id stringlengths 1 6 | url stringlengths 16 1.82k | content stringlengths 37 9.64M |
|---|---|---|
9900 | https://www.additudemag.com/medication/guanfacine/?srsltid=AfmBOorraw__1srWeuLI7CwvMcD94pO6UDXrL_8W41ooeCR9ruJlHrcY | Guanfacine ADHD Non-Stimulant Medication: Side Effects, Uses, Dosages, Warnings
Join ADDitude | Sign In
Search ADDitude
SUBSCRIBE
ADHD
What Is ADHD?
The ADHD Brain
ADHD Symptoms
ADHD in Children
ADHD in Adults
ADHD in Women
Find ADHD Specialists
Symptoms & Tests
Symptom Checker Tool
Symptom Tests
Anxiety
Autism
Bipolar
Depression
ODD
RSD
More in Mental Health
Treatment
ADHD Medications
Medication Reviews
Natural Approaches
ADHD Therapies
Managing Treatment
Treating Your Child
Find ADHD Specialists
ADHD Parenting
Positive Parenting
Behavior & Discipline
Schedules & Routines
Organizing Your Child
Health & Nutrition
Teens with ADHD
Friendships & Activities
School & Learning
IEPs & 504 Plans
Accommodations
Homework & Studying
School Behavior
Learning Challenges
For Teachers
High School
ADHD Adults
Do I Have ADD?
Getting Things Done
Time & Productivity
Relationships
Organization
Health & Nutrition
ADHD Resources
Free Webinars
Free Downloads
Newsletters
News & Research
Guest Blogs
Videos
ADHD Directory
The ADDitude Store
ADHD DIRECTORY
Search Listings
Add a Listing
For Professionals
For Clinicians
For Educators
Research Updates
MAGAZINE
New Issue!
SUBSCRIBE
Manage My Subscription
Give a Gift Sub!
Get Back Issues
Digital Issues
Contact Us
Privacy Policy
Terms of Use
Home
Menu
### ADHD Science & Strategies
Join ADDitude
|
Sign In
ADHD
ADHD Symptoms
ADHD in Children
ADHD in Adults
ADHD in Women
The ADHD Brain
Find ADHD Specialists
Symptoms & Tests
Symptom Checker Tool
Symptom Tests
Anxiety
Autism
Depression
RSD
ODD
More in Mental Health
Treatment
ADHD Medications
Medication Reviews
Complements to Meds
ADHD Therapies
Managing Treatment
Treating Your Child
ADHD Parenting
Positive Parenting
Behavior & Discipline
Schedules & Routines
School & Learning
Health & Nutrition
Teens with ADHD
ADHD Adults
Do I Have ADD?
Getting Things Done
Time & Productivity
Organization
Relationships
Health & Nutrition
ADHD RESOURCES
Free Webinars
Free Downloads
Newsletters
Guest Blogs
eBooks + More
ADHD DIRECTORY
Search Listings
Add a Listing
For Professionals
News & Research
For Clinicians
For Educators
ADHD Directory
MAGAZINE
New Issue!
Subscribe
Gift Subscription
Renew My Subscription
Manage My Subscription
Digital Magazine
Advertisement
ADHD Medication & Treatment
ADHD Medication and Treatment Reviews
guanfacine
Guanfacine HCL is a non-stimulant medication used to treat ADHD symptoms in patients 6 to 17 years of age.
Brand Name: Intuniv
By Janice RoddenMedically reviewed by William Dodson, M.D., LF-APA Reviewed on September 21, 2020
Click to Read 19 Comments & Reviews 💬
Print
Facebook
Twitter
Instagram
Pinterest
What Is Guanfacine? How Does Guanfacine Treat ADHD?
Guanfacine (Brand Name: Intuniv) is a non-stimulant prescription medication used to treat attention deficit hyperactivity disorder (ADHD or ADD). Although the FDA requires all medications to be tested and approved as stand-alone monotherapies, guanfacine is so commonly used in combination with stimulant ADHD medications that it carries an additional FDA approval for use in conjunction with stimulant medications.
The first-line stimulant medications are shown to decrease the frequency of distraction for patients with ADHD. Stimulants may not, however, help with other aspects of ADHD such as hyperactivity, impulsivity, and behavioral and emotional self-control.
Guanfacine is in a class of medications called alpha 2a specific adrenergic receptor agonists or, simply, alpha agonists. These medications activate receptors on nerves that use adrenaline as their neurotransmitter. Adrenaline is familiar to everyone who has ever been frightened. It prepares the brain and the body for flight or for fight. The activation of these receptors tricks the adrenaline nerve into being less active, which in turn leads to lower blood pressure but also a significant decrease in hyperactivity, drive, impulsivity, insomnia, and emotional over-reaction. For these reasons, the stimulants and the alpha agonists are near-perfect complements for the management of ADHD at all ages. The other medication in this class is clonidine, which is also FDA-approved for the treatment of ADHD in children.
Non-stimulant ADHD medications like guanfacine can also be helpful for patients who can’t take stimulants due to undesirable side effects or who don’t experience any benefits from them. People with substance abuse problems may also not be good candidates for this medication.
Unlike some stimulant ADHD medications, guanfacine is not a controlled substance and does not have a high risk of abuse or dependency.
Intuniv is an extended-release (ER) formula of guanfacine. Currently, guanfacine ER is FDA approved only for use in children who are 6 through 17 years old. It has not been studied and approved by the FDA for the treatment of ADHD in children younger than the age of 6 or in adults over the age of 17.
The immediate-release formulation of guanfacine is used to treat high blood pressure. Its safety has not been studied in children under age 12. Because guanfacine is FDA-approved for adults for something other than ADHD, most clinicians feel comfortable using guanfacine IR and ER off-label for adults with ADHD.
Is Guanfacine Similar to Adderall?
No. Guanfacine and Adderall are very different ADHD medications. Both must be prescribed by a doctor and can be used to treat ADHD symptoms but have different side effects and work in the body in different ways. Adderall is a popular stimulant medication that can be prescribed to patients 6 years and older and works to control distractibility and impulsive behavior. It treats inattention, impulsivity, lack of focus, disorganization, forgetfulness, or fidgeting.
Adderall is a combination medication made up of amphetamine and dextroamphetamine that is also available in a generic form. The FDA has approved a generic version of both the immediate-release (Adderall IR) and extended-release (Adderall XR) versions.
Guanfacine can be prescribed when stimulant ADHD medication fails either due to too many negative side effects or a lack of symptom improvement. It is, however, most often used as an addition to a stimulant because it can benefit symptoms and impairments from ADHD that the stimulant medications do not mange well, such as hyperactivity, impulsivity, sleep disturbances, and emotional over-reactions.
Adderall has been classified by the Drug Enforcement Agency (DEA) as a “Schedule II” medication. Schedule II medications are drugs with a high potential for abuse. Guanfacine is in a class of drugs that is not known to be habit-forming.
Guanfacine Dosages — What is Best to Treat ADHD Symptoms?
Guanfacine is available in two formulations:
Immediate-Release Tablet: Taken up to twice daily. Available in 1mg or 2mg dosages.
Extended-Release Tablet: (Guanfacine ER, Brand Name: Intuniv) Taken once daily in the morning or evening at approximately the same time each day. Tablets are available in 1mg, 2mg, 3mg, and 4mg dosages.
The optimal dosage varies patient by patient and by the condition being treated. Your doctor may adjust your dosage weekly by 1mg increments until you or your child experience the best response. The right dose is the one with that gives you (or your child) the most improvement in symptoms without side effects (or with the fewest).
Guanfacine should not be taken with a high-fat meal. This can speed up the release of medication, and increase the risk of side effects. Tablets should be swallowed whole with water or other liquids, and never crushed or chewed. If you miss a dose, you should take it as soon as possible, unless it is already time for your next dose. You should not take two doses of guanfacine at the same time. Contact your doctor if you have questions about skipped pills.
Do not drink alcohol while taking this drug.
When discontinuing treatment, or decreasing dosage, patients should work with a doctor to gradually taper the level of medication by no more than 1mg every 3 to 7 days. Stopping guanfacine suddenly can theoretically create withdrawal symptoms including increased heart rate and high blood pressure. Because people who have ADHD are forgetful and may sometimes forget doses of medications like guanfacine, the FDA did sudden-cessation trials to assess the safety of suddenly stopping an alpha agonist. No rebound high blood pressure or fast heart rate were seen. Nonetheless, when a person stops an alpha agonist, it is strongly recommended to taper off the medication as a precaution.
Before starting or refilling a guanfacine prescription, read the medication guide included with your pills, as it may be updated with new information. This guide should not replace a conversation with your doctor. Your doctor can review you or your child’s medical history, other diagnoses, and other prescriptions for possible interactions and take into account other important considerations. If you have questions, ask your doctor or pharmacist before you begin taking the medication.
As with all medications, follow your guanfacine prescription instructions exactly.
Does Guanfacine Cause Side Effects?
Most people taking guanfacine do not experience side effects. That said, the most common guanfacine side effects are:
sedation or tiredness
low blood pressure usually seen as dizziness when standing up quickly
dizziness
dry mouth
irritability seen in 1-2% of people that does not go away with time. If severely irritable after starting guanfacine, the patient should stop the medication and call their physician
Other less common side effects include
vomiting
slow heart rate
nausea
stomach pain
Serious side effects of guanfacine also include low blood pressure or heart rate, fainting, sleepiness, or withdrawal symptoms. Taking guanfacine may impair your or your teenager’s ability to drive, operate machinery, or perform other potentially dangerous tasks. If side effects are bothersome or continue to persist, talk to your doctor.
Guanfacine and Heart or Blood Pressure Related Problems
Report to your doctor any heart-related problems or a family history of heart and blood pressure problems. Patients with structural cardiac abnormalities, renal failure, and serious heart problems could experience complications while taking guanfacine. Physicians should monitor vital signs closely during treatment. Call your doctor immediately if you or your child experience warning signs such as chest pain, shortness of breath, or fainting while taking guanfacine.
The above is not a complete list of potential side effects. If you notice any health changes not listed above, discuss them with your doctor or pharmacist.
What Precautions are Associated with Guanfacine?
You should not take guanfacine if you have an allergy to guanfacine or any of its ingredients.
Proceed with caution if you have kidney problems, a history of fainting, heart problems, or a history of stroke.
Avoid becoming overheated or dehydrated while taking guanfacine.
If you’re thinking of becoming pregnant, discuss the use of guanfacine with your doctor. Animal studies indicate a potential risk of fetal harm.
It is not known if guanfacine is passed through breastmilk, so nursing mothers are usually instructed not to take this medication.
Store guanfacine in a secure place out of the reach of children, and at room temperature.
Do not share your guanfacine prescription with anyone, even another person with ADHD. Sharing prescription medication is illegal, and can cause harm.
What Should I Know Before Taking Guanfacine?
Be sure and discuss all other active prescription medications with your doctor. Guanfacine can cause dizziness and exacerbate the drowsiness created by depressants including alcohol, barbiturates, antihistamines, or other sedatives.
It’s also important to share information about all vitamin or herbal supplements (especially St. John’s wort), and non-prescription medications (acetaminophen, ibuprofen, etc.) you take with the pharmacist when you fill your prescription and let healthcare providers know you are taking guanfacine before undergoing surgery or having lab tests.
Note: The information in this article is not a complete list of all possible drug interactions.
Guanfacine and Other Medications: More Information
Download: Take Charge of Your Child’s Medication
Read: Intuniv – Answers to Your ADHD Medication Questions
Download: How Do We Know the Medication Is Working?
Find: ADHD Specialists and Clinics Near You
Sources:
Guanfacine- guanfacine hydrochloride tablet. Daily Med (2017).
Guanfacine tablet- extended release. Daily Med (2019).
Newcorn, Jeffrey, et al. Extended‐release guanfacine hydrochloride in 6–17‐year olds with ADHD: a randomised‐withdrawal maintenance of efficacy study. The Association for Child and Adolescent Mental Health (2016).
Facebook
Twitter
Instagram
Pinterest
Tags: ADHD Etc., treating adults, treating kids
More ADHD Articles Recommended for You
The Top ADHD Medications for Children — Rated by Readers
Comorbid Q&A: Treating Mood Disorders, Anxiety, or Autism Alongside ADHD
The History of ADHD and Its Treatments
How Cognitive Behavioral Therapy (CBT) Works
ADHD Newsletter
Treating ADHD
Food and nutrition, medication, supplements, natural therapies, and more.
Email Address
Sign Up
Advertisement
Read These ADHD Articles Next
ADHD Medication Overview: Stimulants, Nonstimulants & More
What Are the Best Non-Stimulant ADHD Medications?
ADHD Medications: Comparison Chart of Stimulants & Non-Stimulants
Advertisement
More ADHD Articles Recommended for You
Stimulants vs. Non-Stimulants: Understanding ADHD Medications
Qelbree (viloxazine)
Study: Supplementing Stimulants with Guanfacine May Improve Executive Functions in Children
Advertisement
ADHD Download
#### Stimulants vs. Non-Stimulants: Understanding ADHD Medications
Get this free download
Advertisement
How the ADHD Brain Works ------------------------ $8.95Add to cart
Advertisement
The ADD itude Commitment
ADDitude collaborates closely with leading medical experts to publish accurate, clear, and authoritative content that millions of readers trust and share.
Review Our Editorial ProcessMeet Our Medical Advisory Panel
ADDitude verifies the factual accuracy of all new content.
Sources are cited for all scientific research and findings.
Content is periodically reviewed and updated to reflect new health insights.
We strive to feature diverse voices and experiences.
Since 1998, millions of parents and adults have trusted ADDitude's expert guidance and support for living better with ADHD and its related mental health conditions. Our mission is to be your trusted advisor, an unwavering source of understanding and guidance along the path to wellness.
Follow us:
Facebook
Twitter
Pinterest
Instagram
LinkedIn
TikTok
Newsletters
(Select your free newsletters)
[x] Parenting a Child with ADHD
[x] Living with Adult ADHD
[x] School and Learning Digest
[x] Treatment Digest
[x] Best of the Week
[x] ADDitude for Women
[x] Webinars & Announcements
Email Address
Sign Up
Subscribe
Get a free issue and free ADDitude eBook, plus save 42% off the cover price.
Order Now
Site Map
Symptom Tests
ADHD Treatment Options
ADHD Parenting Help
School & Learning
For Adults with ADHD
For ADHD Professionals
Community Resources
Free ADHD Webinars
Free ADHD Downloads
Guest Blogs
ADDitude eBooks
Find an ADHD Specialist
ADDitudeMag.com
Join | Log In
About ADDitude
Our ADHD Advisory Panel
Contact Us
Advertise
ADDitude Magazine
Subscribe to ADDitude Magazine
Give ADDitude As a Gift
Manage My Subscription
Subscriber Access to Archives
Pay My Bill
FAQ
Copyright © 1998 - 2025 WebMD LLC. All rights reserved. Your use of this site is governed by our Terms of Use and Privacy Policy. ADDitude does not provide medical advice, diagnosis, or treatment. The material on this web site is provided for educational purposes only.
Advertisement |
9901 | https://www.knowgaucherdisease.com/hcp/pathophysiology | Gaucher disease (GD) is a rare, autosomal recessive disorder.1 Although rare, it is
the most common
type of lysosomal storage disease.1
Patients with GD have a deficiency of the enzyme glucocerebrosidase, the enzyme responsible for
catalyzing glucocerebroside.1–3
The progressive accumulation of glucocerebroside throughout
the body leads to a variety of symptoms that present in various organs.1,4–6
Dr. Weinthal
Hematologist
Dr. Cornier
Geneticist
Disease Overview
What is Type 1 Gaucher
Gaucher disease (GD) is a rare, autosomal recessive lysosomal storage disorder, in which deficiency of the enzyme
glucocerebrosidase leads to the accumulation of its substrate glucocerebroside throughout the
body, primarily in the spleen, liver, and bone marrow.1–3 The accumulation of
glucocerebroside in the different areas of the body leads to the progressive, multi-systemic,
and heterogeneous nature of the disease.1,4
There are three types of GD. Type 1 Gaucher disease (GD1) is the most prevalent form of the
disease, accounting for more than 90% of all cases.4
Unlike type 2 and 3 GD, GD1 is nonneuronopathic.4
Individuals with GD1 can begin to exhibit symptoms at any age, including during early
childhood and late adulthood.3
The type and severity of symptoms will also vary among patients, and there can be varying
degrees of involvement presenting in different organs in the same individual.2-4
This can be related to the rate of substrate accumulation.3,5,6
The clinical course and life expectancy for patients with GD1 are also variable. While
untreated patients will generally survive to adulthood, early recognition, diagnosis, and
treatment of GD1 are key for the best possible patient outcomes and quality of life.1,4,7
An early clinical presentation predicts a more severe disease course, highlighting the
importance of recognizing GD1 as soon as possible to help ensure earlier
intervention and inform treatment decisions.7
Disease Overview
Pathophysiology
Healthy Cell
Gaucher Cell
Disease Overview
GD1 is a lysosomal storage disorder characterized by a deficiency of the enzyme
glucocerebrosidase.2 This is caused by mutations of the gene that controls production
of this enzyme.3
Without sufficient glucocerebrosidase, its substrate glucocerebroside accumulates within the
lysosomes of macrophages, leading to the formation of ‘Gaucher cells’.1,4
‘Gaucher cells’ are engorged macrophages, visually characterized by a displaced nucleus and a lysosomal architecture that is distorted from its normally spherical shape.1,4,8
Gaucher cells infiltrate organs and tissues enriched in cells of the mononuclear phagocyte
system, e.g., the spleen, liver, and bone marrow.1,9,10
Over time, they displace normal cells resulting in progressive impairment; the liver and
spleen may become enlarged, which can interfere with normal functioning and cause a painful
and swollen abdomen.1,8,9,11
Displacement of hematopoiesis by Gaucher cell accumulation within the bone marrow can lead
to anemia and thrombocytopenia, which in turn result in bleeding problems and can compromise
the strength of the skeletal system.6,11
Disease Overview
Prevalence, Genetics, and Inheritance
Disease Overview
Prevalence, Genetics, and Inheritance
Both sexes can be affected by GD1, with males and females sharing an equal risk.3
GD1 is a pan-ethnic condition, although it does have a higher prevalence in people of
Ashkenazi Jewish ancestry.5
GD1 affects:
1–9 in 100,000 within the overall population12
~1 in 600 within the Ashkenazi Jewish population3
~1 in 17 within the Ashkenazi Jewish community that are carriers13
an estimated 6,000 individuals in the United States3,14
GD1 is a lysosomal storage disorder and is caused by mutations of the GBA1 gene that
controls the production of the glucocerebrosidase enzyme.14-16
To date, more than 450 mutations in GBA1 have been identified.17
The type of mutation may to some extent determine the type of GD, but there are other
factors that affect the phenotypic expression. This has been demonstrated by cases in which
symptom type, severity, and disease course have varied among siblings, even identical twins.6,18
GD1 is genetic and has an autosomal recessive inheritance pattern.3 If both
parents are carriers, there is a 50% chance of the child being a carrier, a 25% chance of
them having Gaucher disease, and a 25% chance that they will not be affected.3
Pedigree Diagram Illustrating the Autosomal Recessive Inheritance Pattern of Gaucher Disease
Signs and Symptoms
Visceral, Hematological, and Skeletal
Signs and Symptoms
The age of onset and the severity of symptoms of type 1 Gaucher disease (GD1) can vary
considerably.1,2
The main clinical manifestations include:
Visceral
Splenomegaly3,4
Splenomegaly occurs in 95% of patients with GD1.
It may be one of the first and most physically prominent presentations of GD1 in
patients.
Splenomegaly results in abdominal distension and early satiety.4–7
Many patients complain of acute abdominal pain, which can be worsened by splenic
infarction.1,8
Hepatomegaly3,4
Enlargement of the liver occurs in 80% of patients with GD1, also contributing
to abdominal pain and early satiety.1,5,7
Hepatomegaly may progress to liver disease and cirrhosis.1,4
Hematological
Thrombocytopenia3,4
Bruising, bleeding, and nosebleeds occur due to
thrombocytopenia.1,4,6
Anemia3,4
Fatigue is common, occurring in 50% of patients with GD1.6,7
Gammopathy9,10
Gammopathy can be polyclonal or monoclonal, and is detected very often in
patients.9,10 This can lead to long-term hematological complications.9
Hyperferritinemia9,10
Hyperferritinemia is very often detected in patients and can contribute to
long-term hematological complications.10
Further Complications
Long-term hematological complications include a risk of developing
hematological malignancies and solid tumors.10,11
Skeletal
Skeletal involvement occurs in 70–100% of patients with GD1 and includes
abnormal bone shaping, bone pain, bone infections, and fractures.5
Osteonecrosis of the femoral head may lead to joint collapse and osteoarthritis.12
Skeletal manifestations are associated with considerable pain, limitations in mobility,
and a significantly reduced quality of life.11,13
In severe cases, they can lead to potentially irreversible complications or long-term disability.14
This summarizes the key clinical manifestations of GD1; however, a GD1 diagnosis and the
resulting
symptoms patients experience can also lead to further, long-term impacts on patients that
require
ongoing Monitoring and Management. Visit the
Further Impact section to learn more.
Signs and Symptoms
What to Do If You Recognize the Symptoms of GD1
Signs and Symptoms
What to Do If You Recognize the Symptoms of GD1
Laboratory testing for Gaucher disease (GD) should be carried out when a patient
presents with signs indicative of the disease, such as unexplained splenomegaly,
anemia, or thrombocytopenia.15 Suspicion should be higher when the patient is of
Ashkenazi Jewish descent.14,15 If GD is suspected, an enzyme assay as well as genetic
testing should be performed to confirm the diagnosis. Learn more about the definitive tests
for GD1 here.
Diagnosis and Testing
Overview
Diagnosis and Testing
Overview
A diagnosis of type 1 Gaucher disease (GD1) should be confirmed with an enzyme assay to
determine levels of β-glucosidase enzyme activity, and genetic testing to identify mutations
that cause the disease.1
A bone marrow biopsy or smear can detect Gaucher cells, but it is not recommended as a
diagnostic tool, as it is not differential. Furthermore, it is an invasive procedure, and
there
is a risk of complications.1–3
Some states (including Missouri, New York, Illinois, and Tennessee) currently offer genetic testing to
screen newborns for Gaucher disease (GD), with further states currently considering
implementation of this testing.4 Check the current status of newborn screening
for GD in your
state in case your patients have questions or need guidance in this area.
Diagnosis and Testing
Testing
Diagnosis and Testing
Testing
Enzyme Assay
The β-glucosidase enzyme assay measures the levels of the enzyme
glucocerebrosidase.5 A GD
diagnosis is confirmed by establishing glucocerebrosidase enzyme activity in mononuclear
cells,
leukocytes, or cultured fibroblasts. These are obtained by either a skin biopsy or dried
blood
spot test.6–8 Low levels of this enzyme, less than 15% of mean normal activity,
indicate a
diagnosis of GD.6,9 Enzyme activity, however, does not predict disease
severity.1,6
Genetic Testing
Genetic testing requires either a blood or saliva sample, used to extract the patient’s DNA.10
Genetic sequencing can then be used to detect specific mutations in the GBA1 gene that
result in
GD.11,12
Type 1 Gaucher Disease Diagnosis and Testing ALGORITHMS
Adult
This algorithm provides information to assist in the differential diagnosis of GD1 in
patients
with unexplained splenomegaly and/or thrombocytopenia.
This algorithm is not intended to be a diagnostic tool. It does not replace the need for a
complete evaluation of the patient by a healthcare professional.
In cases of splenomegaly and/or thrombocytopenia it is important to include GD1 in
differential
diagnoses in order to avoid potentially harmful splenectomy or irreversible
complications.7,15
Pediatric
This algorithm has been synthesized from two publications on type 1 Gaucher disease
diagnosis:
‘Consensus Conference: A reappraisal of Gaucher disease – diagnosis and disease management
algorithms’ and ‘Presenting signs and patient covariables in Gaucher disease: outcome of the
Gaucher Earlier Diagnosis Consensus (GED-C) Delphi initiative.’3,13
This diagnostic algorithm for pediatric patients has been synthesized from two
publications on
type 1 Gaucher disease diagnosis: ‘The diagnosis and management of Gaucher disease in
pediatric
patients: Where do we go from here?’ and ‘Screening, patient identification, evaluation, and
treatment in patients with Gaucher disease: Results from a Delphi consensus.’6,14
Diagnosis and Testing
Diagnostic Challenges and Screening
Diagnosis and Testing
Diagnostic Challenges and Screening
Diagnostic challenges
Due to the rare nature and heterogeneous presentation of GD1, patients often consult several
different specialists before reaching the correct diagnosis.16,17 Although
clinical
presentation of variable severity occurs in childhood for the majority of patients,
diagnosis is often delayed 10–15 years until patients are in adulthood.15,17
Patients are
often misdiagnosed, which can lead to disease complications, persistence of untreated
symptoms, emotional distress, and frustration and depression at the lack of an explanation
for their symptoms.6
One of the key reasons for misdiagnoses is that GD1 symptoms can overlap with a number of
other disorders, such as:
Visceral: liver cancer6
Hematological: autoimmune hemolytic
anemia; hematological malignancies; bleeding
disorders; idiopathic thrombocytopenia purpura3,4,18,19
Skeletal: synovitis or osteomyelitis;
multiple myeloma6,9
Additional diagnostic challenges of GD1, which may lead to delayed diagnoses or
misdiagnoses, include:
the high phenotypic heterogeneity in the initial presentation, which can make GD1
difficult to recognize6,17
low physician awareness of GD, due to its rarity5,16
the existence/appearance of nonspecific and mild symptoms of GD1, which can mean that
physicians do not always consider it in their differential
diagnoses2,6,15
CARRIER SCREENING
Genetic testing can also be used to identify carriers of GD1. Although carriers do not
experience symptoms of the disease, it is important that they are identified because GD is a
hereditary condition.12
Carrier screening for mutations commonly associated with GD1 is recommended for the
Ashkenazi Jewish population and other individuals who are at high risk for the disease.6,20
Testing must be offered in conjunction with genetic counseling to provide couples at risk,
even asymptomatic individuals, with a description of the range of associated phenotypes and
their options, which include prenatal diagnosis.9
FAMILY PLANNING
Due to its hereditary nature, an additional challenge faced by patients, carriers, and their
families is how to approach family planning. Make sure your patients affected by GD1 know
they are not alone in this and are directed toward any available Resources or referred to
appropriate specialists to help them with these decisions.
Diagnosis and Testing
Importance of Early Recognition and Diagnosis
Diagnosis and Testing
Importance of Early Recognition and Diagnosis
Prompt diagnosis of GD1 is essential because the disease is progressive.5 Delayed
recognition and subsequent delayed diagnosis of the disease lead to late management, which
can lead to irreversible complications, such as avascular necrosis or multiple myeloma.8,15
Therefore, the aim should be to recognize, diagnose, and treat before the development of
such irreversible complications.2 Early recognition can reduce unnecessary
investigations
and the time to diagnosis, and therefore lead to more effective management strategies and
the timely initiation of appropriate treatment.5,16 This can help to alleviate
some of the
anxiety experienced by patients and their families, and is therefore critical to ensuring
the best patient outcomes.16 Regular Monitoring and Management
are required for patients
with GD1 following their diagnosis.7,8
Monitoring, Management, and Treatment Options
Monitoring
Monitoring, Management, and Treatment Options
Monitoring
Patients with type 1 Gaucher disease (GD1) should be regularly monitored using blood tests,
measurement of visceral volumes (spleen and liver), and skeletal scans to detect any
complications that can occur as the disease progresses.1,2 The monitoring
interval,
which is usually every 6–24 months, depends on the recommendations of the healthcare
provider,
the patient’s individual experience with the disease, and/or treatment.1
Blood Tests
Blood tests are often used to monitor disease progression and measure a
patient’s:1,3
Hemoglobin count
Platelet count
Additionally, the serum levels of several biological markers indicate disease severity and
progression:1,3,4
Glucosylsphingosine (lyso-Gb1/lyso-GL1)2
Tartrate-resistant acid phosphatase (TRAP)1
Chitotriosidase (CHIT1)1,3,4
Angiotensin-converting enzyme (ACE)1
Chemokine ligand 18 (CCL-18)2,3
Visceral Volumes
Visceral volumes can be assessed with:1,3
Volumetric MRI (since repeat assessment is routine in Gaucher disease)1,3
CT or ultrasound can be used when MRI is unavailable1,3
Skeletal Scans
Skeletal scans are used to detect bone marrow infiltration and bone disease, and can be
assessed using:1
MRI, which is particularly useful to identify bone infarction and necrosis
Dual-energy X-ray absorptiometry (DEXA) is the gold standard method for assessing bone
mineral density
Monitoring, Management, and Treatment Options
Management and Treatment Options
Monitoring, Management, and Treatment Options
Management and Treatment Options
Management
As GD1 is a multi-systemic and heterogeneous disorder, a multidisciplinary team and
individualized approach is required for optimal management.5,6
First, the team should conduct a comprehensive assessment of all possible disease
manifestations.7
Therapeutic goals should then be established for the individual patient based on each
disease manifestation.7
Achievement and maintenance of these goals rely on regular ongoing monitoring and the
adjustment of therapy when goals are not reached.4,7
Such management is designed to provide personalized supportive care for the patient,
considering the patient’s severity and rate of disease progression, and their quality of
life.6–8
Treatment Options
Healthcare providers can work with patients to help manage their GD1 symptoms. In addition,
treatments are available for GD1:9
Enzyme replacement therapy (ERT) infusion – a functional version of the deficient enzyme is
administered by intravenous infusion to metabolize glucocerebroside10,11
Oral treatment with a substrate reduction therapy (SRT) – inhibits the synthesis of the
substrate glucocerebroside10,11
ERT is the only currently approved treatment for pediatric patients with type 1 Gaucher
disease.5
Regular ERT infusions are recommended for all children and adolescents who are symptomatic
of GD1
and should be initiated as soon as possible after diagnosis to help prevent irreversible
complications.5
Monitoring, Management, and Treatment Options
Supportive Therapies
Monitoring, Management, and Treatment Options
Supportive Therapies
Management of GD1 has, and occasionally continues to be, limited to supportive care with transfusions, splenectomy, and orthopedic surgery.12
Transfusions may be given in cases of severe anemia or bleeding, for example, prior to
invasive procedures, during pregnancy, and postpartum hemorrhage.5,8,11 Some
patients may become transfusion dependent.11
Painful bone crises often require temporary immobilization, and analgesics may be given to
relieve pain.2,8
Bisphosphonates can be used, but patients receiving them should be monitored for
osteonecrosis of the jaw.8
The use of bisphosphonates is controversial in GD1, as the pathophysiology of bone-mass
decline is poorly understood.2
Joint replacement can be considered for restoration of function and relief from chronic
pain.5
Orthopedic surgeries may be required for bone complications, including avascular necrosis
and pathological fractures.2
Historically and even at times today, partial or total splenectomy is performed in the presence
of hypersplenism. While effective, this procedure increases the risk of infections, avascular
necrosis, and pulmonary hypertension. Furthermore, splenectomized patients generally suffer
from increased bone and liver deposits, leading to increased bone complications and
susceptibility to infection. Splenectomy should therefore be avoided and only performed in
exceptional circumstances, such as emergency situations.2,8,9,14
GD1 will affect a patient physically but also can contribute toward emotional and
psychological issues.
For this reason, psychological support should be offered to patients to help reduce this impact of the disease on patients and their families. Learn more about the potential psychological
and
psychosocial impacts of GD1 here.
It is also important that patients and their loved ones be made
aware of patients’ associations that they can connect with.2,5,8 Visit our
Resources page to learn
about organizations currently working to improve understanding of GD1 and provide support to
patients living with this rare condition.
Monitoring, Management, and Treatment Options
FURTHER IMPACT - DISEASE PROGRESSION
Monitoring, Management, and Treatment Options
FURTHER IMPACT - DISEASE PROGRESSION
Puberty and Growth
For younger patients with GD1, delayed growth and puberty can often arise as complications
in the disease course.2,4,5
Around half of younger patients with GD exhibit delayed growth, with approximately 25%
having a
shorter stature than expected when compared with midparental height.7
Patients who exhibit significantly stunted growth during childhood are likely to have severe
visceral involvement and can also experience delayed puberty.7
Growth below the 5th percentile is seen in approximately 34% of children with GD1
specifically.2
Children with a more severe form of GD1 also display delayed milestones and failure to
thrive.11
For children diagnosed with GD1, regular monitoring of growth should be performed at regular
intervals of every 3 or 6 months.11,15
With appropriate treatment and management, children with GD experiencing delayed growth can
achieve normal height, when compared with population averages.7
Life Expectancy
As a result of the variable clinical course seen in GD1 patients, life expectancy can be
highly variable.7
GD1 patients can fall anywhere on a spectrum between a severe form of the disease with onset
in early childhood to an indolent or even asymptomatic disorder presenting in elderly
adults.7
The average life expectancy of GD1 patients has been reported to be approximately a decade
shorter than that of the general population; however,
it has been acknowledged that the positive effects of appropriate, modern management and
treatment have not yet been taken into account.5
Monitoring, Management, and Treatment Options
FURTHER IMPACT - WELLBEING
Monitoring, Management, and Treatment Options
FURTHER IMPACT - WELLBEING
Health-related Quality of Life (HRQoL)
HRQoL of patients can be impacted by many aspects of GD1, including the range and severity
of symptoms,
as well as how they interact with aspects of daily life.16
The HRQoL of younger patients specifically is seen to be negatively impacted by many of the
key
symptoms associated with GD1 – most notably bone, joint, and abdominal pain, as well as
chronic fatigue and mobility limitations.4,5,16
Subjective experiences relating to symptoms, for example feeling ill, sad, and unable to be
involved
in various activities, were also reported by children to impact substantially on their
perceived quality of life,
especially in relation to social and school functioning.16
Studies comparing adult GD patients to the general population have found that HRQoL domains
(e.g., physical functioning, presence of physical limitations, pain, general state of
health,
vitality, and social functioning) are significantly lower in patients.16
With appropriate management, which also takes into account psychological and subjective emotional functioning, patients may perceive
improvements in HRQoL; for example, they may be able to better control their most debilitating symptoms, lead more active and socially
fulfilling lives, and pursue their education or careers.9,16
Psychological and Psychosocial Impact
Receiving a diagnosis of a chronic, lifelong condition and living with its symptoms can lead to
various psychological and psychosocial issues.17 There can be specific
considerations for different age groups, including what each perceives as the most prevalent issues they face when living with GD1.
Younger children:
It is vital to consider the implications of receiving such a significant diagnosis at a very
young age;
children who have grown up with a GD diagnosis may progress into adulthood without a proper
understanding
of the implications of their condition or the importance of treatment.5
Additionally, children may experience considerable psychosocial problems, for example
difficulty
adjusting to life with their condition, as well as feelings of distress, including anger,
denial, fear, insecurity, and isolation.16
Older children and adolescents:
Delayed growth and puberty can have a negative effect on the body image and social
functioning of adolescent patients,
leading to self-stigmatization and decreased self-esteem.16
An increasing awareness of ‘being different’ can be amplified when symptoms (e.g., fatigue,
splenomegaly,
or bone disease) prevent patients from participating in physical or social activities, such
as sports.16
School performance may also be affected by symptoms such as chronic pain and
fatigue.16
Older children and adolescent patients can benefit from sensitive genetic counseling for their
condition
and support for the above issues, as well as support for the transition from pediatric to
adult healthcare.5
Adults:
Some of the domains affected by GD1 in adult patients include difficulties coping with the
diagnosis
and physical functioning-related daily activities at school, home, or work.16
Patients find that their social lives, including capacity for recreational activities are
limited by both a lack of energy and pain.16
Stress associated with treatment burden, time consumption, and cost is often reported.16
Patients perceive GD1 as likely to limit their options in life and impact future plans;
their emotional health is impacted by these limitations, causing feelings of low
self-esteem, anger, anxiety and depression.16
Psychological counseling can therefore be beneficial for all patients with GD and their
families,
given the chronic, lifelong nature of the disease, presence of incapacitating symptoms,
and effects of these on both day-to-day functioning and overall outlook on
life.11,16
Resources
Downloadable Materials
Resources
Downloadable Materials
The following resources might be useful in your clinical practice to provide further
information
on managing patients with type 1 Gaucher disease (GD1):
ADULT DIAGNOSIS AND TESTING ALGORITHM
Use this algorithm to learn more about the common features of GD1 and help diagnose adult
patients with unexplained splenomegaly and/or thrombocytopenia.
Download
Pediatric Diagnosis and Testing Algorithm
Use this algorithm to learn more about the common features of GD1 specific to children and
adolescents and help diagnose pediatric patients with unexplained splenomegaly with
thrombocytopenia and/or anemia.
Download
Gaucher Disease Monitoring Flyer
This flyer provides guidelines for the clinical monitoring of adult patients diagnosed with
type 1 Gaucher disease and highlights the benefits of ongoing monitoring.
Download
GAUCHER DISEASE PEDIATRIC MONITORING FLYER
This flyer provides guidelines for the clinical monitoring of pediatric patients diagnosed with type 1 Gaucher disease and highlights the benefits of ongoing monitoring.
Download
Gaucher Community Awareness Resources Brochure
This brochure outlines three steps toward a better understanding of type 1 Gaucher disease,
with the intention of spreading awareness of GD1.
Download
There are also resources available that you can share with your patients to help them on
their journey with GD1:
CAREGIVER Guide to Type 1 Gaucher Disease
This caregiver guide, for anyone caring for a patient with GD1, is available for
download below in both English and Spanish.
Download ENG
Download ESP
Resources
Other Resources
Resources
Other Resources
There are a number of organizations that specialize in raising GD1 and wider rare-disease
awareness and building communities for patients living with these diseases. We encourage
you to utilize the information and resources offered by these organizations and to make
your patients aware of the additional support that they offer.
National Gaucher Foundation (NGF)
The NGF supports physicians in their efforts to better understand the diagnosis and
treatment of Gaucher disease. Information and resources can be found on their website.
Resources for your patients and their families are also provided.
Visit Site
Gaucher Community Alliance (GCA)
The GCA is a patient organization that aims to support the patient community.
Resources and opportunities to connect with other patients are available for your patients.
Visit Site
National Organization for Rare Disorders (NORD)
NORD provides resources for clinicians about specific rare disorders,
including type 1 Gaucher disease, to facilitate the timely diagnosis and treatment of their
patients.
NORD also provides programs and services to support your patients.
Visit Site
CheckRare
CheckRare is the only publishing and learning platform dedicated exclusively to rare
diseases and cancers.
Visit to find original content from a range of perspectives, including patients and
healthcare professionals,
as well as industry and regulatory bodies.
Visit Site
Stay updated
Thank you
for signing up to the mailing list.
To unsubscribe, please click here.
Stay updated
Sign up for news and information on type 1 Gaucher disease.
Required field
SITEMAP
SITEMAP
©2024 Takeda Pharmaceuticals U.S.A., Inc., 500 Kendall Street, Cambridge, MA 02142.
1‑877‑TAKEDA‑7 (1‑877‑825‑3327).
All rights reserved. Takeda and are registered trademarks of Takeda Pharmaceutical Company Limited.
US‑NON‑8044v2.0 12/24
All images used on this website are stock photography images and do not portray actual patients.
Diagnosis and Testing
Overview
In the video below, specialists discuss the importance of understanding
and diagnosing Gaucher disease – watch the video now.
Video Player is loading.
Current Time 0:00
/
Duration 0:00
Loaded: 0%
Stream Type LIVE
Remaining Time -0:00
1x
Chapters
descriptions off, selected
captions settings, opens captions settings dialog
captions off, selected
This is a modal window.
Beginning of dialog window. Escape will cancel and close the window.
End of dialog window.
This is a modal window. This modal can be closed by pressing the Escape key or activating the close button.
This is a modal window.
Dr. Weinthal
Hematologist
Dr. Cornier
Geneticist
Healthy Cell
Gaucher Cell
Video Player is loading.
Current Time 0:00
/
Duration 0:00
Loaded: 0%
Stream Type LIVE
Remaining Time -0:00
1x
Chapters
descriptions off, selected
captions settings, opens captions settings dialog
captions off, selected
This is a modal window.
Beginning of dialog window. Escape will cancel and close the window.
End of dialog window.
This is a modal window. This modal can be closed by pressing the Escape key or activating the close button.
This is a modal window. |
9902 | https://assets.thermofisher.com/DirectWebViewer/private/results.aspx?page=NewSearch&LANGUAGE=d__EN&SUBFORMAT=d__CGV4&SKU=ALFAA44435&PLANT=d__ALF | Aluminum fluoride, anhydrous
SAFETY DATA SHEET Page 1 / 7
Creation Date 06-Mar-2014
Revision Date 17-Sep-2025
Version 4
ALFAA44435
SECTION 1. IDENTIFICATION OF THE SUBSTANCE/MIXTURE AND OF THE COMPANY/UNDERTAKING
产品说明: 产品说明: 产品说明: 产品说明: 氟化铝 氟化铝 氟化铝 氟化铝
Product Description: Aluminum fluoride, anhydrous
Cat No. : 44435
Synonyms Aluminum trifluoride.
CAS No 7784-18-1
Molecular Formula AlF3
Supplier Avocado Research Chemicals Ltd.
(Part of Thermo Fisher Scientific)
Shore Road, Heysham
Lancashire, LA3 2XY,
United Kingdom
Office Tel: +44 (0) 1524 850506
Office Fax: +44 (0) 1524 850608
Emergency Telephone Number For information US call: 001-800-227-6701 / Europe call: +32 14 57 52 11
Emergency Number US: 001-201-796-7100 / Europe: +32 14 57 52 99
CHEMTREC Tel. No. US: 001-800-424-9300 / Europe: 001-703-527-3887
E-mail address begel.sdsdesk@thermofisher.com
Recommended Use Laboratory chemicals.
Uses advised against No Information available
SECTION 2. HAZARD IDENTIFICATION
Classification of the substance or mixture
Based on available data, the classification criteria are not met
Label Elements
None required
Storage
P403 - Store in a well-ventilated place
Disposal
P501 - Dispose of contents/ container to an approved waste disposal plant
Physical and Chemical Hazards
None identified.
Health Hazards
Emergency Overview
The product contains no substances which at their given concentration are considered to be hazardous to health.
Physical State
Solid
Appearance
White
Odor
Odorless
Page 1Aluminum fluoride, anhydrous
ALFAA44435
SAFETY DATA SHEET Page 2 / 7
Revision Date 17-Sep-2025
The product contains no substances which at their given concentration are considered to be hazardous to health.
Environmental hazards
Contains no substances known to be hazardous to the environment or not degradable in waste water treatment plants. Is not likely
mobile in the environment due its low water solubility. .
This product does not contain any known or suspected endocrine disruptors.
SECTION 3. COMPOSITION/INFORMATION ON INGREDIENTS
Component CAS No Weight %
Aluminum fluoride 7784-18-1 >95
SECTION 4. FIRST AID MEASURES
Eye Contact
Rinse immediately with plenty of water, also under the eyelids, for at least 15 minutes. Get medical attention.
Skin Contact
Wash off immediately with plenty of water for at least 15 minutes. Get medical attention immediately if symptoms occur.
Inhalation
Remove to fresh air. Get medical attention immediately if symptoms occur.
Ingestion
Clean mouth with water and drink afterwards plenty of water. Get medical attention if symptoms occur.
Most important symptoms and effects
None reasonably foreseeable.
Self-Protection of the First Aider
No special precautions required.
Notes to Physician
Treat symptomatically.
SECTION 5. FIRE-FIGHTING MEASURES
Suitable Extinguishing Media
Use extinguishing measures that are appropriate to local circumstances and the surrounding environment.
Extinguishing media which must not be used for safety reasons
No information available.
Specific Hazards Arising from the Chemical
Thermal decomposition can lead to release of irritating gases and vapors.
Protective Equipment and Precautions for Firefighters
As in any fire, wear self-contained breathing apparatus pressure-demand, MSHA/NIOSH (approved or equivalent) and full
protective gear.
SECTION 6. ACCIDENTAL RELEASE MEASURES
Personal Precautions
Ensure adequate ventilation. Use personal protective equipment as required. Avoid dust formation.
Page 2
Environmental Precautions Aluminum fluoride, anhydrous
ALFAA44435
SAFETY DATA SHEET Page 3 / 7
Revision Date 17-Sep-2025
Should not be released into the environment. See Section 12 for additional Ecological Information.
Methods for Containment and Clean Up
Sweep up and shovel into suitable containers for disposal. Avoid dust formation.
Refer to protective measures listed in Sections 8 and 13.
SECTION 7. HANDLING AND STORAGE
Handling
Wear personal protective equipment/face protection. Ensure adequate ventilation. Avoid contact with skin, eyes or clothing. Avoid
ingestion and inhalation. Avoid dust formation.
Storage
Keep containers tightly closed in a dry, cool and well-ventilated place.
Specific Use(s)
Use in laboratories
SECTION 8. EXPOSURE CONTROLS/PERSONAL PROTECTION
Control Parameters
Component China Taiwan Thailand Hong Kong
Aluminum fluoride -TWA: 2.5 mg/m 3TWA: 2.5 mg/m 3-
Component ACGIH TLV OSHA PEL NIOSH The United Kingdom European Union
Aluminum fluoride TWA: 2.5 mg/m 3(Vacated) TWA: 2
mg/m 3(Vacated) TWA:
2.5 mg/m 3
IDLH: 250 mg/m 3
REL = 2 mg/m 3(TWA)
REL = 2.5 mg/m 3
(TWA)
STEL: 6 mg/m 315 min
TWA: 2 mg/m 38 hr
STEL: 7.5 mg/m 315
min
TWA: 2.5 mg/m 38 hr
Legend
ACGIH - American Conference of Governmental Industrial Hygienists
OSHA - Occupational Safety and Health Administration
NIOSH: NIOSH - National Institute for Occupational Safety and Health
Monitoring methods
BS EN 14042:2003 Title Identifier: Workplace atmospheres. Guide for the application and use of procedures for the assessment of
exposure to chemical and biological agents. MDHS14/3 General methods for sampling and gravimetric analysis of respirable and
inhalable dust
Exposure Controls
Engineering Measures
None under normal use conditions. .
Personal protective equipment
Eye Protection Wear safety glasses with side shields (or goggles) (European standard - EN 166)
Hand Protection Protective gloves
Inspect gloves before use.
Glove material Breakthrough time Glove thickness EU standard Glove comments
Natural rubber
Nitrile rubber
Neoprene
PVC
See manufacturers
recommendations
EN 374 (minimum requirement)
Page 3Aluminum fluoride, anhydrous
ALFAA44435
SAFETY DATA SHEET Page 4 / 7
Revision Date 17-Sep-2025
Please observe the instructions regarding permeability and breakthrough time which are provided by the supplier of the gloves.
(Refer to manufacturer/supplier for information)
Ensure gloves are suitable for the task: Chemical compatability, Dexterity, Operational conditions, User susceptibility, e.g.
sensitisation effects, also take into consideration the specific local conditions under which the product is used, such as the danger
of cuts, abrasion.
Remove gloves with care avoiding skin contamination.
Skin and body protection Long sleeved clothing
Respiratory Protection No protective equipment is needed under normal use conditions.
Large scale/emergency use Use a NIOSH/MSHA or European Standard EN 136 approved respirator if exposure limits
are exceeded or if irritation or other symptoms are experienced
Recommended Filter type: Particle filter
Small scale/Laboratory use Maintain adequate ventilation
Hygiene Measures Handle in accordance with good industrial hygiene and safety practice.
Environmental exposure controls No information available.
SECTION 9. PHYSICAL AND CHEMICAL PROPERTIES
Molecular Formula AlF3
Molecular Weight 83.97
SECTION 10. STABILITY AND REACTIVITY
Stability Stable under normal conditions.
Appearance White
Physical State Solid
Odor Odorless
Odor Threshold No data available
pH No information available
Melting Point/Range No data available
Softening Point No data available
Boiling Point/Range 1291 °C / 2355.8 °F @ 760 mmHg
Flash Point Not applicable Method - No information available
Evaporation Rate Not applicable Solid
Flammability (solid,gas) No information available
Explosion Limits No data available
Vapor Pressure No information available
Vapor Density Not applicable Solid
Specific Gravity / Density No data available
Bulk Density No data available
Water Solubility Slightly soluble
Solubility in other solvents No information available
Partition Coefficient (n-octanol/water)
Autoignition Temperature No data available
Decomposition Temperature No data available
Viscosity Not applicable Solid
Explosive Properties No information available
Oxidizing Properties No information available
Page 4Aluminum fluoride, anhydrous
ALFAA44435
SAFETY DATA SHEET Page 5 / 7
Revision Date 17-Sep-2025
Hazardous Reactions None under normal processing.
Hazardous Polymerization Hazardous polymerization does not occur.
Conditions to Avoid Incompatible products. Excess heat. Avoid dust formation.
Materials to avoid No information available.
Hazardous Decomposition Products None under normal use conditions.
SECTION 11. TOXICOLOGICAL INFORMATION
Product Information
(a) acute toxicity;
Component LD50 Oral LD50 Dermal LC50 Inhalation
Aluminum fluoride >2000 mg/kg ( Rat ) LC50 > 0.53 mg/L ( Rat ) 4 h
(b) skin corrosion/irritation; No data available
(c) serious eye damage/irritation; No data available
(d) respiratory or skin sensitization;
Respiratory No data available
Skin No data available
(e) germ cell mutagenicity; No data available
(f) carcinogenicity; No data available
There are no known carcinogenic chemicals in this product
(g) reproductive toxicity; No data available
(h) STOT-single exposure; No data available
(i) STOT-repeated exposure; No data available
Target Organs No information available.
(j) aspiration hazard; Not applicable
Solid
Symptoms / effects,both acute and
delayed
No information available
SECTION 12. ECOLOGICAL INFORMATION
Ecotoxicity effects .
Persistence and Degradability
Page 5Aluminum fluoride, anhydrous
ALFAA44435
SAFETY DATA SHEET Page 6 / 7
Revision Date 17-Sep-2025
Persistence May persist, based on information available.
Degradability Not relevant for inorganic substances.
Bioaccumulative Potential May have some potential to bioaccumulate
Mobility in soil Is not likely mobile in the environment due its low water solubility
Endocrine Disruptor Information This product does not contain any known or suspected endocrine disruptors
Persistent Organic Pollutant This product does not contain any known or suspected substance
Ozone Depletion Potential This product does not contain any known or suspected substance
SECTION 13. DISPOSAL CONSIDERATIONS
Waste from Residues/Unused
Products
Chemical waste generators must determine whether a discarded chemical is classified as a
hazardous waste. Consult local, regional, and national hazardous waste regulations to
ensure complete and accurate classification.
Contaminated Packaging Empty remaining contents. Dispose of in accordance with local regulations. Do not re-use
empty containers.
Other Information Waste codes should be assigned by the user based on the application for which the product
was used.
SECTION 14. TRANSPORT INFORMATION
Road and Rail Transport Not Regulated
IMDG/IMO Not regulated
IATA Not regulated
Special Precautions for User No special precautions required
SECTION 15. REGULATORY INFORMATION
International Inventories
X = listed, China (IECSC), Europe (EINECS/ELINCS/NLP), U.S.A. (TSCA), Canada (DSL/NDSL), Philippines (PICCS), Japan (ENCS), Japan
(ISHL), Australia (AICS), Korea (KECL).
Component The
Inventory of
Hazardous
Chemicals
(2015
Edition)
List of
dangerous
goods GB
12268 -
2012
TCSI IECSC EINECS TSCA DSL PICCS ENCS ISHL AICS KECL
Aluminum fluoride --XX232-051-1 XXXXXXKE-00970
National Regulations
Page 6Aluminum fluoride, anhydrous
ALFAA44435 SAFETY DATA SHEET Page 7 / 7
Revision Date 17-Sep-2025
SECTION 16. OTHER INFORMATION
Prepared By Health, Safety and Environmental Department
Creation Date 06-Mar-2014
Revision Date 17-Sep-2025
Revision Summary Not applicable.
Training Advice
Chemical hazard awareness training, incorporating labelling, Safety Data Sheets (SDS), Personal Protective Equipment (PPE) and
hygiene.
Legend
Key literature references and sources for data
Suppliers safety data sheet, Chemadvisor - LOLI, Merck index, RTECS
Disclaimer
The information provided in this Safety Data Sheet is correct to the best of our knowledge, information and belief at the
date of its publication. The information given is designed only as a guidance for safe handling, use, processing, storage,
transportation, disposal and release and is not to be considered a warranty or quality specification. The information
relates only to the specific material designated and may not be valid for such material used in combination with any other
materials or in any process, unless specified in the text
End of Safety Data Sheet
CAS - Chemical Abstracts Service TSCA - United States Toxic Substances Control Act Section 8(b)
Inventory
EINECS/ELINCS - European Inventory of Existing Commercial Chemical
Substances/EU List of Notified Chemical Substances
DSL/NDSL - Canadian Domestic Substances List/Non-Domestic
Substances List
PICCS - Philippines Inventory of Chemicals and Chemical Substances ENCS - Japanese Existing and New Chemical Substances
IECSC - Chinese Inventory of Existing Chemical Substances AICS - Australian Inventory of Chemical Substances
KECL - Korean Existing and Evaluated Chemical Substances NZIoC - New Zealand Inventory of Chemicals
WEL - Workplace Exposure Limit TWA - Time Weighted Average
ACGIH - American Conference of Governmental Industrial Hygienists IARC - International Agency for Research on Cancer
DNEL - Derived No Effect Level PNEC - Predicted No Effect Concentration
RPE - Respiratory Protective Equipment LD50 - Lethal Dose 50%
LC50 - Lethal Concentration 50% EC50 - Effective Concentration 50%
NOEC - No Observed Effect Concentration POW - Partition coefficient Octanol:Water
PBT - Persistent, Bioaccumulative, Toxic vPvB - very Persistent, very Bioaccumulative
ICAO/IATA - International Civil Aviation Organization/International Air
Transport Association
IMO/IMDG - International Maritime Organization/International Maritime
Dangerous Goods Code
ADR - European Agreement Concerning the International Carriage of
Dangerous Goods by Road
MARPOL - International Convention for the Prevention of Pollution from
Ships
OECD - Organisation for Economic Co-operation and Development ATE - Acute Toxicity Estimate
BCF - Bioconcentration factor VOC - (Volatile Organic Compound)
Page 7 |
9903 | http://maecourses.ucsd.edu/~kseshadr/mae113-s109/homework/MAE113_HW1_solution.pdf | MAE 113, Summer Session 1, 2009 HW #1 1.2, 1.7, 1.14, 2.3, 2.6 1.2 Develop the following analytical expressions for a turbojet engine: a) When m ° f << m ° o, Pe = Pa, and finlet = fnoz = 0, then the installed thrust is given by: T = m ° o gc HVe - V0L From equation 1.5, F = Im ° o + m ° f M Ve - m ° 0 V0 gc + HPe - P0L Ae when we apply m ° f << m ° o and Pe = Pa, we get F = m ° o Ve - m ° 0 V0 gc From equation 1.9, T = FH1 - finlet - fnozL but, finlet = fnoz = 0, so T=F. Thus, T = m ° o gc HVe - V0L b) By the same conditions, show TSFC = Tgcêm ° o + 2 V0 2 hT hPR From equation 1.20, TSFC = V0 hP hT hPR and from equation 1.16, hP = 2 VeêV0+1 from part (a), T = m ° o gc HVe - V0L T gc m ° o = Ve - V0 T gc m ° o V0 = Ve V0 - 1 Ve V0 = T gc m ° o V0 + 1 Printed by Mathematica for Students plugging this into 1.16 hP = 2 T gc m ° o V0 +1+1 hP J T gc m ° o V0 + 2N = 2 hP V0 J T gc m ° o + 2 V0N = 2 hP V0 = 2 T gc m ° o +2 V0 V0 hP = T gc m ° o +2 V0 2 and, finally, this goes into equation 1.20 to get TSFC = T gc m ° o +2 V0 2 hT hPR c) For V0 = 0 and 500 ft/s, plot the preceding equation for TSFC [in (lbm/h)/lbf] vs specific thrust T/m ° o [in lbf/(lbm/s)] for values of specific thrust from 0 to 120. Use hT = 0.4and hPR = 18, 400 Btu/lbm.
It is very easy to mess up the units of this problem. Note that the input variable, T/m ° o, is in lbf/(lbm/sec), but the output variable is in (lbm/hour)/lbf. You must multiply by the number of seconds in an hour to output the correct units. Also, you must use the conversion 1btu=778.16 ft·lbf.
You can set up an equation of the form TSFC = I3600 sec hourM JX lbf lbmêsN J32.174 lbmÿft lbfÿs2 N+2 J0 or 500 ft s N 2 H0.4L J18,400 Btu lbm 778.16 ftÿlbf Btu N , where X = T m ° o is the input variable Here's Matlab code you can use to make the plot: X=0:0.1:120; TSFC0=3600(X32.174+20)/(20.418400778.16); TSFC500=3600(X32.174+2500)/(20.418400778.16); plot(X,TSFC0,X,TSFC500) 0 20 40 60 80 100 120 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Specific Thrust lbf/(lbm/s) TSFC (lbm/hr)/lbf 2 MAE 113 HW1 solution-2.nb Printed by Mathematica for Students The blue line is the V0 = 0 line and the green line is the V0 = 500 ft ê s line.
d) explain the trends You're on your own here. Say something about how TSFC is always higher when the inlet velocity is higher, and that TSFC increases linearly with specific thrust.
MAE 113 HW1 solution-2.nb 3 Printed by Mathematica for Students 1.7 The JT9D high-bypass-ratio turbofan engine with V0 = 0, P0 = 14.696 psia, T0 = 518.7 °R, and m ° C = 247 lbm ê s, m ° B = 1248 lbm ê s, VCe = 1190 ft ê s, VBe = 885 ft ê s, m ° f = 15 750 lbm ê hr. Estimate the following assuming P0 = Pe : a) Thrust From Problem 1.5, we learn that thrust for a bypass engine is equal to the sum of the thrust from the core and the bypass stream.
F = FC + FB FC = 1 gc AIm ° C + m ° f M VCe - m ° C V0E FB = m ° B gc HVBe - V0L Putting all these together, we get an equation for the thrust for a bypass engine F = 1 gc AIm ° C + m ° f M VCe - m ° C V0 + m ° B VBe - m ° B V0E We have values for all the variables on the right hand side, so we can calculate thrust, but again be very careful with units.
F = 1 32.174 lbmÿft lbfÿs2 AI247 lbm s + 15 750 lbm hour hour 3600 s M 1190 ft s - 247 lbm s H0L + 1248 lbm s 885 ft s - 1248 lbm s H0LE F = 43 626 lbf b) Thermal efficiency, hT, with hPR º 18, 400 Btu lbm Equation 1.13 tells us that for a single inlet and single exhaust hT = W ° out Q ° in W ° out, singleexhaust = 1 2 gc AIm ° o + m ° f M Ve 2 - m ° o V0 2E Q ° in = m ° f hPR here, though, we have two inlets and two exhausts, therefore W ° out, with bypass = W ° Cout + W ° Bout W ° out = 1 2 gc AIm ° C + m ° f M VCe 2 - m ° C V0 2E + 1 2 gc AHm ° BL VBe 2 - m ° B V0 2E but V0 = 0, so W ° out = 1 2 gc AIm ° C + m ° f M VCe 2 + m ° B VBe 2 E inserting this into the first equation gives us the result that hT = AIm ° C+m ° f M VCe 2 +m ° B VBe 2 E 2 gc m ° f hPR 4 MAE 113 HW1 solution-2.nb Printed by Mathematica for Students and now we can put in values hT = BJ247 lbm s +15750 lbm hr hr 3600 sN J1190 ft s N 2 +1248 lbm s J885 ft s N 2 F 2 J32.174 lbmÿft lbfÿs2 N J15750 lbm h h 3600 sN J18,400 Btu lbm 778.16 ftÿlbf Btu N hT = 0.3308 hT = 33.08 % c) Propulsive efficiency, hP, and uninstalled thrust specific fuel consumption, S As defined in equation 1.14, hP = T V0 W ° out And since V0 = 0, hP = 0 From equation 1.10, S = m ° f F S = 15750 lbm hr 43626 lbf S = 0.361 lbmêhr lbf = 1.0028 ÿ 10-4 lbmês lbf MAE 113 HW1 solution-2.nb 5 Printed by Mathematica for Students 1.14 An aircraft with wind area 800 ft2 in level flight at maximum CL ê CD. CD0 = 0.02, K2 = 0, K1 = 0.2, find a) The maximum CL ê CD and corresponding CL and CD Equation 1.48 is used here J CL CD N = 1 2 CD0 K1 +K2 J CL CD N = 1 2 H0.02L H0.2L +0 J CL CD N = 7.906 Also, equation 1.47 says that CL = CD0 K1 CL = 0.02 0.2 CL = 0.3162 now we can find CD 0.3162 CD = 7.906 CD = 0.04 b) The flight altitude and drag for aircraft weight of 45,000 lbf and Mach 0.8. Use eqns 1.29 and 1.30b.
Equations 1.29 and 1.30b are L = n W = CL q Sw q = g 2 P M0 2 = g 2 d Pref M0 2 Solving for d and combining the two equations, d = 2 q g Pref M0 2 d = 2 J n W CL Sw N g Pref M0 2 d = 2 n W g Pref M0 2 CL Sw Now we can plug in values d = 2 H1L H45000 lbfL 1.4 J14.7 lbf in2 144 in2 ft2 N H0.8L2 H0.3162L I800 ft2M d = 0.1876 6 MAE 113 HW1 solution-2.nb Printed by Mathematica for Students Use Appendix A to see that this corresponds to an altitude of about 39, 800 ft .
Next, equation 1.31 gives us drag D = CD q Sw D = CD n W CL D = 0.04 H1L H45000 lbfL 0.3162 D = 5692.6 lbf c) Flight altitude and drag for an aircraft of weight 35,000 lbf and Mach 0.8 Analysis is nearly identical to part b, with only a change in weight.
d = 2 n W g Pref M0 2 CL Sw d = 2 H1L H35000 lbfL 1.4 J14.7 lbf in2 144 in2 ft2 N H0.8L2 H0.3162L I800 ft2M d = 0.1459 Again, use Appendix A to see that this corresponds to an altitude of about 45, 000 ft .
Also, D = CD n W CL D = 0.04 H1L H35000 lbfL 0.3162 D = 4427.6 lbf d) Range for an installed engine TSFC rate of 0.8 (lbm/hr)/lbf, if the 10,000-1bf difference in aircraft weight between parts b and c is due only to fuel consumption.
First start with equation 1.43 for range factor RF = CL CD V TSFC gc g0 We need to calculate velocity in ft/s rather than Mach. Use appendix A and the note that speed of sound a=astd q .
V = a M V = 1116 ft s 0.7519 0.8 V = 774 ft s MAE 113 HW1 solution-2.nb 7 Printed by Mathematica for Students Now we can go back to RF RF = 0.3162 0.04 774 ft s 0.8 lbmêhr lbf hr 3600 s 32.174 lbmÿft lbfÿs2 32.174 lbmÿft lbfÿs2 RF = 27 533 115 ft nm 6080 ft RF = 4528.5 nm Next, we use equation 1.45a to find the range, s W f Wi = expI- s RFM s = RF ln J Wi W f N s = 4528.5 nm lnI 45000 lbf 35000 lbf M s = 1138 nm = 1309.6 mi 8 MAE 113 HW1 solution-2.nb Printed by Mathematica for Students 2.3 Consider the flow shown in Figure P2.2. It has radius r0, velocity V1, and pressure P 1. The fluid leaves with a velocity V2 = VmaxB1 - I r r0 M 2F with uniform pressure P2. Use the conservation of mass and momentum equations to show that the force necessary to hold the pipe in place can be written as F = p r0 2 JP1 - P2 + r V1 2 3 gc N Start with equation 2.20 ⁄Fs = 1 gc I dMs dt + M ° out - M ° inM by assuming conservation of mass (since there is no sink or source), dMs dt = 0.
⁄Fs = 1 gc IM ° out - M ° inM The sum of the forces is ⁄Fs = -F + HP1 - P2L A Where F is the force required to keep the pipe in place and A is the area of the pipe.
We can combine these two equations to get ⁄Fs = 1 gc IM ° out - M ° inM = -F + HP1 - P2L A F = 1 gc IM ° in - M ° outM + HP1 - P2L A These equations are wrong.
However, the first part of this equation does not make sense. Force should not be proportional to M ° in - M ° out, because that would imply that if M ° in > M ° out, the force would be postive. However, M ° in > M ° out implies that air is moving from back to front, which should decrease the force required to keep the pipe steady. This means that we have defined our sum of forces to be in a different direction in each of the equations.
Instead, the equation should read: F = 1 gc IM ° out - M ° inM + HP1 - P2L A MAE 113 HW1 solution-2.nb 9 Printed by Mathematica for Students Momentum flux can be obtained by integrating M ° out = M ° 2 = Ÿm ° 2V2 „m ° M ° 2 = Ÿ0 r0r V2 2 2 p r „r M ° 2 = Ÿ0 r0r JVmaxB1 - I r r0 M 2FN 2 2 p r „r M ° 2 = Ÿ0 r0r Vmax2J1 - I r r0 M2N 2 2 p r „r M ° 2 = 2 p r Vmax2 Ÿ0 r0J1 - 2 I r r0 M 2 + I r r0 M 4N r „r M ° 2 = 2 p r Vmax2 Ÿ0 r0Jr - 2 r3 r02 + r5 r04 N „r M ° 2 = 2 p r Vmax2B r2 2 - 2 r4 4 r02 + r6 6 r04 F 0 r0 M ° 2 = 2 p r Vmax2B r02 2 - 2 r04 4 r02 + r06 6 r04 F M ° 2 = 2 p r Vmax2B r02 2 - r02 2 + r02 6 F M ° 2 = p r Vmax 2 r02 3 Similarly, M ° in = M ° 1 = Ÿm ° 1V1 „m ° M ° 1 = Ÿ0 r0r V1 2 2 p r „r M ° 1 = r V1 2 2 p B r2 2 F 0 r0 M ° 1 = r V1 2 p r02 And plugging into to the equation for force above F = 1 gc IM ° out - M ° inM + HP1 - P2L A F = p r02 :P1 - P2 -r gc J Vmax 2 3 - V12 N> Conservation of mass can be used to find Vmax m ° 1 = m ° 2 p r0 2 r V1 = r Ÿ0 r0Vmax :1 - I r r0 M 2> 2 p r „r V1 = 2 Vmax r0 2 Ÿ0 r0:r - r3 r02 > „r V1 = 2 Vmax r0 2 B r2 2 -r4 4 r02 F 0 r0 V1 = 2 Vmax r0 2 B r0 2 2 -r0 4 4 r02 F V1 = Vmax 2 Vmax = 2 V1 10 MAE 113 HW1 solution-2.nb Printed by Mathematica for Students And we arrive at F = p r02 :P1 - P2 + r gc J H2 V1L2 3 - V1 2 N> F = p r02 :P1 - P2 + r V1 2 gc I 4 3 - 1M> F = p r02 :P1 - P2 + r V1 2 gc I 4 3 - 1M> F = p r02 :P1 - P2 + r V1 2 3 gc > MAE 113 HW1 solution-2.nb 11 Printed by Mathematica for Students 2.6 Using Figure P2.5, with 1500 lbm/s of air at 60°F and 14.7 psia entering the engine at a velocity of 450 ft/s and that 1250 lbm/s of bypass air leaves the engine at 60° to the horizontal, at a velocity of 890 ft/s and pres-sure of 14.7 psia. The remaining 250 lbm/s leaves the engine core at a velocity of 1200 ft/s and pressure of 14.7 psia. Determine the force on the strut, Fx. Assume an ambient pressure of 14.7 psia.
By symmetry, there is no net force in the y-direction. The momentum equation in the x-direction is ⁄Fx = Fx = 1 gc IM ° x core out + M ° x fan out - M ° x inM M ° x in = I1500 lbm s M I450 ft s M = 675 000 ftÿlbm s2 M ° x core out = I250 lbm s M I1200 ft s M = 300 000 ftÿlbm s2 M ° x fan out = I-1250 lbm s M I890 ft s M cosH60 °L = -556 250 ftÿlbm s2 Therefore, the force is Fx = 1 32.174 lbmÿft lbfÿs2 B300 000 ftÿlbm s2 + J-556 250 ftÿlbm s2 N - 675 000 ftÿlbm s2 F Fx = -28, 945 lbf This means a force of 28,945 lbf slowing the plane.
12 MAE 113 HW1 solution-2.nb Printed by Mathematica for Students |
9904 | https://arxiv.org/pdf/2002.05680 | An Almost Optimal Bound on the Number of Intersections of Two Simple Polygons
Eyal Ackerman
Department of Mathematics, Physics, and Computer Science, University of Haifa at Oranim, Tivon 36006, Israel ackerman@sci.haifa.ac.il
Balázs Keszegh
Alfréd Rényi Institute of Mathematics, H-1053 Budapest, Hungary MTA-ELTE Lendület Combinatorial Geometry Research Group keszegh@renyi.hu
Günter Rote
Department of Computer Science, Freie Universität Berlin, Takustr. 9, 14195 Berlin, Germany rote@inf.fu-berlin.de
Abstract
What is the maximum number of intersections of the boundaries of a simple m-gon and a simple
n-gon, assuming general position? This is a basic question in combinatorial geometry, and the answer is easy if at least one of m and n is even: If both m and n are even, then every pair of sides may cross and so the answer is mn . If exactly one polygon, say the n-gon, has an odd number of sides, it can intersect each side of the m-gon at most n − 1 times; hence there are at most mn − m
intersections. It is not hard to construct examples that meet these bounds. If both m and n are odd, the best known construction has mn − (m + n) + 3 intersections, and it is conjectured that this is the maximum. However, the best known upper bound is only mn − (m + d n
6
e), for m ≥ n. We prove a new upper bound of mn − (m + n) + C for some constant C, which is optimal apart from the value of C.
2012 ACM Subject Classification Theory of computation → Computational geometry; Mathematics of computing → Combinatorial problems
Keywords and phrases Simple polygon, Ramsey theory, combinatorial geometry
Funding Eyal Ackerman : The main part of this work was performed during a visit to Freie Universität Berlin which was supported by the Freie Universität Alumni Program.
Balázs Keszegh : Research supported by the Lendület program of the Hungarian Academy of Sciences (MTA), under the grant LP2017-19/2017 and by the National Research, Development and Innovation Office – NKFIH under the grant K 116769.
1 Introduction
To determine the union of two or more geometric objects in the plane is one of the basic computational geometric problems. In strong relation to that, determining the maximum complexity of the union of two or more geometric objects is a basic extremal geometric problem. We study this problem when the two objects are simple polygons. Let P and Q be two simple polygons with m and n sides, respectively, where m, n ≥ 3.For simplicity we always assume general position in the sense that no three vertices (of P
and Q combined) lie on a line and no two sides (of P and Q combined) are parallel. We are interested in the maximum number of intersections of the boundaries of P and Q.This naturally gives an upper bound for the complexity of the union of the polygon areas as well. (In the worst case all the m + n vertices of the two polygons contribute to the complexity of the boundary in addition to the intersection points.)
This paper is to appear in the proceedings of the 36th International Symposium on Computational Geometry
(SoCG 2020) in June 2016; Editors: Sergio Cabello and Danny Chen, Leibniz International Proceedings in Informatics.
arXiv:2002.05680v1 [math.CO] 13 Feb 2020
The Number of Intersections Between Two Simple Polygons 2(c) (b) PPQQ(a) PQ
Figure 1 (a) Optimal construction for m=n= 8 , with 8×8 = 64 intersections. (b) Optimal construction for m= 8 ,n= 7 , with 8×6 = 48 intersections. (c) Lower-bound construction for
m= 9 ,n= 7 . There are 8×6 + 2 = 50 intersections.
This problem was first studied in 1993 by Dillencourt, Mount, and Saalfeld [ 2]. The cases when m or n is even are solved there. If m and n are both even, then every pair of sides may cross and so the answer is mn . Figure 1a shows one of many ways to achieve this number. If one polygon, say Q, has an odd number n of sides, no line segment s can be intersected
n times by Q, because otherwise each side of Q would have to flip from one side of s to the other side. Thus, each side of the m-gon P is intersected at most n − 1 times, for a total of at most mn − m intersections. It is easy to see that this bound is tight when P has an even number of sides, see Figure 1b. When both m and n are odd, the situation is more difficult; the bound that is obtained by the above argument remains at mn − max {m, n }, because the set of m intersections that are necessarily “missing” due to the odd parity of n might conceivably overlap with the
n intersections that are “missing” due to the odd parity of m. However, the best known family of examples gives only mn − (m + n) + 3 = ( m − 1)( n − 1) + 2 intersection points, see Figure 1c. Note that in Figure 1, all vertices of the polygons contribute to the boundary of the union of the polygon areas.
I Conjecture 1. Let P and Q be simple polygons with m and n sides, respectively, such that
m, n ≥ 3 are odd numbers. Then there are at most mn − (m + n) + 3 intersection points between sides of P and sides of Q.
In [ 2] an unrecoverable error appears in a claimed proof of Conjecture 1. Another attempted proof [ 5 ] also turned out to have a fault. The only correct improvement over the trivial upper bound is an upper bound of mn − (m + d n
6
e) for m ≥ n, due to Černý, Kára, Král’, Podbrdský, Sotáková, and Šámal . We will briefly discuss their proof in Section 2. We improve the upper bound to mn − (m + n) + O(1) , which is optimal apart from an additional constant:
I Theorem 1. There is an absolute constant C such that the following holds. Suppose that
P and Q are simple polygons with m and n sides, respectively, such that m and n are odd numbers. Then there are at least m + n − C pairs of a side of P and a side of Q that do not intersect. Hence, there are at most mn − (m + n) + C intersections.
The value of the constant C that we obtain in our proof is around 2267
. We did not make a large effort to optimize this value, and obviously, there is ample space for improvement. E. Ackerman, B. Keszegh, and G. Rote 3III II IIV L(e2) = aL(e1) = ∗L(e5) = bL(e4) = aL(e3) = b
Figure 2 The edge-labeled multi-graph G0in Proposition 2.
IbII IaIV L(e2) = aL(e1) = ∗L(e5) = bL(e4) = aL(e3) = bIII aIII b
Figure 3 The unfolded graph G′
0
2 Overview of the Proof
First we establish the crucial statement that the odd parity of m and n allows us to associate
to any two consecutive sides of one polygon a pair of consecutive sides of the other polygon with a restricted intersection pattern among the four involved sides (Lemma 5 and Figure 5). This is the only place where we use the odd parity of the polygons. A simple observation (Observation 3) relates the bound on C in Theorem 1 to the number of connected components of the bipartite “disjointness graph” between the polygon sides of
P and Q. Our goal is therefore to show that there are few connected components. We proceed to consider two pairs of associated pairs of sides (4 consecutive pairs with 8 sides in total). Unless they form a special structure, they cannot belong to four different connected components (Lemma 7). (Four is the maximum number of components that they could conceivably have.) The proof involves a case distinction with a moderate amount of cases. This structural statement allows us to reduce the bound on the number of components by a constant factor, and thereby, we can already improve the best previous result on the number of intersections (Proposition 9 in Section 6). Finally, to get a constant bound on the number of components, our strategy is to use Ramsey-theoretic arguments like the Erdős–Szekeres Theorem on caps and cups or the pigeonhole principle (see Section 7) in order to impose additional structure on the configurations that we have to analyze. This is the place in the argument where we give up control over the constant C in exchange for useful properties that allow us to derive a contradiction. This eventually boils down again to a moderate number of cases (Section 8.2). By contrast, the proof of the bound mn − (m + d n
6
e) for m ≥ n by Černý et al. proceeds in a more local manner. The core of their argument [ 1 , Lemma 3], which is proved by case distinction, is that it is impossible to have 6 consecutive sides of one polygon together with 6 distinct sides of the other polygon forming a perfect matching in the disjointness graph. This statement is used to bound the number of components of the disjointness graph. (Lemma 8 below uses a similar argument.)
3 An Auxiliary Lemma on Closed Odd Walks
We begin with the following seemingly unrelated claim concerning a specific small edge-labeled multigraph. Let G0 = ( V0, E 0) be the undirected multigraph shown in Figure 2. It has four nodes V0 = {I, II , III , IV } and five edges E0 = {e1 = {II , IV }, e 2 = {I, IV }, e 3 = {I, II }, e 4 =
{I, III }, e 5 = {I, III }} . Every edge ei ∈ E0 has a label L(ei) ∈ { a, b, ∗} as follows: L(e1) = ∗,
L(e2) = L(e4) = a, L(e3) = L(e5) = b.
I Proposition 2. If W is a closed walk in G0 of odd length, then W contains two cyclically consecutive edges of labels a and b.The Number of Intersections Between Two Simple Polygons 4
Proof. Suppose for contradiction that W does not contain two consecutive edges of labels a
and b. Since W cannot switch between the a-edges and the b-edges in I or III, we can split I
(resp., III ) into two nodes Ia and III b (resp., III a and III b) such that every a-labeled edge that is incident to I (resp., III ) in G0 becomes incident to Ia (resp., III a) and every b-labeled edge that is incident to I (resp., III ) in G0 becomes incident to Ib (resp., III b). In the resulting graph G′
0
, which is shown in Figure 3, we can find a closed walk W ′ that corresponds to W
and that uses the edges with the same name as W . Since G′
0
is a path, every closed walk has even length. Thus, W cannot have odd length. J
4 General Assumptions and Notations
Let P and Q be two simple polygons with sides p0, p 1, . . . , p m−1 and q0, q 1, . . . , q n−1. We assume that m ≥ 3 and n ≥ 3 are odd numbers. Addition and subtraction of indices is modulo m or n, respectively. We consider the sides pi and qj as closed line segments. The condition that the polygon P is simple means that its edges are pairwise disjoint except for the unavoidable common endpoints between consecutive sides pi and pi+1 . Throughout this paper, unless stated otherwise, we regard a polygon as a piecewise linear closed curve, and we disregard the region that it encloses. Thus, by intersections between P and Q, we mean intersection points between the polygon boundaries.
As mentioned, we assume that the vertices of P and Q are in general position (no three of them on a line), and so every intersection point between P and Q is an interior point of two polygon sides.
The Disjointness Graph. As in [ 1 ], our basic tool of analysis is the disjointness graph of
P and Q, which we denote by GD = ( V D, E D). (Its original name in [ 1] is non-intersection graph .) It is a bipartite graph with node set V D = {p0, p 1, . . . , p m−1} ∪ { q0, q 1, . . . , q n−1} and edge set ED = { (pi, q j ) | pi ∩ qj = ∅ } . (Since we are interested in the situation where almost all pairs of edges intersect, the disjointness graph is more useful than its more commonly used complement, the intersection graph.) Our goal is to bound from above the number of connected components of GD.
I Observation 3. If GD has at most C connected components, then GD has at least m+n−C
edges. Thus, there are at least m + n − C pairs of a side of P and a side of Q that do not intersect, and there are at most mn − (m + n) + C crossings between P and Q. J
Geometric Notions. Let s and s′ be two line segments. We denote by `(s) the line through
s and by I(s, s ′) the intersection of (s) and(s′) see Figure 4. We say that s and s′
are avoiding if neither of them contains I(s, s ′). (This requirement is stronger than just disjointness.) If s and s′ are avoiding or share an endpoint, we denote by ~rs′ (s) the ray from I(s, s ′) to infinity that contains s, and by ~rs(s′) the ray from I(s, s ′) to infinity that contains s′. Moreover, we denote by Cone (s, s ′) the convex cone with apex I(s, s ′) between these two rays.
I Observation 4. If a segment s′′ that does not go through I(s, s ′) has one of its endpoints in the interior of Cone (s, s ′), then s′′ cannot intersect both ~rs′ (s) and ~rs(s′). In particular, it cannot intersect both s and s′. J
For a polygon side s of P or Q, CC (s) denotes the connected component of the disjointness graph GD to which s belongs. E. Ackerman, B. Keszegh, and G. Rote 5
4.1 Associated Pairs of Consecutive Sides
I Lemma 5. Let pa and pb be two sides of P that are either consecutive or avoiding such that CC (pa) 6 = CC (pb). Then there are two consecutive sides qi, q i±1 of Q such that
(pa, q i), (pb, q i±1) ∈ ED and (pa, q i±1), (pb, q i) /∈ ED. Furthermore, I(pa, p b) ∈ Cone (qi, q i±1)
or I(qi, q i±1) ∈ Cone( pa, p b).
The sign ‘ ±’ is needed since we do not know which of the consecutive sides intersects pi
and is disjoint from pi+1 .
Proof. We may assume without loss of generality that I(pa, p b) is the origin, pa lies on the positive x-axis and the interior of pb is above the x-axis. The lines (pa) and(pb) partition the plane into four convex cones (“quadrants”). Denote them in counterclockwise order by
I, II , III , IV , starting with I = Cone (pa, p b), see Figure 4. Every side of Q must intersect papapbI = Cone( pa, p b)II III IV QI(pa, p b)~rpa (pb)ss′I(s, s ′)~rpb (pa)(s)(s′)
Figure 4 How an odd polygon Q can intersect two segments. The segments pa and pb are avoiding, whereas s and s′ are disjoint but non-avoiding.
or pb (maybe both), since CC (pa) 6 = CC (pb). One can now check that traversing the sides of Q in order generates a closed walk W in the graph G0 of Figure 2. For example, a side of Q that we traverse from its endpoint in I to its endpoint in III and that intersects pa
corresponds to traversing the edge e4 = {I, III } from I to III , whose label is L(e4) = a. We do not care which of pa and pb are crossed when we move between II and IV .It follows from Proposition 2 that Q has two consecutive sides qi, q i±1 such that qi
intersects pb and does not intersect pa, while qi±1 intersects pa and does not intersect pb.Hence, (pa, q i), (pb, q i±1) ∈ ED and (pa, q i±1), (pb, q i) /∈ ED. Furthermore, I(qi, q i±1) must be either in I or III as these are the only nodes in G0 that are incident both to an edge labeled a and an edge labeled b. In the latter case I(pa, p b) ∈ Cone (qi, q i±1), and in the former case I(qi, q i±1) ∈ Cone( pa, p b). J
Let pi, p i+1 be two sides of P such that CC (pi) 6 = CC (pi+1 ). Then by Lemma 5 there are sides qj , q j±1 of Q such that (pi, q j ), (pi+1 , q j±1) ∈ ED. We say that the pair qj , q j±1
is associated to pi, p i+1 . By Lemma 5 we have I(qj , q j±1) ∈ Cone (pi, p i+1 ) or I(pi, p i+1 ) ∈
Cone (qj , q j±1). If the first condition holds we say that pi, p i+1 is hooking and qj , q j±1 is
hooked , see Figure 5. In the second case we say that pi, p i+1 is hooked and qj , q j±1 is hooking .Note that it is possible that a pair of consecutive sides is both hooking and hooked (with respect to two different pairs from the other polygon or even with respect to a single pair, as in Figure 5c).
I Observation 6 (The Axis Property) . If the pair pi, p i+1 and the pair qj , q j±1 are associated such that (pi, q j ), (pi+1 , q j±1) ∈ ED, then the line through I(pi, p i+1 ) and I(qj , q j±1) separates
pi and qj±1 on the one side from pi+1 and qj on the other side. JThe Number of Intersections Between Two Simple Polygons 6pipi+1 qjqj±1pipi+1 qjqj±1pipi+1 qjqj±1(a) (b) (c)
Figure 5 Hooking and hooked pairs of consecutive sides. (a) The pair pi, p i+1 is hooking and the associated pair qj , q j±1 is hooked. (b) vice versa. (c) Both pairs are both hooking and hooked.
We call this line the axis of the associated pairs. In our figures it appears as a dotted line when it is shown.
5 The Principal Structure Lemma about Pairs of Associated Pairs
I Lemma 7. Let pi, p i+1 , p j , p j+1 be two pairs of consecutive sides of P that belong to four different connected components of GD. Then it is impossible that both pi, p i+1 and pj , p j+1
are hooked or that both pairs are hooking.
Proof. Suppose first that both pairs pi, p i+1 and pj , p j+1 , are hooking and let qi′ , q i′±1
and qj′ , q j′±1 be their associated (hooked) pairs such that: (pi, q i′ ), (pi+1 , q i′±1) ∈ ED,
(pj , q j′ ), (pj+1 , q j′±1) ∈ ED, I(qi′ , q i′±1) ∈ Cone( pi, p i+1 ) and I(qj′ , q j′±1) ∈ Cone( pj , p j+1 ).For better readability, we rename pi, p i+1 and qi′ , q i′±1 as a, b and A, B , and we rename
pj , p j+1 and qj′ , q j′±1 as a′, b ′ and A′, B ′. The small letters denote sides of P and the capital letters denote sides of Q. In the new notation, a, b are consecutive sides of P with an associated pair A, B of consecutive sides of Q, and a′, b ′ are two other consecutive sides of P with an associated pair A′, B ′ of consecutive sides of Q. The disjointness graph GD
contains the edges (a, A ), (b, B ), (a′, A ′), (b′, B ′). Since a, b, a ′, b ′ belong to different connected components of GD, it follows that the nodes A, B, A ′, B ′, to which they are connected, belong to the same four different connected components. There can be no more edges among these eight nodes, and they induce a matching in GD. One can remember as a rule that every side of P intersects every side of Q among the eight involved sides, except when their names differ only in their capitalization. In particular, each of A′ and B′ intersects each of a and b.and hence they must lie as in Figure 6a. To facilitate the future discussion, we will now normalize the positions of these four sides. We first ensure that the intersection I(A′, b ) is directly adjacent to the two polygon vertices I(a, b ) and I(A′, B ′) in the arrangement of the four sides, as shown in Figure 6b. This can be achieved by swapping the labels a, A with the labels b, B if necessary, and by independently swapping the labels a′, A ′ with b′, B ′ if necessary. Our assumptions are invariant under these swaps. By an affine transformation we may finally assume that I(A′, b ) is the origin; b lies on the
x-axis and is directed to the right; and A′ lies on the y-axis and is directed upwards. Then a
has a positive slope and its interior is in the upper half-plane, and B′ has a positive slope and its interior is to the right of the y-axis, see Figure 6c. The arrangement of the lines through a, b, A ′, B ′ has 11 faces, some of which are marked as F1, . . . , F 6 in Figure 6. Our current assumption is that both a, b and a′, b ′ are hooking: E. Ackerman, B. Keszegh, and G. Rote 7F1F2F3F4F5F6abB′A′baA′B′I(A′, B ′)I(a, b )I(A′, b )baA′B′(a) (b) (c)
Figure 6 Normalizing the position of a, b, A ′, B ′raF1F2F3F4F5F6abB′A′Ba′Arb
Figure 7 Case 1: I(A, B ) ∈ F1, I(a′, b ′) ∈ F2F1F2F3F4F5F6abB′A′Ba′b′Arb Figure 8 Case 2: I(A, B ) ∈ F1, I(a′, b ′) ∈ F4
The hooking of a, b means that I(A, B ) ∈ Cone (a, b ) = F1 ∪ F2 ∪ F3. By the Axis Property (Observation 6), the line through I(A′, B ′) and I(a′, b ′) must separate A′ from B′. Therefore, the vertex I(a′, b ′) can lie only in F2 ∪ F4 ∪ F5 ∪ F6. Thus, based on the faces that contain
I(A, B ) and I(a′, b ′), there are 12 cases to consider. Some of these cases are symmetric, and all can be easily dismissed, as follows. In the figures, the four sides a′, b ′, A ′, B ′, which are associated to the second associated pair are dashed. All dashed sides of one polygon must intersect all solid sides of the other polygon.
I(A, B ) ∈ F1 and I(a′, b ′) ∈ F2, see Figure 7 (symmetric to I(A, B ) ∈ F2 and I(a′, b ′) ∈
F4). Let ra (resp., rb) be the ray on (a) (resp.,(b)) that goes from the right endpoint of
a (resp., b) to the right. Since a′ is not allowed to cross b, the only way for a′ to intersect
A is by crossing rb. Similarly, in order to intersect B, a′ has to cross ra. However, it cannot intersect both ra and rb, by Observation 4. Since we did not use the assumption that A, B are hooked, the analysis holds for the symmetric Case 6, I(A, B ) ∈ F2 and I(a′, b ′) ∈ F4, as well.
I(A, B ) ∈ F1 and I(a′, b ′) ∈ F4, see Figure 8. Since a′ is not allowed to cross b, the only way for a′ to intersect B is by crossing rb. However, in this case a′ cannot intersect A.
I(A, B ) ∈ F1 and I(a′, b ′) ∈ F5, see Figure 9 (symmetric to I(A, B ) ∈ F3 and I(a′, b ′) ∈
F4). Both a′ and b′ must intersect A, and they have to go below the line (b) to do so. However, a′ can only cross(b) to the right of b, and b′ can only cross `(b) to the left of b,The Number of Intersections Between Two Simple Polygons 8F1F2F3F4F5F6abB′A′Ba′Ab′
Figure 9 Case 3: I(A, B ) ∈ F1 and I(a′, b ′) ∈ F5F1F2F3F4F5F6abB′A′Ba′Ab′
(a) At least one of the sides a′ and b′ has an endpoint in F4.F1F2F3F4F5F6abB′A′Ba′Ab′
(b) None of the sides a′ and b′ has an endpoint in F4.
Figure 10 Case 4: I(A, B ) ∈ F1 (or I(A, B ) ∈ F2, which is similar) and I(a′, b ′) ∈ F6.
and therefore they cross A from different sides. This is impossible, because a′ and b′ start from the same point.
I(A, B ) ∈ F1 and I(a′, b ′) ∈ F6. If one of the polygon sides a′ and b′ has an endpoint in F4
(see Figure 10a), then this side cannot intersect B. So assume otherwise, see Figure 10b. The side a′ intersects B′ and is disjoint from A′, while b′ is disjoint from B′ and intersects
A′. (Due to space limitation some line segments are drawn schematically as curves.) Thus, each of a′ and b′ has an endpoint in F2 ∪ F5. But then I(A, B ) ∈ Cone (a′, b ′) and it follows from Observation 4 that neither A nor B can intersect both a′ and b′.
I(A, B ) ∈ F2 and I(a′, b ′) ∈ F2, see Figure 11. Since a′, b ′ is hooking, I(A′, B ′) ∈
Cone (a′, b ′), and the line segments a′, b ′, A ′, b, B ′ enclose a convex pentagon. The polygon side A must intersect b, a′ and b′, but it is restricted to F2 ∪ F4. It follows that A must intersect three sides of the pentagon, which is impossible. (This is in fact the only place where we need the assumption that a′, b ′ is hooking.)
I(A, B ) ∈ F2 and I(a′, b ′) ∈ F4. This is symmetric to Case 1.
I(A, B ) ∈ F2 and I(a′, b ′) ∈ F5, see Figure 12 (symmetric to I(A, B ) ∈ F3 and I(a′, b ′) ∈
F2). Then A is restricted to F2 ∪ F4, while a′ and b′ do not intersect F2 and F4. Therefore E. Ackerman, B. Keszegh, and G. Rote 9F1F2F3F4F5F6abB′A′Ab′a′
Figure 11 Case 5: I(A, B ) ∈ F2, I(a′, b ′) ∈ F2F1F2F3F4F5F6abB′A′BAb′a′ Figure 12 Case 7: I(A, B ) ∈ F2, I(a′, b ′) ∈ F5
A can intersect neither a′ nor b′.
I(A, B ) ∈ F2 and I(a′, b ′) ∈ F6. This case is very similar to Case 4, where I(A, B ) ∈ F1
and I(a′, b ′) ∈ F6, see Figure 10. If one of the polygon sides a′ and b′ has an endpoint in
F4, then it cannot intersect B. Otherwise, I(A, B ) ∈ Cone (a′, b ′) and therefore, neither
A nor B can intersect both a′ and b′.
I(A, B ) ∈ F3 and I(a′, b ′) ∈ F2. This is symmetric to Case 7.
I(A, B ) ∈ F3 and I(a′, b ′) ∈ F4. This is symmetric to Case 3. F1F2F3F4F5F6abB′A′Aa′Bb′
Figure 13 Case 11: I(A, B ) ∈ F3 and I(a′, b ′) ∈ F5
I(A, B ) ∈ F3 and I(a′, b ′) ∈ F5, see Figure 13. Then the intersection of b′ and A can lie only in the lower left quadrant. It follows that the triangle whose vertices are I(a′, b ′),
I(a′, A ) and I(A, b ′) contains a and does not contain I(A, B ). This in turn implies that
B cannot intersect both b′ and a, without intersecting B′.
I(A, B ) ∈ F3 and I(a′, b ′) ∈ F6, see Figure 14. As in Case 4, we may assume that neither
a′ nor b′ has an endpoint in F4, since then this side could not intersect B. We may also assume that I(A, B ) /∈ Cone (a′, b ′) for otherwise neither A nor B intersects both of a′The Number of Intersections Between Two Simple Polygons 10
and b′, according to Observation 4. If a′ has an endpoint in F2, then it cannot intersect
B (see Figure 14a). Otherwise, if a′ has an endpoint in F5, then B cannot intersect b′
(see Figure 14b). F1F2F3F4F5F6abB′A′b′a′B
(a) If a′ has an endpoint in F2, then it cannot intersect B.F1F2F3F4F5F6abB′A′b′a′B
(b) If a′ has an endpoint in F5, then B cannot intersect b′.
Figure 14 Case 12: I(A, B ) ∈ F3 and I(a′, b ′) ∈ F6.
We have finished the case that a, b and a′, b ′ are hooking. Suppose now that a, b and a′, b ′
are hooked, with respect to some pairs A, B and A′, B ′. Then A, B is hooking with respect to a, b and A′, B ′ is hooking with respect to a′, b ′. Recall that A, B, A′ and B′ belong to four different connected components. Hence, this case can be handled as above, after exchanging the capital letters with the small letters (i.e., exchanging P and Q). J
6 A Weaker Bound
The principal structure lemma is already powerful enough to get an improvement over the previous best bound:
I Lemma 8. GD has at most (n + 5) /2 connected components.
Proof. Partition the sides q0, q 1, . . . , q n−1 of Q into (n − 1) /2 disjoint pairs q2i, q 2i+1 , discard-ing the last side qn−1. Let H+ denote the subset of these pairs that are hooked. Suppose first that this set contains some pair q2i0 , q 2i0+1 of sides that are in two different connected com-ponents. Combining q2i0 , q 2i0+1 with any of the remaining pairs q2i, q 2i+1 of H+, Lemma 7 tells us that the sides q2i and q2i+1 must either belong to the same connected component, or one of them must belong to CC (q2i0 ) or CC (q2i0+1 ). In other words, each remaining pair contributes at most one “new” connected component, and it follows that the sides in H+
belong to at most |H+| + 1 connected components. This conclusion holds also in the case that H+ contains no pair q2i0 , q 2i0+1 of sides that are in different connected components. The same argument works for the complementary subset H− of pairs that are not hooked, but hooking. Along with CC (qn−1) there are at most (|H+| + 1) + ( |H−| + 1) + 1 = (n − 1) /2 + 3 = ( n + 5) /2 components. J
Together with Observation 3, this already improves the previous bound mn − (m + d n
6
e)
for a large range of parameters, namely when m ≥ n ≥ 11 :E. Ackerman, B. Keszegh, and G. Rote 11
I Proposition 9. Let P and Q be simple polygons with m and n sides, respectively, such that m and n are odd and m ≥ n ≥ 3. Then there are at most mn − (m + n−52 ) intersection points between P and Q. J
7 Ramsey-Theoretic Tools
We recall some classic results. A tournament is a directed graph that contains between every pair of nodes x, y either the arc (x, y ) or the arc (y, x ) but not both. A tournament is transitive if for every three nodes x, y, z the existence of the arcs (x, y ) and (y, z ) implies the existence of the arc (x, z ).Equivalently, the nodes can be ordered on a line such that all arcs are in the same direction. The following is easy to prove by induction.
I Lemma 10 (Erdős and Moser [ 3]) . Every tournament on a node set V contains a transitive sub-tournament on 1 + blog 2 |V |c nodes.
Proof. Choose v ∈ V arbitrarily, and let N ⊆ V − { v} with |N | ≥ (|V | − 1) /2 be the set of in-neighbors of v or the set of out-neighbors of v, whichever is larger. Then v together with a transitive sub-tournament of N gives a transitive sub-tournament of size one larger. J
A set of points p1, p 2, . . . , p r in the plane sorted by x-coordinates (and with distinct
x-coordinates) forms an r-cup (resp., r-cap ) if pi is below (resp., above) the line through
pi−1 and pi+1 for every 1 < i < r .
I Theorem 11 (Erdős–Szekeres Theorem for caps and cups in point sets [ 4]) . For any two integers r ≥ 2 and s ≥ 2, the value ES (r, s ) := (r+s−4
r−2
) fulfills the following statement :
Suppose that P is a set of ES (r, s ) + 1 points in the plane with distinct x-coordinates such that no three points of P lie on a line. Then P contains an r-cup or an s-cap. Moreover, ES (r, s ) is the smallest value that fulfills the statement. J
A similar statement holds for lines by the standard point-line duality. A set of lines
1, 2, . . . , r sorted by slope forms an r-cup (resp., r-cap ) ifi−1 and i+1 intersect below (resp., above)i for every 1 < i < r .
I Theorem 12 (Erdős–Szekeres Theorem for lines) . For the numbers ES (r, s ) from Theorem 11, the following statement holds for any two integers r ≥ 2 and s ≥ 2:
Suppose that L is a set of ES (r, s ) + 1 non-vertical lines in the plane no two of which are parallel and no three of which intersect at a common point. Then L contains an r-cup or an
s-cap. JI Theorem 13 (Erdős–Szekeres Theorem for monotone subsequences [ 4 ]) . For any integer
r ≥ 0, a sequence of r2 + 1 distinct numbers contains either an increasing subsequence of length r + 1 or a decreasing subsequence of length r + 1 . J
8 Proof of Theorem 1 8.1 Imposing More Structure on the Examples
Going back to the proof of Theorem 1, recall that in light of Observation 3 it is enough to prove that GD, the disjointness graph of P and Q, has at most constantly many connected components. The Number of Intersections Between Two Simple Polygons 12
We will use the following constants: C6 := 6 ; C5 := ( C6)2+1 = 37 ; C4 := ES (C5, C 5)+1 = (70 35
) + 1 = 112 ,186 ,277 ,816 ,662 ,845 ,433 < 270 ; C3 := 2 C4−1; C2 := C3 + 5 ; C1 := 8 C2;
C := C1 − 1 < 2270
.We claim that GD has at most C connected components. Suppose that GD has at least C1 = C + 1 connected components, numbered as 1, 2, . . . , C 1. For each connected component j, we find two consecutive sides qij , q ij +1 of Q such that CC (qij ) = j and
CC (qij +1 ) 6 = j. We call qij the primary side and qij +1 the companion side of the pair. We take these C1 consecutive pairs in their cyclic order along Q and remove every second pair. This ensures that the remaining C1/2 pairs are disjoint, in the sense that no side of Q belongs to two different pairs. We apply Lemma 5 to each of the remaining C1/2 pairs qij , q ij +1 and find an associated pair pkj , p kj ±1 such that (qij , p kj ), (qij +1 , p kj ±1) ∈ ED. Therefore, CC (qij ) = CC (pkj ) and
CC (qij +1 ) = CC (pkj ±1) 6 = CC (qij ). Again, we call pkj the primary side and pkj ±1 the companion side. As before, we delete half of the pairs pkj , p kj ±1 in cyclic order along P ,along with their associated pairs from Q, and thus we ensure that the remaining C1/4 pairs are disjoint also on P .At least C1/8 of the remaining pairs qij , q ij +1 are hooking or at least C1/8 of them are hooked. We may assume that at least C2 = C1/8 of the pairs qij , q ij +1 are hooking with respect to their associated pair, pkj , p kj ±1, for otherwise, pkj , p kj ±1 is hooking with respect to qij , q ij +1 and we may switch the roles of P and Q. Let us denote by Q2 the set of C2
hooking consecutive pairs (qij , q ij ±1) at which we have arrived. (Because of the potential switch, we have to denote the companion side by qij ±1 instead of qij +1 from now on.) By construction, all C2 primary sides qij of these pairs belong to distinct components. We now argue that all C2 adjacent companion sides qij ±1 with at most one exception lie in the same connected component, provided that C2 ≥ 4.We model the problem by a graph whose nodes are the connected components of GD.For each of the C2 pairs qij , q ij ±1, we insert an edge between CC (qij ) and CC (qij ±1). The result is a multigraph with C2 edges and without loops. Two disjoint edges would represent two consecutive pairs of the form (qij , q ij ±1) whose four sides are in four distinct connected components, but this is a contradiction to Lemma 7. Thus, the graph has no two disjoint edges, and such graphs are easily classified: they are the triangle (cycle on three vertices) and the star graphs K1t, possibly with multiple edges. Overall, the graph involves at least
C2 ≥ 4 distinct connected components CC (qij ), and therefore the triangle graph is excluded. Let v be the central vertex of the star. There can be at most one j with CC (qij ) = v, and we discard it. All other sides qij have CC( qij ) 6 = v, and therefore CC( qij ±1) must be the other endpoint of the edge, that is, v.In summary, we have found C2 − 1 adjacent pairs qij , q ij ±1 with the following properties. The primary sides qij belong to C2 − 1 distinct components. All companion sides qij ±1 belong to the same component, distinct from the other C2 − 1
components. All 2C2 − 2 sides of the pairs qij , q ij ±1 are distinct. Each qij , q ij ±1 is hooking with respect to an associated pair pkj , p kj ±1.All 2C2 − 2 sides of the pairs pkj , p kj ±1 are distinct. Let us denote by Q′
2
the set of C2 − 1 sides qij .
I Proposition 14. There are no six distinct sides qa, q b, q c, q d, q e, q f among the C2 − 1 sides
qij ∈ Q′
2
such that qa, q b are avoiding or consecutive, qc, q d are avoiding or consecutive, and
qe, q f are avoiding or consecutive. E. Ackerman, B. Keszegh, and G. Rote 13
Proof. Suppose for contradiction that there are six such sides. It follows from Lemma 5 that there are two consecutive sides pa′ and pb′ of P such that CC (pa′ ) = CC (qa) and
CC( pb′ ) = CC( qb).Similarly, we find a pair of consecutive sides pc′ and pd′ of P such that CC (pc′ ) = CC (qc)
and CC (pd′ ) = CC (qd), and the same story for e and f . By the pigeonhole principle, two of the three consecutive pairs (pa′ , p b′ ), (pc′ , p d′ ), (pe′ , p f ′ ) are hooking or two of them are hooked. This contradicts Lemma 7. J
Define a complete graph whose nodes are the C2 − 1 sides qij ∈ Q′
2
, and color an edge
(qij , q ik ) red if qij and qik are avoiding or consecutive and blue otherwise. Proposition 14 says that this graph contains no red matching of size three. This means that we can get rid of all red edges by removing at most 4 nodes. To see this, pick any red edge and remove its two nodes from the graph. If any red edge remains, remove its two nodes. Then all red edges are gone, because otherwise we would find a matching with three red edges. We conclude that there is a blue clique of size C3 = C2 − 5, i.e., there is a set Q3 ⊂ Q′
2
of C3 polygon sides among the C2 − 1 sides qij ∈ Q′
2
that are pairwise non-avoiding and disjoint, i.e., they do not share a common endpoint. Our next goal is to find a subset of 7 segments in Q3 that are arranged as in Figure 15. To define this precisely, we say for two segments q and q′ that q stabs q′ if I(q, q ′) ∈ q′. Among any two non-avoiding and non-consecutive sides q and q′, either q stabs q′ or q′ stabs q, but not both. Define a tournament T whose nodes are the C3 sides qij ∈ Q3, and the arc between each pair of nodes is oriented towards the stabbed side. It follows from Lemma 10 that T
has a transitive sub-tournament of size 1 + blog 2 C3c = C4.Furthermore, since C4 = ES (C5, C 5) + 1 , it follows from Theorem 12 that there is a subset of C5 sides such that the lines through them form a C5-cup or a C5-cap. By a vertical reflection if needed, we may assume that they form a C5-cup. We now reorder these C5 sides qij of Q in stabbing order, according to the transitive sub-tournament mentioned above. By the Erdős–Szekeres Theorem on monotone subsequences (Theorem 13), there is a subsequence of size C6 + 1 = √C5 − 1 + 1 = 7 such that their slopes form a monotone sequence. By a horizontal reflection if needed, we may assume that they have decreasing slopes. We rename these 7 segments to a0, a 1, . . . , a 6, and we denote the line (ai) byi, see Figure 15. We have achieved the following properties: The lines 0, . . . , 6 form a 7-cup, with decreasing slopes in this order. The segments ai are pairwise disjoint and non-avoiding.
ai stabs aj for every i < j .These properties allow a0 to lie between any two consecutive intersections on 0. There is no such flexibility for the other sides: Every side aj is stabbed by every preceding side ai.For 1 ≤ i < j , ai cannot stab aj from the right, because then a0 would not be able to stab ai.Hence, the arrangement of the sides a1, . . . , a 6 must be exactly as shown in Figure 15, in the sense that the order of endpoints and intersection points along each linei is fixed. We will ignore a0 from now on.
8.2 Finalizing the Analysis
Recall that every ai is the primary side of two consecutive sides ai, b i of Q that are hooking with respect to an associated pair Ai, B i of consecutive sides of P . The sides ai and Ai are the primary sides and bi and Bi are the companion sides. All these 4 × 6 sides are distinct, The Number of Intersections Between Two Simple Polygons 14 234561`0a3a4a1a2a5a6a0
Figure 15 The seven sides a0, a 1, . . . , a 6. The lines 0, . . . , 6 form a 7-cup.
and they intersect as follows: ai intersects Bi and is disjoint from Ai; bi intersects Ai and is disjoint from Bi; and I(Ai, B i) ∈ Cone( ai, b i).Figure 16 summarizes the intersection pattern among these sides. A side Ai must intersect every side aj with j 6 = i and every side bj since CC (Ai) = CC (ai) 6 = CC (aj )
and CC (Ai) = CC (ai) 6 = CC (bi) = CC (bj ). (Recall that all companion sides bi belong to the same component.) Similarly, every side Bi must intersect every side aj . We have no information about the intersection between Bi and bj , as these sides belong to the same connected component. aibiAiBiajbjAjBjPQ
Figure 16 The subgraph of GD induced on two pairs of consecutive sides ai, b i and aj , b j of P
and their associated partner pairs Ai, B i and Aj , B j of Q. Parts of P and Q are shown to indicate consecutive sides. The dashed edges may or may not be present.
We will now derive a contradiction through a series of case distinctions.
Case 1: There are three segments Ai with the property that Ai crosses `i to the left of ai.Without loss of generality, assume that these segments are A1, A 2, A 3, see Figure 17. The segments A1, A 2, A 3 must not cross because P is a simple polygon. Therefore A1 intersects
a2 to the right of I(a1, a 2) because otherwise A1 would cross A2 on the way between its intersections with 2 and with a1. A3 must cross3, a 2, a 1 in this order, as shown. But then
A1 and A3 (and a2) block A2 from intersecting a3.E. Ackerman, B. Keszegh, and G. Rote 15 ??a1a2a3A1A2A313`2
Figure 17 The assumed intersection points between Aiand `iare marked.
Case 2: There at most two segments Ai with the property that Ai crosses `i to the left of ai.In this case, we simply discard these segments. We select four of the remaining segments and renumber them from 1 to 4. From now on, we can make the following assumption:
General Assumption: For every 1 ≤ i ≤ 4, the segment Ai does not cross i at all, or it crossesi to the right of ai.This implies that A3 must intersect the sides a2, a 1, a 4 in this order, and it is determined in which cell of the arrangement of the lines 1, 2, 3, 4 the left endpoint of A3 lies (see Figures 15 and 18). For the right endpoint, we have a choice of two cells, depending on whether A3 intersects `3 or not. We denote by left (s) and right (s) the left and right endpoints of a segment s. We distinguish four cases, based on whether the common endpoint of A3 and B3 lies at left (A3)
or right( A3), and whether the common endpoint of a3 and b3 lies at left( a3) or right( a3).
Case 2.1: I(A3, B 3) = left( A3) and I(a3, b 3) = right( a3), see Figure 18. As indicated in the figure, we leave it open whether and where A3 intersects 3. We know that b3 must lie below3 because I(A3, B 3) ∈ Cone( a3, b 3).We claim that A2 cannot have the required intersections with a1, a3, and b3. Let us first consider a1: It is cut into three pieces by A3 and B3.If A2 intersects the middle piece of a1 in the wedge between A3 and B3, then A2 intersects exactly one of a3 and b3 inside the wedge, because these parts together with a1 are three sides of a convex pentagon. If A2 intersects a3, then it has crossed `3 and it cannot cross b3
thereafter. If A2 intersects b3, it must cross `4 before leaving the wedge, and then it cannot cross a3 thereafter. Suppose now that A2 crosses the bottom piece of a1. Then it cannot go around A3, B 3
to the right in order to reach a3 because it would have to intersect 4 twice. A2 also cannot pass to the left of A3, B 3 because it cannot cross2 through a2 or, by the general assumption, to the left of a2.Suppose finally that A2 crosses the top piece of a1. Then it would have to cross 3 twice before reaching b3.The Number of Intersections Between Two Simple Polygons 16 a4b3a3A3B33a121a2`4A2I(a3, a 4)I(A3, B 3)
Figure 18 Case 2.1, I(A3, B 3) = left (A3) and I(a3, b 3) = right (a3). A hypothetical segment A2
is shown as a dashed curve. The side a2 and the part of `2 to the left of a2 is blocked for A2.
Case 2.2: I(A3, B 3) = left( A3) and I(a3, b 3) = left( a3).If (A3) does not intersect a3, we derive a contradiction as follows, see Figure 19. We know that the sides a2, a 3, a 4 must be arranged as shown. The segment A3 crosses a2 but not a3.Now, the parts of a3 and A3 to the left of2 form two opposite sides of a quadrilateral, as shown in the figure. If this quadrilateral were not convex, then either (A3) would intersect a3,which we have excluded by assumption, or3 would intersect A3 left of a3, contradicting the General Assumption. Thus, the sides a3 and A3 violate the Axis Property (Observation 6), which requires a3 and A3 to lie on different sides of the line through I(A3, B 3) and I(a3, b 3).Looking back at this proof, we have seen that the configuration of the segments a1, a 2, a 3, a 4
according to Figure 15 in connection with the particular case assumptions make the situation sufficiently constrained that the case can be dismissed by looking at the drawing. The treatment of the other cases will be proofs by picture in a similar way, but we will not always spell out the arguments in such detail. a4a3A33a224(A3)
Figure 19 Case 2.2. I(A3, B 3) = left (A3),
I(a3, b 3) = left (a3), (A3) does not intersect a3.a4a3A33a224
Figure 20 Case 2.3. I(A3, B 3) = right (A3),and I(a3, b 3) = right( a3), A3 lies below `3.
If (A3) intersects a3, the situation must be as shown in Figure 21: the pair A3, B 3 is hooked by a3 and b3. The analysis of Case 2.1 (Figure 18) applies verbatim, except that the word “pentagon” must be replaced by “hexagon”. E. Ackerman, B. Keszegh, and G. Rote 17 a4b3a3A3B33a121a24A2(A3)
Figure 21 Case 2.2, I(A3, B 3) = left (A3), I(a3, b 3) = left (a3), and `(A3) intersects A3. A hypo-thetical segment A2 is shown as a dashed curve.
Case 2.3: I(A3, B 3) = right( A3), and I(a3, b 3) = right( a3).If A3 lies entirely below 3, then A3 together with a3 violates the Axis Property (Obser-vation 6), see Figure 20. a2A313b3a3B34a1I(a3, a 4)b3a3a4A3`3B3A3?(b) (a) A2
Figure 22 Case 2.3. A3 intersects `3.
Let us therefore assume that A3 intersects `3 (to the right of a3), and thus right (A3) =
I(A3, B 3) lies above 3, see Figure 22a. Then b3 must also lie above3, because a3, b 3 is supposed to be hooking, that is, I(A3, B 3) ∈ Cone( a3, b 3).It follows that A3 cannot intersect `3 to the right of I(a3, a 4) (the option shown as a dashed curve), because otherwise it would miss b3: b3 is blocked by a4.Therefore, the situation looks as shown in Figure 22a. Figure 22b shows the position of the relevant pieces. The segments a4, B 3, a 3, b 3, A 3 enclose a convex pentagon. Now, the segment A2 should intersect a3, b3, and a4 without crossing A3 and B3, like the dashed curve in the figure. This is impossible. The Number of Intersections Between Two Simple Polygons 18
Case 2.4: I(A3, B 3) = right( A3) and I(a3, b 3) = left( a3).If A3 intersects 3 (to the right of a3), then A3 together with a3 violates the Axis Property (Observation 6), see Figure 23. We thus assume that A3 lies entirely below3.a4a3A3`3
Figure 23 Case 2.4. A3 intersects 3.a2a4A3A213b3a3B34a1I(a3, b 3)a2a4A33a34a1I(a3, b 3)`(A3)(a) (b)
Figure 24 Case 2.4. A3 lies below `3.
If (A3) passes above I(a3, b 3) = left (a3), the sides a3 and A3 violate the Axis Property see Figure 24a. On the other hand, if(A3) passes below I(a3, b 3) = left (a3), as shown in Figure 24b, then b3 must cross `1 to the right of a1 in order to reach A2. Again by the Axis Property, B3 must remain above the dotted axis line through I(A3, B 3) = right (A3) and
I(a3, b 3) = left (a3). On `1, b3 separates a1 from the axis line, and hence a1 lies below the axis line. Therefore B3 and a1 cannot intersect. This concludes the proof of Theorem 1. J
Acknowledgement. We thank the reviewers for helpful suggestions.
References
1 J. Černý, J. Kára, D. Král’, P. Podbrdský, M. Sotáková and R. Šámal, On the number of intersections of two polygons, Comment. Math. Univ. Carolinae 44:2 (2003), 217–228.
2 M. B. Dillencourt, D. M. Mount and A. J. Saalfeld, On the maximum number of intersections of two polyhedra in 2 and 3 dimensions, Proc. Fifth Canadian Conf. on Computational Geometry ,Waterloo, Ontario, August 1993, 49–54.
3 P. Erdős and L. Moser, On the representation of directed graphs as unions of orderings, Magyar Tud. Akad. Mat. Kutató Int. Közl., 9 (1964), 125–132.
4 P. Erdős and G. Szekeres, A combinatorial problem in geometry, Compositio Mathematica 2
(1935), 463–470.
5 Felix Günther, The maximum number of intersections of two polygons. Preprint, arXiv: 1207.0996 [math.CO], July 2012, withdrawn by the author. |
9905 | https://www.car.org/-/media/CAR/Documents/Transaction-Center/PDF/Standard-Forms/December-2018/December-2018---Legal-Forms/AD_12-18_LegalChange_Draft5.pdf?la=en&hash=8F0FC03A2AEC7B6F2A009AF4154D164738D9ADEA | %PDF-1.4 %ãÏÓ 526 0 obj <> endobj xref 526 34 0000000016 00000 n 0000001451 00000 n 0000001598 00000 n 0000002075 00000 n 0000002129 00000 n 0000002243 00000 n 0000003525 00000 n 0000004477 00000 n 0000005103 00000 n 0000005589 00000 n 0000006130 00000 n 0000006242 00000 n 0000006516 00000 n 0000007149 00000 n 0000007421 00000 n 0000008002 00000 n 0000009989 00000 n 0000011852 00000 n 0000012039 00000 n 0000013550 00000 n 0000015228 00000 n 0000017051 00000 n 0000018745 00000 n 0000046831 00000 n 0000068215 00000 n 0000068350 00000 n 0000079224 00000 n 0000079475 00000 n 0000079833 00000 n 0000081585 00000 n 0000081843 00000 n 0000082149 00000 n 0000001260 00000 n 0000000995 00000 n trailer <<8A00803F664853419D78AF09B400B25B>]/Prev 160296/XRefStm 1260>> startxref 0 %%EOF 559 0 obj <>stream hÞbàf5eeHPbd@�A6 æÁ°?È¿RÔ:#Ç~/0A <be@ÀÃÀäj¤åX,bÊ À¸Áå,s3-ãFÆ Ö w\n3 °÷± 1´±nbd{ÅṲ̂åÊpÑm1ÃY¿ÿ�M¿ÆØ ³ÉÓ�H³10ȦÝ$ÎÀTX ro0�!i G endstream endobj 558 0 obj <>/Filter/FlateDecode/Index[24 502]/Length 40/Size 526/Type/XRef/W[1 1 1]>>stream hÞbbgb`Å3Î ÑøÅ£ñ1£Üø�0�CÞ endstream endobj 527 0 obj <>/Metadata 22 0 R/Pages 21 0 R/StructTreeRoot 24 0 R/Type/Catalog/ViewerPreferences<>>> endobj 528 0 obj <>/ExtGState<>/Font<>/ProcSet[/PDF/Text/ImageB]/XObject<>>>/Rotate 0/StructParents 0/TrimBox[0.0 0.0 612.0 864.0]/Type/Page>> endobj 529 0 obj [/Separation/All/DeviceCMYK 550 0 R] endobj 530 0 obj <> endobj 531 0 obj <>stream H¬ÝnÜ6
ïõ|åÎ 9ü\dݹpÑÀ½7 j§µ¢¯ß3vÍå]C2 h½\Iüxf8g¸ýõÇ·nÜ»wÛËݧ Gîýû;×m¿ýr}óõãþÿ¸v.1ö]î;Î>gvR¼jr«-®Pñµ{¸éþp?»íîÜîʱsW»ß:òêþsÁÝ%Þþ¿~{ÄËÈ×¢Ü]WS]äê)¢(îú®»Çì.v»\¦ÞDbCP»Á~¼ë\®·ÝU÷ä£ÔvÖ×Ï&¡úRl¬ÿ0>1Ñþ«´IúU^?Ú\¯.Z¹¹" V³>²8^"µ)¾Ä~ß»?Ç%ê²6óçsêç%·ó·j~îÇðç)»¿Dp#5ûXd [e§»ÕõÚî!½Ås§IqO!îsë¾K!ù3e¨Ó/C0oOe²VÊ5äjM¤&CØ~º#wñ÷ÑjY+Û ÖU©ÉB¼¥¥0äëUl}ò48üNÌÇQÑr6ª^¥ThÍÈP6Ì·õ×Ûñ[®ÕÇ öû¬Å!èöÑ·Gðád°ñ´Ç:YêÆB4½õv½÷Ãõí,bìÙ% =;l´i.6£{BiÈí ÛaèmhÜSNCç^SÕSH®úp{iUïÎ2·Nú;0Ãs;êY¢5Ç tðZt8PP¢)Rtý>RdJ3|±ö¶g 4ñth±Áó)=[m"¦óDkθ+ö"!ÍY[S}ô/õx Hkzyx¹½¼áaC®E ý phÍéí|¨:´ó[g}®fÊÜ XÀ/ jè-=¾ä)Q«µvT{1Cµ9¢5g kë)m½5É/kdV ëkðD=&rÿ 0�»endstream endobj 532 0 obj <>stream HäM#7
÷u ]À(Ry ¬r Á�ót6¹~Ê3c¹eyá®Z 4´ªôÞrÊûçóö¶e ¯{tá$E©µa²zÎZÒ\_\_··Äû(N²§f%q©}@ÿòëvÁûjºþý²USR!Òü²}ÚþÙþÀ穾#Iû}|ärÃéóÊuzéó»PÕ6&�ÞÏÝõuÄTwJMEî ¿l¯3Ä(TC½ÎãTì%3²ðw%ÓëüÚçoAMØ3ëCù@Çÿj)ë µF^xL8f°ÇD¯¡.4,7~T)øÈÈí1Pù@°t2V°ìYËt\_Íw¨ ×y0ªöÝ%ÝËö¤@
vð[BúKòð $ÔóI1þ &ò;"&Ñë@
\û¼"
´T\_·ÌJ»Kò( DL¼ I \_ÈÄî¤<z 6 hà#a^Ó@±D%e>ÁI²´$Þ0] h·Ælx¯¥ö£o·ðx÷PÃ7ªÙÖMÃFlѦøxïÈx½x¥Zy½Ë S®ew à(²-·dܽMñá®QÃz§L×6 )>Þ1L¹Óx¶¯- h>Åû
ù^ì(ºSF¹Èï CMÒb}àÀ-[LñáVÑÓͳ£Ä¬S|O^ÜB×½ô ¦øpY ~ð.]¦PÉs|¤E-0ù!¾À Ã?M¯¸ÅºÄ@6ÄZO0dòdPã³¼a yo Ãø$[8Þ» >ÁÉîH~ÊïIf0L^0´bÞ³|làèo7Ù¹=KýIüf b{ðîà¾IþÛÆ ³á
8· QQß~scýþ÷ïíÏôí5o±\_,áþ³=cܯÚÇÕ¨H}n Hdp\ãÀ@û¡·HxÌñKñCpcDÌL{ua6íW°¢yÿ÷ê:¿õ¶,®Å=qê{aÆ!ÇÒ :õEzô~~+é¿ùqFû=ýjIÿ 0�yI¼ endstream endobj 533 0 obj <> endobj 534 0 obj <> endobj 535 0 obj <> endobj 536 0 obj <> endobj 537 0 obj <> endobj 538 0 obj <>stream H\ÔÍnÚ@à½bÉ"2xîÜ$ ÉDbÑö=PKŶ³àí;Ç'J¥"ìñù¸ÈïûCßÍ.ÿ> Í1ÎîÜõíoÃûÔDw®ÏÖ
k»fþø´¼6×zÌò´øx¿ÍñzèÏCV.ÿÞæéîªv8ÅÇ,ÿ6µqêú{øµ;>ºüø>5ö³[¹íÖµñ.ô¥¿Ö×èòeÙÓ¡MÇ»ùþÖü;ãç}®X>¯i6ÞƺSÝ\_bV®ÒcëÊ·ôØf±oÿ;nÊe§só»²²ÀÉ«UzKÙ3{ägægä óyǼCÞ3ï\_\_ß ô¼¾Çõýy\0Èìõèõ,È9 +³"³!Óéáôtz8}Å\!¿0¿¤,4 BÀ 4 BÀ 4 BÀ 4 BÀ 4 ^Yz97ÁÜsÌM87ÁÜsÌ-pns 4Íæ@s9Ð4Íæ@s9Ð4 e¯¢WÙ«èUö\z½^e¯¢WÙ«èUö\z½^e¯¢Wù)þ/åÜsSÎM17åÜsSÎM17¥Ya6ÎÍ07£ßà7ú ~£ßà7ú ~£ßà7ú ~£ßà7ú ~£ßà7ú þ ºÕßo<3üa¿Ú çTü-é 7ìÇ[7í0îs\_hÞ§)m Ë6´ìغ>~îTã0º´ Ïì¯��pPæ endstream endobj 539 0 obj <> endobj 540 0 obj <>stream H\Í@
÷yZv/üÔ[-ÁÖnp1?3Ò´I(ã·:9ML@ýB¬{>^óí~·úÙä?ãØlNýÐÅpo± æÎýéúvþ¼[ÞÛK3ey:|¸\_çpÙ§1«kÿJ¯s¼M7Ãcÿ]ýp6¶GnÓô.aMaÖkÓ
Sô¾7òåØÓ¾KÏûùþÎüûÆïûLµÜiÇ.\§¦ ±Î!«tMý®uî¿çjyìxjßÕ¾\é#ñ3ù¼"¯À[òüJ~¿SPm9Çb-É%¸"WK¶! ØXÉ öd¦
¥
Ý7
¹\a® W+Èæ r
¹\a® W+Èæ r
¹\y!¿Ù Ùwv%èJØ +Ç®ºrôtðtôtðtôtðtôtðtôtðtôtðttspSö èA9\_1\_9\_1\_9\_1\_9\_1\_9\_1\_9\_1\_Ù¢e.Yì\_Ñ¿²E'ÊN(;Qt¢ìDѲE'xtéïáïéïáïéïáïéïáïéïáïéïáïééṪ(WËR|þú±iÍ×îµ·ÓÚ-«¾ì6׿Á4N&+û+À�e�Ð endstream endobj 541 0 obj <>stream HÜM,¹ ÷u ] ¢þaø <³ÍôíýÕ]Ïðjê tåR"Á YRI¿=þýøçãü¹xúÏãGÉ{õxtÕÔÌóeïQ×ÍU[úøýñGª±ª¦«åYƽ±Ë¬z«7¿?ÆÌkÌ´ ìTKÍÝWÒ²¹vúñ¨EWzÒr³öʽ°ÐF®e¦=òÖÝÊ£íôñ%wkzìlwÕ'Ï]ÍÓ'ý+ú+>lcÍ0lÌÍKÿv᪼h×Kæí5õìk¦Ë1ÃÆu6{\9ëñX/Úðj;[ç˲sÛ×ûXú Woéê=ÚUe(\¾òÌK°tyÚk£ø¦aBÃ:¦úæ£ýÃÜ®3¸ë«vÌäôJ¬ê§¼×Å©swüòËðöØÈÕÄÉuÇycjï £zmð¬Ì<Øö²¼LÍ\äFÃúõݽËw®50/ ÍϪîØèÜwËþW[ÙÄÇ/gZ¶¤Xðõ:$0Ù�yö$n^ Ê:fà½LÎdXM@IÏ®¤§õmð×ÌÞaÄ7#ÎßШæÁU1´¸OÄòb#M¸Ýnì§M^î&èáaóqîA¡ç#W@g.Ö½Ït¡ ïõW¸»wÌ'ÞUðº~wA!±û° CÆ!{7áQ�9~v =+².Ô¢§Vrecb¬ÅKÈÌ>®�»Ç-0'¦8ã¨Qü9 @ï 5eµ?ïM¹¨Ru (ºéUÄÞË-deBKÞò£Òo y½äûÊ L ÜÄçvrÛ|gÃ[ìBÝÚAh)
_ÐûÄè:yly+ÉGènzà .aª¡öaåGÛÜqÊ£«¯§G¿Àùs}Z^8@+O¥ç¦¤-äTñ í¤&êµ'ñ¦¥/'ë
Òè8D]:GuàÍà´DJ9r®YºeYõá08 ¹'&æÜÚ¯æc¯±7æ®A),s±¤Ku«°&èÉÏá"K½ÿC_G7eo¿³y2oAüvéÊÞ ý ÷TI!B<(7Á6ýÅa©o±ú%Z#¯¦û¨Òò¶ù¥'Á#M# Z; 5%¸ºG¢ê¶Ò6I°.HqÅTb¦$Ò¢ÔAM·^ÿxþÒÞþµè·wëÎV©ëiKÎËK
kês¤((à×.T4tŨË|ºÁ Lðs þî[7J4Nì:[ÉɳõÊ{¹+~×GTd6)¸3døÓK©ù¡8¤úéãd%vPªSUã×AåãæÞõøêÃÝìþÕNqG\Tiç5ìÛ¢I¤«ø T<º£ü|¥AÏ�&Ç\¹mZ AùqEC©¿ó 5zÀÐY
oôA¸)4´G»E=ïéÏô÷ðªüù¤¾²¨>83_¾B'y¨¡Cº7?T¨7RФr3Êø ß}wiÏP¶ºôZÖwè©2ÿ(|¼9\8ÊùWÚ¶õÚ=ÙS|U¤bA£Äµ£®Î9E6TÕ¦ÀHѨêmbüS塹P)ÊE 4Pº,Uëå-ºEÕ Èܤ«R®[Ð÷>EkÄH v#Ûé6G½Þ/ùØLx4[rö"õ·ýÔ· J+ݶ80z ºø yK@sI¼øþ$¶QÄÐ(Åã±^Þ¡xw±¥BÑ^ÒóÚ^»åª «gÃÀ ;úy¬.nM#µm«¡îå´õê_à÷g®Ýk qOuþLL©ï1ü&©K \êCm²R¼«~(q?_ ^kQ�¶Új¾2b{<0Ôê"ð|{ØÉëî3úÏ)búD5O¢��ãÈNÔì¢Èÿ_O4dmEoÙ]lØǸTvc/ÓùÖÆÒB zB¯o2· íç]Ìeº¦cnÜcÕ}7K=ãbCºÛZÄ%îÖ®!f1¦uïѬ ÓD|¥û{!©� ÎiTW 5Sz\Wd)ÜC
½ðTM¼]cL@Ó±ñ3ú/Á¨ÓP¬èùng~ñ³K èY¶½_îöC©Èª%-I$ óÉ/³úhö_JÜ-D\vtöL-+ì!ßgräÜ2hHÅAÝ°Ùé©÷R>,ó,]ôî¤V)Q±ZÓH=îvßpnõÐF¨¿DÁªáÛÏçyLE{²ÀèÚ¨1,aÊýnèãíog^1¥¾Zø(£)Ä¿V3Æ(E~h®R|Õ|°mZlÎîh×ÙG.?é¿ �Y endstream endobj 542 0 obj <>stream H¤W[OG~÷¯Çµs¿Hcêjo[U¡Ôq[!iûïû½mZ$<ßîìùÎmÎc¸9¬d7Áµr,p#=E®cREö÷$W1ÁRØ;rÜÇFÀ»±M;úÑ}²$@Û=x%ø[ > ¾ÇßgPÆÓ³dZî½ÑIV.L(¶ú4øÌò\ ÚðyCèå§ä0ó).f{§æÑm"È¡WY©®bø1N'K#wÚpg Ya²' MòÖ(Y\ISÉún2BÇãã7S<þ~ð¦Láköûà¨,Éòà PØH>&g#}EýJªjaGXÍÊObºÃò·Á¬ÜÅjîYïéR²÷Ä/3¿àÒ²rE±¡ÅLV» [¡"±¾+&ã³ùÉæç·óñðòÛ=è>"}"émlªû3»Esák[Òj1Á¶ ãår?ý÷êK]©ß~¾_´ûïêK¥pv|b2óó·ûü>"ä>Ù ßã&¯;1?ÏgÙø¬<_,Ù¨gÜÚç2Ùe.uû}Ѫ±¥
È%?Z È9kYÀB²Óùrrv¾üa1cCcU,f§ãÅtþöô=ðÉ.ÖL9ã¹ ò-Ëq9cãÓÙÛÉÏpÞYÖòùÅ~¥ÜI[ÅPybÛbcÙe1¾gCYl׿l¶ë÷XóPüú7{¸^³Éæëæ¨br÷~}9ÓgK÷SØg²¶=z¨}~иxÎ5¾ÉX"H!w¼zøruS®ÿzx}2;9QBããcHªªIz eQ¤wù~ÐÉôÈÑ°³3>{
û«OÅ Ü\%!ùL½HU2 yWt$JÀÁIRû"góèdIv~«ËáÁDv;ã]=íÍ .¼¨óñ!B ]~/gÓ¢ý ÏHMn<ýÙþåiò¹©ïhѼ¨©Ù§ùÿeqqÈÖ=žG¡SÿÙ³'
kS´»7åÎ)ÓsÐ!q»u#ÎÑÕ ¡#Wé~q»ïS×zË£D[ýÔµ)9)»¾Ønnr%^=¬x]=êÙªuT5 ¹c0ݲ«19nõ?F9:¦ùi,ZØ./H¥DËÞ%WP\+û\\w¸z6 .gº\ïW©¸b2Ä(T\Ýl> ﶷ«êÝøþþnµ¹zØÜÝVOî>¤&j(14ú¢:qh¤!#Kê~^±ùí÷BTÍ Y×äk¨
ãFÐ x@ë°]c9:Ù|ü²]·7éèæjµ~º½úãz³boHz¢ëv̲FsCCôhº¤Hck�\þ Dóe4ÁÜÐîÀ!úîöDÃi8#[#͵Õyk î X^§¹ìó>g)rÖOì¶g©Åäóê,4÷6BaÔÈJr1Ùìèö3 Ì\ Ò¼ë ¢ß¦ëE(&$}VImß¼S\ÒÉ %´/#䣬úݧVµÀ V4ÕùX!H¥®"}×´ djú-8 Ù
Yª«6eËÉ"rj£y0®ò×$ H ±µæ5BÓĪ4mZÍ $W ÉßQûë¯jfTqå-Z W9 :1öø¹°+HSÑÒ%³ZHM¢j@ñÕú]Юýª$6¤«¥íª!¬ bá>ZdN"Ì ¹CM«ÜRÚ5¨æ¬3/Cÿ{!=Úا^à!ÔÊåÛ
Ô-t%{Q½óÖ=Hµ¹="»\_Ó)ɱ"ÊÄ꣬!I7ùØ 2ÔxýØZ\¡½¶êz顺VU£vÒ7Ô >k|£ú4Ú´OQkWiг]á=ØÕuïcÍQðÌE.$~ͩƵÐ0/¯tµKQÈVP|by7å\_À]/Vö��ötÏ endstream endobj 543 0 obj <> endobj 544 0 obj <>stream HÜWK7¾÷¯Ð1 t[RI% kÆ9! »&NþûùªJîÄ&¬MúF¯z|õè9,¾°iI5»9,1¸9.
ã�a)ÜyêÐ/%ü\Ǽþ'/gZ-îaòKu�- à÷uè²ë§ä·»Gw óîÌ|uá¼{mÞ+2óôfz5½wA5.3,µ:®KK%¹3LÀÊÙDÔܽ\júô\¸:ØSrX×\ñzª¹M6 ô¸7,9ÓFº<Ø!-¡%()Ñ m£|÷ú¿Ç{MD·×WäF[#\VIߦqûo8%³X¨äj^oê')«¨C¸ånÝÒq÷/ãº×¿ýôÎ=}úäåóï_8¤Ö³gw/ðû«éî4=9¬;½B§34:}pMUC&VWHúµ;=L\_½ÿúôëäþ�gTû( endstream endobj 545 0 obj <>stream HÌWÛnÛF}ÐØG PiÞIA�ÛqIjEõaE.¥)®@Rrø÷ÙHÓ¥4(P0Er9gÎì\Î>,&ïÉäæqGKòöíÍçûïwïîÞÃó»Åäf±pKùıÏ'g2'üÍIØïÌIì»vÅvòÅ"KëcNÒ KX¶Å×$ÕÀuÅx¹&»JxÆ2K²diÃEI¦[ϼÙJ¦?;vd5-kªÞãóÑmDÁ½kñ²a«¾¥e JFZF+½ÖúÇUäx·sÜéß_'§1? ?!±gbÕçÄsÛõÖtñõ ïô�Çv!Êø?"ÕzróáÑ!ëºÞGvÀ½>ú×4·&-¸ä !ÜÄ ëDvè¼ÀYNgd÷£ÔÜØ\?ê>àùÿØ-³qÁ2 ÛK¿\à
W2ìì{¶ym³G¢ùÅ8¯ÞíþÏ õYØ©ÊjSà¯Ð¢$ãuº¯ë¥®ßUj6Ð5'kZÉb>±¤×fdêÚ®Ä7b=çت\_Í)׾©Y£ÌíËU v>÷ç mÆ_ß..M»cºõ\K[y[+(ú_oøN>J,èpÛA+þòAVêóg^oôÑ =(»sE䨦@cÅG½0ÙcäíÞëÈÅrQÛóÀq½ÇO~_z÷ܼýððÛb êlCE/¡ÀöuoÉ#+ V!T-¡<«ã©íÔã!± ^72qd,èºblkÖ&]ÀI9B F5ÑSBeLß:Á¢,Z\rë8²b«Þ¯ÔÈM¯;ªÀ¹:墨 Õæ8ìÐå\
xVÁÞå9¯¶\2ðX|2ñÍ7îIoúN !»V¶72n·Ú·_x¶O9Zí²o¶fxJ+6#8Ò×oÚÙÀ,¯a j!ZZ4´gÐýkÇr>ð.ò Îð¹Û·làOÝT0 F$ò;¨¥EA5©Ú~ÏaoT:abßXòé AO¬EIWÓKê'^f¯ÁC 'Ê,>¨Áh2:ÃMæ"ìgõhÐN>¯ÏdLÌ}Vû)ÃSnÞÈçk!2SöuÔ¯ÓzµïÛ÷+~I \kbQ@p±-<¹ÄuÃ& eÄËUB&X\¨ü.=LöBH¯> v¼zÖ®é{"¯¡%ôúJ¾õs¼Oëk¼Z\ßÕe¥óªYßè÷´b0 v®±ÌóFßo>wÒñÁXÙ ÷ÆZ«×C1\¯Í<7ö^zYåÀã÷j G(<Ókû}6íÈ /8]5b¦ôtzl>zîÊG)^Ú4pÁq.å\¾ªYuP3^ä3¥O_îh%+Q=¾ÕÊJ¥¢JÑ(e£[7¤ÈªØU¹eàRbÊ(vÑ\9¿À%X:ÿºE¶F ç]x)Û"¸j(/Ù±Q!W!S~É fF0räåAzN©Î« ôÕ SHug&6#ÇÈnÊ/v<ðç¶ï¸qµ6" Ãðj)&xd0Rêî¿:%
¸¤¤Ó½Ø¢@ÚºÊuèÄçý¿ÿ(s SdÏ}'îCÉ)gËdrýy-y!ãÈçnÔübéÎR8HWç &>Jk8Ø¡$ÔÙµmUÓ)á§&¾Ò|eô¨y³WºiÖ º>;eÐËêÑB;+á<6Øΰ£,gJ¾$<'«tR·shK[ͦb)ÃIÅvÇà\ )b/)PÅPG³l¦únÐÔò= E!b½c k¡E I]´Ñõ6 z4ÕHÅèë-ÍE§õÕÖ^÷UÃð¯½ÑEÑú£¯E£úÉÿ[½+ÑÆRtQîÞï»ÑE¹ºÔÿ@î� Tt> endstream endobj 546 0 obj <>stream HWÛnÛF}7ÐØGPiÞ/EÀNÔEúPÄ}XKik+ì.íèï;{¡H%$¤8sfÎ9»þëb¹¾X~z.®VÜ 7o®>½»}½}{ó®ß¬/®Ökùh]]äÈ9{©ëGh½¿øê<8Kt.÷ÜÄ)[yD¬B;Ö!nJTaÊ»»×´Ù7ß usËXiosrç^þ»Ì+{)¯(æUè¼tF¡¢fèhëZ§BÜB ôØ°çF='wá-i¤~0uöXNáñ£~~YU¤ zô ×-¼s&ë|4kÏõ´.P}dnÅá�Ã8 ohMÅåú¿ ©7$ÄÀ?+tË4@Á1Wí ä5LÎ Mñ$TðçùFnÅñ«©ÿ x ¡?|u»çó d ¥|zÑF7zêc)á?ÊÍT¶?aýU té¹±®ÊsIpÑuc.P¡¢®$ÛÀ+¥'OÀtõ@Põ ã ÖT´TAá¹b¡¢MtVËÄõ<¨AöiS1¾·86Óög{]6ÈK÷·dDXÊ+P´ybõ±£ª¢ª¨~ëéT "»f\îÞ°'27Èiz-è±¥Claf±%!ðÅ»þ°¼[£ûåûå >ÝÞ}Ø@'çæóúOÕÔY-?~\Þërçúî½iÙÍßÿ,ïçWÏ ¢h4©(wó.)è86z; {ÀNfQ¶< بGI9èÈ pSî8k·;ªi
{Æg%\ �bä¹~ê¿ Û«+mÅX~êÆiÁ¸&\_°FØò©±Øpö¨Ò ÅLRÙP<ÍG\<(É] A¼±&Ûên%ç@g²ð§ s5ÖÌ°z¸éÐS²"uMºí¥j¡.Þ´G¸Rm%ÇP]eÍmZ m¬;íTª£¥®&å \ b ÑþyÀ¹Ê&÷{±8Á z¸·æY¸ãmÑaT¶Ø ë7ÅpGî°Ü©X]³çÁÿA8¬XRÃ3µA\G\M©ÌÕ zÁXè~ýÁÁ¬N åøͨ31Ü301¬¾ô[-5¼°ææ¨wF͸¥ÓíãsUÍ;üßôƲÇ/¹2ÏSÖì³kµÚé#ܾ±ÎiM·Ïê¸RöUñ@=bG:¹ÈG¹æg\ÕËÄVKTL©»&¾g¡)PÐbV'²_P?÷Ü0N:©v1YþJu ü MÁÎÞÿXÜd¬½iÑÈ7U)@Õ÷Tãlt·ÍÓ.ÀÚM¶L1>a¦ùõíevªqÓÊý×w8AP²îÃ~uÎü×À(©/ê¶F W&¨¦{ÚÏòæ°ÀIÍ çZóí!¡vº!¢ùj&ª¢ n øÀ@SuÔ=P£
³TÍ'5> Fkö 7oÁªûÈ¢5zãÃxBÅ0º?�ftà´ ¿ Ê4d<¬ö5uv=²9Xá«ýB¤ g0EAçM 8 0j>stream HÔWmÚFþÔÿ0R¿éÎØÆ6¶EÀIH/Ç ÜFQ©¢Å,àÄØÄ»>ßÙ]çZUUËIõÌ<ó<óàÃ�Zé¤ðêUçÃ4Ó°àõë»!þp¶:ah á²eègC¸ üó¡ëéaÐ5 ݲñC¸iý®e»Ó2ÎthzOûÀÖY,ÄwWHNEì!§d$I¬rJ74å x±"§À×;¯éÈ~+§x}Þ¢û¬ÈÍ"]Ðq.tmS·´lhi ')#³ú#|ß VP¾{ }ϬCoùð¾/%Ø1NEÛ·"¸CÌÀCÌR@I¼)¦ãVäG ^<ÅßÉ<+xÕ£%$t
Ú·¦njY<Ë»"\l/(sº¸(KYpte-åRØæÙ³
ØIòôv½�æ¡VWþw!\_µ:o§¬Øyj¬ZaX=·" DlJ @~ò4R¾ß{ßT¬É:ã öaQð½}¦ytH§a0£dPKjKCÌF\\ÑdyªBª°òn<Ç×[¢4 wå^øDÈúmLS+ïáºìÖq´S¾¦¢n0~¤xóu\/$^ 2UJÌ"£lÎhþD¤ºcu\_«cU»Ëø¦4IP¼e
OI;Ô5ãy®°2
þb,E8Ũrz-åT«,ólÔK¶üqN#Rü,)::þ.RuWVpØÞËa[<ìB4(l0ä ÑS½ÌQH }®ÆXÌ¿e"ÛѤ+ÚLWH \_EehzNmZ^
ÍpgÅj-.u%» î(÷\!È Ù0sÑøém©&Bb
Æ,J2Ù¸Y¾¹Á´¥ª[´$KÍ\hc)z£höÐý«¨§éU3~\_OU�gs Ä Ù±}+°2 DÔ¦,6Q°[£DE-É@ÇFKXQ&±VYloU¢LX÷KkXÒÍ°M£6Z7ºõ#Èð+\_ÐP;MÞ @AsÅ©Ià<ÞàD«÷¨öts,q\YiÇ3ÜsÊw\ñÉ«§ÁØ߼ƺU+oÓ÷«´KÚMmi6ªÒ.¡4OAób²N!)ÏìuuÛÅ}é<P2,Sá7lvHødkjQyÎ? \DÅ Wl/r±R6C{¾nv
¨¤qùݲËÁÆ°,oûäêÓ²{ dÈL¿ºå §8ÞÙªpåÊ1DcÁo °tuu¢§Î@\l:è~y¼oôu6~£´×ÎÆÔWuÙÕw£) GÓÁýxúë$ãÉÖúCu,|4çÀ¾Jt¯6 ^uÙ{OÑøaZöSLB9ý6º/{ë< àq2zayiü�åQW»Ã ÀLO@ýI? ÇþÛÖnfýbÔè@hп\5Z¿-
ÔgÇfGîµ7tî± $Y¾è4%X ^í[[kú]½06&}(ÚÉ÷&Rõ²±oÚÿ 0Þ5ÚÜ´ÎVd³üqx?zÍ~ª�1Å8ÄZ¼#~®3ßèêЯÈÀqYAeÛ8¼ýÓïÿöí ²ê×OæÇÃbD\·w -"--¨nËwyö7³ö&Î7³v³ëúý5ÆY íÿ¦Ò0@ÛeERkʲ®mTØ7ÃÔûI[6"ð,§{�ý5ôJ«{ óÚµm¹{ CýòQ¼º©½¡}j$À[õáÓõußÁñ^1ð̶ÓÍÅêöçfóWM}Ýûÿ²o¢¤~ÎëËÒ1|ïÈLÚ3\ ²TqÞöË¢µÑè÷ìm;BߨOñß
|%ógÇCP
{¬cÇ æ²DLÒ}£kÿ¥ÏªÑ£S§+v{§êUµ«Z?�#K_² endstream endobj 548 0 obj <>stream H XTÇÀÿ3óÞ^jÀk}ïÁ[ëQ5µQÑzÄX¯XADlðLÛI¼c<¼1 ´Æñ yî®j4³;ákì÷ïëì7ó?æ?Çþç·;� 1ºG¯ì?9Üs׸x{ÛÛWÎ (V=æÍHeè@Öɼ¿×4{Raü¦¹½
ÇÌNJ~{Ú½l¿�¡Á�[³§'ÆO½uOxààvßg:w,äö7Ü2gAÛßÛ^aÉ© ñé p)ÛíSØÍëF�Ü{ÇK³S¿¾y÷.·GûÛSgÏÓ5uÀã¬~{Z¢½Àºá·�à×P& |Ï=Ã%T a&À~¼ äB»åNù,&^áÕ¡ÃG@ ×#1?fÃ! ¯÷!ÐC,kXa(Û??®a+°¥Á ©çIA<ë_/# Ú@ÒðA\HLh6y £tXдe£½ø>àb\KñQ\ár´e HÑ&\Àw(ü;Á~àÏOºBC´ä»n ÏA(Ah í =t@eøCt ¯'Óq:Ãp¡Já(Á1(/ à øSpá | gápÎCª@ûÐq¼ï¿ôßó,&Ã|Xà+¸µP.¸?}ø ÜÈY?j$º¡®hz½£SdÞP%:Î ðüÎÃx1ið>@ütÄÈR2¼NF#?Ãóµo¢¾h3ÚHBÈX¼?¢Ý¨åðS² æÁbÔ}>EyhÚrI ÐTØK�ïFÛÑdGiè4:Ìy) f~Ö #
þ0ÆÃD ©ç«Ìlȵ¨~/KðR¼/Ãàåx%~~Ô¸©HýW¶¥£,¾æ~tyq;~n§p%®#b&áÄFV$#È!Pè!& \aP#Oÿ2øÚ:úÆR ^c5ÀªZ¬)ÖÅÖµÖÿJF)Pj-Y%E²I=¥¤ÞR?)Jb%»´\\vJE2ÈXe£ì'·²"GÈÝåÞòåy¼CÞ%.Êåòe³§\-ÖJ¨ÒVEQ®Jw¥¥ôW)/)CaÊ%QIVf+ï(ËÊ&e»r\qãðàp9¼Køð JÄÀ\¨XõWÔVjÚ^P»©jjW3ÔEê2u¥ºVÍSÔOÕCj±zTýB=©UÏ«5ê h ´
Ø¢l1¶¶8[mmf§Wê/ÕS¯Õ»êoÔÿø(ç£íC£h4@cé:¡ãé$Oé4
¾EçÒ·é;ô]ú>]LÒåôúwºæÐt3ÝJóévºî¥ÓOèZMkéUz:éuúõº{¹cÝSÝIîn»{©{û¬yL O[O§§¿ç¾ç±ÇÍÞcضd«XËfØ&Ë Ø.VÈØ>¶}ƳRVÎN°Óì,«bX5»Êì\_ì{výÀÜ^׫ÿ púr95¹WzÃ85'95Õ|Ô,Ô¬"ùä+rY�!Hè)¦iB&§f¿X"V?4Q#b q»á§ÆßabM¶fX³¬¹©¥\I:5½85}}ÔÌò95
>j,r ÜZóQ3QÎ×û¨)«85X§&X§¦ÒA§¦§&ÒGÍNM2U¡ØyJ£SS¡885AáRxg51Ϥf¤L5W-T¨95¥j9§æ §¦Z§¦§¦DÛ &jnqj¾¥@;о:æÔ¦céD:&Ð$:¦ÒÙt>ý+MÕt-]O?ÔäqjvÒ=:5¢æ[êq?ïä£fN õ�O¨GötöDÿ5+85uìC¶
mc;|ÔY1+cì$;Ãα ì «es±:5÷uj÷~X(ÞSÞïQ.W{Wy{íÞ$�6éI?YfÝçs?ïàÆ/¢Ò¾¦\Å7{7jÎTWÍõ?\æÚäë¿uº^罿K;]iÎ<×,W²ëM×tW3Ý5Å5¡¡G ç]ç W³VcóeþpÜvvÕ=QÎ~¼ý3ØQí¸¸è¸à¨rw8×uìp8r9ʬÛó¥¿k®Õ{Q+Öhûµ½Z¡¶S+Ð6kë´ÕÚ2m®6MKдIÚm¸6¸a6PoùKF{^ë¬u¨óÖ=¬{Pw®î˺3u{ëöÔÖí¼VêW¹h)²rÞ³±I~ò¬ëÜ\c¾lþÚ\i¾m®iöYf6ÏaÉ·ð7å Ë8ËË$ËËÝÇk°%ÐáY´1jÆJS½ésA4 &h®Í¥ÁÛØóìb0øÛK߸'þ[»B¡HØ'4e¸Y6ëB
pKcD=cñ/b×2Ä,±@¬nkùÿMáOF Àch561¾fÔÏÝùÆZ5Ïj òiþzÛëYû5YM¿°"L¿ñé][S7]vÆxÉd3E6é1zÝ(ÿßbjÑ°kSßð õ×ñQ¬ç±S)Ö4k3-4epiÊþúº~®êL8þxN "EÛb X!HBî,@ ÅÝC±E[p °»í.ma ^ÜÜ íL§3ý ö{ç¼ç=sÎܹgæù|½yÞ"o©·Ò[ëmð¶x;¼ÝÞ{Zêo;ro ·×KL=ø·kö¤¬½÷¦~ÿì%zG½SÞïêÿü3îãþ/·wÙ»îÝòîýµã½GÞ3÷%{}É×|Óø¦óÍèÅ7GʾoÞÔËþ¿}Ê8?E´¶Ñ|<@ã.w
<<0x 5áÔP~ºP ¡>24@¨Ð £4E\_h~Ð ÓC4~ m1#´ÆÌÐ?ö:V̱:ãçÐs@'Ì ]ð èê¦LwÌÝ0æ
Øa Æ¢0¥(Æc9ø+À×1 &a8LÁa:ViXfb øëÀ<¬ó±>,À°B65ØÖc;Ø[°lÃ.°;ce%a 6
ÒnF w,T¬rß?->Ź|"ͦ9x2cÞÛxP:<ñÅcø+ÇxOái<gñÇ xÃKx¯5ç®xï/¥¡´ô eÄøã|Ïð9¾ÀWøßà[LÆwø?àGç{ $"&%#ò£ >¥Lø¯GÁB¡TÊQyª@aN)\Q$¦ÊT
ºQwêA=©õ¦>ÔúQ@©+ ¢ÁTR5FÕi8¡¯¥³t®ÒMzHOé%½þúJ?Wd Á2DÊ72J¾|\_ Ja)\%$HJIi r.¤²TjRCjI©' ¥4(i!m]tX!#e¬2U¦¹&#se¡ÄÉ2Y%ñ²Q¶ÊvI]òùQ~}r@Èq9mQr^.É5I»òPÊ+y#ï¹µ°Vm3iµ³öÖÁúX\_ëgì¡=²ÇöÄÚ3{n/쥽²×öÆÇÏbTÔÓ4Vý4½fÐÏ4»~¡yÔ\_4PKh°jÐ\ZCëhm¤M4JkKÖ6£±ÚI»h7í®=µ·ö×:Xë(¨SuÎ×eºJãu£nÕW]?êOºOè=®çõ^Ó$½«õ©¾Ô·úÁÐMa7I-½e²¬ÎØùÌ߬°Z1+aAl!jå¬EX¤U³VÇX#\_k¢íµ¯½µd{gïí}t5>èC:NghO´\Àh|0«Ã\l«1ÖbØ]!ÈܼpòÀ)ø ÎB>8þp Ãe( ®B¸EáÔC:näWw ¯ ¼\ð\;ïU
·Ð3@ faXbqè 7æ~�}± ôÇB0K¹ QXFb0ÆXM6eØÌK¬À°ÃJl ñØ6cGر°Û»Ù ܸ®YÁÝq.ǸqÍÅU¸]ã¤CtÐQ:F'è<]ÀQtnÐMJ¢[tîÐ]ºG÷e m¥xF×CÊ><'ñ,WD³yOãé<ãx/æ
¼qcNËÉQÜ[p+æ¶Ãí¸wäÜý]OàÀ
¸0áb\¹(q)æÒe8Ër{åLóJ~ίù'ó ~Ã/ù-äß}Ò"ZBËqöÃþ8á0#q4Ãñ8ñdIK)Vð^Íkx=o༷ñvÞájn-Çó:ÞÄy 'ðNÞ
=°öÁØ[ÊÈpé(µ%¯@ æ®ùFKKi%¥äÜG¾ E¤²R^\Ju©)u¥¾4¦ÒL¢%Æ£¿TpÕØIsëKh!-¬¥´¤éçCsjq-ªÅ4fѬ®)7éfÝ;uVÖJé<¦áZ[kj-¯uµîÕõ½ªõ>ÑGúXïè-½mé ý¬ å·Ã²Yv=¥'ô¤Õ¶VKÞJ²+Ô0 ·EvÓ,ÉÚpeclM´©6ÍFØgm}c£m¬}kì;n#m·)2[fÊ,Y ód¾,ŲDVÊrYá\»ZÖ8Ù®õ²Åùv³ü ÿÉù·üGö:×þ"ò_Ù/å _å¨SrBNÊ99#gå7¹ åª\+rS®Ë ¹#·ä¶<øîªêª÷î{o7»Ý$a_6 n±!¡I¤PmÉ"H,¤5±ØD°cÚiJZKE 2J;aìêÐMþ8¢8ÕÁüQZ;¬q¦|äáïÜûÎ&0ýÝs?Þ½çãwÎ5?4?2¯WÍOÌÿÿ6'ÌÛæ¤éx÷y÷{x_ò¾ì=èÝëý)"1-ý¨ Χij@B,q6e¢y/l¬3)ª¬{¬QñEÜÕ«2ßÚö+)ÊÈÛýÑôaÔKƦoÌÏY5w6¥e;:jæÙ@Fu¬%m$Z¾ÖOÅc÷oµÄ¾õÍ-i3¡~1Ð1JÆÒbu['þ¯i³Ó ©hv¤RK±IûjÁvèrwèR;IL²ªWÅÒFekÛWÛÒ}MÑtCS\jÛ±æôéÖ¶ô馨Ja7{Rü>ÑYêÙ3{çäè]V·¥¢iÔ½¸în?#Nß%nA+&hG#Ñ}j¨/nGI·ã6Îj·ýÕ«V·5ã¤vj>Rü¥P¿"¯\TʨJfÄgѪBÆoÅØf&DE¸ð 0T²ÑB (D+§g«½#j7CM0±¼ôb£\·TÈÄ>¥Ø'LJ^BhB8{LÍÌÁ@¿åc(zEs[FT&£Ã(@gÔ§\A \?+H A:i¥H¸'
«\AmÍ$ù!¥c$±Ùj$JHíì×kH\tjôWÐ<Ñ©@^¡êuIàì.2 1 "òÎ ±ÍZ¨[IÊÚÅKêe]me¼Ü뫪D½¾|a;¬Ú[Û·uMõrÎ?ðØ£{k¿üÀLùìƯo^¾:0p-ÞýíûWìøú#¡ìÞ÷xãÖM\_yÔÙmþëfÕvc9ì {á!¬%ÊæAóm¸egyOIÄã wõáðS2LOÙpÁºÀ ,å9H¬9ðt,î¸p¾ý~´}hFº½»ËÃz\_K¼pþ
ù³fpjYrKbÐÀôäy!ÚynUZQ/ =îg°ó@s²äC-±ð®ñã-æ×G]Áóv3(a°XaÑ8~WÐ~0¬;ô¥?�IX¨jxq¹óçyð¼ Î1X¯çXxmpàvÕÿ·¸®�Ò
Ａ>àëÔ¡É~nð!.¼¤¿27gXè¾ äëôÀ¥°´Z¨ÓC\õR\h1@ Ú:4sS'\_ÎJ4¥¥ë4Bû §/=F$!U|ľi¬bÀÆ\4«+¹Gº¢\ à8õHmøÍ£(õf1x^/°Ä2yÞ½¼Í=<ïÏ£ïãyçxÞ ×ÌbHúHÄÒ54 ©åí¤\Øú n#×.(Sø+[/Åàe] .kÖ0Äç ô¨å®Ä¯%~Åíø°È?÷ ss dYÀÏp=µ=u%KV°aËd¶'¥î/8ª¸\$ãÑ?nþãÞð̽ä)ѳÁ³{²7í¬±Fo,' WÁ!ð>x8&ÁùÈ�E:·!·ïû²v4Å!�¹MLöL@ÒM¡öÑerɬ$i\y~;¾¥Î}^ùXÏ+óÕÊüа(ÀäMä^ò¨©¬Z"T» !c+u40xÁ2§Ì! O_ÎYÎaµ+ÀÑ!È÷Å´èÈ'A¾R:¹:PiR¥täyHAZ
ï@«h5h hÖ6}j¾ÄõP,ëÏ ?©Cõ+_
-:àl;Ð=Ã×ÁîÒÁ>±^U D�lBgËH±·Ø Szµë¼êé}oóCO:_ø|íðá_wGö>wÌÙg¾}lóÏç×Ø=:qüÅ]OìúÙößQl¹þö¸µÝú@y>!É\RÁ0êûz}\pºYò.KÚ5 ~Í
Êr¡\¨,aÕ©ZËP®§¦gPÃÈõ4ò ¯ò\ <
HµeìyQv¨Öë)óh½+Ý0ÄÒwý¡Ì\_2ÍB$i¶, Q('ML:Z(DE¼[Çt¿²ßZ"BÊ#ò(ðNC@dÏá5yPfÎÍå!4p-4yÉchÖ¢Kô5 àÖ¼mYRWq½0Ü�ñ¨NQ®µp\jáRÆè\ã¬ÙË4Á gW?ühÈ£¥®îbС@sVóc7 뽺�v¢©½,õB,\äü]ÈÜ[ÈÃÅ.æáb.æÏEø+.ØÉ È ëWÁÞcð 6=Ð#üåµL~NÆËE8T¸há%<"j>$lDºyÌ~çµqçÓÿ<|©ê7UÎÐ-ççìuæúÙ¦¹ÿsÆpÎ9:ÆùÀ¹ Èn¹Yr>Wu)ññCàãt=n&Y9|dfr0 ¬^%"¦RÞ§«Zå« N1( Ïd6±¡ Mµ¡éÆæ$U©ånÈ=Æ!Äà@ñS\ø§Ja :&ûçdI/ÀîØÅ pWªµÄwa¶§µ¿Y0Do×ùÖ2¬Õ>GÃ.Ðsw�ôëI^Xyªì[Yo®[å¿{ïï=g=k¤-¿ä ;&OG" H¬ðw¶;o:'vìd¾;eeFȤ¨ Ô' Ù·¡¥%²6£[b\ÓEOÍmGëæ)G¹¬éæº7¤ÖÓOLéæBªIíèŦÙ8OëÉC]ÿõ;ë@uÀ þ¡%o5Æz°ÄÑIÌ &ãV+iß½Þæ ËÖÕ»b[}ÐFñ¬Ùäê|¤3-ͷضÚXE°ö§M?ÖÞßJQíÖÏìêýuBôÜÛ?Úbåç?Ûûx÷mgd¿e÷DGÙJJúEt0:²ìÐïÔÇÝÑÎÉ\nzò=rnGîõ.Y£ Ý¿ZØ5ìþ¿Ñ,ÆÑ,GïøkÙóù&8Ñ Ö:DU//{}e.Ç"q3Iã4N±!UÙð:/}ĸ¯eD©w2·4K.ÑO®ÅÐÄR¨Î2RVWÖBÿ¶¸ LÌ ^íöm |²ÝtÏ1³ÌßñetTnÍÒQhé¾Ñ}¸Q\ 1}=¥k!;HG¶»¹üîñëUÛÝ×Òvu=%«ÉpêÖl{AOüWéǬB×TX ~p0 Ã?91\ÔSk¥&ïVí�íHT>8®�\_¢ËÁRËc�«�¦ÊæÙ\Ù Ý�Ë¿�¦ bÌ �wÐÏLpFÆwQlQ ËxSèz àM�~Á\_(àÜ&Ia I,KJ¼-Ñ+ö¿ÊÚJàN(ë&t�§ÀÑÐ
¨ :Yþ¡ìǶ¼ °JwáÛ-ñÒO^Ëù6 xìS°$ay�é"Íèfä+°Ëúc»LòP=T\¤ãÕsïLøµGNHêu¼îë�¦ýxý¸Ntßïh"¿é\_À«ÃÑ4X¦¨
7MЦ ]l.Æ}qÏ!]ý¾úÉU(Á<pfNêÉõHô¯u¸ö¤J7EËмýÒõC¼Õd<ãÏaùÀ,tÝWTÞ²åÀÜb¨ <ò2!{w^äÞ¸
Õ��üM9³Àr¾H¶nõ aëÝ0gxý(ü§|ïXMª\_:q>ä\ß p3ÿè±þ6FáÛg�\ð1òÅô:,Yj-ÕµÈ)ÛEs{fÒ3ÏGÖì\¤ KóѤy^%ÖJJX+u]ÞÐ,WéØ rÌ%Xy1ÖY,ëØ´©»Æ¥MK³É¤Mm cú×UG~:e¯úä#[ýV´:Ú½±OÚÙv¾í ÙÀvF/GQ4½fï´åÑëÑövi÷D738Éd;@ìe!nôyR@X¬ HT±ÐÍ¿¥/ª"¤Ìi9eeFVÅFU?Elø Ô3ìÛµ TÝ´áµÓfYbËu#RåÈHGêTu¶.äÛbPz4ÒEÖÅÌXÓ>"¥ù øÑÀ&e<«c3O¹ÅÀë :nEiµÄ6ühmðÁiÖúëø>|.h)gÇOH>45ÆÍe¤]{ãøvÍ˶$¡kO³]ýáÑ=[{jÏr;Þ|´zø¯Û÷®ÝVÛñ»±LP)ÄÅ{~¿þÖ ®peeÿýÿyÕá=OÞvó ×vAëëú÷æ rq"2gjJêI¢1ÒS»~èî.øNfépZÅ2| ÁéOÝlÎãTç?X{!ò¿èàG+öîÛ·;zw0Ç×"1ö×ãxì/?Ò5~là(ÁBË2/a' èVy+{8½Ä¦#ÉW OíÊH,ÈtA@7noÀ °ÀfÚLPun¡è-yË/¸p²Mçð%çäîØA9~ÞiDàÞX6òím]Ì»(º�| ÏÉh»ï~xz .tl5,Ú¶ à % $æA'Ukï?hQ+Û- +/Wn×ǪÈ\ô}V[V1#E0US9:µ ßD3NܨH"SUq¾y~èÏ!XLð=kÍó#U³ ÈQdpEG£²ÉØßη¬R ïofaÊó¿
[[ää©»x÷r£¿T´-°öxÞë'üýä;¯=qòÜÐÜä¹gf8®EMïîuéd÷ã¯#_×ÌhX
îÑTdv¶D¤ÜÙ· ÑØb°ÞÝÐ{¶Æ'3vÎN©ËR´ t&è\%?df5À:û!ìø Äa¿Ôq²\_ íÙt÷+Ï}Þ¼7{ý.bï"XIÐDp)5OÓ"#qZ6aÃ6aÃ6,B# §Ã?aþ¼ã�ÊöÙÿÄù ªcàwQl4íשÔèaÀjØ:Þ)¤ð
)<¦øhjG¾²=kç"IôáAÖqNR2Å]§Ã%vÀØ:6u'Á.½®^¼Ý}E~ ?4Õ\_xvçÕۦΠ;Ofä« Y@NHÜw�®·¾¥ín½ûì/°½0¾Y÷¼k°»£ rXïêô\ëp®
~F¾ü|I-÷Syñu¥ÑRêÒëlD¼Êë<¼Êë<¼Êë¼xÎóîJÑ/sí)
EG·ÈXFÉËWÉkGç&ó<ê<Ú/H/ç(¨¢f\jÌ6ð1!
¥¼²ËiÑúmY¥ÀN#Lü¨ d>I¾#-m½ ÔùJrÔ dN÷µi5KÒk´ $9U XIpÀ@ê~ ð0ì ùÒå±4Æ>o� XKôñð@Ô-3ʵ,\ËÚ¢0Ù%É5gî~fé^îyp?ÁRI³ØjÖèo@_y®Úº!ÆØçÞcñ
0M¿×ÁNɨ86Ó"÷uVLÐ Àô[Lî }ß °$ÝÇÑNù)PþT 6©;H±ë 6\fe4Áòû#KÕ ,U¹ÓꮽKYª�R¸ºGô)¼J]bDåyïÛ[#¶XÆËÆêð:ûìv£¸D¹n¾ÊêI|er¤ªaÒEU=úC«;õ¶µbû¿ %¼õ¥Úm¯ãàCÁ8hxvËþÚ6Ôñ? Íwà�Ï^ 2ËDÚh&J»Jfß2ÿsÚÞ»Ö#9!Ï#]EÐÇýnxü ð�ÛÉÇ $û Øs¢Ða8U@'t{¥ÛD%ÚG°sàã¨Óe\¹#÷ïÉè ½Ê>ÁéUö°?àö\v!nòÊýY>{W¿J0SK9r!ÁT¥1pËÀ«½æHLdK {zÁa3:¨bMA{
0ÈÄÔ¡ +?=ÿª¥#o$x 7@mHÄj3ÑZÊF!aÑ £S[Ä#(ê2¢EÈEnÊ Z×Ñ &¢EQÄÊBGµuRaQT18ûν÷ئTÒ%Ïÿÿ÷{ß9çYöì« ã®f%ÀOë¸yôß�_«Ç.øÃ)¢Xij2[y»¿10ØÇügPÇà·<&ô�\à9äS)ä³pÇuBÛ)¡íßÕÇgS¾Wöà&?áGëI,©ëæäC¬%ktÿ#~lòãóãf fÎùU,£Çþi016ÖÛ^[þ~ô ÖA,¿no_¶,nþ&%fÐmúðt£óB½R,SX#�ÝÎ>" S¶�8p Qwãì«-Ä©-jS9èzE$ 'ï¡Ìɽ�zBlQ(C#¾D_ÉxË Þ'kh54®{ KÌÜ
,£$§TàÕ2ÕLÇÒ¬R)³Ïm#R±<(¦ ÇÏôÝ¢x^ç½b¶·dð[=¤¾é8
¥Í ´ôBÑÌq[¥Ö²Å3Lø vi1;0gêRP1à(?Ê 4(e°AeÖShêÝQ }¨.³ÈS³§³}×Y»uìÍ0«b5¯Ô$§±íæ2X̺¼àA X»uÒ2ÏpùP°(QUSm,0UE!0Q5'äMVAÿUÏ®3µ¦zö´Ø§Ë¶voÞðÃïüxl½shàÔçU½²Ñìü×Tw±Å¶íÍ[½f,Tò=÷ôs?jÉØünOçþ±Ãµî]Áºå§ßy}Ïé@E~ËòÆ"sf7r&Aó(5^ARÎÄÈôÏ\õcKÙd5ðîÙ÷ñã!eAÚ!?ÓoPãò!/hbÃÜ)£J@²R-ÃvºpLÊ ÿ| îü ¥$(ÍÉ "J°¨g1)-¹ÉíÐE&"B>;Ä=Sì¢záý@±^Q¢Ü¡DG62CβßytÚÈܺ/2@³$c-Ù ®&meÀà ?áOÔ$ª!ËM©
l[ìÙ{ §µkÌ´oúðã¯î ýá»ì5?¹ÒßóIú¤ã±/N¤¿L¤Ïk6ǯۮ #æ²ZªÉ·¹}¨ãyn;KêYÒÍ#!ÏrDZ¸L1¿î¨ÃZEî%ûÆG¨}~¬HÜQTø+yMZÛg0@ û;DΠ帷Ê!ìl²7Mö¦j/ díÖÞt³7ÝÒ^Êgz^ò
Ôc5aÙèÎû�º±h dÞÄ¥Ë"6[ÇíÜ] " gÌ÷.qH¼µ.qÏú÷7 Yçæõ³Í:Å^îó+æò
¸ò=Íûò;e<5p8l8Ì9ì)S%¡D@¢<êr&jpA"®(nqEedS9,ÕÄ¥ ò5ÊÍé
å÷òV6|?MeÌOoyIàU·V0oÒ¼ÎòVÃó·|8Ó¯4ü»°Q^ÔõLØ"°¶EEª23ªRÉbdĪc(P¦v QãOÞ·vm;¹÷Àæè)ûÚ+C#Ãÿ´Å4£S=7nó´1æƳ^³DúmÈ^Y¿¦OÕZR¯J u³ T~g¡ßR±jåm±¯.2Ô«Ö+ÚGBS.³TP
%&òÉ�Í^e]8îsÐ7 \_L˪Ûô�µ²·ì
6ü²KUôqá>.u&4XjªìBèê©w õ)(µÉÑ®L¥"ö²Éh"þYCC8½JÁÍìù5ÈoC½Æ»at~Ï@ÉË0(¦´nõ¬gÙÕx{ ¬bJë¢ß¹ÀKqoáÉE+'vóHÓLäÒ%tMMÖ U52TQ +AàDÑ@mÇüÒ!0[B¶Ûï¾ØüFw2iõ|ÏEóÜΠɱf§S¿jîiçÁ'6®ã)áP`RsÀ¢iQw@u»´ªf[¸ã8¥©µúAËû.¾ÕæÈZÐA²»íêñÔ>@V'äw'q¯áË¿öÙãò.ãO/Ów±Ä=ä×| pYw,ÕsôQ®ã@ý×sMC GsMBYÍM9»HàR4EL ¹,sî/ª¥½iï@ÆSÇî@?\«µ�[-À½9´w ©XQMön^2¹çÔø8Ýþ¿Ç:¶8»^aO©ü² ê¿ma2'ÅÆ~²$æqäqû§¦5Ï3|¬l"3TNYÍt/üø>7|ÿð÷ÃJk°¦Ua=ADÅ¿ #Öî u¨92¡FíS6ÄÜP!Ýòò zÄ&¬]X²®+ Ã=8ÜÃ=8ÜÃ=8¼r·ð¨¼zF5#liÜ¢üRÎÿã¼Z¼®ðý¶cb'vlÇ¡1 ÆéRÙLJîak´BÒH´QÛÐ ¶Ñ¬°Ä£QU£"À¼ôaº!Õ @ ¡Nbv÷ØL¢«|>÷ÞÿÞ{ßé³e®ô4+($ s|CyÖ7´g}ÃF=aèsÜê?#Z@©º|TÀ 2ÕJ�´£à³Ô&©}¿PÏEÐ©ß 5^îVã ´Äu&·èÑÄK4X%
´úÐÛ-5ÕàP�t ¯ÈUN ©Ãös.ð
lv(ÅÍY;0 éO6KÜ"à SmÕfD-A[b³ù_Æ åG#aØêJúøÞ£G³£éÍ)éÕîöéV]«t.Ù £àB~1ÊÔ 8»1öç5�¶?Ó9N? ºd<'tv¸ßÇ9x_Ô¸RQ¾OT2[î
Ì¥Ì6b8Qp²º¥æÍÄ ~ JÁ.ø¦e°²WËÊbZYÄïlV£û¥ÔÈ?¯¥:Ã:&0&©&&U/^Ììb !ü¯Àýµ!gQDs¬ e9Ç)ð0A!ýGÍPZìèsAÖÆvËÓÐ jBBª ĹÌEø
ÓÔô 6pHÚ;§íö·«rq|TO RþõÜSãtôØmgþ°ôû]«w $[µzWrÙÝI;¾¯ïèsûNÉÍz×5GW¬>ux,¥2\ ü)ü8ÃÅîãYù|ì gkÉĤÌRÿ;ëLRnÎ/kòÙ¼<²Ïã2¬íH|"Éзãè}Áã>ÿ7}kz}«ÉÐ7K½E|ÁK©Úcl¡ c[½©íP²Çnì0wïßêK¡1Ä×¾wï=ú¿°ñ|ÖRGôäá ¬¨lØ é½oÍz V§¯ST¬ÁO¡ð@'m²ø3º:5¥á:pV åîr4~3ÓI¦'=)À'k ¸HàÌ,ÐÄí¡t
Åì ÙPu¢zÑO½¼°Nz Ý0·yrÍV¯©PPŦ)Ã>´§ûàÁîÎ#)cÒè5£ }ûúåô-sò+7ôüóBê± Ý÷ëé¤ñsðt£ F×S"6½ÀùeD@鼡bgÊ9aräZpBÃuÑ :,-dËÍ V!æ©ç¨Øª;eºËþA9 Ê8 ðà è û¡Lù©õbêáG SwRhXt~jYê!·7ÁÐ?JáG)üȵÅÁYòÀÃcþ8²ùý»vwîêí±×=wãÖåþí¤1ÚûÁ
ç./íëKÿùZúvúX×Ø´¶ñ ½qÒ¯WôëCa ñ¡¼l(&<¨J¶¸Wy8¿ÌkaÄ}È9q²\OfQ®òmù�ËEÐ Ci»ä2½(ð²)þõÞ"ë3ÚÞÊ/½Q ¬Ål«:°Sت±iþ(G ΢d±�º%Ó?»ªf¢&a-éÜõæþî G{öË}¾¸ù§3ëìcôóË©óKU»
FÁ5cfß>%µÑÜ ¶R'¨.º2uq KW¹~|R9ÔTrÀÇàÄu|ù0pöÃã]úг¸NÜhE0,Ìõí�:áÎújTªI8j Ùc¯KM)ùv[·?>ì|ÿðF¸÷S÷.YIxcµùS$O)õzã1èt]D2®t ð;á=©~DIÕnßò¡/ Bî
8§ØsÕd.1ÒE~º¸¡ ÿ$Ñ\꥿nVeÝ®Û ñ¿~5vù8T° %Ì,ÅB§?pU¦ ?ØB ¥¦ÀÃÜC° ²¼ÇÍÍÌÈ·Aú8,uºa[± I( /ª6-0,n2ÈG¨z#ÝóÃö� ,Ñ� Æ:4iÕïU£M:A Zú®¤1þ©§h©É&¤ïf§º§ß@ÓsÁÉ5_äu_? ÃdéTªi¦!·-nçÈ1ÑBã+Gp¥QaÀ_ªNTã0[3ÙQؼlomkÜ\¿éø¡W.]1Ú;v´n[Ö>ÜÞö÷¾ÓçfºyégæΩ9½ýï ¾ÿÁk¯6-®}zJbj{Ë;©ws|ß»izMhb"m{@ÛZ
Au8HRxÈ´#D¬yA´-F^$;9iA.áÿÅ!ð¿dh>6@¥¬×�´cÁí'£3»®RÖÊHsJPP#(@AGA Q=.[�ÈkKµÉ}ñZ!dCà^#k7BþÚ2ª?Xúu#v¨ñÇj~ø Í3çÖ½ðhë+;w&{î±ïLñfÑ«mæïö3Òî½nòW;=E hª8n¡!naïn!n¡nV7ä; eR¥!5(|éÞíÃ^ÜÈ{_»iÏkÖë�48äÒû ÿJ&üЩCý¥¬WÝÌ
lÉÐàü¯.w¾à»E¯K_àN¦zÒZj1d:d¾�De2ÿÍ~¦)XM5ksò/íxRþÊu¶ãcµë× 0Þ~Û>yä0û¿z¿ë'zÅéSîiñ)]$ÅRå�øåÒïuKxUÀ VÐq(7"ÀA56~ìYñÊFòJT1íø½n¥xVY\¼å72þÓ)à�YÄä4¨#¿Êf ,.»NXjñá þX¤½ êS�£Bº·íº^^½^Ô².c$ àFªL¿2n6¶¿3lÇïîGÛTÖzÁ6ÅÖGº Ëérúóy}" Øh/ú§\ëÌ!À®i¢þ'K«ó øt44¶.B¦5Q´ö\_êT& ¡éÑ|Ô+÷Ëüò�¿XÊ/ðçXjÚaRa(pæ[j"Æ4j e)/Ûòçs6ù�§LձġÞ0ÈìÓRH$VR\²À¯hI?hæ{ÆV:¯w½É«EL¦?é�³q¥Æ¯À× èëEÙ¯¸ ;Jÿ(Åù(Å]�·f°Í¯kw�+'
ÄKê(JÉë5¨©õ�Ø×бËeÝ û0ù ÀU FñÚ·dºÀ38ÝJÔɹÄÅ·ßGétÁ.9ÒÅ_hÊ=k¶^a^qn¯\ ðÊ%¼¼¡V@½rÙyQÜt±¸i¥H¸Ç 趨ui< aÐFÌTúÌé³ö¹3Ãöùãé3¸¿ÛgÕã¸3l}ssXçæBûf,¸v!åp°>nPji'}0k; T9):[%:_î÷4¥Ô!´F¦«,Cäß²ÓP;qz K]@Û !½BÆ» í) ø/áUUuÅï}oß¾ÝÍÇnv³Yµf·[I¥&! ©ÁÊCÁÀ@ÍP
±8Kg°Í�¢EP>-SéL©bÇÉ°hNë¸fl üÑqpD©- ^ú;ç¾³IDÑG~{ß½÷ýî¹çãw è,TùÉ1"¾ó´üXÀ\óÊ
Kk5 ÏTjÖ±×ðÚ0;ÅóóT¬T aç¦:S#µ°:¯3@)àãPÍ6T³m>¹H°§^�¸"ÅÒ§ï4ÀAÞ#{ÙFK¸$w°Z@vu~Zw lA?øàÀl/+¯¤¶^WÊÈjKäÕs2r¯V¯JAìÌE2rTFÞ°WêËBn)?äõ²ENÀ%~eC1x/JP¾ÏÒ0 küߢ±?þÀö¼÷ôoÿ
áOÏyûô.oUo
¼^½ñÆ0Vƽûí÷¨?°ÖQWÓIÚ&Ú&4 rÏ !ápÏÒmê ÅíWá£�3µÙU �wáM¥¦ýWÙê¾g\½ ·éÞ$l϶FÏúS"fÞÏ°
IÎ&8¨}¨Àê<)høÒðôÅ1iUì72®yq|\_ÃAUEÕ©Ü ¨¸98J¾±iV(×Ь$vd®w¸çåïXh:v$ý ðÝ pȼ¡üGþÃãt'.9²qX¾gmz¶ZQêwS[D¬F)ýè ,\_°Íx ]AtLÛgrFÒ¯Ëf<4èëÜëÛ¨ejg ^û×<øÄöû[¦Í6Õ{ì ]KrümÕ¼;à×¹F7£.÷ÁwÃú%~jÄUT¨Lÿd/7 W"VHè|SÀÌ÷Ç·J쾾䪨ɵGæUylebd²¼^'à'^1À1¸ÀÓh°SÀnÁí\FZÞ9�ËDr¬Ø$Sv[´Õ·dî$©"Z"2Ú Hëùò¨±[ /A_xÏ[$+ãH(øàQQó¨[§zë2©<9ÌÁQgH¯F0Pqjjî$9Lkb6mÓÙ¶lRg]×7Y?½ñ¼Ý±wïv»lÏ.Öm/ygt¯Äì»)ä¨Ã_6èg=DUåkpæ 55··$+nÛ¶ÜKO<¸víà©}¿(ùNù_FFÌþ¬U¯wÃg© ~F)ÇÇÇï6|h|lÕLµ 8BP±?ØE> 4²¹¶Ý;00Á¥ú5ßÚ
}¦÷±Y²Å->)®ö_;b´ÉUDå ³>î05ßè«ú\?sÜjÐK7ahïNvvn_3$ñ)×)Ñèãw g8;\e)+½À óð½\[ëöd0Ù¢¬\·f}äÓÏ¥»¢¥¯¬µûô ç8æ/£ûb~¨Å�2jd" üÍC4e ?µ2ߨR!¾Qåßh¢x£kº¿¿fs\8àîæÙ'¬õ8>pZiWÐil0¢ZÒ&ùJ<VÇ5i}»Ýg-ÇêôA^pÌúRÌ,õÏY¯ôÈùAkC\vÙ´SNd5B±ã^æÀsÆEg·ÌÙÇ^ÔÔBóóð6=Ü[,5}ÍSÞbk¥s\mGò#üÑ棦¸ ò¯c¼,ß\£ËÆ|k´bá#:YOäê¬Öx»µ]÷üððÑßö9ç½?y«®^òyu^ÝÐoxÄe&¸Ô2n¿PÇüB¤ -�%}X'µÑÎ(C23LÔè¶VËn©WTZ®=Ao±©~õ²>¤k½ÞeÏkóðÍ#Á¸jÜ@³~¹\_5NÆèÞ¹Ó þkCÎÚ¨BàXJºÏïð .O´ª«¸5isòÙø¦S,¦ÍÁ¼)¼½£¨cAÉá½w2[L¢£Oc:¨üHöó§Zjêð±Ià&î¶ô1y +ê¬Ñ$ËXÁÔ\#ݪxZuqZ5¦Õáõ fÛ\Á¨õ"éÑÙëÙ3õqo÷yÓ)>E ASdÔ$ë~õX1ÝØY¯KãÓê ?ày4ñ©´)ì FNqÔÜ\_«kù+r3ú;&½àkð«$ûZ M/Ñ?ì·à 1 &Ö z´G%rt (g/(ÇÍWHïbõiÜÝY<¡PÁ\_;£)w¿ûÀËFÙÚ:pVT{^F®ñHcè´I½¶:'àC×À8iû«&bêÎ1aTOQS?&²45aT³Ý^ [ñÅï Ôv QQ¶@rÆïEò½ãK>c%\´áøéZ°EÅX-SßÆÃiE>ð¼ÆVICÓ±<¥µî]sÙL²ÒMäìܹ·½ý¡¿×ß«éïûh¿÷µ[ùµ¾ c÷±±7yYïq½yÚ½ÚÔT§uôÉ°îLJS\ÙÀwIIóà \ÖGªøÖXÅ'$\_{ ãC²,$ËB²,$ËB²,Ä΢(Á"òHMuRêA :ë6ø¹Î\_Ù>¢IÇÈ>00�3·r\"°Ã¨4-1¯¨BJì()3yy²¬ £y¼«,î,ÙM ÙM ÙM1Ù øà VMÄBç9MP6ªåªaÚ\_G¦¥8·a?mFV¬>k1·©jôc0D1218VbÎÇZÌWÌפ sËFÆûZѱ¶ÕU)OAëÌHÁ(²²QIå³k¹ë;\_¥þ4!í/¸6¨¥¶³iKdí\"×ÞOÍ£G¿Õ U}¼·cDuôvI©??ÓñäTo궩@ xìÞy»~Dñè#=:×c· Ç;ßCrYÍ Ô;£Sz1+ërv%à]yX¿úð$SÉÿ³]50U^gøïç^¸"÷r ^àÞP¬Á¹-µ[ÕFgiEf¤ÌªL 1áÏ1üaþP¤Ä¶Äb©Y2Óv³m¶2"Æ8t³Î4m¶Æ8{× {ß÷|/÷Ts¸Ïw¾óï|ç<ïû©½´¾³ùÈaMﲤãþ»ÃÏÿbwWß°LÝtü;Û·ïuªã ßÌiÍàõѱD>ç|fG>×)ùò®Ú±©;ÀtÄh2ÓJ^½òI»ÌØdÈ 3/\Æü\_$¼\ÿï£üÀ°ÓV<À)BqNeaGG[dqÄébFJmì,NÑA Ááð#)øH#½Àb";LW|Zs94ð¸NÈ
pÜ
ðBb¬ÞVHÒ3sö¨Í2!iÄhªDC Á!zAª{mA-GP£(XM93ôijdæ ê©%ÊG êoÀÇi;.ÒU´ÕsmRÏÛR®ÜªØÔV(B@¬FÔ+�ÐíµpRõðG XÕÙ.(#¾;ð²x¾^#ù¶SÏ×x¬¥©}¿ñö?¸xßzðpØ÷ ]Ç~ÕÿFQ^bÔíÚØ|¶íÎWWûn®2^°ëÒÕ{¢{wóMÞ@-þë"U>Y " }Ä+³y¶ãxÖ¿g0ýÓa} àoêB],ÿf°L «PÇãöÑmå(þ¢õ1bèäë¾e¯Kͱ"ô¯©·ò.d¸+òìp(ʧàd6ñ5ªæüS×ïüR'Á Áah(¬ºø«ºc�oNQÒÈ1¥YÖbX>arV¯;ÏNúÒP¼dIZIà¼b þv²Uëøõ59®µMYgäúýèD%íøþ£1#d6Úzåx¦alåêuÅ Þd°�XT©!6± V2¨g°
A9txDÛ< j ÀJH?ðxô¤(íy¾ÿ¼åå,�YËÛ7¹§Á|+{VsÏágRÌ°\_Ö=(¦©cd¿¥/Q"(&&B詳«ç2ÃúòËk1ÿPjgËþ¾óm=?yu«ö¯ÉÓú²JËþ»uçÊ»µ íGZtí¬nèçLòëùè2r1«9,9ìÎsèÀr¦4pÊ ¬VÔ 0ÈC´ØÁ5Áàðc×qÏÔªt£¤X¥LZ�p~ ájdcG6RCX}E6r ÊÄ�ìl@9÷¹�¶^£\¤?/Ûg>{ ~%¡Úpf�Í ¢Hóðüô >{°éÔÖ=Z©¿aKï9ÍØy÷úHÜ·&¤Kºúï¹äPÛù MçúäضíÖ¥«ÆÛ÷oYwåfù#¨FÎÇ\¹ôª 8ÿ:1ÍPw@¨;ÔêÜ|ðÏ<Î J Ðn÷P(%©ô ô 5¨§dÊ5%:é øab©ÊG4ú1L"B¸l%8õ
áúÝ«¿ø£ùÍÏͦî«ß².@vÙB×qUu¯h1Q ? À¤p8Ó8yHôQ,0 ãÎw,¦s£§£ôiQ2pÐp/\_»&nÒc#JDÅ=ÓíxÚL6§Yü,.�£B®gÆGxæ
»5+aûò°#KY"Ï.}(a¶;u°;u('BdßKá{)t/m½,L¨,¸zaËÈL±¬N&Û8'iÔ@n]ÍéÓ8õ¦ñil%|°:ÌéSAN54á\öÔò6ÖÅM¥]_4ÐÞ ²=!Ló\<¡¸NÌP ã0àOÏ ¿2¥ß#~°Æb?¹¯/nܲ~Oã¦-u¯vîhêînÞqR¶Ü¨Xôûêñ±±W}Ü÷Ïë(ÀÝÒ� «A)&¦E8#ÌÓçå'ú³;B'8Eá@?°î©Eq3BrÈía²iÄ ¹ ó"ż ËÄLæjËÐ[¬èW0áÅf³ë+¶uõƹù TLÕðÜ3?2a@.ó×ÔCLbLbLW,IZe- ú¸²§©®�P MS©ßl÷ mÒU:Ô³B´ gfû¾¨Cdí!A5¤xY¯d{³XpB0¿]«A¸,y¼T»ðÉfЪ�~cV)ÃjQ �Z±´³ ~ÉãÔzÁ© x�&§õ:´ÒΣGövny8~ý¡Ö¸õõ\_>y'iﶮ£Vkî=³ÿÜ@Ç3Öµë2hÿ½Vñ§][eº4~A~[<2~ 1:ji~íË mT¨4$GçÎ ÆÎÍÕ³Bjí|jû»ÿÁ$Q¼?§y~ïüêùzkIò!¸LFQO%o¼¢ÐfÕL^r-¡ÞÎeF&èPMÆ0+DÐÜ(zÍ?Jó2ü^ãO¢×ñ$\^Í×ÿg½üc«¬Î8þ}ÏûãÞ®aYk-]uô¥ÞI-Thj)?ÖaãjĦA&ÈXÁë Ap8R0G&ëºX§1¶LkG©LÝ,ØÒ6ؾû>ç¾÷öíµÅ"üñÉsóë=ç9ÏyóÞ\éc-cýöé¤ ¡Ø¾ÇûXW
Ö Îu�Ûí©Øt{&RíIf]åxùÎ%Ud Qô= Ç0Röv´»øiº«»½(O],ÛÈÑH=òIñ´§ê³k?´]gÄÎQÄÎ_½àµ¹9Ã(<Ëóõ¸¤ê³¾#¾G|f0Äüès½gÓíÖ(²ê¶î@õJ¬¨\®w%%mû}_ìËz¶GÆÕ³ÿ6J:Ç¥X£Ö¼»£YÄòó"2õwú¸7NÿúÐ{Ã^Î=ɾv¢Pöæãá8}h]ü,qßÔöX ÆI÷¢qaí^TôáL$©{Ýã£íFq{"l®ô»uR§þHjuûjãcÑóú°PíÞÇ}qúu±§ ÛBB Ì{ÖÀ54 ¬3åp=ï£8EÎ<ýÏúíCAßòc¹×{6[d'ÉfO´@zªóÔf¾æO!z¶ñ « KÆ#]pÏ«(30ϹRÏv7d{(_@Q
¯¨6ÌQÏ!hþ #ÔKøË êGo,ÄÍÏ£BxÝsÓ'ª 5!ì'óoØÏx¶¬C¨'Õì7ä²¼r2Ùë}ûQ!÷#5Ѽåx=¥ÈÌ1[W3Ö ñCÅänÚÎJÿ_ù10.jÆÄBäÆHÌÑñX8û4Ú 1ZçÅClïa^¸:ó£ÕŹ$/JüJd¿T$/:+Pëà1ùSñ
ñæTLä36dÈ@òßrmëQG©ýÿ«m¾B¹yˬu(6ºé#9ä)ÜͶJè#ïѤcàC:õ½(2O2U!;ÜK+£í3ôÕ9 ¼?ò=~Cæ7ÎòÍĪÜíÒs#-ô[{ù¸1f«cÜ+m,;÷òë-d«g+Ïöº$WëÜòÚãMÆ:y;dý, Ec ý7ó]dÝýºYPÊsÑóOç÷ç³ß·8OµÎû:oy¹×|¬Æè Ç;S¸æÝìòÔGp]²5´ÄãZÚ ±-D[îÀ(N²¬$äiû ùún&òüÃÖ /ë<¢sÙr¯í½Üóµ\ÇÇ"'¹gc bîjf]É{(KOjt4Ó#ñX¯ãnÎÏ¥ãk%9èÔEúÈzä{<{;ÛÚGYË}¤ AÖ¦øþh5îqd ÙBêI© K½~×ÞOù¦ÝEùÊûI£j2ªÉ½@.ôÆÞU¹³ÎǨÞÙE$W¤ó_4hZÑìLE)ï^déd1É'-"mÐr8i" .#Suëþ2®¼KrHoÙ^È2O/¢}U=²) ̤)î¶.&ejë(Í<O0Fø¦¯< Æk(!ßP¿Æ,ÁÜÀ·v
ÛmÇ4ó»|G| ÃúÝÝzòíÍøÈCÓHYÅw/ËÐ[èC#Lòd<×w_#cýAä"=xv ¼>ùíöÑùccôRÊ
ÀD¬¥\0}~èÇÉÀ²=¸ûÈvÆÿ½j;~þTê5äkÉùgÝ,2Þ'+5ߢeÉf6jTJÍK(5X©áÇÔgÛ¨±N3Ç\p·;xKä40¦ÕaïWy¨ nà[»ïðV´R_SÆx×Ýöë8ÀýUXKÇ×7i;^¶ZÔ%¢3÷V:p»u~åJÑ5+§õ^VGµ±/×·I°O¨ïGà»ç.úi ÷'æ]jVDúë©i<ö¿F¤.·Åä94 VÛÍõ~9B´ìñ+"m"ÛÖÿ:¤Ì³îß>ßó ÎnÆÈ]hPsQ)8ç©×à=þéña?ncÁóÚPÏGùÿaSÇQ)aL£|O°¯bË1¹#µýé$I\×äQ«~t'õixÌz¹?ö1×ü衺Ȼa¾Íó%vsíV¼p-¯òÞ¬ÊÜ+Z>
\µÑ½Bmæ:6 Æ9lS3Ü+"Í^ÎA¬QÛÃ&|ÓÖïñ×Õì$r|+÷Gô¼p ?gÿ7¸Fa3Û~ÀüyùTxÅÇX¾1ûKø)cÊjÚõ<¿K¬Øvõ¨ñ¿ð-A;1Êü7É6ó_fksÇ%4Ü¿ãÏõzLñø¶ô½ìµ¦ý&R?§¢«ãRï~Ë÷ïy§o¡Ò,@¥ô×±Iö¼5Æ÷mò±ÅÝot»GY¾b¬Aý2õn3Ì\2qhòøάà©PsÛùÏ×n õUÆTØÃQ«?Àò,R1RßÐW¯[¹ó ËËÈDæçÏÑÍa¿å1YÆõñaN'ß)ìÛæ¶;µ|oÅéÁ<Ö=ív8·»ÓÔ§ô×ùî.±þç¾cm%GQbgøôSvºíæî{ærÌ×®«û»ä¶Ûû¹¿
|_Q7¤ã˹æ,?CzXîbçá{îYÇý-&Ò6(îßn]Í÷Y¾ öI4ÏRîGÏgSä;ñ:ÏöOäÏö<Û7Xn »©¿L}#Ëðå0ÔïP¥vÐöL$~y¥ß~ûû:ÄûôëøÃú¯ûþs³úgüoè²²råÁü÷fåçùÿÍÞøö½·ZÞ7Ñõ}ÓíPßý³'£<ðK,(îý5·=À;ßÎ<ØO¿¿ñý¯ÛuëõÏÄÔããèQ_==êÜSæ<¯Çß#c»´äï¼3ïǼ3s;÷±´i¿BñbÓ~Sl"bØD4Dgüøbþ°b×lPxè'¾Ä¿¾û ûQåµOadßNÏÿ$¹ïòÇV|¼
¼|5ùsúìßX¯Iæbºu÷sD/åmõí³ùí×ÙÎ,ðy1çÄ÷¢¢¿>~Bgÿsví7Èks§XÓ¥ìfÚ/Òî÷KÄ»¸ûÙý ýK3W¶o6õ«]ÃaìÅXþcÝG»yÏÍÎ æNã-8q8Chbq/1MÊ,aÓ®-´É¼Ïäú3æ½øO8Jx¹'ð°¹{H¡9þýoóþ÷2¦,þ3äÈ|ß!\¢LLFWãëèE7¹¼Gy}ØHþ¸à²Á¶ì\_è÷òR« kõSÖê´yïßrîùr~¦Ï$ø¯ëå¼s²¸
¸ÐflÞÅ^ØOû+~ëù5oÆÐu»ÿ6à\kÈ«È«ëèûúÓÀ Õnü5°ö~¦à1ÜãÀÏ�); ôß °ÃsÀ<×sª¯%öÝ}>Eÿ>ɽ·ÇïSÜ×{ÿ à;eûqô{À½ ði³/�ã¹ÿiý^_G«ÊÏøÚ¡nü) :¹±HIú8ò[àè_S¤øÿÆ{ ¾"~ÑÁ±½< L§øzæãÁÕË�ÏîîL"Åñ^À#Ç//¾Ê÷õ£_¾¶=ùûò ×y}æïÇ|w?ù&ðï�'FS¤øÆ7S¤H"E)R¤H"E pwnwt?ï²è~ yÛ¾ Ó¢GæyX1uluä<å¿Ü
uÙ÷\À·²·rd&¿Ïg%r[º9kº~.rú¹1yl)¬¹òv ä®°1ùj¢U6¨~CÔn¿& kcC}ªÕjQÝ°VT;@YÃXE:ÖÑöê@ÕU9¬'ËIË´ª9vjLs# ½¦«½>U =¿ìÉóc7cí©0RIäÔã²"6[~R JªZéà FU+òD×Ó « 'PayqªE5¤ ø®£cG½5[ñy¯Æη=JÄzk;¯¦ãÖÆ̽j3èd¹Ñûd6¦MðmüÀV:°Yh[N=Ìü#[ ³¢ÿnz VîUì¼edzo3r&áøNû#Ev¬gs1u m>÷Í\ü2-]¥£e÷oup¬ÿí¿ÄºìÚ[Ç~ö4má|dmwÔQçÙY3G×>7Ë}8aìÄ=¬iï;dÄÎ$ e?ÇÊ6Ù2³\ÜZ¾+ay%N9ÅUzöìå:kE^Zì§ýcþ/ñy¬ÿ /D9d{33¼õ¬0,
í¸ ÅM¸Ãº{gÙqÖ/Òtq±º5¢)ð<pûM7³ÛÆ7Ç-b
Y»Ë¹à(ËT·2Ùç591Ó0ÇôUèeêGý5¥ö¾xýW[n+ÃÉ ±Jôï³¼GøaöÏ ÀuV?+¼Ç°µïðc/ ^
ìøí{ ~IøUásmaë~× oÞ#|p(<+|NøļɧGòé>stream H XTÇÀÿ3ï½Ý·ËQãúÞcßZª±h¬¥F£õ¨±^±Ýà@ïXoD x!c Ö¯çîZù5mfw¶³Xc¿Ï|_g¿ÿ1ÿ9ö?¿Ý@�ÀÁÈ£»÷;ß«f¬&$¥%ÚÛÞ¿~�E´ì4®dø·Ô UëfOI3ßö³²9)©¦Å�e)9Ó§~;ï°èï=9ZÌ fú¢'̶NO»°èNU&À;í�ÚwMý³{¤%.´¶/8²ÅK³Ó+Gfv6Éb=gîÜ.³wéë·§'Ûã¸} �Ç�ÇïÀe �òPó mPÃG0 Ïð9ÐùûJÈ lU6dø�¼¡pé¦.òzÙ¾ù8¡·o(ëËKËÓ° Øìs hêyVƺáKãHmáÒñQ\Àh÷6A¥è4gFëÑf´ ÀE¸Ã%¸ár\ãJ´màB¹. mÃùlÀȾ Ì�ì¤!B! ¡Ûu+x ZC´¶Ð^ö¨ãln:Î"(cP¥PåPÇ¡N@Sð78 gà,j8àïp.A ªBÑI¼
í¿#tßA,¤ÈËð\P.x�?À#øÜ(�Q F²¢H¤¢®¨ ^GÃÐpt
÷ãBt¡j4?Á¹8/åÚA|ÈMà9 ×æsC¸7¸ÑÈÃMÃG°
¯Q´må¹±øÁýíC(l0¢¾è3ôÊE;ÑÃsM
à}hÚì(EÅ|ÎÄκ?t!ÐÆÁxca6ØÙ³dÁFÔ¿àex9^
WàðJ¼�|x8;ê(ìc\Jÿ½2жæ!tyq;vngð9\Ï!ãL\$gãVq«¹\î"w»ÎðÝùüd>ǯçëc)á[áGC ¡¡¡¯a!ÍpÍà5&Y,ª%ÉfYjÙhùdB¤VER$ÔCúÔKê+EK1R¼dVJùÒ©PËläòK²EVd«ÜMî%ÿYηȻå½òr\)\_S@1)JÒBi¥´VÚ\¢(ªÒEé¦ôV¢~Ê�e 2DªLQTeò®²RÙªlSv)'g$ #;Gö ²Ö8ë)T¬ª¡jK5B}Yµª]Õ(5Zµ«êu
ºZݨæª
êjZ¢«'ÔÓêõZ§Þµ ¶[¸-Úk³%ØlÓl3;¾Þpµá Z«ánÃO³RÞ$Äþ$ £È2L"$¼EÒÈÛdYDÞ%ïÉR²¬$¿õ$l%ÛÉGv}ä�ù|N®ZrÜ$·Ü&߯»§;Þ=Õᶻ»×»/¸©GôzÚz¬®~GÝK h!=HÑ#´ÑJz¥h ½JkéMê ÿ¢ßÑô!u{=^¯þ/ÄèËaÔä0^]è'Á¨9ͨ©åÀOÍFÍ.û»ÆÊ÷àGñSøt~£æPÔ¨ ±ÝðQh±Z¦XR- FM ÔBj-I:5=5}üÔÌò5~jÌrÜJðS3Q^'gû©©k5X§&L§¦Ò^§¦#£&ÊOÍFM2U¡ØùJ¦¥SS¥85¡Rd'?5±/¤f¤ujZ V2jÊÔJFM5£¦V§&QÓ×OM²mNM]5÷5wö¤£& 5£ÉX2L&I$
Ì$³É²ü
dîÿr{½ëÞ-ïÞ\_;îØ{ä=sïX²÷Ñ|Í7o:ß¾Y|s¤ìûæM½ìÿÛ§óStIkiÍÇ4ÎérÆsFüÞIs<÷àOÜÝO#yÓçn§º}ëXØ0Î92ÌùS[Ê97I7Ò1úO#îpîíêÄ0ý ÌJÕ\X XÙiøóðy'ëpÛø3qÓ8§¸3q>§b'"Îy¶qÛ×Ϊá"¸N+ I÷ ܲðBáÇO¡|ðjÁïPªB2ÔCú(Ð�¢A#ô&è ÍÑZbzÁO¡ fVÚágг@ÌÙ ³Cgü:bè9¡ ~]Ýéy æ¼°»Ã ÁX±eÃx,ßbLp 9ÃaFÀÉX¦cUU&Vï±6Ìú0ëÁ¬ ±Äa#X=¶°;Á6ì[±3VáX(ífTq¸ÁpÆb(LÅ)÷ýÓSË'Òl7)3&á-¼÷ð¥ÃCxàQ<¿q<'ñÆ3xÏáy¼ñ7¼ñ^sîºwð>ùRJKPF|ð1>Á§ø ã |
¯ñ ¾Åd|ïñ~t¾B"bR2ò!?Ê@ééSÊ/ñ yL!Je©§0 §ª@T¢¨4U¢ÊÔºSêI½¨7õ¡¾ÔúÓ�H]i ¦4ªÒ0ªFé }-¥tnÒCzJ/éíô×Wú¹r e !2T¾Qò¥äüRP KQ)!ARJJK¨©(¤TêRSjK]i ¥±DKsiã¤ÄÊ)ceL©2Í5É+ %NɲU¶KìÈòìrDËiórI®IÜòT^Éyoͬ¹µ´keIÛZ;ko¬õµ~öÀÚ#{lOì©=³çö^Ú+{mo|ü,VAE=M£iÕOÓkýL³ëGý5@µk¨i¤VÖêZ[ëkCm¬ÑÚL[h¶Ö¶«´vÓîÚS{k¨u¸Ò:Uçè|]¦«4^7êVMpÕõ£þ¤ûôÑãz^/é5MÒ»úPêK}« ÝvÔÒ[&ËêÏü-À [ ³dÁb¡VÎ,Ò¢¬ªU·ÚVßúøZcm¯}í%Û;{o죫9ðAÒq:Cã|Òø¤µO 1zðFÁ ¬«°%¬ÆXa;v
g´pòiȧà+8 ùà øÃ9( ¡\@¸ Eà
kP j»_ ÞA¼ðèä¼WÞBkÌ�m1 Ã0C,�½1?ôÃ�è¡?XÊÕlÒ0a4Àbl °!,æ®X¢a6åØ Vb Ç6°;&
ØÞµÌvLà.¸ÄuÍ îËp9®Äý»h.®Õèç ¢ÃtÒ1:Açé¢ëtnRÝ¢ÛtîÒ=º/ch+Å3ºRöá<g¹"ÍSyOç¹Ç x1/ä%¼qZä(æfÜ[r ·á¶Ü;pGÎÍyØßõT.È\ s.ÆÅ9r.Í!\C¹,·çXÎ$y9¯äçü\_q2¿à7üßòGþݧ-¢%´§a?ìqÃ8Gã8¸)Oæ´h¯Õ¼×óÞÈ[yoç®æÖr<¯ãM¼·pïä]Ø{aì½¥ RKòJ! p(iæo´´ÒJZK.É-yä+) RDII )+å¥TRGêI#i"M%FÚºbô0W4·~©¹´ÒZJKj~®94§×¢ZL³iÍêrnÖݺSwi%¨QÎᡵ´ÖÔzZGëê^ýYÑ«zY¯è}¤õÞÒÛÎÉÐÏ Z~+9,e×SzBOZ-«a5å$»B ·[d7-Îl¨ ·Q6Æ&ØDjÓl ¶q6ÙÙ76ÚÆÚ·6ɾ³é6ÒØx"³e¦Ì2OæËRY,Kd¥,ε«eí:Y/[o7ËòOùìËd¯sí/(ÿýrXÊ!ùUÊ19%'ä¤3rV~ rQ®Êe¹"7åºÜ;rKnËWklT×>ç¾Ö»~°k¼Áöb¯Iéæ!PëØ£u]§T S(6 yÐDE±Á!
�-ª"Ô\ Q7)4Ëm mh)EDI~4¶/ýæ;kè¬wî93sgæ̹f]µþcÝ°¾°®[7ÛÖ-«Çºcõ[³ÕÙælw~îüÙálqÞB%~÷²ØZ
êù»xI=íǤò2&ª©Þ£¢Ø¨#¬HØÛD}X|Ó\,"ö"ʼfNFïIùZ#ïlÈÆÌ<>.#dM,ÖÔÚKðÔ1Ö2kb³Òf|Ö#ÉÊT¬3Öù';c³bOýðÉ´Wñbig\KùÉVü¿ é¦ëSå9¸4=é±Î4¬ð5¬P BvÍXÚ¬nN~/^ßX®oL»n¬)}´9>ÚXî¦Rrrïe¾ÍØìÈÓZæ'ÓõåiêìÔOnz}ggy'üð3è= )îeÔû D4ñ¦\߬^¯tËQéVº°3Õ½5sæ'©!ÿ,ú&¾Ñ¢ZÊ1¸4&ñ.üêÊB2.ª"ÅÓð¡ 6FB!AFûA¬öd6S XX^ÖÝ"+ÊlxA R dD>öÃ^î�Í^DD+È$©(R#�ñb[Ù¥´TÌ0FáEGwaMɨNÄ8;¬.å3ªQ%äqbÄA~V¸ï';½M³ñ(Ë ´#1�íQÔEdW7^4î% £= ³£b¤ÒÞj¸9KÇOÞÛÁ.Ñ\& ÚiîRr#� ´P Æ'×FrÒ©urò¤êÊÑNL¬X-qE2\#nDÑñ¬2'76Ä[ê+XV´}õÊ-;\¾Ü·ß|¿÷ÊõMnT¶ýhfÃ1³~yøÍ[kXõíÆy+½Ö½CìäW32ôíqyök\òÍÏÞi¸ø(?!@ßl¹üê} Xæ?ò«±ÌÙÃm J@mu![pßzdì-�%bhW²>Xʯl lè Ý´¸´\P�8:º²ZÀ̸MÌö¨RÅAAM
Úi-Tn
Ö@ÿÓz×�&�&�&0É°~Ó °Ô¦ÞdhÜ%Uî8ÈRþlªp@1×@ö�ba!ø=Hé©ßØÕxø²�άöQ?\Ýù>ZðÑ|´à£-øhÁG >ZðÑÒ>H7Y·[;'æØÒsàs;xÑlDÍ ÚȪä³TDH\)Hþ·¸P<ð è\´??¨äèÕb¦ø%lfN¡æb¦ ¶DÎÅc @óº,u¬m¤ÒÖ-Y!awvg"Fù¾/X³û£¬ý¶äP[r5.ö Æ$FP7®|J|4Ñ¥ÓÆÖéAê8¡#Þ«¹¦2A¹ñ sÖ1ίZÊ5+¥K?Ó5¢òÙ,o³wÈ(O÷þãýqƬÞWíÃ\_Í Aqöô<´À,ÌX|$ÎÑÏD ÇäÓCÖcÃë« á48Åï2xAX}8"/ i7(.
,Î!JU Á.=Jíh½ö!8üÅÜAá#ÆÇD±ñqh¨c ZóÑÃÎúT£#¢-ÅrêÕn8hØÒHI§?ɧZÛ¯óvÊçÚúÌrïyï¬qÜýg¾ú×Cýo?{tÞ¼æ¾o-kd9õÌ;æ.òn&ßÞ8¶øÎY\þ.áh8§Ïy! Á>YÓ�Êù7 ØCG(À!,¡ a18%¢(wÒ@½¡UäÇj%B }FÎFL«ZRCvÕîh#ZR\ê:Ŧÿï³fMónýæøÑweA×ûwÍû§÷\!§Ë¡r~Ô<çy7n^ô>Ñ{Êírtíó>>3(g×ú9»GGÏÏØ¢QSe,ÅIpù¹ðNIZe©´©«Ó±#9ÛPÉOwø@�#ÄP�ig WÐ×ø qwb±J¢¸R~(|;ý«ÝÞ)ï´±I5Þ½sýÌ»üù~ïEïoÞn/IÉK>[7ás¾¸JûR9G¸£ 0(¦:ªÞÖQaP±ÞÁ:e6ó!ÎBÃJÀD^ Àjìö¿P¯ßÉ Ì Aý]Ý1D Í@3Ôi+U ª2Ô2G
L¢òøgêë1Nö×Þ_[Ó¬¬÷¯ÏëðúÔ°£ãó[à ØËGQa°TS©"2¢ÍOù&wE°÷çï{f#Æ Ï1öÌk/²äòQM ÝkÆOx@²7ò#ï¤UÚ\_alìÞ> Ox½ÊZê ûÑ\dþÎamJ;ªî0XÕaõ3Z"ÝéÓË. EVC)AÕjÕ; ªWt«7^Á6æ¤à0×àÇI » FÈ\7q°ù0% uÎsaH@bh®?WtS¡Ó.Z©ã ´÷²R÷S¦ºæË» «E¾ë]èúqwùv÷ß{ý\²§SÑyèÐ=#¡BvJ\¤2t^ÏðNuSn«;¢aZá
JC'×À]úBêR"=¿cp«í®[«DþuÐ×È'Fþ×M'~«ÏZÛoåª$BìPH9dêr¨Æ=Ý;Ùÿ郎5êh0¶ô¯²÷6Æ ®Gùº{l¦$T3Ô.ÑàÞêÌUæN\_×ÀÆ ¡°õ¹ROÜv:8öÿWº©ödí@ß°÷Ñ·6]ßί5&y\Φm±ÅtA§Mì =ÅùV>Èv¯ÎU b/@t¤2jtø÷v««Öm¾aÑÀƬïwô�½\_/c°ÁdV±9s.ú·5rNëgxVÍà�µ Ðh¯tæ±Î<¾õøËÌ9¦Có´¡UJ.K§È£QØÎ]ë æk»Òªùb¸9¿.å× <Í7I}s£,´MÛôÒµÿc¼j«,¯ðû~?÷»÷¶·÷ööRÌn{騱m1[Fu,MêP±º©2\0ÁJ&¸¡76%ËÌØ2a.,ÙÈ°nþTó:S°Îiä\_òuÏ9ç;ímn¹yoïýÞïý9ï9Ïy¬ñ1ñ×ùËezÇW'TÜÞ6о£Ö&ZJ½¡ã:uÜ2®PvÙÕD0EÌôoæ:)gõáS@ÌzVªx,F¼cøj©1ÆêÏÉ1Õ åý¤\'11Ôho¾l!ê" r¦ÛGÎmuî
³EøosºÏ=9üsgíFÍÉ¿FlùæNÎÇ[®&+I¼Îh\ÑA!N<ªVGµoPr-ÝWr-cyFDê»Ëyóì1Ûfá)ì÷»¨Ò]A¼ÎY4VWµ§^ÁfJØfGæ OÁ <r%ÙNW6Är¼£´Z¦77èå=ñI®RØB¸ïr<¬E£¢µ"ª#û9Å° ëY ðªT4TùÓ�uåìÊíÓqÇ©Ü >CHì¥?Y Xê°:äa7väá('Äx#+t Wí|H
9NÀéhR e0ãøæ§h[ źo|aËUç+mwÜ®Lý¾>¥¼ ÃW|ÓL¼L«_¡~XÇÎÔwÕ_f+ùfÀ"dêÝ¥¯Ï¨(H�@üi$ h«äÃ2\^ 3?CÛFb°W'¹^Ýõ¢ |Pi ¶IIzà$»b�5ò)ëø\7Ô§ÐÃU� äÁ5ýª,ßøXÉGËËuÈÍ(@Sõ®U0 wçãZÓØ^_@f6=rm¶»0ÜïÅÃ×Ãî·ÏÙDÌúöwÿûgÛ¼Ýï×ùÏ]é=¨¹z.åjÇ#ÅCÌi¯A°I9%òsÍçHlH~FÞ@ê"kS¦#ÛþöJÑ.îÊÄI.°4( [¦hCZ'Ú|4Rè;h'¹,ã$ë< ©t´¢£WsäüUB(>1ý Ü÷®0ý"Í?÷)HjÚùºöLJ£ÍWUx»fëTàéeÜ#²ý^a}Z8Û<¤V²·¬@[§&éK>ö°nî{<9±¼mOòmt«ë|Ù#IêÕ=ì?;l8£ÌG]µuUɧ²/+?K1u·þÝÌÌ1ÍV¢ßH»O/æ pøýÛ�G¨¢YÈó ·Wo-MW.¹¤,ud%[¬ÆÓÔÝZn<-ÒMyÎÍl28ÊgRG»\ÈçH #ûýÛ·¿õæ{ª×Ûømï \_ Ã;}#& mòôidÇÎØUöápe¸&\.µ÷½(wµ8"ü(1|Át¯ûLÌé³]º$×õ>°\_D©4^¤¬/Å\ßsVóV~¶Äÿ'T4 êH»e¨ü¡² c\ÊTû\Î}Ì|î6O, ¦Ïèhon(ÔM¯ÉV; çÐò¥}«ÃÃÝÎ/þùѱáf7³äî{z®á#p³7 ß»ù·²Ýê-v^/v FS/d0aeÐèGv©Ç êÆÙ%
'ÊmqóBªÅ»ÚVõ½^qE8UåÇT=åÔÛE¦³]uÔPE uÔHGbÔq|TÇÇ[ÙHq¥ù!öó¼,Å WvçÁ5ÚÒ!aÄL·Ùê\¶ÐÜÑ~©EwÁ½xåò%áw¯;¼Ñ¼Ï4¿+\tËmá áëîÌ\ÿp
óàñ§FÌM>>ðbgªå~iç£|@1BL·Ø7vTk¤éª>ëà vÓµ´R¬sNÈHþ F#Ô+~+U ¶j|TªX¥)òy ÀF\P .ʲ¯ ÎÔ8}Ö¼àWÜÓJ.h¤j¤F¡nÊolXÓÊç2¼zcfAÞС 2C{4<; Q8>èíç¤ßȬztîj0)s2åe�<Á+ öøÕUh ÐÜ^ò¸DåTü#ÌÿHÁ4y&¾Êsçh42ÖË;çQ ¨µÔzZÎXØÄbZsù¶¨ö ñQVù®á¢÷
±VÓáTR> rNPaw©pw8×{ü×o½p£8k£ºÉ îÊM^=¿®yNï IgÈ(/e´ÐzîRð¡ ñÆJ(ì=ZË,U0¾X èÁ§+¥åb¹°=§Kzñ·ÊßAúZx|ÑÎþáÚuÙË'<²itÿ>|Å£?üû7ò À÷ ä+ö+QlÒÈÛ KAFdG2?óI³Vº>\S8QØSðè}K´Æ/²j�tÞ¡
æ,ÓR¥VtbË ®<ëËÕ Õà Q}êb= ¶+xôÓ}m>-£1 u¥ôË¡ÔèÿÚÜvNX
{µ½u '9NAÁ°}ÖÙqÿv.½F´lêSÐ¥õSN{ü÷³Cü)ß«&¹{p"_Z˹h\68A4ôJ%Z¿QÐà®ÒÈ×}ø|µxE @´7T°dM5s5e§g�åÈÊ1H,+¡Iëìr)Àýe¡ú
\rt5y ÓεR+RëÛУEó(gÐ:©f½ò:a®X¶ m=M°%YT¼vi9[§ £¯îç3Ñ>ÅMÇ. Þ$·¿C÷:×'éôñ¾Ø¸ ®=ÚÃü 8rì{Ö£O fZô�Ïn ШۤîÙÅösPF7Yçbwã¹5Nßð"ªõ¾8rÀ»Ö» x=¶ÓLnæÎ8,<¡Ðç-ºÐzÐîBópEq¾r/¼®KXF§T§ÿr^5ÀQ]UøÞ÷ö/dÝ$ $Ù4¡)H¨ÎX S¨ Щò#?) í@Ú[ %Õ:bgJÚ!)¥±\¦ÑBFG°ÑH-1 jÛ-:òwνw÷Î8Ì%gÏ{ïÞ{þ¾ó1QZKY}Ö%ªñ·»J}¾\_LRçÕ^#41#ä D ¹¹F$n1RÄT Rôª Äü ¦p¸E ÷oS!è¤/\ÁÔ¯\_ãj!NIé¨@ ª°ê±fc-¢Ðw!|&Ëû8xµ¿EB9"ÀØ)ÝÏ'¦ÕÑ[6BfÀRZH£Ò~±¬]X A·òݸX>¥ ?EEiäói0#êã¨ùl¼¯2!mÚôªÛ$Mpi¿6Y!Ϊß×2á'×?'K+ÝÞxÏÛϼÚ~@>yIã²Å¹î{ó¿¾dE« óçÏ9ë¡%/ß¹(4aú¼g¿ÿ=GÇ>V9cæw<úü>ʽ=ȽI¹¢TÞ0¨Ü#R)8ÉhØÉQìP§^§¢�äV?Ç#UE$4
+EÓHRh¯|Ö%µ z¡ã !õ äta,¤KQJ¥<Å®;N6 ¤Hz²Ðß$åù\=$P÷ÕîÝ�²®ý1ÁÍã¨ÉºiuSòcQ«2=¾2AþñG6=¼uWóªn¥½æ÷'/ùãrÛý«,³m/þèÐÓ~M=~{ÕtÚ± Î@Ý8L$Å ¡]ûCTÛB¦?©)ÔøPQH2Q ÆLµî9 ÔÇÇ©à(bó!4bmàqE§T[!ì§úRòÔyüôxØË)ó¨ïr¤óji%Þ¯ÝÕÔù\D»·ÝÄÇAþ&YK¥W¦E}TÔM±³&¼<ÝZìºo~äÉØËpÌ}+ºôóë\y÷|òø{2=$'ÊÉøí{ìí}ºésÈmòé:Û5]"^K)ÃÃú8;,Õ~$ 3d¦ MMëe7Y?:§ËLÎ0è,¿Arc<íYDÕV#
a̲ÞD=~æèÔÖJÐÎJjP9l9lGø¤Jz-ì7B¡ñw2
6º8FK¡µO°ÐFtÎÇÉ 9ã«F°1ÌqãÙÓѽ «Hm ^Ü©8=Í1E|FIæ:c2Ú±%pljðõµ·=¶aÙ6ë#÷÷ﯹòÖÜãçì/ô½1,O»Ï¶R»:ûnkg|tðÌ¿Þ>²oýûß´Èñr jçcÏSÿß#½½ª@ï¡Z\pÁx7í1E#DHPÚÔ#JI ªÀð8ô<ÄU9¥I]Mµ×Çg%Ô>À6d]9ä׳.biTN¨lTù-£è±I\q®§àtЯ#6,ÕÎs ,ã�a¸R4 5W#V°W¸×nvdý·õZíËï]?ÔêºòÉvUgðµ<ìô"è~ÑIl}ÃÎ>øĨùIaé%ûx< \;ÏB¸eÑ2µôC EAåón'~×øÄ"-Oü¤ð{|¢XvèÏ4Aä%¬M?ytì\_ÈÛóòò1ù©ÿ)l]HÝNí¬
óÜÿ°Ù× ìÀ²ÔëçO® ÁZ¿ àO§äDÊDFP¹Ù%dÖ�8\_c ³0+@û9bMôC0Fèw{ ôñãn KÆ¢®Î¢<ãò<my¦Wçý·,Ò81²yG®%Ý�puv.þ±A"NÀÀSÁ©¶XÅ{n¶ vS{U«8f~ñmé:¹(X§±.ýË¡7|&+Õ«&´ä¤Ìæ øÁ/&j±ü!üÒxÞlSo.Dé0O !ÐÙ°jÙ6åiH
fÜç,« � ©Ê ´).>¹J~ÙÝ.w»Ý«:Ç®^Ϻ CËåë ÝÃ=Çà·G¹¬¿ siQUæp¹]T¼ÔAz8Iu!ÞnÞ%ØÃïÞ0E²ÁíACnÄwóµ[Ë-Ä}°qf¸d.Y Ú!ÕE+6ÄÕ"ûùî¾êm L q%°[>svªw^èì÷ jèLÓÊàVû±8¢3Ì{ Ðg>¸ã8ù î)$)ÅXRõ(òIï)¤)B7hÊcG6e%|ת0÷éBÒæ+éÚmÝ jÝEܾ}íîÀg-ZåÜÝÐp/bW/íÿÕÒuG¾n=Ó¸iÍÚ¡UÖÁk× ²gµýìÈã/f¸b[ ïòp§¶QÜ)C2$Js§ëP¦ óº;Ñf!óÅõIT7ÐL¼7ÞJå0£8UáTm7æTc(c<áû¿9Õ»¦ë5Á°¬ðõYV"˲îY~ÿVÐý~á´ÕÞ¸ú±æ¡V
fá¿Øûy))Ũ:¹ ÓiµÐÊ}»±Ö + ù3V9G¥\½B#Q>æ|åµaÀ�çùQ¢M¦Bnª4²¯XµAÍTWIÔUR}{ÜLå$4''Èc O#¤(&²ÓÊ\=D²ÓÊn¯�oå7#ª¢gQ7å?|WkpUÕ>û<îMàûJyLÄ$(7!ܤ K(Ô)T EgH
A¡ÆD$ t¢á 0�EÄÅÊL S)édGéðHÁòh
!'ýÖÚg%!Q'³s¿³÷9{¯½ßZË°jj×,ß°þÅÚ5ªÐmúßM÷»yG¹¯zó½û¾áÞvT±JRA5ÌýÜ«YÝöÖg¦Ê¨7Òc]uk:TñÓÙ ÓCìä´¦ÝpÔ×5è]î23ùø"¾.Ëìþ~µ%fl%?Kx)«þ2H]}õ×=ái8ا½4\ÞWÃÒhCæTÚ©OÛ'w
ÖͼÁ
yä¿êk¯ªýõ Ë)u½õïslxqÄLØôöǯìy÷ÈÜÓî]·ÓýpGsEëþgÐýO÷Hr®«É SàZb"5¥ÐJ" ¤zWTwLj÷a&/é¡zêÖ&u"Ý ê¡ÝLÈü!÷rVw¤¯ºõUw'èî¡ûÕMåB/o°EÀè%qµìÅ¥«·°jÙá·g~}æƽóWÚfÇÎÛv½{Hm¼;ÐíUΫ~jò'{4[t2Þ'fcU
ø/C¦ë .y[Y/aôò]�!x¹Ë9½2¡OÃõ]ÑÞ�ûû¼¨ìO§%IE9I\Êí £àÈ1 ·{¤¹9üÕ
ÉcÖ¯A·6þã
m÷+»vmÇ J:ã:ÿa?ûæÛiõFv�PT\(ªN@¹ àÌlpH@¿£»;fÞ³¨]'3HR f ¢¨Æà¢,Ê˳h/ µ Ø'àWÒÍ°N¥�¹%ÒÅC@6 ȧ©jØ0 ÄRrýùú]Ú+½¼Ùdþ$W7Ûó=#-H¦Ì¬Çà#\ѧà ·ñ@£ tjɹóµ,¥�å,îV}úÄÍ}<LÉ6 %J X(F,ìá·É+XÑnÊÖ®Ö&u°làCK1O }¶¬fL,§¸'!î2�ôÓ3~c¤¡óÜ, Kw·.Þþª?´Bé�b1éh�øã2Æ >«áïÁ['ä¹É¤ðO ¨F6EÀzeúþÆó�4»cÆQ GÛÁÛå9ùæ-zÉo�<1 /lò+ 0¼è;jøËjñy8]=
©c2ꥰzc µ>Ô'ôgzKÊ10,½Û@½ímTI SºHÍR)ÉØ\áÝa\>ª²Tóß 35}µ/wv´uDUÁ¢%¿7©åêöK¾nO2oMüù¨Æ~4òµªÕ7Î62Vxøîª=?ÌpZç s3z½ÊÜëtq¯G¢!�vñ,¥ j{d,ÜÎlÕ¿ThÖaï7ÚÌîªï#ð$שک:%Ç<×kG{øzÒöêÑØ1_?1ô-jgigûZ-U¤XÀórq²ÖnIJ^ÔYBªó%(N>+.Î;ç7«l{|SÇΡéöé~Ϭ7·6©èë·¨qçØãÑwô§¢Ûc]B)×0Uä¤òYÄêXj?4�B9äðÄueÌ8»¤QÈÖ}Å=ÒKû½=×!ô)º'p¾³hàxp8îίêyZ¹aí¥á{{Í^Ôvà¸Ê~jöÓ+UöO7:æ̺ö7f5ùz¸ èÒ¯VzýÛ[ÒÕ IÿÖ&3c¥-]íÿçN8ö2ÞÍc¾ÓôþÄ�ß6'óÇ
ãA}§ bx°�ÀÆöfYr5r¨ ìjùûQãè³3�W(ÇÑf'$þ ¨Ð&©�C8j\1£I61¨éi5¯RSÙ.á6M{ÌxªÈx\7ZÍ5Üw¹{ξ¤íñ÷¾²ò æLw=v(AöÀí%ÎÈ]ª/^m @e \æ\_xt¤êß$RãTU^)ÄÛ À6ZÍX''\AFXGÉm<ðû4ù¥Ì=ÊÒ1þ ©ª8×&÷Pm).©&Ã4î1aÓM´®Iº$Ã&éTÆLÇ°éZ &h\ÀZ½q¿EÅù¨aSèyÖ8÷ÇßøÓQÙ±1S2ÜgÕ/;m¹´c¢ÿ=b±]Õmµî~uT
DÕ º~N�$EElðdW÷l¦±\¹÷;=9¼ )ääÅKËÐE©£î)so{ªdûÇÑÏ>ü)ëìDf¨KvY 6¨ÍÆpóÉyÇ2F¿2÷;¼ìÄ5SNêö["vòK>¹I/ºçb«N½ê|;ám?íBæPMi6ÓaXÀ=_¬$À6rðiú¢b&w9ô¢©{/Ó_Qî¿^Û$pÕ=;zÁíÊ&:¿èwn¶SóTÉ¿Qû¦ØêÁ¨q+Ì#1ÊtÙÜÀaLzðy¡ÆµWð>ðü¼Ïia:Má¡ê¬»MÍoqÛU²[¡Õ=ütëhïhw§iß,,µ,ËT2�¦å!3,?¤¥²û¸¦ºåðËí£ç··¨ùî6qÀL6Õûî×Æß#8¤¢óK{s;§×ÁÈçqÖºHí@é4-TTc &-º8'àérÖ - SÀr4 ß÷ÁO}:ðs8'PQêõÝzU}j×ESh"Eìò,q £Ø
ö^ð&¦ º Cí [
R!êLU(R ÌÜÁ¦ZYñ'fÕ¬x=sÇUÆ·ßvî³Ñ½ØÞ®2ÔgϺjµZànq׸+aå°ånðҡÇDãA¸Q6ne¼úk.×_Ô¼®|IqIõªÅìeín%x %Ód1
µá@´áàG´á@ôF�AÍg$p�À][Qå
swyûp¿«î9w}xò»[µ¯NB=2÷ËPÿ%°ë ܲR{²÷&KhͼÕb ÈNf8¢b�iBZü¶¯ëÐ@3{d@{ÛEÛ¦Þ4 8µ©/y7ê¨/ò@ u¿!GfôL,cõ0Uw[ʦ²\KLeÁTLEa©¨v°ØT(tuKhìÀÃ;4ù îWHE4Ýêbª°ÓÄÑ\Vó"üUþxqá+z|;Yèx2Ò¼z22JÛh·ü?7%·4R6OÊle}qVE«^¯dVß˲~ê3KæÎ]d½§Letüö꺢÷¯3fñ2^fðtÆKƲªUlDhÍRH] @©YBl²°H¨Å ¢FJ-Á )Uq庱èT,biCK[@E�&þÓsßûo½qÅSã¡=ÃSÚ;mCL±òä¡9r2e\4ÜVeÌWÆsÊ, W1·!Ål
2Æ)£UÃ1|&&Ô f±8ÉWöbíÍFÇæÜ4\WÝcP¬»È½)TE| 4!)ýÆgI|ñ°Cäçåzò¢F^.<3¼Xqõ×´7.ùÎ ³ç;<ºqÌy7X]=¹qé¦óo nÁìÍ[qß^8iñ²Õ;.jiZñoüÖiçÁí¯nX[_?D2ÚS81)SË]WÒ7F% »7Y«$른ï¥ä¨ ¦ =¯[óñ-Ôn;ò½8ÖaúRÕ® °:Rêìgýi¤zy"ÕÕø©~ê©~êW¤úÅqI%^y¤é !î l,ÊÛ©W@Ô±©f²z׳ê?²M§Ùè¿Ó¹ÞyD/ÐÛ©k^maçkÎAmôcù§ß£®üÛMoF´^äùàãÕzà�¼ä}hEFùTþøW}b>,§�ÌgçÑÜkb3/]'dD¤s¶JçlõlòÙÕl°-ØÉÎÊç²c¹ÿÅîv§$!BÖ@Æ>û?!Ø
r9ébrB=³Å
@P
@PHA!ÕLÀgH°¥clqh+ò˺çTUq! ä¸ØXЬ�¼ÍípzV.´zKÅ%Ïô-ÎW9ç!¾ZãþF³ûNÇiçÅò[çh¿iDæÑÎ_ 'w}ß9ç÷Çw,_Þ¼¨·ûXæR 8ÿªè÷J´¯°ø¸a&ª¼,$¤Ò)®³C¤dnxÙ>0\�]ªjQQª\_¨k,×2ýðZîØnXÿÞà{Ü^¤Igi¥ñæp·x·x)Þbà-¦x·à-èjvÅYý³¦îÄ$k}�8³ð!+¥Ô?¡uL£�lå°!^©xH~m>wÕº¹3ôÉk¾mN=rä®óø¢ó±ÔZæÛ· µs¯¬Û°bÍfý£&§Î\å|zà¦sQ[ -Ó6iõï9 íõj½¹³)ÉZÎü2%3a¸ùd(~ØÕt~¸-"\_ܲ®E4fEL-£L½H¹ÊxKay0ánæóbLtï"Øx\ÙÅ{XSÊè#¥èÑ»z@ <ŪÃw¬Å9Zñðbs¢S4ĹI
æÙ éÌ\_üXð\r)ª²-ªÚ('\#v0JyÁYEá{6m EcaLð:ãÀ=n ù][ÊïʳWúê'dªµeqf(FY!O¦Ò,)¥8X«4Ô¸H³L
"Rg±+øøÛºb"/Ñ.b1Ó]©Î4Õräæ§ç^\6ÃF &õÒ¯O>cæ´i/¾¤.;wé²9sk};7ÿéÜþ˾§OkÚÛÙuÕÕ^8a±3¦vCãw_~uÓcÎí»ù5§öþõO-'fÍzù<§JÄO\ÁU2 n¼UQ «\ Í
áx½ÈÀ ¼ z?¨
A:aTSDaP|,Å:È,Ú,d¿,_ñ,Ô½µìb²'JRLFÀdLFÀdD2ù,IJÄëa¬\ "T"þ)ßÌHJ0ÕÌd%X\ QõÁ[nªrÙÁ¼(·#ûÚ¥% Ç JØ}µ%/Ïl¬_e~ôK:ØG÷i^lcaÖݯÝprÙÇÇ»üý¹x ѳ oûÀeO~Ó!¸þ °/õ tpÌ< Ké¼>2·RÙAMæJo>¶ÓX}e·\_§&}¤e®¶§ÏbÞü�÷b|Lã~\_¡\}
±Çæb ÛÕô³{^I³7 ÿBòl÷ëÏOEÄ'÷eõ{ëÖÐíúû%×ìË ~Ù]Ó·åå4 æú³É¬ã1ù.ûUùÖ¨F»>¯
ÿ®»¾vao¦:á÷v¿þï°Îô ÜÌðl\§=AoĽÑFc
sÁýgõ gpì¤CÄÐÓÀ±ûn\2'\_§LÏÄòÕ®ÑÚO3ôªÑÓ�£7 ÐÏQ¹¶2 vj´%~DÚf żay?ÆL =ÆܨLÛOAu´:Æ9ÊÖP¾ã:\Ôêh¢À>\Ò¿LëVë½µ ½
F ì¡£@ ÃÜKGÍóÔh£y =ÖuÆiZ ÔÁnØ÷ZÒa¾ÐúÒä%}"í7ݺe»ðº°hÃ(¢îÚÖÃú#«F}7ÆÝX;4Ù(Ô=YC^Ö£RUÍÐ7asmü;4zo/¥)¬ã¬ßüªVê×Ä°Ql|µõ«¢ö!Åu9Þd7RuòíuÉ;yÒØ<Éyf?H´%Oò;c&M¶<³GKÎÓ=X\_9Öõ3ïÑóVMÓ§!Nö§ SIðì×þÃ~¹WU]qøwÞ7QT�+RQ0HåUÓD¤APªÁG" SÐH¬R[DfcÇQAK;¶V루íਨ-¡ÐÎtFꬴåôÛç×DÛþÓ/ký¾û±Ö:¼qóV®¥ì_Ã"ô2ú_¡î5Ä_Oi¢ÌcýëµÄÿ91Ì|îzæ|3®} 1¬)\çî¡/ýr¶ìõÚhî·&{å!~±WÁ^ï}Mz»¨3< k´W¯&¹CnÕÑ>õÔ ÏLîð!åW«&اlï¿£~Á Uù ò9Åæ]IîRO»ñ{ͽH\»,>'³¯&ÞfB Ìl ÍÍÌu;mÏ{ûFöÀ¬?|1{Ráþû¹·w<ãá/9¯î\ÎÏ£ \SgñÍ}7#Ƚq rÞc¿Pµ&~pÍn#{þg¿¾5ÜfÊGän4»5ÊÌtný±Y»5ÍÃgE>L, 㼸YOäG^AÎÓYÞPæÿ>Mñ½± wò¾v ßÑNädOä'ÈbhÞ°
¶5Èåq;íö½
Ë?D
ÿ¾»AÝeÖóf¹7[é Èà&m6øU¿;·VÝa ½ .¬Gü%GÁÝìv sæíÎÓ±§Á½0 ¦äQHS\_ JôiÎ$|C;§è|çt ´ßÿáǾϿ¶«ÌH§2ÒÏL§çíÇÚp6kfpî 5_¾é´él§NÝ©:æpçír»·Óî}sý75ØoÔà ·¸¹DÅèUTs>WªçXGUÏ(ÈB:ëc8'áÅ´¯8cvþºþ@$m¾ïn´²1¹ùÞ¼ñ·çó¦ è¡ùïÚÔçãÃgÓÌx¶Nvç
×y3ø3³ñcüYfNËÉûÝsé{Ë~Ù¯6ÌãnÆds:qrsn´«cª7u§ó"q¶µIu^1ë>Q´LÇ+ô |Ò4Î4ò.ª4ßÏjg[¯Þ¤MØ5ê¬ÀÞàm¾y®0qÃߪÓü»UƺÚÝö±ÌÛÄÞZ_áî(½ÖØ".Öúõø»ÑoØ_G[wV¼vC o¡ø»p¥SbOe~¹ï¦ÜHë"êe¤?·_¾¥û îÆp7ë=2&§çËB¬:# ùu]-?ò×atÎú³õÓ)«ËÙ;³ÖÇ.VXõëôzBkïZ?Ãhox»�ìa4ÉüÓÕ'Ø®ZCFÖXîÈÕbZ?o$}Îa /Ë/ôb]³ßêKܯÓã·¥¥ ×øcÉsúóðÞïýøøþ1N¹®ëÜ2î,x!à ¬áwVó»§ó-ò³ðHþ<®-üܪ
w«Á.Õ û>ê.¹Dc'x=Éùüf khñçÐÿ\Í0qR]3ɧj"õ-r® äbûFëðÄ\×Ü+¹ï°îËÉGWëáf_GGq¿yF ¬HdûS ·ÚôÕní ·Àò¸\e LÛö¤n³ýNÅ~Ë®QÕæd¶}2}£¸NNlýA³mÚßÜÎo¬YÖêðYØbmEnE. ÷ïEßcmC#1lµN²ÚÉ7+Uio ³Äø,9J5å5´«ôøvHʽ¼ÙAçgÿ Ê m÷7:Ó»±Þ
-w(\½Ü~\åûe ëëäwMðö2öa6(ÕB;ó&ø[¬ßf3×åÙ bÛûʽfÚfô=V¨ÜJ.1'Ì:w÷º½un¡í@¬]ýí¡î MÊÙäÞÈCËÝØb»¬ÿÄîªó¢ì.böKê¾ nB¿/·ßgß»4ãë\_£þ·ÀçΨðl:Uø<ÃÙ>/ ¯{ÐW}èosÆýeñ-Yco&\_ú>(ÿU:3/ßWê|ÿÛvWïϵ;Ü¿r2ròAê tv¿¨<Ôý?ì÷±àÀï£Ó÷ò9ßG»Ï!êawxo¶÷¹Yòþ¼§ùáÛ(èËï¡>XÉ·gA}¡]ø~»TÿTÁxÿ»ÿè¢]è\_
åµs>é=ìþ|·Ú
ï?ù.¤Íýsá\Êc9.osvÁªOãÑáÊBèWi¢w;® O¨±9û6ë÷ªÃ,©îL)|õ Ëà¸\a=m×òÚz$².Uü×GkM¹ímWU$!ó¿ÐÅK!';)/Ú-1RÑ'èCNvRN®21¢8TEÞSªÑÈ¡H yR®ÜÇ~MÔd¯õäUM@rôk¡Ü- ßÆÐû¸Uá'ycÊ1È!
sDã±Ì8f®õ+ê¼\_¸yTÇ~axð~Ñ|è×É|ÉþäÎ0èIµH©;ôrZ¾N«Þ50»yÌó$4ÁL[ª3tì£oC=dóûØÝ4ÏàLWߺÖ"¨ukùÞ¡Y& îá»ð5 ±ÖðÚMKê¬Ô5Næ&·ÛÉËÛgíj,< k\é'gqZÎo° O cøν|òòɹ=a¯e¯2qßørîüp<ܧ$×É{¹J
:y¹èJ÷jUÄÏ(lå¾Üã#e)ÃÆb)Z×Ûc? _» vðh×X:ùhÏ=)Us̯¥c«×ÑÒ _zWêóhL_¾ÀN¢} ëïÿci ö)1§^ò¿að<k¥!ÜÕ¡¥3Gñ<O#(;»U:»:rtÞ0x'úôÕ2il³4.CkÆs×'î&s§Ðï¢ã>¥üÑo0ÖvÏÁó]§ØSùzJJÊÿ U'ÇÒ4ÞþôùÒ7_f:ѶÓÒrèðKûÙD·4¿9Ñmu÷Itò'ÝEßè¾F''z¦\_P7wAýìY×4üت[ã'iC¶¡Pô´Ð°ë±ÒÐRhp²Å I(Ô¹ ãfGYíhÑhvGÈ#ø:W¾\½ÙGÈ#ôüî§3ÊÄvK iwñ|ÒÑùùÎ'WÏÅÕÕëb9À-q\_ÖoOXl û¢iÄDSÙI"ÖX §\åfJ&b³ÐÈméEesߤNIº-'¶ü¦ÎÊ:S²/&Vê\ÔUfl¥¤°NxU®Ã´Ñ¾°µ¾PB=¢J%§½W¥H§H5K°ùQª\3S^g©�4djºqp\-ÙRn¤¸:OÒ\+tkë}ÉL]é2ÛèL/GÅqIÒRrsÜØHW^Òºr>>©®¤Î¼¶eµ ñy©ÀÍB>]¢Ôùƶ»UéQ)6|-µ ÁJÈÖk¢¯2RÉ{7ÜÝìu<²ÎÈå;ÖÈÓ¬^WÑ:+ÞB¬·Ùî¥)m#M#\È çø»A«ø^Çh¹Ýè>üJÄxÄLIÁ²NÚ¢!õ1nø9æPÙ(h «øªbÅ3TÀ=7±ªy5GtÉÜ\rJ9F k\"Öà¢b¾)¹jT9ûì,9tg\%+Ë$l±ù²°æÌÍÅÕì,r{ö+¸ª¢§\5ôR±cOKyÕÕ cÃuòU : a1¬\_HuI;«ÖdÜe qW « ·¾xUÆz/AËr Õû± J Íõ +m
b]Z=BÿÕ;úo§$(\íÜ«q\_Y¶53ÊØáßÎ c\_É\.ù=7/ëç̸Jk·ñô
lí>¤¿áñ û2འ±ÇVjæ»ÊOÔa¯£}¶ûzÌøyº.ïÑCT÷wzo8!¸ÈeºqýÎ ×nOá¿ù?N|x[ ®V¼£³ù®Þ©wþüÿÔoøüEP~ûÎ9:Û;3ÇEæB,0ø¾¡ÏéÎ÷¸+}wÌr~¢ú.ãÖZ.ÏÖéGº[kù¿×Àè3º-ï/ø͸LåcºM7áõug¹È[.ᤠè.3-@z%¼ÎBÈ«ô%}Úzïµòï¿VþúR½,=ÐÉ=#~ýC\_Ìi9ÏTsTmñ·~Ñ[|q/ûÀ ÄíG<Èvòµ¸ñYĽKt%Öoq'"W"îvÄ»÷"."þq?}Áø¾|»yîÿ�ÍÚã) endstream endobj 550 0 obj <>stream {dup dup dup } endstream endobj 551 0 obj <>stream HVXTWþÏ}o ,øfà E°QDì]°¡D\E%\6lDÍAÌ%v£Á%Ý®®kV#êºFg£ëý³÷|÷{sÎ=÷~ÿ¹÷ÿïZÈÀáí<¦^î^É#÷ Äðägx�Ä]Û0"-UÉN: Øôä°èäÄW/ÆÞj±?0&!#úH¥C:Ïé±Qá;õ-Ôx^¯S,j®Ì~ûα©é+·Þaÿ$P#$!)",À]Ç~Xbxz²MüðÎæ|ebxbT~ìó|ùfrÒäÔÉƤk@·Ìªñä¨d×i¯²¿ñI ªiÖj<9ãð»/]E´¨GBY$¢j æá7mpÀ�øCy-4>æ{䥽'rx}Ùuë°�Y¿ö¼küp×}¿ó¿%Y£µ©V½FMÛZµëØÕWß¾AÃF4mÖܱ
¢Ó;9«W·Z·iÛν½g:wñòöñíê×Í¿{½z÷éÛ¯ÿ :løA£Fù$8dl踰p81)yRÊäÔ)iSÓ32§MÿtÆ̬ìY³çÌýlÞü9 -^²4wÙò+Wå}¾zÍÚ/þ²îËõù 7~õ×ME·lÝöõö;wíÞ³wß7û|[|ðÐá#G¿;vüû'OþáÌÙsç,¹pñ.ÿtåêµë7nÞºm4Cßp¥ÅÌ&-¼1fÒr²Ó0I3¤iT ȶr"9Îu|¦4P½bPÜOÅ[ñUü^ÊLe£²IÙ¦Óèêëêô:®.T·JW¤z¾¾¾¾¾
ÞMßJßO®rNvN:ªPmU;Õ^m¬6SÕÖjÕWMP³Ô9ê|uºB-P·©»ÕbõzB=§^Poª÷ ¾CC!m-,.!±¿§û Æ>±/ ekËCåP9×1Ëñ©R_i¤(Vì×ì
¿Ã¬Ëý½.cwÐ;¾Ç¦´bW>=ðö\5_ÝòûYÆ~±{Àe{Må.±·LÞÚoòÍÏÌOÍæ'æ#æÃæCææ
w×ÝmÿåsËó\^Ïx\ñ¨AÅv b~yÝVÍÊÊ2\4å7ËÊóJóJ7.J7Yg5\T:=÷2ÿ2Ï2gSSo¯ÉËÔÉäir7¹ô¦¦&{6>0Þ3Þ1^3ÏùjÙeÜl\gìkìct3º F'ÍQ«xXÅò®ÿq"V¤ý.÷±|ëhUÃÞ{Cþ,÷OViòÑ¡m(ÄÌyÈÃßñ!Öa36!oÙXJ<Åb|ù$Á'ø[ðÏð/lÀ×8ÓØñ@."qQøgqçñ#Jð�Ѹ¸Á/X«ø Wø' D$"òIHF &c R©øÒ Lçλ\Á\ÍÆÏxbÒl¨U§x3Õ$[ªEµñªCvT@õ¨>ÙSjH¨19PjJͨ9þäH-H!éÉI%¹+¹QKjE© ~Å5jKíÈÚyRêH¨3u!/ò&òEîPWò£näOÝ©õ¤^ÔúP\_êGýi� ÄNì¢A4P�ÒPFÃiÄ+¼Æ]üh¦1ô S¥PGaNã)")¢)biÅQ<R%ÒDJ=ܧd\G9ná6¿ze¸RìÆ¡ ¾Á~.Ø}8Y8yx!Úb ã{|
å¢îøVÌÔøòô NU]ª 'V¡é¿ÝüÜòã Þªüð;¡±ý±¾ÖfùHûÿýMÞÚ&IlG®pÁF~§:'ZcVPutE×-x4
¹ÔGÚû«¡r¼-<Êó÷²ïÌ}D´°ÁuÄP&~ ®Ð#Ö¶«yeÖýJ·;¿~èÑÌÖ鼿È^JYèÎ f~,ÁDÍE q :´Õ½àËsz uÄ,åøϪy<
Ò(WøÉ"SJZÉ^NÒdY&òî5ºóªUsÇ,ÏNd¥d³6׳w0;P¥Ó¡oZÓæCED-HrR8^ÄÉ\¤¸½H µ¬(÷ Ål'ÉCi1áp¹¤6û S2zåoúé>©<?ªl-òEßíRCSY¯îphò@Z&'%åF¯ü¤<³boóðû#ií+a4¾ÖçýóÈ) ÃÛ|©á´ ìÄ%^1p8]± àØîÇ+:e[DE=ÆˬÐ+Ìe! µ v]«ßXÿ |a$-¸äC&΢O_ïóýÏ´ùqÃECÕ#A§ÿ±0½5WpE«7~¾ÌäÁ´ MÝÊmºv >_Ïd ÜvÅÅT¼
ïX(õ¸fòn¾C¨$V2Áü¦,{½rk«L ,ë©#m¾99%çx9.CiX$m÷Ë].%~¥8ef~sfkyVDý R³Zdµ{ë[ÅY«Åí2¶þééy ÈUTSa0í,,:ÜÙ¡IÉ
¤À' Ùdi}~Jà9¡PL¥ ÇÅìæJë-:ää$sÙ ~C«ûÒ¤6$AÚ+¤ÆÒݸm¼ª¥ Uªy�uªFuQ÷ÉøBþµ{Ü=Á=Ð0ºM.£K
~àÉÛ?Ä)(ºxÕmD¬ åÌ8¢õ&bzvÙöå1EÒlz9°ûÔôÉtª÷èÓ³9˧b'V³.=ÊS©|ÞrDB"'¤(YHRÊðòõ±a)}/£1ÔãQ_·h£>µÙb%Ýê¼=ß'KyÏOÕ8çngÚùó='ã=¾~'üIø_a\Á_qÿÑM¸KXÛRWSD9ÆÖP°Ëíqû øÇc £5hmqA¨ ïjP×Ø&¬#ú:Áë2µ¼>o6R»ôí&ÉëC/:«aZÒi¶XèZ#ÓÒ°klÑ«ÒݬExYK¶Ù,pcÝï>}éç;4ú]̸vÇz'ÃðjÌ0ÕþÃ7¥ê½éùw÷÷M
¬MÿçÖ~SéY-Mi0I$¦ zx¹ XkJǾgLÕë×Êoq¬
¦t,8J) h8DSÆ?>7éÆßÜïÙÓÀ¤õ{ª¿S ðGx&^TjRK×ñÛÈz+M××[µR´Jͬ©´þMÛûzþê/{8I<ÎØðÌÕ_^ûòְ訢íÌïzTêfïø¯î-ÌÿéÎyiïÖïS¼Z\E¥¸gôî/¤wùÜn>'çnç>Í}cø\GÈåÁ ø
+«ÝMaúS©¹í^-4 ÔÌù¹(GËÜxã¢Ç¸In£9.$;¸ÖÆb¹owp¨®Íñ0ñ:~\¼pÕ6´¤f رs+³Rò�Ò<6U©O Nr¬oÕ½ òxõö¼mvg,E¯ÙÙn¦ MÕéÓÿRü4Äoí=ptÓÀV«#u·dRLý ò oÕc¤Íµz©ºÜ ©QÙa�³>}ß¡ï"'ã¼MÉö0)ïD #!>ç@Ãø;þ¦S©1,ë7Sï¦8äÈOR¿ÇGØÔGÓùmÓ£çO¥&~5 ¦É°ë Ó®Cþ´xpj£¼\n"¦ÅµJ}¼\éíÒÛ"ÁÅÇg\_£Nì SE%¬U&&<
®T©P»Ùåþ·aäS(÷w_!ãËn3B ìòµÈæIàñ¢È,£eÙkqºr© |»×- ÖÆZÞð¶7½wñ£ kk¿ßZ/ó¹î| yWnr»£9°Ù8¯E-Ãí¿)tx´&¸sblåÝOÜß/ à¤<Q¤!9%ôEæ7QÁõø~íæþîçùîѦëïn85Ixq,ñÇø§ÓUzíLã×úë-¶Ý /zô6ÑãUj^-ý°t©1Ì V±gô "é®ÉN÷/k¬K¡F®~Èþ5Rnªúrºöíu Rm�yø¸Ø�ÝÀ>çyÑ=þ¢Uì°Ù;2~ÕA=ø}3Ü7c>oæklÄÙy1òçËE@dRSߢç¨Ûø/Q\.ßÁ\9
b&JüFS¤õTQYÄÏv2Ä@KÒD0XC ÓK¶~þ̧÷¾ñüP'ëìÿÇc'ßÿF?|eÛ¥
\´Þ·ðúdáä¦Ý{ì;ðØg¯ÌFí÷?ØßøK¶g;7%õ o¤Èþùwö:9þì½ÝÅ]Ëåß:$
ÿnãê'l¨!êêwiW0Õ6Mc ¶wÒÁízGk3¥ç«6gôM$ô6óz³Æç=n.óoæ-úoÇÀîñw!äÐú4]Û«Õþ ÝÔ¶KWA£^(üµ@ö¥|W}ðªøÀrÚ.P= çl¾Ø]°.l%T&ùÙ?%!;1Fô"æåT÷ÿ%è 6 >·¶q%2X¶UÀ ¡a#ÕáÊ(DØ-^¨~¾\_
i(ÌÙ-õdñªØ¬cPÙº°ìÐ2uªÈÖþuÅ$òO �êÎË´¡]oèæÅ\_ Çú;ó\_�à+K&uÓØ7½aÿv7 >\là]Hbfv
m}O¤Ýï©CYjÛàRÔ~3aS5ÉUp à¤[:{¨Ä¬ÏuJMw®ÒoïÜ}¨±ñÝ£O.yèôóO<Ú]ªyvØôkÙÛ·¬:°¼-y¬©·ÚúÃÉmËZä»Y6¯ÔÇ;ZZ\³ãÉøÀéþpq©/;o¨§ÎéÈñ}eóæÎÇÚc]]oÌïZ~çoü¢Ð i\_·?çFÚBægUëµ z++;zï[P×;GÊq!¢u½ W®[làË:¾¬åK/kø²/+ù²Üu'º8n;Ú¹öt8Í!!Öõ÷Ç(ª"Ëkí²£ÖkÖõå,µêMðµuÑûYÓ×p?ż±hL/ÕÇÆ^ÉM1m} ½ËIþp D8)S icßFÆgÿ3¼òF"?§Æ|ìezäÅVKTª½{jkÎþð.^Ejª©Ì÷;
Ê
4« QjO´Hè!]íÖêPY1_z\¾¼@Vø(¼÷ù¬ÜÏys]z0'c!àV§ÓíòýËÚݪ9¥Eͬ|÷SûÚ ,Èòhõn\_tI)îÚòµpsiKzýòGya+ç']è%£·I¿·\n½íräÖæȪ²ÙÃÛ,+z QïRH©
9Ò´¼®@ÍÈ+bÏjíð|½Q3 wdIºÛþÐ0(,ê¤úý롪ªq\_[篫ª« u%¨~¸^+¯¯×êë[:ËeeÞ¦ÖÄÚ"ï¸yc ñgçT$f¢ÖR즦
¡º®ªÚ{jjJ\JíV;ÒµðþHÄϸVuUTø)%/tg£%C~áa{{Ø}Ôýû¬û÷\_ÝÎÕîgÜ»n7svd·³oêôƹZý븢¢C$2a¨3ÜàJÙÚ°G-é ø5Y>¯ÖWQ8O"h}²±¡×ܤUe ªæ+ò¼ h 57ð'OeæüÆìuØëzÖ¢÷ÌéäñññäN½¯ÿ ùöñ1Ñtìh;|{vR<þì·¶?|îvì)Æ¿ë1t%mﺣrÁLðÀ×Ý^k HO?uJÜ;~RÄO,:üÕËÇ¿yõUÌõ²h=6¹¿¯güm³ßÑä¯ËeeÅÅ9ÝZö²9Éý_XÛéålHÚ¶mógÞëÏ|JsÄ¿z^ä7túÐßÿ=èqLãÑ
þÜf/9Ï'§×#ªõ¦vÄÌÜ¥!ÎÒczqÚ¨RÜ¥eƵiqú·'õjÇûgÑ·W¦ÈG$Ñ^ý,-¾,�VÀ+°à>ëÑ°sÎ@»H G6öeyæJa\0¢´PS;îûÍ�
´) kcöã"Ý)G)Û#ÞQÚ~·ã~³ATOÑ yѾnQ%æüÄÀnn²(djaÎZØìd\_ÕÆè-¬½Sh§hïb©Ò&Ø"íA1Maþ³Ô
ïU ½èô´óTb=°×x.ÓÏÚW¸ý±>F%ÏûµèSb®$1 =õ×uó3kbýY§ÿ èv+Ä={®ß¤Û@éu3ô0mÁ«×÷Ö»ôQû ì ÛßÍpLQëc[1GSöHZÏ
Ì àìqu;büàQR ì=«h.Û>lc´ ú¬½ÊÀ/ñX9@ká»$Ï.Üwn@ó3G3¶ï:©~]ÉþsðSX¹v�ÛäVÔ¯BØuÃÉiøã%'ã�ú°^y ôa|3LÝê;ðÅ (þÝÖ!ŵ4ØþXãa°í¥rÈ÷�¼ SäCmdÎ2oÌæx+ôuç50Çgi;¿q{F©{¬sD8f·¬ç¹Á/æIF2Ç9ÓüÁ}¯ùuCÆí+JÞ2VÅgºXG¶z÷?dF§ÏHÄ6ÇWJڲΰáù´¼³) óØÆö<Ïå|À1 s^àؼYêÚf9.n°ç&ÃF®ú%Ú8 Nn[woûf$ç°ñéçgpýÿo!wdüáFFj_¥U?q,r<¥u] ¾Iè6ßsÎ5Þ´'Ì=ö>�µ'>{¥´Ïä]¶Á1ûRöÏÇvg«¸Kç\c'rs}^%Qëý>êÎùÖÌ!·ÙDÛy<·ëß´Zÿ 5IØ+NqnÃó}¹õ÷é>õÞê£?M«U¿ï"·öSì ºIêýeä±w«®P·ö!îÏAþ Ï<ç'x z©!ß=5iÿYµóHnSy6úƨZq}mðÊñ}ÙñÖ:rÖÅÚ5ømu ¶b{ñØýÏå ÉUT {´2Ô½ÒväñÊVøªOÓNÛʹ£Zo¥³ ýç9 ù[è¿Äyï ç¯TêÑDÞx|à¼[Åö ÀëÈÇÐA<\FÎø)¹Uçþ,Rõ¢µ2KÀ%Ô&õÎ=Ö!kíʤǧ/¢»ÌvHè§ÆÎb¿Àß«Ü=ª\s}àéqw£/ëÀyõ¹¼ÿy3¾Uf8n"q>ü-ÈDGmÑ¿fÿý²ã8ãø3{/»Ç9H $ÞÅØ!ÎÙAîáÅ{lܳch\bspîr·Ø&R ¢Xä¦:Åã|H¬+Õôe}®Öú%jU9])À¨¯)ªªôSåþg÷¨í¤jZUUÅ¡ßüyfö·Ýá7|ÿÆ;]ÈÏ(Ì÷gÀº
$/Æý zû_ ûßÓàyÐ
3㾿7îÒóñvd)ïþçØß4Ï&û æp¼sû·wS+öj½µÊ¸rVý«þ/ð!ö Zãy#üãðû¢ø{VEy:ÇxõMspsãnøÅ»}xÁ$à÷¥-ðeåìÄzd±×ð3l?èø¯ãg¹«åÿ·¾k]ÞĹ_û®ÇîwÜ¢àwDëòîjT|èïñbýÉ5ÖXc5ÖXc ÃMîuú3í¦_\Ëjé Ü}þJ6^JtÎTþm£"ä,Û{>g;a/äl¢ô bwá±iö|Îf´QXÈÙþ³m´Ñ³í°·æl'ì}9[¤láÈD¼)¡ Tú£±xdpHSFÇ¢ÚD,¬tNÄ¢ñlhbr@ );víÚ±ÉÎÅ72¢µJ<ÇC5wÜG£c¡J$¡-
GñcJtà³£ô)£Á ¥/ £w1¥?×Ðáñx$ôkèX¢æpK{×þ6ÏJcý/÷Z(B@/RB¤ òAXýXÔMPܬ5¯Báü1<¡P§iEQ#§b¨7AÛà?:<ÞÚ
¿´=gí¤x}4?eUì CÃÐãfojîÛ>´ÃÛ5KBôø"æó¼×YBÍQ³Þ1ø¢4ð
ú~æa66í3ä=4[ÕÌþZsA|>oÜÃçÐÊÓ¸9Îêðh|¶yOßa¬A;uÑ~j#Ï=#{Æle3Åý_÷ß)ýÿcµ£Ý@ïó˵éHµ¶^5ȯ6À|4æ²%}²Aö Ûj´! A ÕfS7§u^2¢\Ñûô0åôWå û8½|¹GÒ=!ÈÃêúrùHB¯÷]]×m } ±Ýöc¡Ý/ åìVú¸¬\cRPýùÒ ò'æ»Ü 7¾bæ£,ËÀNn¤FÎc£$ó©ÕÌ=§Ì©sísSsËsNÅ4¹Å92ÇhɽԾ´¸´¼äÈ°õjÁ×ëå×Ákà¯5!B5ñCZ æÍÜ$kR0%L©ùÔrÊA)wª.¥¦ÚS½)'åÊy©¤ÀÑ öòiö%c:;-Ð{¦nÆF3oÍÌÏØܾX%Z«ÄÙ0
QûÉ ìÀQ@bTË¿õì¥Y]]µÓlݬÀ-á}ö1û;Ìßík#áìÍ°Ës\ß
6¾}¶+³¨®?0g/ÔË»åH ^n|!ðJòUÔ{.P.7fMjIøQ9ª½'±1¦ïå¡JÕXùÆ>¸HwëBsRK7ô¬>¯;zõ>¥'u»@y9Ò#yº-_^WQ hDØ7¥�3ÙÀ|1°pLîkØ=¼è+7±À7±À±xK 1¶êNTæÝ.w¾»Pq)ùJ¡SkEAt>.{µÎ£ÎIç ÎÛN±ÖyÈ)0ÊHtE&¡Lv0Ó²Z%{e[M° xp¾-\þ$8hï^|%ÅÚ¦dX"ÝVµdÈAK,)·d%¥XRlI%ù8-±«Ð@¼¾.�¼ Îo³àUð8 ^�10" @·wØ ²¤×,)³äAK,ɳÄaM.ðkp ü|L¶Õ»]I_»J½i¦{Ì´VUÅäOÅäbò¼üýbò¨ì7K$EÚ(=,KeR©ô T"¹¥uR'IS²KD1¿í'b|ÚQ
®~ÖpT53£ÄOþÎæ2f<ó þf£ÑãÇ\>cx=~ÃÕè¾ÄØ7zà53:»1dî:]a<ÕýÞ¢øéW\¸&O¿ÒÓSê¹÷W¶:Ãüíïá;vEoò¿®$w%oIÓUÆÒ! ùgz¡{²û4rwÖHG3Úê¾$QsÏSÏYzYÈÏC¯{+\{Kݱ'Í!4Uª¸j'vò==FAU³QxÑvßv/ÃÖÁ]+\;ÕTYq]̹á.Æ|Rã¡q³ ¤¶aÜf Íã98ëá¸öû³Ð´qSƵÄ8êCã<ûLidØZ2Ö!C=Ók(U-Èe9ǦªÌ¦YswIÕ[{q¾´ARÕ²ºÅ¿ 0�¹R|Õ endstream endobj 552 0 obj <> endobj 553 0 obj <>stream H\Ûà ï}¹l/9 (Ùr±6»ê¤6F½ÈÛ逸 +è|2óã/o»ÎhüÃͲG£6Êá2ßD¸àU ´ô[<å4XÆIܯǩ3ãÌ�þIÉÅ»6G5\_pËø»Sè´¹æ»í·Àû»µ?8¡ñAÓ½ömxí:Eyí×iþ\¾VPÄ{³ÅÝ®ÈDF«q¦Õ04ê_>/ì2ÊÛà(BqQ¢Ê#S .ëÄuà}}àCCà6qøøøuqê\_þu¸">&-
0ücÊð rÉ»sdOüèKpD|þ-\lö+À�0³ endstream endobj 554 0 obj <>stream HäVipSUþÞ»éÊBYJ-¾$- --"B±[Z
B©mRH´Mm\Q ¨ à®)-àè #¨ücÔÑ. #\5~÷åAÇQÑqÓ~ï|g¹ç{Þ2 >X ÊUyc]GV¢åQQçsµ~<h "Þ©k i¦�j�õ,ýæúÖßåwÎDoÚv44ß\±½ Û)×7z\î·êÄ2 \Æ7Òl@}<õÌF\_hÑÌÎt©3§×ì¯s%e&IíÜo£Ïµ¨U=)XK/îÅåó¬Ü·ú4úÏ´ú¡PªP,ýOëgO¬(¤^K=5}¤FùþAvNÖ«4Ʊ.J&\çýjJgV 5MÝ1« #ÝÊh2e(Á£)Æ\Eü6t§0ÅÅ'$&%÷êݧo?sJÿ© 64ý¢ak«-3køÊÎ7æÒ±»|üWL?åÊ¢{iÙUWOV>}FÅÌÊkfU;öºj³æú97Ì7ß
Û,]¶²ýÁuk~ôõ7=¶yËÖÇ·=±ýÉO=óô Ï¿¿ÎÝàC>/4öUhêzuMmØðÜ¢ÚfÉo¼ æÅ÷÷þ¬¼xt"n½óåwÜu÷=+î½á5¬¦©
X(ý&tñÚf^żcx\_á,~¶¦E£òÞ àm|Sè¶èÇ@t7±+Ú ¼½ÞD31Mß±V¿ Ê:¸YE=:ÐFxÑÑðq{?Oò:±�YåBìBnËx
íÆJq/îÃýhÇÞÁÕXð öbö³J7Æ5è£vò �úò9å]Hªtt(J»³K®èòÝH7wtM+õÚ#Ê|j £,d"G+¬²YS ká©î°V¦5ºÜS.éðyZU/¯³H¡3ýõ8Ç$óô=¿£Ê #ô4A.ÞM¼FHùlê1XÅ@òAÄê;)Ïg!CTÓ¿úvBÊØZõ[Ú+U$ÿªÛ©þD~øø6åKÊDQDtr}"ÕÜ>1CL d=S ^dØ \ÿ)ñ.ñ
¾Ö¿BHbìý:ññ ±öID{ î·ª/²Þ.ò½&«@þf¬"ºôçûÑÀ÷1ÅäÌ<çË|Ìl&îÑaU72~#÷9A©a"aíäô>AÖV1ëÕ�}©çõæßÀ¢8ÃNRÊú¥\Gð\¢8ÖóéXKl%¶ñÌÇ(Ý@] z Ñ|F|ÖÏì2ä"¹ú>rO=öOêVSn1pX¯×ª¬§½¨0ä]¼GS¦Æ¸îb 5'ýøïüFüïþäî¨rõó«ú5U¨îMæìÓóÌBÏ`(0ZÈó O^%'ISäÜÓdp¹ib\E_|dpAû×7!SlÇO¯ñø[´ Õp
¼þ\¨¹Y xCA-à zm÷¬Êê²ì»·¥¡Ö áq{úþó¿W gºÊ�g0yuqþòêö\z8£5Shs[óÔ<Ê6^Ý
Jæ(A5ÊͼvF·0¾VÏÄhÌÐã¼ì|öзy]ÌðW×þsÑÆÄÝñ¿÷¢ìÕ>.ÙÙûyoks,!É4ô£¶! Ù¿0�B endstream endobj 555 0 obj <> endobj 556 0 obj <>stream H\ÁjÃ0 ï~ ÛC±Ë.&0:9¬Ëö�dÅ6sÈÛOvKØBüúÄ/ÉsÿÒA¾S´f|pkÜÈ"8û ·ù^Õß.& Éð°¯>LQh òÅ5ÓgG< y%äà ¯óp9l)ýà!®z3ébY±SïX÷y?1ó×ñ¹'¶ÖÍÍ×d, 3 8:ЯÀàþéí'ûmHè§{TÈVÕS!ï=e¯ v#boõÕT±ã>Nb¦Ê¿ �Xqß endstream endobj 557 0 obj <>stream ÿØÿî�Adobe�d�����ÿÛ�C� ÿÝ��OÿÀ� »r��ÿÄ�Ò������������ �������!1"A#2Qq3 $%&'()456789:BCDEFGHIJRSTUVWXYZabcdefghijrstuvwxyz
¡¢£¤¥¦§¨©ª±²³´µ¶·¸¹ºÁÃÄÅÆÇÈÉÊÑÒÓÔÕÖ×ØÙÚáãäåæçèéêðñòóôõö÷øùúÿÚ����?�ßÃbt""""""""""""sMö[¹3m%YÕ? b+HSd39'VÚ«± ?=þ jÎ'å¸hli;Ø¢pW%ÎJNÃwÝ.Ðäó%Hr ¼_ "Å&NùRoA
³íÏ91Ù(þr_I_¤v@ÿÑ߸tÇHKÙ1¨m&ÁX½Ôã%$OÇ,|¥KHK×AË}ô ýÇ-;bpørÒĵ®¹6ß{ %s±a´¦!ä®J>ØlAÖ´UÎÑ&;æ9ké®Áj~+õ)å�ÿÒß"""""""""""pÙv³§2PsÞQw³1Þ§ÔÉ,35-a4Cöú!½ka§ÓéxG1$¥¯KF@~íáÄãÉ»¸b$BJ(¯Ck#±°î&8ñàçDn»ØÿÓßÃf\Ç{9Ø»;§3¥3¦ÓÁÞÜ-ÄHk·3 #N
ä¹Ù'³f ê°òGtSôPIÕ::¶ê9 ¼÷ 1ûtÈ6\Ù�ÿÕß: «áS\ÇHÇ Ã4ü"6Èý¹
©bLÄs8m®<kÅBMûÉ(b'&qØr#ñ×ÖÍraÆË×-8ÐgXªN.ù+åÉò8¤êñÊ;}Y®ú3\Ç7ìÿÖß(ð~N¹0jB""rJ%Ò"""'&5Ë;
¸é>IÄ^û§iØÈJGU-|¤|·ê|¤¥QZvM"'b5=øò ¸¸ô^:æØËÛRE´ÙªÈùP¹Ä㸵K_"b0ìzj ¦zE0lTËA±ïPÿ×ßqÞ¥^""""r#q× rMìZ|9ëé[Û].L6ßo'VîÌaÐî%IE'6#vió¦1ÍM\µQf:ÿÐß¼tîØíÒ""""""#f0ÔfDyZ
)©Õ³Ez4´K×Ô¨ùo&u%%7]´"½u¤¬IäC6ÔÜ4í)ol2¼¢ÍéÄäÆ·=°êÞaË[M¬ã¶r#±Mz»HÙXäHlÓ¨ìgcpÌÿÑß2d±'ò"""#N2eÇt¦uÚFõ8¿ TVøGºe¬\Aá׬bz¦Ç'èÎþÜ´cµ¨Ê+ÛÚÝ1ãÊÎ
uï²w ;ÊÈÐ>D6(òdÉC¤ùKiÓxçÒÚÒ GyAkc±:èÿÒßÃ&u"#§áñxÐøc E>MÝrcRØÿ�f5/wMÄÆDôÇ8÷ÔRéaíVnÐ)þrV$DH/PóSqÆs¯g"²ìÁÕð'2§$óQÛCì¹7A{#Ø:�ÿÓß©IÉs³½'B"suÒ"#liðé93¦Æîm <ø3«ì íV¸cò474b>æ@)òé9¸å!cOJëºÊءТ¥ÖÈ3LrKÓºI.àºb+F"íÑ3¨æJx<ÿÔß"28Ãêqé%¥1¦ÌaaDiäDH®©O;iÕ±·®¹ÄVv G ¶IYw®»¢¼i³ñÌ٠ש³8o§}ui6iùpÅëË5^H²B"6jn90QSæ!Î#Ä»H¹\cL£»4ðdÉÀÿÕß"""#ªq³ìºæF:DDäÆ¡´a¾Ðɧ"¯´\?1=£MÃáÔv±¾0e+ÔÎZìhRs¬ 㡼û!¦Ën à¼ÛÀ"rcMâ1ïïåZÙ5e5m®ÛÈy3Ù±1Ìl±Pü�ÿÖß"r]"wÇHØ8´!EÄ\;Ú)ð¾}¥:é"¹¥Ì»®Lb¦ÏpÌ$¯�°¡çÜ Ø¦bO ëûaÕåãäµ¦\ÉòåDÒ{ £· ÉlAw E5Ør#�ÿ×ß""'b4ò""""""'$ÚDNLj;TÞ¸OÜv¹.Z"¨¦f]é8)ÉEs0êÆÕØrÕìäPË\_¯'d¡ym[°cvÞñ |5pY!åñì¤É-75'ñiBfÙ¹-ÙZÊ)7·W;d¤ìÿÐßÃf¦ÑáØtÜ4ò"'$ÚDDNa½"M¿Ó´Ô¥ÚÆ'Ó°^ºUðtxGËMGOÍÊÔ×>QC^çE´ø¨¯ö,üGRXlÆj^:Lµ|äÏbù%óµÑÌëv�ÿÑß'0ºFÆF:DDDDDDDäÆc¤DDDDæiW&" Ýv'k}(8Ùª,ÇbRǶãÉ0u4è)±--×Ø;"2ÑTëtÓM)A10|F¤§Ñr×å£CÉàÿÒß""""""""""""""""" ×®AG¯n®R>ªytöç\;Ó¨¹91% çQMMpD×ÀNÞ$I½Lßå?õãåªtÞ.dSÄv¦¹Ó ´a¤ÿÓß"""""""""""""""""'&5×û gcó©[Ë´<ÌÓgÁå>Úº
£}Spè;qȳág®NL§7RùÐó٠ƹkuðl|XGÖ29"\XzÌ;òëËé Óª.ÿÔß""""""""""""""""""6ѨNªLÓb4øtÃͳGjnDfik³ÁõäzM?ñ ¹'dÌ(ðScëþÍDZ~aÕt]ù�ÿÕß""""""""""""""""""$SC¢¸y> |µPÌ#Ìv§aôGh7t£Ã»¥0dIGÒM6È 3Ìë×Ï:\Ó|ÁÕ¡ñ¸ÿÖß""""""""""""""""""""""""""6gÌh[×»®Ú0Ú[$ûÇ3RâǧÅ#êÉÛ|CSåÈ?c±:Xÿ×ß""""""""""'&5Ò""""""""""""""rQO',ünÖ¸ÊIXìlÆSéßMwqåÁLx(Ó4è®3Ìi³n¥2ÿÐß""""""""'7 ']""""""""""""""""rL×öõ@tN1®Ìi³áÆ2dÝ ÊJùrÕÃ1,U$F»?EÌA¶äßHÊ@ÿÑß""""""""""""""""""""""""'$¢uÒ6gÝpÛ> ùqc:¥êc¸Ö3Ú öuLé±T¦vêdÎW=×:ÿÒß""""""'=:DDDDDDDDDDDDDDDDDFÌc¥3Li¸î1¨-n|ز[ÒmqWK:·QÛÁµÃñÝ6ÓÕE\&SGÃÚkð¿f#ä@ÿÓß""""""""""""""""""""""""'&q×7 Ó¢uÒ#lº|óÏÑñ>Bú;ÌW×4ÙNäK\_
û4ÿ�&§Å"6l]ÿÔß""""""rcJ<:DDDDDDDDDDDDã¤DN¤NDaLi~NºHo] jêïÄNA·Ìñ(9)Õ¤v¶®Ìéõ\V:ÁÄlÉ®Lk@ÿÕß""""""""""""""""""s]"""""'&qØ"s6"ÍGjÁDÄ ø("3»AòÌDø ®ÏzXÁßÃöcѳäÉ7±îhÿÖß"""""""""""(òDDDDDDFáÎiDë¤DDF{;©c¤DDDH9-¦Ò;&´kCæ<mã÷&4WÃoFÙo|;¸ÿ×ß"""""Ó{CÎ.93³'ÞRÄc¤Aù8{z2¿> êqE!kîam¥pÿÐß""""#fÇHÉr$"""""""#uØHÇ&ÎWÊ;øDDDDòaºï´frJÀÿÑß""""""r""""""""""""""-""""""#lniGb;&""""r< ´á½×b\Ô|¨úÿÒß""""""""""-""""""""""",Ò"s\ì]±v8×b0±Ò""6$#gÀ^Oe3fÚ椤;É´ÿÓß""""""""""s""""""""""""$s8æ¹K³ºÌÒÌtfÆö>,ïg&źÚIktCÚdÿÔ߸tÛ;®²q6ѦG¤ÔéÑåÒ""")ð^Ã[ÃQii³l2läÊ7ñGdwÈààÿÕ߸t½9]'"t¡Ò"""""""#eÈwJQÄ#ëäéR"'&J']2ldNØËGT.Ë(¦Å>ÈäúW°ªÓ OnÿÖßä×\téɵØ"êt}úéùDæ¹2i93ÅÓÕÛO¤äÌKÓ¸"ë䧰y(;�ÿ×Þã÷ÖÝ"""""'êq9]öï´s\IÔ±Ò'&pØ×-&G&uÉ: Tzså£Òo%ÑÁD§�ÿÐß)Es¸oÔi®8äI¶¶ö'HÙ]×,]ª)IRÇ.lI]>OöwÙ¦1:<ÿÑß)òìS\õIÓêqc\éõ8Ó$ÍòtÎ¥cpèG.D¸¿ïN¬c³äÊùKH2|J)¶öÿÒß¡x+äí8ØÙÙ5ÙDDDd£¦¥¬aÄDFösÙ³8lC$DQå ɺƹÔÒ¬%'eÏE;¥ÿÓß¼s«¨úIr-§×fpÍÓáÒ2]8È(tòìFiLá
X+Iðé¶(8Ù³ÈÉ;Óتi;g4Ûx(±äØÌÓJÛÍ#e$Ö(ox�ÿÔßjòj]Ýs6rcM¬c¶(ìQmrg]¬DDDn»é#µ±Ø¶»3
^ÎÃf®u<É7N×Mñ%¥ÙÍ)vdÔvw¶é{hêPíRG»uvlyX~lHKrr9<ølüDê^l~HÿÕßרkZeÚÚRÚúím©¡ÉðÖ(¾'7bN6Ñ®O\å¹:ú6Ýs´ÊJÙ$ÎìDæ²ô]§ÉëS8çbìZQgfqÎÎL3Äãsþc¹¥ØµÉL»µwKuÎÅtÇd9úL¹g ÛÛQëm¦»®ö�ÿÖßZ[±4üÄv©ZýGs]G5Ãë&Ìc2¶vQúCo\Mõ +a³dòÇ\Xv1§ÖsQÍù
.çE´ç5É&Å3Ìéx2Ör·Á1})EßÏ"JN»ÙÝG>O¯w û ^;XÓmlE3®i³cgÉuC®éÀÿ×ß ýê¼&4î?çéNLMÐl$¬×LÄslòt\£$î¦Íº¥ië�fu''Q±m9É7±ÚocÌ
M\ô[¨í8o®MRDLtK½pþÐe¬nÖ--w¾;<³&ö1-mÕ&뾬>Mض6Äå×/lC®#C3ÙØIQDDDÿÐßZÕü£¯±äø÷¨ñòÌ Db3Ø|¡½QúsNôϱJH¿W/nèb§õIT¥ó¶ i[60îÉtÈ×bqÝ)òm¬eg ´¹2evNÌ6|F^9§æ¦ÌÅPL£ÝKYÓf0ÛÆ'¼bHæØ[ÙgHb0ÛA½ÑÌã¦Ì»o²6dúùIwì'RÉb2Où(f4ø3È <¤íÔÉ´®¥ãðg7ÝuÈðu¹&Pºã³:ã²àÿÒß¹EhÃb©1'³S'÷VÝ¡,NcbØ"'QZ¸¯DáÙú?¶ÃÚK±È,uy14rLrë¶äv94þÈçÏ¡¹'L´DúG~Yêy\µ¶ rc ë®á½JÐlBjÐÉ®X?ÒH6µä¾"·Ù-ÔGx kéGwî»&»6\òË= JdÙòi68æO<öå«2Æ»íÚÞjdÆ\_'ô;1Þ±%ÔÝÄغäK½ÐßÏ ÔÒ|#¼ãÏÓ;äN¯E3LëDDD@ÿÔߨ¥b¬c¢u¨ÙVwª~?&; (ÛêJ)6LL9é/Ô2$¢qÓlÙñá<Æm¬1¥
0oM?¢>YGòÜãìReØã:8SÅ"#Ã¥3Oha§ ×s º)ðg£uü&Ozë·ðuMÅäÛf£"""-"�ÿÕß«´Ì§íëÉó®O5J×Dî¢Çvs]v¾SãáκäÆ×»íÒÌsÁ¶tN6s]êæ9jëtÎ"ï¦XÌ×ÛS\©Ý»9^°¡\_ÙDDfÉÓQ:èÿÖßòïT| ~F¦6Ѩf¨ÜìdÎÇçó\>#OVg|;5Øv|õE{ÍcØ;®|K_¤í£ rg O))ø:$¦ØòXä× µðÉåìdLmÔb ä\\_©Ö¿s"©>f7ìe®<l:' ùÇ,Z%Ï|ÉGº|OoSCf0þ3oìÔdì8løv #�ÿ×ߪ6õC2Ùí'Ìö ö3Tsuúfë³àØòu§É¤DDI©ÉögÍ4Ú×[Ù¤ì¸DDDÿÐß$<Lb/bkÉqòÄ$Eé¨CzkÛÁVÔ/\z¤zêî ·b «|ò¢ÞRDD:DFÚ6�æHBCæpÉs1G}\f5ÊÉÖG§M°é¶÷ê×ìlN8w§$¢q"""$ä:¼Ú5³áÿÑßrgLvGz u3Rq$Æ;ÇÍÆ{Gj£äén£+mIìÌÃyÚß³³lÔºèýrLó%?¦Sw.îÍ¥m6k©)~álN;2g ìí-¨¯|EÙÝ6´Ø»®Owa°RìK:N8UÙãÄD±°m«ÔÓdãbtàÿÒ߸QÝÊF»0o%4«
QSü¼r9zýÚ½à «ÔgzñRæ¹1¡æ½NÝ6°V6¶ø>¼S(íÇÕÈã>8µ7z¥Äpí>#ÄwG.OÑA{7ìæë´¶zï2dÛ´Ûùô¡-g·q^¨æt¢p°°øòm§M"""'!×-MKxÈE>c ö üðàÿÔÝx¥äK¤DîxpñÎ%¼jΣ"""""""IO^½L""""""""""""/nãDDD"+¢k\srÖvaȸhÿÕÝ{9[avK¦Çkmvg'bK">ù/ü={ÏøòöáRÍÜÇ ÙsÑñå§M""'b5%'3¾Ý·,q·'Ì:ÜÔ8f¤éÝ�ÿÖÝÈ:ÌìæK´öåÄë¤DtF¹1Ö5ÛÎDDDDN¸DD\&x¡~¼öÕêi°á;YÇâºä×ða¾Ê %"tëWé©:d}±¸vÎÄc#\Ô¦ÝKdÿ×Ý]Ä»¾ ¯Ñ1ØðtFbHtúËNäFDDDDDDDI-~ÆGܱF±8ïRêímëhêlÕêéfÓ]»¸·dèäéìÎÚáñäèq&\_);�ÿÐÝT#Ê'f1Ø?5ßµv#"v!<|qcy%¯¯ooI¤á²áA;á;ãæÕÍ +x°k'b4°¢5ÇpêÊxÉ3]7©Ð¿Ë31×÷qÙ>µÓ¾DÿÑÝÆÚuÏgdÌ90Ó®&áGyòix"wÐß°¼¯0_àÏ@æSLrÕÃ]ÔÓàËWÒ4èð&+Ý ëÄL9Û$èf1û®×H©ï·ùðìËXóÄaGàêJ蹿>@ÿÒÝi£Äwmv²vDDDìN%¬b)pÙÇÔýºBÞj?z£§Ìj+ä ·AÕëm®ÅÍÓ·êDë
ì´øä>'±òi¦f®Z±ÖR}ËFI2ì{êö|AÑå-)£HÉFDDì¸DDDDDDvÍEq|F{MñvíÒ""'u']:aÆ9è¢4ìÇmÈéóéÇúÛ}¤ë®f9%Sà¦uËxEúqèÊqbY·PSáÿÖÝ2g};ÄW]×úNçÑÍqØòá®I:á ûÆ×s<¹.{:\Y©}Á¾aÞë¾|¹3¦LcìC¶1ØMYØIì±È #O¯ÓZ>ÉëðééÇ:Öê{ê>J!2c�ÿ×Ý<ðG¢!àÆ µ%c³X¦1»;¢8þyðN{"váf;±%ÇÙ)ðlNx5Ø,5&c°}'Å"è¦J+2¼x"ìÕqh¤Tãm8þL¥f©íä-£LµqÀÿÐݪÆDdÛ""""$ü\RG/åéÙÉv8¦¢é¨ðÖìiB^¸#^vM"'}F1
ÜÏx ÏÅòåJÆ.Ö5$WEo.DiÔðu§'ú#f :|
O.Ö<\;ÙÑ ©Å2"sTö¦¹?""rcM;Ã-^ÿÑÝ'$ ¢uÒ""#ºät@: RôEËO îYÄ$DQäï'Ûe ;L7à'üU¢ÞøváLéWÊ\s\'}£Ï&æ;äy6jtxrÐåká;ÅÃöb¢1Ò'&ÑÇ$IúohrÜR§$\_ÑäNa©Ç¦| ÿÒÝÇqD¥IØ:óØ$«ô¯aã«~#}öç§b½´aÄI¶B#¦¸i±Y8§¬Í(HíÜôtLXî±
XQ0^è§a|s}>KôW^ót3i'üù{Ì#µ¦Dq\_¨éåI®�ÿÓÝÃd ãz»±ä¦1I\_)-äúqLãêÞz¡wòlñÖ ùf ¸ù" ÑÉé F;ÓNë®têç·0Ò"#«ìûJ-Óõ5¥iÕ-%v#O'qÞ¦Ì9ãäH¦b¸©Õ193
ºè?·Yû7ftd"¼v\×SGÑ+rLóÇÄÎ¥úÑ)«ñõJ~'ìúÒÿÔÝ'"1Ò"vIØØ®ÑÇb1Ô<Ñ@ñ]½¥xä)äè ºò|DTÿ�ø¾ à¤Y7°Þ¤£)p.Äé®ÀªÇhG0ê9bn©+Sº(] ªHã×gJ¼íìp?f)ìÌA±Ë~¹3«òͤøäÇÉ63<tÏ)¸øû$¶Q¼£µ°¢4²$ÜÇjy�ÿÕÝ""">+§E4þº]ÇÔù´S̤hÝÉñd´ØÄs¦½ %¸æëDNÄhºmQ\_É]7ì.IuqÜ(pØÅl>ðÑ#ı]O^)K^<>(RuÍÇLïïx.ð}B Ú¹Ni{̸t@®\$jWND"í´|ϵAhÃ&5ÊÎ;rãÁf¸ÿÖÞ.¸k;®Zêx:!?"StÏEiëì-û"öê]u%a7áüú¡w¤&JJ@Ï�í²ìYÙ2äs{q×ñ@AJê×R×MiK鶼WØküOÙù³'G$üä]Ênn¦û:¾NéÙñFø6ÕÇ"©éÇØ])ÔËy\ÐÙ35Æ;[uMkEÚQL;1¥íj§W@ÿ×ÞÊñì~ÇϽ#¸µvbWSVMåç5/ÄO"±]w1Ë#;Ùñ|Þ^ñØzÇ ¦¨åÕJdÉ0ç§f É/I:êJøhÓkÆïÊMN,IÔ¢§#½ªþ[Dú[aÝ)òìX¢NÒÝtøpêTëñvr8"µêqïlEì¢QLj9ÄÑnÞ% éÔÜ6!-e%æeÎÿÐÞXøô¼r»<ÜoÉ/ïË2h}#A·äªy'P¤õ¾9×±³^¨=7º¤¿Ò 1äÁékߧQjh2ÒUqÞºäAu;©½Nıº+
³h¨.n²nIt×ì|[J'z/E¥Ó¦>,³§] o&ÖDnSÄäRRÇû2x]I^?®»ÔÿÑÝëi¯¡ø«T\_ÄG"s$±ËUΩbpç´¡2|;ìÆ#ìæ¸s§VEýDSu:\¼0QTµraÞ§ÅêtqL»òe$ðJJ±¦X×0êør<¦Ò:§Bz;±.Ö»é-(Þº#Ôªg0¢1Òs\9¥ØÿÒÝ"ÜK^kU<|\z|LXLLIÑR°ÓªG5Ò«³õÓ&0×}v#qÇhE0lAñT¿/À~¥Ôêñ̤§Tá\ñ¥ùNÇ&EkÁÑ1.ôÉs´ æçè½õÊÆ^<9éØðsvá±)1ØtèDðÓ¨6lïgH,ÆIBJÎ&'"uÒr|Nõ�ÿÓÝ""v'̧<Ô\_yä"ý¼ c³zéDvI#ÚIã>¼X|V é®Êbac¸^\_¨{|F¡úÑ0Ö {¹³©ñR>%Ë|G6ne¦MÝ2xÊ#´i±]7ÙaÑ],â5Þ®ú:#Ov#]L§''&qmm½¹1¨¯J©buÒ"/fuÇ|ÿÔÝ"""v<á¶;záÈoÌÓQéøtFãÁé³xíØ''.µõl0£Ã±].©+°/'\å¯Xðc\cËva¥Çp¡É;1ØòtJñö>8ôê~vwu$³ôtN¥/^ÆÇHtÙv'J']" ÿÕÝ""":³h$ÔÜêQ©àÙ"4î©kÅ2#äqTS8ey"ÎÛ\óÒ)ø´øÈù/©8ì"ÓѦL#µ]Rôî;!5}¢ç:SäFf»î9×N9Ä8ñ8õߧ|£
$Ö8áìm ë#tÛBbpéwpØtÇcúZs±tßö7³±¶ôµÃȵ{9Jõ:g 'Jgs]r#""2'ìÿÐÝ'p¦uÏnÌc§ÖFpòa×ÉØØòlñA³ÅÖ+ýh~\%ÎNYnÌé¸uec°ªäª#º:!?b[w2±¢º¦;5v/4KVåôomC3®Ö\ö¼ÏÌÙÙ¶Û -®vIÙÎÎDzbjlRdïtqÄ$Ì"""2gJ' t¢qÙbuÒ"#m8g·;�ÿÑÝÇb4îL9
T]r\#RñÓV1ê+òEPròîSå¦ñ,b9'aaÅòÓu9ÜÔµã -ðx0£ÅqË!+#:§Ob4Mx× é¤IEqÕÖQIºäÆ8wI>8¥/ÈM^>©ªmgKÑiÎÙÒ×;m£PnºÿÒÝî²´ÃêvPþ®u>\°¥zó8g%å\_<æÚeÍ׬ ãã?åiGÑH#"""""""7±³l:-¥&·$ɺìÁc»®¹®»1i(º~]2Q8v6)±EÎHÿÐÝÚƤõ2gJ©DoXK[v¹1¦91®Xè¶Ø"G2oc·LÔL<É)rdÓ5ËN£>/Þù�4èÇ&æEÝìdÙØ
®z)EiëÉÔuy{»
ß]7³©ÜHÿÑÞJÄQiÑI><I[h® sp#sK©ÎöÃÈÉG3®Åtø¦é1¥ÕÎ¥ãíx©µä3,pÞÅuOqÉsMÓkÇ"xíÐþçÖbNrÔ´vn]ÝîïcÍ�ÿÒÞ(úGGÑ:ex>+ØwcH5Þ§Dc¦u.2æ¹kb(ä;æÎÃ~z)æ&ó&M¤ÝKHÄS;
æºéá2lì[DëºQ)¸ãÒ'KpúØýwHµ±ÿÓÞxI0g]IN=N(ªc&tè6\"v#NôyqÍÄV0"""""rgs±³mr& ±³°'&pÁv´DY,ñÉwJcJ Ý2\"wcczÇw×5ß;Dë®Äc¤´Ã±3'nÿÔÞJ}$\T1)íÙ6öQäã{AM¢ÚÝÑ1é³%ÙÝE6-3]6Ó¦væPZ0Ót»ìþ:öJgK³¹¥93®z'7]:Óôèò8ñ4F$6gOqÌæHR0S.Z4ØuÝ�ÿÕÞZW[RññESºu!ÓzWfqí¨ÎöMx0Í×HÉG3
¨eÎÅ]öìÔËTÇ,|918|"2't· ]©®L]:§O±ÆY " >/nc¦ÚÅ uÌ7ÏG¹ ÿÖß.>!9LuNöDDÔÊ>)æ.Ô"#{;¦L¸g ì0½;ØËJ+Saºçb¬ùM8äÎ3Å×\6' 4ÙKÇ+YbH¢N©Ä[
îºå£]¥¥®kæÿ×ß½6g\#ªq93ûiõ8×\ÇbÚèóJ#ôte9.>Ìl\ùüÉø->cLv')v\RÌ:,å¥Í.ÇÄH(¹½´»9©"N×bp¢JdÝp·¯dÀÿÐß&pÝ=!nÎ!ìµÃÇLÒìFÚ DäÎÉL£í®£6ÎDá[Êd½ ´ì w±¹¥> v Ò±D¹®X£Ã°tK±WHÍt諺å$½>+cfÎhÿÑßiötU!³.Äc¤Ó§V1#äN%ºDdÛ ³§Í¡ gÁGwðnÃ.9ÛM°è/\0´gYÄ3dÁLá{ÄÜuÃár üqÇ]ììNîë³áØIØI
éËɽÜ3uÞÆÌã®ø�ÿÒßh®ZaäDHmË×,y9¥¦^È«ôÕÙ7"#l]¯)wÝEäøvcLôd3Ï6ò^ûX2ØChÓf²çBÉ"©kSôÙ,)3±y-Ó¢Ú{VØSç¢öGD¹Rc¦|2[nl.Çj=;°ÿÓß"'&p7'&ú 6Óø°st¦pÉ»:Rtèòì×]õÍv'"7±³(&øµýH×3ÃgÁñ(¬l2ä> ê|Tw"鸹²|>Óù±óǯÅæî|ımÒ¦La¤~õ>Ævr-÷QÉÐéW-lGM¦Ùrc\w±9ØÿÔß" \ܽ!Ç$ØHG¯Cm&S4ºW0öl¥§i³:nk!IÇá çÊZئÆv7Ùó8dɲic\·Ñ$BûÌTÊüLÀ¯1éòQ|9¸m¦ÙÎÎÄif¥&8ð³¦ÚM;§Ôøðò%n\IÚoÕ(òon§uÿÕßr'La[é@zqyò¤9Ò$ø¬BH&tÓ©Çvv'\ìlnÔÜÃfMÝrgÍ×bQØêÎ3®°XlÎ6Ó;ɶ³="ÅãÚ:ɤ}¿S:á³:I«:ìÃÊG,çWcx\G.ñyÛ¦A{veÉE% ÿÖßÃläNùAkáýìä¹³N75Øt¤r$ÙbÚäÕ§ 2n¥¯-S4í:<Æ;S-|32^>.ä¸áÿ�® ç®Hr(K6í>·Ãç%¯ÙÚ ÚùKVº-qØòJIòe,lBY@@ÿ×ßÎiïc&4¦që¤t;©=èó{;®1É"¯.;o(&0Ý×N¬áÓ:|_Ĥ6POQ²á¯¿Tÿ�^z ÏhÈè&¸á^ñ%×o±EM;w# ]´àÿÐß.Åk§ë¾ éÑ'»¥³®Ìá½µÇw )s(ï&vvÓ¨½;æ;/GlXìe£ IY'PZølN>M"""ÅI·¤¼ÖPÌ÷A1}ì|DDDDDÓß>äÖeÍm4ÙÿÑÞ¨w¶º<µ»8<S8äN2tJYnéÝìQ%2MÍ7ÄS;YÒÍ'7ÔS.Eõ$Ø,ƹ®ý
ÉpØYü5z øºgq8591R4þJ'u""#·ÃûL-§Z5cDDGÖ4îÿÒÞv¼£'bq,K:lNe¼^O³:\×7 ×bÎM<t;F1¦áÄã(íbjh6|h(¥øsqÞÎLëwÔ.§V#ñóy;ã-± á~~¬ý©&ÖÝMÑ_ Åó5gòáÏ�ÿÐÝØ>c©½¤Ò";]©wØn;Z6)wDDD+áx£|¬)Ã~¿,T/>$DDD;B"ÔƼnÍ¢ZØÙÑd}qr;IM²)¸ï4±pW¦´#'¦1OÇÓéÉ5M3kË\WsZ&|9Å>Fòå} [½´XüäÿÑÝîÎ{'Ê<Ç ÙbpêØÉv-LëDI¸y>\®+nÀ9k2\r4ûW©¤DI¸ü»0ídúÓ¿/XÀ³:íªèMHÑÔ¦ ×Mf4ÞºQà»+ÉNÀíPLT©jJWSÞb5Ú×\ô|Î4Üq|9Ûµ°9×áf&ÿÒÝ""":²S;>\""/cø¾e4|,"ø8W|ìÓ8é; x«2.8wÛ©>Ìëi:Ðá$ÔµÅ^¦WÙª³LêS]>±wCrU à &
ÑØãIJËhÄ+};xlÓQõÃz»X:Ö,×o! °´ ÿÓÝ"""#¦M"""""ðÜ}çÊ×´AäÃ>i¶å ~M 'VpD!äFDuiÙkÃÇ×ÚäÏJý;v0v''q=j.X.°exðþ¨¦±góbë®Å"Oé©iL+ñè;W©µò¼UóDñ 'çJ/0¿1í'ÏKoH'νQÅpêóä;¸(°øz²�ÿÔÝ"""'\'&qÒ"7¶{87 ¦VtÛÝ62cL»Ðú9hÔ3T¢zñvÅÃÔÞnX;®ÍpêÔîìñuö}i¢øøúû_.ÌõG^NÇöE5ìo\,>´K;ñĵ0''øfmï¨ÈÚÚj9Rñwzö.æ¸h|U-+ÙaZõ8ÿÕÝ"""""'1Ò:kLiC§Ö;áºÁÙÔ4ÎÉvÄ3qLùãr)®|¼2c]ùIøøþÍ4Þ¼A9¾+· z5ÆF»Ñ57?¹ÅÑȯ0g$l�K2}m£,åëÓÁäNLi¶¦å¦aÑm EéÉ×HqrtÏP|V-Ï I#S¢B ÿÖÝ""''äDDDtÛL×KÐþ|Á>a$ôg´¥¯a¼}RÄDgT."K8ÕÖå2g!¼YNäãz ÁUãÔñ\2-ÓôôV¶»|Nºxüòðß³Ôû^ <óbxå6-§GÜ ZõAon93Lv;ÑN9¿¡| ök ÆDD@ÿ×Ý"rL¢G'ÁC¤DDDì¸K tÓ³N¯ ¬E|G'Ù¯$×˶ç åû2·5cêåäõ!^¦Xw<¤>y1Ø·g-Ë5ôlòÌù69DMæ©øù¢ÞÄdK¾$~ôäêxC1.òYZ]ïi}[iáëìPó^ ¯>Lrc]""Yü4Ä k±:q�ÿÐÝ"#-ãÑv:'J<ºDNÄk}aÙ=àøh¿Yz'3CÏC9-2#Où[×=¡ä·Ù"#û}ÕÉM×~YñÀíø´êm²BÞ ¢hÞÓR'è§#mnªþOr¼Ód¶´.Ý90¦Ý1 ¼êÎ^X¼Ù<@>\_�¦º«áÇkSXà¾ê.'ªjN¤""s@ÿÑÝ""""":±¥YÇv<§´N9òµ8ê>MÏ"Úmùbõ'¥"¾ÝhüEºB)±rþúoÉÙÃwDõÔìÁÜwÙNºùïf³å'%¶ÖÚ}c]ÍwÌ>M´;+
?K3MbÈ-ÁöAïðóGÃM~~£¿Yæ¿QB)¯ÚDR½RV´ù8[¦Ì8ÿÒÝv\""'b5ÏGG¢Më¥cRq{]ÄaE[IõÙIÛòsDúAÛÇùëÍòsÍŦ÷ÙM#sññvêlß©Ñnèù-õIÞ½'ãòóCYo²ÃQ9m¨¨tÖ-b¹>PòGSbu,N;ÇX²KqEß=ã¤BN®] Új9í¥µ×1Òs"#/8o&Ýý;�ÿÓÞIãìi;ã±í«Ôµ®zòÎéÍ<ÐȼuÉTÊÖeØsæuAçOù0ìÒ· .fíc×Kå;1ZÝ|
ìí6,w£±5-¸Ñü¤}b¥^I<aWó%%o5ò>'æA³lÇT}Ê%7&&iMÇÉØFlÆ ¿¹ü¸9&Ól<Ûùzê�ÿÔÝêÞj/Øùãrh|p^ §dèøWmáöoÖýÑzò;5¬R,e¸ñßdña¹ÕcKv;êSHü·z~iEnI²7RÄ$Ó£O4±XWiN§ê¼}G:aÑxªJÞ.xÊx¦Ö"ëDv¼Ê/ÎE9oó"�Åø8/ÖFÚº7Ú^=ôå¹A$¥ú·.ǵ·hrQÌàh¼9.º7SS8±ÚÿÕÝÈN<ÇAoôß2l|T??U-æ©kÆ'¯Pó|KS-»b\_ü¸e¹&ȱ7þíìDÛ í÷Ä:~¢9äÔV¿cvð<¹%¸'f¾ÅÏ4K¿bôìV (2xqÈ×cá´W Ô¦§Ö)Ç ï
^H¾%ÐÿÖÞÇOnüÓ'dé¼Ý½JNON/³MÍëÇ)ëx´$3(o/Î>×!ù&|À®R®Ú!$[RRq.á³Êtr£[ çÉ[ ¾3®Ä¥ö<ùÇ:
¼;Ïcp¡É®3¿2Ìu7Û±X"©&"ݽîow~aÝÅÒvE5ìäñÄ$uÒ""""6ÔõCÀÿ×ÝûÛ±}'MMëm¢ Üw¤ã®xz7·@=LÌBzòõGÛ±üvé>¿,ØÔ|à;»DêJÔùìtr!µ±i:KM×ì90æ '-fAÌ6Þ ÝqgðÑáÄquô2þÑu¦dVr9sÄ4û"éiH²ivåêysÈ]:µêpÿÐÝÇs$ñnÈsDײOV[=óݧÊJÒ ÛòD8í ðÛLBr"× '|Ã'Zúäyw®1ö'&¸ãDìFøkyXzè+WÞÝPû$µ®økaÿ�# µtþ:'Äs}EÖÇÉÔúý?ik?d|1¨«nãÌ#䯵u²b""rg""#{�ÿÑÝÃîtN"M#åÄób§,e¹fèÈHëÇ\>Î{:(ä׬$rÓ]wq$ý.úvÒOv4øzdÌ2ò¯ä°IqØO «8&mïgЮ¢Øb4NuIDæÌ#%·ÔIZÒÍtÉA ÿÒÞnN.²òúW²ZqJ SPø®á3¦Õ¶vÒígO4É,T«5Ù(>³¥èwB¯û>Óô¶Ê׸Áo|{j"Õ)Ô¬jKA¸í-ÖK;N'<1Mvi|Cñ¥Ju!KIÓ¡ÒGîR9Ï""�ÿÔß¼Y/ X7^¦úe¥Õrg\Gs1Òv#KÒOe§M'b0§ÉöiÓ> vu;ÙúäÎ"'¬4×b¼õDVSoä-¨üz(g¤¸udþÚm¬d%¤uòìÞldL¦LiÜtÙK5ì\|ù8]ãNDDDDHïyD@ÿÕÞ®+N¬+å\_zΡÃ&0ÛW #³Á¤äƺDìNθDGÇ\Ý(>iõÑ\)G[c».ýɬÊzXý×bøH¸s©¨{ø0çñ&r<93®Ö|9¥ f¨úÊn¢!ØvÉC>ÆDë¤DDDDäÎ9Ø ÿÖÞZøXñK¼µØ0îÇv®#w""uØsp¦u'}9©~V¥u2×CfÎMÑ¥¬¥¬v71ØK0íiØ]8º¦6moÉí:ó&{§ÁÓlÎB3 ¼áD¤añàì>#\r%ÞØÇg9r+©áÒ""""Ehr!38ì�ÿ×ÞFù¼¥�oàÉ2aÍÔv®Ç&Z¸i;§uÇ"pì0°±:+2µgiìù:˱ì«)bÝ]9Ai\:g ääCÛ±¾ÝvK4ñ=as]N17dKèç&p¸ÎÃ-êIðÖv#Rõw #"""'&Øyb $Ä}µ(ÿÐÞ¾BÌ쮸ç´å® Çeþ ¡Ü±ØINÍCf6 ¼i&"¦ÅOùu±±Á§Ú#ÚݳÈf0¡9Ù¦Ì#ëëµòÒÎ8a\+[¨[ÛPØØ:#.{KmçL¬iôðÙ5$xveȶ3ÙÚÏT&ÍÑÚè»àÚÛ´h;ë¤DNLávtÍr[\IÇd« �ÿÑß.ZÄÆúÎ;:Á÷ãîü}M.¸ku<»xÔ]%ØãS'×\µ"÷3ËöH÷\>³Ï4¢×Mi6+ìùw]rl|àøy·±³Á b1ÎÈîk-DS\>±Ù>xºY¬X=ë3ÅÞúV,}èäSRqÖ"|µ¢ÛüTÏ£&ýÞ£É$¤ úÏ0çÉ´}C'gê%ãøig[á¦=;? 2~dOÃEE#öèµÞÎû-æÚÓ¨ml2³ÚVsá íæMw³½±/}>-ºdø1Û[L3¶øðñäy¶òåkÌDZdñRÇbtÛ´éVtMgN½àÿÓÝ'{,¾·z÷ËÃÑ ÚøOkÉ5/ùòv.ÑMö¢a$Wª«cAýRu&jX:<¤Çu2|£é©¸é¢b"$\C¤DDFDáMÐ̺|%EÉJH&u» ÿÕÝ#£Á{wì+çß.cBÛáóÌøfÀù)ïáN;ÄÔæÁvîż9+ÖLv#RVL ø~ó#ÜÄ¿cé CsOì~Qtù'ü5ÖFË|9(ØìÙÈ2|OdÌÓÄùí¼¿c½½ÍÃ0Z|òE|B;ײ<rg\"u7vÏà¢uÍ×!Øöï³m 4½¢ÓXðÛ[¨{sº±Ò»""1Ò"ôeË®ZB¤ÍÔµôÿÓÝv#SGaôé ¢¨íÁCÉðÿ�dgÖV$mLÙÞÎLê.èm|vo¦W$[]ÒÝr¶ßZp<ÛJ´ò"¾´êé86m!oÉõ!9iÔSá;ÈG[uûØÙsqØs³±áÙIT°ä¢fqÊ+èàü·'£´åÙZgcfMw ·¦ÉaZÝQüÐÿÔÝÇb4ø«,z3ºMÒ#lk±8tƽºßÏ. RÓmÊoù[¥ÓbÆ^Jm®lJhàÄS(æ¬ÅRr_a_)î9íÈ©NGñÆÉ÷²/nÖØ#ìFɳlG3§VÇZØýE°¦æÄëï;JÒò[õ$ù}CÌkµ£øçRdqáÙQ:äÉÑ:ç%ú
I2Om:hÿÕÝĵKÑ4à_mKÅi®M±Ø6ÞEñ-o¥ÉÇÜ4C7GXû7m'Rõ(<Íw4¤ÏI"SÖúeåd|ÄÊkFâq¥hÓ=¹Õ%8ÒÚÌëìN'ë¦Z5°íãrd|¤; ããèíç=g}Dä¶ ×ðzNf÷Ò².ØtN9-ÚQkýDÜíìÞk ÿÖÞIÇ"0_ r;X=¼»N=Ç')³µ³¶°kê"¿ìÞ~>ÞU×6©¢vg :-
Ú%Ø´K±§z7RvEÚ5$Te¸;BºÑéÛ®[Á±:mi
o ø¤þ^NÀf0³K ñØëWuÇbÞe8¦ÃÃGJqà¨)CMànóåMÑÇ ¯ G´ÑêiwcRSÉÜÊ9¶9N?)¯on�ÿ×Þ^Û'LxYV|9é ×°®m^Ý]rdØAkhßäíBóW¬£¬hÃw×6Øð|ÉkS|)6('TN· 3¬ÁL#êfZQYÊÚ×3J U+Ýß'ýf-Å0mg È·c-=RqùJñ�ûÊXÉ#·´Ñ8qk:S8ÚMöì%¬ôòÜU(®¤tÿÐÞj¤¬v>#A\BBEiÄ6V¶ØÛJ;IC¸§¤§ç¦ùZH,v#MDø6$]ôd×Oc£À.>ËðDe©ößöHçRÓ >ùá>´ëáþÈ/É6¤Ã±Q|\]ÄÇÃÄ%æºlÒ%Ø2x¢OÄ\Ó1�ÚfN$g\_igɸï$ïHæ{"䦴a°t;É>C³àúFËɾK@ÿÑÞ\N7øí:²±TEo!mt6E6ï¥b µDýÌcéÑlιäÆê~ëäÎr'RRT§H;\'9¥®Í"Mù0ØMa³¼VÓý#C¢ÚÇ\tÙ+mrN9\_'Òºï/f1mg@Ó:ækQåÿÒÝÝKZ§¯±Ë#$L:äÎ ½ôøGdSuj©s ?|ËÇ:gÓ&5ÍrgNÍ,3Ññ]Xü=:ûe¼DãlG6v,s±8Sg&Mݵ°¢îO)äAkéôwdÄÌÑÞ§&L):yIÇjääÁbØ6éíç±:< ](¬XÉú>T ÿÓÝÄ´6nn¦á§ÌÅc«®Z4Á¨{VSU$ØÉ6²°ãúp¨Y(uÉG®ÄêBû'&q|xxðA9ctÛ#"âRDØÞÚo£$G%ô4ruGn]3¾»áÒ"/NL8¾3®DjjÆDeÃ÷ÔS(åWDZ2ZØM{õ ìNZkÄz7æÈÿÔÝÄÕ¨HgÁÇ¡#µÊH{N¹}jɶ7;àßå«ç§Y Ç&uÍI5ðnM3±ÞÝ¥%oÇ;XÞÎÌäv±Ú v.c&q ¬ò¶B+Ê'q5j;Mã&O'$5¦nKéÎÕ&áÙ¥2çcm:|3¦ÄÛª>Ýò#ÞÝÔãKÓÚûá�ÿÕÞ¬údðÍPÏÃYÃÄP\NºÐñ¡¦qéÑ'CÈdݸìN,8þ/|H^?\sÆæB<Ãacðü]½EN'&qÎéÎÆDê9ß=ø)º9¶&f§SBÔõøwzÅ=¤àäí"ÌuÉðlΦ¡B]>#OEk®Å´?ÑQmq£gë0»)-(þ^¤¯$µ¢íy±@ÿÖÞC{~nXuÑX鳯PGß!Û\a³áv#W<>½"þÀßtêSäÙG©¸x§mZô8ÄVG®[Ê<6g Go¤Q|ñ)&7¦Óðnðï¡äçúiìU·UoÚDë¤äùCX½POu6c²N>N.êÞn.ðTÞ~^Þ{6²QN6k§Äà�ÿÐÝÄü\3Ȳ8ã·®¡¥z#-Ú0ÎÓ bÅ?I éØ=¸¤$Õ áäÅ¥-Øð|ôЩðñ"+X¯£ÞÎMZÈ;¹ìÙ;3®EtExóÙó9éØ¿DVåìÑثǧM$Ù©¹ÅpîøQdµ¾ û¥6d"G{õ ìø2ÖÓ"èµØI0ucSkE}ååAöî�Lé9» ÿÑÝÚÆ®xkÖÞè»ã¯W9%³-y2¼³TÏ©5 ¸»µzÉ®ãNÇÝ=ÒÔáÙMZõ8ÔS(&Øn]×4ÙsRÓl>Q~Þ˾;s~^9íoÕ7h².KD¡j[ ·}Ãx<èø©äFÚÝCZ5Ë-ɦYMtv¼Ö0ä<Ñ°|ODæ;îcÿÒß./$å¶GuÉ¢j ÄwúH»uAkj9E"8ùsØòBM0¾¡¤X}̧áæÏ2']'&pɲ;XÎÈy%guÎÎë±\öçç¤3·MNö\ô'LÒÇ ÍsQÍLN;6û'"FpYfDHmMAkÚì^Ë^d¢4ÿ�0ç´ô®«³8d¨oà!¦�ÿÓß®Á¥ØªH'Ãm)²)(GØɲjst;çEý§ÊÈNNsqt¢ç±a°¸6ó -¼ßfó²i®i�ÿÔßyGìÙ2y1Û\cdv6 g¢¤aºum0>1g/Ç4 >QÍRñpÍTá¦"Ez Gkb=rhoDµðäÌ»ò15)voÓÙALkå$ ¦BÉ; \_{ ñ"5Ò"E~>ÀoY? X¾û6i0#R»¹·Ä·$ØÏdDäÎ:Dm§�ÿÕßSt=¯¦9&éºå§n¸}á¤Iøë+Í×Ò[|T\_ÁVtª.##-c¢t¦4?%z´¡ÙÇ8ßM=8¢Ìá¾!ÔWD#QÌä2é³E;Ù¶ 'J/c¸¼ §Eɱ)8ùgê"#m^ÎVB|Nê:Fu2ku
4¸NË{)GvÄÚL ÿÖÞÛòuGóÍìïcf4ÎXÍÒm6c:+§ÚÙá àøyäö:>!5kÔãl×C7vܺIÍÃ×Ðx¢I?¨ªÙi>QO!¶£Ó«|:ÖÔRtÎ9&¹6sæ"× Î(MÞ;;ÇÇÈÚÞ-è$w¯cc$F¨ç´àg&pض^ÉGÇkW-&Ôÿ×Þi¶¢CdÙ-I[ëdDuk§tKQÚØØ.)äZÖ:Áv;Ñ&Ï"®Õü¹KtNzö òñªí\g Q^Ó(¦qÉÑröw·&4£Ë³8ÖÄ^Èøªk\_ªnì3¦¥2ÖÄ\ü!U[ÍVÜß59kaÐé+¾3®ÖrgÁG±åÞ¿h|NF"´aV5.�ÿÐÝûIlÎu|93®J;nM3¶¥¯ Ü\_g±±^¦^È»L6ngd7«¿7²9×+²°ãH6O&r\_ª/.L Ocz»
6Ç>f$Ä2äß±"¾^ñÛ¯¥¥ ý Î9ìø^òOÃ\_¤ý+ÁÑl<ÿ�ǾUV®tÎ¥V-ðí¹> ÎÇDaÀÿÑÞ £""""ö¡³8lxvµz]q׬$2èÔs(æqÈE?E~~;÷&©ÁÊó;->M""""""IÇòY®«è7'o x£!Sf»nnþ)ê§u ²È~S/ìýèlOtø· |ÜÿÒÝëN¹®8tF¥ëì6×zjË &q䢴á½Ì³n i£8y:¬{lKÊ86úFýÞÒd¸DDDDDNúsH讬~&p¸öEV2×öa'à·YYwı:ìÆFÎ)9oÞQHÞÌá3ì^ÜÇ3RQßNZº]@ÿÓÝ:Ñ»©+Å©g!Ñx)>´O{ÅÁi̵°Ù&M2lgMr#êe¦! ì"x2IO' ùºØ±ÒäɤDDNDë¤DNË}#Vü1ãMÉÉ\ÎN)}}-ÜOµ)É^ñ;W uÜró¦VÚtÆÔÚø>c]HÿÔÝÇSR½½úÏuxÌýͤÏiúWá#)ðÅW)èòrcQZ0ÍGkb>æÌé³8æiõ©·äA0!%(:ü} ¡?&´õc äªb4Û:áÕ]qOn>Ò¼qEoÄòwkÍ1ìùíåELî##Øì¨']6"bBüWLjªc]:ÖÓpɯÙéÒsêºDÿÕÝÙ/gvê3óqÑHr)? 7±v(9íÕ ¹{¢bÓf]éñTäú,ìY4 ã<Oí¼Áýº¡ùíü¢ídDìθDáéó%ëÒL""'cÉY'j¹¥vV½Uÿ�Í> é%ãä%åãØØ B¾¬ùZÄýWHÛW·
/'IG$tú%Ý"$¯a\_YÀÿÖÝé¢0êÓºÖ?NZè{áG,Áõæìm¹qÈæINhK)äÉ~Áý!½°É±VébÄ[ßQå¤ã»jþNI´ðÒ-§îi{pñr|¿(ínÓ®¤éôäFçøñ é"$µ¿ÔG1®x¥f'ÑÄ kõ@MLÓ¯¥jéH°gÃSDüêÿ×Þ:¿E4ò¡µÍ7âÎMõÊËg{ã¿;SDë\ s«Wc=(o=<@B$±iÜgóF¥Ö�ïYS\vûÏôa 䬹o¦´dNn©";£!¿ ɳ$o�¾DÔËG@ÓäÌtSé ¼|«ÀÿÐÞi6èðæ2dI¿TèF®Þz£ìiÑyÿ�Z7Û±\_èé¬SÚ/äóÌ¡'s]Ò¯Ø ×ÏïÍ ËÖ"LÄkD©fº c"",Ò"""">µÑd/',ÞQ{"r ¡Ò;ìFc®ÄÇ¢|Hïh9i©";@?¯
qq¿§iþ/·:ÿÑÞk>Ms¶EJi®MMaÜsÕ-ëóô4êÚjäÌËcMLq̧¯94Ç"ÓMòi¾Þúìkºá;1
©iÅ2/±Oh~"ÝýzDá{tv$ø®¹;/æú\ô+ôUäÎ:Dµá¬µ»N£ÀÿÒÞ8:i¦uÇ:ÊjñvöîANAl±&D:éÜQ&^ ·§>LÒm±½¦ o%I^C¦È{$¥õ,òj ¹ºá2ì[J|¹è"""ÃÃÝ-¸£Ø+rÌÚRÇ.¥ù}ÐÛÍ& '&4£ÁLiDbJ[¬»ÉïQÔ·Y[ðòwM1¦Þ½ã¯O¹ZéVº7¤D� ÿÓÞ¨h+ÍÖO¯$_ظÕ_ËÎ,Sy;ÉÑwkÉ)«Ë5O^^þ$5¼%ãç4\Óo(§eÎòç¿nÅÏöìF:[¤íg«·F
®uÒÁÎc°DDE¢xø¦DÛäø÷åS+Äo$b8:·¯$Z ·T6ÔqÄlÒ"IJ ø{Òöë8þ"'Ø2ÜKNºøé\_¨Ý¤9EÿÔݸ·g\Ãfè¦tpD1ãBÉkó} ófØì[M¯Õ<ÇHÝvg Û¤bdêDDDDukØtpÇl?Zú1ÚO¥púý#¾NjõA6ïÌéõ妧>{w¼\ÉãÄSÇ0çYv)2V¥aXû4s\ÁägÄÎB¦dè¿Ùhû"ëµÑVú0g$Ïj8pÿÕÝG{IMµ"""KYÑS\¨)5Ü#£e\;\Ø®H=Ò'&5Ì9}:Þ¬éÈöN¬aÄäDò">´,>¯´\_}ñ«Å6OGÏÀx¯ËKÚPrN¢"˶{w´IÛ9«|ê(æ6ß|saÜîc¥íkðüqH;½¡[~).Mã°ááÔ·´LøO ¤jL�ÿÖÝB""""$c'jòäFÉLêZÆ#µ´Ò¢äƽXI1®äÛ2/®Dù.»'VuØ:#S×>¨(©§À?ögM~tP5ùæÎny>y» ×Êä"$tW\ gJ#SÌj~+}IMn8>V,©9í0¯½µipêN-8§,ÏP6.ù»ÿ×ÝBöDE¸DDIi\ÇO�í&DØ:o¦{KX:g"s ´³:$rg µÍà ,W èqîDaÑ6QQëæDÃî_ßm6sÛI>^F£2nd;ÑÉIXCèØIkaÑnÉ©d¼|iZTÞJw
W/p
׬!½ô=ãL¯ìæ�ÿÐÝB""""'bp_õEî¸DNÄiÕJDã¦t?½ ÒíÇbº#sÔõáÑ\å>D{:9bçÛ( À:³KtI³& q$Û ÊÇÙ¹®ØÒwÙI$¤èÜ_ÃóLÎ,óÔ ãPòqÝ4ÿÑÝB""""":ë.ÞÏ»®ñM;¡ô¦¥ùhÊLÌòq93ÍðÉÅÔ«¥R×l ?UËÖ8Püù>lý½¦~ªjsbh|¢'³\"$q:Qhº2:aÉsÚMëö®Å~ÌÊùLm
ü$ùYòߤ#ã¼sGÜSPVHÕGÑÛG@ÿÒÝB"""""$óìÛFF¥Ç&ºnFÚîlrLmrf~Ìø¦pÎæͦM"v'MáÎúík"4EkZ{K?êÛ¦Á¯;õM¬aäukØ]§å@9p½±.Ìa¤äÆ "'bpø¯&uÇJg\Ovû1ß \_"þ"à¸ï5y#0/P¤NK®Lj�ÿÓÝB'pw1"":ϵ4ÿ�×Óv\""v#R^q±ïYvkV4Óë]OZö>Ó~MüäW×:})7ýûDGÒ5-ãPLcjiaÉϨäN8E§Ü©Y§Ádz9¯= à¼Dùñ<}ãªc"$S¦�ÿÔÝB""""";¬ÊZW¨¾xÅné'(ú95Ãb4êƦEyÓkD±]+\_§ ó:ucO¬éÇü¹M\_ª ¾Ñ\_8¼@I¾¹N;¦#´ EÍãDD} ëWNÃ{ÝaðVó(~#Øi;Òßáù¾g<ß ïÅÕì91¨ºÈÿÕÝB""""":-M·Ð.9äC~ª ":çSüTٴɺ³¥ ¬$×&b\é +Cæ"[ÌëÒÒ×EÓ¼b+|LG¨ nY±±yû4<ÒÒ0ò"!Ü<ݼI÷§@71àÍ)òá;¡Ô6¶¹õY|}¢øvs±:'Füù3ÍÖ|¹ÍJÛW " ÿÖÝB""""""ÁõåérZ/XULl»Ðú× ml2¶ÃHù¯W#ÃìWDV½Zùá 7¤¤§y´q9ÓínÞÑÍðè(G?IÜÈvµÃÈ_7hÉGn>¨$^0";Ô±àËÛê##ëåsÃ_]VrXÉv#|%|Ê>iòç,Ñ?$EãÀ/DDÿ×ÝB""""""I7Ñ\?O+qM'Ê9¾6ÓHêMÚÇÙ»´ÔÖ´R2=ó©")~\WÜD\ܱLéñrÍÍ(sRv<¤ä¬¯ôàë¦1ÉC1bHÎÖ¸lÜ$¼t,9Ö"�Z4ç"|5ºÉþÐ z¢v±®ÍdÿÐÝB""""""'kn¾üÿ�!gÁj §Ã÷:~§³/lô)6:f)ö¨ûösuÏ&^Y·vÑ{/dÃN/Ïi~";GV0Þ¨æ0ÒMX¼¡'f0¢5oÃX Ë~ÒÈv#ðÆÉÜÈqߤ¦e~$5¶ÜÿÑÝB"""""""#nÆÕ¶ §4×Ñ[mí5f+ôÌòî:FÞ2Di;õ]'DDyrjç]IéÎÅYÓTøãE~~QtHI8ïÒFrNÆ rdúÑ�ÿÕÝXM¡ÇØpo$[ÄåDï§mm4ßDìF¤q5+«>:øIê£æ'oÕ;ÔèÚÆDDD³øHü}rS¨h¬ÊZS|^¿>zÍË7AéZÚѨý'7"ÄL#8!]ií(yøN(fó[´éÉtw4ÿÖÝDOCtDDDDDDDGRq/%|jw³D±"4wëÔýPóGi w£ûAÑ.øn^"AÍNÁkH5¹>\ïdbv# 2òT;G4Á#þ'Ò¨ËXÕ
§åDw\"%ÃúýÀ$b=�ÿ×ÝC«l?¹Û®Ï£9:Ì8DDDDìÆ9¤'ëß ù~åy>d¢ kJêTÿ�]ÄFødxU¹ÑÌÒ'Õ4%i©-XGÖW4·h=ËF/"PrùWZ®Ê;øEíãMKE´á\ñ3áPäÆkùãb$/²"#«=|øòÿÐÝn²äÆ8I«Í5vyû{6/¥Ü¦;£ãDDIb»Ã±Øð&ü6æ:äÆ£Pè¡ÇDjÛåÀ6á'é¡Ò½¢#£ÏGÃ¥GEfJ+©¸¸ö|åÊ£&"EpOQdDñØéf0ÛF ôe§¢M"ÙoðõÀ~À$¶øCÕÇ;(¹£°l�ÿÑÝnòiz:-Z»éÎlÆ8ìWRE ¤FMñi©~©¦zùCóC»örov·eØJm«ØÚÓ©ÓUo7"ÃbuÜKN9»¤Õ%-
0'É··iEckQ|uKïDe©¡'=£J;Çklv²mQuYëUM¦LáJjý2Öst"$¼>%7,uÿÒÝB""#cÁÄt[qÄ[Ù[Q|¬7¼Ay6K´»ÃgÉMíÈõ.²VóõÜ&¡äDcêªmP×$ÁÉ2x£È^NEZã®O)"Ñ)émï1áæ%jä:DäÆ1\NLi:i[Ö:¡ölñÿF<\¤ÓEýãáFM×HrñvVóqÍÒ]aqÓ¼ç�ôPÚtÒwÙñl?t¦5É8¢pGÏ�ÿÔÝB"""""";´cOq¸DGÒÓñÌa$èÿ�ãî#&|9ÅÃ'Ò&mçТ"W|d¾g ÒIN9²Ûå¦7©u.ºá¦gÁ±ðó )âb±Gg!$Ocm¼8;eéöë4CÇ5ËIðYä϶:±¦ßʦ" 'Å4ËF;XÔ¹®¹3
¦÷�ÿÕÝGaÌ""#}»®Zò}Ó]2Ôû$dGÓØzøÐucVôúºyYÒzHtÇTÿ�ÉÞCN>9¸ÊpES÷=siñy×IÉVòüOgmõMí+¼6mlrÄuÌCxr³5öu¹§S.tÖ,ìFçàù·ÖLj;DAÄizM1®VÚm§Á°Ñ®¹Ò�ÿÖÝB>·Ç:ÌNc®c¤DDDæ;[k³hfa'´÷+ác´ùñ<»á Ú±§GHØòìÆ«¯ði(ø<÷(çgÀ§D>üz£±SÇfz¿Å}8æ~"ù ÛNEñ×DñÛ%a¹¢a9éÏj_3çê/2ù§ü©QhJyæ²b.¿¥_'ñÙ+W Ì;ÛOq'ô+Ë[ì�ÿ×ÝB$0ÉpVÿ�fÙ''ÃÌé#x2#$ëlWV Å:<ø~!ðUÇ9>J# ä©ÚÄë½\ê|F$tÜÃfÑUwÄs÷ñÃnÈð¢0Mg V<<ü%´j£øæHȬZ5Sf1N ØÉÃ}QÇ$§çíí5øãÚ³á¦Li¸-m®g{ÁºìÛÇiàÿÐÝXpè®DDDDDDDDum4ô³æù±å ¾Ô7¦òFKåc!¸"Ò# ½°øðé³%S.SRºêîÎ ¿>z ÉÏ ¦Ó&ÕðYÚM$:ɸ¯ÒVq:©øtñ-Ç&Õ3OÁi¬ã[hYä#tñÉiñ1ËMæøo«�&0É®¹1gcK{>+÷�ÿÑÝióéÚÒlìß<#""""""""$ü\¿,WÎ{XÓ0©MÇÃX:sÛ¹ù%ÊyI
3M¢O¾Ó<�ÿÒÝGw""""""""#¢º)rÃáù÷ò"8O¢'&qÒH|FÖ×/æ9¦éñä?%pÚ×°CÌitzáÑñÞ²t
¿YXÍ;½1;ÔðoPö®+á}Ïwx¸óÊ ùG5Ã0½;Nï}µuU~>Á§©³SÚ{;ïX�ÿÓÝG5Ò"'s]é¤DDDDI)\)¾ÅÃqè¯YDDDDæãÚ .L¿· :'\½5اW;;
ìÎõÇ1ØrõÔ1=VüíòS´+çÀ¿à=«EuÔ<רïãú¦sWd©µöåöGzbN>M¬[ߢ|JN^-
dÎ{GLiq#ÿÔÝIðDDDumÔ¦Ó&°²$´ãÕß:CID¤DDNL?gæ¹yÃ"ºe¼´ë³lF\b+f]:+¥u Ø®S¾Z ÕSFp[ ô66#ë®ü£c
Ë|>§¥z42ÔÔYªH¯h2õì üsDX¬IÅÏ4^1?Ã} ¬ÃNàÿÕÝB""/DDDDDNæC±]KÇÃÝgÅÅYÔ3±ëég"4ó-Y9gPõ,Ó¢ºÓM3¦;8îá{3z¦ý©ÆÞºäq¥µ£øäð$ ùf$î>Mq:é.Kb\óÇÕM[ÅÎã!ËqÄ«ù~cÐI¬ÃRf'îí¹8"ÞÜT1à¦�ÿÖÝ[FKÅt¦1ÏDäƺNLk¤DD}!ɶ}k£x¸wØÚÒ\ï´>\¥»$""'$Ø=É´@krÊ;^\æ®E54qbNbaØmH"ëí]\×N´LÌ-;Z;¢Í%qKv¼ô æý#RÒýÎ <!¦b çuô.>\×,\_Z"jÏÕrx«t\_YKåERÖe3^ÊÞn¡¢1Í�ÿ×Ý{ù0½Pé;N9ôøy¼öã«Sh-nÐw¤DäÎ:mAÉÊm'Dö:Ð}k©«Gbb ·5kõJ¼±gohÀÓx"úwm>%¯>r9¼Ø¢ÚlË°íl~hßõo§DV·a\á9¶Ëmxà / äËY1lOÅÍ5XG$}NºîáÄäÆ«¹xð?PÿÐÝkFDNf:DDDD]
Ó#Úh>ÄéK±äø´T,ÏdÔõ8±Æ[;rÀò~ã¢oShã ËñQw¾´KSls¼ÉvE6Çbb]È2ck\KÛ|//νõ ý!<ÑvÙ Ë@Ifæ0þ9ÉËM È=Liµ®¬¢»í92RÇ-eÜú°dÄ[Ó¿DqøO¦Ï$T»ÇÙÿÒÞ{|6yzê.²æa±¼_f)MêΤês¹§Gy-&ãðåï®ñÈøtÇH?=°¼§ã®¥#ºõ2üßd×r-®MuØðîé±áÓt'&qÊÆlÎ#Çkôãôäqh® |µ´qL¤ÆñXxG7¥=¥»¡æ¬À]ÔS]¥Ø´Yh~vø{è~yÉAó$ÿÓÞÎ.nñ õêa9Í4Þõé""äÎÙ7SãÁC½ñ%-©«V·@ónÆöM_Hf «tÐN_¤8äU<$j>[ÅUæìîÉ#ÉÅ$NvÇJè7Itx6lÚn¤ö3þh~Jh\Ãf wÃ9ï_�-^¡íàÌIJã·B7áüñÐ=dÿÔÞÛ©±Ë³í©¾aJM4É®;Ó¢¸{#Ü6m¦éõHÏgÌàkNB"vÚã9.hGku<¿2| |¸WòòvRîiYINf¹> $µÓ{aäNÄj;Î"ZÌ;§Ä|¿f$�í}C1O&1ÚÔTFBzO&±ÈÖÌaDktúδEàÿÕݸ2éxùz¦Ø»±ÇÕ!lÆõ7 Ñó8äÉc½tG0é6Ó:#DDxErãð>"rMÑ 71Ø:ðlyK¾Ï§,BNM2<Hò°H»öjþk|É .\MpØ?ÙßP¢RAæɤeEJl]qÙ'cNª_gÒ4N¼y ÉÖ ËÛÀÉòÔÓ>TMJoA7nT´3»DñSSóGXWF>0 ϾøAO£äÙ¾Dug :®ÏììIÓ.DÉ2áC#RëIè"'BÞM(9xþÉSF|=úûìß Ðܲ4êÆf%©üs'ÁÑæ®DbÿÒÞÃllPì×\%z»0ï¢úÑ)¢Ñsº;4r8ÝéäTé2c î8Ï&|Oº¶ã¯XHDNO\_ÄÕ¾ "'a%£LqØÒÓÐdÁ®ËI«ìN¸ÚjN&¬¡¾J8I ñ¡í3IYðrwè|2m¨&5¶6,ìN¥¬\O[vLD@ÿÓÞ¨;4ÞJ¡»~;ZjYïËöH¦uÇbqaÁx±ryA<G)gkd-íÊõãXíàLðücÄÆÈ.IFDDî¸tNDNëì;iõ³°ïT ¯i;§ÇbuÚÎ$ÅÍz$á¿Y["ä:ä!²)&]°ìÚ]¥aY×Lµ®ö�ÿÔÞ¬vÓÕéýLô#.SËñ�/}dDýdE®Ó¨õÙv³¢"tÊqÍ¥4òÏñ "b}¬GÔ8sN«®æ8DD|WMÇp%gO¬iÉúäjNJg MH>ü}¡_5êµø{öÚ4=£Z êäÓNõ:©fºíkëÌà|Ôý@ÿÕÞ¹/=¯dJ&¢p§ËDB«å×LaA\uÑQ^ ùg´ ÌÃÇÝchÐö Jb""IÆå """""KJ§õ¤C³:aD§ô&Ð|¼}øyò°/hm-MGÐø¡½tê×uv#ªX¶>\s°ÿÖÞ¹ðui®òDD B2k54ëDGD73RÅáÛDIV30¹97(¡G§TÄDDgxrÖË dÒ"$rÆ5ÃHÑ4§S5 )©Ø®;1¡/^®{EÉMjõ4aðóÞ1%:¡öì05jd¦uÚéþaÞÉ«r}}µãbRÕ7à¿a¹#Qz3ÙݺåìÞº_(¶»Jä¹É%ÅæD9ÿÑÞçï[xDDDìFpG ¡øéüsTxÕÐÖãÄ~#G|=~YüÞQ×JüæÚ³æ9ßiùr±ÜI[Ĺ1¾
q× $× |9oH¿Åw5±Ìı0ÔíéÓRzMI[k³o]3b¨ûo'q×�ÿÒÝ""#}»æ;"³O5b>±Ç.[òiõ¾<vrb,åÌJÌ˾P;èW to\ îEý×QVMᶼÊ,Ó£iá5¢CÄiòå|;6pè¶ ã°Aj£¤¥SDc¤@ÿÓÝ""s\¬a)æ ÓNêU®%ÄÔã¯TïA§´øK^¾FñBiEi¡N¢kxo8³®Û!bï2l}2ígJÞ Öñ©0Ó:ÿÔÝ"""$¼^ÅMõêtWDWDÖ½N Ïyú¸íÖ|¬«;90å§ìy8Q:èkðvôë&åø~ñ,GW;jrËè4ø:#]Ã-ìL@î/GZÁ ¨ÚøI( îºå¥Ù)Mê´Uj55d"""�ÿÕÝ"""#é9+:Ïg-Mv¼I«LHHmÉíi²8®,¸ã!!ï¢"2Ðä\_ìîÆû ÉÃä$¢¹c µ{9× ê}¼qàè¤ÉÓÆdV ³¿£¦9n,r¼H¹gÄiݱÏikMMG>MÉéæ:DDDæÿÖÝ"'b5"ÕÃÎÇL¿$P©~ÇLh>^O¨l3;{DDDDDDNDë¤FÍtÞí ºì¿iz¹iêmo¦sÒ)¶,§u
)=#ï2dÎ$áöuõÚNnTÉ62c]Á1ØLWÚåh«±qæÀÿ×Ý#¥Ì:°u É®hKÅÇ«:S£³ÉtOe.èô÷ꮼ鴯e+Ö'&5Ò""""'=:DFÚÚ³ÙìEgL´"ù}O�KñL"ÄÕúHúÌBËáá/5ºÀ:Ó¹Ó^°É8íï=QM¾L¿ÑG©jå£EqÆ &5ΧÿÐÝùbw\sIJöOGÃD!äáê'¤ñë ©´Áô祾o±5¯P¼óä-ðTy3uÒ"rcQOX/eMu é|¨i>³Á}gc°# ¼j:ñõ Ë}8}%'%»!KB䦹]|KiÜÉMN4Õ¯W|ÉÄDDã¤DDÿÑÞI1
Ä:qaYÓ×AóÒe]4r E&j±ïÖ ëó#Ê&ÄÄDDDDDDFL½RûqÒ"$ÔZìy8rp´c±)mIEuV´Í"½NMÄñתB""'"uÌ:mDë¤DÿÒÞ¯0ïCã\'#
-xfÖ¸qü=Ú"SrÛäí\¼d¬aÄDDDDDDFXÜJDä¹q½K«ìAÎO;äNºäÎä¬IãÏb0ØäSVýJbl¼gÏÇ"tÙãÓúºDDÿÓßÃzÁÄDDDDCñÖk"~Ws>LÐDSO""""'5Ò"#&tètÈ)(tvÏGêÊ7$E!©]ùi~çmv ""rLÃ0¦p¢t»:D@ÿÔÞú|^ǺìNºDl×Nr]"""7±DéÄlC±éz7³:ìFM°¢Úvc¤DDDDDäƦ¦Íp«]0æè´³\êqhÒ¯¼"©;�ÿÕÝÈK̬¢Ó}gN'&pØÒÃÈÉG>J'ª_däÎ96r''bÃw\6|#HÛN""""""'&4¢1×&pßgnäø(uÌÀöKÇÿÖÞ)ãê« ØIXÃÄQmwuÏ'Ò0ò"""""G2lɹ§ÓèûJdθ^Ѥ¤$"3z£µÚ{0"']"""""""""""""'4¦4¢1Òs#mn¦|Ø�ÿ×ÞxëÐE òrQt:µHDDDDDdØòî%Æ»ë
+ùC2åk2+SàȺO|N¡·.^H±ñ:é¸qáÊDlÛ»ØÞèg]wöoÉ�ÿÐßSÅu+±.ë¤DDDFÍts©c8ÜrdÉkO³?ÑÑm«¨ílxIù)®D'¢R"""""rg BæÉ É1!9ºé'+²åþ]ĺÿÑßZ¥©ÄäN¸ïg6#ªq8écfº|FÆ£´aÑ8ìùCo.Ö¸í§Òùx}k{GhÇkäCºìVò'çg·c%ØævodÚ6XÞV¸ZÿÒßxw¶2lÉv'}>\Ô)"pò""""""s]""""#z¹568ñÙð|VÚtÎ:FOöy"¼¡òì×PMqËÁ "qÈ(´¹©©uG1�ÿÓßY±VuÛImtíӣû¤lÁÑàÙðìy>']"""""""""""""#&°ïWsr']#f \DDDDDHm:äÆb2ÐfiÕ+Õ;ì(#ìwNÅò ÿÔß)IöRty9ÚDXtøsÇHÈuÌ"Ò"RG&5È®DDDDeïÑÊòÅ ¸wm)òlNìÌ> ¦5Év#Ô±WzñèÿÕß¼6Õ oÄáûiÞÈ´ê¦n¤©äDDDDDDDDDDDDDDSàÞ§l×J!'¢"#{»®.ZöXÃêÇAÎäAÄä; ò"�ÿÖß-z¹OÝuØlæXÓå£OÃÈѱ:Å aȱÄEc¤D^h¢'có8êFã¾ÓD:D@ÿ×ßIbÇLj3°Ýt÷N«³lF8é³îºDDDDNLk]×2 & µÄ¦©ÄDDlΨÉ2Sn®î»1f5ÉzƤ«:îcITD@ÿÐßãÁÐXá¹DBw$è(tHìlÛ ¹v§Ôµ,IÄDNf£º Ã/m6Á¶B-){91Ø¢p½Û±³îÃøù;&7³±:èÿÑÞí(«cý¹viÕ¡Þ鶲-IºáÝ©ý]""A{Ìz3æK½µ ¼!Ãéå%®lJL2\í¥²uàرݴËC±©+ôSÍìtIÝÿÒÞ½v<§Å µpÚÆwN) Ì×H'7Ëv9êÚ'ÑäDDNLã¤DDDDæiÚ :<;0nõ:²tØètɶºÅÝËN£ÔµG2}dFiE1ÔMLµ0ØòJ8ôî"ºQBëìß�ÿÓÞø¢jVT÷&4°Ì:±¼T-¾kÎ$ønIXÏ.[Å×H6dò""#f]ÒÃ;9í~¦Ç,#·&§ã¿h.ºSl:©ÉC{£Ô|'\´Eu'c}¦¬GÔL\´ôض£Ì³DÍ ücx¾"kí7Ö Üów"@ÿÔßmÔ±É9 ùG6Ã~ĵêtÎáXé³8ì¤s}NsKuÒ$dV²?FO.!½»"´+Y/ °tÉ_)HÄrî©CR¼iRøQ4³cJ|¢ btØm;éر³(æq£QiõOÿÕ߸®:DDDDDDDDHù->j Vv±q;èugJ]6líé´Ù±klI»vîäÛ=±&ÛÙ¾ #.ZøXDö©ñèno² 3Ù/ ¾ Äö0ìÔq fpÏ2+2-÷ÓfØeLö2Ó1Éò|ì|}»�ÿÖß"""""""" \ö ©+g&5Ø7:$êòu[Å#/(f©}2dÍ)¶¤ãµËæEQ_²5Ýr|ºîê9ã:°°á.L[+¬Qí<»"¼já¾¥tî®ÌÄ$¬éÄc¤úÕÎæÌéÕ?6IA¶r#Ãàÿ×ßÇ&p¢uÈÄäÎNºDDDDç°<áè±Æëºìɽ¶ê6Óy/Ô<Û\?ù)múwì&w63 I8mE2'RVGÊW«³hE1®rÇsÌ;ÔïE6v2É:b)+L;ÑÙ¦z>0ÌÓíb ×};ÙÉr<2vëæ�ÿÐßÃz»ZÝ\´Ä¾|HÎ);ÚýuÒ"' J\µº«9m¨¦Å3¨ãتÎB)ðäX»c&É)|K7ìn¹®D¦_³G<ÅÛN>g Ù±¯lGó´ë±|WRR¢©Æ&+¹~"+èù¦Æ×j:>&/\ÍGá¹"¨99¤ÊóÍMù-5?TDÛkÎ\Gó;ëåÙ.E±ÛI'¹{ù>³§{. ÿÑßÄvä¡úÐýÑwÕ´[ØÚW©ÓäÆ;³8lò;òèKXس:nµÍÄSìVK$Ùv³bO(í8lL¥6({p¦BLë³ögÔ\_tTGP|ÿ�×Ä+7Hþó¢ÚäÙØóàËS w]¬ÇÝζB?~×"¸ ÿ�ÔÉçCnNÆÖEE{9é·PÉ5-dï ÿÒß sL¡Ó\6ÑUeÅmK3G×IÍ'@rs·6£í¦ñ=\í®Íï.÷½¢ÆtÑÍæµ$ø:#ìÊÑÅrg#äBD±àgû3p®æøÞLé¾áÑ:S8g±¤NêY¦tÓ159z®ï ¼3Ù&Ä¹í £ $äê\-lÎ!äE§ÍRfÿÓÞÉGHÙ")Ä'¸ìù:±!¿&^MI/j~OhȳNêäθDDI)dª Üwû1fÜ0Ù6¾´#ÇÑ|4ÙX^ñÇ ®nyå3MþÉÉ1-µµÓ;Öu¸Iø#ddri»ÑÌé¤uk§ïT[®ëVSWÉ�ÿÔÞÃ%Ãzã1
¦Ä$|âúz«ýgÉõ·T´ºxw=3È|,v2pØsÑ;ºá;®[ÍEWéÓÍJ&% Á¨þ.>¾+,gÚ0ÇuÄ9È¡}«öIx>sÙæQJy,Ç/&;¾ìÿÕÞ»:hÈ¥>\:#ËOªò;T·ÛJxðÌôV¼æúÇêÛªXçªïh³qÙ6´ÇÄ_ÍÖ23s%">ó¦Óiv.²¥¢ÅoÍ°{KK¬HRá;®&ÍÿÖÞ2N¯ ¯áÃÝ$Þƪë{+ÊA¹~¢"""r""v""#lå$e'a$&ãöñ kÈge"""$é®ôÁå=ÓqÉ4øL\$±èϪ&L�ÿÙ endstream endobj 1 0 obj <>/ExtGState<>/Font<>/ProcSet[/PDF/Text]/Properties<>>>/Rotate 0/StructParents 1/TrimBox[0.0 0.0 612.0 864.0]/Type/Page>> endobj 2 0 obj <>stream HÜWio[ÇýÎ_1¥|} Ú£ÖTMPØý ÐåF1%Fòó{î½3o!iÇUÕ%
rß¼;w¿gËn?¨/¯/NqZ«ÙÑj¦ÿuN}7[/µz÷Ól±ZieÔêÛY6],¡¨¹±¡ËZGÕ/Ú#cKWöß¼Z}?Û;¹9<[í¯þ>;]ÍN_á¨ÅfK#ª t§uQ«_TáM8/ä.Ç îêiÇ__¼¤Óør¸þùãíÃêî×_YóÁÁ÷ëEJÙ.û�©«·qurú¯÷§Ç««Ë¥ÚÒ
=«Sé¬';·÷F~0Q~H{×ק7Ëý¿þøOëk;]WOPwoõÕS¬Ü;»¹º\½Ùÿ\Dí(¢øÙDñ35ÊaêB2NØÁ3.ËãWË¿Ü@,B3=Ù
a ÕmæMb9YDNí5Ï>VÏ$W!MRL. 4ͧ4q¯9v
3¥=¨úV"PP=K^êµ4\Kb:aä®l¶8sÍÒhø�
Z%¢ Ç$j{Qaý[4ÚHR®n´ê¬1p-<à ®¹d]ìQ;³eC^(]Ìp𶺲٦¡²y@º+FìEãvhʸJ5×]ï© ;Vç¦b$é^j$ e2t#x¿iL@SM%SÙBæ!ôM&n Ö¬jË¢ç¾zdÛ§õÎ0 ±ª5®SHû,u¸~1rrÍ 4DgÇ'°7jìßMÍA+Òa×Ĭ{ͩɵлW²®® fªC& B÷VÉnJÀL³.¸GæãWæSióñAóv:ý¥êJNìPj:+Éåhò\ÔBÓå-úØ^]¬[÷®x>yg¾!p>9m>U¤-¶JDôLMÅü^Jt]Ð(,á<|OïdISTóE
¦áÈ%×�%Bò¬hNOÜ»«eÌ &GÉbù}ɬ·KCb#OÛ%Q6D~c=M,ÇÙ;W×k® àd~¯k[x� §Z¥a´b1¬6Écr]ó w¾Ö²iKòÆzVMdX&ÍÃPclö°Ø^][¦ OòxÚ;ëëÄÝÃTxó5¦Q ĵ¥8ÆJ_!±Ï>q#}B ÊQÆLv_ þAÒìZkÃ2L[°s£Õ|:ÆóʶJë6Ï<ÚmÜ;H¢V)ÁL%× ¡#Vɬ©T ix Í-kC\25å©v èÈ^¯]C¼× ×"LM%vÓÚs"(,عm õ¼0õTïc² ÜTeYZhÝ�5÷dg¤ø-÷Oo½\9RC<¼+ýS䱤kÖÒÜÛyoÎB5¹Ütn#¥+»X\Ë<¤yͳkɲ Ö1ÞãMÜú&S7ÈÔÕe4v
Ô÷sWñà ÿÁëÆ×M¯{àíßoONûÒÁ ×úF.Ö.]lC¨¥^(TèÑÖ±v HA¯g=¤YçnÚëY? Nèè«ô¶2µÄÛÅI#' }Jöqë¯@N¬ÒLo¾¤qpt=OàÃôøí´LwhÿÏë³üí³@mCg)?ã:j¾Ñæ®1o½ä?_ÊÏÌ¡R!§°+- Ô´¤Ñ¨äÒ9SÚ:G/IÉADf"ç²oÞP åbz#0ñÌDδé--LX¢BHç0°´®w[y[»é:Ö86!'7¿µg"nÿÏ´ý¸L#%%CÍlÌeNj±Ér=¤0Á/,6Ó9
äbúC°¿:�ÛÆ \·-ß.69oû¶êãDÐ2í+ÃJ)nãf!¯¶p¿¿¥Íîäk¡ /@¥vió;¸ÚùgºÚÚ(m/ü+ÔÛM©·ûõN[p´¯qx$]/©¹äÕ¤$¶®¸VÄ~;ïÒ6~¦¼ÏgÁk¾¸yº¤å#N¹ÒHåoÔëI~Ò¨!v\×ImÔ3Ñ9ÝÎp]q7aÛL×cKçKÞý4áªÃb6îF7 'Féðui@Ø#ÅÛzÓM5¡PX©áÑý Êj20r<[~h8u)sH ´y%wpຨWÜÞ®G7ÿ�£u5õO3ºÊ%õr<^©£tõúÿ7¦nËÜvøLJq5µ-±?ãÀPDÅ 2
à
ýD]mR B¶7¾:Vò±8{ÿîããzñbñêøDÔÁÁÑ =º~¸]ß½=¼ýáþýZá¹Vü_$[°×˨µ-Êyt»ÞÅ1ô=^µ<¾${&渹Eàï9DÇ"éÆ·ÃS)Zf,åôôÔJ0vjí"÷Èǵún¡0+jÄ&¼Æúkæ[WCpxn&Ò| 9õkϪ!à ^3¿¤4gzO-$:&òÉE麲|¨ ØZ>8¦á\_Haþ ð¥ ]¡5Ó¦¨XßJK4ÃÉ=¿eËD7jø\ «0=±1ó p| ¼Êm%9¶ÇxÚWØ©ãµgªÜP dYbZ
%¯×Í·©Çè§aéíú°=^¦A]VF'óÖùm,.öobKÓÃ¥ÛC°mü-¬sg««-S;¢QáìWKoãȾï¯è£ x¼ýb?<&ÝÍ!J¦-îÈ¢CJvüïSn´%îLæÈEêÈ®ª¯^_ÁtתÑý® ´ÜJýÙøÀùHÃè)ÚÞpaÒ,,9÷ð¸N9Êa6Ä%íI$Þ¬,¦ÉZhIåa,G=0¨Qc¸"§�ÙÎÖ¦dLÙa(à b£Òp[àÙH~¬ ò3~ÿn6©¤Ì/ðÓYÃÜ÷B(Òyýcö&FrÕ[$B׸$áVé¤Ó%J{©0_±P1c:/9ÊTͯ¢]X¤Ã/ ÈcÌÖà6(Á",j^¶� à]{AjZØPè1Ey·�0A1{s#¦ý
@v¹GóÙhC¯°ä¤¡(
8u:Sàü )6bìÒhm6$iM¡=q¥ì¶|BÝÚ,ôª£ ÏÚóC=ü§V0¹:Á;ZzZ¡°P¿2éF®$Z^CÚ)[|È/&)ôò&.J4Q@]v8P»åù «ÏMFE4Á@øíhÞà2K«Jcq@Ç°')¤ ÁÅ÷ÄÉ"c<ÁË2:îÒÚ Iê$!$ó²;ÏGýRLÍWÀGÐ!©3ªç ¯6Ôÿ52a½zéGfä;Þd°uØóF©DÐ.¹Ä§2!ÿ§Yh E'ð¶d, Úx°'ür|©uY¢Vç?!Í4m§¸(CÔÄr±Ì{¡¦Îd%)Þ1oplö^Z"j!èý¿F1»x²ÔN&mÎÚÔºÒ,A>æ¶Ö)W!ÚÕÿ#ðßÀáJp88YGÔÃÖ8Ð1Lt¥£YEÖ)bLq:JÉIhÀÙ/£Ê¦®N%©y[Cî+z0 ø
&Ñ^z§¯-ÇÌò
JQ×4î(¬
¥qªô9Ï&Ú@"6¸äá'L>tÚ,'è ²_÷Y: |¶5½½ñäMnõ»ÜËѹ¡¢[{ß0~3¶ >ä]¹v¯D25=xàÕ7&Û:ÐWÁÆay&:ÉÖæ=s×l(/¹T$4Ö$íýO²3bL¢WÔ×¥NYN.ÕPæ\ähn¤z².Ùk%ý[x²WKÓ£ïëWl~ï³ý}#ehÎv ¡7=PÙt®ÈÎí
ìIÔD[Ò[XúNÄÚg¬GaÀ¹}ðËåðÁgë>àik£bÑÀÖî÷2î±iò§L=îSiA½¸gN ìtùHåi.IìG¢
µ9·ApTE¨bý&A'OÙ¤W¢cÅ+ö8]ÓDªInHñø/Ûä"äð¥48:õ´Ðstäf¿ÎÅ á¬CQóÈ@ "nt 1ÔëÙ²¹¿¯6K|ç©p',®4í¥h«r-à2P´}Ûôì)ëA¼ ϪÇ'Çúö$ë!ù \ýìiÕºëzYm0Ê.ÛYuÛr[EÛ|f,A\_¬ÊmÆÇQ ß :e<TF �!ýþ±l·5ïjWØÜlN0"o!¯6U×Ñp#Ãû>+>¶Í-¨ Ø^47Ũl{¨«2> Ì+J�úWµ¤aÕ´ð~ /A
<ìÚåªìð³ØàÉÍo%¦::N za^òX18}ÈmûXTeÛòî¯õºFí©ÐÄvRïpZDÐ#út¢½C2 d£¶(ÿY´\· ¦îDWB{äDåõ3°+%cÉó'õª´v8ykSó¯2À³+ßÈDÅ6B÷X(©o ÑYHéK¿¿®ª?¡ù6(m]Ëà ÙY&À¾pÒ\{8±&̵4KQÙV3¬(¾0ØÒ ½Z Tb ¢"§ªo7;l ä3ÔtÓú¡\b¦»gðä®Z¯7>¸i ïõ}Ltï-¦0©ÇÞðÏ]ýøC¥OB}t¦Ç5\è>ÌàOÒ[ñtS£DnDÿ?ØEå¡LM¦-û>;lñXýt�@îûG¦KÌ[kq°ÃÜ$/Û444²mʲð]\34g&nw¦HgS|±\_
:RWVë3c¥p=[\Lhvò+ë@nYcFú! ¶ôë
ØÍC¦L%ç? ¼ªß
2§Í,Õ|·¦/ôW¥GLõ÷8¥´ÁN3º;< ÷
íûÛo[Hürܶ«¶ÙÝØS¸ú;5ÔÑUÕçnÚm}ÌmýªÉ¡Û°¹öe5ØUí#0ý$¼EwøÌlbO2¹KÝ7mµ§ãh/ q¶,»v5~p¬·MûL®<´Õº¾¯7¸ó!K¿/7ª=eRÛsûfñ;º~Þúª%F´ÁÚ0ëKøªmól2s¬3úQ»Õ¶n¶Ï»Äm¦ç ³ßHt,Ô=qÜ ºZªoÇÔoª²¿xf¾ÄEÞè¬0Ù¿ûúþ5Òhôq¤véÎåÿÈ¿C 3�· #M]^çÓ¿Y¯åÆy#øy¹UZü9æÛrHURÙÜVZ,&©½ÎÓ§� °,û²öJ41éîéæ1ÒRó9ª~PÊi±w\Í>{=ܲ^%«jnðMÇÅJ!kÚɳÛ\$Y-k,rvÓ¶ÏJ?°ZæÍ\¢=)$»çJ~?§²zsÍ-]Ûa¼$,6&ÒM¹:aXö6Îóø´A7>§ål}îîrtgWg1JÍéãØßûWl±²Ö¢üX["©=6ÊUË8$ûDZ8C=î©ox\_ȹýRçPÇáö©Ö¿Êí.4+DlM?© 1!ÏaWìFF(¥ðN r«MW[ïUO ³\_ÆR{e¾½<5̵ëÄ5OO«²a>ÐÊÐw<ͳ r#õ¶ äaO+Êù1º«Õí¥Ø0ìr[_ ßÜ\T²±R¶ë±ÜU 8sÔ QdWLcÑÃù®Wþ¤9pä6éWX÷(GOSZù©ný½ùÙ»úº°�!/A"§'¹àZï9,cÉøM£iþåª]À´óÔ+ïû ´ñë=² ¾úk"ݦ+[µÊqºÈ6t_¬@upìDä£ØÄÓ~^¤~7ɯƵ· °CvuR$¡%¦ô¦¦!}Z8P) ä]g!èÐÜöêÃ'rÏ^Ù'ÄRöÏÄPKÀQ.5ò/ÝÔJ À2MRÿÉIN$¥÷cVÄqñÄF®GÔú[ÙKØ|É %9¤¡xn¥<õ5þÔyÖWøMýã«×¶¼\dí8;ÿãßÙôPÀÛ55 KÙÜW)k'psö9à, ÒQÖ¡ëþü!ê NìæÚ·IÓ#÷ú¼ñÞ¼W¬ë£#Ä}GhÆõ§¸^M[þ{fJW ¡ÁÕ»h«·ÍQ§IÀË|O¿6O¢¬Gßá:ô296URà äpW]÷DÑVÖ{ÙÒoZhl_ÊìX¾ê¦.ÉÈgîKùhúþ¦îz/]ö"ÝÊÿ^±ÊT¿-
fþùÚvWòdæ+g/ĪX,ùÐ$DN!GãÁ(GàüÆõ®É8@>¨uÞÖ?8e·²ë¦IgßTUÑöÏxîÊCiH饳©eºáàײªôó,ü~è 0nSvo ý~G£xÈ¿ÀdÐE6P\ßAÍ1Çè~¥ÑíIG¶fr®§7§K%yøý´XÊ1ëðtÃV<¢ È"Êí[ºIÖ6μ ÉR ¾\åß½[g$Hî$ÁJ05ÞH+ësUzvúC¦çx|ÿ>×2@ÍT@6`P20ȨxøMµ£ yJp1-0\Ý0zw;-À(Ï°ug'Xx+YH×¹ÝÐFÓ¡+;I?f'z&L²i¿åîz ʺÀE/nßÞõ£= fUuGíIß9voXìmeAg7½5ùàr,.ØÕ°½DÕ>Õ:ê!ÌÙåt:²Å$³Q$Ô¢iv±©Kkèz)v;y¾õNÒ,Ç-ç¦û ¥æ ¤ybZ?8öo'ÇØ#RÍÓ(HDbääتTÏûÏÓ$H¹¸qìóß ©ß¼!óУ:cÝõi_¾¬·õÙö¾Ïé÷SîȺµ&¼@EÒÑpµé KAã©cÌÝ(EÚh &étJ~?§²z²»rÎh¶e¬7¦S§ku)¯uyY|t\ÜåèVskgCïM ¶±#Þ2pmõùûW©·ÕÙm#åñõkª)À½çÓoýÈÉ=~¯»y½! r@ôHjGËcë$çVU}aDÃQY" ø Ìm/u£Ïyãg&¡>Æ 0V·3éh(
ØÕg4¼Ã®æp)ϼwhÓ$ZjÍNÉ6Ø|ùªñÛTfñ
Oòòe='½ÆP|rõn¸ûoSÍÀ ÷ųµ'K§9 Ï,ø)íìÌDiå¶ÄIµqÉÈú. 't#Ï3±1Óï¹Ü÷Á9É)hà µRîû3@ãZzo²h"Þö×V½'pö'ÿ .£q,ÃGö8®f:ÈE!~¶ÄãåX]7 m¿Z1¡)íȾ0òÎÚ§O)«ðaQ6¯5ðþzlßðWAEÖAÎeæ"¸8Ä,³Ü¾"üí<á ø£ö¸Êʸ{ Éca\_~ ±Ûq¤?iôsdÑ,·aÝÊ3§SzÑvbA\'º¸\d"^¤nDi]r(|'S0:v¾,SÓ¦Ø�ÛbØÿY¯ºÞ8%ú~¥ü±d#ºi É>9wµR¤»Ù«$o±t
Ã̯ÿýên ùêÛygº«N:u\$à0È0$@Ôv/\_Û´sÙðo4!o\Uoë-ÇcJM©ó§pI³åé¹»§£´{eiÎM»ë¤MÇ[¶è%¼=F|^ÑU#mÃÁ=&íàvíHkG÷ÈD¶æ}7Àï:GXJGRÛPQëÖ»æåwJExóÁ]núpc=üº h^u6þmn;öÍra13ïüaH¯ÞÚF×h¦ÎDF FÐ.)9ô[ÝvÀ´Oªì «¼]ià8JT¸Þæ¥~ $ü"}ªÀnõ¬-½
N�¼ûÁ8V+]µ¢sÛmÈA»!E4WÙF¾Çh;_¡dãwÓªI?ï-wÆëU̧d\£Á¶·µ]ÍA5äAuê¨=ä²ô±cnþ#²A=Ô WsÚïá xÈÆͼ dÄðII/UvHkáa Üý¼·@_a«BF<î³a%m[çÅê'sCÁ{e4Un7RÙ×NyjT=e8§L'ùéñEcÍü5Vx\ £³t%Ù×jï,4$à}:¼·UÞÆÞÚñE\JÚÈ(~EWeÑ8Lc2̵¹\ÊÊKFP~[>¢Z\_ñFö¿üZ3+r´ÁV>@xÝÃ¥ø§i\_«óRÔµpóJbÆf¨ÍÝõô.}i·§}sÅÆoî^vøÎ]å+�[ |]DD¡y³Û+i«k%×zëY¸ÆÉ:\_ps>¶8]ª&¥8ðb ¬qÇ÷sþHï#Y^ecݤåÜìçCÁ2?R4ÄþR±Á5ï1zÙé·À°¾¤ð ¼3\ÙÜb¯©&ÞA<0ð8x×è²
m³õf²IG
ñ�å©®>°K¸¹ßwR»ª3¬ÈQs¸u§Ü9;PÝk©.üs7(È´JÕ&Ðþc^¥nÍdGZxÖjFëû#&£þl�Åë8ÐÜ©q(úWvmwÉ¢CÿvHÌ"9¸ÿ]nϯÏV3ÛXî¶
Ég¥æ5ò}o;)ùÖóöªIü÷ð3
U!ä�Võ¬ë3Px1²f9ÖÕøÓc\ ;4¡Ü}>øµ¨C)¦Ü$u¤
f0í,ªæÅï¼àßa)'ËR(duÒ²NÞJW5m Ú¼äú:» §Bg«,¿ðã©9´¦%2M Ñù
ç|Ç6©D!£>ägè£p²S¹Ï«£²Í«¡GÁO òÏÊ¢½tõÕ5Gz@ÙãÔÕµ!}e�®U½¾HQFÄF-d¨1 19ùÖc;mûz«êj¡Ä÷j¯½ÇîÀ8ó&&+{mÃî0CH@F} «¡¡QSM¤¶ð#>û?ñU²8DïÑÿPGak¨Ò~¤±ÜM<;ƾ°[Ó,Û㿵HDÏ¡çd,²2_¾Ì|/òÑqUòJ7ýÖøÔªsD¨HDd0^"ªî)§|´¾é¾µbK¾Ð?(T-W5q M²½þ&2o¶Y±,ól® ±r,hÏ1áp[8ðó
FÍ°Mh¢Ñ× Á
öL>´&Öå"hÚ4©Cbáz¶È}TCG§Y.N#'Ö32r²yÙIgÿä-Ìħþ¡¬²¢+Aͪ mNKiq½¨^õ2©°çp"OùªDwôén¯K¨ª´1¤yvãA©+ß]k^ó8ÄÁÞ ©=·'ıóqäHìÙ¹YUù<#µêìÉÌ\°2m:a OFÚÈÉû^ÎÞÏÖÉ] ®![í¨¬ÒW{²34PìÝðÐÙ³u²;Ç®\_7TaLxÝ:ª@j@ÏVF}ÁQèq¡U\_h\j>jf²´Àp\_ÏeÔë¦È^Ym0ôÕȹ0Zt²}DÒÙ¨v óåªÖT²gIiؼ²°©�yWôH.£óÕ3Ç°ÎgÙSÒ§8Ü0§ мR»RÍNf³ìyÇ0
å×lYµ0\²FV'ôÎR-5õÕu ¿°"Ò<[¨cFDÚK î 6Éí¨úN¥"Ù6»vGÑQð�2ÄN[óe0·Axz
ÇC(�¬Þ [$0.´VÎ7Zï¤Ä, 7j;¯· ùËîMË0-ýÑÖÁkhewRÔÖz¢Þ Jã©+2ú×+ 8bóäia<æeÿ?Ý1e}É$Ëܺ6¼Ï?twäh a3|rúDGÊ¡Ú û:;hÒÙ³åË<+ô6g¹>^,D)Úlûý˨ͺ£N6Çä =ô £>iâ>7ÚÒ"g?èR~pi:BD®'è�ÃÅb²,6lbï |=\áF^[ìû#»Ò3õ&7º©ª¢ Zº¢�iñÕ{/=m ª)]ªÜ/-D'§Csñ~5¶i8ZßU&kóãQ®Ze\_ßYÝàg¦w¯yùëÄß÷ÉV¥¦Yè8ùø-Ìh]M~dfÑP«íQÉ/¥�P¹°zf;'�ð xÖG§Z9Ìì�Õ �Mn· ìi¶{Í25»ª¦K»á¾F9«Íë)Ò ÞÉ × }ñ1&è£Ù@hóOù3#HÑÒwù 0[<|ÚlÕa+ìÈ}rè Ùº×$j¦P\_r¨ÜÈ/z¼þMcÅRÎ+ʹcر×m¾Û' G7~~ëÜÞ\wÚh}O¯ØgÇ÷[÷Ìñx5þ²®ÃCéÜôJCý^IK¼5 º½e·Æ½A ®£æm´ëëpîvÒþØI"¦nnë\¥¬ôWûCæÖ@þëK÷nRÁÎ>óÝ=ü¿W½ñ½s)|¼¶Ã{jäÜ;ÛU\¢;?OÀ¿n÷nh~Jà:ý{VÇ®õF2¸¾|7ºU®w{×½®t>iþOÁ$2çÁO "S:J^¤ÐqÊ>ë;ÿH£6 ¨nSM>øðçrÒÚCÚéT\¦u®èWL§ ÇeJ/eÜ¥x§ÛTPb}%x/NÁè¼7HXFê§áUï91}sï ±õ]¥£Þ0ÕZã ògh¥nýËT±>AôÝ´¬zÜtWÒ 4g˺öª$¨æÕu:¢Ìwn¬],nV3bù ýypû º«óºÇ¥ÇßnoÒqjQOG²{uè~2è#,ÃÏ{«wÍ®ò.tEÈÚþïw¼ ´wT ¿Þ»f%wìöÛ_àÝ¿ �ß9× endstream endobj 3 0 obj <>stream {0 0 0 4 -1 roll } endstream endobj 4 0 obj <>stream HTT PWíf¦{?ãBÓÄÌètÇ T@(å'¯Á ¨3 83#QQÁùøÏU
Ôøÿ¡m@låo+S¦"ºÆ°F½=>ØÚ7hmÕVWÝ÷Þí{Ï»ïûIÈÝ$ù¤ùsÄkÈ+6XÙº´T«>0Á³úQIïá$ )yË¥1#X4U¿þìµ×£¤PÓcF÷$Ü\HrÒã£BÙÈO >5ÀeÃ\6|Ê ²o<á|ðäÉÁü,}AOµY¬| gÊ.0uV>ǧ¸à,|Áb0»ÿ«7Zxo5ëô|9/Xÿõ¼,9Çæç²sóuìFÁÄÏ à «³ó,ÆbCÏ3fL·®0å¬à¦«Ð'YfÙÆÇägÅð:Ï×Ùx\¥ÙcÄuqÑÄgÌVW½1Ûj,0Y&E§¦¹@By½a9Aø#FÄ(ð ¼ BMãä?AÄ$B!rb6IÌ¥ùH©r" ü0©Øñ1a&>':Èq¤|F¹E»¹ËÉ6ɾ=OWË5J Jh:îP¼£U\TJe²PyVùÄ=ÞÝî~Ñý·aSj\/HO/Øú\UÊ¥ÍΤÍ4T# Û)è£.±àZ lS ÐKæ0á5-]d]3äZ©TÉW²4=½»®©¶CYÕXÙØ þ!:ä»uüjÝʵZWLÏ Lð~¢P¥¯mwjÛÉ6²fQÖ aõ $AÌ«¯¡}êniÅ5,l?ØÛ Ã·áÒÐDz°Ü^(ÕªÒ+»îíŤ+¦ïY»EØß²{·òùs ¦ªA5»ç}-Ó÷{-w
®þµþ7 ×)¯JR#Ĭñã´L_dÄ24\ÿéªPònä.rßU(eû ETÍùHø|.°0< F"m®nC~!ç¨ÚEX»P3½lu>ù×£Îjþlè\_8ÎgºùúÍrÎ íÒÈvÏcN(°ÊÍþ±y}Y<ÉßwÍÿ¸³ÿk#ÿ<àç×4 þ[ë¾SÍ\=·ìõéÚ
ëó5ÌÓ9
'¯q´\_w£ -èmÐ ÜVzrñ¿¿¥^˺ȿ1ðfTÕÔC© ê,ñÜ-bzÀN°DiUE?E¡±EvªîR VÓ¨ºF£L§RÕT Ð!@àyEMR×aÒ%òÁz¤A^qyHÁE@-È¡¡KÁH¿÷ÚÑ042¦è=Î4ÁðiÀ4AÊÃC¸\qgßþïª/¶eÑÉNª]¨AÓí´ªÓ&8ÏnÎI"Ó'Ímì¤Íf¤Br ß Å èß«!cúûoåDWH£AJ ¡J´ò9¦üEð¯Íª³Rpªòm¢ Æ:cÙèÓU]UÅÊg È?@=¹
ûfjKK+k>¤ãYð¥Áã~#°g¸æû ¨ïüÓNíµgÖNÓ 5;ó-3äuQ&MöÐ!èjUCìPß ]¡ª/vÚKÈ{x\_o§=ÞåyÎØòN±CîÎÊÓÞ@¥®pVd¯ÓÎF ÚEÉszW) yкah¦np´ãº8'aÞä=oxôLñÝ?w?Ç:RwùN µè]Ì°þ~î Fê°5H©M¥±Ä;à©�¨_w=Md$Hì»}·q5iäµ?Ays½|o-v©x¼þzè·J,å{uÍ¿ÜVC@à14VÆãRá% äé_0!7¤Fn8DãMÞ{ìj»ºÌmÈÁÍ{Àràº÷T÷|}><½ë¦aXõûÈ]T³§añGÜZÒÍ1¿þbìµ!J<ÙGÿÓã-Úy4zwó-«w~îÅuLøaävß@L{¬@B ¦$z0fCmQBÞá(» ,Û[eÚ©?VÚ?ð½bª9ÈQÎZéx<&ºòÁ<\_IùL!äzߦa¾ 4Ó¢)"T ׶-9g)êÚ<H:hL6«|±e
Ö¼¦°Ô I\s¢]ü²¡£;úMëWGëh6(Øó;\.õjÚöÜÎÕX«2ËË>·gX?)Í°\ómæ©C£ûÏÞsìv ÍDn]U±¶é %³¸¾¼¡bïvå
±bSí!5Xê\_ízÚ·Hæ6H(ñ¼\Yοx[!úæ rpÛ"¢"(Qÿ5[Eg6µ´kÀoÄìÅZø
úbs@y2Í|6hÿ|ê-²QÚ;¬E|ÔÌ(¿dütw Ì7dXÝbZúíR ò[æ"íDq0ñÒÉúÓøÀÀéÓÕtBYÌo#¿ SÁ xȾÈOä
6¢8IoÃüu¼¯·s\Ù35Þ 3ÿKcÅ4EQ a¦é8DK2¸¡ÆàBH>TÔ¸vR5¨Å¬ÕXPqµV¬J\C uBÕúc4jbôã>ò&ÆÛÿoî½ç¼sÎ}/e(åX©&>9á¾d|ª¤ªñÚ¬==ѪÝÚlNøÒÅ?.¬îuIìYgZlIë@pLÑ J£úõ±ï Ññ�K0£%(Y° ã{ºaÎ/ø}¯9{C5%eÖ8ÆbÓ ¡¹©ìæ ¹î:ÏclnÛ%×]
Tò"«/¢ßA¸,|§õ5LKýuÈ@àC¢ißV}«ÎÓçY8_Ö:o(¾³£Fr4¬<·Ú®ÚL Å?O±× ýÆÞg bl4õº~v ûçÿÇÓ{ìñ¤)07ßi|B4Q|ÔÑJÞH[ é£ Ç¿UîÌZºÚ¹MR9¡ÅMÿ�%ä%xpa{¾ÏÃ<¢CvÃ6w²q, -a3aª\¤,ÌÓÞ¿ñ@os3óä"ÄBåX½Öiì&B½Ð QÔ+.Á!ðýöáù¤JüücY»r°N6Öt¤H}°¯qçSl,^gcLÊ&Û"Iø8+»åùÙŵ[зNßÒ4¶s'«<çr³Éwµö^gMqK¸þ%ÍðjÌrKüi9¢ fjAOí¶¸÷hk¯Ç?¡^«4ÑÄm:$+,yyÚ]=\_vÚ¡åIUë·r�lË!¡Ü�Õguõ\¨6c(-ÿ 0�iÚã endstream endobj 5 0 obj <> endobj 6 0 obj <>stream H\ËjÃ0D÷þ -EqbKW @H[È¢ö[I lg¿ïOi¡DÇX3: o¶÷ÛØ&M}³ £9t±MáÜ\_RÌ>»- ÓvÍøó6ý7§zÈr ï®ç1¶ñÐgUeò7ýxÓÕܬÛ~fYþÚºx47ÝÌä»Ë0|
S£ÕÊ´á EOõð\ɧØí¶ÕïÝx½ÕÌß÷ëL1½/(Óôm8uR!«æú¬Lõ¨Ï\ ±ý÷½\2¶?4uʪçs]7ä øü \rO=eA.ÀìÀKò¼&¯íbb]µÈÚ\-ÙÙiÑiÙiÑiéfáféfáæèæàæØïÐïØïÐïØïÐïØïÐï,ÊBO§°GÐ#Ì ²¬ +ÌÊõd¾#ßé/ðú ü
þy$ëeUçzëéïáïéàáàéàáàéàáày.¸ôÛÅõëßÙj.)éXM£<Í&©áwÚ~0/û�¬qÈ] endstream endobj 7 0 obj <>stream HT XWÇgæJ ;vgV¨ JµZ
¢/Ð@BÁÊK´¨ò_ k©eK Ñíú@·´ºµ+Z÷ûp·îzéîÞõûöË{çÞsîïüÏrP4MOZ³>zÅzß0uYmÒ$#WÇT3"5É)¦%z!Íyl¢ÈÑ¢§[îh8a4àÖ_;Õ1"5FôwfÈÅÄéÏ·Dn58Í
Y¡sýãç:Ë5¹Æá'¡~B@@ °H¥OP ±V£I5 áºD½!]oPÔª¢´4aÓQX¥6ª fçÃ×´Æ((A©Rk}ÙÓ¨Ôi jC²Ú ,5d&~¬US4:µNXô¾ ÎJLË4jÌê4«¦ITëjJ1è3SHNo²¦«É$Á 4X
÷µ a~R§´J«@( êd á4#NHTLJòiÐUDF¯3Îô_»Úéd¶ R'QM>ÔDLQS(jEPTZÌPá4µ¢¢hVJ§¨#Å©p£þNO çÓU±"I«t´ËuËsûÀ$3»&æ(å¡ÇîþîF÷² ¸Uz^|vibÑôPØNr{èGXÀYp.v»¡xײFÖ9ÃÎ\Ã3dÛÅ1}¥J1CÑ ÉÐÆú!Láڡ1JT²xè·|æ d3
p6s)ó9X¬öÁ@; î×að¦TÌݬ¿ic7vÜñ _CrU³sùnÃñÿÆ7cÊã ͼ´Lÿ&ÃøC¼¼Ô;4Úé±PxVÚ|uh?YìD]ÐÈÀ1äÜ!QDb5f¡oHÅÉ|Ã^î¿ÆvÃ{¤pØÊB²RaNsmPÓ VØk¬]XuDÝWÁ�Õx2ÀÙë±Kü28 rDÎÒ½ðÌMvÑëjÒêÅ/Ð/Ø··ùlå·ÖñU²oQÙÞÔúàUþ¥·Yݲв2÷ïk:Vß¿Á§ð[~{ºÝÌ\_à1û=?çÃnFÜêÉãîõ£×!¡§èññ¸]yÜz¼ú¶¬¯¶ëêî¹ÎVº/1Äå·äå®1Ê^cÐê7äDÈf[÷.à5ç:êNµò;b~FaΡO¯ØX\_ØXrpÏÞ6¯íï>)SÔ)÷þ±¼%ë÷¬Ä8Î7ÆÅÛm°Fßê»T?²¿4Ïa¿xîòZ¼frÀ\îhàcQxVÆPR±ga!Q\Þac»4Ãè¸'[°û²¼\h0Å6ð±~Ô \_)z{=íè
XÌÜDûlÌr¤è?puÿµÆ{EÞ6,-7¨2'zÅD£¼µ+"Ðbà)¼áùX\Æ´loîäîo[ñ ?wÈDò@¼§µÜÞ[Zã÷o\ÚX¼Î¢I£ PJ"nN yJ÷A ';»#:ÄÖ?eÅÏut0ßBÎi]çuöʱx7!Iôe¯ì¿qÖÎÝݾ62zàô0Ùpn)·fCzB)E˾ÎWWj.Po.
0Æáá#µ0ûÙWãxl:ǧ§%0«°±xT¼ÛWÃßnßq0Jýü\«Ï" @¾ÖaÅã&\V´¼ÏÄvõ«[{á?¢¥öoí÷\_öܱÙC4zY»kïØê ÿo¨Fò¶OÃSd1¹VrKÍgnTðÑhbæ©h ¸ouþ£? #µI©2®8MZN羦¦?qãbòø.ô´aó%?nÊh\R íb¾Oòô5]h-Ê)Ïø|×±]uE\_¯^g=,@ÍBtJÁ!Ì:gG³få~¸ËÐnõ®ûìËò2"^¶3Lïx^¬~ãGqqÅK·[Nøó#1Ä+ù1ÁYåK8ûùÚö3ÎO'Þ¾A/S.¿c;µæÁ,Òþ ßþ'ÿ�9�°pûP¥±¢¹È±ç¦Å÷øµPûküSxøSN¥ü¾°{izhvkpc^høkyº÷Þµ÷/³÷»ã¼ùC»ø!y÷g÷e÷D÷n¹÷l¹û÷EûgûhûeûEûl]ûn]÷ûD÷hºûNû ÷/÷T´÷S´÷J÷.÷N÷M÷ û.ûSbûTbûKû/ûM÷÷ÞØÍ«á°²oohgûN-üL¿¶÷\_ëÚû\_Éûx÷¬÷/÷ÀÃG}2x0Oyùûl÷8ûSxùbûj÷\÷¢ü ¾ Å øP÷Ý �u¶ã# endstream endobj 8 0 obj <> endobj 9 0 obj <>stream H\ÏjÃ0 Æï~ ×CI6v !P²rØíR[écô·dfHô3þ¤OXNªú¹vý,÷0fÙõÎÆ[0 /píØdÒöf¾ïß &7Ë4ÃP»nE!<æ°È§£/°É[°zwO\_U³Isóþp³LeYJ ziýk;LbÚº¶xÞÏËsþÅý1£
É·Bë® W)3®R³ÿÎ3ÅiÎ|·AÓ(vÈ·Ì[9'®+33y2çæïwÈë+ª¯X¯H¯X¯H¯X¯¢}ù\Ŭ÷Ì{ó{SÔ:1¹OE}jöÕä«ÙW¯f\_M¾}5ùj®¯©þkb ½ß ]1¾ù¹
£Ï%ΦÕ;x¼(?zYô\_�F>ª= endstream endobj 10 0 obj <>stream application/postscript Adobe Illustrator CS3 2009-04-07T12:17:50-07:00 2009-04-07T12:24:08-07:00 2009-04-07T12:24:08-07:00 uuid:BB91FC0724EE11DEBCA1D0B5C54A0FBE uuid:BB91FC0924EE11DEBCA1D0B5C54A0FBE 1 False True 2.000000 3.000000 Inches Black Default Swatch Group 0 White CMYK PROCESS 0.000000 0.000000 0.000000 0.000000 Black CMYK PROCESS 0.000000 0.000000 0.000000 100.000000 endstream endobj 11 0 obj <> endobj 12 0 obj [/Separation/Black/DeviceCMYK 3 0 R] endobj 13 0 obj <> endobj 14 0 obj <> endobj 15 0 obj <>stream hÞì\_o[ÇÅ¿Ê>¶ÑÝ3û0Èi¶qÁr à TFJKEÑ·ïÙ{aJiÄMS /VZwvvîùÍðÒ¢óÎ7gÙµ>×buV]+ÙYãµ\ Ѹ¹.çÄÏ#çF»äb ÙqÄZ ÅÅêBuÉs£Ð\B_ì9.KÖgîKÆ1r.ÍÅäRä¿]JÎ gÇRåuúË .qäBgÝçÔqntÆQ¹O¨Ü'q4ÐGKtÆÑ5=Ë}T¸ÜËÈÁõa>ºÌa+sXÍ.s]湺>¢§3Èà ýñ CÌ÷Ï#Ì÷Ï3¢+ô\¡¿Ü²c¹ððJwYª+Ý%ãäÊàý5×Cl<¼JÇQ¼EWéÒ3¸JðÙ1Ä^¥KÔê8Q\a1×b�\1ððC -8)në!2¸ÖC3
þ¯Ñ_jÌúËGX2§Ãbà z,<¾b¥úÐó/(0x¥Ö}ôÚ=ÒÒx ÔÕûÊÕÇ
Ö3Lwæ!Ðo ôûg2Õ(5x¢Õ+À#LF0æx è52#Á¿©7x¬51^='f$wM<Ò®|ÍFÏü«¢¿z5}»YÝßÿ}u×¹ñîýô]'§¿8_m×7»ÛõºCôôwëvßMïo7ëyuM><Þ§Ýöár¶{{»;=å.gÿÞ^߬ÿ²¾¾úqçü¾Þ®oh2áÁóëöôz{v9>½\]NO/gËóÕéðòùêyôv°~úÀÃ8Û\_ÝLßòxÖÛyÆVüíá~wýÃã²äÅ].v«íî¯7ÿä,G¼ò3'ó'²ø<Ù«½àt6/îz·º\ýÀ»rá?í9{£q÷ FãN¿"¬ÿvüëEjÿsf>HA'èíêæêëoþqñÇéÜY?¼þt{ùð
«fëw_î?úÞÍxÑ¡·3¾ø´¶Ý]_nÖ}ÞÝn¿¬6Ó9»Ý>NëËÝôùÝíç»Õvuµ]Ýýøù~÷¸YÎOO?ewòìÍõº¸ÌEsÐ5'ÍÐl«æ%¼d²3]7ùYî½6k®u!yÍ+æ YëÖ§¢yñcmù<6Åáu_^ñzÅã×}yÅëuÿ¾ÈÏ~®÷þ£æ¤9h^üßíoòcÏök½²<Ú~½öÑ´è¢%²ýueýÛ×ò^ûº(ê¢è~o/?ûìõó>>~ï÷¿ÿûþqN¼
}QW/Z&Uv5°A\ SQ¯2£ê¤¨Jªò¬ê®f¥´LêjêjêjrAUÁ¼xQ)U%U!ÍUÒ¼xQ.glÕo{U{{Õzzuºx©µuºxQ©5 ºxQoZ¾8µ-µ2u65<õ;µAuG5Gõ\¨çB=ê¹PÏ
z.Tµ ªU-,Uk~\gU?¨úAÕª~PÕª&TM¡ê U[¨úBÕªÞPõª7T½¡ê Uo¨ @]�û¯bêPnCÉ e7ÞP~C e8PCIe9æPCe:êP®CÉe;îP¾C e<òPÎCIe=öPÞCe>úPîCÉe?þPþC�@@@ @@@@@ @$@(@,@0@4@8@@@@D@H@L@P@T@X@\@@d@h@l@p@t@x@|˜ø0ñaÃć‰&>L|˜ø0ñaÃć‰&>L|˜ø0ñaÃć‰&>L|˜ø0ñaÃć‰&>L|˜ø0ñaÃć‰&>L|˜ø0ñaÃć‰&>L|˜ø0ñaÃć‰&>L|˜ø0ñaûgñaÃć‰&>L|˜ø0ñaÃć‰&>L|˜ø0ñaÃć‰&>L|˜ø0ñaÃć‰[øøÔ?šÎ<\_Yùø¸–ü™Î¯ôàÊGõ›ùÁ”†6¢.ü’!fC›=úãÃlˆã†qÔ01<ãßë×·?}L¡ôß™£? ¡ÿ¼[I¯¢%Ÿ4V¹O\_8߬.×ý~z½¹½ü×éé¼Qîùíg÷øæúêa»~ºUÈ'µ0Á#Nre"gã–mÙ\X=ºU9ºo?ͦu¾}ûåÛ¿Ü=¬6ý·ŒWoþüæ÷þìôëêvdõÞ0úß²MÄè6vl›ìo 4ŽšŽ–Ù0φí¸a×<˜?nXG Û aò£†5´QÃ0jG Ó¨a5U&\“F•É£ÊäQeò¨2yT™<ªLU&\“G•É£ÊäQeʨ2eT™2ªLU¦Œ\SF•)£Ê”Qeʨ2eT™:ªLU¦Ž\SG•©£ÊÔQeê¨2uT™:ªLU¦\ÓF•i£Ê´QeÚ¨2mT™6ªLU¦\Ó•á³ílˆã†5´QÃ0jG Ó¨a5,£†uÔ° bTŒ\ƒQe0ª F•Á¨2U£ÊTŒ42c£Êب2ö¢2ÿ�êA»Ô endstream endobj 16 0 obj <>stream hÞŒ–;n1 E·¢<ñ/†Àe#^AªtAö\_år¦IÌ-$GwHõµ—h,¹Ž(ü ¿³$g/ݱÄöROœ²ô€1]6ŒÙ²lœŽ¤Òq–ËyV+Àˆ…<ë•‚<ß+yøt6ò\W·U…,2’±V³æ(Å B0ãe ’–H1>;ÓVÔ»Žñã2B Ü·ŒÃKôööúx}«×÷ùÓësMÕ{ýx}þZ÷åëõõççï÷÷AgÁÁdÁbÁÂM‚¶YPXPY5c¬cÍkÆX3Æš±ÛŒ¾YPXPYÐXÐY0X0Y°Xð° k&X3ÁÛ=ÆÎÁÉÅs³ ° k&Y3ÉIÖL²f5¬dÍk¦X3Å)ÖL±f5S¬bÍk¦X35sn3þ : Þf,¼Íä3Ø$Ø
Y3M±½YPXPYÐXÐY0X0Y0£x©g°IPn3ç¼Íô3¨,h,è,,x±ý ^fLÁÃMºYPXð2cÿù÷ú¯��-WSg endstream endobj 17 0 obj <>stream hÞXÝj7|½@½ýB0$
Þ¤¤¦¹ Á,Á¤¡N\·ïHµWºçóîh¾£3çh´ !:ïBH.!»C0Æê#csÁ'È
18#&9ËÄX$ 1F¶Deù¬¸DL°êR!5A¾è]Näã«s#_ ®¢¹RÈ#ÓéïL®fw7O¾X#&ÄêZ%_l[Éó!o<)Sàq!ISì#kJ|(¤e 07qßFHܸ52'2Gn'd2ÇBfæ2g2'C&sjdf©X.2g2çJf¦ÈÉ\ ÌÕõBåÆB!sdf\h,}(dnÞ¹Íss¡podî©õ
Ð
ìë µ Ûì©EuAíÕ£|Ü\¨£|]Ø^>1NXû&Ù'¡/÷¶JO¬§ÒÆ[Á÷²4öïù²|6º¨%gÉsSauÉLmªU>þIsVX#Î ûd|[¬æbWÙZ{2z&eþf>äKåCOÉÅÞ×ÆÆ]XøÐs¡äÑØ;2GÖÇX¾¹9CïàÜÈù½qo1ç¾Ì}KO-Kß%3Hè[¡ä)±CûO¼xqøõð2^÷WÎ8Þýy¸úè0ÞÞ~=~¹¼üwiweX0®m�Óh~]ØÚXp×À´ Ì»À² ¬»À©LY£ßb8©k íã.0íó.pÓÖÀº lÀ4áºbv¶ »À´ Ì»À² ¬»À¶ Ì»Êä]eò®2yW¼«LÊÌ»À² ¬»À¶ ,~]àT&¬¶ »À©yXvu 8ÁX2k va8AY2ÈkàPi ìÊô«öóÀ«#ì÷ò÷ñ º~swýõxüxüú×õ?ßoo®Ç¢×OãÖ.λûïóûwý ÏOÇ ~FSI1+Ū87÷¡M1\¯ÖMÞy]pÓhÇE~ÆÓßý¾o#k/ï>}¸½ûxɯ^Ýýû.]øÜøýÆûzñ"þãÏg~ðþðÇá÷ã÷»o«ÛãÏ7\_¯nï>ü}yÉjùêDR¨zÙo>~»¿QÍfV¼ìϼ"³üi±¶¼\_@\_ûϪfV5³ª¢ªU·¬ºeU¹H"¾"¾"¾"¾"¾rRE|E|E|U|U|U|U|U|U|U|õ$³øN]Ðħn4u£©MÝhêFS7ºÑÔ¦nêƨnêƨn³-|·ø3cóÔnÖ.'eÁ\_É¥bà{ ¹,ºmü[¶[¿á~õ\|¿ZϦ67µ¹$RHA"$RHA"ñøL|&>ÏÄg;§ÃÄÅÅÅ7Ýéبç EOFðì÷iQm«®DX¼ïýï»ÃøñP º½¹»ÿ|¼}D
´Ä9@§Ìud?³è43h85ÍÉ;{ÖÅxËÏy§)úÇóʵè]Ðátx
jO/ÛeÿavzuÚ_ÅýEñqí,bqMx¾~'ìûÿ¨ÌÅ9oQæ%Ó¼o)b"ÎË¥9·imNOw"È, ³Ì2 È, ³Ì2 È, ³Ì2 È, ³Ì: 2 È, ³µ½ºê)è)èD½/ê}hZ qæsÎæÑ)h¢ f \hª ±æ ,h² Ñf .hº ñæ 0h f 2hÊ 1æ 4hÒ Qf 6T<\_ÿ�@Ô endstream endobj 18 0 obj <>stream hÞ¬XÛ\5üÿ�Ý\_$´Ò\/ D!"¼Eh5a)lVË"¿§ú"fËÃl{ËÝåê>ãöªX©EÅË^Ä×Q´Å÷Y´Ç÷UtᣵXà´o VOÕÒ%¬¾°VáNÃö2k¬eÎÀϲ¶ç«´V1èACä)Ò´ÆÅ cYi&ñÄKsW @ÓxØ(#¨Ú,mH<çÑñÄáyÖ«Ó2Þ´»?©¶Ñ ÒÓvÚAã/ñZ½\_§ÀíçÜÖðÛÈ·o#ßF¾'ãd¼I¿ÜP?¡~"õ¨P4P¸!á &@f6àºpE<åR ¬X-p%\_%\_%\_¥ºr~||LëÈר¤ò5&Ükn¯S?''''§>ÎxÎxûïô×é¯Ó\_§¿î9^ñ:ãuÆë7o0ÞHê=ÈgÏHòä3ÈgÏ8\_×ø<{swúòôæöõ#:Ïëáëóãûw¿<^>¿ºûátoO¿>^¿}s{wxg§íê ¯È^ê[ÄÊ7ÏßÝÜ·Çû×7??¾{ºùÀ?^+Ýë dÿIýó6¯±Kô>áu/ÁÃ[/Ø÷Ç»( NYpºÜÖíVhöà )k~¾1gìÈFÎ4reRÓÈFJ¹£òuT¾êó£ºãrQ\[ÄÓçç³½#-ô%¹ÿdlý¨M=\_ M½³Wpö Î^ÁÙ+8{g¯àì¼Ú§ t³knkfðÀ}|õìïa--¿è¶mnõ/nÏþ ìä-²Y%³vÙßEô½+ïÄ$JÍËJù'ÿÙh
Vs +@)¶)°Ü 3Ø¡pÛuùl¥dòxÐZO&½ñ 7ôÖ©rë#,6ÆÆØXXo¹dù¸$Kß.ûccÞsIòú<ÍêóT5[¹ß"¶©¦Tm¨± 5cvØfÛLcil3ÍÎgç|süþtý#ú¢lQLGêQÒanQÿRjy%0^ L/¿æýÜd´;&XëEgrÙdþ7g¼Ú¯v&ܶ0å¼of&É o+õ3´ß4ÿuÆvãð;eyñ¶:e\_ÿ!7ªÌ ½5Mú¡úl6ÿ°z©öÆT¢Bmï\ÎwMF:óÐî\ÛÕ TòPÍC-õ4´í HF¬î^÷´ t»ñ¬nר¹òìw¾ß{Գк]¤g + m5mç¡¿ 0�G: endstream endobj 19 0 obj <>stream hÞÍnÜ8\_
o0dÿ± =XÉ1ØO{Ù?ìû¶J ÀÚ'ÒÓKÝÕM{dk½x3Å-ËlCÖlc ÖÕdèM<±¦}¦fXµiN¬¶1¦yN£ùàg³¹3Ítuµ0·Þ"ñXì'30i3À!ìà NÎ[.rѵm¶5É!¿>.l@wl 1pÆgH±Y ªC O$ô=ÚP!5CÙ:a(³> (ó' (;«(;] ({0å A9!(Ç Ê ÑÓÊ3qjB9a©M('}w#iÔò¢SÊVMoÒéÕ lhA·f²KÙ.úèÐ0)Ä-ú-Qö8 Éþ-2pR,¡ìtéÓÅI@.¶ tcA9è, Ayb õˤ? ÊIðIú³ ôÕÊt(/T@µÃ1ï²Ç¼cÌFwl #ìØ$áhh áÍ$Mu^ØðøèVL#eC|@ cPÔmmÓª'qÊøòÔòÄ9´t¢('jw¤« ] ¼8î\]èçO\_/òø!Ç+>tÜ»¯×ßÁä¶ýöøöÏÛ\_?Q£VF{l¨VÐYG÷Fõ::¢ÅTê¨VÑ9 䪣÷²èæ?nä-;Yèf\Urõ29ʤI-V&½LF$dgNÇ!|£¡gv\_÷u{rî³R¸©sUÉìer¾O/ø7×#P÷Cø:¨Î4"á˧Ū³×È«¼¦Ô5æYÆqßhÄ×þìB
;Û¤M»'}¯ pË|É,«JF/kîÓ-F ®ðuV# õ3~\_kÄÙm·<o4Îj±8o4ÎîÙ!|éíB
aÛGáFè6iQ¸Ö«¤îÎfRwÿ ¡tVIÙÉ$JÉU%µÉQ&¥Lj´2éòõm<¾|ç[BÛKÖßpPÖÿßc~ýûß?ßþø×'oÏÕçÆþeð/|-¼ÿ"R©HóøÆy z\(÷ À×·Îo¶ï\¬ïÛKÍço 'Þþó5zk§<ÇaûgÕ³ï{1¦²Y!7#¼zMþ'À�Ý Jn endstream endobj 20 0 obj <>stream hÞ252Q0P052U01R°±ÑwÖw4Ñ÷V04Ñ�åôÒL 03X?¸ 1ÏÎI¥!X¥.��ëË¢ endstream endobj 21 0 obj <> endobj 22 0 obj <>stream 2018-10-29T16:58:29-07:00 2018-10-29T16:58:34-07:00 2018-10-29T16:58:34-07:00 Adobe InDesign CS6 (Windows) uuid:bf88bce9-7f64-4335-95a0-ac8b83665ee7 xmp.did:C5950823B7206811871FF9B96FD29D29 xmp.id:3AA4A580D6DBE81195F4C9B811799A3B proof:pdf xmp.iid:39A4A580D6DBE81195F4C9B811799A3B xmp.did:39A4A580D6DBE81195F4C9B811799A3B xmp.did:C5950823B7206811871FF9B96FD29D29 default converted from application/x-indesign to application/pdf Adobe InDesign CS6 (Windows) / 2018-10-29T16:58:29-07:00 application/pdf Adobe PDF Library 10.0.1 False endstream endobj 23 0 obj <> endobj xref 0 526 0000000000 65535 f 0000124182 00000 n 0000124592 00000 n 0000138008 00000 n 0000138145 00000 n 0000141121 00000 n 0000141522 00000 n 0000141995 00000 n 0000144485 00000 n 0000144850 00000 n 0000145269 00000 n 0000148755 00000 n 0000148792 00000 n 0000148845 00000 n 0000149368 00000 n 0000149808 00000 n 0000152022 00000 n 0000152882 00000 n 0000154599 00000 n 0000156135 00000 n 0000157157 00000 n 0000157333 00000 n 0000157393 00000 n 0000160115 00000 n 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f 0000000000 65535 f trailer <<8A00803F664853419D78AF09B400B25B>]>> startxref 116 %%EOF~~~~~~~~ |
9906 | https://www.cecm.sfu.ca/~nbruin/nn2.pdf | Some ternary Diophantine equations of signature (n, n, 2) Nils Bruin1 Simon Fraser University. Email: bruin@member.ams.org Summary. In this article, we will determine the primitive integral solutions x, y, z to equations of the form xn + yn = Dz2 with n = 4, 5, 6, 7, 9, 11, 13, 17 and D ∈{2, 3, 5, 6, 10, 11, 13, 17}.
These equations form the small exponent cases of the equations considered by Bennett and Skinner in , where their modular techniques do not apply.
The computations necessary form a nice showcase of the arithmetic geometric functionality in the Magma computer algebra system. We will show how to construct curves, how to test curves for local solubility, how to analyse elliptic curves over number fields and how to use Chabauty-techniques to determine the rational points on a curve.
1 Introduction The following result is stated in the paper by Bennett and Skinner.
Theorem 1. If n ≥4 is an integer and D ∈{2, 3, 5, 6, 10, 11, 13, 17} then the equation xn + yn = Dz2 has no solutions in nonzero coprime integers (x, y, z) with, say, x > y, unless (n, D) = (4, 17) or (n, D, x, y, z) ∈{(5, 2, 3, −1, ±11), (5, 11, 3, 2, ±5)}.
In that paper the authors use techniques based on Galois representations on torsion subgroups of elliptic curves and modular forms to prove a large part of this theorem, but these methods do not apply to all combinations of 1 The research described in this paper is partly funded by NSERC and the Univer-sity of Sydney.
2 Nils Bruin n, D that occur in the statement and for each n ≤17, there are some values of D for which they refer to this paper rather than give a proof. Indeed, here we we will prove the following result.
Theorem 2. If n ∈{5, 6, . . . , 17} and D ∈{2, 3, 5, 6, 10, 11, 13, 17} then the equation xn + yn = Dz2 has no solutions in coprime nonzero integers except those arising from the identities 1n + 1n = 2 · 12, 35 −15 = 2 · 112 and 35 + 25 = 11 · 52.
We will also prove Proposition 1. The equation x4+y4 = Dz2 has no integral solutions for D ∈ {3, 5, 6, 10, 11, 13}. For D = 2, the only integral solutions with gcd(x, y, z) = 1 are (x, y, z) = (±1, ±1, ±1). For D = 17, it has infinitely many integral solutions with distinct values of x/y.
We will use the proofs to introduce the reader to some of the very powerful tools that Magma offers for solving arithmetic geometric questions. The article is laid out in the following way.
As an introduction we give an easy proof to Proposition 1. It shows the basic mechanisms that are available in Magma to define arithmetic geometric objects and answer questions about them. We try to point out that many questions can be formulated and answered using Magma in a language that is very close to the one that mathematicians are used to.
Next, we review some mathematical concepts and constructions that will prove indispensible in the rest of the paper. We recall a theorem from that translates questions like the one in Theorem 2 to questions about rational points on some algebraic curves.
In Section 4, we apply those results to x5 + y5 = Dz2 and, using Magma, obtain some curves that parametrise the primitive solutions to the equation under consideration. We then construct elliptic subcovers of those curves such that an application to them of the methods from yields the rational points on the original curves. We defer the actual application to Section 7.
We trust that after this demonstration of the problem solving capability of Magma, the reader will be interested in knowing some of the algorithms employed. In Section 5 we give a full account of the algorithms the author has implemented in Magma to test schemes for local solvability. A highlight is an algorithm that decides local solvability of hyperelliptic curves in time that is essentially independent of the size of the residue class field in the odd residue characteristic case.
In Section 6, we explain how 2-Selmer groups and 2-isogeny Selmer groups of elliptic curves over number fields can be computed, how they can be used to bound the free ranks of Mordell-Weil groups and how they can be used to find generators for Mordell-Weil groups. We also explain how one can do this in Magma using the implementation of the author, based on .
Some ternary Diophantine equations of signature (n, n, 2) 3 In Section 7, we explain how the techniques first introduced in can be applied to use elliptic curves over number fields to find the rational points on curves. We outline how one can use the implementation by the author to prove the result in Theorem 2 for n = 5.
In the last section, we give an outline of successfull strategies to solve the remaining cases from Theorem 2. For full details and a transcript of the Magma session that obtains all computational results in this paper, we refer the reader to the electronic resource .
2 Proof of Proposition 1 To get a taste for things to come, we first prove Proposition 1. It is very straightforward. First note that any solution to x4 + y4 = Dz2 corresponds to a rational point (u, v) = (x/y, Dz/y) on the curve D(u4 + 1) = v2.
We can simply ask Magma to compute for each desired value of D, whether this curve has any points over, say, Q2 (see Section 5.4).
_:=PolynomialRing(Rationals()); > Dset:={2,3,5,6,10,11,13,17}; > {D:D in Dset| IsLocallySolvable(HyperellipticCurve(D(x^4+1)),2)}; { 2, 17 } So just by testing local solvability at 2, we have already proved the lemma for all values of D except 2 and 17. Let’s first consider D = 2. Clearly, the curve 2u4 + 2 = v2 has a rational point (u, v) = (1, 2), so it is isomorphic to an elliptic curve E. The rational points of an elliptic curve form a finitely generated group. Magma can compute an upper bound on the free rank of that group (see Section 6) and as it turns out, it is 0.
C2:=HyperellipticCurve(2(x^4+1)); > p0:=C2![1,2]; > E,C2toE:=EllipticCurve(C2,p0); > RankBound(E); 0 > #TorsionSubgroup(E); 4 We find an upper bound of 0 on the free rank, so E(Q) consists entirely of torsion points, of which there are 4. Indeed, there are 4 obvious points: (u, v) = (1, 2), (−1, 2), (1, −2), (−1, −2).
and these all correspond to solutions with x = ±y.
For D = 17 we proceed similarly. We find the point (u, v) = (2, 17) on the curve 17u4 +17 = v2. This time we find an upper bound of 2 on the free rank.
4 Nils Bruin > C17:=HyperellipticCurve(17(x^4+1)); > p0:=C17![2,17]; > E,C17toE:=EllipticCurve(C17,p0); > RankBound(E); 2 In fact, Magma can find two independent points on E (note that the command MordellWeilGroup in principle could return a group of smaller rank, so one should always check that the rank of the returned group corresponds to the expected rank).
G,GtoE:=MordellWeilGroup(E); > G; Abelian Group isomorphic to Z/2 + Z/2 + Z + Z Defined on 4 generators Relations: 2G.1 = 0 2G.2 = 0 > [Inverse(C17toE)(GtoE(g)):g in OrderedGenerators(G)]; [ (-1 : 17 : 2), (-2 : -17 : 1), (13 : 697 : 2), (314 : 3097553 : 863) ] The last solution corresponds to the primitive solution: (2 · 157)4 + 8634 = 17 · (182209)2 and, using the group law on E, arbitrarily many can be constructed.
3 Construction of parametrising curves In this section, we recall a result from in an explicit form, occurring in , which relates integer solutions (x, y, z) of equations like xn + yn = Dzm with gcd(x, y, z) = 1 to rational points on some algebraic curves.
First we need some notation.
Let f(x, y) ∈Z[x, y] be a square-free homogeneous form of degree n and assume for simplicity that f(x, 1) is monic of degree n, i.e, f(x, y) = xn + y(· · · ). We construct the algebra A = Q[θ] = Q[x]/f(x, 1). This allow us to express f as a norm form f(x, y) = NA[x,y]/Qx,y.
Let S be a finite set of rational primes and let K be a number field. For a prime p of K we write p ∤S is p does not extend any prime in S to K.
Following , we define K(p, S) := {a ∈K∗: vp(a) ≡0 (mod p) for all primes p ∤S}/K∗p.
Following Magma’s terminology, we refer to this set as the (p, S)-Selmer group of K. It is a finite, effectively computable group. An algorithm for computing Some ternary Diophantine equations of signature (n, n, 2) 5 it is described in and the implementation and optizations in MAGMA are due to Fieker .
Since f is square-free, the algebra A is isomorphic to a direct product of number fields K1 × · · · × Kr. We generalise the notation above to A(p, S) := K1(p, S) × · · · × Kr(p, S) ⊂A∗/A∗p.
Furthermore, we will identify elements of A(p, S) with some set of represen-tatives in A∗.
Since {1, θ, . . . , θn−1} forms a Q[Z0, . . . , Zn−1]-basis of the vector space A[Z0, . . . , Zn−1], for any δ ∈A∗there are unique homogeneous forms Qi = Qδ,i ∈Q[Z0, . . . , Zn−1] of degree m such that Q0 + Q1θ + · · · + Qn−1θn−1 = δ(Z0 + Z1θ + · · · + Zn−1θn−1)m.
We define the projective curve Cδ := {Q2 = Q3 = · · · = Qn−1 = 0} ⊂Pn−1 and the map φδ : Cδ →P1 defined by φδ : (Z0 : · · · : Zd−1) 7→−Q0(Z0, . . . , Zn−1) Q1(Z0, . . . , Zn−1).
From [2, Theorem 3.1.1], it follows that Cδ is absolutely irreducible of genus 1 + mn−2( 1 2n(m −1) −m) and that φδ is a Galois cover with Galois-group (Z/mZ)n−1, ramified exactly at {(x : y) ∈P1(Q) : f(x, y) = 0}.
For given nonzero integer D and m ≥1, we consider the equation f(x, y) = Dzm.
Let S be a finite set of primes containing the prime divisors of DDisc(f).
∆:= {δ ∈A(m, S) : NA/Q(δ)/D ∈Q∗m}.
A consequence of [2, Lemma 3.1.2] is Theorem 3. Let f, D, m, θ, A, ∆be defined as above. Then n (x : y) : x, y, z ∈Z, f(x, y) = Dzm, gcd(x, y, z) = 1 o ⊂ [ δ∈∆ φδ(Cδ(Q))} One can easily recover (x, y, z) from (x : y) in the following way. Let (x0 : y0) ∈P1(Q). If (x, y, z) is a solution with (x : y) = (x0 : y0), then there is a λ ∈Q∗such that x = λx0, y = λy0, z = m r λnf(x0, y0) D .
6 Nils Bruin Given x0, y0, it is straightforward to determine for which values of λ the above system has solutions with gcd(x, y, z) = 1 and what those solutions are.
The strategy for proving Theorem 2 is to determine the relevant curve Cδ for each of the combinations (n, D) and to find the rational points on Cδ. In some cases we get away with only constructing a subcover of Cδ/P1.
In the particular, when m divides n, then the weighted projective equiva-lence classes of solutions to f(x, y) = Dzm are in bijection with the rational points of the curve given by the weighted projective model C′ : f(x, y) = Dzm, where (x, y, z) have weights (1, 1, n/m). In the special case that m = 2, we see that C′ is a double cover of a projective line. The curves Cδ are twists of the unramified cover of C′ obtained by embedding C′ in its jacobian Jac(C′) and taking the pullback along the multiplication-by-2 map on Jac(C′). These properties are explained and exploited in and in a trivial way in Section 8.
In the particular case that m = 2 and n = 4, we recover the multiplication-by-2 covers of curves of genus 1 that play a role in 2-descents and 4-descents. If f has a rational root, then C′ is isomorphic to its jacobian and the Cδ are the homogeneous spaces that play a role in 2-descents as described in Section 6.
If f does not have a rational root, then C′ can still be expressed as a Z/2Z × Z/2Z cover of its jacobian. The Cδ are then homogeneous spaces associated to a 4-descent on the jacobian. See and .
4 The equation x5 + y5 = Dz2 We begin putting the construction from Section 3 into Magma. In our case f = x5 + y5. We model this by defining a univariate polynomial in Magma.
This allows us to construct the algebra straight away.
:=PolynomialRing(Rationals()); > f:=x^5+1; > A:=quo; Next we construct the rings A[Z0, . . . , Z4], Q[Z0, . . . , Z4] and the correspond-ing projective space. Note that, while mathematically Q[θ][Z0, . . . , Z4] ≃Q[Z0, . . . , Z4][θ] in a canonical way, this is not the case in a computer algebra system. We first construct the left hand side (PA) and then obtain the right hand side (AP) using SwapExtension. We also get the appropriate isomorphism swap.
We then extract Q[Z0, . . . , Z4] as the base ring of AP.
PA:=PolynomialRing(A,5); > AP,swap:=SwapExtension(PA); > :=BaseRing(AP); > P4:=Proj(BaseRing(AP)); > P1:=ProjectiveSpace(Rationals(),1); Some ternary Diophantine equations of signature (n, n, 2) 7 Given δ ∈A, it is now straightforward to construct Cδ and φδ. We present a Magma routine that takes δ as input and returns both Cδ and φδ. Note that Coefficients on an element of AP returns a sequence of coefficients with respect to the power basis [1, θ, . . . , θ4], i.e., the Qi for us.
function Cdelta(delta) function> g:=delta(&+[PA.itheta^(i-1): i in [1..5]])^2; function> Q:=Coefficients(swap(g)); function> Crv:=Scheme(P4,[Q,Q,Q]); function> phi:=mapP1|[Q,-Q]>; function> return Crv,phi; function> end function; Given D, we can compute the set ∆as well. For that, we need to represent Q as a number field Q and A as an algebra over Q. We also compute the decom-position of A = Q × K, where K = Q(ζ), the field generated by a primitive 5th root of unity. By giving an explicit representation to AbsoluteAlgebra, we make sure the system uses that representation.
Q:=NumberField(x-1:DoLinearExtension); > OQ:=IntegerRing(Q); > Qx:=PolynomialRing(Q); > AQ:=quo; > AQtoA:=homA|[theta]>; > K:=NumberField(x^4 - x^3 + x^2 - x + 1); > OK:=IntegerRing(K); > Aa,toAa:=AbsoluteAlgebra(AQ:Fields:={Q,K}); We can now compute ∆in the following way. We take S to be the set of primes that divide DDisc(f). We compute the subgroup of A(2, S) that has square norm and we translate it over the class of D in A(2, S).
function DeltaForD(D) function> S:=Support(DDiscriminant(f)OQ); function> slmA,slmAmap:=pSelmerGroup(AQ,2,S); function> slmQ,slmQmap:=pSelmerGroup(2,S); function> slmNorm:=mapslmQ|a:->slmQmap(Norm(a@@slmAmap))>; function> slmSquareNorm:=Kernel(homslmQ| function> [slmNorm(a):a in OrderedGenerators(slmA)]>); function> classD:=slmAmap(D); function> return {AQtoA( (d-classD)@@slmAmap):d in slmSquareNorm}; function> end function; We gather all δs that are relevant together (remember we can always recover the corresponding D from NA/Q(δ)) and throw out any for which Cδ is not locally solvable at 2, 5 or 11 (other primes turn out to make no further con-tributions).
8 Nils Bruin > Dset:={2,3,5,6,10,11,13,17}; > BigDelta:=&join{DeltaForD(D):D in Dset}; > Delta:=BigDelta; > Delta:={delta:delta in Delta| IsLocallySolvable(Cdelta(delta),2: > AssumeIrreducible,AssumeNonsingular)}; > Delta:={delta:delta in Delta| IsLocallySolvable(Cdelta(delta),5: > AssumeIrreducible,AssumeNonsingular)}; > Delta:={delta:delta in Delta| IsLocallySolvable(Cdelta(delta),11: > AssumeIrreducible,AssumeNonsingular)}; This eliminates already 4/5th of the parametrising curves. Now we construct a subcover, derived from the ring homomorphism m1 : A →Q given by θ 7→−1.
From x −θy = δc2 0, it follows that f(x, y) = NA/Q(δ)NA/Q(c0)2. Hence, x4 −x3y + x2y2 −xy3 + y4 = f(x, y)/m1(x −θy) = N(δ)/m1(δ)N(c0)2.
Putting d = N(δ)/m1(δ), it follows that Cδ covers Ed : u4 −u3 + u2 −u + 1 = dv2.
We compute which curves occur.
m1:=homRationals()|-1>; > {PowerFreePart(Norm(delta)/m1(delta),2):delta in Delta}; { 1, 5, 55 } Unfortunately, Ed has infinitely many rational points for d = 1, 55. For d = 5 we do get some useful information.
E5:=HyperellipticCurve(5(x^4-x^3+x^2-x+1)); > p0:=E5![-1,5]; > ell:=EllipticCurve(E5,p0); > RankBound(ell); 0 > #TorsionSubgroup(ell); 2 We see that E5 only has two rational points (−1, ±5). It is straightfor-ward to check that these correspond to the obvious solutions (x, y, z) = (1, −1, 0), (−1, 1, 0) for x5 + y5 = Dz2. If we take heed of these solutions, we can discard any δ for which Cδ covers E5.
Delta:={delta:delta in Delta|PowerFreePart(Norm(delta)/m1(delta),2) ne 5}; > #Delta; 16 For these remaining values, we use the same idea as above, but now we use the map m2 : A →K given by θ 7→ζ. We define d = N(δ)/m2(δ) and we obtain the following subcover of Cδ/P1.
Some ternary Diophantine equations of signature (n, n, 2) 9 Ed : u4 + ζu3 + ζ2u2 + ζ3u + ζ4 = dv2 where the cover φδ : Cδ →P1 induces the cover u : Ed →P1. Since the value of d only matters up to squares, we take unique representatives by mapping through K(2, S′) for an appropriate S′.
m2:=homK|zeta>; > slmK,slmKmap:=pSelmerGroup(2,Support(235111317OK)); > dset:={K|(slmKmap(Norm(delta)/m2(delta)))@@slmKmap:delta in Delta}; > dset; { 2zeta^3 - 2zeta^2 - 2, 1, 15zeta^3 - 5zeta^2 + 8zeta - 17, -3zeta^3 - 7zeta^2 - 8zeta - 9, -2zeta^2 - 2 } For our subsequent operations, it is beneficial to compute a Weierstrass model of Ed using the point (u, v) = (−1, 0). We express the function u in the coordinates of that model. We obtain Ed : Y 2 = X3 −d(3ζ3 + ζ −1)X2 −d2(ζ2 + ζ + 1)X and u = −X + d(ζ3 −1) X −d(ζ3 + ζ) .
Using Magma, one can verify this using a few lines of code. Notice that the elliptic curve is represented as a projective curve.
Kd:=RationalFunctionField(K); > KdX:=PolynomialRing(Kd); > FEd1:=(X^4+zetaX^3+zeta^2X^2+zeta^3X+zeta^4)/d; > Ed1:=HyperellipticCurve(FEd1); > Ed2,toEd2:=EllipticCurve(Ed1,Ed1![-1,0]); > umap:=mapP1|[Ed1.1,Ed1.3]>; > FEd:=X^3+(-3zeta^3-zeta+1)dX^2+(-zeta^2-zeta-1)d^2X; > Ed:=EllipticCurve(FEd); > bl,toEd:=IsIsomorphic(Ed2,Ed); > u:=Expand(Inverse(toEd2toEd)umap); > u:Minimal; (xE : yE : zE) -> ((-2zeta^3 + 3zeta^2 - 3zeta + 2)/d^2xE + (-zeta^3 - 2zeta + 2)/d (2zeta^3 - 3zeta^2 + 3zeta - 2)/d^2xE + (2zeta^2 - zeta + 2)/dzE) We have the following diagram of covers.
10 Nils Bruin Cδ π ( R R R R R R φδ Ed u wnnnnnn P1 /Q Clearly, φδ(Cδ(Q)) ⊂u(Ed(K)) ∩P1(Q).
We can now complete the proof of Theorem 2 for n = 5 by computing the right hand side of the above inclusion. By techniques we will explain in Section 7, we find the following table.
d u(Ed(K)) ∩P1(Q) 1 {−1, 0, ∞} −2ζ2 −2 {−1, 1} 2ζ3 −2ζ2 −2 {−3, −1, −1/3} −3ζ3 −7ζ2 −8ζ −9 {−1, 3/2} 15ζ3 −5ζ2 + 8ζ −17 {−1, 2/3} It is straightforward to check that all of these values for x/y lead to solutions with xyz = 0 or solutions that are mentioned in Theorem 2.
5 Deciding local solvability As we have seen in Sections 2 and 4, the first step in solving arithmetic geo-metric questions often involves deciding if, for a projective variety X over a number field K, the set X(Kp) is empty for some prime p. In this section, we outline several algorithms that have been implemented in Magma by the au-thor to test local solvability. They include tools for determining the Kp-points of separated 0-dimensional schemes, Kp-solvability of complete intersections, Kp-solvability of smooth projective curves, given by a possibly singular planar model and Kp-solvability of hyperelliptic curves.
For the rest of this section, O will be a complete local ring of characteristic 0 with maximal ideal p and finite residue field O/p. We write π for a generator of p and L for the field of fractions of O. We use ν : L∗→Z to denote the normalised valuation, i.e., ν(πe) = e and use the customary extension ν(0) = ∞.
For any object f (vector, matrix, polynomial) defined over O, we write f for the corresponding reduced object over O/π. We also write ν(f) for the minimum of ν(c), where c runs through the coefficients of f.
5.1 Determining X(L) for a reduced 0-dimensional projective scheme Let Pn be n-dimensional projective space over K with variables (X0 : . . . : Xn) and let X be a reduced 0-dimensional projective scheme, defined by Some ternary Diophantine equations of signature (n, n, 2) 11 f1 = · · · = fm = 0, where fi ∈K[X0, . . . , Xm]. Without loss of generality, we can assume fi ∈ O[X0, . . . , Xm].
Note that any point in Pn(L) has a representative (x0 : · · · : xn) such that for some 0 ≤N ≤n, we have (x1, . . . , xN−1, xN, xN+1, . . . , xn) ∈O × · · · × O × {1} × p × · · · × p Hence, it is sufficient to solve the problem of finding O-integral points on an affine separated 0-dimensional scheme Y , given by equations fi ∈O[y] = O[y1, . . . , yn].
In principle, one could solve the problem in the same way as one does for exactly representable fields like Q, Fq and number fields, by using resultants and univariate factorisation. In practice, however, the objects considered are not exactly represented and it is almost impossible to make such algorithms numerically stable. Therefore, we will present an algorithm here that simply builds solutions one π-adic digit at the time, until the solution is verifyably separated and Hensel liftable. We simply reduce the system of equations to O/p, determine the solutions over that finite field and interpret what these solutions mean over O.
The first step is to pick gi ∈O[y] such that (g1, . . . , gm)O[y] is as close as possible to I = (f1, . . . , fm)K[y] ∩O[y].
Let Mf = ∂ ∂yj fi i,j ∈Om×n Algorithm Saturate(f1, . . . , fm): 1. REPEAT 2.
Let T ∈GLm(O) such that T(Mf(0, . . . , 0)) is in row echelon form.
3.
(f1, . . . , fm)t ←T(f1, . . . , fm)t.
4.
FOR i ∈{1, . . . , m}: 5.
fi ←fi/πν(fi) 6. UNTIL in step 5 no fi was changed.
7. RETURN (f1, . . . , fm) This algorithm does not always find generators of I. However, if (0, . . . , 0) is sufficiently close to a non-singular point of Y , then for all f ∈I, the minimal valuation of the coefficients of f is attained by the coefficient of a linear term.
It is clear that in this situation, the algorithm will find gi that generate I.
12 Nils Bruin In practice, the coefficients of fi are only given up to finite precision. Hence, in step 5, it might happen that a coefficient has no precision left. In that case, an error should be generated.
It is now straightforward to determine the integer points of Y up to some precision bound r: Algorithm IntegerPoints(r, f1, . . . , fm): 1. (f1, . . . , fm) ←Saturate(f1, . . . , fm).
2. Let V ⊂On be a set of representatives of the solutions of f1 = · · · = fm = 0 in O/p.
3. Let V0 ⊂V represent the points over O/p with 0-dimensional tangent space.
4. W ←{}; V1 = V \ V0 5. FOR v ∈V0: 6.
v ←Hensel lift of v to precision r using a suitable subset {fi1, . . . , fin} 7.
IF for all i we have ν(fi(v)) ≥r THEN 8.
Add v to W 9.
ELSE 10.
Discard v 11. FOR v ∈V1: 12.
gi ←fi(v1 + πy1, . . . , vn + πyn) for i = 1, . . . , m 13.
FOR w ∈IntegerPoints(r −1, g1, . . . , gm): 14.
Add (v1 + πw1, . . . , vn + πwn) to W 15. RETURN W Obviously, if Y has some higher multiplicity O-point, then succesive ap-proximations to it will be in V1 and never in V0. The algorithm recurses infinitely. Therefore, if IntegerPoints gets called with r ≤0, an error should be generated, indicating that the scheme Y has points that do not separate below the requested precision level. An alternative is to return such points as non-separating approximations to possible solutions and return them. These give the user neighbourhoods that could not be resolved at the requested precision.
It should also be clear, and this is an essential problem, that step 7 only tests that v is approximately on Y . While v can be uniquely lifted to arbitrary precision r′ using {fi1, . . . , fin} (provided the fi themselves are given to suf-ficient precision), it may be that this lift does not satisfy the other equations to precision r′, but that it did to precision r. Obviously, if Y is presented as a complete intersection and n = m, then this problem will not arise. Otherwise, the best one can do is to assume that the user supplies a sufficiently high r to begin with.
5.2 Determining solvability of complete intersections Let X ⊂Pn be a complete intersection defined over a number field K of dimension d. We assume that X is equidimensional, which means that its Some ternary Diophantine equations of signature (n, n, 2) 13 maximal components are all of dimension d. This condition is certainly met if X is irreducible. Let X′ denote the reduced singular subscheme.
Let L be the completion of K at some finite prime. First we assume X′(L) is empty. In this case, our task is to decide if X(L) contains any non-singular points. We follow the same approach as in Section 5.1 and note that, again, it is sufficient to solve the problem for integral points on affine complete intersections Y : Algorithm HasNSIntegralPoints(f1, . . . , fn−d): 1. (f1, . . . , fn−d) ←Saturate(f1, . . . , fn−d).
2. Let V ⊂On be a set of representatives of the solutions of f1 = · · · = fn−d = 0 in O/p.
3. IF any of the points in V represent a point over O/p with a d-dimensional tangent space: 4.
RETURN true 5. FOR v ∈V : 6.
gi ←fi(v1 + πy1, . . . , vn + πyn) for i = 1, . . . , n −d 7.
IF HasNSIntegralPoints(g1, . . . , gn−d): 8.
RETURN true 9. RETURN false Since we assume that Y is a complete intersection, the problem of step 7 in Section 5.1 does not arise.
The strategy to determine if X(L) is empty is now straightforward: Algorithm CIHasPoints(X, L): 1. if X′(L) is nonempty, then X(L) is nonempty 2. otherwise, use HasNSIntegralPoints on the affine patches of X to decide if X(L) is nonempty.
Obviously, step 1 can only be decided if X′ is of one of the types we have considered before, i.e., X′ is empty, dim X′ = 0 or X′ is a complete intersec-tion. One may be able to show that X′(L) is empty by showing that some complete intersection containing X′ has no points over L, but the converse does not hold.
In order to compute X′, it is essential that X is represented exactly over some field allowing exact arithmetic, because only then do Groebner basis algorithms allow for the computation of the radical of an ideal.
5.3 Solvability of smooth curves In this section we consider a reduced scheme X ⊂P2 given by a single equation f(x, y, z) = 0. We present an algorithm to determine the local solvability of the desingularisation ˜ X of X. As in Section 5.2, we note that it is sufficient to solve the problem for integral points on affine curves Y : f(x, y) = 0 and to a large extent, the algorithm is the same. The only difference occurs 14 Nils Bruin in how singular points in Y (O) are treated. Instead of considering them as rational points, we blow up Y at the singularity and remove the exceptional component. We then look for L-valued points on the resulting scheme.
First, let us study how to blow up an affine curve in (0, 0). Therefore, let f ∈O[x, y] describe a curve in A2 with a singularity at (0, 0) and no other O-valued singularities. We resolve the singularity by blowing up A2 at (0, 0).
This means we take the inverse image under β : {xv = yu} ⊂P2 × A2 → A2 (u : v; x, y) 7→(x, y) Note that any point (u : v; x, y) that has an image (x, y) ∈A2(O) under β has a representative of one of the forms (u : 1; x, y) or (1 : πv; x, y) with u, v ∈O.
Hence, any integral point on β−1Y is covered by an integral point on one of β−1 1 Y or β−1 2 Y , where β1 : A2 → A2, (u, y) →(uy, y) β2 : A2 → A2, (v, x) →(x, πxv) To remove the exceptional component from β−1 1 Y : f(uy, y) = 0, compute Y1 : f1(u, y) = f(uy, y)/u(highest possible power).
The curve Y1 may have new singularities, but since Y1 is isomorphic to Y outside u = 0, any integral-valued singularities will have u = 0. The singular points of Y1 can be easily described as Y ′ 1(O) = (u, 0) : u ∈O and f1(u, 0) = ∂f1 ∂u (u, 0) = ∂f1 ∂y (u, 0) = 0 and can be computed using univariate root finding for polynomials over O.
Of course, in practice, an expression like “= 0” should be interpreted as “is indistinguishable from 0 at the given precision”. For β−1 2 Y we can proceed similarly.
Given a list S of integral-valued singularities, one can check the desingu-larisation of Y : f(x, y) = 0 for integral points: Algorithm HasSmoothIntegralPoints(f, S): 1. f ←f/πν(f) 2. IF S = {(x0, y0)}: 3.
f ←f(x + x0, y + y0) 4.
f1 ←f(uy, y)/u(highest possible power) 5.
S1 ← n (u, 0) : u ∈O and f1(u, 0) = ∂f1 ∂u (u, 0) = ∂f1 ∂y (u, 0) = 0 o 6.
IF HasSmoothIntegralPoints(f1, S1): RETURN true 7.
f2 ←f(x, πxv)/v(highest possible power) Some ternary Diophantine equations of signature (n, n, 2) 15 8.
S2 ← n (v, 0) : v ∈O and f2(v, 0) = ∂f2 ∂v (v, 0) = ∂f2 ∂x (v, 0) = 0 o 9.
IF HasSmoothIntegralPoints(f2, S2): RETURN true 10. ELSE 11.
Let V ⊂O2 be a set of representatives of the solutions of f = 0 in O/p.
12.
IF a point in V represents a nonsingular point over O/p: RETURN true 13.
FOR (x0, y0) ∈V : 14.
g ←f(x0 + πx, y0 + πy) 15.
S′ ←{((x1 −x0)/π, (y1 −y0)/π) : (x1, y1) ∈S} 16.
Remove any non-integral entries from S′ 17.
IF HasSmoothIntegralPoints(g, S′): RETURN true 18. RETURN false To determine local solvability of the desingularisation of a reduced projec-tive plane curve X ⊂P2, one can determine the reduced singular subscheme X′ of X, find X′(L) using Section 5.1 and apply HasSmoothIntegralPoints to each affine patch of X, using X′(L) to initialise S.
5.4 Solvability of hyperelliptic curves In this section, we adopt Magma’s terminology and understand hyperelliptic curve to mean nonsingular double cover of P1. Some geometric hyperelliptic curves can be represented in this category (but not the ones that have a twisted P1 as a canonical model). Conics and some curves of genus 1 also fit in this category.
We represent such curves as a nonsingular curve in weighted projective space P(1,d,1) with coordinates (x, y, z) and a model of the form C : y2 + h(x, z)y = f(x, z).
Over fields of odd characteristic we can complete the square and without loss of generality, we can assume h = 0. In this case, the nonsingularity of C means that f(x, z) is a square-free form of degree 2d and a simple application of Riemann-Hurwitz shows that C is of genus d −1.
Of course, to decide if a hyperelliptic curve has points over L, one could cover it with two non-singular affine patches and use Section 5.3. One can also use [2, Appendix A.2], which is slightly more efficient. Both these algorithms are essentially polynomial in #O/p, though. We can do better if O/p is of odd characteristic and satisfies (#O/p) −2(d −1) p #O/p > 0.
We generalise an algorithm that is presented for d = 2 in and . It is based on the fact that a curve defined over a finite field of large cardinality compared to the genera of the components, must be very singular not to have any nonsingular rational points.
16 Nils Bruin Since multiplying f with an even power of π does not change the local solvability of y2 = f(x, z), we can assume f ∈O[x, z] with 0 ≤ν(f) ≤1.
Note that if ν(f) = 0, any point (x : y : z) ∈P(1,d,1)(O/p) satisfying y2 = f(x, z) with y ̸= 0 or (x : z) a zero of f of multiplicity 1 is a nonsingular point on C and hence Hensel-lifts to a point in C(L). If we can show such a point exists, then C(L) is not empty. We distinguish the following cases.
1. ν(f) = 1. If x, z ∈O such that f(x, z) is a square, then in particular, ν(f(x, z)) ≡0 mod 2. Hence, (x : z) must be a root of f/π in O/p. For any such root we take a representative (x0, z0) ∈sO2 and we test y2 = f(x0+πx, z0+πz)/π2 for local solvability. If any of those cases is solvable, then so is the original equation. If none is, or if no roots are avaliable, then y2 = f(x, z) has no solutions.
2. ν(f) = 0 and f = α(g(x, z))2 with α a non-square in O/p. If x, z ∈O such that f(x, z) is a square, then g(x, z) = 0. Hence we take representatives (x0, z0) ∈O2 for the roots of g and test y2 = f(x0 + πx, z0 + πz) for local solvability. If any of those cases is solvable, then so is the original equation. If none is, or if no roots are avaliable, then y2 = f(x, z) has no solutions.
3. ν(f) = 0 and f = α(g(x, z))2 with α a nonzero square in O/p. We take (x0, z0) ∈O to represent a non-root of g in P1(O/p). Note that g has at most d roots, while #P1(O/p) = #O/p + 1 > 2(d −1) points, so this is easy. Then f(x0, z0) is a square, because it represents a non-zero square in O/p. Therefore, the original equation is solvable.
4. In all other cases, f = g1(x, z)(g2(x, z))2, where g1 is square-free and deg(g1) + 2 deg(g2) = 2d. The curve D : y2 1 = g1(x, z) is a hyperelliptic curve over O/p of genus (deg(g1) −2)/2 and hence, by the Hasse-Weil bounds, has at least (#O/p) −2(d −1) p #O/p + (2 deg g2 + 2) p #O/p points. It follows that D must have points (x, y1, z) with g2(x, z) ̸= 0. Since the higher multiplicity roots of f are exactly the roots of g2, it follows that y2 = f(x, z) has a non-singular point, which is Hensel-liftable. It follows that C(L) is non-empty.
These cases lead directly to a recursive algorithm, where the most difficult operation is factorisation of univariate polynomials of degree at most 2d over a finite field. The branching degree of the algorithm is bounded by 2d as well and not (as is the algorithm in Section 5.3), essentially by #O/p.
Some ternary Diophantine equations of signature (n, n, 2) 17 6 Mordell-Weil groups of Elliptic curves The Mordell-Weil group of an abelian variety A over a number field K is the set of K-rational points A(K). The abelian variety structure of A induces a group structure on A(K). A celebrated theorem of Weil, which for elliptic curves over Q was already proved by Mordell, states that A(K) is a finitely generated commutative group. Actually determining A(K), even in the case where A is an elliptic curve and K = Q, is still more an art than a science.
However, even an artist works better if he has proper tools available. In this chapter, we introduce the tools that Magma offers to determine Mordell-Weil groups of elliptic curves over number fields. The Magma implementation is based on .
First, we review some of the fundamental definitions connected to the subject. We do not give much detail, since many other excellent descriptions already exist (see for instance ). Computational concerns that arise specif-ically when applying the methods outlined here to elliptic curves over number fields are addressed in .
Let E be an elliptic curve over a number field K. In order to bound the free rank of E(K), we bound the size of E(K)/2E(K). For this, we use the 2-Selmer group of E over K. From the exact Galois-cohomology sequence 0 →E(K)/2E(K) →H1(K, E) →H1(K, E) we derive a set that approximates the image of E(K)/2E(K) in H1(K, E) everywhere locally. We define S(2)(E/K) to be the intersection of the kernels of H1(K, E) →H1(Kp, E) for all primes p of K: 0 →S(2)(E/K) →H1(K, E) → Y p H1(Kp, E).
Clearly, S(2)(E/K) provides a sharp bound, unless H1(K, E) maps to any cocycle in H1(K, E) that trivialises under all restrictions Gal(Kp) ⊂Gal(K).
The group consisting of such cocycles is called the Shafarevich-Tate group III(E/K) and we have the exact sequence 0 →E(K)/2E(K) →S(2)(E/K) →III(E/K) →0.
The group S(2)(E/K), as a Galois-module, can be represented in the fol-lowing way (see ). For an elliptic curve E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6, we define the following algebra: A[θ] = K[X]/(X3 + a2X2 + a4X + a6 + (a1X + a3)2/4).
The Galois-module H1(K, E) can be identified with the subgroup of A∗/A∗2 consisting of the elements of square norm and for some suitable, effectively 18 Nils Bruin computable, set S of primes of K, we have S(2)(E/K) ⊂A(2, S). The map µ : E(K) →S(2)(E/K) is induced by (x, y) 7→x −θ where x −θ ∈A∗.
Magma computes S(2)(E/K) by computing the local images of E(Kp) →A∗/A∗2 ⊗Kp and computing the elements from A(2, S) of square norm that land in these local images. Wherever possible, elements of A(2, S) are left in product rep-resentation, to avoid coefficient blowup.
As an example, we compute S(2)(Ed/K) for d = 2ζ3 −2ζ2 −2, as defined in Section 4.
_:=PolynomialRing(Rationals()); > K:=NumberField(x^4-x^3+x^2-x+1); > OK:=IntegerRing(K); > d:=2zeta^3-2zeta^2-2; > E:=EllipticCurve([0,(-3zeta^3-zeta+1)d,0,(-zeta^2-zeta-1)d^2,0]); > two:=MultiplicationByMMap(E,2); > mu,tor:=IsogenyMu(two); > S2E,toS2E:=SelmerGroup(two);S2E; Abelian Group isomorphic to Z/2 + Z/2 + Z/2 + Z/2 Defined on 4 generators in supergroup: S2E.1 = $.1 + $.2 + $.6 + $.7 + $.8 + $.9 S2E.2 = $.2 + $.4 + $.7 + $.8 S2E.3 = $.1 + $.2 + $.5 + $.7 S2E.4 = $.3 + $.9 Relations: 2S2E.1 = 0 2S2E.2 = 0 2S2E.3 = 0 2S2E.4 = 0 So we see that E(K)/2E(K) ⊂(Z/2Z)4. Part of this corresponds to the image of the torsion subgroup of E.
Etors,EtorsMap:=TorsionSubgroup(E); > sub; Abelian Group isomorphic to Z/2 + Z/2 Defined on 2 generators in supergroup S2E: $.1 = S2E.3 + S2E.4 $.2 = S2E.1 + S2E.4 Relations: 2$.1 = 0 2$.2 = 0 Mapping from: Abelian Group isomorphic to Z/2 + Z/2 Defined on 2 generators in supergroup S2E: $.1 = S2E.3 + S2E.4 Some ternary Diophantine equations of signature (n, n, 2) 19 $.2 = S2E.1 + S2E.4 Relations: 2$.1 = 0 2$.2 = 0 to GrpAb: S2E We conclude that the free rank of E(K) is at most 2. We look for rational points on E, up to some tiny bound and we see that the found points already generate E(K)/2E(K).
V:=MyRationalPoints(E,5); > sub eq S2E; true We then select some minimal subset of V that generates E(K)/2E(K) and construct a group homomorphism from an abstract abelian group into G.
gs:=[E![0,0], > E![-2zeta^3 - 2zeta + 2,0], > E![-2zeta^3,-4zeta^2], > E![-2zeta^3 - 4zeta + 4,-4zeta^3 + 4zeta]]; > assert S2E eq sub; > G:=AbelianGroup([2,2,0,0]); > mwmap:=mapE|g:->&+[c[i]gs[i]:i in [1..#gs]] where c:=Eltseq(g)>; In fact, we could have left this all to the system and just executed: > success,G,mwmap:=PseudoMordellWeilGroup(E); > assert success; Here, it is of the utmost importance to check that success is true. Only then is there a guarantee that the returned group is of finite (odd) index in E(K).
If the value false is returned, then only a subgroup is returned that will itself be 2-saturated in E(K) (meaning that, if 2P ∈G and P ∈E(K) then P ∈G as well), but need not be of finite index.
In fact, the computation done by PseudoMordellWeilGroup is not com-pletely equivalent to the computation we did above. By default, if possible, PseudoMordellWeilGroup uses a 2-isogeny descent (see ). For any non-trivial element of E2, there is an associated 2-isogeny φ : E →E′, together with a dual isogeny ˆ φ : E′ →E, such that ˆ φ ◦φ = 2|E. In complete analogy to the 2-Selmer group, we define the φ-Selmer group by considering the exact sequence 0 →E′(K)/φE(K) →H1(K, E[φ]) →H1(K, E) and we define S(φ)(E/K) by insisting on exactness of 0 →S(φ)(E/K) →H1(K, E[φ]) → Y p H1(Kp, E).
20 Nils Bruin From the exact sequence 0 →Eφ →E2 →E′ˆ φ → E′(K)/φE(K) →E(K)/2E(K) →E(K)/ˆ φE′(K) →0 it follows that 4#E(K)/2E(K) = #E′(K)/φE(K) · #E(K)/ˆ φE′(K) · #E2.
Therefore, we can use φ-Selmer groups to bound the free rank of E(K) as well. One can compute φ-Selmer groups in the same way as 2-Selmer groups.
phi:=TwoIsogeny(E![0,0]); > Sphi,toSphi:=SelmerGroup(phi); > phihat:=DualIsogeny(phi); > Sphihat,toSphihat:=SelmerGroup(phihat); > 4#S2E, #Sphi, #Sphihat, #TwoTorsionSubgroup(E); 64 2 8 4 Apart from providing an upper bound on the rank of E(K), Selmer groups also contain information about possible generators of E(K). To access this information, it is useful to interpret S(2)(E/K) ⊂H1(K, E) as a set of twists of the cover E 2 →E. The second return value of IsogenyMu gives a map that computes such a cover from an element of H1(K, E). The covering space is represented as an intersection X of two quadrics in P3, with a map φ : X →E. If the cover represents an element from S(2)(E/K), however, one can construct a model of X of the form C : v2 = f0u4 + · · · + f4. A call to Quartic realises this.
delta:=S2E.2; > psi:=tor(delta@@toS2E); > XX:=Domain(psi); > C,CtoXX:=Quartic(XX); One can then search for points on C, which can be mapped back to E.
V:=MyRationalPoints(C,10); > assert #V gt 0; > P:=psi(CtoXX(Rep(V)));P; (-zeta^3 - 4zeta^2 - 2zeta - 2 : -14zeta^3 + zeta^2 - 6zeta + 10 : 1) > assert delta eq toS2E(mu(P)); Note, however, that it is a rarity for it to make sense to search for points on C as computed. The model computed for C generallly does not have particu-lary small coefficients and there is no reason to expect that the point we are looking for will be easier to find on C than on E. Over Q, a rather satisfactory solution to this problem has been found in the form of a proper minimization and reduction theory , . For other number fields, a satisfactory theory Some ternary Diophantine equations of signature (n, n, 2) 21 is woefully lacking and Magma leaves it to the art and ingenuity of the user to find a suitable model from the returned one.
The same functionality is available for 2-isogenies as well. Here, the cover corresponding to an element in the Selmer group naturally has a model of the form C : v2 = f0u4 + f2u2 + f4, and therefore it does make sense to look for rational points on the covering curve. Therefore, the routine PseudoMordellWeilGroup uses the following default strategy: 1. If a 2-isogeny is available, this is chosen as isogeny φ, Otherwise full multiplication-by-2 is used as φ.
2. The φ-Selmer group is computed and, if φ ̸= 2, then also the ˆ φ-Selmer group is computed.
3. The image of the torsion subgroup is determined in the computed Selmer groups.
4. The elliptic curve is searched for rational points up to a preset bound and, if relevant, also the 2-isogenous curve is searched. If the found points already generate the Selmer group(s), we are done.
5. Otherwise, if φ is a 2-isogeny or if the elliptic curve is defined over Q, the covers corresponding to elements of the Selmer group that are not repre-sented by rational points are constructed (and, if reduction is available, reduced) and searched for points.
6. If this still leaves some elements of the Selmer group(s) not corresponding to found rational points, then false is returned, together with the group generated by the found points. Otherwise, true is returned.
One can override the default choice of isogeny and whether or not homoge-neous spaces should be used for searching for rational points.
If III(E/K) is nontrivial, then obviously neither a 2-descent nor a 2-isogeny descent will provide a sharp bound on E(K)/2E(K). In this situ-ation, a 4-descent may give more information ( and ). For K = Q, Tom Womack has implemented routines to perform such a computation in MAGMA. Another option consists of using the Cassels-Tate pairing to obtain more information (see ).
Alternatively, one may use visualisation (see ) to obtain more informa-tion. See for an explicit approach using MAGMA.
7 Chabauty methods using elliptic curves In this section, we show how, given an elliptic curve E over a number field K and a map u : E →P1, one can try to determine {p ∈E(K) : u(p) ∈ P1(Q)}. The method is an adaptation of Chabauty’s partial proof of Mordell’s conjecture and is described in and . A similar method applied to bielliptic genus 2 curves is described in . We quickly review the theory here.
22 Nils Bruin We write O for the ring of integers of K. and we fix models of E and P1 over O. We choose a prime p such that the primes p1, . . . , pt of K over p are unramified and such that the cover u : E →P1 has good reduction at each pi, as a scheme morphism over O.
Let p be a rational prime which is unramified in K. Let p1, . . . , pt be the primes of K over p. Suppose that u : E →P1 has good reduction at all pi. We write O for the ring of integers of K, E(O/pi) for the points in the special fibre of E, considered as a scheme over Opi and E(1)(Kpi) for the kernel of reduction: 0 →E(1)(Kpi) →E(Kpi) ρi →E(O/pi) →0 Let g1, . . . , gr ∈E(K) be generators of the free part of E(K). Then if P0 = T + n1g1 + · · · + nrgr ∈E(K) has u(P0) ∈P1(Q), then certainly (abusing notation), u(ρi(P0)) ∈P1(Fp) and in fact u(ρi(P0)) = u(ρj(P0)). The points P0 ∈E(K) define a collection of cosets of Λp = t \ i=1 E(K) ∩E(1)(Kpi) Let Vp be this coset collection and let b1, . . . , br be generators of Λp. In Magma, both Vp and Λp are easily computed.. We take the the same elliptic curve as in the previous chapter, together with its (finite index subgroup of the) Mordell-Weil group and the cover suggested in Section 4.
P1:=ProjectiveSpace(Rationals(),1); > u:=mapP1|[-X + (zeta^3 - 1)dZ,X+(-zeta^3-zeta)dZ]>; > V3:=RelevantCosets(mwmap,u,Support(3OK)); > Lambda3:=Kernel(V3); > GmodLambda3:=Codomain(V3); > V3; As is clear, the coset data is returned as a tuple consisting of the map G → G/Λp, together with the collection of cosets, represented as elements of G/Λp.
We can compute a similar coset collection Vq and intersect it with Vp. This gives a new coset collection mod Λp + Λq. Alternatively, one could project Vp ∩Vq down to get again a coset collection modulo Λp. This is what in Magma is called a Weak coset intersection.
Some ternary Diophantine equations of signature (n, n, 2) 23 > V11:=RelevantCosets(mwmap,u,Support(11OK)); > V3i11:=CosetIntersection(V3,V11:Weak); > V3i11; In order to bound the number of points P ∈E(K) with u(P) ∈P1(Q), we make use of the formal group description of the group structure on E. Let b1, . . . , br be generators of Λp ⊂E(K). In terms of formal power series, there are isomorphisms ExpE : K →E(K), LogE : E(K) →K, where z is a local coordinate on E around the origin. These power series converge on E(1)(Kp) for unramified primes of odd residue characteristic and establish an isomorphism E(1)(Kp) ≃pOp. Therefore, for each prime pi we obtain a power series θP0,i(n1, . . . , nr) = u P0 + ExpE r X j=1 njLogE(bj) ∈Opi.
If u(P0+n1b1+· · ·+nrbr), then θP0,i(n1, . . . , nr) ∈Qp and θP0,i(n1, . . . , nr) = θP0,j(n1, . . . , nr). Using that Opi is a finite Zp-module, we can decompose Opi = ⊕Zp and express the above equations as [K : Q] −1 equations in Zp.
We can do this in Magma: > P0:=mwmap(G.3+G.4); > u(P0); (-3 : 1) > theta:=ChabautyEquations(P0,u,mwmap,Support(3OK)); > PrintToPrecision(theta,1);"";PrintToPrecision(theta,1);"";PrintToPrecision(theta O(3^5) - (3 + O(3^5))$.1 + (3^210 + O(3^5))$.2 O(3^5) - (329 + O(3^5))$.1 - (332 + O(3^5))$.2 O(3^5) - (3^25 + O(3^5))$.1 - (3^4 + O(3^5))$.2> A consequence of the shape of ExpE(z) is that the power series returned by ChabautyEquations have the property that the coefficient c of a monomial of total degree d satisfies ordp(c) ≥d −⌊ordp(d!)⌋. In particular, from the 24 Nils Bruin power series above, one can see that any integral solution (n1, n2) must satisfy n1 ≡n2 ≡0 mod 3 and, since det −1 0 −29 −32 ̸≡0 mod 3, by Hensel’s lemma such an integral solution lifts uniquely to (0, 0). In other words, there is at most one point in the coset G3 +G4 +Λ3 that has a rational image under u. One can do similar arguments for the other fibres of reduction: > N,V,R,C:=Chabauty(mwmap,u,3:Aux:={7}); > assert N eq #V; > assert #C eq 0; > R; 4 > V; { 0, G.3 - G.4, -G.3 + G.4, G.3 + G.4, -G.3 - G.4 } > {EvaluateByPowerSeries(u,mwmap(P)):P in V}; { (-1 : 1), (-1/3 : 1), (-3 : 1) } To interpret the above results, consider that in the previous computations, we have not really used that we have generators of E(K). In fact, for this particular example, we don’t know we have. We only know we have generators of some finite odd index subgroup G. For the finite field arguments, we only need that the [E(K) : G] is prime to [E(O/pi) : ρi(G)] for each of the i.
Since the power series argument works for n1, n2 ∈Zp, we only need that that [E(K) : G] is prime to p as well. However, when computing LogE(bj), we can often already deduce that p ∤[E(K) : G].
We only need G to be q-saturated in E(K) for finitely many l. The l that are encountered during the computations, are collected as prime divisors of R. In our case, this is only 2 and since we already know G to be 2-saturated in E(K), any conclusions we draw from G will also be valid for E(K). The interpretation of the other return values can be stated as follows.
{P ∈E(K) : u(P) ∈P1(Q)} ≤N V ⊂{P ∈E(K) : u(P) ∈P1(Q)} ⊂V ∪C Here, C is a coset collection of the type we described before. Note that if #V = N then all inequalities above are identities.
The routine Chabauty only tries a limited number of techniques to de-termine p-adic solution and only with finite precision. It uses an adaptation Some ternary Diophantine equations of signature (n, n, 2) 25 of the algorithm in Section 5.1 to find solutions of multiplicity 1 and it uses a generalisation of [2, Lemma 4.5.1] to test if the solution (0, . . . , 0) is the only integral solution of possibly higher multiplicity. It may therefore fail to produce a finite bound at all. In that case, N = ∞is returned.
As an advanced example, we also give the computation for d = −3ζ3 − 7ζ2 −8ζ −9.
d:=-3zeta^3-7zeta^2-8zeta-9; > E:=EllipticCurve([0,(-3zeta^3-zeta+1)d,0,(-zeta^2-zeta-1)d^2,0]); > P1:=ProjectiveSpace(Rationals(),1); > u:=mapP1|[-X + (zeta^3 - 1)dZ,X+(-zeta^3-zeta)dZ]>; > success,G,mwmap:=PseudoMordellWeilGroup(E); > assert success; > [mwmap(P):P in OrderedGenerators(G)]; [ (-41zeta^3 + 18zeta^2 - 14zeta + 42 : 0 : 1), (0 : 0 : 1), (-5zeta^3 + 6zeta^2 + 9 -69zeta^3 + 17zeta^2 - 34zeta + 60 : 1), (-6zeta^3 + 3zeta^2 - zeta + 6 : 57zet 16zeta^2 + 27zeta - 51 : 1), (-36zeta^3 + 8zeta^2 - 20zeta + 32 : 10zeta^3 + 72 60zeta + 62 : 1) ] > N,V,R,C:=Chabauty(mwmap,u,3); > C31,R31:=RelevantCosets(mwmap,u,Support(31OK)); > R:=LCM(R,R31); > Cnew:=CosetIntersection(C,C31:Weak); > assert #Cnew eq 0; > R; 2 > V; { 0, G.4 - G.5, -G.4 + G.5 } > {EvaluateByPowerSeries(u,mwmap(P)):P in V}; { (3/2 : 1), (-1 : 1) } An interesting feature of this example is, that the 3-adic argument by itself is not sufficient. We see that there are two 3-adic “ghost” solutions. The 3-adic computation did come up with a rather precise 3-adic approximation of these putative solutions. The cosets are disjoint from V31, so we proved that they indeed only correspond to Z3-solutions and not rational ones.
Incidentally, specifying 31 as an “auxiliary” prime, such that V3 and V31 get intersected before the 3-adic argument, would have solved this particular equation as well, as would 191 by itself.
The other 3 values of d mentioned in Section 4 can be solved in a similar way, either with p = 31 or p = 191.
26 Nils Bruin 8 The equations xn + yn = Dz2 for n = 6, 7, 9, 11, 13, 17 The proof of Theorem 2 for the remaining cases is straightforward and, in many cases, easier than for n = 5, because there are no non-trivial solutions.
For each n, we outline a successful strategy. For full details, we refer the reader to the accompanying electronic resource .
x6 + y6 = Dz2: Since 6 is even, we can reduce the genus (and the number) of the curves to consider tremendously. Note that a solution with y ̸= 0 corresponds to a rational point on the genus 2 curve Y 2 = DX6 + D. For D ∈{2, 3, 5, 6, 10, 11, 13, 17}, we conclude that only for D = 2 does this curve have points over Q2 and Q7. Following the same approach as in , we write 2X6 + 2 = (2X2 + 2)(X4 −X2 + 1) and we conclude that any point (X, Y ) corresponds to a solution (X, Y1, Y2) of dY 2 1 = 2X2 + 2, dY 2 2 = X4 −X2 + 1 for d ∈Q(2, {2, 3}). Only for d = 1 does this system of equations have solutions over Q2. The curve Y 2 2 = X4 −X2 + 1 only has rational points with X ∈{−1, 0, 1, ∞}.
x7 + y7 = Dz2: We note that any solution corresponds to a solution to Cd : Y 2 = d(X6 −X5 + X4 −X3 + X2 −X + 1) for some d ∈Q(2, S), where S contains the prime divisors of 7D. For the relevant values of D, only d = 1, 7 yield curves with points over R, Q2, Q7, Q11.
With it is straightforward to check that Jac(C7)(Q) is of free rank 1 and using Stoll’s implementation of , (3-adically), one finds that all rational points have X = −1.
For C1, one uses and the techniques outlined in Section 7 to show that all rational points have X ∈{−1, 0, 1, ∞} x9 + y9 = Dz2: We factor: y2 1 = d1(x6 −x3z3 + z6) y2 2 = d2(x2 −xz + z2) y2 3 = Dd1d2(x + 1) and note that any primitive solution (x, y, z) gives rise to a solution of the system above for d1, d2 ∈Q(2, S), where S contains the prime divisors of 3D. Furthermore, because gcd(x, z) = 1, we have gcd(d1, d2) | 3.
We can dehomogenize the first two equations and if we test them for simultaneous solvability over R, Q2, Q3, Q5, we are only left with d1 ∈ {1, 3}. The first equation gives rise to a curve of genus 2 with a Mordell-Weil group of free rank 1. Again, yields that all points have X ∈ {−1, 1, 0, ∞}.
Some ternary Diophantine equations of signature (n, n, 2) 27 x11 + y11 = Dz2: Using the same argument as for n = 7, we find all that solutions correspond to rational points on Cd : Y 2 = d(X10 −X9 + · · · −X + 1) for d ∈{1, 11}. Rather than applying to Cd directly, we substitute (U, V ) = ((X2 + 1)/X, Y + Y/X) to find the covered curve Dd : V 2 = d(U 6 + U 5 −6U 4 −5U 3 + 9U 2 + 5U −2).
For d = 1 the free rank of the Mordell-Weil group is bounded above by 2 and for d = 11 the free rank is bounded by 1, but we could not find a generator. Using the techniques from , we find that U(D1(Q)) ⊂ {−2, −1, 2, ∞} and that U(D11)(Q)) ⊂{−2, −1, 1, 2, ∞}. From this, it follows easily that Cd only has rational points above X ∈{−1, 0, 1, ∞} for d = 1, 11.
x13 + y13 = Dz2: Using the same argument as for n = 7, we find that all solutions correspond to rational points on Cd : Y 2 = d(X12 −X9 + · · · −X + 1) for d ∈{1, 13}. For d = 13 we substitute (U, V ) = ((X2 + 1)/X, Y/X3) to obtain Dd : V 2 = 13(U 6 −U 5 −5U 4 + 4U 3 + 6U 2 −3U −1).
Following yields that U(D13(Q)) ⊂{−2}, which corresponds to X = −1.
For d = 1 we get a bound on the Mordell-Weil rank of 2, so we use that over Q(β) with β3 −β2 −4β + 1 = 0, we have a quartic factor X4 + βX3 + (β2 + β −1)X2 + βX + 1 of (X13 + 1)/(X + 1). Using and we find X(C1(Q)) ⊂{0, 1, ∞}.
x17 + y17 = Dz2: Using the same argument as for n = 7, we find that all solutions correspond to rational points on Cd : Y 2 = d(X16 −X9 + · · · −X + 1) for d ∈{1, 17}. Over K = Q(β) with β4 + β3 −6β2 −β + 1 = 0 we have R(X) := X4 + βX3 + 1/2(−β3 + 6β + 1)X2 + βX + 1 with NK/QR(X) = (X17 + 1)/(X + 1). Hence, any rational point on Cd has an X-coordinate corresponding to a rational point on Dδ : V 2 = δR(X).
28 Nils Bruin for some δ ∈K(2, S) with dNK/Q(δ) a square, where S contains the primes above 2 · 17 · δ.
Local arguments show that only δ = 1, β3+2β2−3β+1 need consideration.
The techniques from Section 7 then show that X(Cd(Q)) ⊂{−1, 0, 1, ∞} for d = 1, 17.
References 1. Michael A. Bennett and Chris M. Skinner. Ternary Diophantine equations via Galois representations and modular forms. Canad. J. Math., 56(1):23–54, 2004.
2. N. R. Bruin. Chabauty methods and covering techniques applied to generalized Fermat equations, volume 133 of CWI Tract. Stichting Mathematisch Centrum Centrum voor Wiskunde en Informatica, Amsterdam, 2002. Dissertation, Uni-versity of Leiden, Leiden, 1999.
3. Nils Bruin. Algae, a program for 2-selmer groups of elliptic curves over number fields. see
4. Nils Bruin.
The diophantine equations x2 ± y4 = ±z6 and x2 + y8 = z3.
Compositio Math., 118:305–321, 1999.
5. Nils Bruin. Chabauty methods using elliptic curves. J. Reine Angew. Math., 562:27–49, 2003.
6. Nils Bruin.
Transcript of computations.
available from 2003.
7. Nils Bruin. Visualising Sha in abelian surfaces. Math. Comp., 73(247):1459– 1476 (electronic), 2004.
8. Nils Bruin and E.
Victor Flynn.
Towers of 2-covers of hy-perelliptic curves.
Technical Report PIMS-01-12, PIMS, 2001.
9. J.W.S. Cassels. Diophantine equations with special reference to elliptic curves.
J. London Math. Soc., 41:193–291, 1966.
10. J.W.S. Cassels. Second descents for elliptic curves. J. Reine Angew. Math., 494:101–127, 1998.
Dedicated to Martin Kneser on the occasion of his 70th birthday.
11. Claude Chabauty.
Sur les points rationnels des vari´ et´ es alg´ ebriques dont l’irr´ egularit´ e est sup´ erieure ` a la dimension. C. R. Acad. Sci. Paris, 212:1022– 1024, 1941.
12. J. E. Cremona. Reduction of binary cubic and quartic forms. LMS J. Comput.
Math., 2:64–94 (electronic), 1999.
13. John E. Cremona and Barry Mazur. Visualizing elements in the Shafarevich-Tate group. Experiment. Math., 9(1):13–28, 2000.
14. Henri Darmon and Andrew Granville.
On the equations zm = F(x, y) and Axp + Byq = Czr. Bull. London Math. Soc., 27(6):513–543, 1995.
15. Claus Fieker. p-Selmer groups of number fields. Private communication.
16. E. Victor Flynn and Joseph L. Wetherell. Finding rational points on bielliptic genus 2 curves. Manuscripta Math., 100(4):519–533, 1999.
17. E.V. Flynn. A flexible method for applying chabauty’s theorem. Compositio Mathematica, 105:79–94, 1997.
18. J. R. Merriman, S. Siksek, and N. P. Smart. Explicit 4-descents on an elliptic curve. Acta Arith., 77(4):385–404, 1996.
Some ternary Diophantine equations of signature (n, n, 2) 29 19. Bjorn Poonen and Edward F. Schaefer. Explicit descent for jacobians of cyclic covers of the projective line. J. reine angew. Math., 488:141–188, 1997.
20. Samir Siksek. Descent on curves of genus 1. PhD thesis, University of Exeter, 1995.
21. Joseph H. Silverman. The Arithmetic of Elliptic Curves. GTM 106. Springer-Verlag, 1986.
22. Denis Simon. Computing the rank of elliptic curves over number fields. LMS J.
Comput. Math., 5:7–17 (electronic), 2002.
23. Michael Stoll.
Implementing 2-descent for Jacobians of hyperelliptic curves.
Acta Arith., 98(3):245–277, 2001.
24. Michael Stoll and John E. Cremona. Minimal models for 2-coverings of elliptic curves. LMS J. Comput. Math., 5:220–243 (electronic), 2002.
25. Tow Womack. Four descent on elliptic curves over Q. PhD thesis, University of Nottingham, 2003.
Index AbsoluteAlgebra, 7 ChabautyEquations, 23 Chabauty, 24 IsogenyMu, 20 MordellWeilGroup, 4 PseudoMordellWeilGroup, 19 Quartic, 20 SwapExtension, 6 abelian, 17, 19 affine, 11, 13–15 algebra, 2, 4–7, 17 algebraic, 2, 4 algorithm, 2, 4, 10–13, 15, 16, 25 arithmetic, 2, 10, 13 Bennett, 1 bijection, 6 blowup, 18 branching, 16 cardinality, 15 Cassels, 21 Chabauty, 21 characteristic, 2, 10, 15, 23 cocycle, 17 cohomology, 17 commutative, 17 completion, 13 component, 13–15 Conics, 15 connected, 17 coset, 22, 24, 25 cover, 2, 5–9, 14, 15, 20–22, 27 covering, 20, 21 covers, 2, 6, 8, 9, 21 cremona, 20 curve, 1–6, 8–10, 13–17, 21, 22, 26, 27 curves, 1–4, 6, 8, 10, 13, 15, 17, 21, 26 decomposition, 7 descent, 6, 19, 21, 26 descents, 6 desingularisation, 13–15 dimension, 10–13 divisors, 5, 24, 26 dual, 19 echelon, 11 Elliptic, 17 elliptic, 1–3, 9, 10, 15–17, 21, 22 embedding, 6 equidimensional, 12 exact, 5, 11, 13, 16, 17, 19, 20 exactness, 19 extension, 10 factorisation, 11, 16 Fermat, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 Fieker, 5 formal, 23 Galois, 1, 5, 17 genus, 1, 5, 6, 15, 16, 26 geometric, 2, 10, 15 Groebner, 13 Hasse, 16 32 Index Hensel, 11, 12, 16, 24 homogeneous, 4–6, 21 homomorphism, 8, 19 Hurwitz, 15 hyperelliptic, 2, 10, 15, 16 intersection, 10, 12, 13, 17, 20, 22 isogenies, 21 isogenous, 21 isogeny, 2, 19, 21 isomorphism, 6, 23 isomorphisms, 23 jacobian, 6 kernel, 17, 22 kernels, 17 lift, 11, 12, 16, 24 liftable, 11, 16 lifted, 12 lifts, 16, 24 mapped, 20 mapping, 9 maps, 5, 8, 14, 17, 18 maximal, 10, 13 minimal, 11, 19, 20 minimization, 20 minimum, 10 model, 6, 9, 10, 15, 20–22 models, 22 modular, 1 Mordell, 2, 4, 16, 17, 19, 21, 22, 26, 27 morphism, 6, 8, 19, 22, 23 neighbourhoods, 12 nonsingular, 15, 16 nonsingularity, 15 norm, 1, 4, 7, 10, 17, 18 normalised, 10 pairing, 21 parametrise, 2 parametrising, 4, 8 planar, 10 plane, 15 primitive, 2, 4, 7, 26 pullback, 6 quadrics, 20 quartic, 27 radical, 13 ramified, 5, 6, 22, 23 rank, 2–4, 17, 19, 20, 26, 27 ranks, 2, 17 reduce, 10, 11, 13, 15, 21, 26 reduced, 10, 13, 15, 21 reduction, 20–22, 24 represent, 1, 5, 7, 9, 11–18, 20–22 representable, 11 representation, 1, 7, 18 representations, 1 representative, 5, 9, 11–16 representatives, 5, 9, 12, 13, 15, 16 represented, 9, 11, 13, 15, 17, 20–22 represents, 15, 16, 20 residue, 2, 10, 23 restrictions, 17 resultants, 11 Riemann, 15 Saturate, 11–13 saturated, 19, 24 scheme, 2, 10–15, 22 schemes, 2, 10 Selmer, 2, 4, 17, 19–21 Shafarevich, 17 siksek, 15 singularities, 14 singularity, 14, 15 Skinner, 1 smooth, 10, 13, 15, 16 solution, 1–6, 8, 10–13, 15, 16, 20, 24–27 solutions, 1–4, 6, 8, 10–13, 15, 16, 25–27 Solvability, 13, 15 solvability, 2, 3, 10, 12, 13, 15, 16, 26 solvable, 7, 16 Stoll, 26 subcover, 2, 6, 8 subcovers, 2 subgroup, 1, 7, 17–19, 21, 22, 24 subgroups, 1 subscheme, 13, 15 Supp, 1, 22 Twist, 1 twisted, 15 Index 33 twists, 6, 20 unramified, 6, 22, 23 visualisation, 21 Weak, 22 Weierstrass, 9 weighted, 6, 15 Weil, 2, 4, 16, 17, 19, 21, 22, 26, 27 Womack, 21 words, 24 |
9907 | https://oehha.ca.gov/proposition-65/methanol-fact-sheet | Methanol - A Fact Sheet - OEHHA
Skip to main content
Skip to Main Content
CA.gov CA.gov
Official website of the State of California
Reset
High contrast
Increase font size Font increase
Decrease font size Font decrease
Dyslexic font
State of California OEHHA
Search Menu
Menu
Custom Google Search Submit
Close
Environmental Topics
All Environmental Topics
Air
Climate Change
Fish
Pesticides
Water
CalEnviroScreen
About
About OEHHA
What We Do
Jobs
Programs and Reports
Contact Us
OEHHA Laws and Regulations
Join the Listserv
Meet the Executive Office
Organizational Chart
Proposition 65
Proposition 65 Overview
About Proposition 65
The Proposition 65 List
Meetings, Hearings and Workshops
Notices
Law and Regulations
Warnings
News and Events
Latest News
Events
Public Comments
Library
Library Overview
Chemical Databases
Documents
Maps
Prop 65 Chemical List
Videos
Home
Proposition 65
Methanol - A Fact Sheet
Methanol - A Fact Sheet
June 12, 2012
Developmental Toxicity - In 2012 Methanol was added to the Proposition 65 list because the National Toxicology Program found that animal studies showed birth defects at high doses.
Other known health effects -Methanol poisoning can cause headaches and muscle pain, and at high enough doses,blindness or death.
What is Methanol?
Methanol is a clear and colorless liquid alcohol sometimes called “wood alcohol.” Methanol is not the kind of alcohol that people drink. Methanol’s main use is in the production of chemicals. For example, 30 percent of methanol in commerce is used to produce formaldehyde.
2. ### What products contain methanol?
Methanol occurs naturally at low levels in fruits, vegetables, juices and other foods prepared from fruits and vegetables. It also can be present in some adhesives, paints, varnishes and other consumer products. Cigarette smoke also contains methanol.
Methanol is used as a fuel in race cars and may have some use in alternative fuels for transportation.
3. ### What are the health effects of methanol exposure?
Scientists at the National Toxicology Program recently determined that methanol causes birth defects in animals. Based on this finding, methanol was added to the Proposition 65 list of chemicals that cause developmental toxicity.
Methanol can be toxic in other ways. Methanol poisoning happens when people are exposed to more methanol than their bodies can handle. You should never drink pure methanol. Drinking a few teaspoons of undiluted methanol can lead to blindness and can even be fatal. Less-severe symptoms of high methanol exposure include headaches, blurred vision, and muscle pain.
4. ### What is Proposition 65?
Proposition 65 is a law passed by California voters in 1986. It requires the Governor to maintain a list of carcinogens and reproductive toxicants.
5. ### What does it mean that methanol appears on California’s Proposition 65 list?
California added methanol to the list of chemicals known to cause reproductive toxicity in 2012 after federal scientists at the National Toxicology Program found that the chemical caused birth defects in laboratory animals. As of March 2013, businesses with more than 10 employees must provide a warning if their operations or products cause sufficiently high exposures to methanol. Additionally, as of November 2013, methanol may not be discharged in significant levels into a source of drinking water. Businesses that fail to comply with these requirements could face civil lawsuits brought by state or local prosecutors or members of the public.
OEHHA has proposed two “maximum allowable dose levels” that identify levels of exposure to methanol that require warnings and prohibit discharges to sources of drinking water. The proposed levels are 47,000 micrograms per day for inhalation, 23,000 micrograms for ingestion of methanol.
Fruits and vegetables that naturally contain methanol do not require Proposition 65 warnings. Similarly, warnings are not required for juices and other products prepared from fruits and vegetables with naturally occurring methanol.
6. ### How are most people exposed to methanol?
Generally, people may be exposed to low amounts of methanol by touching or breathing it in from certain consumer items. Items that may contain methanol include varnishes, shellacs, paints, windshield washer fluid, antifreeze, tobacco smoke and adhesives. Improperly produced home-made distilled spirits may contain dangerously high levels of methanol.
People can be exposed to methanol in workplaces where methanol is manufactured or used. People who work with methanol are exposed to higher levels than the average person.
7. ### How can I reduce my exposure to methanol?
Wear appropriate protective gear when using paints, adhesives, and varnishes. Washing your hands after using these items also reduces exposure. If you work with methanol, follow the protection requirements in your workplace.
Cutting back or quitting smoking will significantly decrease your exposure to methanol and many other toxic substances.
8. ### Where can I find out more about methanol and the risk of reproductive toxicity?
To learn more about the basis for the National Toxicology Program’s findings that methanol caused birth defects in laboratory animals, please see:
Downloads
Notice of Proposed Rulemaking MADL for Methanol
March 1, 2012
Interpretive Guideline No. 2012-001 Consumption of Methanol Resulting from Pectin that Occurs Naturally In Fruits And Vegetables
March 16, 2012
Methanol - A Fact Sheet
June 12, 2012
Chemical Reference
Methanol
Cal EPA
Air Resources BoardCal RecycleDepartment of Pesticide RegulationDepartment of Toxic Substances ControlState Water Resources Control Board
Alerts
Amber AlertCal AlertsMy Hazards
About
GovernorLt. GovernorCalifornia Data
CA.gov
Conditions of use/Privacy Policy
Accessibility
Help
Sitemap
Facebook
Instagram
LinkIn
Twitter/X
YouTube
Copyright © 2025 State of California
Back to top |
9908 | https://www.youtube.com/watch?v=n3FbQGGKVfc | Art of Problem Solving: Simplifying Linear Expressions
Art of Problem Solving
103000 subscribers
623 likes
Description
89820 views
Posted: 24 Dec 2011
Art of Problem Solving's Richard Rusczyk explains how to simplify one-variable expressions.
This video is part of our AoPS Prealgebra curriculum. Take your math skills to the next level with our Prealgebra materials:
📚 AoPS Prealgebra Textbook:
🖥️ AoPS Prealgebra 1 Course:
🔔 Subscribe to our channel for more engaging math videos and updates
Transcript:
This is an expression. "Expression" is a fancy math word for a bunch
of stuff multiplied, subtracted, added or divided together. So that is an expression. Now, if we wanted to evaluate this expression
-- you know, pick different values of x and stick them in there and figure out what this
expression equals for different values of x — we'd have to do a bunch of multiplications,
bunch of additions and subtractions. Let's simplify this first. And by "simplify," that means make this into
a simpler expression. That means we won't have to do as many multiplications,
additions or subtractions to figure out what the expression equals for each value of x. Now, we see right away one way we can simplify
this. Since all we're doing here is adding and subtracting
a bunch of stuff, we can rearrange it and put the 7 and the 8 together, because we know
how to combine those two. You know, x plus 2x minus 5x - notice I'm
careful to keep the minus with the 5x - plus 7 plus 8. So, I put these two over here, these two terms,
what we call these, each of these is a term. 8, 7, 5x, 2x, x, they're all terms. We put these two terms that don't have an
x in them, we put them together, because we can add those very easily. We can add 7 and 8 and we get 15. And we call these terms that don't have variables
in them, we call these "constants," because 7, well, it's constant, no matter how large
or how small you draw that 7, it's still 7. These over here have variables in them; x
is a variable. It can vary. So we've simplified the two constant terms
by combining them. 7 and 8 is 15. Can we simplify this out here, the x plus
2x minus 5x? We use the distributive property. We can factor the x out. But before we do that, I'm going to put in
a 1 out here. It's a little imaginary 1. Now, we usually don't include the 1's in these
terms, because 1 times x, well, that's just x. But we're going to include it here so that,
when we factor the x out, we remember it. We remember that there's a 1 out here. We just factor the x out. See, when we multiply this back out, we get
1x plus 2x minus 5x, plus 15. And now we know how to combine all this. 1 plus 2 is 3. 3 minus 5, that's negative 2, so we end up
with negative 2x plus 15. This is a lot easier to deal with than all
this mess up here. So if you're going to pick a value of x, you'd
rather put it in this, than put it in that mess up there. All we mean when we say simplify here is,
we combine the terms with variables in them and we combine the terms without variables
in them. So we combine the constant terms 7 and 8,
gave us 15. The variable terms - x plus 2x is 3x, minus
5x is negative 2x. Let's try one more. All right, here we have a different variable,
we're not using x anymore, we're going to use c. And we can start off here just by expanding
these, again with the distributive property. 3 times 2 plus c, that gives me 3 times 2
plus 3 times c. Now here, when I distribute, again, I have
to be very careful about my signs. I'm going to remember that there's a negative
out here. This is negative 2 times c and negative 2
times negative 5. Notice how careful I am to keep, keep my signs. It stays a negative 2, negative 5. Have to pay attention to your signs. Most of the time when you make mistakes simplifying
expressions, it's usually a sign error. Now, 3 times 2 is 6. And 3 times c, well, that's just 3c. Minus 2 times c is minus 2c. Minus 2 times minus 5 is 10. Negative times a negative is positive. And now we know what to do here. We combine the variable terms. We'll put those two together, 3c minus 2c. And we combine the constant terms, the 6 and
the 10. And again, we can do the distributive property
here, factor out the c, add the two constant terms. Now usually, I don't bother writing this step
in the middle. I just realize, 3c minus 2c, it's going to
be 3 minus 2, which is 1c. We don't need to write the 1. 1 times c is just c. And we get c plus 16. And we're done. |
9909 | https://www.chegg.com/homework-help/questions-and-answers/find-parabola-form-y-x-2-b-best-fits-points-1-0-4-4-5-8-minimizing-sum-s-squares-vertical--q111151581 | Solved Find the parabola of the form y=ax2+b which best fits | Chegg.com
Skip to main content
Books
Rent/Buy
Read
Return
Sell
Study
Tasks
Homework help
Understand a topic
Writing & citations
Tools
Expert Q&A
Math Solver
Citations
Plagiarism checker
Grammar checker
Expert proofreading
Career
For educators
Help
Sign in
Paste
Copy
Cut
Options
Upload Image
Math Mode
÷
≤
≥
o
π
∞
∩
∪
√
∫
Math
Math
Geometry
Physics
Greek Alphabet
Math
Calculus
Calculus questions and answers
Find the parabola of the form y=ax2+b which best fits the points (1,0),(4,4),(5,8) by minimizing the sum, S, of squares of the vertical distances from the above points to the parabola given by S=(a+b)2+(16a+b−4)2+(25a+b−8)2.y= NOTE. You need to minimize S as a function of the variables a,b and then write the expression for y:y=ax2+b with a and b that you
Your solution’s ready to go!
Our expert help has broken down your problem into an easy-to-learn solution you can count on.
See Answer See Answer See Answer done loading
Question: Find the parabola of the form y=ax2+b which best fits the points (1,0),(4,4),(5,8) by minimizing the sum, S, of squares of the vertical distances from the above points to the parabola given by S=(a+b)2+(16a+b−4)2+(25a+b−8)2.y= NOTE. You need to minimize S as a function of the variables a,b and then write the expression for y:y=ax2+b with a and b that you
Show transcribed image text
There are 2 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 The given sum function is S=(a+b)2+(16 a+b−4)2+(25 a+b−8)2…(1) Now for minimize (1) at first find the critical point . View the full answer Step 2 UnlockAnswer Unlock Previous questionNext question
Transcribed image text:
Find the parabola of the form y=a x 2+b which best fits the points (1,0),(4,4),(5,8) by minimizing the sum, S, of squares of the vertical distances from the above points to the parabola given by S=(a+b)2+(16 a+b−4)2+(25 a+b−8)2.y= NOTE. You need to minimize S as a function of the variables a,b and then write the expression for y:y=a x 2+b with a and b that you have found. Find three positive real numbers whose sum is 39 and whose product is a maximum. List them in ascending order. HINT. Introduce three variables x,y,z. Express one of them (say, z ) in terms of the two others (using the fact that their sum is given). The product x yz becomes then a function of those two variables. You need to maximize it.
Not the question you’re looking for?
Post any question and get expert help quickly.
Start learning
Chegg Products & Services
Chegg Study Help
Citation Generator
Grammar Checker
Math Solver
Mobile Apps
Plagiarism Checker
Chegg Perks
Company
Company
About Chegg
Chegg For Good
Advertise with us
Investor Relations
Jobs
Join Our Affiliate Program
Media Center
Chegg Network
Chegg Network
Busuu
Citation Machine
EasyBib
Mathway
Customer Service
Customer Service
Give Us Feedback
Customer Service
Manage Subscription
Educators
Educators
Academic Integrity
Honor Shield
Institute of Digital Learning
© 2003-2025 Chegg Inc. All rights reserved.
Cookie NoticeYour Privacy ChoicesDo Not Sell My Personal InformationGeneral PoliciesPrivacy PolicyHonor CodeIP Rights
Do Not Sell My Personal Information
When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Sale of Personal Data
[x] Sale of Personal Data
Under the California Consumer Privacy Act, you have the right to opt-out of the sale of your personal information to third parties. These cookies collect information for analytics and to personalize your experience with targeted ads. You may exercise your right to opt out of the sale of personal information by using this toggle switch. If you opt out we will not be able to offer you personalised ads and will not hand over your personal information to any third parties. Additionally, you may contact our legal department for further clarification about your rights as a California consumer by using this Exercise My Rights link.
If you have enabled privacy controls on your browser (such as a plugin), we have to take that as a valid request to opt-out. Therefore we would not be able to track your activity through the web. This may affect our ability to personalize ads according to your preferences.
Targeting Cookies
[x] Switch Label label
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices
mmmmmmmmmmlli mmmmmmmmmmlli mmmmmmmmmmlli |
9910 | https://www.tutorialspoint.com/difference-between-asa-and-aas | Difference Between ASA and AAS
The study of geometry is enjoyable. Sizes, distances, and angles are the primary focus of this branch of mathematics known as geometry. Shapes are the focus of geometry, a branch of mathematics. It's not hard to understand how geometry may be used to solve problems in the actual world. It finds application in a wide range of fields, including engineering, architecture, the arts, sports, and more.
Today, we'll talk about a special topic in triangle geometry called congruence. But first, let's define congruence so we may use it. Whenever one figure can be superimposed over the other in such a way that all of its elements match up, we say that the two figures are congruent. That is, if two figures share the same dimensions and shape, we say that they are congruent. If you take a look at two congruent figures, you'll see that they are the same shape at two distinct locations.
True, triangle congruence serves as the cornerstone of many geometrical theorems and proofs. The notion of triangle congruence is central to the study of geometry in high school. The idea of sufficiency, that is, determining the criteria which fulfil that two triangles are congruent, is often disregarded while teaching and learning about triangle congruence.
We will just cover two of the five possible methods for checking congruence between two triangles (the ASA and AAS methods). To clarify, "Angle, Angle, Side" (AAS) is the opposite of "Angle, Side, Angle" (ASA). Let's check out how you may utilise the two to figure out if a pair of triangles is indeed congruent.
ASA Triangle Congruence
ASA stands for Angle-Side-Angle. This criterion states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent. In other words, if we know that two triangles have two angles and one side in common, then we can conclude that they are congruent.
AAS Triangle Congruence
AAS stands for Angle-Angle-Side. This criterion states that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent. In other words, if we know that two triangles have two angles and one non-included side in common, then we can conclude that they are congruent.
Differences: ASA and AAS
The main difference between ASA and AAS is the order in which the angles and sides are congruent. In ASA, the included side is between the two congruent angles, while in AAS, the non-included side is opposite to one of the congruent angles. This means that in ASA, we have two angles and one side, while in AAS, we have two angles and two sides.
Another difference between ASA and AAS is the number of sides that are congruent. In ASA, we only have one side that is congruent, while in AAS, we have two sides that are congruent. This means that AAS is a stronger criterion than ASA, as it requires more information to prove congruence.
When to Use ASA and AAS?
ASA and AAS criteria are used to prove congruence between two triangles. However, they are not interchangeable, and it is important to use the correct criterion for the given situation.
ASA criterion is used when we have two angles and the included side in common. This is useful in situations where we are given the length of one side and two angles, and we need to find the length of another side.
AAS criterion is used when we have two angles and one non-included side in common. This is useful in situations where we are given the length of two sides and one angle, and we need to find the length of another side.
The following table highlights the major differences between ASA and AAS Triangle Congruence ?
| Characteristics | ASA | AAS |
---
| Terminology | ASA stands for "Angle, Side, Angle". ASA refers to any two angles and the included side. | AAS means "Angle, Angle, Side". AAS refers to the two corresponding angles and the non-included side./p> |
| Congruence | According to ASA congruence, two triangles are congruent if they have an equal side contained between corresponding equal angles. In other words, if two angles and an included side of one triangle are equal to the corresponding angles and the included side of the second triangle, then the two triangles are called congruent, according to the ASA rule. | The AAS rule, on the other hand, states that if the vertices of two triangles are in one-to-one correspondence such that two angles and the side opposite to one of them in one triangle are equal to the corresponding angles and the non- included side of the second triangle, then the triangles are congruent. |
Characteristics
ASA
AAS
Terminology
ASA stands for "Angle, Side, Angle". ASA refers to any two angles and the included side.
AAS means "Angle, Angle, Side". AAS refers to the two corresponding angles and the non-included side./p>
Congruence
According to ASA congruence, two triangles are congruent if they have an equal side contained between corresponding equal angles.
In other words, if two angles and an included side of one triangle are equal to the corresponding angles and the included side of the second triangle, then the two triangles are called congruent, according to the ASA rule.
The AAS rule, on the other hand, states that if the vertices of two triangles are in one-to-one correspondence such that two angles and the side opposite to one of them in one triangle are equal to the corresponding angles and the non- included side of the second triangle, then the triangles are congruent.
Conclusion
In summary, ASA and AAS are two criteria used in geometry to determine when two triangles are congruent. ASA requires two angles and the included side to be congruent, while AAS requires two angles and one non-included side to be congruent. It is important to use the correct criterion for the given situation to prove congruence between two triangles.
6K+ Views
Kickstart Your Career
Get certified by completing the course
TOP TUTORIALS
TRENDING TECHNOLOGIES
CERTIFICATIONS
COMPILERS & EDITORS
Tutorials Point is a leading Ed Tech company striving to provide the best learning material on technical and non-technical subjects.
Tutorials Point is a leading Ed Tech company striving to provide the best learning material on technical and non-technical subjects.
© Copyright 2025. All Rights Reserved. |
9911 | https://spaceplace.nasa.gov/moon-distance/ | How Far Away Is the Moon? | NASA Space Place – NASA Science for Kids
moon-distance
Vea en Español
Earth Sun Solar System Universe Science and Tech Educators
How Far Away Is the Moon?
The Short Answer:
The Moon is an average of 238,855 miles away from Earth, which is about 30 Earths away.
You might be surprised.
Often when we see drawings of the Earth and the Moon, they look really close together.
Don’t be fooled! They’re actually really far apart. The Moon is an average of 238,855 miles (384,400 km) away. How far away is that? That’s 30 Earths.
Average distance?
Why mention the average distance? Well, the Moon is not always the same distance away from Earth. The orbit is not a perfect circle.
When the Moon is the farthest away, it’s 252,088 miles away. That’s almost 32 Earths. When it's closest, the Moon is 225,623 miles away. That’s between 28 and 29 Earths.
So far apart!
The Moon definitely seems close because we can see it so well without a telescope,
but remember, it’s farther away than most people realize!
Related Resources for Educators
Distance to the Moon (Educator Guide)
Launchpad: Moon Magic
article last updated July 23, 2021
How far apart?
To see for yourself how far apart Earth and the moon are, try this:
Materials
Basketball
Tennis ball
Measuring tape
If Earth was the size of a basketball, the moon would be the size of a tennis ball. This can give you an idea of how they compare.
Directions
Place the basketball on the ground. This represents Earth.
Use the measuring tape to find a spot 23 feet 9 inches away from the middle of the basketball. Place the tennis ball there. The tennis ball is the moon.
Are the basketball Earth and tennis ball moon farther apart than you expected? Take a picture and show your friends and family how far apart the moon and Earth really are.
Extra!
Use nearby objects to remember how far 23 feet 9 inches is from the basketball. Ask others to place the tennis ball where they think the moon is.
They will be very surprised when you show them how far away it needs to be!
If you liked this, you may like:
What Are the Moon’s Phases?
Lunar Eclipses and Solar Eclipses
How Is the Sun Completely Blocked in an Eclipse?
Games Crafts Activities Videos Glossary Mystery
About UsPrivacy PolicyImage UseAccessibility
Contact NASA Space Place Last Updated: September 9th, 2025
MoreLess
By Subject
Space
Sun
Earth
Solar System
People & Technology
Parents & Educators
By Type
Explore
Do
Play
MoreLess
How far apart?
To see for yourself how far apart Earth and the moon are, try this:
Materials
Basketball
Tennis ball
Measuring tape
If Earth was the size of a basketball, the moon would be the size of a tennis ball. This can give you an idea of how they compare.
Directions
Place the basketball on the ground. This represents Earth.
Use the measuring tape to find a spot 23 feet 9 inches away from the middle of the basketball. Place the tennis ball there. The tennis ball is the moon.
Are the basketball Earth and tennis ball moon farther apart than you expected? Take a picture and show your friends and family how far apart the moon and Earth really are.
Extra!
Use nearby objects to remember how far 23 feet 9 inches is from the basketball. Ask others to place the tennis ball where they think the moon is.
They will be very surprised when you show them how far away it needs to be! |
9912 | https://aquatherm.com/tech-bulletins/equivalent-length-of-fittings | EQUIVALENT LENGTH OF FITTINGS - Aquatherm
LOGIN
REGISTER
CATALOG
FIELDPRO
QUICK ESTIMATION
PRESSURE TEST
CONTACT US
Search
PRODUCTS arrow_drop_down
aquatherm blue
aquatherm services
Other Products
Product Features
APPLICATIONS arrow_drop_down
Heating & Cooling
Brewery
Fire Protection
Industrial
Data Centers
CONTRACTORS arrow_drop_down
Fabrication Services
Training
Hangers
Ancillary Products
Find Local Support
Installation Equipment
ENGINEERS arrow_drop_down
Linear Thermal Expansion Calculator
Head Loss Calculator
Download CAD and Revit Files
Chemical Compatibility
RESOURCES arrow_drop_down
Design Tools
Downloads
Videos
Case Studies
Technical Bulletins
Training Resources
White Papers
FAQ
ABOUT arrow_drop_down
Our Story
Our Benefits
Installation
Longevity
Reliability
Chemical Purity
Sustainability
Search
CATALOG
QUICK ESTIMATION
PRESSURE TEST
PRODUCTS arrow_drop_down
aquatherm blue
aquatherm services
Other Products
Product Features
APPLICATIONS arrow_drop_down
Heating & Cooling
Brewery
Fire Protection
Industrial
Data Centers
CONTRACTOR SERVICES arrow_drop_down
Fabrication Services
Training
Hangers
Ancillary Products
Find Local Support
Installation Equipment
ENGINEERING arrow_drop_down
Linear Thermal Expansion Calculator
Head Loss Calculator
Download CAD and Revit Files
Chemical Compatibility
RESOURCES arrow_drop_down
Design Tools
Downloads
Videos
Case Studies
Technical Bulletins
Training Resources
White Papers
FAQ
ABOUT arrow_drop_down
Our Story
Our Benefits
Installation
Longevity
Reliability
Chemical Purity
Sustainability
menu
EQUIVALENT LENGTH OF FITTINGS
The flow rates, friction loss data and velocities for a few given flow rates are shown in the catalog. This data are the result of calculations made with the Hazen-Williams formula which is also given in the catalog. The data shows the results of the calculations showing the friction loss of the pipe in feet per hundred feet (ft./100 ft.) of pipe. Note that the pipe data in the catalog includes the losses associated with one joint (socket or butt-weld) per length of pipe, so these joints do not need to be added separately to the friction loss calculations. The velocity of the fluid flowing in the pipe per pipe size and gpm flow rate is also calculated.
Friction loss data for the fittings is found on the pages of the catalog directly following the friction loss data for the pipe. These pages show the friction loss of the fittings in equivalent lengths of straight pipe.
For example an 8” 90 deg. elbow SDR 17.6 shows an equivalent length of 14.0 ft. This means that for every 8” 90 deg. elbow you have in your system you need to add 14.0 ft. equivalent length of pipe to the total length of pipe in your system when calculating the total friction loss.
Engineers and system designers need to know equivalent lengths of fittings and the total length of pipe in a system in order to size the system pump and ensure proper flows and pressures at fixtures and equipment.
Consider the sketch below of a pumping/piping system. It can be a cooling system or a heating system or both. If we need to determine the pump size, we need to know the length of pipe involved in the system, the resulting friction head and the flow rate in gallons per minute.
We can set up a table that looks like this:
With the total system pipe length determined (725.6 ft.) and the flow rate in gpm (525 gpm) we can size the pump. Using the Hazen-Williams formula we find that 6” pipe at 525 gpm has a friction head of 2.5 ft./100 ft. In other words for every 100 feet of pipe we will lose 2.5 ft. of head.
Another way of looking at it is friction loss in psi. In our case, we lose 1.1 psi per 100 ft. of pipe; so for every 100 feet of pipe we lose 1.1 psi of head.
With the friction head of 2.5 ft./100 ft. of pipe we can multiply this friction head by the total length of pipe, including the equivalent lengths of pipe for the fittings:
2.5 ft/100 ft. x 725.6 ft. =18.14 ft.
Our pump will need to have a head capacity of at least 18.14 ft. The velocity in the system will be 6.9 ft. per second.
cloud_download Tech Bulletin
arrow_upward
About Us
FAQ
Contact
Terms of Use
Privacy Policy
Congtogel adalah agen toto macau sekaligus situs togel terlengkap yang menyediakan bonus menarik bagi seluruh penggemar bandar toto togel terpercaya.
;)Facebook
;)Twitter
;)LinkedIn
825 West 600 North
Lindon, Utah 84042
(801) 805-6657
© 2025 aquatherm |
9913 | https://pounds-to-ounces.appspot.com/40-pounds-to-ounces.html | 40 Pounds To Ounces Converter | 40 lbs To oz Converter
Pounds To Ounces
40 lbs to oz
40 Pound to Ounces
40 Pound to Ounce converter
lbs
=
oz
How to convert 40 pound to ounces?
40 lbs 16.0 oz=640.0 oz
1 lbs
A common question isHow many pound in 40 ounce?And the answer is 2.5 lbs in 40 oz. Likewise the question how many ounce in 40 pound has the answer of 640.0 oz in 40 lbs.
How much are 40 pounds in ounces?
40 pounds equal 640.0 ounces (40lbs = 640.0oz). Converting 40 lb to oz is easy. Simply use our calculator above, or apply the formula to change the length 40 lbs to oz.
Convert 40 lbs to common mass
| Unit | Mass |
--- |
| Microgram | 18143694800.0 µg |
| Milligram | 18143694.8 mg |
| Gram | 18143.6948 g |
| Ounce | 640.0 oz |
| Pound | 40.0 lbs |
| Kilogram | 18.1436948 kg |
| Stone | 2.8571428571 st |
| US ton | 0.02 ton |
| Tonne | 0.0181436948 t |
| Imperial ton | 0.0178571429 Long tons |
What is 40 pounds in oz?
To convert 40 lbs to oz multiply the mass in pounds by 16.0. The 40 lbs in oz formula is [oz] = 40 16.0. Thus, for 40 pounds in ounce we get 640.0 oz.
40 Pound Conversion Table
Further pounds to ounces calculations
39 Pounds to oz
39.1 lbs to oz
39.2 Pounds to oz
39.3 lbs to Ounce
39.4 Pounds to oz
39.5 lbs to Ounce
39.6 Pounds to oz
39.7 Pounds to oz
39.8 Pounds to Ounces
39.9 lbs to oz
40 lbs to oz
40.1 lbs to Ounce
40.2 lbs to oz
40.3 lbs to Ounces
40.4 Pounds to Ounce
40.5 Pounds to oz
40.6 Pounds to oz
40.7 lbs to oz
40.8 lbs to Ounce
40.9 lbs to oz
41 lbs to Ounce
Alternative spelling
40 lbs to oz, 40 lbs to Ounce, 40 lbs in Ounce, 40 lb to Ounces, 40 lb in Ounces, 40 lb to oz, 40 lb in oz, 40 Pounds to Ounces, 40 Pound to Ounce, 40 Pound to oz, 40 Pound in oz, 40 lb to Ounce, 40 Pounds to oz,
Further Languages
40 Pounds To Ounces
40 Фунт в Унция
40 Libra Na Unce
40 Pund Til Unse
40 Pfund In Unze
40 λίμπρα σε Ουγγιά
40 Libra En Onza
40 Nael Et Unts
40 Pauna Unssi
40 Livre En Once
40 Funta U Unca
40 Font Uncia
40 Libbra In Oncia
40 Svaras Iki Uncija
40 Lira Fil Uqija
40 Pond Naar Ons
40 Funt Na Uncja
40 Libra Em Onça
40 livră în Uncie
40 Libra Na Unca
40 Pund Till Uns
40 Pond In ons
40 أونصة إلى رطل
40 Narınlamaq Unsiyası
40 পাউন্ড মধ্যে আউন্স
40 Lliura A Unça
40 पाउण्ड से औंस
40 Pon Ke Ons
40 オンスにポンド
40 파운드 온스
40 Libra Til Unse
40 Фунт в Унция
40 Funt V Unča
40 Funta Në Dërhemi
40 ปอนด์เพื่อออนซ์
40 પાઉન્ડ ઔંસ
40 Pound Ons
40 Фунт в Унція
40 Pound Sang Ounce
40 磅为盎司
40 磅至盎司
40 Pound To Ounce
Sitemap 0.1 - 100
Sitemap 101 - 1000
Sitemap 1001 - 2000
Sitemap 2001 - 3000
Sitemap 3001 - 4000
Sitemap 4001 - 5000
Sitemap 5005 - 10000
Sitemap 10100 - 100000
Impressum
© Meta Technologies GmbH |
9914 | https://foodsafety.institute/research-methodology/role-of-hypotheses-in-scientific-research/ | Research Methodology
The Role of Hypotheses in Scientific Research
A hypothesis serves as the backbone of scientific research, providing a tentative explanation or prediction about the relationship between variables. This educated guess guides researchers through the complex journey of discovery, helping them formulate clear research questions and design appropriate methodologies. Without a well-crafted hypothesis, researchers risk wandering aimlessly through data collection and analysis, potentially wasting valuable time and resources on inconclusive results.
Table of Contents
What is a hypothesis?
Types of hypotheses in research
Research and null hypotheses
Directional vs. non-directional hypotheses
Simple vs. complex hypotheses
The process of hypothesis formulation
Starting with observation and curiosity
Reviewing existing literature
Identifying variables
Stating the hypothesis formally
Testing hypotheses: From theory to evidence
Designing appropriate research methods
Statistical testing and analysis
Drawing conclusions
Common pitfalls in hypothesis formulation and testing
Confirmation bias
Overgeneralization
Confusing correlation with causation
p-hacking and data dredging
The value of hypotheses in the scientific method
Providing direction and focus
Facilitating communication
Enabling cumulative knowledge
Bridging theory and practice
Evolving approaches to hypotheses in modern research
Exploratory vs. confirmatory research
Bayesian approaches
Big data and machine learning
What is a hypothesis?
At its core, a hypothesis is a proposed explanation for a phenomenon based on limited evidence, serving as a starting point for further investigation. It represents the researcher’s best guess about how variables might interact or what outcome might result from specific conditions. Unlike casual assumptions we make in everyday life, scientific hypotheses are formulated based on existing knowledge, preliminary observations, and theoretical frameworks.
A well-formulated hypothesis has several key characteristics:
Testability: It must be possible to evaluate the hypothesis through observation or experimentation
Falsifiability: There must be a potential observation or experiment that could prove the hypothesis false
Clarity: It should be stated in clear, precise language that avoids ambiguity
Relevance: It should address the research question and be connected to existing knowledge
Specificity: It should identify the variables involved and their proposed relationship
Types of hypotheses in research
Understanding the different types of hypotheses can help researchers select the most appropriate form for their specific research questions.
Research and null hypotheses
The most fundamental distinction in hypothesis formulation is between research (alternative) and null hypotheses:
Research hypothesis (H₁): This states that there is a relationship between variables or a difference between groups. For example, “Students who study for more than 5 hours score higher on exams than those who study less.”
Null hypothesis (H₀): This states that there is no relationship between variables or no difference between groups. It’s what researchers attempt to reject through statistical testing. Following our example: “There is no difference in exam scores between students who study more than 5 hours and those who study less.”
Directional vs. non-directional hypotheses
Research hypotheses can be further classified based on whether they specify the direction of the relationship:
Directional hypothesis: Predicts the specific direction of the relationship or effect. For instance, “Increased consumption of vitamin C reduces the duration of common cold symptoms.”
Non-directional hypothesis: Suggests a relationship exists but doesn’t specify its direction. For example, “There is a relationship between vitamin C consumption and duration of common cold symptoms.”
Simple vs. complex hypotheses
Hypotheses can also vary in complexity:
Simple hypothesis: Suggests a relationship between two variables. For example, “Exercise frequency is related to reduced stress levels.”
Complex hypothesis: Proposes a relationship between multiple variables. For instance, “Exercise frequency, sleep quality, and dietary habits together influence stress levels.”
The process of hypothesis formulation
Developing a robust hypothesis is not a random process but follows a systematic approach that enhances its scientific value.
Starting with observation and curiosity
Most hypotheses begin with observation of a phenomenon that sparks curiosity. For example, a researcher might notice that students who participate in class discussions seem to perform better on exams. This observation leads to questions: Is there a relationship between class participation and exam performance? What factors might influence this relationship?
Reviewing existing literature
Before formulating a hypothesis, researchers must conduct a thorough review of existing literature. This helps them understand what is already known about the topic, identify gaps in knowledge, and build upon previous findings. A literature review might reveal that while several studies have examined the relationship between class participation and exam performance, few have considered the mediating role of learning styles.
Identifying variables
Clear identification of variables is crucial for hypothesis formulation. Researchers must specify:
Independent variables: The factors manipulated or selected by the researcher
Dependent variables: The outcomes or effects measured in the study
Control variables: Additional factors that might influence the relationship and need to be controlled
In our class participation example, the independent variable might be the frequency of participation, the dependent variable could be exam scores, and control variables might include prior academic performance, attendance, and study habits.
Stating the hypothesis formally
Once variables are identified, the hypothesis can be stated formally. A well-structured hypothesis typically follows the “If… then…” format or clearly states the expected relationship between variables. For example: “Students who participate in class discussions at least three times per week will achieve higher exam scores compared to those who participate less frequently.”
Testing hypotheses: From theory to evidence
The formulation of a hypothesis is just the beginning. Its true value lies in how it guides the testing process and leads to meaningful conclusions.
Designing appropriate research methods
The hypothesis directly influences the research design. Different types of hypotheses may require different methodological approaches:
Experimental designs: Best suited for testing causal relationships between variables
Correlational designs: Appropriate for examining relationships without manipulating variables
Qualitative approaches: Useful for exploring complex phenomena where variables cannot be easily isolated
The research design must enable collection of data that can either support or refute the hypothesis. For our class participation example, a researcher might design a study comparing exam scores between two groups of students with different participation rates, while controlling for other relevant factors.
Statistical testing and analysis
Once data is collected, statistical tests are used to determine whether the results support or reject the hypothesis. The choice of statistical test depends on the type of data and the nature of the hypothesis:
Parametric tests: Such as t-tests, ANOVA, or regression analysis for normally distributed data
Non-parametric tests: Such as chi-square, Mann-Whitney U, or Kruskal-Wallis for data that doesn’t meet parametric assumptions
Statistical significance is typically set at p < 0.05, meaning that there is less than a 5% probability that the observed results occurred by chance. If the p-value is below this threshold, researchers typically reject the null hypothesis and accept the alternative (research) hypothesis.
Drawing conclusions
Based on statistical analysis, researchers draw conclusions about their hypotheses:
Rejecting the null hypothesis: If statistical tests show significant results, researchers reject the null hypothesis and accept the alternative hypothesis
Failing to reject the null hypothesis: If results are not statistically significant, researchers cannot reject the null hypothesis (though this doesn’t necessarily mean the null hypothesis is true)
It’s important to note that hypotheses are never “proven” in an absolute sense. Rather, they are supported or not supported by evidence. Scientific knowledge advances through the accumulation of supporting evidence over time and across multiple studies.
Common pitfalls in hypothesis formulation and testing
Even experienced researchers can fall into traps when formulating and testing hypotheses. Being aware of these common pitfalls can help avoid them:
Confirmation bias
Researchers may unconsciously favor evidence that confirms their hypotheses while dismissing contradictory findings. To counter this, researchers should actively seek evidence that might disprove their hypotheses and be open to unexpected results.
Overgeneralization
Drawing broad conclusions from limited data can lead to invalid generalizations. Researchers must be cautious about the scope of their conclusions and acknowledge the limitations of their studies.
Confusing correlation with causation
Observing a relationship between variables doesn’t necessarily mean one causes the other. Additional evidence and specific research designs are required to establish causality.
p-hacking and data dredging
These practices involve manipulating data or analyses until statistically significant results emerge. Such approaches undermine the scientific process and can lead to false conclusions. Pre-registering hypotheses and analysis plans before data collection can help prevent these issues.
The value of hypotheses in the scientific method
Despite potential pitfalls, well-formulated hypotheses remain essential to scientific progress for several reasons:
Providing direction and focus
Hypotheses narrow the scope of investigation, allowing researchers to focus their efforts on specific aspects of complex phenomena. This focused approach leads to more efficient use of resources and more meaningful results.
Facilitating communication
Clear hypotheses make it easier for researchers to communicate their ideas and findings to colleagues, reviewers, and the broader scientific community. This facilitates collaboration and builds upon existing knowledge.
Enabling cumulative knowledge
The systematic testing of hypotheses contributes to the accumulation of scientific knowledge over time. Even rejected hypotheses provide valuable information that helps refine theories and generate new hypotheses.
Bridging theory and practice
Hypotheses connect abstract theoretical ideas with concrete observations and measurements, allowing researchers to test and refine theoretical frameworks based on empirical evidence.
Evolving approaches to hypotheses in modern research
As research methodologies continue to evolve, so too do approaches to hypothesis formulation and testing:
Exploratory vs. confirmatory research
While traditional hypothesis testing follows a confirmatory approach, modern research often incorporates exploratory phases where patterns and relationships are examined without rigid hypotheses. This can lead to new insights that inform subsequent hypothesis-driven research.
Bayesian approaches
Bayesian statistics provide an alternative to traditional null hypothesis significance testing. Rather than simply rejecting or failing to reject hypotheses, Bayesian approaches update the probability of hypotheses based on new evidence, allowing for more nuanced conclusions.
Big data and machine learning
With the advent of big data and machine learning algorithms, researchers can now identify patterns and relationships in vast datasets that might not have been hypothesized in advance. This has led to debates about the role of traditional hypothesis testing in data-driven research.
What do you think? Is the traditional approach to hypothesis testing still relevant in the age of big data and artificial intelligence? How might researchers balance the benefits of exploratory data analysis with the rigor of hypothesis-driven research?
How useful was this post?
Click on a star to rate it!
Average rating 0 / 5. Vote count: 0
No votes so far! Be the first to rate this post.
We are sorry that this post was not useful for you!
Let us improve this post!
Tell us how we can improve this post?
Comments
Leave a Reply Cancel reply
Research Methodology
1 Selection of Research Problem
Science and Characteristics of Scientific Knowledge
Need for Scientific Methodology
Identification of Research Problem
Statement of the Problem and Objectives
2 Review of Literature
Review of Literature: Sources and Classification
Uses of Review of Literature
Steps in Review of Literature
Writing Review of Literature and Theoretical Orientation
Citation
Writing Bibliographical Details of a Reference
3 Concept and Variables, Formulation and Testing of Hypothesis
Concept, Construct and Variables
Types of Variables
Hypothesis
Types and Forms of Hypothesis
Characteristics, Function and Testing of Hypothesis
4 Research Design
Characteristics of Research Design
Criteria of a Research Design
Max-Min-Con Principle
Classification of Research Design
Experimental Research Design
Descriptive Research Design
5 Descriptive and Survey Research Design
Characteristics of Descriptive Research Design
Steps in Descriptive Research
Aims of Descriptive Research Design
Types of Descriptive Research Design
Case Studies
Observational Studies
Historical Studies
Field Studies
Diagnostic Studies
Explorative Studies
Longitudinal Studies
Correlational Studies
Cross-Sectional Studies
Action Research
Evaluation Research
Survey Research
6 Experimental Research
Testing of hypothesis
t-test
χ2-test
F-test
Principles of Experimental Designs
Completely Randomised Designs
Randomized Complete Block Design
Latin Square Design
Factorial Experiments
2n factorial experiment
3n factorial experiment
7 Levels of Measurement
Concept of Measurement
Postulates of Measurement
Nominal Scale
Ordinal Scale
Interval Scale
Ratio Scale
8 Knowledge Test Constructions
Knowledge Test
Characteristics of a Good Test
Steps in Standardised Test Construction
Item Analysis
Writing Test Items
Preliminary Administration
Reliability of the Final Test
Validity of the Final Test
Norms of the Final Test
Item Difficulty and Discrimination
9 Data Collection
Secondary Data Sources
Instruments Used for Collecting Primary Data
Validity, Data Editing, and Coding
Data Tabulation and Presentation
10 Sampling Technique
Importance of Sampling
Types of Sampling Techniques
Probability based Sampling Techniques
Non-Probability based Sampling Techniques
Sample Size Determination
Sampling and Non-Sampling Errors
11 Quantitative Techniques
Frequency Distribution
Measures of Central Tendency
Measures of Dispersion
Correlation
Regression
Multiple Regressions
Dummy Variable Analysis
Discriminant Function Analysis
Factor Analysis
Principal Component Analysis
12 Qualitative Techniques
Observation Method
Interview Method
Questionnaire Method
Case Study Method
Projective Techniques
13 Statistical Analysis and Packages
χ2- test
t-test
F-test
Basic Experimental Designs
Factorial Experiments
Non-Parametric Tests
Run Test
Sign Test
Wilcoxon Signed Rank Test
Mann-Whitney U-Test
Kruskal-Wallis One-way Analysis of Variance
Friedman Two-way Analysis of Variance
14 Report Writing
Research Report
Steps in Preparing the Report: Preliminary Considerations
Main Components of a Research Report
Diagrammatic Presentation
Common Weaknesses in Research Report Writing
Share This
Share on Mastodon |
9915 | https://www.wikiwand.com/en/articles/Palindromic_polynomial | Reciprocal polynomial
Polynomial with reversed root positions From Wikipedia, the free encyclopedia
In algebra, given a polynomial
| | |
--- |
| | This article needs additional citations for verification. (April 2021) |
| | |
--- |
| | This article needs editing to comply with Wikipedia's Manual of Style. In particular, it has problems with MOS:FORMULA - avoid mixing <math>...</math> and {{math}} in the same expression. (July 2025) |
<math>...</math>
with coefficients from an arbitrary field, its reciprocal polynomial or reflected polynomial, denoted by p∗ or pR, is the polynomial
That is, the coefficients of p∗ are the coefficients of p in reverse order. Reciprocal polynomials arise naturally in linear algebra as the characteristic polynomial of the inverse of a matrix.
In the special case where the field is the complex numbers, when
the conjugate reciprocal polynomial, denoted p†, is defined by,
where
a
i
¯
{\displaystyle {\overline {a_{i}}}}
denotes the complex conjugate of
a
i
{\displaystyle a_{i}}
, and is also called the reciprocal polynomial when no confusion can arise.
A polynomial p is called self-reciprocal or palindromic if p(x) = p∗(x).
The coefficients of a self-reciprocal polynomial satisfy ai = an−i for all i.
Properties
Reciprocal polynomials have several connections with their original polynomials, including:
Other properties of reciprocal polynomials may be obtained, for instance:
Palindromic and antipalindromic polynomials
A self-reciprocal polynomial is also called palindromic because its coefficients, when the polynomial is written in the order of ascending or descending powers, form a palindrome. That is, if
is a polynomial of degree n, then P is palindromic if ai = an−i for i = 0, 1, ..., n.
Similarly, a polynomial P of degree n is called antipalindromic if ai = −an−i for i = 0, 1, ..., n. That is, a polynomial P is antipalindromic if P(x) = –P∗(x).
Examples
From the properties of the binomial coefficients, it follows that the polynomials P(x) = (x + 1)n are palindromic for all positive integers n, while the polynomials Q(x) = (x – 1)n are palindromic when n is even and antipalindromic when n is odd.
Other examples of palindromic polynomials include cyclotomic polynomials and Eulerian polynomials.
Properties
Real coefficients
A polynomial with real coefficients all of whose complex roots lie on the unit circle in the complex plane (that is, all the roots have modulus 1) is either palindromic or antipalindromic.
Conjugate reciprocal polynomials
A polynomial is conjugate reciprocal if
p
(
x
)
≡
p
†
(
x
)
{\displaystyle p(x)\equiv p^{\dagger }(x)}
and self-inversive if
p
(
x
)
=
ω
p
†
(
x
)
{\displaystyle p(x)=\omega p^{\dagger }(x)}
for a scale factor ω on the unit circle.
If p(z) is the minimal polynomial of z0 with |z0| = 1, z0 ≠ 1, and p(z) has real coefficients, then p(z) is self-reciprocal. This follows because
So z0 is a root of the polynomial
z
n
p
(
z
¯
−
1
)
¯
{\displaystyle z^{n}{\overline {p({\bar {z}}^{-1})}}}
which has degree n. But, the minimal polynomial is unique, hence
for some constant c, i.e.
c
a
i
a
n
−
i
¯
=
a
n
−
i
{\displaystyle ca_{i}={\overline {a_{n-i}}}=a_{n-i}}
. Sum from i = 0 to n and note that 1 is not a root of p. We conclude that c = 1.
A consequence is that the cyclotomic polynomials Φn are self-reciprocal for n > 1. This is used in the special number field sieve to allow numbers of the form x11 ± 1, x13 ± 1, x15 ± 1 and x21 ± 1 to be factored taking advantage of the algebraic factors by using polynomials of degree 5, 6, 4 and 6 respectively – note that φ (Euler's totient function) of the exponents are 10, 12, 8 and 12.[citation needed]
Per Cohn's theorem, a self-inversive polynomial has as many roots in the unit disk
{
z
∈
C
:
|
z
|
<
1
}
{\displaystyle {z\in \mathbb {C} :|z|<1}}
as the reciprocal polynomial of its derivative.
Application in coding theory
The reciprocal polynomial finds a use in the theory of cyclic error correcting codes. Suppose xn − 1 can be factored into the product of two polynomials, say xn − 1 = g(x)p(x). When g(x) generates a cyclic code C, then the reciprocal polynomial p∗ generates C⊥, the orthogonal complement of C.
Also, C is self-orthogonal (that is, C ⊆ C⊥), if and only if p∗ divides g(x).
See also
Notes
References
External links
Wikiwand - on
Seamless Wikipedia browsing. On steroids.
Wikiwand ❤️ Wikipedia |
9916 | https://or.stackexchange.com/questions/7394/can-you-calculate-the-mean-of-some-mip-variables-using-linear-constraints | Skip to main content
Operations Research
Can you calculate the mean of some MIP variables using linear constraints?
Ask Question
Asked
Modified
3 years, 8 months ago
Viewed
529 times
This question shows research effort; it is useful and clear
4
Save this question.
Show activity on this post.
got a lingering question from a graduate course in integer programming that's been bugging me ever since.
Is it possible to find the mean of some variables in a MIP without resorting to quadratic constraints?
Here's an example of what I mean:
Suppose there are N cities you could travel to and you've given each city a rating, r. In some goal program style MIP you create a binary variable, Xi, to indicate whether or not you will be travelling to city i.
Finding the average rating you've attributed to a city is simple enough:
r¯=1N∑iri
But what if you wanted to find the average rating of the cities the program has selected, i.e., those whos Xi=1?
This is where I am stuck - the calculations I have to find this new average are pretty simple:
R¯=1∑iXi∑iriXi
But because I am multiplying R¯ and ∑iXi, both decision variables that the model calculates based on its solution, the constraint used to find R¯ is quadratic.
For clarification - in this problem, you're trying to find a subtour of the N cities to travel to, so it cannot be assumed that ∑iXi=N. The number of cities you travel to, ∑iXi, is in itself a decision variable.
Is there any way to make this kind of calculation linear?
The aforementioned graduate class seemed to imply such a thing could be done, but I could have just misinterpreted what the lecture was saying at that time.
As of yet, I haven't been able to come up with a way to do this, so I'm wondering if the smart people of the internet might be able to tell me for certain if its impossible or not.
mixed-integer-programming
linear-programming
linearization
quadratic-programming
indicator-constraints
Share
CC BY-SA 4.0
Improve this question
Follow this question to receive notifications
edited Dec 8, 2021 at 23:22
SecretAgentMan
1,95522 gold badges1616 silver badges4242 bronze badges
asked Dec 7, 2021 at 19:16
gjgutier545gjgutier545
16099 bronze badges
4
2
What role does the average play in the model? Are you trying to optimize it or maybe bound it?
RobPratt
–
RobPratt
12/07/2021 19:41:26
Commented
Dec 7, 2021 at 19:41
The actual problem I've pulled this example from is a larger goal program where I'm solving a TSP, but instead of minimizing the distance traveled while hitting every city I'm minimizing the negative deviation from the original average rating (among other things). So, the TSP formulation gets the subset of cities to travel to and calculates our new average. I want this new average to deviate from the original as minimally as possible. Hope that explains it well enough!
gjgutier545
–
gjgutier545
12/07/2021 19:49:28
Commented
Dec 7, 2021 at 19:49
1
For TSP, every city is visited, so ∑iXi=N, a constant.
RobPratt
–
RobPratt
12/07/2021 19:53:07
Commented
Dec 7, 2021 at 19:53
Aye, but this isn't an exact TSP formulation. Subtours are encouraged in this formulation so ∑iXi doesn't necessarily have to equal N in this case.
gjgutier545
–
gjgutier545
12/07/2021 19:54:54
Commented
Dec 7, 2021 at 19:54
Add a comment
|
1 Answer 1
Reset to default
This answer is useful
3
Save this answer.
Show activity on this post.
You can do it if you have a fairly high pain tolerance. :-)
Let's start by introducing binary variables Z1,…,ZK (where K is the dimension of X) and making N a variable. Add constraints
∑j=1KZj=1,
N=∑j=1Kj⋅Zj
and
∑j=1KXj=N.
I'm assuming here that at least one of the X variables will be 1, since N=0 makes the mean a bit undefined.
Now comes the fun part. If variable Y is supposed to be the mean, we want to set
Y⋅N=∑j=1Krj⋅Xj.
The right side is linear, so that's no problem. The left side expands to ∑Kj=1j⋅Zj⋅Y. Each term is a product of a general variable (Y) and a binary variable (Zj). Rob's answer to Mixed-integer optimization with bilinear constraint shows how to linearize those by adding more variables and constraints, assuming that you can bound Y.
Share
CC BY-SA 4.0
Improve this answer
Follow this answer to receive notifications
answered Dec 7, 2021 at 20:11
prubin♦prubin
42.1k33 gold badges4343 silver badges114114 bronze badges
2
With this and Rob's answer you linked to, it looks like everything is falling into place! Took a bit of finagling, but I managed to work this approach into my problem and all appears to be working(: Very much appreciated!
gjgutier545
–
gjgutier545
12/08/2021 01:50:41
Commented
Dec 8, 2021 at 1:50
Glad it worked. Mathematically it's correct, but "big M" models actually solving in reasonable time is somewhat hit and miss.
prubin
–
prubin
♦
12/08/2021 17:27:53
Commented
Dec 8, 2021 at 17:27
Add a comment
|
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
mixed-integer-programming
linear-programming
linearization
quadratic-programming
indicator-constraints
See similar questions with these tags.
Linked
7
Mixed-integer optimization with bilinear constraint
Related
9
Static stochastic knapsack problem: unbounded version
7
Why is there not a feasible solution for a MIP?
5
Linearizing objective function with variables inside an indicator function
4
Name for subclass of ILP without any inequality constraints (including constraints on x)
11
Efficient way to solve "easy" quadratic optimization problem
5
What is the go-to practical method for optimizing separable quadratic programs?
2
Finding a maximum violated infeasible subset
0
How can I calculate the UB if I'm using Fischetti's Alternative Cut Generation for Benders'?
Hot Network Questions
What does it mean to "rotate the knight"?
How does the Linux kernel decide whether to deny memory allocation or invoke the OOM killer when a cgroup exceeds its memory limit?
LuaLaTeX Accessible PDF not validated by Yuja Panorama
Chinese periodic table of elements (元素周期表)
Strange or normal ohmeter measurements of IRF630 N-channel mosfets
Why does a new Windows 11 PC have repeated paths in %PATH%?
What's a plausible way for a creature to be petrified into a statue-like form without being harmed?
Fiery tsunami on a sunny day
What will happen when a pull request includes changes to ignored .gitignore?
Alter table alter column does not work with user defined types
Is the gravitational field around a black hole identical to space being "denser?"
What are the years N in the Gregorian Calendar, such that duplicating any digit leaves the calendar unchanged?
Salesforce CLI not working, not able to retrieve files to the VS code
Float vs Decimal
How can I force reevaluation of Association values?
Does this video portray a river being drained into the ocean after a beachgoer digs a trench with a shovel?
A single man to save mankind
How the inductor current doesn't become immediately zero even when there is no path to flow the current?
Package csvsimple inserts a spurious greek letter capital psi when in tab separated mode. How do I not get that?
垂れ幕 pitch accent
Name for moduli space of torsion-free sheaves in Nakajima’s “Lectures on Hilbert Schemes…”
P channel MOSFET for Reverse polarity Protection
Could the UN General Assembly meet in Vienna, Geneva, or Nairobi?
How do I prevent bacteria and fungi from growing on PVA hydrogel?
Question feed
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. |
9917 | https://www.teachervision.com/back-school-headquarters/middle-school-math-diagnostic-pre-assessment | Math Diagnostic Assessment for Middle School - Identify Learning Gaps - TeacherVision
Skip to main content
Site search input
Sign In
Sign Up
SubjectsArt Health and Safety Language Arts and Writing Languages Mathematics Music Physical Education Library Reading and Literature Geography Science Government Social Studies and History 21st Century Skills-SEL All Subjects
GradesEarly Learning Pre-K Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade 9th Grade 10th Grade 11th Grade 12th Grade All Grades
Report Card Comments50 Quick Report Card Comments Academic Achievement & Improvement Advice for Parents All Subjects Character and Social Traits Citizenship Distance Learning and Homeschooling ESL Students End of Year Essentials Kit Handwriting Kindergarten Language Arts and Reading Math Personality & Attitude Positive Descriptions of Student Behavior Preschool Science Social Studies Work and Study Habits All Report Card Comments
Warm Ups2nd Grade Hispanic Heritage Month 4th Grade 5th Grade 6th Grade 7th Grade Animal Habitats High School Grammar Packet Language Arts and Writing Math Middle School Grammar Packet Mountains Numbers and Numeration Reading All Warm Ups
Graphic OrganizersTop 10 Most Popular Analogy Organizer Argumentative Writing Cause and Effect Chart, Three-Column Citing My Sources Five-Paragraph Essay General KWL Chart Literature Math Research Paper Outline Setting SMART Goals Using "KWL" In Your Classroom Strategy Writing Writing Research Papers All Graphic Organizers
Blog
Site search input
SubjectsArtHealth and SafetyLanguage Arts and WritingLanguagesMathematicsMusicPhysical EducationLibraryReading and LiteratureGeographyScienceGovernmentSocial Studies and History21st Century Skills-SEL
GradesEarly LearningPre-KKindergarten1st Grade2nd Grade3rd Grade4th Grade5th Grade6th Grade7th Grade8th Grade9th Grade10th Grade11th Grade12th Grade
Report Card Comments50 Quick Report Card CommentsAcademic Achievement & ImprovementAdvice for ParentsAll SubjectsCharacter and Social TraitsCitizenshipDistance Learning and HomeschoolingESL StudentsEnd of YearEssentials KitHandwritingKindergartenLanguage Arts and ReadingMathPersonality & AttitudePositive Descriptions of Student BehaviorPreschoolScienceSocial StudiesWork and Study Habits
Warm Ups2nd GradeHispanic Heritage Month4th Grade5th Grade6th Grade7th GradeAnimal HabitatsHigh School Grammar PacketLanguage Arts and WritingMathMiddle School Grammar PacketMountainsNumbers and NumerationReading
Graphic OrganizersTop 10 Most PopularAnalogy OrganizerArgumentative WritingCause and EffectChart, Three-ColumnCiting My SourcesFive-Paragraph EssayGeneral KWL ChartLiteratureMathResearch Paper OutlineSetting SMART GoalsUsing "KWL" In Your Classroom StrategyWritingWriting Research Papers
Blog
PlanningAwards & CertificatesBack-to-School HeadquartersBlog ArticlesBulletin BoardsClassroom ActivitiesClassroom FormsClassroom Passes & NotesEducator’s CalendarGamesGraphic OrganizersGraphs & ChartsLesson PlansLiterature GuidesMini-LessonsProject-based LearningRubricsThemed PacketsVideos & ActivitiesWorksheets
StrategiesAssessmentBehavior ManagementClassroom ManagementClassroom OrganizationEducational TechnologyEnglish-Language LearnersGetting to Know Your StudentsIcebreakersNew Teacher ResourcesOpen House ResourcesProfessional DevelopmentRemote LearningReport Card Comments & PhrasesSpecial NeedsStudy Skills and Test PrepTeacher-Parent CollaborationTeaching Tips and Advice
HolidaysAsian Pacific American Heritage MonthBlack History MonthChinese New YearChristmasColombus DayDay of the DeadEarth DayEasterFather’s DayFlag DayFourth of JulyGroundhog DayHalloweenHanukkahHispanic Heritage MonthKwanzaaLabor DayMartin Luther King DayMemorial DayMother’s DayNative American Heritage MonthNew Year’s DayPassoverPresident’s DaySt Partick’s DayTeacher Appreication WeekThanksgivingValentine’s DayVeteran’s DayWomen’s History Month
ThemesAll Kinds of WeatherAnimalsAutumnBack to SchoolBiographiesCommunity Helpers and JobsDinosaursFamous ExplorersFamilyGreen ActivitiesHeroesHuman and Animal HomesHealth & NutritionInsects and BugsInventors and InventionsOceansPlantsSpaceSpringSummerWinterWomen in STEM
Home Learning
Self-Care
Daily Warm-Ups
Premium Resources
For Schools
Games
Sign up
Sign in
Assessment
Mathematics
Middle School Math Diagnostic Pre-Assessment
Middle School Math Diagnostic Pre-Assessment
Download
Save to Drive
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
SHARE
Authored by: Dwayne Vega
Last edited: August 18, 2021
Grade:
6
7
8
Subjects:
Mathematics
show tags
Simplified Content Type
Assessment
Holidays:
Back to School Headquarters
Teaching Strategies:
Assessment
This middle school diagnostic math assessment for back to school is designed to help you benchmark your students' skills at the beginning of the year and plan for intervention and remediation of learning gaps resulting from remote, hybrid, and quarantine instruction, or summer learning loss.
This assessment for middle school grades (6-8) includes:
15 mixed problems focused on standards-aligned skills, divided into 4 sections:
Ratios and Proportional Relationships
The Number System
Expressions and Equations
Statistics and Probability
An answer key and detailed recommendation rubric
A scoring guide with suggestions for skills to review based on student performance on the assessment.
Skills measured in this assessment include:
Analyze proportional relationships and use them to solve real-world and mathematical problems
Apply and extend previous understandings of operations with fractions to add subtract, multiply, and divide rational numbers.
Solve real-life and mathematical problems using numerical and algebraic expressions and equations
Use random sampling to draw inferences
Featured Middle School Resources
ACTIVITIES
Test Prep Strategies, Tools, and Practice Questions
Help set your students up for academic success with this packet of test-taking tips, test preparation strategies, and pr...
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
ASSESSMENT
Exit Tickets for Middle School Classrooms
Exit tickets are a form of assessment that are often used informally to assess how well students grasped a lesson, what ...
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
CHOICE BOARDS
Geography Activities for Middle School
Help students explore the world and develop their geography knowledge with this packet of geography activities for middl...
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
Related Resources
ACTIVITIES
Getting to Know You Activities for Middle School
Get to know your new batch of middle grade students with fun cross-curricular activities! These get to know you questio...
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
WORKBOOKS
A Grading And Assessment Workbook For Teachers
TeacherVision contributor and instructional coach Julie Mason designed this workbook in order to share best practices fo...
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
ASSESSMENT
Middle School Reading Diagnostic Pre-Assessment
This middle school diagnostic reading assessment for back to school is designed to help you benchmark reading assessment...
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
ASSESSMENT
Middle School Writing Diagnostic Pre-Assessment
This middle school diagnostic writing assessment for back to school is designed to help you benchmark writing assessment...
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
TEACHING RESOURCE
Math Differentiation Strategies and Activities for Middle School
Close learning gaps this Fall with differentiated math instruction This resource provides 5 strategies and 3 editable t...
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
ASSESSMENT
Middle School Pre-Algebra Diagnostic Pre-Assessment
This middle school diagnostic pre-algebra assessment for back to school is designed to help you benchmark your students'...
ADD TO FAVORITES
Hide lower priority columns
| | Add to Folder |
--- |
| - [x] | creative writing |
| - [x] | children's book |
| - [x] | activities |
| - [x] | classroom tools |
| - [x] | language arts and writing |
| - [x] | vocabulary |
Create new folderCREATE NEW FOLDER
About the author
Dwayne Vega
Contributor
About Dwayne
Whilst serving as a soldier in the US Army, Dwayne Vega (B.A., M.Ed,. M.A.) completed a Bachelor’s Degree in English, before going on to attain a Master’s Degree in… Read more
View Dwayne's profile
ADVERTISEMENT|REMOVE ADS
Advertisement
Exclusive PBL Projects
Project-Based Learning
for busy teachers.
Ready to print.
Easy to implement.
Browse all PBL packets »
ADVERTISEMENT|REMOVE ADS
Register
Search our growing library of professionally created teacher resources.
Site search input
With the help of certified and current classroom teachers, TeacherVision creates and vets classroom resources that are accurate, timely, and reflect what teachers need to best support their students.
Teachervision is part of the Sandbox Learning family of educational and reference sites for parents, teachers and students.
Advertise on this site.
Pricing & Benefits
For Schools
Teacher Blog
Newsletters
About Us
Help
Teacher Advisors & Experts
Terms of Use
Privacy Policy
Cancellation Policy
Contact Us
Advertise with Us
Do Not Sell My Info (for CA Residents)
Publisher Partners
Browse Lesson Plans
Browse Worksheets
Browse Teaching Resources
©2025 Sandbox Networks Inc. All rights reserved. Sandbox Learning is part of Sandbox & Co., a digital learning company.
BDOW! |
9918 | https://www.wikidata.org/wiki/Q7843925 | trirectangular tetrahedron - Wikidata
Jump to content
[x] Main menu
Main menu
move to sidebar hide
Navigation
Main page
Community portal
Project chat
Create a new Item
Recent changes
Random Item
Query Service
Nearby
Help
Special pages
Lexicographical data
Create a new Lexeme
Recent changes
Random Lexeme
Search
Search
English
[x] Appearance
Appearance
move to sidebar hide
Text
Small Standard Large
This page always uses small font size
Width
Standard Wide
The content is as wide as possible for your browser window.
Donate
Create account
Log in
[x] Personal tools
Donate
Create account
Log in
Pages for logged out editors learn more
Contributions
Talk
trirectangular tetrahedron(Q7843925)
Item
Discussion
[x] English
Read
View history
[x] Tools
Tools
move to sidebar hide
Actions
Read
View history
General
What links here
Related changes
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Concept URI
Print/export
Create a book
Download as PDF
Printable version
In other projects
From Wikidata
geometrical shape
edit
In more languages
Configure
| Language | Label | Description | Also known as |
--- --- |
| default for all languages | No label defined | – | |
| English | trirectangular tetrahedron | geometrical shape | |
| Spanish | tetraedro trirectangular | No description defined | |
| Traditional Chinese | 直角四面體 | No description defined | |
| Chinese | 三直角四面體 | No description defined | 直角四面体 |
All entered languages
Statements
subclass of
tetrahedron
edit
0 references
add reference
add value
maintained by WikiProject
WikiProject Mathematics
edit
0 references
add reference
add value
add statement
Identifiers
Freebase ID
/m/0hzryhj
edit
0 references
add reference
add value
MathWorld ID
TrirectangularTetrahedron
subject named as
Trirectangular Tetrahedron
edit
1 reference
imported from Wikimedia project
English Wikipedia
add reference
add value
Microsoft Academic ID
131918839
edit
0 references
add reference
add value
add statement
Sitelinks
Collapse
Wikipedia(5 entries)
edit
✕
bgПравоъгълен тетраедър
enTrirectangular tetrahedron
esTetraedro trirrectangular
frTétraèdre trirectangle
ukПрямокутний тетраедр
Wikibooks(0 entries)
edit
✕
Wikinews(0 entries)
edit
✕
Wikiquote(0 entries)
edit
✕
Wikisource(0 entries)
edit
✕
Wikiversity(0 entries)
edit
✕
Wikivoyage(0 entries)
edit
✕
Wiktionary(0 entries)
edit
✕
Multilingual sites(0 entries)
edit
✕
Retrieved from "
This page was last edited on 4 February 2025, at 06:26.
All structured data from the main, Property, Lexeme, and EntitySchema namespaces is available under the Creative Commons CC0 License; text in the other namespaces is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy.
Privacy policy
About Wikidata
Disclaimers
Code of Conduct
Developers
Statistics
Cookie statement
Mobile view
Data access
Search
Search
trirectangular tetrahedron(Q7843925)
Add topic |
9919 | http://www.360doc.com/document/17/0802/10/15143161_676029365.shtml | 第1讲 等比数列
搜索
我的图书馆
查看信箱
系统消息
官方通知
设置
开始对话
有 11 人和你对话，查看 忽略
历史对话记录
通知设置
发文章
发文工具
撰写网文摘手文档视频思维导图随笔相册原创同步助手
其他工具
图片转文字文件清理AI助手
留言交流
×
微信扫一扫关注
查看更多精彩文章
第1讲 等比数列
吴其明的图书馆 2017-08-02 |104阅读|7 转藏 下载 1
转藏分享
QQ空间QQ好友新浪微博微信
460.7K 大小
/26
按Esc退出全屏模式
第1页
本站是提供个人知识管理的网络存储空间，所有内容均由用户发布，不代表本站观点。请注意甄别内容中的联系方式、诱导购买等信息，谨防诈骗。如发现有害或侵权内容，请点击 一键举报。
转藏分享
微信QQ空间QQ好友新浪微博
献花（0） +1
来自： 吴其明的图书馆>《春季实验班讲座》
举报/认领
上一篇：
第十五讲 期末测试
下一篇：
第11讲 组合数学之基本图论练习题
猜你喜欢
0 条评论
发表
请遵守用户评论公约
查看更多评论
类似文章更多
5-3
1.理解等比数列的概念.2.掌握等比数列的通项公式与前n项和公式.3.能在具体的问题情境中识别数列的等比关系，并能用有关知识解决相应的问题.4.了解等比数列与指数函数的关系.2.等比数列的通项公式。考向...
2020年高考理科数学《数列》题型归纳与训练
（2）如果要证明一个数列是等差数列，则必须用定义法或等差中项法．题组二等比数列的判定与证明例2设数列{an}的前n项和为Sn，已知a1=1，Sn＋1=4an＋2.(1)设bn=an＋1?2an，证明：数列{bn}是等比数列；(2...
2018年 高中数学 必修5 数列 解答题 专项练习（含答案）
2018年高中数学必修5数列解答题专项练习。（1）数列的通项公式；6、在数列中，为常数，，且成公比不等于1的等比数列.（I）证明数列是等比数列，并求数列的通项公式；（1）求证数列是等差数列，并求数列...
高中数学小题专练(九) 数列
高中数学小题专练(九) 数列。肖博数学小题专练(九) 数列。2.(2017·洛阳第一次统考)等差数列{an}为递增数列，若 a.和 Sn 等于( )=2，公比 q=3 的等比数列，其前 n 项和 Sn=解析 依题意，an+1=Sn+1-...
4.3.1等比数列的概念课件(1)优质教学课件PPT
选填
1.3.20等比数列
选填
求数列通项公式的常用方法
求数列通项公式的常用方法。例：在数列{an}中，若a1=1，an+1=an+2(n1)，求该数列的通项公式an。例：已知数列{an}的前n项和Sn满足an=SnSn-1(n2)，且a1=-，求数列{an}的通项公式。解：∵an=SnSn-1(n2)，...
高考数学难点突破_难点13__数列的通项与求和
难点13数列的通项与求和。数列是函数概念的继续和延伸，数列的通项公式及前n项和公式都可以看作项数n的函数，是函数思想在数列中的应用.数列以通项为纲，数列的问题，最终归结为对数列通项的研究，而数...
2012高考复习专题限时集训：等差数列与等比数列
2012高考复习专题限时集训：等差数列与等比数列。1．C 【解析】 已知Sn－Sn－3＝51(n>3)＝an－2＋an－1＋an＝3an－1，由此得an－1＝...
吴其明的图书馆
关注对话
TA的最新馆藏
在北京工作22年的片段 [转]北京鲁迅故居游 [转]颐和园中的六座城关 [转]华为手机怎么截图 华为截屏怎么操作 [转]2016第二十一届华罗庚金杯少年数学邀请赛决赛试题解析(小学高年级) [转]关于的中秋，十首唯美古诗，你看了吗？
喜欢该文的人也喜欢更多
×
¥.00
微信或支付宝扫码支付：
开通即同意《个图VIP服务条款》
正在支付中，请勿关闭二维码！
微信支付后，该微信自动注册为你的个人图书馆账号
付费成功，还是不能使用？
复制成功！
绑定账号，享受 特权
恭喜你成为个图VIP！
在打印前，点击“下一步”观看2个提示
下一步
● 电子书免费读
● 全站无广告
● 全屏阅读
● 高品质朗读
● 批量上传文档
● 可关注600人
● 5千个文件夹
● 专属客服
微信支付查找“商户单号”方法：
1.打开微信app，点击消息列表中和“微信支付”的对话
2.找到扫码支付给360doc个人图书馆的账单，点击“查看账单详情”
3.在“账单详情”页，找到“商户单号”
4.将“商户单号”填入下方输入框，点击“恢复VIP特权”，等待系统校验完成即可。
支付宝查找“商户订单号”方法：
1.打开支付宝app，点击“我的”-“账单”
2.找到扫码支付给个人图书馆的账单，点击进入“账单详情”页
3.在“账单详情”页，找到“商家订单号”
4.将“商家订单号”填入下方输入框，点击“恢复VIP特权”，等待系统校验完成即可。
已经开通VIP，还是不能 打印？
请通过以下步骤，尝试恢复VIP特权
第1步 在下方输入你支付的微信“商户单号”或支付宝“商家订单号”
第2步 点击“恢复VIP特权”，等待系统校验完成即可
如何查找商户单号？
恢复VIP特权
正在查询...
订单号过期！
该订单于2020/09/09 23:59:59支付，VIP有效期：2020/09/09 23:59:59至2020/09/11 23:59:59！如需使用VIP功能，建议重新开通VIP
返回上一页
支付成功！
确定
确定复制刚才选中的内容？
确定
复制
分享文章微信QQ空间QQ好友新浪微博
AI解释
复制
×
复制成功！
¥.00
微信或支付宝扫码支付：
开通即同意《个图VIP服务条款》
正在支付中，请勿关闭二维码！
自动续费¥18/月，可随时取消
开通即同意《自动续费服务协议》|《个图VIP服务条款》
如何开发票？
全部>>
● 电子书免费读
● 全站无广告
● 全屏阅读
● 高品质朗读
● 批量上传文档
● 可关注600人
● 5千个文件夹
● 专属客服
×
支付确认
请在手机上打开的页面进行支付；
如支付完成，请点击“支付完成”。
支付完成取消支付 |
9920 | https://bio.libretexts.org/Bookshelves/Microbiology/Microbiology_(OpenStax)/10%3A_Biochemistry_of_the_Genome/10.02%3A_Structure_and_Function_of_DNA | Last updated Apr 21, 2024
Save as PDF
OpenStax
OpenStax
Learning Objectives
Describe the biochemical structure of
deoxyribonucleotides
Identify the base pairs used in the synthesis of
deoxyribonucleotides
Explain why the double helix of
DNA
is described as
antiparallel
In Microbial Metabolism, we discussed three classes of macromolecules: proteins, lipids, and carbohydrates. In this chapter, we will discuss a fourth class of macromolecules: nucleic acids. Like other macromolecules, nucleic acids are composed of monomers, called nucleotides, which are polymerized to form large strands. Each
nucleic acid
strand contains certain nucleotides that appear in a certain order within the strand, called its
base sequence
. The
base sequence
of
deoxyribonucleic acid (DNA)
is responsible for carrying and retaining the hereditary information in a cell. In Mechanisms of Microbial Genetics, we will discuss in detail the ways in which
DNA
uses its own
base sequence
to direct its own synthesis, as well as the synthesis of RNA and proteins, which, in turn, gives rise to products with diverse structure and function. In this section, we will discuss the basic structure and function of
DNA
.
DNA
Nucleotides
The building blocks of nucleic acids are nucleotides. Nucleotides that compose
DNA
are called
deoxyribonucleotides
. The three components of a deoxyribonucleotide are a five-carbon sugar called deoxyribose, a phosphate group, and a
nitrogenous base
, a nitrogen-containing ring structure that is responsible for complementary base pairing between
nucleic acid
strands (Figure 10.2.1). The carbon atoms of the five-carbon deoxyribose are numbered 1ʹ, 2ʹ, 3ʹ, 4ʹ, and 5ʹ (1ʹ is read as “one prime”). A nucleoside comprises the five-carbon sugar and
nitrogenous base
.
Figure 10.2.1: (a) Each deoxyribonucleotide is made up of a sugar called deoxyribose, a phosphate group, and a
nitrogenous base
—in this case,
adenine
. (b) The five carbons within deoxyribose are designated as 1ʹ, 2ʹ, 3ʹ, 4ʹ, and 5ʹ.
The deoxyribonucleotide is named according to the nitrogenous bases (Figure 10.2.2). The nitrogenous bases
adenine
(A) and
guanine
(G) are the
purines
; they have a double-ring structure with a six-carbon ring fused to a five-carbon ring. The
pyrimidines
,
cytosine
(C) and
thymine
(T), are smaller nitrogenous bases that have only a six-carbon ring structure.
Figure 10.2.2: Nitrogenous bases within
DNA
are categorized into the two-ringed
purines
adenine
and
guanine
and the single-ringed
pyrimidines
cytosine
and
thymine
.
Thymine
is unique to
DNA
.
Individual nucleoside triphosphates combine with each other by covalent bonds known as 5ʹ-3ʹ
phosphodiester bonds
, or linkages whereby the phosphate group attached to the 5ʹ carbon of the sugar of one
nucleotide
bonds to the hydroxyl group of the 3ʹ carbon of the sugar of the next
nucleotide
. Phosphodiester bonding between nucleotides forms the
sugar-phosphate backbone
, the alternating sugar-phosphate structure composing the framework of a
nucleic acid
strand (Figure 10.2.3). During the polymerization process, deoxynucleotide triphosphates (dNTP) are used. To construct the
sugar-phosphate backbone
, the two terminal phosphates are released from the dNTP as a pyrophosphate. The resulting strand of
nucleic acid
has a free phosphate group at the 5ʹ carbon end and a free hydroxyl group at the 3ʹ carbon end. The two unused phosphate groups from the
nucleotide
triphosphate are released as pyrophosphate during phosphodiester bond formation. Pyrophosphate is subsequently hydrolyzed, releasing the energy used to drive
nucleotide
polymerization.
Figure 10.2.3:
Phosphodiester bonds
form between the phosphate group attached to the 5ʹ carbon of one
nucleotide
and the hydroxyl group of the 3ʹ carbon in the next
nucleotide
, bringing about polymerization of nucleotides in to
nucleic acid
strands. Note the 5ʹ and 3ʹ ends of this
nucleic acid
strand.
Exercise 10.2.1
What is meant by the 5ʹ and 3ʹ ends of a
nucleic acid
strand?
Discovering the Double Helix
By the early 1950s, considerable evidence had accumulated indicating that
DNA
was the genetic material of cells, and now the race was on to discover its three-dimensional structure. Around this time, Austrian biochemist Erwin Chargaff 1(1905–2002) examined the content of
DNA
in different species and discovered that
adenine
,
thymine
,
guanine
, and
cytosine
were not found in equal quantities, and that it varied from species to species, but not between individuals of the same species. He found that the amount of
adenine
was very close to equaling the amount of
thymine
, and the amount of
cytosine
was very close to equaling the amount of
guanine
, or A = T and G = C. These relationships are also known as Chargaff’s rules.
Other scientists were also actively exploring this field during the mid-20th century. In 1952, American scientist Linus Pauling (1901–1994) was the world’s leading structural chemist and odds-on favorite to solve the structure of
DNA
. Pauling had earlier discovered the structure of protein α helices, using X-ray
diffraction
, and, based upon X-ray
diffraction
images of
DNA
made in his laboratory, he proposed a triple-stranded model of
DNA
.2 At the same time, British researchers Rosalind Franklin (1920–1958) and her graduate student R.G. Gosling were also using X-ray
diffraction
to understand the structure of
DNA
(Figure 10.2.4). It was Franklin’s scientific expertise that resulted in the production of more well-defined X-ray
diffraction
images of
DNA
that would clearly show the overall double-helix structure of
DNA
.
Figure 10.2.4: The X-ray
diffraction
pattern of
DNA
shows its helical nature. (credit: National Institutes of Health)
James Watson (1928–), an American scientist, and Francis Crick (1916–2004), a British scientist, were working together in the 1950s to discover
DNA
’s structure. They used Chargaff’s rules and Franklin and Wilkins’ X-ray diffractionimages of
DNA
fibers to piece together the purine-pyrimidine pairing of the double helical
DNA
molecule (Figure 10.2.5). In April 1953, Watson and Crick published their model of the
DNA
double helix in _Nature_.3 The same issue additionally included papers by Wilkins and colleagues,4 as well as by Franklin and Gosling,5 each describing different aspects of the molecular structure of
DNA
. In 1962, James Watson, Francis Crick, and Maurice Wilkins were awarded the Nobel Prize in Physiology and Medicine. Unfortunately, by then Franklin had died, and Nobel prizes at the time were not awarded posthumously. Work continued, however, on learning about the structure of
DNA
. In 1973, Alexander Rich(1924–2015) and colleagues were able to analyze
DNA
crystals to confirm and further elucidate
DNA
structure.6
Figure 10.2.5: In 1953, James Watson and Francis Crick built this model of the structure of
DNA
, shown here on display at the Science Museum in London.
Exercise 10.2.2
Which scientists are given most of the credit for describing the molecular structure of
DNA
?
DNA
Structure
Watson and Crick proposed that
DNA
is made up of two strands that are twisted around each other to form a right-handed helix. The two
DNA
strands are
antiparallel
, such that the 3ʹ end of one strand faces the 5ʹ end of the other (Figure 10.2.6). The 3ʹ end of each strand has a free hydroxyl group, while the 5ʹ end of each strand has a free phosphate group. The sugar and phosphate of the polymerized nucleotides form the backbone of the structure, whereas the nitrogenous bases are stacked inside. These nitrogenous bases on the interior of the molecule interact with each other, base pairing.
Analysis of the
diffraction
patterns of
DNA
has determined that there are approximately 10 bases per turn in
DNA
. The asymmetrical spacing of the sugar-phosphate backbones generates major grooves (where the backbone is far apart) and minor grooves (where the backbone is close together) (Figure 10.2.6). These grooves are locations where proteins can bind to
DNA
. The binding of these proteins can alter the structure of
DNA
, regulate
replication
, or regulate
transcription
of
DNA
into RNA.
Figure 10.2.6: Watson and Crick proposed the double helix model for
DNA
. (a) The sugar-phosphate backbones are on the outside of the double helix and
purines
and
pyrimidines
form the “rungs” of the
DNA
helix ladder. (b) The two
DNA
strands are
antiparallel
to each other. (c) The direction of each strand is identified by numbering the carbons (1 through 5) in each sugar molecule. The 5ʹ end is the one where carbon #5 is not bound to another
nucleotide
; the 3ʹ end is the one where carbon #3 is not bound to another
nucleotide
.
Base pairing takes place between a purine and pyrimidine. In
DNA
,
adenine
(A) and
thymine
(T) are
complementary base pairs
, and
cytosine
(C) and
guanine
(G) are also
complementary base pairs
, explaining Chargaff’s rules (Figure 10.2.7). The base pairs are stabilized by hydrogen bonds;
adenine
and
thymine
form two hydrogen bonds between them, whereas
cytosine
and
guanine
form three hydrogen bonds between them.
Figure 10.2.7: Hydrogen bonds form between complementary nitrogenous bases on the interior of
DNA
.
In the laboratory, exposing the two
DNA
strands of the double helix to high temperatures or to certain chemicals can break the hydrogen bonds between complementary bases, thus separating the strands into two separate single strands of
DNA
(single-stranded
DNA
[ssDNA]). This process is called
DNA
denaturation and is analogous to protein denaturation, as described in Proteins. The ssDNA strands can also be put back together as double-stranded
DNA
(dsDNA), through reannealing or renaturing by cooling or removing the chemical denaturants, allowing these hydrogen bonds to reform. The ability to artificially manipulate
DNA
in this way is the basis for several important techniques in
biotechnology
(Figure 10.2.8). Because of the additional hydrogen bonding between the C = G base pair,
DNA
with a high GC content is more difficult to denature than
DNA
with a lower GC content.
Figure 10.2.8: In the laboratory, the double helix can be denatured to single-stranded
DNA
through
exposure
to heat or chemicals, and then renatured through cooling or removal of chemical denaturants to allow the
DNA
strands to reanneal. (credit: modification of work by Hernández-Lemus E, Nicasio-Collazo LA, Castañeda-Priego R)
Link to Learning
View an animation on
DNA
structure from the
DNA
Learning Center to learn more.
Exercise 10.2.3
What are the two
complementary base pairs
of
DNA
and how are they bonded together?
DNA
Function
DNA
stores the information needed to build and control the cell. The transmission of this information from mother to daughter cells is called
vertical gene transfer
and it occurs through the process of
DNA
replication
.
DNA
is replicated when a cell makes a duplicate copy of its
DNA
, then the cell divides, resulting in the correct distribution of one
DNA
copy to each resulting cell.
DNA
can also be enzymatically degraded and used as a source of nucleosides and nucleotides for the cell. Unlike other macromolecules,
DNA
does not serve a structural role in cells.
Exercise 10.2.4
How does
DNA
transmit genetic information to offspring?
Paving the Way for Women in Science and Health Professions
Historically, women have been underrepresented in the sciences and in medicine, and often their pioneering contributions have gone relatively unnoticed. For example, although Rosalind Franklin performed the X-ray
diffraction
studies demonstrating the double helical structure of
DNA
, it is Watson and Crick who became famous for this discovery, building on her data. There still remains great controversy over whether their acquisition of her data was appropriate and whether personality conflicts and gender bias contributed to the delayed recognition of her significant contributions. Similarly, Barbara McClintock did pioneering work in maize (corn) genetics from the 1930s through 1950s, discovering transposons (jumping
genes
), but she was not recognized until much later, receiving a Nobel Prize in Physiology or Medicine in 1983 (Figure 10.2.9).
Today, women still remain underrepresented in many fields of science and medicine. While more than half of the undergraduate degrees in science are awarded to women, only 46% of doctoral degrees in science are awarded to women. In academia, the number of women at each level of career advancement continues to decrease, with women holding less than one-third of the positions of Ph.D.-level scientists in tenure-track positions, and less than one-quarter of the full professorships at 4-year colleges and universities.7 Even in the health professions, like nearly all other fields, women are often underrepresented in many medical careers and earn significantly less than their male counterparts, as shown in a 2013 study published by the _Journal of the American Medical Association_.8
Why do such disparities continue to exist and how do we break these cycles? The situation is complex and likely results from the combination of various factors, including how society conditions the behaviors of girls from a young age and supports their interests, both professionally and personally. Some have suggested that women do not belong in the laboratory, including Nobel Prize winner Tim Hunt, whose 2015 public comments suggesting that women are too emotional for science 9 were met with widespread condemnation.
Perhaps girls should be supported more from a young age in the areas of science and math (Figure 10.2.9). Science, technology, engineering, and mathematics (STEM) programs sponsored by the American Association of University Women (AAUW)10 and National Aeronautics and Space Administration (NASA)11 are excellent examples of programs that offer such support. Contributions by women in science should be made known more widely to the public, and marketing targeted to young girls should include more images of historically and professionally successful female scientists and medical professionals, encouraging all bright young minds, including girls and women, to pursue careers in science and medicine.
Figure 10.2.9: (a) Barbara McClintock’s work on maize genetics in the 1930s through 1950s resulted in the discovery of transposons, but its significance was not recognized at the time. (b) Efforts to appropriately mentor and to provide continued societal support for women in science and medicine may someday help alleviate some of the issues preventing gender equality at all levels in science and medicine. (credit a: modification of work by Smithsonian Institution; credit b: modification of work by Haynie SL, Hinkle AS, Jones NL, Martin CA, Olsiewski PJ, Roberts MF)
Clinical Focus: Part 2
Based upon his symptoms, Alex’s physician suspects that he is suffering from a foodborne illness that he acquired during his travels. Possibilities include bacterial
infection
(e.g., enterotoxigenic _E. coli_, _Vibrio cholerae_, _Campylobacter jejuni_, _Salmonella_), viral
infection
(rotavirus or norovirus), or
protozoan
infection
(_Giardia lamblia_, _Cryptosporidium parvum_, or _Entamoeba histolytica_).
His physician orders a stool sample to identify possible causative agents (e.g.,
bacteria
,
cysts
) and to look for the presence of blood because certain types of infectious agents (like _C. jejuni_, _Salmonella_, and _E. histolytica_) are associated with the production of bloody stools.
Alex’s stool sample showed neither blood nor
cysts
. Following analysis of his stool sample and based upon his recent travel history, the hospital physician suspected that Alex was suffering from traveler’s diarrhea caused by enterotoxigenic _E. coli_ (ETEC), the
causative agent
of most traveler’s diarrhea. To verify the diagnosis and rule out other possibilities, Alex’s physician ordered a diagnostic lab test of his stool sample to look for
DNA
sequences encoding specific
virulence
factors of ETEC. The physician instructed Alex to drink lots of fluids to replace what he was losing and discharged him from the hospital.
ETEC produces several
plasmid
-encoded
virulence
factors that make it pathogenic compared with typical _E. coli_. These include the secreted toxins heat-labile
enterotoxin
(LT) and heat-stabile
enterotoxin
(ST), as well as colonization factor (CF). Both LT and ST cause the excretion of chloride ions from intestinal cells to the intestinal
lumen
, causing a consequent loss of water from intestinal cells, resulting in diarrhea. CF encodes a bacterial protein that aids in allowing the bacterium to adhere to the lining of the small intestine.
Exercise 10.2.5
Why did Alex’s physician use genetic analysis instead of either isolation of
bacteria
from the stool sample or direct Gram stain of the stool sample alone?
Key Concepts and Summary
Nucleic acids are composed of nucleotides, each of which contains a pentose sugar, a phosphate group, and a nitrogenous base. Deoxyribonucleotides within
DNA
contain deoxyribose as the pentose sugar.
DNA
contains the pyrimidines
cytosine and thymine, and the purinesadenine and guanine.
Nucleotides are linked together by
phosphodiester bonds
between the 5ʹ phosphate group of one
nucleotide
and the 3ʹ hydroxyl group of another. A nucleic acid
strand has a free phosphate group at the 5ʹ end and a free hydroxyl group at the 3ʹ end.
Chargaff discovered that the amount of adenine is approximately equal to the amount of thymine in
DNA
, and that the amount of the guanine is approximately equal to cytosine. These relationships were later determined to be due to complementary base pairing.
Watson and Crick, building on the work of Chargaff, Franklin and Gosling, and Wilkins, proposed the double helix model and base pairing for
DNA
structure.
DNA
is composed of two complementary strands oriented antiparallel to each other with the phosphodiester backbones on the exterior of the molecule. The nitrogenous bases of each strand face each other and complementary bases hydrogen bond to each other, stabilizing the double helix.
Heat or chemicals can break the hydrogen bonds between complementary bases, denaturing
DNA
. Cooling or removing chemicals can lead to renaturation or reannealing of
DNA
by allowing hydrogen bonds to reform between complementary bases.
DNA
stores the instructions needed to build and control the cell. This information is transmitted from parent to offspring through vertical gene transfer.
Footnotes
1 N. Kresge et al. “Chargaff's Rules: The Work of Erwin Chargaff.” _Journal of Biological Chemistry_ 280 (2005):e21.
2 L. Pauling, “A Proposed Structure for the Nucleic Acids.” _Proceedings of the National Academy of Science of the United States of America_ 39 no. 2 (1953):84–97.
3 J.D. Watson, F.H.C. Crick. “A Structure for Deoxyribose
Nucleic Acid
.” _Nature_ 171 no. 4356 (1953):737–738.
4 M.H.F. Wilkins et al. “Molecular Structure of Deoxypentose Nucleic Acids.” _Nature_ 171 no. 4356 (1953):738–740.
5 R. Franklin, R.G. Gosling. “Molecular Configuration in Sodium Thymonucleate.” _Nature_ 171 no. 4356 (1953):740–741.
6 R.O. Day et al. “A Crystalline Fragment of the Double Helix: The Structure of the Dinucleoside Phosphate Guanylyl-3',5'-Cytidine.” _Proceedings of the National Academy of Sciences of the United States of America_ 70 no. 3 (1973):849–853.
7 N.H. Wolfinger “For Female Scientists, There's No Good Time to Have Children.” _The Atlantic_ July 29, 2013. www.theatlantic.com/sexes/arc...ildren/278165/.
8 S.A. Seabury et al. “Trends in the Earnings of Male and Female Health Care Professionals in the United States, 1987 to 2010.” _Journal of the American Medical Association Internal Medicine_ 173 no. 18 (2013):1748–1750.
9 E. Chung. “Tim Hunt, Sexism and Science: The Real 'Trouble With Girls' in Labs.” _CBC News Technology and Science_, June 12, 2015. Accessed 8/4/2016.
10 American Association of University Women. “Building a STEM Pipeline for Girls and Women.” www.aauw.org/what-we-do/stem-education/. Accessed June 10, 2016.
11 National Aeronautics and Space Administration. “Outreach Programs: Women and Girls Initiative.” Accessed June 10, 2016. |
9921 | https://www.ejpam.com/index.php/ejpam/article/view/5390/1766 | EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 4, 2024, 3399-3414 ISSN 1307-5543 – ejpam.com Published by New York Business Global
Nonlinear Mixed λ-Jordan Triple Derivation on
∗-algebras
Amal S. Alali 1, Junaid Nisar 2, ∗, Nadeem ur Rehman 3, Hafedh M. Alnoghashi 4
1
Department of Mathematical Sciences, College of Science, Princess Nourah bint Abdulrahman University, P. O. Box 84428, Riyadh 11671, Saudi Arabia
2
Department of Applied Sciences, Symbiosis Institute of Technology, Symbiosis International (Deemed) University, Lavale, Pune,India
3
Department of Mathematics, Aligarh Muslim University, Aligarh-202002 India
4
Department of Computer Science, College of Engineering and Information Technology, Amran University, Amran, Yemen.
Abstract. Let A be a ∗-algebra with unit I and P1 and P2 = I − P1 includes a non-trivial projections, and let λ ∈ C \ { 0, −1}. In this paper, we aim to study the characterization of nonlinear mixed λ-Jordan triple derivation on ∗-algebras. As an application, we can also apply our results on prime ∗-algebras, factor von-Neumann algebras and standard operator algebras.
2020 Mathematics Subject Classifications : 16W10, 47B47, 46K15.
Key Words and Phrases : λ-Mixed Jordan triple derivation, ∗-derivation, ∗- algebra.
Introduction
Consider an ∗-algebra A defined over the complex field C. Introducing the λ-Jordan product U ♢λV = U V + λV U and the skew Lie product [ U, V ]∗ = U V − V U ∗ for nonzero scalar λ, these algebraic structures have gained significant attention in various research do-mains, as evidenced by studies such as [1–5, 7, 8, 12]. In the context of additive mappings, an additive derivation is characterized by Π( U V ) = Π( U )V + U Π( V ) for all U, V ∈ A . If the additional condition Π( U ∗) = Π( U )∗ holds for all U ∈ A , then Π is termed an additive
∗-derivation. Now, let Π : A → A be a map without assuming additivity. The concept of a nonlinear skew Lie derivation is introduced, defined by the relation Π([ U, V ]∗) = [Π( U ), V ]∗ + [ U, Π( V )] ∗
for all U, V ∈ A . Notably, Kong and Zhang established the result that every nonlinear skew Lie derivation is, in fact, an additive ∗-derivation.
∗Corresponding author. DOI:
Email addresses: asalali@pnu.edu.sa (A. Alali), junaidnisar73@gmail.com (J. Nisar),
nu.rehman.mm@amu.ac.in (N. Rehman), halnoghashi@gmail.com (H. Alnoghashi) 3399 Copyright: ©2024 The Author(s). (CC BY-NC 4.0) . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3400
Similarly, a map Π : A → A is termed a nonlinear skew Lie triple derivation if it satisfies the equation Π([[ U, V ]∗, W ]∗) = [[Π( U ), V ]∗, W ]∗ + [[ U, Π( V )] ∗, W ]∗ + [[ U, V ]∗, Π( W )] ∗
for all U, V, W ∈ A .Several recent studies have delved into the exploration of derivations and isomorphisms associated with innovative products resulting from the combination of Lie and skew Lie products, as evidenced by works such as [6, 9, 11]. Notably, Zhou et al. established the result that every nonlinear mixed Lie triple derivation on a prime ∗-algebra is, in fact, an additive ∗-derivation. Additionally, Pang et al. demonstrated that every second nonlinear mixed Jordan triple derivable mapping on factor von Neumann algebras also an additive ∗-derivation. Motivated by these previous works, our paper introduces the λ-Jordan product defined as U ♢λV = U V +λV U . We specifically focus on the derivation corresponding to the novel product obtained by combining the skew Lie product and the λ-Jordan product. In this context, we define a map Π : A → A as a mixed λ-Jordan triple derivation if it satisfies the equation Π([ U, V ]∗♢λW ) = [Π( U ), V ]∗ ⋄Λ W + [ U, Π( V )] ∗♢λW + [ U, V ]∗♢λΠ( W )for all U, V, W ∈ A . Our main result establishes that Π is a nonlinear mixed λ-Jordan triple derivation on ∗-algebras if and only if Π is an additive ∗-derivation.
Main Result
Theorem 2.1. Let A be a unital ∗-algebra with unity I containing a non-trivial projection
P satisfies
XAP = 0 = ⇒ X = 0 (▲)
and
XA(I − P ) = 0 = ⇒ X = 0 . (▼)
Define a map Π : A → A such that
Π([ U, V ]∗♢λW ) = [Π( U ), V ]∗♢λW ) + [ U, Π( V )] ∗♢λW + [ U, V ]∗♢λΠ( W ).
Then Π is an additive ∗-derivation.
Consider a non-trivial projection P = P1 in the algebra A, and let P2 = I − P1, where
I is the unity element of the algebra. Utilizing the Peirce decomposition of A, we express
A as the direct sum A = P1AP1 ⊕ P1AP2 ⊕ P2AP1 ⊕ P2AP2. Denoting the corresponding subspaces as A11 = P1AP1, A12 = P1AP2, A21 = P2AP1, and A22 = P2AP2, we can represent any element U ∈ A as the sum U = U11 + U12 + U21 + U22 , where Uij ∈ A ij
and U ∗
ij
∈ A ji for i, j = 1 , 2. Before proving Theorem 2.1, we need several lemmas and remarks. . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3401
Lemma 2.1. Π(0) = 0 .
Proof. It is obvious that Π(0) = Π([0 , 0] ∗♢λ0) = [Π(0) , 0] ∗♢λ0 + [0 , Π(0)] ∗♢λ0 + [0 , 0] ∗♢λΠ(0) = 0 .
Lemma 2.2. Let U12 ∈ A 12 and U21 ∈ A 21 . Then Π( U12 + U21 ) = Π( U12 ) + Π( U21 ).
Proof. Let T = Π( U12 + U21 ) − Π( U12 ) − Π( U21 ). Since [ U12 , P 1]∗♢λP2 = 0 and by using Lemma 2 .1, we have Π([ U12 + V21 , P 1]∗♢λP2) = Π([ U12 , P 1]∗♢λP2) + Π([ V21 , P 1]∗♢λP2)= [Π( U12 ), P 1]∗♢λP2 + [ U12 , Π( P1)] ∗♢λP2 + [ U12 , P 1]∗♢λΠ( P2)+[Π( V21 ), P 1]∗♢λP2 + [ V21 , Π( P1)] ∗♢λP2 + [ V21 , P 1]∗♢λΠ( P2).
On the other hand, we find Π([ U12 + V21 , P 1]∗♢λP2 = [Π( U12 + V21 ), P 1]∗♢λP2 + [ U12 + V21 , Π( P1)] ∗♢λP2
+[ U12 + V21 , P 1]∗♢λΠ( P2).
From the above two equations, we get [ T, P 1]∗♢λP2 = 0. That means −P1T ∗P2+λP 2T P 1 =0. Multiplying by P2 from the left and since λ̸ = 0, we get P2T P 1 = 0. Similarly, one can show that P1T P 2 = 0. Now, for every X21 ∈ A 21 , it follows from [ X21 , U 12 ]∗♢λP1 = 0 and using Lemma 2.1 that Π([ X21 , U 12 + V21 ]∗♢λP1) = Π([ X21 , U 12 ]∗♢λP1) + Π([ X21 , V 21 ]∗♢λP1)= [Π( X21 ), U 12 ]∗♢λP1 + [ X21 , Π( U12 )] ∗♢λP1
+[ X21 , U 12 ]∗♢λΠ( P1) + [Π( X21 ), V 21 ]∗♢λP1
+[ X21 , Π( V21 )] ∗♢λP1 + [ X21 , V 21 ]∗♢λΠ( P1).
On the other hand, we have Π([ X21 , U 12 + V21 ]∗♢λP1) = [Π( X21 ), U 12 + V21 ]∗♢λP1 + [ X21 , Π( U12 + V21 )] ∗♢λP1
+[ X21 , U 12 + V21 ]∗♢λΠ( P1).
From the above expressions, we find that [ X21 , T ]∗♢λP1 = 0. That means X21 T P 1 −
λP 1T X ∗
21
= 0. Multiplying both sides by P1 from right, we get X21 T P 1 = 0 . By using ( ▲)and ( ▼), we have P1T P 1 = 0 . Similarly, we can show that P2T P 2 = 0. Hence, T = 0 i.e., Π( U12 + U21 ) = Π( U12 ) + Π( U21 ).
Lemma 2.3. For any Uij ∈ A ij , 1 ≤ i, j ≤ 2, we have
Π(
2
X
i,j =1
Uij ) =
2
X
i,j =1
Π( Uij ).. Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3402
Proof. Let T = Π( U11 + U12 + U21 + U22 ) − Π( U11 ) − Π( U12 ) − Π( U21 ) − Π( U22 ). For every X12 ∈ A 12 , also [ P1, U 11 ]∗♢λX12 = [ P1, U 22 ]∗♢λX12 = 0 and using Lemmas 2.1 and 2.2, we get Π([ P1, U 11 + U12 + U21 + U22 ]∗♢λX12 ) = Π([ P1, U 11 ]∗♢λX12 ) + Π([ P1, U 12 ]∗♢λX12 )+Π([ P1, U 21 ]∗♢λX12 ) + Π([ P1, U 22 ]∗♢λX12 )= [Π( P1), U 11 ]∗♢λX12 + [ P1, Π( U11 )] ∗♢λX12
+[ P1, U 11 ]∗♢λΠ( X12 ) + [Π( P1), U 12 ]∗♢λX12
+[ P1, Π( U12 )] ∗♢λX12 + [ P1, U 12 ]∗♢λΠ( X12 )+[Π( P1), U 21 ]∗♢λX12 + [ P1, Π( U21 )] ∗♢λX12
+[ P1, U 21 ]∗♢λΠ( X12 ) + [Π( P1), U 22 ]∗♢λX12
+[ P1, Π( U22 )] ∗♢λX12 + [ P1, U 22 ]∗♢λΠ( X12 ).
On the other hand, we have Π([ P1, U 11 + U12 + U21 + U22 ]∗♢λX12 ) = [Π( P1), U 11 + U12 + U21 + U22 ]∗♢λX12
+[ P1, Π( U11 + U12 + U21 + U22 )] ∗♢λX12
+[ P1, U 11 + U12 + U21 + U22 ]∗♢λΠ( X12 ).
By comparing the above two equations, we get [ P1, M ]∗♢λX12 = 0 from which we obtain
P1T X 12 − T X 12 − λX 12 T P 1 = 0 . Multiplying P2 from left and right, we get P2T X 12 = 0 .
By using ( ▲) and ( ▼), we have P2T P 1 = 0 . Similarly, we can show that P1T P 2 = 0. Again for X12 ∈ A 12 , it follows from [ X12 , U 11 ]∗♢λP2 = [ X12 , U 12 ]∗♢λP2 = [ X12 , U 12 ]∗♢λP2 =0 that Π([ X12 , U 11 + U12 + U21 + U22 ]∗♢λP2) = Π([ X12 , U 11 ]∗♢λP2) + Π([ X12 , U 12 ]∗♢λP2)+Π([ X12 , U 21 ]∗♢λP2) + Π([ X12 , U 22 ]∗♢λP2)= [Π( X12 ), U 11 ]∗♢λP2 + [ X12 , Π( U11 )] ∗♢λP2
+[ X12 , U 11 ]∗♢λΠ( P2) + [Π( X12 ), U 12 ]∗♢λP2
+[ X12 , Π( U12 )] ∗♢λP2 + [ X12 , U 12 ]∗♢λΠ( P2)+[Π( X12 ), U 21 ]∗♢λP2 + [ X12 , Π( U21 )] ∗♢λP2
+[ X12 , U 21 ]∗♢λΠ( P2) + [Π( X12 ), U 22 ]∗♢λP2
+[ X12 , Π( U22 )] ∗♢λP2 + [ X12 , U 22 ]∗♢λΠ( P2).
On the other hand, we get Π([ X12 , U 11 + U12 + U21 + U22 ]∗♢λP2) = [Π( X12 ), U 11 + U12 + U21 + U22 ]∗♢λP2
+[ X12 , Π( U11 + U12 + U21 + U22 )] ∗♢λP2
+[ X12 , U 11 + U12 + U21 + U22 ]∗♢λΠ( P2).
By the above two equations, we get [ X12 , T ]∗♢λP2 = 0. That means that X12 T P 2 −
λP 2T X ∗
12
= 0 . When we multiply both sides by P1 on the left, the result is X12 T P 2 = 0. . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3403
By using ( ▲) and ( ▼), we have P2T P 2 = 0. Similarly, P1T P 1 = 0. Hence, T = 0 i.e., Π( U11 + U12 + U21 + U22 ) = Π( U11 ) + Π( U12 ) + Π( U21 ) + Π( U22 ).
Lemma 2.4. For any Uij , V ij ∈ A ij with (1 ≤ i̸ = j ≤ 2) , we have
Π( Uij + Vij ) = Π( Uij ) + Π( Vij ).
Proof. Initially, we establish the result for i = 1 and j = 2. Let T = Π( U12 + V12 ) −
Π( U12 ) − Π( V12 ). Since [ X12 , U 12 ]∗♢λP2 = 0, and using Lemma 2.1, we get Π([ X12 , U 12 + V12 ]∗♢λP2) = Π([ X12 , U 12 ]∗♢λP2) + Π([ X12 , V 12 ]∗♢λP2)= [Π( X12 ), U 12 ]∗♢λP2 + [ X12 , Π( U12 )] ∗♢λP2
+[ X12 , U 12 ]∗♢λΠ( P2) + [Π( X12 ), V 12 ]∗♢λP2
+[ X12 , Π( V12 )] ∗♢λP2 + [ X12 , V 12 ]∗♢λΠ( P2).
On the other hand, we have Π([ X12 , U 12 + V12 ]∗♢λP2) = [Π( X12 ), U 12 + V12 ]∗♢λP2 + [ X12 , Π( U12 + V12 )] ∗♢λP2
+[ X12 , U 12 + V12 ]∗♢λΠ( P2).
By comparing the last two expressions, we get [ X12 , T ]∗♢λP2 = 0. That means X12 T P 2 −
P2T X ∗
12
= 0 . By left-multiplying both sides of the preceding equation by P1 and utilizing (▲) and ( ▼), we obtain P2T P 2 = 0. Similarly, we can show that P1T P 1 = 0. Now, again for any X12 ∈ A 12 . Since [ P1, U 12 ]∗♢λX12 = 0 and using Lemma 2.1, we have Π([ P1, U 12 + V12 ]∗♢λX12 ) = Π([ P1, U 12 ]∗♢λX12 ) + Π([ P1, V 12 ]∗♢λX12 )= [Π( P1), U 12 ]∗♢λX12 + [ P1, Π( U12 )] ∗♢λX12
+[ P1, U 12 ]∗♢λΠ( X12 ) + [Π( P1), V 12 ]∗♢λX12
+[ P1, Π( V12 )] ∗♢λX12 + [ P1, U 12 ]∗♢λΠ( X12 ).
On the other hand, we find Π([ P1, U 12 + V12 ]∗♢λX12 ) = [Π( P1), U 12 + V12 ]∗♢λX12 + [ P1, Π( U12 + V12 )] ∗♢λX12
+[ P1, U 12 + V12 ]∗♢λΠ( X12 ).
From the last two expressions, we find [ P1, T ]∗♢λX12 = 0. That means P1T X 12 − T X 12 −
λX 12 T P 1 = 0. Multiplying both sides by P1 from right and since λ̸ = 0, we have X12 T P 1 =0. Thus, P2T P 1 = 0 follows from ( ▲) and ( ▼). Similarly, we can show that P1T P 2 = 0 .
Hence, T = 0 i.e., Π( U12 + V12 ) = Π( U12 ) + Π( V12 ).
By using the same technique as above, one can show that Π( U21 + V21 ) = Π( U21 ) + Π( V21 ).. Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3404
Lemma 2.5. For any U11 , V 11 ∈ A 11 and U22 , V 22 ∈ A 22 , we have (i) Π( U11 + V11 ) = Π( U11 ) + Π( V11 ).
(ii) Π( U22 + V22 ) = Π( U22 ) + Π( V22 ).
Proof. Let T = Π( U11 + V11 ) − Π( U11 ) − Π( V11 ). On the one hand, we have Π([ P2, U 11 + V11 ]∗♢λP1) = [Π( P2), U 11 + V11 ]∗♢λP1 + [ P2, Π( U11 + V11 )] ∗♢λP1
+[ P2, U 11 + V11 ]∗♢λΠ( P1).
On the other hand, it follows from [ P2, U 11 ]∗♢λP1 = 0 that Π([ P2, U 11 + V11 ]∗♢λP1) = Π([ P2, U 11 ]∗♢λP1) + Π([ P2, V 11 ]∗♢λP1)= [Π( P2), U 11 ]∗♢λP1 + [ P2, Π( U11 )] ∗♢λP1 + [ P2, U 11 ]∗♢λΠ( P1)+[Π( P2), V 11 ]∗♢λP1 + [ P2, Π( V11 )] ∗♢λP1 + [ P2, V 11 ]∗♢λΠ( P1).
By comparing the last two equations, we find [ P2, T ]∗♢λP1 = 0. This gives P2T P 1 −
λP 1T P 2 = 0 and hence, P2T P 1 = P1T P 2 = 0 .
Again for any X12 ∈ A 12 and since [ X12 , U 11 ]∗♢λP2 = 0, we find Π([ X12 , U 11 + V11 ]∗♢λP2) = Π([ X12 , U 11 ]∗♢λP2) + Π([ X12 , V 11 ]∗♢λP2)= [Π( X12 ), U 11 ]∗♢λP2 + [ X12 , Π( U11 )] ∗♢λP2
+[ X12 , U 11 ]∗♢λΠ( P2) + [Π( X12 ), V 11 ]∗♢λP2
+[ X12 , Π( V11 )] ∗♢λP2 + [ X12 , V 11 ]∗♢λΠ( P2).
From the other side, we get Π([ X12 , U 11 + V11 ]∗♢λP2) = [Π( X12 ), U 11 + V11 ]∗♢λP2 + [ X12 , Π( U11 + V11 )] ∗♢λP2
+[ X12 , U 11 + V11 ]∗♢λΠ( P2).
From the last two equations, we get [ X12 , T ]∗♢λP2 = 0 . That means X12 T P 2 −λP 2T X ∗
12
=0. Thus, X12 T P 2 = 0. By using ( ▲) and ( ▼), we get P2T P 2 = 0 . Similarly, we can show that P1T P 1 = 0 . Hence, T = 0 i,e. Π( U11 + V11 ) = Π( U11 ) + Π( V11 ).
(ii ). By using the same argument as in ( i), one can show that Π( U22 + V22 ) = Π( U22 ) + Π( V22 ).
Lemma 2.6. Π is an additive map. . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3405
Proof. For any U, V ∈ A , we write U = U11 + U12 + U21 + U22 and V = V11 + V12 +
V21 + V22 . By using Lemmas 2.3 - 2.5, we get Π( U + V ) = Π( U11 + U12 + U21 + U22 + V11 + V12 + V21 + V22 )= Π( U11 + V11 ) + Π( U12 + V12 ) + Π( U21 + V21 ) + Π( U22 + V22 )= Π( U11 ) + Π( V11 ) + Π( U12 ) + Π( V12 ) + Π( U21 ) + Π( V21 ) + Π( U22 ) + Π( V22 )= Π( U11 + U12 + U21 + U22 ) + Π( V11 + V12 + V21 + V22 )= Π( U ) + Π( V ).
Lemma 2.7. (i) Π( P1)∗ = Π( P1), Π( P2)∗ = Π( P2).
(ii) P1Π( P1)P2 = −P1Π( P2)P2.
(iii) P2Π( P2)P1 = −P2Π( P1)P1.
Proof. (i) In the view of [ P1, P 1]∗♢λI = 0 and using Lemma 2.1, we have 0 = Π([ P1, P 1]∗♢λI)= [Π( P1), P 1]∗♢λI + [ P1, Π( P1)] ∗♢λI + [ P1, P 1]∗♢λΠ( I)= (1 + λ)( P1Π( P1)∗ − P1Π( P1)) .
Since λ̸ = −1, we have P1Π( P1)∗ = P1Π( P1). That means
P1Π( P1)∗P1 = P1Π( P1)P1. (2.1)
P1Π( P1)∗P2 = P1Π( P1)P2 (2.2)
P2Π( P1)∗P1 = P2Π( P1)P1 (2.3) Also, [ P1, P 2]∗♢λI = 0 and Π(0) = 0, we have 0 = Π([ P1, P 2]∗♢λI = 0) = [Π( P1), P 2]∗♢λI + [ P1, Π( P2)] ∗♢λI + [ P1, P 2]∗♢λΠ( I)= (1 + λ)( P2Π( P1)P2 − P2Π( P1)∗P2).
That is,
P2Π( P1)∗P2 = P2Π( P1)P2. (2.4) From equations (2.1)-(2.4), we conclude Π( P1)∗ = Π( P1).
Similarly, by using the same technique as above, we get Π( P2)∗ = Π( P2).
(ii) Again [ P1, P 2]∗♢λP2 = 0 and Π(0) = 0, we find 0 = Π([ P1, P 2]∗♢λP2). Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3406
= [Π( P1), P 2]∗♢λP2 + [ P1, Π( P2)] ∗♢λP2 + [ P1, P 2]∗♢λΠ( P2)= Π( P1)P2 − P2Π( P1)∗P2 + λP 2Π( P1)P2 − λP 2Π( P1)∗ + P1Π( P2)P2 − λP 2Π( P2)P1.
Multiplying above equation by P1 from left, we get P1Π( P1)P2 = −P1Π( P2)P2.(iii) By using the same technique as in (ii), we can show that P2Π( P2)P1 = −P2Π( P1)P1.
Lemma 2.8. For every Uij ∈ A ij (1 ≤ i̸ = j ≤ 2) , we have
Pj Π( λU ij )Pi = 0 .
Proof. To begin, we establish the result for i = 1 and j = 2. For any U12 ∈ A 12 , we get Π( λU 12 ) = Π([ P1, λU 12 ]∗♢λP2)= [Π( P1), λU 12 ]∗♢λP2 + [ P1, Π( λU 12 )] ∗♢λP2 + [ P1, λU 12 ]∗♢λΠ( P2)= λΠ( P1)U12 − λU 12 Π( P1)P2 + λ2P2Π( P1)U12 + P1Π( λU 12 )P2
−λP 2Π( λU 12 )P1 + λU 12 Π( P2) + λ2Π( P2)U12 .
By left-multiplying the above equation with P2 and right-multiplying with P1, we obtain (1 + λ)P2Π( λU 12 )P1 = 0 . (2.5) Since λ̸ = −1, we find P2Π( λU 12 )P1 = 0 . Similarly, by using the same technique for
i = 2 , j = 1 , one can show that P1Π( λU 21 )P2 = 0 .
Lemma 2.9. (i) P1Π( P2)P1 = P2Π( P1)P2 = 0 .
(ii) P1Π( P1)P1 = P2Π( P2)P2 = 0 .
Proof. (i) For every X12 ∈ A 12 , it follows from [ X12 , P 1]∗♢λP1 = 0 that 0 = Π([ X12 , P 1]∗♢λP1)= [Π( X12 ), P 1]∗♢λP1 + [ X12 , Π( P1)] ∗♢λP1 + [ X12 , P 1]∗♢λΠ( P1)= Π( X12 )P1 − P1Π( X12 )∗P1 + λP 1Π( X12 )P1 − λP 1Π( X12 )∗
+X12 Π( P1)P1 − Π( P1)X∗
12
λX 12 Π( P1) − λP 1Π( P1)X∗
12
.
By left-multiplying the preceding equation by P1 and right-multiplying by P2, and con-sidering the fact that λ̸ = 0, we obtain
X12 Π( P1)P2 = P1Π( X12 )∗P2. (2.6) That means,
P2Π( P1)X∗
12
= P2Π( X12 )P1.. Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3407
Also, [ P1, X 12 ]∗♢λX12 = 0 and using Lemma 2.1, we get 0 = Π([ P1, X 12 ]∗♢λX12 )= [Π( P1), X 12 ]∗♢λX12 + [ P1, Π( X12 )] ∗♢λX12 + [ P1, X 12 ]∗♢λΠ( X12 )= −X12 Π( P1)∗X12 + λX 12 Π( P1)X12 + P1Π( X12 )X12 − Π( X12 )X12 − λX 12 Π( X12 )P1
+X12 Π( X12 ) + λX 12 Π( X12 ).
Multiplying the above equation by P1 from left and right, we find X12 Π( X12 )P1 = 0 . By using ( ▲) and ( ▼), we get P2Π( X12 )P1 = 0. That means
P1Π( X12 )∗P2 = 0 .
From Equation (2.6),( ▲) and ( ▼), we have P2Π( P1)P2 = 0. Similarly, we can show that P1Π( P2)P1 = 0. (ii) For any X21 ∈ A 21 and using Lemma 2.6, we have Π([ X21 , P 1]∗♢λP1) = Π( X21 ) − Π( λX ∗
21
).
On the other hand, we have Π([ X21 , P 1]∗♢λP1) = [Π( X21 ), P 1]∗♢λP1 + [ X21 , Π( P1)] ∗♢λP1 + [ X21 , P 1]∗♢λΠ( P1)= Π( X21 )P1 − P1Π( X21 )∗P1 + λP 1Π( X21 )P1 − λP 1Π( X21 )∗
+X21 Π( P1)P1 − λP 1Π( P1)X∗
21
X21 Π( P1) + λΠ( P1)X21 .
By comparing the aforementioned two equations and subsequently left-multiplying by P2
and right-multiplying by P1, we obtain 2X21 Π( P1)P1 + λP 2Π( P1)X21 + P2Π( λX ∗
21
)P1 = 0 Now, by using Lemma 2.8 and Lemma 2.9(i), we get X21 Π( P1)P1 = 0. Hence, P1Π( P1)P1 =0. Similarly, we can show that P2Π( P2)P2 = 0 .
Now, let M = P1Π( P1)P2 − P2Π( P1)P1. Then M = −M ∗. We define a mapping ∆ : A → A as ∆( U ) = Π( U ) − (U M − M U ) for all U ∈ A . It can be easily verified that for all U, V, W ∈ A , ∆([ U, V ]∗♢λW ) = [∆( U ), V ]∗♢λW ) + [ U, ∆( V )] ∗♢λW + [ U, V ]∗♢λ∆( W ).
Remark 2.1. The mapping ∆ possesses the following properties: (i) ∆ is additive. (ii) ∆( P1) = ∆( P2) = 0 .
(iii) ∆( I) = 0
(iv) For every Uij ∈ A ij (1 ≤ i̸ = j ≤ 2) , we have
Pj ∆( λU ij )Pi = 0 . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3408
(v) ∆ is a ∗-derivation if and only if Π is an ∗ -derivation. Proof. (i) Since [ U, M ] is additive and also using Lemma 2.6, it is clear that ∆ is additve.
(ii) By using Lemma 2.9, we have ∆( P1) = Π( P1) − P1Π( P1)P2 − P2Π( P1)P1
= P1Π( P1)P2 + P2Π( P1)P1 − P1Π( P1)P2 − P2Π( P1)P1
= 0.
Similarly, we can show that ∆( P2) = 0.
(iii) By using additivity of ∆, we have ∆( I) = ∆( P1 + P2) = ∆( P1) + ∆( P2) = 0 .
(iv) For i = 1 and j = 2, it can be inferred from Lemma 2.8 that
P2∆( λU 12 )P1 = P2(Π( λU 12 ) − λU 12 M + M λU 12 )P1
= 0.
In the same way, one can show for i = 2 , j = 1 , i.e., P1∆( λU 21 )P2 = 0 .
(v) Since [ U, M ] = U M − M U is an additive ∗-derivation. Therefore, ∆ qualifies as a
∗-derivation if and only if Π is a ∗-derivation.
Lemma 2.10. ∆( Uij ) ⊆ Uij , i, j = 1 , 2.
Proof. First, we prove for i = 1 , j = 1 . For every U11 ∈ A 11 , it follows from Remark 2.1 that 0 = ∆([ P1, U 11 ]∗♢λP1)= [P1, ∆( U11 )] ∗♢λP1
= P1∆( U11 )P1 − ∆( U11 )P1 + λP 1∆( U11 ) − λP 1∆( U11 )P1.
Left multiplying the above equation by P2, we find
P2∆( U11 )P1 = 0 . (2.7) Similarly, again by using Remark 2.1, we have 0 = ∆([ P2, U 11 ]∗♢λP1)= [P2, ∆( U11 )] ∗♢λP1
= P2∆( U11 )P1 − λP 1∆( U11 )P2.
Multiplying P2 on the right and since λ̸ = 0 , we get
P1∆( U11 )P2 = 0 (2.8) . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3409
Now, for every X12 ∈ A 12 and ∆( P2) = 0, it follows that 0 = ∆([ X12 , U 11 ]∗♢λP2)= [∆( X12 ), U 11 ]∗♢λP2 + [ X12 , ∆( U11 )] ∗♢λP2
= −U11 ∆( X12 )∗P2 + λP 2∆( X12 )U11 + X12 ∆( U11 )P2
−λP 2∆( U11 )X∗
12
.
Multiplying P1 from left and P2 from right and using Lemma 2.8, we get
X12 ∆( U11 )P2 = U11 ∆( X12 )∗P2 = U11 (P2∆( X12 )P1)∗ = 0 .
Thus, X12 ∆( U11 )P2 = 0. By using ( ▲) and ( ▼), we have
P2∆( U11 )P2 = 0 . (2.9) From Equations (2.7)-(2.9), we have ∆( U11 ) ⊆ U11 . Similarly, we can show that ∆( U22 ) ⊆
U22 .
Next, we establish the result for i = 1 , j = 2. Additionally, for any U12 ∈ A 12 and ∆( P2) = 0, we have ∆( U12 ) = ∆([ P1, U 12 ]∗♢λP2)= [P1, ∆( U12 )] ∗♢λP2
= P1∆( U12 )P2 − λP 2∆( U12 )P1.
By left and right multiplying the above equation by P1, we obtain
P1∆( U12 )P1 = 0 . (2.10) Additionally, by left and right-multiplying by P2, we find
P2∆( U12 )P2 = 0 . (2.11) Similarly, multiplying P2 from left and P1 from right and since λ̸ = −1, we get
P2∆( U12 )P1 = 0 (2.12) From equations (2.10)- (2.12), we get ∆( U12 ) ⊆ U12 .
Similarly, by using the same technique as above, one can prove that ∆( U21 ) ⊆ U21 .
Lemma 2.11. For any Ui,j , V i,j ∈ A ij , 1 ≤ i, j ≤ 2, we have (i) ∆( U11 V12 ) = ∆( U11 )V12 + U11 ∆( V12 ) and ∆( U22 V21 ) = ∆( U22 )V21 + U22 ∆( V21 ).
(ii) ∆( U12 V21 ) = ∆( U12 )V21 + U12 ∆( V21 ) and ∆( U21 V12 ) = ∆( U21 )V12 + U21 ∆( V12 ).
(iii) ∆( U11 V11 ) = ∆( U11 )V11 + U11 ∆( V11 ) and ∆( U22 V22 ) = ∆( U22 )V22 + U22 ∆( V22 ).. Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3410
(iv) ∆( U12 V22 ) = ∆( U12 )V22 + U12 ∆( V22 ) and ∆( U21 V11 ) = ∆( U21 )V11 + U21 ∆( V11 ).
Proof. (i) Using Lemma 2.10 and ∆( P2) = 0, we get ∆([ U11 , V 12 ]∗♢λP2) = [∆( U11 ), V 12 ]∗♢λP2 + [ U11 , ∆( V12 )] ∗♢λP2
= ∆( U11 )V12 + U11 ∆( V12 ).
On the other side, we get ∆([ U11 , V 12 ]∗♢λP2) = ∆( U11 V12 ).
By comparing the above two equations, we get ∆( U11 V12 ) = ∆( U11 )V12 + U11 ∆( V12 ). Similarly, we can show that ∆( U22 V21 ) = ∆( U22 )V21 + U22 ∆( V21 ).
(ii) For any X12 ∈ A 12 and by using Lemma 2.11 (1), we have ∆([ U12 , V 21 ]∗♢λX12 ) = ∆( U12 V21 X21 )= ∆( U12 V21 )X12 + U12 V21 ∆( X12 ).
On the other hand, we have ∆([ U12 , V 21 ]∗♢λX12 ) = [∆( U12 ), V 21 ]∗♢λX12 + [ U12 , ∆( V21 )] ∗♢λX12
+[ U12 , V 21 ]∗♢λ∆( X12 )= ∆( U12 )V21 X12 + U12 ∆( V21 )X12 + U12 V21 ∆( X12 ).
From the above two expressions, we get (∆( U12 V21 ) − ∆( U12 )V21 − U12 ∆( V21 )) X12 = 0 .
Thus, by ( ▲) and ( ▼), ∆( U12 V21 ) = ∆( U12 )V21 + U12 ∆( V21 ). Similarly, we can show that ∆( U21 V12 ) = ∆( U21 )V12 + U21 ∆( V12 ).
(iii) For any X12 ∈ A 12 , it follows from Lemma 2.11(i) that ∆( U11 V11 X12 ) = ∆( U11 V11 )X12 + U11 V11 ∆( X12 ).
Again using Lemma 2.11 from the other side, we have ∆( U11 V11 X12 ) = ∆( U11 )V11 X12 + U11 ∆( V11 X12 )= ∆( U11 )V11 X12 + U11 ∆( V11 )X12 + U11 V11 ∆( X12 ).
By comparing the aforementioned two equations, we obtain (∆( U11 V11 ) − ∆( U11 )V11 −
U11 ∆( V11 )) X12 = 0. Therefore, utilizing ( ▲) and ( ▼), we conclude that ∆( U11 V11 ) = ∆( U11 )V11 + U11 ∆( V11 ). Similarly, one can show that ∆( U22 V22 ) = ∆( U22 )V22 + U22 ∆( V22 ).
(iv) For any X21 ∈ A 21 . It follows from Lemma 2.11(2) that ∆( U12 V22 X21 ) = ∆( U12 V22 )X21 + U12 V22 ∆( X21 ).. Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3411
Again on the other side, it follows from Lemma 2.11(i) and Lemma 2.11(ii) that ∆( U12 V22 X21 ) = ∆( U12 )V22 X21 + U12 ∆( V22 X21 )= ∆( U12 )V22 X21 + U12 ∆( V22 )X21 + U12 V22 ∆( X21 ).
From the above two equations and using ( ▲) and ( ▼), we find ∆( U12 V22 ) = ∆( U12 )V22 +
U12 ∆( V22 ). Similarly, one can show that ∆( U21 V11 ) = ∆( U21 )V11 + U21 ∆( V11 ).
Lemma 2.12. ∆( U ∗) = ∆( U )∗ for all U ∈ A .
Proof. For any X12 ∈ A 12 , it follows from Remark 2.1 and Lemma 2.11(i) that ∆([ U11 , P 1]∗♢λX12 ) = ∆( U11 X12 ) − ∆( U ∗
11
X12 )= ∆( U11 )X12 + U11 ∆( X12 ) − ∆( U ∗
11
)X12 − U ∗
11
∆( X12 ).
Alternatively, it can be deduced from ∆( P1) = 0 that ∆([ U11 , P 1]∗♢λX12 ) = [∆( U11 ), P 1]∗♢λX12 + [ U11 , P 1]∗♢λ∆( X12 )= ∆( U11 )X12 − ∆( U11 )∗X12 + U11 ∆( X12 ) − U ∗
11
∆( X12 ).
From the above two equations, we have (∆( U11 )∗ − ∆( U ∗
11
)) X12 = 0. Now, by using ( ▲)and ( ▼), we get ∆( U11 )∗ = ∆( U ∗
11
). (2.13) Similarly, by using the same technique, one can show that ∆( U22 )∗ = ∆( U ∗
22
). (2.14) Again, it follows from Lemma 2.10 and ∆( P2) = 0 that ∆([ U12 , P 2]∗♢λX12 ) = [∆( U12 ), P 2]∗♢λX12 + [ U12 , P 2]∗♢λ∆( X12 )= −∆( U12 )∗X12 − λX 12 ∆( U12 )∗ − U ∗
12
∆( X12 ) − λ∆( X12 )U ∗
12
.
On the other hand, using Lemma 2.11, we have ∆([ U12 , P 2]∗♢λX12 ) = ∆( −U ∗
12
X12 − λX 12 U ∗
12
)= −∆( U ∗
12
X12 ) − ∆( X12 λU ∗
12
)= −∆( U ∗
12
)X12 − U ∗
12
∆( X12 ) − λ∆( X12 )U ∗
12
− X12 ∆( λU ∗
12
).
By comparing the above two equations, we get (∆( U12 )∗ − ∆( U ∗
12
)) X12 + λX 12 ∆( U12 )∗ − X12 ∆( λU ∗
12
) = 0 .
By left-multiplying both sides by P2 and utilizing ( ▲) and ( ▼), we obtain ∆( U12 )∗ = ∆( U ∗
12
). (2.15) . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3412
Similarly, we can show that ∆( U21 )∗ = ∆( U ∗
21
). (2.16) From equations (2.13)-(2.16) and using additivity of ∆, we get ∆( U ∗) = ∆( U )∗.
Proof of Theorem 2.1 For every U, V ∈ A , we can write U = U11 + U12 + U21 + U22 and
V = V11 + V12 + V21 + V22 . Since, ∆ is additive and using Lemma 2.11, we get ∆( U V ) = ∆( U11 V11 + U11 V12 + U12 V21 + U12 V22
+U21 V11 + U21 V12 + U22 V21 + U22 V22 )= ∆( U11 V11 ) + ∆( U11 V12 ) + ∆( U12 V21 ) + ∆( U12 V22 )+∆( U21 V11 ) + ∆( U21 V12 ) + ∆( U22 V21 ) + ∆( U22 V22 )= ∆( U11 + U12 + U21 + U22 )( V11 + V12 + V21 + V22 )+( U11 + U12 + U21 + U22 ) ∆( V11 + V12 + V21 + V22 )= ∆( U )V + U ∆( V ).
So, ∆ is a derivation. By using Lemma 2.12, ∆ is an additive ∗-derivation. Hence, by Remark 2.1, Π is an additive ∗-derivation. This completes the proof of Theorem 2.1. The corollaries following directly from Theorem 2.1 are as follows:
Corollary 2.1. Let A be a standard operator algebra on an infinite dimensional complex Hilbert space H containing identity operator I. Suppose that A is closed under adjoint operation. Define Π : A → A such that
Π([ U, V ]∗♢λW ) = [Π( U ), V ]∗♢λW ) + [ U, Π( V )] ∗♢λW + [ U, V ]∗♢λΠ( W )
for all U, V, W ∈ A . Then Π is an additive ∗-derivation.
Corollary 2.2. Let A ba a factor von Neumann algebra with dim M ≥ 2. Define Π :
M → M such that
Π([ U, V ]∗♢λW ) = [Π( U ), V ]∗♢λW ) + [ U, Π( V )] ∗♢λW + [ U, V ]∗♢λΠ( W )
for all U, V, W ∈ A . Then Π is an additive ∗-derivation.
Corollary 2.3. Let A be a prime ∗-algebra with unit I containing non-trivial projection
P . A map Π : A → A satisfies
Π([ U, V ]∗♢λW ) = [Π( U ), V ]∗♢λW ) + [ U, Π( V )] ∗♢λW + [ U, V ]∗♢λΠ( W )
for all U, V, W ∈ A . Then Π is an additive ∗-derivation. REFERENCES 3413
Acknowledgements
The authors extended their appreciation to Princess Nourah bint Abdulrahman Uni-versity for funding this research under Researchers Supporting Project number (PNURSP2024R231), Princess Nourah bint Abdulrahman University, Riyadh Saudi Arabia.
Conflicts of Interest: The authors declare no conflict of interest.
References
M. Ashraf, Md. Shamim Akhter, and M. Ansari. Nonlinear bi-skew jordan-type derivations on factor von neumann algebras. Filomat , 37(17):5591–5599, 2023. D. Huo, B. Zheng, J. Xu, and H. Liu. Nonlinear mappings preserving jordan multiple– product on factor von neumann algebras. Linear and Multilinear Algebra , 63(5):1026– 1036, 2015. L. Kong and J. Zhang. Nonlinear skew lie derivations on prime-rings. Indian Journal of Pure and Applied Mathematics , 54(2):475–484, 2023. C. Li, Q. Chen, and T. Wang. Nonlinear maps preserving the jordan triple-product on factor von neumann algebras. Chinese Annals of Mathematics, Series B, 39(4):633–642, 2018. C. Li and F. Lu. Nonlinear maps preserving the jordan triple 1--product on von neumann algebras. Complex Analysis and Operator Theory , 11:109–117, 2017. C. Li and D. Zhang. Nonlinear mixed jordan triple-derivations on-algebras. Siberian Mathematical Journal , 63(4):735–742, 2022. C. Li, F. Zhao, and Q. Chen. Nonlinear skew lie triple derivations between factors.
Acta Mathematica Sinica, English Series , 32(7):821–830, 2016. C. Li, Y. Zhao, and F. Zhao. Nonlinear-jordan-type derivations on-algebras. Rocky Mountain Journal of Mathematics , 51(2):601–612, 2021. Y. Liang and J. Zhang. Nonlinear mixed lie triple derivations on factor von neumann algebras. Acta Math Sci Chinese Series , 62:1–13, 2019. Y. Pang, D. Zhang, and D. Ma. The second nonlinear mixed jordan triple deriv-able mapping on factor von neumann algebras. Bulletin of the Iranian Mathematical Society , 48(3):951–962, 2022. N. Rehman, J. Nisar, and M. Nazim. A note on nonlinear mixed jordan triple deriva-tion on-algebras. Communications in Algebra , 51(4):1334–1343, 2023. REFERENCES 3414
F. Zhang. Nonlinear η-jordan triple ∗-derivation on prime ∗-algebras. Rocky Mountain J. Math , 52:323–333, 2022. Z. Zhou, Y. Zhujun, and Z Jianhua. Nonlinear mixed lie triple derivations on prime -algebras. Communications in Algebra , 47(11):4791–4796, 2019. |
9922 | https://www.lao.ca.gov/sections/health/agency_reports_06-07/epsdt.pdf | Review of Sampling and Extrapolation Methodologies, Early and Periodic Screening, Diagnosis and Treatment Claims Audits Prepared for the California Department of Mental Health, Medi-Cal, Epidemiology, and Forecasting Unit October 2006 By Randall MacIntosh, Ph.D., Principal Investigator Senior Faculty Associate & Professor of Sociology Institute for Social Research California State University, Sacramento Table of Contents Page Introduction.......................................................................................................................1 Issue 1: Does sampling work? ..........................................................................................2 Issue 2: What is sampling error and what can be done to reduce it?...............................2 Issue 3: How much sampling error is present under the current process? ......................3 Issue 4: Is the sample size sufficient for extrapolation? ...................................................6 Issue 5: How much precision is there in the extrapolation?.............................................7 Issue 6: How much larger should the sample be if more precision is desired?................8 Issue 7: Does it matter if more than one claim is selected from a client’s chart?..........10 Issue 8: Is the Raosoft calculator appropriate to determine sample size? .....................12 Issue 9: What are the technical advantages and disadvantages of probe samples? ......12 Issue 10: What factors could invalidate the legal entity selection process?...................12 References.......................................................................................................................14 Appendix.........................................................................................................................15 EPSDT Sampling Study, Institute for Social Research October 2006 Page 1 Introduction This report is prepared at the request of the California Department of Mental Health, Medi-Cal, Epidemiology, and Forecasting unit on the basis of recent legislation. The DMH requested that the review cover: • The statistical validity of the proposed sampling methodology that addresses the issues of stratified random selection, confidence intervals and related parameters, and sample size; • the statistical validity of the extrapolation methodology; • the adequacy of the statistical software programs used for the above processes; • the consequences to the State and Service Providers due to sampling error; and, • suggestions for improving the sampling and extrapolation methodologies. On the basis of that scope of work, interviews were conducted with DMH staff and representatives of providers. Ten technical issues were identified in those conversations. We have sought to explain and evaluate these issues to promote a constructive dialogue between providers and the agency without an overwhelming morass of technical details. As this is a technical report, however, we have laid out some procedural alternatives for policymakers to consider. We have sought to make it explicit in several places that we wish to avoid making any policy recommendations. General conclusion Our general conclusion is that most aspects of the sampling as implemented are at random and scientifically defensible. The precision of the estimates generated by the process, however, appears to be poor. We offer several recommendations for improvement. Primary recommendations Our primary recommendation is for policymakers to consider establishing an acceptable precision range. Sample size would then be determined by the level of precision desired for recoupment amounts. Samples could be drawn in an iterative process until a desired precision is achieved. Under the present system, the sample size is fixed for each audit, regardless of the precision it affords. We also recommend the use of stratification on the basis of dollar amounts. Similarly, if there are several sites that deliver service to different types of clients, it may be desirable to stratify on the basis of location, as well. Other stratification variables may also be tested to reduce sampling error We also recommend that sample characteristics be compared to population values as part of an on-going monitoring of the sample selection process. Policymakers may consider reporting these values to other stakeholders to promote dialogue and transparency. EPSDT Sampling Study, Institute for Social Research October 2006 Page 2 It may be more appropriate to use a variable sample calculator, such as the one found in RAT-STATS. An alternative is to use the formulae found in statistical auditing textbooks. (References are provided on page 14.) These computations are straightforward and can be carried out in a spreadsheet. Issues for Review Issue 1: Does sampling work? Most modern science relies on the use of random samples as a means of estimating unknown values in a population. It is possible, for example, to tell what’s going on in your body from a few milliliters of blood. It is not necessary to analyze the entire 5 liters of blood that we carry within us. That’s because the small sample is representative of the entire supply from which it was randomly selected. There are many mathematical proofs that demonstrate that the mean (or average) from a properly drawn random sample will be a close estimate of the population average. While the sample mean will be close, it will not be precisely the population value. The discrepancy between the population average and the sample average is because of “sampling error.” Issue 2: What is sampling error and what can be done to reduce it? Sampling error is the difference that arises when using a sample to estimate an unknown population characteristic. It is, for example, the difference between the population mean and the sample mean. Several factors determine the amount of error present in a sample. One factor is the amount of variation (or differences in magnitudes) in the population. A second factor is the size of the sample. Sample size can be adjusted to reduce sampling error and obtain a desired level of precision. There is, of course, a trade-off in the cost of the study when sample size is increased. And the increase in precision obtained from a larger sample is not a straight line. In other words, increasing the sample size will only reduce the error up to a certain point. After that point, the increased cost of additional samples greatly exceeds the amount of additional information they would provide. This is shown below in Figure 1. EPSDT Sampling Study, Institute for Social Research October 2006 Page 3 0% 5% 10% 15% 20% 25% 30% 35% 10 50 100 200 1,000 3,000 Sample Size Amount of Sampling Error Fig. 1. Relationship between sample size and sampling error There is a given amount of variation in the population and this cannot be manipulated. There are, however, sampling strategies that can be applied to address this problem. If there is a large amount of variation in the population, it is termed “heterogeneous.” Breaking the population into more homogeneous sub-groups is called “stratifying.” Sub-samples are then drawn from each stratum (group). This strategy typically results in a substantial reduction in sampling error. If there is a large variation in the dollar value of claims, for example, it might be advisable to stratify on this basis and draw proportionate sub-samples from a pool of small claims and a pool of large claims. Similarly, if there are several sites that deliver service to different types of clients, it may be desirable to stratify on the basis of location. This idea has already been implemented, in part, by stratifying by service type and restricting most audits to one type. This eliminates the heterogeneity with regard to service function. As outlined above, other stratification variables also may be tested to reduce sampling error. Issue 3: How much sampling error is present under the current process? An analysis of a recently completed audit of a legal entity (labeled for purposes of this report LE00A) provides an illustration of sampling error. As shown in Table 1, the mean value of the universe (population) of all 5,839 claims filed within the mental health services category is $272.86. That is to be contrasted with the mean of $296.75 derived from a sample of 190 claims. The sample mean overstates the actual population mean by approximately $24. As will be shown later, this sampling error has implications for extrapolation. Claims Mean Claims universe (population) $272.86 Total sample of 190 claims $296.75 Disallowed claims only $323.80 Table 1. Averages (means) from the LE00A audit. EPSDT Sampling Study, Institute for Social Research October 2006 Page 4 Two points should be noted. First, if the process was repeated and another audit sample drawn, it would be expected to be as likely to under-estimate the population mean by a similar amount. Additional study would be required to determine of this over-estimate is simply an unlucky occurrence for the legal entity1 because of random error, or a systematic defect. The analysis of one audit sample is insufficient to draw any conclusion with regard to the validity of the process. Table 1 also shows the average amount for the eight disallowed claims ($323.80) is approximately $50 greater than the population average. Additional study would also be necessary to reach any conclusion as to whether auditors are giving greater scrutiny to larger claims or whether larger claims are more complex and therefore more error-prone. As the three means shown above are simple to obtain, DMH may wish to consider regular reporting of this information on an audit-by-audit basis to stakeholders to facilitate on-going monitoring of the process. Table 2 shows the values for the eight disallowed claims. There is substantial variation in these amounts. As will be demonstrated later, this wide variation has an important impact on extrapolation precision Disallowed claims 1. $818.12 5.
$81.36 2. $649.80 6.
$54.72 3. $576.30 7.
$41.04 4. $335.16 8.
$33.90 Table 2. Values of the eight disallowed claims 1 A “legal entity” is a county or a corporation that is the subject of the audit. EPSDT Sampling Study, Institute for Social Research October 2006 Page 5 The population and sample means for upcoming audits are shown in Table 3. Audit Sample Size LE Claims Population Average LE Claims Sample Average Difference Coefficient of Variation1 1 194 $ 92.35 $ 89.27 -$3.08 74% 2 191 $118.94 $118.94 $0.00 0% 3 192 $203.93 $192.30 -$11.63 73% 4 18 $149.76 $149.76 $0.00 61% 5 194 $111.50 $108.92 -$2.58 52% 6 195 $156.93 $145.60 -$11.33 72% 7 191 $141.92 $141.84 -$0.08 54% 8 195 $ 81.14 $ 88.49 $7.35 75% 9 190 $179.71 $192.39 $12.68 75% 10 195 $135.33 $143.22 $7.89 61% 11 195 $ 83.61 $ 88.60 $4.99 110% 12 195 $131.32 $125.66 -$5.66 87% 13 187 $132.79 $141.26 $8.47 69% 14 195 $164.33 $172.58 $8.25 70% 15 193 $197.76 $196.70 -$1.06 48% 16 195 $147.92 $165.13 $17.21 58% 17 195 $154.28 $149.12 -$5.16 57% 18 192 $156.78 $161.11 $4.33 82% 19 194 $115.07 $106.87 -$8.20 57% 20 195 $292.65 $298.40 $5.75 69% 22 195 $185.00 $164.92 -$20.08 101% 23 194 $130.60 $140.75 $10.15 77% 24 194 $ 97.22 $ 98.83 $1.61 64% 25 194 $152.01 $146.74 -$5.27 75% 26 61 $ 58.39 $ 61.37 $2.98 103% 27 188 $339.52 $363.38 $23.86 64% Overall Avereage $150.41 $152.01 $1.59 69% 1 Coefficient of variation is the standard deviation divided by the population mean. Higher percentages means less consistency among the claim values and this greater variation could be expected to introduce more sampling error. (Standard deviation values are not shown in this table.) Table 3. Sample and population means for planned audits Table 3 shows the 27 sample averages slightly overstate the overall population average by about 1 percent. This is well within the expected range of sampling error. Notice, however, that the over- and under-estimate is substantially greater for some individual audit subjects. This is because the amount of variation within the claims is greater. The “coefficient of variation” column gives a measure of how much variation there is among claim values for each legal entity. As these values vary widely among the LEs, the number of sampled units would also vary to provide similar levels of precision. Resources could be better managed by reallocating portions of samples from entities with a smaller coefficient of variation to those with more variability. EPSDT Sampling Study, Institute for Social Research October 2006 Page 6 Issue 4: Is the sample size sufficient for extrapolation? A review of reports issued by Office of Audit Services within the Inspector General’s Office of the U.S. Department of Health and Human Services (DHHS) shows the proposed California DMH sample size of 200 to 250 is at the upper end of the range used by the federal agency in its audits. Some of the federal audit reports show the DHHS extrapolation procedure differs from that of CA DMH. The DHHS has used the lower limit of a 90 percent confidence interval as the repayment amount. The current DMH method does not take sampling error into account and therefore does not use any confidence interval when establishing repayment amounts through extrapolation. Audit Sample Size PCH Health Systems outpatient physical and occupational therapy 114 Hospital of the University of Pennsylvania air ambulance services 100 Regent Care Center skilled services 50 Independent Diagnostic Testing Facilities 230 New Jersey school-based health services 150 Table 4. Sample sizes used in U.S. DHHS audits. The level of desired precision and the equity of the extrapolation methodology are policy matters and beyond the scope of this review. Consequently, we make no recommendations in these areas. EPSDT Sampling Study, Institute for Social Research October 2006 Page 7 Issue 5: How much precision is there in the extrapolation? One method of evaluating the extrapolation precision is to use the audit sample to estimate the total value of claims in the mental health services category for LE00A (again for confidentiality this is a hypothetical number). This is useful because we know what the true value is and it can be compared to the extrapolated value. Panel A of Table 5 shows the extrapolated value and the true value, which differ by $139,524. The extrapolation overestimates the true value because of the sampling error discussed above under Issue 3 and is reflected in the discrepancy between the sample and population means shown in Table 1. A. Value Extrapolated estimate of total claim value $1,732,728 True total claim value $1,593,204 Difference $139,524 B. Extrapolated ranges Confidence Level 80% Lower limit $1,641,779 Upper limit $1,823,677 90% Lower limit $1,615,831 Upper limit $1,849,625 95% Lower limit $1,593,227 Upper limit $1,872,230 Table 5. Extrapolation of total claims value based on the audit sample. The amount of sampling error can be taken into account by constructing a confidence interval around the extrapolated value. It gives a range of plausible values based on the amount of variation in the sample. Panel B of Table 5 shows the confidence intervals for the extrapolated amount obtained by adding and subtracting a “margin of error.” The margin of error is computed based on the sampling error. Several confidence intervals are available depending on the level of confidence desired. The wider the interval, the greater the confidence we have that the true value will fall within it. Notice that only the lower limit of the 95 percent confidence interval ($1,593,227) approaches the true value of $1,593,204. The upper limit substantially overstates the true value and note that the other more narrow confidence intervals do not include the true value. EPSDT Sampling Study, Institute for Social Research October 2006 Page 8 The same procedure, with the same level of confidence, can be applied to the disallowed values. Doing so produces a wide range between the upper and lower limits. This is because there is a wide disparity in the disallowed values (shown previously in Table 2). When the value of the eight claims disallowed in the audit ($2,590.40) is extrapolated to the universe of 5,839 mental health services claims, the recoupment amount is estimated to be $79,6072. Taking sampling error into account and computing a margin of error reveals the imprecision of the estimate, as shown in Table 6. A 95 percent confidence interval is obtained for the extrapolation of disallowed values. The range for the true value is between a lower limit of $6,067 and an upper limit of $153,147. Recall that is was the lower limit that approached the true value for the estimate of all claims shown in Table 5. But the lower limit in the table below is 1/13th of the extrapolated disallowed value. The difference between these two values is not inconsequential. The level of precision can be summarized. The summaries are part of the standard output of RAT-STATS, a program specifically designed to analyze audit data written by the DHHS Office of Audit Services and distributed free. For each level of confidence, RAT-STATS computes a margin of error (labeled “Precision Amount”) and computes the size of the margin of error relative to the estimate (labeled “Precision Percent”). The margin of error for the 95 percent confidence interval for the disallowed extrapolation is $73,540 or an amount that is 92.4% of the extrapolated estimate of $79,607 (shown in Table 6). Put more simply, the margin of error is almost equal to the recoupment estimate. (The complete RAT-STATS output is contained in the Appendix.) Value Extrapolated disallowed estimate $79,607 Margin of error $73,540 Confidence Level 95% Lower limit $6,067 Upper limit $153,147 Table 6. Extrapolated disallowed amount based on audit sample average. Issue 6: How much larger should the sample be if more precision is desired? Keep in mind that sample stratification would reduce the sampling error and provide some additional precision at little additional cost. Additional research could reveal which stratification variables would be the most useful. If claims are stratified on value, Roberts (1978) recommends that the top stratum be sampled on a 100 percent basis. For LE00A (again, number 2 This value is for exposition purposes only. Because the value of the eight disallowed claims is 4.6 percent of the total sample value, no extrapolation is being applied in this audit. (Another $230 in disallowed claims would have triggered extrapolation.) The values discussed above are to demonstrate how the process would work if the disallowed amount exceeded the 5 percent threshold. EPSDT Sampling Study, Institute for Social Research October 2006 Page 9 is hypothetical for reasons of confidentiality), there are 138 claims larger than $600. The virtue of examining all large claims is that the sample error for this stratum is reduced to zero.3 Beyond stratification, increasing the sample size would also deliver estimates that are more precise. Each audit will be different because the variation in the disallowed claims is expected to differ. With that caveat in mind, presented below is a simulation of how various sample sizes would affect the extrapolation precision for the LE00A audit. This analysis is based on two assumptions. First, it is assumed that larger samples would uncover the same proportion of disallowed claims. Second, it is also assumed that the variation of those additional disallowed claims would approximate the variation shown for the disallowed claims reported in Table 2. Sample Size Extrapolation Margin of Error1 Precision Percentage2 Expected Disallowed Claims 190 $73,540 92% 8 200 $71,620 90% 8 225 $67,370 85% 9 250 $63,770 80% 11 275 $60,670 76% 12 300 $57,950 73% 13 325 $55,560 70% 14 350 $53,410 67% 15 375 $51,480 65% 16 400 $49,740 62% 17 425 $48,140 60% 18 450 $46,670 59% 19 475 $45,320 57% 20 500 $44,070 55% 21 525 $42,910 54% 22 550 $41,830 53% 23 575 $40,810 51% 24 600 $39,850 50% 25 1 Margin of error is the “Precision Amount” in RAT-STATS output. It is the standard error of the estimate multiplied by a t-value of 1.97. This value, when added to and subtracted from the extrapolated estimate, provides a 95 percent confidence interval. 2 The precision percentage represents the size of the margin of error relative to the estimated recoupment amount. Smaller values in this column represent more precision. Table 7. Extrapolation precision estimates for various sample sizes As shown in Table 7, doubling the sample size to 400 would be expected to reduce the margin of error for the extrapolation by about one-third (92% versus 62%). A sample of 600 claims would be expected to reduce the margin of error to approximately 50 percent of the extrapolated 3 Details on determining stratum boundaries as recommended by Gunning, Horgan and Yancey (2004) are found in the Appendix. EPSDT Sampling Study, Institute for Social Research October 2006 Page 10 recoupment amount. The cost of increasing the sample size could be offset by conducting fewer audits. Samples could be drawn in an iterative process until a desired precision is achieved. Issue 7: Does it matter if more than one claim is selected from a client’s chart? All of the precision computations presented and discussed in previous sections of this report are based on the assumption of simple random sampling (SRS). Under SRS, each selected claim is assumed unrelated to the others in the sample. If, however, this is not the case, the computations for variance (and precision) estimates become much more complicated. If the claims selected in the sample are somehow related to each other, then the variance is understated and the precision is overstated. If some of the sample claims come from one client’s record, then they are said to be “clustered.” Additional research could describe the exact nature of how this clustering affects the claim contents. But it is likely, for example, that multiple claims from one client’s chart are processed by the same clerical persons and documented by the same professionals. If one claim in this file is problematic, then it is likely others could be as well. In statistical terms, the claims’ quality within a client’s file are said to be “correlated.” It is this correlation among sampled claims that leads to the understatement of variance and overstatement of precision. Client Claims Client Claims Client Claims 8323 13 6532 3 6740 2 8311 12 8462 3 1297 1 8388 10 2616 3 5066 1 7413 10 3256 3 8164 1 0887 9 3388 3 2767 1 6091 7 9345 3 7122 1 0556 6 5862 3 3024 1 9703 6 0568 2 4614 1 8071 5 5753 2 1870 1 0491 5 0585 2 8006 1 1613 4 4864 2 5724 1 9466 4 3576 2 9901 1 0880 4 5348 2 8378 1 3077 4 6173 2 8614 1 5591 4 6405 2 9661 1 4777 4 6445 2 9663 1 6115 4 6909 2 3402 1 6281 4 4582 2 4223 1 0508 4 1424 2 7435 1 2204 3 1526 2 9263 1 Table 8. Distribution of sampled claims by client. Table 8 shows that nearly one-third of the audit sample claims came from six clients and only 19 of the 190 sampled claims fit the SRS assumption. EPSDT Sampling Study, Institute for Social Research October 2006 Page 11 Client Disallowed claims Claims in the sample 0887 2 9 8311 1 12 1613 1 4 5591 1 4 0508 1 4 6405 1 2 1424 1 2 Table 9. Disallowed claims and number of claims sampled by client. Table 9 shows that all the disallowed claims came from charts where more than one claim was sampled. Two of the eight disallowed claims came from one chart. None of the disallowed claims came from the 19 client records that met the SRS assumption. Table 10 illustrates the effect clustering has on the sample variance estimate. Under the assumption of SRS, the variance estimate is 12.3. When taking clustering into account, the variance is inflated by about one-fourth4 and is estimated to be 15.5. (This inflation is called the “design effect.”) The practical import of this issue is that the variance computations are the basis for precision estimates. If the variance is understated, the precision will be overstated. Sample mean Standard error assuming SRS Standard error assuming clustering Design Effect 296.75 12.3 15.5 1.64 Table 10. Effect of clustering on sample variance estimates. Should policymakers determine that some precision standard is desired, the continued use of clustered data would make subsequent analysis tasks more complex. The computational difficulties associated with clustered data, however, could be avoided by limiting the sample selection to one claim per client record. If there are an insufficient number of clients to achieve this and clustered samples are unavoidable, then complex sample analysis routines could be utilized, such as PROC SURVEYMEANS and SURVEYREG in SAS, to obtain appropriate variance estimates. The clustering has a smaller effect on the variance estimates of the disallowed claims because there is only one instance where more than one disallowed claim came from the same client’s record (as shown in Table 10). This may not be the case, however, in other audits. 4 The square root of the design effect gives the amount that clustering inflates the variance estimate. In this instance 1.64.5=1.28 or 28%. EPSDT Sampling Study, Institute for Social Research October 2006 Page 12 Issue 8: Is the Raosoft calculator the appropriate software to determine sample size? Preliminarily, a distinction must be made between attribute sampling and variable sampling. Attribute sampling is used when the goal of the study is to simply estimate the proportion of disallowed claims, where each claim is evaluated on a “pass/fail” basis. Variable sampling, on the other hand, is used when the goal is to estimate a dollar value for disallowed claims. The Raosoft sample size calculator implements standard statistical formulae to determine sample sizes for a given confidence level for attribute sampling.5 As the goal of the audits is to estimate the value of disallowed claims, it may be more appropriate to use a variable sample calculator. RAT-STATS has such a module that will give sample sizes for various precision levels. One limitation of the RAT-STATS module is that the user cannot specify additional alternative levels of desired precision. The computations for sample sizes based on desired precision levels are presented in textbooks on audit sampling, such as Roberts (1978) or Guy, Carmichael and Whittington (1998). As with the Raosoft calculator, the RAT-STATS sample calculator requires the input of population sizes and several assumed values: strata means and standard deviations. These values can be computed using the claims database. It will be difficult, however, to calculate sample sizes to arrive at a desired precision based on the mean and standard deviation of disallowed claims. The only way to estimate these values is to use a probe sample to uncover some disallowances. Issue 9: What are the advantages and disadvantages of probe samples? Providers expressed great interest in using probe samples. From a technical point of view, a probe sample would allow for overall sample sizes to be determined for a desired level of precision (as discussed above), which is a departure from current practice. The use of probe samples would also allow DMH to more efficiently allocate its resources in that portions of samples from providers with low disallowance rates could be reallocated to providers with higher disallowance rates to obtain more precise recoupment extrapolation. DMH expressed concern about adding to the project management complexity if probe samples were used. The primary concerns expressed were in the area of audit notification to the providers and timeframe management. As these are policy matters, they are beyond the scope of this report and we make no recommendation. Issue 10: What factors could invalidate the legal entity selection process? A two-stage sampling process is currently used by DMH (a detailed description is found in the Appendix). In the first stage, legal entities are selected to be audited. Briefly, entities are stratified by the claim dollar amounts and the number of entities selected for an audit in each stratum is roughly proportional to the total claims values within the service category. Not all 5 The parameters used by DMH are a 95% confidence interval and an assumed failure (disallowance) proportion of 15 percent. Details are found in the Appendix. EPSDT Sampling Study, Institute for Social Research October 2006 Page 13 entities are subject to audit selection, but DMH does not want to disclose which legal entities fall into this category. For the legal entities subject to selection, a random number is assigned as a means of carrying out selection. Deviations could invalidate the random selection process, such as repeatedly drawing the sample of selected audit subjects and choosing a particular subject sample based on undocumented selection criteria. EPSDT Sampling Study, Institute for Social Research October 2006 Page 14 References Gunning, Patricia, Horgan, Jane M., and Yancey, W. 2004. A Simple Method for Setting Stratum Boundaries for Statistical Auditing, Proc. Of Amer, Acc, Association Southern Region Meeting, Austin Texas, March Guy, Dan M., Douglas R Carmichael and O. Ray Whittington. 1998. Audit Sampling: An Introduction. Wiley: New York. Roberts, Donald M. 1978. Statistical Auditing. American Institute of Certified Public Accountants: New York. EPSDT Sampling Study, Institute for Social Research October 2006 Page 15 APPENDIX EPSDT Sampling Study, Institute for Social Research October 2006 Page 16 RAT-STATS OUTPUT DEPARTMENT OF HEALTH & HUMAN SERVICES OIG - OFFICE OF AUDIT SERVICES Date: 9/4/2006 VARIABLE UNRESTRICTED APPRAISAL Time: 21:20 AUDIT/REVIEW: LE00A (fictitious number for confidentiality) DATA FILE USED: C:\My Documents\auditing\ExaminedAuditedLE00A.xls SAMPLE EXAMINED NONZERO TOTAL OF TOTAL OF SIZE VALUE DIFFS DIFF VALUES AUD VALUES 190 56,382.66 8 2,590.40 53,792.26 ----------------------- E X A M I N E D ------------------------ MEAN / UNIVERSE 296.75 5,839 STANDARD DEVIATION 169.73 STANDARD ERROR 12.11 SKEWNESS .59 KURTOSIS 3.05 POINT ESTIMATE 1,732,728 CONFIDENCE LIMITS 80% CONFIDENCE LEVEL LOWER LIMIT 1,641,779 UPPER LIMIT 1,823,677 PRECISION AMOUNT 90,949 PRECISION PERCENT 5.25% T-VALUE USED 1.286046870294 90% CONFIDENCE LEVEL LOWER LIMIT 1,615,831 UPPER LIMIT 1,849,625 PRECISION AMOUNT 116,897 PRECISION PERCENT 6.75% T-VALUE USED 1.652955801726 95% CONFIDENCE LEVEL LOWER LIMIT 1,593,227 UPPER LIMIT 1,872,230 PRECISION AMOUNT 139,502 PRECISION PERCENT 8.05% T-VALUE USED 1.972595079100 ----------------------- A U D I T E D -------------------------- MEAN / UNIVERSE 283.12 5,839 STANDARD DEVIATION 169.01 STANDARD ERROR 12.06 SKEWNESS .45 KURTOSIS 2.89 POINT ESTIMATE 1,653,121 CONFIDENCE LIMITS 80% CONFIDENCE LEVEL LOWER LIMIT 1,562,560 UPPER LIMIT 1,743,682 PRECISION AMOUNT 90,561 PRECISION PERCENT 5.48% T-VALUE USED 1.286046870294 90% CONFIDENCE LEVEL LOWER LIMIT 1,536,723 UPPER LIMIT 1,769,519 PRECISION AMOUNT 116,398 PRECISION PERCENT 7.04% T-VALUE USED 1.652955801726 95% CONFIDENCE LEVEL LOWER LIMIT 1,514,215 UPPER LIMIT 1,792,027 PRECISION AMOUNT 138,906 PRECISION PERCENT 8.40% T-VALUE USED 1.972595079100 EPSDT Sampling Study, Institute for Social Research October 2006 Page 17 --------------------- D I F F E R E N C E ---------------------- MEAN / UNIVERSE 13.63 5,839 STANDARD DEVIATION 89.48 STANDARD ERROR 6.38 SKEWNESS 7.33 KURTOSIS 57.56 POINT ESTIMATE 79,607 CONFIDENCE LIMITS 80% CONFIDENCE LEVEL LOWER LIMIT 31,662 UPPER LIMIT 127,552 PRECISION AMOUNT 47,945 PRECISION PERCENT 60.23% T-VALUE USED 1.286046870294 90% CONFIDENCE LEVEL LOWER LIMIT 17,983 UPPER LIMIT 141,231 PRECISION AMOUNT 61,624 PRECISION PERCENT 77.41% T-VALUE USED 1.652955801726 95% CONFIDENCE LEVEL LOWER LIMIT 6,067 UPPER LIMIT 153,147 PRECISION AMOUNT 73,540 PRECISION PERCENT 92.38% T-VALUE USED 1.972595079100 EPSDT Sampling Study, Institute for Social Research October 2006 Page 18 Defining Stratum Boundaries (from Gunning, Horgan and Yancey 2004) The following assumes the top stratum (greater than $600) is sampled on a 100 percent basis. 1. Arrange the stratification variable X in ascending order; 2. Take the minimum value as the first term, and the maximum value as the last term of the geometric series with L+1 terms; 3. Calculate the common ratio: r = (max/min)1/L ; 4. Take the boundaries of each stratum to be the X values corresponding to the terms in the geometric progression with this common ratio: Minimum k0= a, ar, ar2 ….. arL = maximum kL. L=3, k0=11,….. ,k3=600: thus r = (600/11) 1/3 = 54.55 1/ 3 = 3.89, and kh=113.89 h (h=0,1,2,3). Sample sizes should correspond to the percent of the population within each stratum. (Research has shown up to five strata provide the most cost-effective stratification.) Stratum Range Population Percent 1 $11-$41.70 2.4 2 $41.71-$158.19 22.3 3 $158.20-$600 75.3 4 $600-$1,100 Table 11. Stratified sample distribution for LE00A All claims greater than $600 are examined. This threshold could be adjusted upward for LEs with a large number of high-value claims. EPSDT Sampling Study, Institute for Social Research October 2006 Page 19 Fig. 2. Histogram of all claim values for LE00A. 1000 800 600 400 200 0 Claim value 600 500 400 300 200 100 0 Frequency Skew: .72 Kurtosis: .98. EPSDT Sampling Study, Institute for Social Research October 2006 Page 20 Fig. 3. Non-normal distribution of all claim values LE00A 1.0 0.8 0.6 0.4 0.2 0.0 Observed Cum Prob 1.0 0.8 0.6 0.4 0.2 0.0 Expected Cum Prob Normal P-P Plot of Claim Values EPSDT Sampling Study, Institute for Social Research October 2006 Page 21 DMH Documents Sampling Process 1. A Legal Entity is a corporation or a county. A corporation may have one or more providers. Counties usually have multiple providers, but small counties may have only one provider. 2. Legal Entities are assigned into three categories based on their total approved amount during the review fiscal year. The data source is the SD/MC approved claims file. 3. The LEs are assigned two random numbers. The first number determines which LEs will be reviewed during the review period. The second number determines when the selected LEs will be reviewed. 4. About a month prior to the review, several SAS programs are run to obtain the sample. • Claims are selected for a specific LE within county and saved in a dataset. • During the selection those claims with a disallowed indicator are deleted and excluded from the dataset. • Claims in the dataset are assigned a random number, using the SAS software random number generator. • The RaoSoft software is used to determine the sample size based on the number of claims for the LE within county and type of service for the review time period. • A maximum of 250 claims will be selected, plus a maximum of 20 replacement claims per service category. These claims are selected based on the random number. • The replacement claims are used if the LE disallowed a claim prior to notification of the review and the LE has a receipt for the disallowed claim. • Usually a LE will have one or more types of service reviewed to a maximum of 250 claims. • The sample claims and replacement claims are unduplicated by client and provider to obtain a list of clients and charts that will be used in the audit. EPSDT Sampling Study, Institute for Social Research October 2006 Page 22 Sample Size Calculation 1) The sample size will be determined by using the Raosoft web site at www.raosoft.com. 2) Where there is more than one county per Legal Entity (LE), each county will have their own sample size calculated. 3) The sample Size will be calculated for each type of service within county within LE. 4) Where there is more than one county per LE any disallowance will be extrapolated to each county using the county’s sample of claims. The parameters for the Raosoft product will be the following: Error level will be set to 5% The confidence level will be set to 95% The distribution level will be set to 85%. This level was selected based on past reviews of charts. It is expected that at least 85% of the claims will not have a disallowance. EPSDT Sampling Study, Institute for Social Research October 2006 Page 23 Enabling Legislation SEC. 81. The State Department of Mental Health shall revise its method for auditing entities that provide specialty mental health services under the Early and Periodic Screening, Diagnosis, and Treatment Program, and its method for extrapolating data obtained from those audits, pursuant to this section. Commencing July 1, 2006, and continuing thereafter, the following provisions shall apply: (a) The department shall select statistically valid stratified samples by service function for each entity to be audited. (b) The department shall not extrapolate the results of any audit to the full audited service function unless the error rate determined by the audit is five percent or greater. If the error rate is less than five percent, the department shall disallow only the specific claims found to be in error. (c) The department, in consultation with stakeholders, shall select an independent statistician to review the sampling methodology and extrapolation methodology used by the department. No later than October 1, 2006, the statistician shall prepare a public report on the statistical validity of those methodologies. If the statistician determines either methodology to be invalid, the department shall adopt a new methodology, which shall be used by the department only after its validity is verified by the statistician. |
9923 | https://www.stemlynsblog.org/podcast-ppv-npv/ | Positive and Negative Predictive Values
Skip to content
St.Emlyn's
Emergency Medicine #FOAMed
AboutMenu Toggle
Our TeamMenu Toggle
Professor Simon Carley
Professor Richard Body
Dr Iain Beardsell
Dr Liz Crowe
Prof Daniel Horner
Dr Natalie May
Dr Trudie Pestell
Administration and BackgroundMenu Toggle
Research
Cite St.Emlyn’s.
Caveats and Caution
Disclaimer
Case presentations on St Emlyn’s
Privacy policy. St Emlyn’s
Who funds St.Emlyn’s?
St Emlyn’s Lesson PlansMenu Toggle
Induction
SHO Teaching
HST Teaching
Medical School
EM CurriculaMenu Toggle
RCEM Curriculum 2021
Undergraduate Acute and Emergency Care Curriculum
ACCS Curriculum (2015)
CT3 Curriculum (2015)
HST Curriculum (2015)
ACP Curriculum
Revision GuidesMenu Toggle
FRCEM Revision GuideMenu Toggle
FRCEM Revision Guide – Forward
FRCEM Revision Guide – Contents
Chapter 1 – The FRCEM Short Answer Question (SAQ)/Single Best Answer (SBA)
Chapter 2: The FRCEM Objective Structured Clinical Examination (OSCE)
Chapter 3: History taking stations
Chapter 4 – Communication Stations
Chapter 5 – Psychiatry, substance misuse and safeguarding stations
Chapter 6 – System examination stations
Chapter 7 – Musculoskeletal stations
Chapter 8 – Resuscitation Stations
Chapter 9 – Procedure Stations
Chapter 10 – Teaching stations
MLA Revision GuideMenu Toggle
MLA Revision Guide – Forward
Medical Licensing Assessment Revision- Contents
Chapter 1 – The Applied Knowledge Test (MCQ/SBA)
Chapter 2 – The MLA Clinical and Professional Skills Assessment
Chapter 3 – History taking CPSA Stations
Chapter 4 – Communication Stations
Chapter 5 – Psychiatry, substance misuse and safeguarding CPSA stations
Chapter 6 – System Examination CPSA Stations
Chapter 7 – Musculoskeletal CPSA Stations
Chapter 8 – Resuscitation CPSA Stations
Chapter 9 – Procedure CPSA Stations
Chapter 10 – Prescribing CPSA Station
e-BooksMenu Toggle
Journal Club 2019 E-book
Journal Club 2018 E-book
The Resuscitationist’s Guide to Health and Wellbeing
Understanding Risk and Probability in Emergency Medicine
101 Lessons from Sydney HEMS
A beginner’s guide to using social media and #FOAMed resources to teach and learn in postgraduate medical education programs.
Contact
ECG Library
Search for:
SpotifyiTunesYouTubeInstagramBlueskyTwitterFacebook
St.Emlyn's
Emergency Medicine #FOAMed
Main Menu
Search for:
AboutMenu Toggle
Our TeamMenu Toggle
Professor Simon Carley
Professor Richard Body
Dr Iain Beardsell
Dr Liz Crowe
Prof Daniel Horner
Dr Natalie May
Dr Trudie Pestell
Administration and BackgroundMenu Toggle
Research
Cite St.Emlyn’s.
Caveats and Caution
Disclaimer
Case presentations on St Emlyn’s
Privacy policy. St Emlyn’s
Who funds St.Emlyn’s?
St Emlyn’s Lesson PlansMenu Toggle
Induction
SHO Teaching
HST Teaching
Medical School
EM CurriculaMenu Toggle
RCEM Curriculum 2021
Undergraduate Acute and Emergency Care Curriculum
ACCS Curriculum (2015)
CT3 Curriculum (2015)
HST Curriculum (2015)
ACP Curriculum
Revision GuidesMenu Toggle
FRCEM Revision GuideMenu Toggle
FRCEM Revision Guide – Forward
FRCEM Revision Guide – Contents
Chapter 1 – The FRCEM Short Answer Question (SAQ)/Single Best Answer (SBA)
Chapter 2: The FRCEM Objective Structured Clinical Examination (OSCE)
Chapter 3: History taking stations
Chapter 4 – Communication Stations
Chapter 5 – Psychiatry, substance misuse and safeguarding stations
Chapter 6 – System examination stations
Chapter 7 – Musculoskeletal stations
Chapter 8 – Resuscitation Stations
Chapter 9 – Procedure Stations
Chapter 10 – Teaching stations
MLA Revision GuideMenu Toggle
MLA Revision Guide – Forward
Medical Licensing Assessment Revision- Contents
Chapter 1 – The Applied Knowledge Test (MCQ/SBA)
Chapter 2 – The MLA Clinical and Professional Skills Assessment
Chapter 3 – History taking CPSA Stations
Chapter 4 – Communication Stations
Chapter 5 – Psychiatry, substance misuse and safeguarding CPSA stations
Chapter 6 – System Examination CPSA Stations
Chapter 7 – Musculoskeletal CPSA Stations
Chapter 8 – Resuscitation CPSA Stations
Chapter 9 – Procedure CPSA Stations
Chapter 10 – Prescribing CPSA Station
e-BooksMenu Toggle
Journal Club 2019 E-book
Journal Club 2018 E-book
The Resuscitationist’s Guide to Health and Wellbeing
Understanding Risk and Probability in Emergency Medicine
101 Lessons from Sydney HEMS
A beginner’s guide to using social media and #FOAMed resources to teach and learn in postgraduate medical education programs.
Contact
ECG Library
Podcast – Positive and Negative Predictive Values: Critical Appraisal Nugget 11
By Rick Body / July 24, 2024 / #FOAMed, Emergency Medicine, Evidence Based Medicine, Journal Club, Podcast
Welcome back to another instalment of our Critical Appraisal Nugget series with Rick Body and Greg Yates here at St Emlyn’s. In our previous podcast, we delved into the concepts of sensitivity and specificity, laying a strong foundation for understanding diagnostic tests. Today, we’re building on that knowledge by exploring positive predictive value (PPV) and negative predictive value (NPV).
Listening time – 11:16
Revisiting sensitivity and specificity
Before we dive into positive and negative predictive values, let’s briefly revisit sensitivity and specificity. Sensitivity is the ability of a test to correctly identify those with the disease (true positives), while specificity is the ability of a test to correctly identify those without the disease (true negatives). Both metrics are intrinsic properties of the test, meaning they don’t change regardless of the prevalence of the disease in the population.
Positive predictive value (PPV)
Positive Predictive Value (PPV) is the probability that a person has the disease given that they have tested positive. It answers the question, “If the test result is positive, what are the chances that the patient actually has the disease?” We can calculate the PPV as follows: true positives / (true positives + false positives)
Negative predictive value (NPV)
Negative Predictive Value (NPV) is subtly different: it is the probability that a person does not have the disease given that they have tested negative. It answers the question, “If the test result is negative, what are the chances that the patient does not have the disease?” We can calculate the NPV as follows: true negatives / (true negatives + false negatives)
The importance of prevalence
One of the key differences between sensitivity/specificity and PPV/NPV is that positive and negative predictive values are heavily influenced by the prevalenceof the disease in the population being tested. As prevalence increases, PPV increases and NPV decreases, and vice versa. This dependency on prevalence makes PPV and NPV highly relevant in clinical practice, where the pre-test probability (or prevalence) can vary significantly across different populations and settings. However, it also gives us reason for caution: if the prevalence is very low then we would expect to see a high NPV even if the test is not very good. That being so, while the NPV and PPV help to give us a practical idea of the post-test probability of disease in a given cohort, we also need to interpret them alongside the sensitivity and specificity, which are less dependent on prevalence and more dependent on the ability of the diagnostic test to differentiate between those who have the disease (or condition) and those who don’t.
Practical Example
Let’s consider a diagnostic test for a disease with a prevalence of 10% in a given population. Suppose the test has a sensitivity of 90% and a specificity of 90%. The 2×2 table below illustrates this:
Disease present Disease absent
Test positive 90 90
Test negative 10 810
2×2 table for a test with 90% sensitivity and 90% specificity with 10% prevalence
In this example, the PPV is equal to 90 / (90 + 90) = 50%. The NPV is equal to 810 / (10 + 810) = 98.8%. So we have a pretty good rule-out test but not such a good rule-in test. Let’s now change the prevalence to 70% – a really high-risk cohort – but let’s keep the sensitivity and specificity of the test the same, at 90% each. Let’s see how the 2×2 table looks:
Disease present (n=100)Disease absent (n=42)
Test positive 90 4
Test negative 10 38
2×2 table for a test with 90% sensitivity and 90% specificity with 70% prevalence
From this table, you can work out that the sensitivity and specificity are still both equal to 90% – so the test is working just as well. However, the PPV and NPV have changed drastically. The PPV = 90 / (90 + 4) = 95.7%, whereas the NPV = 38 / (10 + 38) = 79.2%. Suddenly, we have a pretty good rule-in test but not such a good rule-out test! So, you can see why we need to take account of both sensitivity and specificity, and the PPV and NPV.
Why PPV and NPV matter
In clinical practice, PPV and NPV are crucial because they provide direct information about the test’s performance in a real-world setting. They help clinicians make more informed decisions about patient care based on the likelihood of disease presence or absence after testing.
Summary
Understanding PPV and NPV is a really important skill for emergency physicians, whether it be for critical appraisal of a paper to decide if you should use a new test in practice, to guide your own research or to inform your daily practice as you apply diagnostic tests in everyday patient care. Remember to consider the prevalence of disease when interpreting PPV and NPV, and be sure to also look at the sensitivity and specificity too. While the PPV and NPV give us a practical idea of the post-test probability of disease, the sensitivity and specificity will also help to reassure us that the test is doing something more than just rolling a dice! We hope you find this CAN podcast helpful. Stay tuned for more critical appraisal nuggets at St Emlyn’s.
Podcast Transcript
Hello and welcome to the St. Emlyn’s podcast. I’m Rick Body. And I’m Greg Yates. It’s been a little while since we did our last critical appraisal nugget podcast, but here we are going to talk about positive predictive value and negative predictive value. This continues the theme of diagnostic test accuracy. If you remember, last time we talked about sensitivity and specificity. Now, these two concepts, positive and negative predictive value, are related to sensitivity and specificity, but they’re different in an important way. It’s really vital that if you’re preparing for exams or if you’re just thinking about how you can best apply diagnostic tests in your practice, these are really important concepts to get your head around.
Positive predictive value and negative predictive value are arguably a more clinically helpful way of thinking about what our different tests can do. Although, as we’re going to talk about today, there are probably some caveats to that statement. So, Rick, what do we actually mean when we say positive predictive value and negative predictive value?
By positive predictive value, we are asking what is the probability that a patient with a positive test actually has the condition that we’re trying to diagnose? A negative predictive value is looking at the probability that a patient with a negative test doesn’t have the disease. If that’s not already boggled, it might be helpful to think about a clinical example. So, last time we did one of these critical appraisal nuggets, we thought about border force officer Rick Body. Now this time, it’s going to be ward cover FYI Rick Body.
So, cast your mind back into the past. You’re covering the wards at night. It’s midnight. You’ve just prescribed your 19th bag of maintenance dextrose and you’re fed up. Your bleep has gone off again and it’s the vascular ward calling. They want you to come and see a patient there who’s just developed some chest pain. Very helpfully, the nurses have already done an ECG and they’ve sent off a series of blood tests, including a high-sensitivity troponin. So far, so good. Unfortunately, before you can see this patient, you get caught up with a cardiac arrest. Sorting that out and documenting everything that happened takes a long while, as it does. By the time you finish and finally get to the vascular ward, most of those blood tests have come back.
This patient with chest pain has had their blood test resulted and the high-sensitivity troponin is positive. So, what’s going through your mind right now? You’ve got a positive troponin, chest pain, vascular ward. Okay, so, I mean, clearly, I’m going to want to know the clinical context of this particular patient as we always should. And then having got that, I’m going to want to know what’s the probability that a patient with a positive troponin is actually having a heart attack. That’s what I’m trying to diagnose here. The troponin isn’t the diagnosis. The diagnosis I’m trying to make is a myocardial infarction. I know that troponin will be perfect, but I want to know that probability because it’s going to affect what I do. I will want to prescribe some treatments for the patient that are having a myocardial infarction, which might include aspirin or other antiplatelet medications. I want to get the cardiologists involved. But I don’t want to do that if the patient is unlikely to be having a myocardial infarction. So, it’s really important now to know that probability.
And I’d add to that, even a negative test is very helpful, isn’t it? Because you might consider an alternate diagnosis. You might think about where you could best spend your time with another patient, for example. So, there’s lots of value to knowing how likely a negative test is to reflect somebody who doesn’t have the disease.
Exactly. And had the test been negative, I’d similarly want to know, well, what’s the chance now that the patient’s got a myocardial infarction? Because I have to appreciate not all tests are black and white. It’s not going to be a 0% chance of a myocardial infarction. So, I want to know what’s the chance of missing that diagnosis. What you’ve basically described there are the two facts, or best attempt at the facts, that we get from positive and negative predictive values. Positive predictive value, as a sort of recap, is going to tell us the proportion of people with a positive test result who truly have the disease.
So, applying to ward cover Rick, it’s if I have a troponin that’s come back elevated, what percentage of patients with this elevated result are actually having a heart attack? Now, if that troponin had come back negative, again, our question would be about negative predictive value. If I have a troponin that has come back normal, what percentage of patients with this result are truly disease-free, i.e., not having a heart attack? You can see how at least a rough understanding of the positive and negative predictive values of your diagnostic tests can be pretty helpful in practice, particularly if you’re a junior doctor and you’re still getting to grips with these tests.
So, I can assume positive predictive value, negative predictive value, perfect statistics, why would we ever use anything else? Podcast is over.
It would be tempting to think that’s really all we need to know when we are treating patients, positive and negative predictive value, because that’s the practical knowledge we need. You then might ask, well, why did I need to listen to your last podcast about sensitivity and specificity? Because actually, it only matters to me if I know whether the patient’s got the disease. On the ward, when I’m seeing a patient, I don’t actually know. I don’t know the test results. So, the PPV and NPV are the ones that are most meaningful for me. However, there is a really important caveat on this. You need both, and the reason you need both sensitivity and specificity and positive and negative predictive values is because the positive and negative predictive values are heavily affected by the prevalence of the condition that we’re trying to diagnose. If we have a really rare condition, let’s say only one percent of the patients we’re testing have the condition, the positive predictive value is going to tend to be lower.
So, our test is basically going to shift that probability up if it’s positive in a sort of fixed proportion. The higher the prevalence, the higher the positive predictive value; the lower the prevalence, the lower the positive predictive value. And that works the same for negative predictive values too. So, if you have a high prevalence, you tend to find it difficult to get a low negative predictive value. And if you start with a low prevalence, you can achieve a high negative predictive value quite easily, even with a test that isn’t doing any work. It might be a useless test of very poor sensitivity, but you still get a high negative predictive value because the prevalence of the condition is so low. So, we need to know both sensitivity, specificity, and positive and negative predictive values to make sense of how good a test is.
Okay, so that’s disappointing. So, they’re not perfect statistics. We actually still need to think. When we get a test result back and we’re thinking about our NPV and our PPV, we actually need to think, okay, these statistics have been calibrated to a particular population with a particular prevalence. We need to think about the risk profile of the patient in front of us. If we’re in an environment where that disease that we’re trying to test for is very prevalent, that’s going to affect our PPV and NPV. Whereas, if it’s not very prevalent, that’s also going to affect our PPV and NPV. Is there anything else that we actually need to think about when it comes to these statistics?
Yeah, a couple of things. Just give you a couple of practical examples about how we might interpret these values in practice. Let’s imagine that we’re trying to diagnose a myocardial infarction now, but we want to diagnose it in everybody who attends the emergency department, no matter what they’re coming with. It might be a stubbed toe, but we’re interested in making that diagnosis. Now, imagine that the test we are applying is actually useless. It doesn’t detect anyone with a myocardial infarction. Its sensitivity is zero percent. But we’re going to apply this in a population where the prevalence of myocardial infarction is really low because we’ve got so many people with other conditions. The negative predictive value of that test, which has zero percent sensitivity, could be really high, like 99.99 percent potentially, if we’ve got a population without many myocardial infarctions in there. It would sound like a really good test, which is actually perfectly useless as a test. So, that’s why we need to know both.
Let’s give you one more example to turn things around and show how we need both concepts. Let’s think about the ECG. Think about looking for an ST depression on an ECG. Now, ST depression is quite a specific change for myocardial infarction in someone with chest pain. The specificity might be, let’s say, 95 percent. Sounds really high. Now, from the last podcast, you’ll remember, snuff it out and spin it in. So, high sensitivity is what we need to rule out a condition, high specificity is what we need to rule in. With a specificity as high as 95 percent, you’d think, ah, I’ve ruled in a myocardial infarction. Sounds great, actually, doesn’t it? But if the prevalence is low, let’s say we have only five percent or ten percent of the people with a myocardial infarction, the positive predictive value could be as low as 20 percent, even with that high specificity. So, we absolutely need to know both. I hope we’ve hammered that home enough. Do you think we have?
I think we probably have hammered that home to death. You need to know the prevalence of the actual disease in the group of patients that you’re seeing to correctly use these statistics. Let’s maybe finish with one final tip. Negative predictive value, the probability of the patient not having the condition if they have a negative test. A bit complicated, right? Yeah. Let’s think about when you see a negative predictive value, do 100 minus the negative predictive value. So, instead of, let’s say you see a negative predictive value of 99 percent, turn that round to one percent. That’s the probability of having the disease if you have a negative test. That’s much more meaningful for us. It’s the post-test probability.
Okay, that makes sense. So, to wrap up then, quick recaps. Positive predictive value is the likelihood that if you have a positive test, you’re actually going to encounter a patient who has the disease. Whereas the negative predictive value is when you have a negative test, what’s the percentage likelihood of a patient who truly doesn’t have the disease? Think about the prevalence of the disease and how that affects these statistics. Still think about your sensitivity and specificity, and think about the population that the test was calibrated in.
Cite this article as: Rick Body, "Podcast – Positive and Negative Predictive Values: Critical Appraisal Nugget 11," in St.Emlyn's, July 24, 2024,
Click to share on X (Opens in new window)X
Click to share on Facebook (Opens in new window)Facebook
Click to share on LinkedIn (Opens in new window)LinkedIn
Click to share on Reddit (Opens in new window)Reddit
Click to share on Pocket (Opens in new window)Pocket
Click to share on WhatsApp (Opens in new window)WhatsApp
Click to email a link to a friend (Opens in new window)Email
Click to print (Opens in new window)Print
Podcast – Sensitivity and Specificity: Critical Appraisal Nugget 10
This episode ofr our CAN podcasts introduces two core concepts: sensitivity and specificity. These are two ways of thinking about the accuracy of a diagnostic test. Knowing the sensitivity and specificity of an investigation will give you a decent idea of how it should be used in the emergency department.
Podcast – Likelihood Ratios: Critical Appraisal Nugget 12
Mastering likelihood ratios can transform your diagnostic skills. In this podcast with Rick and Greg, discover how likelihood ratios can enhance decision-making and improve patient care.
CAN 8: p-values
We've not done a critical appraisal nugget for a while now. No excuses, we've just been busy with other stuff, but we need to rectify that now. This week I sat down with Rick Body and talked through some of the issues around the use and abuse of p-values in…
About The Author
#### Rick Body
Professor Richard Body MB ChB, FRCEM, PhD is Professor of Emergency Medicine at the University of Manchester. He is honorary Consultant in Emergency Medicine at Manchester Foundation trust. He is also the Group Director of Research & Innovation at Manchester University NHS Foundation Trust, which is the largest NHS trust in England. His research interests include diagnostics, cardiac disease and the philosophy of emergency medicine. He is an acclaimed international speaker on cardiac diagnostics . He can be found on Twitter as @richardbody
← Previous Post
Next Post →
2 thoughts on “Podcast – Positive and Negative Predictive Values: Critical Appraisal Nugget 11”
Sahba Sabeti September 5, 2024 at 3:19 am
“Sensitivity is the ability of a test to correctly identify those with the disease (true positives), while specificity is the ability of a test to correctly identify those without the disease (true negatives).”
Should these definitions be reversed?
Reply
1. richardbody September 18, 2024 at 2:06 pm
Actually, that’s exactly right! You might wonder how that fits with SpIN (specificity rules in) and SnOUT (sensitivity rules out), when sensitivity is about patients with disease. The point is that if a test accurately identifies almost everyone with the disease then there are very few false negatives – patients who will be missed. And if we achieve that (a high sensitivity), then we often say it’s a ‘rule-out’. So the definitions are correct, even though it might seem counter-intuitive at first.
Reply
Thanks so much for following. Viva la #FOAMedCancel reply
Type your email…
Subscribe
Join 10.2K other subscribers
Our Latest Podcast
Recent Posts
Stay or go? Should we Move Patients with Refractory Out-of-Hospital Cardiac Arrest to Hospital? The EVIDENCE Trial
JC: Ketamine or Midazolam as first choice for Paediatric Status Epilepticus? St Emlyn’s
Hydrofluoric acid burns: stay on brand!
GECCo’s Global Health & Emergency Care Research Summit, July 2025
Proudly supported by
Quick Links
Social Media
Copyright © 2025 St.Emlyn's | Powered by Medmastery
Contact Us
FOAMed Medical Education Resources bySt Emlyn’sis licensed under aCreative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Based on a work atLife in the Fast Lane
Scroll to Top |
9924 | http://aleph0.clarku.edu/~djoyce/elements/bookI/propI30.html | Euclid's Elements, Book I, Proposition 30
Euclid's Elements =================Book I ======
Proposition 30
Straight lines parallel to the same straight line are also parallel to one another.
Let each of the straight lines AB and CD be parallel to EF.
I say that AB is also parallel to CD.
I.29
Let the straight line GK fall upon them. Since the straight line GK falls on the parallel straight lines AB and EF, therefore the angle AGK equals the angle GHF.
I.29
Again, since the straight line GK falls on the parallel straight lines EF and CD, therefore the angle GHF equals the angle GKD.
C.N.1
But the angle AGK was also proved equal to the angle GHF. Therefore the angle AGK also equals the angle GKD, and they are alternate.
Therefore AB is parallel to CD.
Therefore straight lines parallel to the same straight line are also parallel to one another.
Q.E.D.
Guide ----- For this proposition it is supposed that the three lines lie in one plane. Proposition XI.9 applies to the case where the three lines do not lie in a plane.
Playfair’s axiom
A number of the propositions in the Elements are equivalent to the parallel postulate Post.5 in the sense that if the rest of the postulates are assumed and any one of these propositions is assumed, then the parallel postulate can be proved as a proposition. This one I.30, the last I.29, either part of I.32, and almost any later one. Thus, Euclid had many statements to choose from to take as a postulate.
In many modern expositions of synthetic geometry, Playfair’s axiom (John Playfair, 1748–1819) is chosen as that postulate instead of Euclid’s parallel postulate Post.5. Playfair’s axiom states that there is at most one line parallel to a given line passing through a given point. (That there is at least one follows from the next proposition I.31, which doesn’t depend on the parallel postulate.)
Two advantages of Playfair’s axiom over Euclid’s parallel postulate are that it is a simpler statement, and it emphasizes the distinction between Euclidean and hyperbolic geometry.
Two disadvantages are that it does not have the historical importance of Euclid’s parallel postulate, and the proof of the parallel postulate from Playfair’s axiom is nonconstructive. That proof is a proof by contradiction that begins assuming that a point does not exist, deriving a contradiction, and concluding that the point must exist, but does not construct it. It may well be that Euclid chose to make the construction an assumption of his parallel postulate rather rather than choosing some other equivalent statement for his postulate.
Elegance in mathematics
Euclid’s Elements form one of the most beautiful works of science in the history of humankind. This beauty lies more in the logical development of geometry rather than in geometry itself. It is not the diagrams that excite our interest; rather it is the concepts, the way the concepts interconnect, and the way Euclid selected and presented these concepts and their interconnections. The Elements are elegant. Elegance in mathematics is characterized by simplicity and clarity. An elegant presentation is easy for the reader to follow. But elegance is not only in the presentation, it is in the selection of definitions and proofs. The elegant definition is the one that makes the rest of the theory easy. The elegant proof is the one that is easiest to follow, one that is designed just right to fit the goal. Extraneous concepts should not be involved. Even the goals need to be adjusted to the right level of generality to cover the concepts, but not so abstract that the abstraction itself obscures the goal.
One of the criticisms of Euclid’s parallel postulate was that it isn’t simple.The statement of this proposition, I.30, is much simpler, and Playfair’s axiom is much simpler. As they’re each logically equivalent to Euclid’s parallel postulate, if elegance were the primary goal, then Euclid would have chosen one of them in place of his postulate. Perhaps the reasons mentioned above explain why Euclid used Post.5 instead.
Use of Proposition 30
This proposition is used in I.45 and IV.7.
Next: I.31
Previous: I.29
Book I
©1996, 1997, 2003
David E. Joyce
Department of Mathematics and Computer Science
Clark University
Worcester, MA 01610 |
9925 | https://stackoverflow.com/questions/34794664/how-should-i-calculate-the-average-speed-by-road-segment-for-multiple-segments | sql - How should I calculate the average speed by road segment for multiple segments? - Stack Overflow
Join Stack Overflow
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
Sign up with GitHub
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Overflow
1. About
2. Products
3. For Teams
Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers
Advertising Reach devs & technologists worldwide about your product, service or employer brand
Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models
Labs The future of collective knowledge sharing
About the companyVisit the blog
Loading…
current community
Stack Overflow helpchat
Meta Stack Overflow
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Let's set up your homepage Select a few topics you're interested in:
python javascript c#reactjs java android html flutter c++node.js typescript css r php angular next.js spring-boot machine-learning sql excel ios azure docker
Or search from our full list:
javascript
python
java
c#
php
android
html
jquery
c++
css
ios
sql
mysql
r
reactjs
node.js
arrays
c
asp.net
json
python-3.x
.net
ruby-on-rails
sql-server
swift
django
angular
objective-c
excel
pandas
angularjs
regex
typescript
ruby
linux
ajax
iphone
vba
xml
laravel
spring
asp.net-mvc
database
wordpress
string
flutter
postgresql
mongodb
wpf
windows
xcode
amazon-web-services
bash
git
oracle-database
spring-boot
dataframe
azure
firebase
list
multithreading
docker
vb.net
react-native
eclipse
algorithm
powershell
macos
visual-studio
numpy
image
forms
scala
function
vue.js
performance
twitter-bootstrap
selenium
winforms
kotlin
loops
express
dart
hibernate
sqlite
matlab
python-2.7
shell
rest
apache
entity-framework
android-studio
csv
maven
linq
qt
dictionary
unit-testing
asp.net-core
facebook
apache-spark
tensorflow
file
swing
class
unity-game-engine
sorting
date
authentication
go
symfony
t-sql
opencv
matplotlib
.htaccess
google-chrome
for-loop
datetime
codeigniter
perl
http
validation
sockets
google-maps
object
uitableview
xaml
oop
visual-studio-code
if-statement
cordova
ubuntu
web-services
email
android-layout
github
spring-mvc
elasticsearch
kubernetes
selenium-webdriver
ms-access
ggplot2
user-interface
parsing
pointers
c++11
google-sheets
security
machine-learning
google-apps-script
ruby-on-rails-3
templates
flask
nginx
variables
exception
sql-server-2008
gradle
debugging
tkinter
delphi
listview
jpa
asynchronous
web-scraping
haskell
pdf
jsp
ssl
amazon-s3
google-cloud-platform
jenkins
testing
xamarin
wcf
batch-file
generics
npm
ionic-framework
network-programming
unix
recursion
google-app-engine
mongoose
visual-studio-2010
.net-core
android-fragments
assembly
animation
math
svg
session
intellij-idea
hadoop
rust
next.js
curl
join
winapi
django-models
laravel-5
url
heroku
http-redirect
tomcat
google-cloud-firestore
inheritance
webpack
image-processing
gcc
keras
swiftui
asp.net-mvc-4
logging
dom
matrix
pyspark
actionscript-3
button
post
optimization
firebase-realtime-database
web
jquery-ui
cocoa
xpath
iis
d3.js
javafx
firefox
xslt
internet-explorer
caching
select
asp.net-mvc-3
opengl
events
asp.net-web-api
plot
dplyr
encryption
magento
stored-procedures
search
amazon-ec2
ruby-on-rails-4
memory
canvas
audio
multidimensional-array
random
jsf
vector
redux
cookies
input
facebook-graph-api
flash
indexing
xamarin.forms
arraylist
ipad
cocoa-touch
data-structures
video
azure-devops
model-view-controller
apache-kafka
serialization
jdbc
woocommerce
razor
routes
awk
servlets
mod-rewrite
excel-formula
beautifulsoup
filter
docker-compose
iframe
aws-lambda
design-patterns
text
visual-c++
django-rest-framework
cakephp
mobile
android-intent
struct
react-hooks
methods
groovy
mvvm
ssh
lambda
checkbox
time
ecmascript-6
grails
google-chrome-extension
installation
cmake
sharepoint
shiny
spring-security
jakarta-ee
plsql
android-recyclerview
core-data
types
sed
meteor
android-activity
activerecord
bootstrap-4
websocket
graph
replace
scikit-learn
group-by
vim
file-upload
junit
boost
memory-management
sass
import
async-await
deep-learning
error-handling
eloquent
dynamic
soap
dependency-injection
silverlight
layout
apache-spark-sql
charts
deployment
browser
gridview
svn
while-loop
google-bigquery
vuejs2
dll
highcharts
ffmpeg
view
foreach
makefile
plugins
redis
c#-4.0
reporting-services
jupyter-notebook
unicode
merge
reflection
https
server
google-maps-api-3
twitter
oauth-2.0
extjs
terminal
axios
pip
split
cmd
pytorch
encoding
django-views
collections
database-design
hash
netbeans
automation
data-binding
ember.js
build
tcp
pdo
sqlalchemy
apache-flex
mysqli
entity-framework-core
concurrency
command-line
spring-data-jpa
printing
react-redux
java-8
lua
html-table
ansible
jestjs
neo4j
service
parameters
enums
material-ui
flexbox
module
promise
visual-studio-2012
outlook
firebase-authentication
web-applications
webview
uwp
jquery-mobile
utf-8
datatable
python-requests
parallel-processing
colors
drop-down-menu
scipy
scroll
tfs
hive
count
syntax
ms-word
twitter-bootstrap-3
ssis
fonts
rxjs
constructor
google-analytics
file-io
three.js
paypal
powerbi
graphql
cassandra
discord
graphics
compiler-errors
gwt
socket.io
react-router
solr
backbone.js
memory-leaks
url-rewriting
datatables
nlp
oauth
terraform
datagridview
drupal
oracle11g
zend-framework
knockout.js
triggers
neural-network
interface
django-forms
angular-material
casting
jmeter
google-api
linked-list
path
timer
django-templates
arduino
proxy
orm
directory
windows-phone-7
parse-platform
visual-studio-2015
cron
conditional-statements
push-notification
functional-programming
primefaces
pagination
model
jar
xamarin.android
hyperlink
uiview
visual-studio-2013
vbscript
google-cloud-functions
gitlab
azure-active-directory
jwt
download
swift3
sql-server-2005
configuration
process
rspec
pygame
properties
combobox
callback
windows-phone-8
linux-kernel
safari
scrapy
permissions
emacs
scripting
raspberry-pi
clojure
x86
scope
io
expo
azure-functions
compilation
responsive-design
mongodb-query
nhibernate
angularjs-directive
request
bluetooth
reference
binding
dns
architecture
3d
playframework
pyqt
version-control
discord.js
doctrine-orm
package
f#
rubygems
get
sql-server-2012
autocomplete
tree
openssl
datepicker
kendo-ui
jackson
yii
controller
grep
nested
xamarin.ios
static
null
statistics
transactions
active-directory
datagrid
dockerfile
uiviewcontroller
webforms
discord.py
phpmyadmin
sas
computer-vision
notifications
duplicates
mocking
youtube
pycharm
nullpointerexception
yaml
menu
blazor
sum
plotly
bitmap
asp.net-mvc-5
visual-studio-2008
yii2
electron
floating-point
css-selectors
stl
jsf-2
android-listview
time-series
cryptography
ant
hashmap
character-encoding
stream
msbuild
asp.net-core-mvc
sdk
google-drive-api
jboss
selenium-chromedriver
joomla
devise
cors
navigation
anaconda
cuda
background
frontend
binary
multiprocessing
pyqt5
camera
iterator
linq-to-sql
mariadb
onclick
android-jetpack-compose
ios7
microsoft-graph-api
rabbitmq
android-asynctask
tabs
laravel-4
environment-variables
amazon-dynamodb
insert
uicollectionview
linker
xsd
coldfusion
console
continuous-integration
upload
textview
ftp
opengl-es
macros
operating-system
mockito
localization
formatting
xml-parsing
vuejs3
json.net
type-conversion
data.table
kivy
timestamp
integer
calendar
segmentation-fault
android-ndk
prolog
drag-and-drop
char
crash
jasmine
dependencies
automated-tests
geometry
azure-pipelines
android-gradle-plugin
itext
fortran
sprite-kit
header
mfc
firebase-cloud-messaging
attributes
nosql
format
nuxt.js
odoo
db2
jquery-plugins
event-handling
jenkins-pipeline
nestjs
leaflet
julia
annotations
flutter-layout
keyboard
postman
textbox
arm
visual-studio-2017
gulp
stripe-payments
libgdx
synchronization
timezone
uikit
azure-web-app-service
dom-events
xampp
wso2
crystal-reports
namespaces
swagger
android-emulator
aggregation-framework
uiscrollview
jvm
google-sheets-formula
sequelize.js
com
chart.js
snowflake-cloud-data-platform
subprocess
geolocation
webdriver
html5-canvas
centos
garbage-collection
dialog
sql-update
widget
numbers
concatenation
qml
tuples
set
java-stream
smtp
mapreduce
ionic2
windows-10
rotation
android-edittext
modal-dialog
spring-data
nuget
doctrine
radio-button
http-headers
grid
sonarqube
lucene
xmlhttprequest
listbox
switch-statement
initialization
internationalization
components
apache-camel
boolean
google-play
serial-port
gdb
ios5
ldap
youtube-api
return
eclipse-plugin
pivot
latex
frameworks
tags
containers
github-actions
c++17
subquery
dataset
asp-classic
foreign-keys
label
embedded
uinavigationcontroller
copy
delegates
struts2
google-cloud-storage
migration
protractor
base64
queue
find
uibutton
sql-server-2008-r2
arguments
composer-php
append
jaxb
zip
stack
tailwind-css
cucumber
autolayout
ide
entity-framework-6
iteration
popup
r-markdown
windows-7
airflow
vb6
g++
ssl-certificate
hover
clang
jqgrid
range
gmail
Next You’ll be prompted to create an account to view your personalized homepage.
Home
Questions
AI Assist Labs
Tags
Challenges
Chat
Articles
Users
Jobs
Companies
Collectives
Communities for your favorite technologies. Explore all Collectives
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Collectives™ on Stack Overflow
Find centralized, trusted content and collaborate around the technologies you use most.
Learn more about Collectives
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
How should I calculate the average speed by road segment for multiple segments?
Ask Question
Asked 9 years, 8 months ago
Modified9 years, 8 months ago
Viewed 2k times
This question shows research effort; it is useful and clear
-6
Save this question.
Show activity on this post.
I have a table of driver speeds and road segments:
sql
driver_lpr | segment | speed
0000001 | A | 30
0000002 | B | 60
0000003 | A | 50
0000004 | A | 100
0000005 | B | 60
And I want to have a table of average speed per segment
sql
segment | average speed
A | 47.368
B | 60
How can this be done in SQL ?
sql
sql-server
mean
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Improve this question
Follow
Follow this question to receive notifications
edited Jan 16, 2016 at 18:54
Shog9
160k 36 36 gold badges 237 237 silver badges 242 242 bronze badges
asked Jan 14, 2016 at 16:26
Alejandro RodriguezAlejandro Rodriguez
252 6 6 silver badges 15 15 bronze badges
3
3 Under meta discussion How can I demonstrate to voters that a question is less trivial than it seems?Bhargav Rao –Bhargav Rao 2016-01-15 09:33:37 +00:00 Commented Jan 15, 2016 at 9:33
2 It probably would have helped if the OP had mentioned that the 'average' he expects is not 'your everyday average'. TBH, I had never even heard of the term Harmonic mean and it took me a while to realise his example result was 'strange' =)deroby –deroby 2016-01-17 11:28:30 +00:00 Commented Jan 17, 2016 at 11:28
2 He may not have known, @deroby (Alejandro, can you clarify?) - wouldn't be the first time someone's used arithmetic mean for averaging rates and gotten incorrect results without realizing why.Shog9 –Shog9 2016-01-18 20:17:15 +00:00 Commented Jan 18, 2016 at 20:17
Add a comment|
2 Answers 2
Sorted by: Reset to default
This answer is useful
21
Save this answer.
+50
This answer has been awarded bounties worth 50 reputation by Alejandro Rodriguez
Show activity on this post.
When averaging speeds, the harmonic mean is in need.
The straight forward AVG() approach is wrong, the arithmetic mean yields the wrong result for average velocity.
There is no predefined function for the harmonic mean, but it could be achieved with this query:
sql
SELECT segment,
COUNT()/SUM(1e0/speed) AS avg_speed
FROM T
GROUP BY segment
SQL Fiddle
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Improve this answer
Follow
Follow this answer to receive notifications
edited Jan 18, 2016 at 21:10
Martin Smith
456k 97 97 gold badges 776 776 silver badges 882 882 bronze badges
answered Jan 14, 2016 at 17:22
Uri GorenUri Goren
13.8k 8 8 gold badges 62 62 silver badges 113 113 bronze badges
1 Comment
Add a comment
Steve Ford
Steve FordOver a year ago
This is true if the segments are the same distance, if the time interval is the same then a standard arithmetic mean suffices, if neither then a weighted harmonic or arithmetic mean is required.
2016-01-19T11:44:44.173Z+00:00
0
Reply
Copy link
This answer is useful
3
Save this answer.
Show activity on this post.
This is little bit complex, but will take care of 0 speed also. I have 2 similar queries to do it based on different scenarios.
Assume your source table like below.
sql
+-------------+----------+-------+
| driver_lpr | segment | speed |
+-------------+----------+-------+
| 0000001 | A | 30 |
| 0000002 | B | 60 |
| 0000003 | A | 50 |
| 0000004 | A | 100 |
| 0000005 | B | 60 |
| 0000006 | B | 0 |
| 0000007 | C | 0 |
+-------------+----------+-------+
I have added 2 new rows with 0 speed.
Case 1:
Addition of a 0 speed in segment B, will give average speed as 40``(602+01)/(2+1).
Addition of a new segment, C with 0 speed will give average speed as 0
So output would be
sql
+----------+-------------------+
| segment | average_speed |
+----------+-------------------+
| A | 47.36842105263158 |
| B | 40 |
| C | 0 |
+----------+-------------------+
SQLFiddle Demo CASE 1
Case 2:
There will be no change in average speed of B with addition of a 0.
However, a new segment C will have 0 average speed.
Output would be
sql
+----------+-------------------+
| segment | average_speed |
+----------+-------------------+
| A | 47.36842105263158 |
| B | 60 |
| C | 0 |
+----------+-------------------+
SQLFiddle Demo CASE 2
Query for Case 1:
sql
with tbl1 as
(SELECT segment,
case when speed = 0 then cast(0 as float) else
cast(1 as float)/cast(speed as float)
end as newspeed
FROM T
),
tbl2 as
(
select segment,cast(1 as float)/avg(newspeed) as avgspeed,count() as cnt
from tbl1
where newspeed <> 0
group by segment
union
select segment,0 as avgspeed,count() as cnt
from tbl1
where newspeed =0
group by segment
)
select segment,
sum(avgspeedcnt)/sum(cnt) as "average_speed"
from tbl2
group by segment
Query for Case2
sql
with tbl1 as
(SELECT segment,
case when speed = 0 then cast(0 as float) else
cast(1 as float)/cast(speed as float)
end as newspeed
FROM T
),
tbl2 as
(
select segment,cast(1 as float)/avg(newspeed) as avgspeed,count() as cnt
from tbl1
where newspeed <> 0
group by segment
union
select segment,0 as avgspeed,count() as cnt
from tbl1
where newspeed =0
group by segment
)
select segment,
sum(avgspeed) as "average_speed"
from tbl2
group by segment
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Improve this answer
Follow
Follow this answer to receive notifications
edited Jan 23, 2016 at 16:50
answered Jan 23, 2016 at 14:09
UtsavUtsav
8,153 2 2 gold badges 22 22 silver badges 40 40 bronze badges
2 Comments
Add a comment
Martin Smith
Martin SmithOver a year ago
How can there be a speed of 0? That would mean they are stationary and therefore could not complete any road segments.
2016-01-23T20:00:25.25Z+00:00
1
Reply
Copy link
Utsav
UtsavOver a year ago
Just wanted to cover all possible scenarios.
2016-01-23T22:50:08.123Z+00:00
0
Reply
Copy link
Your Answer
Thanks for contributing an answer to Stack Overflow!
Please be sure to answer the question. Provide details and share your research!
But avoid …
Asking for help, clarification, or responding to other answers.
Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Draft saved
Draft discarded
Sign up or log in
Sign up using Google
Sign up using Email and Password
Submit
Post as a guest
Name
Email
Required, but never shown
Post Your Answer Discard
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
sql
sql-server
mean
See similar questions with these tags.
The Overflow Blog
The history and future of software development (part 1)
Getting Backstage in front of a shifting dev experience
Featured on Meta
Spevacus has joined us as a Community Manager
Introducing a new proactive anti-spam measure
New and improved coding challenges
New comment UI experiment graduation
Policy: Generative AI (e.g., ChatGPT) is banned
7 people chatting
Community activity
Last 1 hr
Users online activity 7833 users online
3 questions
16 answers
52 comments
144 upvotes
Popular tags
phpiospythonjavac++
Popular unanswered question
Merge columns into multi line cells
latexr-markdowngt
jfronc
1
2 hours ago
Linked
8Calculating bytes per second (the smooth way)
-1Calculate average speed over x-distance
Related
1073How can I get column names from a table in SQL Server?
1062Reset identity seed after deleting records in SQL Server
535How can I update top 100 records in SQL server?
844SQL update query using joins
1621What are the options for storing hierarchical data in a relational database?
963How to Join to first row
1328What do Clustered and Non-Clustered index actually mean?
1103How can I do 'insert if not exists' in MySQL?
798Get top 1 row of each group
Hot Network Questions
Copy command with cs names
Discussing strategy reduces winning chances of everyone!
Why include unadjusted estimates in a study when reporting adjusted estimates?
Overfilled my oil
Matthew 24:5 Many will come in my name!
On being a Maître de conférence (France): Importance of Postdoc
Exchange a file in a zip file quickly
Repetition is the mother of learning
Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward"
Are there any world leaders who are/were good at chess?
Checking model assumptions at cluster level vs global level?
Triangle with Interlacing Rows Inequality [Programming]
Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations?
Alternatives to Test-Driven Grading in an LLM world
Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road?
Do we need the author's permission for reference
I have a lot of PTO to take, which will make the deadline impossible
how do I remove a item from the applications menu
Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term?
в ответе meaning in context
What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel?
How can the problem of a warlock with two spell slots be solved?
Can you formalize the definition of infinitely divisible in FOL?
What happens if you miss cruise ship deadline at private island?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
lang-sql
Why are you flagging this comment?
Probable spam.
This comment promotes a product, service or website while failing to disclose the author's affiliation.
Unfriendly or contains harassment/bigotry/abuse.
This comment is unkind, insulting or attacks another person or group. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Stack Overflow
Questions
Help
Chat
Products
Teams
Advertising
Talent
Company
About
Press
Work Here
Legal
Privacy Policy
Terms of Service
Contact Us
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings |
9926 | https://math.stackexchange.com/questions/108509/calculate-perimeter-from-parametric-form-with-an-ellipse | calculus - Calculate perimeter from parametric form with an ellipse? - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Calculate perimeter from parametric form with an ellipse?
Ask Question
Asked 13 years, 7 months ago
Modified13 years, 7 months ago
Viewed 2k times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
Suppose I have a thing such as an ellipse:
(x a)2+(y b)2=1(x a)2+(y b)2=1
now we can define it so that x a=c o s(θ)x a=c o s(θ) and y b=s i n(θ)y b=s i n(θ). I know the perimeter formula
μ(S)=∫1+(f′(x))2−−−−−−−−−−√d x.μ(S)=∫1+(f′(x))2 d x.
It is easy to paramerize the ellipse but how can I parametrize the perimeter formula so that I can easily calculate the perimeter?
I find that I am doing things the hard way like this:
y=±b 1−(x a)2−−−−−−−−√y=±b 1−(x a)2
now if I plug in the y into the formula of perimeter, it is messy. Can I do it elegantly with parametric form somehow?
calculus
geometry
conic-sections
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Feb 12, 2012 at 15:22
hhhhhh
asked Feb 12, 2012 at 14:07
hhhhhh
5,605 9 9 gold badges 60 60 silver badges 115 115 bronze badges
3
I gave a mention of, and an algorithm for, the complete elliptic integral in this answer.J. M. ain't a mathematician –J. M. ain't a mathematician 2012-02-12 15:11:55 +00:00 Commented Feb 12, 2012 at 15:11
possible duplicate of why we only have a approximation for every circumference for ellipse but not define a special formula for each ellipseAmérico Tavares –Américo Tavares 2012-02-12 15:14:59 +00:00 Commented Feb 12, 2012 at 15:14
1 If there would be a simple formula for the perimeter of an ellipse you would have met it in high school ……Christian Blatter –Christian Blatter 2012-02-12 16:23:28 +00:00 Commented Feb 12, 2012 at 16:23
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
4
Save this answer.
Show activity on this post.
I would write the following
d s=(d x d θ)2+(d y d θ)2−−−−−−−−−−−−−−−√d θ d s=(d x d θ)2+(d y d θ)2 d θ
and so
P=∫2 π 0 a 2 sin 2 θ+b 2 cos 2 θ−−−−−−−−−−−−−−−√d θ P=∫0 2 π a 2 sin 2θ+b 2 cos 2θ d θ
and this is just an elliptic integral. The final result takes the form (b>a b>a)
P=4 b E(e)P=4 b E(e)
being
E(e)=∫π 2 0 1−e 2 sin 2 θ−−−−−−−−−−√d θ E(e)=∫0 π 2 1−e 2 sin 2θ d θ
and e 2=1−a 2 b 2 e 2=1−a 2 b 2 the eccentricity.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Feb 12, 2012 at 17:46
answered Feb 12, 2012 at 14:54
JonJon
5,742 1 1 gold badge 20 20 silver badges 27 27 bronze badges
12
...I have never really understoond what the S S here means? In physics, it sometimes denotes some surface S S but here it is? Do you change the coordinates somehow and use the Jacobian matrix or something like that, sorry thinking alound (maybe totally skewed wrong ideas)...hhh –hhh 2012-02-12 15:23:52 +00:00 Commented Feb 12, 2012 at 15:23
1 @hhh: it's the differential corresponding to arclength.J. M. ain't a mathematician –J. M. ain't a mathematician 2012-02-12 15:27:41 +00:00 Commented Feb 12, 2012 at 15:27
@J.M.: yes but what is is S S? It is a partial derivative apparently, there is something I cannot grap in the very beginning of this answer -- chain rule ...have to calculate... thinking still aloud...hhh –hhh 2012-02-12 15:32:57 +00:00 Commented Feb 12, 2012 at 15:32
1 If we have x=a cos θ x=a cosθ and y=b sin θ y=b sinθ as in OP and a≥b a≥b (major axis on x x axis, length 2 a 2 a), then e 2=1−(b a)2 e 2=1−(b a)2 and P=4 a E(e)P=4 a E(e); cf. en.wikipedia.org/wiki/Ellipse#Eccentricity&en.wikipedia.org/wiki/Ellipse#Circumference. So in general, e 2=|a 2−b 2|max(a,b)2 e 2=|a 2−b 2|max(a,b)2 and P=4 max(a,b)E(e)P=4 max(a,b)E(e). @hhh: s s is arc length, i.e. the length traversed along the portion of curve parametrized in the integral.bgins –bgins 2012-02-12 17:34:44 +00:00 Commented Feb 12, 2012 at 17:34
1 @hhh: Indeed, the initial integral goes from 0 0 to 2 π 2 π but the elliptic integral is given between 0 0 and π 2 π 2. The factor 4 comes out as you need four times the elliptic integral.Jon –Jon 2012-02-12 20:25:52 +00:00 Commented Feb 12, 2012 at 20:25
|Show 7 more comments
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
calculus
geometry
conic-sections
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Linked
0why we only have a approximation for every circumference for ellipse but not define a special formula for each ellipse
Related
0Convert ellipse parameter from General parametric form to General polar form
1Calculate tangent point on ellipse
0differential perimeter of an ellipse
9Converting a rotated ellipse in parametric form to cartesian form
0Calculate ellipse parameters from cone second degree equation.
4Extrema of ellipse from parametric form
11Finding out the perimeter of the ellipse?
0How to find center of ellipse form a given edge point
Hot Network Questions
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
The rule of necessitation seems utterly unreasonable
Is it ok to place components "inside" the PCB
How to start explorer with C: drive selected and shown in folder list?
Bypassing C64's PETSCII to screen code mapping
Stress in "agentic"
Calculating the node voltage
Checking model assumptions at cluster level vs global level?
Lingering odor presumably from bad chicken
How to home-make rubber feet stoppers for table legs?
For every second-order formula, is there a first-order formula equivalent to it by reification?
Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term?
alignment in a table with custom separator
Why multiply energies when calculating the formation energy of butadiene's π-electron system?
Alternatives to Test-Driven Grading in an LLM world
Another way to draw RegionDifference of a cylinder and Cuboid
What were "milk bars" in 1920s Japan?
Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations?
Gluteus medius inactivity while riding
How to convert this extremely large group in GAP into a permutation group.
Does the curvature engine's wake really last forever?
Drawing the structure of a matrix
Xubuntu 24.04 - Libreoffice
Proof of every Highly Abundant Number greater than 3 is Even
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
9927 | https://www.cs.umd.edu/users/gasarch/COURSES/250/S22/slides/COMB/gridcoltalk.pdf | Grid Coloring 250H 3 x n Grids ● No matter how you 2-color a 3 x 9 grid there exists a mono rectangle ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ 3 x n Grids ● No matter how you 2-color a 3 x 9 grid there exists a mono rectangle ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ 3 x n Grids ● No matter how you 2-color a 3 x 9 grid there exists a mono rectangle ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ 3 x n Grids ● No matter how you 2-color a 3 x 9 grid there exists a mono rectangle ● What about a ○ 3 x 8 ○ 3 x 7 ○ 3 x 6 ○ ect?
3 x n Grids ● No matter how you 2-color a 3 x 9 grid there exists a mono rectangle ● What about a ○ 3 x 8 ○ 3 x 7 ○ 3 x 6 ○ ect?
● I know you are all super excited to talk about this with your classmates 3 x n Grids ● No matter how you 2-color a 3 x 9 grid there exists a mono rectangle ● What about a ○ 3 x 8 ○ 3 x 7 ○ 3 x 6 ○ ect?
⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ Let’s Look Closer A 2-coloring of a 3x7 grid can be viewed as an 8-coloring of the rows, so if there are 9 rows, two are the same.
⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ What about?
● 4x4 ● 4x5 ● 4x6 ● 5x5 ● 5x6 What about?
● 4x4 ● 4x5 ● 4x6 ● 5x5 ● 5x6 ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ ⬤ n x m grid 2-Coloring Theorem Theorem: n x m is 2-colorable without a monochromatic rectangle if and only if it does not contain a 3 x 7, 7 x 3, or 5 x 5 grid.
What if we have 3 colors?
We can expand this to look at what grids we can color for 3 colors |
9928 | https://arxiv.org/pdf/2505.00922 | Cluster deletion and clique partitioning in graphs with bounded clique number
Nicola Galesi # Ñ
Department of Computer, Control and Management Engineering. Sapienza University Rome, Italy
Tony Huynh # Ñ
Discrete Mathematics Group, Institute for Basic Science (IBS), Daejeon, South Korea
Fariba Ranjbar
Department of AI, Data and Decision Sciences, Luiss Guido Carli, Italy
Abstract
The Cluster Deletion problem takes a graph G as input and asks for a minimum size set of edges X such that G − X is the disjoint union of complete graphs. An equivalent formulation is the Clique Partition problem, which asks to find a partition of V (G) into cliques such that the number of edges in the cliques is maximized. We begin by giving a much simpler proof of a theorem of Gao, Hare, and Nastos [ 15 ] that
Cluster Deletion is efficiently solvable on the class of cographs. We then investigate Cluster Deletion and Clique Partition on permutation graphs, which are a superclass of cographs. Our findings suggest that Cluster Deletion may be NP-hard on permutation graphs. Finally, we prove that for graphs with clique number at most c, there is a 2(c
2
)(c
2
)+1 -approximation algorithm for Clique Partition . This is the first polynomial time algorithm which achieves an approximation ratio better than 2 for graphs with bounded clique number. More generally, our algorithm runs in polynomial time on any graph class for which Max Clique can be computed in polynomial time. We also provide a class of examples which shows that our approximation ratio is best possible.
2012 ACM Subject Classification G.2.2, F.2.0
Keywords and phrases Cluster deletion, Clique partition, Graph modification, Edge-deletion, Cographs, Permutation graphs, Approximation algorithms
1 Introduction
Graph clustering is a fundamental problem in graph theory with numerous real world applications. To name just a few, graph clustering has applications in computational biology where it helps to identify functionally related genes [ 5], social network analysis where it aids in detecting tightly-knit communities [ 14 ], image segmentation [ 28 ], and of course increasingly in machine learning [ 3, 23 ]. The goal of graph clustering is to partition the vertices of a graph into clusters , such that there are many edges between vertices in the same cluster, and only few edges between different clusters. Since a complete graph is the densest possible graph, the following approach to graph clustering has become extremely popular.
Cluster Deletion :
Input: A graph G.
Output: A minimum size set of edges X ⊆ E(G) such that G − X is a disjoint union of complete graphs. A disjoint union of complete graphs is also called a cluster graph . Given a graph G as input, the Cluster Deletion problem is equivalent to maximizing the number of edges in a spanning subgraph H of G which is a cluster graph. We call the equivalent ‘dual’ version the Clique Partition problem.
arXiv:2505.00922v2 [cs.DS] 25 Sep 2025
2 Cluster deletion and clique partitioning
Clique Partition :
Input: A graph G.
Output: A partition (X1, . . . , X s) of V (G) such that each Xi is a clique of G and ∑si=1 |E(Xi)|
is maximum.
Previous work. Shamir et al. [ 27 ] proved that Cluster Deletion is NP-hard for general graphs. Despite its computational difficulty, the problem has been extensively explored within the realms of parameterized complexity [18, 9, 7, 15, 4, 20], approximation algorithms [8, 10, 25, 31, 30, 2], and restricted graph classes. For instance, Cluster Deletion is polynomial time solvable on cographs [ 15 ], split graphs [ 6, 22 ], (proper) interval graphs [ 6, 22 ], P4- and paw-free graphs [ 19 ] and graphs with maximum degree at most three [ 21 ]. On the other hand, NP-hardness persists on C4-free graphs with maximum degree four [ 21 ], planar graphs [ 16 ], chordal graphs [ 6], C5-free graphs and (2 K2, 3 K1)-free graphs . Dessmark et. al [ 10 ] introduced a simple greedy strategy (which we call Greedy ) of repeatedly choosing maximum cliques, and proved that Greedy is a 2-approximation for both Cluster Deletion and Clique Partition . However, Greedy does not necessarily run in polynomial time, since it requires Max Clique as a subroutine. However, Gao et. al [ 15 ] proved that Greedy optimally solves Clique Partition (and hence also Cluster Deletion ) on cographs in polynomial time. Interestingly, the weighted version of Cluster Deletion often exhibits greater computational hardness even on instances where the unweighted version is tractable. For instance, while unweighted Cluster Deletion is tractable on split graphs and cographs, the weighted version remains NP-hard on these classes [ 6]. There are also other popular generalizations of Cluster Deletion , including correlation clustering [ 3, 27 ], where one is allowed to delete or add edges. In this work, we focus on the unweighted variant of Cluster Deletion .From the perspective of approximation algorithms, there have been a series of results for
Cluster Deletion with progressively improved guarantees. The first approximation algorithm for Cluster Deletion achieved an approximation ratio of 4 using a linear programming (LP) relaxation [ 8 ]. Very recent work of Balmaseda et. al [ 2] has refined the analysis of two earlier algorithms, leading to an improvement of the approximation ratio from 4 to 3 [ 2]. The state of the art is the 2-approximation of Veldt et. al [ 31 ]. However, the LP-based approaches are computationally expensive due to the large number of constraints involved. To address this, recent work has introduced faster algorithms by leveraging alternative LP formulations such as Strong Triadic Closure (STC) labeling [ 29 ], resulting in practical 4-approximations [ 30 ]. Veldt [ 30 ] also proposed the first combinatorial algorithm, MatchFlipPivot , which achieves similar approximation guarantees at significantly reduced runtime . Note that the approximation status of the vertex version of Cluster Deletion , (where one deletes vertices instead of edges) has been completely settled by Aprile et. al [ 1], who gave a 2-approximation algorithm (which is best possible assuming the Unique games conjecture ).
Our contributions. We begin by giving a much simpler proof of correctness of the main result from .
▶ Theorem 1.1 ([ 15 ]) . For every cograph G, Greedy outputs an optimal clique partition of
G.
The proof in [ 15 ] uses a quite subtle induction, which requires carefully extending a given clique partition while controlling the number of parts of the clique partition. We instead use N. Galesi, T. Huynh, and F. Ranjbar 3
the well-known fact that every cograph is either the disjoint union of two smaller cographs or the join of two smaller cographs. This gives a very clean inductive proof of Theorem 1.1. We next analyze Clique Partition and Cluster Deletion on permutation graphs ,which are a superclass of cographs. Unfortunately, for permutation graphs, Greedy does not necessarily solve Clique Partition optimally. One reason for this is that Greedy does not include a tie-breaking rule in case there are two different maximum cliques to choose from. We introduce a very natural tie-breaking rule, which is to always choose a locally optimal maximum clique. Here, locally optimal means the number of edges with exactly one end in X is smallest among all maximum cliques. We call this refinement of the greedy algorithm Smart Greedy . Note that Smart Greedy runs in polynomial time on any graph class G for which all maximum cliques of G can be computed in polynomial time for all
G ∈ G . Permutation graphs are one example of such a graph class, since given a permutation representation of an n-vertex permutation graph G, one can find all maximum cliques of
G in O(n2) time via dynamic programming. Unfortunately, we show that Smart Greedy
also does not solve Clique Partition optimally on permutation graphs. Finally, we introduce another variant of Greedy . For a graph G, we let ω(G) and ω′(G)
be the number of vertices and edges, respectively, in a maximum clique of G. The main idea is very simple: if ω(G) ≤ 2, then a maximum clique partition is just a maximum matching, and hence can be found in polynomial-time via Edmonds’ matching algorithm [ 12 ]. Therefore, we propose the following modification of Greedy . We run Greedy as long as ω(G) ≥ 3. Once
ω(G) ≤ 2, we instead use Edmonds’ matching algorithm on the remaining graph and add the maximum matching to the clique partition. We call this algorithm Greedy Edmonds .Our main result is that for every graph G, Greedy Edmonds is a 2ω′(G)
ω′(G)+1
-approximation algorithm for Clique Partition .
▶ Theorem 1.2. For every graph G, Greedy Edmonds returns a clique partition X such that
OPT ≤ 2ω′(G)
ω′(G) + 1 |E(X )|,
where OPT is the number of edges in an optimal clique partition of G.
Although optimal solutions to Clique Partition correspond to optimal solutions to
Cluster Deletion , it is not necessarily true that an α-approximate solution to Clique Partition corresponds to an α-approximate solution to Cluster Deletion . Indeed, for general n-vertex graphs, Dessmark et al. [ 10 ] proved that there is no polynomial time
n1−O( 1(log n)γ )-approximation algorithm for Clique Partition (for some constant γ), unless
NP ⊆ ZPTIME (2 (log n)O(1)
). On the other hand, as already mentioned, there are constant factor approximations for Cluster Deletion . We do not know if our approximation guarantee from Theorem 6.2 also holds for Cluster Deletion .Note that Greedy Edmonds runs in polynomial time on any graph class for which a maximum clique can be found in polynomial time. Examples of such graph classes include graphs of bounded treewidth, perfect graphs (which include interval graphs and chordal graphs), distance-hereditary graphs, and graphs with bounded clique number. As far as we can tell, for graphs of bounded clique number, Greedy Edmonds is the first polynomial time algorithm which achieves an approximation ratio better than 2 for Clique Partition .For general graphs G, the best approximation algorithm for Clique Partition is the
ω(G)-approximation algorithm from . For permutation graphs, we also obtain the following corollary to Theorem 6.2. 4 Cluster deletion and clique partitioning
▶ Corollary 1.3. Given an n-vertex permutation graph G as input, Greedy Edmonds
returns a clique partition X such that
OPT ≤ 2ω′(G)
ω′(G) + 1 |E(X )|,
where OPT is the number of edges in an optimal clique partition of G in O(n2 log n) time.
Proof. Let G be an n-vertex permutation graph. It is well-known that a permutation representation of G can be found in O(n) time. Since a maximum clique in a permutation graph can be found in O(n log n) time, the greedy portion of Greedy Edmonds can be made to run in O(n2 log n) time, Finally, Rhee and Liang [ 26 ] proved that a maximum matching of an n-vertex permutation graph can be found in O(n log log n)-time. Thus, the total running time is O(n2 log n). ◀
We also found an example of a permutation graph for which neither Smart Greedy nor
Greedy Edmonds computes an optimal clique partition. This provides some evidence that
Cluster Deletion may be NP-hard on permutation graphs. Finally, we show that the approximation ratio in Theorem 6.2 is best possible.
▶ Theorem 1.4. For each ℓ ≥ 3, there exists a graph Gℓ with ω(G) = ℓ and a clique partition
Xℓ output by Greedy Edmonds such that
lim
ℓ→∞
(ℓ
2
) + 1 2(ℓ
2
) OPT ℓ
|E(Xℓ)| = 1 ,
where OPT ℓ is the maximum number of edges in a clique partition of Gℓ.
2 Preliminaries
All graphs in this paper are assumed to be simple and undirected. We use standard graph theoretical notation and terminology (see [ 11 ]). For example, we say that a graph G is H-free ,if it does not contain H as an induced subgraph. Note that G is a cluster graph if and only if G is P3-free, where P3 is a 3-vertex path. A graph G is a cograph if it is P4-free. A clique in a graph G is a set of pairwise adjacent vertices of G. A clique partition of G is a partition of V (G) into cliques of G. If X is a clique partition of G, we let E(X ) and E(X)
denote the set of edges of G contained (respectively, not contained) in the cliques of X .For each n ∈ N, we let [n] := {1, . . . , n } and Sn denote the set of all permutations of [n].Throughout this paper, permutations will be represented as follows. If π : [ n] → [n] is a permutation, it will be denoted as π = [ π(1) , . . . , π (n)] . The permutation graph associated with a permutation π ∈ Sn, denoted by Gπ , is defined as follows. The vertex set is
V (Gπ ) := [ n], and the edge set is
E(Gπ ) := {{ i, j } : i, j ∈ [n], i < j and π(i) > π (j)} .
In other words, an edge exists between vertices i and j if i < j but i appears after j in the permutation sequence. Permutation graphs were originally introduced by [ 13 ] and [ 24 ]. A comprehensive overview of their structural properties and associated algorithms is provided in . Let α ≥ 1. An α-approximation algorithm is an algorithm which returns a solution which is within a multiplicative factor of α to an optimal solution. N. Galesi, T. Huynh, and F. Ranjbar 5
3 A Greedy Algorithm
In this section we review the algorithm for Clique Partition introduced by Dessmark et al. [ 10 ]. The algorithm requires Max Clique as a subroutine, and therefore does not run in polynomial time for general graphs. However, in every graph class for which Max Clique
can be computed in polynomial time, the algorithm runs in polynomial time. We call the algorithm Greedy , since it proceeds greedily. The description of Greedy is as follows. Let
G be a graph and X be a maximum clique of G. Greedy adds X to the clique partition and then recurses on G − X.
Algorithm 1 Greedy
Input : a graph G
Output : a clique partition X for G
X = ∅
while G̸ = ( ∅, ∅)
X = Max Clique (G)
X = X ∪ { X}
G = G − X
Output : X
Dessmark et al. [ 10 ] proved that Greedy is a 2-approximation algorithm for both
Cluster Deletion and Clique Partition .
▶ Theorem 3.1. For every graph G, Greedy returns a clique partition X such that
2|E(X )| ≥ OPT , where OPT is the number of edges in an optimal clique partition of
G.
▶ Theorem 3.2. For every graph G, Greedy returns a clique partition X such that
|E(X )| ≤ 2OPT , where OPT is the number of edges in an optimal solution to Cluster Deletion .
For completeness 1, we include proofs of both Theorem 3.1 and Theorem 3.2 in Section A.
4 Cluster Deletion in Cographs
In this section, we give a simpler proof that Greedy computes an optimal clique partition in cographs, which was first proved in [ 15 ]. We require the following basic property that all cographs satisfy. The join of two graphs G1 and G2 denoted by G1 ∨ G2 is the graph obtained from the disjoint union G1 ⊔ G2 by adding all edges between V (G1) and V (G2).
▶ Lemma 4.1. Every cograph G with at least two vertices has a partition (A, B ) of V (G)
such that either G = G[A] ⊔ G[B] or G = G[A] ∨ G[B].
The other ingredient we need is the following partial order on the set of clique partitions of a fixed graph, introduced in [ 15 ]. Let X := {X1, . . . , X s} and Y = {Y1, . . . , Y t} be clique partitions of G such that |X1| ≤ · · · ≤ | Xs| and |Y1| ≤ . . . ≤ | Yt|. We write X ⪯ Y if
∑ji=1 |Xi| ≤ ∑ji=1 |Yi| for all 1 ≤ j ≤ max {s, t }.
1The proofs presented in are correct but quite terse.
6 Cluster deletion and clique partitioning
▶ Lemma 4.2 ([ 15 ]) . If X is a clique partition of a graph G which is maximal under ⪯, then
X is an optimal clique partition of G.
By Lemma 4.2, it suffices to prove the following.
▶ Theorem 4.3. Let G be a cograph. Then Greedy always outputs a clique partition X of
G which is maximal under ⪯.
Proof. Let G be a cograph. We proceed by induction on |V (G)|. Let X := {X1, . . . , X s} be a clique partition of G output by Greedy and let Y := {Y1, . . . , Y t} be an arbitrary clique partition of G with |X1| ≤ · · · ≤ | Xs| and |Y1| ≤ · · · ≤ | Yt|. By Lemma 4.1, V (G) has a partition (A, B ) such that either G = G[A] ⊔ G[B] or G = G[A] ∨ G[B].First suppose G = G[A]∨G[B]. Towards a contradiction, suppose ∑
i∈[j]
|Xi| < ∑
i∈[j]
|Yi|
for some j ≤ max {s, t }. By symmetry, we may assume ∑
i∈[j]
|Xi ∩ A| < ∑
i∈[j]
|Yi ∩ A|.
Let m be the largest index such that Xm ∩ A̸ = ∅. Note that X | A := {Xi ∩ A : i ∈ [m]} is a greedy clique partition of G[A] with |X1 ∩ A| ≤ · · · ≤ | Xm ∩ A|. By induction, X | A is a maximal clique partition of G[A] under ⪯. Thus, ∑
i∈[j]
|Xi ∩ A| ≥ ∑
i∈[j]
|Yi ∩ A|, which is a contradiction. Next suppose that G = G[A] ⊔ G[B]. Thus, every clique of G is either contained in A or contained in B. Let XA := {Xi : Xi ⊆ A}, XB := {Xi : Xi ⊆ B}, YA := {Yi : Yi ⊆ A}, and
YB := {Yi : Yi ⊆ B}. Note that XA and XB are greedy clique partitions of G[A] and G[B],respectively. Let IA := {i ∈ [s] : Xi ⊆ A}, IB := {i ∈ [s] : Xi ⊆ B}, JA := {i ∈ [t] : Yi ⊆ A},and JB := {i ∈ [t] : Yi ⊆ B}. By symmetry, we may assume that |IA ∩ [ℓ]| ≥ | JA ∩ [ℓ]| := a.Choose an arbitrary ℓ ≤ max {s, t }. Let I′
A
be the a largest elements of IA and let I′
B
be the
ℓ − a largest elements of IB , where I′
B
= IB if ℓ − a ≥ | IB |. Then,
∑
i∈[ℓ]
|Xi| = ∑
i∈IA∩[ℓ]
|Xi| + ∑
i∈IB∩[ℓ]
|Xi|≥ ∑
i∈I′
A
|Xi| + ∑
i∈I′
B
|Xi|≥ ∑
i∈JA∩[ℓ]
|Yi| + ∑
i∈JB∩[ℓ]
|Yi|
= ∑
i∈[ℓ]
|Yi|,
where the second to last line follows by induction. Thus, X ⪯ Y , as required. ◀
5 Cluster Deletion in Permutation Graphs
The question of whether Cluster Deletion can be solved in polynomial time on the class of permutation graphs is listed among the open problems in [ 22 ]. Moreover, since every cograph is a permutation graph, a positive answer would extend the main result in [ 15 ]. An important observation is that Max Clique can be solved efficiently on permutation graphs. Indeed, given a permutation [π(1) , . . . , π (n)] , a clique in Gπ is simply a decreasing subsequence of
[π(1) , . . . , π (n)] . Hence, all maximum cliques of Gπ can be efficiently generated via dynamic programming. Could it be that Greedy always finds optimal clique partitions for permutations graphs? The next very simple counterexample shows that this is not the case. Let G be the 4-vertex path P depicted below, which is a permutation graph represented by the permutation N. Galesi, T. Huynh, and F. Ranjbar 7
[3 , 1, 4, 2] .
1 3 2 4
The optimal clique partition for G is Y = {{ 1, 3}, {2, 4}} . On the other hand,
{1, 3}, {2, 3}, {2, 4} are the maximum cliques of G and therefore it is possible that
Greedy chooses {2, 3} in its first iteration. This would result in the clique partition
X = {{ 1}, {2, 3}, {4}} , which is not optimal. In searching for polynomial time algorithms for Cluster Deletion on permutation graphs this example immediately suggests a modification of Greedy where a tie-breaking rule is used in case there are multiple maximum cliques Greedy can choose from. Here is a very natural candidate for a tie-breaking rule. For X ⊆ V (G), we let δ(G) be the set of edges of G with exactly one end in X. Among all maximum cliques, we add to X the maximum clique X with |δ(X)| minimum. This is a locally optimal strategy, since it minimizes the number of edges deleted at each step (subject to always choosing a maximum clique). We call this algorithm Smart Greedy . In Smart Greedy we use a subroutine Clique List (G)
which takes a graph G as input and outputs the list of all the maximum cliques of G. For permutation graphs, this subroutine can be implemented in polynomial time via dynamic programming.
Algorithm 2 Smart Greedy
Input : a graph G
Output : a clique partition X for G
X = ∅
while G̸ = ( ∅, ∅)
L = Clique List (G)
Let X be a clique in L with |δ(X)| minimum 5. X = X ∪ { X}
G = G − X
Output : X
Smart Greedy computes an optimal clique partition of the 4-vertex path example above, and in fact performs well for many examples of permutation graphs. However, we found a permutation graph for which Smart Greedy does not solve Clique Partition
optimally. Consider the following graph G
214376
5
G is a permutation graph since it is generated by the permutation [2 , 5, 4, 1, 7, 3, 6] . The optimal clique partition of G is X = {{ 3, 4, 5}, {1, 2}, {6, 7}} , obtained by deleting 3 edges from G. Smart Greedy obtains this partition if and only if it chooses the 3-clique {3, 4, 5}
during the first iteration. The maximum cliques of G are X1 := {1, 4, 5} and X2 := {3, 4, 5}.Since |δ(X1)| = |δ(X2)| = 3 , Smart Greedy could choose {1, 4, 5} during the first iteration, in which case it will not output an optimal clique partition. This example suggests that all local tie-breaking rules will fail, since the correct first choice depends on a global parity phenomenon. 8 Cluster deletion and clique partitioning
Another potential approach is to use the two orderings of the vertices provided by the permutation. That is, given a permutation [π(1) , . . . , π (n)] we can either process the vertices of Gπ in order 1, 2, . . . , n or in order π(1) , π (2) , . . . , π (n). One natural rule would be to always add the newest vertex in the ordering to the largest clique possible. However, for the permutation [2 , 4, . . . , 2k, 2k − 1, . . . , 3, 1] , if we process the vertices in order
1, 2, . . . , n , then this rule will output the clique partition {{ 1, 2}, {3, 4}, . . . , {2k − 1, 2k}} .This clique partition only contains k edges, while the optimal clique partition contains (k+1 2
) edges. Similarly, for the permutation [2 k, 1, 2k − 1, 2, 2k − 2, 3, . . . , k + 1 , k ], if we process the vertices in order π(1) , π (2) , . . . , π (n), then we will output the clique partition
{{ 1, 2k}, {2, 2k − 1}, . . . , {k, k + 1 }} . Again, this clique partition only contains k edges, while the optimal clique partition contains (k+1 2
) edges. Thus, the two orderings provided by π
must be used in a more sophisticated manner.
6 An improved approximation algorithm
In this section we present an improved approximation algorithm for Clique Partition .The description of the algorithm is simple. If ω(G) ≥ 3, we compute a maximum clique X
and put X into the clique partition. If ω(G) = 2 , then we compute a maximum matching M
and put all edges of M into the clique partition. We call the algorithm Greedy Edmonds .
Algorithm 3 Greedy Edmonds
Input : a graph G
Output : a clique partition X for G
X = ∅
while G̸ = ( ∅, ∅)
if ω(G) = 2
then compute a maximum matching M of G
X = X ∪ M
G = G − V (M)
else X = Greedy (G)
X = X ∪ X
G = G − X
Output : X
Let G be a graph, Y be a clique partition of G, and C ⊆ V (G). We let Y − C be the nonempty sets of the form Y − C, where Y ∈ Y . Note that Y − C is a clique partition of
G − C. Recall that ω′(G) is the number of edges in a maximum clique of G.We now show that Greedy Edmonds is a 2ω′(G)
w′(G)+1
-approximation algorithm for every graph G. We first require the following lemma.
▶ Lemma 6.1. Let G be a graph, Y be a clique partition of G, λ(k) := min {| E(Y − X)X| = k}, and C be a subset of V (G) of size k. If |E(Y − C)| = λ(k), then there exists an ordering c1, . . . , c k of C such that ci is in a maximum clique of Y − { c1, . . . , c i−1} for all
i ∈ [k].
Proof. We proceed by induction on k. The lemma clearly holds if k = 1 , so we may assume
k ≥ 2. Let C be a subset of V (G) of size k such that |E(Y − C)| = λ(k). We claim that C
contains a vertex c1 such that c1 is in a maximum clique of Y. If not, let y ∈ V (G) be in a N. Galesi, T. Huynh, and F. Ranjbar 9
maximum clique of Y, x ∈ C, and C′ := ( C \x)∪{ y}. Then, |E(Y − C′)| < |E(Y − C)|, which is a contradiction. Thus, c1 exists. Let C1 := C \ { c1} and Y1 := Y − { c1}. By induction,
C \ { c1} has an ordering c2, . . . , c k such that ci is in a maximum clique of Y1 − { c2, . . . , c i−1}
for all i ∈ { 2, . . . , k }. Thus, c1, c 2, . . . , c k is the required ordering of C. ◀▶ Theorem 6.2. For every graph G, Greedy Edmonds returns a clique partition X such that
OPT ≤ 2ω′(G)
ω′(G) + 1 |E(X )|,
where OPT is the number of edges in an optimal clique partition of G.
Proof. We proceed by induction on |V (G)|. Let ℓ := ω(G). If ℓ ≤ 2, then we are done since
Greedy Edmonds computes an optimal clique partition of G. Thus, we may assume ℓ ≥ 3.Let Y be an optimal clique partition of G. For each i ∈ [ℓ] let yi be the number of cliques of size i in Y and let xi be the number of cliques of size i in X .Let A1, . . . , A xℓ be the cliques of size ℓ in X and let B1, . . . , B yℓ be the cliques of size
ℓ in Y. Let A := ⋃xℓ
i=1
V (Ai) and B := ⋃yℓ
i=1
V (Bi). Let G′ := G − A, Y′ := Y − A, and
X ′ := X − A.Let a := ℓx ℓ and c := min {| E(Y − X)X| = a}. Let C be a subset of V (G) of size a
such that |E(Y − C)| = c. For each i ∈ [ℓ] let zi be the number of cliques of size i in Y − C.By Lemma 6.1, there is an ordering c1, . . . , c a of C such that ci is in a maximum clique of
Y −{ c1, . . . , c i−1} for all i ∈ [a]. By the definition of Greedy Edmonds , ω(G′) ≤ ℓ−1. Thus,
A contains at least one vertex from each clique of Y of size ℓ, which implies |C| = |A| ≥ yℓ.Let c1, . . . , c a be the ordering of C given by Lemma 6.1. By Lemma 6.1, c1, . . . , c yℓ are each in a different maximum clique of Y. Thus, zℓ = 0 . Let k be the largest index such that
zk > 0.Again by Lemma 6.1,
a ≥
ℓ
∑
i=k
(i − k)yi,zk =
ℓ
∑
i=k
yi −
(
a −
ℓ
∑
i=k
(i − k)yi
)
=
ℓ
∑
i=k
(i − k + 1) yi − a, zk−1 = yk−1 +
(
a −
ℓ
∑
i=k
(i − k)yi
)
, and
zi = yi, for all i ∈ [k − 2] .Thus,
|E(Y − C)| =
(k
2
) ( ℓ∑
i=k
(i − k + 1) yi − a
)
+
(k − 12
) (
a −
ℓ
∑
i=k
(i − k)yi
)
+
k−1
∑
i=1
(i
2
)
yi
= a(1 − k) +
ℓ
∑
i=k
(2 i − k)( k − 1) 2 yi +
k−1
∑
i=1
(i
2
)
yi.10 Cluster deletion and clique partitioning
Next note that
|E(X )| =
(ℓ
2
)
xℓ + |E(X ′)|≥
(ℓ
2
)
xℓ +
(ℓ−12
) + 1 2(ℓ−12
) |E(Y′)| (by induction)
≥
(ℓ
2
)
xℓ +
(ℓ−12
) + 1 2(ℓ−12
) |E(Y − C)|.
Recall that our goal is to show that |E(X )| ≥ (ℓ
2
)+1 2(ℓ
2
) |E(Y)|. Substituting the value of
|E(Y − C)| into the above inequality and noting that (ℓ−12 )+1 2(ℓ−12 ) > (ℓ
2
)+1 2(ℓ
2
) , it suffices to show that
(ℓ
2
)
xℓ +
(ℓ−12
) + 1 2(ℓ−12
)(
a(1 − k) +
ℓ
∑
i=k
(2 i − k)( k − 1) 2 yi
)
≥
(ℓ
2
) + 1 2(ℓ
2
)
ℓ
∑
i=k
(i
2
)
yi.
Observe that,
(ℓ
2
)
xℓ +
(ℓ−12
) + 1 2(ℓ−12
) a(1 − k) = ℓ − 12 a + (1 − k)( (ℓ−12
) + 1) 2(ℓ−12
) a
= (ℓ − k)(ℓ−12
) − k + 1 2(ℓ−12
) a
≥ (ℓ − k)(ℓ−12
) − k + 1 2(ℓ−12
)
ℓ
∑
i=k
(i − k)yi
Thus, it suffices to show
(ℓ − k)(ℓ−12
) − k + 1 2(ℓ−12
)
ℓ
∑
i=k
(i−k)yi +
(ℓ−12
) + 1 2(ℓ−12
)
ℓ
∑
i=k
(2 i − k)( k − 1) 2 yi −
(ℓ
2
) + 1 2(ℓ
2
)
ℓ
∑
i=k
(i
2
)
yi ≥ 0.
(1) Let
f (i, k, ℓ ) := (ℓ − k)(ℓ−12
) − k + 1 2(ℓ−12
) (i − k) +
(ℓ−12
) + 1 2(ℓ−12
) (2 i − k)( k − 1) 2 −
(ℓ
2
) + 1 2(ℓ
2
)(i
2
)
.
Then (1) can be rewritten as
ℓ
∑
i=k
f (i, k, ℓ )yi ≥ 0.
Therefore, it suffices to show that f (i, k, ℓ ) ≥ 0 for all positive integers k ≤ i ≤ ℓ with ℓ ≥ 3
and k ≤ ℓ − 1. Expanding and simplifying, we have that f (i, k, ℓ ) is equal to
(−ℓ3 + 3 ℓ2 − 4ℓ + 4) i2 + (2 ℓ4 − 7ℓ3 + 7 ℓ2 − 4) i + ℓk (−2ℓ3 + ℓ2k + 7 ℓ2 − 3ℓk − 7ℓ + 4 k)4ℓ(ℓ − 1)( ℓ − 2) .
Since ℓ ≥ 3, the denominator 4ℓ(ℓ − 1)( ℓ − 2) is positive. Therefore, it suffices to show that the numerator
g(i, k, ℓ ) := ( −ℓ3 +3 ℓ2 −4ℓ+4) i2 +(2 ℓ4 −7ℓ3 +7 ℓ2 −4) i+ℓk (−2ℓ3 +ℓ2k +7 ℓ2 −3ℓk −7ℓ+4 k)N. Galesi, T. Huynh, and F. Ranjbar 11
Figure 1 The graph G6, with k1= 6 , k 2= 5 , k 3= 5 , k 4= 4 , k 5= 3 , k 6= 3 . Vertices of the same (non-white) color form a clique in the graph.
is nonnegative for all positive integers k ≤ i ≤ ℓ with ℓ ≥ 3 and k ≤ ℓ − 1. Regarding, k and ℓ
as constants, g(i, k, ℓ ) is a quadratic function of i with leading coefficient −ℓ3 +3 ℓ2 −4ℓ+4 < 0
(since ℓ ≥ 3). Therefore, to show that g(i, k, ℓ ) ≥ 0 for all i ∈ [k, ℓ ], it suffices to show that
g(k, k, ℓ ) ≥ 0 and g(ℓ, k, ℓ ) ≥ 0. It is easy to check that
g(k, k, ℓ ) = 4 k(k − 1) ≥ 0,
since k ≥ 1. Next note that
g(ℓ, k, ℓ ) = ( ℓ2 − 3ℓ + 4) k2 + ( −2ℓ3 + 7 ℓ2 − 7ℓ)k + ( ℓ4 − 4ℓ3 + 3 ℓ2 + 4 ℓ − 4) .
This is a quadratic in k with positive leading coefficient and two roots k1 := ℓ − 1 and
k2 := (ℓ−2) 2(ℓ+1)
ℓ2−3ℓ+4
. It is easy to check that k2 > k 1 for all ℓ ≥ 6 and that ℓ − 2 ≤ k2 < k 1 for
ℓ ∈ { 3, 4, 5}. Since g(ℓ, x, ℓ ) < 0 if and only if k1 < x < k 2, it follows that in either case,
g(ℓ, k, ℓ ) ≥ 0 for all positive integers k and ℓ with 1 ≤ k ≤ ℓ − 1 and ℓ ≥ 3. This completes the proof. ◀
Note that the permutation graph at the end of Section 5 also shows that Greedy Edmonds does not always output an optimal clique partition of a permutation graph. We now give a family of graphs which shows that the approximation ratio from Theorem 6.2 is best possible.
▶ Theorem 6.3. For each ℓ ≥ 3, there exists a graph Gℓ with ω(G) = ℓ and a clique partition
Xℓ output by Greedy Edmonds such that
lim
ℓ→∞
(ℓ
2
) + 1 2(ℓ
2
) OPT ℓ
|E(Xℓ)| = 1 ,
where OPT ℓ is the maximum number of edges in a clique partition of Gℓ.
Proof. Let ℓ ≥ 3. We first define a sequence of positive integers ki as follows. Let k1 := ℓ
and inductively define ki to be the smallest integer ki such that ℓ2 − ∑i−1
i=1
ki ≤ ℓk i. Let t
be the largest index such that kt ≥ 3. Let Gℓ be the graph on vertex set [ℓ] × [ℓ] consisting of ℓ disjoint ‘horizontal’ cliques of size ℓ and disjoint cliques of size k1, . . . , k t which are sequentially placed as ‘leftmost’ as possible in [ℓ] × [ℓ] (see Figure 1). Clearly ω(Gℓ) = ℓ and the ‘horizontal’ cliques form an optimal clique partition of
Gℓ. Thus, an optimal clique partition of Gℓ has ℓ(ℓ
2
) edges. On the other hand, 12 Cluster deletion and clique partitioning
Greedy Edmonds can output a clique partition with clique sizes k1, . . . , k t together with ℓ(ℓ − 1) − ∑mi=1 ki cliques of size 2. Thus, Greedy Edmonds can output a clique partition with f (ℓ) := ∑mi=1
(ki
2
) + ( ℓ)( ℓ − 1) − ∑mi=1 ki edges. It is easy to check that
lim ℓ→∞
(ℓ
2
)+1 2(ℓ
2
)
ℓ
(ℓ
2
)
f(ℓ)
= 1 . ◀
Acknowledgements. Tony Huynh is supported by the Institute for Basic Science (IBS-R029-C1).
References
1 Manuel Aprile, Matthew Drescher, Samuel Fiorini, and Tony Huynh. A tight approximation algorithm for the cluster vertex deletion problem. Math. Program. , 197(2):1069–1091, 2023.
doi:10.1007/s10107-021-01744-w .
2 Vicente Balmaseda, Ying Xu, Yixin Cao, and Nate Veldt. Combinatorial approximations for cluster deletion: Simpler, faster, and better. arXiv preprint arXiv:2404.16131 , 2024.
3 Nikhil Bansal, Avrim Blum, and Shuchi Chawla. Correlation clustering. Mach. Learn. ,56(1–3):89–113, 2004. doi:10.1023/B:MACH.0000033116.57574.95 .
4 Gabriel Bathie, Nicolas Bousquet, Yixin Cao, Yuping Ke, and Théo Pierron. (sub)linear kernels for edge modification problems toward structured graph classes. Algorithmica , 84(11):3338– 3364, 2022. doi:10.1007/s00453-022-00969-1 .
5 Amir Ben-Dor, Ron Shamir, and Zohar Yakhini. Clustering gene expression patterns. Journal of Computational Biology , 6(3-4):281–297, 1999. doi:10.1089/106652799318274 .
6 Flavia Bonomo, Guillermo Durán, and Mario Valencia-Pabon. Complexity of the cluster deletion problem on subclasses of chordal graphs. Theoretical Computer Science , 600:59–69, 2015. doi:10.1016/j.tcs.2015.07.001 .
7 Sebastian Böcker and Peter Damaschke. Even faster parameterized cluster deletion and cluster editing. Information Processing Letters , 111(14):717–721, 2011. doi:10.1016/j.ipl.2011.05. 003 .
8 Moses Charikar, Venkatesan Guruswami, and Anthony Wirth. Clustering with qualitative information. Journal of Computer and System Sciences , 71(3):360–383, 2005. doi:10.1016/j. jcss.2004.10.012 .
9 Peter Damaschke. Bounded-degree techniques accelerate some parameterized graph algorithms. In Parameterized and Exact Computation: 4th International Workshop, IWPEC 2009 , pages 98–109, 2009.
10 Anders Dessmark, Jesper Jansson, Andrezej Lingas, Eva-Marta Lundell, and Mia Person. On the approximability of maximum and minimum edge clique partition problems.
International Journal of Foundations of Computer Science , 18(02):217–226, 2007. doi: 10.1142/S0129054107004656 .
11 Reinhard Diestel. Graph theory , volume 173 of Graduate Texts in Mathematics . Springer, Berlin, sixth edition, ©2025.
12 Jack Edmonds. Paths, trees, and flowers. Canadian Journal of Mathematics , 17:449–467, 1965.
doi:10.4153/CJM-1965-045-4 .
13 Shimon Even, Amir Pnueli, and Abraham Lempel. Permutation graphs and transitive graphs.
J. ACM , 19(3):400–410, 1972. doi:10.1145/321707.321710 .
14 Santo Fortunato. Community detection in graphs. Physics Reports , 486(3):75–174, 2010.
doi:10.1016/j.physrep.2009.11.002 .
15 Yong Gao, Donovan R. Hare, and James Nastos. The cluster deletion problem for cographs.
Discrete Math. , 313(23):2763–2771, 2013. doi:10.1016/j.disc.2013.08.017 .
16 Petr A. Golovach, Pinar Heggernes, Athanasios L. Konstantinidis, Paloma T. Lima, and Charis Papadopoulos. Parameterized Aspects of Strong Subgraph Closure. In 16th Scandinavian Symposium and Workshops on Algorithm Theory (SWAT 2018) , volume 101, pages 23:1–23:13, 2018. doi:10.4230/LIPIcs.SWAT.2018.23 .N. Galesi, T. Huynh, and F. Ranjbar 13
17 Martin Charles Golumbic. Algorithmic graph theory and perfect graphs , volume 57. Elsevier, 2004.
18 Jens Gramm, Jiong Guo, Falk Hüffner, and Rolf Niedermeier. Graph-modeled data clustering: Exact algorithms for clique generation. Theory of Computing Systems , 38(4):373–392, 2005.
doi:10.1007/s00224-004-1178-y .
19 Niels Grüttemeier and Christian Komusiewicz. On the relation of strong triadic closure and cluster deletion. Algorithmica , 82:853–880, 2020. doi:10.1007/s00453-019-00617-1 .
20 Giuseppe F. Italiano, Athanasios L. Konstantinidis, and Charis Papadopoulos. Structural parameterization of cluster deletion. Algorithmica , pages 1–21, 2025.
21 Christian Komusiewicz and Johannes Uhlmann. Cluster editing with locally bounded modifications. Discrete Applied Mathematics , 160(15):2259–2270, 2012. doi:10.1016/j. dam.2012.05.019 .
22 Athanasios L. Konstantinidis and Charis Papadopoulos. Cluster deletion on interval graphs and split related graphs. Algorithmica , 83(7):2018–2046, 2021. doi:10.1007/S00453-021-00817-8 .
23 Pan Li, Gregory J. Puleo, and Olgica Milenkovic. Motif and hypergraph correlation clustering.
IEEE Transactions on Information Theory , 66(5):3065–3078, 2020. doi:10.1109/TIT.2019. 2940246 .
24 Amir Pnueli, Abraham Lempel, and Shimon Even. Transitive orientation of graphs and identification of permutation graphs. Canadian Journal of Mathematics , 23(1):160–175, 1971.
doi:10.4153/CJM-1971-016-5 .
25 Gregory J. Puleo and Olgica Milenkovic. Correlation clustering with constrained cluster sizes and extended weights bounds. SIAM Journal on Optimization , 25(3):1857–1872, 2015.
doi:10.1137/140994198 .
26 Chongkye Rhee and Y. Daniel Liang. Finding a maximum matching in a permutation graph.
Acta Inform. , 32(8):779–792, 1995. doi:10.1007/s002360050033 .
27 Ron Shamir, Roded Sharan, and Dekel Tsur. Cluster graph modification problems. Discrete Applied Mathematics , 144(1):173–182, 2004. doi:10.1016/j.dam.2004.01.007 .
28 Jianbo Shi and J. Malik. Normalized cuts and image segmentation. IEEE Transactions on Pattern Analysis and Machine Intelligence , 22(8):888–905, 2000. doi:10.1109/34.868688 .
29 Stavros Sintos and Panayiotis Tsaparas. Using strong triadic closure to characterize ties in social networks. In Proceedings of the 20th ACM SIGKDD International Conference on Knowledge Discovery and Data Mining , KDD ’14, page 1466–1475, 2014. doi:10.1145/2623330.2623664 .
30 Nate Veldt. Correlation clustering via strong triadic closure labeling: Fast approximation algorithms and practical lower bounds. In Proceedings of the 39th International Conference on Machine Learning , volume 162, pages 22060–22083, 2022.
31 Nate Veldt, David F. Gleich, and Anthony Wirth. A correlation clustering framework for community detection. In Proceedings of the 2018 World Wide Web Conference , WWW ’18, page 439–448, 2018. doi:10.1145/3178876.3186110 .
A Proofs of Theorem 3.1 and Theorem 3.2
Proof of Theorem 3.1. Let X := ( X1, . . . , X ℓ) be the sequence of cliques constructed by
Greedy , where X1 is a maximum clique of G. We proceed by induction on ℓ. If ℓ = 1 , then
|E(X )| = OPT . Thus, we may assume ℓ ≥ 2. Let Y := ( Y1, . . . , Y m) be an optimal clique partition of G. Let G′ := G−X1, X ′ := ( X2, . . . , X ℓ), Y′ := ( Y1 \X1, . . . , Y m \X1), and OPT ′
be the number of edges in an optimal clique partition of G′. Let k := |X1|. For each vertex
x in X1, let Y(x) be the unique cluster Y of Y such that x ∈ Y . Define y(x) := |Y (x)| − 1.Since X1 is a maximum size clique of G, y(x) ≤ k − 1 for all x ∈ X1. Therefore,
|E(Y)| − | E(X ′)| ≤ ∑
x∈X1
y(x) ≤ ∑
x∈X1
k − 1 = k(k − 1) .14 Cluster deletion and clique partitioning
We also clearly have
|E(X )| − | E(X ′)| = k(k − 1) 2 .
By induction, |E(X ′)| ≥ OPT ′
2
≥ |E(Y′)|
2
. Therefore, |E(X )| ≥ |E(Y)|
2
, as required. ◀
Proof of Theorem 3.2. Let X := ( X1, . . . , X ℓ) be the sequence of cliques constructed by
Greedy , where X1 is a maximum clique of G. We proceed by induction on ℓ. If ℓ = 1 , then
|E(X )| = 0 = OPT . Thus, we may assume ℓ ≥ 2. Let Y := ( Y1, . . . , Y m) be an optimal clique partition of G. Let Z := ( X1, Y 1 \ X1, . . . , Y m \ X1), G′ := G − X1, X ′ := ( X2, . . . , X ℓ),
Y′ := ( Y1 \ X1, . . . , Y m \ X1) and OPT ′ be the minimum size of a set of edges Z such that
G′ − Z is a cluster graph. Let k := |X1|. For each i ∈ [m], let xi := |X1 ∩ Yi|.Note that
|E(Z) \ E(Y)| = ∑
i∈[m]
xi(|Yi| − xi) ≤ ∑
i∈[m]
xi(k − xi) = ∑
i∈[m]
xi
∑
j̸=i
xj = 2 ∑
i̸=j
xixj
On the other hand,
|E(Y) \ E(Z)| = ∑
i̸=j
xixj .
Let δ(X1) be the set of edges of G with exactly one end in X1. By induction, |E(X ′)| ≤
2OPT ′ ≤ 2|E(Y′)|. Therefore,
|E(X )| = |δ(X1)| + |E(X ′)|≤ | δ(X1)| + 2 |E(Y′)|
= |E(Z)| + |E(Y′)|
= |E(Z) \ E(Y)| + 2 |E(Y′)|≤ 2 ∑
i̸=j
xixj + 2 |E(Y′)|
= 2 |E(Y) \ E(Z)| + 2 |E(Y′)|
= 2 |E(Y)|
= 2 OPT ,
as required. ◀ |
9929 | https://explainingscience.org/2019/05/28/stellar-aberration/ | Skip to content
Stellar aberration
This post is a bit more technical than usual and discusses stellar aberration, the apparent movement of stars as a result of the motion of the Earth. It is an interesting case of the application of the scientific method. Observations were made to test a scientific theory, but unexpected results were found, which in turn led to new discoveries.
Discovery of stellar aberration
In 1727 the English astronomer James Bradley had been taking measurements of the position of Eltanin, the brightest star in the constellation Draco.
James Bradley (1693 – 1762) – image from Wikimedia Commons
Bradley was looking for a small shift in the position of the star, known as the parallax.
The expected shift in the position of a star during the year due to parallax
In Bradley’s time the heliocentric theory, in which the Earth and all the planets orbit the Sun, was generally accepted by astronomers. However, it had not yet been accepted by the Catholic church and many thinkers outside the field of astronomy still believed the older geocentric theory Finding the parallax of a star would confirm that the heliocentric theory was correct and would also allow the distance to the star to be calculated.
The distance to star (Dss) is equal to the distance between the Earth and the Sun (Des) divided by the tangent of the parallax angle (A).
However, Bradley found that the position of Eltanin did not change as predicted. Although he found a small change in its position over the course of the year, the position shift was different from what it would have been if parallax were the cause.
For clarity. the change in position has been greatly exaggerated. The maximum shift of Eltanin around its mean position is only 0.0056 of a degree, roughly 1% of the diameter of the Moon.
In trying to explain his observations Bradley discovered an entirely different effect which came to be known as stellar aberration. His discovery not only confirmed the heliocentric theory but allowed an accurate measurement of the speed of light.
To understand Bradley’s explanation of the cause of stellar aberration, consider a rainy day where there is no wind – rain falls vertically from the sky. However, if you are moving, rain no longer falls vertically but is tilted in your direction of travel. Two common examples of this are: anyone carrying an umbrella and walking quickly will need to tilt their umbrella forward to shield from the rain and when a train is moving, rain appears to streak down its windows diagonally.
Bradley reasoned that the change in the star’s position which became known as stellar aberration was due to the Earth’s motion around the Sun. Bradley believed that light consisted of small particles which travelled at a finite speed, and the light particles appear displaced in the direction the Earth was moving in the same way raindrops are.
As the Earth goes around its orbit, the direction in which it is moving moves continually changes and so the displacement of the star from its mean position continually changes too.
Bradley’s explanation not only proved that the Earth revolved around the Sun, but he was able to work out the speed of light. To do this he showed that the maximum angle θ, by which any star is displaced from its mean position, is given by the formula:
θ = v/c
v is the Earth’s velocity around the Sun
c is the speed of light
θ the angle the star is displaced, measured in radians (2π radians equals 360 degrees, so one radian is approximately 57.3 degrees).
From this we have
c = v/ θ
Using this, Bradley was able to estimate the speed of light as 301,000 km/s in modern units (Walker 2008). This is within 0.5% of the correct value.
Limitations of the classical theory
Bradley, like the majority of eighteenth-century physicists, believed in a particle theory of light. In the nineteenth century a wave theory of light became generally accepted. The wave theory was better able to explain phenomena such as diffraction and interference patterns. However, when this theory was used to try and explain stellar aberration, it ran into a couple of difficulties.
Firstly, the aberration angle (v/c) should varies with the speed of light (c). In the wave theory the speed of light depends on the medium through which light is passing. Light travels more slowly in air than it does in a vacuum, but the difference is small. In a denser medium, the slowing of light is more significant. For example, when light passes through water it travels 1.3 times more slowly than it does in a vacuum. So, if we filled a telescope with water, the aberration should be 1.3 times larger than if it were filled with air. This experiment was first tried by the English astronomer George Airy in 1871 and no change in the aberration angle was detected,
Secondly, the nineteenth-century wave theory of light required that it travelled in a medium called ‘the ether’. This was an invisible material which had no interaction with physical objects. The speed of light is a constant when measured with respect to this mysterious ether. To explain the results of a historic experiment called the Michelson-Morley experiment, it was proposed that the ether was dragged along by the motion of the Earth when it goes around the Sun, i.e. the relative motion of the Earth against the background ether is zero. However, if this were the case stellar aberration would not exist
The modern explanation of stellar aberration
Today the generally accepted explanation of stellar aberration is by Einstein’s theory of special relativity. In special relativity there is no absolute space and no absolute motion. As the Earth moves around the Sun in its near-circular orbit, we are observing a star in different reference frames, in which the relative direction of motion of the Earth when measured with respect to the star is different. Using the equations of special relativity the star has a different position when we transform its coordinates from one reference frame to another.
Despite this Bradley’s explanation gives excellent agreement with measurements made through an air filled telescope
What about parallax?
The shift in position of a nearby star caused by parallax proved to be very much smaller than the position shift due to stellar aberration, which unlike parallax does not vary with a star’s distance. Even for nearby stars the parallax is so small that it wasn’t successfully measured until 1838, when the German astronomer Friedrich Wilhelm Bessel detected it for the star 61 Cygni. The parallax he measured was 0.314 arc seconds, which is around 65 times smaller than the shift due to stellar aberration. The shift in position of 61 Cygni due to parallax is equivalent to a width of a 2 cm at a distance of 12 km. Because parallax was so difficult to detect, by the year 1900 only 60 nearby stars had had their parallax measured. It wasn’t until the development of machines to accurately measure the position of stars on photographic plates in the twentieth century that a large number of stellar parallaxes were calculated
The table below summarises the differences between parallax and stellar aberration.
Bradley was the first person to accurately measure and attempt explain the cause of stellar aberration. A small shift in position of stars over the 12-month cycle been reported by earlier astronomers but it had not been accurately measured or accounted for.
I hope you have enjoyed this post, for those of you who want a little more detail I’ve put some further notes below.
Update 24 August 2020
A video containing some of the material in this post can be viewed on the Explaining Science YouTube channel. This channel is a new venture and over the next few months I plan to upload additional videos.
Technical notes
How the amount of aberration varies with the position of a star
In the Sun’s frame of reference, consider a beam of light from a distant star, which lies at an angle A to the direction of the Earth’s motion. The speed of light is c and the light beam has x and y velocity components ux and uy, where
ux is its velocity in the direction of the Earth’s motion.
uy is its velocity in a direction perpendicular to x.
Therefore
ux = cCOS(A)
uy = cSIN(A)
Because the star is at an angle A relative to the direction of the Earth’s motion,
tan(A) = uy/ux = sin(A)/cos(A).
If the Earth is moving at velocity v in the x direction relative to the Sun, then in the Earth’s frame of reference:
the x component of the beam’s velocity is ux’ = ux +v.
the y component of the velocity is unchanged.
From Earth the star appears at an angle A’ to the direction of the Earth’s motion.
tan A’ = uy/ux’
= sin A / (cos A +v/c))
We want to find the displacement of the star due to the Earth’s motion. If we give this the symbol θ, then θ = A’ – A.
If we take A as equal to 90 degrees, this means that the direction of the star is perpendicular to the direction in which the Earth is moving. So, sin(A) = 1 which means tan(θ) = – v/c
However, because the angle θ is so small, to a good approximation θ = tan(θ). So θ = -v/c
If we take A as equal to 0 degrees, this means that the star lies in the same direction as the Earth is moving. So, sin(A) = 0, which means θ = 0 and the star is unchanged in position.
If we take A as equal to 180 degrees, this means that the star lies in the opposite direction from that in which the Earth is moving. So, sin(A) = 0, which means θ = 0 and the star is unchanged in position.
Full relativistic calculation
The calculation of the angle of aberration θ shown above is an oversimplification. When we are adding velocities, we should really use the relativistic addition formula from Einstein’s theory of special relativity to preserve the fact that nothing can travel faster than the speed of light. However, because the Earth’s velocity around the Sun is so small compared to the speed of light, a more complicated relativistic formula gives virtually the same results as the simple formula derived above.
Examples
For simplicity, if we assume the Earth moves in a perfect circle around the Sun at a constant velocity of 29.8 km/s and the speed of light is 299,792 km/s then the table below gives the amount of aberration for different angles to the Earth’s direction of motion:
If we consider a star which is positioned at a right angle to the Earth’s orbit, then as the Earth moves around its orbit the Earth would always be moving at 90 degrees to the direction of the star. The star would appear to move in a small circle of radius 20.5 arc seconds.
In reality, the Earth moves in a slightly elliptical orbit. It travels at 30.3 km/s when it is closest to the Sun in January and 29.3 km/s when it is furthest away in July. This causes the aberration to vary between 20.2 arcsec in July and 20.8 arcsec in January and the star appears to move in an slightly elliptical path rather than in a perfect circle.
If we consider a star which in the same plane as the Earth’s orbit then, assuming it lies directly opposite the Sun in June, its change in position during the year is shown in the diagram below.
In June the Earth is moving at 90 degrees to the direction of the star. So, the star is displaced an angle of v/c in the direction of the Earth’s motion.
In September the Earth is moving directly away from the star. Because the angle between the star and the direction of the earth’s motion is zero the star is not displaced
In December the Earth is moving at 90 degrees to the direction of the star (but in the opposition direction compared to June). So, the star is displaced an angle of v/c in the direction of the Earth’s motion.
In March the Earth is moving directly towards the star. So, the angle between the star and the direction of the earth’s motion is zero and the star is not displaced.
As seen from Earth, over the course of the year a star in the plane of the Earth’s orbit would appear to trace out a straight line centred around its mean position. Stars at angles between zero and 90 degrees to plane of the Earth’s orbit would trace out ellipses centred around their mean position.
Reference
Walker B H (2008) Optical engineering fundamentals, second edition,Available at: (Accessed: 31 May 2019).
Share this:
Click to email a link to a friend (Opens in new window)
Email
Click to share on LinkedIn (Opens in new window)
LinkedIn
Click to share on X (Opens in new window)
X
Click to share on Facebook (Opens in new window)
Facebook
Click to share on Pinterest (Opens in new window)
Pinterest
Like Loading...
Related
Galileo and the telescope
Revised 18 September 2022 Telescopes are instruments which use multiple lenses to produce magnified images of distant objects. It is unclear who invented the first telescope: lenses had been widely used in Europe to correct poor eyesight since the fourteenth century and I expect that, over time, the telescope was…
In "General"
Copernicus
In this post I'll talk about Nicolas Copernicus (1473 - 1543) and the heliocentric theory. The move away from the prevailing Earth-centred theory of the Universe to the heliocentric theory represents one of the greatest advances in astronomy ever made. Nicolas Copernicus - Image from Wikimedia Commons Background - the…
In "Cosmology"
The discovery of pulsars 1967
In 1967 Jocelyn Bell, a 24-year-old student from Cambridge University, was doing the research for her PhD. She was using a radio telescope to study radio waves emitted from compact astronomical objects known as quasars, and when she analysed the data she had collected, she noticed a signal which appeared…
In "Space"
Published by Steve Hurley
Hi I am Steve Hurley. I work in the IT industry. I studied for a PhD in astronomy in the 1980s. Outside work my real passion is explaining scientific concepts to a non-scientific audience. My blog (explainingscience.org) covers various scientific topics, but primarily astronomy. It is written in a style that it is easily understandable to the non scientist.
Publications and videos
For links to my books and videos please visit www.explainingscience.org View all posts by Steve Hurley
23 thoughts on “Stellar aberration”
[…] the Church had been wrong in 1633. The shift in the science was effectively complete by the time of Bradley’s observation of stellar aberration in 1729. It would take much longer for a shift in Church […]
LikeLike
2. […] Credit: Steve Hurley/Explaining Science […]
LikeLike
3. […] Credit: Steve Hurley/Explaining Science […]
LikeLike
4. I have the same question as the one posted by Muhammad in April 2019 and not yet answered here: why was the aberration angle not affected when the light from a star passed through a medium with a substantially higher refractive index than space (or air)? If it’s because in that experiment the light traveled through the alternative medium only within the telescope, i.e. only for a tiny fraction of its path from the star, then why was the experiment done at all?
LikeLiked by 1 person
Apologies for not coming back you earlier, Over the last 12 months I have been focussing on building up my YouTube channel ( and I have not have had the time to review all comment on my blog.
IF Bradley’s explanation were correct (which it isn’t although it gives very accurate agreement with the results when the telescope is filled with a medium of low refractive index such as air) the formula tan(ϴ) = -v/c sin A, would mean that the telescope would have to be tilted at a greater angle if filled with water, because of the lower value of the speed of light in water
However, Bradley’s explanation is not correct, the “correct explanation” is given by Einstein’s theory of special relativity, which is little more complicated. I will write a post about this sometime, but in the meantime, I have put a link on my references pages ( )to a good paper which gives the relativistic explanation of stellar aberration
LikeLike
Thanks for replying and for supplying the very useful reference.
LikeLiked by 1 person
5. I suppose that since light is composed of individual photons, it can be compared to falling raindrops ………
LikeLiked by 1 person
6. Greeting: Thank you for all the good presentations. The graphics is easy to understand. However, the reasoning by Bradley is contestable. The same results can be produced as observational errors because the 20.5 arc seconds is also can be found in a physics lab and without looking at the skies doing gravity experiments and electrostatics experiments and many other experiments suggesting something totally different and it’s not the light coming from the stars but a moving frame of reference or the sun’s motion. Id the sun is moving it produces the stellar aberration as error and produces all of relativity special and general as errors of a moving frame of reference of the sun. This fact I can prove and this fact the French mathematician Fourier worked on, and this fact was mentioned in 10th century and the fact is the moving stun as reference produces visual effects = 500 years of all wrong astronomy of a solar system.
LikeLike
7. “So, if we filled a telescope with water, the aberration should be 3 times larger than if it were filled with air. This experiment was first tried by the English astronomer George Airy in 1871 and no change in the aberration angle was detected.” The question is Why? Why no change was detected?
LikeLike
Apologies for not coming back you earlier, Over the last 12 months I have been focussing on building up my YouTube channel ( and I have not have had the time to review all comment on my blog.
IF Bradley’s explanation were correct (which it isn’t although it gives very accurate agreement with the results when the telescope is filled with a medium of low refractive index such as air) the formula tan(ϴ) = -v/c sin A, would mean that the telescope would have to be tilted at a greater angle if filled with water, because of the lower value of the speed of light in water
However, Bradley’s explanation is not correct, the “correct explanation” is given by Einstein’s theory of special relativity, which is little more complicated. I will write a post about this sometime, but in the meantime, I have put a link on my references pages ( )to a good paper which gives the relativistic explanation of stellar aberration
LikeLike
8. Thank you very much for your vivid, concise and understandable description of the situation of stellar parallax / stellar aberration.
LikeLike
Thank you and thank you your postive email about my video
BTW I am currently doing a video about special relativity,
LikeLike
9. I was sort of curious about the algebra of the expressions for A – A’ where you employ the difference of tangents. Is there any way you could show the steps for how you got the result? I have tried a few times now and am missing something
LikeLike
Thanks for your comment
I assume that you mean that one or more of the steps under the phrase
“We want to find the displacement of the star due to the Earth’s motion. If we give this the symbol θ, then θ = A’ – A”.
If you can indicate which of these steps are unclear I will add some additional explanation
Steve
LikeLike
10. yes, indeed – and nearly 300 years on!
I checked through the University notes I have and can’t find it mentioned there either. I will post a note to them about it.
LikeLike
11. Although familiar with stellar parallax, I hadn’t appreciated stellar aberration until I saw your article. The final section, with the last picture and 4 bullet points were very helpful; I then went back to the maths and it all “clicked”.
LikeLiked by 1 person
Thank you for your comment Barbara,
It is interesting that although stellar aberration causes a much larger moving in stars’ positions than parallax, and was important in the development of modern astronomy, knowledge of it is much less common than parallax.
LikeLike
12. […] Source link […]
LikeLike
13. Sounds great fun!
Do you happen to remember what value you got ?
LikeLike
14. I remember calculating the speed of light using Bradley’s method on Optics 101. The bonus points were for calculating the aberration, I believe…a lot of fun, indeed.
LikeLike
15. […] Source link […]
LikeLike
16. Cool despite not really having a clue what you’re on about 🙂
LikeLiked by 1 person
LikeLike
Leave a comment Cancel reply
Comment
Reblog
Subscribe
Subscribed
Explaining Science Already have a WordPress.com account? Log in now.
Explaining Science
Subscribe
Subscribed
Sign up
Log in
Copy shortlink
Report this content
View post in Reader
Manage subscriptions
Collapse this bar
Original text
Rate this translation
Your feedback will be used to help improve Google Translate |
9930 | https://www.expii.com/t/parallel-transversal-angle-relationships-5513 | Parallel-Transversal Angle Relationships - Expii
Expii
Parallel-Transversal Angle Relationships - Expii
Using common angle pairs, you can find and relate the different angles formed by parallel lines and transversals. As an application, we could prove that the sum of angles in a triangle is 180 degrees.
HomeLog inSign up
Search for math and science topics
Search for topics
Log InSign Up
GeometryTransversal Proofs in Geometry
Parallel-Transversal Angle Relationships
Using common angle pairs, you can find and relate the different angles formed by parallel lines and transversals. As an application, we could prove that the sum of angles in a triangle is 180 degrees.
Parallel-Transversal Angle Relationships
Go to Topic
Explanations (1)
DAN THE MAN
Text
1
So parallel lines and transversals. When a tranversal intersects two parallel lines, 8 angles are formed. There are 4 types of angles formed. One is corresponding angles, angles on the same side of the transversal, but are not adjacent or both exterior/ both interior. One must be exterior and the other interior for the angles to be corresponding. In parallel lines, these two angles will be congruent. Another pair of angles formed is alternate interior. There are the angles within the parallel lines, but are on opposite sides of the transversal. In parallel lines, these angles will also be congruent. Another is consecutive interior angles. These are angles within the parallel lines, but are also on the same side of the transversal. These angles will be supplementary in parallel lines. The final types of angles formed are the alternate exterior angles. These are angles formed on the outside of the parallel lines, that are on opposite sides of the transversal. These will be congruent in parallel lines.
ReportShare1 Like
Related Lessons
Parallel Lines Create Congruent Angles
Rotation Around a Point
Congruent Angles Prove Parallelism
Reflection over a Line
View All Related Lessons
You've reached the end
TOP
ABOUT US
AboutJobsContact
EXPLORE
Daily ChallengeSolveModerate Zoom ChatLive Math Courses
INITIATIVES
SparkRamanujanShowcase
INFO
Terms of ServiceCode of ConductPrivacy NoticeLabor Condition Applications
© 2019 Expii, Inc.
How can we improve?
close
Feedback Type
General
Bug
Feature
Message Email (optional) Send
Send Feedback |
9931 | https://www.tiger-algebra.com/drill/3x_4y_5z=35;2x_5y_3z=27;2x_y_z=13/ | Copyright Ⓒ 2013-2025
tiger-algebra.com
This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here.
Solution - Linear equations with three unknowns
Other Ways to Solve
Step by Step Solution
System of Linear Equations entered :
Solve by Substitution :
// Solve equation for the variable z
// Plug this in for variable z in equation
// Plug this in for variable z in equation
// Solve equation for the variable y
// Plug this in for variable y in equation
// Solve equation for the variable x
// By now we know this much :
// Use the x value to solve for y
// Use the x and y values to solve for z
z = -2(4)-(2)+13 = 3
Solution :
How did we do?
Why learn this
Terms and topics
Related links
Latest Related Drills Solved
Copyright Ⓒ 2013-2025
tiger-algebra.com |
9932 | https://study.com/skill/learn/using-the-area-of-similar-triangles-theorem-to-solve-for-area-explanation.html | Using the Area of Similar Triangles Theorem to Solve for Area | Geometry | Study.com
Log In
Sign Up
Menu
Plans
Courses
By Subject
College Courses
High School Courses
Middle School Courses
Elementary School Courses
By Subject
Arts
Business
Computer Science
Education & Teaching
English (ELA)
Foreign Language
Health & Medicine
History
Humanities
Math
Psychology
Science
Social Science
Subjects
Art
Business
Computer Science
Education & Teaching
English
Health & Medicine
History
Humanities
Math
Psychology
Science
Social Science
Art
Architecture
Art History
Design
Performing Arts
Visual Arts
Business
Accounting
Business Administration
Business Communication
Business Ethics
Business Intelligence
Business Law
Economics
Finance
Healthcare Administration
Human Resources
Information Technology
International Business
Operations Management
Real Estate
Sales & Marketing
Computer Science
Computer Engineering
Computer Programming
Cybersecurity
Data Science
Software
Education & Teaching
Education Law & Policy
Pedagogy & Teaching Strategies
Special & Specialized Education
Student Support in Education
Teaching English Language Learners
English
Grammar
Literature
Public Speaking
Reading
Vocabulary
Writing & Composition
Health & Medicine
Counseling & Therapy
Health
Medicine
Nursing
Nutrition
History
US History
World History
Humanities
Communication
Ethics
Foreign Languages
Philosophy
Religious Studies
Math
Algebra
Basic Math
Calculus
Geometry
Statistics
Trigonometry
Psychology
Clinical & Abnormal Psychology
Cognitive Science
Developmental Psychology
Educational Psychology
Organizational Psychology
Social Psychology
Science
Anatomy & Physiology
Astronomy
Biology
Chemistry
Earth Science
Engineering
Environmental Science
Physics
Scientific Research
Social Science
Anthropology
Criminal Justice
Geography
Law
Linguistics
Political Science
Sociology
Teachers
Teacher Certification
Teaching Resources and Curriculum
Skills Practice
Lesson Plans
Teacher Professional Development
For schools & districts
Certifications
Teacher Certification Exams
Nursing Exams
Real Estate Exams
Military Exams
Finance Exams
Human Resources Exams
Counseling & Social Work Exams
Allied Health & Medicine Exams
All Test Prep
Teacher Certification Exams
Praxis Test Prep
FTCE Test Prep
TExES Test Prep
CSET & CBEST Test Prep
All Teacher Certification Test Prep
Nursing Exams
NCLEX Test Prep
TEAS Test Prep
HESI Test Prep
All Nursing Test Prep
Real Estate Exams
Real Estate Sales
Real Estate Brokers
Real Estate Appraisals
All Real Estate Test Prep
Military Exams
ASVAB Test Prep
AFOQT Test Prep
All Military Test Prep
Finance Exams
SIE Test Prep
Series 6 Test Prep
Series 65 Test Prep
Series 66 Test Prep
Series 7 Test Prep
CPP Test Prep
CMA Test Prep
All Finance Test Prep
Human Resources Exams
SHRM Test Prep
PHR Test Prep
aPHR Test Prep
PHRi Test Prep
SPHR Test Prep
All HR Test Prep
Counseling & Social Work Exams
NCE Test Prep
NCMHCE Test Prep
CPCE Test Prep
ASWB Test Prep
CRC Test Prep
All Counseling & Social Work Test Prep
Allied Health & Medicine Exams
ASCP Test Prep
CNA Test Prep
CNS Test Prep
All Medical Test Prep
College Degrees
College Credit Courses
Partner Schools
Success Stories
Earn credit
Sign Up
Copyright
Using the Area of Similar Triangles Theorem to Solve for Area
Florida Math Standards (MAFS) - Geometry Skills Practice
Click for sound
6:50
You must c C reate an account to continue watching
Register to access this and thousands of other videos
Are you a student or a teacher?
I am a student I am a teacher
Try Study.com, risk-free
As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed.
Get unlimited access to over 88,000 lessons.
Try it risk-free
It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time.
Already registered? Log in here for access
Back
What teachers are saying about Study.com
Try it risk-free for 30 days
Already registered? Log in here for access
00:04 Using the area of…
02:44 Using the area of…
05:02 Using the area of…
Jump to a specific example
Speed Normal
0.5x
Normal
1.25x
1.5x
1.75x
2x
Speed
Lalitha Kannan, David Harmon
Instructors Lalitha Kannan
Lalitha Kannan has taught math at all grade levels for over 15 years. She has a Master's degree in Mathematics degree from the University of Madras and a Master's degree in Computer Science from Oakland University.
View bio
David Harmon
David Harmon has taught middle school math for over 12 years. They have a B.A. in Middle Grades Education in Math and Science from University of North Carolina at Wilmington. They also have a certification in Special Education.
View bio
Example SolutionsPractice Questions
Steps to Solve for Area Using the Area of Similar Triangles Theorem
Step 1: Measure the corresponding sides of the similar triangles.
Square the ratio of the corresponding sides of both the triangles.
Step 2: Take the ratio of the areas of both the triangles. Use Area of Similar Triangles theorem and
equate the ratio of areas of both the triangles to the square of the ratio of the corresponding sides of the two triangles.
Step 3: Solve for the missing area of the triangle.
Equations & Definitions to Solve for Area Using the Area of Similar Triangles Theorem
Similar Triangles: Two triangles are similar if all the corresponding sides are proportional to each other, and
their corresponding angles are also congruent.
Area of Similar Triangles Theorem: Given two similar triangles, the ratio of the areas of the triangles is equal to
the square of the ratio of any of their corresponding sides.
Formula for Area of Similar Triangles Theorem: Given two similar triangles Δ A B C and Δ D E F , then the
A r e a o f Δ A B C A r e a o f Δ D E F=(A B D E)2=(B C E F)2=(A C D F)2
Example Problem 1: Solve for Area Using the Area of Similar Triangles Theorem
Examine the figure given below and solve for the area of Δ A B C.
Area of Similar Triangles
Step 1: Measure the corresponding sides of the similar triangles. Square the ratio of the
corresponding sides of both the triangles.
From the diagram above, we see that
In Δ A B C
A C=20 m
and in Δ D E F , the corresponding side
D F=10 m
Ratio of A C and D F is given by
A C D F=20 10=2
Square the ratio (A C D F)2=2 2=4
Step 2: Take the ratio of the areas of both the triangles. Use Area of Similar Triangles theorem and
equate the ratio of areas of both the triangles to the square of the ratio of the corresponding sides of the two triangles.
From the figure, we can see that the area of Δ D E F is given as
A r e a o f Δ D E F=15 m 2
A r e a o f Δ A B C is missing. Let the missing area be X
Ratio of A r e a o f Δ A B C a n d A r e a o f Δ D E F is given by
A r e a o f Δ A B C A r e a o f Δ D E F=X 15
Using Area of Similar Triangles Theorem we can equate
A r e a o f Δ A B C A r e a o f Δ D E F=(A C D F)2
Step 3: Solve for X , the missing area of the triangle Δ A B C.
From the above steps, we see that
X 15=4 X=15×4=60 m 2
Therefore, the area of the triangle Δ A B C=60 m 2
Example Problem 2: Solve for Area Using the Area of Similar Triangles Theorem
Examine the figure given below and solve for the area of Δ D E F.
Area of Similar Triangles
Step 1: Measure the corresponding sides of the similar triangles. Square the ratio of the
corresponding sides of both the triangles.
From the diagram above, we see that
In Δ A B C
A C=15 c m
and in Δ D E F , the corresponding side
D F=3 c m
Ratio of A C and D F is given by
A C D F=15 3=5
Square the ratio (A C D F)2=5 2=25
Step 2: Take the ratio of the areas of both the triangles. Use Area of Similar Triangles theorem and
equate the ratio of areas of both the triangles to the square of the ratio of the corresponding sides of the two triangles.
From the figure, we can see that the area of Δ A B C is given as
A r e a o f Δ A B C=187.5 c m 2
A r e a o f Δ D E F is missing. Let the area be X
Ratio of A r e a o f Δ A B C a n d A r e a o f Δ D E F is given by
A r e a o f Δ A B C A r e a o f Δ D E F=187.5 X
Using Area of Similar Triangles Theorem we can equate
A r e a o f Δ A B C A r e a o f Δ D E F=(A C D F)2
Step 3: Solve for X, the missing area of the triangle Δ D E F.
From the above steps, we see that
187.5 X=25 25 X=187.5 X=187.5 25=7.5 c m 2
Therefore, the missing area of the triangle Δ D E F=7.5 c m 2.
Get access to thousands of practice questions and explanations!
Create an account
Table of Contents
Steps to Solve for Area Using the Area of Similar Triangles Theorem
Equations & Definitions to Solve for Area Using the Area of Similar Triangles Theorem
Example Problem 1
Example Problem 2
Test your current knowledge
Practice
Using the Area of Similar Triangles Theorem to Solve for Area
Recently updated on Study.com
Videos
Courses
Lessons
Articles
Quizzes
Concepts
Teacher Resources
The First & Second Balkan Wars | Background & Consequences
Ethnic Groups in Indonesia | Demographics & People
Gods of the Winter Solstice
Holes by Louis Sachar | Themes, Quotes & Analysis
Aurangzeb | Empire, Achievements & Failures
Libya Ethnic Groups | Demographics, Population & Cultures
Cold War Lesson for Kids: Facts & Timeline
The Chronicles of Narnia Series by C.S. Lewis | Overview...
Isotherms Definition, Maps & Types
Physical Development | Overview & Examples
Introduction to Renaissance Literature: Characterizing...
General Studies Literature: Help & Review
GACE 713 Study Guide - English (6-12) Exam Prep
GACE 711 Study Guide - Mathematics (6-12) Exam Prep
GACE 706 Study Guide - Middle Grades Language Arts (4-8)...
GACE 704 Study Guide - Middle Grades Math (4-8) Exam Prep
GACE 735 Study Guide - Special Education General...
Physics 112L: Physics II with Lab
Science 101L: Intro to Life Sciences with Lab
Pennsylvania Biology Keystone Study Guide and Exam Prep
NES Elementary Education Subtest II (103) Study Guide and...
Essentials of Organizational Behavior
AQA A-level Anthropology: Practice & Study Guide
Student Resources
Praxis 5531 Study Guide - Pre-Kindergarten Education Exam...
Instructional Strategies for Teaching Online Courses
GACE History Test II (035) Study Guide and Test Prep
Time Management for Sales Professionals
How to be an Effective Manager
Types of Curriculum | Null Curriculum
Sedation Definition, Levels & Drugs
Rock in Performing Arts | History, Characteristics & Facts
Reproductive Endocrinology Meaning & Specialization
Drug Rehabilitation Types, Effects & Importance
Cardiology Definition, Tests & Specializations
Neurology | Definition, Importance & Disorders
Anaerobic Bacteria Definition, Types & Examples
Urban Renewal: Definition, Projects & Strategy
Surface Area Affecting Friction | Equation, Process &...
Squanto | Biography, Facts & Thanksgiving
Finding the Inverse of a 3x3 Matrix | Overview & Formula
Collective Security vs. Self-Help in Resolving...
Jainism vs. Islam
Oceania Theory: Overview
History of the Americas | Timeline & Facts
Aboriginal Dance: History & Ceremonies
Clark v. Arizona | Case Brief & Famous Diminished...
How to Use & Cite AI Tools in College Saver Course...
Understanding Generative AI as a Student: Uses, Benefits...
WEST Prep Product Comparison
What Are the Features of My Institutional Student Account...
How to Pass the SLLA Exam
How to Pass the TExES Exams
How to Activate a Study.com Group Plan Account
Multiplication Math Games
Math Card Games for Kids
Registering for the C1 Advanced & What to Bring
Can You Use a Calculator on the GMAT?
Quiz & Worksheet - Legacy of the Neo-Babylonian Empire
Quiz & Worksheet - Aslan in The Lion, the Witch & the...
Quiz & Worksheet - The Mule in Their Eyes Were Watching God
Quiz & Worksheet - Who Dies in Bridge to Terabithia?
Quiz & Worksheet - Barbados Ethnic Groups
Quiz & Worksheet - Periodic Functions
Quiz & Worksheet - Sensory & Motor Changes in Late Adulthood
Math
Social Sciences
Science
Business
Humanities
Education
Art and Design
History
Tech and Engineering
Health and Medicine
Plans
Study help
Test prep
College credit
Teacher resources
Working Scholars®
School group plans
Online tutoring
About us
Blog
Careers
Teach for us
Press Center
Ambassador
Scholarships
Support
Contact support
FAQ
Site feedback
Resources and Guides
Download the app
Study.com on Facebook Study.com on YouTube Study.com on Instagram Study.com on Twitter Study.com on LinkedIn
© Copyright 2025 Study.com. All other trademarks and copyrights are the property of their respective owners. All rights reserved.
Contact us by phone at (877)266-4919, or by mail at 100 View Street#202, Mountain View, CA 94041.
About Us
Terms of Use
Privacy Policy
DMCA Notice
ADA Compliance
Honor Code For Students
Support
× |
9933 | https://www.lastminutelecture.com/2025/06/thermal-properties-materials-chapter-19-callister-materials-science.html | Thermal Properties of Materials — Chapter 19 Summary from Callister’s Materials Science and Engineering
Skip to main content
Search This Blog
Last Minute Lecture
Thermal Properties of Materials — Chapter 19 Summary from Callister’s Materials Science and Engineering
Get link
Facebook
X
Pinterest
Email
Other Apps
June 08, 2025
Thermal Properties of Materials — Chapter 19 Summary from Callister’s Materials Science and Engineering
Chapter 19 of Materials Science and Engineering by William D. Callister, Jr. and David G. Rethwisch explores the fundamental ways materials respond to heat, focusing on critical topics such as heat capacity, thermal expansion, thermal conductivity, and the development of thermal stresses. These properties are crucial for engineering safe, durable, and efficient products across industries.
Watch the full podcast-style summary below, and subscribe to Last Minute Lecture for clear, chapter-by-chapter study guides to foundational engineering textbooks!
Fundamental Thermal Properties
Heat Capacity (C): The quantity of heat required to change a material’s temperature, often measured as specific heat (c) per unit mass. In solids, heat is absorbed mainly through atomic vibrations (phonons), with additional minor contributions from electronic and magnetic effects under certain conditions.
Thermal Expansion: Most materials expand upon heating and contract when cooled. This is described by the linear coefficient of thermal expansion (αl) and the volume coefficient (αv). Materials with stronger atomic bonds generally have lower thermal expansion.
Thermal Conductivity (k): A measure of a material’s ability to transfer heat. In metals, free electrons are the primary heat carriers; in ceramics and polymers, phonons (quantized lattice vibrations) are dominant.
Thermal Stresses: Develop when thermal expansion is constrained or non-uniform, potentially leading to cracking or failure—especially relevant for brittle materials like ceramics.
Thermal Behavior by Material Class
Metals: High thermal conductivity (20–400 W/m·K) due to free electrons, moderate thermal expansion (5–25 × 10⁻⁶ (°C)⁻¹). Specialized alloys like Invar and Kovar are engineered for minimal expansion.
Ceramics: Low thermal expansion (0.5–15 × 10⁻⁶ (°C)⁻¹) and moderate thermal conductivity (2–50 W/m·K). Their resistance to thermal shock is a function of low expansion, higher conductivity, and fracture strength.
Polymers: Low thermal conductivity (~0.3 W/m·K), but high thermal expansion (50–400 × 10⁻⁶ (°C)⁻¹). Performance can be improved with crosslinking or foaming to reduce expansion and enhance insulation.
Key Mechanisms and Engineering Strategies
Heat Capacity Mechanisms: Vibrational energy (phonons) dominates, with electronic/magnetic effects only important in special cases. The Debye temperature (ΘD) marks where heat capacity levels off.
Thermal Expansion Mechanism: Asymmetric interatomic potential-energy curves cause atoms to shift further apart as temperature rises; weaker bonds lead to greater expansion.
Thermal Conductivity:
Metals: Electron movement dominates; described by the Wiedemann–Franz Law linking thermal and electrical conductivities.
Ceramics: Phonon transport is hindered by defects and scattering.
Polymers: Transfer heat via molecular vibrations and rotations, with more crystalline polymers conducting heat better.
Thermal Stress Management: Minimized by selecting materials with compatible expansion, improving design to avoid constraint, and using heat treatments like annealing to relieve internal stresses.
Glossary of Key Terms
Debye Temperature (ΘD): The temperature above which a material's heat capacity becomes nearly independent of temperature.
Phonon: Quantum of vibrational energy in a solid, analogous to a photon in electromagnetic waves.
Specific Heat (c): Heat capacity per unit mass.
Thermal Shock Resistance (TSR): Ability of a material to withstand sudden temperature changes without cracking, estimated as TSR ∝ σf k / (E αl).
Thermal Stress (σ): Stress induced by temperature changes, calculated by σ = EαlΔT.
Pilling-Bedworth Ratio: Ratio that predicts whether oxide layers will protect metals from further oxidation.
Conclusion: Managing Heat for Material Reliability
Understanding thermal properties is essential for material selection, design, and failure prevention in engineering systems subject to temperature fluctuations. Effective management of heat capacity, expansion, and stresses ensures reliability in everything from electronics to engines to construction. For a thorough and accessible explanation, watch the podcast above and subscribe to Last Minute Lecture for expertly summarized textbook chapters.
If you found this breakdown helpful, be sure to subscribe to Last Minute Lecture for more chapter-by-chapter textbook summaries and academic study guides.
callisterheat capacitymaterials sciencestudy guidethermal conductivitythermal expansionthermal propertiesthermal shockthermal stresses
Get link
Facebook
X
Pinterest
Email
Other Apps
Comments
Post a Comment
Popular posts from this blog
Behavior Therapies & Evidence-Based Practice — Chapter 9 Summary from Systems of Psychotherapy
June 07, 2025
Behavior Therapies & Evidence-Based Practice — Chapter 9 Summary from Systems of Psychotherapy Chapter 9 of Systems of Psychotherapy: A Transtheoretical Analysis explores the robust field of behavior therapies, which focus on changing maladaptive behaviors through practical, empirical methods. These approaches, grounded in learning theory and continuous assessment, have become foundational in treating a range of psychological conditions. This expanded summary, based on our podcast-style breakdown of Chapter 9 , highlights the key techniques, theoretical foundations, and clinical applications of behavior therapy. Want a concise video guide? Watch the chapter summary above and subscribe to Last Minute Lecture for more psychology study resources! The Foundations of Behavior Therapy Behavior therapy focuses on observable, measurable changes in behavior rather than probing unconscious processes. Using empirical methods, therapists continuously assess and adjust treatmen...
Read more
Cognitive & Rational-Emotive Therapies — Chapter 10 Summary from Systems of Psychotherapy
June 07, 2025
Cognitive & Rational-Emotive Therapies — Chapter 10 Summary from Systems of Psychotherapy Chapter 10 of Systems of Psychotherapy: A Transtheoretical Analysis explores the evolution and techniques of Cognitive Therapy (CT) and Rational-Emotive Behavior Therapy (REBT). Developed by Aaron Beck and Albert Ellis, these therapies target the root of emotional distress by challenging and changing maladaptive thought patterns. This in-depth summary, based on our podcast breakdown of Chapter 10 , explains the major concepts, therapeutic techniques, and clinical impact of these cognitive-based approaches. Watch the chapter summary above and subscribe to Last Minute Lecture for expert textbook guides and psychology study resources! The Foundations of Cognitive Therapy & REBT Cognitive therapies emerged in response to limitations of psychoanalysis and behaviorism, emphasizing the role of thoughts in shaping emotions and behaviors. Aaron Beck’s Cognitive Therapy and Albert E...
Read more
A Day in the Life of Adam and Eve — Hunter-Gatherer Society, Diet, and Spirituality Explained | Chapter 3 from Sapiens by Yuval Noah Harari
July 04, 2025
A Day in the Life of Adam and Eve — Hunter-Gatherer Society, Diet, and Spirituality Explained | Chapter 3 from Sapiens by Yuval Noah Harari What was daily life really like for our prehistoric ancestors? In Chapter 3 of Sapiens: A Brief History of Humankind , Yuval Noah Harari takes us inside the complex, often misunderstood world of hunter-gatherer societies. To fully understand our modern minds and social habits, Harari argues, we must examine the foraging communities in which Homo sapiens evolved for 95% of our history. Watch the full video summary below, and subscribe to Last Minute Lecture for more insightful breakdowns of Sapiens and other essential books! The Forager Legacy: How Hunter-Gatherers Shaped Humanity For most of human existence, people lived in small, mobile groups—relying on hunting, gathering, and foraging. Harari dispels the myth that these societies were either idyllic Edens or brutal battlegrounds. Instead, they were diverse, adaptable, and surpri...
Read more
Powered by Blogger
Theme images by Veronica Olson
Request a Book
Last Minute Lecture
I'm the creator of Last Minute Lecture — a blog and YouTube channel dedicated to helping students, educators, and lifelong learners better understand academic concepts, one chapter at a time. Our content focuses on detailed, easy-to-digest summaries of major college textbooks in psychology, biology, medicine, and related fields. Whether you're preparing for exams, trying to catch up on reading, or just love to learn, this space is designed to simplify complex ideas and support academic success. Every blog post is paired with a YouTube video that brings the chapter to life through podcast-style narration and visual reinforcement. I'm passionate about making high-quality educational content free and accessible for everyone. With a background in psychology and science communication, I created this platform to help others study smarter, learn faster, and stay motivated. Thanks for being here — let's keep learning.
Visit profile
All Books
Atomic Habits
Brock Biology of Microorganisms
Campbell Biology – Chapter Summaries
Certification & Recertification Exam Review for Physician Assistants
Child Psychopathology
Clinical Psychopharmacology
Cognitive Psychology
Developing Person Through the Life Span
Dopamine Nation
Earth: Portrait of a Planet
Essentials of Abnormal Psychology
Financial Accounting
Introduction to Algorithms
Learn Computer Forensics
Man and His Symbols
Materials Science and Engineering
Neuroscience For Dummies
Personality Psychology
Precarious Life
Principles of Microeconomics
Principles of Psychology
Research Methods for the Behavioral Sciences
Sapiens
Saunders NCLEX-PN Review
Scattered Minds
Systems of Psychotherapy A Transtheoretical Analysis
The Mountain Is You
Think and Grow Rich
The 48 Laws of Power
University Physics
Archive
September 2025 16
August 2025 65
July 2025 98
June 2025 445
May 2025 53 |
9934 | https://pubmed.ncbi.nlm.nih.gov/35073425/ | The role of dermoscopy in the diagnosis of subungual glomus tumors - PubMed
Clipboard, Search History, and several other advanced features are temporarily unavailable.
Skip to main page content
An official website of the United States government
Here's how you know
The .gov means it’s official.
Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site.
The site is secure.
The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely.
Log inShow account info
Close
Account
Logged in as:
username
Dashboard
Publications
Account settings
Log out
Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation
Search: Search
AdvancedClipboard
User Guide
Save Email
Send to
Clipboard
My Bibliography
Collections
Citation manager
Display options
Display options
Format
Save citation to file
Format:
Create file Cancel
Email citation
Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page.
To:
Subject:
Body:
Format:
[x] MeSH and other data
Send email Cancel
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Add to My Bibliography
My Bibliography
Unable to load your delegates due to an error
Please try again
Add Cancel
Your saved search
Name of saved search:
Search terms:
Test search terms
Would you like email updates of new search results? Saved Search Alert Radio Buttons
Yes
No
Email: (change)
Frequency:
Which day?
Which day?
Report format:
Send at most:
[x] Send even when there aren't any new results
Optional text in email:
Save Cancel
Create a file for external citation management software
Create file Cancel
Your RSS Feed
Name of RSS Feed:
Number of items displayed:
Create RSS Cancel
RSS Link Copy
Full text links
Wiley
Full text links
Actions
Cite
Collections
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Permalink
Permalink
Copy
Display options
Display options
Format
Page navigation
Title & authors
Abstract
Similar articles
Cited by
References
MeSH terms
Related information
LinkOut - more resources
Int J Dermatol
Actions
Search in PubMed
Search in NLM Catalog
Add to Search
. 2022 Jul;61(7):826-832.
doi: 10.1111/ijd.16042. Epub 2022 Jan 24.
The role of dermoscopy in the diagnosis of subungual glomus tumors
Fatih Göktay1,Gaye Güldiken1,Zeynep Altan Ferhatoğlu2,Pembegül Güneş3,Güldehan Atış1,Eckart Haneke456
Affiliations Expand
Affiliations
1 Department of Dermatology, Hamidiye Medical Faculty, Haydarpaşa Numune Training and Research Hospital, University of Health Sciences Turkey, İstanbul, Turkey.
2 Department of Dermatology, Cerrahpaşa Medical Faculty, İstanbul University-Cerrahpaşa, İstanbul, Turkey.
3 Department of Pathology, Hamidiye Medical Faculty, Haydarpaşa Numune Training and Research Hospital, University of Health Sciences Turkey, İstanbul, Turkey.
4 Department of Dermatology, Inselspital, Bern University Hospital, University of Bern, Bern, Switzerland.
5 Centro de Dermatología Epidermis, Instituto CUF, Porto, Portugal.
6 Dermatology Practice Dermaticum, Freiburg, Germany.
PMID: 35073425
DOI: 10.1111/ijd.16042
Item in Clipboard
The role of dermoscopy in the diagnosis of subungual glomus tumors
Fatih Göktay et al. Int J Dermatol.2022 Jul.
Show details
Display options
Display options
Format
Int J Dermatol
Actions
Search in PubMed
Search in NLM Catalog
Add to Search
. 2022 Jul;61(7):826-832.
doi: 10.1111/ijd.16042. Epub 2022 Jan 24.
Authors
Fatih Göktay1,Gaye Güldiken1,Zeynep Altan Ferhatoğlu2,Pembegül Güneş3,Güldehan Atış1,Eckart Haneke456
Affiliations
1 Department of Dermatology, Hamidiye Medical Faculty, Haydarpaşa Numune Training and Research Hospital, University of Health Sciences Turkey, İstanbul, Turkey.
2 Department of Dermatology, Cerrahpaşa Medical Faculty, İstanbul University-Cerrahpaşa, İstanbul, Turkey.
3 Department of Pathology, Hamidiye Medical Faculty, Haydarpaşa Numune Training and Research Hospital, University of Health Sciences Turkey, İstanbul, Turkey.
4 Department of Dermatology, Inselspital, Bern University Hospital, University of Bern, Bern, Switzerland.
5 Centro de Dermatología Epidermis, Instituto CUF, Porto, Portugal.
6 Dermatology Practice Dermaticum, Freiburg, Germany.
PMID: 35073425
DOI: 10.1111/ijd.16042
Item in Clipboard
Full text links Cite
Display options
Display options
Format
Abstract
Background and objectives: The dermoscopic features of glomus tumors have only been described in a few case reports. The aim of this research was to define the clinical and dermoscopic features of subungual glomus tumors.
Methods: Thirty-two patients with subungual glomus tumors were evaluated retrospectively.
Results: On the photographs, longitudinal erythronychia, longitudinal leukonychia, punctate leukonychia, splinter hemorrhage, isolated capillaries, distal notching, distal subungual hyperkeratosis, onycholysis, and onychoschizia were found. There was no statistical difference between the rates of detection of these findings by evaluation from clinical photographs alone and from both clinical and dermoscopic photographs. While ramified vessels with bluish spots could be detected in only five of 26 cases with bluish spots in their clinical photographs, these ramified vessels were seen in 14 cases in bluish spots in dermoscopic photographs (P = 0.004). Compared to clinical examination, dermoscopy was able to detect blue spots in three more cases. Lesion duration was higher in the cases with ramified vessels (P = 0.018).
Conclusions: Dermoscopy seems to contribute to the clinical examination in displaying only ramified vessels located in bluish spots and in determining the localization of the subungual tumors. The presence of ramified vessel in the bluish spots is strongly related to lesion duration.
Keywords: bluish spot mixed with ramified vessels; dermoscopy; homogenous bluish spot; nail; subungual glomus tumor.
© 2022 the International Society of Dermatology.
PubMed Disclaimer
Similar articles
Dermoscopy of glomus tumour: a cross-sectional study of 86 cases.Álvarez-Salafranca M, Bañuls J, Thomas L, Hirata SH, Argenziano G, Medina C, Lacarrubba F, Del Pozo LJ, Ara M, Zaballos P.Álvarez-Salafranca M, et al.J Eur Acad Dermatol Venereol. 2022 Nov;36(11):2016-2024. doi: 10.1111/jdv.18432. Epub 2022 Aug 5.J Eur Acad Dermatol Venereol. 2022.PMID: 35841303
Clinical and onychoscopic characteristics of subungual glomus tumor: a cross-sectional study.Grover C, Jayasree P, Kaliyadan F.Grover C, et al.Int J Dermatol. 2021 Jun;60(6):693-702. doi: 10.1111/ijd.15358. Epub 2020 Dec 17.Int J Dermatol. 2021.PMID: 33332614 Review.
Clinical, dermoscopic, and pathologic features of onychopapilloma: A review of 47 cases.Tosti A, Schneider SL, Ramirez-Quizon MN, Zaiac M, Miteva M.Tosti A, et al.J Am Acad Dermatol. 2016 Mar;74(3):521-6. doi: 10.1016/j.jaad.2015.08.053. Epub 2015 Oct 27.J Am Acad Dermatol. 2016.PMID: 26518173 Review.
Intraoperative dermoscopy features of subungual squamous cell carcinoma: a study of 53 cases.Carlioz V, Perier-Muzet M, Debarbieux S, Amini-Adle M, Dalle S, Duru G, Thomas L.Carlioz V, et al.Clin Exp Dermatol. 2021 Jan;46(1):82-88. doi: 10.1111/ced.14345. Epub 2020 Sep 12.Clin Exp Dermatol. 2021.PMID: 32569407
Dermoscopy of Onychopapilloma: A Benign Mimic of Subungual Malignancy.Woodie BR, Subhadarshani S.Woodie BR, et al.Cureus. 2025 Jan 27;17(1):e78103. doi: 10.7759/cureus.78103. eCollection 2025 Jan.Cureus. 2025.PMID: 40018485 Free PMC article.
See all similar articles
Cited by
Surgical Management of Difficult Nail Glomus Tumors.Gioia Di Chiacchio N, Di Chiacchio N.Gioia Di Chiacchio N, et al.Skin Appendage Disord. 2024 Aug;10(4):254-261. doi: 10.1159/000538028. Epub 2024 Mar 25.Skin Appendage Disord. 2024.PMID: 39108551 Free PMC article.Review.
Blue Nail Discoloration: Literature Review and Diagnostic Algorithms.Hwang JK, Lipner SR.Hwang JK, et al.Am J Clin Dermatol. 2023 May;24(3):419-441. doi: 10.1007/s40257-023-00768-6. Epub 2023 Mar 27.Am J Clin Dermatol. 2023.PMID: 36971947 Review.
Modified nail folding approach: A novel strategy for the treatment of subungual glomus tumors-A case series study.Ahangar P, Akbaribazm M, Rahimi M, Pirmohamadi H.Ahangar P, et al.Clin Case Rep. 2023 Oct 30;11(11):e8129. doi: 10.1002/ccr3.8129. eCollection 2023 Nov.Clin Case Rep. 2023.PMID: 37915733 Free PMC article.
Nail Unit Arteriovenous Hemangioma Presenting as Longitudinal Erythronychia.Hwang JK, Magro CM, Lipner SR.Hwang JK, et al.Skin Appendage Disord. 2023 Aug;9(4):300-305. doi: 10.1159/000530739. Epub 2023 Jun 16.Skin Appendage Disord. 2023.PMID: 37588479 Free PMC article.
References
Pasch M, Haneke E, Baran R, et al. Tumors of the nail apparatus and adjacent tissues. In: Baran R, de Berker DAR, Holzberg M, Piraccini BM, Richert B, Thomas L, eds. Baran & Dawber’s Diseases of the Nails and their Management, 5th edn. West Sussex: Wiley-Blackwell, 2018: 675-824.
McGrath JA. The structure and function of skin. In: Calonje JE, Brenn T, Lazar A, Billings S, eds. In: McKee’s Pathology of the Skin with Clinical Correlations, 5th edn. China: Elsevier, 2020: 1-35.
Piraccini BM. Nail disorders A practical guide to diagnosis and management, 1st edn. Milano: Springer, 2014: 125-153.
Senhaji G, Gallouj S, El Jouari O, et al. Rare tumor in unusual location - Glomus tumor of the finger pulp (clinical and dermoscopic features): a case report. J Med Case Rep 2018; 12: 196.
Turan H, Uslu E, Erdem H, et al. What is your diagnosis 1? Turkderm - Turk Arch Dermatol Venereol 2015; 49: 232-233.
Show all 27 references
MeSH terms
Dermoscopy
Actions
Search in PubMed
Search in MeSH
Add to Search
Glomus Tumor / diagnostic imaging
Actions
Search in PubMed
Search in MeSH
Add to Search
Humans
Actions
Search in PubMed
Search in MeSH
Add to Search
Nail Diseases / diagnosis
Actions
Search in PubMed
Search in MeSH
Add to Search
Retrospective Studies
Actions
Search in PubMed
Search in MeSH
Add to Search
Skin Neoplasms / diagnostic imaging
Actions
Search in PubMed
Search in MeSH
Add to Search
Skin Neoplasms / pathology
Actions
Search in PubMed
Search in MeSH
Add to Search
Related information
Cited in Books
MedGen
LinkOut - more resources
Full Text Sources
Wiley
Full text links[x]
Wiley
[x]
Cite
Copy Download .nbib.nbib
Format:
Send To
Clipboard
Email
Save
My Bibliography
Collections
Citation Manager
[x]
NCBI Literature Resources
MeSHPMCBookshelfDisclaimer
The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited.
Follow NCBI
Connect with NLM
National Library of Medicine
8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov |
9935 | https://www.statology.org/r-cor-function/ | How to Use cor() to Calculate Correlation Coefficients in R
You can use the cor() function in R to calculate correlation coefficients between variables.
Here are the most common ways to use this function:
Method 1: Calculate Pearson Correlation Coefficient Between Two Variables
Use the Pearson correlation coefficient when calculating the correlation between two continuous variables. (e.g. height and weight)
Method 2: Calculate Pearson Correlation Coefficient Between All Numeric Variables in Data Frame
This method will return a correlation matrix that contains the Pearson correlation coefficient between each pairwise combination of numeric variables in a data frame.
Method 3: Calculate Spearman Correlation Coefficient Between Two Variables
Use the Spearman correlation coefficient when calculating the correlation between two ranked variables. (e.g. rank of a student’s math exam score vs. rank of their science exam score in a class)
Method 4: Calculate Kendall’s Correlation Coefficient Between Two Variables
Use the Kendall correlation coefficient when when you wish to use Spearman Correlation but the sample size is small and there are many tied ranks.
The following examples show how to use each method in practice with the following data frame in R that shows the number of hours spent studying, number of practice exams taken, and final exam score for eight different students:
Example 1: Calculate Pearson Correlation Coefficient Between Two Variables
The following code shows how to use the cor() function to calculate the Pearson correlation coefficient between the hours and score variables:
The Pearson correlation coefficient between hours and score turns out to be 0.86.
Note that if there are NA values in your data frame, you can use the argument use=’complete.obs’ to only use the rows where there are no NA values:
Example 2: Calculate Pearson Correlation Coefficient Between All Numeric Variables
The following code shows how to use the cor() function to create a correlation matrix that contains the Pearson correlation coefficient between all numeric variables in the data frame:
Here’s how to interpret the output:
Note: The Pearson correlation coefficient between each individual variable and itself is always 1, which is why each value along the diagonal of the correlation matrix is 1.
Example 3: Calculate Spearman Correlation Coefficient Between Two Variables
The following code shows how to use the cor() function to calculate the Spearman correlation coefficient between the hours and prac_exams variables:
The Spearman correlation coefficient between hours and prac_exams turns out to be -.125.
Example 4: Calculate Kendall’s Correlation Coefficient Between Two Variables
The following code shows how to use the cor() function to calculate Kendall’s correlation coefficient between the hours and prac_exams variables:
Kendall’s correlation coefficient between hours and prac_exams turns out to be -.123.
Additional Resources
The following tutorials explain how to perform other common tasks in R:
How to Calculate Rolling Correlation in R
How to Calculate Autocorrelation in R
How to Calculate Partial Correlation in R
Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.
Post navigation
Leave a Reply Cancel reply
Your email address will not be published. Required fields are marked
Comment
Name
Email
Δ
Search
ABOUT STATOLOGY
Statology makes learning statistics easy by explaining topics in simple and straightforward ways. Our team of writers have over 40 years of experience in the fields of Machine Learning, AI and Statistics. Learn more about our team here.
Featured Posts
Statology Study
Statology Study is the ultimate online statistics study guide that helps you study and practice all of the core concepts taught in any elementary statistics course and makes your life so much easier as a student.
Introduction to Statistics Course
Introduction to Statistics is our premier online video course that teaches you all of the topics covered in introductory statistics. Get started with our course today.
You Might Also Like
Join the Statology Community
Sign up to receive Statology's exclusive study resource: 100 practice problems with step-by-step solutions. Plus, get our latest insights, tutorials, and data analysis tips straight to your inbox!
By subscribing you accept Statology's Privacy Policy. |
9936 | https://www.khanacademy.org/science/kn-class-10-science/xf0d7317fb5b753a0:life-processes/xf0d7317fb5b753a0:untitled-492/v/mechanism-of-breathing-and-gas-exchange-in | Mechanism of Breathing and Gas Exchange (video) | Khan Academy
Skip to main content
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked.
Explore
Browse By Standards
Explore Khanmigo
Math: Pre-K - 8th grade
Math: High school & college
Math: Multiple grades
Math: Illustrative Math-aligned
Math: Eureka Math-aligned
Math: Get ready courses
Test prep
Science
Economics
Reading & language arts
Computing
Life skills
Social studies
Partner courses
Khan for educators
Select a category to view its courses
Search
AI for Teachers FreeDonateLog inSign up
Search for courses, skills, and videos
Help us do more
We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever.
Select gift frequency
One time
Recurring
Monthly
Yearly
Select amount
$10
$20
$30
$40
Other
Give now
By donating, you agree to our terms of service and privacy policy.
Skip to lesson content
KA Science Class 10
Course: KA Science Class 10>Unit 5
Lesson 3: Respiration
Aerobic & anaerobic respiration
Respiration site & ATP
Types of respiration
Meet the lungs
The lungs and pulmonary system
How does the volume of the Lungs change?
Breathing in animals
Fermentation and anaerobic respiration
The respiratory system review
Respiration in Plants and Animals
Respiration in Plants and Animals
Upper Respiratory System
Lower Respiratory System
Human Respiratory System
Mechanism of Breathing and Gas Exchange
Mechanism of Breathing and Gas Exchange
Science>
KA Science Class 10>
Life processes>
Respiration
© 2025 Khan Academy
Terms of usePrivacy PolicyCookie NoticeAccessibility Statement
Mechanism of Breathing and Gas Exchange
Google Classroom
Microsoft Teams
About About this video
How does breathing work in our body? What makes air enter and leave our body? Inside the body, how does oxygen know where to go? Watch this video to understand the mechanism of breathing and gas exchange.Created by Nivedhitha Suresh.
Skip to end of discussions
Questions Tips & Thanks
Want to join the conversation?
Log in
Sort by:
Top Voted
Video transcript
Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube
Up next: exercise
Use of cookies
Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy
Accept All Cookies Strictly Necessary Only
Cookies Settings
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session.
Functional Cookies
[x] Functional Cookies
These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences.
Targeting Cookies
[x] Targeting Cookies
These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service.
Performance Cookies
[x] Performance Cookies
These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
9937 | https://www.wikiwand.com/en/articles/Incenter%E2%80%93excenter_lemma | Incenter–excenter lemma - Wikiwand
English
Sign in
Top Qs
Timeline
Chat
Perspective
Top Qs
Timeline
Chat
Perspective
AllArticlesDictionaryQuotesMap
Incenter–excenter lemma
A statement about properties of inscribed and circumscribed circles From Wikipedia, the free encyclopedia
Discussion (2)
Remove ads
Remove ads
Incenter–excenter lemma
•
•
•
StatementProofApplication to triangle reconstructionGeneralizationSee alsoReferences
In geometry, the incenter–excenter lemma is the theorem that the line segment between the incenter and any excenter of a triangle, or between two excenters, is the diameter of a circle (an incenter–excenter or excenter–excenter circle) also passing through two triangle vertices with its center on the circumcircle. This theorem is best known in Russia, where it is called the trillium theorem (теорема трилистника) or trident lemma (лемма о трезубце), based on the geometric figure's resemblance to a trillium flower or trident; these names have sometimes also been adopted in English.
These relationships arise because the incenter and excenters of any triangle form an orthocentric system whose nine-point circle is the circumcircle of the original triangle. The theorem is helpful for solving competitive Euclidean geometry problems, and can be used to reconstruct a triangle starting from one vertex, the incenter, and the circumcenter.
Remove ads
Statement
incenter–excenter lemma with incenter I and excenter E
Let ABC be an arbitrary triangle. Let I be its incenter and let D be the point where line BI (the angle bisector of ∠ABC) crosses the circumcircle of ABC. Then, the theorem states that D is equidistant from A, C, and I. Equivalently:
The circle through A, C, and I has its center at D. In particular, this implies that the center of this circle lies on the circumcircle.
The three triangles AID, CID, and ACD are isosceles, with D as their apex.
A fourth point E, the excenter of ABC relative to B, also lies at the same distance from D, diametrically opposite from I.
Remove ads
Proof
Summarize
Perspective
By the inscribed angle theorem, ∠I B A=∠D C A,∠I B C=∠D A C.{\displaystyle \angle IBA=\angle DCA,\ \angle IBC=\angle DAC.}
Since B I{\displaystyle BI} is an angle bisector, ∠D C A=∠D A C⟹A D=C D.{\displaystyle \angle DCA=\angle DAC\implies AD=CD.}
We also get
∠D I A=180∘−∠A I B=180∘−(180∘−∠I A B−∠I B A)=∠I A B+∠I B A=∠I A C+∠D A C=∠I A D⟹A D=D I.{\displaystyle {\begin{aligned}\angle DIA&=180^{\circ }-\angle AIB\&=180^{\circ }-(180^{\circ }-\angle IAB-\angle IBA)\&=\angle IAB+\angle IBA\&=\angle IAC+\angle DAC\&=\angle IAD\\implies AD&=DI.\end{aligned}}}
Remove ads
Application to triangle reconstruction
This theorem can be used to reconstruct a triangle starting from the locations only of one vertex, the incenter, and the circumcenter of the triangle. For, let B be the given vertex, I be the incenter, and O be the circumcenter. This information allows the successive construction of:
the circumcircle of the given triangle, as the circle with center O and radius OB,
point D as the intersection of the circumcircle with line BI,
the circle of the theorem, with center D and radius DI, and
vertices A and C as the intersection points of the two circles.
However, for some triples of points B, I, and O, this construction may fail, either because line IB is tangent to the circumcircle or because the two circles do not have two crossing points. It may also produce a triangle for which the given point I is an excenter rather than the incenter. In these cases, there can be no triangle having B as vertex, I as incenter, and O as circumcenter.
Other triangle reconstruction problems, such as the reconstruction of a triangle from a vertex, incenter, and center of its nine-point circle, can be solved by reducing the problem to the case of a vertex, incenter, and circumcenter.
Generalization
Let I and J be any two of the four points given by the incenter and the three excenters of a triangle ABC. Then I and J are collinear with one of the three triangle vertices. The circle with IJ as diameter passes through the other two vertices and is centered on the circumcircle of ABC. When one of I or J is the incenter, this is the trillium theorem, with line IJ as the (internal) angle bisector of one of the triangle's angles. However, it is also true when I and J are both excenters; in this case, line IJ is the external angle bisector of one of the triangle's angles.
Remove ads
See also
Angle bisector theorem
References
Loading content...
Edit in WikipediaRevision historyRead in Wikipedia
Related topics
Triangle Shape with three sides Bisection Division of something into two equal or congruent parts Pedal triangle Triangle found by projecting a point onto the sides of another triangle Nine-point circle Circle constructed from a triangle Orthocenter Intersection of triangle altitudes
Show more
Wikiwand - on chrome
Seamless Wikipedia browsing. On steroids.
Install on Chrome
Remove ads
Wikiwand ❤️ Wikipedia
PrivacyTerms
Remove ads |
9938 | https://www.youtube.com/watch?v=LNsBvs_Wv44 | OpenStax Calculus Exercise 4.1 Problem 19 | Related Rate | Surface area of a sphere
Ardi Satriawan
32900 subscribers
Description
9 views
Posted: 28 Jul 2025
The radius of a sphere decreases at a rate of 3 m/sec. Find the rate at which the surface area decreases when the radius is 10 m.
The radius of a sphere decreases at a rate of 3 m/sec. Find the rate at which the surface area decreases when the radius is 10 m.
The radius of a sphere decreases at a rate of 3 m/sec. Find the rate at which the surface area decreases when the radius is 10 m.
The radius of a sphere decreases at a rate of 3 m/sec. Find the rate at which the surface area decreases when the radius is 10 m.
The radius of a sphere decreases at a rate of 3 m/sec. Find the rate at which the surface area decreases when the radius is 10 m.
Transcript:
The radius of a sphere decreases at a rate of 3 m per second. Find the rate at which the surface area decreases when the radius is 10 m. Suppose that we have a very big ball like this with a radius of 10 m. Okay. So we have the radius here is decreasing at a rate of 3 m per second. So the ball is deflating here. So we will have the rate of the ball deflating is drt and that will equal to 3 m/s. Oke but we need to find out the rate of the surface area decreasing. Oke. What is the surface area of a sphere or a ball? We know that the area of a square that will be 4π multiplied by r square. Oke. But then we need to find out the rate of change of the area or we have it as dt. However, here on the right hand side we only have R here. So let's differentiate DA with respect to R first. But to get DA over DT we multiply this by DR DT which is not a problem because we know drdt. The only problem is we don't know yet but we can calculate that from this function here using power row. So here we will have dt and that will equal to what is the differentiation bit from this function with respect to r I think we will have 4π as a constant and the differentiation of r squ with respect to r is 2r so i have 2r then multiplied by drdt ok we can simplify this write that so have dt is equal to 8π multip by ruled by drdt oke and we have the most important equation in this problem and all that we need to do is to plug in the numerical values when the radius is 10 met so we will need to plug in when R is equal to 10 m. Oke, so we will have dt and then when r is equal to 10 so we will have 8π multip by 10 and then multiply that by drdt and that will be 3. So we will have 240 pi and the unit here will be dr ya that will be met squ per second and this is the exact values but if you want to make it an approximate value with decimal number we can just plug this into calculator. So we will have 240 multiplied by pi. So we will have 753.98. 753.98 and the unit here will be met squ per second and let's highlight that because this two is the final answer for this problem. You can answer it as 240 pi or 753.98. And hopefully I did not make any mistakes during my calculation. Thank you for watching and see you in the next video. |
9939 | https://quantummicrowave.com/product/wr-19-waveguide-to-1-85mm-female-adapter-40-to-60-ghz/ | QMC-WC19-1.85FR385 ==> WR-19 Waveguide to 1.85mm Female Adapter – Quantum Microwave
Skip to content
Millimeter Wave & Microwave Suppliers
Products Available
Online Store
Tech Tools
About
Contacts
General Policies
Cart
Request a Quote
Search for:
Products Available
Online Store
Tech Tools
About
Contacts
General Policies
Cart
Request a Quote
QMC-WC19-1.85FR385 ==> WR-19 Waveguide to 1.85mm Female Adapter
$875.00
In stock
#### Product Overview
#### Request a Quote
#### Product Overview
QMC-WC19-1.85FR385
• Frequency: 40 to 60 GHz
• Max VSWR: 1.35:1
• Connector: V-Female (1.85mm)
• Length: 1.00’’
• Waveguide WR-19: UG-385/U-M (Custom 0.750’’ Ø)
• Copper Waveguide with Brass Flanges
• Finish: Gold Plated
View Datasheet
GO BACK
#### Request a Quote
Your Name (required)
Your Email (required)
Your Phone
Part Number
Quantity
Connector 1 (ex: SMA F)
Connector 2 (ex: SMA M)
Quote Request Please List any other part number(s) & quantity you want included in your quote
×
In stock
QMC-WC19-1.85FR385 ==> WR-19 Waveguide to 1.85mm Female Adapter quantity
Add to cart
SKU: QMC-WC19-1.85FR385 Categories: MMwave Products, QM-Store-Adapters, QM-Waveguide to CoaxTag: WR-19
Related products
### QMC-WB10-H90 ==> WR-10 Waveguide 90 Degree Bend | H-Plane Bend | Gold Plated
$170.00 Add to cartDetails
### WR-15 Waveguide Twist | 90º Degree | QMC-TWST15-90
$140.00 Add to cartDetails
### QMC-MX4-15F13 50 to 75 GHz WR-15 Active Multiplier
$3,575.00 Add to cartDetails
### QMC-AMP12-78151621 71 to 86 GHz WR-12 Power Amplifier Pout +21 dBm
Details
Quantum Microwave
Experts in Cryogenic, Microwave & Millimeter Wave Solutions, when only the best performance matters Quantum Microwave is here for your requirement.
The Lowest Noise Figure, Low Power Consumption, Broadband performance, Small size, Products that deliver both performance and value.
Toggle Navigation
Low Noise Factory
Raytheon BBN
Millimeter Wave Products
Online Store
Contact Us
Request a Quote
Checkout
Copyright © 1995-2020 Quantum Microwave Inc. All Rights Reserved. Privacy Policy
Page load link |
9940 | https://www.youtube.com/watch?v=ouDS2tZU94g | Naming Alkynes - IUPAC Nomenclature & Common Names
The Organic Chemistry Tutor
9880000 subscribers
4864 likes
Description
369092 views
Posted: 29 Apr 2018
This organic chemistry video tutorial provides a basic introduction into naming alkynes using iupac nomenclature and writing the common names of alkynes.
Alkyne Synthesis Test Question:
Organic Chemistry - Video Lessons:
Final Exam and Test Prep Videos:
Alkyne Reactions Review:
Alkene Reaction - Stereochemistry Test Question:
Stereochemistry R/S Configuration:
Optical Activity & Specific Rotation:
SN1, SN2, E1, E2 Reaction Mechanisms:
SN1, SN2, E1, E2 - Practice Test:
Alkene Reactions Review:
Organic Chemistry PDF Worksheets:
Organic Chemistry 1 Exam 2 Playlist:
Organic Chemistry 1 Final Exam Review:
Full-Length Videos and Worksheets:
123 comments
Transcript:
in this video we're gonna focus on naming alkynes so let's begin with the common names how can we name this particular alkyne this is C - H - and its common name is acetylene now this is also called a sign because it's an alkyne so it ends with the suffix yne but the common name is acetylene now let's try another example what do you think the common name is for this molecule so here we have the acetylene portion of the molecule which is the C triple bond C part and notice that we have a methyl group in front so therefore the common name for this alkyne is simply methyl acetylene so based on that example go ahead and try this one so write the common name for this molecule so let's focus on the acetylene portion of the molecule and notice that we have a methyl group on the left and on the right so therefore this is going to be called dimethyl excitedly dimethyl acetylene and so hopefully you're getting the hang of how to write the common names of our kinds now let's move on to our next example which is going to be ch3 ch2 C triple bond CH so go ahead and write the common name for that alkyne and also write thigh you PAC name since we're here already let's go ahead and do that let's start with that you PAC name actually so this is carbon one two three four so four is associated with butane but instead of writing butane we're gonna write butane now the triple bond is on carbon one so this is called one view time now what is the common name of this particular alkyne so we can see that we have our acetylene group and this is an ethyl group attached to it so the common name is going to be ethyl acetylene and so that's it for this example here's another one so for this example right the IU PAC name and also the common name of this particular alkyne so let's start with the IU PAC name we need some number in such a way that the alkyne has the lowest possible number so this is going to carbon 1 2 3 4 5 so that's a pentane but instead of saying pentane it's gonna be pent on now the triple bond is between carbons 2 & 3 so we have to pick the lower of those two numbers so this is going to be called 2 pen time now let's try the common name so here is our acetylene functional group and on the left side we have a methyl attached to it and on the right side we have an ethyl group attached to it so we need to put it in alphabetical order so the common name is ethyl methyl acetylene now for this next example all I want you to do is to write the common name so let's say that we have a benzene ring attached to the carbon-carbon triple bond how can we name it so notice that whenever you have a benzene ring as a substituent it's called phenyl so to name this it's simply going to be phenol acetylene and for the common name you need to write it as one word so if you see me put a space it really should be together as one word so just keep that in mind now let's work on another example so for this one right the common name and the I you pack name as well so let's start with the I you pack name this is going to be carbon one two three four five six so we have a methyl on carbons two and five so it's going to be two five dimethyl and four six carbons that's hexane but we're going to use hex I and the triple bonds on carbon three so it's gonna be - three - hex sign and so that's how you can name this particular our time now what is the common name for this particular molecule so notice that we have an isopropyl group on the left side and on the right side so this is going to be called dye isopropyl acetylene now let's work on one final example what's the common name for in this molecule so we have two phenyl groups attached to the acetylene part so it's simply going to be diphenyl acetylene now let's work on some more examples with our u-pack nomenclature so go ahead and write the IU pack name for this particular alkyne now should we number it from left to right or right to left if we go from right to left the triple bond will be on carbon dream and if we go from left to right the triple bond will still be on carbon three so which way should we go going from left to right we're going to have the name 2-chloro and then we have a methyl group on carbon 5 so it's going to be 5 methyl and we have the alkyne on carbon 3 so that's going to be 3 hex on because we have 6 carbons in the parent chain now if we go in the other direction this will be 1 2 3 4 5 6 so chloro still comes before methyl we still need to put it in alphabetical order but it's gonna be 5 chloro - 2 methyl - 3 hack sign and so the first one is gonna win because you want the first of excuse me the first substituent to have the lower number and so this is the right answer now let's move on to our next example so it's going to be ch3 ch2 - CH with a ch3 attached to it and then we have our carbon-carbon triple bond and another triple bond at the end so go ahead and name that molecule so because we have the chill bond at the end here we want to start counting from right to left we want to give the triple bond the lower values so we have a nine carbon chain that's the longest chain and so that's gonna be non name but instead of nonane is gonna be none nine we do have two alkynes or two triple bonds so it's gonna be a dye I so let's put it all together we have a methyl group on carbon 7 so it's seven methyl and then we have a triple bond by carbon one and five so it's gonna be 1 comma 5 Nana and then dye I because we have two triple bond functional groups and so this is how we could name this particular molecule now what do you do when you have a molecule that contains an alkyne functional group and an alkyne functional group which one has more priority should we place the one at the alkyne or at the alkyne in the situation such as this the alkyne has more priority so we're gonna count given the alkyne the lower number so we have six carbons so this is hexane but it's gonna be hexene instead now the way you name it it's gonna be one heck seen because the alkyne is a carbon one and then - five iron because the triple bond is on carbon five and so that's how we can name a molecule that has an alkene and an alkyne functional group now what if we have an alkyne with an alcohol functional group which one has more priority the alcohol or the alkyne it turns out that that alcohol has a higher priority than the alkyne so we're gonna give the O age group the lower number so this is called 3eu time cuz the alkyne is on carbon 3 and then - 2 all because the alcohol is on carbon 2 and so that's how we can name this molecule now let's try one more example go ahead and name this particular molecule so how should we count it should we count it like this 1 2 3 4 5 6 7 8 or should we count it like this 1 2 3 4 5 6 7 when you count the longest chain it has to include the alkyne functional group so we need to use the numbers in green so this is carbon one two three four five six and seven now we have a propyl group on carbon 3 so this is going to be called 3 propyl - one peptide since we have a seven carbon chain with the alkyne functional group and the triple bond is that carbon one and so that's it for this video so now you know how to name alkynes using IU pack nomenclature and you know how to write the common names for Al Quaeda molecules thanks for watching |
9941 | https://redcliffelabs.com/myhealth/kidney/bun-creatinine-ratio-understanding-its-importance-and-normal-range/ | BUN Creatinine Ratio: Understanding Its Importance & Normal Range
Kidney
BUN Creatinine Ratio: Understanding Its Importance & Normal Range
Medically Reviewed By
Dr. Geetanjali Gupta
Written By Muskan Taneja
on Dec 17, 2024
Last Edit Made By Muskan Taneja
on Aug 16, 2025
share
The kidney is an important organ in the human body. Its key functions include filtration of toxins and waste products.
Any abnormality may lead to kidney disease or kidney failure. A study found that around 17% of Indians have chronic kidney disease (CKD). However, the number varies in every region.
This ratio can be curbed with early detection of kidney functionality. Early detection helps plan effective disease management and treatment.
BUN creatinine ratio is a calculated formula that determines kidneys' functionality in filtering waste products. Today’s blog will provide blood urea nitrogen meaning, its normal range, and BUN creatinine ratio.
Understanding BUN Creatinine Ratio
When your kidney fails to work properly, your healthcare provider may recommend a creatinine test and a blood urea nitrogen (BUN) test. These two tests are essential for monitoring your kidney function. A BUN creatinine ratio is a critical marker for kidney health.
Experts recommend that you should take tests from a reliable diagnostic service provider. Redcliffe Labs is a leading omnichannel PAN India diagnostic laboratory, offering over 3600 quality diagnostic tests at affordable prices. To make diagnostics even more accessible, they provide a home sample collection service. After booking tests, their DMLT-certified phlebotomist will visit your home for sample collection.
They offer the blood urea nitrogen (BUN) and creatinine tests at INR 149 and 140, respectively.
BUN Test
BUN, or blood urea nitrogen test, is a blood test. The test measures the amount of urea in your blood. Urea nitrogen is the waste compound your liver utilizes from the protein in your foods. This protein collapses to create BUN.
A doctor may recommend the BUN test:
For assessing dehydration
For monitoring, medication effects
For evaluating liver function
As a part of a routine health check-up
To check your kidney functioning
For evaluating symptoms of kidney disease
Symptoms of Abnormal BUN Levels
The symptoms of abnormal BUN levels differ depending on whether the level is high or low. Here are some common symptoms related to abnormal BUN levels.
High BUN Levels
Swelling in ankles or legs
Fatigue
Changes in urination
Itchy skin
Nausea and vomiting
Low BUN Levels
Weakness
Swelling or Edema
Nutritional deficiencies
Mental confusion
Gastrointestinal Issues
Creatinine Test
A creatinine test is a simple blood test that measures creatinine levels and assesses kidney functioning. Creatinine is a protein that produces energy for muscle contraction. Your kidney filters it out from the blood and then eliminates it through urine.
Why can your doctor recommend the Creatinine Test?
A doctor can recommend a creatinine test to screen for kidney disease. They can also recommend the test if you have diabetes, high blood pressure, or other medical conditions.
Symptoms of Abnormal Creatinine Levels
The symptoms of abnormal creatinine and BUN levels are similar, which can be associated with other medical conditions.
Swelling
Fatigue
Nausea
Vomiting
Shortness of breath
Also, read
What is the Blood Test BUN Creatinine Ratio?
A blood test BUN creatinine ratio compares blood urea nitrogen (BUN) levels to creatinine in the blood. It is a routine screening test that helps detect kidney disease and other underlying health issues.
BUN test results can be normal, lower, or higher. The low and high ranges indicate underlying health conditions.
The normal range of BUN creatinine ratio is between 10:1 and 20:1. It is a better indicator of kidney health than BUN or creatinine levels alone.
The high BUN ratio indicates congestive heart failure, gastrointestinal bleeding, and dehydration. Meanwhile, the low BUN ratio indicates malnutrition or liver disease.
What is the BUN Creatinine Ratio Calculator?
The BUN creatinine ratio calculator assesses kidney function by comparing the levels of creatinine and BUN in the blood.
Through this method, doctors can determine the efficiency of your kidney to filter out waste products from the blood. The BUN creatinine ratio also helps identify any underlying abnormalities.
The steps to assess the BUN creatinine ratio are simple.
Get a BUN and creatinine blood test and their values in your blood.
Ensure that the measurements of the BUN and creatinine units match. The unit is milligrams per deciliter (mg/dL).
Divide the BUN value by the creatinine value with the BUN creatinine ratio formula.
The formula is:
BUN Creatinine Ratio = BUN (mg/dL) / Creatinine (mg/dL)
For example:
If your creatinine value is 1 mg/dL and BUN value is 20 mg/dL. The calculation would be:
BUN creatinine ratio = 20 mg/dL / 1 mg/dL = 20.
So, your BUN creatinine ratio is 20.
BUN Creatinine Ratio Normal Range
A normal bun creatinine ratio typically falls between 10:1 and 20:1. The BUN creatinine ratio differs by age and gender. Here is a detailed BUN creatinine ratio chart for every age and gender.
| | | |
---
| Blood Urea Nitrogen (BUN) Levels | | |
| Age | Male (mg/dL) | Female (mg/dL) |
| < 1 month | 4-12 | 3-17 |
| 1 month to 11 months | 2-13 | 4-14 |
| 1 year to 3 years | 3-12 | 3-14 |
| 4 years to 19 years | 7-20 | 7-20 |
| 20 years and above | 7-25 | 7-25 |
| Creatinine Levels | | |
| Age | Male (mg/dL) | Female (mg/dL) |
| 2 days or below | 0.79- 1.58 | 0.79- 1.58 |
| 3-27 days | 0.35-1.23 | 0.35-1.23 |
| 1 month to 9 years | 0.20-0.73 | 0.20-0.73 |
| 10 years to 12 years | 0.30-0.78 | 0.30-0.78 |
| 13 years to 15 years | 0.40-1.05 | 0.40-1.00 |
| 16 years to 17 years | 0.60-1.20 | 0.50-1.00 |
| 18 years to 19 years | 0.60-1.26 | 0.50-1.00 |
| 20 years to 49 years | 0.60-1.35 | 0.50-1.10 |
| 50 years to 59 years | 0.70-1.25 | 0.50-0.99 |
| 60 years to 69 years | 0.70-1.25 | 0.50-0.99 |
| 70 years to 79 years | 0.70-1.18 | 0.60-0.93 |
| 80 years or above | 0.70-1.11 | 0.60-0.88 |
Factors Affecting BUN to Creatinine Ratio
The BUN creatinine calculator easily measures the BUN creatinine ratio. However, certain factors before the BUN and creatinine tests can affect the ratio. These include:
Your diet
Medications such as corticosteroids, antibiotics, and diuretics
Liver disease
Sickle cell anemia
Kidney damage
Pregnancy
Metabolic disorders
Rhabdomyolysis
Your age
Hydration level
Causes of BUN Creatinine Ratio Low
A low BUN creatinine ratio can occur due to various health conditions. These include low protein intake, malnutrition, certain conditions affecting muscle mass, overhydration, and liver disease.
Causes of BUN Creatinine Ratio High
A high BUN creatinine indicates kidney damage, disease, or dysfunction. If your BUN creatinine levels are elevated, it can be a sign of low kidney filtration ability or impaired blood loss in the kidney. To clearly understand, your doctor may recommend further diagnostic tests to identify the underlying health issue and plan the management and treatment accordingly.
10 other causes of the high BUN creatinine ratio include:
Stress
Burns
Dehydration
High-protein diet
Pregnancy
Heart attack
Inability to pee because of a blockage in the urinary tract
Gastrointestinal bleeding
Age
Certain medicines, including steroids and antibiotics
Also, read
What Do Low BUN Levels Mean?
Low BUN levels are uncommon and, in most cases, aren't a cause for concern. However, severe health conditions can cause BUN levels to be unusually low.
A low BUN level may be normal during the second or third trimester of pregnancy. They could also indicate:
Overhydration
Liver failure
Severe lack of protein
Malnutrition
What Do High BUN Levels Mean?
A BUN level over 20 mg/dL is considered abnormal and can point to some harmful conditions. But a dangerously high BUN level, often starting at 50 mg/dL, indicates kidney damage that needs immediate medical attention.
A patient with a BUN between 100 mg/dL and 250 mg/dL may have severe kidney dysfunction, resulting from a problem with the organ's ability to filter waste products out of the bloodstream.
A high BUN level typically indicates some dysfunction with a patient's kidneys. This can be caused by:
Certain medications
Dehydration
Urinary tract obstruction
Congestive heart failure
Recent heart attack
Gastrointestinal bleeding
Shock
What level of BUN indicates kidney failure?
Health care providers typically don't rely solely on blood urea nitrogen (BUN) levels to diagnose kidney failure.
However, if a patient's BUN levels exceed their normal baseline and their creatinine levels are elevated, it indicates potential kidney failure.
A BUN level higher than 20 mg/dL suggests that the kidneys may not be functioning at optimal capacity.
Key Takeaways
The BUN creatinine ratio is essential for diagnosing and managing various liver and kidney diseases. The BUN test and creatinine test both equally support kidney health. Take the blood urea nitrogen (BUN) and creatinine tests from Redcliffe Labs. Measure the values and take preventive measures to ensure overall well-being.
Frequently Asked Questions (FAQs)
Q1. What is BUN in a blood test used for?
Ans: It helps assess kidney and liver function, hydration status, and protein metabolism.
Q2. What does a high BUN Creatinine Ratio mean?
Ans: It may indicate dehydration, internal bleeding, or heart failure, even if creatinine levels are normal.
Q3. What causes a low BUN Creatinine Ratio?
Ans: It may be caused by liver disease, a low-protein diet, or excess fluid intake.
Q4. Can medications affect BUN and creatinine levels?
Ans: Yes. Diuretics, NSAIDs, ACE inhibitors, and some antibiotics can impact kidney filtration.
Q5. Should I fast before a BUN blood test?
Ans: Fasting isn't usually required unless combined with other tests. Always follow your doctor's instructions.
Q6. Can diet alone fix an abnormal ratio?
Ans: Diet plays a role, but it's essential to identify and treat the underlying cause with a doctor's help.
Leave a comment
6 Comments
Robin Kelly
May 13, 2025 at 4:33 AM.
My bun is 29/ Creatine ratio is 1
Myhealth Team
May 13, 2025 at 6:25 AM.
Your BUN is 29 and creatinine is 1, giving a high BUN/Creatinine ratio of 29. This could be due to dehydration, high protein intake, or stress. Drink more water, avoid excess protein or salt, and consult a doctor if symptoms like fatigue or swelling appear.
Montez
Apr 22, 2025 at 6:15 PM.
My Bun/creatine ratio 5600 is 28.33 mg/dL is that bad
Myhealth Team
Apr 23, 2025 at 5:26 AM.
A BUN/Creatinine ratio of 28.33 is slightly high and may suggest dehydration, high protein intake, or possible kidney-related issues. It's best to consult your doctor for a detailed evaluation, especially if you have other symptoms.
Patricia Davis
Apr 4, 2025 at 5:25 PM.
My bun creatinine levels are low, should I be concerned?
Myhealth Team
Apr 7, 2025 at 2:16 AM.
Low BUN and creatinine levels are usually not serious and may result from low protein intake, overhydration, or low muscle mass. If you feel fine and other tests are normal, it’s likely not a concern. Still, check with your doctor if you have symptoms or health issues.
Jeanne Tuttle
Mar 4, 2025 at 1:13 PM.
77 year old female with a high bun ratio
Of 28. Any suggestions?
Myhealth Team
Mar 6, 2025 at 12:43 PM.
A BUN ratio of 28 in a 77-year-old female could indicate dehydration, kidney dysfunction, or increased protein breakdown. Ensure proper hydration, reduce excess protein intake, and consult a doctor to assess kidney function and underlying causes. Regular monitoring of creatinine, electrolytes, and GFR is recommended.
Pavika
Mar 2, 2025 at 7:19 AM.
My grand daughter aged 2.5 years BUN and Creatinine ratio is 53.5 . Kindly suggest causes and remedy
Myhealth Team
Mar 4, 2025 at 8:00 AM.
A BUN/Creatinine ratio of 53.5 in a 2.5-year-old may indicate dehydration, kidney issues, or high protein intake. Ensure she drinks plenty of fluids and repeat the test. Watch for symptoms like swelling, fatigue, or reduced urination. Consult a pediatric nephrologist for further evaluation if levels remain high.
PARDEEP KUMAR PUNIA
Jan 19, 2025 at 9:06 AM.
My BUN/Creatinine is 65.29 mg/dl & urea is 139.81 mg/dl. How can i improve.
Myhealth Team
Jan 20, 2025 at 6:53 AM.
High BUN/Creatinine (65.29) and urea (139.81) may indicate kidney strain or dehydration. Consult a doctor, stay hydrated, limit protein intake, and manage underlying conditions like diabetes or high blood pressure. Follow a low-sodium, kidney-friendly diet and monitor levels regularly.
Related Posts
Consult Now |
9942 | https://ashfordstpeters.net/Guidelines_Maternity/APH-Guideline-Sep-2021.pdf | MANAGEMENT OF ANTEPARTUM HAEMORRHAGE WOMEN'S HEALTH AND PAEDIATRICS MATERNITY UNIT Amendments Version Date Comments Approved by 1 August 2021 New guideline Perinatal Guidelines Group Compiled by: Obstetrics Registrar Irene Ray In consultation with: Perinatal Governance Group Ratified by: Perinatal Governance Group Date ratified: September 2021 Next review date: September 2025, or if legislation, national guidance or lessons learnt indicate an earlier review Target audience: All health professionals within the maternity services Equality impact assessment: Perinatal Governance Group Comments on this document to: Perinatal Governance Guideline Group Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 1 of 11 Contents 1.0 Definition: Bleeding from or into the genital tract, occurring after 24+0 weeks of pregnancy till the birth of the baby. ....................................................................................................................... 6 2.0 Incidence: 3-5%...................................................................................................................6 3.0 Etiology: .............................................................................................................................. 6 Placenta praevia......................................................................................................... 6 Show .......................................................................................................................... 6 Cervical cause e.g. erosion/ ectropion........................................................................ 6 Vasa praevia .............................................................................................................. 6 Unexplained APH ....................................................................................................... 6 4.0 Classification of APH (RCOG): ............................................................................................ 6 5.0 APH prediction..................................................................................................................... 6 6.0 Risk factors for APH ............................................................................................................ 7 7.0 Complications of APH:......................................................................................................... 7 8.0 APH management: .............................................................................................................. 8 9. References ............................................................................................................................... 10 Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 2 of 11 Fig 1. Management of spotting in pregnancy Spotting Any fresh red per vaginal bleeding in pregnancy needs face to face clinical assessment regardless of gestation and the amount of bleeding Observations and fetal heart assessment (cCTG from 26/40) History and examination to establish the cause for APH Abdominal palpation for contractions/ uterine tenderness – labour or, abruption Speculum examination to check lower genital cause for APH e.g. cervical erosion or SROM Check USS report for placental localization, if not present, to arrange USS for placental localization when bleeding stops (≥20/40) FBC and G&S, consider anti-D if appropriate. Kleihauer if indicated Can be sent home with safety netting if no active PV bleed and placenta praevia has been excluded As imminent delivery unlikely, corticosteroids are unlikely of benefit but can still be considered Single / recurrent PV spotting due to cervical ectropion should not alter subsequent antenatal care Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 3 of 11 Fig 2. Management of minor APH or major APH in pregnancy (no hypovolemic shock) Minor APH (<50ml blood loss, settled) or, Major APH (50-1000ml blood loss, no sign of hypovolemic shock) Normal CTG Abnormal CTG Initial assessment as per spotting. Admission till 24-48 hours bleed-free/ longer if needed tailored to suit clinical situation. IV access and send blood for FBC and G&S, Kleihauer (if indicated). A coagulation screen is not indicated unless the platelet count is abnormal. Anti-D if Rh negative. Check smear history. Speculum examination to identify cervical dilation or a lower genital tract cause for the APH (cervical erosion/ ectropion) Consider steroids if 23+0 - 34+6 Check placental site on USS. If placenta praevia: send a group and save sample every 72 hours throughout admission, discuss and document the risks of caesarean and additional procedures that may become necessary to control bleeding. Women with atypical antibodies form a particularly high-risk group and discussions in these cases should involve the local haematologist and blood bank. If over 37 weeks and not placenta praevia, induction of labour should be considered. Encourage mobility, TEDs and hydration to reduce risk of venous thromboembolism. Complete the VTE assessment. Avoid/ with-hold Clexane/ Aspirin until bleeding stops. Recurrent APH should be re-classified as consultant-led care. USS for fetal growth should be performed if repeated bleeds/ abruption/ un-explained APH. APH is a relative contra-indication for tocolysis and is contra indicated in major APH. Inform NICU of admission Observations and CTG while admitted will depend on amount and cause of bleed. Exclude SROM (vasa-praevia) Determine cause – abruption / placenta praevia Intrauterine resuscitation Delivery by the quickest means possible Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 4 of 11 Fig 3. Management of massive APH in pregnancy +/- hypovolemic shock Massive APH (blood loss >1000mls) and / or signs of hypovolemic shock Call for help – dial 2222 and state ‘massive obstetric haemorrhage’. Inform consultant obstetrician and consultant anaesthetist. Alert others in MDT protocol– midwifery, hematologist, porters, laboratory, NICU, theatre staff. Follow ABCDE approach for resuscitation Temperature should be taken every 15mins Administer oxygen 10-15L/min via face mask Left lateral tilt IV access (at least 2 x grey cannulas) Send ‘URGENT’ blood sample for FBC, cross match 4 units, clotting, fibrinogen, U&Es, LFTs, Kleihauer Follow MOH protocol (separate guideline) at 1L of APH and on-going bleeding. Ensure on-going effective maternal resuscitation Foley’s catheter and strict fluid balance Administer cortico-steroids Tocolysis contra-indicated Painless bleed Painful bleed Suspect placenta praevia especially if malpresentation Avoid digital examination Check placental site on USS, anticipate placenta accreta if previous LSCS and anterior placenta) Emergency LSCS with senior staff Anticipate PPH and hysterectomy Suspect placental abruption Live fetus: emergency LSCS IUD: consider vaginal delivery if maternal condition is stable Anticipate PPH Suspect uterine rupture especially if uterus scarred or multi-parous Laparotomy Anticipate PPH and hysterectomy Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 5 of 11 Antepartum Haemorrhage 1.0 Definition: Bleeding from or into the genital tract, occurring after 24+0 weeks of pregnancy till the birth of the baby. 2.0 Incidence: 3-5%. 3.0 Etiology: Main causes Other causes Placenta praevia Placental abruption or abruptio placentae Show Cervical cause e.g. erosion/ ectropion Vasa praevia Unexplained APH 4.0 Classification of APH (RCOG): Spotting Minor APH - <50mls, settled Major APH – 50-1000mls, no hypovolemic shock Massive APH - >1000mls, +/- hypovolemic shock Recurrent APH: >1 APH episodes 5.0 APH prediction Cannot be reliably predicted as heterogenous pathophysiology. 70 % of abruption occurs in low-risk pregnancy. Some modifiable risk factors can however be changed such as, smoking and drug misuse. Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 6 of 11 6.0 Risk factors for APH Placenta praevia Placental abruption Previous placenta praevia Previous caesarean section (risk increases with increasing number of CS) Previous termination of pregnancy Multiparity Advanced maternal age Multiple pregnancy Smoking Deficient endometrium due to h/o: o uterine scar o endometritis o manual removal of placenta o curettage o submucous fibroid Assisted conception Previous abruption (risk higher with increasing number of abruptions) Pre-eclampsia Fetal growth restriction Non-vertex presentation Polyhydramnios Advanced maternal age Multiparity Low BMI Assisted reproductive techniques Intra-uterine infections PROM Abdominal trauma (accident/DV) Smoking Drug misuse in pregnancy (cocaine, amphetamine) 7.0 Complications of APH: Maternal complications Fetal complications Anaemia Infection Maternal shock Renal tubular necrosis Consumptive coagulopathy Postpartum haemorrhage Prolonged hospital stay Psychological sequelae Blood transfusion complications Fetal hypoxia Small of gestational age and growth restriction Prematurity (iatrogenic and spontaneous) Fetal death Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 7 of 11 8.0 APH management: 8.1All APH should be assessed in a hospital. 8.2Multi-disciplinary team approach (midwife and obstetric staff, laboratory access, theatre, neonatal and anaesthetic services). 8.3History taking: A through history should be taken (maternal condition permitting) to ascertain cause of bleed. Placenta praevia is usually associated with painless PV bleed. Causes for painful PV bleed could be – labour or abruption. Vasa praevia should be considered in case of PV bleed following SROM and subsequent fetal compromise. Cervical erosion or ectropion can present with bleed following sexual intercourse or h/o abnormal smears in the past can point towards a cervical neoplastic cause. Domestic violence should be always ruled out incase of presentation with APH. Enquiry should be made about fetal movements. Current medications. If on anti-coagulants, will need to with-hold, may need IV unfractionated heparin till bleeding controlled. 8.4Examination to assess the amount and cause of APH: Observations and amount of blood loss. Observed loss may be underestimated as haemorrhage may be concealed. Observations (including temperature) should be repeated every 15 minutes in case of major haemorrhage with ongoing bleed / hypovolemic shock. Auscultation for fetal heart sound +/- cCTG (from 26+0). Vasa praevia should be suspected where APH is following spontaneous or artificial rupture of membranes and associated with CTG abnormality. Abdominal palpation: to check for contractions / uterine tenderness (woody hard feel in case of abruption). Speculum examination: to check cervix/ to look for rupture of membranes. Digital vaginal examination should not be performed unless placenta praevia has been ruled out. Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 8 of 11 8.5Investigations: Blood tests: 1. Full blood count 2. Group and save 3. Kleihauer, if rhesus negative or IUD to quantify feto-maternal haemorrhage 4. Coagulation profile, if major/ massive blood loss or abnormal platelet count. 5. Anti-D (1500 IU IM) to be given to all non-sensitised Rhesus negative women, irrespective of routine antenatal anti-D prophylaxis and at 6 weekly intervals after 20+0 weeks if recurrent APH. Ultrasound scan is not routinely indicated at first presentation where a previous departmental scan has confirmed placental location e.g. anomaly scan. USS for assessing growth in case of recurrent APH (not for cervical erosion). USS has poor sensitivity for diagnosis of abruption which is mainly a clinical diagnosis. 8.6Management of APH based on the amount of bleeding - see flowcharts above. 8.7Timing and mode of delivery: Any woman admitted with an APH will have a consultant review and individual plan of care within 14 hours of admission Placenta not low lying: >39 weeks Any APH would prompt consideration of delivery usually by induction of labour if no contraindication. This decision can be made at the time of presentation by the obstetric registrar with input from the on-call consultant as needed. 37-39 weeks Individual decision considering degree of bleeding, previous history, and assessment of maternal and fetal wellbeing. <37 weeks In the absence of any objective evidence of fetal or maternal compromise expectant management would usually be advised to optimise perinatal outcomes with advancing gestation. A diagnosis of abruption or clinical concern about maternal or fetal wellbeing would usually prompt consideration of delivery. The mode of birth to be decided by the consultant on-call. At extremes of gestation expectant management may be advised, taking on board the wishes of the family. Placenta praevia / low lying placenta (within 20mm of cervical os): Delivery would normally be indicated at >36 weeks with any degree of APH. At gestations less than 36 weeks a consultant plan should be made taking in to account the relative risks and benefits of expectant management. Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 9 of 11 8.8 Intra-partum monitoring: APH is an indication for undertaking continuous fetal monitoring in labour. Please refer to intra-partum fetal monitoring guidelines for further information on this. 8.9 Obstetric anaesthesia: The decision about mode of anaesthesia should be a multi-disciplinary decision taking into consideration the seniority of staff and maternal and fetal wellbeing. Where in doubt, early consultant involvement is advised. 8.10 Post-partum management: Anticipate post-partum haemorrhage Active management of third stage of labour – 5U oxytocin or syntometrine + 40U oxytocin infusion Consider syntometrine, unless hypertensive. Risk assess for post-natal thromboprophylaxis Debriefing Clinical incident reporting 9. References Royal College of Obstetricians and Gynaecologists. Green Top guideline No.63 - Antepartum Haemorrhage. 2011 Royal College of Obstetricians and Gynaecologists. Green Top guideline No 27a - Placenta praevia, placenta praevia accreta diagnosis and management. 2018 Royal College of Obstetricians and Gynaecologists. Green Top guideline No 27b - Vasa praevia: diagnosis and management. 2018 Royal College of Obstetricians and Gynaecologists. Green Top guideline No 52 Postpartum Haemorrhage, Prevention and Management. 2016 MBRRACEUK report. Saving Lives, Improving Mothers' Care 2020: Lessons to inform maternity care from the UK and Ireland Confidential Enquiries in Maternal Death and Morbidity 2016-18. 10. Abbreviations: APH: Antepartum haemorrhage MOH: Massive obstetric haemorrhage PPH: Post-partum haemorrhage CTG: cardiotocography FBC: Full blood count MDT: Multidisciplinary team G&S: Group and save VTE: venous thromboembolism Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 10 of 11 Section 1 Organisational Policy Current Version is held on the Intranet First ratified: September 2021 Review date: September 2025 Version Page 11 of 11 |
9943 | https://projecteuclid.org/journals/involve-a-journal-of-mathematics/volume-12/issue-1/Patterns-in-colored-circular-permutations/10.2140/involve.2019.12.157.pdf | inv lve a journal of mathematics msp Patterns in colored circular permutations Daniel Gray, Charles Lanning and Hua Wang 2019 vol. 12, no. 1 msp INVOLVE 12:1 (2019) dx.doi.org/10.2140/involve.2019.12.157 Patterns in colored circular permutations Daniel Gray, Charles Lanning and Hua Wang (Communicated by Joshua Cooper) Pattern containment and avoidance have been extensively studied in permutations.
Recently, analogous questions have been examined for colored permutations and circular permutations. In this note, we explore these problems in colored circular permutations. We present some interesting observations, some of which are direct generalizations of previously established results. We also raise some questions and propose directions for future study.
1. Background Patterns are essentially subpermutations of a bigger permutation. For two permuta-tions and of lengths n and k with n k, we say that contains as a pattern if there is a subsequence of entries of , .i1; i2; i3; : : : ; ik/, which is order isomorphic to ; i.e., is it if and only if s t. Such a subsequence is called an occurrence of in . If no occurrence of is present in , we say that avoids .
Most of the earlier work on patterns concerns pattern avoidance; see [Bóna 2012] for a nice introduction. A more comprehensive study of pattern containment was first proposed by H. Wilf in 1992 [Liendo 2012]. There are two natural questions one might ask regarding pattern containment. First, what is the shortest permutation that contains every element in some set of permutations? Second, for a given pattern, in what permutation does this pattern occur the most? The former deals with superpatterns, whereas the latter concerns pattern packing.
Superpatterns. For a set P of permutations we say that a permutation is a P-superpattern if it contains at least one occurrence of every 2 P. We also define sp.P/ D minfn W there is a P -superpattern of length ng and sp.k/ D sp.P/ when P is the set of all permutations of length k.
For results on the bounds of sp.k/, see [Arratia 1999; Eriksson et al. 2007; Miller 2009]. Bounds of sp.P / have also been studied for layered permutations [Gray MSC2010: primary 05A05; secondary 05A15, 05A16.
Keywords: Circular permutations, patterns.
This work was partially supported by a grant from the Simons Foundation (#245307).
157 158 DANIEL GRAY, CHARLES LANNING AND HUA WANG 2015], 321-avoiding permutations [Bannister et al. 2014], m-colored permutations [Gray and Wang 2016], and words [Burstein et al. 2002/03].
Pattern packing. Letting f .; / be the number of occurrences of in , we define g.n; / D maxff .; / W is a permutation of length ng: and the packing density of as ı. / D lim n!1 g.n; / n k : A permutation (of length n) with f .; / D g.n; / is called -optimal.
For packing densities of length-3 and length-4 patterns, see [Albert et al. 2002; Price 1997; Stromquist 1993]. There are three length-4 patterns whose packing densities remain open, as are any longer nonlayered patterns.
Pattern avoidance. Pattern avoidance has been well-studied for permutations; see [Bóna 2012] for details. In the case of colored permutations, [Mansour 2001] pro-vides a formula for the number of permutations avoiding all length-2 permutations whose entries are colorable in r ways. For circular permutations, [Callan 2002] counts the number of circular permutations avoiding 1324, 1342, and 1234. Both topics are relatively new, and there are still many open questions.
Our contribution. There are two natural variations of permutations, colored permu-tations and circular permutations, where the first one assigns colors to each entry and the second arranges entries around a circle. In colored permutations, superpatterns [Gray and Wang 2016], pattern packing [Just and Wang 2016], and pattern avoidance [Mansour 2001] have been considered. Noncolored pattern containment [Gray et al.
2017] and pattern avoidance [Callan 2002] have been studied for circular patterns.
In this paper, we will consider the combination of these two variations, the colored circular permutations. First, we will introduce the necessary terminology and notation in Section 2. We then discuss “supercolored circular permutations” in Section 3, where we point out that many of the results in [Gray and Wang 2016] can be directly generalized to the colored circular permutations. In Section 4, we discuss pattern packing in colored circular permutations, including some generalizations of results in [Just and Wang 2016]. Lastly, in Section 5 we consider pattern avoidance in colored circular permutations. We conclude our work by commenting on the many remaining problems for future work in Section 6.
2. Terminologies in colored and circular permutations We start with some formal terminologies and notations for colored permutations and patterns.
PATTERNS IN COLORED CIRCULAR PERMUTATIONS 159 Definition 2.1. Let k and m be any positive integers. An m-colored permutation of length k is any permutation of length k where each entry is colored one of m given colors; we allow distinct entries of the permutation to be colored differently.
We denote the set of all permutations of length k in m colors by Sk;m.
In the case that there are only two or three colors, we will color the entries of a permutation “red”, “green”, or “blue”; thus, we may have the colored permutation 2r1b3r, which denotes the permutation 213 whose first and third entries are colored “red” and whose second entry is colored “blue”. If more than three colors are allowed, we will just label the colors with natural numbers; hence, the colored permutation 1134415322 is the permutation 13452 whose first and third entries are colored 1, fifth entry is colored 2, fourth entry is colored 3, and second entry is colored 4.
Definition 2.2. Let k and m be any positive integers. A monochromatic m-colored permutation of length k is any m-colored permutation for which every entry is colored the same color. We denote the set of all monochromatic m-colored permu-tations of length k by Mk;m.
Definition 2.3. Let k and m be any positive integers. A nonmonochromatic m-colored permutation of length k is any m-colored permutation for which there exist at least two distinct entries that are colored differently. We denote the set of all nonmonochromatic m-colored permutations of length k by Nk;m.
The union of Nk;m and Mk;m is Sk;m, the set of all m-colored permutations of length k. For example, 1r2b3b is nonmonochromatic since the first entry and second entry are colored differently, while 1r2r3r and 1b2b3b are both monochromatic.
For comparison, we list S2;2, N2;2, and M2;2 below: S2;2 D f1r2r; 1r2b; 1b2r; 1b2b; 2r1r; 2r1b; 2b1r; 2b1bg; N2;2 D f1r2b; 1b2r; 2r1b; 2b1rg; M2;2 D f1r2r; 1b2b; 2r1r; 2b1bg: Definition 2.4. For colored permutations p and q we say that p contains q as a colored pattern if there is some subsequence of p, say P, which satisfies the following two conditions: The i-th entry of P is the same color as the i-th entry of q for all i.
P is order isomorphic to q.
If there is no such P satisfying both conditions, we say that p avoids q as a colored pattern.
We will usually drop the phrase “as a colored pattern” and just say “p contains q” when it is obvious that we are dealing with colored permutations. For instance, if 160 DANIEL GRAY, CHARLES LANNING AND HUA WANG p D 1r3b2r and q D 2b1r, we see that p contains q since the subsequence .3b; 2r/ of p satisfies both of the conditions above. However, if q D 2r1b then p avoids q since there is no subsequence of p simultaneously satisfying both conditions.
Similar to the noncolored case, for a collection P of colored permutations we define the P-superpattern and sp.P/ accordingly. Note that the permutation p D 1r2b6r5b4r3b contains every colored permutation in S2;2. Hence, p is an S2;2-superpattern. Brute force shows that there is no shorter S2;2-superpattern; therefore sp.S2;2/ D 6.
Next we formalize the concept of pattern containment/avoidance in circular permutations. Note that the following definition also applies to colored permutations.
Definition 2.5. Let p D p1p2 pn be a permutation of length n. The circular shift of p, denoted S.p/, is given by S.p/ D pnp1p2 pn 1: If we take a permutation, , and wrap its entries clockwise around a circle, equally spread out within one revolution, then we have created a circular permutation, c.
We say that c D c if is just a cyclic shift of , i.e., Si. / D for some i.
Definition 2.6. For colored permutations p and q we say that p contains q circularly if p contains Si.q/ as a colored pattern for some nonnegative integer i.
Definition 2.7. Let P be any collection of permutations. A circular P-superpattern is a permutation which contains every p 2 P as a circular pattern. We let spc.P/ denote the length of the shortest circular P-superpattern. When P is the set of all circular patterns of length k we simply write spc.k/.
A useful concept in the study of pattern packing in colored permutations will be “colored blocks”, which we define below.
Definition 2.8. In a colored permutation , a colored block is a maximal monochro-matic segment .a/ i in which every entry in this segment has color a and every entry not in this segment is either larger or smaller than each entry in .a/ i .
For example, the permutation D 1r2r6b5b3b4r has four colored blocks. From left to right, they are .r/ 1 D 1r2r, .b/ 2 D 6b5b, .b/ 3 D 3b, .r/ 4 D 4r. Indeed every colored permutation has a unique decomposition into colored blocks: Given a colored permutation, it can first be decomposed into maximal monochromatic subsequences and it is easy to see that there is a unique way to do this. Within each monochromatic subsequence there is a unique way to separate the entries according to their numerical values.
When comparing the numerical values between different blocks, we say that .r/ i < .b/ j when all entries of .r/ i are less than all entries of .b/ j . It is easy to see that this concept generalizes naturally to the circular case.
PATTERNS IN COLORED CIRCULAR PERMUTATIONS 161 For colored patterns and permutations we define fc.; / to be the number of occurrences of in wrapped around a circle; i.e., fc.; / D f .; / C f .; S. // C f .; S2. // C C f .; Sk 1. //: Then, similar to before, gc.n; / D maxffc.; / W is a permutation of length ng: If is of length n and fc.; / D gc.n; /, then we say that is circular -optimal.
Definition 2.9. Let be a colored permutation of length k. The circular packing density of , denoted by ıc. /, is defined by ıc. / D lim n!1 gc.n; / n k : 3. Superpatterns In this section we consider questions related to superpatterns in colored circular per-mutations. We note that some of the results in this section are direct generalizations from those in [Gray and Wang 2016]. For this reason some details will be omitted.
Theorem 3.1. For any positive integers k and m, we have that spc.Sk;m/ D m spc.k/: Proof. Let p0 be a circular Sk;m-superpattern and p0 i be the longest monochromatic subsequence in p0 in color i. It follows that p0 is a circular k-superpattern and consequently jp0 ij spc.k/ for any 1 i m. Hence jp0j D m X iD1 jp0 ij m spc.k/: Now, let p be a circular permutation of length spc.k/ that contains all noncolored patterns of length k. Consider the m-colored circular permutation p00, constructed from p by replacing each 1 j spc.k/ in p with the sequence sj D Œm.j 1/ C 11Œm.j 1/ C 22 Œm.j 1/ C mm: It is easy to see that jp00j D mjpj D m spc.k/ and that p00is a Sk;m-superpattern. Thus, spc.Sk;m/ jp00j D m spc.k/: With Theorem 3.1, we can use previously established results [Gray et al. 2017] on spc.k/ to bound spc.Sk;m/.
162 DANIEL GRAY, CHARLES LANNING AND HUA WANG Corollary 3.2. For positive integers k and m we have spc.Sk;m/ D m spc.k/ mg.k/k2 e2 ; where g.k/ ! 1 as k ! 1, and spc.Sk;m/ D m spc.k/ m.sp.k 1/ C 1/ m 1 2k.k 1/ C 1 : Consequently, mg.k/k2 e2 spc.Sk;m/ m 1 2k.k 1/ C 1 ; where g.k/ ! 1 as k ! 1.
As expected, the bounds for spc.Sk;m/ are simply m times the bounds for spc.k/.
Next, we restrict our attention to only monochromatic or nonmonochromatic patterns.
First we note the following facts on the sizes of Mk;m and Nk;m: jSk;mj D mkkŠ .
jMk;mj D mkŠ .
jNk;mj D jSk;mj jMk;mj D .mk 1 1/jMk;mj.
We now establish a lower bound for spc.Nk;m/.
Theorem 3.3. For positive integers k and m, spc.Nk;m/ mg.k; m/k2 e2 ; where g.k; m/ ! 1 as k ! 1.
Proof. Let n D spc.Nk;m/, and our Nk;m-superpattern of length n must contain a circular shift of every permutation in Nk;m. Note that at most k such permutations can be circular shifts of each other; hence at least jNk;mj=k permutations from Nk;m must be contained in the superpattern. Consequently n k jNk;mj k D .mk m/kŠ k D .mk m/.k 1/Š : By the fact nk=kŠ n k and Stirling’s approximation kŠ p 2k.kk=ek/, we have nk kŠ .mk m/.k 1/Š and hence n .mk m/.kŠ/2 k 1=k .mk m/2 k2k e2k 1=k D m .1 m kC1/2 1=k k2 e2 D mg.k;m/k2 e2 with g.k; m/ D .1 m kC1/2 1=k ! 1 as k ! 1.
PATTERNS IN COLORED CIRCULAR PERMUTATIONS 163 It is interesting to note that this lower bound is similar to that found for spc.Sk;m/ in Corollary 3.2. To bound spc.Nk;m/ from above, first note that a circular Mk;m-superpattern must have m copies of a circular k-superpattern, one for each color.
Then, we have spc.Nk;m/ spc.Sk;m/ D m spc.k/ D spc.Mk;m/: Given the fact that jNk;mj D .mk 1 1/jMk;mj, it is rather surprising that the shortest Nk;m-superpattern is not longer than the shortest Mk;m-superpattern. The following further analyzes the relationship between them.
Theorem 3.4. For any positive integers k 2 and m, we have spc.Mk 1;m/ spc.Nk;m/ spc.Mk;m/: Proof. The second inequality follows from the discussion above.
On the other hand, let q be an m-colored pattern of length k with all but one entry of color i. For some circular shift of q to be contained in a circular Nk;m-superpattern, a circular shift of the length-.k 1/ monochromatic pattern in color i must be contained in the superpattern. Hence all length-.k 1/ monochro-matic patterns (of any color) must occur in a circular Nk;m-superpattern, and spc.Mk 1;m/ spc.Nk;m/.
4. Pattern packing Our results in this section mainly concern the characteristics of the optimal colored circular permutations when the pattern under consideration is described through colored blocks. Again, some of our results here are direct generalizations of those in noncircular case [Just and Wang 2016], for which reason we skip some details.
In the case of having only two colored blocks, we can see that a pattern must be of the form D 12 with 1 in red and 2 in blue. We will assume, without loss of generality, that 1 < 2. In this case, we may simply say that the pattern is of the form rb with r < b, and similarly for patterns with more colored blocks.
Theorem 4.1. For a pattern with two colored blocks of the form rb with r < b, there is an optimal circular permutation of the form RB with R < B.
Proof. Let be a -optimal permutation of length n with colored blocks 12 k.
We can assume without loss of generality that 1 is red.
Now, let us take all the red blocks r1r2 rs and blue blocks b1b2 bt, and form a new circular permutation 0 D r1 rsb1 bt. It is easy to see that any occurrence of in is also in 0.
Next, since is of the form rb with r < b, we claim that, in our optimal permutation 0, every red entry must be (numerically) less than every blue entry.
Otherwise, one may always “rearrange” the numerical values so that the numerical 164 DANIEL GRAY, CHARLES LANNING AND HUA WANG ordering stays the same among entries of the same color, and so that all red entries are smaller than the blue ones. The resulting permutation can only contain more occurrences of .
Consequently, all red blocks together simply form a single block in and so do the blue blocks. Our conclusion, then, follows.
Next, we consider patterns with three colored blocks. Note that for circular patterns with three colored blocks and two colors, rb1b2 with b1 < r < b2 is the only case that we needed to investigate: With three colored blocks and two colors there are always one block with one color (say red) and two blocks with the other (say blue). One of the circular shifts of this pattern must be of the form rb1b2. By taking a circular shift of the reversed pattern (i.e., rb2b1) if necessary, we may also assume that b1 < r < b2.
Theorem 4.2. For a pattern with three colored blocks of the form rb1b2 and b1 < r < b2, there is an optimal circular permutation of the same form.
Proof. Let be a -optimal circular permutation of length n. First, we will show that we can put all blue blocks in increasing order of their numerical values and next to each other. Let r1; r2; : : : ; rs be the red blocks of .
Now, for an occurrence of in , suppose R (in ) is the part corresponding to r (in ). Let bR be the set of all blue blocks greater than R. Then, any occurrence of rb1b2 with r R must have b1 occurring in bR. The maximum number of such occurrences (i.e., the maximum possible contribution of R to fc.; /) is f .bR; b2/: As far as the ordering of the blue blocks is concerned, arranging the blue blocks in increasing order achieves the above maximum. At this point it is also easy to see that putting blocks of the same color together will not reduce the number of occurrences of . Denote such an optimal permutation by 0 D r1 rsb1 bt, with bi < biC1 for any 1 i t 1.
Next, we show that all red entries form a single block, or equivalently, the numerical value of any red block is between those of bj0 1 and bj0 for some fixed j0. Let bj be the collection of blue blocks bj ; bj C1; : : : ; bt, and let b<j be the collection of blocks b1; : : : ; bj . Then, there must exist some j0 that maximizes the occurrences of b1 in b<j and b2 in bj . In other words, f .b<j0; b1/f .bj0; b2/ f .b<j ; b1/f .bj ; b2/ for any 1 < j t. So, fc.0; / f .r1 rs; r/f .b<j0; b1/f .bj0; b2/ PATTERNS IN COLORED CIRCULAR PERMUTATIONS 165 with equality when bj0 1 < ri < bj0 for any 1 i s. From this it follows that there are exactly one single red block and two blue blocks in D RB1B2 with B1 < R < B2.
It remains to consider the case when we have three colored blocks in three different colors, i.e., the pattern rbg with r < b < g.
Theorem 4.3. For a pattern of the form rbg with r < b < g, there is an optimal circular permutation of the same form.
Proof. Let be a -optimal circular permutation of length n with R0, B0 and G0 being the collections of all red blocks (in their original order), blue blocks and green blocks respectively. An occurrence of in must consist of an occurrence of r in R0, an occurrence of b in B0, and an occurrence of g in G0. Hence f .; / f .R0; r/f .B0; b/f .G0; g/ with equality if each of R0, B0 and G0 is a single block, arranged in this order, and R0 < B0 < G0.
To summarize the above observations, we have the following.
Corollary 4.4. For any circular pattern with two or three colored blocks, there is a corresponding optimal circular permutation of the same form.
Remark 4.5. After seeing the above results on patterns with two or three colored blocks, it is natural to guess that the same holds for patterns with more blocks.
Consequently one can ask if there is always an optimal circular permutation of the same form as the pattern. We have not been able to prove either way.
On the other hand, considering D 1r2b, it is not hard to check that D .1r3b2r4b/c is an optimal length-4 circular permutation for . Thus, there does exist an optimal permutation that is not of the form RB with R < B. Evidence seems to suggest that this is the only such case.
5. Pattern avoidance The numbers of m-colored (noncircular) permutations that avoid one or two 2-letter patterns were presented in [Mansour 2001], together with some discussion of the connection between pattern avoidance in noncolored permutations and colored permutations. In [Callan 2002], pattern avoidance in circular permutations was studied. It was pointed out that, when considered as circular permutations, none avoid any 2-letter patterns and the identity (reverse identity) is the only one avoiding the pattern 132 (123). In this section we extend this study to colored circular permutations, generalizing a little of both [Callan 2002] and [Mansour 2001].
166 DANIEL GRAY, CHARLES LANNING AND HUA WANG Avoiding a monochromatic length-2 pattern in Sk;m. Without loss of generality, we may assume the monochromatic length-2 pattern is 1121. Then, to avoid such a pattern, our permutation in Sk;m can contain at most one entry of color 1.
Consider two cases: There is no entry with color 1 in . Then, there are a total of .k 1/Š noncolored circular permutations of length k, and each of the k entries has m 1 choices of colors (i.e., the colors 2; 3; : : : ; m); thus the number of such permutations is .k 1/Š .m 1/k: There is exactly one entry with color 1. Out of the k entries 1; 2; : : : ; k there are k choices for this particular entry of color 1. There are still .k 1/Š ways to wrap the k entries (regardless of their colors) around a circle. Now for each of the remaining k 1 entries that are not of color 1, there are m 1 choices of colors.
Hence the number of such permutations is kŠ .m 1/k 1: Consequently, we have the following.
Theorem 5.1. The number of circular permutations in Sk;m that avoid a given monochromatic length-2 pattern is .k 1/Š .m 1/k C kŠ .m 1/k 1 D .k 1/Š .m 1/k 1.k C m/: Avoiding nonmonochromatic length-2 pattern in Sk;m. Again, without loss of generality, let us assume this pattern to be 1122. For a circular permutation in Sk;m, let E1 and E2 be the sets of entries in that are colored 1 and 2 respectively.
It is easy to see that a 1122 pattern will occur if there is any entry in E1 that is of smaller numerical value than one in E2. Thus, all entries in E1 are larger than those in E2. Suppose jE1 [E2j D i for some 0 i k. Then, there are i C1 ways to partition the entries into E1 and E2 (i.e., to find a j D 0; 1 : : : ; i such that the smallest j entries are colored 2 and the rest are colored 1).
Thus, still with .k 1/Š ways to wrap all entries around a circle, there are k i ways to choose entries of E1 [ E2. After identifying j there are .m 2/k i ways to color the remaining entries. Consequently the number of such permutations is .k 1/Š k X iD0 .m 2/k i.i C1/ k i D .k 1/Š k X iD1 .m 2/k ii k i C.k 1/Š k X iD0 .m 2/k ik i D .k 1/Š k X iD1 .m 2/.k 1/ .i 1/k k 1 i 1 C.k 1/Š k X iD0 .m 2/k i k k i PATTERNS IN COLORED CIRCULAR PERMUTATIONS 167 D kŠ ..m 2/C1/k 1C.k 1/Š ..m 2/C1/k D .k 1/Š .m 1/kCkŠ .m 1/k 1 D .k 1/Š .m 1/k 1.kCm/: As a result we have the following.
Theorem 5.2. The number of circular permutations in Sk;m that avoid a given nonmonochromatic length-2 pattern is .k 1/Š .m 1/k 1.k C m/: Avoiding monochromatic patterns of length 3 in Sk;m. All circular permutations of length 3 are equivalent under circular shift and reverse. So we will only consider, without loss of generality, the pattern 113121. It is known that in the noncolored case the only circular permutation that avoids 132 is the identity permutation.
Let be a permutation that avoids 113121 in Sk;m, and let E1 (of cardinality i D 0; 1; : : : ; k) be the set of entries of color 1, then: If i 3, there is only one way to order the entries in E1 (i.e., in increasing order). Starting with .k 1/Š ways to wrap the k entries (regardless of color) around a circle, only one of the .i 1/Š orderings of the entries in E1 can be chosen.
Noting that there are k i ways to pick the numerical values of the entries in E1 and .m 1/k i ways to assign colors to the remaining entries, we have the number of such permutations as k X iD3 k i .k 1/Š .i 1/Š .m 1/k i : If i 2, then there are k i ways to pick these i entries, .k 1/Š ways to wrap all the entries around the circle and .m 1/k i ways to color the other entries. The number of such permutations is 2 X iD0 k i .k 1/Š.m 1/k i : We may combine the above two formulas and conclude the following.
Theorem 5.3. The number of circular permutations in Sk;m that avoid a given monochromatic length-3 pattern is .k 1/Š .m 1/k C k X iD1 k i .k 1/Š .i 1/Š .m 1/k i : Wilf classes. Theorems 5.1 and 5.2 imply that for any m-colored pattern of length 2, say , the number of -avoiding circular permutations in Sk;m is .k 1/Š.m 1/k 1.k C m/. This interesting (and perhaps a little surprising) observation is analogous to the findings in [Mansour 2001] in noncircular case.
168 DANIEL GRAY, CHARLES LANNING AND HUA WANG This also implies that there is only one Wilf class of colored circular permutations when restricted by one pattern of length 2.
6. Concluding remarks and additional questions In this short note, we considered questions related to superpatterns, pattern pack-ing, and pattern avoidance in colored circular permutations. We presented some elementary observations, especially those generalized from previously established results on colored (but not circular) permutations, on each of these three questions.
Many interesting questions remain to be further explored.
In Section 3, we introduced generalizations of a few facts on colored superpatterns.
The arguments of these generalizations follow from direct adjustment of those in [Gray and Wang 2016]. There are, however, also some constructive proofs that cannot be directly generalized to circular cases. It would be interesting to further investigate them.
The several theorems in Section 4 claim that for patterns with two or three colored blocks their corresponding optimal colored circular permutations include at least one with exactly the same format (in terms of the colored blocks). With these facts, one may easily calculate the packing densities of various patterns. It is not clear whether this is true for more colored blocks. It is also mentioned that, for the pattern D 1r2b, there exist optimal permutations that have a different format. It seems likely that this is the only such case, though we do not have a proof yet.
The numbers of permutations avoiding various given patterns, as studied in Section 5, lead to an interesting statement on the Wilf classes of colored circular permutations restricted by 2-letter patterns. It appears to be much more complicated to examine the same problem for colored circular permutations restricted by longer patterns or more than one 2-letter patterns.
References [Albert et al. 2002] M. H. Albert, M. D. Atkinson, C. C. Handley, D. A. Holton, and W. Stromquist, “On packing densities of permutations”, Electron. J. Combin. 9:1 (2002), art. id. R5. MR Zbl [Arratia 1999] R. Arratia, “On the Stanley–Wilf conjecture for the number of permutations avoiding a given pattern”, Electron. J. Combin. 6 (1999), art. id. N1. MR Zbl [Bannister et al. 2014] M. J. Bannister, W. E. Devanny, and D. Eppstein, “Small superpatterns for dominance drawing”, pp. 92–103 in 2014 Proceedings of the Eleventh Workshop on Analytic Algo-rithmics and Combinatorics .ANALCO/, edited by M. Drmota and M. D. Ward, SIAM, Philadelphia, PA, 2014. MR [Bóna 2012] M. Bóna, Combinatorics of permutations, 2nd ed., CRC Press, Boca Raton, FL, 2012.
MR Zbl [Burstein et al. 2002/03] A. Burstein, P. Hästö, and T. Mansour, “Packing patterns into words”, Electron. J. Combin. 9:2 (2002/03), art. id. R20. MR Zbl [Callan 2002] D. Callan, “Pattern avoidance in circular permutations”, preprint, 2002. arXiv PATTERNS IN COLORED CIRCULAR PERMUTATIONS 169 [Eriksson et al. 2007] H. Eriksson, K. Eriksson, S. Linusson, and J. Wästlund, “Dense packing of patterns in a permutation”, Ann. Comb. 11:3-4 (2007), 459–470. MR Zbl [Gray 2015] D. Gray, “Bounds on superpatterns containing all layered permutations”, Graphs Combin.
31:4 (2015), 941–952. MR Zbl [Gray and Wang 2016] D. Gray and H. Wang, “Note on superpatterns”, Involve 9:5 (2016), 797–804.
MR Zbl [Gray et al. 2017] D. Gray, C. Lanning, and H. Wang, “Pattern containment in circular permutations”, preprint, 2017. To appear in Integers.
[Just and Wang 2016] M. Just and H. Wang, “Note on packing patterns in colored permutations”, Online J. Anal. Comb. 11 (2016), art. id. 4. MR Zbl [Liendo 2012] M. L. Liendo, Preferential arrangement containment in strict superpatterns, master’s thesis, East Tennessee State University, 2012, available at https:/ /dc.etsu.edu/etd/1428/.
[Mansour 2001] T. Mansour, “Pattern avoidance in coloured permutations”, Sém. Lothar. Combin. 46 (2001), art. id. B46g. MR Zbl [Miller 2009] A. Miller, “Asymptotic bounds for permutations containing many different patterns”, J. Combin. Theory Ser. A 116:1 (2009), 92–108. MR Zbl [Price 1997] A. L. Price, Packing densities of layered patterns, Ph.D. thesis, University of Pennsylva-nia, 1997, available at https:/ /search.proquest.com/docview/304421853. MR [Stromquist 1993] W. Stromquist, “Packing layered posets into posets”, preprint, 1993, available at http:/ /walterstromquist.com/papers/POSETS.DOC.
Received: 2017-11-29 Revised: 2018-01-21 Accepted: 2018-02-14 dagray@georgiasouthern.edu Department of Mathematical Sciences, Georgia Southern University, Statesboro, GA, United States lannin3@clemson.edu Department of Mathematical Sciences, Clemson University, Clemson, SC, United States hwang@georgiasouthern.edu Department of Mathematical Sciences, Georgia Southern University, Statesboro, GA, United States mathematical sciences publishers msp involve msp.org/involve INVOLVE YOUR STUDENTS IN RESEARCH Involve showcases and encourages high-quality mathematical research involving students from all academic levels. The editorial board consists of mathematical scientists committed to nurturing student participation in research. Bridging the gap between the extremes of purely undergraduate research journals and mainstream research journals, Involve provides a venue to mathematicians wishing to encourage the creative involvement of students.
MANAGING EDITOR Kenneth S. Berenhaut Wake Forest University, USA BOARD OF EDITORS Colin Adams Williams College, USA John V. Baxley Wake Forest University, NC, USA Arthur T. Benjamin Harvey Mudd College, USA Martin Bohner Missouri U of Science and Technology, USA Nigel Boston University of Wisconsin, USA Amarjit S. Budhiraja U of North Carolina, Chapel Hill, USA Pietro Cerone La Trobe University, Australia Scott Chapman Sam Houston State University, USA Joshua N. Cooper University of South Carolina, USA Jem N. Corcoran University of Colorado, USA Toka Diagana Howard University, USA Michael Dorff Brigham Young University, USA Sever S. Dragomir Victoria University, Australia Behrouz Emamizadeh The Petroleum Institute, UAE Joel Foisy SUNY Potsdam, USA Errin W. Fulp Wake Forest University, USA Joseph Gallian University of Minnesota Duluth, USA Stephan R. Garcia Pomona College, USA Anant Godbole East Tennessee State University, USA Ron Gould Emory University, USA Andrew Granville Université Montréal, Canada Jerrold Griggs University of South Carolina, USA Sat Gupta U of North Carolina, Greensboro, USA Jim Haglund University of Pennsylvania, USA Johnny Henderson Baylor University, USA Jim Hoste Pitzer College, USA Natalia Hritonenko Prairie View A&M University, USA Glenn H. Hurlbert Arizona State University,USA Charles R. Johnson College of William and Mary, USA K. B. Kulasekera Clemson University, USA Gerry Ladas University of Rhode Island, USA Suzanne Lenhart University of Tennessee, USA Chi-Kwong Li College of William and Mary, USA Robert B. Lund Clemson University, USA Gaven J. Martin Massey University, New Zealand Mary Meyer Colorado State University, USA Emil Minchev Ruse, Bulgaria Frank Morgan Williams College, USA Mohammad Sal Moslehian Ferdowsi University of Mashhad, Iran Zuhair Nashed University of Central Florida, USA Ken Ono Emory University, USA Timothy E. O’Brien Loyola University Chicago, USA Joseph O’Rourke Smith College, USA Yuval Peres Microsoft Research, USA Y.-F. S. Pétermann Université de Genève, Switzerland Robert J. Plemmons Wake Forest University, USA Carl B. Pomerance Dartmouth College, USA Vadim Ponomarenko San Diego State University, USA Bjorn Poonen UC Berkeley, USA James Propp U Mass Lowell, USA Józeph H. Przytycki George Washington University, USA Richard Rebarber University of Nebraska, USA Robert W. Robinson University of Georgia, USA Filip Saidak U of North Carolina, Greensboro, USA James A. Sellers Penn State University, USA Andrew J. Sterge Honorary Editor Ann Trenk Wellesley College, USA Ravi Vakil Stanford University, USA Antonia Vecchio Consiglio Nazionale delle Ricerche, Italy Ram U. Verma University of Toledo, USA John C. Wierman Johns Hopkins University, USA Michael E. Zieve University of Michigan, USA PRODUCTION Silvio Levy, Scientific Editor Cover: Alex Scorpan See inside back cover or msp.org/involve for submission instructions. The subscription price for 2019 is US $/year for the electronic version, and $/year (+$, if shipping outside the US) for print and electronic. Subscriptions, requests for back issues and changes of subscriber address should be sent to MSP.
Involve (ISSN 1944-4184 electronic, 1944-4176 printed) at Mathematical Sciences Publishers, 798 Evans Hall #3840, c/o University of California, Berkeley, CA 94720-3840, is published continuously online. Periodical rate postage paid at Berkeley, CA 94704, and additional mailing offices.
Involve peer review and production are managed by EditFLOW® from Mathematical Sciences Publishers.
PUBLISHED BY mathematical sciences publishers nonprofit scientific publishing © 2019 Mathematical Sciences Publishers involve 2019 vol. 12 no. 1 1 Optimal transportation with constant constraint WYATT BOYER, BRYAN BROWN, ALYSSA LOVING AND SARAH TAMMEN 13 Fair choice sequences WILLIAM J. KEITH AND SEAN GRINDATTI 31 Intersecting geodesics and centrality in graphs EMILY CARTER, BRYAN EK, DANIELLE GONZALEZ, RIGOBERTO FLÓREZ AND DARREN A. NARAYAN 45 The length spectrum of the sub-Riemannian three-sphere DAVID KLAPHECK AND MICHAEL VANVALKENBURGH 63 Statistics for fixed points of the self-power map MATTHEW FRIEDRICHSEN AND JOSHUA HOLDEN 79 Analytical solution of a one-dimensional thermistor problem with Robin boundary condition VOLODYMYR HRYNKIV AND ALICE TURCHANINOVA 89 On the covering number of S14 RYAN OPPENHEIM AND ERIC SWARTZ 97 Upper and lower bounds on the speed of a one-dimensional excited random walk ERIN MADDEN, BRIAN KIDD, OWEN LEVIN, JONATHON PETERSON, JACOB SMITH AND KEVIN M. STANGL 117 Classifying linear operators over the octonions ALEX PUTNAM AND TEVIAN DRAY 125 Spectrum of the Kohn Laplacian on the Rossi sphere TAWFIK ABBAS, MADELYNE M. BROWN, RAVIKUMAR RAMASAMI AND YUNUS E. ZEYTUNCU 141 On the complexity of detecting positive eigenvectors of nonlinear cone maps BAS LEMMENS AND LEWIS WHITE 151 Antiderivatives and linear differential equations using matrices YOTSANAN MEEMARK AND SONGPON SRIWONGSA 157 Patterns in colored circular permutations DANIEL GRAY, CHARLES LANNING AND HUA WANG 171 Solutions of boundary value problems at resonance with periodic and antiperiodic boundary conditions ALDO E. GARCIA AND JEFFREY T. NEUGEBAUER |
9944 | https://corestandards.org/ | Common Core State Standards Initiative – Preparing America's Students for College & Career
Skip to content
Common Core State Standards
Common Core State Standards
Disclaimer:On this page you can find the Common Core State Standards in Mathematics and English Language Arts. The standards are temporarily located on this page while we work to address technical challenges with CoreStandards.org. Thank you for your patience and we will share an update on CoreStandards.org when we have one.
English Language Arts Standards
English Language Arts Standards (ADA Compliant)
Mathematics Standards
Mathematics Standards (ADA Compliant)
© 2025 Common Core State Standards Initiative
Common Core State Standards |
9945 | https://www.doubtnut.com/qna/20476941 | Explain why cations are smaller and anions larger in radii than their parent atoms?
More from this Exercise
A cation has a fewer number of electrons than its parent atom, while its nuclear charge remains the same. As a result, the attraction of electrons to the nucleus is more in a cation than in its parent atom. Therefore, a cation is smaller in size than its parent atom. On the other hand, an anion has one or more electrons than its parent atom, resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. As a result, the distance between the valence electrons and the nucleus is more in anions than in it’s the parent atom. Hence, an anion is larger in radius than its parent atom.
Topper's Solved these Questions
Explore 40 Videos
Explore 40 Videos
Explore 20 Videos
Similar Questions
Explain why callions are smaller and anions larger in radii than their parent atoms?
Explain why cations are smaller and anions larger in radii than their parent atoms?
Knowledge Check
The cation is ________ and the anion is ________ than the parent atom.
Consider the following statements : (I) The radius of an anion is larger than that of the parent atom (II) The ionization energy generally increases with increasing atomic number in a period. (III) The electronegativity of an element is the tendency of an isolated atom to at tract an electron. Which of the above statements is/are correct?
Consider the following statements: I. The radius of an anion is larger than that of parent atom II. The I.E. increases from left to right in a period generally III. The electronegativity of an element is the tendency of an isolated atom to attract an electron The Correct statements are-
Cations are smaller than their parent atom whereas anions are larger in size than their parent atom. Explain.
Why are cations smaller than neutral atom?
Anion size is always _______ than the parent atom.
Give reason for the following.
a) The electron gain enthalpy of O is less negative than that of S.
b) The size of anion is larger than the parent atom.
Why Ca has larger atomic radius than Al ?
NCERT-CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES-EXERCISE
What do you understand by isoelectronic species? Name a species that w...
Consider the following species: N^(3-),O^(2-),F^(ө),Na^(o+),Mg^(2+) ...
Explain why cations are smaller and anions larger in radii than their ...
What is the significance of the terms-'isolated gaseous atom' and 'gro...
Energy of an electron in the ground state of the hydrogen atom is -2.1...
Among the second period elements the actual ionisation enthalpies are ...
How would you explain the fact that the first ionisation enthalpy of s...
What are the various factors due to which the ionisation enthalpy of t...
The first ionisation enthalpy of group 13 elements are : Explain ...
Which of the following pairs of elements would have a more negative el...
Would you expect the second electron gain enthalpy of O as positive, m...
What is the basic difference between the terms electron gain enthalpy ...
How would you react to the statement that the electronegativity of N o...
Describe the theory associated with the radius of an atom as it a. g...
Would you expect the first ionisation enthalpies for two isotopes of t...
What are the major differences between metals and non-metals?
Use the periodic table to answer the following questions. a. Identif...
The increasing order of reactivity among group 1 elements is LiltNaltK...
Write the general outer electronic configuration of s-,p-,d- and f-blo...
Assign the position of the element having outer electronic configurati...
Exams
Free Textbook Solutions
Free Ncert Solutions English Medium
Free Ncert Solutions Hindi Medium
Boards
Resources
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams.
Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation
Contact Us |
9946 | https://dictionary.cambridge.org/us/dictionary/english/incongruity | Cambridge Dictionary +Plus
My profile
+Plus help
Log out
{{userName}}
Cambridge Dictionary +Plus
My profile
+Plus help
Log out
Log in
/
Sign up
English (US)
Meaning of incongruity in English
Add to word list
Add to word list
the fact that something is incongruous
SMART Vocabulary: related words and phrases
Faults and mistakes
(that's) your hard luck idiom
aberration
Achilles heel
adrift
be at fault for something/doing something
black mark
electrical fault
errata
erratum
error
failing
make a gaffe
malaprop
miscue
misfunction
misperception
systematic bias
that's/it's your funeral! idiom
tragic flaw
typo
See more results »
(Definition of incongruity from the Cambridge Advanced Learner's Dictionary & Thesaurus © Cambridge University Press)
Examples of incongruity
incongruity
This was an incongruity, as someone rich enough to buy such a home would surely want modern communications -- unless they were hiding.
From VentureBeat
Though she views such incongruities as "a bit muddy," she said she wishes the auction agency and seller luck.
From CNN
The result created an incongruity of the breast as a source of infant nutrition.
From Slate Magazine
This incongruity over the group's designation is more than skin-deep; it reflects some confusion over the group's original aims.
From International Business Times
Happily for the network, it's an incongruity that seems to leave viewers unbothered.
From Washington Post
Not that the incongruity always works out for the band -- or the audience.
From NPR
The incongruity is even starker outside the upper chamber, he said.
From Huffington Post
When usual good guys go bad at the movies, the incongruity often results in some of their most memorable performances.
From The Atlantic
The venture capital gap is real, and it is an agonizing incongruity.
From CNBC
Nobody's pondering the difference between the relief and incongruity theories of humor.
From Slate Magazine
There's the incongruity of the context -- what cop could summon this much alacrity in a facility this drab?
From New York Times
One approach is the assumption that humor comes from an incongruity between two things.
From CNN
It's an incongruity, a failed law.
From NPR
These examples are from corpora and from sources on the web. Any opinions in the examples do not represent the opinion of the Cambridge Dictionary editors or of Cambridge University Press or its licensors.
What is the pronunciation of incongruity?
Translations of incongruity
in Chinese (Traditional)
不協調, 不合適…
See more
in Chinese (Simplified)
不协调, 不合适…
See more
in Spanish
incongruencia…
See more
in Portuguese
incongruência…
See more
in more languages
in French
in Turkish
in Dutch
in Czech
in Danish
in Indonesian
in Thai
in Vietnamese
in Polish
in Swedish
in Malay
in German
in Norwegian
in Ukrainian
incongruité…
See more
uymazlık, uygunsuzluk…
See more
ongerijmdheid, misplaatstheid…
See more
neshoda, nesoulad…
See more
uoverensstemmelse…
See more
keanehan…
See more
ความไม่สอดคล้องกัน…
See more
sự không thích hợp…
See more
niestosowność…
See more
oförenlighet, inkongruens…
See more
ketidaksesuaian…
See more
die Nichtübereinstimmung…
See more
uoverensstemmelse, selvmotsigelse…
See more
невідповідність, недоречність…
See more
Need a translator?
Get a quick, free translation!
Translator tool
Browse
inconformity
incongruence
incongruent
incongruently
incongruity
incongruous
incongruously
inconsecutive
inconsequent
Cambridge
Dictionary +Plus
Learn more with +Plus
Create word lists and quizzes for free
Sign up
or
Log in
Word of the Day
take something back
to admit that something you said was wrong
About this
Blog
Calm and collected (The language of staying calm in a crisis)
Read More
New Words
vibe coding
More new words
has been added to list
To top
Contents
EnglishExamplesTranslations
Cambridge Dictionary +Plus
My profile
+Plus help
Log out
English (US)
Change
English (UK)
English (US)
Español
Português
中文 (简体)
正體中文 (繁體)
Dansk
Deutsch
Français
Italiano
Nederlands
Norsk
Polski
Русский
Türkçe
Tiếng Việt
Svenska
Українська
日本語
한국어
ગુજરાતી
தமிழ்
తెలుగు
বাঙ্গালি
मराठी
हिंदी
Follow us
Choose a dictionary
Recent and Recommended
English
Grammar
English–Spanish
Spanish–English
Definitions
Clear explanations of natural written and spoken English
English
Learner’s Dictionary
Essential British English
Essential American English
Grammar and thesaurus
Usage explanations of natural written and spoken English
Grammar
Thesaurus
Pronunciation
British and American pronunciations with audio
English Pronunciation
Translation
Click on the arrows to change the translation direction.
Bilingual Dictionaries
English–Chinese (Simplified)
Chinese (Simplified)–English
English–Chinese (Traditional)
Chinese (Traditional)–English
English–Dutch
Dutch–English
English–French
French–English
English–German
German–English
English–Indonesian
Indonesian–English
English–Italian
Italian–English
English–Japanese
Japanese–English
English–Norwegian
Norwegian–English
English–Polish
Polish–English
English–Portuguese
Portuguese–English
English–Spanish
Spanish–English
English–Swedish
Swedish–English
Semi-bilingual Dictionaries
English–Arabic
English–Bengali
English–Catalan
English–Czech
English–Danish
English–Gujarati
English–Hindi
English–Korean
English–Malay
English–Marathi
English–Russian
English–Tamil
English–Telugu
English–Thai
English–Turkish
English–Ukrainian
English–Urdu
English–Vietnamese
Dictionary +Plus
Word Lists
Contents
English
Examples
Translations
Grammar
All translations
My word lists
To add incongruity to a word list please sign up or log in.
Sign up
or
Log in
My word lists
Add incongruity to one of your lists below, or create a new one.
{{name}}
Go to your word lists
Tell us about this example sentence: |
9947 | https://www.cuemath.com/geometry/unit-circle/ | Unit Circle
A unit circle from the name itself defines a circle of unit radius. A circle is a closed geometric figure without any sides or angles. The unit circle has all the properties of a circle, and its equation is also derived from the equation of a circle. Further, a unit circle is useful to derive the standard angle values of all the trigonometric ratios.
Here we shall learn the equation of the unit circle, and understand how to represent each of the points on the circumference of the unit circle, with the help of trigonometric ratios of cosθ and sinθ.
| | |
--- |
| 1. | What is a Unit Circle? |
| 2. | Finding Trigonometric Functions Using a Unit Circle |
| 3. | Unit Circle with Sine Cos and Tan |
| 4. | Unit Circle Chart |
| 5. | Unit Circle and Trigonometric Identities |
| 6. | Unit Circle Pythagorean Identities |
| 7. | Unit Circle and Trigonometric Values |
| 8. | Unit Circle in Complex Plane |
| 9. | FAQs on Unit Circle |
What is Unit Circle?
A unit circle is a circle with a radius measuring 1 unit. The unit circle is generally represented in the cartesian coordinate plane. The unit circle is algebraically represented using the second-degree equation with two variables x and y. The unit circle has applications in trigonometry and is helpful to find the values of the trigonometric ratios sine, cosine, tangent.
Unit Circle Definition
The locus of a point which is at a distance of one unit from a fixed point is called a unit circle.
Equation of a Unit Circle
The general equation of a circle is (x - a)2 + (y - b)2 = r2, which represents a circle having the center (a, b) and the radius r. This equation of a circle is simplified to represent the equation of a unit circle. A unit circle is formed with its center at the point(0, 0), which is the origin of the coordinate axes. and a radius of 1 unit. Hence the equation of the unit circle is (x - 0)2 + (y - 0)2 = 12. This is simplified to obtain the equation of a unit circle.
Equation of a Unit Circle: x2 + y2 = 1
Here for the unit circle, the center lies at (0,0) and the radius is 1 unit. The above equation satisfies all the points lying on the circle across the four quadrants.
Finding Trigonometric Functions Using a Unit Circle
We can calculate the trigonometric functions of sine, cosine, and tangent using a unit circle. Let us apply the Pythagoras theorem in a unit circle to understand the trigonometric functions. Consider a right triangle placed in a unit circle in the cartesian coordinate plane. The radius of the circle represents the hypotenuse of the right triangle. The radius vector makes an angle θ with the positive x-axis and the coordinates of the endpoint of the radius vector is (x, y). Here the values of x and y are the lengths of the base and the altitude of the right triangle. Now we have a right angle triangle with the sides 1, x, y. Applying this in trigonometry, we can find the values of the trigonometric ratio, as follows:
sinθ = Altitude/Hypoteuse = y/1
cosθ = Base/Hypotenuse = x/1
We now have sinθ = y, cosθ = x, and using this we now have tanθ = y/x. Similarly, we can obtain the values of the other trigonometric ratios using the right-angled triangle within the unit circle. Also by changing the θ values we can obtain the principal values of these trigonometric ratios.
Unit Circle with Sin Cos and Tan
Any point on the unit circle has coordinates(x, y), which are equal to the trigonometric identities of (cosθ, sinθ). For any values of θ made by the radius line with the positive x-axis, the coordinates of the endpoint of the radius represent the cosine and the sine of the θ values. Here we have cosθ = x, and sinθ = y, and these values are helpful to compute the other trigonometric ratio values. Applying this further we have tanθ = sinθ/cosθ or tanθ = y/x.
Another important point to be understood is that the sinθ and cosθ values always lie between 1 and -1, and the radius value is 1, and it has a value of -1 on the negative x-axis. The entire circle represents a complete angle of 360º and the four quadrant lines of the circle make angles of 90º, 180º, 270º, 360º(0º). At 90º and at 270º the cosθ value is equal to 0 and hence the tan values at these angles are undefined.
Example: Find the value of tan 45º using sin and cos values from the unit circle.
Solution:
We know that, tan 45° = sin 45°/cos 45°
Using the unit circle chart:
sin 45° = 1/√2
cos 45° = 1/√2
Therefore, tan 45° = sin 45°/cos 45°
= (1/√2)/(1/√2)
= 1
Answer: Therefore, tan 45° = 1
Unit Circle Chart in Radians
The unit circle represents a complete angle of 2π radians. And the unit circle is divided into four quadrants at angles of π/2, π. 3π/2, and 2π respectively. Further within the first quadrant at the angles of 0, π/6, π/4, π/3, π/2 are the standard values, which are applicable to the trigonometric ratios. The points on the unit circle for these angles represent the standard angle values of the cosine and sine ratios. On close observation of the below figure the values are repeated across the four quadrants, but with a change in sign. This change in sign is because of the reference x-axis and y-axis, which are positive on one side and negative on the other side of the origin. Now with the help of this, we can easily find the trigonometric ratio values of standard angles, across the four quadrants of the unit circle.
Unit Circle and Trigonometric Identities
The unit circle identities of sine, cosecant, and tangent can be further used to obtain the other trigonometric identities such as cotangent, secant, and cosecant. The unit circle identities such as cosecant, secant, cotangent are the respective reciprocal of the sine, cosine, tangent. Further, we can obtain the value of tanθ by dividing sinθ with cosθ, and we can obtain the value of cotθ by dividing cosθ with sinθ.
For a right triangle placed in a unit circle in the cartesian coordinate plane, with hypotenuse, base, and altitude measuring 1, x, y units respectively, the unit circle identities can be given as,
sinθ = y/1
cosθ = x/1
tanθ = sinθ/cosθ = y/x
sec(θ = 1/x
csc(θ) = 1/y
cot(θ) = cosθ/sinθ = x/y
Unit Circle Pythagorean Identities
The three important Pythagorean identities of trigonometric ratios can be easily understood and proved with the unit circle. The Pythagoras theorem states that in a right-angled triangle the square of the hypotenuse is equal to the sum of the square of the other two sides. The three Pythagorean identities in trigonometry are as follows.
sin2θ + cos2θ = 1
1 + tan2θ = sec2θ
1 + cot2θ = cosec2θ
Here we shall try to prove the first identity with the help of the Pythagoras theorem. Let us take x and y as the legs of the right-angled triangle having a hypotenuse 1 unit. Applying Pythagoras theorem we have x2 + y2 = 1 which represents the equation of a unit circle. Also in a unit circle, we have, x = cosθ, and y = sinθ, and applying this in the above statement of the Pythagoras theorem, we have, cos2θ + sin2θ = 1. Thus we have successfully proved the first identity using the Pythagoras theorem. Further within the unit circle, we can also prove the other two Pythagorean identities.
Unit Circle and Trigonometric Values
The various trigonometric identities and their principal angle values can be calculated through the use of a unit circle. In the unit circle, we have cosine as the x-coordinate and sine as the y-coordinate. Let us now find their respective values for θ = 0°, and θ = 90º.
For θ = 0°, the x-coordinate is 1 and the y-coordinate is 0. Therefore, we have cos0º = 1, and sin0º = 0. Let us look at another angle of 90º. Here the value of cos90º = 1, and sin90º = 1. Further, let us use this unit circle and find the important trigonometric function values of θ such as 30º, 45º, 60º. Also, we can also measure these θ values in radians. We know that 360° = 2π radians. We can now convert the angular measures to radian measures and express them in terms of the radians.
Unit Circle Table:
The unit circle table is used to list the coordinates of the points on the unit circle that corresond to common angles with the help of trigonometric ratios.
| Angle θ | Radians | Sinθ | Cosθ | Tanθ = Sinθ/Cosθ | Coordinates |
--- --- --- |
| 0° | 0 | 0 | 1 | 0 | (1, 0) |
| 30° | π/6 | 1/2 | √3/2 | 1/√3 | (√3/2, 1/2) |
| 45° | π/4 | 1/√2 | 1/√2 | 1 | (1/√2, 1/√2) |
| 60° | π/3 | √3/2 | 1/2 | √3 | (1/2, √3/2) |
| 90° | π/2 | 1 | 0 | undefined | (0,1) |
We can find the secant, cosecant, and cotangent functions also using these formulas:
secθ = 1/cosθ
cosecθ = 1/sinθ
cotθ = 1/tanθ
We have discussed the unit circle for the first quadrant. Similarly, we can extend and find the radians for all the unit circle quadrants. The numbers 1/2, 1/√2, √3/2, 0, 1 repeat along with the sign in all 4 quadrants.
Unit Circle in Complex Plane
A unit circle consists of all complex numbers of absolute value as 1. Therefore, it has the equation of |z| = 1. Any complex number z = x + iy will lie on the unit circle with equation given as x2 + y2 = 1.
The unit circle can be considered as unit complex numbers in a complex plane, i.e., the set of complex numbers z given by the form,
z = eit = cos t + i sin t = cis(t)
The relation given above represents Euler's formula.
Read More
Download FREE Study Materials
Unit Circle Worksheet
Unit Circle Worksheet
Worksheet on Trigonometry Level-1
Unit Circle Examples
Example 1: Does the point P (1/2, 1/2) lie on the unit circle?
Solution:
We know that equation of a unit circle is:
x2 + y2 = 1
Substituting x = 1/2 and y = 1/2, we get:
= x2 + y2
= (1/2)2 + (1/2)2
= 1/4 + 1/4
= 1/2
≠ 1
Since, x2 + y2 ≠ 1, the point P (1/2, 1/2) does not lie on the unit circle.
Answer: Therefore (1/2, 1/2) doesn't lie on the unit circle.
2. Example 2: Find the exact value of tan 210° using the unit circle.
Solution:
We know that
tan 210° = sin 210°/cos210°
Using the unit circle chart:
sin 210° = -1/2
cos 210° = -√3/2
Therefore,
tan 210° = sin 210°/cos 210°
= (-1/2)/(-√3/2)
= 1/√3
Answer: Therefore, tan 210° = 1/√3
3. Example 3: Find the value of sin 900° using unit circle chart.
Solution:
Since, the unit circle has 0°- 360°, let us represent 900° in terms of 360°.
900° is 2 full rotations of 360° and an additional rotation of 180°.
Hence, 900° will have the same trigonometric ratio as 180°.
Therefore,
sin 900° = sin 180°
From the unit circle chart, we know that:
sin 180° = 0
Answer: sin 900° = 0
View More >
Breakdown tough concepts through simple visuals.
Math will no longer be a tough subject, especially when you understand the concepts through visualizations.
Book a Free Trial Class
Practice Questions on Unit Circle
Try These! >
FAQs on Unit Circle
What is Unit Circle in Math?
A unit circle is a circle with a radius of one unit. Generally, a unit circle is represented in the coordinate plane with its center at the origin. The equation of the unit circle of radius one unit and having the center at (0, 0) is x2 + y2 = 1. Further, the unit circle has applications in trigonometry and is used to find the principal values of sine and cosine trigonometric ratios.
How Do you Find Sin and Cos Using the Unit Circle?
The unit circle can be used to find the values of sinθ and Cosθ. In a unit circle of radius 1 unit and having the center at (0, 0), let us take a radius inclined to the positive x-axis at an angle θ, and the endpoint of the radius as (x, y). Draw a perpendicular from the end of the radius to the x-axis and it forms a right-angled triangle with the radius as the hypotenuse. The adjacent side of this triangle is the x value, the opposite side of the triangle is the y value and the hypotenuse is of 1 unit. Further using the trigonometric ratio formula we have sinθ = Opp/Hyp = y/1, and cosθ = Adj/Hyp = x/1. Thus we have sinθ = y, and cosθ = x.
What is the Unit Circle Definition of Trig Functions?
The trigonometric function can be calculated for the principal values using the unit circle. For a unit circle having the center at the origin(0, 0), the radius of 1 unit, if the radius is inclined at an angle θ and the endpoint of the radius vector is (x, y), then cosθ = x and sinθ = y. Further, all the other trigonometric ratios can be calculated from these two values. Also, the principal values can be computed by changing the θ value.
How to Find Terminal Point on Unit Circle?
The terminal point on a unit circle can be found with the help of the equation of the unit circle x2 + y2 = 1. If the given point satisfies this equation then it is a point lying on the unit circle. Further, the terminal point on the unit value can be found for the θ value, by finding the values of cosθ and sinθ.
What is the Equation of Unit Circle?
The equation of a unit circle is x2 + y2 = 1. Here it is considered that the unit circle has its center at the origin(0, 0) of the coordinate axes, and has a radius of 1 unit. This equation of unit circle has been derived using the help of the distance formula.
How Do you Derive the Equation of a Unit Circle?
The equation of a unit circle can be calculated using the distance formula of coordinate geometry. For a circle having the center at the origin(0, 0), the radius of 1 unit, any point on the circle can be taken as (x, y). Applying the definition of a circle, and using the distance formula we have (x - 0)2 + (y - 0)2 = 1, which can be simplified as x2 + y2 = 1.
When is Tan Undefined on the Unit Circle?
The unit circle having an equation of x2 + y2 = 1 is helpful to find the trigonometric ratios of sinθ = y and cosθ = x. Using these values we can conveniently find the value of tanθ = sinθ/cosθ = y/x. Tanθ will be undefined for cosθ = 0, i.e., when θ is equal to 90° and 270°.
What is the Connection Between Right Triangles and the Unit Circle?
The right triangles and a unit circle are uniquely connected. Any point on the unit circle can be visualized as a right triangle with radius as the hypotenuse of the right triangle and the coordinates of the point as the other two sides of the right triangle. The equation of a circle x2 + y2 = 1 completely satisfies the Pythagoras theorem related to the right triangle. Also, the right triangle within the unit circle is helpful to derive the trigonometric ratio values.
What is the Unit Circle Used for?
The unit circle is prominently useful in trigonometry. For the trigonometric ratios of sinθ, cosθ, tanθ, their principal angle values of 0º, 30º, 45º, 60º, 90º can be easily calculated using the unit circle. Additionally, the unit circle is useful to represent complex numbers in the argand plane.
What are the Quadrants of the Unit Circle?
The unit circle has four quadrants similar to the quadrants in the coordinate system. The four quadrants are of equal area and they represent one-fourth of the area of the circle. Each of the quadrants subtends an angle of 90º or a right angle at the center of the circle.
How Do you Describe a Unit Circle in Terms of Complex Numbers?
A unit circle consists of all complex numbers of absolute value as 1, thus any complex number z = x + iy will lie on the unit circle with equation given as x2 + y2 = 1. Therefore, the equation of the unit circle can be given as |z| = 1.
Q1: What is tan 45 degrees?
Q2: What is tan 90 degrees in radians using unit circle?
Q3: What is the general form of unit circle?
Q4: What is the value of 4cos0°−4sin0°+sec0°?
Q5: What is the relationship between the sides of the sine function?
Math worksheets and
visual curriculum
FOLLOW CUEMATH
Facebook
Youtube
Instagram
Twitter
LinkedIn
Tiktok
MATH PROGRAM
Online math classes
Online Math Courses
online math tutoring
Online Math Program
After School Tutoring
Private math tutor
Summer Math Programs
Math Tutors Near Me
Math Tuition
Homeschool Math Online
Solve Math Online
Curriculum
NEW OFFERINGS
Coding
SAT
Science
English
MATH ONLINE CLASSES
1st Grade Math
2nd Grade Math
3rd Grade Math
4th Grade Math
5th Grade Math
6th Grade Math
7th Grade Math
8th Grade Math
ABOUT US
Our Mission
Our Journey
Our Team
MATH TOPICS
Algebra 1
Algebra 2
Geometry
Calculus math
Pre-calculus math
Math olympiad
QUICK LINKS
Maths Games
Maths Puzzles
Our Pricing
Math Questions
Events
MATH WORKSHEETS
Kindergarten Worksheets
1st Grade Worksheets
2nd Grade Worksheets
3rd Grade Worksheets
4th Grade Worksheets
5th Grade Worksheets
6th Grade Worksheets
7th Grade Worksheets
8th Grade Worksheets
9th Grade Worksheets
10th Grade Worksheets
Terms and ConditionsPrivacy Policy |
9948 | https://mathworld.wolfram.com/FerrersDiagram.html | Ferrers Diagram -- from Wolfram MathWorld
TOPICS
AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld
Discrete Mathematics
Combinatorics
Partitions
History and Terminology
Wolfram Function Repository
Ferrers Diagram
A Ferrers diagram represents partitions as patterns of dots, with the th row having the same number of dots as the th term in the partition. The spelling "Ferrars" (Skiena 1990, pp.53 and 78) is sometimes also used, and the diagram is sometimes called a graphical representation or Ferrers graph (Andrews 1998, p.6). A Ferrers diagram of the partition
for a list , , ..., of positive integers with is therefore the arrangement of dots or square boxes in rows, such that the dots or boxes are left-justified, the first row is of length , the second row is of length , and so on, with the th row of length . The above diagram corresponds to one of the possible partitions of 100.
The Ferrers diagram of a given partition is implemented in the Wolfram Function Repository as ResourceFunction["FerrersDiagram"][n].
The partitions of integers less than or equal to in which there are at most parts and in which no part is larger than correspond (1) to Young tableaux which fit inside an rectangle and (2) to lattice paths which travel from the upper right corner of the rectangle to the lower left in leftward and downward steps. The number of Young diagrams fitting inside an rectangle is given by the binomial coefficient. The above example shows the
Young diagrams.
See also
Conjugate Partition, Durfee Square, Self-Conjugate Partition, Staircase Walk
Explore with Wolfram|Alpha
More things to try:
partitions
Baudet's conjecture
8:5 odds, bet 97 euros
References
Andrews, G.E. The Theory of Partitions. Cambridge, England: Cambridge University Press, pp.6-7, 1998.Comtet, L. "Ferrers Diagrams." §2.4 in Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. enl. ed. Dordrecht, Netherlands: Reidel, pp.98-102, 1974.Liu, C.L. Introduction to Combinatorial Mathematics. New York: McGraw-Hill, 1968.MacMahon, P.A. Combinatory Analysis, Vol.2. New York: Chelsea, pp.3-4, 1960.Propp, J. "Some Variants of Ferrers Diagrams." J. Combin. Th. A52, 98-128, 1989.Riordan, J. An Introduction to Combinatorial Analysis. New York: Wiley, pp.108-109, 1980.Skiena, S. "Ferrers Diagrams." §2.1.2 in Implementing Discrete Mathematics: Combinatorics and Graph Theory with Mathematica. Reading, MA: Addison-Wesley, pp.53-55, 1990.Stanley, R.P. Enumerative Combinatorics, Vol.1. Cambridge, England: Cambridge University Press, 1999.Stanton, D. and White, D. Constructive Combinatorics. New York: Springer-Verlag, 1986.
Referenced on Wolfram|Alpha
Ferrers Diagram
Cite this as:
Weisstein, Eric W. "Ferrers Diagram." From MathWorld--A Wolfram Resource.
Subject classifications
Discrete Mathematics
Combinatorics
Partitions
History and Terminology
Wolfram Function Repository
About MathWorld
MathWorld Classroom
Contribute
MathWorld Book
wolfram.com
13,278 Entries
Last Updated: Sun Sep 28 2025
©1999–2025 Wolfram Research, Inc.
Terms of Use
wolfram.com
Wolfram for Education
Created, developed and nurtured by Eric Weisstein at Wolfram Research
Created, developed and nurtured by Eric Weisstein at Wolfram Research |
9949 | https://www.merriam-webster.com/sentences/hobnob | Examples of 'HOBNOB' in a Sentence | Merriam-Webster
Chatbot
Chatbot
Games
Word of the Day
Grammar
Word Finder
Slang
NewNewsletters
Wordplay
Rhymes
Thesaurus
Join MWU
More
Games
Word of the Day
Grammar
Wordplay
Slang
Rhymes
Word Finder
Newsletters New
Thesaurus
Join MWU
Shop
Books
Merch
Log In
Username
My Words
Recents
Account
Log Out
Est. 1828
Example Sentenceshobnob
verb
How to Use hobnob in a Sentence
hobnob
verb
Definition of hobnob
Synonyms for hobnob
He loves to hobnob with celebrities.
Please drop by to see me from 10 a.m. to 1 p.m. to hobnobor share one of your precious keepsakes.
— Steven Wayne Yvaska, The Mercury News, 4 July 2019
Throughout the festivities, the A-list celebrities made the rounds and hobnobbed with friends in the massive white tent.
— Paul Chi, Vogue, 23 Feb. 2025
Throughout the festivities, the A-list celebrities made the rounds and hobnobbedwith friends in the massive white tent.
— Paul Chi, Vogue, 23 Feb. 2025
When not hobnobbingwith the hip, Cameron tends to play-act a bit onstage and on record as a showbiz failure.
— Milwaukee, Milwaukee Journal Sentinel, 11 Jan. 2018
For his part, Musk has been busy hobnobbingwith heads of state and the future president.
— Anna Kaufman, USA TODAY, 26 Nov. 2024
The movie contains lots of footage from Stars’ heyday, which shows Tower hobnobbing with the politicians and celebrities of the era.
— Mick Lasalle, kansascity.com, 24 May 2017
The movie contains lots of footage from Stars’ heyday, which shows Tower hobnobbing with the politicians and celebrities of the era.
— Mick Lasalle, Orange County Register, 25 Apr. 2017
Emma Stone seemed to be hobnobbingmore than most A-listers.
— Nate Jones, Vulture, 17 Feb. 2024
The tall, burly Mr. Gilmer frequently hobnobbedwith the world’s elite wine collectors and growers.
— James B. Stewart, New York Times, 25 May 2023
This networking event is sure to give you some insight – and the chance to hobnobwith folks who build festival lineups for a living.
— Detroit Free Press Staff, Detroit Free Press, 14 Apr. 2022
The number of times up-and-coming artists get to hobnob with music-industry bigwigs?
— Brendan Kiley, The Seattle Times, 4 May 2017
The lobby captures the art deco glamor of the Jazz Age, when flappers hobnobbedwith royalty.
— Cnt Editors, Condé Nast Traveler, 13 Dec. 2023
The other women, while not hobnobbingas much with the A-lister, had floor seats and were very defensive about that level of access.
— Shamira Ibrahim, Vulture, 3 June 2024
Anyone who’s anyone seemed to flock to Kives’ estate to save the world while hobnobbing, and many appeared to depart with a major takeaway: a Skims swag bag.
— Emily St. Martin, Los Angeles Times, 27 May 2023
The politicians, pundits and reporters caught in an endless cycle get together to hobnoband talk about the importance of freedom of the press.
— Staff, cleveland.com, 30 Apr. 2018
But the optics are tricky since Youngkin will spend precious time in the state election’s homestretch hobnobbingwith out-of-state donors, some with no clear interest in Virginia.
— Laura Vozzella, Washington Post, 30 Sep. 2023
For Iowans, the growing prominence of the caucuses brought attention, money, and a chance to hobnobwith the political class.
— Sam Walker, WSJ, 7 Feb. 2020
The next time members will be able to hobnobwon't be until the end of October, when the winter season kicks off again with the club's annual Halloween party.
— Kate Bennett, CNN, 19 May 2021
There, Delvey had been hoping to establish a members-only social and cultural club where the crème de la crème could hobnob.
— Tori Latham, Robb Report, 3 Nov. 2022
Gone are the opportunities to hobnobat private views or first-night parties.
—The Economist, 2 June 2020
Anyone who is—or wants to be—a hot ticket rocks up on the VIP days to see art and to be seen seeing art, before hobnobbingat the evenings’ gallery dinners, parties, and after-parties.
— George Nelson, ARTnews.com, 12 Oct. 2024
In the four-month tour that followed, the prince rode elephants, hobnobbedwith maharajahs, visited the Taj Mahal by moonlight—and found plenty of treasures.
— J.s. Marcus, WSJ, 18 May 2018
The long-married Muriel is nervous about seeing her old boyfriend, who now hobnobs with the elite of Hollywood, and is uncertain about her feelings toward him.
— Punch Shaw, star-telegram.com, 16 May 2017
Jeffrey Epstein has hobnobbedwith some of the world's most powerful people during his jet-setting life.
—CBS News, 9 July 2019
Jeffrey Epstein has hobnobbedwith some of the world's most powerful people during his jet-setting life.
—CBS News, 9 July 2019
Jeffrey Epstein has hobnobbedwith some of the world's most powerful people during his jet-setting life.
— Jill Colvin, sun-sentinel.com, 8 July 2019
Jeffrey Epstein has hobnobbedwith some of the world's most powerful people during his jet-setting life.
— Jill Colvin, chicagotribune.com, 8 July 2019
Jeffrey Epstein has hobnobbedwith some of the world's most powerful people during his jet-setting life.
—CBS News, 9 July 2019
That’s not to say Coyne wasn’t proud of his role in music history, which included perks of hobnobbingwith industry’s elite.
— Rob Ledonne, Billboard, 12 Apr. 2018
Some of these examples are programmatically compiled from various online sources to illustrate current usage of the word 'hobnob.' Any opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback about these examples.
Last Updated: 21 Aug 2025
More from Merriam-Webster
### Can you solve 4 words at once? Play Play ### Can you solve 4 words at once? Play Play
Word of the Day
obliterate
See Definitions and Examples »
Get Word of the Day daily email!
Merriam Webster
Learn a new word every day. Delivered to your inbox!
Help
About Us
Advertising Info
Contact Us
Privacy Policy
Terms of Use
Facebook
Twitter
YouTube
Instagram
© 2025 Merriam-Webster, Incorporated
✕
Do not sell or share my personal information.
You have chosen to opt-out of the sale or sharing of your information from this site and any of its affiliates. To opt back in please click the "Customize my ad experience" link.
This site collects information through the use of cookies and other tracking tools. Cookies and these tools do not contain any information that personally identifies a user, but personal information that would be stored about you may be linked to the information stored in and obtained from them. This information would be used and shared for Analytics, Ad Serving, Interest Based Advertising, among other purposes.
For more information please visit this site's Privacy Policy.
CANCEL
CONTINUE
Information from your device can be used to personalize your ad experience.
Do not sell or share my personal information. |
9950 | https://testbook.com/question-answer/the-dimensions-of-emf-are--5f64b48ac55dbf275d952c1e | [Solved] The dimensions of EMF are
Get Started
ExamsSuperCoachingTest SeriesSkill Academy
More
Pass
Skill Academy
Free Live Classes
Free Live Tests & Quizzes
Previous Year Papers
Doubts
Practice
Refer & Earn
All Exams
Our Selections
Careers
English
Hindi
Home
Physics
Units, Dimensions and Measurements
Dimensions of physical quantities
Question
Download Solution PDF
The dimensions of EMF are
This question was previously asked in
BSF RM Official Paper (Conducted on 22-Sep-2019)
Download PDFAttempt Online
View all BSF RO Papers >
ML 2 T-3 A-1
ML 2 T 2 A 3
M-1 T 3
ML 3 T 1 A 3
Answer (Detailed Solution Below)
Option 1 : ML 2 T-3 A-1
Crack with India's Super Teachers FREE Demo Classes Available Explore Supercoaching For FREE
Free Tests
View all Free tests >
Free
General Knowledge for All Defence Exams (शूरवीर): Special Live Test
31.9 K Users
20 Questions 20 Marks 16 Mins
Start Now
Detailed Solution
Download Solution PDF
CONCEPT:
Emf of the cell (E):
The potential difference across the terminals of a cell when it is not supplying any current is called its emf.
Equation of cell can be written as
⇒ E = V + Ir
Where E = emf of the cell, V =Potential difference, I = current and r = internal resistance
EXPLANATION:
When no current is passing through the cell, then I = 0 A, hence
⇒ E = V
As we knowelectric potential can be is written as
⇒E=V=W o r k d o n e(W)C h a r g e(q)
The dimension of work done (W) = [ML 2 T-2]
The dimension of charge (q) = [IT]
∴ The dimension of emf is given by
⇒E=M L 2 T−2 A T=[M L 2 A−1 T−3]
Important Points
Quantity Unit Dimension
Pressure Pascal[ML-1 T-2]
Stress Pascal[ML-1 T-2]
Young Modulus Pascal[ML-1 T-2]
Speed m/s[LT-1]
Momentum kg⋅m/s[MLT-1]
Potential difference Volt[ML 2 T-3 A-1]
Torque Newton meter[M 1 L 2 T-2]
Work Joule[ML 2 T-2]
Energy Joule[ML 2 T-2]
Weight Newton[MLT-2]
Capacitance (C)Coulomb/volt or Farad[M-1 L 2 T 4 A 2]
Resistivity or Specific resistance (ρ)Ohm-meter[ML 3 T-3 A-2]
Electric current (I)Ampere[A]
Download Solution PDFShare on Whatsapp
Latest BSF RO Updates
Last updated on Aug 20, 2025
->BSF RO Notification 2025 has been released for 910 vacancies.
->Interested candidates can apply between 24th August to 23rd September 2025.
->Candidates selected as BSF Radio Operators will receive a salary between Rs 25,500 to Rs. 81,100.
->The candidates can check theBSF RO Previous Year Paperswhich helps to give clarity on the difficulty level of the exam.
India’s #1 Learning Platform
Start Complete Exam Preparation
Daily Live MasterClasses
Practice Question Bank
Mock Tests & Quizzes
Get Started for Free
Trusted by 7.6 Crore+ Students
More Dimensions of physical quantities Questions
Q1.Which one of the following is dimensionless quantity?
Q2.Which of the following pairs of physical quantities has the same dimensions?
Q3.Which of the following is dimensionless?
Q4.What is the dimension of gravitational constant?
Q5.Which of the following pairs of physical quantities does have same dimensional formula?
Q6.Find the dimension of the ratio of the pressure to the stress.
Q7.The dimensional formula of force:
Q8.What is the dimension of coefficient of friction?
Q9.If the product of 'muscular strength' and speed is equal to dimensions of power, then the dimensions of 'muscular strength' is:
More Units, Dimensions and Measurements Questions
Q1.Which one of the following is dimensionless quantity?
Q2.A physical quantity P is related to four observations a, b, c and d as follows: P = a³b² / c√d The percentage errors of measurement in a, b, c and d are 1%, 3%, 2%, and 4% respectively. The percentage error in the quantity P is:
Q3.Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at x = 0.1 cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is M = 5 cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction is:
Q4.A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a gas of density ρ and takes a spherical shape of radius R. When the gas is allowed to flow freely out of it, its radius r changes from R to 0 (zero) in time T. If the speed v(r) of gas coming out of the balloon depends on r as ra and T ∝ Sα Aβ ργ Rδ
Q5.The unit of charge in SI system is
Q6.The number of significant figures in 1.73 seconds is _____.
Q7.Ev is the unit of
Q8.The dimensions of Plank’s Constant equals to that of
Q9.An object has mass 'M' on Earth. The mass of the object at Moon = ?
Q10.The dimensions of electrical conductivity is
Crack Super Pass Live with India's Super Teachers Ananya Singh Testbook Lalit Kumar Testbook Explore Supercoaching For FREE
Suggested Test Series
View All >
BSF Head Constable (RO/RM) Mock Test Series 2025
190 Total Tests with 1 Free Tests
Start Free Test
General Knowledge for All Defence Exams
317 Total Tests with 1 Free Tests
Start Free Test
Suggested Exams
Indian Coast Guard Navik GD Army Technical Agniveer Navy SSR Agniveer Indian Navy Agniveer SSR
Indian Coast Guard Navik GD Important Links
More Physics Questions
Q1.Flying of bird is a proof of Newton's
Q2.For a hydrogen atom the potential energy of the electron in the field of the nucleus is given by:
Q3.A system that does NOT allow exchange of heat with its surrounding is called
Q4.Which one of the following is dimensionless quantity?
Q5.If the length of a copper wire is increased by twice, then its resistivity will be
Q6.What happens when the sunlight travels through the Earth’s atmosphere?
Q7.Which one of the following is NOT a basic property of electric charge?
Q8.The length of a simple pendulum is increased four times to its previous value while the mass is doubled. What is the ratio of the new and previous time period of the pendulum?
Q9.The frequency (f), wavelength (λ) and speed (v) of a sound wave are related as
Q10.Two bodies of unequal masses are dropped from a tower. At any instant, they have equal
Important Exams
SSC CGLSSC CHSLSSC JESSC CPO
IBPS POIBPS ClerkIBPS RRB POIBPS RRB Clerk
IBPS SOSBI POSBI ClerkCUET
UGC NETRBI Grade BRBI AssistantUPSC IAS
UPSC CAPF ACUPSC CDSUPSC IESUPSC NDA
RRB NTPCRRB Group DRRB JERRB SSE
LIC AAOLIC AssistantNABARD Development AssistantSEBI Grade A
Super Coaching
UPSC CSE CoachingBPSC CoachingAE JE electrical CoachingAE JE mechanical Coaching
AE JE civil Coachingbihar govt job CoachingGATE mechanical CoachingSSC Coaching
CUET CoachingGATE electrical CoachingRailway CoachingGATE civil Coaching
Bank Exams CoachingCDS CAPF AFCAT CoachingGATE cse CoachingGATE ece Coaching
CTET State TET CoachingCTET CoachingUPTET CoachingREET Coaching
MPTET CoachingJTET Coaching
Exams
Indian Coast GuardNavy MRIndian Navy TradesmanIndian Coast Guard Assistant Commandant
Navy AAIndian Coast Guard YantrikIndian Coast Guard Navik GDIndian Navy SSC Officer
Navy Tradesman MateIndian Coast Guard Navik DBIndian Coast Guard Group CIndian Navy Agniveer
Naval Ship Repair Yard Motor DriverIndian Coast Guard Group BIndian Navy Civilian PersonnelIndian Navy Driver
Indian Navy Chargeman
Test Series
Indian Coast Guard Assistant Commandant (GD/CPL) Mock TestAgniveer Navy SSR Mock TestCoast Guard Navik GD Mock TestNavy Tradesman Mate Mock Test
Coast Guard Navik DB (Domestic Branch) Mock Test
Previous Year Paper
Indian Coast Guard Previous Year PapersNavy MR Previous Year PapersIndian Coast Guard Assistant Commandant Previous Year PapersIndian Coast Guard Navik GD Previous Year Papers
Indian Coast Guard Navik DB Previous Year PapersIndian Navy Agniveer Previous Year PapersIndian Navy SSR Agniveer Previous Year Paper
Objective Question
Electric Drives MCQFourier Transform MCQGravitation MCQInternational Marketing MCQ
Law Of Motion MCQLinear Programming MCQMoving Charges And Magnetism MCQMysql MCQ
Notice MCQOne Word Substitution MCQP Block MCQPhotosynthesis MCQ
Profit And Loss MCQRespiration In Plant MCQRural Development MCQcurrent affairs MCQ
computers MCQms word MCQconstitution of india MCQenvironmental issues MCQ
co ordinate geometry MCQbig data MCQstates of matter MCQacids bases and salts MCQ
Testbook Edu Solutions Pvt. Ltd.
1st & 2nd Floor, Zion Building,
Plot No. 273, Sector 10, Kharghar,
Navi Mumbai - 410210
support@testbook.com
Toll Free:1800 833 0800
Office Hours: 10 AM to 7 PM (all 7 days)
Company
About usCareers We are hiringTeach Online on TestbookPartnersMediaSitemap
Products
Test SeriesTestbook PassOnline CoursesOnline VideosPracticeBlogRefer & EarnBooks
Our Apps
Testbook App Download now
Current Affairs Download now
Follow us on
Copyright © 2014-2022 Testbook Edu Solutions Pvt. Ltd.: All rights reserved
User PolicyTermsPrivacy
Sign Up Now &
Daily Live Classes
250+ Test series
Study Material & PDF
Quizzes With Detailed Analytics
More Benefits
Get Free Access Now |
9951 | https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality?srsltid=AfmBOoq-rbHvBJnovu-Iz7haYuNNWiZ8dlqhPlKJKn0TfGKjakMSRUuq | Art of Problem Solving
Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality - AoPS Wiki
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical CampusVisit the Virtual Campus
Sign In
Register
online school
Class ScheduleRecommendationsOlympiad CoursesFree Sessions
books tore
AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates
community
ForumsContestsSearchHelp
resources
math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten
contests on aopsPractice Math ContestsUSABO
newsAoPS BlogWebinars
view all 0
Sign In
Register
AoPS Wiki
ResourcesAops Wiki Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality
Page
ArticleDiscussionView sourceHistory
Toolbox
Recent changesRandom pageHelpWhat links hereSpecial pages
Search
Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality
The Root-Mean Power-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (RMP-AM-GM-HM) or Exponential Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (EM-AM-GM-HM) or Quadratic Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (QM-AM-GM-HM), is an inequality of the root-mean power, arithmetic mean, geometric mean, and harmonic mean of a set of positivereal numbers that says:
, where , and is the .
The geometric mean is the theoretical existence if the root mean power equals 0, which we couldn't calculate using radicals because the 0th root of any number is undefined when the number's absolute value is greater than or equal to 1. This creates the indeterminate form of . Then, we can say that the limit as x goes to 0 is the geometric mean of the numbers.
The quadratic mean's root mean power is 2 and the arithmetic mean's root mean power is 1, as and the harmonic mean's root mean power is -1 as . Similarly, there is a root mean cube (or cubic mean), whose root mean power equals 3.
When the root mean power approaches , the mean approaches the highest number. When the root mean power reaches , the mean approaches the lowest number.
with equality if and only if . This inequality can be expanded to the power mean inequality, and is also known as the Mean Inequality Chain.
As a consequence, we can have the following inequality: If are positive reals, then with equality if and only if ; which follows directly by cross multiplication from the AM-HM inequality. This is extremely useful in problem-solving.
The Root Mean Power of 2 is also known as the quadratic mean, and the inequality is therefore sometimes known as the QM-AM-GM-HM Inequality.
Proof
The inequality is a direct consequence of the Cauchy-Schwarz Inequality; Alternatively, the RMS-AM can be proved using Jensen's inequality: Suppose we let (We know that is convex because and therefore ). We have: Factoring out the yields: Taking the square root to both sides (remember that both are positive):
The inequality is called the AM-GM inequality, and proofs can be found here.
The inequality is a direct consequence of AM-GM; , so , so .
Therefore, the original inequality is true.
Geometric Proofs
The inequality is clearly shown in this diagram for
Desmos SlidersDesmos Equation NOTE: The Desmos equation will not show the line when the numbers are negative.
(Note how the RMS is "sandwiched" between the minimum and the maximum)
Retrieved from "
Categories:
Algebra
Inequalities
Art of Problem Solving is an
ACS WASC Accredited School
aops programs
AoPS Online
Beast Academy
AoPS Academy
About
About AoPS
Our Team
Our History
Jobs
AoPS Blog
Site Info
Terms
Privacy
Contact Us
follow us
Subscribe for news and updates
© 2025 AoPS Incorporated
© 2025 Art of Problem Solving
About Us•Contact Us•Terms•Privacy
Copyright © 2025 Art of Problem Solving
Something appears to not have loaded correctly.
Click to refresh. |
9952 | https://mobile.fpnotebook.com/ENT/Mouth/PrtnslrAbscs.htm | Peritonsillar Abscess
Mouth
Peritonsillar Abscess
search
Peritonsillar Abscess, Peritonsillitis, Peritonsillar Cellulitis, Quinsy Sore Throat
See Also
Peritonsillar Abscess Needle Aspiration
Pharyngitis
Pharyngitis Causes
Dysphagia
Tonsillitis
Group A Streptococcal Pharyngitis
Retropharyngeal Abscess
Lemierre Syndrome
Diphtheria
Chronic Pharyngeal Carriage of Streptococcus pyogenes
Tonsillectomy Indications
Epidemiology
Highest Incidence ages 20-40 years old
Accounts for 30% of head and neck abscesses
Definitions
Peritonsillar Abscess
Suppurative fluid collection within the peritonsillar space (between Tonsillar capsule and superior constrictor Muscle)
Pathophysiology
Sequelae of Acute Tonsillitis or Tonsillopharyngitis
Abscess forms between lateral Tonsil and pharyngeal constrictor Muscles
Progression from exudative Tonsillitis to Peritonsillar Cellulitis to Peritonsillar Abscess
Weber's Glands (mucous Salivary Glands within Soft Palate)
Reside in Soft Palate, superior to Tonsil
Duct between Weber Gland and Tonsil
Cellulitis develops within the Weber Gland
Weber Gland duct obstructs and abscess forms
Risk Factors
Exudative Tonsillitis
Periodontal Disease
Tobacco Abuse
Causes
Group A Streptococcal Pharyngitis complication
Streptococcus Pyogenes (most common aerobic organism)
Mixed oropharyngeal flora
Staphylococcus Aureus
Corynebacterium
Streptococcus Milleri (S. intermedius, S. anginosus, S. Constellatus)
Haemophilus Influenzae
Neisseria
Anaerobic Bacteria
Fusobacterium
Peptostreptococcus
Prevotella
Bacteroides
Porphyromonas
Symptoms
Fever
Temp over 39.4 F suggests more serious infection (Parapharyngeal Space Infection, Sepsis)
Severe, unilateral throat pain
Dysphagia and Odynophagia (difficult and painful Swallowing)
Halitosis
Malaise
Otalgia (ipsilateral to abscess)
Signs
General
Ill appearance
Muffled ("hot potato") voice
Trismus
Drooling
Tender Cervical Lymphadenopathy
Oropharynx
Uvula deviates away from abscess to the opposite side
Localized swelling of Soft Palate over affected Tonsil
Swollen Tonsil (usually superior pole)
Indurated, fluctuant mass
Exudate may be present
Erythematous peritonsillar area
Usually unilateral
Labs
Complete Blood Count (CBC)
Throat Culture
Streptococcal Rapid Antigen Test
Monospot
Imaging
Indications
Uncomplicated Peritonsillar Abscess is a clinical diagnosis that may often be managed without imaging
Confirm Peritonsillar Abscess
Diagnosis is uncertain
Failed aspiration (Ultrasound)
Evaluate contiguous soft tissues and vessels (CT or MRI)
Significant Trismus
Suspected deep space infection
Neck Ultrasound
Preferred imaging modality for diagnosis and aspiration guidance of Peritonsillar Abscess
Endocavitary probe transducer intraoral (preferred)
Alternatively, may attempt visualization over Submandibular Gland
Abscess is echo-free with irregular border
CT Neck with contrast
Abscess appears with low attenuation
High False Positive Rate for Peritonsillar Abscess
Shows contiguous spread of infection to deep neck tissue
MRI neck
Evaluate for deep neck infections (better than CT without inonizing radiation)
Evaluate Internal Jugular Vein Thrombosis and Carotid Artery sheath erosion
Differential Diagnosis
Peritonsillar Cellulitis (no pus in capsule)
Retropharyngeal Abscess
Dental Infection (e.g abscessed tooth, Retromolar abscess)
Lemierre Syndrome
Epiglottitis
Mononucleosis (up to 6% coinfection, esp. in teens and young adults)
Cervical adenitis
Sialolithiasis or Sialadenitis
Mastoiditis
Internal cartoid artery aneurysm
Malignancy (e.g. Lymphoma)
Management
Needle aspiration
See Peritonsillar Abscess Needle Aspiration
Be prepared for airway emergency
Observe patient for several hours after observation and confirm able to tolerate liquids
Failed aspiration of pus
May be consistent with Peritonsillar Cellulitis
Consider imaging soft tissue for deep space infection
If no serious findings, may discharge home with close follow-up on oral Antibiotics and steroids
Medical intervention (Antibiotics, steroids) alone without aspiration has similar outcomes in uncomplicated PTA
Medical management without aspiration has a similar failure rate as with aspiration (5 to 8%)
Forner (2020) Otolaryngol Head Neck Surg 163(5):915-922 +PMID: 32482146 [PubMed]
Zebolsky (2021) Am J Otolaryngol 42(4):102954 +PMID: 33581462 [PubMed]
Battaglia (2018) Otolaryngol Head Neck Surg 158(2):280-286 +PMID: 29110574 [PubMed]
Disposition: Indications for inpatient management (typically 2-4 day stays)
Children
Dehydration
Toxic appearance
Persistent significant Trismus or Dysphagia (refractory to aspiration)
Airway compromise risk (e.g. "kissing" Tonsils)
Disposition: Outpatient Management
Observe after aspiration for several hours before discharge (confirm tolerating liquids)
Prescribe Antibiotics, Corticosteroids (typically) and Analgesics
Close interval follow-up at 24-36 hours
Antibiotics for 10-14 days
Broad spectrum Antibiotics are typically needed (polymicrobial infections, often with resistance)
May adjust Antibiotic based on needle aspiration sample
Parenteral
Combination
Penicillin G 10 MU IV every 6 hours and
Metronidazole 1.0 g load, and then 500 mg IV every 6 hours
Piperacillin/Tazobactam (Zosyn) 3.375 mg every 6 hours
Ampicillin with Sulbactam (Unasyn) 3 grams every 6 hours
Ceftriaxone 1 g every 12 hours AND Metronidazole
Clindamycin 900 mg IV every 8 hours (if Penicillin allergic)
Consider Vancomycin AND Flagyl if MRSA concern
Oral agents
Clindamycin 300 to 450 mg orally every 8 hours
Cefdinir (Omnicef) 300 mg every 12 hours AND Metronidazole
Augmentin 875 mg orally twice daily
Combination
Penicillin VK 500 mg orally every 6 hours AND
Metronidazole 500 mg orally every 6 hours
Corticosteroids as adjunct to Antibiotics
Dexamethasone 10 mg orally for 1 dose
[OBrien (1993) Ann Emerg Med 22(2): 212-5 [PubMed]](
Depo Medrol 2-3 mg/kg up to 250 mg IV for 1 dose
Patients improved faster when adjunctive steroids were used
Ozbek (2004) J Laryngol Otol 118:439-42 [PubMed]
Efficacy
Decreased pain and improved oral intake within 12-24 hours
Faster recovery and shorter hospital stays
Lee (2016) Clin Exp Otorhinolaryngol 9(2): 89-97 [PubMed]
Complications
Airway obstruction
Lung infection (Aspiration Pneumonia or Lung Abscess) from Peritonsillar Abscess rupture
Erosion into Carotid Artery sheath (uniformly fatal)
Internal jugular vein Thrombophlebitis
Deep neck or mediastinal infection from contiguous spread
Lemierre Syndrome
Follow-up
Consider Tonsillectomy 3-6 months after Peritonsillar Abscess (40% recurrence rate)
References
Anderson (2019) Crit Dec Emerg Med 33(9): 3-10
Guess and Pittman (2022) Crit Dec Emerg Med 36(7): 12-4
Roberts (1998) Procedures Emergency Medicine, p. 1122-6
Swadron and Finley in Herbert (2018) EM:Rap 18(7): 3-4
Brook (2004) J Oral Maxillofac Surg 62:1545-50 [PubMed]
Galioto (2017) Am Fam Physician 95(8): 501-6 [PubMed]
Kieff (1999) Otolaryngol Head Neck Surg 120(1):57-61 [PubMed]
Steyer (2002) Am Fam Physician 65(1):93-96 [PubMed]
Type your search phrase here
clear text
topENTMouth
Please see the Terms and Conditions.
This page was written by Scott Moses, MD. This page was last revised on 6/21/2024 and last published on 9/1/2025.
©2025, Family Practice Notebook, LLC
loading |
9953 | https://math.stackexchange.com/questions/578943/generalized-pigeonhole-principle | combinatorics - Generalized Pigeonhole Principle - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Generalized Pigeonhole Principle
Ask Question
Asked 11 years, 10 months ago
Modified11 years, 10 months ago
Viewed 14k times
This question shows research effort; it is useful and clear
2
Save this question.
Show activity on this post.
Can somebody explain this to me? I am very confused.
I have a question that says "What is the minimum number of students required in a discrete mathematics class to be sure that at least six will receive the same grade, if there are five possible grades, A, B, C, D, and F?"
and the solution is:
The minimum number of students needed to ensure that at least six students receive the same grade is the smallest integer N such that [N/5] = 6. The smallest such integer is N = 5 5 + 1 = 26. If you have only 25 students, it is possible for there to be five who have received each grade so that no six students have received the same grade. Thus, 26 is the minimum number of students needed to ensure that at least six students will receive the same grade. The bolded part is the part that confuses me the most. Unless I haven't understood math in the past 20 years, N/5 = 6 would mean that N would have to be 30. So how exactly am I supposed to approach this question to understand what it is I'm trying to understand?
combinatorics
pigeonhole-principle
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
asked Nov 24, 2013 at 6:39
FrostyStrawFrostyStraw
1,069 5 5 gold badges 16 16 silver badges 24 24 bronze badges
1
4 I suppose they must be taking [−][−] to be the ceiling function.Casteels –Casteels 2013-11-24 06:55:43 +00:00 Commented Nov 24, 2013 at 6:55
Add a comment|
3 Answers 3
Sorted by: Reset to default
This answer is useful
1
Save this answer.
Show activity on this post.
When there are 25 people, clearly the required condition is not met as it is possible for 5 groups of 5 people each to have gotten each of the different grades. (this is the only case with 25 people when no 6 people have got a common grade)
When you have one person extra, irrespective of his grade, there will be a grade now that is repeated 6 times. Hence, the minimum number of people is is 26.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Nov 24, 2013 at 9:31
AksharaAkshara
80 5 5 bronze badges
Add a comment|
This answer is useful
1
Save this answer.
Show activity on this post.
We can use the Pigeonhole Principle (Strong form). That is q 1+q 2+⋯+q n−n+1 q 1+q 2+⋯+q n−n+1 where q i q i is the number of desired test scores of a certain type, (6 6) and n n is the number of score types in general, (5 5). So 6+6+6+6+6−5+1=26 6+6+6+6+6−5+1=26.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Nov 24, 2013 at 17:31
1233dfv1233dfv
5,773 1 1 gold badge 28 28 silver badges 43 43 bronze badges
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
Are you sure that it doesn’t say ⌈N/5⌉⌈N/5⌉ or ⌈N 5⌉⌈N 5⌉ and not [N/5][N/5] or [N 5][N 5]?
By definition ⌈x⌉⌈x⌉, the ceiling of x x, is the smallest integer n n such that x≤n x≤n. This means that it’s the unique integer n n such that n−1<x≤n n−1<x≤n. Thus, ⌈N/5⌉=6⌈N/5⌉=6 if and only if
5<N 5≤6,5<N 5≤6,
which is true if and only if 25<N≤30 25<N≤30. Since N N here is an integer, this means that ⌈N/5⌉=6⌈N/5⌉=6 if and only if N N is 26,27,28,29 26,27,28,29, or 30 30. With fewer than 26 26 students, it’s possible that no more than 5 5 received each grade; with more than 30 30 students, at least one of the grades must have been obtained by 7 7 or more students.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Nov 24, 2013 at 21:24
Brian M. ScottBrian M. Scott
633k 57 57 gold badges 824 824 silver badges 1.4k 1.4k bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
combinatorics
pigeonhole-principle
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
0Pigeonhole principle in practice?
7Generalized pigeonhole principle: 15 workstations and 10 servers
0Pigeon Hole Priciple and Genaralized Pigeon Hole Principle Question
2The Generalized Pigeonhole Principle
2Counting number of possible grade assignments
0Pigeonhole Principle: splitting into groups
0understanding a pigeonhole principle problem
2Problem about the generalized pigeonhole principle
5Why must we consider seemingly unnecessary cases for combinatorics problems?
Hot Network Questions
Is it safe to route top layer traces under header pins, SMD IC?
What happens if you miss cruise ship deadline at private island?
Can a cleric gain the intended benefit from the Extra Spell feat?
Is there a way to defend from Spot kick?
Does a Linux console change color when it crashes?
Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish?
Lingering odor presumably from bad chicken
Storing a session token in localstorage
What is a "non-reversible filter"?
Does the curvature engine's wake really last forever?
Can you formalize the definition of infinitely divisible in FOL?
Implications of using a stream cipher as KDF
Passengers on a flight vote on the destination, "It's democracy!"
Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation
Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done?
Another way to draw RegionDifference of a cylinder and Cuboid
Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term?
Checking model assumptions at cluster level vs global level?
Why do universities push for high impact journal publications?
Is encrypting the login keyring necessary if you have full disk encryption?
Childhood book with a girl obsessessed with homonyms who adopts a stray dog but gives it back to its owners
Why are LDS temple garments secret?
в ответе meaning in context
Analog story - nuclear bombs used to neutralize global warming
more hot questions
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
9954 | https://www.youtube.com/watch?v=eMmaQH8xUow | Quadratic Equation Pitfalls: Avoid This Common Mistake!
Algebra-1 with Mr. Peters
11100 subscribers
17 likes
Description
290 views
Posted: 23 Sep 2023
Welcome To My Channel Algebra-1 with Mr. Peters
Subscribe Here:
Welcome back to Algebra-1 with Mr. Peters! Today, we're diving into a common stumbling block for many students - solving radical equations. But don't worry, we've got you covered! In this video, we'll first identify the common mistakes students make when tackling these problems. Then, we'll walk you through two methods to correctly solve radical equations. By the end of this video, you'll be able to confidently solve these equations and avoid the common pitfalls. Remember, practice makes perfect! So, don't forget to try out some problems on your own after watching the video. If you find this video helpful, please give it a thumbs up share it with your classmates, and subscribe to our channel for more algebra tips and tutorials. We upload new videos every Monday, Tuesday, and Saturday, and we host live streams every other Wednesday. Happy learning, and see you in the next video!
🎥Watch My Recent Videos🎥
💠Key Steps to Simplifying Negative Exponents with Fractions!
✔️
💠How to Solve & identify Extraneous Solutions
✔️
💠Algebra 1 Final Exam Review. Exponents, Equations, and Radicals examples
✔️
💠Simplifying Radical Expressions : Learn How to Add Subtract & Distribute Radicals
✔️
💠How to Rationalize & Simplify Denominator's with Radicals
✔️
usingthezeroproductproperty
binomialsquared
problemsinvolvingquadraticequations
5 comments
Transcript:
what's good everybody welcome back to another show with Alger one with Mr Peters today's video I'm going to show you the biggest mistake students make when they're solving quadratic equations like this then I'm going to show you guys two methods you can use to solve this problem so you avoid that common mistake let's jump into today's video the biggest mistake I see students make guys is incorrectly squaring the binomial so they'll apply this exponent and what they get is x squared minus 16 is equal to 9. then they start solving like a regular equation right we add 16 on both sides we have x squared is equal to 25 we take the square root of both sides to get rid of that exponent and now we'll get X is equal to positive or negative five but the issue with this answer is that when we plug it back in it's not going to give us the right answer so 5 minus 4 squared is not going to give us 9. this is going to be 1 squared is equal to nine we know one is not equal to nine that is wrong now if we did the opposite right negative 5 minus four and we square that answer we would have negative 9 squared is equal to 9 and we know 81 is not equal to 9. so this is the incorrect way to solve so the correct way to solve I'm going to show you guys that right now so the exponent is important what we're going to do is expand the polynomial and I'm going to use the box method in case you guys are familiar with it or if you want to chat out yourself so what we're doing here is x minus 4 multiply by x minus 4. and I'm going to use the box method to show you how to do this so we put our Factor on the top x minus 4 negative 4 and then we put it on the side x minus 4. we have to keep what's in parentheses grouped together so when I multiply I'll get x squared x times x x times negative 4X gives me negative 4X we go down now we do the same thing with negative four negative 4 times x is be negative 4X negative 4 times 4 gives me positive 16. I combine my like terms those negative four X's so after I expand this binomial to the second power I should have x squared minus eight X plus 16 is equal to 9. now remember when we're solving equations even quadratic equations we have to combine our like terms so I'm going to bring 9 over and subtract right because most of our quadratic equations are set or equal to zero right so now we have x squared minus 8x plus 7 is equal to zero now we have set the quadratic equation equal to zero we could factor and break it down luckily for us the only factors are for seven are one and seven so once we draw parentheses I have X in the first one X in the second one I know it has to be this combination so I have seven and I have one and for me to get a positive 7 when I multiply or both those signs either have to be positive or negative but this negative 8X when we add lets me know they're both going to be negative so at this step guys we found the factors the numbers that will satisfy this quadratic equation what we have to do now is set both of them equal to zero x minus seven is equal to zero so we know X is equal to seven and then we go to the second one x minus 1 is equal to zero and we say that X is equal to one so at this step now let's call parenting raise the box method and let's double check to see if this answer actually satisfies the equation so let's try one so I have one minus four squared is equal to nine once we do the subtraction we're going to get negative three to the second power is equal to nine once we simplify nine is equal to nine so we know the factor of positive one is correct now when I plug in my second Factor seven I have seven minus 4 squared is equal to nine once we simplify we'll get 3 to the second power is equal to nine so nine is equal to nine both of these factors work now before I show you the second method to solve this quadratic equation guys just remember if one of these factors did not satisfy the equation that is what we would call the extraneous solution the extra solution that does not satisfy the equation and with that being said we're going to jump to the last part of this video and show you the second method guys we really hope you found value in our show and that you like the way that we're teaching this math then we want to make it easier for you our show will be nothing without you guys so the second method that I really like to show students and I don't do this often because students will confuse when and when they cannot use this and if you comment down below or just like this video I'll make sure they're going to extra practice to show you when you can use this method but let's do it so in this method we're going to get rid of the exponent right we know we can get rid of the exponents by what exactly taking the square root so when I take the square root of this side oops and I take the square root of this side what is left is x minus 4 is equal to three we took the square root of nine and we know the square root cancels out V exponent now I could solve like a regular equation and before I forget anything this is so important with the square root we have to have positive or negative three so let's split this into two different equations x minus four is equal to positive three and x minus 4 is equal to negative oops so now when we simplify I'm going to get X is equal to 7 which we just had in our previous example the first method I showed you so we know this is right and then when I go to the second one and I do the same thing by solving for x and adding 4 on both sides I'm gonna get X is equal to positive one and voila we got the same exact answers so like I said if you found this method helpful comment down below like this video so we know hey we need to make another video and with that being said guys thank you so much for joining our show algebra one with Mr Peters today hope you enjoyed this video you found it helpful smash the like button subscribe to the channel and leave comments for future videos you guys like to see on the channel or if you had questions from today's video |
9955 | https://www.doubtnut.com/qna/8489876 | My Profile
Maths
Physics
Chemistry
Biology
English
Maths
Physics
Chemistry
Biology
English
Maths
Physics
Chemistry
Biology
English
Maths
Physics
Chemistry
Biology
English
Reasoning
Maths
Physics
Chemistry
Biology
English
Maths
Physics
Chemistry
Biology
English
Class 11
Class 12
Class 12+
Class 11
Class 12
Class 12+
Class 9
Class 10
Class 11
Class 12
Class 9
Class 10
Class 11
Class 12
Class 9
Class 10
Class 11
Class 12
Class 9
Class 10
Class 11
Class 12
Class 9
Class 10
Class 11
Class 12
Class 9
Class 10
Class 11
Class 12
Class 9
Class 10
Class 11
Class 12
Class 9
Class 10
Class 11
Class 12
Class 9
Class 10
Class 11
Class 12
Class 9
Class 10
Class 11
Class 12
IIT-JEE
NEET
CBSE
UP Board
Bihar Board
JEE
NEET
Class 6-10
Class 12
Class 11
Class 10
Class 9
Class 8
Class 7
Class 6
Online Class
Blog
Select Theme
Prove that the tangent and normal at any point of an ellipse bisect the external and internal angles between the focal distances of the point
Video Solution
Know where you stand among peers with
ALLEN's JEE Enthusiast Online Test Series
|
Answer
Step by step video & image solution for Prove that the tangent and normal at any point of an ellipse bisect the external and internal angles between the focal distances of the point by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
Class 12MATHSCONIC SECTIONS - FOR COMPETITION
Similar Questions
Prove that the lengths of tangents drawn from an external point to a circle are equal.
View Solution
Tangent Normal (Angle between Curves)
View Solution
Knowledge Check
Statement 1 The tangent and normal at any point P on a ellipse bisect the external and internal angles between the focal distance of P. Statement 2 The straight line joining the foci of the ellipse subtends of a right angle at P.
AStatement I is true, statement II is true: statement II is a correct explanation for statement I
BStatement I is true, statement II is true, statement II is not a correct explanation for statement I
Cstatement I is true, statement II is false
Dstatement I is false, statement II is true
## If (−2,5) and (3,7) are the points of intersection of the tangent and normal at a point on a parabola with the axis of the parabola, then the focal distance of that point is
A√292
B52
C√29
D25
## What is the sum of focal radii of any point on an ellipse equal to ?
ALength of latusrectum
BLength of major-axis
CLength of minor-axis
DLength of semi-latusretum
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
View Solution
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.
View Solution
Prove that the tangent at any point of circle is perpendicular to the radius through the point of contact.
View Solution
If the tangent at any point of an ellipse x2a2+y2b2=1 makes an angle α with the major axis and an angle β with the focal radius of the point of contact then show that the eccentricity ‘e’ of the ellipse is given by the absolute value of cosβcosα
View Solution
If the normal at any point P on the ellipse cuts the major and minor axes in G and g respectively and C be the centre of the ellipse, then
View Solution
Recommended Questions
Prove that the tangent and normal at any point of an ellipse bisect th...
10:59
2. Prove that the angle between the two tangents drawn from an external p...
05:02
3. Statement 1 : If the line x+y=3 is a tangent to an ellipse with focie ...
04:04
4. Prove that the tangent and normal at any point of an ellipse bisect th...
10:59
5. Prove that form any external point two tangents can be drawn to circle...
03:20
6. Prove that the internal bisector of the angle between two tangents dra...
7. Prove that the internal angle between two tangents drawn from an exter...
8. सिद्ध कीजिये की किसी बाह्म बिंदु से वृत्त पर खींची गई दो स्पर्श रेखाओं...
9. सिद्ध करो कि किसी बाह्य बिंदु से किसी वृत्त पर खींची गई स्पर्श...
Text Solution |
9956 | https://codesignal.com/learn/courses/maximizing-efficiency-in-problem-solving-techniques-in-java/lessons/java-combinatorial-pair-counting? | Hello, coding enthusiast! In our journey to master coding and problem-solving, we've arrived at an interesting challenge today. We're going to focus heavily on combinatorial problems in practice. Specifically, we're examining combinatorial problems that involve working with large datasets and multiple pairs of numbers. We'll learn how to solve significant problems efficiently by implementing smart use of data structures like HashMap and sidestepping expensive operations like iterating over large arrays. Are you ready? Let's dive in!
HashMap
In this unit's task, you'll be given a large ArrayList composed of pairs of distinct, positive integers, including up to 1,000,000 elements. Your challenge is to write a Java function to count the number of indices (i, j) (i≠ji \ne ji=) where the pair does not share a common element with the pair. A crucial point to remember is that a pair is considered identical to , meaning the order of elements in a pair is irrelevant in this case. It is guaranteed that no two pairs are element-wise equal.
ArrayList
(i, j)
i-th
j-th
(a, b)
(b, a)
For example, given the array {{2, 5}, {1, 6}, {3, 2}, {4, 2}, {5, 1}, {6, 3}}, the output should be 8. The required index pairs are the following: (0, 1) (i.e., the pair {2, 5} does not share a common element with the pair {1, 6}), (0, 5) ({2, 5} does not share a common element with {6, 3}), (1, 2), (1, 3), (2, 4), (3, 4), (3, 5), (4, 5).
{{2, 5}, {1, 6}, {3, 2}, {4, 2}, {5, 1}, {6, 3}}
8
(0, 1)
{2, 5}
{1, 6}
(0, 5)
{2, 5}
{6, 3}
(1, 2)
(1, 3)
(2, 4)
(3, 4)
(3, 5)
(4, 5)
At the core of our solution, we're going to leverage the power of combinatorics and a smart way of keeping track of occurrences to solve this problem efficiently.
The central idea is to calculate the total number of pairs and then subtract from this total the number of pairs that share a common element. This will leave us with the count of pairs that do not share a common element, which is what we're after.
Firstly, we will calculate the total number of pairs possible in the array. In a set of n numbers, the number of pairs is given by the formula n⋅(n−1)/2n \cdot (n - 1) / 2n⋅(n−1)/2. This is because each element in the set can pair with every other element, but we divide by 2 because the order of pairs doesn't matter (i.e., pair (a, b) is identical to pair (b, a)).
n
(a, b)
(b, a)
Secondly, we'll count the number of pairs that have at least one common element. To do this, we will use a HashMap to track each number's appearance in the pairs. For each number, we calculate how many pairs it appears in and sum these numbers up.
HashMap
Let's begin with the initial steps of our solution. The first thing we need is a convenient place to store the occurrence of each number in the pairs. Here, Java's data structure, HashMap, shines. It enables us to efficiently track the number and its corresponding occurrences.
HashMap
Next, we calculate the total number of pairs using the formula n⋅(n−1)/2n \cdot (n - 1) / 2n⋅(n−1)/2. We'll need this for our final calculation.
Let's initialize an empty HashMap and calculate the total pairs.
HashMap
With the first step completed, our next move is to populate the HashMap by iterating over the array of pairs. For each pair, we'll examine its two elements and either append the current index to the list of indices for this number (if it’s already in the HashMap) or start a new list for it (if it isn't).
HashMap
HashMap
Here's how we modify our function to carry out this operation:
Finally, with all the data in place, we arrive at our final step of calculation. We need to calculate the total pairs of indices that share at least one common element. For that, we'll consider each number in the array and count the number of times those numbers occur in different pairs. We'll use the same formula as before.
Finally, we subtract these common pairs from the total pairs to get our answer — the count of pairs without a common number.
Adding this last part to our function gives us the solution:
Great job! Today's challenge was certainly a tough one, but you managed to navigate through it successfully. You utilized a HashMap to efficiently track occurrences within a large dataset and applied combinatorial reasoning to subtract the opposite case from the total possibilities. Consequently, you came up with a solution that operates in an efficient time frame.
HashMap
This knowledge will serve you well in solving similar complex problems in the future. Remember, the best way to handle large data is to apply clever techniques that sidestep unnecessary computations, just like we did today.
Now, it's time to solidify your understanding. Up next are practice problems related to today's lesson. Start working on them to reinforce these concepts. Happy coding! |
9957 | https://www.solitaryroad.com/c1039.html | Electrostatics. Static electricity. Structure of matter. Pith ball and leaf electroscopes. Proof plane. Insulators and conductors. Charging by induction. Discharging effect of points.
SolitaryRoad.comWebsite owner: James Miller
[Home] [Up] [Info] [Mail]
Electrostatics. Static electricity. Structure of matter. Pith ball and leaf electroscopes. Proof plane. Insulators and conductors. Charging by induction. Discharging effect of points.
Structure of matter. The current view of matter is as follows: All matter consists of atoms. Each atom contains a nucleus which consists of several kinds of particles, the main ones being protons and neutrons. About the nucleus, electrons orbit in a way similar to that in which planets orbit the sun. Protons carry a unit of positive charge, neutrons carry no charge, and electrons carry a unit of negative charge. Each atom contains the same number of electrons as protons. Each atom is thus electrically neutral. The number of protons in an atom is called its atomic number. It is the atomic number that distinguishes between elements, that identifies an element. A lithium atom is shown in Fig. 1. It has an atomic number of 3.
Almost the entire mass of an atom comes from its protons and neutrons. The electrons contribute almost nothing to the mass. An electron weighs about 1/1837 of a proton. Protons and neutrons weigh the same. There is a weight scale devised for atomic size weights. In this system the weight of a proton is one atomic weight unit (a.w.u.).
Size of an atom. A helium atom has a diameter of about 1 Angstrom (10-10 meters) while its nucleus has a diameter of only 1 femtometer (10-15 meters). Thus the diameter of the atom is 100,000 times that of the nucleus. If you made a scale model of an aluminum atom with a nucleus the size of a marble, the outermost electrons would fly around the nucleus at a distance of 150 feet. An atom is thus mostly empty space.
Atoms are extremely small, so small it is hard to imagine a particle so small. For example, if a single drop of water were magnified to the size of the earth, the individual atoms would be about the size of tennis balls.
Current view of the nature of electricity. The electrons in the outer orbits of some elements (particularly metals) are more loosely held than in others. Electrons can be stripped from an atom leaving it with a positive charge. For example, if we rub a hard rubber rod with wool flannel the rod will acquire a negative charge and the wool will acquire a positive charge. The conclusion is that electrons have been stripped from the wool atoms by the rubbing process and transferred to the rubber rod. If, on the other hand, we rub a glass rod with silk the rod will acquire a positive charge and the silk will acquire a negative charge. Here electrons have been stripped from the glass rod and transferred to the silk. For another example, the flow of electric current in a conductor is regarded as a flow of free electrons through the conductor.
Familiar electrical phenomena.
Walking across a carpet on a cold day and receiving a shock when touching a doorknob. Here a charge was transferred from the carpet to you and then was discharged at the doorknob.
Lightening. A cloud containing moisture becomes oppositely charged with respect to another cloud or the earth. When the electrical pressure between the two becomes great enough, the air, normally an insulator, breaks down and a lightening flash occurs.
The crackling noise heard when dry hair is brushed.
The sticking of pieces of paper together due to static electricity.
Discovery of static electricity. The Greek philosopher Thales of Miletus, 6 th century BC, observed that rubbing an amber rod with cats’s fur would give the rod the power to attract light objects such as feathers in a way similar to the way lodestone will attract iron. William Gilbert (1540 - 1603), an early English experimenter, on investigating this phenomenon, discovered that many kinds of materials behave the same way as amber when they are rubbed: they acquire the ability to pick up light objects.
Detection of static electricity. The presence of an electrostatic charge can be detected by using an electroscope. The simplest type of electroscope consists of a pith ball suspended on a silk thread. The electroscope can be made more sensitive by coating the ball with metallic paint. See Fig. 2.
Two kinds of electrical charge. Suppose we charge a hard rubber rod by rubbing it with wool flannel and hold it near the pith ball of an electroscope. The ball is first attracted to the rod and then, if the ball touches the rod, it immediately rebounds. The ball is first attracted to the rod and then, on touching it, is repelled from it. What has happened is that some of the charge on the rod has been transferred to the ball making both ball and rod similarly charged and repulsion takes place. See Fig. 2. The same thing occurs if we charge a glass rod by rubbing it with silk and hold it near the ball. We also find that, after the experiments, the wool flannel and silk show signs of being charged when tested with an electroscope.
Let us now do a second experiment. Let us deliberately put a charge on the electroscope pith ball by rubbing a hard rubber rod with wool flannel and touching it to the ball. After touching, the ball is repelled by the rod. Now let us rub a glass rod with silk and hold it near the pith ball. We discover that it is attracted to the ball. The ball responds differently to the glass rod than it does to the rubber rod. This tells us that we are dealing here with two different kinds of electricity. Charles du Fay was the first to realize that there were two kinds of electricity. He called them “vitreous” and “resinous” electricity. That was in a 1733 paper titled, “Two kinds of electrical fluid: vitreous and resinous”. Later Benjamin Franklin suggested the names “positive” and “negative” electricity.
Experimentation gives the following rule:
First law of electrostatics. Like charges repel each other; unlike charges attract each other.
Two negatively charged pith balls (balls that have been given a negative charge by contact with a charged rubber rod) will repel each other. Two positively charged pith balls (balls that have been given a positive charge by contact with a charged glass rod) also repel each other. However, a positively charged ball and a negatively charged ball will attract each other.
Why will an electrically charged rod attract neutral objects? If an amber or hard rubber rod is charged by rubbing it with wool flannel, it will attract light objects such as pieces of paper, feathers, etc. Similarly, a glass rod that has been charged by rubbing with silk will attract light objects. Why? We know that a charged rod will attract objects of opposite charge. But why will a charged rod attract electrically neutral objects? Answer: The current theory on this question involves the effect of an electric field on dielectrics (i.e. nonconducting materials), the polarization of molecules by an electric field, and induced charges on a dielectric. These topics will be discussed later. However, if a negatively charged rod, for example, is brought near a neutral object, the electric field of that rod will cause induced charges of opposite sign to appear on the near side of the object and induced charges of the same sign to appear on the opposite side of the object as shown in Fig. 3. Because the force of attraction is greater for close charges than for far ones, the object will be pulled towards the rod. The logic is the same for the case of a positively charged rod.
Leaf electroscope. An electroscope that is more sensitive than the pith-ball electroscope is shown in Fig. 4. It consists of very thin leaves of gold leaf or aluminum foil hanging from a
metal stem with a metal knob at the top. The leaves are protected by a glass flask or by a metal case with glass observation windows. The metal stem is passed into the case through an insulating bushing. If an electric charge is applied to the metal knob, the charge distributes itself over the top, stem and leaves, charging both leaves with electricity of the same kind, and causing them to repel each another. The angle of divergence of the leaves can be used as a measure of the amount of charge on them.
Proof planes. If one applies charges that too large to a leaf electroscope, the leaves are so strongly repelled that they are likely to be damaged. To prevent this from happening, proof planes are often used. A proof plane is a small metal disk with an insulating handle. One can easily make one by cementing a penny to a glass rod. To use such a proof plane, we first touch the penny to the charged object and then to the knob of the electroscope.
Conductors and insulators. In Fig. 5 a brass ball B is suspended by a silk thread and connected to the knob A of an electroscope by a copper wire. If we place a charge on ball B, the leaves of the electroscope will diverge. This shows that the charge that was placed on B traveled through the copper wire to the electroscope. Now let us replace the copper wire by a silk thread and repeat the experiment. This time when we place a charge on ball B, the leaves of the electroscope do not diverge. The charge did not travel through the silk thread.
Def. Conductor. A conductor is a material through which an electric charge can readily travel. Metals are good conductors. Silver is the best known conductor. Copper and aluminum are also very good conductors.
Def. Insulator (or dielectric). An insulator (or dielectric) is a material through which an electric charge will not readily pass. Some of the best insulators are mica, rubber, Bakelite, paraffin, shellac, oils, silk, wool, sulfur, dry air and many plastics.
The discovery of how electrical “fluid” will travel through metal wire was made by the English experimenter Stephen Gray around 1732.
Charging by induction. If a negatively charged rubber rod is brought near a conductor that is insulated from its surroundings, the electrons in the conductor will be repelled; the far side of the conductor will become negatively charged and the near side of the conductor will become positively charged. See Fig. 6(a). One can use this fact to place a charge on a conductor. To place a charge on a conductor, use the following procedure:
Bring a charged rod (or other charged object) near the conductor.
Ground the conductor by touching it with your hand or with a ground wire.
Remove your hand or ground wire.
Remove the charged rod.
The conductor will now have a charge on it opposite in sign to that on the rod.
If a negatively charged rod is brought close to the conductor, grounding it will cause electrons to flow into the ground, leaving it with a net positive charge. If a positively charged rod is brought near the conductor, grounding it will cause electrons to flow from the ground into the conductor, giving it a net negative charge.
Charging a leaf electroscope by induction. Although a leaf electroscope can be charged by transferring charges to it by contact with a charged body, electroscopes are often charged by induction. One simply brings a charged body near the knob and then grounds the electroscope by touching the knob with your finger. See Fig. 6(b). On removing your finger, the electroscope will have a charge on it opposite in sign to that on the rod. Induction is the proper way to charge a very delicate electroscope.
A straightforward application of the method of charging by induction is found in the electrophorus.
The electrophorus. The electrophorus is a simple device for creating an electrostatic charge. It consists of a flat bed made of an insulating material such as hard rubber, sealing wax, or resin, and a metal disk with an insulating handle. See Fig. 7(a). Using it involves the following procedure:
Rub the bed A with fur or wool. This places a negative charge on the bed. See Fig. 7(b).
Place the metal disk B on the bed. Induction will cause the electrons in disk B to be repelled to its upper surface, leaving its lower surface positively charged. Very little charge is transferred from the bed A to the disk B because contact is made at very few points. The process is thus primarily induction.
Ground disk B by touching it with your hand or a grounded wire. This will leave disk B positively charged.
Lift the disk. It now has a positive charge available for use. If, for example, you discharged it through a tube containing neon, the tube would emit the light characteristic of neon.
We note that almost no charge has been removed from the bed. Thus you can repeat the process over and over without recharging the bed.
Method for determining the type of charge (i.e. positive or negative) on a charged body. A charged electroscope can be used to determine the sign of the charge on a body. Suppose the electroscope is positively charged and the leaves diverge at a particular angle. If a positively charged body is brought near the spectroscope, the angle of divergence of the leaves will be increased. This is because electrons will be attracted up from the leaves by the charged body, thus increasing the positive charge on the leaves. If a negatively charged body is brought near a positively charged electroscope, the angle of divergence of the leaves will be decreased.
Location of charge on bodies. Michael Faraday performed several experiments in investigation of the question of where charges were located on bodies. In one experiment he used a conical shaped silk bag to demonstrate that electric charges always lie on the outside. See Fig. 8. He gave the bag a charge and tested it and found that there was a charge on the outside but no charge on the inside. He then pulled the bag inside out and tested it again. Again there was a charge on the outside but none on the inside.
When electric charges are established on conductors, they always reside on the outer surface of the conductors as a consequence of the mutual repulsion of like charges and the freedom with which charges move in conductors. If a charge is placed on the inside surface of a hollow conductor it will not remain there but will immediately travel through the conductor to the outer surface. Consider the apparatus shown in Fig. 9. A hollow metal cylinder is mounted on an insulated stand. Two pith balls are connected electrically to the exterior of the cylinder and another two pith balls are connected electrically to the interior of the cylinder. The cylinder is connected by a chain or wire to a static machine. When the machine charges the cylinder, we find that the pith balls connected to the exterior diverge but the pith balls connected to the inside are not affected.
Effect of the shape of a conductor on charge distribution. Suppose we electrify an egg-shaped conductor such as the one shown in Fig. 10. If we test it with a proof plane, we find that the charge density is not the same everywhere. It is greater at the small end than at the large end. If we increase the curvature at the small end by successively making it more and more pointed, the density of the charge increases, also. This illustrates the following rule:
Rule: The electrical density, or quantity of charge per unit area, varies with the curvature of the surface and is greatest at the point of greatest curvature.
The discharging effect of points. The high electrical density at points on a conductor tends to cause the air surrounding the point to become ionized. Gas molecules in the air can lose electrons to form positive ions if the conductor is positively charged; or they can gain electrons to form negative ions if the conductor is negatively charged. (An ion is an atom or molecule that has a positive or negative charge due to the loss or gain of one or more electrons.) In the first case, if the conductor is positively charged, the gas near the point becomes positively charged, too. The strong positive charge of the conductor strips electrons from the gas molecules causing them to become positively charged ions. These positively charged gas ions are then repelled from the conductor with sufficient velocity to produce an electrical wind. The electrons torn from the molecules by the positively charged conductor are then attracted to the conductor resulting in a neutralization of its charge. As a consequence, a pointed conductor loses its charge rapidly. The leakage of electricity from points on a charged object is called the discharging effect of points.
A charged electroscope can be discharged rapidly if one holds a needle a few inches from the knob. The point of the needle becomes charged by induction and creates ions that neutralize the charge on the electroscope.
Why does a humid environment cause charge to be dissipated from a charged body? To answer this question we note that if one holds a charged rod near a hanging pith ball, the pith ball will first be attracted to it, steal some of its charge, and then be repelled away. Water droplets in the air act in the same way. They are first attracted to the body, steal some charge, and then are repelled away. In no time the body has lost all its charge.
References
Dull, Metcalfe, Brooks. Modern Physics.
Sears, Zemansky. University Physics
Semat, Katz. Physics.
Bennet. Physics.
Freeman. Physics Made Simple.
More from SolitaryRoad.com:
The Way of Truth and Life
God's message to the world
Jesus Christ and His Teachings
Words of Wisdom
Way of enlightenment, wisdom, and understanding
Way of true Christianity
America, a corrupt, depraved, shameless country
On integrity and the lack of it
The test of a person's Christianity is what he is
Who will go to heaven?
The superior person
On faith and works
Ninety five percent of the problems that most people have come from personal foolishness
Liberalism, socialism and the modern welfare state
The desire to harm, a motivation for conduct
The teaching is:
On modern intellectualism
On Homosexuality
On Self-sufficient Country Living, Homesteading
Principles for Living Life
Topically Arranged Proverbs, Precepts, Quotations. Common Sayings. Poor Richard's Almanac.
America has lost her way
The really big sins
Theory on the Formation of Character
Moral Perversion
You are what you eat
People are like radio tuners --- they pick out and listen to one wavelength and ignore the rest
Cause of Character Traits --- According to Aristotle
These things go together
Television
We are what we eat --- living under the discipline of a diet
Avoiding problems and trouble in life
Role of habit in formation of character
The True Christian
What is true Christianity?
Personal attributes of the true Christian
What determines a person's character?
Love of God and love of virtue are closely united
Walking a solitary road
Intellectual disparities among people and the power in good habits
Tools of Satan. Tactics and Tricks used by the Devil.
On responding to wrongs
Real Christian Faith
The Natural Way -- The Unnatural Way
Wisdom, Reason and Virtue are closely related
Knowledge is one thing, wisdom is another
My views on Christianity in America
The most important thing in life is understanding
Sizing up people
We are all examples --- for good or for bad
Television --- spiritual poison
The Prime Mover that decides "What We Are"
Where do our outlooks, attitudes and values come from?
Sin is serious business. The punishment for it is real. Hell is real.
Self-imposed discipline and regimentation
Achieving happiness in life --- a matter of the right strategies
Self-discipline
Self-control, self-restraint, self-discipline basic to so much in life
We are our habits
What creates moral character?
[Home] [Up] [Info] [Mail] |
9958 | https://math.stackexchange.com/questions/254561/proof-of-arctanx-arcsinx-sqrt1x2 | trigonometry - Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$ - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Proof of arctan(x)=arcsin(x/1+x 2−−−−−√)arctan(x)=arcsin(x/1+x 2)
Ask Question
Asked 12 years, 9 months ago
Modified8 years, 8 months ago
Viewed 38k times
This question shows research effort; it is useful and clear
9
Save this question.
Show activity on this post.
I've tried following this way, but I haven't succeeded.
Thank you!
trigonometry
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Dec 9, 2012 at 15:16
levap
67.8k 5 5 gold badges 86 86 silver badges 132 132 bronze badges
asked Dec 9, 2012 at 13:53
AlexAlex
251 2 2 gold badges 4 4 silver badges 7 7 bronze badges
1
2 Please add the condition: x x is real, since this equation is false in general for complex numbers x x.GEdgar –GEdgar 2012-12-09 14:14:26 +00:00 Commented Dec 9, 2012 at 14:14
Add a comment|
7 Answers 7
Sorted by: Reset to default
This answer is useful
19
Save this answer.
Show activity on this post.
Consider the right angled triangle with sides 1,x,1+x 2−−−−−√1,x,1+x 2
Let ϕ ϕ be the angle opposite to the side of length x x.
We find that:
ϕ=arcsin(x/1+x 2−−−−−√)ϕ=arcsin(x/1+x 2)
ϕ=arctan(x/1)ϕ=arctan(x/1)
Thus:
arcsin(x/1+x 2−−−−−√)=arctan(x)arcsin(x/1+x 2)=arctan(x)
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Dec 9, 2012 at 13:56
AmrAmr
20.5k 5 5 gold badges 60 60 silver badges 125 125 bronze badges
0
Add a comment|
This answer is useful
8
Save this answer.
Show activity on this post.
Calculate the derivative of both sides:
(arctan x)′=1 1+x 2(arctanx)′=1 1+x 2
(arcsin x 1+x 2−−−−−√)′=1+x 2−−−−−√−x 2 1+x 2√1+x 2⋅1 1−x 2 1+x 2−−−−−−−√=(arcsinx 1+x 2)′=1+x 2−x 2 1+x 2 1+x 2⋅1 1−x 2 1+x 2=
=1(1+x 2)1+x 2−−−−−√⋅1+x 2−−−−−√1–√=1 1+x 2=1(1+x 2)1+x 2⋅1+x 2 1=1 1+x 2
Since both derivatives are equal the functions are the same up to the sum of a constant:
arctan x=arcsin x 1−x 2−−−−−√+C,C=a constant arctanx=arcsinx 1−x 2+C,C=a constant
Finally, to find what C C is you can, for example, input x=0 x=0 in the above...
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Jan 3, 2017 at 21:00
answered Dec 9, 2012 at 15:17
DonAntonioDonAntonio
215k 19 19 gold badges 143 143 silver badges 291 291 bronze badges
2
This is fine, but I think drawing the picture that gives you the answer without calculus is something that should be known by anyone who asks a question like this. See my answer below.Michael Hardy –Michael Hardy 2012-12-09 17:42:17 +00:00 Commented Dec 9, 2012 at 17:42
I know, @MichaelHardy. In fact, your answer is exactly the same as amr's, which I upvoted at once and which I'd have accepted as the best one. Perhaps the OP is now in calculus I precisely seeing derivatives and stuff and my answer appealed to him better...DonAntonio –DonAntonio 2012-12-09 18:18:34 +00:00 Commented Dec 9, 2012 at 18:18
Add a comment|
This answer is useful
5
Save this answer.
Show activity on this post.
As soon as you see arctan x arctanx, draw a right triangle in which the "opposite" side has length x x and the "adjacent" side has length 1 1. Then the angle to which those are "opposite" and "adjacent" is arctan x arctanx.
The Pythagorean theorem then tells you the length of the hypotenuse.
That gives you the sine of the angle, since sin=o p p h y p sin=o p p h y p.
That tells you what the angle in question is the arcsine of.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Dec 9, 2012 at 21:08
answered Dec 9, 2012 at 17:40
Michael HardyMichael Hardy
1
Add a comment|
This answer is useful
1
Save this answer.
Show activity on this post.
Let arctan x=y⇔x=tan y arctanx=y⇔x=tany. Then,
sin 2 y+cos 2 y=1⇔tan 2 y+1=1 cos 2 y⇔1 x 2+1=1−sin 2 y⇔sin 2 y=x 2 x 2+1 sin 2y+cos 2y=1⇔tan 2y+1=1 cos 2y⇔1 x 2+1=1−sin 2y⇔sin 2y=x 2 x 2+1
and so
sin y=x 1+x 2−−−−−√⇒arctan x=y=arcsin x 1+x 2−−−−−√siny=x 1+x 2⇒arctanx=y=arcsinx 1+x 2
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Dec 9, 2012 at 14:01
NamelessNameless
13.7k 3 3 gold badges 39 39 silver badges 59 59 bronze badges
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
put
x=tan(θ)x=tan(θ)
Now rewrite the formula in θ θ instead of x x. All you need, really, are these:
tan(x)=sin(x)/cos(x)tan(x)=sin(x)/cos(x)
sin 2(x)+cos 2(x)=1 sin 2(x)+cos 2(x)=1
should I be more explicit?
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Dec 9, 2012 at 17:37
Michael Hardy
1
answered Dec 9, 2012 at 14:00
mousomermousomer
395 2 2 silver badges 7 7 bronze badges
2
Yes, please....Alex –Alex 2012-12-09 14:02:10 +00:00 Commented Dec 9, 2012 at 14:02
nameless just did...mousomer –mousomer 2012-12-09 14:17:48 +00:00 Commented Dec 9, 2012 at 14:17
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
Let arctan x=y arctanx=y
⟹(i)tan y=x⟹(i)tany=x
and (i i)−π 2≤y≤π 2(i i)−π 2≤y≤π 2 (using the definition of principal value)
⟹cos y≥0⟹cosy≥0
We have
sin y x=cos y 1=±sin 2 y+cos 2 y x 2+1 2−−−−−−−−−−−−√=±1 x 2+1−−−−−√siny x=cosy 1=±sin 2y+cos 2y x 2+1 2=±1 x 2+1
⟹cos y=+1 x 2+1−−−−−√⟹cosy=+1 x 2+1 and sin y=x x 2+1−−−−−√siny=x x 2+1
So, arctan x=y=arcsin x x 2+1−−−−−√=arccos 1 x 2+1−−−−−√arctanx=y=arcsinx x 2+1=arccos1 x 2+1
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Nov 23, 2013 at 14:59
answered Dec 9, 2012 at 14:06
lab bhattacharjeelab bhattacharjee
279k 20 20 gold badges 213 213 silver badges 337 337 bronze badges
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
To prove that
sin(arctan x)=x(1+x 2)1 2 sin(arctanx)=x(1+x 2)1 2
let a=arctan x a=arctanx. Then
s=sin a=tan a⋅cos a=tan a(1−(sin a)2)1 2 s=sina=tana⋅cosa=tana(1−(sina)2)1 2
s 2=(x 2)⋅(1−s 2)s 2=(x 2)⋅(1−s 2)
s 2=x 2 1+x 2 s 2=x 2 1+x 2
s=sin a=sin a(arctan x)=x(1+x 2)1 2 s=sina=sina(arctanx)=x(1+x 2)1 2
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Sep 14, 2014 at 12:33
Joonas Ilmavirta
26.4k 10 10 gold badges 62 62 silver badges 112 112 bronze badges
answered Sep 14, 2014 at 12:05
alexalex
1
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
trigonometry
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Linked
3Evaluating ∫sin θ+cos θ sin 2 θ√d θ∫sinθ+cosθ sin2 θ d θ
3Find the limit: limit x tends to zero [arcsin(x)-arctan(x)]/(x^3)
2Definite integrals-is the question wrong or my method?
0Trigonometric function problem
0How would I use calculus to prove arcsin x=arctan(x/1−x 2−−−−−√)arcsinx=arctan(x/1−x 2)?
Related
2Trig Question - arcsin(2–√/2)arcsin(2/2) and arc trig functions in general
0Integration of arcsin polynomial
2Simplify sin(arctan 2−arcsin 1 10√).sin(arctan2−arcsin1 10).
1Evaluation of :lim n→∞(1+1 arctan(n))arctan(n)lim n→∞(1+1 arctan(n))arctan(n)
0arcsin(x)=arctan(2 x)arcsin(x)=arctan(2 x)
2Learning arcsin,arccos,arctan arcsin,arccos,arctan - how to?
2Solving trigonometric equation α=arctan(o/q)−arcsin(b/q)α=arctan(o/q)−arcsin(b/q)
1Evaluating lim x→0 1 x 2(arctan(1+x 2)−arcsin cos x 2√)lim x→0 1 x 2(arctan(1+x 2)−arcsincosx 2)
Hot Network Questions
Hypernym for different publication types and different outlets in which they are published
What happens when a creature with the See Invisibility spell is blinded?
Trying to understand the fundamental behavior of parallel connections
Stirring an Alphabet Soup
Compute the integer hull of a polyhedral set in Mathematica
New in my job, made a mistake with crediting a colleague – how should I handle this?
Word choice errors in submitted thesis
Ordinary insertion sort vs. straight insertion sort in Java (benchmark)
Why smoothness is not homeomorphism-invariant
What does the SH pin on HDMI do?
What is the purpose of the option delete readonly = yes available for the configuration of a Samba service?
two fish hiding something
What would be the impact of a being that is so hot it melts metal from 1km away. How far away would be habitable? (On Earth btw)
How to justify a mass hierarchal administrative demotion without causing a coup/civil war
Why aren't these functions o-minimal?
Maximizing profit given a PMF
How to model a twisted vase geometry with vertical ridges using Geometry Nodes?
Pressure treated or regular wood for temporary ramp?
Why it is that a 64-bit EXE file (PE64) can contain a custom MS-DOS stub (a MZ executable), but not a custom 32-bit Windows stub?
What is the legal ramification, if any, of this traffic sign in Sofia, Bulgaria?
Book about two ships that leave Earth decades apart but the later ship has FTL technology
What is the probability that a plane through three random points on the surface of the Earth intersects the Moon?
CSV TimeSheet Calculator
What are possible options to disembark for health issues if your airline keeps you on board at the tarmac for several hours?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.17.34093
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
9959 | https://math.stackexchange.com/questions/4662205/if-b-n-1n-fracn2019-find-the-sum-of-the-first-2019-terms | sequences and series - If $b_n=(-1)^n+\frac{n}{2019}$, find the sum of the first $2019$ terms - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
If b n=(−1)n+n 2019 b n=(−1)n+n 2019, find the sum of the first 2019 2019 terms
Ask Question
Asked 2 years, 6 months ago
Modified2 years, 6 months ago
Viewed 89 times
This question shows research effort; it is useful and clear
0
Save this question.
Show activity on this post.
The common term of the sequence b 1,b 2,...,b n,...b 1,b 2,...,b n,... is b n=(−1)n+n 2019,n=1,2,...b n=(−1)n+n 2019,n=1,2,... What is the sum of the first 2019 2019 terms of the sequence?
I have not found a way to solve the problem, as I do not see how to find a formula for the sum of the first n n terms of the given sequence (as we would have if the numbers were in a.p. or g.p.).
Finding the first few elements didn't help me much even on an intuitive level:
b 1=−2018 2019,b 2=2021 2019,b 3=−2016 2019,b 4=2023 2019,...b 1=−2018 2019,b 2=2021 2019,b 3=−2016 2019,b 4=2023 2019,...
Are we supposed to find such a formula, or is there something special about the sum of the first 2019 2019 terms of the sequence?
sequences-and-series
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this question to receive notifications
asked Mar 19, 2023 at 14:57
yinivem462yinivem462
644 3 3 silver badges 8 8 bronze badges
1
6 Can you compute ∑2019 n=1(−1)n∑n=1 2019(−1)n and ∑2019 n=1 n 2019∑n=1 2019 n 2019?Martin R –Martin R 2023-03-19 15:04:23 +00:00 Commented Mar 19, 2023 at 15:04
Add a comment|
2 Answers 2
Sorted by: Reset to default
This answer is useful
2
Save this answer.
Show activity on this post.
b k=(−1)k+1 2019 k b k=(−1)k+1 2019 k
∑k=1 n b k=∑k=1 n(−1)k+1 2019∑k=1 n k∑k=1 n b k=(−1)n−1 2+1 2019 n(n+1)2∑k=1 2019 b k=(−1)2019−1 2+1 2019 2019(2019+1)2=−1+2020 2=1009.∑k=1 n b k=∑k=1 n(−1)k+1 2019∑k=1 n k∑k=1 n b k=(−1)n−1 2+1 2019 n(n+1)2∑k=1 2019 b k=(−1)2019−1 2+1 2019 2019(2019+1)2=−1+2020 2=1009.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
answered Mar 19, 2023 at 15:06
K.defaoiteK.defaoite
14.3k 4 4 gold badges 24 24 silver badges 54 54 bronze badges
Add a comment|
This answer is useful
2
Save this answer.
Show activity on this post.
For a way to do it in your head: Calculate
b 1+b 2018 b 1+b 2018
b 2+b 2017 b 2+b 2017
......
b 1008+b 1011 b 1008+b 1011
b 1009+b 1010 b 1009+b 1010
Then calculate b 2019 b 2019, the one missing term from the sum, and add...
(Thank you, Gauss)
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
answered Mar 19, 2023 at 20:25
DJohnMDJohnM
3,630 1 1 gold badge 14 14 silver badges 18 18 bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
sequences-and-series
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
4Is the sequence 1 arctan(−n)⋅3 n−2 n 2+n+10 1 arctan(−n)⋅3 n−2 n 2+n+10 increasing or decreasing?
1Determine the arithmetic and geometric sequence given the relationships betweeen the first four terms
0Guess and Prove by induction a formula for the n n-th element in a sequence b n b n
0Finding out whether a progression is A.P. or G.P. and find the first four terms of the series.
2About Proving Convergence of Sequence
3Write b n+1 b n+1 in terms of b n b n, where b n=2 a 0 a 1 a 2...a n−1 a n∀n≥1 b n=2 a 0 a 1 a 2...a n−1 a n∀n≥1
1Assuming all the terms of A.P. [a n][a n] are integers with a 1=2019 a 1=2019. Find number of such sequence satisfying given condition
3If a 1,a 2,a 3 a 1,a 2,a 3 is geometric sequence such that a 1+a 2+a 3=91 a 1+a 2+a 3=91 and a 1,a 2,(a 3−13)a 1,a 2,(a 3−13) is arithmetic sequence, what the value of a 1 a 1?
Hot Network Questions
Direct train from Rotterdam to Lille Europe
How do you emphasize the verb "to be" with do/does?
Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator
Cannot build the font table of Miama via nfssfont.tex
What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left?
In Dwarf Fortress, why can't I farm any crops?
How to rsync a large file by comparing earlier versions on the sending end?
On Chern roots and the Chern polynomial
Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward"
Implications of using a stream cipher as KDF
What is a "non-reversible filter"?
What's the expectation around asking to be invited to invitation-only workshops?
Drawing the structure of a matrix
The rule of necessitation seems utterly unreasonable
Passengers on a flight vote on the destination, "It's democracy!"
Making sense of perturbation theory in many-body physics
в ответе meaning in context
Does the Mishna or Gemara ever explicitly mention the second day of Shavuot?
On being a Maître de conférence (France): Importance of Postdoc
Another way to draw RegionDifference of a cylinder and Cuboid
ICC in Hague not prosecuting an individual brought before them in a questionable manner?
Proof of every Highly Abundant Number greater than 3 is Even
Is it safe to route top layer traces under header pins, SMD IC?
Languages in the former Yugoslavia
more hot questions
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
9960 | https://www.khanacademy.org/math/ka-math-class-12/x6f66cbee9a2f805b:probability-ncert-new/x6f66cbee9a2f805b:independent-events/v/independent-events-dice-thrown-and-balls-picked | Independent events (dice thrown and balls picked) (video) | Khan Academy
Skip to main content
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked.
Explore
Browse By Standards
Explore Khanmigo
Math: Pre-K - 8th grade
Math: High school & college
Math: Multiple grades
Math: Illustrative Math-aligned
Math: Eureka Math-aligned
Math: Get ready courses
Test prep
Science
Economics
Reading & language arts
Computing
Life skills
Social studies
Partner courses
Khan for educators
Select a category to view its courses
Search
AI for Teachers FreeDonateLog inSign up
Search for courses, skills, and videos
Help us do more
We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever.
Select gift frequency
One time
Recurring
Monthly
Yearly
Select amount
$10
$20
$30
$40
Other
Give now
By donating, you agree to our terms of service and privacy policy.
Skip to lesson content
KA Math Class 12
Course: KA Math Class 12>Unit 13
Lesson 5: Independent events
Independent & dependent probability
Independent events example: test taking
Interpreting general multiplication rule
Independent probability
Testing for independence
Testing for independence
Finding probabilities given independence
Finding probabilities given independence
Proving statements for independent events
Independent events (dice thrown and balls picked)
Independent events (other scenarios)
Finding probabilities of independent events
Math>
KA Math Class 12>
Probability>
Independent events
© 2025 Khan Academy
Terms of usePrivacy PolicyCookie NoticeAccessibility Statement
Independent events (dice thrown and balls picked)
Google Classroom
Microsoft Teams
About About this video
In this video, we apply the concept of independent events to solve practical probability problems. We'll first tackle a problem involving tossing a die three times and finding the probability of getting an odd number at least once, by considering the complementary event. Then, we'll analyze a scenario where two balls are drawn at random with replacement from a box containing black and red balls, and calculate probabilities for different outcomes like both balls being red, or one being black and the other red.Created by Ashish Gupta.
Skip to end of discussions
Questions Tips & Thanks
Want to join the conversation?
Log in
Sort by:
Top Voted
Video transcript
Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube
Up next: video
Use of cookies
Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy
Accept All Cookies Strictly Necessary Only
Cookies Settings
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session.
Functional Cookies
[x] Functional Cookies
These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences.
Targeting Cookies
[x] Targeting Cookies
These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service.
Performance Cookies
[x] Performance Cookies
These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
9961 | https://www.albert.io/blog/combinations-of-functions-and-composite-functions/ | Published Time: 2025-02-03T18:30:27+00:00
Combinations of Functions and Composite Functions | Albert Resources
Skip to content
Subjects
Classes
Pricing
Schedule a Demo
Resources
Help Center
Albert Academy
Try Albert
School Success
Free Resources
Subjects
Classes
Pricing
Schedule a Demo
Resources
Help Center
Albert Academy
Try Albert
School Success
Free Resources
Log In
Sign Up
➜
AP® Precalculus
RESOURCES
PRACTICE
FREE RESPONSE
ASSESSMENTS
OVERVIEW
RESOURCES
PRACTICE
FREE RESPONSE
ASSESSMENTS
OVERVIEW
Combinations of Functions and Composite Functions
The Albert Team
Last Updated On: May 5, 2025
What We Review
Toggle - [x]
Introduction
What are Functions?
Understanding Combinations of Functions
Exploring Composite Functions
Domain of Composite Functions
Commutativity of Functions
Identity Function
Constructing Composite Functions
Conclusion
Quick Reference Chart
Need help preparing for your AP® Precalculus exam?
Introduction
Combinations of functions and composite functions are key concepts in precalculus and appear frequently in the AP® Precalculus exam. These foundational ideas help students to better understand how different mathematical operations interact. As mathematics involves building on previous knowledge, grasping these concepts is crucial. This guide encourages engagement with the material to ensure success on the exam.
Start practicing AP® Precalculus on Albert now!
What are Functions?
A function is a special relationship where each input has exactly one corresponding output. Think of it as a machine: you put something in (an input), and it processes to give you something out (an output).
Example:
Given f(x)=2 x+3 f(x) = 2x + 3 f(x)=2 x+3:
Input value: x x x
Output value: The result of 2 x+3 2x + 3 2 x+3 after substituting for x x x
When x=1 x = 1 x=1:f(1)=2(1)+3=5 f(1) = 2(1) + 3 = 5 f(1)=2(1)+3=5
The input is 1 1 1, and the output is 5 5 5.
Understanding Combinations of Functions
Combinations of functions involve combining two or more functions through basic operations: addition, subtraction, multiplication, and division.
Addition: (f+g)(x)=f(x)+g(x) (f + g)(x) = f(x) + g(x) (f+g)(x)=f(x)+g(x)
Subtraction: (f−g)(x)=f(x)−g(x) (f - g)(x) = f(x) - g(x) (f−g)(x)=f(x)−g(x)
Multiplication: (f⋅g)(x)=f(x)⋅g(x) (f \cdot g)(x) = f(x) \cdot g(x) (f⋅g)(x)=f(x)⋅g(x)
Division: (f/g)(x)=f(x)/g(x) (f / g)(x) = f(x) / g(x) (f/g)(x)=f(x)/g(x) (provided g(x)≠0 g(x) \neq 0 g(x)=0)
Example:
Given f(x)=x 2 f(x) = x^2 f(x)=x 2 and g(x)=3 x g(x) = 3x g(x)=3 x:
Calculate (f+g)(x)=f(x)+g(x)=x 2+3 x (f + g)(x) = f(x) + g(x) = x^2 + 3x (f+g)(x)=f(x)+g(x)=x 2+3 x:
Try substituting x=2 x = 2 x=2:
(f+g)(2)=2 2+3(2)=4+6=10 (f + g)(2) = 2^2 + 3(2) = 4 + 6 = 10 (f+g)(2)=2 2+3(2)=4+6=1 0
Ready to boost your AP® scores? Explore our plans and pricing here!
Exploring Composite Functions
Composite functions involve applying one function to the result of another. The notation f(g(x)) f(g(x)) f(g(x)) or f∘g(x) f \circ g(x) f∘g(x) is used to represent this composition.
Example:
Given f(x)=x+1 f(x) = x + 1 f(x)=x+1 and g(x)=2 x g(x) = 2x g(x)=2 x:
Calculate f(g(x)) f(g(x)) f(g(x)):
Start with g(x) g(x) g(x): g(x)=2 x g(x) = 2x g(x)=2 x
Place the result into f(x) f(x) f(x): f(g(x))=(2 x)+1 f(g(x)) = (2x) + 1 f(g(x))=(2 x)+1
For x=3 x = 3 x=3: g(3)=2⋅3=6→f(g(3))=6+1=7 g(3) = 2 \cdot 3 = 6 \rightarrow f(g(3)) = 6 + 1 = 7 g(3)=2⋅3=6→f(g(3))=6+1=7
The original uploader was H Padleckas at English Wikibooks.,CC BY-SA 3.0, via Wikimedia Commons
Domain of Composite Functions
The domain of a composite function is critical, as it dictates which input values are valid. It must satisfy both functions.
Example:
For f(x)=x f(x) = \sqrt{x} f(x)=x and g(x)=x−4 g(x) = x - 4 g(x)=x−4:
Determine the domain of f(g(x)) f(g(x)) f(g(x)):
g(x)=x−4 g(x) = x - 4 g(x)=x−4, so g(x)≥0 g(x) \geq 0 g(x)≥0 when: x−4≥0⇒x≥4 x - 4 \geq 0 \Rightarrow x \geq 4 x−4≥0⇒x≥4
Domain of f(g(x)) f(g(x)) f(g(x)) is x≥4 x \geq 4 x≥4 because it must ensure the square root has a non-negative input.
Commutativity of Functions
Function composition is not generally commutative. Let’s see why by comparing f(g(x)) f(g(x)) f(g(x)) and g(f(x)) g(f(x)) g(f(x)).
Example:
Given f(x)=x+2 f(x) = x + 2 f(x)=x+2 and g(x)=x 2 g(x) = x^2 g(x)=x 2:
f(g(x))=(x 2)+2 f(g(x)) = (x^2) + 2 f(g(x))=(x 2)+2
g(f(x))=(x+2)2 g(f(x)) = (x + 2)^2 g(f(x))=(x+2)2
Since:
f(g(3))=3 2+2=11 and g(f(3))=(3+2)2=25 f(g(3)) = 3^2 + 2 = 11 \text{ and } g(f(3)) = (3 + 2)^2 = 25 f(g(3))=3 2+2=1 1 and g(f(3))=(3+2)2=2 5
They are different, showing composition isn’t commutative.
Identity Function
The identity function is f(x)=x f(x) = x f(x)=x. Composing a function with the identity function will leave the function unchanged.
Example:
If f(x)=x f(x) = x f(x)=x then for any g(x) g(x) g(x):
f(g(x))=g(x) f(g(x)) = g(x) f(g(x))=g(x)
For g(x)=2 x+1 g(x) = 2x + 1 g(x)=2 x+1:
f(g(x))=2 x+1 f(g(x)) = 2x + 1 f(g(x))=2 x+1
Constructing Composite Functions
Creating a composite function can be simplified with a step-by-step approach. Break it down into simpler functions whenever possible.
Example:
Decompose f(x)=(3 x+6)2 f(x) = (3x + 6)^2 f(x)=(3 x+6)2:
Let h(x)=3 x+6 h(x) = 3x + 6 h(x)=3 x+6
Then f(x)=h(x)2 f(x) = h(x)^2 f(x)=h(x)2
Verify:
f(x)=(3 x+6)2=h(x)2 f(x) = (3x + 6)^2 = h(x)^2 f(x)=(3 x+6)2=h(x)2
This approach reinforces understanding through composition skills.
Conclusion
Combinations and composite functions form a significant part of precalculus studies. Mastery of these topics is vital for exam success. Practice with available resources can solidify these concepts, making complex operations much less daunting.
Quick Reference Chart
TermDefinition
Function A relationship where each input has exactly one output.
Composite Function A function where the output of one function is used as the input of another.
Domain The set of possible input values for a function.
Identity Function A function that returns the same value for any input: f(x)=x f(x) = x f(x)=x.
Commutative Property A property that does not generally apply to function composition.
Sharpen Your Skills for AP® Precalculus
Are you preparing for the AP® Precalculus exam? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world math problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now!
2.5 Exponential Function Context and Data Modeling
2.6 Competing Function Model Validation
2.8 Inverse Functions
Need help preparing for your AP® Precalculusexam?
Albert has hundreds of AP® Precalculus practice questions, free responses, and an AP® Precalculus practice test to try out.
Start practicing AP® Precalculus on Albert now!
Interested in a school license?
Bring Albert to your school and empower all teachers with the world's best question bank for:
➜ SAT® & ACT®
➜ AP®
➜ ELA, Math, Science, & Social Studies
➜ State assessments
Options for teachers, schools, and districts.
EXPLORE OPTIONS
Popular Posts
AP® Score Calculators
Simulate how different MCQ and FRQ scores translate into AP® scores
AP® Review Guides
The ultimate review guides for AP® subjects to help you plan and structure your prep.
Core Subject Review Guides
Review the most important topics in Physics and Algebra 1.
SAT® Score Calculator
See how scores on each section impacts your overall SAT® score
ACT® Score Calculator
See how scores on each section impacts your overall ACT® score
Grammar Review Hub
Comprehensive review of grammar skills
AP® Posters
Download updated posters summarizing the main topics and structure for each AP® exam.
Get Started
Log In
Sign Up
Course Library
Schedule a Demo
Company
Our Values
Our Team
Careers
Content
High School
Middle School
ELA
Literature
Math
Social Studies
Science
Advanced Placement®
SAT® and ACT®
Resources
Help Center
Adoption Guides
Albert Academy
Pilot Program
Our Services
Blog
Contact Us
Support
hello@albert.io
(312) 470-2290 ext. 0
Sales
schools@albert.io
(312) 470-2290 ext. 1
© 2025 Learn By Doing, Inc.
AP® and Advanced Placement® are registered trademarks of the College Board, which was not involved in the production of, and does not endorse, this product. SAT® is a registered trademark of the College Board, which was not involved in the production of, and does not endorse this product. ACT® is a registered trademark of ACT®, Inc., which was not involved in the production of, and does not endorse, this product. GMAT® is a registered trademark of the Graduate Management Admission Council®, which was not involved in the production of, and does not endorse, this product. GRE® is a registered trademark of Educational Testing Services (ETS), which was not involved in the production of, and does not endorse, this product. |
9962 | https://www.embibe.com/books/S-L-Loney-The-Elements-of-COORDINATE-GEOMETRY-Part-1-Cartesian-Coordinates/The-Parabola-%28Continued%29/EXAMPLES-XXIX/kve12433823-1 | S L Loney solutions for The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates The Parabola (Continued) S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 1: EXAMPLES XXIX
Harvest Smarter Results! Celebrate Baisakhi with smarter learning and steady progress.
Unlock discounts on all plans and grow your way to success!
Subscribe NowScan to download the App
E M B I B E
STUDY MATERIAL
EXAMS
PRODUCTS
ABOUT US
Free Sign Up
The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 9 - The Parabola (Continued)>EXAMPLES XXIX
Share
S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 1: EXAMPLES XXIX
Author:S L Loney
S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 1: EXAMPLES XXIX
Attempt the practice questions on Chapter 9: The Parabola (Continued), Exercise 1: EXAMPLES XXIX with hints and solutions to strengthen your understanding. The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates solutions are prepared by Experienced Embibe Experts.
Questions from S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 1: EXAMPLES XXIX with Hints & Solutions
MEDIUM
JEE Advanced
IMPORTANT
The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 9 - The Parabola (Continued)>EXAMPLES XXIX>Q 20
Find the locus of the middle point of chord of the parabola y 2=4 a x which passes through the fixed point h,k.
HARD
JEE Advanced
IMPORTANT
The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 9 - The Parabola (Continued)>EXAMPLES XXIX>Q 21
Find the locus of the middle points of the chords of the parabola which are normal to the curve.
HARD
JEE Advanced
IMPORTANT
The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 9 - The Parabola (Continued)>EXAMPLES XXIX>Q 22
Find the locus of the middle points of the chords of the parabola which subtend a constant angle θ at the vertex.
HARD
JEE Advanced
IMPORTANT
The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 9 - The Parabola (Continued)>EXAMPLES XXIX>Q 23
Find the locus of the middle points of chords of the parabola y 2=4 a x which are of given length l.
HARD
JEE Advanced
IMPORTANT
The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 9 - The Parabola (Continued)>EXAMPLES XXIX>Q 24
Find the locus of the middle points of chords of the parabola which are such that the normal at their extremities meet on the parabola.
HARD
JEE Advanced
IMPORTANT
The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 9 - The Parabola (Continued)>EXAMPLES XXIX>Q 25
Through each point of the straight line x=m y+h a chord of the parabola y 2=4 a x is drawn, which is bisected at the point. Prove that it always touches the parabola y+2 a m 2=8 a x-h.
HARD
JEE Advanced
IMPORTANT
The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 9 - The Parabola (Continued)>EXAMPLES XXIX>Q 26
Two parabolas have the same axis and tangents are drawn to the second from points on the first. Prove that the locus of the middle points of the chords of contact with the second parabola all lie on a fixed parabola.
HARD
JEE Advanced
IMPORTANT
The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 9 - The Parabola (Continued)>EXAMPLES XXIX>Q 27
Prove that the locus of the feet of the perpendicular drawn from the vertex of the parabola y 2=4 a x upon chords, which subtend an angle 45° at the vertex, is the curve r 2-24 a r cos θ+16 a 2 cos 2 θ=0.(where r is the perpendicular distance of the chord from the vertex)
Free Sign UpAsk a DoubtGet Free App
Learn and Prepare for any exam you want !
NCERT Books
NCERT Class 12 BooksNCERT Class 11 BooksNCERT Class 10 BooksNCERT Class 9 BooksNCERT Class 8 BooksNCERT Class 7 BooksNCERT Class 6 Books
NCERT Solutions
NCERT Solutions for Class 12NCERT Solutions for Class 11NCERT Solutions for Class 10NCERT Solutions for Class 9NCERT Solutions for Class 8NCERT Solutions for Class 7NCERT Solutions for Class 6
School Boards
CBSE Class 6CBSE Class 7CBSE Class 8CBSE Class 9CBSE Class 10CBSE Class 11CBSE Class 12Tamil Nadu Board Class 10Karnataka Board Class 10ICSE Class 9
Entrance Exams
NEETJEE MainMHT CETJEE AdvancedAP EAMCETTS EAMCETKCETKVPYBITSATCLAT
VITEEEKEAMAP EAPCETNATAGUJCETSRMJEEEGCETCOMEDK UGETCUSAT CATASSAM CEE
Government Exams
CATSSC GD ConstableCisf Head ConstableXATUP Police ConstableUPSC CDSRajasthan Police ConstableSBI CBORRB NTPC CBT 2IBPS PO Mains
SSC CGL Tier 2UGC NET Paper 1IBPS Clerk MainsSSC CGL Tier-IHSSC ClerkIDBI Assistant ManagerRBI Office AttendantIBPS PO PrelimsIPMAT IndoreCTET Paper 1
UP Police SISBI ApprenticeCTET Paper 2SBI PO MainsHaryana Police ConstableIBPS Clerk PrelimsSBI Clerk PrelimsSBI PO PrelimsDelhi Police MTSESIC Stenographer
Rajasthan PatwariSBI Clerk MainsRBI Assistant PrelimsSSC CHSL Tier-IRBI Assistant MainsHaryana Police SISSC MTS Tier IRRB NTPC CBT 1South Indian Bank ClerkUPTET Paper 1
UPTET Paper 2IB ACIO Tier - IIBPS RRB Office Assistant PrelimsIBPS RRB Office Assistant MainsRRB ALP CBT 1IBPS RRB Officer Scale-I PrelimsIBPS RRB Officer Scale-I MainsDelhi Police Constable CBTNDA & NA EEIIFT
Our Partners-State Governments
Tamil NaduHimachal PradeshRajasthanBiharJammukashmir
MaharashtraMadhya PradeshPunjabNagalandJharkhand
KarnatakaTelanganaAndhra PradeshOdishaHaryana
West BengalAssamChhattisgarhGoaMeghalaya
UttarakhandUttar PradeshArunachal PradeshLadakhSikkim
Help
Notices
Privacy Policy
VDP
Contact Us
Terms & Conditions
Join Us
Investor Relations
TakeDown Policy
© Copyright 2025, Embibe. All rights reserved |
9963 | https://fiveable.me/key-terms/hs-physical-science/avogadros-law | Avogadro's Law - (Physical Science) - Vocab, Definition, Explanations | Fiveable | Fiveable
new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom
ap study content toolsprintablespricing
my subjectsupgrade
All Key Terms
Physical Science
Avogadro's Law
🫴physical science review
key term - Avogadro's Law
Citation:
MLA
Definition
Avogadro's Law states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules. This principle is significant because it connects the volume of a gas to the number of particles it contains, establishing a direct relationship that is fundamental in understanding how gases behave under various conditions and also plays a key role in stoichiometry by linking gas volume to the amounts of reactants and products in chemical reactions.
5 Must Know Facts For Your Next Test
Avogadro's Law implies that if you double the volume of a gas at constant temperature and pressure, you also double the number of molecules present.
The law applies specifically to gases and assumes ideal behavior, meaning it works best under low pressure and high temperature conditions.
Avogadro's number, approximately 6.022 x 10^23, represents the number of molecules in one mole of a substance and is fundamental in relating moles to particles.
Avogadro's Law can be mathematically expressed as V1/n1 = V2/n2, where V is volume and n is the number of moles for two different states of the same gas.
This law allows chemists to predict how gases will react with each other based on their volumes when conducting experiments.
Review Questions
How does Avogadro's Law facilitate our understanding of gas behavior in reactions?
Avogadro's Law allows us to relate the volume of a gas to the number of molecules it contains when temperature and pressure are constant. This helps predict how much of a gaseous reactant will be needed to produce a certain amount of product in a chemical reaction. By knowing that equal volumes contain equal numbers of molecules, chemists can use this relationship to balance equations involving gases effectively.
Discuss how Avogadro's Law interacts with other gas laws like Boyle's and Charles' Laws.
Avogadro's Law complements other gas laws such as Boyle's and Charles' Laws by providing a fuller understanding of how gases behave. While Boyle's Law focuses on the relationship between pressure and volume at constant temperature, and Charles' Law examines volume and temperature at constant pressure, Avogadro's Law adds the dimension of particle quantity to these relationships. Together, they form the Ideal Gas Law, which integrates these principles into a comprehensive equation that describes ideal gas behavior.
Evaluate the implications of Avogadro's Law for stoichiometric calculations in chemical reactions involving gases.
Avogadro's Law has significant implications for stoichiometry since it allows chemists to use volumes of gases directly in calculations related to reactions. When balancing chemical equations involving gaseous reactants and products, knowing that equal volumes contain equal numbers of molecules simplifies calculations. This means that if one knows the volume of one gaseous reactant at standard conditions, they can directly calculate the required volumes of other reactants or expected products without converting to moles first. This efficiency enhances accuracy in quantitative analysis and helps streamline experimental procedures.
Related terms
Molar Volume:The volume occupied by one mole of a substance, typically measured at standard temperature and pressure (STP), which is approximately 22.4 liters for an ideal gas.
Ideal Gas Law:
A relation between pressure, volume, temperature, and the number of moles of a gas, represented by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
Stoichiometry:The branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction.
"Avogadro's Law" also found in:
Subjects (3)
AP Chemistry
College Physics I – Introduction
Intro to Chemistry
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
every AP exam is fiveable
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
every AP exam is fiveable
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
every AP exam is fiveable
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
every AP exam is fiveable
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
0 |
9964 | https://chem.libretexts.org/Courses/Brevard_College/CHE_104%3A_Principles_of_Chemistry_II/02%3A_The_States_of_Matter/2.03%3A_Gas_Laws | 2.3: Gas Laws - Chemistry LibreTexts
Skip to main content
Table of Contents menu
search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode
school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Search
Search this book
Submit Search
x
Text Color
Reset
Bright Blues Gray Inverted
Text Size
Reset
+-
Margin Size
Reset
+-
Font Type
Enable Dyslexic Font - [x]
Downloads expand_more
Download Page (PDF)
Download Full Book (PDF)
Resources expand_more
Periodic Table
Physics Constants
Scientific Calculator
Reference expand_more
Reference & Cite
Tools expand_more
Help expand_more
Get Help
Feedback
Readability
x
selected template will load here
Error
This action is not available.
chrome_reader_mode Enter Reader Mode
2: The States of Matter
CHE 184: Principles of Chemistry II
{ }
{ "2.01:_States_of_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.02:_Kinetic-Molecular_Theory" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.03:_Gas_Laws" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.04:_Diffusion_and_Effusion-_Graham\'s_Law" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.05:_Changes_in_State_and_Heating_Curves" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
{ "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Electrochemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_The_States_of_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Solutions_and_Colloids" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Thermochemistry_and_Thermodynamics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Reaction_Rates" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Chemical_Equilibrium" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Acid_and_Base_Equilibria" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Radioactivity_and_Nuclear_Processes" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
Fri, 21 May 2021 18:11:49 GMT
2.3: Gas Laws
221449
221449
Hernan Biava
{ }
Anonymous
Anonymous User
2
false
false
[ "article:topic", "showtoc:no", "license:ccbyncsa", "source-chem-16103", "source-chem-16103", "licenseversion:40" ]
[ "article:topic", "showtoc:no", "license:ccbyncsa", "source-chem-16103", "source-chem-16103", "licenseversion:40" ]
Search site Search Search Go back to previous article
Sign in
Username Password Sign in
Sign in
Sign in
Forgot password
Expand/collapse global hierarchy
1. Home
2. Campus Bookshelves
3. Brevard College
4. CHE 184: Principles of Chemistry II
5. 2: The States of Matter
6. 2.3: Gas Laws
Expand/collapse global location
2.3: Gas Laws
Last updated May 21, 2021
Save as PDF
2.2: Kinetic-Molecular Theory
2.4: Diffusion and Effusion- Graham's Law
Page ID 221449
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Boyle’s Law
2. Charles’s Law
3. Gay-Lussac's Law
4. Combined Gas Law
5. Avogadro´s law
6. The Ideal Gas Law
7. Dalton's Law of Partial Pressures
8. Application of Dalton's Law: Collecting Gases over Water
9. Henry’s Law
1. Application of Henry's Law: Scuba diving
Concept Review Exercises
Answers
Key Takeaway
Exercises
Answers
Attribution and citations
Learning Objectives
To predict the properties of gases using the gas laws.
Experience has shown that several properties of a gas can be related to each other under certain conditions. The properties are pressure (P), volume (V), temperature (T, in kelvins), and amount of material expressed in moles (n). What we find is that a sample of gas cannot have any random values for these properties. Instead, only certain values, dictated by some simple mathematical relationships, will occur.
Boyle’s Law
The first simple relationship, referred to as a gas law, is between the pressure of a gas and its volume. If the amount of gas in a sample and its temperature are kept constant, then as the pressure of a gas is increased, the volume of the gas decreases proportionately. Mathematically, this is written as
(2.3.1)P∝1 V
where the “∝” symbol means “is proportional to.” This is one form of Boyle’s law, which relates the pressure of a gas to its volume.
A more useful form of Boyle’s law involves a change in conditions of a gas. For a given amount of gas at a constant temperature, if we know the initial pressure and volume of a gas sample and the pressure or volume changes, we can calculate what the new volume or pressure will be. That form of Boyle’s law is written
(2.3.2)P iV i=P fV f
where the subscript i refers to initial conditions and the subscript f refers to final conditions.
To use 2.3.2, you need to know any three of the variables so that you can algebraically calculate the fourth variable. Also, the pressure quantities must have the same units, as must the two volume quantities. If the two similar variables don’t have the same variables, one value must be converted to the other value’s unit.
Example 2.3.1: Increasing Pressure in a Gas
What happens to the volume of a gas if its pressure is increased? Assume all other conditions remain the same.
Solution
If the pressure of a gas is increased, the volume decreases in response.
Exercise 2.3.1: Increasing Volume in a Gas
What happens to the pressure of a gas if its volume is increased? Assume all other conditions remain the same.
Answer
If the volume of a gas is increased, the pressure decreases.
Example 2.3.2: Gas Compression
If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is reduced to 0.987 atm? Assume that the amount and the temperature of the gas remain constant.
Solution
The key in problems like this is to be able to identify which quantities represent which variables from the relevant equation. The way the question is worded, you should be able to tell that 1.56 atm is P i, 7.02 L is V i, and 0.987 atm is P f. What we are looking for is the final volume—V f. Therefore, substituting these values into P i V i = P f V f:
(1.56 atm)(7.02 L) = (0.987 atm) × V f
The expression has atmospheres on both sides of the equation, so they cancel algebraically:
(1.56)(7.02 L) = (0.987) × V f
Now we divide both sides of the expression by 0.987 to isolate V f, the quantity we are seeking:
(1.56)(7.02 L)0.987=V f
Performing the multiplication and division, we get the value of V f, which is 11.1 L. The volume increases. This should make sense because the pressure decreases, so pressure and volume are inversely related.
Exercise 2.3.2
If a sample of gas has an initial pressure of 3.66 atm and an initial volume of 11.8 L, what is the final pressure if the volume is reduced to 5.09 L? Assume that the amount and the temperature of the gas remain constant.
Answer
8.48 atm
If the units of similar quantities are not the same, one of them must be converted to the other quantity’s units for the calculation to work out properly. It does not matter which quantity is converted to a different unit; the only thing that matters is that the conversion and subsequent algebra are performed properly. The following example illustrates this process.
Example 2.3.3
If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is changed to 1,775 torr? Does the answer make sense? Assume that the amount and the temperature of the gas remain constant.
Solution
This example is similar to Example 2.3.2, except now the final pressure is expressed in torr. For the math to work out properly, one of the pressure values must be converted to the other unit. Let us change the initial pressure to torr:
1.56 atm×760 torr 1 atm=1,190 torr
Now we can use Boyle’s law:
(1,190 torr)(7.02 L) = (1,775 torr) × V f
Torr cancels algebraically from both sides of the equation, leaving
(1,190)(7.02 L) = (1,775) × V f
Now we divide both sides of the equation by 1,775 to isolate V f on one side. Solving for the final volume,
V f=(1,190)(7.02 L)1,775=4.71 L
Because the pressure increases, it makes sense that the volume decreases.
The answer for the final volume is essentially the same if we converted the 1,775 torr to atmospheres: 1,775 torr×1 atm 760 torr=2.336 atm. Using Boyle’s law: (1.56 atm)(7.02 L) = (2.335 atm) × V f; V f=(1.56 atm)(7.02 L)2.336 atm=4.69 L.
Exercise 2.3.3
If a sample of gas has an initial pressure of 375 torr and an initial volume of 7.02 L, what is the final pressure if the volume is changed to 4,577 mL? Does the answer make sense? Assume that amount and the temperature of the gas remain constant.
Answer
575 torr
To Your Health: Breathing
Breathing certainly is a major contribution to your health! Without breathing, we could not survive. Curiously, the act of breathing itself is little more than an application of Boyle’s law.
The lungs are a series of ever-narrowing tubes that end in a myriad of tiny sacs called alveoli. It is in the alveoli that oxygen from the air transfers to the bloodstream and carbon dioxide from the bloodstream transfers to the lungs for exhalation. For air to move in and out of the lungs, the pressure inside the lungs must change, forcing the lungs to change volume—just as predicted by Boyle’s law.
The pressure change is caused by the diaphragm, a muscle that covers the bottom of the lungs. When the diaphragm moves down, it expands the size of our lungs. When this happens, the air pressure inside our lungs decreases slightly. This causes new air to rush in, and we inhale. The pressure decrease is slight—only 3 torr, or about 0.4% of an atmosphere. We inhale only 0.5–1.0 L of air per normal breath.
Exhaling air requires that we relax the diaphragm, which pushes against the lungs and slightly decreases the volume of the lungs. This slightly increases the pressure of the air in the lungs, and air is forced out; we exhale. Only 1–2 torr of extra pressure is needed to exhale. So with every breath, our own bodies are performing an experimental test of Boyle’s law.
Charles’s Law
Another simple gas law relates the volume of a gas to its temperature. Experiments indicate that as the temperature of a gas sample is increased, its volume increases as long as the pressure and the amount of gas remain constant. The way to write this mathematically is
(2.3.3)V∝T
At this point, the concept of temperature must be clarified. Although the Kelvin scale is the preferred temperature scale, the Celsius scale is also a common temperature scale used in science. The Celsius scale is based on the melting and boiling points of water and is actually the common temperature scale used by most countries around the world (except for the United States, which still uses the Fahrenheit scale). The value of a Celsius temperature is directly related to its Kelvin value by a simple expression:
Kelvin temperature = Celsius temperature + 273
Thus, it is easy to convert from one temperature scale to another.
The Kelvin scale is sometimes referred to as the absolute scale because the zero point on the Kelvin scale is at absolute zero, the coldest possible temperature. On the other temperature scales, absolute zero is −260°C or −459°F.
The expression relating a gas volume to its temperature begs the following question: to which temperature scale is the volume of a gas related? The answer is that gas volumes are directly related to the Kelvin temperature. Therefore, the temperature of a gas sample should always be expressed in (or converted to) a Kelvin temperature.
Example 2.3.4: Increasing Temperature
What happens to the volume of a gas if its temperature is decreased? Assume that all other conditions remain constant.
Solution
If the temperature of a gas sample is decreased, the volume decreases as well.
Exercise 2.3.4
What happens to the temperature of a gas if its volume is increased? Assume that all other conditions remain constant.
Answer
The temperature increases.
As with Boyle’s law, the relationship between volume and temperature can be expressed in terms of initial and final values of volume and temperature, as follows:
(2.3.4)V i T i=V f T f
where V i and T i are the initial volume and temperature, and V f and T f are the final volume and temperature. This is Charles’s law. The restriction on its use is that the pressure of the gas and the amount of gas must remain constant.
Example 2.3.5
A gas sample at 20°C has an initial volume of 20.0 L. What is its volume if the temperature is changed to 60°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant.
Solution
Although the temperatures are given in degrees Celsius, we must convert them to the kelvins before we can use Charles’s law. Thus,
20°C + 273 = 293 K = T i 60°C + 273 = 333 K = T f
Now we can substitute these values into Charles’s law, along with the initial volume of 20.0 L:
20.0 L 293 K=V f 333 K
Multiplying the 333 K to the other side of the equation, we see that our temperature units will cancel:
(333 K)(20.0 L)293 K=V f
Solving for the final volume, V f = 22.7 L. So, as the temperature is increased, the volume increases. This makes sense because volume is directly proportional to the absolute temperature (as long as the pressure and the amount of the remain constant).
Exercise 2.3.5
A gas sample at 35°C has an initial volume of 5.06 L. What is its volume if the temperature is changed to −35°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant.
Answer
3.91 L
Gay-Lussac's Law
Another simple gas law relates the pressure of a gas to its temperature. Experiments indicate that as the temperature of a gas sample is increased in a container at constant volume, its pressure increases. The way to write this mathematically is
(2.3.5)P∝T
As with Boyle’s law, the relationship between volume and temperature can be expressed in terms of initial and final values of volume and temperature, as follows:
(2.3.6)P i T i=P f T f
where P i and T i are the initial pressure and temperature, and P f and T f are the final pressure and temperature. The restriction on its use is that the volume of the gas and the amount of gas must remain constant. Just like in Charles'law,the temperature of a gas sample should always be expressed in a Kelvin temperature when applying Gay-Lussac's Law.
Combined Gas Law
The combined gas law brings Boyle’s, Charles’s land Gay-Lussac´s laws together to relate pressure, volume, and temperature changes of a gas sample:
\mathrm{\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}}
To apply this gas law, the amount of gas should remain constant. As with the other gas laws, the temperature must be expressed in kelvins, and the units on the similar quantities should be the same. Because of the dependence on three quantities at the same time, it is difficult to tell in advance what will happen to one property of a gas sample as two other properties change. The best way to know is to work it out mathematically.
Example 2.3.6
A sample of gas has P i = 1.50 atm, V i = 10.5 L, and T i = 300 K. What is the final volume if P f = 0.750 atm and T f = 350 K?
Solution
Using the combined gas law, substitute for five of the quantities:
(1.50 atm)(10.5 L)300 K=(0.750 atm)(V f)350 K
We algebraically rearrange this expression to isolate V f on one side of the equation:
V f=(1.50 atm)(10.5 L)(350 K)(300 K)(0.750 atm)=24.5 L
Note how all the units cancel except the unit for volume.
Exercise 2.3.6
A sample of gas has P i = 0.768 atm, V i = 10.5 L, and T i = 300 K. What is the final pressure if V f = 7.85 L and T f = 250 K?
Answer
0.856 atm
Example 2.3.7
A balloon containing a sample of gas has a temperature of 22°C and a pressure of 1.09 atm in an airport in Cleveland. The balloon has a volume of 1,070 mL. The balloon is transported by plane to Denver, where the temperature is 11°C and the pressure is 655 torr. What is the new volume of the balloon?
Solution
The first task is to convert all quantities to the proper and consistent units. The temperatures must be expressed in kelvins, and the pressure units are different so one of the quantities must be converted. Let us convert the atmospheres to torr:
22°C + 273 = 295 K = T i
11°C + 273 = 284 K = T f
1.09 atm×760 torr 1 atm=828 torr=P i
Now we can substitute the quantities into the combined has law:
(828 torr)(1,070 mL)295 K=(655 torr)×V f 284 K
To solve for V f, we multiply the 284 K in the denominator of the right side into the numerator on the left, and we divide 655 torr in the numerator of the right side into the denominator on the left:
(828 torr)(1,070 mL)(284 K)(295 K)(655 torr)=V f
Notice that torr and kelvins cancel, as they are found in both the numerator and denominator. The only unit that remains is milliliters, which is a unit of volume. So V f = 1,300 mL. The overall change is that the volume of the balloon has increased by 230 mL.
Exercise 2.3.7
A balloon used to lift weather instruments into the atmosphere contains gas having a volume of 1,150 L on the ground, where the pressure is 0.977 atm and the temperature is 18°C. Aloft, this gas has a pressure of 6.88 torr and a temperature of −15°C. What is the new volume of the gas?
Answer
110,038 L
Avogadro´s law
Avogadro's law relates the volume of a gas to the amount of gas present.The law says that, at constant temperature and pressure, the volume of a gas is directly proportional to the amount of moles (or grams) of a gas:
(2.3.7)V∝n
In this equation,V is volume and n is amount of moles.For comparing the same substance under two different sets of conditions, the law can be usefully expressed as follows:
(2.3.8)V i n i=V f n f
where V i and n i are the initial volume and number of moles, and V f and n f are the final volume and number of moles.
The Ideal Gas Law
So far, the gas laws we have used have focused on changing one or more properties of the gas, such as its volume, pressure, or temperature. There is one gas law that relates all the independent properties of a gas under any particular condition, rather than a change in conditions. This gas law is called the ideal gas law. The formula of this law is as follows:
(2.3.9)PV=nRT
In this equation, P is pressure, V is volume, n is amount of moles, and T is temperature. R is called the ideal gas law constant and is a proportionality constant that relates the values of pressure, volume, amount, and temperature of a gas sample. The variables in this equation do not have the subscripts i and f to indicate an initial condition and a final condition. The ideal gas law relates the four independent properties of a gas under any conditions.
The value of R depends on what units are used to express the other quantities. If volume is expressed in liters and pressure in atmospheres, then the proper value of R is as follows:
(2.3.10)R=0.08205 L⋅atm mol⋅K
This may seem like a strange unit, but that is what is required for the units to work out algebraically.
Example 2.3.8
What is the volume in liters of 1.45 mol of N 2 gas at 298 K and 3.995 atm?
Solution
Using the ideal gas law where P = 3.995 atm, n = 1.45, and T = 298,
(3.995 atm)×V=(1.45 mol)(0.08205 L⋅atm mol⋅K)(298 K)
On the right side, the moles and kelvins cancel. Also, because atmospheres appear in the numerator on both sides of the equation, they also cancel. The only remaining unit is liters, a unit of volume. So
3.995 × V = (1.45)(0.08205)(298) L
Dividing both sides of the equation by 3.995 and evaluating, we get V = 8.87 L. Note that the conditions of the gas are not changing. Rather, the ideal gas law allows us to determine what the fourth property of a gas (here, volume) must be if three other properties (here, amount, pressure, and temperature) are known.
Exercise 2.3.8
What is the pressure of a sample of CO 2 gas if 0.557 mol is held in a 20.0 L container at 451 K?
Answer
1.03 atm
For convenience, scientists have selected 273 K (0°C) and 1.00 atm pressure as a set of standard conditions for gases. This combination of conditions is called standard temperature and pressure (STP). Under these conditions, 1 mol of any gas has about the same volume. We can use the ideal gas law to determine the volume of 1 mol of gas at STP:
(2.3.11)(1.00 atm)×V=(1.00 mol)(0.08205 L⋅atm mol⋅K)(273 K)
This volume is 22.4 L. Because this volume is independent of the identity of a gas, the idea that 1 mol of gas has a volume of 22.4 L at STP makes a convenient conversion factor:
1 mol gas = 22.4 L (at STP)
Example 2.3.9
Cyclopropane (C 3 H 6) is a gas that formerly was used as an anesthetic. How many moles of gas are there in a 100.0 L sample if the gas is at STP?
Solution
We can set up a simple, one-step conversion that relates moles and liters:
100.0 L C 3H 6×1 mol 22.4 L=4.46 mol C 3H 6
There are almost 4.5 mol of gas in 100.0 L.
Note: Because of its flammability, cyclopropane is no longer used as an anesthetic gas.
Exercise 2.3.9
Freon is a trade name for a series of fluorine- and chlorine-containing gases that formerly were used in refrigeration systems. What volume does 8.75 mol of Freon have at STP?
Note: Many gases known as Freon are no longer used because their presence in the atmosphere destroys the ozone layer, which protects us from ultraviolet light from the sun.
Answer
196 L
Airbags
Airbags (Figure 2.3.3) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN 3. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN 3 to initiate its decomposition:
(2.3.12)2 NaNA 3(s)→3 NA 2(g)+2 Na(s)
This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN 3 will generate approximately 50 L of N 2.
Figure 2.3.3: Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)
Dalton's Law of Partial Pressures
The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a gas mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments. The only new concept we need in order to deal with gas mixtures is the partial pressure, a concept invented by the famous English chemist John Dalton (1766-1844). Dalton reasoned that the low density and high compressibility of gases indicates that they consist mostly of empty space; from this it follows that when two or more different gases occupy the same volume, they behave entirely independently. The contribution that each component of a gaseous mixture makes to the total pressure of the gas is known as the partial pressure of that gas.
The definition of Dalton's Law of Partial Pressures that address this is:
The total pressure of a gas is the sum of the partial pressures of its components
which is expressed algebraically as
(2.3.13)P total=P 1+P 2+P 3...=∑i P i
or, equivalently
(2.3.14)P total=RT V∑i n i
There is also a similar relationship based on volume fractions, known as Amagat's law of partial volumes. It is exactly analogous to Dalton's law, in that it states that the total volume of a mixture is just the sum of the partial volumes of its components. But there are two important differences: Amagat's law holds only for ideal gases which must all be at the same temperature and pressure. Dalton's law has neither of these restrictions. Although Amagat's law seems intuitively obvious, it sometimes proves useful in chemical engineering applications. We will make no use of it in this course.
Example 2.3.10
Three flasks having different volumes and containing different gases at various pressures are connected by stopcocks as shown. When the stopcocks are opened,
What will be the pressure in the system?
Which gas will be most abundant in the mixture?
Assume that the temperature is uniform and that the volume of the connecting tubes is negligible.
Solution
The trick here is to note that the total number of moles n T and the temperature remain unchanged, so we can make use of Boyle's law PV = constant. We will work out the details for CO 2 only, denoted by subscripts a.
For CO 2,
(2.3.15)P aV a=(2.13 atm)(1.50 L)=3.19 L⋅atm
Adding the PV products for each separate container, we obtain
(2.3.16)∑i P iV i=6.36 L⋅atm=n TRT
We will call this sum P 1 V 1. After the stopcocks have been opened and the gases mix, the new conditions are denoted by P 2 V 2.
From Boyle's law (2.3.2,
(2.3.17)P 1V 1=P 2V 2=6.36 L⋅atm
(2.3.18)V 2=∑i V i=4.50 L
Solving for the final pressure P 2 we obtain (6.36 L-atm)/(4.50 L) = 1.41 atm.
For part (b), note that the number of moles of each gas is n = PV/RT. The mole fraction of any one gas is X i = n i /n T. For CO 2, this works out to (3.19/RT) / (6.36/RT) = 0.501. Because this exceeds 0.5, we know that this is the most abundant gas in the final mixture.
Dalton’s law states that in a gas mixture (P total) each gas will exert a pressure independent of the other gases (P n) and each gas will behave as if it alone occupies the total volume. By extension, the partial pressure of each gas can be calculated by multiplying the total pressure (P total) by the gas percentage (%).
(2.3.19)P Total=P 1+P 2+P 3+P 4+...+P n
or
(2.3.20)P n=n P Total
Table 2.3.1: Partial Pressures for the gases in air on a typical day| Gas | Partial Pressure (mm Hg) | Percentage (%) |
---
| Nitrogen, (N_2) | P N 2 = 594 | 78 |
| Oxygen, O 2 | P O 2= 160 | 21 |
| Carbon Dioxide, CO 2 | P CO 2 = 0.25 | 0.033 |
| Water Vapor, H 2O | P H 2O = 5.7 | 0.75 |
| Other trace gases | P Other = 0.05 | 0.22 |
| Total air | P Total = 760 | 1 |
Application of Dalton's Law: Collecting Gases over Water
A common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. This arrangement is called a pneumatic trough, and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and becomes trapped in the upper part.
Figure 2.3.1:An Apparatus for Collecting Gases by the Displacement of Water
The volume of the gas can be observed by means of a calibrated scale on the bottle, but what about its pressure? The total pressure confining the gas is just that of the atmosphere transmitting its force through the water. (An exact calculation would also have to take into account the height of the water column in the inverted tube.) But liquid water itself is always in equilibrium with its vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected, and gaseous H 2 O. The partial pressure of H 2 O is known as the vapor pressure of water and it depends on the temperature. In order to determine the quantity of gas we have collected, we must use Dalton's Law to find the partial pressure of that gas.
Example 2.3.11
Oxygen gas was collected over water as shown above. The atmospheric pressure was 754 torr, the temperature was 22°C, and the volume of the gas was 155 mL. The vapor pressure of water at 22°C is 19.8 torr. Use this information to estimate the number of moles of O 2 produced.
Solution
From Dalton's law,
P O 2=P total–P H 2O=754–19.8=734 torr=0.966 atm
Now use the Ideal Gas Law to convert to moles
n=PV RT=(0.966 atm)(0.155 L)(0.082 Latmmol−1K−1)(295 K)=0.00619 mol
Henry’s Law
Henry's law is one of the gas laws formulated by William Henry in 1803. It states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
To explain this law, Henry derived the equation:
(2.3.21)C=kP gas
where
Henry’s Law tells us that the greater the pressure of gas above the surface of a liquid, the higher the concentration of the gas in the liquid. Also, Henry’s law tells us that gases diffuse from areas of high gas concentration to areas of low gas concentration.
Applicability of Henry's Law
Henry's law only works if the molecules are at equilibrium.
Henry's law does not work for gases at high pressures (e.g., N 2(g) at high pressure becomes very soluble and harmful when in the blood supply).
Henry's law does not work if there is a chemical reaction between the solute and solvent (e.g., HCl(g) reacts with water by a dissociation reaction to generate H 3O+ and Cl− ions).
Application of Henry's Law: Scuba diving
Our respiratory systems are designed to maintain the proper oxygen concentration in the blood when the partial pressure of O 2 is 0.21 atm, its normal sea-level value. Below the water surface, the pressure increases by 1 atm for each 10.3 m increase in depth; thus a scuba diver at 10.3 m experiences a total of 2 atm pressure pressing on the body. In order to prevent the lungs from collapsing, the air the diver breathes should also be at about the same pressure.
Figure 2.3.2: Scuba Dviging actively takes into account both Henry's and Dalton's Laws
But at a total pressure of 2 atm, the partial pressure of O 2 in ordinary air would be 0.42 atm; at a depth of 100 ft (about 30 m), the O 2 pressure of 0.8 atm would be far too high for health. For this reason, the air mixture in the pressurized tanks that scuba divers wear must contain a smaller fraction of O 2. This can be achieved most simply by raising the nitrogen content, but high partial pressures of N 2 can also be dangerous, resulting in a condition known as nitrogen narcosis. The preferred diluting agent for sustained deep diving is helium, which has very little tendency to dissolve in the blood even at high pressures.
Career Focus: Respiratory Therapist
Certain diseases—such as emphysema, lung cancer, and severe asthma—primarily affect the lungs. Respiratory therapists help patients with breathing-related problems. They can evaluate, help diagnose, and treat breathing disorders and even help provide emergency assistance in acute illness where breathing is compromised.
Most respiratory therapists must complete at least two years of college and earn an associate’s degree, although therapists can assume more responsibility if they have a college degree. Therapists must also pass state or national certification exams. Once certified, respiratory therapists can work in hospitals, doctor’s offices, nursing homes, or patient’s homes. Therapists work with equipment such as oxygen tanks and respirators, may sometimes dispense medication to aid in breathing, perform tests, and educate patients in breathing exercises and other therapy.
Because respiratory therapists work directly with patients, the ability to work well with others is a must for this career. It is an important job because it deals with one of the most crucial functions of the body.
Concept Review Exercises
What properties do the gas laws help us predict?
What makes the ideal gas law different from the other gas laws?
Answers
Gas laws relate four properties: pressure, volume, temperature, and number of moles.
The ideal gas law does not require that the properties of a gas change.
Key Takeaway
The physical properties of gases are predictable using mathematical formulas known as gas laws.
C is the solubility of a gas at a fixed temperature in a particular solvent (in units of M or mL gas/L)
k is Henry's law constant (often in units of M/atm)
P gas is the partial pressure of the gas (often in units of Atm)
Exercises
What conditions of a gas sample should remain constant for Boyle’s law to be used?
What conditions of a gas sample should remain constant for Charles’s law to be used?
Does the identity of a gas matter when using Boyle’s law? Why or why not?
Does the identity of a gas matter when using Charles’s law? Why or why not?
A sample of nitrogen gas is confined to a balloon that has a volume of 1.88 L and a pressure of 1.334 atm. What will be the volume of the balloon if the pressure is changed to 0.662 atm? Assume that the temperature and the amount of the gas remain constant.
A sample of helium gas in a piston has a volume of 86.4 mL under a pressure of 447 torr. What will be the volume of the helium if the pressure on the piston is increased to 1,240 torr? Assume that the temperature and the amount of the gas remain constant.
If a gas has an initial pressure of 24,650 Pa and an initial volume of 376 mL, what is the final volume if the pressure of the gas is changed to 775 torr? Assume that the amount and the temperature of the gas remain constant.
A gas sample has an initial volume of 0.9550 L and an initial pressure of 564.5 torr. What would the final pressure of the gas be if the volume is changed to 587.0 mL? Assume that the amount and the temperature of the gas remain constant.
A person draws a normal breath of about 1.00 L. If the initial temperature of the air is 18°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and amount of the gas remain constant.
A person draws a normal breath of about 1.00 L. If the initial temperature of the air is −10°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and the amount of the gas remain constant.
An air/gas vapor mix in an automobile cylinder has an initial temperature of 450 K and a volume of 12.7 cm 3. The gas mix is heated to 565°C. If pressure and amount are held constant, what is the final volume of the gas in cubic centimeters?
Given the following conditions for a gas: V i = 0.665 L, T i = 23.6°C, V f = 1.034 L. What is T f in degrees Celsius and kelvins?
Assuming the amount remains the same, what must be the final volume of a gas that has an initial volume of 387 mL, an initial pressure of 456 torr, an initial temperature of 65.0°C, a final pressure of 1.00 atm, and a final temperature of 300 K?
When the nozzle of a spray can is depressed, 0.15 mL of gas expands to 0.44 mL, and its pressure drops from 788 torr to 1.00 atm. If the initial temperature of the gas is 22.0°C, what is the final temperature of the gas?
Use the ideal gas law to show that 1 mol of a gas at STP has a volume of about 22.4 L.
Use a standard conversion factor to determine a value of the ideal gas law constant R that has units of L•torr/mol•K.
How many moles of gas are there in a 27.6 L sample at 298 K and a pressure of 1.44 atm?
How many moles of gas are there in a 0.066 L sample at 298 K and a pressure of 0.154 atm?
A 0.334 mol sample of carbon dioxide gas is confined to a volume of 20.0 L and has a pressure of 0.555 atm. What is the temperature of the carbon dioxide in kelvins and degrees Celsius?
What must V be for a gas sample if n = 4.55 mol, P = 7.32 atm, and T = 285 K?
What is the pressure of 0.0456 mol of Ne gas contained in a 7.50 L volume at 29°C?
What is the pressure of 1.00 mol of Ar gas that has a volume of 843.0 mL and a temperature of −86.0°C?
A mixture of the gases N 2, O 2, and Ar has a total pressure of 760 mm Hg. If the partial pressure of N 2 is 220 mm Hg and of O 2 is 470 mm Hg, What is the partial pressure of Ar?
What percent of the gas above is Ar?
Apply Henry’s Law to the diagram below to explain:
why oxygen diffuses from the alveoli of the lungs into the blood and from the blood into the tissues of the body. why carbon dioxide diffuses from the tissues into the blood and from the blood into the alveoli and then finally out into the atmosphere.
Answers
temperature and amount of the gas
pressure and amount of the gas
The identity does not matter because the variables of Boyle’s law do not identify the gas.
The identity does not matter because the variables of Charles law do not identify the gas.
3.79 L
31.1 mL
92.1 mL
918.4 torr
1.07 L
1.18 L
23.7 cm 3
461 K; 188 0 C
206 mL
835 K; 562 0 C
The ideal gas law confirms that 22.4 L equals 1 mol.
760 torr 1 atm
1.63 mol
4.2 x 10-4 mol
405 K; 132°C
14.5 L
0.151 atm
18.2 atm
70 mm Hg
9.2%
Gases diffuse from high concentration to low concentration (Henry's Law). The partial pressure of oxygen is high in the alveoliand low in the blood of the pulmonary capillaries. As a result, oxygen diffuses across the respiratory membrane from the alveoli into the blood. It's also higher partial pressure in the blood than in the tissues, hence it transfers to the tissues. On the other hand, carbon dioxide diffuses from the tissues (highest CO 2 partial pressure) and across the respiratory membrane from the blood into the alveoliand outto the atmosphere.
Attribution and citations
Gas Laws. (2020, November 4). Retrieved May 21, 2021, from
Wikipedia contributors. (2021, May 6). Avogadro's law. In Wikipedia, The Free Encyclopedia. Retrieved 18:11, May 21, 2021, from
2.3: Gas Laws is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
Back to top
2.2: Kinetic-Molecular Theory
2.4: Diffusion and Effusion- Graham's Law
Was this article helpful?
Yes
No
Recommended articles
8.5: Gas LawsThe physical properties of gases are predictable using mathematical formulas known as gas laws.
8.5: Gas LawsThe physical properties of gases are predictable using mathematical formulas known as gas laws.
13.5: Gas LawsThe physical properties of gases are predictable using mathematical formulas known as gas laws.
2.1: States of MatterAnother way that we can describe the properties of matter is the state (also called phase). The amount of energy in molecules of matter determines the...
2.2: Kinetic-Molecular TheoryThe kinetic-molecular theory is a theory that explains the states of matter and is based on the idea that matter is composed of tiny particles that ar...
Article typeSection or PageLicenseCC BY-NC-SALicense Version4.0Show Page TOCno on page
Tags
source-chem-16103
source-chem-16103
© Copyright 2025 Chemistry LibreTexts
Powered by CXone Expert ®
?
The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org.
Support Center
How can we help?
Contact Support Search the Insight Knowledge Base Check System Status× |
9965 | https://askfilo.com/user-question-answers-smart-solutions/using-the-factor-method-x-square-x-12-0-3136393630353830 | Question asked by Filo student
Using the factor method x square - X + 12 = 0
Views: 5,205 students
Updated on: Dec 28, 2024
Text SolutionText solutionverified iconVerified
Concepts:
Quadratic equations, Factoring
Explanation:
To solve the quadratic equation x2−x+12=0 using the factor method, we need to find two numbers that multiply to the constant term (12) and add up to the coefficient of the linear term (-1).
Step by Step Solution:
Step 1
Rewrite the quadratic equation: x2−x+12=0.
Step 2
Identify the coefficients: a=1, b=−1, and c=12.
Step 3
Find two numbers that multiply to ac=1⋅12=12 and add up to b=−1. However, there are no such pairs of integers.
Step 4
Since the quadratic equation cannot be factored using integers, use the quadratic formula: x=2a−b±b2−4ac.
Final Answer:
The quadratic equation x2−x+12=0 cannot be factored using integers. Use the quadratic formula to find the roots.
Students who ask this question also asked
Views: 5,490
Topic:
Smart Solutions
View solution
Views: 5,325
Topic:
Smart Solutions
View solution
Views: 5,194
Topic:
Smart Solutions
View solution
Views: 5,951
Topic:
Smart Solutions
View solution
Stuck on the question or explanation?
Connect with our tutors online and get step by step solution of this question.
| | |
--- |
| Question Text | Using the factor method x square - X + 12 = 0 |
| Updated On | Dec 28, 2024 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 10 |
| Answer Type | Text solution:1 |
Are you ready to take control of your learning?
Download Filo and start learning with your favorite tutors right away!
Questions from top courses
Explore Tutors by Cities
Blog
Knowledge
© Copyright Filo EdTech INC. 2025 |
9966 | https://www.chegg.com/homework-help/questions-and-answers/arccos-x-0-y-2-use-figure-write-expression-algebraic-form-given-y-cot-y-q25658921 | Solved arccos(x), where 0 < y < π/2. Use the figure to write | Chegg.com
Skip to main content
Books
Rent/Buy
Read
Return
Sell
Study
Tasks
Homework help
Understand a topic
Writing & citations
Tools
Expert Q&A
Math Solver
Citations
Plagiarism checker
Grammar checker
Expert proofreading
Career
For educators
Help
Sign in
Paste
Copy
Cut
Options
Upload Image
Math Mode
÷
≤
≥
o
π
∞
∩
∪
√
∫
Math
Math
Geometry
Physics
Greek Alphabet
Math
Calculus
Calculus questions and answers
arccos(x), where 0 < y < π/2. Use the figure to write the expression in algebraic form given y cot(y) =
Your solution’s ready to go!
Our expert help has broken down your problem into an easy-to-learn solution you can count on.
See Answer See Answer See Answer done loading
Question: arccos(x), where 0 < y < π/2. Use the figure to write the expression in algebraic form given y cot(y) =
Please help solve the following problem Show transcribed image text
There are 3 steps to solve this one.Solution 100%(5 ratings) Share Share Share done loading Copy link Step 1 Solution:
Given:
y=arccos(x)or y=cos−1(x) then find cot(y)=? View the full answer Step 2 UnlockStep 3 UnlockAnswer Unlock Previous questionNext question
Transcribed image text:
arccos(x), where 0 < y < π/2. Use the figure to write the expression in algebraic form given y cot(y) =
Not the question you’re looking for?
Post any question and get expert help quickly.
Start learning
Chegg Products & Services
Chegg Study Help
Citation Generator
Grammar Checker
Math Solver
Mobile Apps
Plagiarism Checker
Chegg Perks
Company
Company
About Chegg
Chegg For Good
Advertise with us
Investor Relations
Jobs
Join Our Affiliate Program
Media Center
Chegg Network
Chegg Network
Busuu
Citation Machine
EasyBib
Mathway
Customer Service
Customer Service
Give Us Feedback
Customer Service
Manage Subscription
Educators
Educators
Academic Integrity
Honor Shield
Institute of Digital Learning
© 2003-2025 Chegg Inc. All rights reserved.
Cookie NoticeYour Privacy ChoicesDo Not Sell My Personal InformationGeneral PoliciesPrivacy PolicyHonor CodeIP Rights
Do Not Sell My Personal Information
When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Sale of Personal Data
[x] Sale of Personal Data
Under the California Consumer Privacy Act, you have the right to opt-out of the sale of your personal information to third parties. These cookies collect information for analytics and to personalize your experience with targeted ads. You may exercise your right to opt out of the sale of personal information by using this toggle switch. If you opt out we will not be able to offer you personalised ads and will not hand over your personal information to any third parties. Additionally, you may contact our legal department for further clarification about your rights as a California consumer by using this Exercise My Rights link.
If you have enabled privacy controls on your browser (such as a plugin), we have to take that as a valid request to opt-out. Therefore we would not be able to track your activity through the web. This may affect our ability to personalize ads according to your preferences.
Targeting Cookies
[x] Switch Label label
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices
mmmmmmmmmmlli mmmmmmmmmmlli mmmmmmmmmmlli |
9967 | https://online.stat.psu.edu/stat415/lesson/10/10.2 | Skip to main content
Keyboard Shortcuts
Help
: F1 or
?
Previous Page
: ← +
CTRL (Windows)
: ← +
⌘ (Mac)
Next Page
: → +
CTRL (Windows)
: → +
⌘ (Mac)
Search Site
: CTRL +
SHIFT +
F (Windows)
: ⌘ +
⇧ +
F (Mac)
Close Message
: ESC
10.2 - T-Test: When Population Variance is Unknown
Now that, for purely pedagogical reasons, we have the unrealistic situation (of a known population variance) behind us, let's turn our attention to the realistic situation in which both the population mean and population variance are unknown.
Example 10-2 Section
It is assumed that the mean systolic blood pressure is = 120 mm Hg. In the Honolulu Heart Study, a sample of people had an average systolic blood pressure of 130.1 mm Hg with a standard deviation of 21.21 mm Hg. Is the group significantly different (with respect to systolic blood pressure!) from the regular population?
Answer
The null hypothesis is , and because there is no specific direction implied, the alternative hypothesis is . In general, we know that if the data are normally distributed, then:
follows a -distribution with degrees of freedom. Therefore, it seems reasonable to use the test statistic:
for testing the null hypothesis against any of the possible alternative hypotheses , , and . For the example in hand, the value of the test statistic is:
The critical region approach tells us to reject the null hypothesis at the level if or if . Therefore, we reject the null hypothesis because , and therefore falls in the rejection region:
Again, as always, we draw the same conclusion by using the -value approach. The -value approach tells us to reject the null hypothesis at the level if the -value . In this case, the -value is :
As expected, we reject the null hypothesis because -value .
Again, we'll learn how to ask Minitab to conduct the t-test for a mean in a bit, but this is what the Minitab output for this example looks like:
Test of mu = 120 vs not = 120
| N | Mean | StDev | SE Mean | 95% CI | T | P |
| 100 | 130.100 | 21.210 | 2.121 | (125.891, 134.309) | 4.76 | 0.000 |
By the way, the decision to reject the null hypothesis is consistent with the one you would make using a 95% confidence interval. Using the data, a 95% confidence interval for the mean is:
which simplifies to . That is, we can be 95% confident that the mean systolic blood pressure of the Honolulu population is between 125.89 and 134.31 mm Hg. How can a population living in a climate with consistently sunny 80 degree days have elevated blood pressure?!
Anyway, the critical region approach for the hypothesis test tells us to reject the null hypothesis that :
if or if
which is equivalent to rejecting:
if or if
which is equivalent to rejecting:
if or if
which, upon inserting the data for this particular example, is equivalent to rejecting:
if or if
which just happen to be (!) the endpoints of the 95% confidence interval for the mean. Indeed, the results are consistent! |
9968 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5198232/ | Progression of Barrett’s esophagus toward esophageal adenocarcinoma: an overview - PMC
Skip to main content
An official website of the United States government
Here's how you know
Here's how you know
Official websites use .gov
A .gov website belongs to an official government organization in the United States.
Secure .gov websites use HTTPS
A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites.
Search
Log in
Dashboard
Publications
Account settings
Log out
Search… Search NCBI
Primary site navigation
Search
Logged in as:
Dashboard
Publications
Account settings
Log in
Search PMC Full-Text Archive
Search in PMC
Journal List
User Guide
View on publisher site
Download PDF
Add to Collections
Cite
Permalink PERMALINK
Copy
As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health.
Learn more: PMC Disclaimer | PMC Copyright Notice
Ann Gastroenterol
. 2016 Sep 30;30(1):1–6. doi: 10.20524/aog.2016.0091
Search in PMC
Search in PubMed
View in NLM Catalog
Add to search
Progression of Barrett’s esophagus toward esophageal adenocarcinoma: an overview
Nele Schoofs
Nele Schoofs
1 Department of Gastroenterology, University Hospitals Leuven and Department of Oncology, KU Leuven, Belgium
Find articles by Nele Schoofs
1, Raf Bisschops
Raf Bisschops
1 Department of Gastroenterology, University Hospitals Leuven and Department of Oncology, KU Leuven, Belgium
Find articles by Raf Bisschops
1, Hans Prenen
Hans Prenen
1 Department of Gastroenterology, University Hospitals Leuven and Department of Oncology, KU Leuven, Belgium
Find articles by Hans Prenen
1,✉
Author information
Article notes
Copyright and License information
1 Department of Gastroenterology, University Hospitals Leuven and Department of Oncology, KU Leuven, Belgium
✉
Correspondence to: Hans Prenen, MD, PhD, University Hospitals Leuven, Department of Gastroenterology, Digestive Oncology Unit, Herestraat 49, B3000 Leuven, Belgium, Tel.: +32 16 34 42 18, Fax: +32 16 34 44 19, e-mail: hans.prenen@uzleuven.be
Received 2016 Jun 10; Accepted 2016 Sep 12; Issue date 2017.
Copyright: © Hellenic Society of Gastroenterology
This is an open-access article distributed under the terms of the Creative Commons Attribution-Noncommercial-Share Alike 3.0 Unported, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC5198232 PMID: 28042232
Abstract
In Barrett’s esophagus, normal squamous epithelium is replaced by a metaplastic columnar epithelium as a consequence of chronic gastroesophageal reflux disease. There is a strong association with esophageal adenocarcinoma. In view of the increasing incidence of esophageal adenocarcinoma in the western world, it is important that more attention be paid to the progression of Barrett’s esophagus toward esophageal adenocarcinoma. Recently, several molecular factors have been identified that contribute to the sequence towards adenocarcinoma. This might help identify patients at risk and detect new targets for the prevention and treatment of esophageal adenocarcinoma in the future.
Keywords: Barrett’s esophagus, esophageal adenocarcinoma, biomarkers
Introduction
In view of the increasing incidence of esophageal adenocarcinoma (EAC) in the Western world, it is important to better understand the process of neoplastic progression of Barrett’s esophagus (BE) toward EAC. In this review we will focus on the known risk factors for this progression, as well as the molecular pathways involved. We searched PubMed for articles published in English from 2000 onwards and used the search terms “esophageal cancer”, “Barrett’s esophagus”, “etiology”, “pathology”, “molecular pathogenesis”, “genetics”, “pathophysiology”, “diagnosis”, “epidemiology”, and “chemoprevention”.
Definition of BE
BE is most commonly seen as the condition in which a metaplastic columnar epithelium replaces the stratified squamous epithelium that normally lines the distal esophagus [1,2]. The metaplastic epithelium is acquired as a consequence of chronic gastroesophageal reflux disease (GERD), and is a predisposing factor for the development of adenocarcinoma of the esophagus.
There are many theories concerning the origin of a BE and no consensus has been reached, as was stated by the authors of a recent review . They concluded that Barrett glands are more complex and possibly unique within the human gastrointestinal epithelium. Barrett epithelium shows multilinear gastric and intestinal differentiation, as documented by the findings of gene expression arrays of both gastric and intestinal epithelium and the different types (complete versus incomplete) of intestinal metaplasia. This could lead to different hypotheses concerning the development of BE and the possible progression to EAC.
Epidemiology
It is difficult to determine the epidemiology of BE because there are many affected individuals who are asymptomatic and remain undiagnosed. Most prevalence data have been derived from BE diagnoses made during esophagogastroduodenoscopy performed to investigate symptoms of dyspepsia. BE is more common in developed countries, affecting 2% of the general adult population . This is most likely to be attributable to the higher incidence of GERD in this population, since this is a well-known causal factor for the development of BE.
The incidence of BE on diagnostic endoscopy is rising independently of an increase in the number of endoscopies carried out, suggesting a true increase in incidence rather than a higher detection rate . In the literature, several publications have studied the prevalence of BE in unselected populations. Rex et al evaluated a cohort of 961 patients undergoing colonoscopy who were offered an additional endoscopy, and found an overall prevalence of 6.8%, with 5.5% for short-segment BE in persons aged 40 years or older . In another similar colonoscopy-based study of 300 patients over the age of 65 years, the prevalence was 4% and 15% for long- and short-segment BE, respectively . In other population-based studies, the prevalence of BE in the general population ranged between 1.3%, 1.6% and 1.9% [8-10].
The incidence of esophageal cancer is highly variable, depending on the region, with “‘the esophageal cancer belt’” as the highest-risk area for the development of esophageal cancer, stretching from northern Iran through the central Asian republics to north-central China. In this area however, 90% of cases are squamous cell carcinomas. Over the past decades, the frequency of adenocarcinoma of the esophagus has increased dramatically, certainly in western countries (“low-risk areas”). Several hypotheses have been proposed concerning the increasing incidence of BE and consequently of EAC. The first hypothesis is that the lower prevalence of Helicobacter pylori (H. pylori) infection, particularly the cagA+ strain, is associated inversely with BE . A second hypothesis is the increase in the known risk factors of overweight and obesity [12,13].
Diagnosis
A diagnosis of BE is made endoscopically by visualization of a columnar lined epithelium arising circumferentially at least 1 cm above the gastroesophageal junction, with intestinal metaplasia on histological investigation of biopsies. Endoscopic reporting can be difficult and should be done using uniform criteria. The Prague classification consists of criteria to uniformly describe the length of a Barrett’s segment during endoscopy. In this classification, the circumferential extent and maximum length of the Barrett epithelium are described in cm (Fig. 1). This classification was recently validated .
Figure 1.
Open in a new tab
The Prague classification of Barrett’s esophagus. The C-value is used for the circumferential pattern (C) and the M value for the maximum length (M) GEJ, gastroesophageal junction
For a definitive diagnosis of BE, histological diagnosis is necessary. The presence of specialized columnar epithelium, characterized by acid mucin-containing goblet cells, in a biopsy specimen of the esophagus has been accepted as diagnostic of BE . Endoscopic surveillance of BE with systematic biopsies is necessary to exclude the presence of dysplasia.
BE can be divided into short- and long-segment BE. Short-segment BE has a maximal length of less than 3 cm, whereas long-segment has a length of more than 3 cm. Long-segment BE has a higher risk for development of EAC. The length of the BE segment is known to be associated with risk of progression to neoplasia, as discussed below.
Risk for progression to EAC
Incidence rates of EAC and high-grade dysplasia (HGD) among patients with BE are variable. Seven systematic reviews have been published on the cancer risk in patients with BE. The annual incidence of EAC among BE patients varied from 0.3% to 0.6%, and the combined incidence of HGD and EAC from 0.9% to 1.0% [16-22]. These incidence rates are still referred to in recent guidelines. One study, however, revealed a publication bias in the reporting of progression rates in patients with BE . Sikkema et al analyzed data derived from high-quality studies to obtain more accurate data on the risk of EAC in BE patients . The annual EAC incidence rate in BE cohorts with less than 2000 patient years of follow up was 0-3.55%, compared to 0.07-0.82% in cohorts with more than 2000 patient years . In two population-based BE follow-up studies, absolute annual risks of 0.12-0.14% were reported, considerably lower than was assumed until now [23,24].
Considering the known risk for the progression of BE toward EAC, current guidelines advise patients with BE to be enrolled in endoscopic surveillance programs in order to detect HGD or EAC. Endoscopic surveillance consists of extensive biopsy sampling, known as the Seattle biopsy protocol. This protocol consists of four quadrant biopsies (every 1-2 cm) with biopsies of mucosal abnormalities. The surveillance interval is determined by histology results. In the absence of dysplasia, the American College of Gastroenterology recommends surveillance endoscopy at 3-year intervals. For patients with low-grade dysplasia, an annual endoscopy is recommended, and for those with HGD who receive no invasive therapy, endoscopic exams should be performed every three months. In a recent review , the authors discuss the current guidelines and the dilemmas in endoscopic surveillance. Several studies suggest a modest effect of surveillance programs on EAC mortality in patients with BE [25-27]. This outcome is probably influenced by the estimates of annual incidence of EAC among patients with BE, considering the fact that the incidence employed in current surveillance guidelines is still 0.4-0.5%, which is much lower than discussed above. Another dilemma in surveillance strategies for BE is the possibility of sampling error and poor adherence to the Seattle protocol, which has been reported to be as low as 30% . To improve the success of endoscopic surveillance, advanced endoscopic techniques, such as chromoendoscopy, narrow-band imaging (NBI) and autofluorescence endoscopy, have attracted major interest during the past decade. In 2012, a prospective, randomized, controlled trial compared the standard endoscopic surveillance strategy with NBI-targeted biopsies, with similar results in diagnosing intestinal metaplasia and even detecting more areas with dysplasia . When using NBI-targeted strategies, fewer biopsies are necessary, which could increase cost-effectiveness. Some authors conclude that this technique should, therefore, be considered in patients with longer segments of BE, considering the poorer adherence in these cases . This is not supported in recent British Society of Gastroenterology (BSG) guidelines, considering the conflicting results in the literature and the lack of hard evidence. According to the BSG guidelines, however, advanced imaging modalities such as chromoendoscopy or “‘virtual chromoendoscopy”‘ are not superior to standard white light endoscopy in BE surveillance and are therefore not recommended for routine use . There is also a bias concerning the incidence rate of EAC in these studies, as stated previously.
The grading of dysplasia in these guidelines is also a concern in view of the poor inter- and intra-observer histological agreement as to the diagnosis of Barrett’s dysplasia , as an inaccurate grading of dysplasia can impact the frequency of endoscopic surveillance.
Ross-Innes et al (BEST2 Study Group) described the use of Cytosponge as a minimally invasive technique for screening of BE . They used Trefoil factor 3 (TFF3) as a marker of intestinal metaplasia in the Cytosponge samples, as it is the subtype most strongly associated with a risk of progression. In a study with 463 controls and 467 BE patients, the Cytosponge test appeared to be a safe and well-tolerated BE screening method that can be carried out in a primary care setting. The prevalence of BE of 3.0% reported in this study is comparable with other studies. The sensitivity of the combination test (Cytosponge + TFF3) is approximately 80% for diagnosing BE; this increases with BE segment length and is not compromised in the presence of dysplasia. The specificity of the test was 92%. In a microsimulation model, both endoscopy and Cytosponge were compared, with no screening, in 50-year-old men with symptoms of GERD . This already showed cost-effectiveness and reduced mortality from EAC for the Cytosponge test compared with no screening.
Molecular pathways
The progression from BE to EAC was first documented in the 1970’s, providing targets for the screening, monitoring, and management of early-stage neoplasia . Currently, there are several risk factors described as possible predictors of progression. The Prasad’s group published a review in 2010, summarizing the current evidence on risk factors for progression . An attempt was made to create a potential risk stratification to optimize the management of patients with BE and make it more cost-effective. A progression score for BE could contain a group of clinical factors and a biomarker panel. In the recent literature, there are certain factors that are suggested as being important risk factors. The clinical panel consists of age, (male) sex, and length of Barrett segment, with increasing length giving an increasing risk but without an evident cut off length. Biomarkers included are aneuploidy/polysomy, p53 loss of heterozygosity (LOH) and p16 LOH. This possible score, however, certainly needs validation in a prospective study of a large cohort of patients before it can be used in clinical practice. If validated, this score could be important for determining which patients are at higher risk of developing an adenocarcinoma and would therefore benefit the most from inclusion in a program of intensive surveillance. A familial aggregation of BE has also been described .
Obesity is implicated as a risk factor for EAC and BE, independently of GERD . Thrift et al described a possible role for high levels of circulating pro-inflammatory cytokines and leptin in the mechanism towards progression . In another analysis, an inverse association between high molecular weight adiponectin levels and risk of progression to EAC is reported .
A possible sequence for progression is described in a hypothesis by Chandrasoma et al , derived from the idea of BE as a complex, multilinear epithelium with different types of metaplasia (cf. definition). The authors suggest an evolution from the esophageal squamous epithelium to cardiac-type glands and further into intestinal metaplastic glands, which can progress to neoplasia. A concept of Barrett glands evolving from metaplasia of the stem cells of the proximal columnar gastric or cardiac epithelium has been stated .
In the recent literature, much attention has been given to understanding the molecular pathways leading to the progression from BE to adenocarcinoma of the esophagus, in order to specify possible therapeutic targets in the prevention and treatment of this type of (early) cancer (Fig. 2). Several embryological signaling pathways are described in the development and malignant transformation of BE . In embryology, the esophagus is derived from the foregut, whose lumen divides along the sagittal axis into the trachea, with columnar epithelium, and the esophagus, with squamous epithelium. Studies of transgenic mouse models have identified four main signaling pathways active in the differentiation of the embryological foregut: the bone morphogenetic protein (BMP), hedgehog (Hh), wingless-type MMTV integration site family (WNT), and retinoic acid (RA) signaling pathways . The three key transcription factors expressed by these pathways for the regulation of differentiation of foregut epithelium toward a squamous or columnar type are NKX2.1, SOX2 and p63. SOX2 and p63 induce squamous differentiation, while NKX2.1 expression is required for columnar differentiation of the foregut epithelium. These pathways and transcription factors not only play an important role in foregut embryology, but may also be important in the development of BE and its progression toward EAC. In normal squamous epithelium, the BMP pathway is not activated, but in the case of inflammation (such as that caused by GERD) the BMP pathway is activated with stromal BMP4 expression, which contributes to a columnar transdifferentiation of squamous esophageal epithelium. The role of this BMP pathway in the progression toward adenocarcinoma requires further research. The Hh pathway is also involved in the development of BE by stimulating the BMP pathway, but it can also act by inducing a transcription factor, epithelial SOX9 expression. Its role in malignant progression is less clear. In contrast, WNT signaling is not involved in the development of BE, but is an important factor in its progression toward EAC. This is indicated by the progressive increase in WNT signaling in the metaplasia–dysplasia–carcinoma sequence. In the development of BE, activity of the RA signaling pathway is increased. In contrast, activation of this pathway reduces the development of EAC, which could be seen as a possible protective effect in progression.
Figure 2.
Open in a new tab
Molecular pathways leading to the progression of Barrett’s esophagus to adenocarcinoma. Blue arrows indicate activation, red arrows therapeutic inhibition, BMP, bone morphogenetic protein; GERD, gastro-esophageal reflux disease; RA, retinoic acid signaling pathway; PPI, proton pump inhibitor; HH, hedgehog signaling pathway; WNT, wingless-type MMTV integration site family; UDCA, ursodeoxycholic acid; NSAIDs, non-steroid anti-inflammatory drugs
Recently, two papers were published in Nature Genetics regarding the genetic changes involved in the progression from BE to EAC. In both studies, the authors used techniques such as whole-exome and whole-genome sequencing to provide more insight into the transition towards cancer. Ross-Innes, et al conducted whole-genome sequencing of paired EAC and BE samples with limited overlap between mutations . The mutational signatures, however, were similar in the different lesions. This could be indicative of a comparable mutagenic process. Another whole-exome sequencing study found similar mutational signatures for EAC and BE . Analysis of mutations in TP53 and cyclin-dependent kinase inhibitor 2A showed less prevalent oncogene activation events. They concluded that this probably occurred later in tumor progression. A model is presented whereby EAC can progress via two separate pathways: TP53 mutation followed by genome doubling versus progressive loss of tumor suppressors in those that do not undergo genome doubling.
Chemoprevention
To improve the survival of patients with EAC, the best strategy remains early diagnosis; however, the cancer often spreads before symptoms occur. BE is a known precursor to EAC, yet an ideal management strategy remains elusive and the utility of current endoscopic surveillance is controversial and costly. For these reasons, the use of chemoprevention strategies has gained considerable interest in recent studies, which aim to prevent the progression of BE towards EAC with pharmacological strategies.
Possible therapeutic targets according to the different molecular pathways involved in the progression of BE toward EAC can be summarized as downregulation of the Hh and WNT signaling pathways, and upregulation of the RA and (probably) the BMP signaling pathways. In the Hh signaling pathway, a possible target is the suppression of zinc-finger transcription factors GLI using a combination of ursodeoxycholic acid and aspirin. This significantly decreased the incidence of EAC in a rat model . A second possibility is the use of WNT antagonists, such as non-steroid anti-inflammatory drugs (NSAIDs), to decrease the WNT signaling pathway and thereby reduce the risk of BE dysplasia and adenocarcinoma. The RA signaling pathway is a third potential therapeutic strategy; however, a clear target has not yet been described.
Several studies have reported a possible reduction in the risk of EAC among patients with BE with the use of NSAIDs, low-dose aspirin, statins and proton pump inhibitors (PPIs) [46-54]. These were all based on small, selected groups of patients with EAC. A meta-analysis, however, showed a significantly lower risk of esophageal cancer in patients who frequently used NSAIDs or aspirin . This was confirmed for the risk of both EAC and HGD in another meta-analysis . The effect of PPIs in the malignant transformation is still a matter for debate. PPIs are known as a treatment for symptom relief in BE, which suggests a decrease in the risk of progression. Nevertheless, certain studies show an increase in the risk of progression [48,56]. This could be explained by the use of PPIs in the treatment of GERD, a possible risk factor for EAC. According to recent studies, statins may also have potential for chemoprevention. In 2013, a systematic review and meta-analysis showed a reduction in the risk of EAC among patients with BE who were taking statins . In 2014, Choi et al evaluated the effectiveness and cost-effectiveness of aspirin, statin, and combination chemoprevention for BE management . They suggested that, among the four treatment strategies analyzed (endoscopic surveillance alone, aspirin chemoprevention, statin chemoprevention and the combination of aspirin and statin chemoprevention), aspirin therapy is the most cost-effective chemoprevention strategy for patients with BE. A combination of aspirin and statins could potentially be cost-effective in those patients with BE who have a higher risk of progression to EAC. A recent matched case-control study evaluated the risk of EAC among patients with BE associated with the use of NSAIDs, low-dose aspirin, statins and PPIs . In this study, a non-significant reduction in the risk of HGD and EAC was only present when PPIs were used at the highest dose in patients with BE. The use of NSAIDs or low-dose aspirin was also not associated with a decrease in the risk of EAC. For statin use, a non-significant dose-duration response was seen. No statistically significant argument was found to indicate chemoprevention in daily practice for patients with BE.
To further investigate the possibility for chemoprevention, a large study centered in the United Kingdom, the Aspirin and Esomeprazole Chemoprevention in Barrett’s Metaplasia (AspECT) study ( is evaluating the impact of low- and high-dose aspirin on BE progression rates to cancer. However, no similar trial that evaluates the impact of statins and the combination of a statin and aspirin is currently ongoing.
Concluding remarks
In view of the increasing incidence of EAC, it is important that more attention be paid to the prevention of this premalignant condition. However, the rate of progression towards adenocarcinoma is very small in comparison to what was thought previously. For this reason, it is important that risk factors for progression are identified to identify patients at risk.
Recently, a molecular revolution has occurred, with the identification of several possible factors contributing to the sequence towards adenocarcinoma. On this basis it may be possible to identify new targets for the treatment and prevention of adenocarcinoma.
In the future, a better understanding of the evolutionary dynamics of Barrett’s clones is crucial for understanding the process of neoplastic progression. This will have important implications for the clinical management of the disease, as currently no good markers exist for the prediction of neoplastic progression. The ultimate goal should be to accurately identify on the one hand the low-risk Barrett cases and on the other hand the ones that are “born to be bad”. Surveillance programs or chemoprevention can be applied more specifically to these groups of patients, but more research and improvement of current surveillance programs are still necessary.
Acknowledgment
Editorial support was provided by Mark English.
Biography
University Hospitals Leuven, Leuven, Belgium
Footnotes
Conflict of Interest: None
References
1.Spechler SJ. Barrett’s esophagus. Semin Gastrointest Dis. 1996;7:51–60. [PubMed] [Google Scholar]
2.Spechler SJ, Fitzgerald RC, Prasad GA, Wang KK. History, molecular mechanisms, and endoscopic treatment of Barrett’s esophagus. Gastroenterology. 2010;138:854–869. doi: 10.1053/j.gastro.2010.01.002. [DOI] [PMC free article] [PubMed] [Google Scholar]
3.Naini BV, Souza RF, Otze RD. Barrett’s esophagus: a comprehensive and contemporary review for pathologists. Am J Surg Pathol. 2016;40:e45–66. doi: 10.1097/PAS.0000000000000598. [DOI] [PMC free article] [PubMed] [Google Scholar]
4.Bennett C, Moayyedi P, Corley DA, et al. BOB CAT Consortium. BOB CAT: A large-scale review and Delphi consensus for management of Barrett’s Esophagus with no dysplasia, indefinite for, or low-grade dysplasia. Am J Gastroenterol. 2015;110:662–682. doi: 10.1038/ajg.2015.55. [DOI] [PMC free article] [PubMed] [Google Scholar]
5.Jankowski J, Barr H. Improving surveillance for Barrett’s oesophagus: AspECT and BOSS trials provide an evidence base. BMJ. 2006;332:1512. doi: 10.1136/bmj.332.7556.1512. [DOI] [PMC free article] [PubMed] [Google Scholar]
6.Rex DK, Cummings OW, Shaw M, et al. Screening for Barrett’s esophagus in colonoscopy patients with and without heartburn. Gastroenterology. 2003;125:1670–1677. doi: 10.1053/j.gastro.2003.09.030. [DOI] [PubMed] [Google Scholar]
7.Ward EM, Wolfsen HC, Achem SR, et al. Barrett’s esophagus is common in older men and women undergoing screening colonoscopy regardless of reflux symptoms. Am J Gastroenterol. 2006;101:12–17. doi: 10.1111/j.1572-0241.2006.00379.x. [DOI] [PubMed] [Google Scholar]
8.Ronkainen J, Aro P, Storskrubb T, et al. Prevalence of Barrett’s esophagus in the general population: an endoscopic study. Gastroenterology. 2005;129:1825–1831. doi: 10.1053/j.gastro.2005.08.053. [DOI] [PubMed] [Google Scholar]
9.Zagari RM, Fuccio L, Wallander MA, et al. Gastro-oesophageal reflux symptoms, oesophagitis and Barrett’s oesophagus in the general population: the Loiano-Monghidoro study. Gut. 2008;57:1354–1359. doi: 10.1136/gut.2007.145177. [DOI] [PubMed] [Google Scholar]
10.Zou D, He J, Ma X, et al. Epidemiology of symptom-defined gastroesophageal reflux disease and reflux esophagitis: the systematic investigation of gastrointestinal diseases in China (SILC) Scand J Gastroenterol. 2011;46:133–141. doi: 10.3109/00365521.2010.521888. [DOI] [PubMed] [Google Scholar]
11.Rubenstein JH, Inadomi JM, Scheiman J, et al. Association between Helicobacter pylori and Barrett’s esophagus, erosive esophagitis, and gastroesophageal reflux symptoms. Clin Gastroenterol Hepatol. 2014;12:239–245. doi: 10.1016/j.cgh.2013.08.029. [DOI] [PMC free article] [PubMed] [Google Scholar]
12.de Jonge PJ, van Blankenstein M, Grady WM, Kuipers EJ. Barrett’s oesophagus:epidemiology, cancer risk and implications for management. Gut. 2014;63:191–202. doi: 10.1136/gutjnl-2013-305490. [DOI] [PMC free article] [PubMed] [Google Scholar]
13.Cook MB, Chow WH, Devesa SS. Oesophageal cancer incidence in the United States by race, sex, and histologic type 1977-2005. Br J Cancer. 2009;101:855–859. doi: 10.1038/sj.bjc.6605246. [DOI] [PMC free article] [PubMed] [Google Scholar]
14.Alvarez Herrero L, Curvers WL, van Vilsteren FG, et al. Validation of the Prague C&M classification of Barrett’s esophagus in clinical practice. Endoscopy. 2013;45:876–882. doi: 10.1055/s-0033-1344952. [DOI] [PubMed] [Google Scholar]
15.Salemme M, Villanacci V, Cengia G, Cestari R, Missale G, Bassotti G. Intestinal metaplasia in Barrett’s oesophagus: An essential factor to predict the risk of dysplasia and cancer development. Dig Liver Dis. 2016;48:144–147. doi: 10.1016/j.dld.2015.10.021. [DOI] [PubMed] [Google Scholar]
16.Shaheen NJ, Crosby MA, Bozymski EM, Sandler RS. Is there publication bias in the reporting of cancer risk in Barrett’s esophagus?Gastroenterology. 2000;119:333–338. doi: 10.1053/gast.2000.9302. [DOI] [PubMed] [Google Scholar]
17.Chang EY, Morris CD, Seltman AK, et al. The effect of antireflux surgery on esophageal carcinogenesis in patients with Barrett esophagus: a systematic review. Ann Surg. 2007;246:11–21. doi: 10.1097/01.sla.0000261459.10565.e9. [DOI] [PMC free article] [PubMed] [Google Scholar]
18.Thomas T, Abrams KR, De Caestecker JS, Robinson RJ. Meta analysis: Cancer risk in Barrett’s oesophagus. Aliment Pharmacol Ther. 2007;26:1465–1477. doi: 10.1111/j.1365-2036.2007.03528.x. [DOI] [PubMed] [Google Scholar]
19.Yousef F, Cardwell C, Cantwell MM, Galway K, Johnston BT, Murray L. The incidence of esophageal cancer and high-grade dysplasia in Barrett’s esophagus: a systematic review and meta-analysis. Am J Epidemiol. 2008;168:237–249. doi: 10.1093/aje/kwn121. [DOI] [PubMed] [Google Scholar]
20.Sikkema M, de Jonge PJ, Steyerberg EW, Kuipers EJ. Risk of esophageal adenocarcinoma and mortality in patients with Barrett’s esophagus: a systematic review and meta-analysis. Clin Gastroenterol Hepatol. 2010;8:235–244. doi: 10.1016/j.cgh.2009.10.010. [DOI] [PubMed] [Google Scholar]
21.Wani S, Puli SR, Shaheen NJ, et al. Esophageal adenocarcinoma in Barrett’s esophagus after endoscopic ablative therapy: a meta-analysis and systematic review. Am J Gastroenterol. 2009;104:502–513. doi: 10.1038/ajg.2008.31. [DOI] [PubMed] [Google Scholar]
22.Desai TK, Krishnan K, Samala N, et al. The incidence of oesophageal adenocarcinoma in non-dysplastic Barrett’s oesophagus: a meta-analysis. Gut. 2012;61:970–976. doi: 10.1136/gutjnl-2011-300730. [DOI] [PubMed] [Google Scholar]
23.de Jonge PJ, van Blankenstein M, Looman CW, Casparie MK, Meijer GA, Kuipers EJ. Risk of malignant progression in patients with Barrett’s oesophagus: a Dutch nationwide cohort study. Gut. 2010;59:1030–1036. doi: 10.1136/gut.2009.176701. [DOI] [PubMed] [Google Scholar]
24.Hvid-Jensen F, Pedersen L, Drewes AM, Sørensen HT, Funch-Jensen P. Incidence of adenocarcinoma among patients with Barrett’s esophagus. N Engl J Med. 2011;365:1375–1383. doi: 10.1056/NEJMoa1103042. [DOI] [PubMed] [Google Scholar]
25.van der Burgh A, Dees J, Hop WC, van Blankenstein M. Oesophageal cancer is an uncommon cause of death in patients with Barrett’s oesophagus. Gut. 1996;39:5–8. doi: 10.1136/gut.39.1.5. [DOI] [PMC free article] [PubMed] [Google Scholar]
26.Macdonald CE, Wicks AC, Playford RJ. Final results from 10 year cohort of patients undergoing surveillance for Barrett’s oesophagus: observational study. BMJ. 2000;321:1252–1255. doi: 10.1136/bmj.321.7271.1252. [DOI] [PMC free article] [PubMed] [Google Scholar]
27.Dulai GS, Guha S, Kahn KL, Gornbein J, Weinstein WM. Preoperative prevalence of Barrett’s esophagus in esophageal adenocarcinoma: a systematic review. Gastroenterology. 2002;122:26–33. doi: 10.1053/gast.2002.30297. [DOI] [PubMed] [Google Scholar]
28.Curvers WL, Peters FP, Elzer B, et al. Quality of Barrett’s surveillance in The Netherlands: a standardized review of endoscopy and pathology reports. Eur J Gastroenterol Hepatol. 2008;20:601–607. doi: 10.1097/MEG.0b013e3282f8295d. [DOI] [PubMed] [Google Scholar]
29.Sharma P, Hawes RH, Bansal A, et al. Standard endoscopy with random biopsies versus narrow band imaging targeted biopsies in Barrett’s oesophagus: a prospective, international, randomised controlled trial. Gut. 2013;62:15–21. doi: 10.1136/gutjnl-2011-300962. [DOI] [PubMed] [Google Scholar]
30.Fitzgerald RC, di Pietro M, Ragunath K, et al. British Society of Gastroenterology. British Society of Gastroenterology guidelines on the diagnosis and management of Barrett’s oesophagus. Gut. 2014;63:7–42. doi: 10.1136/gutjnl-2013-305372. [DOI] [PubMed] [Google Scholar]
31.Kerkhof M, van Dekken H, Steyerberg EW, et al. CYBAR study group. Grading of dysplasia in Barrett’s oesophagus: substantial interobserver variation between general and gastrointestinal pathologists. Histopathology. 2007;50:920–927. doi: 10.1111/j.1365-2559.2007.02706.x. [DOI] [PubMed] [Google Scholar]
32.Ross-Innes CS, Debiram-Beecham I, O’Donovan M, et al. Evaluation of a minimally invasive cell sampling device coupled with assessment of trefoil factor 3 expression for diagnosing Barrett’s esophagus: a multi-center case-control study. PLoS Med. 2015;12:e1001780. doi: 10.1371/journal.pmed.1001780. [DOI] [PMC free article] [PubMed] [Google Scholar]
33.Benaglia T, Sharples LD, Fitzgerald RC, Lyratzopoulos G. Health benefits and cost effectiveness of endoscopic and nonendoscopic cytosponge screening for Barrett’s esophagus. Gastroenterology. 2013;144:62–73. doi: 10.1053/j.gastro.2012.09.060. [DOI] [PubMed] [Google Scholar]
34.Chevaux JB, Piessevaux H, Jouret-Mourin A, Yeung R, Danse E, Deprez PH. Clinical outcome in patients treated with endoscopic submucosal dissection for superficial Barrett’s neoplasia. Endoscopy. 2015;47:103–112. doi: 10.1055/s-0034-1390982. [DOI] [PubMed] [Google Scholar]
35.Prasad GA, Bansal A, Sharma P, Wang KK. Predictors of progression in Barrett’s esophagus: current knowledge and future directions. Am J Gastroenterol. 2010;105:1490–1502. doi: 10.1038/ajg.2010.2. [DOI] [PMC free article] [PubMed] [Google Scholar]
36.Verbeek RE, Spittuler LF, Peute A, et al. Familial clustering of Barrett’s esophagus and esophageal adenocarcinoma in a European cohort. Clin Gastroenterol Hepatol. 2014;12:1656–1663. e1. doi: 10.1016/j.cgh.2014.01.028. [DOI] [PubMed] [Google Scholar]
37.Singh S, Sharma AN, Murad MH, et al. Central adiposity is associated with increased risk of esophageal inflammation, metaplasia, and adenocarcinoma: a systematic review and meta-analysis. Clin Gastroenterol Hepatol. 2013;11:1399–1412. doi: 10.1016/j.cgh.2013.05.009. [DOI] [PMC free article] [PubMed] [Google Scholar]
38.Thrift AP, Garcia JM, El-Serag HB. A multibiomarker risk score helps predict risk for Barrett’s esophagus. Clin Gastroenterol Hepatol. 2014;12:1267–1271. doi: 10.1016/j.cgh.2013.12.014. [DOI] [PMC free article] [PubMed] [Google Scholar]
39.Duggan C, Onstad L, Hardikar S, Blount PL, Reid BJ, Vaughan TL. Association between markers of obesity and progression from Barrett’s esophagus to esophageal adenocarcinoma. Clin Gastroenterol Hepatol. 2013;11:934–943. doi: 10.1016/j.cgh.2013.02.017. [DOI] [PMC free article] [PubMed] [Google Scholar]
40.Chandrasoma PT, Der R, Dalton P, et al. Distribution and significance of epithelial types in columnar-lined esophagus. Am J Surg Pathol. 2001;25:1188–1193. doi: 10.1097/00000478-200109000-00010. [DOI] [PubMed] [Google Scholar]
41.McDonald SAC, Lavery D, Wright NA, Jansen M. Barrett esophagus: lessons on its origins from the lesion itself. Nat Rev Gastroenterol Hepatol. 2015;12:50–60. doi: 10.1038/nrgastro.2014.181. [DOI] [PubMed] [Google Scholar]
42.Pavlov K, Meijer C, van den Berg A, Peters FT, Kruyt FA, Kleibeuker JH. Embryological signaling pathways in Barrett’s metaplasia development and malignant transformation;mechanisms and therapeutic opportunities. Crit Rev Oncol Hematol. 2014;92:25–37. doi: 10.1016/j.critrevonc.2014.05.002. [DOI] [PubMed] [Google Scholar]
43.Ross-Innes CS, Becq J, Warren A, et al. Oesophageal Cancer Clinical and Molecular Stratification OCCAMS Study Group. Whole-genome sequencing provides new insights into the clonal architecture of Barrett’s esophagus and esophageal adenocarcinoma. Nat Genet. 2015;47:1038–1046. doi: 10.1038/ng.3357. [DOI] [PMC free article] [PubMed] [Google Scholar]
44.Stachler MD, Taylor-Weiner A, Peng S, et al. Paired exome analysis of Barrett’s esophagus and adenocarcinoma. Nat Genet. 2015;47:1047–1055. doi: 10.1038/ng.3343. [DOI] [PMC free article] [PubMed] [Google Scholar]
45.Rizvi S, Demars CJ, Comba A, et al. Combinatorial chemoprevention reveals a novel smoothened-independent role of GLI1 in esophageal carcinogenesis. Cancer Res. 2010;70:6787–6796. doi: 10.1158/0008-5472.CAN-10-0197. [DOI] [PMC free article] [PubMed] [Google Scholar]
46.Beales IL, Vardi I, Dearman L. Regular statin and aspirin use in patients with Barrett’s oesophagus is associated with a reduced incidence of oesophageal adenocarcinoma. Eur J Gastroenterol Hepatol. 2012;24:917–923. doi: 10.1097/MEG.0b013e3283543f01. [DOI] [PubMed] [Google Scholar]
47.Corley DA, Kerlikowske K, Verma R, Buffler P. Protective association of aspirin/NSAIDs and esophageal cancer: a systematic review and meta-analysis. Gastroenterology. 2003;124:47–56. doi: 10.1053/gast.2003.50008. [DOI] [PubMed] [Google Scholar]
48.Nguyen DM, Richardson P, El-Serag HB. Medications (NSAIDs, statins, proton pump inhibitors) and the risk of esophageal adenocarcinoma in patients with Barrett’s esophagus. Gastroenterology. 2010;138:2260–2266. doi: 10.1053/j.gastro.2010.02.045. [DOI] [PMC free article] [PubMed] [Google Scholar]
49.Kantor ED, Onstad L, Blount PL, Reid BJ, Vaughan TL. Use of statin medications and risk of esophageal adenocarcinoma in persons with Barrett’s esophagus. Cancer Epidemiol Biomarkers Prev. 2012;21:456–461. doi: 10.1158/1055-9965.EPI-11-1014. [DOI] [PMC free article] [PubMed] [Google Scholar]
50.Kastelein F, Spaander MC, Biermann K, Steyerberg EW, Kuipers EJ, Bruno MJ Probar-study Group. Nonsteroidal anti-inflammatory drugs and statins have chemopreventative effects in patients with Barrett’s esophagus. Gastroenterology. 2011;141:2000–2008. doi: 10.1053/j.gastro.2011.08.036. quiz e13-e14. [DOI] [PubMed] [Google Scholar]
51.Abnet CC, Freedman ND, Kamangar F, Leitzmann MF, Hollenbeck AR, Schatzkin A. Non-steroidal anti-inflammatory drugs and risk of gastric and oesophageal adenocarcinomas: results from a cohort study and a meta-analysis. Br J Cancer. 2009;100:551–557. doi: 10.1038/sj.bjc.6604880. [DOI] [PMC free article] [PubMed] [Google Scholar]
52.Rothwell PM, Fowkes FG, Belch JF, Ogawa H, Warlow CP, Meade TW. Effect of daily aspirin on long-term risk of death due to cancer: analysis of individual patient data from randomised trials. Lancet. 2011;377:31–41. doi: 10.1016/S0140-6736(10)62110-1. [DOI] [PubMed] [Google Scholar]
53.Singh S, Garg SK, Singh PP, Iyer PG, El-Serag HB. Acid-suppressive medications and risk of oesophageal adenocarcinoma in patients with Barrett’s oesophagus: a systematic review and meta-analysis. Gut. 2014;63:1229–1237. doi: 10.1136/gutjnl-2013-305997. [DOI] [PMC free article] [PubMed] [Google Scholar]
54.Sivarasan N, Smith G. Role of aspirin in chemoprevention of esophageal adenocarcinoma: a meta-analysis. J Dig Dis. 2013;14:222–230. doi: 10.1111/1751-2980.12047. [DOI] [PubMed] [Google Scholar]
55.Zhang S, Zhang XQ, Ding XW, et al. Cyclooxygenase inhibitors use is associated with reduced risk of esophageal adenocarcinoma in patients with Barrett’s esophagus: a meta-analysis. Br J Cancer. 2014;110:2378–2388. doi: 10.1038/bjc.2014.127. [DOI] [PMC free article] [PubMed] [Google Scholar]
56.García Rodríguez LA, Lagergren J, Lindblad M. Gastric acid suppression and risk of oesophageal and gastric adenocarcinoma: a nested case control study in the UK. Gut. 2006;55:1538–1544. doi: 10.1136/gut.2005.086579. [DOI] [PMC free article] [PubMed] [Google Scholar]
57.Singh S, Singh AG, Singh PP, Murad MH, Iyer PG. Statins are associated with reduced risk of esophageal cancer, particularly in patients with Barrett’s esophagus: a systematic review and meta-analysis. Clin Gastroenterol Hepatol. 2013;11:620–629. doi: 10.1016/j.cgh.2012.12.036. [DOI] [PMC free article] [PubMed] [Google Scholar]
58.Choi SE, Perzan KE, Tramontano AC, Kong CY, Hur C. Statins and aspirin for chemoprevention in Barrett’s oesophagus: results of a cost-effectiveness analysis. Cancer Prev Res (Phila) 2014;7:341–350. doi: 10.1158/1940-6207.CAPR-13-0191-T. [DOI] [PMC free article] [PubMed] [Google Scholar]
59.Masclee GM, Coloma PM, Spaander MC, Kuipers EJ, Sturkenboom MC. NSAIDs, statins, low-dose aspirin and PPIs, and the risk of oesophageal adenocarcinoma among patients with Barrett’s oesophagus: a population-based case-control study. BMJ Open. 2015;5:e006640. doi: 10.1136/bmjopen-2014-006640. [DOI] [PMC free article] [PubMed] [Google Scholar]
Articles from Annals of Gastroenterology : Quarterly Publication of the Hellenic Society of Gastroenterology are provided here courtesy of The Hellenic Society of Gastroenterology
ACTIONS
View on publisher site
PDF (411.3 KB)
Cite
Collections
Permalink PERMALINK
Copy
RESOURCES
Similar articles
Cited by other articles
Links to NCBI Databases
On this page
Abstract
Introduction
Acknowledgment
Biography
Footnotes
References
Cite
Copy
Download .nbib.nbib
Format:
Add to Collections
Create a new collection
Add to an existing collection
Name your collection
Choose a collection
Unable to load your collection due to an error
Please try again
Add Cancel
Follow NCBI
NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed
Connect with NLM
NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube
National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov
Back to Top |
9969 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8123957/ | Evaluating Guanfacine Hydrochloride in the Treatment of Attention Deficit Hyperactivity Disorder (ADHD) in Adult Patients: Design, Development and Place in Therapy - PMC
Skip to main content
An official website of the United States government
Here's how you know
Here's how you know
Official websites use .gov
A .gov website belongs to an official government organization in the United States.
Secure .gov websites use HTTPS
A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites.
Search
Log in
Dashboard
Publications
Account settings
Log out
Search… Search NCBI
Primary site navigation
Search
Logged in as:
Dashboard
Publications
Account settings
Log in
Search PMC Full-Text Archive
Search in PMC
Journal List
User Guide
New Try this search in PMC Beta Search
View on publisher site
Download PDF
Add to Collections
Cite
Permalink PERMALINK
Copy
As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health.
Learn more: PMC Disclaimer | PMC Copyright Notice
Drug Des Devel Ther
. 2021 May 11;15:1965–1969. doi: 10.2147/DDDT.S221126
Search in PMC
Search in PubMed
View in NLM Catalog
Add to search
Evaluating Guanfacine Hydrochloride in the Treatment of Attention Deficit Hyperactivity Disorder (ADHD) in Adult Patients: Design, Development and Place in Therapy
Toyosaku Ota
Toyosaku Ota
1 Department of Psychiatry, Nara Medical University, Kashihara, Japan
2 Faculty of Nursing, School of Medicine, Nara Medical University, Kashihara, Japan
Find articles by Toyosaku Ota
1,2,✉, Kazuhiko Yamamuro
Kazuhiko Yamamuro
1 Department of Psychiatry, Nara Medical University, Kashihara, Japan
Find articles by Kazuhiko Yamamuro
1, Kosuke Okazaki
Kosuke Okazaki
1 Department of Psychiatry, Nara Medical University, Kashihara, Japan
Find articles by Kosuke Okazaki
1, Toshifumi Kishimoto
Toshifumi Kishimoto
1 Department of Psychiatry, Nara Medical University, Kashihara, Japan
Find articles by Toshifumi Kishimoto
1
Author information
Article notes
Copyright and License information
1 Department of Psychiatry, Nara Medical University, Kashihara, Japan
2 Faculty of Nursing, School of Medicine, Nara Medical University, Kashihara, Japan
✉
Correspondence: Toyosaku Ota Department of Psychiatry, Nara Medical University, 840 Shijyo-Cho, Kashihara, Nara, 634-8522, Japan, Phone: Tel +81-742-22-3051, Fax +81-742-22-3854 Email toyosaku@naramed-u.ac.jp
Received 2021 Mar 1; Accepted 2021 May 1; Collection date 2021.
© 2021 Ota et al.
This work is published and licensed by Dove Medical Press Limited. The full terms of this license are available at and incorporate the Creative Commons Attribution – Non Commercial (unported, v3.0) License ( By accessing the work you hereby accept the Terms. Non-commercial uses of the work are permitted without any further permission from Dove Medical Press Limited, provided the work is properly attributed. For permission for commercial use of this work, please see paragraphs 4.2 and 5 of our Terms (
PMC Copyright notice
PMCID: PMC8123957 PMID: 34007156
Abstract
Attention-deficit/hyperactivity disorder (ADHD) is characterized by age-inappropriate and impairing levels of inattention, hyperactivity, or impulsivity, or a combination of these characteristics. It is estimated to affect around 4% of adults worldwide. In the past few decades, prescriptions for ADHD drugs (psychostimulants and non-psychostimulants) have increased significantly. However, the efficacy and safety of adult ADHD medications remains controversial. Guanfacine extended-release (GXR) is a non-psychostimulant ADHD drug that is a selective α2A-adrenergic receptor agonist, first approved for treatment of adult ADHD in Japan in June 2019. Our aim was to provide an overview of GXR pharmacology and review the studies on efficacy and safety that have been conducted in adults with ADHD. The beneficial actions of guanfacine are thought to be attributed to the strengthening of prefrontal cortical network connections, which regulate attention, emotion, and behavior via the activity at post-synaptic α2A receptors. Current evidence of GXR efficacy and safety suggests that GXR is an effective monotherapy treatment option for adults with ADHD.
Keywords: attention-deficit/hyperactivity disorder, pharmacotherapy, guanfacine, adults, nonstimulant
Introduction
Attention-deficit/hyperactivity disorder (ADHD) is characterized by age-inappropriate and impairing levels of inattention, hyperactivity, or impulsivity, or a combination of these characteristics.1 It is estimated to affect 7–10% of children2 and about 4% of adults.3 Available pharmacological treatments for ADHD include psychostimulant (eg methylphenidate and amphetamine) and non-psychostimulant medications (eg atomoxetine and α2-agonists). In the past few decades, prescriptions for ADHD drugs have significantly increased globally.4,5 Despite clinical guidelines recommending the use of these drugs for treating ADHD,6–9 the treatment guidelines for adult ADHD are developing.
Guanfacine extended-release (GXR) is a nonstimulant, selective, α2A-adrenergic receptor agonist. The guanfacine extended-release formulation has been available in the United States since 2009. The GXR is approved by the US Food and Drug Administration for the treatment of ADHD in patients aged 6–17 years, either as monotherapy or an adjunctive therapy for stimulants. However, there have not been sufficient examinations of efficacy and safety of this drug in adults with ADHD. In fact, GXR for adults was not included in the comprehensive 2018 systematic review and meta-analysis of medications for ADHD;10 moreover, it has not been included in the current international guidelines.11
In Japan, GXR was first approved for treatment of adults with ADHD in June 2019. Therefore, we sought to provide an overview of GXR pharmacology and review studies that have evaluated the efficacy and safety of GXR for the treatment of adults with ADHD.
Pharmacology of GXR
Guanfacine differs from clonidine in that it is a phenylacetylguanidine derivative12 and is more selective in its α2 agonism.13 Like the action of clonidine, it lowers blood pressure by activating brainstem receptors, which leads to decreased sympathetic tone.14 Guanfacine’s beneficial effects on ADHD are thought to occur via its agonism at α2A-adrenoceptors in the prefrontal cortex. The prefrontal cortex is the primary brain area implicated in ADHD, and the prefrontal dysfunction may underlie symptoms of inattentiveness, impulsivity, and excessive motor activity.15 Furthermore, the prefrontal cortex is important for executive functions (eg planning, decision making, judgment, and response inhibition).
Both norepinephrine and dopamine are crucial for prefrontal cortex functions.16 Sufficient levels of norepinephrine are required for optimal working memory and attention; however, excessive norepinephrine release (eg during stress) can also impair cognitive functions.15 Optimal levels of norepinephrine activate α2A-adrenoceptors, whereas excessive levels lead to stimulation of lower affinity α1-adrenoceptors. Experimentally, blocking prefrontal α2A-adrenoreceptors induces motor hyperactivity in primates. Guanfacine has been shown to enhance attention and hyperactivity in animal models of ADHD.17 Guanfacine has also been shown to extend the activity of the frontal cortex in both rats18 and humans.19
The GXR tablet is formulated as a single daily extended-release matrix tablet,20 that should be swallowed whole, instead of chewed, crushed, or split, to increase the rate of guanfacine release. Guanfacine is absorbed effectively from GXR tablets that are taken orally;21 the time to peak plasma concentration is approximately 5 hours in children and adolescents with ADHD.21 However, consumption of a food has an effect on exposure to guanfacine via GXR tablets, with a 75% increase in peak concentration of plasma and 40% increase in area under the concentration–time curve of plasma when exposed to a high-fat breakfast.21 Therefore, exposure to a high-fat food should be shunned.
In plasma, guanfacine is approximately 70% bound to plasma proteins.21 Guanfacine is mainly metabolized by hepatic cytochrome p450 (CYP) 3A4 microsomal enzymes, and exposure to guanfacine has the potential to be increased by the simultaneous use of CYP 3A4 inhibitors (eg itraconazole, clarithromycin, fluvoxamine, and grapefruit juice) or decreased by the simultaneous use of CYP 3A4 inducers (eg rifampin, carbamazepine, phenytoin, and St. John’s Wort).21 However, CYP enzymes are not inhibited or induced by guanfacine.21 The elimination half-life of guanfacine following GXR administration is 16–17 hours,20 which enables a single daily administration of GXR. Furthermore, 50% of one GXR dose is excreted unaltered within the urine.20 In addition, it is suggested that guanfacine may affect the pharmacokinetics of valproic acid and lead to a significant increase in valproate plasma levels when used concurrently with this agent.
Guanfacine reduces sympathetic nerve impulses, which leads to reduced sympathetic outflow and follow-on decrease in vasomotor tone and heart rate. Additionally, guanfacine selectively binds to postsynaptic α2A-adrenoreceptors in the prefrontal cortex and has been theorized to enhance delay-related neural firing at the prefrontal cortex. This affects behavioral inhibition and working memory, which improves symptoms associated with ADHD. Similar to norepinephrine, guanfacine acts on α2A-receptors within the cortex to reinforce the signal-to-noise ratio of surrounding stimuli, which is assumed to enhance the power to focus on a specific stimulus “above the noise” during times of low arousal.22
GXR for Treating Adult ADHD
Butterfield et al studied 26 adults with a diagnosis of ADHD who were 19–62 years of age.23 The study’s primary outcome measures were the ADHD Rating Scale (ADHD-RS) and the Clinical Global Impression-Severity (Table 1). Patients were randomly assigned to two treatment groups, where their existing psychostimulant treatment regimen was supplemented with either a titrated dose (1–6 mg) of GXR or placebo over a 10-week trial period. The data were analyzed using a standard mixed-model analysis of variance procedure, and patients in both groups showed significant improvement in symptoms and functioning during the trial. There was no difference between the treatments in terms of their efficacy, safety, or tolerability.
Table 1.
Characteristics of GXR Study for Adult ADHD
| Author, Year | Duration | Diagnosis of ADHD | Mean Age (SD) | Male % | Intervention (No. Dose) | Primary Outcome Measures |
--- --- ---
| Butterfield, 2016 | 10 weeks | DSM-IV-TR | 37.5 (12.2) | 46.2% | Placebo | ADHD-RS, CGI-S |
| GXR 1–6mg |
| Total: 26 |
| Iwanami, 2020a | 12 weeks | DSM-5 | 31.1 (8.1) | 66.0% | Placebo (100) | ADHD-RS |
| GXR (101) 4–6mg |
| Iwanami, 2020b | 50 weeks | DSM-5 | 33.1 (9.4) | 68.1% | GXR (191) 4–6mg | ADHD-RS |
Open in a new tab
Iwanami et al assessed GXR efficacy and safety in adults with ADHD in a Phase 3, double-blind, placebo-controlled study that included Japanese patients with ADHD aged ≥ 18 years.24 Patients were administered GXR (n = 101) or placebo (n = 100), titrated from 2 mg/day to 4–6 mg/day with dose optimization during 5 weeks, followed by 4–6 mg/day in maintaining the dose during 5 weeks, then tapered to 2 mg/day for 2 weeks (Table 1). Compared with placebo, GXR treatment showed a significant decrease in ADHD-RS total score. Additionally, treatment with GXR showed significantly greater improvements in two subscale (inattention and hyperactivity-impulsivity) scores of the ADHD-RS, Clinical Global Impression-Improvement scale scores, and Patient Global Impression-Improvement scale scores. However, there were more participants in the GXR group compared with participants in the placebo group who reported treatment-emergent adverse events and who discontinued GXR because of treatment-emergent adverse events. The main adverse events in the GXR group were somnolentia, mouth dryness, reduced blood pressure, nasopharyngitis, dizziness upon standing, and astriction, although most treatment-emergent adverse events were mild to moderate in regard to severity. Therefore, the study concluded that GXR monotherapy improves ADHD symptoms in Japanese adults with ADHD without major safety concerns.
In another study, the safety and efficacy of prolonged pharmaceutical therapy of GXR in adults with ADHD were assessed by Iwanami et al.25 In this open-label, prolonged period, phase 3, extension study conducted in Japan, 150 participants who had continued from a double-blind period, and 41 newly attended participants were administered single daily GXR from 2 mg/day during 50 weeks (maintenance dose 4–6 mg/day) (Table 1). One hundred and eighty of 191 participants reported more than one treatment-emergent adverse event and 38 of 191 participants discontinued because of treatment-emergent adverse events. Most treatment-emergent adverse events were mild to moderate with regard to severity; two were considered serious treatment-emergent adverse events but there were no deaths. Treatment-emergent adverse events that were reported by more than 10% of participants were somnolentia, mouth dryness, nasopharyngitis, decreased blood pressure, dizziness upon standing, cardiac slowing, lassitude, astriction, and dizziness. The ADHD-RS scores (total, inattention subscale, and hyperactivity-impulsivity subscale), Adult ADHD Quality of Life Questionnaire total scores, and executive functioning showed significant improvement in participants in double-blind trial and newly enrolled participants at the final point, and there was also an increase in the proportion of patients who showed considerable improvement in Clinical Global Impression-Improvement scale scores and Patient Global Impression-Improvement scale scores. Therefore, results demonstrated that there were no major safety issues for prolonged pharmaceutical therapy of GXR for adults with ADHD, and patients showed significant ameliorations in ADHD symptoms, quality of life, and executive function following long-term GXR monotherapy.
Taking it into consideration, rebound hypertension is a potential concern with sudden discontinuation of alpha-2 agonists. Persistent blood pressure increases of up to 10 mmHg have been observed in a few individuals at 30 days post-discontinuation.26 All reviewed trials had a dose tapering period following the maintenance phase, in keeping with the manufacturer’s recommendations for gradual dosage decrements of no more than 1 mg every 3–7 days when tapering off GXR.26
GXR for Treating Comorbid ASD and Tic Disorder
Scahill et al assessed GXR efficacy and safety in autism spectrum disorder with hyperactivity.27 This study was a five-site randomized, double-blind placebo-controlled, fixed-flexible dose clinical trial conducted by the Research Units on Pediatric Psychopharmacology Autism Network. Sixty-two subjects were randomly assigned to GXR (n=30) or placebo (n=32) for 8 weeks. The GXR group showed a 43.6% decline in scores on the Aberrant Behavior Checklist-hyperactivity subscale compared with a 13.2% decrease in the placebo group. The most common adverse events included drowsiness, fatigue, and decreased appetite. There were no significant changes on ECG in either group. For subjects in the GXR group, blood pressure declined in the first 4 weeks, with return nearly to baseline by endpoint (week 8). Pulse rate showed a similar pattern but remained lower than baseline at endpoint. Therefore, the study concluded that GXR appears to be safe and effective for reducing hyperactivity, impulsiveness, and distractibility in children with autism spectrum disorder although there was no effects for autism spectrum symptoms.
In the meta-analysis of treatment for children with ADHD with comorbid tic disorder, methylphenidate, alpha-2 agonists, desipramine, and atomoxetine demonstrated efficacy in improving ADHD symptoms in children with comorbid tics. In addition, alpha-2 agonists and atomoxetine significantly improved comorbid tic symptoms.28 In the current systematic review, methylphenidate, clonidine, guanfacine, desipramine, and atomoxetine appear to reduce ADHD symptoms in children with tics, though the quality of the available evidence is low to very low.29 Therefore, GXR is not recommended for the treatment of ADHD in adults with comorbid tic disorder as first line therapeutic agent.
Comparison Between ADHD Drugs
Medications used to treat youths with ADHD have been studied extensively; however, direct comparisons between medications have not been investigated in head-to-head clinical trials. Faraone searched literature to identify double-blind, placebo-controlled trials in young people with ADHD from 1979 onwards that described the variability of drug-placebo effect sizes.30 Faraone conducted meta-analysis regression to assess the impact of medication type on pharmacological effects. The effect sizes for immediate-release (effect size 0.99) and long-acting stimulants (effect size 0.95) were broadly similar. On the other hand, the effect size for nonstimulants (effect size 0.57) was smaller than the effect sizes for stimulants. Similar investigations are needed for pharmacotherapy studies in adults with ADHD.
The abuse potential of stimulants is still controversially discussed. However, GXR has no abuse potential. This characteristic is one of the strengths of this medicine.
Conclusion
Guanfacine is an agonist that acts on α2A-adrenoreceptors, that are highly concentrated in the locus coeruleus and prefrontal cortex. The beneficial actions arise from its ability to strengthen prefrontal cortical network connections to regulate attention, emotion, and behavior via its activity at postsynaptic α2A receptors. The current efficacy and safety evidence indicates that GXR as monotherapy is an effective treatment option for adults with ADHD.
Acknowledgments
We thank Sarina Iwabuchi, PhD, from Edanz Group ( for editing a draft of this manuscript.
Abbreviations
ADHD, attention-deficit/hyperactivity disorder; GXR, guanfacine extended-release; CYP, cytochrome p450; ADHD-RS, ADHD Rating Scale.
Disclosure
The authors declare that they have no conflicts of interest.
References
1.American Psychiatric Association. Diagnostic and Statistical Manual of Mental Disorders. 5th ed. Arlington: American Psychiatric Association; 2013. [Google Scholar]
2.Xu G, Strathearn L, Liu B, Yang B, Bao W. Twenty–year trends in diagnosed attention-deficit/hyperactivity disorder among US children and adolescents, 1997–2016. JAMA Netw Open. 2018;1(4):e181471. doi: 10.1001/jamanetworkopen.2018.1471 [DOI] [PMC free article] [PubMed] [Google Scholar]
3.Song P, Zha M, Yang Q, Zhang Y, Li X, Rudan I. The prevalence of adult attention-deficit hyperactivity disorder: a global systematic review and meta-analysis. J Glob Health. 2021;11:04009. doi: 10.7189/jogh.11.04009 [DOI] [PMC free article] [PubMed] [Google Scholar]
4.Chai G, Governale L, McMahon AW, Trinidad JP, Staffa J, Murphy D. Trends of outpatient prescription drug utilization in US children, 2002–2010. Pediatrics. 2012;130(1):23–31. doi: 10.1542/peds.2011-2879 [DOI] [PubMed] [Google Scholar]
5.Renoux C, Shin JY, Dell’Aniello S, Fergusson E, Suissa S. Prescribing trends of attention-deficit hyperactivity disorder (ADHD) medications in UK primary care, 1995–2015. Br J Clin Pharmacol. 2016;82(3):858–868. doi: 10.1111/bcp.13000 [DOI] [PMC free article] [PubMed] [Google Scholar]
6.Pliszka S; AACAP Work Group on Quality Issues. Practice parameter for the assessment and treatment of children and adolescents with attention-deficit/hyperactivity disorder. J Am Acad Child Adolesc Psychiatry. 2007;46(7):894–921. doi: 10.1097/chi.0b013e318054e724 [DOI] [PubMed] [Google Scholar]
7.Kooij SJ, Bejerot S, Blackwell A, et al. European consensus statement on diagnosis and treatment of adult ADHD: the European Network Adult ADHD. BMC Psychiatry. 2010;10:67. [DOI] [PMC free article] [PubMed] [Google Scholar]
8.Wolraich M, Brown L, Brown RT, et al.; Subcommittee on Attention-Deficit/Hyperactivity Disorder; Steering Committee on Quality Improvement and Management. ADHD: clinical practice guideline for the diagnosis, evaluation, and treatment of attention-deficit/hyperactivity disorder in children and adolescents. Pediatrics. 2011;128(5):1007–1022. [DOI] [PMC free article] [PubMed] [Google Scholar]
9.Bolea-Alamañac B, Nutt DJ, Adamou M, et al.; British Association for Psychopharmacology. Evidence-based guidelines for the pharmacological management of attention deficit hyperactivity disorder: update on recommendations from the British Association for Psychopharmacology. J Psychopharmacol. 2014;28(3):179–203. doi: 10.1177/0269881113519509. [DOI] [PubMed] [Google Scholar]
10.Cortese S, Adamo N, Del Giovane C, et al. Comparative efficacy and tolerability of medications for attention-deficit hyperactivity disorder in children, adolescents, and adults: a systematic review and network meta-analysis. Lancet Psychiatry. 2018;5(9):727–738. doi: 10.1016/S2215-0366(18)30269-4 [DOI] [PMC free article] [PubMed] [Google Scholar]
11.National Institute for Health and Care Excellence. Attention deficit hyperactivity disorder: diagnosis and management. NICE guideline [NG87]; 2018. Available from: Accessed January16, 2021. [PubMed]
12.Bream JB, Lauener H, Picard CW, Scholtysik G, White TG. Substituted phenyl-acetyl guanidine: a new class of antihypertensive agents. Arzeimittel-forschung. 1975;25(10):1477–1482. [PubMed] [Google Scholar]
13.Doxey JC. Pre- and postsynaptic effects of alpha-agonists in the anococcygeus muscle of the pithed rat. Eur J Pharmacol. 1979;54(1–2):185–189. doi: 10.1016/0014-2999(79)90423-0 [DOI] [PubMed] [Google Scholar]
14.Westfall TC, Westfall DP. Adrenergic agonists and antagonists. In: Brunton LL, editor. Goodman & Gilman’s the Pharmacological Basis of Therapeutics. New York: McGraw-Hill; 2011:277–334. [Google Scholar]
15.Arnsten AF, Li BM. Neurobiology of executive functions: catecholamine influences on prefrontal cortical functions. Biol Psychiatry. 2005;57(11):1377–1384. doi: 10.1016/j.biopsych.2004.08.019 [DOI] [PubMed] [Google Scholar]
16.Brozoski T, Brown RM, Rosvold HE, Goldman PS. Cognitive deficit caused by regional depletion of dopamine in prefrontal cortex of rhesus monkey. Science. 1979;205(4409):929–932. doi: 10.1126/science.112679 [DOI] [PubMed] [Google Scholar]
17.Sagvolden T. The alpha-2A adrenoceptor agonist guanfacine improves sustained attention and reduces overactivity and impulsiveness in an animal model of attention-deficit/hyperactivity disorder (ADHD). Behav Brain Funct. 2006;2(1):41. doi: 10.1186/1744-9081-2-41 [DOI] [PMC free article] [PubMed] [Google Scholar]
18.Easton N, Shah YB, Marshall FH, Fone KC, Marsden CA. Guanfacine produces differential effects in the frontal cortex compared with striatum: assessed by phMRI BOLD contrast. Psychopharmacology. 2006;189(3):369–385. doi: 10.1007/s00213-006-0558-1 [DOI] [PubMed] [Google Scholar]
19.Swartz BE, Kovalik K, Thomas K, Torgersen D, Mandelkern MA. The effects of an alpha2 adrenergic agonist, guanfacine, on rCBF in human cortex in normal controls and subjects with focal epilepsy. Neuropsychopharmacology. 2000;23(3):263–275. doi: 10.1016/S0893-133X(00)00101-9 [DOI] [PubMed] [Google Scholar]
20.Cruz MP. Guanfacine extended-release tablets (Intuniv), a nonstimulant selective alpha2a-adrenergic receptor agonist for attention-deficit/hyperactivity disorder. P T. 2010;35(8):448–451. [PMC free article] [PubMed] [Google Scholar]
21.Bukstein OG, Head J. Guanfacine ER for the treatment of adolescent attention-deficit/hyperactivity disorder. Expert Opin Pharmacother. 2012;13(15):2207–2213. doi: 10.1517/14656566.2012.721778 [DOI] [PubMed] [Google Scholar]
22.Stahl SM. Stahl’s Essential Psychopharmacology: Neuroscientific Basis and Practical Applications. New York: Cambridge University Press; 2008. [Google Scholar]
23.Butterfield ME, Saal J, Young B, Young JL. Supplementary guanfacine hydrochloride as a treatment of attention deficit hyperactivity disorder in adults: a double blind, placebo-controlled study. Psychiatry Res. 2016;236:136–141. doi: 10.1016/j.psychres.2015.12.017 [DOI] [PubMed] [Google Scholar]
24.Iwanami A, Saito K, Fujiwara M, Okutsu D, Ichikawa H. Efficacy and safety of guanfacine extended-release in the treatment of attention-deficit/hyperactivity disorder in adults: results of a randomized, double-blind, placebo-controlled Study. J Clin Psychiatry. 2020a;81(3):19m12979. doi: 10.4088/JCP.19m12979 [DOI] [PubMed] [Google Scholar]
25.Iwanami A, Saito K, Fujiwara M, Okutsu D, Ichikawa H. Safety and efficacy of guanfacine extended-release in adults with attention-deficit/hyperactivity disorder: an open-label, long-term, phase 3 extension study. BMC Psychiatry. 2020b;20(1):485. doi: 10.1186/s12888-020-02867-8 [DOI] [PMC free article] [PubMed] [Google Scholar]
26.Shire Canada Inc. Intuniv XR product monograph. Ville Saint-Laurent, QC: Shire Canada Inc. 2013. [Google Scholar]
27.Scahill L, McCracken JT, King BH, et al. Extended-release guanfacine for hyperactivity in children with autism spectrum disorder. Am J Psychiatry. 2015;172(12):1197–1206. doi: 10.1176/appi.ajp.2015.15010055 [DOI] [PubMed] [Google Scholar]
28.Bloch MH, Panza KE, Landeros-Weisenberger A, Leckman JF. Meta-analysis: treatment of attention-deficit/hyperactivity disorder in children with comorbid tic disorders. J Am Acad Child Adolesc Psychiatry. 2009;48(9):884–893. [DOI] [PMC free article] [PubMed] [Google Scholar]
29.Osland ST, Steeves TDL, Pringsheim T. Pharmacological treatment for attention deficit hyperactivity disorder (ADHD) in children with comorbid tic disorders. Cochrane Database Syst Rev. 2018;2018(6):CD007990. [DOI] [PMC free article] [PubMed] [Google Scholar]
30.Faraone SV. Using meta-analysis to compare the efficacy of medications for attention-deficit/hyperactivity disorder in youths. P T. 2009;34(12):678–694. [PMC free article] [PubMed] [Google Scholar]
Articles from Drug Design, Development and Therapy are provided here courtesy of Dove Press
ACTIONS
View on publisher site
PDF (227.9 KB)
Cite
Collections
Permalink PERMALINK
Copy
RESOURCES
Similar articles
Cited by other articles
Links to NCBI Databases
On this page
Abstract
Introduction
Pharmacology of GXR
GXR for Treating Adult ADHD
GXR for Treating Comorbid ASD and Tic Disorder
Comparison Between ADHD Drugs
Conclusion
Acknowledgments
Abbreviations
Disclosure
References
Cite
Copy
Download .nbib.nbib
Format:
Add to Collections
Create a new collection
Add to an existing collection
Name your collection
Choose a collection
Unable to load your collection due to an error
Please try again
Add Cancel
Follow NCBI
NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed
Connect with NLM
NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube
National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov
Back to Top |
9970 | https://www.wyzant.com/resources/answers/629/for_all_real_numbers_a_and_b_a_b_is_equivalent_to_b_a_is_the_statement_true_or_false | Log in Sign up
Search
Search Find an Online Tutor Now Ask
Ask a Question For Free
Login
Algebra 1
Keonya D.
asked • 09/18/12
for all real numbers a and b, ab is equivalent to ba. is the statement true or false ?
for all real numbers a and b, a(bc) = ab ac
Follow • 5
Comment • 1
More
Report
Ted S.
This statement is true...
ab = ba
use 2=a and 3=b
2 3 = 3 2
Report
09/26/12
5 Answers By Expert Tutors
By:
Jehsuamo C. answered • 09/20/12
Tutor
5.0 (51)
A Mathemagician, at your service!
See tutors like this
See tutors like this
This statement is true for all real numbers and is known as the Commutative Property. It not only applies to multiplication, but to addition as well: a + b = b + aThe other case you mentioned a(bc) = abac is false.There are certain other properties that are central to understanding algebra and working with real numbers that you may also learn depending on your level:Assiociative property: Given real numbers a, b, and c: (a +b) + c = a + (b +c) also, (ab)c = a(bc)Distributive property: Given real numbers a, b, and c: a(b + c) = ab + acThis property can be shown by applying the definition/concept of multiplication:a(b + c) means that we have "a" groups of (b + c) hence,a(b + c) = (b + c) + ... + (b + c) <- a times, we are adding an "a" number of (b + c)'s.Next we ask the questions how many b's do we have and how many c's do we have? Well applying basic counting we know that there an "a" number of b's, and an "a" number of c's. in other words,a(b + c) = (b + c) + ... + (b + c) [a times] = ab + acIdentity: Pretty states that there is a real number for addition and there is a real number for multiplication that doesn't interact with any other number for the respective operation. For addition the number is 0. 0 is know as the additive identity. For multiplication the number is 1, 1 is known as the multiplicative identity.Additive inverse: The number that you add to a given real number to get the additive identity 0. For example -1 is the additive inverse of 1, and 5/2 is the additive inverse of -5/2. 0 is the additive inverse of 0.Multiplicative inverse: The number that you multiply to a given real number to get the multiplicative identity. In almost every case it is the reciprical of that number. For example:1/2 is the multiplicative inverse of 2, -3 is the multiplicative inverse of -1/3, as before 1 is the multiplicative inverse of 1. HOWEVER, 0 is the only real number with out a multiplicative inverse. There is no real number you can multiply by 0 to get 1. This is why we can't divide by 0, and this is why we say once we understand left and right cancelation that when zero is in the denominator the fraction is "undefined".Left Cancelation: Given that an operation of addition or multiplication is taking place you can "undo" by using its inverse.Examples: a + 4 = 10a + 4 + -4 = 10 + -4a + 0 = 6a = 6b 7 = 21b 7 (1/7) = 21 (1/7)b 1 = 3b = 3The same can pretty much be said without lost of generality for right cancelation when we are subtracting and dividing on both sides we are really adding and dividing by the additive and multiplicative inverse. Trichotemy Law: given two arbitrary real numbers a and b, one out of the following three relationships must take place: either a < b, a = b, a > b.If you are not majoring in mathematics, the first five is pretty much all you need to know.
Upvote • 1 Downvote
Add comment
More
Report
Jaison N. answered • 09/18/12
Tutor
4.9 (150)
A PhD to Teach You Math
See tutors like this
See tutors like this
Plug in numbers for a, b, c. Let a = 2, b = 1, and c = 3 and see for yourself! When you are trying to determine the validity of a mathematical proposition, create a few concrete examples as a check. Use the simplest numbers whenever possible.
Upvote • 1 Downvote
Add comment
More
Report
Major S. answered • 09/19/12
Tutor
4.9 (522)
College Graduate Wishing To Help All Achieve Success
About this tutor ›
About this tutor ›
The commutative property of multiplication does state that ab = ba for all real numbers.
However, a(bc) is not equal to ab ac for all real numbers. This is not an example of the distributive property. Distributive property a(b+c) = ab + ac or a(b-c) = ab - ac.
a(bc) = abc which is not equal to abac which is what ab ac represents. This is not true for all real numbers. I say this because it is true for certain values of a, b, c, but not all values.
Upvote • 0 Downvote
Add comment
More
Report
Nicole C. answered • 09/18/12
Tutor
4.9 (167)
Increase confidence in math and science, chemistry, algebra II
See tutors like this
See tutors like this
True. All real numbers is any positive or negative number, it just can't be a number with the i in it (imaginary number).
a has to be the same on both sides of the equation, so does b and c
both sides are the same, just written differently. If you distribute the left side you get the right side.
So, if the numbers are the same on both sides, and the equations are the same on both sides, the answer will always be true
Upvote • 0 Downvote
Add comment
More
Report
Lois P. answered • 09/18/12
Tutor
New to Wyzant
Miss Lois -- Elementary School Tutor
See tutors like this
See tutors like this
Ror all real numbers a and b, ab is equivalent to ba. This is indeed TRUE! Google Commutative Property for more details, but the basics are this:
Commutative property: When two numbers are multiplied together, the product is the same regardless of the order of the multiplicands. For example 4 2 = 2 4
As for the sub question for all real numbers a and b, a(bc) = ab ac, this is also TRUE. Google Distributive property for more details, again the basics are as follows:
Distributive property: The sum of two numbers times a third number is equal to the sum of each addend times the third number. For example 4 (6 + 3) = 46 + 43
Upvote • 0 Downvote
Comments • 3
More
Report
Matt L.
Lois, you write: "As for the sub question for all real numbers a and b, a(bc) = ab ac, this is also TRUE."
This is actually FALSE, as many others have pointed out. The distributive property for the real numbers holds only for addition (i.e. a(b+c)=ab+ac), not for multiplication. Be careful!
Matt
B.S. mathematics, MIT
Report
09/20/12
Lois P.
Yes, I see that you are right. I will edit my answer so as not to confuse anyone. Thank you.
Report
09/21/12
Lois P.
The system will not allow me to edit my response.
Report
09/21/12
Still looking for help? Get the right answer, fast.
Ask a question for free
Get a free answer to a quick problem.Most questions answered within 4 hours.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
RELATED TOPICS
Math Algebra 2 Calculus Geometry Algebra Inequalities Word Problem Algebra Question Equations Linear Equations ... Functions Word Problems Algebra Help Algebra Equation Math Help Algebra Word Problem Mathematics Math Word Problem Algebra 1 Word Problem Math Help For College
RELATED QUESTIONS
##### how to do I make a graph?
Answers · 8
##### How do I know my answer?
Answers · 5
##### how do you solve 8x+32/x^2-16
Answers · 15
##### 12p+15c>360
Answers · 5
##### 12p+15c>360
Answers · 3
RECOMMENDED TUTORS
Ingrid M.
5.0 (1,173) Deborah T.
5.0 (455) Priti S.
5.0 (737) See more tutors
find an online tutor
Algebra 1 tutors
Algebra 2 tutors
Algebra tutors
College Algebra tutors
Hesi tutors
CBEST tutors
PTCB tutors
Precalculus tutors |
9971 | https://www.frontiersin.org/journals/pediatrics/articles/10.3389/fped.2021.727954/full | Frontiers | Biopsy or Biomarker? Children With Minimal Change Disease Have a Distinct Profile of Urinary Epidermal Growth Factor
Frontiers in Pediatrics
About us
About us
Who we are
Mission and values
History
Leadership
Awards
Impact and progress
Frontiers' impact
Our annual reports
Publishing model
How we publish
Open access
Peer review
Research integrity
Research Topics
FAIR² Data Management
Fee policy
Services
Societies
National consortia
Institutional partnerships
Collaborators
More from Frontiers
Frontiers Forum
Frontiers Planet Prize
Press office
Sustainability
Career opportunities
Contact us
All journalsAll articlesSubmit your researchSearchLogin
Frontiers in Pediatrics
Sections
Sections
Children and Health
General Pediatrics and Pediatric Emergency Care
Genetics of Common and Rare Diseases
Neonatology
Obstetric and Pediatric Pharmacology
Pediatric Cardiology
Pediatric Critical Care
Pediatric Endocrinology
Pediatric Gastroenterology, Hepatology and Nutrition
Pediatric Hematology and Hematological Malignancies
Pediatric Immunology
Pediatric Infectious Diseases
Pediatric Nephrology
Pediatric Neurology
Pediatric Obesity
Pediatric Occupational Therapy
Pediatric Oncology
Pediatric Orthopedics
Pediatric Otolaryngology
Pediatric Pulmonology
Pediatric Rheumatology
Pediatric Surgery
Pediatric Urology
Social Pediatrics
ArticlesResearch TopicsEditorial board
About journal
About journal
Scope
Field chief editors
Mission & scope
Facts
Journal sections
Open access statement
Copyright statement
Quality
For authors
Why submit?
Article types
Author guidelines
Editor guidelines
Publishing fees
Submission checklist
Contact editorial office
About us
About us
Who we are
Mission and values
History
Leadership
Awards
Impact and progress
Frontiers' impact
Our annual reports
Publishing model
How we publish
Open access
Peer review
Research integrity
Research Topics
FAIR² Data Management
Fee policy
Services
Societies
National consortia
Institutional partnerships
Collaborators
More from Frontiers
Frontiers Forum
Frontiers Planet Prize
Press office
Sustainability
Career opportunities
Contact us
All journalsAll articlesSubmit your research
Frontiers in Pediatrics
Sections
Sections
Children and Health
General Pediatrics and Pediatric Emergency Care
Genetics of Common and Rare Diseases
Neonatology
Obstetric and Pediatric Pharmacology
Pediatric Cardiology
Pediatric Critical Care
Pediatric Endocrinology
Pediatric Gastroenterology, Hepatology and Nutrition
Pediatric Hematology and Hematological Malignancies
Pediatric Immunology
Pediatric Infectious Diseases
Pediatric Nephrology
Pediatric Neurology
Pediatric Obesity
Pediatric Occupational Therapy
Pediatric Oncology
Pediatric Orthopedics
Pediatric Otolaryngology
Pediatric Pulmonology
Pediatric Rheumatology
Pediatric Surgery
Pediatric Urology
Social Pediatrics
ArticlesResearch TopicsEditorial board
About journal
About journal
Scope
Field chief editors
Mission & scope
Facts
Journal sections
Open access statement
Copyright statement
Quality
For authors
Why submit?
Article types
Author guidelines
Editor guidelines
Publishing fees
Submission checklist
Contact editorial office
Frontiers in Pediatrics
Sections
Sections
Children and Health
General Pediatrics and Pediatric Emergency Care
Genetics of Common and Rare Diseases
Neonatology
Obstetric and Pediatric Pharmacology
Pediatric Cardiology
Pediatric Critical Care
Pediatric Endocrinology
Pediatric Gastroenterology, Hepatology and Nutrition
Pediatric Hematology and Hematological Malignancies
Pediatric Immunology
Pediatric Infectious Diseases
Pediatric Nephrology
Pediatric Neurology
Pediatric Obesity
Pediatric Occupational Therapy
Pediatric Oncology
Pediatric Orthopedics
Pediatric Otolaryngology
Pediatric Pulmonology
Pediatric Rheumatology
Pediatric Surgery
Pediatric Urology
Social Pediatrics
ArticlesResearch TopicsEditorial board
About journal
About journal
Scope
Field chief editors
Mission & scope
Facts
Journal sections
Open access statement
Copyright statement
Quality
For authors
Why submit?
Article types
Author guidelines
Editor guidelines
Publishing fees
Submission checklist
Contact editorial office
Submit your researchSearchLogin
Your new experience awaits. Try the new design now and help us make it even better
Switch to the new experience
ORIGINAL RESEARCH article
Front. Pediatr., 25 November 2021
Sec. Pediatric Nephrology
Volume 9 - 2021 |
Biopsy or Biomarker? Children With Minimal Change Disease Have a Distinct Profile of Urinary Epidermal Growth Factor
Niels Lodeweyckx 1Kristien Wouters2Kristien J. Ledeganck3Dominique Trouet1,3
1 Department of Pediatric Nephrology, Antwerp University Hospital, Edegem, Belgium
2 Clinical Trial Center (CTC), Clinical Research Center (CRC) Antwerp, Antwerp University Hospital, University of Antwerp, Edegem, Belgium
3 Laboratory of Experimental Medicine and Pediatrics and Member of the Infla-Med Centre of Excellence, University of Antwerp, Edegem, Belgium
Background: In this study, the profile of urinary EGF excretion (uEGF/uCreat) was mapped in children presenting with prolonged proteinuria or with nephrotic syndrome refractory to or dependent of steroids. We investigated whether uEGF/uCreat could be linked to the underlying biopsy result, taking into account its response to immunosuppressive medication and to ACE inhibition, as well as genetic predisposition.
Methods: Ninety-eight pediatric patients with initial presentation of nephrotic syndrome or prolonged proteinuria were included in this study, along with 49 healthy controls and 20 pediatric Alport patients. All patients had a normal kidney function and were normotensive during the course of the study, whether or not under ACE inhibition. In repeated urine samples, uEGF was measured and concentration was normalized by urine creatinine. In order to compare diagnosis on kidney biopsy, genetic predisposition and response of uEGF/uCreat to immunosuppression and to ACE inhibition, uEGF/uCreat is studied in a linear mixed effects model.
Results: Patients with Minimal Change Disease (MCD) showed a significantly different profile of uEGF/uCreat in comparison to healthy children, as well as compared to patients with Focal Segmental Glomerulosclerosis (FSGS) or another glomerulopathy on kidney biopsy. The response of uEGF/uCreat to ACE inhibition was absent in minimal change disease and contrasted with an impressive beneficial effect of ACE inhibition on uEGF/uCreat in FSGS and other proteinuric glomerulopathies. Absence of a genetic predisposition was also associated with a significantly lower uEGF/uCreat.
Conclusions: Despite preserved kidney function, children with a proteinuric or nephrotic glomerular disease on kidney biopsy show a significantly lower uEGF/uCreat, indicative of early tubulo-interstitial damage, which appears reversible under ACE inhibition in any underlying glomerulopathy except in minimal change disease. In view of the distinct profile of uEGF/uCreat in minimal change disease compared to other glomerulopathies, and the link between genetic predisposition and uEGF/uCreat, our study suggests that uEGF/uCreat can be a helpful tool to decide on the need for a renal biopsy in order to differentiate minimal change disease from other proteinuric glomerular diseases.
Introduction
Children presenting with a first episode of nephrotic-range proteinuria without kidney failure are prompt treated with high-dose steroids, assuming an underlying minimal change disease (MCD) favorably responding to a single cure of steroids (1–3). However, when this idiopathic nephrotic syndrome acquires a frequent relapsing, steroid-dependent, or steroid-resistant character, the initiation of other immunosuppressive therapy is most often guided by the findings on kidney biopsy (4, 5). Intriguingly, in this subgroup of refractory nephrotic children, up to half of the biopsy results reveal normal findings on light microscopy with no evidence of glomerular disease on immunofluorescence (4, 6). Although ultrastructural abnormalities are not obligatory present, the effacement of the podocytes' foot processes on electron microscopy are the only hint to a suggested underlying podocyte disease called minimal change disease (7). Those minimal morphological features may even disappear after immunosuppressive treatment, and are indicative of a reversible fragility of the podocytes to exogenous factors in MCD. In contrast, the sclerotic lesions on light microscopy in Focal Segmental Glomerulosclerosis (FSGS) encountered in a minority of the nephrotic patients are irreversible and point out to a more severe, genetically driven podocyte disease translated into the clinical features of an often steroid-resistant nephrotic syndrome with high chance of evolution toward chronic kidney failure (8–10).
Despite the overlap in clinical presentation, and in spite of similar therapeutical responses on T-cell or B-cell inhibition, MCD is considered a more benign podocyte disease than FSGS, with a different underlying genetic predisposition and pathophysiology (7). Of course, it should be kept in mind that the limited number of glomeruli within one biopsy sample could falsely identify an intrinsic FSGS as MCD, sometimes requiring repeated biopsies to unmask a “fake” MCD (7, 8). Moreover, recent research in adults provides increasing evidence that MCD is a distinct independent podocyte disease.
Comparative differential proteomic analyses recently gave evidence for a different profile of urinary biomarkers in MCD compared to FSGS (11).
In case of prolonged unexplained non-nephrotic proteinuria, either or not associated with hematuria or hypertension, children are also referred for a kidney biopsy to reveal the underlying glomerulopathy, in order to start tailored immunosuppressive treatment (12). In this group of pediatric patients besides MCD or FSGS, kidney biopsy often reveals an underlying IgA nephropathy, and less frequently C3 glomerulopathy, membranous glomerulonephritis, or other glomerulopathies rarely encountered at pediatric age (13, 14). The degree of genetic predisposition in this group of patients still remains unclear (15).
Due to its unique origin, mostly restricted to kidney tissue (16), urinary Epidermal Growth Factor (uEGF) functions as a predictive marker for progression in renal decline, and directly indicates the mitogenic function and interstitial regenerative capacity of the kidney (17–19). uEGF levels correlate with intrarenal mRNA expression of EGF on kidney biopsy and inversely correlate with the degree of interstitial fibrosis and tubular atrophy (16, 19, 20). Theoretically, already at an early stage in the disease, a decrease in uEGF could thus precede glomerular deterioration visualized on kidney biopsy (15, 19, 21). Tubular injury is closely linked to progression in glomerular nephropathies (22). Therefore, tubular proteins such as uEGF are potentially better predictive markers for progression of glomerular damage and renal deterioration than proteinuria and albuminuria (17). Particularly for uEGF, over the last few years, increasing evidence highlights its added value in predicting kidney deterioration in many disease entities such as systemic lupus disease (21), diabetes mellitus (23), but also in IgA nephropathy (15, 24), and other chronic kidney diseases (17) in adults. Specifically in children, recent reports illustrated the prognostic capacities of uEGF regarding renal decline in pediatric patients with chronic kidney disease (25), Alport syndrome (20), and nephrotic syndrome (19).
In this present study, we formulated three subsequent research questions. First, we explored whether uEGF could be linked to biopsy result (MCD or other proteinuric glomerulopathies) in children presenting with either prolonged non-nephrotic proteinuria or displaying a steroid-resistant or steroid-dependent nephrotic syndrome. Second, we investigated the impact of immunosuppression and ACE inhibition on the profile of uEGF in these patients. Third, we investigated if uEGF is related to a genetic predisposition of proteinuric glomerulopathies in this pediatric study population.
Patients and Methods
Study Design
In this longitudinal ambispective observational clinical trial, 98 pediatric patients with the initial presentation of nephrotic or non-nephrotic proteinuria were included between March 2016 and April 2021 at the Antwerp University Hospital. At three time points, with an interval of at least 1 month, urine samples were collected from each patient to determine creatinine, protein, and uEGF.
Study Patients
Prior to the design and initiation of this study, the patients had already undergone a renal needle biopsy at some point of the disease course, and were on a biopsy-guided immunosuppressive treatment at the start of the study. As a consequence, children with steroid-resistant, steroid-dependent, or frequent-relapsing nephrotic syndrome were included, as well as children without nephrotic syndrome but with unexplained prolonged proteinuria more than 0.5 g/g creatinine, either or not associated with hematuria.
In every patient presenting with hypertension or residual proteinuria after initiation of immunosuppression, ACE inhibition had been started, irrespective of and prior to the initiation of the study. Patients were in remission under treatment at the time the urine samples for the study were taken, thereby displaying not more than 0.3 g/g protein/creatinine in the urine.
Included patients had to be normotensive and have a normal kidney function [estimated glomerular filtration rate (eGFR) > 90 ml/min/1.73 m 2] at the start of and during the course of the study.
Exclusion criteria were the use of diuretics or aminoglycosides, an active urinary tract infection, or severe co-morbidity.
Healthy Control Patients
A recently published healthy control group (n = 49) functioned as reference for normal, age-specific uEGF values (26). Another group of patients was also included as a second, separate control group, in order to determine the effect of ACE inhibition without immunosuppression on uEGF. These children had initially presented with isolated hematuria (persistent microscopic hematuria with or without intermittent macroscopic hematuria) (n = 20). Diagnosis of Alport syndrome in these patients had been made by genetic testing and/or by kidney biopsy. All but three patients were on preventive monotherapy with ACE inhibition as recommended (27), but were not treated with any immunosuppression. Every Alport patient was normotensive, had a normal kidney function, and showed no proteinuria under ACE inhibition during the course of the study.
Ethics
The study was conducted in accordance with the Declaration of Helsinki and the principles of Good Clinical Practice. The protocol was approved by the Ethics Committee of the Antwerp University Hospital (file number 9/44/231). All patients and their parents and/or legal guardians gave a written informed consent.
Determination of Urinary EGF
Urinary EGF was measured using an EGF human Elisa kit (Invitrogen, Waltham, MA, USA) according to the manufacturer's guidelines. The detection limit of this assay was 3.9 pg/ml. The intra-and intervariability of the EGF human Elisa kit was excellent (28).
In order to avoid bias from differences in urinary concentration, uEGF was expressed as uEGF/U creatinine ratio. In view of the age-specific exponential decline of normal uEGF values, all analyses regarding uEGF were adjusted for age.
Genetic Screening
The gene panel used for screening was developed by Dahan et al. (Institut de Pathologie et de Génétique de Gosselies, Belgium) and comprised the following genes: ACTN4, ANLN, APOL1, ARHGDIA, CD151, CD2AP, COL4A3, COL4A4, COL4A5, COQ2, COQ6, COQ8A, COQ8B, CRB2, DGKE, EMP2, GLA, INF2, LAMB2, LMX1B, MAGI2, MYH9, MYO1E, NPHS1, NPHS2, PAX2, PDSS2, PLCE1, PTPRO, TRPC6, TTC21B, WDR73, and WT1.
Statistical Analysis
Patient characteristics were presented as number and percentages for categorical data, and mean (standard deviation) or median (min–max) otherwise. The relation between categorical characteristics was investigated by Fisher exact test and visualized in mosaic plots.
The log-transformed ratio uEGF/uCreat was studied in a linear mixed effects model with random subject effect and fixed effect age, supplemented with different patient and treatment characteristics and combinations of these. By including a random intercept per patient, dependency among observations of the same individual is taken into account in the analysis. Including all characteristics in one big model was not feasible due to the fact that patient and treatment characteristics are highly entangled; therefore, separate models were fitted to highlight different aspects of the associations.
Post-hoc comparisons were made based on these linear mixed effects models, and after back-transformation of the logarithm, results could be expressed as percentage difference, with 95% confidence interval. Bonferroni–Holm correction was applied to adjust for multiple testing.
Results
Descriptives
As shown in Table 1, a total of 118 patients were included, of which 62% are boys. From 91 patients 1 to 3 urine samples were collected for measurement of uEGF/uCreat. Median age of this group was 11 years (3–19 years). One hundred and ten patients had undergone a kidney biopsy prior to and independent of this study. In 87, patients a genetic screening had been performed.
TABLE 1
Table 1. Overview of patients' characteristics, clinical presentation, results of histology, and genetic screening.
The clustered group of “other glomerulopathy” comprised two patients with membranoproliferative lupus nephritis, three subjects with C3 membranonoproliferative glomerulonephritis, one child with diffuse mesangial sclerosis, and one with C1q nephropathy.
Link Between Initial Presentation and Biopsy Results
Analysis of the 110 patients who had undergone a kidney biopsy, showed a significant link between diagnosis on biopsy and the initial presentation (p< 0.001, Fisher exact).
The diagnosis on biopsy in children with nephrotic syndrome was significantly different compared to the patients displaying non-nephrotic proteinuria (p< 0.001). All but one patients with Minimal Change Disease initially presented with nephrotic syndrome. FSGS was only diagnosed in case of nephrotic syndrome, while other glomerulopathies (IgA nephropathy, Lupus membranoproliferative glomerulonephritis, C3 glomerulopathy, etc.) were mostly present in children presenting with non-nephrotic proteinuria, whether or not associated with hematuria, and only in a minority of nephrotic children (Figure 1A).
FIGURE 1
Figure 1. Mosaic plots visualizing the relation between (A) presentation and biopsy, (B) presentation and genetics, and (C) biopsy and genetics.
When disregarding the patients with nephrotic syndrome, and comparing mutually the groups presenting with isolated non-nephrotic proteinuria, combined proteinuria and hematuria, and isolated hematuria, again a significant link was seen with the results on biopsy (p = 0.013, Fisher exact) (Figure 1A).
Link Between Initial Presentation, Genetics, and Biopsy
Analysis of the 87 patients in whom genetic screening had been performed showed a significant link between the genetic result and the initial presentation (p = 0.044, Fisher exact), probably a statistical reflection of the fact that in the majority of patients with nephrotic syndrome, genetic screening was not able to show any genetic cause (Figure 1B).
However, there was no significant link between genetic results and diagnosis on biopsy (Figure 1C).
Link Between Initial Presentation, Biopsy, Genetics, and uEGF
As illustrated in Table 2, patients with nephrotic syndrome had a 46% lower uEGF/uCreat compared to the healthy controls (p< 0.001), while no significant difference could be seen for patients with initial non-nephrotic proteinuria (p = 0.73). Compared to patients with initial non-nephrotic proteinuria, the subjects with nephrotic syndrome displayed a 32% lower uEGF/uCreat (p< 0.001).
TABLE 2
Table 2. Comparison of uEGF/uCreat between different subgroups.
Additionally, uEGF/uCreat was 42% lower in patients without genetic cause compared to healthy controls (p< 0.001) (Table 2C).
Immunosuppressive Medication and Its Influence on uEGF
Calcineurin-inhibitors (CNI) cause a decrease in uEGF in children (28). As shown in Table 3, this finding was confirmed in our present study, as CNI-treated patients had a significantly lower uEGF/uCreat compared to those patients not receiving CNI (p< 0.001). No significant difference in uEGF/uCreat was seen for patients treated with B-cell inhibitory drugs Mycophenolate (MMF) and Rituximab (p = 0.27).
TABLE 3
Table 3. Comparisons based on linear mixed effects models with outcome log(uEGF/uCreat), random intercept, fixed effect for age, Cyclosporine/Tacrolimus, and MMF/Rituximab.
When making a distinction based on biopsy results, between minimal change disease on the one hand and every other glomerulopathy identified by aberrant light microscopy and immunofluorescence on the other hand (“FSGS/IgAN/other GN”), no significant difference could be withheld in the number of patients treated with CNIs, or for the use of B-cell inhibitors (Table 4). However, the use of CNI did differ significantly when comparing the nephrotic patients with the non-nephrotic proteinuric group. A difference in uEGF levels between nephrotic and proteinuric patients could therefore be explained merely by this difference in immunosuppressive treatment (Table 5).
TABLE 4
Table 4. Biopsy results in relation to the treatment regimen.
TABLE 5
Table 5. Nephrotic patients and non-nephrotic proteinuria in relation to the treatment regimen.
Influence of ACE Inhibition on Urinary EGF
A significant beneficial effect of ACE inhibition on urinary excretion of EGF was noticed. Irrespective of the initial presentation or biopsy result, patients receiving ACE inhibition had a 30% higher uEGF/uCreat compared to the patients treated with ACE inhibitors (p = 0.007) (Table 6A).
TABLE 6
Table 6. The effect of ACE inhibition on uEGF/uCreat according to the biopsy diagnosis.
To go further into depth on this finding, the biopsy results were subdivided into two types: minimal changes (MCD) vs. obvious glomerular changes visible on light microscopy or immunofluorescence, either FSGS, IgA nephropathy, or any other glomerulopathy (“FSGS/IgAN/other GN”) (Figure 2).
FIGURE 2
Figure 2. Graphical compilation visualizing the absolute values of U EGF/uCreat for age, according to the different clusters of biopsy results, comparing patients treated with ACE inhibitor to those without ACE inhibition.
This revealed that the significant beneficial ACE-I effect on uEGF/uCreat was due to a spectacular difference in uEGF/uCreat in the “FSGS/IgAN/GN” biopsy group, where patients treated with ACE inhibitors showed a 78% higher uEGF/uCreat than those without ACE inhibition. ACE inhibition had no significant effect on uEGF/uCreat in the minimal change disease biopsy type (Table 6A).
Patients Without ACE Inhibition
Compared to the healthy controls, uEGF/uCreat was significantly lower in both the MCD-group (38% lower than healthy, p = 0.002) and in the children with an aberrant “FSGS/IgAN/GN” biopsy type (61% lower than healthy, p< 0.001) (Table 6B).
More importantly, a significant difference in uEGF/uCreat was seen between the MCD patients and the patients with “FSGS/IgA/GN.” MCD patients had a 57% higher uEGF/uCreat than the children with FSGS, IgA nephropathy, or another glomerulopathy (p = 0.026) (Table 6B).
Patients Treated With ACE Inhibition
When focusing on all subjects receiving ACE inhibition, a separate group of Alport patients under ACE inhibition which functioned as a second control showed no significant difference in uEGF/uCreat compared to the healthy controls (Table 6C).
Under ACE inhibition, both in the minimal change disease and in the “FSGS/IgAN/ GN,” a significant difference in uEGF/uCreat was still seen compared to healthy controls as well as in comparison to Alport syndrome (Table 6C).
However, under ACE inhibition, minimal change disease patients no longer had a significantly different uEGF/uCreat compared to the “FSGS/IgAN, GN” group, due to a pronounced increase in uEGF/uCreat in the latter biopsy group.
Nephrotic Syndrome and uEGF/uCreat
Similar findings resulted from narrowing the focus toward the patients with nephrotic syndrome (Table 7).
TABLE 7
Table 7. Comparisons based on linear mixed effects models with outcome log(uEGF/uCreat), random intercept, fixed effect for age, and combination of biopsy result and ACE inhibition.
Discussion
In this study, the urinary profile of EGF excretion was mapped in children presenting with prolonged proteinuria or nephrotic syndrome refractory to or dependent of steroids. We investigated whether, taking into account genetic predisposition, response to immunosuppressive medication and influence of ACE inhibition, uEGF/uCreat could be linked to the underlying biopsy result.
uEGF/uCreat Is Significantly Lower in Children With a Proteinuric or Nephrotic Glomerular Disease, Despite Preserved Kidney Function
The additive value of uEGF in predicting renal decline in adults and children with progressive kidney disease or with chronic systemic diseases has been elaborated extensively over the 5 years (25, 28, 29). A similar predictive capacity of uEGF regarding renal decline in nephrotic pediatric patients was suggested a year ago by Gipson et al. (19), a multicenter study in which the initial value of uEGF/uCreat in children with minimal change disease or with FSGS was linked to their eGFR.
However, a comparison with a healthy control group was lacking in that study. Normal values of uEGF are very age-dependent (26) and have an exponential decline with age. So, specifically in a pediatric population this major influence of age on urinary excretion of EGF should be reckoned in order to avoid bias by age. Moreover, the impact of immunosuppressive medication or ACE inhibition on uEGF/uCreat was not elaborated either in the study by Gipson et al.
Our present study expands and finetunes the insights published by Gipson et al. We confirmed that children displaying a “difficult to treat” nephrotic syndrome (refractory to or dependent of steroids), but with preserved kidney function, have significantly lower uEGF/uCreat levels compared to healthy controls. As uEGF is a barometer for renal interstitial resilience, this indicates that in our specific pediatric population with normal eGFR, glomerular damage of any degree (even minimal changes) is translated into early secondary interstitial and peritubular changes. The strength of our study compared to the that of Gipson et al. (19) is the fact that we performed repeated measurements of uEGF/uCreat in the majority of the patients (thereby strengthening the consistency of our findings), that the age-specific values of uEGF/uCreat were respected in the statistical analysis, and that these UEGF/uCreat levels were compared with those of healthy controls. Moreover, the impact of immunosuppressive medication and ACE inhibition was extensively elaborated in our study. As we observed a major effect of CNI on uEGF/uCreat, and a pronounced influence of ACE inhibition on uEGF/uCreat, one could wonder if the data described by Gipson et al. would not be a reflection of the use of CNI and ACE inhibition, rather than an intrinsic disease-induced decrease in eGFR.
The fact that also glomerulopathies other than FSGS were included in our study, and displayed a similar pattern of uEGF/uCreat as in FSGS, seems to confirm once more that any immune- imbalance induced glomerular damage (also IgA nephropathy, Lupus nephritis, C3 glomerulonephritis, etc.) has a prompt repercussion on the interstitial “health” of the kidney, irrespective of the underlying pathophysiology causing the glomerular damage, long before a significant decrease in e GFR would be observed.
Minimal Change Disease Has a Different Urinary Profile of EGF Compared to Other Proteinuric Glomerulopathies: Implications for uEGF as Diagnostic Tool?
The diagnostic and prognostic limitations of a kidney biopsy in children with refractory or steroid-dependent nephrotic syndrome are well-known, since the often-encountered verdict of “Minimal Change Disease” does not evitate the clinical need for strong immunosuppressive treatment, nor does it guarantee less relapses or a better outcome at long term.
In our study, the profile of uEGF/uCreat was significantly different in children with biopsy-proven MCD compared to pediatric patients having any other glomerular diagnosis on kidney biopsy. These findings suggest that uEGF/uCreat could have diagnostic potentials to distinguish minimal change disease from any other glomerular diagnosis, thereby possibly diminishing the indication for a kidney biopsy.
It is reasonable to hypothesize that, compared to MCD, urinary EGF expression decreases more in FSGS, IgAN, and other glomerulopathies due to more pronounced glomerular damage and podocyte dropout. Although our data are perhaps not strong enough to suggest that urinary EGF could completely replace an initial biopsy, it is realistic that monitoring the profile of U EGF over time would help to minimize the need for repeat biopsies. Pairing U EGF/uCreat assessment with biopsy in case of steroid-resistant nephrotic syndrome would, for example, be useful to identify false-negative biopsies that miss the FSGS lesion.
Obviously, to confirm this hypothesis, a study on a larger scale is mandatory, with an expansion of the number of patients as well as with long-term follow-up of UEGF/uCreat in each patient.
The Role of ACE Inhibition: Emphasizing the Difference Between Minimal Change Disease and Other Glomerulopathies
Urinary EGF is a biomarker for tubulo-interstitial damage and, as such, a hallmark of kidney disease progression, including deterioration of any glomerular disease (17, 18, 23). Besides its direct glomerular effect, ACE inhibition is known to have a major influence on repair and regeneration of peritubular fibrosis secondary to glomerular injury (19, 30, 31).
So, not surprisingly, a significant beneficial effect on uEGF/uCreat was observed in our study population, as uEGF/uCreat was significantly higher in the patients receiving ACE inhibition, irrespective of the initial presentation or biopsy result.
Strikingly however, the rise in uEGF/uCreat was selectively more pronounced in patients with an obvious glomerulopathy on biopsy, compared to children with minimal change disease. Theoretically, this implicates that measuring uEGF/uCreat in nephrotic children before and during ACE inhibition, might help to predict the probability that minimal change disease is the underlying diagnosis on kidney biopsy.
This impressive difference for ACE inhibition reinforces the hypothesis that minimal change disease has a distinct underlying pathophysiology, with less impact of its glomerular damage on the renal interstitial compartment, and as a consequence less sensitivity to ACE inhibition.
Finally, also the observed link between uEGF/uCreat and the absence of genetic predisposition in our study seems to strengthen the idea that, in general, minimal change disease should be considered a separate entity rather than an early precursor of FSGS.
The overall beneficial effect of ACE inhibition in our study population, was also confirmed in the control group of pediatric Alport patients, which under ACE inhibition displayed no significant difference in uEGF/uCreat compared to the healthy controls. Of course, a limitation of this finding was the lack of comparison with Alport patients not receiving ACE inhibition. However, a recent paper on pediatric Alport patients by Li et al. (20) demonstrated that at any age, and irrespective of the presence of proteinuria of kidney function, children with Alport syndrome (not treated with ACE inhibition) display a significantly lower uCreat than healthy controls.
Our study has limitations. The relatively small sample size and the fact that every subject in the study was under immunosuppressive treatment at the time the urine samples were taken are considerable restrictions. Although calcineurin inhibitors are known to influence Ucr, the profile of immunosuppressive treatment in the subgroups of our study did not differ significantly. Also, individual response of uEGF/uCreat to ACE inhibition is lacking, as patients were either already on ACE inhibition at the start of the study, or not receiving ACE inhibition at all during the course of the study.
On the other hand, the fact that all treatment options were taken into account in our study provided extra information on their respective effect on uEGF/uCreat, and even shed a new light on the findings of Gipson et al. (19).
Upgrading the study to a larger scale by increasing the number of patients, would allow to statistically calculate the diagnostic value of uEGF/uCreat in predicting the presence of an underlying minimal change disease.
This study thus opens the road for further investigating the predictive role of uEGF/uCreat in pediatric glomerular diseases, to further unravel their underlying pathophysiology and the mode of action of immunosuppressive drugs, in a larger pediatric population with or without nephrotic range proteinuria.
In conclusion, our study provides several new insights:
Despite preserved kidney function, children with a proteinuric or nephrotic glomerular disease on kidney biopsy show a significantly lower uEGF/uCreat, indicative of early tubulo-interstitial damage, which appears partially reversible under ACE inhibition in a degree dependent of the underlying biopsy type.
Our data suggest that uEGF/uCreat might be a helpful, non-invasive tool to distinguish minimal change disease from other proteinuric glomerular diseases, by linking the percentage of decrease in uEGF/uCreat to its response on ACE inhibition and to the absence of genetic predisposition.
The beneficial effect of ACE inhibition on uEGF/uCreat differs significantly in children with minimal change disease compared to children with obvious signs of glomerular changes on kidney biopsy. This reinforces the hypothesis that a distinct underlying pathophysiology lies at the origin of minimal change disease.
Data Availability Statement
The original contributions presented in the study are included in the article/supplementary materials, further inquiries can be directed to the corresponding author/s.
Ethics Statement
The studies involving human participants were reviewed and approved by Ethical Committee of the Antwerp University Hospital. Written informed consent to participate in this study was provided by the participants' legal guardian/next of kin.
Author Contributions
DT designed the study and wrote the major part of manuscript. NL collected the clinical data into the database and helped writing the manuscript. KW performed all statistical analyses. KL was responsible for the analyses of U EGF by ELISA in the laboratory LEMP. All authors were actively involved in interpreting the results and contributed to the conclusions in the discussion.
Conflict of Interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher's Note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
Acknowledgments
We thank R. Hellemans (UZA, Edegem, Belgium) and A. De Guchtenaere (AZ Damiaan, Oostende, Belgium) for their willingness to carefully read the manuscript and to provide constructive comments. We are grateful to K. Van Hoeck (UZA, Edegem, Belgium) for informing part of his patients about the possibility to participate in our study. We also thank K. Dahan, V. Benoit, and P. Hilbert (IPG), Gosselies, Belgium for performing the genetic screening. Finally, we extend our gratitude to all the patients who participated in this study.
Abbreviations
MCD, Minimal change disease; FSGS, Focal segmental glomerulosclerosis; EGF, Epidermal growth factor; uEGF, Urinary epidermal growth factor; eGFR, Estimated glomerular filtration rate; ACE, Angiotensin converting enzyme; CNI, Calcineurin inhibitor; MMF, Mycophenolate mofetil; IgAN, IgA nephropathy; GN, Glomerulonephritis; uCreat, Urinary creatinine.
References
1. Gipson DS, Massengill SF, Yao L, Nagaraj S, Smoyer WE, Mahan JD, et al. Management of childhood onset nephrotic syndrome. Pediatrics. (2009) 124:747–57. doi: 10.1542/peds.2008-1559
PubMed Abstract | CrossRef Full Text | Google Scholar
2. MacHardy N, Miles PV, Massengill SF, Smoyer WE, Mahan JD, Greenbaum L, et al. Management patterns of childhood-onset nephrotic syndrome. Pediatr Nephrol. (2009) 24:2193–201. doi: 10.1007/s00467-009-1282-y
PubMed Abstract | CrossRef Full Text | Google Scholar
3. The primary nephrotic syndrome in children. Identification of patients with minimal change nephrotic syndrome from initial response to prednisone. A report of the International Study of Kidney Disease in Children. J Pediatr. (1981) 98:561–4. doi: 10.1016/S0022-3476(81)80760-3
PubMed Abstract | CrossRef Full Text | Google Scholar
4. Nammalwar BR, Vijayakumar M, Prahlad N. Experience of renal biopsy in children with nephrotic syndrome. Pediatr Nephrol. (2006) 21:286–8. doi: 10.1007/s00467-005-2084-5
PubMed Abstract | CrossRef Full Text | Google Scholar
5. Samuel S, Morgan CJ, Bitzan M, Mammen C, Dart AB, Manns BJ, et al. Substantial practice variation exists in the management of childhood nephrotic syndrome. Pediatr Nephrol. (2013) 28:2289–98. doi: 10.1007/s00467-013-2546-0
PubMed Abstract | CrossRef Full Text | Google Scholar
6. Kyrieleis HA, Löwik MM, Pronk I, Cruysberg HRM, Kremer JAM, Oyen WJG, et al. Long-term outcome of biopsy-proven, frequently relapsing minimal-change nephrotic syndrome in children. Clin J Am Soc Nephrol. (2009) 4:1593–600. doi: 10.2215/CJN.05691108
PubMed Abstract | CrossRef Full Text | Google Scholar
7. Saleem MA, Kobayashi Y. Cell biology and genetics of minimal change disease. F1000Res. (2016) 5:1. doi: 10.12688/f1000research.7300.1
PubMed Abstract | CrossRef Full Text | Google Scholar
8. Suzuki T, Kohatsu K, Han W, Watanabe S, Yahagi K, Nakata M, et al. Morphological features of minimal change disease and focal segmental glomerulosclerosis using repeat biopsy and parietal epithelial cell marker. Kidney Dis. (2020) 6:119–24. doi: 10.1159/000505125
PubMed Abstract | CrossRef Full Text | Google Scholar
9. Lovric S, Ashraf S, Tan W, Hildebrandt F. Genetic testing in steroid-resistant nephrotic syndrome: when and how? Nephrol Dial Transplant. (2016) 31:1802–13. doi: 10.1093/ndt/gfv355
PubMed Abstract | CrossRef Full Text | Google Scholar
10. Troyanov S, Wall CA, Miller JA, Scholey JW, Cattran DC, Group TGR. Focal and segmental glomerulosclerosis: definition and relevance of a partial remission. J Am Soc Nephrol. (2005) 16:1061–8. doi: 10.1681/ASN.2004070593
PubMed Abstract | CrossRef Full Text | Google Scholar
11. Pérez V, López D, Boixadera E, Ibernón M, Espinal A, Bonet J, et al. Comparative differential proteomic analysis of minimal change disease and focal segmental glomerulosclerosis. BMC Nephrol. (2017) 18:49. doi: 10.1186/s12882-017-0452-6
PubMed Abstract | CrossRef Full Text | Google Scholar
12. Kidney Disease Improving Global Outcomes (KDIGO) Glomerulonephritis Work Group. KDIGO clinical practice guideline for glomerulonephritis. Kidney Int. (2012) 2021(Suppl.2):139–274. doi: 10.1038/kisup.2012.9
CrossRef Full Text | Google Scholar
13. Coppo R, Gianoglio B, Porcellini MG, Maringhini S. Frequency of renal diseases and clinical indications for renal biopsy in children (report of the Italian National Registry of Renal Biopsies in Children). Group of Renal Immunopathology of the Italian Society of Pediatric Nephrology and Group of Renal Immunopathology of the Italian Society of Nephrology. Nephrol Dial Transplant. (1998) 13:293–7. doi: 10.1093/ndt/13.2.291
PubMed Abstract | CrossRef Full Text | Google Scholar
14. Hama T, Nakanishi K, Shima Y, Mukaiyama H, Togawa H, Tanaka R, et al. Renal biopsy criterion in children with asymptomatic constant isolated proteinuria. Nephrol Dial Transplant. (2012) 27:3186–90. doi: 10.1093/ndt/gfr750
PubMed Abstract | CrossRef Full Text | Google Scholar
15. Maixnerova D, Reily C, Bian Q, Neprasova M, Novak J, Tesar V. Markers for the progression of IgA nephropathy. J Nephrol. (2016) 29:535–41. doi: 10.1007/s40620-016-0299-0
PubMed Abstract | CrossRef Full Text | Google Scholar
16. Mattila AL, Viinikka L, Saario I, Perheentupa J. Human epidermal growth factor: renal production and absence from plasma. Regul Pept. (1988) 23:89–93. doi: 10.1016/0167-0115(88)90424-7
PubMed Abstract | CrossRef Full Text | Google Scholar
17. Ju W, Nair V, Smith S, Zhu L, Shedden K, Song PXK, et al. Tissue transcriptome-driven identification of epidermal growth factor as a chronic kidney disease biomarker. Sci Transl Med. (2015) 7:316ra193. doi: 10.1126/scitranslmed.aac7071
PubMed Abstract | CrossRef Full Text | Google Scholar
18. Humes HD, Cieslinski DA, Coimbra TM, Messana JM, Galvao C. Epidermal growth factor enhances renal tubule cell regeneration and repair and accelerates the recovery of renal function in postischemic acute renal failure. J Clin Invest. (1989) 84:1757–61. doi: 10.1172/JCI114359
PubMed Abstract | CrossRef Full Text | Google Scholar
19. Gipson DS, Trachtman H, Waldo A, Gibson KL, Eddy S, Dell KM, et al. Urinary epidermal growth factor as a marker of disease progression in children with nephrotic syndrome. Kidney Int Rep. (2020) 5:414–25. doi: 10.1016/j.ekir.2019.11.018
PubMed Abstract | CrossRef Full Text | Google Scholar
20. Li B, Zhang Y, Wang F, Nair V, Ding F, Xiao H, et al. Urinary epidermal growth factor as a prognostic marker for the progression of Alport syndrome in children. Pediatr Nephrol. (2018) 33:1731–9. doi: 10.1007/s00467-018-3988-1
PubMed Abstract | CrossRef Full Text | Google Scholar
21. Mejia-Vilet JM, Shapiro JP, Zhang XL, Cruz C, Zimmerman G, Méndez-Pérez RA, et al. Association Between Urinary Epidermal Growth Factor and Renal Prognosis in Lupus Nephritis. Arthritis Rheumatol. (2021) 73:244–54. doi: 10.1002/art.41507
PubMed Abstract | CrossRef Full Text | Google Scholar
22. Mariani LH, Martini S, Barisoni L, Canetta PA, Troost JP, Hodgin JB, et al. Interstitial fibrosis scored on whole-slide digital imaging of kidney biopsies is a predictor of outcome in proteinuric glomerulopathies. Nephrol Dial Transplant. (2018) 33:310–8. doi: 10.1093/ndt/gfw443
PubMed Abstract | CrossRef Full Text | Google Scholar
23. Betz BB, Jenks SJ, Cronshaw AD, Lamont DJ, Cairns C, Manning JR, et al. Urinary peptidomics in a rodent model of diabetic nephropathy highlights epidermal growth factor as a biomarker for renal deterioration in patients with type 2 diabetes. Kidney Int. (2016) 89:1125–35. doi: 10.1016/j.kint.2016.01.015
PubMed Abstract | CrossRef Full Text | Google Scholar
24. Stangou M, Alexopoulos E, Papagianni A, Pantzaki A, Bantis C, Dovas S, et al. Urinary levels of epidermal growth factor, interleukin-6 and monocyte chemoattractant protein-1 may act as predictor markers of renal function outcome in immunoglobulin A nephropathy. Nephrology. (2009) 14:613–20. doi: 10.1111/j.1440-1797.2008.01051.x
PubMed Abstract | CrossRef Full Text | Google Scholar
25. Tsau Y, Chen C. Urinary epidermal growth factor excretion in children with chronic renal failure. Am J Nephrol. (1999) 19:400–4. doi: 10.1159/000013485
PubMed Abstract | CrossRef Full Text | Google Scholar
26. Meybosch S, De Monie A, Anné C, Bruyndonckx L, Jürgens A, De Winter BY, et al. Epidermal growth factor and its influencing variables in healthy children and adults. PLoS ONE. (2019) 14:e0211212. doi: 10.1371/journal.pone.0211212
PubMed Abstract | CrossRef Full Text | Google Scholar
27. Kashtan CE, Ding J, Gregory M, Gross O, Heidet L, Knebelmann B, et al. Clinical practice recommendations for the treatment of Alport syndrome: a statement of the Alport Syndrome Research Collaborative. Pediatr Nephrol. (2013) 28:5–11. doi: 10.1007/s00467-012-2138-4
PubMed Abstract | CrossRef Full Text | Google Scholar
28. Ledeganck KJ, Anné C, De Monie A, Meybosch S, Verpooten GA, Vinckx M, et al. Longitudinal study of the role of epidermal growth factor on the fractional excretion of magnesium in children: effect of calcineurin inhibitors. Nutrients. (2018) 10:60677. doi: 10.3390/nu10060677
PubMed Abstract | CrossRef Full Text | Google Scholar
29. Kim JS, Bellew CA, Silverstein DM, Aviles DH, Boineau FG, Vehaskari VM. High incidence of initial and late steroid resistance in childhood nephrotic syndrome. Kidney Int. (2005) 68:1275–81. doi: 10.1111/j.1523-1755.2005.00524.x
PubMed Abstract | CrossRef Full Text | Google Scholar
30. van den Belt SM, Heerspink HJL, Gracchi V, de Zeeuw D, Wühl E, Schaefer F, et al. Early proteinuria lowering by angiotensin-converting enzyme inhibition predicts renal survival in children with CKD. J Am Soc Nephrol. (2018) 29:2225–33. doi: 10.1681/ASN.2018010036
PubMed Abstract | CrossRef Full Text | Google Scholar
31. Ruggenenti P, Mosconi L, Vendramin G, Moriggi M, Remuzzi A, Sangalli F, et al. ACE inhibition improves glomerular size selectivity in patients with idiopathic membranous nephropathy and persistent nephrotic syndrome. Am J Kidney Dis. (2000) 35:381–91. doi: 10.1016/S0272-6386(00)70190-9
PubMed Abstract | CrossRef Full Text | Google Scholar
Keywords: minimal change disease, urinary epidermal growth factor, ACE-inhibition, diagnostic biomarker, genetic predisposition
Citation: Lodeweyckx N, Wouters K, Ledeganck KJ and Trouet D (2021) Biopsy or Biomarker? Children With Minimal Change Disease Have a Distinct Profile of Urinary Epidermal Growth Factor. Front. Pediatr. 9:727954. doi: 10.3389/fped.2021.727954
Received: 20 June 2021; Accepted: 05 October 2021;
Published: 25 November 2021.
Edited by:
Constantinos J. Stefanidis, “Mitera” Children's Hospital, Greece
Reviewed by:
Hamidreza Badeli, Gilan University of Medical Sciences, Iran
Ali Asghar Anwar Lanewala, Sindh Institute of Urology and Transplantation, Pakistan
Copyright © 2021 Lodeweyckx, Wouters, Ledeganck and Trouet. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Dominique Trouet, dominique.trouet@uza.be
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
Frontiers' impact
Articles published with Frontiers have received 12 million total citations
Your research is the real superpower - learn how we maximise its impact through our leading community journals
Explore our impact metrics
Download article
Download PDF
ReadCube
EPUB
XML
Share on
Export citation
EndNote
Reference Manager
Simple Text file
BibTex
2,8K
Total views
775
Downloads
2
Citations
Citation numbers are available from Dimensions
View article impact
View altmetric score
Share on
Edited by
Constantinos J Stefanidis "Mitera" Children's Hospital, Greece
Reviewed by
Ali Asghar Anwar Lanewala Sindh Institute of Urology and Transplantation, Pakistan Hamidreza Badeli Gilan University of Medical Sciences, Iran
Table of contents
Abstract
Introduction
Patients and Methods
Results
Discussion
Data Availability Statement
Ethics Statement
Author Contributions
Conflict of Interest
Publisher's Note
Acknowledgments
Abbreviations
References
Export citation
EndNote
Reference Manager
Simple Text file
BibTex
Check for updates
People also looked at
Use of Eculizumab in Pediatric Patients With Transplant Associated Thrombotic Microangiopathy
Laura Gomez-Ganda, Maria Isabel Benitez-Carabante, Aurora Fernandez-Polo, Marina Muñoz-Lopez, Berta Renedo-Miro, Gema Ariceta and Cristina Diaz De Heredia
Recent Advances in Kidney Bioengineering
Nina Cintron Pregosin, Robert Bronstein and Sandeep K. Mallipattu
The Prevalence of Insufficient Sleep and Bedtime Delay Among Kindergarten Children Aged 3 to 6 Years in a Rural Area of Shanghai: A Cross-Sectional Study
Xi Chen, Yan Qiang, Xia Liu, Qing Yang, Qingqing Zhu, Bin Li and Ruiping Wang
Catheter-Obtained Urine Culture Contamination Among Young Infants: A Prospective Cohort Study
Hilla Bahat, Revital Apelman Cipele, Tali Maymon, Ilan Youngster and Michael Goldman
Utilization Pattern for Eculizumab Among Children With Hemolytic Uremic Syndrome
Saritha Ranabothu, Clare C. Brown, Richard Blaszak, Rachel Millner, Kristen Rice Moore and Parthak Prodhan
Guidelines
Author guidelines
Services for authors
Policies and publication ethics
Editor guidelines
Fee policy
Explore
Articles
Research Topics
Journals
How we publish
Outreach
Frontiers Forum
Frontiers Policy Labs
Frontiers for Young Minds
Frontiers Planet Prize
Connect
Help center
Emails and alerts
Contact us
Submit
Career opportunities
Follow us
© 2025 Frontiers Media S.A. All rights reserved
Privacy policy|Terms and conditions
Download article
Download
Download PDF
ReadCube
EPUB
XML
X (6)
Mendeley (11)
See more details |
9972 | http://verso.mat.uam.es/~dragan.vukotic/grado/var_com_I-18-19/varcomI_2018-19_apuntes_sucesiones-series.pdf | Variable Compleja I (CURSO 2018-19, Universidad Autónoma de Madrid) Apuntes breves y ejercicios resueltos Las convergencias puntual y uniforme de sucesiones y series de funciones Definición. Decimos que la sucesión de funciones (fn) converge uniformemente a la función f en en un con-junto A ⊂C (y escribimos fn ⇒A f ) si ∀ε > 0 ∃N ∈N ∀n ≥N ∀z ∈A |fn(z)−f (z)| < ε.
Es irrelevante si ponemos < ε ó ≤ε en la definición arriba. La formulación con ≤ε es equivalente a ∀ε > 0 ∃N ∈N ∀n ≥N sup z∈A |fn(z)−f (z)| ≤ε.
Es decir, fn ⇒f en A si y sólo si supz∈A |fn(z)−f (z)| →0 cuando n →∞.
Ejercicio 1 Demuestre que la sucesión de funciones complejas fn(z) = zn: (a) converge a cero si |z| < 1; (b) converge a 0 uniformemente en cualquier subconjunto compacto del disco unidad D = {z ∈C : |z| < 1}; (c) no converge uniformemente en D.
(d) diverge si |z| ≥1 y z ̸= 1.
SOLUCIÓN. (a) Sabemos de Cálculo I que si q ∈R y |q| < 1 entonces limn→∞qn = 0. Por tanto, si |z| < 1, tomando como q = |z|, vemos que |zn| = |z|n →0, lo cual significa que limn→∞zn = 0.
(b) Sea K un subconjunto compacto de D (notación frecuente: K ⋐D). Puesto que K es cerrado y acotado, es fácil ver (y lo hemos justificado con detalle en clase) que existe un número R, 0 < R < 1, tal que para todo z ∈K se cumple |z| ≤R. Si z ∈K entonces |zn| = |z|n ≤Rn así que sup z∈K |zn −0| ≤Rn →0, n →∞.
Por lo tanto, zn ⇒0 en K .
(c) La sucesión fn(z) = zn →0 para cada z ∈D así que sólo tenemos que examinar si supz∈D |fn(z)| →0 o no cuando n →∞. Resulta que sup z∈D |fn(z)| = sup z∈D |z|n ≥ ( 1−1 n )n .
Puesto que ( 1−1 n )n →1 e > 0 cuando n →∞, se sigue que supz∈D |fn(z)| ̸→0, así que la convergencia no es uniforme en D.
(d) Sabemos de Cálculo I que limn→∞qn = +∞cuando q ∈R y q > 1. Si |z| > 1 entonces |zn| = |z|n →+∞, lo cual significa por definición que limn→∞zn = ∞(en el plano complejo extendido).
1 Queda por comprobar que zn diverge si |z| = 1 y z ̸= 1. Equivalentemente, veremos que |z| = 1 junto con la convergencia de zn a un valor implica que z = 1.
Sea z = eit. Por la fórmula de A. de Moivre, zn = eint = cosnt +i sennt. Si esta sucesión compleja converge, entonces también convergen las sucesiones reales xn = cosnt e yn = sennt. Veremos que esto sólo es posible cuando eit = 1. Sean x = limn→∞xn e y = limn→∞yn. Entonces también x = limn→∞x2n e y = limn→∞y2n. Eso significa que x2n = cos2nt = 2cos2 nt −1 = 2x2 n −1, y2n = sen2nt = 2sennt cosnt = 2xnyn .
Pasando al límite cuando n →∞, obtenemos que x = 2x2 −1, y = 2xy .
De la segunda igualdad se sigue que o bien y = 0 o bien x = 1/2. Como el valor x = 1/2 no satisface la condición x = 2x2 −1, se sigue que y = 0. Puesto que la identidad básica cos2 nt + sen2 nt = 1 implica x2 + y2 = 1, con-cluimos que x = 1 ó x = −1. De nuevo, x = −1 incumple x = 2x2 −1, así que x = 1. Por lo tanto, hemos llegado a la conclusión de que cosnt →1, sennt →0 cuando n →∞.
Usando esta última conclusión y tomando el límite cuando n →∞en la fórmula xn+1 = cos(nt + t) = cosnt cost −sennt sent obtenemos 1 = cost, mientras que tomando el límite cuando n →∞en la fórmula yn+1 = sen(nt + t) = sennt cost +cosnt sent obtenemos 0 = sent y, por tanto, z = eit = 1, que es lo que queríamos demostrar.
Definición. La serie compleja ∑∞ n=1 zn converge absolutamente si converge la serie asociada de números posi-tivos ∑∞ n=1 |zn|.
Al igual que en Cálculo I (para las series de números reales), si una serie converge absolutamente entonces converge y su suma cumple ¯ ¯∑∞ n=1 zn ¯ ¯ ≤∑∞ n=1 |zn|.
Definición. Diremos que la serie funcional ∑∞ n=1 fn(z) converge uniformemente en A ⊂C a la suma S(z) si las sumas parciales SN(z) = ∑N n=1 fn(z) convergen uniformemente a S(z) en A ⊂C.
Teorema. (Criterio de Weierstrass) Si para todo n ∈N (o para todo n ≥N0 para un N0 ∈N fijo) y para todo z ∈A se cumple |fn(z)| ≤Mn y la serie de números positivos ∑∞ n=1 Mn converge, entonces la serie funcional ∑∞ n=1 fn(z) converge uniformemente en A y absolutamente para todo z ∈A y su suma satisface la desigualdad ¯ ¯∑∞ n=1 fn(z) ¯ ¯ ≤∑∞ n=1 |fn(z)|.
Obsérvese que el criterio de Weierstrass sólo nos permite concluir que la serie converge pero no nos dice nada acerca del valor de la suma.
Ejercicio 2 ¿Para qué valores de z converge la serie ∑∞ n=0 zn? ¿Qué podemos afirmar acerca del tipo de conver-gencia para esos valores?
2 SOLUCIÓN. Si la serie converge para un valor de z, entonces el término general zn →0 cuando n →∞. Según el resultado del anterior ejercicio, para que esto ocurra se tiene que cumplir la condición |z| < 1. Veamos ahora que la serie converge en el disco unidad D = {z : |z| < 1}.
De hecho, veremos más: la convergencia es uniforme en cada subconjunto compacto K de D. Las sumas parciales de la serie son conocidas y convergen puntualmente: N ∑ n=0 zn = 1−zN+1 1−z → 1 1−z , N →∞.
Si fijamos un subconjunto compacto K ⋐D, existe un número r ∈(0,1) tal que para todo z ∈K se cumple |z| ≤r. Entonces, por la desigualdad triangular inversa, |1−z| ≥1−|z| ≥1−r > 0 y luego podemos estimar la diferencia de las sumas parciales y la suma de la serie: ¯ ¯ ¯ ¯ ¯ N ∑ n=0 zn − 1 1−z ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 1−zN+1 1−z − 1 1−z ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ zN+1 1−z ¯ ¯ ¯ ¯ ≤r N+1 |1−z| ≤r N+1 1−r .
La última cantidad tiende a cero cuando N →∞(independientemente de z ∈K ya que r es fijo); es decir, sup z∈K ¯ ¯ ¯ ¯ ¯ N ∑ n=0 zn − 1 1−z ¯ ¯ ¯ ¯ ¯ ≤r N+1 1−r →0, N →∞.
Por definición, esto significa que las sumas parciales ∑N n=0 zn convergen a 1 1−z uniformemente en K .
Ejercicio 3 ¿Para qué valores de z converge la serie ∑∞ n=0 ( 1+z 1−z )n?
SOLUCIÓN. Después del cambio de variable w = 1+ z 1−z , la serie se convierte en la serie geométrica de la variable w, que -después de los cálculos pertinentes- se puede volver a transformar en una función de z: ∞ ∑ n=0 (1+ z 1−z )n = ∞ ∑ n=0 wn = 1 1−w = 1 1−1+z 1−z = 1−z −2z = 1 2 −1 2z .
Puesto que la serie geométrica sólo converge cuando |w| < 1, nuestra serie será convergente sólo para los z que cumplan |1+ z| < |1−z|. Es decir, cuando |z −(−1)| < |z −1|.
Geométricamente, dicho conjunto representa el lugar geométrico de los puntos que están más cerca de −1 que de 1, que es el semiplano izquierdo abierto.
Esto mismo también se puede ver algebraicamente: el conjunto indicado es el conjunto de los puntos para los que |1+ z|2 < |1−z|2 o, equivalentemente, 1+|z|2 +2Rez < 1+|z|2 −2Rez .
Esto significa que Rez < 0, lo cual viene a decir que z está el semiplano izquierdo abierto.
3 Definición. La función compleja coseno se define en términos de la función exponencial como sigue: cosz = eiz +e−iz 2 como generalización de la función real coseno vista en los cursos de Cálculo.
Ejercicio 4 Demuestre que las series complejas ∞ ∑ n=1 e−n cos(nz), ∞ ∑ n=1 ne−nsen(nz) convergen absoluta y uniformemente en cada banda horizontal cerrada de la forma Ωε = {z ∈C: |Im z| ≤1−ε}, ε > 0.
SOLUCIÓN. Cuando z = x +i y ∈Ωε, tenemos |y| ≤1−ε y, por tanto, e±y ≤e1−ε. Luego (aplicando la definición de la función exponencial, la desigualdad triangular y esta última desigualdad para la exponencial) |e−n cosnz| = ¯ ¯ ¯ ¯e−n · einz +e−inz 2 ¯ ¯ ¯ ¯ = e−n 2 · ¯ ¯ ¯einxe−ny +e−inxeny¯ ¯ ¯ ≤ e−n 2 · ( e−ny +eny) ≤e−n 2 ·2·en(1−ε) = e−nε .
La serie numérica ∑∞ n=1 e−nε es sumable por ser una serie geométrica cuya razón es q = e−ε < 1. El criterio de Weierstrass implica que la serie ∑∞ n=1 e−n cosnz converge absoluta y uniformemente en cada Ωε.
Preparado por Dragan Vukotić, profesor de un grupo de la asignatura en 2014-15, 2015-16 y 2018-19 4 |
9973 | https://www.twoodo.com/tools/ev-to-j/ | EV to J - Electron-volt to Joule Converter
About Electron-volts and Joules
An electron-volt (eV) is a unit of energy equal to approximately 1.602176634 × 10⁻¹⁹ joules. It represents the amount of kinetic energy gained by a single electron accelerating from rest through an electric potential difference of one volt in vacuum.
A joule (J) is the derived unit of energy in the International System of Units (SI). It is equal to the energy transferred to an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one meter.
Conversion Table: eV to Joules
| Electron-volts (eV) | Joules (J) |
--- |
| 1 eV | 1.602176634 × 10⁻¹⁹ J |
| 1 keV | 1.602176634 × 10⁻¹⁶ J |
| 1 MeV | 1.602176634 × 10⁻¹³ J |
| 1 GeV | 1.602176634 × 10⁻¹⁰ J |
| 1 TeV | 1.602176634 × 10⁻⁷ J |
How to Convert Electron-volts to Joules
To convert electron-volts to joules, multiply the eV value by 1.602176634 × 10⁻¹⁹. This conversion factor represents the exact relationship between the two units.
Example Calculation:
Let's convert 5 eV to joules:
5 eV = 5 × (1.602176634 × 10⁻¹⁹) J
5 eV = 8.01088317 × 10⁻¹⁹ J
Common Applications
Electron-volts are commonly used in:
Atomic and nuclear physics
Particle physics
Semiconductor physics
Quantum mechanics |
9974 | https://www.merckmillipore.com/IN/en/product/Sodium-carbonate-decahydrate,MDA_CHEM-106391 | Merck | Israel | Life Science Products & Service Solutions
Skip to Content
Products
Cart 0
IL EN
Products
Products Applications Services Documents Support
Analytical ChemistryCell Culture & AnalysisChemistry & BiochemicalsClinical & DiagnosticsFiltration
Greener Alternative Products
Industrial MicrobiologyLab AutomationLabwareMaterials ScienceMolecular Biology & Functional GenomicsmAbs Development & ManufacturingmRNA Development & ManufacturingPharma & Biopharma ManufacturingProtein BiologyWater Purification
Analytical ChemistryCell Culture & AnalysisChemistry & SynthesisClinical & DiagnosticsEnvironmental & Cannabis TestingFood & Beverage Testing & ManufacturingGenomicsMaterials Science & EngineeringMicrobiological TestingmAbs Development & ManufacturingmRNA Development & ManufacturingPharma & Biopharma ManufacturingProtein BiologyResearch & Disease AreasWater Purification
Contract ManufacturingContract TestingCustom ProductsDigital Solutions for Life ScienceIVD Development & ManufacturingProduct ServicesSupport
Safety Data Sheets (SDS)
Certificates of Analysis (COA)
Certificates of Origin (COO)
Certificates of Quality (COQ)
Customer Support
Contact Us
Get Site Smart
FAQ
Quality & Regulatory
Calculators & Apps
Webinars
Login
Order Lookup
Quick Order
Cart 0
All In 1 Cart Premier brands partner with us to benefit you Explore Now
Online Custom Media & Chemical Requests With mConfig™ Configurator get feasibility checks & expert guidance Get Access
Unlock Your Yields Discover our new CHOZN® Elite cell line, a next generation CHO host to boost your titer up to 7-13 g/L! Explore Here
Be the Base Editor Boss PURedit® Cas9 Cytosine Base Editors are your streamlined gene editing solution Get Editing
Welcome to Merck
New Arrivals
Slide 1 of 4
Sigma-Aldrich 935042 AC1 Screen Printed Electrodes
Quick View
Sigma-Aldrich 934968 Platinum Microelectrode
Quick View
Sigma-Aldrich SML3846 Motixafortide trifluoroacetate
Quick View
Supelco P-179 Phosphatidylethanol 16:0/18:1 (PEth 16:0/18:1) solution
Quick View
Special Offers & Activities
View our list of seasonal offers and specific product promotions for various research areas.
Learn More
Need Support?
Call 08-9484222 / 08-9484100
Contact Customer Service
Shop Top Categories
Slide 1 of 12
Certified Reference Materials
Microbial Filtration
Solvents
Microbial Culture Media
Cell Separation Media for Research, Diagnostic & Pharmaceutical Applications
Millex® Syringe Filters
Glassware
Pathogen Screening Kits, Instruments & Disposables
Amicon® Centrifugals
Millipore® Filter Membranes
Milli-Q® Lab Water Purification Consumables
Greener Alternative Products
Sign In to Receive Customized Recommendations Based on Your Preferences
Best Sellers
Popular Products
New Arrivals
Slide 1 of 7
1 of 2
Sigma-Aldrich D2650 Dimethyl sulfoxide
Quick View
USP 1044651 Atovaquone
Quick View
Sigma-Aldrich T0440 3,3′,5,5′-Tetramethylbenzidine (TMB) Liquid Substrate System for ELISA
Quick View
Sigma-Aldrich M2933 MES hydrate
Quick View
SAFC 24366C EX-CELL® Advanced CHO Fed-batch Medium
Quick View
Sigma-Aldrich F7524 Fetal Bovine Serum
Quick View
Sigma-Aldrich T9531 D-(+)-Trehalose dihydrate
Quick View
Slide 1 of 10
1 of 3
Sigma-Aldrich D2650 Dimethyl sulfoxide
Quick View
Sigma-Aldrich 34860 Methanol
Quick View
Sigma-Aldrich 258148 Hydrochloric acid
Quick View
Sigma-Aldrich S9888 Sodium chloride
Quick View
Sigma-Aldrich D8418 Dimethyl sulfoxide
Quick View
Sigma-Aldrich 34851 Acetonitrile
Quick View
Sigma-Aldrich 179124 Acetone
Quick View
Sigma-Aldrich 34863 2-Propanol
Quick View
Sigma-Aldrich 459844 Ethyl alcohol, Pure
Quick View
Sigma-Aldrich 472301 Dimethyl sulfoxide
Quick View
Slide 1 of 4
1 of 1
Sigma-Aldrich 935042 AC1 Screen Printed Electrodes
Quick View
Sigma-Aldrich 934968 Platinum Microelectrode
Quick View
Sigma-Aldrich SML3846 Motixafortide trifluoroacetate
Quick View
Supelco P-179 Phosphatidylethanol 16:0/18:1 (PEth 16:0/18:1) solution
Quick View
Local Office Information
Meet the team.
Learn More
Virtual Seminars
We are committed to sharing our technical expertise and supporting your science through our Virtual Seminar Program.
Learn More
CRISPR
We are trusted worldwide for innovative solutions in CRISPR/Cas9 products and services.
Learn More
Get Site Smart
Discover our latest website features and enhance your experience with Merck.
Learn More
Discover Our Brands
Merck offers consultative services to help customers optimize their workflows and overcome complex challenges in areas of life science research, biomanufacturing, and analytical testing.
Millipore®
Preparation, Separation, Filtration & Monitoring Products
Sigma-Aldrich®
Lab & Production Materials
SAFC®
Pharma & Biopharma Raw Material Solutions
Supelco®
Analytical Products
Milli-Q®
Lab Water Solutions
BioReliance®
Pharma & Biopharma Manufacturing & Testing Services
Popular Searches
EthanolDimethyl sulfoxideMethanolAcetoneSodium hydroxideAcetic acidHydrochloric acidEthyl acetateSodium chlorideAcetonitrileTolueneDichloromethaneSulfuric acidWaterHexane
Top
Sign In To Continue
To continue reading please sign in or create an account.
Sign InDon't Have An Account?Register
SupportCustomer SupportContact UsFAQSafety Data Sheets (SDS)Certificates (COA/COO)Quality & RegulatoryCalculators & AppsWebinars
OrdersQuick OrderCustom ProductseCommerce Solutions
CompanyAbout UsResponsibilityEventsPress ReleasesProgramsCareersOffices
Social Media
Merck
Research. Development. Production.
We are a leading supplier to the global Life Science industry with solutions and services for research, biotechnology development and production, and pharmaceutical drug therapy development and production.
Sigma-Aldrich® SolutionsBioReliance® SolutionsMillipore® SolutionsSAFC® SolutionsMilli-Q® SolutionsSupelco® Solutions
© 2025 Merck KGaA, Darmstadt, Germany and/or its affiliates. All Rights Reserved, including Text and Data Mining for AI training and similar technologies.
Reproduction of any materials from the site is strictly forbidden without permission.
Site Use Terms|Privacy Policy|General Terms and Conditions of Sale|Copyright Consent|Site Map|Cookie Settings
English - EN
Learn More
Cookie Disclaimer
We use cookies in our website to give you the most relevant experience. By clicking or navigating the site, you are accepting the use of all our cookies according to our Cookie Policy / Privacy Statement. You are free to decide in the Cookie Settings which categories you would like to allow. Please note that depending on what you select, the full functionality of the website may no longer be available. You may review and change your choices at any time. Further information can be found in our Privacy Policy
Cookie Settings Reject Unnecessary Cookies Accept Cookies
We recognize your browser's Opt-Out Preference Signal.
To honor your signal, Targeting & Social Media Cookies from our websites have been disabled. But you may choose to enable them by adjusting your preferences below.
Cookie Preferences
Necessary
Functional
Targeting
Social Media Cookies
Cookie Preferences
When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about how we use and share your data, please follow this link.
Privacy Statement
Necessary
Always Active
These cookies are necessary to provide and safeguard the functionalities of our services and for the website to operate. We set these cookies as required and in response to your actions, and they depend on how you choose to access and interact with our website as well as specific service(s) requested.
Functional
[x] Functional
These cookies enable the provision of advanced functionalities and are used for personalizing your browsing experience on our websites. We set these cookies in response to your actions, and they depend on your interactions with our website as well as any specific requests or preference(s) settings you have elected.
Targeting
[x] Targeting
These cookies may be set to learn more about your interests through your interaction with our website and are analyzed to provide better website performance and experiences, but also to enhance our advertising activities. We may use service providers to support our usage of these cookies, which work by uniquely identifying your browser and/or device. By integrating these cookies, we aim to learn more about your interests and surfing behavior to place our advertising in a more effective and relevant manner.
Social Media Cookies
[x] Social Media Cookies
These cookies may be set to learn more about your interests through your interactions with our website, as well as your interactions with third parties’ websites our external marketing agencies are partnered with to show you relevant ads throughout your overall internet browsing experience. These cookies work by uniquely identifying your browser(s) and/or device(s) and sharing your browsing activity with third party advertisement companies. By integrating these cookies, we aim to learn more about our customers interests and overall internet browsing behavior to be able to place our advertising in a more effective and relevant manner.
Cookie List
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Clear
[x] checkbox label label
Apply Cancel
Confirm My Choices
Disable Unnecessary Cookies Enable All
Feedback |
9975 | https://en.wikipedia.org/wiki/Sheehan%27s_syndrome | Contents
Sheehan's syndrome
Sheehan's syndrome
Other names | Simmond's syndrome, postpartum hypopituitarism, postpartum pituitary gland necrosis
Anatomy of normal pituitary gland and surrounding structures
Empty sella turcica on MRI as seen in severe cases of Sheehan's syndrome
Specialty | Endocrinology,obstetrics and gynaecology
Sheehan's syndrome, also known as postpartum pituitary gland necrosis, occurs when the pituitary gland is damaged due to significant blood loss and hypovolemic shock (ischemic necrosis) or stroke, originally described during or after childbirth leading to decreased functioning of the pituitary gland (hypopituitarism). Classically, in the milder partial form, the mother is unable to breastfeed her baby, due to failure of the pituitary to secrete the hormone prolactin, and also has no more periods, because FSH (Follicle Stimulating Hormone) and LH (Luteinising Hormone) are not secreted. Although postmenopausal, the mother with this milder form of Sheehan's syndrome does not experience hot flushes, because the pituitary fails to secrete FSH (high levels of FSH, secreted by the pituitary in healthy postmenopausal women is an attempt to trigger ovulation, and these high levels of FSH cause hot the flushes). The failure to breastfeed and amenorrhea no more periods, were seen as the syndrome (a collection of symptoms), but we now view Sheehan's as the pituitary failing to secrete 1-5 of the 9 hormones that it normally produces (the anterior (front) lobe of the pituitary produces FSH, LH, prolactin, ACTH (Adreno-cortico-trophic hormone),TSH (Thyroid Stimulating Hormone) and GH (Growth Hormone); the posterior (the lobe at the back) pituitary produces ADH (Anti-Diuretic Hormone) and Oxytocin, i.e. the pituitary is involved in the regulation of many hormones. It is very important to recognise Sheehan' stroke as, the ACTH deficiency Sheehan's in the presence of the stress of a bacterial infection, such as a urine infection, will result in death of the mother from Addisonian crisis. This gland is located on the under-surface of the brain, the shape of a cherry and the size of a chickpea and sits in a pit or depression of the sphenoid bone known as the sella turcica (the Turk's saddle). The pituitary gland works in conjunction with the hypothalamus, and other endocrine organs to modulate numerous bodily functions including growth, metabolism, menstruation, lactation, and even the "fight-or-flight" response. These endocrine organs, (like the thyroid gland in the neck, or adrenals on the upper pole of the kidneys), release hormones in very specific pathways, known as hormonal axes. For example, the release of a hormone in the hypothalamus will target the pituitary to trigger the release thyroid stimulating hormone (TSH), and the pituitary's released hormone (TSH) will target the next organ in the pathway i.e. the thyroid to release thyroxin. Hence, damage to the pituitary gland can have downstream effects on any of the aforementioned bodily functions.
Signs and symptoms
The various signs and symptoms in Sheehan's syndrome are caused by damage to the pituitary, thereby causing a decrease in one or more of the hormones it normally secretes. Since the pituitary controls many glands in the endocrine system, partial or complete loss of a variety of functions may result. Many of the signs and symptoms of Sheehan's are considered "nonspecific" in the medical community; in other words these signs and symptoms are seen in a number of different disease processes, and are not specific to a singular disease or syndrome.
In some cases, a woman with Sheehan syndrome may be relatively asymptomatic initially; therefore, the diagnosis would not be made until years later when features of hypopituitarism become evident. In rare instances this syndrome can present acutely with unstable vital signs, dangerously low blood glucose levels, heart failure, or even psychosis. Hypopituitarism can lead to an interruption in any of the following hormone pathways: thyroid disorder (secondary hypothyroidism), adrenal gland (adrenal insufficiency due to glucocorticoid deficiency), sex hormone (gonadotropin deficiency), prolactin (a hormone responsible for lactation), growth hormone, or rarely anti-diuretic hormone deficiency (central diabetes insipidus). Since damage to the pituitary can cause a deficiency in more than one of these hormone pathways simultaneously, it is possible to have a mix of any of the signs or symptoms listed below.
Sheehan's syndrome's most common initial symptoms are difficulties with or total absence of lactation (agalactorrhea). Another common sign is infrequent menstrual cycles (oligomenorrhea) or absent menstrual cycles (amenorrhea) following delivery. In addition to menstrual irregularities other signs of sex hormone deficiency are hot flashes, decreased libido, and breast involution. Symptoms and signs of thyroid disorder are tiredness, intolerance to cold, constipation, weight gain, hair loss, slowed thinking, as well as a slowed heart rate and low blood pressure. Adrenal gland malfunction can present acutely or chronically. In a more chronic case, it is similar to Addison's disease with symptoms including fatigue, weight loss, hypoglycemia (low blood sugar levels), low hemoglobin levels (anemia) and hyponatremia (low sodium levels) that develop over several months or years. Acute adrenal insufficiency is referred to as an adrenal crisis, which can be life-threatening, and occurs very shortly after the inciting event i.e. significant blood loss post-partum in the context of Sheehan's syndrome. Adrenal crisis signs and symptoms include hypoglycemia, hypotension, weakness, fatigue, and seizures from severe hyponatremia.
Growth hormone deficiency is one of the most common hormone deficiencies of hypopituitarism seen in Sheehan's syndrome. Low levels of growth hormone may present with low energy, body aches, or subtle wrinkling of the skin around the eyes or mouth. The symptoms of anti-diuretic hormone deficiency are increased thirst, excessive urination, headache, and fatigue. Hematological changes might be seen as well such as anemia or low platelets (thrombocytopenia). Hyponatremia is seen in many cases of Sheehan's syndrome because it can result from multiple etiologies. Drops in thyroid hormones and glucocorticoid/adrenal hormones can indirectly lead to hyponatremia through water retention, while blood loss can trigger hyponatremia through ADH secretion. The development of Syndrome of Inappropriate Anti-Diuretic Hormone in patients with Sheehan's syndrome has been documented in the literature, although the mechanism is not well understood.
Causes
As stated, Sheehan's syndrome is caused by damage to the pituitary, thereby causing a decrease in one or more of the hormones it normally secretes. Sheehan's syndrome typically occurs because of excessive blood loss after delivery (post-partum hemorrhage), although there are several risk factors that may contribute to its development. This syndrome does not appear to be exclusively linked to childbirth, as Sheehan's syndrome has been reported in pregnant patients that experienced massive hemorrhage from non-obstetrical causes. The pituitary gland grows and has a higher metabolic demand during pregnancy because the pituitary needs to rev up the production of certain hormones associated with pregnancy. This higher metabolic demand, in turn, leads to higher demand for blood flow.
Thus, if the body enters a state of shock from excessive blood loss in post-partum delivery, the pituitary gland is more susceptible to injury. Although the vast majority of cases of Sheehan's syndrome occur in the setting of massive blood loss, cases have been documented of acute Sheehan's syndrome occurring with blood loss volumes that are not considered "massive". Some possible predisposing factors to Sheehan's syndrome may include: disseminated blood coagulation (DIC), hypotension, small sella turcica size, and blood clots from a pre-existing hypercoagulable disorder. Atony of the uterus is a leading cause of post-partum hemorrhage, therefore uterine atony could induce Sheehan's syndrome.
Pathophysiology
This syndrome seems to arise when certain factors compound each other to cause pituitary injury. The physiologic enlargement of the pituitary gland in conjunction with an interference in its blood supply, such as episiotomy progressing to anal tearing, ultimately result in pituitary ischemia and necrosis. One cause of pituitary growth associated with the risk of Sheehan's syndrome is the hyperplasia of lactotrophs which produce prolactin, the hormone responsible for milk production. Other hormone-secreting cells of the pituitary undergo rapid growth in pregnant women as well, which contribute to the gland's enlargement.
The anterior pituitary is supplied by a low pressure portal venous system. The anterior pituitary is more commonly affected in Sheehan's syndrome because of the structure of the portal venous system. Posterior pituitary involvement leading to central diabetes insipidus is much rarer, and typically reflects more extensive damage to the organ and more severe disease. It has been suggested that the arrangement of the pituitary's blood supply contribute to its susceptibility for injury. "The highly vascularized pituitary tissue involves one of the most rapid blood flow in the human body and probably, therefore, has a tendency to infarction because even small degrees of change in the pituitary intravascular pressure cause an arrest of blood flow". Ischemia may occur as a result of vasospasm from shock, hypotension, thrombosis, or direct vascular compression of the hypophyseal artery from the enlarged pituitary gland itself. The presence of disseminated intravascular coagulation (i.e., in amniotic fluid embolism or HELLP syndrome) also appears to be a factor in its development.
Diagnosis
Typically an important clue that leads to a diagnosis of Sheehan's syndrome is identifying a deficiency in one or more of the hormones produced directly, or indirectly, by the pituitary gland. The extent of hormone deficiency, and which hormones are affected depends on the extent of the damage to the pituitary. Hormonal assays measure the levels of these hormones which include but are not limited to T4, TSH, estrogen, gonadotropin, cortisol, and ACTH. It might be difficult to detect damage to these hormone pathways if hormone levels are at the borderline of the abnormal range. In this case, stimulation tests will be done to determine if the pituitary is responsive to hypothalamic hormones.
MRI is useful in diagnosing Sheehan's syndrome since it examines the structure of the pituitary and may identify any anatomical damage. MRI findings will vary based on how early or late in the disease process the test is being conducted. If an MRI is conducted early enough in the disease process the pituitary may appear larger than normal, and show changes that are consistent with damage from lack of blood supply. Later in the disease process of this syndrome the damage imposed on the pituitary gland will cause it to shrink, and leave a partially empty or totally empty sella turcica on MRI.
Treatment
The mainstay of treatment is hormone replacement therapy for the hormones that are missing. Treatment plans and dosages should be individualized by an endocrinologist. Glucocorticoids may be administered to address or prevent an adrenal crisis, a potential serious complication of Sheehan's syndrome. Hormone replacement is vital in reducing the morbidity and mortality of this syndrome.
Epidemiology
The exact prevalence of this syndrome is difficult to define because the incidence varies so much from country to country. Sheehan syndrome is more prevalent in developing countries than developed countries. In a study from the United Kingdom in 2001 only 1.4% of patients with hypopituitarism were diagnosed with Sheehan's syndrome. Just a few years earlier in 1996 the World Health Organization estimated that 3 million women were effected by Sheehan's syndrome. In a study of 1,034 symptomatic adults, Sheehan's syndrome was found to be the sixth-most frequent etiology of growth hormone deficiency, being responsible for 3.1% of cases (versus 53.9% due to a pituitary tumor). Additionally, it was found that the majority of women who experienced Sheehan syndrome gave birth at home rather than in a hospital.
History
The specific association with postpartum shock or hemorrhage was described in 1937 by the British pathologist Harold Leeming Sheehan (1900–1988). The initial distinction was made in the research article "Post-Partum Necrosis of the Anterior Pituitary". In his research, Dr. Sheehan reviewed (through autopsy) the effects of pituitary necrosis on 12 cases of patients that experienced postpartum necrosis. He observed cases where lesions and death occurred during or after pregnancy, as well as cases where death occurred in the late stage of necrosis (years later). This started the initial distinction of Sheehan's syndrome from Simmonds' disease (also known as hypopituitarism). Sheehan noted that significant feature of these patients' cases was hemorrhaging, which in his experience was most commonly caused by either: placenta Previa (low placenta), uterine rupture, cervical or uterine tears, post-partum atony, or retained placenta. Simmonds' disease, however, occurs in either sex due to causes unrelated to pregnancy.
However, in his 1939 publication, "Simmonds' Disease due to Post-partum Necrosis of the Anterior Pituitary", Sheehan displays post-partum necrosis as a cause of Simmonds' disease, thus establishing the relationship between the two conditions. According to Sheehan in 1939 approximately 41% of survivors of severe postpartum hemorrhage (PPH) and/or hypovolemic shock experienced severe or partial hypopituitarism.
Society and culture
In the developed world Sheehan's Syndrome is a rare complication of pregnancy; although this syndrome is more prevalent in developing countries it continues to effect women around the world. A retrospective study in Turkey found that the prevalence of Sheehan's syndrome was directly proportional to the amount of at-home deliveries each decade. This may be due to previously limited obstetric techniques present in a home environment. Blood loss associated with episiotomy and forceps exacerbating blood loss when the placenta separates from the wall of the uterus, particularly in mothers with low blood pressure, even in obstetric setting, namely hospital, caused a more subtle Sheehan's syndrome of Growth Hormone, Anti-Duretic Hormone, ACTH deficiency, which may be life threatening if missed; PubMed.gov Sheehan's in modern times:a nationwide retrospective study Iceland 2011, where every mother gives birth in hospital, with full obstetric care available.
Research
At present, the part that autoimmunity plays in the development of Sheehan's syndrome is uncertain. Several case reports have identified anti-pituitary antibodies in patients diagnosed with Sheehan's. Some patients also tested positive for anti-hypothalamus antibodies. Given that many patients that have developed Sheehan's syndrome do not have detectable levels of these antibodies, it is unclear whether these antibodies cause this syndrome or result from it.
References
External links
Classification | DICD-10:E23.0ICD-9-CM:253.2MeSH:D007018DiseasesDB:11998
External resources | MedlinePlus:001175eMedicine:med/1914
vtePituitary disease
Hyperpituitarism | AnteriorAcromegalyHyperprolactinaemiaPituitary ACTH hypersecretionPosteriorSIADHGeneralNelson's syndromeHypophysitis | Anterior | AcromegalyHyperprolactinaemiaPituitary ACTH hypersecretion | Posterior | SIADH | General | Nelson's syndromeHypophysitis
Anterior | AcromegalyHyperprolactinaemiaPituitary ACTH hypersecretion
Posterior | SIADH
General | Nelson's syndromeHypophysitis
Hypopituitarism | AnteriorKallmann syndromeGrowth hormone deficiencyIsolated growth hormone deficiencyHypoprolactinemiaACTH deficiency/Secondary adrenal insufficiencyGnRH insensitivityFSH insensitivityLH/hCG insensitivityPosteriorCentral diabetes insipidusGeneralEmpty sella syndromePituitary apoplexySheehan's syndromeLymphocytic hypophysitisPituitary adenoma | Anterior | Kallmann syndromeGrowth hormone deficiencyIsolated growth hormone deficiencyHypoprolactinemiaACTH deficiency/Secondary adrenal insufficiencyGnRH insensitivityFSH insensitivityLH/hCG insensitivity | Posterior | Central diabetes insipidus | General | Empty sella syndromePituitary apoplexySheehan's syndromeLymphocytic hypophysitisPituitary adenoma
Anterior | Kallmann syndromeGrowth hormone deficiencyIsolated growth hormone deficiencyHypoprolactinemiaACTH deficiency/Secondary adrenal insufficiencyGnRH insensitivityFSH insensitivityLH/hCG insensitivity
Posterior | Central diabetes insipidus
General | Empty sella syndromePituitary apoplexySheehan's syndromeLymphocytic hypophysitisPituitary adenoma
Anterior | AcromegalyHyperprolactinaemiaPituitary ACTH hypersecretion
Posterior | SIADH
General | Nelson's syndromeHypophysitis
Anterior | Kallmann syndromeGrowth hormone deficiencyIsolated growth hormone deficiencyHypoprolactinemiaACTH deficiency/Secondary adrenal insufficiencyGnRH insensitivityFSH insensitivityLH/hCG insensitivity
Posterior | Central diabetes insipidus
General | Empty sella syndromePituitary apoplexySheehan's syndromeLymphocytic hypophysitisPituitary adenoma |
9976 | https://stackoverflow.com/questions/2815083/efficient-data-structure-for-word-lookup-with-wildcards | Stack Overflow
About
For Teams
Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers
Advertising Reach devs & technologists worldwide about your product, service or employer brand
Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models
Labs The future of collective knowledge sharing
About the company Visit the blog
Collectives¢ on Stack Overflow
Find centralized, trusted content and collaborate around the technologies you use most.
Learn more about Collectives
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Efficient data structure for word lookup with wildcards
Ask Question
Asked
Modified 8 years, 3 months ago
Viewed 11k times
25
I need to match a series of user inputed words against a large dictionary of words (to ensure the entered value exists).
So if the user entered:
"orange" it should match an entry "orange' in the dictionary.
Now the catch is that the user can also enter a wildcard or series of wildcard characters like say
"or__ge" which would also match "orange"
The key requirements are:
this should be as fast as possible. use the smallest amount of memory to achieve it.
If the size of the word list was small I could use a string containing all the words and use regular expressions.
however given that the word list could contain potentially hundreds of thousands of enteries I'm assuming this wouldn't work.
So is some sort of 'tree' be the way to go for this...?
Any thoughts or suggestions on this would be totally appreciated!
Thanks in advance, Matt
performance
algorithm
data-structures
memory-management
Share
Improve this question
edited May 12, 2010 at 22:23
Norman Ramsey
203k6262 gold badges374374 silver badges541541 bronze badges
asked May 11, 2010 at 23:12
SwaySway
1,65744 gold badges1717 silver badges1919 bronze badges
4
1
I'm not sure, but I think a Suffix Tree could be what you're looking for - en.wikipedia.org/wiki/Suffix_tree
Rubys
– Rubys
2010-05-11 23:15:51 +00:00
Commented May 11, 2010 at 23:15
1
Do you have to support all grep style wildcards or just the ? (underscore _ in your case)?
Michael Dorgan
– Michael Dorgan
2010-05-11 23:25:33 +00:00
Commented May 11, 2010 at 23:25
Do the wildcards match only a single character or can they match a string of arbitrary length?
drawnonward
– drawnonward
2010-05-11 23:29:16 +00:00
Commented May 11, 2010 at 23:29
Just the underscore, each underscore would represent a single character.
Sway
– Sway
2010-05-11 23:54:58 +00:00
Commented May 11, 2010 at 23:54
Add a comment |
6 Answers 6
Reset to default
16
Put your word list in a DAWG (directed acyclic word graph) as described in Appel and Jacobsen's paper on the World's Fastest Scrabble Program (free copy at Columbia). For your search you will traverse this graph maintaining a set of pointers: on a letter, you make a deterministic transition to children with that letter; on a wildcard, you add all children to the set.
The efficiency will be roughly the same as Thompson's NFA interpretation for grep (they are the same algorithm). The DAWG structure is extremely space-efficient—far more so than just storing the words themselves. And it is easy to implement.
Worst-case cost will be the size of the alphabet (26?) raised to the power of the number of wildcards. But unless your query begins with N wildcards, a simple left-to-right search will work well in practice. I'd suggest forbidding a query to begin with too many wildcards, or else create multiple dawgs, e.g., dawg for mirror image, dawg for rotated left three characters, and so on.
Matching an arbitrary sequence of wildcards, e.g., ______ is always going to be expensive because there are combinatorially many solutions. The dawg will enumerate all solutions very quickly.
Share
Improve this answer
edited May 12, 2010 at 22:25
answered May 12, 2010 at 1:01
Norman RamseyNorman Ramsey
203k6262 gold badges374374 silver badges541541 bronze badges
4 Comments
Matthieu M.
Matthieu M.
Since I don't have access to the publications, I'm wondering one thing: do they build one DAWG for each different length or not ? I think it could considerably speed up the search, since in this case we know beforehand how many letters the word we seek has.
Norman Ramsey
Norman Ramsey
@Matthieu: Google will get you the paper, but I've also added a (possibly ephemeral) link. As for one DAWG per length, you can do this, but it's a time-space tradeoff. The DAWG will store a long word list very effectively with lots of sharing. With one DAWG per length you will lose that sharing. As for speedup it's an experimental question, and experiments may come out differently depending on the machine's cache.
user1502040
user1502040
@Norman Ramsey I've been working on a similar problem (more than 10 years later!), and two good solutions to this that I've found are to keep a bitset of all suffix lengths at each node, or to have a DAWG for each length but share nodes across different lengths. Both work well, but I ended up with the second solution (only 30% larger than a single DAWG, with my implementation).
user1502040
user1502040
@NormanRamsey And for some problems you can get a lot of pruning by, for each node, maintaining a bitset of all characters that appear in any suffix of that node.
4
I would first test the regex solution and see whether it is fast enough - you might be surprised! :-)
However if that wasn't good enough I would probably use a prefix tree for this.
The basic structure is a tree where:
The nodes at the top level are all the possible first letters (i.e. probably 26 nodes from a-z assuming you are using a full dictionary...).
The next level down contains all the possible second letters for each given first letter
And so on until you reach an "end of word" marker for each word
Testing whether a given string with wildcards is contained in your dictionary is then just a simple recursive algorithm where you either have a direct match for each character position, or in the case of the wildcard you check each of the possible branches.
In the worst case (all wildcards but only one word with the right number of letters right at the end of the dictionary), you would traverse the entire tree but this is still only O(n) in the size of the dictionary so no worse than a full regex scan. In most cases it would take very few operations to either find a match or confirm that no such match exists since large branches of the search tree are "pruned" with each successive letter.
Share
Improve this answer
answered May 12, 2010 at 0:01
mikeramikera
107k2828 gold badges264264 silver badges427427 bronze badges
Comments
3
No matter which algorithm you choose, you have a tradeoff between speed and memory consumption.
If you can afford ~ O(NL) memory (where N is the size of your dictionary and L is the average length of a word), you can try this very fast algorithm. For simplicity, will assume latin alphabet with 26 letters and MAX_LEN as the max length of word.
Create a 2D array of sets of integers, set table[MAX_LEN].
For each word in you dictionary, add the word index to the sets in the positions corresponding to each of the letters of the word. For example, if "orange" is the 12345-th word in the dictionary, you add 12345 to the sets corresponding to [o], [r], [a], [n], [g], [e].
Then, to retrieve words corresponding to "or..ge", you find the intersection of the sets at [o], [r], [g], [e].
Share
Improve this answer
answered May 12, 2010 at 0:36
Igor KrivokonIgor Krivokon
10.3k11 gold badge4040 silver badges4141 bronze badges
Comments
1
You can try a string-matrix:
0,1: A 1,5: APPLE 2,5: AXELS 3,5: EAGLE 4,5: HELLO 5,5: WORLD 6,6: ORANGE 7,8: LONGWORD 8,13:SUPERLONGWORD
Let's call this a ragged index-matrix, to spare some memory. Order it on length, and then on alphabetical order. To address a character I use the notation x,y:z: x is the index, y is the length of the entry, z is the position. The length of your string is f and g is the number of entries in the dictionary.
Create list m, which contains potential match indexes x.
Iterate on z from 0 to f.
Is it a wildcard and not the latest character of the search string?
Continue loop (all match).
Is m empty?
Search through all x from 0 to g for y that matches length. !!A!!
Does the z character matches with search string at that z? Save x in m.
Is m empty? Break loop (no match).
Is m not empty?
Search through all elements of m. !!B!!
Does not match with search? Remove from m.
Is m empty? Break loop (no match).
A wildcard will always pass the "Match with search string?". And m is equally ordered as the matrix.
!!A!!: Binary search on length of the search string. O(log n) !!B!!: Binary search on alphabetical ordering. O(log n)
The reason for using a string-matrix is that you already store the length of each string (because it makes it search faster), but it also gives you the length of each entry (assuming other constant fields), such that you can easily find the next entry in the matrix, for fast iterating. Ordering the matrix isn't a problem: since this has only be done once the dictionary updates, and not during search-time.
Share
Improve this answer
edited May 12, 2010 at 0:54
answered May 12, 2010 at 0:44
PindatjuhPindatjuh
10.6k11 gold badge4444 silver badges6969 bronze badges
Comments
0
If you are allowed to ignore case, which I assume, then make all the words in your dictionary and all the search terms the same case before anything else. Upper or lower case makes no difference. If you have some words that are case sensitive and others that are not, break the words into two groups and search each separately.
You are only matching words, so you can break the dictionary into an array of strings. Since you are only doing an exact match against a known length, break the word array into a separate array for each word length. So byLength is the array off all words with length 3. Each word array should be sorted.
Now you have an array of words and a word with potential wild cards to find. Depending on wether and where the wildcards are, there are a few approaches.
If the search term has no wild cards, then do a binary search in your sorted array. You could do a hash at this point, which would be faster but not much. If the vast majority of your search terms have no wildcards, then consider a hash table or an associative array keyed by hash.
If the search term has wildcards after some literal characters, then do a binary search in the sorted array to find an upper and lower bound, then do a linear search in that bound. If the wildcards are all trailing then finding a non empty range is sufficient.
If the search term starts with wild cards, then the sorted array is no help and you would need to do a linear search unless you keep a copy of the array sorted by backwards strings. If you make such an array, then choose it any time there are more trailing than leading literals. If you do not allow leading wildcards then there is no need.
If the search term both starts and ends with wildcards, then you are stuck with a linear search within the words with equal length.
So an array of arrays of strings. Each array of strings is sorted, and contains strings of equal length. Optionally duplicate the whole structure with the sorting based on backwards strings for the case of leading wildcards.
The overall space is one or two pointers per word, plus the words. You should be able to store all the words in a single buffer if your language permits. Of course, if your language does not permit, grep is probably faster anyway. For a million words, that is 4-16MB for the arrays and similar for the actual words.
For a search term with no wildcards, performance would be very good. With wildcards, there will occasionally be linear searches across large groups of words. With the breakdown by length and a single leading character, you should never need to search more than a few percent of the total dictionary even in the worst case. Comparing only whole words of known length will always be faster than generic string matching.
Share
Improve this answer
answered May 12, 2010 at 0:46
drawnonwarddrawnonward
53.7k1616 gold badges109109 silver badges112112 bronze badges
1 Comment
Pindatjuh
Pindatjuh
"If the search term both starts and ends with wildcards, then you are stuck with a linear search within the words with equal length." Check out my answer: I skip the wildcards only if it's not the latest in the search string (in case of a full wildcards only search, which is linear), which forces it to make use of the binary search, no matter if it's wildcarded.
0
Try to build a Generalized Suffix Tree if the dictionary will be matched by sequence of queries. There is linear time algorithm that can be used to build such tree (Ukkonen Suffix Tree Construction).
You can easily match (it's O(k), where k is the size of the query) each query by traversing from the root node, and use the wildcard character to match any character like typical pattern finding in suffix tree.
Share
Improve this answer
answered May 12, 2010 at 3:48
monnmonn
73711 gold badge55 silver badges1313 bronze badges
Comments
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
performance
algorithm
data-structures
memory-management
See similar questions with these tags.
The Overflow Blog
The history and future of software development (part 1)
Getting Backstage in front of a shifting dev experience
Featured on Meta
Spevacus has joined us as a Community Manager
Introducing a new proactive anti-spam measure
New and improved coding challenges
New comment UI experiment graduation
Policy: Generative AI (e.g., ChatGPT) is banned
Linked
Data Structure for representing patterns in strings
Efficient data structure to hold strings with wildcards
0 How to effectively use the Levenshtein algorithm for text auto-completion
1 Efficient ID lookup using wildcard strings as keys
Related
7 most efficient data structure for a read-only list of strings (about 100,000) with fast prefix search
0 Table data structure for quick searching with wildcards
3 A memory-efficient large array of words
3 python: fast dictionary word lookup with wildcards
Efficient Data Structure For Substring Search?
1 Memory and speed efficient search on Strings
2 Efficient data structure to hold strings with wildcards
Data structure for fast full text search
1 Look for a data structure to match words by letters
2 Best suited data structure for prefix matching search
Hot Network Questions
What is this chess h4 sac known as?
Is there a specific term to describe someone who is religious but does not necessarily believe everything that their religion teaches, and uses logic?
how do I remove a item from the applications menu
How can the problem of a warlock with two spell slots be solved?
Switch between math versions but without math versions
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man
How to sample curves more densely (by arc-length) when their trajectory is more volatile, and less so when the trajectory is more constant
Is it safe to route top layer traces under header pins, SMD IC?
Do sum of natural numbers and sum of their squares represent uniquely the summands?
What meal can come next?
Can a cleric gain the intended benefit from the Extra Spell feat?
Analog story - nuclear bombs used to neutralize global warming
How to design a circuit that outputs the binary position of the 3rd set bit from the right in an 8-bit input?
Can Monks use their Dex modifier to determine jump distance?
How to rsync a large file by comparing earlier versions on the sending end?
How do trees drop their leaves?
Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator
How to use cursed items without upsetting the player?
Can a GeoTIFF have 2 separate NoData values?
Where is the first repetition in the cumulative hierarchy up to elementary equivalence?
What can be said?
в ответе meaning in context
Storing a session token in localstorage
more hot questions
Question feed |
9977 | https://vixra.org/pdf/1306.0113v1.pdf | A real explanation for imaginary eigenvalues and complex eigenvectors by Eckhard MS Hitzer Department of Mechanical Engineering Faculty of Engineering, Fukui University 3-9-1 Bunkyo, 910-8507 Fukui, Japan Email: hitzer@mech.fukui-u.ac.jp March 1-5, 2001, Nagpur
, India Abstract This paper first reviews how anti-symmetric matrices in two dimensions yield imaginary eigenvalues and complex eigenvectors. It is shown how this carries on to rotations by means of the Cayley transformation. Then the necessary tools from real geometric algebra are introduced and a real geometric interpretation is given to the eigenvalues and eigenvectors. The latter are seen to be two component eigenspinors which can be further reduced to underlying vector duplets. The eigenvalues are interpreted as rotors, which rotate the underlying vector duplets. The second part of this paper extends and generalizes the treatment to three dimensions. The final part shows how all entities and relations can be obtained in a constructive way, purely assuming the geometric algebras of 2-space and 3-space. I. Introduction … for geometry, you know, is the gate of science, and the gate is so low and small that one can only enter it as a little child. William K. Clifford But the gate to life is narrow and the way that leads to it is hard, and there are few people who find it. …I assure you that unless you change and become like children, you will never enter the Kingdom of heaven. Jesus Christ The motivation for this article appears somehow accidental. I had to make linear algebra problems for students about eigenvectors of matrices and their Cayley transformations. The textbook already had the problem to show that the (real) eigenvector of a three-dimensional anti-symmetric matrix was also an eigenvector of its Cayley transformation. I thought somehow why restrict it to the one real eigenvector, Accepted for publication for the Proceedings of the National Symposium on Mathematical Sciences, March 1-5, Nagpur, India. E. Hitzer, A real explanation for imaginary eigenvalues and complex eigenvectors, (Nat. Symp. on Math. Sc., 2001.3, Nagpur, India.) Proc. Einst. Foundation Int. 1, pp. 1-26 (2001).
why not let the students work on a problem with the two complex eigenvectors as well? But somehow the question came back to me like a boomerang, and I asked myself, what does it really mean to have complex eigenvalues and complex eigenvectors. I thought there must somehow be some real geometric meaning to this and began to explore the simple two-dimensional case. I was already familiar with geometric algebra [3,4,5,7] and was somehow convinced to get an answer applying it. Geometric algebra in a way completes our knowledge about how to properly multiply vectors, by adding the inner product and a dimension independent outer product to one new invertible, associative and distributive geometric vector product. In two dimensions it contains an even sub-algebra isomorphic to complex numbers. This paper therefore follows a line of argument first presenting the usual problem that occurs in matrix algebra and then showing how to shift the interpretation to a completely real interpretation in terms of geometric algebra. For this purpose I briefly introduce the basics of geometric algebra and show how it helps to fill out the gaps of our understanding. Having achieved the task in two dimensions it is only natural to try it in three dimensions as well, learn thereby something about three-dimensional geometric algebra and with hindsight get even a better understanding of what happened in two dimensions. This will also show what is particular in two dimensions and needs to be refined for solving the three-dimensional problem. After I mastered all this in two and three dimensions, using geometric algebra to provide a real geometric understanding for what I did in complex matrix algebra, I desired to turn the strategy around: I wanted to know if it was feasible to pretend not to know about anti-symmetric matrices, their imaginary eigenvalues and complex eigenvectors in the first place, but arrive at all these entities and their relationships in a synthetic way. In the last section of this article I therefore start with pure geometric algebra in two and three dimensions and show how in a natural way relationships arise which can be put into a form completely resembling the relationships of complex matrix algebra, yet equipped with clear and well defined real-geometric meanings. The first quotation stems from Clifford himself, who initially was a theologian and then became an atheist. But somehow his view of science was strongly colored by what Jesus taught as the Gospel about the Kingdom of God. To agree or disagree on what Clifford believed is a matter of faith and not of science. But I think the point made by him, that geometry is like a gateway to a new understanding of science is quite worthwhile to reflect upon. I hope that the reading of this article may help the reader to appreciate Clifford’s opinion to some degree. II.
Two real dimensions II.1 Complex treatment Any anti-symmetric matrix in two real dimensions is proportional to ¸ ¸ ¹ · ¨ ¨ © § 0 1 1 0 U .
The characteristic polynomial equation of the matrix U is 0 1 1 1 det 2 ¸ ¸ ¹ · ¨ ¨ © § O O O OE U or 1 2 O . (J) The classical way to solve this equation is to postulate an imaginary entity j to be the root of –1: 1 j . This leads to many interesting consequences, yet any real geometric meaning of this imaginary quantity is left obscure. The two eigenvalues are therefore the imaginary unit j and – j .
j 1 O , j 2 O The corresponding complex eigenvectors x x1 and x x2 are ¸ ¸ ¹ · ¨ ¨ © § o j j U 1 1 1 1 1 1 x x x x O (Evc) ¸ ¸ ¹ · ¨ ¨ © § o j j U 1 2 2 2 2 2 x x x x O The Cayley transformation C(–kU), with --sin cos 1 k ¸ ¸ ¹ · ¨ ¨ © § ----cos sin sin cos ) ) ( ( 1 2 )) ( ( )) ( ( ) ( 2 2 1 kU kU k E kU E kU E kU C (C) allows to describe two dimensional rotations. The third expression of equation (C) shows that U and C(–kU) must have the same eigenvectors x x1 and x x2. The corresponding eigenvalues of C(–kU) can now easily be calculated from (C) as 2 1 1 1 1 1 2 1 k k k c O O O , 2 2 2 2 1 1 2 1 k k k c O O O .
Inserting and we obtain the complex eigenvalues of the two-dimensional rotation C(–kU) as j 1 O j 2 O --O sin cos 1 j c , --O sin cos 2 j c We now face the question what the imaginary and complex eigenvalues and the complex eigenvectors of U and the rotation C(kU) mean in terms of purely real geometry. In order to do this let us turn to the real geometric algebra of a real two-dimensional vector space.
II.2 Real two-dimensional geometric algebra The theory developed in this section is not limited to two dimensions. In the case of higher dimensions we can always deal with the two-dimensional subspace spanned by two vectors involved, etc. II.2.1 The geometric product Let us start with the real two-dimensional vector space R2. It is well known that vectors can be multiplied by the inner product which corresponds to a mutual projection of one vector a a onto another vector b b and yields a scalar: the projected length a cosT times the vector length b, i.e. --cos cos ab b a b a .
The projected vector a aц itself can then be written as 2 /b bb a a|| , (P1) where I use the convention that inner and outer products have preference to geometric products.
In 1844 the German mathematician H. Grassmann introduced another general (dimension independent) vector product: the anti-symmetric exterior product. This product yields the size of the area of the parallelogram spanned by the two vectors together with an orientation, depending on the sense of following the contour line (e.g.
clockwise and anticlockwise), a b b a .
Grassmann later on unified the inner product and the exterior product to yield the extensive product, or how it was later called by W. Clifford, the geome ric product of vectors: t b a b a ab . (GP) We now demand (nontrivial!) this geometric product to be associative, i.e. (a ab)c c = a(b bc) and distributive, i.e. .
ac ab c) a(b Let us now work out the consequences of these definitions in the two-dimensional real vector space R2. We choose an orthonormal basis {VV V}. This means that 1 V V V V V , , . (Unit) 1 V V V V V 0 V V V V Please note that e.g. in (Unit) we don’t simply multiply the coordinate representations of the basis vectors, we multiply the vectors themselves. We are therefore still free to make a certain choice of the basis vectors, i.e. we work coordinate free! The product of the two basis vectors gives i { V V V V V V V V V V V V V V V V (I) the real oriented area element, which I call i i. It is important that you beware of confusing this real area element i i with the imaginary unit j mentioned in the last section!
But what is then the square of i i?
V V V V V V V V V V V V V V ) )( ( ) )( ( ii i2 (II) The square of the oriented real unit area element of i i is therefore i i2 = 1 ! This is the same value as the square of the imaginary unit j. The big difference however is, that j is postulated just so that the equation (J) can be solved, whereas for i i we followed a constructive approach: We just performed the geometric product repeatedly on the basis vectors of a real two-dimensional vector space!
So far we have geometrically multiplied vectors with vectors and area elements with area elements. But what happens when we multiply vectors and area elements geometrically?
II.2.2 Rotations, vector inverse and spinors We demonstrate this by calculating both Vi and Vi: Vi = V(VV) = (VV) V V Vi = V(VV) = (VV) V V This is precisely a 90 degree anticlockwise (mathematically positive) rotation of the two basis vectors and therefore of all vectors by linearity. From this we immediately conclude that multiplying a vector twice with the oriented unit area element i i constitutes a rotation by 180 degree. Consequently, the square i i2 = 1 geometrically means just to rotate vectors by 180 degree. I emphasize again that j and i i need to be thoroughly kept apart. j also generates a rotation by 90 degree, but this is in the plane of complex numbers commonly referred to as the Gaussian plane. It is not to be confused with the 90 degree real rotation i i of real vectors in the two-dimensional real vector space.
i also generates all real rotations with arbitrary angles. To see this let a a and b b be unit vectors. Then I calculate: a(a ab)= (a aa)b b = b (r) Multiplying a a with the product a ab therefore rotates a a into b b. Rab=a ab is therefore the “rotor” that rotates (even all!) vectors by the angle between a a and b b. What this has to do with i i? Performing the geometric product a ab explicitely yields: ab = cosT +sinT i (Please keep in mind that here a a2 = b b2 =1 and that the area of the parallelogram spanned by a a and b b is precisely sinTab, which explains the second term.) This can formally be written by using the exponential function as: Rab= a ab = exp(i i Tab) = cosTab +sinTab i (R) We can therefore conclude that the oriented unit area element i i generates indeed all rotations of vectors in the real two-dimensional vector space.
Another important facet of the geometric product is that it allows to universally define the inverse of a vector with respect to (geometric) multiplication as: 2 x x x x def 1 1 , .
x x xx x2 That this is indeed the inverse can be seen by calculating 1 1 1 2 x xx x x xx .
Using the inverse b b-1 of the vector b b, we can rewrite the projection of a a unto b b simply as -1 || bb a a . (P2) It proves sometimes useful to also define an inverse for area elements A = ±|A|i i: A-1 = A/A2 = A/(|A|2) = A/|A|2, where |A| is the scalar size of the area and one of the signs stands for the orientation of A relative to i i. We can see that this is really the inverse by calculating AA-1 = A-1A = AA/A2 = A2/A2 = |A|2/(|A|2) = 1. By now we also know that performing the geometric product of vectors of a real two-dimensional vector space will only lead to (real) scalar multiples and linear combinations of scalars (grade 0), vectors (grade 1) and oriented area elements (grade 2). In algebraic theory one assigns grades to each of these. All these entities which are generated such form the real geometric algebra of a real two-dimensional vector space, designated with R2 (note that the index is now a lower index). R2 can be generated through (real scalar) linear combinations of the following list of 22=4 elements {1, VV Vi}.
This list is said to form the basis of R2. When analyzing any algebra it is always very interesting to know if there are any subsets of an algebra which stay closed when performing both linear combinations and the geometric product. Indeed it is not difficult to see that the subset {1,i} is closed, because 1i = i and ii = 1. This sub-algebra is in one-to-one correspondence with the complex numbers C. We thus see that we can “transfer” all relationships of complex numbers, etc. to the real two-dimensional geometric algebra R2. We suffer therefore no disadvantage by refraining from the use of complex numbers altogether. The important operation of complex conjugation (replacing j by j in a complex number) corresponds to rever ion in geometric algebra, that is the order of all vectors in a product is reversed: s ba ab = and therefore i . i 2 1 1 2 2 1 V V V V V V = = In mathematics the geometric product of two vectors (compare e.g. (GP),(Unit),(II),(R)) is also termed a spinor. In physics use of spinors is frequently considered to be confined to quantum mechanics, but as we have just seen in (R), spinors describe every elementary rotation in two dimensions.
(Spinors describe rotations in higher dimensions as well, since rotations are always performed in plane two-dimensional subspaces, e.g. in three dimensions the planes perpendicular to the axis of rotation.) By now we have accumulated enough real geometric tools in order to work out the real explanation for the imaginary and complex eigenvalues and –vectors of section 1. II.3 Real explanation Let us skip back to the characteristic polynomial equation of the matrix U in section 1: O Instead of postulating the imaginary unit j we now turn to the real two-dimensional algebra R2 and set the eigenvalues Oand Oto simply be: O i, O i.
The corresponding “eigenvectors” x x1 and x x2 will then be: ¸ ¸ ¹ · ¨ ¨ © § i x 1 1 , .
¸ ¸ ¹ · ¨ ¨ © § i x 1 2 As in section 1, the eigenvectors of the Cayley transformation C(kU) will be the same.
And the eigenvalues of C(kU) now become: , . (LC) --O sin cos 1 i c --O sin cos 2 i c We can now take the first step in our real explanation and identify Oand Oas the real oriented unit area element with both orientations (). We can further identify the two “eigenvectors” x x1 and x x2 as two-component spinors with the entries: x11=1, x12= i and x21=1, x22= i i. Finally the eigenvalues Ocand Oc of the Cayley transformation C(kU) are seen to simply be rotors (compare (R)), i.e. operators which rotate vectors by Tand T, respectively.
Now we want to better understand what the real-oriented-unit-area-element eigenvalues O i, O i as well as and do when multiplied with the two-component eigen-spinors (previously termed “eigenvectors”) x x --O sin cos 1 i c --O sin cos 2 i c 1 and x x2 . In the last section on the real two-dimensional geometric algebra, we already learnt that every spinor can be understood to be the geometric product of two vectors. We therefore choose an arbitrary, but fixed reference vector z z from the vector space R2. For simplicity let us take z z to be z z =V. We can then factorize the spinor components of the eigen-spinors x x1 and x x2 to: x11=1=VV, x12= i VV VV and x21=1=VV, x22= i i VV VV The eigen-spinor x x1 is thus seen to correspond (modulus the geometric multiplication from the right with z z =V) to the real vector pair (VV V), whereas x x2 corresponds to the real vector pair (VV V). Multiplication with Ofrom the left as in ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § 12 11 12 11 1 1 1 x x x x U i i i x x O results in 1 2 1 1 1 2 1 1 2 1 11 11 ) ( ) ( ) )( ( 1 V V V V V V V V V V o i ix x .
That is the multiplication with O = i i from the left transforms the first entry in the vector pair (VV V), which corresponds to x x1, to V V. Performing the same calculation for the second entry V in (VV V) yields V. So the whole vector pair (VV V) is transformed to the new pair (V VV V). Which upon a close look is seen to be a simple rotation by –90 degree. Here the mathematically speaking non-commutative nature of the geometric product comes into play. That is the order of the factors in a geometric product does really matter. The multiplication of a vector with i i from the right gives a rotation by +90 degree, whilst the multiplication with i i from the left yields a rotation by –90 degree. Analogous calculations for Ox2 = ix2 show that the pair (VV V), which corresponds to x2 is transformed to (VV V), i.e. it is rotated by +90 degree. Let me summarize therefore that the multiplications Ox1 and Ox2 are thus understood to rotate the underlying vector pairs (corresponding to x x1 and x x2, respectively) by –90 and +90 degrees, respectively! In analogy to the treatment of Ox1 and Ox2, I will now treat Oc1x1 and Oc2x2.
Comparing (LC) and (R) one may already suspect that Oc1 and Oc2 may simply rotate the vector pairs corresponding to x x1 and x2 by T and +Trespectively. (As for the sign of T, the non-commutative nature of the geometric product needs again to be taken into account.) But let us prove this now explicitly. o 1 1 1 1 1 1 1 1 11 1 11 ) sin (cos ) sin (cos ) )( sin (cos ı i ı ı ı iı ı ı ı i x x ------Oc 1 1 1 1 1 1 )) ( ( )) sin( ) (cos( ) sin (cos ı ı ı i ı ı i ı ----- R .
After the third equation sign we have used the fact that the oriented unit area element i anti-commutes with all vectors in the plane characterized by i i. I will show this explicitly for V: i i 1 2 1 1 1 2 1 1 2 1 1 ) ( ) ( ) ( V V V V V V V V V V V . (ac) The multiplication Oc1x1 is therefore shown to rotate the first vector V, in the vector pair that corresponds to the eigen-spinor x x1, into V>VR(T). According to (r) and (R) this is a rotation of Vby T. Performing the same calculation for Oc1 x12 we find that the second vector Vin the vector pair that corresponds to x x1 is rotated likewise: V>VR(T).
In the very same way it can be proven explicitly that Oc2x2 rotates the vector pair (VV V), which corresponds to x x2 into (VR(T),V VR(T)).
We have therefore confirmed that the multiplications Oc1x1 and Oc2x2 geometrically mean to rotate the vector pairs corresponding to x x1 and x2 by T and +Trespectively.
Summarizing we see that the complex eigenvectorsx1 and x2 may rightfully be interpreted as two-component eigen-spinors with underlying vector pairs.
The multiplication of these eigen-spinors with the unit-oriented-area-element eigenvalues O and Omeans a real rotation of the underlying vector pairs by –90 and +90 degrees, respectively. Whereas the multiplication with Oc1 and Oc2means a real rotation of the underlying vector pairs by T and +Trespectively.
I concede that despite of the strong case for a real explanation of both imaginary (and complex) eigenvalues and complex eigenvectors, one may at first sight wonder (1) how to extend this explanation to higher dimensions and (2) whether the treatment of higher dimensions might not become too complicated. In order to show that the extension to higher dimensions is fairly easy, straight forward and not at all complicated, I will now discuss the same problem for the case of three dimensions. III. Three real dimensions III.1 Complex treatment of three dimensions Any anti-symmetric matrix in 3 dimensions is proportional to a matrix of the form U 㧩 0 -c b c 0 -a -b a 0 with a2+b2+c2=1. The characteristic polynomial equation of the matrix U is 0 ) ( det 2 2 2 2 ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § c b a b a b c a c E U O O O O O O .
If we use the condition that a2+b2+c2=1, this simplifies and breaks up into the two equations 1 2 2 , 1 O (J3) and .
0 3 O That means we have one eigenvalue O equal to zero and for the other two eigenvalues OOwe have the same condition as in the two-dimensional case for the matrix 0 1 -1 0 .
It is therefore clear that in the conventional treatment one would again assign O jand O j . The corresponding eigenvectors are: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § 2 2 2 1 1 1 c ja bc jb ac ja bc b jc ab jb ac jc ab a 1 x ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § 2 2 2 1 1 1 c ja bc jb ac ja bc b jc ab jb ac jc ab a 2 x . (CEV) The sign expresses that all three given forms are equivalent up to the multiplication with a scalar (complex) constant. The eigenvector that corresponds to O ѳ simply is: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § c b a 3 x .
The fact that O simply means that the matrix U projects out any component of a vector parallel to x x3 . U maps the three-dimensional vector space therefore to a plane perpendicular to x x3 containing the origin. The Cayley transformation C(kU) with --sin cos 1 k now describes rotations in three dimensions: ) ) ( ( ) ( 1 2 )) ( ( )) ( ( ) ( 2 2 2 2 2 1 kU kU c b a k E kU E kU E kU C ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ) 1 )( cos 1 ( 1 ) cos 1 ( sin ) cos 1 ( sin ) cos 1 ( sin ) 1 )( cos 1 ( 1 ) cos 1 ( sin ) cos 1 ( sin ) cos 1 ( sin ) 1 )( cos 1 ( 1 2 2 2 c bc a ac b bc a b ab c ac b ab c a ---------------. (C3) The vector x x3 plays here the role of the rotation axis. If e.g. we set a=b=0, c=1, we get the usual rotation around the z-axis: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § 1 0 0 0 cos sin 0 sin cos )] 1 , 0 , 0 ( )[ ( 3 ----x kU C .
The expression for C(kU) after the second equal sign clearly shows that the eigenvectors of U and C(kU) agree in three dimensions as well. The general formula for calculating the eigenvalues Oc of C(kU) from the eigenvalues O of U reads as follows: 2 1 2 2 2 2 1 1 2 1 ) ( 1 1 2 1 2 2 2 k k k c b a k k k c b a c O O O O O Inserting OOand Oin this formula yields: --O sin cos 1 j c , , . --O sin cos 2 j c 1 3 c O We now already see that the most interesting differences to the two-dimensional case lie in the (complex) eigenvectors, besides the fact that the expression for the rotation matrix C(kU) looks rather more complicated.
III.2 Real three-dimensional geometric algebra We begin with a real three-dimensional vector space R3. In R3 we introduce an orthonormal set of basis vectors {VV VV V}, that is VmкVn for n=m and VmкVn for n ҁ m, {n,m=1,2,3}. The basic 23 = 8 geometric entities we can form with these basis vectors are: 1, {VV VV V}, {i3= VV Vi1= VV Vi2= VV V}, i= VV VV scalar vectors oriented unit real area oriented real volume (grade 0) (grade 1) elements (grade 2) element (grade 3) We now have three real oriented unit area elements { i1i2i3} corresponding to the three plane area elements of a cube oriented with its edges along VV VandV VThis set of eight elements {1,VV VV V, i i1i2i3, i } forms the real geometric algebra R3 of the three dimensional vector space R3. By looking at the subsets {1,VV Vi3}, {1,VV Vi1} and {1,VV Vi2} we see that R3 comprises three plane geometric sub-algebras, as we have studied them in section I.2. In general, by taking any two unit vectors {u u, v} which are perpendicular to each other, we can generate new two-dimensional plane geometric sub-algebras of R3 with the unit area element i i = u uv.
As in the two-dimensional case we have i i12 =i22 =i32 = i i2 = –1. And we have i 2 = i i = VV VVVV VV V VV VVV V VV V VV VV V VV V V V V VV VV V VV V V V VV V V VV V VV V Each permutation after the third, fourth and fifth equal sign introduced a factor of –1 as in (I). The square of the oriented three-dimensional volume element is therefore also i 2=–1.
In three dimensions the vector a a unto b b projection formula (P2) does not change, since it relates only entities in the a a,b b plane. But beyond that we can also project vectors a a onto i planes, by characterizing a plane by its oriented unit area element i i. In this context it proves useful to generalize the definition of scalar product to elements of higher grades : a a a r r r r B B B 1 ) 1 ( 2 1 , where r denotes the grade of the algebraic element Br. For Br = b (r=1) we have as usual ba ab b a 2 1 , but for Br = i i (example with grade r=2) we have ia ai i a 2 1 .
We can calculate for example 0 2 1 1 3 2 3 2 1 1 V V V V V V V 1 i 3 2 3 2 3 2 2 2 2 1 V V V V V V V V 1 i 2 3 3 2 3 2 3 3 2 1 V V V V V V V V 1 i If we now rotate V (and V V) with i1-1 = i1 from the right by –90 degree in the VV V plane, we obtain 2 3 2 3 3 2 V V V V V V 1 -1 1 1 i i i , , 3 3 2 2 2 3 V V V V V V 1 -1 1 1 i i i respectively.
The projection of any vector a a unto the VV V plane is therefore given by -1 1 1 || i i a a .
We say therefore instead of VV V plane also simply i i1-plane. And in general the projection of a vector a a unto any i i-plane is then given by -1 || i i a a , which is in perfect analogy to the vector unto vector projection in formula (P2). There is more[4,5] to be said about R3, but the above may suffice for our present purposes.
III.3 Real explanations for three dimensions If we follow the treatment of the two-dimensional case given in section II.3, then we need to replace the imaginary unit j in the eigenvalues OOand in the eigenvectors x x1, x2 by an element of the real three-dimensional geometric algebra R3. In principle there are two different choices: The volume element i or any two-dimensional unit area element like e.g. i i1i2i3.
Let me argue for the second possibility: We have seen in section III.1 that the multiplication of U with a vector always projects out the component of this vector parallel to x x3 so that the y y on the right hand side of equations like Ux x = y y is necessarily a vector in the two-dimensional plane perpendicular to x x3 containing the origin. O x1 and Ox2 are precisely such vectors, since they arise from Ux x1 and Ux x2, respectively. O x1 and Ox2 are therefore recognized as vectors belonging to the two-dimensional plane perpendicular to x x3 containing the origin. Thus it seems only natural to interpret the squareroot of –1 in the solution of equation (J3) to be the oriented unit area element i characteristic for the plane perpendicular to x x3 containing the origin as opposed to the volume element element i or any other two-dimensional unit area element.
.
The oriented unit area elements i i1of the VV Vplane, i i2 of the VV Vplane, and i i3 of the VV Vplane, are perpendicular to the vectors VV Vand Vrespectively. The unit area element i of the two-dimensional plane perpendicular to x x3 will therefore simply be: i = a i1bi2c i3 = a VVb VVc VV I therefore consequently set O = i iO = --i and ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § 2 2 2 1 1 1 c a bc b ac a bc b c ab b ac c ab a i i i i i i x1 ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § 2 2 2 1 1 1 c a bc b ac a bc b c ab b ac c ab a i i i i i i x2 (EV3) As in the two-dimensional case I interpret the three components of each “eigenvector” as spinorial components, i.e. elementary geometric products of two vectors. (In the following we will therefore use the expression three component eigenspinor instead of “eigenvector”.) I again arbitrarily fix one vector from the i i plane (the plane perpendicular to x x3) as a reference vector z z with respect to which I will factorize the three component eigenspinors x x1 and x x2. With respect to the first representation of the eigenspinors x x1 and x x2 we choose to set .
V V V V V ac ab a __ ) 1 ( 2 -1 ii z The square of z z gives the first component spinor of x x1 and x x2: 2 2 2 2 2 2 1 ) 1 ( a c a b a a z z zz z2 , where we make use of the condition a2 +b2 +c2=1.
The inverse of z z will therefore be: z–1 = z z/zz = Vab/(a) Vac/(a) V. In order to now solve the equations for the two other vectors underlying the eigenspinor components –ab–i ic and –ac–i ib of x x1: n2z = –ab–i ic and n n3z = –ac–i ib, we can simply multiply with z z–1 from the right: n2 = n2zz–1 = (–ab–i ic) z–1 = (–ab– c {a i1bi2c i3})(Vab/(a) Vac/(a) V = (–abV+ a2b2/(a)Va2bc/(a)V– ca i1Vbci2Vci3V+ a2bc/(a) i i1V2 ab2c/(a) i i2V2 abc2/(a) i3V2 + a2c2/(a) i i1V3 abc2/(a) i i2V3ac3/(a) i3V3) In order to simplify the above equation we first calculate i1V1 = VVV i i2V1 = VVV V i3V1 VVV V VVV V i1V V VVV V VVV V i2V V VVV V VVV i i3V= VVV V V i1V3 = VVV V V i2V3 = VVV V VVV V V i3V3 V VVV i After reordering everything we get: n2 = (b2)V2 ab V1 bc V3 = Vц. In the very same way we can now calculate n3 = n3zz–1 = (c2)V3 ac V1 bc V2 = Vц.
Summarizing these calculations we have (setting n n1 = z z= Vц): ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § __ __ __ __ __ __ V V V V V V z n z n z n i i x 3 2 1 1 b ac c ab a2 1 So all we need to give a real geometric interpretation for the three-component eigenspinor Z are geometric products of the projections of the three basis vectors V V andV onto the i plane. The other two equivalent representations of x x1 given in (EV3) can be written as: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # __ __ __ __ __ __ V V V V V V a bc b c ab i i x1 2 1 and x .
¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # __ __ __ __ __ __ V V V V V V 2 1 c a bc b ac i i 1 We see that this simply corresponds to a different choice of the reference vector z z, as z=Vц and as z z=Vц, respectively. In general all possible ways to write x x1 correspond to different choices of z z from the i i plane. The geometric product Rzz’ = z zz’ of any two such reference vectors z z and z z’ gives the rotation operation to rotate one x x1(z z) choice into the other x x1(z z’)= x x1(z z) Rzz’.
Comparing the (complex) “eigenvectors” x x1 and x x2 in (CEV) we see that they are related to each other by complex conjugation. Since the corresponding operation in geometric algebra is the reversion of the order of vectors it is no wonder that the components of the eigenspinor x x2 are formed by taking the reverse order of vector factors appearing in the factorization of x x1: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § # ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ V V V V V V V V V V V V V V V V V V 3 2 1 2 zn zn zn i i x b ac c ab a 2 1 .
Naturally nothing can hinder us to factorize out the reference vector z z in for x x2 also to the right, we just end up with somewhat less handy expressions. Let us now turn to the interpretation of the eigenvalues Oc1 and Oc2 of the Cayley transformation C(kU). (We will see that the eigenvalues O1 and O2 may indeed be understood as special cases of Oc1 and Oc2 by simply setting T degree.) We now write them – replacing j by i – as: Oc1 = cos T i sin Tand Oc2= cos T i sin T The action of Oc1 on the three-component eigenspinor x x1 is: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § z n z n z n z n z n z n x x 3 1 2 1 1 1 3 2 1 1 1 1 1 ) ( c c c c c kU C O O O O O . (Lc3.1) As in the two-dimensional treatment of section II.2, the products Oc1n1,Oc1n2,Oc1n3 can now be understood as a real two-dimensional rotation of the underlying vector triplet {n n1,n2,n3} by the angle Tin the plane perpendicular to x x3 Since all triplet vectors are elements of the plane perpendicular to x x3, and i is the oriented unit area element of precisely this plane, the two-dimensional treatment fully applies.
In line with this, the action of Oc2 on the three-component eigenspinor x x2: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § 2 3 2 2 2 1 3 2 2 2 1 2 3 2 1 2 2 2 2 ) ( c c c c c c c c kU C O O O O O O O O zn zn zn zn zn zn zn zn zn x x (Lc3.2) may now be interpreted as the same two-dimensional rotation of the vector triplet {n n1,n2,n3} by the angle T1ote that here we have Oc2 to the right of each triplet vector, and not to the left as in (Lc3.1), which explains the same sign of the angle. For the third equation sign we have used the fact that spinors commute, since 1 (one) and i i commute with each other. Note also that the resulting spinor components continue to be different, because the different order of vector factors in the spinorial components of x x1 and x x2.
As for O1 and O2 we now set T degree and we conclude from the above discussion that O1 rotates the vector triplet {n n1,n2,n3} belonging to x x1 by –90 degree. If we look at the last expression in (Lc3.2) O2 can be interpreted to also rotate the vector triplet {n n1,n2,n3} by –90 degree. (Note again the order of vector factors in (Lc3.2)!) As already mentioned the fact that O3=0 means a projecting out of any component parallel to x x3. The third eigenvalue of the Cayley transformation C(kU) is Oc3=1, which means that any component parallel to x x3 will be invariant under multiplication with C(kU).
Let me also remark that instead of interpreting the action of the eigenvalues O1, O2, Oc1 and Oc2 as triplet rotations, we may just as well interpret them as rotating the second factor in the vector factorization of the eigenspinor components, i.e. the reference vector z itself. Yet this interpretation has the drawback that we would need to abandon the fact that z z did not change so far, which makes comparisons in the plane perpendicular to x3 more straight forward. Summarizing the real situation in three dimensions, we see that the projecting out of the components parallel to x x3 lead to the use of the oriented unit area element i, which characterizes the plane perpendicular to x x3. i i thus replaces the imaginary unit j, which lacked any real geometric interpretation. Subsequently O1 and O2 (Oc1 and Oc2) were seen to act as rotation operators on the vector triplet {n n1,n2,n3} underlying the three-component eigenspinors x x1 and x x2. The underlying triplet vectors themselves (and the reference vector z z) are all elements of the plane perpendicular to x x3 and continue to be so, when being rotated. IV. Bottom up real explanation It is interesting to consider whether geometric algebra can only serve as a tool of real analysis of the complex situation, or if we may even completely forget about matrices and complex numbers and let geometric algebra generate expressions that are finally interpretable in terms of matrices and complex numbers. IV.1 Bottom up for two dimensions We therefore now only assume the real two-dimensional vector space R2 and its geometric algebra R2 as described in section II.2. In section II.3 we learnt that the geometric multiplications of the two basis vectorsV Vand V with the plane oriented unit area element i from the left yield: i V V i V V V If we simply write this in matrix form we obtain ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § 1 2 2 1 0 1 1 0 V V V V . (B2) And we therefore see that the matrix corresponding to the geometric multiplication with i from the left is exactly the two-dimensional anti-symmetric matrix U with which we started off in section II.1. Multiplying both sides of (B2) with Vfrom the right, we obtain the two-component spinorial form of the equation: ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § 1 1 0 1 1 0 0 1 1 0 1 1 1 2 1 2 1 1 i i V V V V V V V V . (B2s) Please note that, because of the existence of the vector inverse in section II.2.2, this is an invertible operation! The second and fourth expressions in equation (B2s) correspond exactly to the equation Ux x1=Ox1 ix1 of section II.3. By applying the reversion operation of the order of vectors in all geometric products involved, as defined in section II.2.2, we get: ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § 1 1 0 1 1 0 0 1 1 0 1 1 2 1 2 1 1 1 i i V V V V V V V V (B2r), which corresponds exactly to the equation Ux x2=Ox2 ix2 explained in section II.3. What we just did with the unit oriented area element i i, we will now do with the rotor R(T)=cosT + i i sinT. It transforms two basis vectorsV Vand V to: R(T)V= (cosT + i i sinT V= cosT V V+ i iV sinT cosT V VsinTV V R(T)V= (cosT + i i sinT V= cosT V V+ i iV sinT sinTV V + cosT V V Rewriting this in Matrix form we obtain: ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § 2 1 2 1 2 1 cos sin sin cos cos sin sin cos V V V V V V --------. (BR2) The matrix that corresponds to the rotor R(T) operating on the basis vectors V Vand V is therefore exactly the Cayley transformation C(kU). Geometrically multiplying both sides of equation (BR2) with V Vfrom the right we obtain: ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § i i i 1 ) sin (cos 1 cos sin sin cos cos sin sin cos 1 2 1 1 ----------V V V V ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § 1 2 1 1 1 2 1 1 ) ( ) ( )) sin( ) (cos( )) sin( ) (cos( V V V V V V V V ------R R i i . (BR2s) We now see a complete correspondence of equation (BR2s) with the eigenspinor equation C(kU)x x1=Ocx1=(cosT + i isinT) x1, with x x1=(VVV VV)=(1, i), as explained in section II.3. Reversing the order of all elementary geometric vector products in equation (BR2s) yields: ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § i i i 1 ) sin (cos 1 cos sin sin cos cos sin sin cos 2 1 1 1 ----------V V V V ¸ ¸ ¹ · ¨ ¨ © § ¸ ¸ ¹ · ¨ ¨ © § 2 1 1 2 1 1 ) ( ) ( ) sin (cos ) sin (cos V V V V V V V V ------R R i i .
This reversed form of (BR2s) is therefore seen to perfectly correspond to the second eigenspinor equation of the Cayley transformation C(kU)x x2=Ocx2=(cosT isinT) x2.
It is therefore demonstrated that by purely assuming the real two-dimensional geometric algebra R2 all expressions which are known from complex matrix algebra are generated automatically. The invertible transformation of the vector equations into spinor equations through the geometric multiplication with V Vfrom the right makes the correspondence explicite.
IV.2 Bottom up for three dimensions That what we have shown in two dimensions is not only due to the simplicity of two-dimensional algebra, will now be explicitly demonstrated in three dimensions as well.
Here we start in the real three-dimensional geometric algebra R3 by considering the multiplication of the plane area element i=a i1 +b i2 +c i3 with the components of the three basis vectors VVandV Vparallel to the plane characterized by i i, i.e. the plane perpendicular to (a,b,c): iV= c Vb V iV= c Va V iV= b Va V iB3) It is made use of the condition that a2+b2+c2=1. These calculations can either be performed by hand as in the previous sections, or one may simply use geometric algebra capable mathematical software, like as MAPLE V with the Cambridge GA package, etc. Rewriting the three equations in matrix form yields: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¸ ¹ · ¨ ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § __ __ __ __ __ __ __ __ __ V V V V V V V V V V V V V V V V V V 0 0 0 0 0 0 a b a c b c a b a c b c a b a c b c a b a c b c . % We now recognize the three-dimensional anti-symmetric matrix U of section III.1. The third equation (B3) may simply be calculated from the second one, the fourth expression is just rewriting the third in matrix form as well. The distinction between the basis vectors and their i i-plane projections seems at first sight unnecessary, but the operation of i on the basis vectors themselves does not yield the equations (iB3). Geometrically multiplying both sides of (B3) with V from the rightwe obtain the three-component spinorial form: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¸ ¹ · ¨ ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § __ __ __ __ __ __ __ __ __ __ __ __ i i i i i i i i i i i b ac c ab a b ac c ab a a b a c b c 2 2 1 ) ( ) ( ) 1 ( 0 0 0 V V V V V V V V V V V V . (B3s) This exactly corresponds to the equation Ux x1=O1x1= ix1 of section III.3. Note that the multiplication with V is invertible. Multiplying with V or V instead yields the equivalent forms of x x1 also mentioned in section III.3. Reversing all elementary geometric vector products involved in (B3s) yields: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¸ ¹ · ¨ ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § __ __ __ __ __ __ __ __ __ __ __ __ i i i i i i i i i i i b ac c ab a b ac c ab a a b a c b c 2 2 1 ) ( ) ( ) 1 ( 0 0 0 V V V V V V V V V V V V . (B3r) This is seen to be Ux x2=O2x2= ix2 of section III.3. As in the bottom up explanation for two dimensions, we will now look at the action of the rotor R(T)=cosT + i i sinT, with i=a i1 +b i2 +c i3. As before, we could look straight away for what R(T) does with the three basis vectors projected onto the i i-plane, i.e. with VVandV V. But a better way is to use the fact that i i anti-commutes with all vectors in its plane [see section II.3, equ. (ac)] and commutes with vectors perpendicular to its plane: i( a V +b V +c V)=( a V +b V +c V) i. (com) In three dimensions a V +b V +c V is the only vector perpendicular to the i i-plane. We can therefore rewrite e.g. the rotation of V as: Oc1V=R(T)V R(T)R(T)V R(T)V__R(T) (DSR) Moving a R(T) factor to the right causes the angle to change sign, because of the anti-commutativity of i i with VIf we apply the same formula to the vector a V +b V +c V we obtain: R(T)(a V +b V +c V)R(T) R(T)R(T) (a V +b V +c V) = a V +b V +c V, because the vector a V +b V +c V is perpendicular to the i i-plane, i.e. it doesn’t change the sign of the angle in the factor R(T), if the factor is moved back to the left. That is to say the double sided description of the rotation as in (DSR) does exactly what a rotation around the axis a V +b V +c V is expected to do, it leaves the axis invariant and rotates only vectors in the i i-plane. The formula (DSR) automatically takes care of the necessary projections. This corresponds well with the Cayley transformation C(kU), which also leaves the vector a V +b V +c V invariant. Instead of looking at Oc1V=R(T)V__R(T)Oc1V=R(T)VR(T)and Oc1V=R(T)VR(T)(LRh) we will look straight away at: R(T)VR(T)={1+(cosT)(a)} V+ {c sinT+ab(cosT)} V + {b sinT+ac(cosT)} V R(T)VR(T)={c sinT+ab(cosT)} V + {1+(cosT)(b)} V+ { a sinT+bc(cosT)} V R(T)VR(T)={b sinT+ac(cosT)} V+ { a sinT+bc(cosT)} V+ {1+(cosT)(c)} V.
Rewriting these three equations in matrix form gives: ¸ ¹ · ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¹ · ¨ © § 2 R 2 R -- V V V ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § V V V ) 1 )( cos 1 ( 1 ) cos 1 ( sin ) cos 1 ( sin ) cos 1 ( sin ) 1 )( cos 1 ( 1 ) cos 1 ( sin ) cos 1 ( sin ) cos 1 ( sin ) 1 )( cos 1 ( 1 2 2 2 c bc a ac b bc a b ab c ac b ab c a ---------------.
(BR3) The matrix obtained is exactly C(kU), as in section III.1. Both sides of equation (BR3) leave components parallel to a V +b V +c V invariant. They can therefore be subtracted on both sides without changing the form of the equation: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¹ · ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¹ · ¨ © § __ __ __ __ __ __ __ __ __ V V V V V V V V V ) ( R 2 R 2 R kU C ---.
According to (DSR) I simplified the first expression, to have only R(T)=Oc1 to the left.
Multiplying by V__from the right we obtain: ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § ¸ ¸ ¸ ¹ · ¨ ¨ ¨ © § __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ i i i i b ac c ab a kU C kU C b ac c ab a c c 2 2 1 1 1 ) ( ) ( 1 R V V V V V V V V V V V V V V V V V V O O -.
(BR3s) This is seen to exactly be the Cayley transformation eigenvector equation C(kU)=Oc1x1=(cosT + i i sinT) x x1, i.e. equ. (Lc3.1) in section III.3. Using V__or V instead of V__ from the right, we obtain the other forms of x x1 mentioned in section III.3. Reversing the order of all geometric products of vectors involved in (BR3s) yields equation (Lc3.2) of section III.3. Using the laws governing the real geometric algebra R3, and properly rewriting the action of the eigenvalues Oc1 and Oc2 as in equ. (DSR), we have therefore obtained all relationships for the three-dimensional anti-symmetric matrix U and its Cayley transform C(kU) in a purely constructive manner.
V. Conclusion In this paper the eigenvalues and eigenvectors of anti-symmetric matrices in two and three dimensions and of their Cayley transformations were briefly reviewed as described by standard linear algebra. Traditionally the imaginary and complex eigenvalues and the complex eigenvectors have no direct real geometric interpretation. In an effort to unravel the real geometric interpretation I used the geometric algebras of the plane and Eulidean three-space respectively. All relationships involving imaginary and complex values or components found an explanation. The imaginary and complex eigenvalues were found to act like rotation operators on duplets and triplets of vectors. In three-space the triplets of vectors were found to be the projections of the coordinate vectors onto the plane of rotation, unique up to an arbitrary constant rotation in the plane. Geometric multiplication of this triplet of vectors with any other vector (termed reference vector) in the plane of rotation gives the three components equivalent to a complex three-component eigenvector in the standard treatment. The imaginary eigenvalues corresponded to real space +90 degree or –90 degree rotation operators and the complex eigenvalues of the Cayley transformations to arbitrary real space rotations in the plane of rotation, depending on the scalar parameter k of the Cayley transformation. Finally, for both two and three dimensions a bottom up explanation was found in a constructive way, only assuming the two- and three-dimensional geometric algebras. This explanation yielded expressions and equations that completely resemble the eigenvalue and eigenvector relationships for anti-symmetric matrices and their Cayley transformations.
The extension to Euclidean and Minkowskian four-space might be a natural next step. This way the interpretation could be further refined and new insights in areas of physics whose theories are based on such spaces might be gained. I especially think of the special theory of relativity, electrodynamics and relativistic quantum mechanics. Some preliminary calculations seem to indicate that the extension to four-spaces will not be trivial. Independent of the question for higher dimensions, the two and three dimensional treatment has already yielded a definitive and very satisfactory picture of how imaginary and complex eigenvalues and complex eigenvectors should truly be understood in terms of real geometry. Something undergraduate students, exposed to linear algebra for the first time, as well as scientists who teach and apply it will certainly appreciate. VI. Acknowledgements I first of all thank God for the joy of studying his creation: “…since the creation of the world God’s invisible qualities – his eternal power and divine nature – have been clearly seen, being understood from what has been made…”. I thank my wife for encouraging my research in many ways. I further thank my very helpful and teaching assistant T. Ido for the good cooperation in teaching linear algebra. Dr. H. Ishi helped me through discussion and comments on this work. I thank Prof. O. Giering and Prof. J.S.R. Chisholm for attracting my interest to geometry as such. Fukui University provided a good research environment for conducting this research. I thank K. Shinoda (Kyoto) for his prayerful support and encouragement for me. References ਛጟ⒤ޔㇱ᥏ᄦޔ✢ᒻઍᢙ㐷ޔ♿દ࿖ደᦠᐫޔ᧲੩㧝㧥㧤㧢㧚 H. Grassmann, (F. Engel, editor) Die Ausdehnungslehre von 1844 und die Geometrische Analyse, vol. 1, part 1, Teubner, Leipzig, 1894. D. Hestenes, New Foundations for Classical Mechanics, Kluwer Academic Publishers, Dordrecht, 1999. S. Gull, A. Lasenby, C. Doran, Imaginary Numbers are not Real. - the Geometric Algebra of Spacetime, Found. Phys. 2 23 (9), 1175 (1993). D. Hestenes, Spacetime Calculus, Arizona State University, download version: Cambridge Maple GA package, D. Hestenes, G. Sobczyk, Clifford Algebra to Geometric Calculus, Kluwer Academic Publishers, Dordrecht, 1984. Apostle Paul, Letter to the Romans, chapter 1, verse 20, The Holy Bible New International Version, International Bible Society, Colorado, 1973. Jesus Christ, in the Gospel according to Matthew, Bible, Today’s English Version. 㩷 E. Hitzer, A real explanation for imaginary eigenvalues and complex eigenvectors, (Nat. Symp. on Math. Sc., 2001.3, Nagpur, India.) Proc. Einst. Foundation Int. 1, pp. 1-26 (2001). |
9978 | https://www.webmd.com/drugs/2/drug-7212/labetalol-oral/details | Skip to main content
Labetalol (Normodyne, Trandate) - Uses, Side Effects, and More
Written by Alex Poppen, PharmD | Medically Reviewed by Kimberly Rath, PharmD on Jun 13, 2024
Common Brand Name(s): Normodyne, Trandate
Common Generic Name(s): labetalol, labetalol hydrochloride
Pronunciation: la-BET-a-loe
Drug Classes: beta-adrenergic blocking agent
Availability: prescription only, generic available
How is it used? liquid that is injected into a blood vessel, tablet that is swallowed
Uses
Side Effects
Warnings & Precautions
Interactions
Overdose/Missed Dose
Reviews (163)
6 mins read
What is labetalol used for?
Labetalol is commonly used to lower high blood pressure (hypertension).
Labetalol may also be used for other conditions as determined by your healthcare provider.
How does labetalol work (mechanism of action)?
Labetalol blocks beta-1 receptors in your heart and alpha-1 receptors in your blood vessels. These receptors normally bind to hormones called catecholamines. When catecholamines bind to these receptors, it causes your heart to beat harder and faster and your blood vessels to tighten. Blocking these receptors helps your heart beat more slowly and your blood vessels relax. These effects can help lower blood pressure, and may reduce your risk for heart attack and stroke.
How is labetalol supplied (dosage forms)?
Labetalol is available in the following dosage forms that are taken by mouth.
100 mg oral tablets
200 mg oral tablets
300 mg oral tablets
Labetalol is also available in injectable forms.
How should I store labetalol?
Labetalol tablets should be stored at room temperature, between 68 F to 77 F (20 C to 25 C). It can be exposed to temperatures between 59 F to 86 F (15 C to 30 C), for shorter periods of time, such as when transporting it. Store in a cool, dry place.
Side Effects
What are the most common side effects of labetalol?
The most common side effects of labetalol are listed below. Tell your healthcare provider if you have any of these side effects that bother you.
Low blood pressure (see below)
Dizziness
Nausea
Feeling unusually weak or tired
There may be other side effects of labetalol that are not listed here. Contact your healthcare provider if you think you are having a side effect of a medicine. In the U.S., you can report side effects to the FDA at www.fda.gov/medwatch or by calling 800-FDA-1088. In Canada, you can report side effects to Health Canada at www.health.gc.ca/medeffect or by calling 866-234-2345.
What are the serious side effects of labetalol?
While less common, the most serious side effects of labetalol are described below, along with what to do if they happen.
Heart Rate Changes. Labetalol can make your heart beat slower than normal (bradycardia). Call your healthcare provider right away if you have any of the following symptoms of bradycardia.
Feeling lightheaded, dizziness, or fainting
Confusion
Feeling weak or easily tired
Chest pain
Shortness of breath
Decreased Blood Pressure (Hypotension). Labetalol may cause low blood pressure. If you feel faint or dizzy, lie down. Tell your healthcare provider right away if you have any of the following symptoms of low blood pressure.
Dizziness, feeling lightheaded, or fainting
Confusion
Feeling weak or tired
New or Worsening Heart Failure. Labetalol may worsen existing heart failure or cause fluid buildup in your body. For individuals with underlying heart disease or heart damage, labetalol can potentially lead to heart failure over time. Tell your healthcare provider right away if you have any of the following symptoms.
Shortness of breath or trouble breathing, especially while lying down
Swelling in your feet, ankles, or legs
Unusually fast weight gain
Unusual tiredness
Breathing Problems. Labetalol may cause the muscles around the airways to tighten (bronchospasm) instead of relaxing, which can be serious. Stop using carvedilol and contact your healthcare provider right away if you have trouble breathing.
Liver Damage. Liver damage, also called hepatotoxicity, can happen when taking labetalol. Call your healthcare provider right away if you have any of the following symptoms of liver damage.
Nausea or vomiting
Stomach or belly pain
Fever
Weakness or unusual tiredness
Itching
Loss of appetite
Light-colored poop
Dark-colored urine
Your skin or the whites of your eyes turning yellowish in color (also called jaundice)
Severe Allergic Reactions. Labetalol may cause allergic reactions, which can be serious. Stop using labetalol and get help right away if you have any of the following symptoms of a serious allergic reaction.
Breathing problems or wheezing
Racing heart
Fever or general ill feeling
Swollen lymph nodes
Swelling of the face, lips, mouth, tongue, or throat
Trouble swallowing or throat tightness
Itching, skin rash, or pale red bumps on the skin called hives
Nausea or vomiting
Dizziness, feeling lightheaded, or fainting
Stomach cramps
Joint pain
Warnings & Precautions
Who should not use labetalol?
Allergies to Ingredients. People who are allergic to any of the following should not use labetalol.
Normodyne
Trandate
Labetalol
Any of the ingredients in the specific product dispensed
Your pharmacist can tell you all of the ingredients in the specific labetalol products they stock.
Asthma. Labetalol should not be used by those with asthma or similar breathing problems.
Bradycardia. Labetalol should not be used if your heart is beating too slow (bradycardia).
Heart Block. Labetalol should not be used if the electrical signals in your heart move too slow (heart block).
Severe or Worsening Heart Failure. Labetalol should not be used if your heart failure is severe or getting worse, causing you to feel very tired, have trouble breathing, or swelling in your legs.
Cardiogenic Shock. Labetalol should not be used if you have a life-threatening condition called cardiogenic shock where your heart cannot pump enough blood to the rest of your body.
Drug Interactions. Labetalol should not be taken while you are using certain other medicines. Before taking labetalol, tell your healthcare provider about any prescription or over-the-counter (OTC) medicines, vitamins/minerals, herbal products, and other supplements you are using. See the Interactions section for more details.
What should I know about labetalol before using it?
Do not take labetalol unless it has been prescribed to you by a healthcare provider. Take it as prescribed.
Do not share labetalol with other people, even if they have the same condition as you. It may harm them.
Keep labetalol out of the reach of children.
Labetalol can affect your alertness or coordination. Do not drive or do other activities that require alertness or coordination until you know how labetalol affects you.
Stopping labetalol suddenly may lead to chest pain or heart attack in people with heart disease. Do not stop taking this medication without talking to your healthcare provider first.
What should I tell my healthcare provider before using labetalol?
Tell your healthcare provider about all of your health conditions and any prescription or over-the-counter (OTC) medicines, vitamins/minerals, herbal products, and other supplements you are using. This will help them determine if labetalol is right for you.
In particular, make sure that you discuss any of the following.
Current and Past Health Conditions. Tell your healthcare provider if you have any of the following.
Heart or circulation problems
Breathing problems
Liver problems
Pheochromocytoma
Diabetes. Labetalol can decrease early signs of low blood sugar, such as fast heartbeat, and make it more likely for the low blood sugar to be serious, especially for people with diabetes.
Cataract Surgery. Labetalol may cause a condition called Intraoperative Floppy Iris Syndrome (IFIS) during cataract surgery, which can make the surgery more difficult. Tell your eye doctor about all the medications you take, including labetalol, before your cataract surgery.
Pregnancy. It is not known if or how labetolol could affect pregnancy or harm an unborn baby. Tell your healthcare provider if you are or plan to become pregnant.
Breastfeeding. Labetalol passes into breast milk. Tell your healthcare provider if you are breastfeeding or plan to breastfeed.
Interactions
Does labetalol interact with foods or drinks?
There are no known interactions between labetalol and foods or drinks.
It is unknown if drinking alcohol will affect labetalol. The risk of dizziness may be increased if you drink alcohol while taking labetalol.
Does labetalol interact with other medicines (drug interactions)?
Always tell your healthcare provider about any prescription or over-the-counter (OTC) medicines, vitamins/minerals, herbal products, and other supplements you are using.
In particular, make sure that you discuss if you are using any of the following before taking labetalol.
Diltiazem and verapamil, which are medicines used to treat chest pain and high blood pressure
Digoxin (Lanoxin), which is a medicine used to treat irregular heartbeat and some types of heart failure
Cimetidine (Tagamet HB), which is a medicine used for heartburn
Beta-receptors agonists (e.g., albuterol, formoterol), which are used to treat asthma and other breathing problems
Nitroglycerin, which is a medicine used to treat chest pain and lower blood pressure
A tricyclic antidepressant, such as amitriptyline, nortriptyline, or protriptyline
Any medicine for irregular heart rate or rhythm
Any medicine for high blood pressure
Any medicine for diabetes
Does Labetalol (Normodyne, Trandate) interact with other drugs you are taking?
Enter your medication into the WebMD interaction checker
Overdose/Missed Dose
What should I do if I accidentally use too much labetalol?
If you or someone else has used too much labetalol, get medical help right away, call 911, or contact a Poison Control center at 800-222-1222.
What should I do if I miss a dose of labetalol?
If you miss a dose, take it as soon as you remember. If it is almost time for your next dose, skip the missed dose and only take the next dose. Do not take double or extra doses.
FEATURED RESULTS
Questions about this drug?
Talk to a physician near you.
See All
Carol Bland, MD
Family Medicine
9 Ratings
View Profile
Kellie Greene, NP
Internal Medicine
10 Ratings
View Profile
Girraj Bansal, MD
Internal Medicine
36 Ratings
View Profile
Chelsea Willis, DO
Internal Medicine
1 Rating
View Profile
Tony Buchta, R.Ph., MBA
Internal Medicine
0 Rating
View Profile
Save this page
Save Drug
Labetalol (Par Pharmaceutical) US Prescribing Information, March 2017.
Labetalol hydrochloride (Almaject) US Prescribing Information, May 2019.
Labetalol hydrochloride injection (Hikma Pharmaceuticals USA) US Prescribing Information, March 2022.
You Might Also Like
How PAH Is Diagnosed
Common Searches
Adderall
Celexa
Cipro
Cymbalta
Flexeril
Hydrocodone
Prilosec
Prozac
Seroquel
Synthroid
Tramadol
Trazodone
Lexapro
Lisinopril
Mobic
Naproxen
Neurontin
Pradaxa
Prednisone
Vicodin
Warfarin
Wellbutrin
Xanax
Zocor
Zoloft
Show MoreShow Less
Drug Survey
Are you currently using Labetalol (Normodyne, Trandate)?
This survey is being conducted by the WebMD marketing sciences department.
Top doctors in Columbus, OH
From our sponsor
TN
Dr. Thomas Nguyen, MD
Family Medicine, Preventative Medicine, Emergency Medicine, Critical Care Medicine
5.0
2 Ratings
(833) 603-8376View Full Profile
KG
Kellie Greene, NP
Internal Medicine, Family Medicine, Nurse Practitioner, Preventative Medicine, Emergency Medicine, Critical Care Medicine
4.6
10 Ratings
(833) 603-9362View Full Profile
Dr. Chelsea Willis, DO
Internal Medicine, Family Medicine, Preventative Medicine, Emergency Medicine, Critical Care Medicine
5.0
1 Rating
(833) 813-6533View Full Profile
Related Links
Find a Drug
Pill Identifier
Interaction Checker
Drugs & Medications A-Z
Drugs & Medical Conditions
Drug and Medication Updates
Find a Vitamin
First Aid
Pain Management
Commonly Abused Drugs
OTC Pain Relief
How to Read Drug Labels |
9979 | http://physics.bu.edu/~duffy/semester2/c03_parallel_plate.html | J2S._Canvas2D (eField4.EField) "EFieldApp"[x]
Two parallel plates
Consider two large flat plates placed near one another. The plates are parallel, and have equal and opposite charges uniformly distributed. This configuration is known as a parallel-plate capacitor.
The net field is the vector sum of the individual fields, which have the same magnitude. In this situation there is a uniform field between the plates, and zero field everywhere else. The field between the plates is twice the field from one plate, so the net field is:
E = σ/ε o
A parallel-plate capacitor is a great way to create a uniform field. The field is perfectly uniform only when the plates are infinite in size, but as long as the width of the plates is large compared to the distance between the plates the field is fairly uniform.
EFieldApp ready |
9980 | https://www.youtube.com/watch?v=CXGRYOrOUJ0 | Discrete mathematics : - ( Function ; Equal functions ) - 31.
Ameen Academy
6310 subscribers
5 likes
Description
464 views
Posted: 23 Jan 2018
Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous.
Function is a relation between a set of inputs and set of permissible outputs with the property that input is related to exactly one output.
Equal Functions :
Domain of f = Domain of g ,
Range of x = Range of g.
f(x) = g(x) for every x belonging to their common domain.
For more videos please visit : www.ameenacademy. com
Please subscribe our YouTube channel for you to get latest Uploads.
Transcript:
welcome to Amin academy.com in discrete mathematics regarding functions let us discuss about equal functions two functions are said to be equal to functions or said to be equal if and only if the domain of F the first function f is equal the domain of the second function G and the comb the codomain of f is equal to first 1 the domain of F is equal to the domain of G and the co domain of F is equaled the core domain of G and f of X equal G of X for every X belong to there common domain these are the condition for the functions to be equal now let us do an example to understand this concept example if X equal set of 1 & 2 y is 3 and 6 and f of X gives Y means f of X equal to Y given by f of X equal x square plus 2 and G of X equal to 3x then show that f of X equal G of X means we have to show both functions are equal now f of X equal given x squared plus 2 and G of X equal 3 X is given now here we have to substitute the domain domain value of the function function that is f of 1 first one that is what the first function is XY plus 1 means 1 square plus 2 which is equivalent square is 1 so is 3 like G of 1 equal 3 into 1 equal 3 F of 2 they call 2 square plus 2 2 square is 4 4 plus 2 is 4 plus 2 is 6 here G of 2 equal 3 into 2 equals 6 hence we get f of X equal G of X so both functions are equal now let us do one more problem a function f is defined by f of X equal to x squared plus 5x minus 5 for X is greater than zero find the value of x which is unchanged by the mapping now since that it's given find the value of x which is unchanged by the mapping which means we can say f of X equal x now substitute f of X that is given x squared plus 5x mod X square minus plus five X minus five is equal X which gives X square plus Phi X minus X minus y equal to zero by simplification you get X square plus 4x minus 5 equals 0 by factorizing and solving we get X plus 5 and X minus 1 equal to 0 from that we get x equals minus y and x equals 1 the possible value of therefore can say the possible value of x is 100 the - we cannot take because the domain is given the value of X is greater than zero so we got the value of therefore we got the value of x is one they have done |
9981 | https://www.youtube.com/watch?v=YAk0htu7FnQ | How can we do that? 5 equal-area rectangles in a square problem. Reddit geometry r/theydidthemath
bprp math basics
234000 subscribers
2578 likes
Description
135817 views
Posted: 24 Aug 2024
Here's a fun "find the area" problem from Reddit. We are given a square that has been equally divided into 5 parts. We are only given one side is 2 and asked to find the area of the square. Is this question even possible?
I help students master the basics of math. You can show your support and help me create even better content by becoming a patron on Patreon 👉 Every bit of support means the world to me and motivates me to keep bringing you the best math lessons! Thank you!
reddit #mathbasics #geometry
263 comments
Transcript:
request how can we do that the question is we are
going to find the area of this square and we are given that the rectangles inside they all have
the same area so now let's take a look we only know this right here is two and then we have to
find the area of the square well to find the area of the square we must have the side times the side
right and because I don't know what the side right here is so let me just put down X right here and
of of course that will also be X and just a small note real quick once that's X the whole area
is just going to be x X and now it's x² so X second power all right now here's the deal 1 2 3
4 five we have five rectangles and they all have the same area the total area is x² so we just have
to divide it by five that means this right here must be x^2 over 5 likewise this likewise this
likewise this and also likewise that x² over 5 awesome now what though hm here's the deal because
we have the number right here right this is the only given number so we must somehow use it to
figure out what x is and uh we know this area is x² over 5 this is 2 so if we can figure pick out
this side from here to here then we can set up an equation but how though now check this out this
little strip right here right this it's just like you know the driveway of the building something
like that it's X squ over five right 1/5 of the whole Square so that means the whole thing left
right here must be what just 1 2 3 4 four of this together so the red part is 4x2 over 5 yeah now
check this out because we know from here to here is X and this right here is 4x² over 5 so from
here to here must be equal to what well just do x² / X we have X and then we still multiply by 4
over 5 so this part must be 4 over 5x just a quick check if we take this this X we do get 4x2 over
5 for the red part and now we have this that must be equal to this so we can set an equation
2 4 over 5x must be equal to X SAR over 5 and now we can just work this out so this is like 2
over 1 4 over 5 so this is 8 over 5x = x² over 5 multiply both sides by five now just let me show
you so they cancel and we have 8X = X2 and then we can put uh this to the other side but let me write
this down first so X2 minus 8x = Z and a factor on X so that means X is equal to 0 or x - 8 is
equal to 0 so X is equal to 8 of course we have a geometry equation if x is equal Zer
we don't have a square so not this we want X is equal to 8 so the area must be equal
to 8 squared so the area is 64 so that's it |
9982 | https://www.pearson.com/channels/calculus/asset/664a9781/the-accompanying-figure-shows-the-velocity-v-dsdt-ft-msec-of-a-body-moving-along | Calculus
Improve your experience by picking them
Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m/sec. Because the acceleration of gravity at the planet’s surface was gₛ m/sec², the explorers expected the ball bearing to reach a height of s = 15t − (1/2)gₛt² m t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of gₛ?
Motion Along a Line The graphs in Exercises 105 and 106 show the position s=f(t) of an object moving up and down on a coordinate line. At approximately what times is the (c) Acceleration equal to zero?
Motion Along a Line The graphs in Exercises 105 and 106 show the position s=f(t) of an object moving up and down on a coordinate line. At approximately what times is the (d) When is the acceleration positive? Negative?
Motion Along a Line The graphs in Exercises 105 and 106 show the position s=f(t) of an object moving up and down on a coordinate line. At approximately what times is the (b) velocity equal to zero?
Suppose the position of a particle moving along a straight line is given by the graph below. At time t=2 seconds, estimate the value of the velocity and acceleration of the particle using the graph.
Given the vector-valued function r(t)=(t2,23t3,t), find the unit tangent vector T, the unit normal vector N, and the unit binormal vector B at the point (4,-163,-2). Which of the following correctly gives the unit tangent vector T at that point?
Which of the following best describes the gradient vector field of the function f(x,y)=x2+y2?
Given the position equation s(t)s\left(t\right)s(t), calculate the average velocity (in meters per second) based on the given interval, and the instantaneous velocity (in meters per second) at the end of the time interval.
s(t)=9t−t2s\left(t\right)=9t-t^2s(t)=9t−t2, 0≤t≤30\le t\le30≤t≤3 |
9983 | https://www.youtube.com/watch?v=s8x59_iXDz4 | Sale Arithmetic: Relationship between total price, unit price and number of quantities
Miss Hamilton's Mathematics Class
611 subscribers
24 likes
Description
346 views
Posted: 21 Mar 2021
In this lesson you will solve problems that involve finding:
1. number of quantities given the total price and unit price
2. unit price given the total price and number of quantities
3. total price given unit price and number of quantities
5 comments
Transcript:
[Applause] [Music] [Applause] [Music] [Applause] [Music] [Applause] [Music] student student students welcome back to another lesson today we're looking at sales arithmetic specifically we're going to learn how to calculate total price when we're given the quantity and the unit price we're going to compute the unit price when we're given the quantity and total price and we're going to calculate the quantity when we have the total price and the unit price let's first let's start off with some definition and definitions and formulas right so unit price unit means one and we learned that from when we were doing um best buy see unit price cost of one item and that's the formula for calculating the price of one item we take our total price and we divide it by the number of quantities or the number of things that we've purchased quantity is the item being sold and to find the quantity we take our u total price and we divide it by unit price and finally total price is the number of quantities multiplied by the unit price let's look at this example so you went to your corner shop and these are some of the um the items on the on the in the store all right first question what is the cost of two things of tuna how do we do that we want to know how much two things are tuna costs remember total price is number of quantities times unit price so there are two tunas and the unit price is 180 we multiply that and we get 360 dollars awesome next question what is the cost of three spice bones and two slices of cheese we're going to use the same formula there are three spice bonds all right so we're going to multiply that by 150 the unit price and we get that 450. there are two sizes of trees so we're going to multiply that by the unit price which is 75 and get 150. then we're going to add it all right at those prices now we get 600 dollars let's look on this example a box of donor costs four hundred dollars if one donor costs seventy five dollars how many donors are there in the box so an entire box costs 450 and one out of the box cost them to five so how many donuts are in the box who are we going to figure that out to find the quantity we take our total price and divide it by our unit price like so and the answer that we get is six so 450 divided by 75 is six try this final example a case of soda costs 960 dollars if there are one dozen sodas in the case how much would one soda cost what formula are we going to use here right we need to find the unit price so we're going to take the total price that we're provided with are going to divide it by the number of quantities and how many items make up a dozen 12 so we're going to take our 960 divided by 12 and that's our answer we get 80 dollars all right no recap what's the unit price how do we find unit price how do we find number of quantities how do we find the total sales all right once you know that then calculating sale items will be easy thanks for watching this video |
9984 | https://www.osmosis.org/answers/condylomata-lata | Condylomata Lata: What Is It, Cause, Presentation, and More | Osmosis
Plans
Medicine (USMLE)Medicine (non-USMLE)Medicine (Osteopathy)Physician AssistantNurse PractitionerRegistered NursingLicensed Practical NursingDentistryPharmacyHealth Professional
Library
Medicine (USMLE)Medicine (non-USMLE)Medicine (Osteopathy)Physician AssistantNurse PractitionerRegistered NursingLicensed Practical NursingDentistryPharmacyHealth Professional
Resources
EventsBlogQuestions & answer topicsRaise the Line PodcastSpreadjoy e-cardsOsmosis around the world
InstitutionsSign inTry it free
Fall in love with Osmosis at 20% off! Save now until September 30 at 11:59 PM PT.
Learn more
Skip to the first question
Condylomata Lata
What Is It, Cause, Presentation, and More
Author: Nikol Natalia Armata, MD
Editor: Alyssa Haag, MD
Editor: Emily Miao, MD, PharmD
Editor: Kelsey LaFayette, DNP, ARNP, FNP-C
Illustrator: Jillian Dunbar
Copyeditor: David G. Walker
Modified: May 09, 2025
What is condylomata lata?
Condylomata lata, also known as condyloma latum, refers to a benign and painless cutaneous manifestation of secondary syphilis. They are hypopigmented, gray or white moist plaques located in the genital area, inner thighs, axillae, or within the oral mucosa. Condylomata lata are reported in about 9% to 44% of individuals who have syphilis.
Syphilis is one of the most commonsexually transmitted infectionscaused by the spirochete bacteria, Treponema pallidum. It progresses through three distince stages: primary, secondary, and tertiary. If left untreated, secondary syphilis can develop 2 to 8 weeks after the initial infection, and can be associated with systemic symptoms such as fatigue, fever, lymphadenopathy.
Conditions that can mimic condylomata lata include genital warts (i.e., condyloma acuminatum) fromhuman papillomavirus (HPV)infection and malignancy (e.g., squamous cell carcinoma).
What causes condylomata lata?
Condylomata lata are caused by syphilis; they are characteristic of secondary stage syphilis. Secondary syphilis can occur approximately 2 to 8 weeks after the initial presentation of primary syphilis lesions, if left untreated. During the secondary phase, the infection spreads throughout the body, and the infected individual is highly contagious.
What are the signs and symptoms of condylomata lata?
Condylomata lata appear as smooth, soft, flat skin growths that may vary in shape and size, and range in color from white or gray but can be pink and flesh colored. They typically develop in warm, moist regions, such as the genitals (e.g., penis, labia), anus, or mouth. Less frequently, they can also develop in the axilla, palms, face, umbilicus, and between the toes. In individuals who are immunocompromised, such as those withHIV/AIDS, transplanted organs, or undergoing chemotherapy, condylomata lata can multiply into large clusters.
What are the differential diagnoses for condylomata lata?
Differential diagnoses involve considering various possible conditions that could be causing symptoms and then ruling out each one through use of history, clinical evaluation, diagnostic tests, and critical thinking. This process helps to narrow down the list of potential diagnoses to determine the most likely cause of the symptoms.
Differential diagnoses can be broken down into four categories:most likely,less likely,least likely, andcan’t miss.Most likelydiagnoses are conditions most probable based on symptoms and clinical presentation.Less likelydiagnoses are not as probable but should still be considered. On the other hand,least likelydiagnoses can be considered if other, more probable conditions are excluded. Finally,can’t missdiagnoses are less common but critical to promptly identify and treat as they can lead to severe consequences.
Differential diagnoses for condylomata lata include:
Most likely:
Condyloma acuminata: Also known as genital warts, caused by human papillomavirus (HPV). These are similar in appearance but are typically cauliflower-like and dry, unlike the moist, flat lesions of condylomata lata.
Molluscum contagiosum: Caused by a poxvirus, presenting as small, firm, dome-shaped papules with a central dimple. These can be mistaken for condylomata lata but are usually smaller and less moist.
Genital herpes: Caused by herpes simplex virus, presenting with painful vesicles and ulcers. While different in appearance, it can be confused with condylomata lata in early stages.
Less likely:
Lichen planus: An inflammatory condition causing flat-topped, purple lesions. It can affect the genital area but is usually more itchy and less moist than condylomata lata.
Least likely:
Fixed drug eruption: A reaction to medication causing round, red patches that recur at the same site. These can appear on the genitals but are usually solitary and well-defined.
Pemphigus vulgaris: An autoimmune blistering disorder that can affect mucous membranes, including the genitals, but presents with painful blisters rather than wart-like lesions.
Behçet disease: A rare inflammatory disorder causing ulcers in the mouth and genitals, but these are typically painful and not wart-like.
Can’t miss:
Syphilis (primary and tertiary):Primary syphilis presents with a painless chancre, while tertiary syphilis can cause gummas and systemic symptoms. Both require prompt diagnosis and treatment to prevent complications.
HIV/AIDS:Immunosuppression can lead to various skin manifestations, including atypical presentations of infections like syphilis.
Squamous cell carcinoma: A type of skin cancer that can present as a persistent, non-healing lesion on the genitals. It requires prompt diagnosis and treatment to prevent progression and spread.
How is condylomata lata diagnosed and treated?
Diagnosis of condylomata lata begins with a history and physical examination. A healthcare provider may ask about the individual’s sexual history and any prior episodes of sexually transmitted diseases. In most cases, condylomata lata can be easily identified and diagnosed through physical examination of the affected area(s). Examination of the lesion under magnification using a colposcope may also be used if a more detailed evaluation is required.
Oftentimes, diagnosis can be confirmed by additional diagnostic testing, including serologic tests, which are blood tests detecting the presence of antibodies against Treponema pallidum. Dark-field microscopy, a specialized microscope, can be used to directly visualize the spirochetes. A biopsy may be needed to better differentiate syphilitic condylomata from HPV-induced lesions (e.g. genital warts) or from cancerous lesions. The visualization of numerous spirochetes by immunostaining, which is the process of selectively identifying proteins in cells, can also confirm the diagnosis of condylomata lata.
Treatment of condylomata lata requires treatment of the underlying syphilis infection. The first-choice treatment for all manifestations of syphilis is an intramuscular injection of the antibiotic, penicillin. For individuals allergic to penicillin, alternatives, such as doxycycline, ceftriaxone, or penicillin desensitization, where induced tolerance to penicillin occurs, may also be administered. It may take a few months for condyloma lata to resolve completely.
What are the most important facts to know about condylomata lata?
Condylomata lata refer to benign, painless, wart-like lesions associated with secondary syphilis. They are typically gray-white growths that develop in warm, moist regions, like the mouth, genitals, or anus. Typically, condylomata lata are smooth, soft, flat, and but can vary in shape and size. Diagnosis can be made through a review of medical history and a physical exam of the affected area. When necessary, further serologic testing or even a biopsy may follow. Treatment of condylomata lata is focused on treating the underlying syphilis infection, typically with the administration of penicillin.
Related videos and concepts
13:49 Cell wall synthesis inhibitors: Penicillins 10:01 Syphilis: Nursing 13:58 Treponema pallidum (Syphilis) 22:39 Anatomy of the male urogenital triangle 9:01 Genital warts: Nursing
References
Barei F, Murgia G, Stefano Ramoni, Cusini M, Marzano AV. Secondary syphilis with extra-genital condyloma lata: A case report and review of the literature.International Journal of STD & AIDS. 2022;33(12):1022-1028. doi:
Herzum A, Burlando M, Micalizzi C, Parodi A. Condylomata lata and papular rash of secondary syphilis.Actas Dermo-Sifiliograficas. 2023;114(5):447. doi:
Pourang A, Fung MA, Tartar D, Brassard A. Condyloma lata in secondary syphilis.JAAD Case Rep. 2021;10:18-21. Published 2021 Feb 9. doi:10.1016/j.jdcr.2021.01.025
Tang B, Chang X. Condylomata lata. New England Journal of Medicine. 2023;388(2):e3. doi:
Company
About usCareersLibraryEditorial BoardBlogBecome An Osmosis Student Leader
Support
Help centerContact usFAQ
For Institutions
Medicine (DO)Medicine (MD)Nurse Practitioner (NP)Physician Assistant (PA)DentistryNursing (RN)
More
PlansiOS appAndroid appSwagCreate custom contentRaise the Line Podcast
Copyright © 2025 Elsevier, its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies.
Cookies are used by this site.
Cookie settings | Your Privacy Choices
Terms and Conditions
Privacy Policy
Cookie Notice
Sitemap
USMLE® is a joint program of the Federation of State Medical Boards (FSMB) and the National Board of Medical Examiners (NBME). COMLEX-USA® is a registered trademark of The National Board of Osteopathic Medical Examiners, Inc. NCLEX-RN® is a registered trademark of the National Council of State Boards of Nursing, Inc. Test names and other trademarks are the property of the respective trademark holders. None of the trademark holders are endorsed by nor affiliated with Osmosis or this website.
We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. For more information, see ourCookie Policy
Cookie Settings Accept all cookies
Cookie Preference Center
We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors.
You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings.
You may also be able to exercise your privacy choices as described in our Privacy Policy
Allow all
Manage Consent Preferences
Strictly Necessary Cookies
Always active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookie Details List
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookie Details List
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site.
Cookie Details List
Targeting Cookies
[x] Targeting Cookies
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. If you do not allow these cookies, you will experience less targeted advertising.
Cookie Details List
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Confirm my choices |
9985 | https://ell.stackexchange.com/questions/65667/is-the-word-meander-the-same-as-wander | Is the word 'meander' the same as 'wander'? - English Language Learners Stack Exchange
Join English Language Learners
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
English Language Learners helpchat
English Language Learners Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Is the word 'meander' the same as 'wander'?
Ask Question
Asked 10 years, 1 month ago
Modified10 years, 1 month ago
Viewed 1k times
This question shows research effort; it is useful and clear
4
Save this question.
Show activity on this post.
In a dictionary, I found the word 'meander', which means 'wander'. I've heard a lot about the phrase including 'wander', but never about meander. So is it ok to say, "I meandered through the forest."? Is it the same as 'wandered'?
word-difference
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Improve this question
Follow
Follow this question to receive notifications
asked Aug 29, 2015 at 7:59
AkihiroAkihiro
531 4 4 silver badges 13 13 bronze badges
2
2 While it does strictly mean the same as "wander", "meander" is most often encountered in reference to a river winding across a landscape. Saying that you "meander through a forest" would be interpreted as saying that you wandered through it in a particular way, performing a sort of slow left-right slalom.IanF1 –IanF1 2015-08-29 08:18:54 +00:00 Commented Aug 29, 2015 at 8:18
Yep. Even when the dictionary defines it simply as identical in actual meaning, 'meander' carries the connotation of referring to a frequently changing direction, whereas 'wander' could refer to not only a back-and-forth path, but also a fairly straight (or, realistically, slightly curved in one direction) path through the forest, just as long as there's not a preplanned route or destination involved.Dan Henderson –Dan Henderson 2015-08-29 19:12:26 +00:00 Commented Aug 29, 2015 at 19:12
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
4
Save this answer.
Show activity on this post.
You could use "meander"; however, it is more specific than "wander". (See IanF1's comment.)
meanderverb
to proceed by or take a winding or indirect course:
the stream meandered through the valley
to wander aimlessly; ramble:
the talk meandered on
Source: Dictionary.com
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Improve this answer
Follow
Follow this answer to receive notifications
answered Aug 29, 2015 at 9:12
Dog LoverDog Lover
1,696 10 10 silver badges 17 17 bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
word-difference
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Related
2Difference between nonconformity and inconformity?
3Go over/through vs look over/through
4Private Entrepreneur vs Sole proprietorship (Sole Trader)
1Borderline between "translucent" and "transparent"
1Autumn vs Fall - difference in meaning?
1To turn (around/round) the corner
0Do 'make a joke of' and 'joke around' have the same meaning?
4"arrive" Vs. "reach" what are the differences if any?
1Is there a specific word for financial records?
0Are marketing ploys the same as marketing gimmicks?
Hot Network Questions
How to rsync a large file by comparing earlier versions on the sending end?
My dissertation is wrong, but I already defended. How to remedy?
Analog story - nuclear bombs used to neutralize global warming
Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results
What is this chess h4 sac known as?
Clinical-tone story about Earth making people violent
Is existence always locational?
Spectral Leakage & Phase Discontinuites
Passengers on a flight vote on the destination, "It's democracy!"
How to start explorer with C: drive selected and shown in folder list?
What were "milk bars" in 1920s Japan?
Is encrypting the login keyring necessary if you have full disk encryption?
Numbers Interpreted in Smallest Valid Base
Making sense of perturbation theory in many-body physics
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
Can a cleric gain the intended benefit from the Extra Spell feat?
Suspicious of theorem 36.2 in Munkres “Analysis on Manifolds”
Calculating the node voltage
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
Cannot build the font table of Miama via nfssfont.tex
Overfilled my oil
Are there any world leaders who are/were good at chess?
Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward"
Change default Firefox open file directory
more hot questions
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
English Language Learners
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
9986 | https://teachers.institute/assessment-for-learning/comparing-mean-median-mode-assessments/ | Assessment for Learning
Comparing Mean, Median, and Mode in Educational Assessments
When we think of assessing students’ performance or understanding the distribution of data in any educational context, one of the first statistical tools that comes to mind is the concept of “central tendency.” Central tendency refers to the central or typical value in a dataset, and it helps in summarizing large sets of data in a meaningful way. The three most commonly used measures of central tendency are the mean, median, and mode. However, each of these has its strengths and limitations depending on the type of data you are dealing with. In this blog post, we will explore and compare the mean, median, and mode, highlighting their respective advantages, limitations, and how they can be applied to different educational assessments. By the end, you will have a better understanding of how to select the most appropriate measure of central tendency based on the nature of the data and the specific goals of your assessment.
Table of Contents
Understanding the Three Measures of Central Tendency
What is the Mean?
What is the Median?
What is the Mode?
Comparing Mean, Median, and Mode
Advantages and Limitations of the Mean
Advantages and Limitations of the Median
Advantages and Limitations of the Mode
Choosing the Right Measure for Educational Assessments
When to Use the Mean
When to Use the Median
When to Use the Mode
Conclusion
Understanding the Three Measures of Central Tendency
The mean, median, and mode are essential tools in descriptive statistics. They offer insights into the central location of a dataset, but each one serves a different purpose. Let’s break them down to understand their unique characteristics.
What is the Mean?
The mean, often referred to as the average, is calculated by adding up all the values in a dataset and then dividing by the total number of values. For example, if a class of students scored 60, 70, 80, and 90 in an exam, the mean score would be:
Mean = (60 + 70 + 80 + 90) / 4 = 75
The mean is particularly useful when the data is spread evenly, and every data point holds similar importance. It’s highly sensitive to extreme values (outliers), which can skew the result significantly. For example, if one student in the class scored 10 instead of 60, the mean would drastically change. This makes the mean less reliable for data sets with outliers.
What is the Median?
The median represents the middle value in a dataset when the numbers are arranged in ascending or descending order. If the number of data points is odd, the median is the middle number; if even, it is the average of the two middle numbers. For example, if five students scored the following marks: 30, 40, 60, 80, and 100, the median score would be 60 because it is the middle value when the scores are arranged in order. However, if the dataset had an even number of scores, say 30, 40, 60, 70, 80, and 100, the median would be calculated as:
Median = (60 + 70) / 2 = 65
The median is resistant to outliers, making it a better measure when the dataset contains extreme values. For example, if one student scored 5, the median would remain unaffected, whereas the mean would decrease.
What is the Mode?
The mode refers to the value that occurs most frequently in a dataset. A dataset may have one mode (unimodal), more than one mode (bimodal or multimodal), or no mode at all if all the values occur with equal frequency. For instance, if the scores of students are 50, 60, 70, 70, 80, and 90, the mode would be 70 because it occurs twice, while the others appear only once. If every score in a dataset appears once, the dataset has no mode.
The mode is particularly useful for categorical data where we want to know which category or value is most popular. It is simple and easy to identify, but like the median, it can be less informative when dealing with continuous or numerical data.
Comparing Mean, Median, and Mode
While all three measures of central tendency aim to represent the “center” of a dataset, they each do so in different ways and are suited to different types of data. Let’s examine their advantages and limitations in more detail.
Advantages and Limitations of the Mean
Advantages:
Provides a precise measure when the data are continuous and symmetrical.
Useful for further statistical analysis (e.g., variance, standard deviation).
Reflects the influence of every data point in the dataset.
Limitations:
Sensitive to outliers, which can distort the results if there are extreme values.
Not suitable for skewed distributions, as it may not represent the “typical” value.
Advantages and Limitations of the Median
Advantages:
Not affected by outliers, making it a robust measure for skewed data.
Useful when the data is ordinal or when dealing with non-symmetrical distributions.
Limitations:
Does not take into account the exact values of all the data points—only the middle value.
Less precise than the mean for symmetric, continuous datasets.
Advantages and Limitations of the Mode
Advantages:
Simple and easy to identify, especially for categorical data.
Useful when dealing with nominal data where other measures like mean and median don’t apply.
Can be used for any type of data (nominal, ordinal, interval, and ratio).
Limitations:
May not exist or be useful for datasets with no repeating values.
Does not provide information about the spread or dispersion of the data.
Choosing the Right Measure for Educational Assessments
In the context of educational assessments, selecting the right measure of central tendency is crucial for making informed decisions about students’ performance. The choice between the mean, median, and mode depends largely on the type of data you are dealing with and the insights you are looking for. Let’s explore some scenarios where each measure is best suited.
When to Use the Mean
The mean is ideal for numerical, continuous data, particularly when you want to understand the overall trend of performance. For example, in a large class of students taking a math exam, the mean score gives an accurate picture of the average performance. However, you should be cautious if there are extreme scores (outliers), as these can distort the average. For instance, if most students scored between 70 and 90, but one student scored 10, the mean would not truly reflect the typical student’s performance.
When to Use the Median
The median is particularly useful in cases where the data is skewed or contains outliers. For instance, in a class where most students performed well on a test but a few students had poor results, the median would give a more accurate representation of the “typical” student’s performance. Similarly, when dealing with ordinal data, such as survey responses on a Likert scale (strongly agree, agree, neutral, disagree, strongly disagree), the median can help capture the middle preference or opinion.
When to Use the Mode
The mode is useful in cases where you want to know the most frequent value or category in your data. For example, in a classroom where students choose their preferred study material from a list of options, the mode can reveal the most popular choice. The mode is also helpful when dealing with categorical data, such as identifying the most common grade, or when analyzing attendance patterns (e.g., the most common day of the week for school absences).
Conclusion
In summary, each measure of central tendency—mean, median, and mode—has its own strengths and is useful in different educational assessment scenarios. The mean is suitable for continuous data but can be influenced by outliers. The median is robust and works well for skewed distributions or ordinal data. The mode, while simple, is great for categorical data and identifying the most common values.
As an educator or someone involved in educational assessments, understanding when to use each measure will help you make more accurate and meaningful interpretations of your students’ performance. Consider the nature of your data—whether it’s continuous, ordinal, or nominal—and your specific assessment goals when choosing the most appropriate measure of central tendency.
What do you think? Have you ever faced a situation where the choice of central tendency measure significantly affected the interpretation of your data? How did you handle it?
How useful was this post?
Click on a star to rate it!
Average rating 0 / 5. Vote count: 0
No votes so far! Be the first to rate this post.
We are sorry that this post was not useful for you!
Let us improve this post!
Tell us how we can improve this post?
PDF 📄
Comments
Leave a Reply Cancel reply
Assessment for Learning
1 Concept and Purpose of Evaluation
Basic Concepts
Relationships among Measurement, Assessment, and Evaluation
Teaching-Learning Process and Evaluation
Assessment for Enhancing Learning
Other Terms Related to Assessment and Evaluation
2 Perspectives of Assessment
Behaviourist Perspective of Assessment
Cognitive Perspective of Assessment
Constructivist Perspective of Assessment
Assessment of Learning and Assessment for Learning
3 Approaches to Evaluation
Approaches to Evaluation: Placement Formative Diagnostic and Summative
Distinction between Formative and Summative Evaluation
External and Internal Evaluation
Norm-referenced and Criterion-referenced Evaluation
Construction of Criterion-referenced Tests
4 Issues, Concerns and Trends in Assessment and Evaluation
What is to be Assessed?
Criteria to be used to Assess the Process and Product
Who will Apply the Assessment Criteria and Determine Marks or Grades?
How will the Scores or Grades be Interpreted?
Sources of Error in Examination
Learner-centered Assessment Strategies
Question Banks
Semester System
Continuous Internal Evaluation
Choice-Based Credit System (CBCS)
Marking versus Grading System
Open Book Examination
ICT Supported Assessment and Evaluation
5 Techniques of Assessment and Evaluation
Concept Tests
Self-report Techniques
Assignments
Observation Technique
Peer Assessment
Sociometric Technique
Portfolios
Project Work
Debate
School Club Activities
6 Criteria of a Good Tool
Evaluation Tools: Types and Differences
Essential Criteria of an Effective Tool of Evaluation
Reliability
Validity
Usability
Objectivity
Norm
7 Tools for Assessment and Evaluation
Paper Pencil Test
Oral Test
Aptitude Test
Achievement Test
Diagnostic–Remedial Test
Intelligence Test
Rating Scales
Questionnaire
Inventories
Checklist
Interview Schedule
Observation Schedule
Anecdotal Records
Learners Portfolios and Rubrics
8 ICT Based Assessment and Evaluation
Importance of ICT in Assessment and Evaluation
Use of ICT in Various Types of Assessment and Evaluation
Role of Teacher in Technology Enabled Assessment and Evaluation
Online and E-examination
Learners’ E-portfolio and E-rubrics
Use of ICT Tools for Preparing Tests and Analyzing Results
9 Teacher Made Achievement Tests
Understanding Teacher Made Achievement Test (TMAT)
Types of Achievement Test Items/Questions
Construction of TMAT
Administration of TMAT
Scoring and Recording of Test Results
Reporting and Interpretation of Test Scores
10 Commonly Used Tests in Schools
Achievement Test
Aptitude Test
Achievement Test Versus Aptitude Test
Performance Based Achievement Test
Diagnostic Testing and Remedial Activities
Question Bank
Oral Test
General Observation Techniques
Practical Test
11 Identification of Learning Gaps and Corrective Measures
Educational Diagnosis
Diagnostic Tests: Characteristics and Functions
Diagnostic Evaluation Vs. Formative and Summative Evaluation
Diagnostic Testing
Achievement Test Vs. Diagnostic Test
Diagnosing and Remedying Learning Difficulties: Steps Involved
Areas and Content of Diagnostic Testing
Remediation
12 Continuous and Comprehensive Evaluation
Continuous and Comprehensive Evaluation: Concepts and Functions
Forms of CCE
Recording and Reporting Students Performance
Students Profile
Cumulative Records
13 Tabulation and Graphical Representation of Data
Use of Educational Statistics in Assessment and Evaluation
Meaning and Nature of Data
Organization/Grouping of Data: Importance of Data Organization and Frequency Distribution Table
Graphical Representation of Data: Types of Graphs and its Use
Scales of Measurement
14 Measures of Central Tendency
Individual and Group Data
Measures of Central Tendency: Scales of Measurement and Measures of Central Tendency
The Mean: Use of Mean
The Median: Use of Median
The Mode: Use of Mode
Comparison of Mean, Median, and Mode
15 Measures of Dispersion
Measures of Dispersion
Standard Deviation
16 Correlation – Importance and Interpretation
The Concept of Correlation
Types of Correlation
Methods of Computing Co-efficient of Correlation (Ungrouped Data)
Interpretation of the Co-efficient of Correlation
17 Nature of Distribution and Its Interpretation
Normal Distribution/Normal Probability Curve
Divergence from Normality
Share on Mastodon |
9987 | https://www.britannica.com/science/significant-figures | SUBSCRIBE
SUBSCRIBE
Home
History & Society
Science & Tech
Biographies
Animals & Nature
Geography & Travel
Arts & Culture
ProCon
Money
Games & Quizzes
Videos
On This Day
One Good Fact
Dictionary
New Articles
History & Society
Lifestyles & Social Issues
Philosophy & Religion
Politics, Law & Government
World History
Science & Tech
Health & Medicine
Science
Technology
Biographies
Browse Biographies
Animals & Nature
Birds, Reptiles & Other Vertebrates
Bugs, Mollusks & Other Invertebrates
Environment
Fossils & Geologic Time
Mammals
Plants
Geography & Travel
Geography & Travel
Arts & Culture
Entertainment & Pop Culture
Literature
Sports & Recreation
Visual Arts
Image Galleries
Podcasts
Summaries
Top Questions
Britannica Kids
Ask the Chatbot Games & Quizzes History & Society Science & Tech Biographies Animals & Nature Geography & Travel Arts & Culture ProCon Money Videos
significant figures
Rules for significant figures
Calculations with significant figures
References & Edit History Quick Facts & Related Topics
significant figures
print Print
Please select which sections you would like to print:
verifiedCite
While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions.
Select Citation Style
Share
Share to social media
Facebook X
URL
Feedback
Thank you for your feedback
Our editors will review what you’ve submitted and determine whether to revise the article.
External Websites
National Center for Biotechnology Information - PubMed Central - Significant Figures
Bellevue College - Department of Physics - Significant Figures
Khan Academy - Intro to significant figures
Indiana University - Meaningful Numbers and �Significant Figures�
Open Library Publishing Platform - Significant Figures
Open Washington Pressbooks - Significant Figures
Chemistry LibreTexts - Significant Figures in Calculations
United States Naval Academy - Significant figures
OpenStax - University Physics Volume 1 - Significant Figures
Oberlin College and Conservatory - Significant Figures (PDF)
UMass Amherst Libraries - Open Books - Physics 132 Lab Manual - Review of Significant Figures
BCcampus Open Publishing - Significant Figures
Also known as: significant digits
Written by
Written by
Ken Stewart
Ken Stewart is a former educator with an honours degree in chemistry, physics, and mathematics.
Ken Stewart
Fact-checked by
Fact-checked by
The Editors of Encyclopaedia Britannica
Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree....
The Editors of Encyclopaedia Britannica
Last Updated: •Article History
Also called:
: significant digits
Related Topics:
: numerals and numeral systems
: number
On the Web:
: OpenStax - University Physics Volume 1 - Significant Figures (Aug. 30, 2025)
See all related content
significant figures, any of the digits of a number beginning with the digit farthest to the left that is not zero and ending with the last digit farthest to the right that is either not zero or that is a zero but is considered to be exact. Significant figures are used to report a value, measured or calculated, to the correct number of decimal places or digits that will reflect the precision of the value.
The number of significant figures a value has depends on how it was measured, or how it was calculated. When a measurement is taken, the precision of that measurement is dependent on the equipment used to take the measurement. The measurement will have some digits that are certain and one digit that is uncertain, or estimated. The digits will be certain to the lowest increment division of the equipment used to take the measurement, and the estimated digit will be based on a best guess when the measurement is between two of the lowest increments of the equipment. If, for example, a measurement is taken with a metric ruler that has millimetre markings as the lowest increment, the number of millimetres of the measurement will be known for certain. One other digit will be estimated, since the measured quantity can be between two millimetre increments. If the item being measured falls exactly on a millimetre marking, then the estimated digit is written as a zero, to indicate that the digit of the measurement beyond the millimetre increments on the ruler is thought to be a zero, and not a nonzero estimated value.
The basic concept of significant figures in measurements relates to the fact that a measured value cannot be more precise than the equipment being used to make the measurement. It should be noted here that the equipment used to take a measurement does not determine the number of significant figures a measured value will have, only where the precision of a value must stop being written. Only one estimated value can ever be included in a measured quantity.
Rules for significant figures
Determining the number of significant figures in measured quantities is essential when reporting the precision of measured values and the precision that can be reported when measured values are used in calculations. The rules for determining the number of significant figures are as follows:
All nonzero digits are significant.
For example, the value 211.8 has four significant figures.
All zeros that are found between nonzero digits are significant.
Thus, the number 20,007, with three 0s between the 2 and 7, has a total of five significant figures.
Leading zeros (to the left of the first nonzero digit) are not significant.
A value such as 0.0085, for example, has two significant figures because the 0s before the 8 are placeholders and are not significant.
Trailing zeros for a whole number that ends with a decimal point are significant.
For example, a value written as 320. shows the decimal point, which indicates that the 0 to the right of the 2 was measured; therefore, the value has a total of three significant figures. If the decimal point was not written, then 320 would have only two significant figures. In general, any confusion this may cause can be avoided by writing values such as these in scientific notation.
Trailing zeros to the right of the decimal place are significant.
This means a value such as 12.000 has a total of five significant figures, since the 0s after the decimal place have been measured to be zeros, indicating they are as significant as any other nonzero digit.
Exact numbers, and irrationally defined numbers like Euler’s number (e) and pi (π), have an infinite number of significant figures.
In a defining expression like 1 metre = 100 centimetres, these values are considered exact and thus have an infinite number of significant figures. While π is usually written as 3.14 for ease of calculation, the π button on the calculator would be used in any calculations, and thus it is considered to be a value with infinite significant figures.
For any value written in scientific notation as A ×10x, the number of significant figures is determined by applying the above rules only to the value of A; the x is considered an exact number and thus has an infinite number of significant figures.
For example, the value 4,500 can be written in scientific notation to reflect two, three, and four significant digits:
4.5 × 103 has two significant figures
4.50 × 103 has three significant figures
4.500 × 103 has four significant figures
Calculations with significant figures
For calculations involving measured quantities, the first step in determining the precision of the answer is to determine the number of significant figures in each of the measured quantities. Once done, the number of significant figures in a calculated value involving measurements is determined based on the mathematical operation being performed.
When two or more measured quantities are added or subtracted, the resulting value will have the same number of decimal places as the value with the fewest number of decimal places (the limiting value). So if the measured values of 22.35 and 47.773 are added, the limiting value of 22.35 has two decimal places, which means that the result of the addition will have only two decimal places.
When two or more measured quantities are being multiplied or divided, the answer will have the same number of total significant figures as the value with the fewest number of significant figures. So if the measured values of 2.445 and 31.7 are being multiplied, the resulting value will have three significant figures, since 2.445 has four significant figures, but 31.7 has only three significant figures.
Access for the whole family!
Bundle Britannica Premium and Kids for the ultimate resource destination.
When a value is to be rounded off, the rules for rounding are:
When the digit to the right of the one being rounded to is less than a 5, the remaining digit remains the same as the value rounds down.
For example, 33.742 is to be rounded to one decimal place. Here, the 7 in the first decimal place is followed by a 4, which is less than 5, which means that 33.742 rounded to one decimal place is 33.7. Note that only the 4 that is to the right of the 7 is looked at here; the 2 in the third decimal place is insignificant when rounding to one decimal place.
When the digit to the right of the one being rounded to is greater than 5, the value rounds up.
For example, 2.8763 is to be rounded to two decimal places. In this case, the 6 in the third decimal place is greater than 5, so the 7 in the second decimal place is rounded up to 8. This means that when rounded to two decimal places, 2.8763 rounds to 2.88. Again, the 3 in the fourth decimal place is insignificant when rounding to two decimal places.
When the digit to the right of the one being rounded to is exactly a 5 (which means no nonzero digit follows it), the value is rounded so that the final digit is an even number. This rule is designed to avoid always rounding up or always rounding down; it creates more balance when rounding.
Thus, 21.45 rounds to one decimal place to 21.4, while 36.75 would round to 36.8.
However, if a value such as 38.25003 is to be rounded to one decimal place, it rounds to 38.3. This is the only type of rounding where a digit farther than immediately to the right of the one being rounded to is ever considered. In this example, the digit looked at when rounding off to one decimal place is a 5. However, farther along the decimal portion of the value there is a nonzero digit. The number being rounded is therefore rounded up, as the 0.00003 indicates that the value of 0.05003 is larger than just the 0.05. For this reason, the value rounded to one decimal place is 38.3, not 38.2.
Ken Stewart |
9988 | https://www.youtube.com/watch?v=vo9HEpwR4_Q | Comparing Fractions - 3rd Grade Math [For TEACHERS & PARENTS]
McCarthy Math Academy
74700 subscribers
157 likes
Description
12736 views
Posted: 16 Nov 2021
If you are a teacher, parent, or anyone else who needs a REFRESHER to help a struggling 3rd, 4th, or 5th grader, this video is for you!
My name is Sarah McCarthy, and my mission is to make MATH make SENSE!
In just 9 minutes, you'll learn:
- Important vocabulary
- How to compare fractions with the same NUMERATOR.
- How to compare fractions with the same DENOMINATOR
- How to compare fractions using a visual model.
- How to check your work using cross multiplication (butterfly method)
- How to help students reason with fractions.
- And much, much more.
Here are the links I refer to in the video:
- Website:
- More Videos like this:
- Multiplication MashUp:
- Email: McCarthyMathAcademy@gmail.com
- Instagram:
- Facebook:
Transcript:
are you helping some third graders compare fractions and you don't even know where to get started well don't worry because in this episode we are gonna break that down what's going on everybody my name is sarah but a lot of third fourth and fifth graders know me as miss mccarthy while i create a ton of math videos and resources for students i thought it might be helpful to create some videos for you for the parents teachers tutors basically anyone who is looking to help third fourth and fifth graders to make math make sense so without further ado let's jump into today's episode so today i'm going to teach you how to compare fractions two different ways one using a visual model which will help to make the math make sense and another way is a way that i teach students to check their work using cross multiplication so for their grade we are comparing two fractions right and usually we either have the same numerator or the same denominator so what i'm going to do is make it the same new denominator first let's say that we have two fractions they both have a denominator of six and they have let's say this one has a denominator of i'm sorry a numerator of three and over here we have a numerator of four the top number is called the numerator the bottom number is called the denominator and have a fancy little song that goes like this you see that number on top that's called the numerator it describes the amount that is being considered or shaded and when you jump down from the fraction bar denominator it's the total number of equal parts sure love me some fractions okay so we have two different fractions right there we have three sixths and four sixths all right six is so difficult for me to say um so to compare two fractions we're going to draw it out first all right so to draw them out first you start with a large rectangle this is how i do it anyway and this is how i teach students to do it draw a large rectangle we have two fractions so we're going to split that rectangle in half and we will place three sixths on top and four sixths on the bottom all right move it up so you can see it okay now we break the parts into the total number of equal parts which is six right which is even and we learned in a previous episode that if a denominator is even you can cut it in half and then we can halves these it will do three on that side and three equal parts on that side to make it six equal parts now notice i went ahead and i split both of these rectangles why because they both have a denominator of six so they should both have equal parts now for the numerator it describes the amount that is being considered so we're gonna shade in those three equal parts one two three on top and we have one two three four on the bottom one because our second fraction is four sixths i usually i wait for a student to ask this because it creates that teachable moment but they always say miss mccarthy do i can i shade in like one two and then three over here and i'll say you absolutely could but when we're comparing fractions let's start just like when we read start from the left and move our way over that way it's easy to compare our fractions okay and here it's it's easy to tell when we compare them that 4 6 is actually shaded in longer right and this was a simple example because we have the same denominator right what we want to get students to is where they can reason by looking at the fractions even and just say hey i know both of these fractions are going to be broken into the same size piece which is into sixths and i know that this one's only gonna have three shaded in out of the six and four sixths will have four shaded in out of the six so therefore this one is greater and we make the sign like that to go greater and a lot of a lot of teachers i know i use like a little alligator and yet chomp the one that is greater so that's how you would model these two fractions with a visual model now let's go ahead and take a look at how you can use cross multiplication um i am only a fan of this once a student understands what's going on here okay once they understand what fractions are how to compare them this way i only use this way that i'm about to teach you as a way to check i want them to understand fractions and sometimes the reason why they would want to do that i'll try to do a different example too to show you this um is when the pieces get really close and it's kind of hard to tell which one is eq if they're equal or if one is greater than the other this is when i would use cross multiplication so what you do is you start from the denominator and i go like this two start from the denominator and cross up two so six times four would be six twelve and eighteen twenty four twenty four and then 6 times 3 would be 6 12 and 18. that's from the multiplication mashup that i have on youtube and 18 and 24 as those products well 24 is greater so we would make a sign like that the same sign that shows that 4 6 is greater okay so we just use cross multiplication to check but what if we were comparing these two fractions and here you can see that our numerators are equal they're the same hang on i'm not going to appreciate that our numerators have the same number but our denominators have different numbers same numerators different denominators let's start by drawing it out make our big rectangle [Music] a lot of times kids when they're doing this they get kind of sloppy on the end and they'll kind of like go like that and that's going to make it real funky so make sure that they're learning to draw it as pretty much as straight as they can okay nice and straight down nice and perpendicular all right we've got two fractions we're gonna do three fourths right here and we're gonna do three fifths on the bottom okay let's break it into the total number of equal parts that's four it's even so i'm gonna do two on this side and two on that side moving on to the second fraction we've got five as a denominator so i know that four parts were on top so now i need to fit five parts on the bottom and the same amount of space so i know my parts need to be a little bit smaller which is what we want students to be thinking and reasoning with okay and then both we are shading in three parts one two three on top and one two three on the bottom and it's clear to see from our visual that three-fourths is greater than three-fifths and we can use cross-multiplication to check okay four times three is twelve four eight twelve and five times three is 5 10 15 and we know that 15 is greater so this is how i teach comparing fractions in third grade and in fourth grade we can have different kind of denominators and numerators which makes it a little bit trickier and again what we want students to be doing is understanding how big these pieces are because a lot of times um a student if they didn't draw it out they might go oh well there's five in the denominator here they both have the same numerator but five is greater than four so it must be this one which is the wrong thinking because we've got five pieces inside the hole which means we're splitting this into smaller pieces and so three of those pieces are going to be three smaller pieces than the three pieces up top so we really want them to be understanding and seeing it so really stick with those visual models you can also do it on a number line too but just today for this episode i was showing you how to do it on an area model all right everybody that's it for today's episode but if you have a specific skill that you'd like for me to present in an upcoming video go ahead and drop that in the comments you can also email me mccarthymathacademy gmail.com links below i'm also on instagram and facebook at mccarthy math academy so you can find me there i love making videos that you need so just drop me a line for third fourth and fifth grade content and like i mentioned at the beginning of this video i create a ton of video lessons for students so if you found value in this video today which is geared towards teachers and parents and tutors and basically adults looking to help students so if this video helped you today imagine how much my other videos could help your students you can check out my website for more details just go to mccarthymathacademy gmail.com which you can find the links below right before we go remember that you were born for a reason you matter and what you choose to do with your life matters too so get out there and change the world in your own special way and i will see you on the next episode |
9989 | https://content.myconnectsuite.com/api/documents/c0536c08c8ac461eb181b1e6b3d0dae7.pdf | 1 8th Grade Common Core State Standards Flip Book This document is intended to show the connections to the Standards of Mathematical Practices and the content standards and to get detailed information at each level. Resources used: CCSS, Arizona DOE, Ohio DOE and North Carolina DOE. This “Flip Book” is intended to help teachers understand what each standard means in terms of what students must know and be able to do. It provides only a sample of instructional strategies and examples. The goal of every teacher should be to guide students in understanding & making sense of mathematics. Construction directions: Print on cardstock. Cut the tabs on each page starting with page 2. Cut the bottom off of this top cover to reveal the tabs for the subsequent pages. Staple or bind the top of all pages to complete your flip book. Compiled by Melisa Hancock (Send feedback to: melisa@ksu.edu) 2 1. Make sense of problems and persevere in solving them. In grade 8, students solve real world problems through the application of algebraic and geometric concepts. Students seek the meaning of a problem and look for efficient ways to represent and solve it. They may check their thinking by asking themselves, ―What is the most efficient way to solve the problem?‖, ―Does this make sense?‖, and ―Can I solve the problem in a different way?‖ 2. Reason abstractly and quantitatively. In grade 8, students represent a wide variety of real world contexts through the use of real numbers and variables in mathematical expressions, equations, and inequalities. They examine patterns in data and assess the degree of linearity of functions. Students contextualize to understand the meaning of the number or variable as related to the problem and decontextualize to manipulate symbolic representations by applying properties of operations. 3. Construct viable arguments and critique the reasoning of others. In grade 8, students construct arguments using verbal or written explanations accompanied by expressions, equations, inequalities, models, and graphs, tables, and other data displays (i.e. box plots, dot plots, histograms, etc.). They further refine their mathematical communication skills through mathematical discussions in which they critically evaluate their own thinking and the thinking of other students. They pose questions like ―How did you get that?‖, ―Why is that true?‖ ―Does that always work?‖ They explain their thinking to others and respond to others‘ thinking. 4. Model with mathematics. In grade 8, students model problem situations symbolically, graphically, tabularly, and contextually. Students form expressions, equations, or inequalities from real world contexts and connect symbolic and graphical representations. Students solve systems of linear equations and compare properties of functions provided in different forms. Students use scatterplots to represent data and describe associations between variables. Students need many opportunities to connect and explain the connections between the different representations. They should be able to use all of these representations as appropriate to a problem context. 5. Use appropriate tools strategically. Students consider available tools (including estimation and technology) when solving a mathematical problem and decide when certain tools might be helpful. For instance, students in grade 8 may translate a set of data given in tabular form to a graphical representation to compare it to another data set. Students might draw pictures, use applets, or write equations to show the relationships between the angles created by a transversal. 6. Attend to precision. In grade 8, students continue to refine their mathematical communication skills by using clear and precise language in their discussions with others and in their own reasoning. Students use appropriate terminology when referring to the number system, functions, geometric figures, and data displays. 7. Look for and make use of structure. (Deductive Reasoning) Students routinely seek patterns or structures to model and solve problems. In grade 8, students apply properties to generate equivalent expressions and solve equations. Students examine patterns in tables and graphs to generate equations and describe relationships. Additionally, students experimentally verify the effects of transformations and describe them in terms of congruence and similarity. 8. Look for and express regularity in repeated reasoning. (Inductive Reasoning) In grade 8, students use repeated reasoning to understand algorithms and make generalizations about patterns. Students use iterative processes to determine more precise rational approximations for irrational numbers. During multiple opportunities to solve and model problems, they notice that the slope of a line and rate of change are the same value. Students flexibly make connections between covariance, rates, and representations showing the relationships between quantities. Mathematical Practice Standards (MP) summary of each standard Mathematics Practice Standards 3 In Grade 8, instructional time should focus on three critical areas: (1) formulating and reasoning about expressions and equations, including modeling an association in bivariate data with a linear equation, and solving linear equations and systems of linear equations; (2) grasping the concept of a function and using functions to describe quantitative relationships; (3) analyzing two- and three-dimensional space and figures using distance, angle, similarity, and congruence, and understanding and applying the Pythagorean Theorem. (1) Students use linear equations and systems of linear equations to represent, analyze, and solve a variety of problems. Students recognize equations for proportions (y/x = m or y = mx) as special linear equations (y = mx + b), understanding that the constant of proportionality (m) is the slope, and the graphs are lines through the origin. They understand that the slope (m) of a line is a constant rate of change, so that if the input or x-coordinate changes by an amount A, the output or y-coordinate changes by the amount mA. Students also use a linear equation to describe the association between two quantities in bivariate data (such as arm span vs. height for students in a classroom). At this grade, fitting the model, and assessing its fit to the data are done informally. Interpreting the model in the context of the data requires students to express a relationship between the two quantities in question and to interpret components of the relationship (such as slope and y-intercept) in terms of the situation. Students strategically choose and efficiently implement procedures to solve linear equations in one variable, understanding that when they use the properties of equality and the concept of logical equivalence, they maintain the solutions of the original equation. Students solve systems of two linear equations in two variables and relate the systems to pairs of lines in the plane; these intersect, are parallel, or are the same line. Students use linear equations, systems of linear equations, linear functions, and their understanding of slope of a line to analyze situations and solve problems. (2) Students grasp the concept of a function as a rule that assigns to each input exactly one output. They understand that functions describe situations where one quantity determines another. They can translate among representations and partial representations of functions (noting that tabular and graphical representations may be partial representations), and they describe how aspects of the function are reflected in the different representations. (3) Students use ideas about distance and angles, how they behave under translations, rotations, reflections, and dilations, and ideas about congruence and similarity to describe and analyze two-dimensional figures and to solve problems. Students show that the sum of the angles in a triangle is the angle formed by a straight line, and that various configurations of lines give rise to similar triangles because of the angles created when a transversal cuts parallel lines. Students understand the statement of the Pythagorean Theorem and its converse, and can explain why the Pythagorean Theorem holds, for example, by decomposing a square in two different ways. They apply the Pythagorean Theorem to find distances between points on the coordinate plane, to find lengths, and to analyze polygons. Students complete their work on volume by solving problems involving cones, cylinders, and spheres. Critical Areas for Mathematics in 8th Grade Critical Areas 4 Domain: The Number System (NS) Cluster: Know that there are numbers that are not rational, and approximate them by rational numbers. Standard: 8.NS.1. Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number. Standards for Mathematical Practice (MP): 8.MP.2. Reason abstractly and quantitatively. 8.MP.6. Attend to precision. 8.MP.7. Look for and make use of structure. Connections: This cluster goes beyond the Grade 8 Critical Areas of Focus to address Working with irrational numbers, integer exponents, and scientific notation. This cluster builds on previous understandings from Grades 6-7, The Number System. Instructional Strategies A rational number is of the form a/b, where a and b are both integers, and b is not 0. In the elementary grades, students learned processes that can be used to locate any rational number on the number line: Divide the interval from 0 to 1 into b equal parts; then, beginning at 0, count out those parts. The surprising fact, now, is that there are numbers on the number line that cannot be expressed as a/b, with a and b both integers, and these are called irrational numbers. Students construct a right isosceles triangle with legs of 1 unit. Using the Pythagorean theorem, they determine that the length of the hypotenuse is √2. In the figure right, they can rotate the hypotenuse back to the original number line to show that indeed √2 is a number on the number line. In the elementary grades, students become familiar with decimal fractions, most often with decimal representations that terminate a few digits to the right of the decimal point. For example, to find the exact decimal representation of 2/7, students might use their calculator to find 2/7 = 0.2857142857…, and they might guess that the digits 285714 repeat. To show that the digits do repeat, students in Grade 7 actually carry out the long division and recognize that the remainders repeat in a predictable pattern—a pattern that creates the repetition in the decimal representation (see 7.NS.2.d). Thinking about long division generally, ask students what will happen if the remainder is 0 at some step. They can reason that the long division is complete, and the decimal representation terminates. If the reminder is never 0, in contrast, then the remainders will repeat in a cyclical pattern because at each step with a given remainder, the process for finding the next remainder is always the same. Thus, the digits in the decimal representation also repeat. When dividing by 7, there are 6 possible nonzero remainders, and students can see that the decimal repeats with a pattern of at most 6 digits. In general, when finding the decimal representation of m/n, students can reason that the repeating portion of decimal will have at most n-1 digits. The important point here is that students can see that the pattern will repeat, so they can imagine the process continuing without actually carrying it out. Continued next page 5 Conversely, given a repeating decimal, students learn strategies for converting the decimal to a fraction. One approach is to notice that rational numbers with denominators of 9 repeat a single digit. With a denominator of 99, two digits repeat; with a denominator of 999, three digits repeat, and so on. 13/99 = 0.13131313… 74/99 = 0.74747474… 237/999 = 0.237237237… 485/999 = 0.485485485… From this pattern, students can go in the other direction, conjecturing, for example, that the repeating decimal 0.285714285714… = 285714/999999. And then they can verify that this fraction is equivalent to 2/7. Once students understand that (1) every rational number has a decimal representation that either terminates or repeats, and (2) every terminating or repeating decimal is a rational number, they can reason that on the number line, irrational numbers (i.e., those that are not rational) must have decimal representations that neither terminate nor repeat. And although students at this grade do not need to be able to prove that √2 is irrational, they need to know that √2 is irrational (see 8.EE.2), which means that its decimal representation neither terminates nor repeats. Nonetheless, they can approximate √2 without using the square root key on the calculator. Students can create tables like those below to approximate √2 to one, two, and then three places to the right of the decimal point: From knowing that 12 = 1 and 22 = 4, or from the picture above, students can reason that there is a number between 1 and 2 whose square is 2. In the first table above, students can see that between 1.4 and 1.5, there is a number whose square is 2. Then in the second table, they locate that number between 1.41 and 1.42. And in the third table they can locate √2 between 1.414 and 1.415. Students can develop more efficient methods for this work. For example, from the picture above, they might have begun the first table with 1.4. And once they see that 1.422 > 2, they do not need to generate the rest of the data in the second table. Use set diagrams to show the relationships among real, rational, irrational numbers, integers, and counting numbers. The diagram should show that the all real numbers (numbers on the number line) are either rational or irrational. Given two distinct numbers, it is possible to find both a rational and an irrational number between them. 6 Explanations and Examples 8.NS.1 Students distinguish between rational and irrational numbers. Any number that can be expressed as a fraction is a rational number. Students recognize that the decimal equivalent of a fraction will either terminate or repeat. Fractions that terminate will have denominators containing only prime factors of 2 and/or 5. This understanding builds on work in 7th grade when students used long division to distinguish between repeating and terminating decimals. Students convert repeating decimals into their fraction equivalent using patterns or algebraic reasoning. One method to find the fraction equivalent to a repeating decimal is shown below. Change 0.4 to a fraction. Let x = 0.444444….. Multiply both sides so that the repeating digits will be in front of the decimal. In this example, one digit repeats so both sides are multiplied by 10, giving 10x = 4.4444444…. Subtract the original equation from the new equation. 10x = 4.4444444…. x = 0.444444….. 9x = 4 Solve the equation to determine the equivalent fraction. 9x = 4 9 9 x = 4 9 Additionally, students can investigate repeating patterns that occur when fractions have a denominator of 9, 99, or 11. For example, 9 is equivalent to 0.4, 9 is equivalent to 0.5, etc. Students can use graphic organizers to show the relationship between the subsets of the real number system. Common Misconceptions Some students are surprised that the decimal representation of pi does not repeat. Some students believe that if only we keep looking at digits farther and farther to the right, eventually a pattern will emerge. A few irrational numbers are given special names (pi and e), and much attention is given to sqrt(2). Because we name so few irrational numbers, students sometimes conclude that irrational numbers are unusual and rare. In fact, irrational numbers are much more plentiful than rational numbers, in the sense that they are ―denser‖ in the real line. 8.NS.1 7 Domain: The Number System (NS) Cluster: Know that there are numbers that are not rational, and approximate them by rational numbers. Standard: 8.NS.2. Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., 2). For example, by truncating the decimal expansion of √2, show that √2 is between 1and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.4. Model with mathematics. MP.7. Look for and make use of structure. MP.8. Look for and express regularity in repeated reasoning. Connections: See 8.NS.1. Instructional Strategies: See 8.NS.1. Explanations and Examples: 8.NS.2 Students locate rational and irrational numbers on the number line. Students compare and order rational and irrational numbers. Additionally, students understand that the value of a square root can be approximated between integers and that non-perfect square roots are irrational. Students also recognize that square roots may be negative and written as - 28. To find an approximation of 28, first determine the perfect squares 28 is between, which would be 25 and 36. The square roots of 25 and 36 are 5 and 6 respectively, so we know that 28 is between 5 and 6. Since 28 is closer to 25, an estimate of the square root would be closer to 5. One method to get an estimate is to divide 3 (the distance between 25 and 28) by 11 (the distance between the perfect squares of 25 and 36) to get 0.27. The estimate of 28 would be 5.27 (the actual is 5.29). Students can approximate square roots by iterative processes. Examples: Approximate the value of to the nearest hundredth. Solution: Students start with a rough estimate based upon perfect squares. falls between 2 and 3 because 5 falls between 22 = 4 and 32 = 9. The value will be closer to 2 than to 3. Students continue the iterative process with the tenths place value. falls between 2.2 and 2.3 because 5 falls between 2.22 = 4.84 and 2.32 = 5.29. The value is closer to 2.2. Further iteration shows that the value of is between 2.23 and 2.24 since 2.232 is 4.9729 and 2.242 is 5.0176. Compare √2 and √3 by estimating their values, plotting them on a number line, and making comparative statements. Solution: Statements for the comparison could include: √2 is approximately 0.3 less than √3 √2 is between the whole numbers 1 and 2 √3 is between 1.7 and 1.8 Common Misconceptions: See 8.NS.1. 5 5 5 5 8.NS.2 8 Domain: Expressions and Equations (EE) Cluster: Work with radicals and integer exponents. Standard: 8.EE.1. Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 323–5 = 3–3 = 1/33 = 1/27. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: This cluster goes beyond the Grade 8 Critical Areas of Focus to address Working with irrational numbers, interger exponents, and Scientific notation. This cluster connects to previous understandings of place value, very large and very small numbers. Instructional Strategies Although students begin using whole-number exponents in Grades 5 and 6, it is in Grade 8 when students are first expected to know and use the properties of exponents and to extend the meaning beyond counting-number exponents. It is no accident that these expectations are simultaneous, because it is the properties of counting-number exponents that provide the rationale for the properties of integer exponents. In other words, students should not be told these properties but rather should derive them through experience and reason. For counting-number exponents (and for nonzero bases), the following properties follow directly from the meaning of exponents. 1. anam = an+m 2. (an)m = anm 3. anbn = (ab)n Students should have experience simplifying numerical expressions with exponents so that these properties become natural and obvious. For example, 23∙25=(2∙2∙2)∙(2∙2∙2∙2∙2)=28 (53)4=(5∙5∙5)∙(5∙5∙5)∙(5∙5∙5)∙(5∙5∙5)=512 (3∙7)4=(3∙7)∙(3∙7)∙(3∙7)∙(3∙7)=(3∙3∙3∙3)∙(7∙7∙7∙7)=34∙74 If students reason about these examples with a sense of generality about the numbers, they begin to articulate the properties. For example, ―I see that 3 twos is being multiplied by 5 twos, and the results is 8 twos being multiplied together, where the 8 is the sum of 3 and 5, the number of twos in each of the original factors. That would work for a base other than two (as long as the bases are the same).‖ Note: When talking about the meaning of an exponential expression, it is easy to say (incorrectly) that ―35 means 3 multiplied by itself 5 times.‖ But by writing out the meaning, 35=3∙3∙3∙3∙3, students can see that there are only 4 multiplications. So a better description is ―35 means 5 3s multiplied together.‖ Students also need to realize that these simple descriptions work only for counting-number exponents. When extending the meaning of exponents to include 0 and negative exponents, these descriptions are limiting: Is it sensible to say ―30 means 0 3s multiplied together‖ or that ―3-2 means -2 3s multiplied together‖? The motivation for the meanings of 0 and negative exponents is the following principle: The properties of counting-number exponents should continue to work for integer exponents. Example next page 9 For example, Property 1 can be used to reason what 30 should be. Consider the following expression and simplification: 30∙35=30+5=35. This computation shows that the when 30 is multiplied by 35, the result (following Property 1) should be 35, which implies that 30 must be 1. Because this reasoning holds for any base other than 0, we can reason that a0 = 1 for any nonzero number a. To make a judgment about the meaning of 3-4, the approach is similar: 3−4∙34=3−4+4=30=1. This computation shows that 3-4 should be the reciprocal of 34, because their product is 1. And again, this reasoning holds for any nonzero base. Thus, we can reason that a−n = 1/an. Putting all of these results together, we now have the properties of integer exponents, shown in the above chart. For mathematical completeness, one might prove that properties 1-3 continue to hold for integer exponents, but that is not necessary at this point. A supplemental strategy for developing meaning for integer exponents is to make use of patterns, as shown in the chart to the right. The meanings of 0 and negative-integer exponents can be further explored in a place-value chart: Thus, integer exponents support writing any decimal in expanded form like the following: 3247.568=3∙103+2∙102+4∙101+7∙100+5∙10−1+6∙10−2+8∙10−3. Expanded form and the connection to place value is important for helping students make sense of scientific notation, which allows very large and very small numbers to be written concisely, enabling easy comparison. To develop familiarity, go back and forth between standard notation and scientific notation for numbers near, for example, 1012 or 10-9. Compare numbers, where one is given in scientific notation and the other is given in standard notation. Real-world problems can help students compare quantities and make sense about their relationship. Provide practical opportunities for students to flexibly move between forms of squared and cubed numbers. For example, If then . This flexibility should be experienced symbolically and verbally. 32=9 9=3 Opportunities for conceptually understanding irrational numbers should be developed. One way is for students to draw a square that is one unit by one unit and find the diagonal using the Pythagorean Theorem. The diagonal drawn has an irrational length of √2. Other irrational lengths can be found using the same strategy by finding diagonal lengths of rectangles with various side lengths. Continued next page 10 Explanations and Examples 8.EE.1 Integer (positive and negative) exponents are further used to generate equivalent numerical expressions when multiplying, dividing or raising a power to a power. Using numerical bases and the laws of exponents, students generate equivalent expressions. Examples: Common Misconceptions Equivalent to or Is x2x3 x5 x6 ? Students may make the relationship that in scientific notation, when a number contains one nonzero digit and a positive exponent, that the number of zeros equals the exponent. This pattern may incorrectly be applied to scientific notation values with negative values or with more than one nonzero digit. Students may mix up the product of powers property and the power of a power property. 25 64 5 4 2 3 256 1 4 1 4 4 4 4 4 4 7 3 7 3 4-3 52 = 4-3 ´ 1 52 = 1 43 ´ 1 52 = 1 64 ´ 1 25 = 1 16,000 8.EE.1 11 Domain: Expressions and Equations Cluster: Work with radicals and integer exponents. Standard: 8.EE.2. Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.EE.1. Explanations and Examples 8.EE.2 Students recognize that squaring a number and taking the square root √ of a number are inverse operations; likewise, cubing a number and taking the cube root 3 are inverse operations. This understanding is used to solve equations containing square or cube numbers. Equations may include rational numbers such as x2 = 1 /4, x2 = 4 /9 or x3 = 1/8 (NOTE: Both the numerator and denominators are perfect squares or perfect cubes.) Students recognize perfect squares and cubes, understanding that non-perfect squares and non-perfect cubes are irrational. Students understand that in geometry a square root is the length of the side of a square and a cube root is the length of the side of a cube. The value of p for square root and cube root equations must be positive. Examples: and and Solve Solution: Solve Solution: Instructional Strategies See 8.EE.1. Common Misconceptions: See 8.EE.1. 9 32 3 9 27 1 3 1 3 1 3 3 3 3 1 27 1 27 1 3 3 3 9 2 x 9 2 x x2 = ± 9 3 x 8 3 x 8 3 x 3 3 3 8 x 2 x 8.EE.2 12 Domain: Expressions and Equations Cluster: Work with radicals and integer exponents. Standard: 8.EE.3. Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. For example, estimate the population of the United States as 3108 and the population of the world as 7109, and determine that the world population is more than 20 times larger. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. Connections: See 8.EE.1. Explanations and Examples 8.EE.3 Students express numbers in scientific notation. Students compare and interpret scientific notation quantities in the context of the situation. If the exponent increases by one, the value increases 10 times. Students understand the magnitude of the number being expressed in scientific notation and choose an appropriate corresponding unit. For example, 3 x 108 is equivalent to 30 million, which represents a large quantity. Therefore, this value will affect the unit chosen. Instructional Strategies See 8.EE.1. Common Misconceptions: See 8.EE.1. 8.EE.3 13 Domain: Expressions and Equations Cluster: Work with radicals and integer exponents. Standard: 8.EE.4. Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. Connections: See 8.EE.1. Explanations and Examples 8.EE.4 Students use laws of exponents to multiply or divide numbers written in scientific notation. Additionally, students understand scientific notation as generated on various calculators or other technology. Students can convert decimal forms to scientific notation and apply rules of exponents to simplify expressions. In working with calculators or spreadsheets, it is important that students recognize scientific notation. Students should recognize that the output of 2.45E+23 is 2.45 x 1023 and 3.5E-4 is 3.5 x 10-4. Students enter scientific notation using E or EE (scientific notation), (multiplication), and ^ (exponent) symbols. Instructional Strategies See 8.EE.1. Common Misconceptions: See 8.EE.1. 8.EE.4 14 Domain: Expressions and Equations (EE) Cluster: Understand the connections between proportional relationships, lines, and linear equations Standard: 8.EE.5. Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed. Standards for Mathematical Practice (MP): MP.1. Make sense of problems and persevere in solving them. MP.2. Reason abstractly and quantitatively. MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. MP.8. Look for and express regularity in repeated reasoning. Connections: This cluster is connected to the Grade 8 Critical Area of Focus #1, Formulating and reasoning about expressions and equations, including modeling an association in bivariate data with a linear equation, and solving linear equations and systems of linear equaitons and Critical Area of Focus #3, Analyzing two- and three-dimensional space and figures using distance, angle, similarity, and congruence, and understanding and applying the Pythagorean Theorem. Explanations and Examples 8.EE.5 Students build on their work with unit rates from 6 grade and proportional relationships in 7 grade to compare graphs, tables and equations of proportional relationships. Students identify the unit rate (or slope) in graphs, tables and equations to compare two or more proportional relationships. Using graphs of experiences that are familiar to students increases accessibility and supports understanding and interpretation of proportional relationship. Students are expected to both sketch and interpret graphs. Example: Compare the scenarios to determine which represents a greater speed. Include a description of each scenario including the unit rates in your explanation. Scenario 1: Scenario 2: y = 50x x is time in hours y is distance in miles 15 Instructional Strategies This cluster focuses on extending the understanding of ratios and proportions. Unit rates have been explored in Grade 6 as the comparison of two different quantities with the second unit a unit of one, (unit rate). In seventh grade unit rates were expanded to complex fractions and percents through solving multistep problems such as: discounts, interest, taxes, tips, and percent of increase or decrease. Proportional relationships were applied in scale drawings, and students should have developed an informal understanding that the steepness of the graph is the slope or unit rate. Now unit rates are addressed formally in graphical representations, algebraic equations, and geometry through similar triangles. Distance time problems are notorious in mathematics. In this cluster, they serve the purpose of illustrating how the rates of two objects can be represented, analyzed and described in different ways: graphically and algebraically. Emphasize the creation of representative graphs and the meaning of various points. Then compare the same information when represented in an equation. By using coordinate grids and various sets of three similar triangles, students can prove that the slopes of the corresponding sides are equal, thus making the unit rate of change equal. After proving with multiple sets of triangles, students can be led to generalize the slope to y = mx for a line through the origin and y = mx + b for a line through the vertical axis at b. Instructional Resources/Tools Carnegie Math™ graphing calculators SMART™ technology with software emulator National Library of Virtual Manipulatives (NLVM)©, The National Council of Teachers of Mathematics, Illuminations Annenberg™ video tutorials, www.nsdl.org 8.EE.5 16 Domain: Expressions and Equations Cluster: Understand the connections between proportional relationships, lines, and linear equations. Standard: 8.EE.6. Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.7. Look for and make use of structure. MP.8. Look for and express regularity in repeated reasoning. Connections: See 8.EE.5. Explanations and Examples 8.EE.6 Triangles are similar when there is a constant rate of proportion between them. Using a graph, students construct triangles between two points on a line and compare the sides to understand that the slope (ratio of rise to run) is the same between any two points on a line. The triangle between A and B has a vertical height of 2 and a horizontal length of 3. The triangle between B and C has a vertical height of 4 and a horizontal length of 6. The simplified ratio of the vertical height to the horizontal length of both triangles is 2 to 3, which also represents a slope of 2/3for the line. Students write equations in the form y = mx for lines going through the origin, recognizing that m represents the slope of the line. Students write equations in the form y = mx + b for lines not passing through the origin, recognizing that m represents the slope and b represents the y-intercept. Example: Explain why is similar to , and deduce that has the same slope as . Express each line as an equation. Instructional Strategies See 8.EE.5. Common Misconceptions: See 8.EE.5. ACB DFE AB BE 8.EE.6 17 Domain: Expressions and Equations (EE) Cluster: Analyze and solve linear equations and pairs of simultaneous linear equations. Standard: 8.EE.7. Solve linear equations in one variable. a. Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers). b. Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: This cluster is connected to the Grade 8 Critical Area of Focus #1, Formulating and reasoning about expressions and equations, including modeling an association in bivariate data with a linear equation, and solving linear equations and systems of linear equaitons. This cluster also builds upon the understandings in Grades 6 and 7 of Expressions and Equations, Ratios and Proportional Relationships, and utilizes the skills developed in the previous grade in The Number System. Explanations and Examples 8.EE.7 Students solve one-variable equations with the variables being on both sides of the equals sign. Students recognize that the solution to the equation is the value(s) of the variable, which make a true equality when substituted back into the equation. Equations shall include rational numbers, distributive property and combining like terms. Equations have one solution when the variables do not cancel out. For example, 10x – 23 = 29 – 3x can be solved to x = 4. This means that when the value of x is 4, both sides will be equal. If each side of the equation were treated as a linear equation and graphed, the solution of the equation represents the coordinates of the point where the two lines would intersect. In this example, the ordered pair would be (4, 17). 10 • 4 – 23 = 29 – 3 • 4 40 – 23 = 29 – 12 17 = 17 Equations having no solution have variables that will cancel out and constants that are not equal. This means that there is not a value that can be substituted for x that will make the sides equal. For example, the equation -x + 7 – 6x = 19 – 7x, can be simplified to -7x + 7 = 19 – 7x. If 7x is added to each side, the resulting equation is 7 = 19, which is not true. No matter what value is substituted for x the final result will be 7 = 19. If each side of the equation were treated as a linear equation and graphed, the lines would be parallel. Continued next page 18 An equation with infinitely many solutions occurs when both sides of the equation are the same. Any value of x will produce a valid equation. For example the following equation, when simplified will give the same values on both sides. –1 /2(36a – 6) = 3 4(4 – 24a) –18a + 3 = 3 – 18a If each side of the equation were treated as a linear equation and graphed, the graph would be the same line. As students transform linear equations in one variable into simpler forms, they discover the equations can have one solution, infinitely many solutions, or no solutions. When the equation has one solution, the variable has one value that makes the equation true as in 12-4y=16. The only value for y that makes this equation true is -1. When the equation has infinitely many solutions, the equation is true for all real numbers as in 7x + 14 = 7 (x+2). As this equation is simplified, the variable terms cancel leaving 14 = 14 or 0 = 0. Since the expressions are equivalent, the value for the two sides of the equation will be the same regardless which real number is used for the substitution. When an equation has no solutions it is also called an inconsistent equation. This is the case when the two expressions are not equivalent as in 5x - 2 = 5(x+1). When simplifying this equation, students will find that the solution appears to be two numbers that are not equal or -2 = 1. In this case, regardless which real number is used for the substitution, the equation is not true and therefore has no solution. Examples: Solve for x: o o o Solve: o Instructional Strategies 8.EE.7-8 In Grade 6, students applied the properties of operations to generate equivalent expressions, and identified when two expressions are equivalent. This cluster extends understanding to the process of solving equations and to their solutions, building on the fact that solutions maintain equality, and that equations may have only one solution, many solutions, or no solution at all. Equations with many solutions may be as simple as 3x = 3x, 3x + 5 = x + 2 + x + x + 3, or 6x + 4x = x(6 + 4), where both sides of the equation are equivalent once each side is simplified. Continued next page 4 ) 7 ( 3 x 8 4 8 3 x x 2 3 5 ) 1 ( 3 x x 7 ) 3 ( 7 m y y 3 1 4 3 3 2 4 1 19 Table 3 on page 90 of CCSS generalizes the properties of operations and serves as a reminder for teachers of what these properties are. Eighth graders should be able to describe these relationships with real numbers and justify their reasoning using words and not necessarily with the algebraic language of Table 3. In other words, students should be able to state that 3(-5) = (-5)3 because multiplication is commutative and it can be performed in any order (it is commutative), or that 9(8) + 9(2) = 9(8 + 2) because the distributive property allows us to distribute multiplication over addition, or determine products and add them. Grade 8 is the beginning of using the generalized properties of operations, but this is not something on which students should be assessed. Pairing contextual situations with equation solving allows students to connect mathematical analysis with real-life events. Students should experience analyzing and representing contextual situations with equations, identify whether there is one, none, or many solutions, and then solve to prove conjectures about the solutions. Through multiple opportunities to analyze and solve equations, students should be able to estimate the number of solutions and possible values(s) of solutions prior to solving. Rich problems, such as computing the number of tiles needed to put a border around a rectangular space or solving proportional problems as in doubling recipes, help ground the abstract symbolism to life. Experiences should move through the stages of concrete, conceptual and algebraic/abstract. Utilize experiences with the pan balance model as a visual tool for maintaining equality (balance) first with simple numbers, then with pictures symbolizing relationships, and finally with rational numbers allows understanding to develop as the complexity of the problems increases. Equation-solving in Grade 8 should involve multistep problems that require the use of the distributive property, collecting like terms, and variables on both sides of the equation. This cluster builds on the informal understanding of slope from graphing unit rates in Grade 6 and graphing proportional relationships in Grade 7 with a stronger, more formal understanding of slope. It extends solving equations to understanding solving systems of equations, or a set of two or more linear equations that contain one or both of the same two variables. Once again the focus is on a solution to the system. Most student experiences should be with numerical and graphical representations of solutions. Beginning work should involve systems of equations with solutions that are ordered pairs of integers, making it easier to locate the point of intersection, simplify the computation and hone in on finding a solution. More complex sytems can be investigated and solve by using graphing technology. Contextual situations relevant to eighth graders will add meaning to the solution to a system of equations. Students should explore many problems for which they must write and graph pairs of equations leading to the generalization that finding one point of intersection is the single solution to the system of equations. Provide opportunities for students to connect the solutions to an equation of a line, or solution to a system of equations, by graphing, using a table and writing an equation. Students should receive opportunities to compare equations and systems of equations, investigate using graphing calculators or graphing utilities, explain differences verbally and in writing, and use models such as equation balances. Continued next page 20 Problems such as, ―Determine the number of movies downloaded in a month that would make the costs for two sites the same, when Site A charges $6 per month and $1.25 for each movie and Site B charges $2 for each movie and no monthly fee.‖ Students write the equations letting y = the total charge and x = the number of movies. Site A: y = 1.25x + 6 Site B: y = 2x Students graph the solutions for each of the equations by finding ordered pairs that are solutions and representing them in a t-chart. Discussion should encompass the realization that the intersection is an ordered pair that satisfies both equations. And finally students should relate the solution to the context of the problem, commenting on the practicality of their solution. Problems should be structured so that students also experience equations that represent parallel lines and equations that are equivalent. This will help them to begin to understand the relationships between different pairs of equations: When the slope of the two lines is the same, the equations are either different equations representing the same line (thus resulting in many solutions), or the equations are different equations representing two not intersecting, parallel, lines that do not have common solutions. System-solving in Grade 8 should include estimating solutions graphically, solving using substitution, and solving using elimination. Students again should gain experience by developing conceptual skills using models that develop into abstract skills of formal solving of equations. Provide opportunities for students to change forms of equations (from a given form to slope-intercept form) in order to compare equations. Common Misconceptions 8.EE.7-8 Students think that only the letters x and y can be used for variables. Students think that you always need a variable = a constant as a solution. The variable is always on the left side of the equation. Equations are not always in the slope intercept form, y=mx+b Students confuse one-variable and two-variable equations. 8.EE.7 21 Domain: Expressions and Equations Cluster: Analyze and solve linear equations and pairs of simultaneous linear equations. Standard: 8.EE.8. Analyze and solve pairs of simultaneous linear equations. a. Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously. b. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6. c. Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair. MP.1. Make sense of problems and persevere in solving them. MP.2. Reason abstractly and quantitatively. MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. MP.8. Look for and express regularity in repeated reasoning. Connections: See 8.EE.7 Explanations and Examples: Systems of linear equations can also have one solution, infinitely many solutions or no solutions. Students will discover these cases as they graph systems of linear equations and solve them algebraically. A system of linear equations whose graphs meet at one point (intersecting lines) has only one solution, the ordered pair representing the point of intersection. A system of linear equations whose graphs do not meet (parallel lines) has no solutions and the slopes of these lines are the same. A system of linear equations whose graphs are coincident (the same line) has infinitely many solutions, the set of ordered pairs representing all the points on the line. By making connections between algebraic and graphical solutions and the context of the system of linear equations, students are able to make sense of their solutions. Students need opportunities to work with equations and context that include whole number and/or decimals/fractions. Examples: • Find x and y using elimination and then using substitution. 3x + 4y = 7 - 2x + 8y = 10 • Plant A and Plant B are on different watering schedules. This affects their rate of growth. Compare the growth of the two plants to determine when their heights will be the same. Let W = number of weeks Let H = height of the plant after W weeks Plant A W H 0 4 (0,4) 1 6 (1,6) 2 8 (2,8) 3 10 (3,10) Plant B W H 0 2 (0,1) 1 6 (1,6) 2 10 (2,10) 3 14 (3,14) 22 Given each set of coordinates, graph their corresponding lines. Solution: Write an equation that represent the growth rate of Plant A and Plant B. Solution: Plant A H = 2W + 4 Plant B H = 4W + 2 • At which week will the plants have the same height? Solution: The plants have the same height after one week. Plant A: H = 2W + 4 Plant B: H = 4W + 2 Plant A: H = 2(1) + 4 Plant B: H = 4(1) + 2 Plant A: H = 6 Plant B: H = 6 After one week, the height of Plant A and Plant B are both 6 inches. 8.EE.8 23 Extended Standards: The Alternate Achievement Standards for Students With the Most Significant Cognitive Disabilities Non-Regulatory Guidance states, “…materials should show a clear link to the content standards for the grade in which the student is enrolled, although the grade-level content may be reduced in complexity or modified to reflect pre-requisite skills.” Throughout the Standards descriptors such as, describe, count, identify, etc, should be interpreted to mean that the students will be taught and tested according to their mode of communication. EXTENDED 8. EE 24 Domain: Functions (F) Cluster: Define, evaluate, and compare functions. Standard: 8.F.1. Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output. (Function notation is not required in Grade 8.) Standards for Mathematical Practice (MP): MP.1. Make sense of problems and persevere in solving them. MP.2. Reason abstractly and quantitatively. MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. MP.8. Look for and express regularity in repeated reasoning. Connections: Connections: This Cluster is connected to the Grade 8 Critical Area of Focus #2, Grasping the concept of a function and using functions to describe quantitative relationships. Expressions and Equations: Linear equations in two variables can be used to define linear functions, and students can use graphs of functions to reason toward solutions to linear equations. Geometry: Similar triangles are used to show that the slope of a line is constant. Statistics and Probability: Bivariate data can often be modeled by a linear function. Explanations and Examples 8.F.1 Students distinguish between functions and non-functions, using equations, graphs, and tables. Non- functions occur when there is more than one y-value is associated with any x-value. Students are not expected to use the function notation f(x) at this level. For example, the rule that takes x as input and gives x2+5x+4 as output is a function. Using y to stand for the output we can represent this function with the equation y = x2+5x+4, and the graph of the equation is the graph of the function. Students are not yet expected use function notation such as f(x) = x2+5x+4. Instructional Strategies In grade 6, students plotted points in all four quadrants of the coordinate plane. They also represented and analyzed quantitative relationships between dependent and independent variables. In Grade 7, students decided whether two quantities are in a proportional relationship. In Grade 8, students begin to call relationships functions when each input is assigned to exactly one output. Also, in Grade 8, students learn that proportional relationships are part of a broader group of linear functions, and they are able to identify whether a relationship is linear. Nonlinear functions are included for comparison. Later, in high school, students use function notation and are able to identify types of nonlinear functions. To determine whether a relationship is a function, students should be expected to reason from a context, a graph, or a table, after first being clear which quantity is considered the input and which is the output. When a relationship is not a function, students should produce a counterexample: an ―input value‖ with at least two ―output values.‖ If the relationship is a function, the students should explain how they verified that for each input there was exactly one output. The ―vertical line test‖ should be avoided because (1) it is too easy to apply without thinking, (2) students do not need an efficient strategy at this point, and (3) it creates 25 misconceptions for later mathematics, when it is useful to think of functions more broadly, such as whether x might be a function of y. ―Function machine‖ pictures are useful for helping students imagine input and output values, with a rule inside the machine by which the output value is determined from the input. Notice that the standards explicitly call for exploring functions numerically, graphically, verbally, and algebraically (symbolically, with letters). This is sometimes called the ―rule of four.‖ For fluency and flexibility in thinking, students need experiences translating among these. In Grade 8, the focus, of course, is on linear functions, and students begin to recognize a linear function from its form y = mx + b. Students also need experiences with nonlinear functions, including functions given by graphs, tables, or verbal descriptions but for which there is no formula for the rule, such as a girl‘s height as a function of her age. In the elementary grades, students explore number and shape patterns (sequences), and they use rules for finding the next term in the sequence. At this point, students describe sequences both by rules relating one term to the next and also by rules for finding the nth term directly. (In high school, students will call these recursive and explicit formulas.) Students express rules in both words and in symbols. Instruction should focus on additive and multiplicative sequences as well as sequences of square and cubic numbers, considered as areas and volumes of cubes, respectively. When plotting points and drawing graphs, students should develop the habit of determining, based upon the context, whether it is reasonable to ―connect the dots‖ on the graph. In some contexts, the inputs are discrete, and connecting the dots can be misleading. For example, if a function is used to model the height of a stack of n paper cups, it does not make sense to have 2.3 cups, and thus there will be no ordered pairs between n = 2 and n = 3. Provide multiple opportunities to examine the graphs of linear functions and use graphing calculators or computer software to analyze or compare at least two functions at the same time. Illustrate with a slope triangle where the run is "1" that slope is the "unit rate of change." Compare this in order to compare two different situations and identify which is increasing/decreasing as a faster rate. Students can compute the area and perimeter of different-size squares and identify that one relationship is linear while the other is not by either looking at a table of value or a graph in which the side length is the independent variable (input) and the area or perimeter is the dependent variable (output). Common Misconceptions Some students will mistakenly think of a straight line as horizontal or vertical only. Some students will mix up x- and y-axes on the coordinate plane, or mix up the ordered pairs. When emphasizing that the first value is plotted on the horizontal axes (usually x, with positive to the right) and the second is the vertical axis (usually called y, with positive up), point out that this is merely a convention: It could have been otherwise, but it is very useful for people to agree on a standard customary practice. 8.F.1 26 Domain: Functions (F) Cluster: Define, evaluate, and compare functions. Standard: 8.F.2. Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a linear function represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change. Standards for Mathematical Practice (MP): MP.1. Make sense of problems and persevere in solving them. MP.2. Reason abstractly and quantitatively. MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. MP.8. Look for and express regularity in repeated reasoning. Connections: See 8.F.1. Explanations and Examples 8.F.2 Students compare functions from different representations. For example, compare the following functions to determine which has the greater rate of change. Function 1: y = 2x + 4 Function 2: x y -1 -6 0 -3 2 3 Examples: Compare the two linear functions listed below and determine which equation represents a greater rate of change. Function 1: Compare the two linear functions listed below and determine which has a negative slope. Function 2: The function whose input x and output y are related by y = 3x + 7 27 Function 1: Gift Card Samantha starts with $20 on a gift card for the book store. She spends $3.50 per week to buy a magazine. Let y be the amount remaining as a function of the number of weeks, x. x y 0 20 1 16.50 2 13.00 3 9.50 4 6.00 Function 2: The school bookstore rents graphing calculators for $5 per month. It also collects a non-refundable fee of $10.00 for the school year. Write the rule for the total cost (c) of renting a calculator as a function of the number of months (m). Solution: Function 1 is an example of a function whose graph has negative slope. Samantha starts with $20 and spends money each week. The amount of money left on the gift card decreases each week. The graph has a negative slope of -3.5, which is the amount the gift card balance decreases with Samantha‘s weekly magazine purchase. Function 2 is an example of a function whose graph has positive slope. Students pay a yearly nonrefundable fee for renting the calculator and pay $5 for each month they rent the calculator. This function has a positive slope of 5 which is the amount of the monthly rental fee. An equation for Example 2 could be c = 5m + 10. Instructional Strategies See 8.F.1. Common Misconceptions: See 8.F.1. 8.F.2 28 Domain: Functions Cluster: Define, evaluate, and compare functions. Standard: 8.F.3. Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.F.1. Explanations and Examples 8.F.3 Students use equations, graphs and tables to categorize functions as linear or non-linear. Students recognize that points on a straight line will have the same rate of change between any two of the points. Example: Determine which of the functions listed below are linear and which are not linear and explain your reasoning. o y = -2x2 + 3 non linear o y = 2x linear o A = πr2 non linear y = 0.25 + 0.5(x – 2) linear Instructional Strategies See 8.F.1. Common Misconceptions: See 8.F.1. 8.F.3 29 Domain: Functions (F) Cluster: Use functions to model relationships between quantities. Standard: 8.F.4. Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values. Standards for Mathematical Practice (MP): MP.1. Make sense of problems and persevere in solving them. MP.2. Reason abstractly and quantitatively. MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. MP.8. Look for and express regularity in repeated reasoning. Connections: Connections: This cluster is connected to the Grade 8 Critical Area of Focus #2, Grasping the concept of a function and using functions to describe quantitative relationships. Expressions and Equations: Linear equations in two variables can be used to define linear functions, and students can use graphs of functions to reason toward solutions to linear equations. Geometry: Similar triangles are used to show that the slope of a line is constant. Statistics and Probability: Bivariate data can often be modeled by a linear function. Explanations and Examples 8.F.4 Students identify the rate of change (slope) and initial value (y-intercept) from tables, graphs, equations or verbal descriptions. Students recognize that in a table the y-intercept is the y-value when x is equal to 0. The slope can the determined by finding the ratio y x between the change in two y-values and the change between the two corresponding x-values. The y-intercept in the table below would be (0, 2). The distance between 8 and -1 is 9 in a negative direction is -9; the distance between -2 and 1 is 3 in a positive direction. The slope is the ratio of rise to run or y /x or 9 /3 = -3. x y -2 8 0 2 1 -1 Using graphs, students identify the y-intercept as the point where the line crosses the y-axis and the slope as the rise. run In a linear equation the coefficient of x is the slope and the constant is the y-intercept. Students need to be given the equations in formats other than y = mx + b, such as y = ax + b (format from graphing calculator), y = b + mx (often the format from contextual situations), etc. Note that point-slope form and standard forms are not expectations at this level. 30 In contextual situations, the y-intercept is generally the starting value or the value in the situation when the independent variable is 0. The slope is the rate of change that occurs in the problem. Rates of change can often occur over years. In these situations it is helpful for the years to be ―converted‖ to 0, 1, 2, etc. For example, the years of 1960, 1970, and 1980 could be represented as 0 (for 1960). Students use the slope and y-intercepts to write a linear function in the form y = mx +b. Situations may be given as a verbal description, two ordered pairs, a table, a graph, or rate of change and another point on the line. Students interpret slope and y-intercept in the context of the given situation. Examples: The table below shows the cost of renting a car. The company charges $45 a day for the car as well as charging a one-time $25 fee for the car‘s navigation system (GPS).Write an expression for the cost in dollars, c, as a function of the number of days, d. Students might write the equation c = 45d + 25 using the verbal description or by first making a table. Days (d) Cost (c) in dollars 1 70 2 115 3 160 4 205 Students should recognize that the rate of change is 45 (the cost of renting the car) and that initial cost (the first day charge) also includes paying for the navigation system. Classroom discussion about one time fees vs. recurrent fees will help students model contextual situations. When scuba divers come back to the surface of the water, they need to be careful not to ascend too quickly. Divers should not come to the surface more quickly than a rate of 0.75 ft per second. If the divers start at a depth of 100 feet, the equation d = 0.75t – 100 shows the relationship between the time of the ascent in seconds (t) and the distance from the surface in feet (d). o Will they be at the surface in 5 minutes? How long will it take the divers to surface from their dive? Make a table of values showing several times and the corresponding distance of the divers from the surface. Explain what your table shows. How do the values in the table relate to your equation? Instructional Strategies In Grade 8, students focus on linear equations and functions. Nonlinear functions are used for comparison. Students will need many opportunities and examples to figure out the meaning of y = mx + b. What does m mean? What does b mean? They should be able to ―see‖ m and b in graphs, tables, and formulas or equations, and they need to be able to interpret those values in contexts. For example, if a function is used to model the height of a stack of n paper cups, then the rate of change, m, which is the slope of the graph, is the height of the ―lip‖ of the cup: the amount each cup sticks above the lower cup in the stack. The ―initial value‖ in this case is not valid in the 31 context because 0 cups would not have a height, and yet a height of 0 would not fit the equation. Nonetheless, the value of b can be interpreted in the context as the height of the ―base‖ of the cup: the height of the whole cup minus its lip. Use graphing calculators and web resources to explore linear and non-linear functions. Provide context as much as possible to build understanding of slope and y-intercept in a graph, especially for those patterns that do not start with an initial value of 0. Give students opportunities to gather their own data or graphs in contexts they understand. Students need to measure, collect data, graph data, and look for patterns, then generalize and symbolically represent the patterns. They also need opportunities to draw graphs (qualitatively, based upon experience) representing real-life situations with which they are familiar. Probe student thinking by asking them to determine which input values make sense in the problem situations. Provide students with a function in symbolic form and ask them to create a problem situation in words to match the function. Given a graph, have students create a scenario that would fit the graph. Ask students to sort a set of "cards" to match a graphs, tables, equations, and problem situations. Have students explain their reasoning to each other. From a variety of representations of functions, students should be able to classify and describe the function as linear or non-linear, increasing or decreasing. Provide opportunities for students to share their ideas with other students and create their own examples for their classmates. Use the slope of the graph and similar triangle arguments to call attention to not just the change in x or y, but also to the rate of change, which is a ratio of the two. Emphasize key vocabulary. Students should be able to explain what key words mean: e.g., model, interpret, initial value, functional relationship, qualitative, linear, non-linear. Use a ―word wall‖ to help reinforce vocabulary. Common Misconceptions Students often confuse a recursive rule with an explicit formula for a function. For example, after identifying that a linear function shows an increase of 2 in the values of the output for every change of 1 in the input, some students will represent the equation as y = x + 2 instead of realizing that this means y = 2x + b. When tables are constructed with increasing consecutive integers for input values, then the distinction between the recursive and explicit formulas is about whether you are reasoning vertically or horizontally in the table. Both types of reasoning—and both types of formulas—are important for developing proficiency with functions. When input values are not increasing consecutive integers (e.g., when the input values are decreasing, when some integers are skipped, or when some input values are not integers), some students have more difficulty identifying the pattern and calculating the slope. It is important that all students have experience with such tables, so as to be sure that they do not overgeneralize from the easier examples. Some students may not pay attention to the scale on a graph, assuming that the scale units are always ―one.‖ When making axes for a graph, some students may not using equal intervals to create the scale. Some students may infer a cause and effect between independent and dependent variables, but this is often not the case. Some students graph incorrectly because they don‘t understand that x usually represents the independent variable and y represents the dependent variable. Emphasize that this is a convention that makes it easier to communicate. 8.F.4 32 Domain: Functions Cluster: Use functions to model relationships between quantities. Standard: 8.F.5. Describe qualitatively the functional relationship between two quantities by analyzing a graph (e.g., where the function is increasing or decreasing, linear or nonlinear). Sketch a graph that exhibits the qualitative features of a function that has been described verbally. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.F.4. Explanations and Examples 8.F.5 Given a verbal description of a situation, students sketch a graph to model that situation. Given a graph of a situation, students provide a verbal description of the situation. Example: The graph below shows a student‘s trip to school. This student walks to his friend‘s house and, together, they ride a bus to school. The bus stops once before arriving at school. Describe how each part A-E of the graph relates to the story. Instructional Strategies See 8.F.4. Common Misconceptions: See 8.F.4. 8.F.5 33 Domain: Geometry (G) Cluster: Understand congruence and similarity using physical models, transparencies, or geometry software. Standard: 8.G.1. Verify experimentally the properties of rotations, reflections, and translations: a. Lines are taken to lines, and line segments to line segments of the same length. b. Angles are taken to angles of the same measure. Parallel lines are taken to parallel lines. Standards for Mathematical Practice (MP): MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. MP.8. Look for and express regularity in repeated reasoning. Connections: This cluster is connected to the Grade 8 Critical Area of Focus #3, Analyzing two- and three-dimensional space and figures using distance, angle, similarity, and congruence, and understanding and applying the Pythagorean Theorem. This cluster builds from Grade 7 Geometry, Ratios and Proportional Relationships, and prepares students for more formal work in high school geometry. Explanations and Examples 8.G.1 In a translation, every point of the pre-image is moved the same distance and in the same direction to form the image. A reflection is the ―flipping‖ of an object over a line, known as the ―line of reflection‖. A rotation is a transformation that is performed by ―spinning‖ the figure around a fixed point known as the center of rotation. The figure may be rotated clockwise or counterclockwise. Students use compasses, protractors and ruler or technology to explore figures created from translations, reflections and rotations. Characteristics of figures, such as lengths of line segments, angle measures and parallel lines are explored before the transformation (pre-image) and after the transformation (image). Students understand that these transformations produce images of exactly the same size and shape as the pre-image and are known as rigid transformations. Students need multiple opportunities to explore the transformation of figures so that they can appreciate that points stay the same distance apart and lines stay at the same angle after they have been rotated, reflected, and/or translated. Students are not expected to work formally with properties of dilations until high school. Instructional Strategies A major focus in Grade 8 is to use knowledge of angles and distance to analyze two- and three-dimensional figures and space in order to solve problems. This cluster interweaves the relationships of symmetry, transformations, and angle relationships to form understandings of similarity and congruence. Inductive and deductive reasoning are utilized as students forge into the world of proofs. Informal arguments are justifications based on known facts and logical reasoning. Students should be able to appropriately label figures, angles, lines, line segments, congruent parts, and images (primes or double primes). Students are expected to use logical thinking, expressed in words using correct terminology. They are NOT expected to use theorems, axioms, postulates or a formal format of proof as in two-column proofs. Transformational geometry is about the effects of rigid motions, rotations, reflections and 34 translations on figures. Initial work should be presented in such a way that students understand the concept of each type of transformation and the effects that each transformation has on an object before working within the coordinate system. For example, when reflecting over a line, each vertex is the same distance from the line as its corresponding vertex. This is easier to visualize when not using regular figures. Time should be allowed for students to cut out and trace the figures for each step in a series of transformations. Discussion should include the description of the relationship between the original figure and its image(s) in regards to their corresponding parts (length of sides and measure of angles) and the descriptionof the movement, including the atributes of transformations (line of symmetry, distance to be moved, center of rotation, angle of rotationand the amount of dilation).The case of distance – preserving transformation leads to the idea of congruence. It is these distance-preserving transformations that lead to the idea of congruence. Work in the coordinate plane should involve the movement of various polygons by addition, subtraction and multiplied changes of the coordinates. For example, add 3 to x, subtract 4 from y, combinations of changes to x and y, multiply coordinates by 2 then by 12. Students should observe and discuss such questions as ‗What happens to the polygon?‘ and ‗What does making the change to all vertices do?‘. Understandings should include generalizations about the changes that maintain size or maintain shape, as well as the changes that create distortions of the polygon (dilations). Example dilations should be analyzed by students to discover the movement from the origin and the subsequent change of edge lengths of the figures. Students should be asked to describe the transformations required to go from an original figure to a transformed figure (image). Provide opportunities for students to discuss the procedure used, whether different procedures can obtain the same results, and if there is a more efficient procedure to obtain the same results. Students need to learn to describe transformations with both words and numbers. Through understanding symmetry and congruence, conclusions can be made about the relationships of line segments and angles with figures. Students should relate rigid motions to the concept of symmetry and to use them to prove congruence or similarity of two figures. Problem situations should require students to use this knowledge to solve for missing measures or to prove relationships. It is an expectation to be able to describe rigid motions with coordinates. Provide opportunities for students to physically manipulate figures to discover properties of similar and congruent figures, for example, the corresponding angles of similar figures are equal. Additionally use drawings of parallel lines cut by a transversal to investigate the relationship among the angles. For example, what information can be obtained by cutting between the two intersections and sliding one onto the other? 35 In Grade 7, students develop an understanding of the special relationships of angles and their measures (complementary, supplementary, adjacent, vertical). Now, the focus is on learning the about the sum of the angles of a triangle and using it to, find the measures of angles formed by transversals (especially with parallel lines), or to find the measures of exterior angles of triangles and to informally prove congruence. By using three copies of the same triangle labeled and placed so that the three different angles form a straight line, students can: • explore the relationships of the angles, • learn the types of angles (interior, exterior, alternate interior, alternate exterior, corresponding, same side interior, same side exterior), and • explore the parallel lines, triangles and parallelograms formed. Further examples can be explored to verify these relationships and demonstrate their relevance in real life. Investigations should also lead to the Angle-Angle criterion for similar triangles. For instance, pairs of students create two different triangles with one given angle measurement, then repeat with two given angle measurements and finally with three given angle measurements. Students observe and describe the relationship of the resulting triangles. As a class, conjectures lead to the generalization of the Angle-Angle criterion. Students should solve mathematical and real-life problems involving understandings from this cluster. Investigation, discussion, justification of their thinking, and application of their learning will assist in the more formal learning of geometry in high school. Common Misconceptions: Students often confuse situations that require adding with multiplicative situations in regard to scale factor. Providing experiences with geometric figures and coordinate grids may help students visualize the difference. 8.G.1 36 Domain: Geometry (G) Cluster: Understand congruence and similarity using physical models, transparencies, or geometry software. Standard: 8.G.2. Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.4. Model with mathematics. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.G.1. Explanations and Examples 8.G.2 This standard is the students‘ introduction to congruency. Congruent figures have the same shape and size. Translations, reflections and rotations are examples of rigid transformations. A rigid transformation is one in which the pre-image and the image both have exactly the same size and shape since the measures of the corresponding angles and corresponding line segments remain equal (are congruent). Students examine two figures to determine congruency by identifying the rigid transformation(s) that produced the figures. Students recognize the symbol for congruency (≅) and write statements of congruency. Examples: Is Figure A congruent to Figure A‘? Explain how you know. Describe the sequence of transformations that results in the transformation of Figure A to Figure A‘. Instructional Strategies See 8.G.1. 8.G.2 37 Domain: Geometry Cluster: Understand congruence and similarity using physical models, transparencies, or geometry software. Standard: 8.G.3. Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates. Standards for Mathematical Practice (MP): MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.G.1. Explanations and Examples 8.G.3 Students identify resulting coordinates from translations, reflections, and rotations (90°, 180°and 270°both clockwise and counterclockwise), recognizing the relationship between the coordinates and the transformation. For example, a translation of 5 left and 2 up would subtract 5 from the x-coordinate and add 2 to the y-coordinate. D (- 4, -3) à D‘ (-9, -1). A reflection across the x-axis would change B (6, -8) to B‘ (6, 8) . Additionally, students recognize the relationship between the coordinates of the pre-image, the image and the scale factor following a dilation from the origin. Dilations are non-rigid transformations that enlarge (scale factors greater than one) or reduce (scale factors less than one) the size of a figure using a scale factor. A dilation is a transformation that moves each point along a ray emanating from a fixed center, and multiplies distances from the center by a common scale factor. In dilated figures, the dilated figure is similar to its pre-image. Translation: A translation is a transformation of an object that moves the object so that every point of the object moves in the same direction as well as the same distance. In a translation, the translated object is congruent to its pre-image. has been translated 7 units to the right and 3 units up. To get from A (1,5) to A‘ (8,8), move A 7 units to the right (from x = 1 to x = 8) and 3 units up (from y = 5 to y = 8). Points B + C also move in the same direction (7 units to the right and 3 units up). Reflection: A reflection is a transformation that flips an object across a line of reflection (in a coordinate grid the line of reflection may be the x or y axis). In a rotation, the rotated object is congruent to its pre-image. When an object is reflected across the y axis, the reflected x coordinate is the opposite of the pre-image x coordinate. ABC 38 Rotation: A rotated figure is a figure that has been turned about a fixed point. This is called the center of rotation. A figure can be rotated up to 360˚. Rotated figures are congruent to their pre-image figures. Consider when is rotated 180˚ clockwise about the origin. The coordinates of are D(2,5), E(2,1), and F(8,1). When rotated 180˚, has new coordinates D‘(-2,-5), E‘(-2,-1) and F‘(-8,-1). Each coordinate is the opposite of its pre-image. Instructional Strategies See 8.G.1. Common Misconceptions: See 8.G.1. DEF DEF ' ' ' F E D 8.G.3 39 Domain: Geometry Cluster: Understand congruence and similarity using physical models, transparencies, or geometry software. Standard: 8.G.4. Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.G.1. Explanations and Examples 8.G.4 This is the students‘ introduction to similarity and similar figures. Students understand similar figures have angles with the same measure and sides that are proportional. Similar figures are produced from dilations. Students describe the sequence that would produce similar figures, including the scale factors. Students understand that a scale factor greater than one will produce an enlargement in the figure, while a scale factor less than one will produce a reduction in size. Examples: Is Figure A similar to Figure A‘? Explain how you know. Describe the sequence of transformations that results in the transformation of Figure A to Figure A‘. Instructional Strategies See 8.G.1. 8.G.4 40 Domain: Geometry Cluster: Understand congruence and similarity using physical models, transparencies, or geometry software. Standard: 8.G.5. Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the sum of the three angles appears to form a line, and give an argument in terms of transversals why this is so. Standards for Mathematical Practice (MP): MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.G.1. Explanations and Examples 8.G.5 Students use exploration and deductive reasoning to determine relationships that exist between a) angle sums and exterior angle sums of triangles, b) angles created when parallel lines are cut by a transversal, and c) the angle- angle criterion for similarity of triangle. Students construct various triangles and find the measures of the interior and exterior angles. Students make conjectures about the relationship between the measure of an exterior angle and the other two angles of a triangle. (The measure of an exterior angle of a triangle is equal to the sum of the measures of the other two interior angles) and the sum of the exterior angles (360°). Using these relationships, students use deductive reasoning to find the measure of missing angles. Students construct parallel lines and a transversal to examine the relationships between the created angles. Students recognize vertical angles, adjacent angles and supplementary angles from 7th grade and build on these relationships to identify other pairs of congruent angles. Using these relationships, students use deductive reasoning to find the measure of missing angles. Students construct various triangles having line segments of different lengths but with two corresponding congruent angles. Comparing ratios of sides will produce a constant scale factor, meaning the triangles are similar. Examples: Students can informally prove relationships with transversals. Show that m 3 + m + m = 180˚ if l and m are parallel lines and t1 & t2 are transversals. + + = 180˚. Angle 1 and Angle 5 are congruent because they are corresponding angles ( ). can be substituted for . Continued next page 4 5 1 2 3 1 5 1 5 41 : because alternate interior angles are congruent. can be substituted for Therefore m + m + m = 180˚ Students can informally conclude that the sum of a triangle is 180º (the angle-sum theorem) by applying their understanding of lines and alternate interior angles. In the figure below, line x is parallel to line yz: Angle a is 35º because it alternates with the angle inside the triangle that measures 35º. Angle c is 80º because it alternates with the angle inside the triangle that measures 80º. Because lines have a measure of 180º, and angles a + b + c form a straight line, then angle b must be 65 º (180 – 35 + 80 = 65). Therefore, the sum of the angles of the triangle are 35º + 65 º + 80 º Instructional Strategies See 8.G.1. Common Misconceptions: See 8.G.1. 2 4 4 2 3 4 5 8.G.5 42 Domain: Geometry (G) Cluster: Understand and apply the Pythagorean Theorem. Standard: 8.G.6. Explain a proof of the Pythagorean Theorem and its converse. Standards for Mathematical Practice (MP): MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: This cluster is connected to the Grade 8 Critical Area of Focus #3, Analyzing two- and three-dimensional space and figures using distance, angle, similarity, and congruence, and understanding and applying the Pythagorean Theorem. Explanations and Examples 8.G.6 Students explain the Pythagorean Theorem as it relates to the area of squares coming off of all sides of a right triangle. Students also understand that given three side lengths with this relationship forms a right triangle. Students should verify, using a model, that the sum of the squares of the legs is equal to the square of the hypotenuse in a right triangle. Students should also understand that if the sum of the squares of the 2 smaller legs of a triangle is equal to the square of the third leg, then the triangle is a right triangle. Instructional Strategies Previous understanding of triangles, such as the sum of two side measures is greater than the third side measure, angles sum, and area of squares, is furthered by the introduction of unique qualities of right triangles. Students should be given the opportunity to explore right triangles to determine the relationships between the measures of the legs and the measure of the hypotenuse. Experiences should involve using grid paper to draw right triangles from given measures and representing and computing the areas of the squares on each side. Data should be recorded in a chart such as the one below, allowing for students to conjecture about the relationship among the areas within each triangle. Students should then test out their conjectures, then explain and discuss their findings. Finally, the Pythagorean Theorem should be introduced and explained as the pattern they have explored. Time should be spent analyzing several proofs of the Pythagorean Theorem to develop a beginning sense of the process of deductive reasoning, the significance of a theorem, and the purpose of a proof. Students should be able to justify a simple proof of the Pythagorean Theorem or its converse. 43 Previously, students have discovered that not every combination of side lengths will create a triangle. Now they need situations that explore using the Pythagorean Theorem to test whether or not side lengths represent right triangles. (Recording could include Side length a, Side length b, Sum of a2 + b2, c2, a2 + b2 = c2, Right triangle? Through these opportunities, students should realize that there are Pythagorean (triangular) triples such as (3, 4, 5), (5, 12, 13), (7, 24, 25), (9, 40, 41) that always create right triangles, and that their multiples also form right triangles. Students should see how similar triangles can be used to find additional triples. Students should be able to explain why a triangle is or is not a right triangle using the Pythagorean Theorem. The Pythagorean Thereom should be applied to finding the lengths of segments on a coordinate grid, especially those segments that do not follow the vertical or horizontal lines, as a means of discussing the determination of distances between points. Contextual situations, created by both the students and the teacher, that apply the Pythagorean theorem and its converse should be provided. For example, apply the concept of similarity to determine the height of a tree using the ratio between the student's height and the length of the student's shadow. From that, determine the distance from the tip of the tree to the end of its shadow and verify by comparing to the computed distance from the top of the student's head to the end of the student's shadow, using the ratio calculated previously. Challenge students to identify additional ways that the Pythagorean Theorem is or can be used in real world situations or mathematical problems, such as finding the height of something that is difficult to physically measure, or the diagonal of a prism. 8.G.6 44 Domain: Geometry (G) Cluster: Understand and apply the Pythagorean Theorem. Standard: 8.G.7. Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions. Standards for Mathematical Practice (MP): MP.1. Make sense of problems and persevere in solving them. MP.2. Reason abstractly and quantitatively. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.G.6. Explanations and Examples 8.G.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions. Through authentic experiences and exploration, students should use the Pythagorean Theorem to solve problems. Problems can include working in both two and three dimensions. Students should be familiar with the common Pythagorean triplets. Instructional Strategies See 8.G.6. 8.G.7 45 Domain: Geometry Cluster: Understand and apply the Pythagorean Theorem. Standard: 8.G.8. Apply the Pythagorean Theorem to find the distance between two points in a coordinate system. Standards for Mathematical Practice (MP): MP.1. Make sense of problems and persevere in solving them. MP.2. Reason abstractly and quantitatively. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.G.6. Explanations and Examples 8.G.8 One application of the Pythagorean Theorem is finding the distance between two points on the coordinate plane. Students build on work from 6th grade (finding vertical and horizontal distances on the coordinate plane) to determine the lengths of the legs of the right triangle drawn connecting the points. Students understand that the line segment between the two points is the length of the hypotenuse. The use of the distance formula is not an expectation. Example: Students will create a right triangle from the two points given (as shown in the diagram below) and then use the Pythagorean Theorem to find the distance between the two given points. Instructional Strategies See 8.G.6. 8.G.8 46 Domain: Geometry (G) Cluster: Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres Standard: 8.G.9. Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. Standards for Mathematical Practice (MP): MP.1. Make sense of problems and persevere in solving them. MP.2. Reason abstractly and quantitatively. MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. MP.8. Look for and express regularity in repeated reasoning. Connections: This cluster is connected to the Grade 8 Critical Area of Focus #3, Analyzing two- and three-dimensional space and figures using distance, angle, similarity, and congruence, and understanding and applying the Pythagorean Theorem. Explanations and Examples 8.G.9 Students build on understandings of circles and volume from 7th grade to find the volume of cylinders, cones and spheres. Students understand the relationship between the volume of a) cylinders and cones and b) cylinders and spheres to the corresponding formulas. Example: James wanted to plant pansies in his new planter. He wondered how much potting soil he should buy to fill it. Use the measurements in the diagram below to determine the planter‘s volume. Instructional Strategies Begin by recalling the formula, and its meaning, for the volume of a right rectangular prism: V = l ×w ×h. Then ask students to consider how this might be used to make a conjecture about the volume formula for a cylinder: continued next page 47 Most students can be readily led to the understanding that the volume of a right rectangular prism can be thought of as the area of a ―base‖ times the height, and so because the area of the base of a cylinder is pi r2 the volume of a cylinder is Vc = π r2h. To motivate the formula for the volume of a cone, use cylinders and cones with the same base and height. Fill the cone with rice or water and pour into the cylinder. Students will discover/experience that 3 cones full are needed to fill the cylinder. This non-mathematical derivation of the formula for the volume of a cone, V = 1/3 π r2h, will help most students remember the formula. In a drawing of a cone inside a cylinder, students might see that that the triangular cross-section of a cone is 1/2 the rectangular cross-section of the cylinder. Ask them to reason why the volume (three dimensions) turns out to be less than 1/2 the volume of the cylinder. It turns out to be 1/3. For the volume of a sphere, it may help to have students visualize a hemisphere ―inside‖ a cylinder with the same height and ―base.‖ The radius of the circular base of the cylinder is also the radius of the sphere and the hemisphere. The area of the ―base‖ of the cylinder and the area of the section created by the division of the sphere into a hemisphere is π r2. The height of the cylinder is also r so the volume of the cylinder is π r3. Students can see that the volume of the hemisphere is less than the volume of the cylinder and more than half the volume of the cylinder. Illustrating this with concrete materials and rice or water will help students see the relative difference in the volumes. At this point, students can reasonably accept that the volume of the hemisphere of radius r is 2/3 π r3 and therefore volume of a sphere with radius r is twice that or 4/3 π r3. There are several websites with explanations for students who wish to pursue the reasons in more detail. (Note that in the pictures above, the hemisphere and the cone together fill the cylinder.) Students should experience many types of real-world applications using these formulas. They should be expected to explain and justify their solutions. Common Misconceptions: A common misconception among middle grade students is that ―volume‖ is a ―number‖ that results from ―substituting‖ other numbers into a formula. For these students there is no recognition that ―volume‖ is a measure – related to the amount of space occupied. If a teacher discovers that students do not have an understanding of volume as a measure of space, it is important to provide opportunities for hands on experiences where students ―fill‖ three dimensional objects. Begin with right‖- rectangular prisms and fill them with cubes will help students understand why the units for volume are cubed. See Cubes 8.G.9 48 Extended Standards: The Alternate Achievement Standards for Students With the Most Significant Cognitive Disabilities Non-Regulatory Guidance states, “…materials should show a clear link to the content standards for the grade in which the student is enrolled, although the grade-level content may be reduced in complexity or modified to reflect pre-requisite skills.” Throughout the Standards descriptors such as, describe, count, identify, etc, should be interpreted to mean that the students will be taught and tested according to their mode of communication. EXTENDED 8.G 49 Domain: Statistics and Probability (SP) Cluster: Investigate patterns of association in bivariate data. Standard: 8.SP.1. Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities. Describe patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear association. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: This Cluster is connected to the grade 8 Critical Area of Focus #1, Formulating and reasoning about expressions and equations, including modeling an association in bivariate data with a linear equation, and solving linear equations and systems of linear equations. Explanations and Examples 8.SP.1 Bivariate data refers to two variable data, one to be graphed on the x-axis and the other on the y-axis. Students represent measurement (numerical) data on a scatter plot, recognizing patterns of association. These patterns may be linear (positive, negative or no association) or non-linear Students build on their previous knowledge of scatter plots examine relationships between variables. They analyze scatterplots to determine positive and negative associations, the degree of association, and type of association. Students examine outliers to determine if data points are valid or represent a recording or measurement error. Students can use tools such as those at the National Center for Educational Statistics to create a graph or generate data sets. ( Examples: Data for 10 students‘ Math and Science scores are provided in the chart. Describe the association between the Math and Science scores. Student 1 2 3 4 5 6 7 8 9 10 Math 64 50 85 34 56 24 72 63 42 93 Science 68 70 83 33 60 27 74 63 40 96 Data for 10 students‘ Math scores and the distance they live from school are provided in the table below. Describe the association between the Math scores and the distance they live from school. Student 1 2 3 4 5 6 7 8 9 10 Math score 64 50 85 34 56 24 72 63 42 93 Dist from school (miles) 0.5 1.8 1 2.3 3.4 0.2 2.5 1.6 0.8 2.5 50 Data from a local fast food restaurant is provided showing the number of staff members and the average time for filling an order are provided in the table below. Describe the association between the number of staff and the average time for filling an order. Number of staff 3 4 5 6 7 8 Average time to fill order (seconds) 180 138 120 108 96 84 The chart below lists the life expectancy in years for people in the United States every five years from 1970 to 2005. What would you expect the life expectancy of a person in the United States to be in 2010, 2015, and 2020 based upon this data? Explain how you determined your values. Date 1970 1975 1980 1985 1990 1995 2000 2005 Life Expectancy (in years) 70.8 72.6 73.7 74.7 75.4 75.8 76.8 77.4 Instructional Strategies Building on the study of statistics using univariate data in Grades 6 and 7, students are now ready to study bivariate data. Students will extend their descriptions and understanding of variation to the graphical displays of bivariate data. Scatter plots are the most common form of displaying bivariate data in Grade 8. Provide scatter plots and have students practice informally finding the line of best fit. Students should create and interpret scatter plots, focusing on outliers, positive or negative association, linearity or curvature. By changing the data slightly, students can have a rich discussion about the effects of the change on the graph. Have students use a graphing calculator to determine a linear regression and discuss how this relates to the graph. Students should informally draw a line of best fit for a scatter plot and informally measure the strength of fit. Discussion should include ―What does it mean to be above the line, below the line?‖ The study of the line of best fit ties directly to the algebraic study of slope and intercept. Students should interpret the slope and intercept of the line of best fit in the context of the data. Then students can make predictions based on the line of best fit. Common Misconceptions: Students may believe: Bivariate data is only displayed in scatter plots. 8.SP.4 in this cluster provides the opportunity to display bivariate, categorical data in a table. In general, students think there is only one correct answer in mathematics. Students may mistakenly think their lines of best fit for the same set of data will be exactly the same. Because students are informally drawing lines of best fit, the lines will vary slightly. To obtain the exact line of best fit, students would use technology to find the line of regression. 8.SP.1 51 Domain: Statistics and Probability Cluster: Investigate patterns of association in bivariate data. Standard: 8.SP.2. Know that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line, and informally assess the model fit by judging the closeness of the data points to the line. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.SP.1. Explanations and Examples 8.SP.2 Students understand that a straight line can represent a scatter plot with linear association. The most appropriate linear model is the line that comes closest to most data points. The use of linear regression is not expected. Examples: The capacity of the fuel tank in a car is 13.5 gallons. The table below shows the number of miles traveled and how many gallons of gas are left in the tank. Describe the relationship between the variables. If the data is linear, determine a line of best fit. Do you think the line represents a good fit for the data set? Why or why not? What is the average fuel efficiency of the car in miles per gallon? Miles Traveled 0 75 120 160 250 300 Gallons Used 0 2.3 4.5 5.7 9.7 10.7 Instructional Strategies See 8.SP.1. 8.SP.2 52 Domain: Statistics and Probability Cluster: Investigate patterns of association in bivariate data. Standard: 8.SP.3. Use the equation of a linear model to solve problems in the context of bivariate measurement data, interpreting the slope and intercept. For example, in a linear model for a biology experiment, interpret a slope of 1.5 cm/hr as meaning that an additional hour of sunlight each day is associated with an additional 1.5 cm in mature plant height. Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.SP.1. Explanations and Examples 8.SP.3 Linear models can be represented with a linear equation. Students interpret the slope and y-intercept of the line in the context of the problem. Examples: 1. Given data from students‘ math scores and absences, make a scatterplot. 2. Draw a line of best fit, paying attention to the closeness of the data points on either side of the line. 3. From the line of best fit, determine an approximate linear equation that models the given data (about y = 95 3 25 x ) 4. Students should recognize that 95 represents the y intercept and 3 25 represents the slope of the line. 53 5. Students can use this linear model to solve problems. For example, through substitution, they can use the equation to determine that a student with 4 absences should expect to receive a math score of about 62. They can then compare this value to their line. Instructional Strategies See 8.SP.1. 8.SP.3 54 Domain: Statistics and Probability Cluster: Investigate patterns of association in bivariate data. Standard: 8.SP.4. Understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. Use relative frequencies calculated for rows or columns to describe possible association between the two variables. For example, collect data from students in your class on whether or not they have a curfew on school nights and whether or not they have assigned chores at home. Is there evidence that those who have a curfew also tend to have chores? Standards for Mathematical Practice (MP): MP.2. Reason abstractly and quantitatively. MP.3. Construct viable arguments and critique the reasoning of others. MP.4. Model with mathematics. MP.5. Use appropriate tools strategically. MP.6. Attend to precision. MP.7. Look for and make use of structure. Connections: See 8.SP.1. Explanations and Examples 8.SP.4 Students recognize that categorical data can also be described numerically through the use of a two-way table. A two-way table is a table that shows categorical data classified in two different ways. The frequency of the occurrences are used to identify possible associations between the variables. For example, a survey was conducted to determine if boys eat breakfast more often than girls. The following table shows the results: Male Female Eat breakfast on a regular basis 190 110 Do not eat breakfast on a regular basis 130 165 Students can use the information from the table to compare the probabilities of males eating breakfast (190 of the 320 males à 59%) and females eating breakfast (110 of the 375 females à29%) to answer the question. From this data, it can be determined that males do eat breakfast more regularly than females. Example: The table illustrates the results when 100 students were asked the survey questions: Do you have a curfew? and Do you have assigned chores? Is there evidence that those who have a curfew also tend to have chores? Solution: Of the students who answered that they had a curfew, 40 had chores and 10 did not. Of the students who answered they did not have a curfew, 10 had chores and 40 did not. From this sample, there appears to be a positive correlation between having a curfew and having chores. 8.SP.4 55 Extended Standards: The Alternate Achievement Standards for Students With the Most Significant Cognitive Disabilities Non-Regulatory Guidance states, “…materials should show a clear link to the content standards for the grade in which the student is enrolled, although the grade-level content may be reduced in complexity or modified to reflect pre-requisite skills.” Throughout the Standards descriptors such as, describe, count, identify, etc, should be interpreted to mean that the students will be taught and tested according to their mode of communication. EXTENDED 8.SP |
9990 | https://www.umt.edu/coop-unit/documents/martin_publications/martin600.pdf | 2505 q 2002 The Society for the Study of Evolution. All rights reserved.
Evolution, 56(12), 2002, pp. 2505–2518 ENVIRONMENTAL INFLUENCES ON THE EVOLUTION OF GROWTH AND DEVELOPMENTAL RATES IN PASSERINES VLADIMı ´R REMES ˇ1,2 AND THOMAS E. MARTIN1,3 1U.S. Geological Survey Biological Resources Division, Montana Cooperative Wildlife Research Unit, Natural Science Building, University of Montana, Missoula, Montana 59812 3E-mail: tmartin@selway.umt.edu Abstract.
The reasons why growth and developmental rates vary widely among species have remained unclear.
Previous examinations of possible environmental influences on growth rates of birds yielded few correlations, leading to suggestions that young may be growing at maximum rates allowed within physiological constraints. However, estimations of growth rates can be confounded by variation in relative developmental stage at fledging. Here, we re-estimate growth rates to control for developmental stage. We used these data to examine the potential covariation of growth and development with environmental variation across a sample of 115 North American passerines. Contrary to previous results, we found that growth rates of altricial nestlings were strongly positively correlated to daily nest predation rates, even after controlling for adult body mass and phylogeny. In addition, nestlings of species under stronger predation pressure remained in the nest for a shorter period, and they left the nest at lower body mass relative to adult body mass. Thus, nestlings both grew faster and left the nest at an earlier developmental stage in species with higher risk of predation. Growth patterns were also related to food, clutch size, and latitude. These results support a view that growth and developmental rates of altricial nestlings are strongly influenced by the environmental conditions experienced by species, and they generally lend support to an adaptive view of interspecific variation in growth and developmental rates.
Key words.
Adaptation, aerial foraging, allometry, comparative analysis, development, independent contrasts, nest predation.
Received April 11, 2002.
Accepted August 20, 2002.
A major question in evolutionary biology is to explain why life-history traits vary among species (Roff 1992, 2002; Stearns 1992; Charnov 1993). Most attention has focused on traits such as fecundity, age at first reproduction, survival, and rate of aging. Growth and developmental rates have re-ceived less attention even though they are integral compo-nents of life-history strategies and vary widely among spe-cies. As a result, our understanding of why growth and de-velopment varies among species remains unclear.
Life-history theory predicts that individuals should grow and develop at an infinite rate to achieve maximum output of offspring (Stearns 1992). However, this ‘‘Darwinian de-mon’’ ideal is not possible because of basic limits set by physiological constraints on growth rates (Ricklefs 1969b; West et al. 2001). Nevertheless, organisms might be expected to grow and develop at the fastest rate possible within these constraints (Ricklefs 1969b, 1979a; Ricklefs et al. 1998).
Such views have been reinforced by an absence of correla-tions between growth rates and environmental variation in past interspecific examinations (see Ricklefs et al. 1998). On the other hand, benefits of maximum growth rates might be compromised by physiological or environmental costs. For example, fast growth can yield costs to reproductive output (Roff 1992; Stearns 1992), antiherbivore defense (Herms and Mattson 1992), starvation, predation, parasitism, disease sus-ceptibility (Nylin and Gotthard 1998; Lankford et al. 2001), physical performance (Billerbeck et al. 2001), and adult body size (Nylin and Gotthard 1998) based on intraspecific studies of growth in plants, insects, fish, frogs, and lizards (sum-marized in Arendt 1997; Metcalfe and Monaghan 2001). Con-2 Present address: Department of Zoology, Faculty of Science, Palacky ´ University, Tr ˇ. Svobody 26, CZ-771 46 Olomouc, Czech Republic; E-mail: remes@prfnw.upol.cz.
sequently, we might expect growth rates to be optimized among species that occupy different environmental condi-tions that vary in such costs and benefits. A critical question then centers on which selection pressures cause variation in growth rates among species.
Birds are ideal for addressing this question for four reasons.
First, data on growth rates are relatively abundant. Second, birds vary widely in growth rates and nestling periods (at least a 30-fold difference from seabirds to small passerines; Ricklefs 1979a, 1983). Third, a rich array of factors has been hypothesized to influence the evolution of growth and de-velopmental rates in birds, with little resolution of the relative importance of these differing factors (see Table 1; summa-rized in Starck and Ricklefs 1998a). Finally, different the-oretical models of avian growth and development differ sub-stantially in their assumptions and predictions. For instance, some predicted a strong relationship between time-dependent juvenile mortality and growth rates (Case 1978), whereas others predicted a very weak one or even the absence of such a relationship (Ricklefs 1969b, 1984). In support of the latter, some comparative studies found no correlation between nest mortality rates and growth rates (Ricklefs 1969b; Ricklefs et al. 1998). In contrast, a negative correlation between nest mortality rates and duration of the nestling period has been consistently observed (Ricklefs 1969b; Bosque and Bosque 1995; Martin 1995; Ricklefs et al. 1998). These contrasting results may suggest that nest mortality rates influence du-ration of the nestling period, but does not influence rate of growth. However, previous analyses may have been com-promised by a confounding between growth rate and duration of the nestling period. In particular, nestlings of species with low mortality remain in the nest longer and generally achieve either a longer period of asymptotic growth or weight re-2506 V. REMES ˇ AND T. E. MARTIN TABLE 1.
Summary of factors that were hypothesized to drive the evolution of growth and development in birds together with the con-ceived direction. A 1 means higher growth rates and longer nestling periods, a 2 means slower growth rates and shorter nestling periods, and 0 means no effect.
Factor Growth Nestling period Body size1 Developmental mode (altricial or precocial)2 Clutch size (energetic limitation)3 Sibling competition4 Time-dependent mortality5 Foraging ecology6 Latitude7 Nutritional limitation8 Growth rate (K)9 2 2 2 1 1 or 0 2 1 2 n/a 1 2 1 2 2 1 2 1 2 1 Ricklefs 1968a; Starck and Ricklefs 1998d.
2 From altricial to precocial species, that is, slower growth and shorter to lacking nestling period; Ricklefs 1973, 1979a; Starck and Ricklefs 1998c.
3 Lack 1968; Ricklefs 1986a.
4 Werschkul and Jackson 1979; Ricklefs 1982; Bortolotti 1986; Royle et al.
1999.
5 Williams 1966; Lack 1968; Ricklefs 1984 (positive effect), 1969b, 1979a (no effect).
6 Unpredictability of food, Case 1978; aerial foragers, O’Connor 1978; sea-birds, Ricklefs 1979a, 1982.
7 Ricklefs 1976; Oniki and Ricklefs 1981.
8 Concerns tropical frugivorous birds, Ricklefs 1983.
9 We can expect that species growing at higher rates (K) will also fledge earlier.
FIG. 1.
Schematic depiction of the influence of the pattern of growth data (either with weight recession in later phases of the growth or without it) on the estimation of parameters of the logistic growth curve. Both datasets have the same initial rate of the weight increase up to day 18. The growth curve parameters for the growth data without recession (full circles with growth curve) are: K 5 0.5, A 5 20, ti 5 6. The same parameters for the data with recession (open circles, lower growth curve) are K 5 0.54, A 5 18.3, ti 5 5.5. The difference is caused by the recession data forcing the asymptote of the best fit curve to be lower and K to be higher.
Consequently, we get two different values of K for the exact same rate of increase because of differing patterns of development after reaching the maximum weight in the nest (see Fig. 2).
cession than for species that leave the nest earlier. These differences can cause overestimation of growth rate for spe-cies that remain in the nest longer (see Fig. 1 and Methods) and, thereby, may obfuscate relationships between growth rates and nest mortality rates.
Here we re-examine the potential relationships between growth rates and nest mortality, as well as other potential environmental selection pressures for a sample of 115 North American passerines. We used original data and recalculated growth rates to standardize for analytical approach and to examine the possible interaction of nestling period on growth rate estimates. We controlled for the influence of develop-mental mode (altricial–precocial spectrum) by including only altricial passerines. Our primary ecological factors of interest were time-dependent nest mortality caused by predation, lat-itude, and foraging ecology. Foraging ecology was suggested to influence growth and developmental strategies in seabirds (Ricklefs 1979a). So, we examined aerial versus nonaerial foragers because aerial foragers differ from other passerine groups by unpredictability of their food supply over evolu-tionary time, which seems to influence life-history strategies (O’Connor 1978; Martin 1995). We also examined clutch size, which may influence the per capita rate of food delivery.
We use a comparative approach, which cannot test causal hypotheses, but can provide insight into possible selective forces by examining which factors explain variation in growth and developmental rates among species.
METHODS Dataset We collected data for as many species of passerines as we could find in the literature (see Appendix 1). Data on growth rates, latitudes, and nest predation rates are from original sources. We began with data on nestling growth listed by Starck and Ricklefs (1998b), but we checked and reanalyzed all growth rate estimates (see below). In addition, we were able to locate additional data, yielding primary growth data for 183 populations of 115 species of North American pas-serines. Of these, we were able to locate data on nest pre-dation rates (proportion of nests taken by predators) for 107 species. Data on the duration of nestling period, clutch size, and foraging category (aerial vs. nonaerial foragers) were from Poole and Gill (1992–2002). Data on adult body mass were from Dunning (1993). Where separate data on adult mass are given for males and females, we took the average.
Time-dependent mortality is the only part of nest mortality that can be expected to favor elevated growth and develop-mental rates (Ricklefs 1969a; Case 1978). Other sources of mortality that do not bear direct relation to the length of nest cycle are not relevant for the evolution of growth rates. Time-dependent mortality of nests can be caused, for example, by inclement weather but its main source is nest predation (Rick-lefs 1969a; Martin 1993). Percent of nests lost to predation was the common form of data available from primary sources and they are a compound result of the rate of time-dependent mortality caused by nest predation and the duration of nesting cycle. Thus, we transformed our data on nest predation to daily nest mortality rates by the formula: dmr 5 2(ln S)/T, where dmr is daily nest mortality rate caused by nest pre-dation, S is proportion of nests that were successful (1— proportion depredated), and T is the duration of the nest cycle (for discussion on this transformation procedure see Ricklefs 1969a; Ricklefs et al. 1998).
Growth rate can be characterized as an increase in body 2507 GROWTH STRATEGIES OF PASSERINES FIG. 2.
Estimates of growth rate constant K (mean 1 SE) for aerial (n 5 20) and nonaerial (n 5 95) foragers for raw data and when adjusted for weight recession. In general, K is higher for unadjusted data than when data were truncated at the highest mass reached by the young. This effect was pronounced only in aerial foragers due to their typical growth pattern with strong weight recession in the nest (see also Fig. 1). Paired t-tests: t1,188 5 24.27, P , 0.0001, n 5 190 for nonaerial foragers; t1,38 5 24.10, P 5 0.0006, n 5 40 for aerial foragers.
mass over time, whereas development is characterized by timing of developmental events during ontogeny of an in-dividual. For example, nestlings of species that fledge earlier develop their body functions more quickly than nestlings in later-fledging species (Ricklefs 1967b, 1979b; Austin and Ricklefs 1977). This different rate of maturation of body functions is certainly connected to the demands of life outside the nest, and we tried to analyze which ecological factors might be responsible for timing of nest leaving (i.e., duration of the nestling period), while controlling for the effect of growth rate. In addition, we analyzed relative fledging mass, defined as the ratio of body mass at fledging to adult body mass as an estimate of the relative stage of development at fledging. Body mass at fledging was defined as the average mass of nestlings at the last day in the nest and was taken from the original studies.
Body mass increases in passerines in an S-shaped function, which is expected on theoretical grounds (West et al. 2001).
Many S-shaped mathematical curves could potentially fit growth data, but the logistic growth curve is traditionally used (Ricklefs 1967a, 1968a). Although Brisbin et al. (1987) criticized its use and suggested the use of more complex curves that are able to describe shape as well as rate, the advantage of the logistic growth curve is that it produces only three parameters that are readily biologically interpret-able. The logistic growth curve has a form of W(t) 5 A/{1 1 e }, where W(t) denotes body mass of a nestling at [2K(t2t )] i time t, A is the asymptotic body mass that the nestling ap-proaches, ti is the inflection point on the time axis in which growth changes from accelerating to decelerating, e is the base of natural logarithm, and K is a constant scaling rate of growth. Because the value of K indexes growth rate inde-pendently of absolute time of growth (in time21), it is a con-venient measure for comparative purposes (Ricklefs 1968a).
Estimates of K (growth rate) are problematic for compar-ative purposes because of a negative correlation between K and A (asymptotic body mass). The most serious problem arises in species with nestlings that remain in the nest longer and experience weight recession in later phases of the nestling period (see Ricklefs 1968b). Here the downward hook of the data, after a maximum value is achieved, forces A to be estimated lower than it would be without this hook, and con-sequently the estimate of K is artificially inflated (see Fig.
1). Consequently, species with the exact same growth during the initial growth period (illustrated by the vertical line in Fig. 1) yield different estimates of K when they remain in the nest for differing periods of time (Fig. 1; V. Remes ˇ, unpubl. data). Recently, a new modification of the logistic growth curve was developed to deal with this problem in seabirds (Huin and Prince 2000). It has two K parameters, one for mass increase (K1) and the other for the rate of mass decrease during weight recession phase (K2). However, it suffers from the same shortcoming as the traditional method.
Although this new approach fits growth data with weight recession well, the value of K1 (which equals K of the tra-ditional model and is of interest to us) still depends on the pattern and extent of weight recession (V. Remes ˇ, unpubl.
data).
To overcome these problems and obtain growth estimates that are comparable across different ecological groups of birds, we used two approaches. First, we fit the traditional logistic curve to the growth data truncated at the highest mass achieved by nestlings in species that remained in the nest past a maximum mass; truncation was necessary for 61 pop-ulations of 39 species, including 15 (of 20) species of aerial foragers, and 24 (of 95) species of nonaerial foragers. The effect of this adjustment on the estimation of K was marked only in aerial foragers, because these have a high incidence of weight recession (see above and Fig. 2). However, even after this adjustment, estimations of K could still be con-founded by differences in ages and relative mass at fledging.
Consequently, our second approach was to fit the logistic curve to the growth series truncated at 70% of adult body mass. This approach completely standardizes the relative nestling mass over which growth rates are estimated. We chose 70% as a compromise between retaining as much of the growth curve as possible and the maximum number of species possible (some species leave the nest at a lighter mass than 70% of adult body mass and consequently had to be excluded). This procedure led to loss of 18 species but still yielded highly standardized data on 97 species. Both these adjustments yield more standardized and appropriate esti-mation of K than in previous analyses. Analyses based on K fit to the growth data without either of the two adjustments produced virtually identical results to adjusted data, with ex-ception that the explanatory power of aerial foraging for K was significantly reduced. Given the redundancy of these analytical results, we report only results of the analyses with the two standardized sets of K.
For our analyses we used average K for species. However, growth rate of nestlings can be adversely affected by poor environmental, especially food, conditions during rearing (e.g., Martin 1987; Gebhardt-Henrich and Richner 1998; 2508 V. REMES ˇ AND T. E. MARTIN TABLE 2.
Multiple regression analyses of the constant K of the logistic growth curve in North American passerines in relation to potential ecological factors and covariates. Sample sizes are numbers of species for raw species data (not corrected for phylogenetic relationships) and of phylogenetically independent contrasts for PICs (corrected for phylogenetic relationships). For partial regression and correlation coefficients from the analysis of raw species data, see Figure 3.
K 1 Raw species data (n 5 107) PICs (n 5 103) Full model t P Best model2 t P Full model t P Best model2 t P Adult body mass Clutch size Dmr3 Foraging mode4 Latitude 26.50 21.14 4.49 21.90 1.69 ,0.0001 0.2587 ,0.0001 0.0604 0.0940 26.54 — 5.11 21.97 — ,0.0001 — ,0.0001 0.0515 — 22.34 20.16 4.53 0.49 0.94 0.0214 0.8764 ,0.0001 0.6257 0.3509 22.39 — 4.48 — — 0.0188 — ,0.0001 — — 1 Fit to the growth data that were truncated at the highest mass reached by the young in the nest.
2 Selected by the backward selection procedure in the multiple regression model of SPSS.
3 Dmr is daily nest mortality rate caused by nest predation.
4 Foraging mode is aerial foragers (coded 1) and nonaerial foragers (coded 0).
Schew and Ricklefs 1998). As a result, we repeated analyses with maximum K for species, which might better reflect evo-lutionary responses of growth rates than average values. Nev-ertheless, results of these two analyses did not differ. Con-sequently, only the results of the analyses with average values of K are reported because we used average values for all other variables.
Phylogenetic Analyses We analyzed raw species data, but also employed the meth-od of phylogenetically independent contrasts to control for possible phylogenetic influences (Felsenstein 1985; Harvey and Pagel 1991) based on the CAIC software package (Purvis and Rambaut 1995). We analyzed the independent contrasts in a phylogenetic regression framework (Grafen 1989), in which contrasts computed by CAIC (CRUNCH algorithm) were analyzed with standard multiple linear regressions forced through the origin (see Garland et al. 1992).
For analyses adjusting for phylogenetic relationships, we used a working phylogeny depicted in Martin and Clobert (1996), which is based on Sibley and Ahlquist’s (1990) DNA-DNA hybridization phylogenetic hypothesis, and supple-mented it with more recent molecular phylogenetic infor-mation (details are available from the authors upon request).
We did not have consistent estimates of branch lengths and so we used equal branch lengths (Garland et al. 1993).
Previous analyses comparing equal branch lengths versus variable ones found little effect on results (Martins and Gar-land 1991). Estimation of branch lengths is an empirical issue and their performance should be statistically tested (Garland et al. 1992). We checked the performance of equal branch lengths by plotting the absolute values of the standardized contrasts against their standard deviations (Garland et al.
1992). In all cases, performance of equal branch lengths was good and much better than that of another option, Grafen’s (1989) branch lengths.
Statistics Logistic growth curves were fit in nonlinear regression in SPSS (1996). We used the Levenberg-Marquardt estimation method, sum of squared residuals loss function, and no pa-rameter constraints. The ability of various factors to explain interspecific variation in the growth rate constant K, duration of nestling period, relative fledging mass, and premature fledging was tested by multiple linear regressions in SPSS.
The best models were selected by the backward selection procedure. Our P-to-enter and P-to-remove values were 0.05 and 0.1, respectively (Sokal and Rohlf 1995). Selection of the final model in backward selection procedure depends in part on these P-values. Thus, we also validated the models by the means of Akaike’s information criterion (AIC), which is computed as nln(SSE/n) 1 2p, where n is the number of observations, SSE is the sum-of-squares error, and p is the number of model parameters. This is a general criterion for choosing the best number of parameters to include in a model.
We chose a model with the minimum number of parameters from the set of models for which the difference between AIC(i) and AIC(min) was lower than two, where AIC(i) is AIC of the particular model and AIC(min) is minimum AIC of all the possible models (see Anderson et al. 2000). Mar-ginal means for raw species data were estimated by AN-COVAs in SPSS. Phylogenetic regressions were performed on independent contrasts as ordinary multiple linear regres-sions forced through the origin.
Relative fledging mass was distributed normally. Other variables were transformed to meet the assumption of normal distribution. Adult body mass, clutch size, K, latitude, and premature fledging were log10 transformed and daily mor-tality rates caused by nest predation were square-root trans-formed.
RESULTS Growth Rate (K) Growth rates (K) fit to the data truncated at the highest mass achieved by nestlings were significantly correlated with adult body mass and daily nest predation rates (full model: R2 adj. 5 0.45, F5,101 5 18.07, P , 0.0001; reduced model: R2 adj. 5 0.44, F3,103 5 28.81, P , 0.0001). The influence of foraging mode was only marginally significant (Table 2, Fig. 3). Results were essentially the same for phylogeneti-cally independent contrasts, with the exception that foraging mode was dropped from the model (Table 2).
2509 GROWTH STRATEGIES OF PASSERINES FIG. 3.
Scattergrams of the standardized residuals from the mul-tiple regression model of raw species data for K fit to the growth data that were truncated at the highest mass reached by the young in the nest (see Table 2). (A) K versus adult body mass (full circles, Corvidae; open circles, others), both corrected for daily predation rates and foraging mode; (B) K versus daily predation rates, both corrected for adult body mass and foraging mode; and (C) marginal means for foraging mode (black bars, all species; white bars, with-out four species of Empidonax), adjusted for adult body mass and daily predation rates. K relates to adult body mass, daily predation rates, and foraging mode (coded as nonaerial foragers 5 0 [n 5 89], aerial foragers 5 1 [n 5 18]) by the equation: log(K [day21]) 5 20.256 (0.040) 2 0.128 (0.020) log(body mass [g]) 1 0.976 (0.191) square root (dpr [day21]) 2 0.039 (0.020) foraging mode.
Parameters are partial regression coefficients (SE). rp-values are partial correlation coefficients.
In the analyses of raw species data, the relationship be-tween K and daily predation rate was curvilinear (Fig. 3B), as indicated by a significant nonlinear quadratic term (t1,104 5 23.44, P 5 0.0008), when effects of adult body mass and foraging mode were controlled. No such effect was apparent in the analyses of phylogenetically independent contrasts when adult body mass was controlled (t1,101 5 20.02, P 5 0.9807).
Although foraging mode was marginally significant in the analysis of raw species data (Table 2), this influence was due to four species of the genus Empidonax (see Fig. 3C), and after adjusting for phylogenetic effects it disappeared (Table 2). A strong phylogenetic effect was also apparent in the scaling of growth rate with adult body mass due to the in-fluence of the family Corvidae (see Fig. 3A). After accounting for the effects of phylogeny, the influence of adult body mass on growth rate was much smaller (Tables 2, 3). Moreover, when the contrast between Corvidae and Vireonidae was ex-cluded, the effect of adult body mass on growth rate com-pletely disappeared (t1,100 5 20.99, P 5 0.3257, rp 5 20.10) and daily predation rate became the best predictor (t1,100 5 4.69, P , 0.0001, rp 5 0.43).
Very similar results were obtained from the analyses of growth rates (K) fit to the data truncated at 70% of adult body mass. The difference was that both clutch size and latitude also entered the model in both raw species data (full model: R2 adj. 5 0.44, F5,83 5 13.23, P , 0.0001; reduced model: R2 adj. 5 0.44, F4,84 5 16.52, P , 0.0001) and phyloge-netically independent contrasts (Table 3).
Nestling Period Adult body mass, daily nest predation rates, foraging mode, and K were significantly related to the length of the nestling period when estimates of K were fit to the data truncated at the highest mass achieved by the nestlings (R2 adj. 5 0.72, F5,101 5 56.15, P , 0.0001; Table 4, Fig. 4). The same results were found when analyses were performed on phylogeneti-cally independent contrasts (Table 4). When the covariate K was based on data truncated at 70% of adult body mass, the analyses of factors influencing length of the nestling period yielded virtually identical results.
Premature Fledging We also examined the extent to which nestlings were will-ing to fledge prematurely, defined as duration of the normal nestling period minus duration of the nestling period when nestlings fledge prematurely. The extent of premature fledg-ing decreased with daily predation rates (t1,30 5 23.93, P 5 0.0005, n 5 33) after accounting for the effect of adult body mass (t1,30 5 3.70, P 5 0.0009, n 5 33; Fig. 5; whole model: R2 5 0.43, F2,30 5 13.16, P , 0.0001). This was the same for phylogenetically independent contrasts. Neither clutch size nor the growth rate constant K entered these models.
Relative Fledging Mass Relative fledging mass (6 SE, n), defined as the ratio of the mass at fledging to adult body mass, averaged 0.819 (6 0.015, 115) over all species. Both daily predation rates 2510 V. REMES ˇ AND T. E. MARTIN TABLE 3.
Multiple regression analyses of the constant K of the logistic growth curve in North American passerines in relation to potential ecological factors and covariates. Sample sizes are numbers of species for raw species data (not corrected for phylogenetic relationships) and of phylogenetically independent contrasts for PICs (corrected for phylogenetic relationships).
K 1 Raw species data (n 5 89) PICs (n 5 85) Full model t P Best model2 t P Full model t P Best model2 t P Adult body mass Clutch size Dmr3 Foraging mode4 Latitude 25.78 22.57 2.41 20.70 3.13 ,0.0001 0.0121 0.0181 0.4885 0.0024 25.76 22.59 2.81 — 3.18 ,0.0001 0.0114 0.0061 — 0.0021 23.86 22.19 2.15 0.08 2.45 0.0002 0.0317 0.0342 0.9374 0.0163 23.90 22.20 2.17 — 2.47 0.0002 0.0307 0.0327 — 0.0157 1 Fit to the growth data that were truncated at 70% of adult body mass.
2 Selected by the backward selection procedure in the multiple regression model of SPSS.
3 Dmr is daily nest mortality rate caused by nest predation.
4 Foraging mode is aerial foragers (coded 1) and nonaerial foragers (coded 0).
TABLE 4.
Multiple regression analyses of the duration of the nestling period in North American passerines in relation to potential ecological factors and covariates. Sample sizes are numbers of species for raw species data (not corrected for phylogenetic relationships) and of phylo-genetically independent contrasts for PICs (corrected for phylogenetic relationships). For partial regression and correlation coefficients from the analysis of raw species data, see Figure 4.
Nestling period Raw species data (n 5 107) PICs (n 5 103) Full model t P Best model1 t P Full model t P Best model1 t P Adult body mass Clutch size Dmr2 Foraging mode3 K 4 3.95 1.92 23.55 6.08 25.74 0.0001 0.0570 0.0006 ,0.0001 ,0.0001 3.73 — 24.34 5.98 25.74 0.0003 — ,0.0001 ,0.0001 ,0.0001 3.91 1.23 22.64 2.54 22.56 0.0002 0.2220 0.0095 0.0126 0.0121 3.80 — 22.88 2.61 22.52 0.0003 — 0.0048 0.0105 0.0134 1 Selected by the backward selection procedure in the multiple regression model of SPSS.
2 Dmr is daily nest mortality rate caused by nest predation.
3 Foraging mode is aerial foragers (coded 1) and nonaerial foragers (coded 0).
4 Fit to the growth data that were truncated at the highest mass reached by the young in the nest.
and foraging mode were significantly related to relative fledging mass, with adult body mass controlled (Table 5, Fig. 6; whole model: R2 adj. 5 0.46, F3,102 5 30.24, P , 0.0001). Examination of phylogenetically independent contrasts, with adult body mass controlled, still yielded significant effects of daily predation rates, but foraging mode was dropped from the stepwise selection procedure (Table 5).
Relative fledging mass was curvilinearly related to daily predation rate, as reflected by a significant nonlinear qua-dratic term (t1,103 5 2.46, P 5 0.0154), when the effects of adult body mass and foraging mode were controlled (see Fig.
6B). However, no such term was apparent in phylogenetically adjusted data, when the effect of adult body mass was taken into account (t1,100 5 0.86, P 5 0.3928). All analyses of the relative fledging mass were performed without the strongly outlying Leucosticte tephrocotis (relative fledging mass 5 1.54; see also Appendix 1).
DISCUSSION Previous comparative studies of variation in growth rates found little in the way of environmental correlates (see Rick-lefs et al. 1998). In direct contrast, we found very strong effects of predation, with smaller but significant contributing effects of foraging mode, clutch size, and latitude on both growth rate and developmental strategies. These results show that growth and development of altricial nestlings are shaped by extrinsic environmental forces.
Daily predation rates scaled positively with growth rates and negatively with duration of the nestling period (after controlling for growth rates; Tables 2–4; Figs. 3B, 4B). Fur-thermore, species with higher nest predation fledged at lighter relative mass (Table 5, Fig. 6B). These results suggest that multidimensional aspects of the growth strategies of altricial nestlings are strongly influenced by risk of nest predation.
Such results are in line with theoretical arguments made by Lack (1968) and Case (1978), and with empirical studies of Kleindorfer et al. (1997) and Hałupka (1998).
However, the benefits of growing faster and shortening the nestling period may reach a point of diminishing returns.
Indeed, the curvilinear relationship between nest mortality versus growth rates (Fig. 3B; Results) and relative fledging mass (Fig. 6B; Results) could arise if costs (ecological: Mar-tin 1992; Martin et al. 2000a,b; physiological: Metcalfe and Monaghan 2001) of even faster growth exceed the benefits.
Alternatively, leveling off of growth rates could be caused by reaching the maximum growth rate possible within cellular and physiological constraints (see Ricklefs 1969b). More-2511 GROWTH STRATEGIES OF PASSERINES FIG. 4.
Scattergrams of the standardized residuals from the multiple regression model of raw species data for the duration of the nestling period, with K fit to the growth data that were truncated at the highest mass reached by the young in the nest (see Table 4). (A) Duration of the nestling period versus adult body mass, both corrected for daily predation rates, foraging mode, and K; (B) nestling period versus daily predation rates, both corrected for adult body mass, foraging mode, and K; (C) nestling period versus K, both corrected for adult body mass, daily predation rates, and foraging mode; (D) marginal means for foraging mode adjusted for adult body mass, daily predation rates, and K. Nestling period relates to adult body mass, daily predation rates, K, and foraging mode (coded as nonaerial foragers 5 0 [n 5 89], aerial foragers 5 1 [n 5 18]), by the equation: log(nestling period [days]) 5 0.862 (0.053) 1 0.097 (0.026) log(body mass [g]) 2 1.038 (0.239) square root (dpr [day21] 2 0.634 (0.110) log(K [day21]) 1 0.135 (0.022) foraging mode. Parameters are partial regression coefficients (SE). rp-values are partial correlation coefficients.
over, because birds that grow quicker also fledge and mature earlier, the trade-off between growth and maturation could play a role (Ricklefs et al. 1994, 1998; but see Krijgsveld et al. 2001).
Several studies have shown that duration of the nestling period scales negatively with nest predation rates (Lack 1968; Martin and Li 1992; Bosque and Bosque 1995; Martin 1995; Yanes and Sua ´rez 1997). We show that this is true even after accounting for pure growth rates (Table 4, Fig. 4B). Thus, shorter nestling periods result not only from faster growth, but also from earlier timing of nest-leaving at an earlier stage of development. The latter is clearly reflected by the lower relative fledging mass in species with higher predation rates (Table 5, Fig. 6B). However, the precocity of fledging is clearly limited. This limit is evident from the negative re-lationship between the extent of premature fledging relative to nest predation rates. Whereas species with high predation already leave at an early stage of development and cannot leave many days earlier, species with low predation stay in the nest to a much later stage of development and have the capability of fledging many days earlier than they do (see Fig. 5). In addition, the reduction in fledging mass when fledging prematurely can be so high and costs connected with it so strong (Linde ´n and Møller 1989; Magrath 1991; Geb-hardt-Henrich and Richner 1998; Lindstro ¨m 1999) that it can be advantageous to keep this strategy as optional for the circumstances of real predation danger.
Predation was not the only environmental correlate of growth and development. Food limitation has long been pro-posed to explain the evolution of growth and development of birds through energetic expensiveness of large clutch sizes (Lack 1968), shorter time available for feeding offspring at lower latitudes (Ricklefs 1976), and unpredictability of food supply (O’Connor 1978). Of these three, growth rates were associated with aerial foraging when using K fit to the growth data truncated at the highest mass (Table 2) and with clutch size and latitude when using K fit to the growth data truncated at 70% of adult body mass (Table 3). Nestling period duration differed strongly between aerial and nonaerial foragers (Table 4, Fig. 4D). In contrast, the influence of aerial foraging on relative fledging mass was rather weak in the raw species data (Table 5, Fig. 6C), and disappeared after adjusting for 2512 V. REMES ˇ AND T. E. MARTIN FIG. 5.
Scattergram of the standardized residuals of premature fledging (reduction of the duration of the nestling period when fledging prematurely as compared to normal fledging, in days) ver-sus daily predation rates, both corrected for adult body mass, for raw species data. Log(premature fledging [day]) 5 0.333 (0.159) 2 3.444 (0.876) square root (dpr [day21]) 1 0.336 (0.091) log(body mass [g]), n 5 33. Parameters are partial regression coefficients (SE). rp-value is partial correlation coefficient.
TABLE 5.
Multiple regression analyses of the relative fledging mass (fledging mass/adult body mass) in North American passerines in relation to potential ecological factors and covariates. Sample sizes are number of species for raw species data (not corrected for phylogenetic rela-tionships) and of phylogenetically independent contrasts for PICs (corrected for phylogenetic relationships). For partial regression and correlation coefficients from the analysis of raw species data, see Figure 6.
Relative fledging mass Raw species data (n 5 106) PICs (n 5 102) Full and best model1 t P Full model t P Best model1 t P Adult body mass Dmr2 Foraging mode3 26.01 25.14 2.62 ,0.0001 ,0.0001 0.0101 24.58 22.92 20.34 ,0.0001 0.0043 0.7383 24.58 22.91 — ,0.0001 0.0044 — 1 Selected by the backward selection procedure in the multiple regression model of SPSS.
2 Dmr is daily nest mortality rate caused by nest predation.
3 Foraging mode is aerial foragers (coded 1) and nonaerial foragers (coded 0).
phylogeny (Table 5). This is not surprising, however, because aerial foragers show a high incidence of weight recession (see Methods), and thus rather long nestling periods do not translate to very high mass of fledglings relative to adult body mass.
Aerial foragers seem to be food limited in general (Martin 1987, 1995). Their food supply is temporally unpredictable and there is an ample evidence that adverse climatic condi-tions can lead on a proximate level to longer incubation and nestling periods and impaired growth, condition, and survival of nestlings (e.g., Bryant 1975; O’Connor 1978; McCarty and Winkler 1999). Nestlings of aerial foragers must survive relatively long periods without parental attendance during inclement weather conditions, whose incidence and duration are unpredictable. Adaptations to sustain these periods in-clude extensive fat deposition and early thermal indepen-dence of the nestlings (Ricklefs 1967b; O’Connor 1978). It is possible to speculate that these adaptations aimed at de-creasing susceptibility to starvation could also have led to or have been connected with adjustments in growth strategies of nestlings (O’Connor 1978), including lower growth rates and longer nestling periods. On the other hand, the strongest result concerning aerial foragers (i.e., longer nestling periods) can be explained alternatively—that they must wait in the nest until the maturation of their flight muscles that are crit-ical for their demanding flight life. Thus, results concerning aerial foragers must be interpreted with caution and more analyses are clearly needed, but the positive results found here with relatively low power hold promise for future anal-yses.
Lack (1968) proposed a trade-off between clutch size and growth rate (see also Ricklefs 1968a). Our data illustrate negative covariation among species; species with larger clutch size show slower growth of nestlings for the same adult body mass, nest predation, and latitude. Such a co-variation could be merely a correlated response to the same factor—lower nest predation risk may simply favor both larg-er clutches (Martin 1995) and slower growth (Case 1978) independently. However, because we control for nest pre-dation by multiple regression, it may be that this is cause and effect (i.e., a trade-off). A positive relationship between growth rate of nestlings and latitude was predicted by Rick-lefs (1976). Although he suggested this effect to explain slower growth of tropical species, we show that this can be true even on a much smaller geographical scale. However, because both the effect of clutch size and latitude were ap-parent only when using K fit to the growth data truncated at 70% of adult body mass, they should be taken with caution and be subject to additional analyses.
Adult body mass can have an allometric influence on life-history traits. Because K indexes growth rate independently of body mass (Ricklefs 1968a; Starck and Ricklefs 1998b), however, there is no a priori reason to expect its negative scaling with growth rate. The negative scaling observed here may suggest basic design constraints of large body size (West et al. 2001). Alternatively, slower growth rates and longer nestling periods of larger birds could be a result of lower parental investment in offspring. Larger birds have higher adult survival rates (Sæther 1987, 1988, 1989; Martin 1995), and species with higher adult survival are expected to invest less in progeny (Charnov and Schaffer 1973; Martin 2002).
Adult body mass could work here just as a reflection of adult survival rate. To test this hypothesis, comparative studies of 2513 GROWTH STRATEGIES OF PASSERINES FIG. 6.
Scattergrams of the standardized residuals from the mul-tiple regression model of raw species data for the relative fledging mass (fledging mass/adult body mass; see Table 5). (A) Relative fledging mass versus adult body mass, both corrected for daily predation rates and foraging mode; (B) relative fledging mass versus daily predation rates, both corrected for adult body mass and for-aging mode; and (C) marginal means for foraging mode, adjusted for adult body mass and daily mortality rates. Relative fledging mass relates to adult body mass, daily predation rates, and foraging mode (coded as nonaerial foragers 5 0 [n 5 88], aerial foragers 5 1 [n 5 18]) by the equation: relative fledging mass 5 1.244 (0.062) 2 0.182 (0.030) log(adult body mass [g]) 2 1.524 (0.296) square root (dpr [day21]) 1 0.080 (0.031) foraging mode. Parameters are partial regression coefficients (SE). rp-values are partial correlation coefficients.
growth strategies including adult survival rates and feeding rates are needed.
In sum, we show that growth rates, duration of nestling period, and relative developmental stage at nest-leaving cov-ary in altricial nestlings with nest predation rates, aerial for-aging, clutch size, and latitude after taking into account adult body mass and phylogenetic effects. This is in line with the view that growth strategies of altricial nestlings are finely tuned to environmental conditions typical of each species (Lack 1968; Kleindorfer et al. 1997) and lends support to an adaptive view of variation in growth and development (Arendt 1997). Moreover, we show that studies addressing evolution of growth strategies should simultaneously ex-amine both pure growth and timing of developmental events (here, fledging), because these two show independent evo-lution in relation to extrinsic selective forces.
ACKNOWLEDGMENTS During the work on this manuscript, VR was supported by a grant from J. W. Fulbright Commission, and TEM by grants from National Science Foundation (DEB-9707598, DEB-9981527). We thank R. E. Ricklefs and an anonymous re-viewer for comments that greatly improved quality of the manuscript.
LITERATURE CITED Anderson, D. R., K. P. Burnham, and W. L. Thompson. 2000. Null hypothesis testing: problems, prevalence, and an alternative. J.
Wildl. Manage. 64:912–923.
Arendt, J. D. 1997. Adaptive intrinsic growth rates: an integration across taxa. Q. Rev. Biol. 72:149–177.
Austin, G. T., and R. E. Ricklefs. 1977. Growth and development of the rufous-winged sparrow (Aimophila carpalis). Condor 79: 37–50.
Billerbeck, J. M., T. E. Lankford Jr.,and D. O. Conover. 2001.
Evolution of intrinsic growth and energy acquisition rates. I.
Trade-offs with swimming performance in Menidia menidia.
Evolution 55:1863–1872.
Bortolotti, G. R. 1986. Evolution of growth-rates in eagles: sibling competition vs. energy considerations. Ecology 67:182–194.
Bosque, C., and M. T. Bosque. 1995. Nest predation as a selective factor in the evolution of developmental rates in altricial birds.
Am. Nat. 145:234–260.
Brisbin, I. L., Jr., C. T. Collins, G. C. White, and D. A. McCallum.
1987. A new paradigm for the analysis and interpretation of growth data: the shape of things to come. Auk 104:552–555.
Bryant, D. M. 1975. Breeding biology of house martins Delichon urbica in relation to aerial insect abundance. Ibis 117:180–216.
Case, T. J. 1978. On the evolution and adaptive significance of postnatal growth rates in the terrestrial vertebrates. Q. Rev. Biol.
53:243–282.
Charnov, E. L. 1993. Life history invariants. Oxford Univ. Press, Oxford, U.K.
Charnov, E. L., and W. M. Schaffer. 1973. Life-history conse-quences of natural selection: Cole’s result revisited. Am. Nat.
107:791–793.
Dunning, J. B., Jr. 1993. Body weights of 686 species of North American birds. Eldon Publishing, Cave Creek, AZ.
Felsenstein, J. 1985. Phylogenies and the comparative method. Am.
Nat. 125:1–15.
Garland, T., P. H. Harvey, and A. R. Ives. 1992. Procedures for the analysis of comparative data using phylogenetically inde-pendent contrasts. Syst. Biol. 41:18–32.
Garland, T., A. W. Dickerman, C. M. Janis, and J. A. Jones. 1993.
Phylogenetic analysis of covariance by computer simulation.
Syst. Biol. 42:265–292.
2514 V. REMES ˇ AND T. E. MARTIN Gebhardt-Henrich, S., and H. Richner. 1998. Causes of growth var-iation and its consequences for fitness. Pp. 324–339 in J. M.
Starck and R. E. Ricklefs, eds. Avian growth and development.
Oxford Univ. Press, Oxford, U.K.
Grafen, A. 1989. The phylogenetic regression. Philos. Trans. R.
Soc. Lond. B Biol. Sci. 326:119–157.
Hałupka, K. 1998. Partial nest predation in an altricial bird selects for the accelerated development of young. J. Avian Biol. 29: 129–133.
Harvey, P. H., and M. D. Pagel. 1991. The comparative method in evolutionary biology. Oxford Univ. Press, Oxford, U.K.
Herms, D. A., and W. J. Mattson. 1992. The dilemma of plants: to grow or defend. Q. Rev. Biol. 67:283–335.
Huin, N., and P. A. Prince. 2000. Chick growth in albatrosses: curve fitting with a twist. J. Avian Biol. 31:418–425.
Kleindorfer, S., H. Hoi, and R. Ille. 1997. Nestling growth patterns and antipredator responses: a comparison between four Acro-cephalus warblers. Biologia (Bratisl.) 52:677–685.
Krijgsveld, K. L., J. M. Olson, and R. E. Ricklefs. 2001. Catabolic capacity of the muscles of shorebird chicks: maturation of func-tion in relation to body size. Physiol. Biochem. Zool. 74: 250–260.
Lack, D. 1968. Ecological adaptations for breeding in birds. Me-thuen, London.
Lankford, T. E., Jr., J. M. Billerbeck, and D. O. Conover. 2001.
Evolution of intrinsic growth and energy acquisition rates. II.
Trade-offs with vulnerability to predation in Menidia menidia.
Evolution 55:1873–1881.
Linden, M., and A. P. Møller. 1989. Cost of reproduction and co-variation of life-history traits in birds. Trends Ecol. Evol. 4: 367–371.
Lindstro ¨m, J. 1999. Early development and fitness in birds and mammals. Trends Ecol. Evol. 14:343–348.
Magrath, R. D. 1991. Nestling weight and juvenile survival in the blackbird, Turdus merula. J. Anim. Ecol. 60:335–351.
Martin, T. E. 1987. Food as a limit on breeding birds—a life-history perspective. Annu. Rev. Ecol. Syst. 18:453–487.
———. 1992. Interaction of nest predation and food limitation in reproductive strategies. Curr. Ornithol. 9:163–197.
———. 1993. Nest predation among vegetation layers and habitat types: revising the dogmas. Am. Nat. 141:897–913.
———. 1995. Avian life-history evolution in relation to nest sites, nest predation, and food. Ecol. Monogr. 65:101–127.
———. 2002. A new view of avian life-history evolution tested on an incubation paradox. Proc. R. Soc. Lond. B Biol. Sci. 269: 309–316.
Martin, T. E., and J. Clobert. 1996. Nest predation and avian life-history evolution in Europe versus North America: a possible role of humans? Am. Nat. 147:1028–1046.
Martin, T. E., and P. J. Li. 1992. Life-history traits of open-nesting vs. cavity-nesting birds. Ecology 73:579–592.
Martin, T. E., P. R. Martin, C. R. Olson, B. J. Heidinger, and J. J.
Fontaine. 2000a. Parental care and clutch sizes in North and South American birds. Science 287:1482–1485.
Martin, T. E., J. Scott, and C. Menge. 2000b. Nest predation in-creases with parental activity: separating nest site and parental activity effects. Proc. R. Soc. Lond. B Biol. Sci. 267:2287–2293.
Martins, E. P., and T. Garland Jr. 1991. Phylogenetic analyses of the correlated evolution of continuous characters: a simulation study. Evolution 45:534–557.
McCarty, J. P., and D. W. Winkler. 1999. Relative importance of environmental variables in determining the growth of nestling tree swallows Tachycineta bicolor. Ibis 141:286–296.
Metcalfe, N. B., and P. Monaghan. 2001. Compensation for a bad start: Grow now, pay later? Trends Ecol. Evol. 16:254–260.
Nylin, S., and K. Gotthard. 1998. Plasticity in life-history traits.
Annu. Rev. Entomol. 43:63–83.
O’Connor, R. J. 1978. Growth strategies in nestling passerines.
Living Bird 16:209–238.
Oniki, Y., and R. E. Ricklefs. 1981. More growth rates of birds in the humid new world tropics. Ibis 123:349–354.
Poole, A., and F. Gill, eds. 1992–2002. The birds of North America.
The Birds of North America, Philadelphia, PA.
Purvis, A., and A. Rambaut. 1995. Comparative analysis by inde-pendent contrasts (CAIC): an Apple-Macintosh application for analyzing comparative data. Comput. Appl. Biosci. 11:247–251.
Ricklefs, R. E. 1967a. A graphical method of fitting equations to growth curves. Ecology 48:978–980.
———. 1967b. Relative growth, body constituents, and energy con-tent of nestling barn swallows and red-winged blackbirds. Auk 84:560–570.
———. 1968a. Patterns of growth in birds. Ibis 110:419–451.
———. 1968b. Weight recession in nestling birds. Auk 85:30–35.
———. 1969a. An analysis of nesting mortality in birds. Smithson.
Contrib. Zool. 9:1–48.
———. 1969b. Preliminary models for growth rates in altricial birds. Ecology 50:1031–1039.
———. 1973. Patterns of growth in birds. II. Growth rate and mode of development. Ibis 115:177–201.
———. 1976. Growth rates of birds in the humid New World trop-ics. Ibis 118:179–207.
———. 1979a. Adaptation, constraint, and compromise in avian postnatal development. Biol. Rev. (Camb.) 54:269–290.
———. 1979b. Patterns of growth in birds. V. A comparative study of development in the starling, common tern, and Japanese quail.
Auk 96:10–30.
———. 1982. Some considerations on sibling competition and avi-an growth rates. Auk 99:141–147.
———. 1983. Avian postnatal development. Pp. 1–83 in D. S.
Farner, ed. Avian biology. Vol. VII. Academic Press, New York.
———. 1984. The optimization of growth rate in altricial birds.
Ecology 65:1602–1616.
Ricklefs, R. E., R. E. Shea, and I. H. Choi. 1994. Inverse relationship between functional maturity and exponential growth rate of avi-an skeletal muscle: a constraint on evolutionary response. Evo-lution 48:1080–1088.
Ricklefs, R. E., J. M. Starck, and M. Konarzewski. 1998. Internal constraints on growth in birds. Pp. 266–287 in J. M. Starck and R. E. Ricklefs, eds.
Roff, D. A. 1992. The evolution of life histories: theory and anal-ysis. Chapman and Hall, New York.
———. 2002. Life history evolution. Sinauer Associates, Sunder-land, MA.
Royle, N. J., I. R. Hartley, I. F. Owens, and G. A. Parker. 1999.
Sibling competition and the evolution of growth rates in birds.
Proc. R. Soc. Lond. B Biol. Sci. 266:923–932.
Sæther, B-E. 1987. The influence of body weight on the covariation between reproductive traits in European birds. Oikos 48:79–88.
———. 1988. Pattern of covariation between life-history traits of European birds. Nature 331:616–617.
———. 1989. Survival rates in relation to body weight in European birds. Ornis Scand. 20:13–21.
Schew, W. A., and R. E. Ricklefs. 1998. Developmental plasticity.
Pp. 288–304 in J. M. Starck and R. E. Ricklefs, eds. Avian growth and development. Oxford Univ. Press, Oxford, U.K.
Sibley, C. G., and J. E. Ahlquist. 1990. Phylogeny and classification of birds. Yale Univ. Press, New Haven, CT.
Sokal, R. R., and F. J. Rohlf. 1995. Biometry. 3rd ed. W. H. Free-man, New York.
SPSS, Inc. 1996. SPSS Base 7.0 for Windows user’s guide. SPSS, Inc., Chicago, IL.
Starck, J. M., and R. E. Ricklefs, eds. 1998a. Avian growth and development. Oxford Univ. Press, Oxford, U.K.
———. 1998b. Data set of avian growth parameters. Pp. 381–423 in J. M. Starck and R. E. Ricklefs, eds. Avian growth and de-velopment. Oxford Univ. Press, Oxford, U.K.
———. 1998c. Patterns of development: the altricial-precocial spectrum. Pp. 3–30 in J. M. Starck and R. E. Ricklefs, eds. Avian growth and development. Oxford Univ. Press, Oxford, U.K.
———. 1998d. Variation, constraint, and phylogeny: comparative analysis of variation in growth. Pp. 247–265 in J. M. Starck and R. E. Ricklefs, eds. Avian growth and development. Oxford Univ. Press, Oxford, U.K.
Stearns, S. C. 1992. The evolution of life histories. Oxford Univ.
Press, Oxford, U.K.
2515 GROWTH STRATEGIES OF PASSERINES APPENDIX 1 Species means used in the analyses for adult body mass (Abm, g), clutch size (Cl, no. of eggs per nest), duration of nestling period (Nstl, days), duration of shortened nestling period when fledging prematurely (Nstl prem, days), growth rate constant K of the logistic curve fit to the growth data truncated at the highest mass achieved by nestlings (K, day21), data truncated at 70% of adult body mass (K 70%, day21), body mass at fledging (Fl mass, g), latitude of the study of growth (Lat, 8N), number of primary growth series for the species (No. stud), foraging mode (Forag; n, nonaerial; a, aerial foragers), proportion of nests taken by predators (Pred, %), and references for primary sources (Ref ).
Species Abm Cl Nstl Nstl prem K K 70% Fl mass Lat No.
stud Forag Pred Ref Agelaius phoeniceus Aimophila botterii Aimophila carpalis Aimophila cassinii Ammodramus bairdii Ammodramus caudacutus Ammodramus maritimus 52.6 19.9 15.3 18.9 17.5 19.3 23.3 3.28 3.65 3.80 3.97 4.33 3.90 3.39 11.5 10.0 8.5 8.0 9.0 9.7 10.0 — 8.0 — — — 8.0 8.0 0.533 0.489 0.555 0.515 0.410 0.564 0.579 0.585 0.484 0.580 — 0.461 0.546 0.565 35.6 14.7 11.9 12.0 13.6 15.4 16.8 41.6 31.5 32.0 31.5 50.8 40.3 40.3 3 1 1 1 1 2 2 n n n n n n n 44.40 29.20 68.00 30.77 43.00 36.20 24.53 1, 4 5, 6 7 6, 8 9, 10 11, 12 11, 13 Ammodramus savannarum Amphispiza bellli Anthus spinoletta Anthus spragueii Aphelocoma coerulescens Auriparus flaviceps Bombycilla cedrorum Calamospiza melanocorys 17.0 18.9 20.9 25.3 83.3 6.8 31.9 37.6 4.30 3.28 4.60 4.50 3.27 3.70 4.06 3.80 8.0 95.
14.1 11.2 18.0 19.0 15.5 8.5 7.5 — 12.0 — 12.0 — — 7.0 0.462 0.492 0.491 0.515 0.302 0.337 0.439 0.456 — 0.490 0.448 — 0.296 0.463 0.479 — 10.5 13.9 18.6 15.5 59.8 6.7 33.2 23.6 45.0 43.8 45.0 50.8 27.3 33.4 41.6 40.5 1 1 1 1 1 1 1 1 n n n n n n n n 58.90 43.20 33.33 32.75 29.50 19.50 21.43 54.14 1, 14 1, 15 16 9, 17 1, 18 19 20 21 Calcarius lapponicus Calcarius mccownii Calcarius ornatus Calcarius pictus 27.3 23.2 18.9 26.4 5.25 3.43 4.29 4.08 9.0 10.0 10.0 8.1 — — — — 0.545 0.480 0.511 0.486 0.535 0.492 0.462 0.594 22.1 18.7 14.9 22.0 69.5 45.7 49.0 58.8 2 4 4 1 n n n n 32.40 38.46 44.74 25.32 1, 22 9, 23, 23a 9, 24 25 Campylorhynchus bruneicapillus Cardinalis cardinalis Carduelis flammea Carduelis pinus Carpodacus mexicanus Catharus bicknelli Catharus fuscescens Catharus guttatus Catharus ustulatus 38.9 44.7 13.6 14.6 21.4 28.1 31.2 31.0 30.8 3.37 3.00 4.20 3.80 4.26 3.53 4.00 3.46 3.57 21.0 9.5 11.0 15.0 15.0 11.6 11.0 12.0 13.0 — 7.0 — 12.0 13.0 — — — 10.0 0.396 0.598 0.435 0.375 0.627 0.519 0.646 0.448 0.510 0.380 — 0.548 0.427 — 0.694 0.632 0.467 0.518 31.3 27.1 9.2 18.6 11.8 24.8 24.5 24.8 28.9 32.2 40.4 59.8 40.8 46.9 44.0 40.9 44.1 44.6 1 2 2 1 1 1 1 2 1 n n n n n n n n n 28.60 54.00 — — 45.80 68.53 55.00 60.10 35.14 1, 26 1, 27, 32 28 29 1, 30 31 2, 27 1, 33 34 Chondestes grammacus Cinclus mexicanus Cistothorus palustris Cistothorus platensis Corvus brachyrhynchos Corvus caurinus Dendroica discolor Dendroica kirtlandii Dendroica petechia 29.0 57.8 11.3 9.0 448.0 391.5 7.7 13.8 9.5 4.09 4.30 4.92 6.59 5.00 4.02 3.89 4.63 4.27 11.5 25.4 14.0 13.0 33.3 31.7 9.4 9.4 8.4 7.0 — 12.5 11.0 — 25.0 — 8.0 — 0.675 0.312 0.466 0.408 0.216 0.264 0.507 0.547 0.579 — 0.242 0.487 0.342 0.236 0.272 0.469 0.398 0.562 14.0 53.0 11.5 7.4 390.0 303.0 6.3 12.7 8.9 34.0 46.9 44.3 44.0 45.9 49.7 39.2 44.0 50.6 1 1 1 1 2 2 1 1 4 n n n n n n n n n 38.70 3.80 41.39 22.70 45.61 49.10 61.80 40.10 34.20 1, 35 36 1, 37 1, 38 39 1, 40 1, 41 1, 42 1, 43 Dendroica striata Dendroica virens Dolichonyx oryzivorus Dumetella carolinensis Empidonax difficilis 13.0 8.8 31.6 36.9 10.9 4.32 4.00 5.00 3.64 3.30 9.5 10.0 10.5 10.5 15.5 — 9.0 8.8 — — 0.538 0.736 0.511 0.516 0.433 0.499 0.566 0.511 0.495 0.413 11.3 7.7 22.1 28.5 10.5 — 45.5 43.3 45.0 36.4 1 1 1 1 2 n n n n n — — 29.80 31.20 58.90 44 45 1, 46 1, 47 1, 48 Empidonax minimus Empidonax oberholseri Empidonax traillii Eremophila alpestris Euphagus cyanocephalus Geothlypis trichas 10.3 10.4 13.4 31.4 62.7 10.1 3.86 3.78 3.48 3.10 5.13 4.00 14.3 17.8 13.5 9.5 13.3 9.8 — — — — — 8.0 0.452 0.434 0.388 0.522 0.501 0.537 0.474 0.394 0.499 0.559 0.519 0.598 10.8 10.1 12.5 22.3 46.0 10.3 50.2 37.9 44.6 48.9 41.7 43.5 1 1 2 5 1 2 a a a n n n 53.30 55.50 44.20 24.90 45.50 14.50 1, 49 1, 50 1, 51 1, 9, 52 1, 53 1, 54 Gymnorhinus cyanoephalus Hirundo pyrrhonota Hirundo rustica 103.0 21.6 18.6 47.4 4.00 3.50 4.53 3.25 21.0 22.7 20.3 13.5 — — — — 0.309 0.442 0.431 0.529 0.355 0.455 0.429 0.550 81.3 20.8 18.0 36.7 35.2 37.9 43.5 40.9 1 2 2 1 n a a n 41.20 13.62 10.35 52.50 1, 55 56 23a, 57 1, 27 Werschkul, D. F., and J. A. Jackson. 1979. Sibling competition and avian growth rates. Ibis 121:97–102.
West, G. B., J. H. Brown, and B. J. Enquist. 2001. A general model for ontogenetic growth. Nature 413:628–631.
Williams, G. C. 1966. Adaptation and natural selection. Princeton Univ. Press, Princeton, NJ.
Yanes, M., and F. Sua ´rez. 1997. Nest predation and reproductive traits in small passerines: a comparative approach. Acta Oecol.
18:413–426.
Corresponding Editor: T. Smith 2516 V. REMES ˇ AND T. E. MARTIN APPENDIX 1 Continued.
Species Abm Cl Nstl Nstl prem K K 70% Fl mass Lat No.
stud Forag Pred Ref Hylocichla mustelina Icteria virens Junco phaeonotus Lanius ludovicianus Leucosticte tephrocotis Melospiza lincolnii Melospiza melodia 25.3 20.4 47.4 26.3 17.4 20.8 3.69 3.54 5.40 4.48 4.24 3.88 8.9 11.5 18.1 18.5 10.5 10.8 — — — — — — 0.528 0.457 0.370 0.363 0.574 0.484 — 0.472 0.416 0.479 0.492 0.480 16.3 17.7 44.3 40.5 14.7 17.8 40.0 31.9 39.5 51.5 40.5 40.0 1 1 2 1 1 1 n n n n n n 52.90 26.00 19.40 16.70 42.00 28.10 1, 58 2, 59 1, 23a, 60 61 2, 62 1, 63 Mimus polyglottos Myioborus pictus Myiodynastes luteiventris Oporornis formosus Oreoscoptes montanus Parus atricapillus Passerculus sandwichensis Passerina amoena Pheuticus melanocephalus Pica pica Pipilo aberti 48.5 7.9 46.3 14.0 43.3 10.8 20.1 15.5 44.5 177.5 46.0 3.70 3.40 3.42 4.12 3.80 7.00 4.13 3.37 3.40 6.48 2.85 12.0 13.0 17.0 8.5 11.9 16.0 9.3 10.0 11.5 28.4 12.5 10.0 10.0 — 7.0 — 12.0 8.0 8.0 — — — 0.514 0.557 0.355 0.680 0.480 0.402 0.519 0.480 0.358 0.223 0.497 0.504 0.478 0.386 0.530 0.463 0.431 0.511 0.604 0.385 0.324 0.497 35.0 8.8 39.0 12.5 38.0 11.3 15.4 12.6 33.0 180.5 31.3 28.2 36.0 32.0 38.0 44.0 42.4 43.3 42.0 37.6 49.7 33.6 1 1 1 1 2 1 4 1 1 2 1 n n a n n n n n n n n 47.10 37.90 — 30.00 32.00 19.70 43.40 35.00 34.00 29.00 63.80 1, 64 65 66 1, 67 68 1, 69 1, 9, 70 71 1, 72 73 1, 74 Pipilo erythrophthalmus Piranga olivacea Piranga rubra Plectrophenax nivalis Pooecetes gramineus Progne subis Protonotaria citrea Quiscalus major Quiscalus mexicanus Quiscalus quiscula 40.5 28.6 29.0 42.2 25.7 49.4 16.2 166.5 149.0 113.5 3.30 3.53 3.20 5.71 4.00 4.52 4.61 2.77 3.50 4.80 10.5 10.0 9.5 12.8 10.5 28.5 10.0 13.5 12.0 13.5 7.5 — — — — — 9.0 — — — 0.519 0.431 0.704 0.569 0.612 0.306 0.654 0.403 0.422 0.458 — 0.477 — 0.556 0.600 0.364 0.520 — — — 0.24.7 20.5 18.2 30.4 17.5 52.0 12.7 77.7 100.5 60.5 39.6 42.5 39.0 71.4 43.1 42.3 42.2 28.0 30.6 40.9 2 1 1 1 2 1 1 1 1 3 n n n n n a n n n n 51.90 32.60 64.00 27.90 52.90 6.25 31.00 41.24 31.08 19.35 1, 27, 75 1, 76 77 1, 78 1, 79 80 1, 81 82 83 84 Riparia riparia Sayornis nigricans Sayornis phoebe Seiurus aurocapillus Seiurus motacilla Setophaga ruticilla Sialia currucoides Sialia mexicana Sialia sialis Spizella arborea 14.6 18.7 19.8 19.4 20.3 8.3 29.0 28.1 31.6 20.1 4.38 4.16 4.58 4.20 5.00 3.92 5.39 5.00 4.40 4.96 19.4 19.5 17.0 8.5 10.0 9.0 19.5 21.4 18.0 9.0 — 14.5 12.0 — 9.0 7.5 — — — — 0.377 0.450 0.425 0.473 0.590 0.613 0.369 0.487 0.463 0.543 0.277 0.416 0.414 0.550 0.575 0.575 0.541 0.476 0.436 0.539 14.5 18.7 17.5 13.5 17.1 7.7 25.8 25.1 27.2 16.7 42.7 36.0 39.0 41.6 42.4 43.5 46.3 46.0 42.5 58.5 2 1 1 2 1 1 3 1 2 1 a a a n n n a a a n — 7.29 15.90 24.50 — 37.80 10.81 29.59 48.60 — 85 86 1, 87 1, 27, 88 89 1, 90 91, 92 92 1, 93 94 Spizella breweri Spizella pallida Spizella passerina Spizella pusilla Stelgidopteryx serripennis Sturnella neglecta Tachycineta bicolor Tachycineta thalassina 10.4 12.0 12.3 12.5 15.9 97.7 20.1 14.2 3.27 4.00 3.60 3.83 6.25 5.10 5.45 4.40 8.00 8.00 10.5 7.5 19.3 11.0 20.0 23.5 — — 8.0 5.0 — 8.0 17.0 — 0.543 0.532 0.558 0.656 0.478 0.469 0.511 0.357 0.631 0.497 0.539 0.677 0.449 — 0.491 0.589 9.6 10.3 10.6 8.9 14.1 40.0 20.6 16.5 42.8 53.5 43.3 41.6 42.2 50.8 45.1 48.8 3 1 4 4 1 1 3 1 n n n n a n a a 20.50 54.20 41.20 60.40 19.00 46.90 29.44 5.70 1, 23a, 95 1, 96 1, 97 1, 98 2, 99 1, 9 1, 100 101 Toxostoma curvirostre Toxostoma longirostre Toxostoma rufum Troglodytes aedon Turdus migratorius Tyrannus forficatus Tyrannus tyrannus Tyrannus verticalis Vermivora peregrina Vireo atricapillus 79.4 69.9 68.8 10.9 77.3 43.2 39.5 39.6 10.0 8.5 3.12 3.75 3.60 6.34 3.30 4.58 3.40 3.89 5.50 4.00 15.0 13.0 10.8 17.0 13.0 15.4 16.5 16.0 11.5 10.5 — — — — — — — — — — 0.403 0.5456 0.471 0.515 0.519 0.382 0.438 0.413 0.654 0.412 — — — 0.471 0.466 0.388 0.457 0.459 0.558 0.357 46.6 37.3 41.5 10.2 56.9 30.1 31.8 36.0 7.3 8.2 30.2 28.2 39.0 49.5 42.4 39.0 40.2 39.0 49.8 32.0 2 1 1 3 2 1 3 1 1 1 n n n n n a a a n n 40.20 62.00 29.00 28.50 40.20 25.95 32.70 37.60 40.00 39.10 1, 102 103 1, 104 1, 105 1, 106 107 1, 87, 108 1, 108 109 110 Vireo belli Vireo griseus Vireo olivaceus 8.5 11.4 16.7 3.40 3.74 3.20 11.3 10.0 11.0 — — — 0.576 0.486 0.554 0.737 0.526 0.559 8.1 10.4 13.8 32.0 36.8 43.5 1 1 1 n n n 38.50 17.02 24.90 1, 110 111 1, 112 Xanthocephalus xanthocephalus Zonotrichia albicollis Zonotrichia atricapilla Zonotrichia leucophrys Zonotrichia querula 64.6 25.9 28.8 28.1 36.3 3.78 4.15 4.48 3.90 3.92 10.5 8.5 9.8 9.3 9.1 — — — 7.8 — 0.492 0.513 0.636 0.564 0.541 0.495 0.506 0.557 0.567 — 45.1 20.3 23.1 20.2 24.5 42.9 45.8 55.2 47.9 63.7 5 1 1 6 1 n n n n n 34.40 41.30 37.00 51.10 30.00 1, 113 1, 114 115 1, 116, 117 1, 117 2517 GROWTH STRATEGIES OF PASSERINES APPENDIX 2 Sources of growth rate (G) and nest predation (P) data in Ap-pendix 1 (for sources of other variables, see Methods).
(1) Martin 1995 (P); (2) Conway and Martin 2000 (P); (3) Starck and Ricklefs 1998b (G); (4) Williams 1940, Holcomb and Twiest 1968, Cronmiller and Thompson 1980 in 3 (G); (5) Webb and Bock 1996, no. 216 in 118 (P); (6) Maurer, B. A., E. A. Webb, and R.
K. Bowers. 1989. Nest characteristics and nestling development of Cassin’s and Botteri’s sparrows in southeastern Arizona. Condor 91:736–738 (G); (7) Lowther et al. 1999, no. 422 in 118 (P); Austin and Ricklefs 1977 in 3 (G); (8) Dunning et al. 1999, no. 471 in 118 (P); (9) Maher, W. J. 1972. Growth of ground-nesting passerine birds at Matador, Saskatchewan, Canada. Pp. 85–102 in S. C. Ken-deigh and J. Pinowski, eds. Productivity, population dynamics and systematics of granivorous birds. Polish Scientific Publishers, War-szawa (G); (10) Davis, S. K., and S. G. Sealy. 1998. Nesting biology of the Baird’s sparrow in southwestern Manitoba. Wilson Bull. 110: 262–270 (P); (11) Woolfenden 1956 in 3 (G); (12) Greenlaw and Rising 1994, no. 112 in 118 (G, P); (13) Post and Greenlaw 1994, no. 127 in 118 (G); Post, W. 1981. The influence of rice rats Ory-zomys palustris on the habitat use of the seaside sparrow Ammospiza maritima. Behav. Ecol. Sociobiol. 9:35–40 (P); Post, W. 1974. Func-tional analysis of space-related behavior in the seaside sparrow.
Ecology 55:564–575 (P); (14) Walkinshaw, L. H. 1940. Some Mich-igan notes on the grasshopper sparrow. Jack-Pine Warbler 18:50– 59 (G); (15) Petersen et al. 1986 in 3 (G); (16) Verbeek, N. A. M.
1970. Breeding ecology of the water pipit. Auk 87:425–451 (G, P); (17) Maher, W. J. 1973. Birds. I. Population dynamics. Canadian Committee for the International Biological Programme (Matador project) Technical report no. 34. Univ. of Saskatchewan, Saskatoon (P); (18) Woolfenden 1978 in 3 (G); (19) Taylor, W. K. 1971. A breeding biology study of the Verdin, Auriparus flaviceps (Sun-devall) in Arizona. Am. Midl. Nat. 85:289–328 (G, P); Austin, G.
T. 1977. Production and survival of the Verdin. Wilson Bull. 89: 572–582 (P); (20) Putnam 1949 in 3 (G); Young, H. 1949. A com-parative study of nesting birds in a five-acre park. Wilson Bull. 61: 36–47 (P); (21) Shane 2000, no. 542 in 118 (G, P); (22) Maher 1964, Fox et al. 1987 in 3 (G); (23) Mickey 1943 in 3 (G); With 1994, no. 96 in 118 (G); (23a) Porter, D. K., and R. A. Ryder.
1974. Avian density and productivity studies and analysis on the pawnee site in 1972. Grassland biome US International Biological Program Technical Report no. 252 (G, P); (24) Hill and Gould 1997, no. 288 in 118 (G); O’Grady, D. R., D. P. Hill, and R. M. R. Barelay.
1996. Nest visitation by humans does not increase pred on chestnut-collared longspur eggs and young. J. Field Ornithol. 67:275–280 (P); (25) Jehl, J. R., Jr. 1968. The breeding biology of Smith’s longspur. Wilson Bull. 80:123–149 (G); Briskie 1993, no. 34 in 118 (P); (26) Anderson and Anderson 1961 in 3 (G); (27) Norris 1947 in 3 (G); (28) Grinnell 1943 in 3 (G); Walkinshaw, L. H.
1948. Nesting of some passerine birds in western Alaska. Condor 50:64–70 (G); (29) Perry 1965 in 3 (G); (30) Badyaev, A. V., and T. E. Martin. 2000. Individual variation in growth trajectories: phe-notypic and genetic correlations in ontogeny of house finch (Car-podacus mexicanus). J. Evol. Biol. 13:290–301 (G); (31) Wallace 1939 in Rimmer, C. C., K. P. McFarland, W. G. Ellison, and J. E.
Goetz. 2001. Bicknell’s thrush (Catharus bicknelli) in A. Poole and F. Gill, eds. The birds of North America. no. 592. The Birds of North America, Inc., Philadelphia, PA (G); BBIRD database, MCWRU, Univ. of Montana, Missoula (P); (32) Eckerle, K. P., and R. Breitwisch. 1997. Reproductive success of the northern cardinal, a large host of brown-headed cowbirds. Condor 99:169–178 (G); (33) Perry, E. M., and W. A. Perry. 1918. Home life of the vesper sparrow and the hermit thrush. Auk 35:310–321 (G); Gross, A. O.
1964. Eastern hermit thrush (Hylocichla guttata faxoni Bangs and Penard). Pp. 143–162 in A. C. Bent. Life histories of North Amer-ican thrushes, kinglets and their allies. Dover Publications, Inc., New York (G); (34) Stanwood 1913 in 3 (G); Evans Mack and Yong 2000, no. 540 in 118 (P); (35) Martin and Parrish 2000, no.
488 in 118 (G); (36) Sullivan, J. O. 1973. Ecology and behavior of the dipper: adaptations of a passerine to an aquatic environment.
Ph.D. diss., University of Montana, Missoula (G, P); (37) Welter, W. A. 1935. The natural history of the long-billed marsh wren.
Wilson Bull. 47:3–34 (G, P); (38) Crawford, R. D. 1977. Polygy-nous breeding of short-billed marsh wrens. Auk 94:359–362 (G); (39) Parmalee 1952 in 3 (G); Ignatiuk, J. B., and R. G. Clark. 1991.
Breeding biology of American crow in Saskatchewan, Canada, parkland habitat. Can. J. Zool. 69:168–175 (G); Caffrey, C. 2000.
Correlates of reproductive success in cooperatively breeding west-ern American crows: if helpers help, it’s not by much. Condor 102: 333–341 (P); (40) Butler et al. 1984 in 3 (G); Verbeek, N. A. M.
1995. Body temperature and growth of nestling northwestern crows, Corvus caurinus. Can. J. Zool. 73:1019–1023 (G); (41) Nolan, V., Jr. 1978. The ecology and behavior of the prairie warbler Dendroica discolor. Ornithological Monographs no. 26. The American Orni-thologists’ Union (G); (42) Mayfield 1960 in 3 (G); (43) Schrantz 1943 in 3 (G); Biermann, G. C., and S. G. Sealy. 1982. Parental feeding of nestling yellow warblers in relation to brood size and prey availability. Auk 99:332–341 (G); Weatherhead, P. J. 1989.
Sex ratios, host-specific reproductive success, and impact of brown-headed cowbirds. Auk 106:358–366 (G); Briskie, J. V. 1995. Nest-ing biology of the yellow warbler at the northern limit of its range.
J. Field Ornithol. 66:531–543 (G); (44) Hunt and Eliason 1999, no.
431 in 118 (G); (45) Pitelka, F. A. 1940. Breeding behavior of the black-throated green warbler. Wilson Bull. 52:3–18 (G); (46) Mar-tin, S. G. 1974. Adaptations for polygynous breeding in the bob-olink, Dolichonyx oryzivorus. Am. Zool. 14:109–119 (G); (47) Zim-merman, J. L. 1963. A nesting study of the catbird in southern Michigan. Jack-Pine Warbler 41:142–160 (G); (48) Davis et al.
1963 in 3 (G); Lowther 2000, no. 556 in 118 (G); (49) Briskie, J.
V., and S. G. Sealy. 1989. Determination of clutch size in the least flycatcher. Auk 106:269–278 (G); (50) Pereyra, M. E., and M. L.
Morton. 2001. Nestling growth and thermoregulatory development in subalpine dusky flycatchers. Auk 118:116–136 (G); (51) King 1955 in 3 (G); Walkinshaw, L. H. 1966. Summer biology of Traill’s flycatcher. Wilson Bull. 78:31–46 (G); (52) Pickwell 1931, Beason and Franks 1973, Cunnings and Threlfall 1981 in 3 (G); Maher, W. J. 1980. Growth of the horned lark at Rankin Inlet, Northwest Territories. Can. Field-Nat. 94:405–410 (G); (53) Balph 1975 in 3 (G); (54) Stewart 1953 in 3 (G); Hofslund, P. B. 1959. A life history study of the yellowthroat, Geothlypis trichas. Proc. Minn. Acad.
Sci. 27:144–174 (G); (55) Bateman and Balda 1973 in 3 (G); (56) Stoner 1945, Chapman and George 1991 in 3 (G); Brown, C. R., and M. B. Brown. 1996. Coloniality in the cliff swallow. University of Chicago Press, Chicago, (P); (57) Stoner 1935 in 3 (G); Shields, W. M., and J. R. Crook. 1987. Barn swallow coloniality: a net cost for group breeding in the Adirondacks? Ecology 68:1373–1386 (G); (58) Eckerle and Thompson 2001, no. 575 in 118 (G); (59) Weath-ers, W. W., and K. A. Sullivan. 1991. Growth and energetics of nestling yellow-eyed juncos. Condor 93:138–146 (G); (60) Miller, A. H. 1931. Systematic revision and natural history of American shrikes (Lanius). Univ. Calif. Publ. Zool. 38:11–242 (G); (61) Yar-brough, C. G. 1970. The development of endothermy in nestling gray-crowned rosy finches, L. t. griseonucha. Comp. Biochem. Phy-siol. 34:917–925 (G); Twining, H. 1940. Foraging behavior and survival in the Sierra Nevada rosy finch. Condor 42:64–72 (P); (62) Ammon 1995, no. 191 in 118 (G); (63) Nice 1937 in 3 (G); (64) Fischer 1983 in 3 (G); (65) Marshall, J., and R. P. Balda. 1974.
The breeding ecology of the painted redstart. Condor 76:89–101 (G, P); Christoferson, L. L., and M. L. Morrison. 2001. Integrating methods to determine breeding and nesting status of 3 western songbirds. Wildl. Soc. Bull. 29:688–696 (P); (66) Ligon, J. D. 1971.
Notes on the breeding of the sulphur-bellied flycatcher in Arizona.
Condor 73:250–252 (G); (67) McDonald 1998, no. 324 in 118 (G); (68) Reynolds et al. 1999, no. 463 in 118 (G); Rotenberry, J. T., and J. A. Wiens. 1989. Reproductive biology of shrubsteppe pas-serine birds: geographical and temporal variation in clutch size, brood size and fledging success. Condor 91:1–14 (P); (69) Kluyver 1961 in 3 (G); (70) Williams and Prints 1986 in 3 (G); Threlfall, W, and R. J. Cannings. 1979. Growth of nestling savannah sparrows.
Bird-Banding 50:164–166 (G); Rogers, C. M. 1985. Growth rate and determinants of fledging weight in Michigan-breeding savannah sparrows. Condor 87:302–303 (G); (71) Greene et al. 1996, no. 232 in 118 (G, P); (72) Weston, H. G., Jr. 1955. Notes on the devel-opment of immature black-headed grosbeaks (Pheuticus melano-cephalus). Proc. Iowa Acad. Sci. 61:636–646 (G); (73) Dhindsa, M. S., and D. A. Boag. 1990. The effect of food supplementation 2518 V. REMES ˇ AND T. E. MARTIN on the reproductive success of black-billed magpies Pica pica. Ibis 132:595–602 (G); Todd, K. S., Jr. 1968. Weights of black-billed magpies from southwestern Montana. Auk 85:508–510 (G); Reese, K. P., and J. A. Kadlec. 1985. Influence of high density and parental age on the habitat selection and reproduction of black-billed mag-pies Pica pica hudsonia. Condor 87:96–105 (P); (74) Finch 1984 in 3 (G); (75) Barbour, R. W. 1950. Growth and feather development of towhee nestlings. Am. Midl. Nat. 44:742–749 (G); (76) Prescott, K. W. 1965. The scarlet tanager Piranga olivacea. NJ State Museum Investigations no. 2. New Jersey State Museum, Trenton (G); (77) Fitch, H. S., and V. R. Fitch. 1955. Observations on the summer tanager in northeastern Kansas. Wilson Bull. 67:45–54 (G); White, D. H., B. R. Chapman, J. H. Brunjes IV, R. V. Raftovich, Jr., and J. T. Segniak. 1999. Abundance and reproduction of songbirds in burned and unburned pine forests of the Georgia Piedmont. J. Field Ornithol. 70:414–424 (P); (78) Maher 1964 in 3 (G); (79) Perry, E. M., and W. A. Perry. 1918. Home life of the vesper sparrow and the hermit thrush. Auk 35:310–321 (G); Dawson and Evans 1960 in 3 (G); (80) Allen, R. W., and M. M. Nice. 1952. A study of the breeding biology of the purple martin (Progne subis). Am. Midl.
Nat. 47:606–665 (G); Stutchbury, B. J. 1991. Coloniality and breed-ing biology of purple martins Progne subis hisperia in saguaro cacti.
Condor 93:666–675 (P); (81) Walkinshaw, L. H. 1938. Nesting studies of the prothonotary warbler. Bird-Banding 9:32–46 (G); (82) Bancroft 1984 in 3 (G); Dunham, M. L. 1990. Nest site selection by boat-tailed grackles. Wilson Bull. 102:702–706 (P); (83) Gotie and Kroll 1973 in 3 (G); McIIhenny, E. A. 1937. Life history of the boat-tailed grackle in Louisiana. Auk 54:274–295 (P); (84) Will-son, M. F., R. D. St. John, R. J. Lederer, and S. J. Muzos. 1971.
Clutch size in grackles. Bird-Banding 42:28–35 (G); Hamel, P. B.
1974. Age and sex determination of nestling common grackles.
Bird-Banding 45:16–23 (G); Peer, B. D., and E. K. Bollinger. 1997.
Explanation for the infrequent cowbird parasitism in common grackles. Condor 99:151–161 (G); Peterson, A., and H. Young.
1950. A nesting study of the bronzed grackle. Auk 67:466–476 (P); (85) Petersen, A. J. 1955. The breeding cycle in the bank swallow.
Wilson Bull. 67:235–286 (G); Marsh, R. L. 1979. Development of endothermy in nestling barn swallows (Riparia riparia). Physiol.
Zool. 52:340–353 (G); (86) Wolf, B. O. 1991. The reproductive biology and natural history of the black phoebe (Sayornis nigricans Swainson) in central California. M. A. thesis, California State Uni-versity San Jose (G); Wolf 1997, no. 268 in 118 (P); (87) Murphy, M. T. 1981. Growth and aging of nestling eastern kingbirds and eastern phoebes. J. Field Ornithol. 52:309–316 (G); (88) Hann 1937 in 3 (G); (89) Eaton 1958 in 3 (G); (90) Baker 1944 in 3 (G); (91) Power, H. W., III. 1966. Biology of the mountain bluebird in Mon-tana. Condor 68:351–371 (G); Herlugson, C. J. 1983. Growth of nestling mountain bluebirds. J. Field Ornithol. 54:259–265 (G); (92) McCluskey, D. C., J. W. Thomas, and E. C. Meslow. 1977. Effects of aerial application of DDT on reproduction in house wrens and mountain and western bluebirds. USDA Forest Service Research Paper PNW-228. Pacific northwest forest and range experimental station. USDA Forest Service, Portland, OR (G, P); (93) Hamilton 1943 in 3 (G); Pinkowski, B. C. 1975. Growth and development of eastern bluebirds. Bird-Banding 46:273–289 (G); (94) Baumgart-ner, A. M. 1938. A study of development of young tree sparrows at Churchill, Manitoba. Bird-Banding 9:69–79 (G); (95) Petersen et al. 1986 in 3 (G); Rotenberry et al. 1999, no. 390 in 118 (G); (96) Salt, W. R. 1966. A nesting study of Spizella pallida. Auk 83: 274–281 (G); (97) Weaver 1937, Walkinshaw 1944, Dawson and Evans 1957, Reynolds 1984 in 3 (G); (98) Walkinshaw 1936, Daw-son and Evans 1957, Best 1977 in 3 (G); Carey, M. 1990. Effects of brood size and nestling age on parental care by male field spar-rows (Spizella pusilla). Auk 107:580–586 (G); (99) Lunk, W. A.
1962. The rough-winged swallow: a study based on its breeding biology in Michigan. Bulletin of the Nuttall Ornithological Club no. 4 (G); (100) Paynter 1954 in 3 (G); Zach, R., and K. R. Mayoh.
1982. Weight and feather growth of nestling tree swallows. Can.
J. Zool. 60:1080–1090 (G); Quinney, T. E., D. J. T. Hussell, and C. D. O. Ankney. 1986. Sources of variation in growth of tree swallows. Auk 103:389–400 (G); Finch, D. M. 1990. Effects of predation and competitor inference on nesting success of house wrens and tree swallows. Condor 92:674–687 (P); (101) Edson 1943 in 3 (G); Brawn, J. D. 1987. Density effects on reproduction of cavity nesters in north Arizona. Auk 104:783–788 (P); (102) Rand 1941, Fischer 1983 in 3 (G); (103) Fischer 1983 in 3 (G); Fischer, D. H. 1980. Breeding biology of curve-billed thrashers Toxostoma curvirostre and long-billed thrasher Toxostoma longirostre in south-ern Texas, USA. Condor 82:392–397 (P); (104) Cavitt and Haas 2000, no. 557 in 118 (G); (105) Edson 1930 in 3 (G); Huggins, S.
E. 1940. Relative growth in the house wren. Growth 4:225–236 (G); Zach, R. 1982. Nestling house wrens: weight and feather growth. Can. J. Zool. 60:1417–1425 (G); (106) Hamilton 1935, Howell 1942 in 3 (G); (107) Murphy 1988 in 3 (G); Nolte, K. R., and T. E. Fulbright. 1996. Nesting ecology of scissor-tailed fly-catchers in South Texas. Wilson Bull. 108:302–316 (P); (108) Mur-phy 1983, 1988 in 3 (G); (109) Holmes, S. B. 1998. Reproduction and nest behaviour of Tennessee warblers Vermivora peregrina in forests treated with Lepidoptera-specific insecticides. J. Appl. Ecol.
35:185–194 (G); Rimmer and McFarland 1998, no. 350 in 118 (P); (110) Graber, J. W. 1961. Distribution, habitat requirements, and life history of the black-capped vireo (Vireo atricapilla). Ecol. Mon-ogr. 31:313–336 (G, P); (111) Hopp et al. 1995, no. 168 in 118 (G, P); (112) Southern, W. E. 1958. Nesting of the red-eyed vireo in the Douglas Lake Region, Michigan. Jack-Pine Warbler 36:185– 207 (G); (113) Fautin 1941, Willson 1966, Richter 1983 in 3 (G); Richter, W. 1984. Nestling survival and growth in the yellow-head-ed blackbird, Xanthocephalus xanthocephalus. Ecology 65:597–608 (G); Ortega, C. P., and A. Cruz. 1992. Differential growth patterns of nestling Brown-headed Cowbirds and Yellow-headed Blackbirds.
Auk 109:368–376 (G); (114) Knapton 1984 in 3 (G); (115) Hendrick 1987 in 3 (G); Norment et al. 1998, no. 352 in 118 (P); (116) Banks 1959, Morton et al. 1972 in 3 (G); King, J. R., and J. D. Hubbard.
1981. Comparative patterns of nestling growth in white-crowned sparrows. Condor 83:362–369 (G); (117) Norment, C. J. 1992. Com-parative breeding biology of Harris’ sparrows and Gambel’s white-crowned sparrows in the northwest territories of Canada. Condor 94:955–975 (G); (118) Poole and Gill 1992–2002. |
9991 | https://zhuanlan.zhihu.com/p/90483310 | 对数函数和有关幂函数、指数函数、对数函数的补充 - 知乎
关注推荐热榜专栏圈子 New付费咨询知学堂
直答
切换模式
登录/注册
对数函数和有关幂函数、指数函数、对数函数的补充
首发于做以数学为主的教育事业
切换模式
对数函数和有关幂函数、指数函数、对数函数的补充
杨树森
数学话题下的优秀答主
收录于 · 做以数学为主的教育事业
22 人赞同了该文章
对数函数是指
f(x)=log ax,a∈(0,1)∪(1,+∞).f(x)=\log_ax,\ a\in(0,1)\cup(1,+\infty).
这里
∀x>0,log ax=b⇔x=a b.\forall x>0,\ \log_ax=b\Leftrightarrow x=a^b.
对数函数的性质主要有
定义域是(0,+∞),(0,+\infty),值域是R.\mathbb R.
当a>1 a>1时单调递增，当0<a<1 0<a<1时单调递减。
过定点(1,0).(1,0).
关于指数和对数的运算，在过去的文章里已经说得很详细。这次我主要想说两个问题，第一个是关于反函数。定义域为D D的函数f(x)f(x)具有反函数的条件是它是单射，也就是
∀x 1,x 2∈D,f(x 1)=f(x 2)⇒x 1=x 2.\forall x_1,x_2\in D,\ f(x_1)=f(x_2)\Rightarrow x_1=x_2.
在这个条件下，定义 f(x)f(x) 的反函数为
f−1(f(x))=x,x∈D.f^{-1}(f(x))=x,\ x\in D.
特别地，连续函数具有反函数的充要条件是单调。这是很直观的结论，首先单调函数肯定是单射，其次如果区间上的连续函数不单调，那么它不是单射。
在平面直角坐标系中，若 f(x)f(x) 有反函数，则图像
y=f(x)y=f(x) 和 y=f−1(x)y=f^{-1}(x)
关于直线 y=x y=x 对称。这一点可以理解为，反函数就是把 x x 和 y y 交换。
在高中我们着重指出的是，指数函数和对数函数
f(x)=a x,g(x)=log ax f(x)=a^x,\ g(x)=\log_a x
互为反函数。
第二个问题是关于幂函数、指数函数和对数函数的增长速率。
当 x 充分大时，幂函数 (\alpha>0), 指数函数和对数函数 (a>1) 都是充分大的。但是我们可以看出它们的增长速率差别很大。
这里的增长速率的比较不能通过直接求差的方式观察，而是通过求比值。如果两个函数的比值当 x 充分大时接近一个正的常数，就说它们的增长速率是相等的。
幂函数 x^\alpha\ (\alpha>0) 的增长速率随着 \alpha 的增大而增大，据此定义幂函数 x^{\alpha}\ (\alpha>0) 的增长速率是 \alpha. 这样我们就发现所有的多项式函数
a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0
的增长速率是 n.
另外我们发现指数函数的增长速率非常大，而对数函数的增长速率非常小。
事实上当 x 充分大时
\frac{a^x}{x^{\alpha}},\ \frac{x^{\alpha}}{\log_ax}\ (a>1,\alpha>0)
总是充分大。这是Excel的计算结果
可以直观地考虑原因是当 x 充分大时，每当 x 再增大，指数函数的增量是幂函数的任意多倍，幂函数的增量是对数函数的任意多倍。这就说明
指数函数 a^x\ (a>1) 的增长速率比任何幂函数都大，所以增长速率是 +\infty,
对数函数 \log_ax\ (a>1) 的增长速率比任何幂函数都小，所以增长速率是无穷小。
编辑于 2019-11-06 09:54
数学
数学分析
高考
赞同 22添加评论
分享
喜欢收藏申请转载
写下你的评论...
还没有评论，发表第一个评论吧
关于作者
杨树森
做以数学为主的教育事业
数学话题 的优秀答主
回答 545文章 755关注者 168,047
关注他发私信
推荐阅读
对数函数 ==== 我们知道，指数函数 y=a^{x},(a>0,a e 1) ，对于每一个确定值x，都有一个y值与它相对应。并且当x取不同值时，得到的函数值y也是不同的。也就是说指数函数的自变量与因变量是一一对应的。… 反低头联盟...发表于反低头联盟...对数函数 ==== 视频链接——对数函数关键词：对数的基本知识，对数的性质与运算法则，对数函数定义域值域分析，对数函数单调性，对数函数神奇结论，对数复合函数，对数函数的渐近线、定点、对称性，对数奇… 振华对数函数的常见性质 ========= 对数函数的常见性质对数函数 f(x)=\log_ax~(a>0,a eq1)有如下性质：定义域：x>0；单调性：当a>1时，单调增加；当 0<a<1 时，单调减少；符号： 当a>1，且x>1时，为正；… GaryG...发表于写给学生的...函数篇：对数函数 ======== 1.对数概念及对数运算如果你想知道 中x的大小，显然x不是特殊值。只能用一种新的方式来表示x，这种表示方式就是对数。即表示为： 对数中，2的位置叫底数，7的位置叫真数。因为对数来源于指… 不止数学 发表于不止数学马...
想来知乎工作？请发送邮件到 jobs@zhihu.com
打开知乎App
在「我的页」右上角打开扫一扫
其他扫码方式：微信
下载知乎App
无障碍模式
验证码登录
密码登录
开通机构号
中国 +86
获取短信验证码
获取语音验证码
登录/注册
其他方式登录
未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》
扫码下载知乎 App
关闭二维码
打开知乎App
在「我的页」右上角打开扫一扫
其他扫码方式：微信
下载知乎App
无障碍模式
验证码登录
密码登录
开通机构号
中国 +86
获取短信验证码
获取语音验证码
登录/注册
其他方式登录
未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》
扫码下载知乎 App
关闭二维码 |
9992 | https://tasks.illustrativemathematics.org/content-standards/tasks/2063 | Illustrative Mathematics
Loading [MathJax]/jax/output/HTML-CSS/config.js
Engage your students with effective distance learning resources. ACCESS RESOURCES>>
Partitioning a Rectangle into Unit Squares
No Tags
Alignments to Content Standards:2.G.A.22.OA.C.4
Student View
Task
Materials
• Copies of a rectangle with edges marked (one for each student/ group, see attached blackline master)
• A straight edge tool
Actions
The teacher should guide students through these actions, as the text in this task is too complex for some second graders.
Draw a grid on the rectangle by connecting each mark to the one directly across from it on the opposite edge.
2. The grid separates the rectangle into many little squares. How many squares are there?
There are five little squares in each row. Count by fives to find how many squares there are in the entire rectangle. Â
What other methods can you think of to quickly count how many squares there are in the entire rectangle?
Write a number in each little square to count them and show that your answers are correct.
One number sentence which shows the total number of squares is 3 + 3 + 3 + 3 + 3 = 15. Â Write another number sentence which shows the total number of squares.
IM Commentary
The purpose of this task is to show the student that a rectangle can be partitioned into unit squares, and that there are a number of reasonable ways to count the resulting squares.Â
The third part encourages the student to count by fives, as called for by 2.NBT.2, and makes the connection between the equal-sized groups and area of a rectangle representations for multiplication, which will be developed in third grade. Counting by numbers other than five is not called for by the standards, but would be a nice additional outcome.
Many extensions are possible:
Inviting students to make their own rectangle with a grid of a size of their choosing.
To learn about world cultures as well as mathematics, one can take the opportunity to explore game boards with this kind of shape. In particular, Snakes and Ladders is a game which has been played around the world. (Commercially, Chutes and Ladders in the U.S. is based on it). See the Wikipedia page on Snakes and Ladders for more information.
Students could make their own game with different-sized rectangular boards and different ways to navigate them, reinforcing the fact that there are different ways to count and all ways give the same answer (connecting 2.G.A.2 with K.CC.4.B). Playing such games can develop skills in estimation, the relationship between multiplication and division, and probability, setting the ground for a wide range of later work.
Investigating which numbers can be represented as the total number of squares in some grid. Even at this grade level, some students may take an interest in numbers which can only be made by a grid with side length one (that is, prime numbers).
Students need to experience partitioning rectangles themselves to understand area. If students have not yet done this, then they should do the partitioning themselves. However, if the students are already proficient with partitioning and understand that the squares taken together constitute the larger rectangle, then they can start with a pre-partitioned rectangle.Â
Solution
For this particular example, there are fifteen little squares.
There are five little squares in each row. To find the total amount, the student can count by fives: five, ten, fifteen. Since the student has already counted that there are a total of fifteen squares, they can feel confident that counting by fives is giving them the correct number of squares. Compare with standard 2.OA.C.4
4. The student might also notice that there are three little squares in each column. To find the total amount, the student can count by threes: three, six, nine, twelve, fifteen. Counting by numbers other than five, while not strictly necessary in a task at this grade level, nonetheless provides a foundation for working on multiplication in third grade.
Students can label the squares in different ways. In particular, it might be helpful for the student to make a few copies of their grid and try a few different labeling schemes.
To reinforce counting by the number of squares in each column, the teacher might encourage the student to label the top row with the numbers 1Â through 5, the second row with the numbers 6Â through 10, and the third row with the numbers 11Â through 15.
To reinforce counting by the number of squares in each column, the teacher might encourage the student to label the leftmost column with the numbers 1 through 3, the second column with the numbers 4Â through 6, and so forth.
Another number sentence which can be used is 5 + 5 + 5 = 15.  While others are possible, including for example 9 + 6 = 15Â(breaking it up into a familiar square an six more blocks) and 1 + 1 + ...Â+ 1 = 15, number sentences which describe breaking up by rows or columns are the desired and most likely responses.
Partitioning a Rectangle into Unit Squares
Materials
• Copies of a rectangle with edges marked (one for each student/ group, see attached blackline master)
• A straight edge tool
Actions
The teacher should guide students through these actions, as the text in this task is too complex for some second graders.
Draw a grid on the rectangle by connecting each mark to the one directly across from it on the opposite edge.
2. The grid separates the rectangle into many little squares. How many squares are there?
There are five little squares in each row. Count by fives to find how many squares there are in the entire rectangle. Â
What other methods can you think of to quickly count how many squares there are in the entire rectangle?
Write a number in each little square to count them and show that your answers are correct.
One number sentence which shows the total number of squares is 3 + 3 + 3 + 3 + 3 = 15. Â Write another number sentence which shows the total number of squares.
Print Task
Typeset May 4, 2016 at 18:58:52. Licensed by Illustrative Mathematics under a
Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. |
9993 | https://www.youtube.com/watch?v=-WFgtTWVyzM | Nucleophilic Addition to the Carbonyl Group || Clayden Series (Part-3) || Basic Organic Chemistry
Chem Explore
110000 subscribers
57 likes
Description
2810 views
Posted: 8 Mar 2024
🔴🔴Play Store ChemExplore App Link: 👇👇👇👇👇👇
🔴FREE Class PDF Link:
🔴FREE CLAYDEN PDF:
🔴Clayden Organic Series (Playlist):
🔴🔴📖📖Admission Open || 📢📢Register Now || 🎫🎟️Special Discount || Learn from IITan
👇👇👇Complete New Batches👇👇👇
🔴CSIR-NET-JUNE-2025 (Live):
🔴CSIR-NET-JUNE-2025 (Recorded):
🔴 GATE-2026 (Hybrid Batch):
🔴CSIR-NET & GATE Combined-Batch (12-Months):
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
👇👇👇👇👇👇👇👇
🔴Subscribe For Regular Videos:
🔴Facebook Group:
🔴Like us on Facebook:
🔴Follow me on Instagram:
🔴Myweb pagee link:
.............................................................................................................
Thanks for watching my videos. Subscribe to Chem Explore and share with your friends.
Keep Sharing, Keep Supporting!!!
Suman Bhowmick ................................................................................................................
nucleophilic #adition #carbonyl Group #grignardreagent #lithiumreagent #alkyllithium #basicchemistry #clayden #organic #organicchemistry #clayden_organic #gate2025 #csirnet #chemicalscience #chemistry #net2024 #csirnet2024 #gatechemistry #iitjam
Disclaimer Copyright Disclaimer Under Section 107 of the Copyright Act 1976, allowance is made for "fair use" for purposes such as comment, news reporting, teaching, scholarship, and research. Fair use is a use permitted by copyright statute that might otherwise be infringing. Non-profit, educational or personal use tips the balance in favor of fair use. The information provided on this channel is a public service with the understanding that Chem Explore CSIR-NET GATE IIT-JAM makes no warranties, either expressed or implied, related to completeness, accuracy, reliability or suitability of the information. The channel disclaims liability for errors and omissions that may have crept in its content.
3 comments
Transcript:
वेलकम टू केम x चैनल इस वीडियो में ना हम लोग न्यूक्लियोफिलिक एडिशन टू द कार्बोनेल ग्रुप्स के ऊपर डिस्कस करेंगे राइट क्लेन सीरीज अ ऑर्गेनिक केमिस्ट्री का एक इंपॉर्टेंट सीरीज़ है इसके ऊपर हम लोग डिस्कस कर रहे हैं नाउ बिफोर गोइंग टू स्टार्ट दिस वीडियो अ यह कंप्लीट बेसिक ऑर्गेनिक केमिस्ट्री का सीरीज़ चल रहा है राइट इफ यू आर प्रिपेरिस सीएसआर यूजीसी नेट गेट आईटी जेम केमिस्ट्री और एनी काइंड ऑफ़ एमएससी एंट्रेंस एग्जाम सेट लेवल एग्जाम असिस्टेंट प्रोफेसर एग्जाम और इफ यू आर लुकिंग फॉर बीएससी कोर्सेस राइट सो यू कैन लुक एट दिस वीडियो सीरीज काफी इंपोर्टेंट है हम लोग यहां पे बेसिक टू एडवांस लेवल तक डिस्कस करें अगर किसी का ऑर्गेनिक थोड़ी बहुत वीक है वो देख सकते हो एंड यू कैन एंजॉय दिस वीडियो कंप्लीट फ्री है एजें पोर्शन है उसमें हम लोग मूव करेंगे सो इट विल बी रियली इंटरेस्टिंग राइट एंड और एक रिक्वेस्ट है आप लोग चैनल को सब्सक्राइब नहीं करते हो राइट तो डेफिनेटली चैनल को सब्सक्राइब करो दैट विल गिव मी अ मोटिवेशन एज वेल ऑलराइट सो देखो स्टार्ट करते हैं तो इस क्लास में मेनली हम लोग स्टार्ट करेंगे द वेरी बेसिक पार्ट द न्यूक्लि फाइल एंड देन वी आर गोइंग टू डू द एडिशन टू योर कार्बोन ग्रुप सो लेट्स टेक अ सिंपल न्यूक्लि फाइल सिंपल न्यूक्लि फाइल कैन बी योर अ कैन बी योर c माइ तो यहां पे यह जो न्यूक्लि फाइल है नी एन मींस योर न्यूक्लि फाइल न्यूक्लि फाइल को हम लोग शॉर्ट में डिफाइन करते हैं ए माइन से ठीक है सो न्यूक्लि फाइल क्या करेगी कार्बोन ग्रुप नाउ व्हेन एवर यू आर टॉकिंग अबाउट योर कार्बोन कार्बोन तो यहां पे ना कार्बोन को हम लोग दो तरीके से डिफाइन करेंगे वन इज योर कीटोन अनदर इज योर एल्डिहाइड सो दिस इज योर एल्डिहाइड राइट एंड देन देन यू हैव और यू कैन राइट डाउन कीटोन दैट मींस आर सी डबल ब ओ आर ड इसको हम लोग क्या बोलते कीटोन क्लियर हो गया नाउ आर कैन बी एनी एल्का ग्रुप ट कैन बी एनी एल्का ग्रुप मिथाइल एल्काइन फिनाइल ठीक है अल्का और य कैन से अल्का मींस मिथाइल इथाइल राइट और यू हैव अ फिनाइल ग्रुप ठीक है एक्ट एरोमेटिक ग्र नाउ द नेक्स्ट पार्टट वी टॉकिंग अबाउट न फिलिक एडिशन तो न्यूक्लि फाइल कैसे कार्बोन में एडिशन होती है कार्बोनिल्स इसका एडिशन होती है ये न्यूक्लि फाइल यहां पे जाके अटैक करेगी ये जो कार्बोन का जो कार्बोन सेंटर है वहां प जाके अटैक करती है सिमिलरली ये जो नेगेटिव चार्ज है वो भी इधर जाके अटैक करेगी एंड इस बॉन्ड को ओपन कर देती है क्लियर हो गया सो दिस इज वेरी बेसिक केमिस्ट्री नाउ इफ आई कैन टेक अ न्यूक्लि फाइल हियर वी कन टॉक अबाउट सो मेनी न्यूक्लि फाइल सी न्यूक्लि फाइल दैट इन दिस सीरीज और इन दिस लेक्चर वी कन टॉक अबाउट सायनो दैट इ इज योर नाइट्राइल इसको हम लोग क्या बोलते हैं नाइट्राइल एंड देन यू हैव h माइ एक्सेट्रा ये सब न्यूक्लि फाइल के ऊपर डिस्कस करेंगे नाउ व्हेन एवर वी आर टॉकिंग अबाउट द c ए माइ व्हाट इज द स्ट्रक्चर मेनली तो ये जो नेगेटिव चार्ज मैंने य पे ड्रॉ किया ना ये किसके ऊपर है ये बेसिकली कार्बन के ऊपर है नाइट्रोजन के ऊपर नहीं है तो य ये जो अटैक होती है ना ये कार्बोन साइड से होती है नाइट्रोजन साइड से नहीं होती है तो वो भी हम लोग डिस्कस करेंगे व्हाट इज द मॉलिक्यूल र् बाइटल वट इज द हाइब्रिडाइज्ड दिस अ बेसिक केमिस्ट्री लेट्स से इफ यू हैव अ इफ यू हैव अ कीटोन सिंपल कीटोन जो भी हम लोग लैब में यूज़ करते हैं एसीटोन राइट एसीटोन सॉल्वेंट राइट सब लोग ऑर्गेनिक लैब में यूज़ करते हैं सो वी हैव अ स दिस ch3 cooch3 वी कॉल इट एज़ ए एसीटोन मॉलिक्यूल नाउ इन दिस एसीटोन मॉलिक्यूल इफ यू ऐड दिस नाइट्राइल ग्रुप सो नाइट्राइल को तो हम लोग ऐसे नहीं डाल सकते इन द फॉर्म ऑफ़ साल्ट सॉल्ट कैन बी योर सोडियम अ दिस nnmv.org.in अगर हम लोग एडिशन करेंगे सो देन यू हैव ए प् स ए माइनस नाउ दिस इज योर न्यूक्लि फाइल सोर्स ऑफ योर न्यूक्लि फाइल स ए माइनस तो वही स ए माइनस जो नाइट्रोजन नहीं कार्बन का जो साइट है वहां से इसका एडिशन हो जाती है कहां पे कार्बोन में नाउ ट सडन य प कार्बोन का जो कार्बन कार्बन है वहां प क्यों एडिशन हो रहा है राइट वो भी हम लोग थोड़ी देर में डिस्कस करेंगे नाउ आफ्टर दैट यू आर गोइंग टू गेट ch3 o माइन राइट एंड o माइ एंड देन यू हैव ए ch3 एंड यू हैव ए नाइट्राइल ग्रुप क्लियर हो गया तो ये जो नेगेटिव चार्ज क्रिएट हुआ है वो बेसिकली यहां पे आफ्टर दैट सोडियम इसके बाद इफ यू नीट इफ यू यूज योर hso4 एंड वाटर देन यहां से जो प्रोटोन रहती है वो इजली ये जो नेगेटिव चार्ज क्रिएट हुआ है ना वो h+ को ले लीए मीडियम से राइट तो ऐसे एक साथ एडिशन नहीं करना है व यू हैव टू क्रिएट दिस दिस इंटरमीडिएट फास्ट फॉलो बाय योर दिस एसिडिफिकेशन ओके नाउ यू आर गोइंग टू गेट दिस नाउ यहां पे और एक पोर्शन को मैं मेंशन कर देता हूं ये जो एसिडिक कंडीशन को यहां पे यूज़ किया गया है ना अ दैट शुड बी वेरी वेरी डाइल्यूट राइट अगर इसको थोड़ी बहुत स्ट्रांग कर देंगे ना तो ये जो नाइट्राइल ग्रुप है वो फादर ऑक्सीडो के सड एज राइट जो एसिड ग्रुप है उसमें कन्वर्ट हो जाती है सो बी केयरफुल सो दैट शुड बी वेरी डाइल्यूट कंडीशन राइट क्लियर हो गया तो ऐसे करके हम लोग इस रिएक्शन को स्टडी कर सकते हैं क्लियर हो गया सो ये जो स्टेप हम लोग कर रहे ये जो स्टेप हम लोग कर रहे हैं वी कॉल इट एज ए प्रोटोनेशन ये जो नेगेटिव चार्ज क्रिएट है वो h+ को मीडियम से ले रहा है ठीक है सो दैट इज व्हाट यू नीड टू अंडरस्टैंड नाउ इस मैकेनिज्म को हम लोग थोड़ी और तरीके से भी कर सकते हैं क्योंकि ये रिएक्शन थोड़ी एसिड कंडीशन में ज्यादा फैसा इल होते है ठीक है तो इसको हम लोग डिफरेंट तरीके से भी कर सकते हैं बट द प्रॉब्लम इज अगर हम लोग ये डिफरेंट तरीके से करेंगे तो तो ये जो मॉलिक्यूल देख रहे हो ये मतलब डिस्ट्रॉय हो सकता है व्हाट आई एम सेइंग दैट अगर इस रिएक्शन में अगर हम लोग एसिड कंडीशन पहले ऐड कर देते हैं ना तो प्रॉब्लम इज दैट ये जो सोडियम नाइट्रेट जो सॉल्ट है एए वो डिस्ट्रॉय हो सकती है राइट सो दैट इज वेरी वेरी इंपोर्टेंट तो दिस इज योर फर्स्ट स्टेप एंड दिस इज योर सेकंड स्टेप क्लियर हो गया नाउ बेस्ड ऑन दिस इंफॉर्मेशन अभी हम लोग ये डिस्कस करेंगे ये जो कार्बोनिल्स होती है ना कार्बोन ग्रुप्स का जो ऑर्बिटल होती है सडन वाई न्यूक्लि फाइल वही कार्बोन में जाके ये जो कार्बन है यहां पे जाके क्यों अटैक कर रही है सो व्हाट इज द कफिट वेदर होमो में अटैक कर रही है और लूमो में अटैक कर रही है तो वो भी हम लोग डिस्कस करेंगे सो जस्ट लुक एट दिस मॉलिक्यूल दिस इज योर वेरी वेरी मोस्ट रिएक्टिव कार्बोनिल्स योर फॉर्मल डिहाइड सो वी कैल कॉल इट एज ए फॉर्मल डिहाइड राइट तो ये जो फॉर्मल डिहाइड है तो यू नो दैट जो डबल बॉन्ड होती है जो डबल बॉन्ड होती है यहां पे एक तो सिग्मा बॉन्ड होती है और एक पाई बॉन्ड होती है सो इन दिस मॉलिक्यूल यू हैव अ सिगमा 1 सिगमा प्स 1 पा बॉन्ड क्लियर हो गया सो दिस इज योर ch2 दिस इज योर ch2 क्लियर हो गया सो यू हैव अ ch2 ग्रुप जहां पे व सिगमा 1 पाई बंड है क्लियर हो गया तो इसको हम लोग देखो एक सिग्मा अगर हम लोग सिग्मा सो नाउ दिस कार्बन इज बेसिकली र ए टू हाइब्रिडाइज्ड एंगल दिस बन एंगल जो बेसिकली होती है दैट इज र 120 डिग्री अगर sp3 हाइब्रिडाइज्ड एंगल 109.5 बॉन्ड है तो सिग्मा बॉन्ड को हम लोग कैसे ड्र कर सकते हैं देखो अगर हम लोग यहां पे सिग्मा बॉन्डिंग की बात करेंगे तो सिग्मा बॉन्डिंग देखने में कैसी होती है ठीक है एंड देन यह जो ऑक्सीजन होती है ना य पर दो लोन पेयर भी होती है ऑक्सीजन का वैलेंसी कितना होता है सिक्स होती है राइट तो इस लोन पेयर को मैं ऐसे करके ड्र कर देती हूं सो दिस इज द सिगमा बॉन्डिंग उसके बाद जो हाइड्रोजन के साथ जो बॉन्डिंग हो रहा है दिस इ र sp2 हाइब्रिड तो ये जो देखो ऑर्बिटल मैंने ड्र किया है दिस इज वन टूथ य तीन जो ऑर्बिटल ट इ sp3 ऑर्बिटल ठीक है अभी देखो ये जो हाइड्रोजन बाबू है ना य प हाइड्रोजन ऐसे करके सिग्मा बंड ऐसे करके वेरी सिंपल स्ट्रक्चर है क्लियर हो गया नाउ दिस इज योर सिग्मा सी बॉन्ड ये जो बंड देख रहे हो मैं एक लाइन ड्र कर देता हूं तो ये जो मैंने य प ड्र किया ट इज योर सिग्मा सीओ सिगमा सी बॉन्ड क्लियर हो गया नेक्स्ट इ दैट योर पाई बडि नेक्स्ट इ योर पाई बॉन्डिंग तो पाई बॉन्डिंग को हम लोग कैसे डिफाइन करेंगे देखो यह जो कार्बन है और यह जो ऑक्सीजन है सिगमा बंड तो छोड़ो अभी पाई बंडिंग की अगर बात करें सो य पर दिस इज योर नॉर्मल हम लोग नॉर्मल जो थी उसके ऊपर डिस्कस करेंगे सो दिस इ अ पाई बॉन्डिंग इंटरेक्शन सो ये सब सेम फेज में होगी ठीक है एंड नाउ य हैव हाइड्रोजन एंड य हैव हाइड्रोजन क्लियर हो गया तो दैट इज योर पाई बॉन्डिंग इंटरेक्शन देखो ऐसे करके पाई बॉन्डिंग करना है क्लियर हो गया तो ऐसे करके हम लोग जो कार्बोनियम पाउंड होती है उसकी पाई बॉन्डिंग को भी डिफाइन कर सकते हैं पा ऑफ स बॉन्ड क्लियर हो गया सो दिस इज वेरी जनरल स्ट्रक्चर उसके बाद हम लोग इसके ऊपर डिस्कस करेंगे दैट सडन कार्बन सेंटर में ये जो मॉलिक्यूल देख रहे हो यहां पे पे कार्बन सेंटर में उसका अटक क्य हो रही है ओके एंड आल्सो ये जो लेक्चर सीरीज चल रही है ना इसका पीडीएफ भी अवेलेबल होती है जैसे इफ यू डोंट वांट टू राइट डाउन जो भी मैं यहां पे लिख रहा हूं राइट स्क्रीन में सो इफ यू डोंट वांट टू राइट इ इफ यू डोंट वांट टू राइट और इफ यू वांट टू सेव योर टाइम तो इसका पीडीएफ भी अवेलेबल है फ्री पीडीएफ है डाउनलोड भी कर सकते हो उसका लिंक भी डिस्क्रिप्शन बॉक्स में दे मैं दे दूंगी ठीक है एंड आल्सो राइट ये क्लेन सीरीज का जो बुक है पीडीएफ वो भी अवेलेबल है फ्रीली अवेलेबल है उसका लिंक भी मैं डिस्क्रिप्शन बॉक्स में दे ी एवरीथिंग अ दैट इज अवेलेबल इन योर केम एक्सप्लोर एप्लीकेशन डाउनलोड भी कर सकते हो ओके नाउ इफ यू वांट जॉइन ठीक है अ हम लोगों के पास एक न्यू बैचेज भी हमने लॉन्च किए सो दिस इज कंप्लीट दिस इज अ ऑन गोइंग बैचेज इफ यू वान टू जॉइन फॉर अपकमिंग सीएसआर यूजीसी ने जून का एग्जाम है अ यू हैव अ दिस रिकॉर्डेड एज वेल एज योर कंप्लीट लाइव बैचेज तो यहां पे देख सकते हो दिस इज लाइव बैचेज एंड रिकॉर्डेड बैचेज इसका प्राइस कुछ ऐसा है काफी सारे ऑफर भी भी चल रहा है ओके एंड देन इफ यू वा जॉइन फॉर एनी अपकमिंग गेट केमिस्ट्री 2025 का एंड आल्सो दिसंबर का जो नेट आने वाली है उसके भी अलग-अलग नेम है सो ये जो गेट का है दैट इज योर गोल्ड बैच जो निकेल का बैच है दैट इज फॉर योर दिसंबर का जो नेट है उसके लिए काफी सारे बैच है तो उसमें भी जॉइन कर सकते हो काफी सारे ऑफर भी चल रहा है ठीक है एंड देन इफ यू वांट टू टेक द रिकॉर्डेड बैच ओनली फॉर दिसंबर नेट एंड गेट 2025 तो वो भी ले सकते हो उसके लिए प्राइस थोड़ी बहुत कम है ठीक है तो देख सकते हो कि व्हाट इज द प्राइस ठीक है काफी सारे ऑफर भी चल रहा है ठीक है तो इसमें सब ऑफ ऑफर भी मिल जाएगी ये जो प्राइस इसके ऊपर भी ₹5000000 कर रही है क्लियर हो गया द कंसेप्ट इज दैट जो भी न्यूक्लि फाइल होती है वो कार्बन में क्यों अटैक कर रही है व्हाई नॉट ऑक्सीजन सो द मेन रीजन इज द इलेक्ट्रोनेगेटिविटी राइट सो इलेक्ट्रोनेगेटिविटी का जो डिफरेंस है सो आई कैन राइट डाउन हियर द इलेक्ट्रो नेगेटिविटी तो ये जो इलेक्ट्रोनेगेटिविटी उसको हम लोग ऐसे काई सिंबल से डिनोट करते हैं ठीक है सो द कार्बन का जो इलेक्ट्रोनेगेटिविटी वैल्यू होती है ना वो बेसिकली अराउंड 2.5 होती है ठीक है एंड ऑक्सीजन का जो होती है वो बेसिकली अराउंड योर 3.5 फ्लोरिन का सबसे ज्यादा होता है फोर अबोव फोर 4.1 समथिंग लाइक दैट सो ड्यू टू दिस इलेक्ट्रोनेगेटिविटी डिफरेंस ऑक्सीजन जो बेसिकली वो डेल्टा नेगेटिव साइड होती है एंड कार्बन वो बेसिकली डेल्टा पॉजिटिव सेंटर होती है क्लियर हो गया ड्यू टू दिस इलेक्ट्रोनेगेटिविटी डिफरेंस ओके सो दैट इज द रीजन नाउ ये जो हम लोग न्यूक्लियोफिल का एडिशन कर रहे कार्बोन में तो ट्राई टू अंडरस्टैंड कौन सी मॉलिक्यूलर ऑर्बिटल इसमें इवॉल्व होती है वेरी बेसिक केमिस्ट्री होते है जैसे कोई भी अगर हम लोग डबल बॉन्ड की बात करते हैं दिस इज योर स कार्बन योर कार्बन एंड ऑक्सीजन बॉन्ड सो अगर कोई भी मैं डबल बॉन्ड की बात करता हू लेट्स से एक सिंपल c डबल बॉन्ड c की अगर बात करेंगे सो यू हैव अ वन 2p ऑर्बिटल अनदर 2p ऑर्बिटल दो 2p ऑर्बिटल एटॉमिक ऑर्बिटल कंबाइन करके मॉलिक्यूलर ऑर्बिटल क्रिएट करेगी सो देन अगर दो एटॉमिक ऑर्बिटल होगी तो दो मॉलिक्यूलर ऑर्बिटल होती है राइट रिमेंबर दो एटॉमिक ऑर्बिटल होती है तो दो मॉलिक्यूलर ऑर्बिटल होती है रिमेंबर हम लोग यहां पे c स की बात कर रहे ठीक है तो ये c1 दिस इज c2 सो ये वन नंबर कार्बन के लिए ये दो नंबर कार्बन के लिए ठीक है ये थोड़ा छोटा लग रही है बट ऐसे नहीं है सब इक्वल है ठीक है एंड सिमिलरली यू हैव अनदर मॉलिक्यूलर ऑर्बिटल तो ये जो मॉलिक्यूलर ऑर्बिटल कैन बी योर ईव एंड देन यू हैव ए शई टू क्लियर हो गया नाउ नेक्स्ट पार्ट इज दैट ये जो ग्राउंड लेवल होती है इसमें नोड जीरो होती है कोई भी नोड नहीं होती नोड का रोल क्या हो है कि जो साइन होती है ना लोप की वो चेंज हो जाती है ऊपर में जब ऊपर एक-एक करके ऊपर जाएंगे तो तो यू हैव अनदर वन नोड n0 n1 n2 ऐसे करके होती है तो n1 मींस यहां पे एक नोड हो जाएगी अच्छा नोड का क्या रोल होती है नोड अगर रहेगी बिफोर n0 में क्या होती है ये जो देखो साइन सेम रहती है ऊपर ही है ना तो अगर यहां पे साइन होगी ड्यू टू दिस नोट ये साइन चेंज हो जाती है एंड सिमिलरली इसका एनर्जी जो है वो ऊपर की तरफ इंक्रीज करर ठीक है नाउ डबल बॉन्ड में कितनी पाई इलेक्ट्रॉन रहती है दो दो इलेक्ट्रॉन तो यहां पे यही होगी योर दो इलेक्ट्रॉन ठीक है तो ऊपर में देखो खाली है राइट क्लियर हो गया ऊपर में देखो खाली है दैट्ची का जो यू नो साइवान है दैट वी कॉल इट एज ए होमो क्लियर हो गया नाउ नाउ जब भी हम लोग न्यूक्लि फाइल की बात करें तो यहां पे स है ठीक है तो इसका ना हम लोग चेंज भी कर सकते हैं ससी को सी भी कर सकते हैं थोड़ी बहुत कफिट चेंज हो जाएगी मतलब ये लोप जो है चेंज हो जाएगी ठीक है नाउ ये जो स की बात कर रहे ठीक है तो स में आई टोल्ड यू दैट न्यूक्लि फाइल कार्बोन में जब अटैक कर रही है राइट तो यहां पे इलेक्ट्रॉन दे रही है राइट सो न्यूक्लि फाइल यहां पे इलेक्ट्रॉन को प्रोवाइड कर रही है सप्लाई कर रही है नेगेटिव चार्ज तो ऐसे करके इलेक्ट्रॉन के तौर पे ऐसे करके आ रही है ना कार्बोन में तो कौन सी मॉलिक्यूलर ऑर्बिटल यहां पे पार्टिसिपेट करेगी डेफिनेटली योर लोएस्ट अन ऑक्यूपाइड मॉलिक्यूल ट क्योंकि यहां पे खाली प्ले प्लेस है टू यू नो एक्सेप्ट टू इलेक्ट्रॉन राइट दैट इज द रीजन नाउ राइट नाउ व्हाट यू नीड टू डू इज दैट ये जो अ योर लूमो है हम लोग उसको अभी ड्रॉ करेंगे देखो कैसे हम लोग ड्रॉ करेंगे इसको ऐसे करके लिखते हैं नाउ इसको हम लोग कैसे ड्रॉ करेंगे लूमो जो होती है ना आई टोल्ड यू लूमो का जो को एफिशिएंट होती है वह कार्बन के केस में थोड़ी बहुत हाई होती है अच्छा इसका जो ऑर्बिटल होती है वह थोड़ी बहुत तो हाई होती है मतलब साइज भी थोड़ी बड़ा होगी ठीक है वही है थोड़ी साइज बड़ा होगी एंड ऑक्सीजन में थोड़ी कम होती है ठीक है तो यहां पे एक नोड है इसलिए बिकॉज़ ऑफ़ दिस नोड बिकॉज़ ऑफ़ दिस नोड यह जो सिंबल होती है वह एक्सचेंज हो जाती है चेंज हो जाती है क्लियर हो गया सो दिस इज योर लोएस्ट अन ऑक्यूपाइड मॉलिक्यूलर ऑर्बिटल ठीक है यू कैन राइट डाउन योर हाइड्रोजन और एनीथिंग ठीक है नाउ ये जो योर न्यूक्लि फाइल है ये जो न्यूक्लि फफा इल है दैट इज योर एय माइनर ठीक है तो ये न्यूक्लि फाइल क्या करेगी अभी न्यूक्लि फाइल का जो होमो होती है वही जाके यहां पे अटैक करेगी क्योंकि होमो जाके होमो तो इलेक्ट्रॉन दे सकते तो वही जाके लूमो में जाके अटैक करेगी सो दिस इज योर न्यूक्लि फाइल ठीक है दिस इज योर न्यूक्लि फाइल विथ होमो सो दिस इ कैन ब र नेगेटिव माइनस सो दिस इ योर होमो एओ एओ एंड दिस इज योर लूमो क्लियर हो गया तो वही जाके अ कर ऐसे करके करना है ठीक है ओके उसके बाद क्या करना है उसके बाद देखो यहां पर क्या हो जाएगी ऐसे करके हम लोग इसको ऐसे ड्र कर सकते सी अच्छा ये बंड ओपन हो जाएगी तो इसको हम ऐसे नहीं लिख सकते o माइनस सो य प हाइड्रोजन हाइड्रोजन एंड देन यू हैव अ न्यूक्लि फाइ ठीक है नाउ ये जो सेंटर है अभी ये बन गई sp3 पहले क्या था इसका हाइब्रिडाइज्ड दैट यू हैव ए sp3 सेंटर क्लियर हो गया सो दिस इज वेरी वेरी इंपोर्टेंट क्लियर हो गया ऐसे करके हम लोग इसका अटक करेंगे नाउ अभी हम लोग रियल एक एग्जांपल के ऊपर डिस्कस करते हैं रियल एग्जांपल कैन बी योर सोडियम नाइट्रेट दैट आई टोल्ड यू दैट वहां पे जब हम लोग इसका अटैक किए थे ना रिमेंबर वहां पे जब हम लोग अटैक किए थे सोडियम नाइटेल के केस में तो इन दैट केस कार्बोनिफरस लूमो पार्टिसिपेट करेगी कार्बन का जो ऑर्बिटल है वो थोड़ा बड़ा होता है इन केस ऑफ लूमो ठीक है कफिट थोड़ा हाई होती है क्योंकि वी आर टॉकिंग अबाउट पॉजिटिव कफिट तो कार्बन का साइड थोड़ा ज्यादा होती है बिकॉज ऑफ द इलेक्ट्रोनेगेटिविटी डिफरेंस ये जो डापोर मूवमेंट है वो ऑक्सीजन की तरफ होती है ठीक है नाउ इन दिस केस ये जो नाइट्राइल है नाइट्राइल देखो सी ट्रिपल बॉन्ड n तो ओबवियसली यह जो लेस इलेक्ट्रोनेगेटिव जो साइड होती है ठीक है लेस इलेक्ट्रोनेगेटिविटी जो साइड होती है दैट इज योर कार्बोन साइड वही यहां पे कार्बोनेल का जो कार्बोन है उसके साथ रिएक्ट करेगी सो दिस इज योर होमो हाईएस्ट ऑक्यूपाइड मॉलिक्यूलर ऑर्बिटल एंड दिस इज योर लूमो लोएस्ट अन ऑक्यूपाइड मॉलिक्यूलर ऑर्बिटल लूमो इज बेसिकली योर पाई स्टार रिमेंबर लूमो इज बेसिकली योर पाई स्टार ठीक है एंड यहां पे और एक चीज मैं मेंशन कर देता हूं इसको व्हाट इज द हाइब्रिडाइज्ड आइज ये नाइट्रोजन भी ए हाइब्रिडाइज्ड तो यहां पे ए हाइब्रिडाइज्ड है कार्बन का जो होमो है वही जाके यहां पे कार्बोनेल का जो लूमो पास्टर कार्बोनी कार्बोन साइड में जाके अटक कर है ठीक है तो ऐसे करके हम लोग इसका मॉलिक्यूल ऑर्बिटल को भी ड्रॉ कर सकते हैं क्लियर हो गया नाउ हम लोग कुछ बेसिक केमिस्ट्री के ऊपर डिस्कस करेंगे जैसे हम लोग कार्बोन में न्यूक्लि फाइल का एडिशन कर रहे हैं ठीक है जैसे हम लोग न्यूक्लि फाइल में न्यूक्लि फाइल का एडिशन कार्बोन में कर रहे हैं सो वी कैन राइट डाउन सम बेसिक एग्जांपल दैट इज अवेलेबल इन योर क्लेन रेफरेंस राट सो ले यव एल्डिहाइड लिखते ठीक एड ओपन तरीके से लिखेंगे एंड नाउ इन दिस मॉलिक्यूल व हैव ओ प्रोटेक्शन फल प्रोटेक्शन ठीक नाउ इन दिस मॉलिक्यूल यू कैन ड ए एंड वव प्स ठीक है तो यह सीक्वेंशियल अगर ऐड करेंगे तो डेफिनेटली c यहां पे जाके अटैक कर देगी फ्रॉम कार्बन साइड o माइनस बन जाएगी वो माइनस यहां से h+ को एक्सेप्ट कर ली राइट क्लियर हो गया तो ऐसे करके स्टेप वाइज मैकेनिज्म को सोचना है बहुत सिंपल है एंड नाउ वी हैव अ o प देन यू हैव अ ओ ए हो जाएगी राइट यू हैव अ हाइड्रोजन एंड देन c ए ठीक है c ए को हम लोग ऐसे भी लिख सकते हैं थोड़ी एलेबोरेट तरीके से c तो दैट मींस कार्बन साइड से इसका अटैक हुआ है क्लियर हो गया तो ऐसे करके इजली हम लोग इसका मैकेनिज्म कर सकते हैं नाउ द नेक्स्ट न्यूक्लि फाइल ओके और एक पॉइंट यहां प मैं मेंशन कर देता हूं जैसे हम लोग जब भी अटैक कर रही है तो जब भी हम लोग अटैक कर रहे हैं ना तो कार्बोन का जो हाइब्रिडाइज्ड इजेशन राइट तो यहां पे देखो ये कार्बोन है नाउ वी हैव अ आर यू हैव आर ड सो व्हाट इज द हाइब्रिड आई टोल्ड यू दिस इज योर 120 डिग्री बन एंगल राइट दिस इ 120 डिग्री बन एंगल सो व्हाट इज द हाइब्रिडाइज्ड ए2 हाइब राइट नाउ इन दिस मॉलिक्यूल इन दिस मॉलिक्यूल इन दिस मॉलिक्यूल इफ यू ड नक् फाइल ट ए ए एंड देन वाटर ए तो य पर क्या बनेगी योर आर डैश ठीक है थोड़ी बन एंगल कम हो जाती है यू हैव ओ एंड देन यू हैव स ए रिमेंबर यू हैव टू यूज वेरी वीक एसिड कंडीशन अदर वाइज ये जो नाइट्राइल है उसका ऑ फरदर ऑक्सीडेशन होके कार्बोसिंथ है ठीक है बी केयरफुल सो इसका बन एंगल क्या होगी 100 9.8 डिग्री क्लियर हो गया तो ऐसे बन एंगल कम हो जाती है बन एंगल कम हो जाती है तो अगर बन एंगल कम हो जाएगी तो ये आर आर ग्रुप में एक थोड़ी बहुत स्टेरी क्राउड आ जाएगी सो इस चीज को ना याद रखना ठीक है क्योंकि हम लोग इसको बाद में यूज भी करेंगे नाउ वी हैव डिस्कस अबाउट योर नाइट्राइल न्यूक्लि फाइल वी हैव डिस्कस अबाउट द नाइट्राइल न्यूक्लि फाइल तो यहां पे और भी न्यूक्लि फाइल हो सकती है जैसे ए ए सोडियम हाइड्रोक्साइड का जो ओ है दिस ओ कैन एक्ट एज एक्ट एज ए बेस एज वेल एज न्यूक्लि फाइल राइट तो देखो यहां पे अगर कोई भी कीटोन है ठीक है कोई भी अगर कीटोन है कोई भी अगर कीटोन है तो कीटोन में और कार्बोनेल कंपाउंड में हम लोग एओ को ऐड कर सकते हैं देखो वी हैव अ r सीड ब दिस आर तो ये जो कंपाउंड देख रहे हो दिस इज योर कंपलीटली कीटोन कंपाउंड तो कीटोन है तो यहां पे अगर ओ माइनस ऐड कर देंगे तो ओ माइनस यहां पे अटैक कर सकती है राइट अंडर स्ट्रांग कंडीशन नाउ वी हैव नाउ वी हैव व्हाट यू नीड टू अंडरस्टैंड दैट और एक चीज मैं मेंशन करने य पे भूल गया इसको हम लोग यहां पे एडिशन नहीं करेंगे ठीक है यहां पे हम लोग इसका एडिशन नहीं करेंगे हम लोग य पे c-n - का एडिशन करेंगे सो वी हैव अ स वी हैव अ o माइनस एंड देन वी हैव ए c ए ठीक है नाउ अंडर आफ्टर सम यू नो इफ यू हैव अ वाटर इन दिस मीडियम और वेरी वीक एसिडिक कंडीशन तो यहां पे हम लोगों को क्या मिल जाएगी o एंड देन यू हैव ए नाइट्राइल एंड देन यू हैव ए r नाउ द पॉइंट इज यह जो कंपाउंड अभी हम लोगों को मिला ना यह उतना स्टेबल होती नहीं है ठीक है यह उतना स्टेबल होती नहीं ये इमीडिएट यू नो ब्रेक भी हो सकती है अंडर बेसिक कंडीशन ये ऐसे करके ना ऐसे करके c माइ रिमूव भी हो सकती है सो बी केयरफुल दैट व ये जो रिएक्शन होती है राइट इसको हम लोग बोलते हैं साइनो हाइड्रिन फॉर्मेशन इसको हम लोग क्या बोलते हैं सायनो हाइड्रिन फॉर्मेशन व्हिच इज योर रिवर्सिबल ये जो फॉर्मेशन होती है ये लिटिल बिट ये रिवर्सिबल होती है एक्चुअली दैट मींस इसको हम लोग कैसे लिख सकते हैं देखो दैट मींस वी कैन राइट डाउन आर ये रिवर्सिबल होती है अंडर बेसिक कंडीशन ठीक है सो लेट्स से यू हैव अ ए ए वही तो होगी सो दिस इज योर k इक्विलियम अच्छा k इक्विलियम का बेसिक व्हाट इज द बेसिक फार्मूला व्हाट इज द बेसिक फार्मूला ऑफ योर k इक्विलियम सो k इक्विलियम वी कैन राइट डाउन दैट दिस इज जस्ट लाइक र प्रोडक्ट कंसंट्रेशन बाय रिएक्टेंट कंसंट्रेशन राइट सो k इक्विलियम वी कैन राइट डाउन k फॉरवर्ड बाय k बैकवर्ड राइट एंड देन योर प्रोडक्ट कंसंट्रेशन एट इक्विलियम एट इक्विलियम कंडीशन व्हाट इज द प्रोडक्ट कंसंट्रेशन बाय इक्विलियम कंडीशन व्हाट इज योर रिएक्टेंट कंसंट्रेशन तो अगर अगर रिएक्टेंट तो अगर इफ यू हैव अ मोर यू नो इफ यू हैव अ हाई के इक्विलियम वैल्यू अगर के इक्विलियम वैल्यू हाई है इसका मतलब क्या है प्रोडक्ट कंसंट्रेशन हाई है र आई कैन राइट डाउन फॉरवर्ड रिएक्शन इज मोर फीजिबल देन र बैकवर्ड रिएक्शन इफ यू हैव अ के इक्विलियम का वैल्यू अगर हाई होगी ठीक है तो इस पॉइंट को भी मेंशन याद रखना ठीक है नाउ द नेक्स्ट पॉइंट जो भी यहां पे मेंशन किया गया दैट इज योर एंगल जिस एंगल में न्यूक्लि फाइल कार्बोन में अप्रोच करती है व्हाट इज द बोन एंगल ठीक है दैट इज वेरी वेरी इंपोर्टेंट सो ट्राई टू अंडरस्टैंड लेट्स से ये जो कार्बोन है ठीक है यह जो कार्बोन है ठीक है इसका जो बन एंगल होती है जिस एंगल में यह कोई भी न्यूक्लि फाइल की अगर बात करें जैसे अगर नाइट्राइल ग्रुप की बात करें और यू कैन राइट डाउन सिंपल न्यू फाइल तो इसका जो एप्रोच होती है ना इस एप्रोच जो होती है दैट इज योर 107 डिग्री इस वैल्यू को ना याद रखना 107 डिग्री जिको हम लोग बोलते हैं बुर्गी कुछ स्पेलिंग है ऐसा ठीक है बुर्गी ट्रेक्टर तो इस एंगल में हम लोग करते हैं ठीक है तो यूजिंग दिस अ एंगल सो ही एंगल में इसका अप्रोच होती है ट्राई टू अंडरस्टैंड इस एंगल में क्लियर हो गया सो क्लियर हो गया ऐसे करके हम लोग इसको करेंगे बट व्हाट इज द न्यूक्लि फाइल देखो न्यूक्लि फाइल का लो ऐसे रहेगी एंड न्यूक्लि फाइल इज योर मेनली होमो एंड ये जो कार्बोन है दिस इज योर लूमो यह थोड़ा बड़ा होती है ठीक है और यह थोड़ा छोटा होती है राइट सो व्हाट इज द एंगल एंगल को हम ऐसे डिफाइन करेंगे सो ये जो एंगल होती है दिस इज बेसिकली 7 डिग्री ठीक है सो दैट इज द एंगल दैट यू नीड टू रिमेंबर ठीक है ये चीज को याद रखना क्योंकि एग्जाम में क्वेश्चन भी पूछा जाता है ओके तो क्लियर हो गया सो वी हैव डिस्कस अबाउट द सायनो और नाइट्राइल सो वी कॉल इट एज ए नाइट्राइल न्यूक्लि फाइल एंड कैसे हम लोग नाइट्राइल न्यूक्लि फाइल को कार्बोनेल में अटैक कर सकते हैं व्हाट इज द बन एंगल व्हाट इज द मॉलिक्यूलर ऑर्बिटल डायग्राम व्हाई सनली कार्बोनेल सेंटर में अटैक कर रही है ड्यू टू दिस पोलराइजेशन डिफरेंस इलेक्ट्रोनेगेटिविटी डिफरेंस राइट नाउ व्हाट इज द सिंपलेस्ट न्यूक्लि फाइल दैट वी नो ट हाइड्राइड तो सबसे सिंपलेस्ट न्यूक्लि फाइल जो भी यनो अवेलेबल होती है ट इ र हाइड्राइट न्यूक्लि फाइल सो द हाइड्राइड और यू कन राइट डाउन ए माइनस एस ए न्यूक्लि फाइल हाइड्राइट और h माइनस एस ए न्यूक्लि फाइल तो ये जो हाइड्राइट न्यूक्लि फाइल की हम लोग बा बा कर रहे ना तो इसको हम लोग कैसे कार्बोन सेंटर में अटैक कर सकते हैं राइट वी नो द य नो हाइड्राइट रिएजेंट होती है सोडियम बोरो हाइड्राइड लिथियम एलम हाइड्राइड राइट तो वहां पे हम लोग इसको कैसे करेंगे ट्राई टू अंडरस्टैंड हाइड्राइड का जो सोर्स होती है इन जनरल वी नो दिस सोडियम हाइड्राइड राइट पोटेशियम हाइड्राइड इसको हम लोग यूज करते हैं ना बट रिमेंबर यह जो सॉल्ट देख रहे हो ना इन दिस सॉल्ट सोडियम हाइड्राइड और पोटेशियम हाइड्राइड ये जो हाइड्राइड होती है वो न ा नहीं होती वो बेस होती है तो यहां पर जो h माइनस होती है फ्रॉम दिस सॉल्ट h माइनस एक्ट्स एज ए बेस नाउ न्यूक्लि फाइल हम लोग कैसे करेंगे न्यूक्लि फाइल का जो न्यूक्लि फाइल जो होती है वो थोड़ी डिफरेंट होती है लेट्स से यू हैव अ बी ए4 माइनस सोडियम बोरो हाइड्राइड जो होती है ना तो सोडियम बोरो हाइड्राइड मतलब na4 सो यहां से हम लोग को क्या मिल जाएगी बी ए4 माइनस नाउ वी हैव लिथियम एलुमिनियम हाइड्राइड य से क्या मिल जाएगी al4 माइनस क्लियर सो इसका स्ट्रक्चर क्या होती है हम लोग इसको ऐसे नहीं ड्र कर सकते बी ए ए ए ए ट इ र माइनस सिमिलरली एलुमिनियम का जो ए ए ए ए ए माइनस ओके नाउ यहां पे देखो ये जो हमने यहां पे ड्रॉ किया ना सोडियम बोरो हाइड्राइड लिथियम एलुमिनियम हाइड्राइड यहां पे ये जो कोई भी एक हाइड्रोजन पिक अप कर लेते दिस हाइड्रोजन और दिस हाइड्रोजन विल एक्ट एज अ न्यूक्लि फाइल क्लियर हो गया तो यहां पे जो h माइ ए दैट एक्ट्स एज ए न्यूक्लि फाइल वेरी वेरी इंपोर्टेंट ठीक है ट एक्ट न्यूक्लि फाइल ठीक है तो उसको अभी हम लोग मैकेनिज्म के तौर पर डिस्कस करेंगे ठीक है देखो तो सोडियम हाइड्रेट जो होती है तो इसका रोल क्या होता है यह बेसिकली बेस के तौर पर काम करती है लेट्स से सनली इफ यू हैव h एक ठीक है कोई भी हैलाइड है तो यहां पे इफ यू ड सोडियम हाइड्राइड तो य प ये जो सोडियम हाइड्रे जो h माइनस है वो ऑलवेज सिमिलर एटम जो है सिमिलर मतलब और एक हाइड्रोजन के साथ ही हाइड्रोजन एटम के साथ ही वो रिएक्ट करेगी ठीक है सो व्ट यू कैन राइट डाउन हियर सो य पे बेसिकली h माइनस ऑलवेज रिएक्ट एस ए बेस क्लियर हो गया ऑलवेज रिएक्ट एस ए बेस क्लियर हो गया एंड नाउ वी हैव समथिंग लाइक x माइनस एंड देन वी हैव हाइड्रोज नाउ इफ यू कैन रिमेंबर कोई भी अगर ओ ग्रुप होती है ना कोई भी अगर ओ ग्रुप है आरओ कैन बी एनीथिंग आर कैन बी एनीथिंग योर योर एनी एलिफेटिक अल्कोहल इसका अगर हम लोग इन दिस केस इफ यू ऐड सोडियम हाइड्राइड ना ये काफी इंपोर्टेंट रिएजेंट होती है जो ओ ग्रुप होती है उसको डी पोर्टनेट करने के लिए और आई कैन से ये बेसिकली ऐसे हाइड्र को ले लेगी और य से o माइनस क्रिएट हो जाती है सो फॉर पर्टिकुलर रीजन हम लोग ये सोडियम हाइड्राइड पोटेशियम हाइड्राइड को यूज कर करते टू ट एक्सेप्ट दिस हाइड्रोजन एंड देन यू हैव o माइनस फ्रीली o माइनस दैट विल बी योर न्यूक्लियोफिल जो बेसिकली कोई भी कंपाउंड में जाके अटैक कर सकती है एक मैं सिंपल कंपाउंड यहां पे लिख देता हूं t एए ट्राई मिथाइल सिलाइल क्लोराइड यहां पे जाके अटक कर देते व्हाट इज द स्ट्रक्चर ऑफ टीएएसी देखो सिलिकॉन ट्राई मिथाइल तीन मिथाइल है ये सब मिथाइल है ठीक है एंड वी हैव अ स सेम ग्रुप है कार्बन सिलिकॉन सेम ही तो ग्रुप है बाबू तो ये o माइ एक कर देगी एंड ये स यहां से रिमूव हो जाएगी सो वी हैव आओ ए आ me3 तो ऐसे करके हम लोग इस रिएक्शन को करते हैं ऑलवेज लैब में हम लोग ये यूज कर देते हैं ठीक है नाउ बेस्ड ऑन दिस कांसेप्ट नाउ सो बेसिकली सोडियम हाइड्रेट पोटेशियम हाइड्रेट जो होती है यहां पे जो h माइनस होती है वो बेसिकली ए बेस के तौर पे काम करती है ट्स ल राइट नाउ बाकी जो रिजन यहां पे हमने ड्र किया था यहां पे इसको हम लोग कैसे ड्र करेंगे ट्राई टू अंडरस्टैंड दैट लेट्स से यू हैव कीटोन कीटोन से और यू कैन टेक एल्डिहाइड व्ट एवर यू वांट टू टेक ठीक है सी आर ये जो आ आड है ये बेसिकली हम लोग फॉर सेक ऑफ सिंपलीसिटी हम लोग कीटोन पकड़ रहे ठीक है नाउ इन दिस कीटोन मॉलिक्यूल ठीक है इन दिस कीटोन मॉलिक्यूल यह जो बी सोडियम बोरो आइट है सो य कैन राइट न बी ए4 माइन रिमेंबर ये माइनस अटक नहीं करती है ट इज वेरी वेरी इंपोर्टेंट जो भी गलत है वो भी बता रहा कि बी माइन अटक नहीं करेगी अगर बी माइनस अटैक कर देगी सो वी आर गोइंग टू गेट लुक एट दिस स्ट्रक्चर प्रॉपर्ली तो बोन ठीक है बोन के पास तो फ लेंसी नहीं हो सकती है बोन के पास डी ऑर्बिटल नहीं है तो यह रिएक्शन स्टेप पॉसिबल नहीं है डोंट डू दिस बट एक्चुअल में होती क्या है देखो इन दिस मॉलिक्यूल य बी ठीक है ये जो ए है वो वो बेसिकली हाइड्राइड के तौर प अटक कर देगी और इसको ओपन कर देगी क्लियर हो गया सो य आर आर ओ यू नो ओ माइनस एंड देन यू हैव हाइड्रोजन क्लियर हो गया ओके देन वी कैन ऐड सम एसिड मीडियम और वर्क अप वी कॉल इट एज ए वर्क अप देखो एक एक चीज ना हम लोग यहां पर डिस्कस कर रहे काफी इंपोर्टेंट है फर्स्ट ऑफ ल फर्स्ट टे में हम लोग सोडियम बो हाइड्रेट ऐड कर रहे सम सॉल्वेंट कंडीशन लोअर टेंपरेचर कंडीशन दिस इज द फर्स्ट स्टेप एंड सेकंड स्टेप में हम लोग ये वर्क अप कर रहे हैं नाउ ये वर्क अप जो है यू कैन ऐड सम वेरी डाइल्यूट एसिड सॉल्यूशन और यू कैन यूज अमोनियम क्लोराइड वाटर इस मिक्सचर को भी हम लोग यूज़ कर सकते हो और यू कैन यूज डाइल्यूट एसिडिक कंडीशन आफ्टर दैट ये जो वर्क अप के बाद ये o माइनस जो बेसिकली ओज में कन्वर्ट हो जाएगी सो ये बेसिकली सेकेंडरी अल्कोहल है सो वी हैव अ सेकेंडरी अल्कोहल इन हैंड क्लियर हो गया तो ऐसे करके हम लोग सेकेंडरी अल्कोहल भी प्रिपेयर कर सकते हैं सिमिलरली हम लोग एल्डिहाइड का भी कर सकते हैं नेक्स्ट स्लाइड में हम लोग करते हैं सो सिमिलरली इफ यू हैव अ एल्डिहाइड इसमें भी हम लोग न्यूक्लि फाइल ऐड कर सकते हैं ऐसे करके ठीक है यू कैन टेक ए एलुमिनियम हाइड्राइड लिथियम एलुमिनियम हाइड्राइड तो ये एलुमिनियम नेगेटिव होती है ठीक है सिमिलरली वही गलती मत करना ये एज ए h माइन के तौर पे य से एज ए h माइनस के तौर पे अटैक करेगी ठीक है एज ए h माइनस हाइड्राइट होती है ना दैट इज योर न्यूक्लियोफिलिक सोर्स तो ऐसे करके अगर ओपन कर देंगे तो दिस इज योर फर्स्ट फर्स्ट स्टेप लिथियम एलुमिनियम तो इसको हम लोग ऐसे भी लिखते लिथियम एलुमिनियम हाइड्रेट को हम लोग ऐसे लिखते ए एज मींस लिथियम एलुमिनियम हाइड्राइड ठीक है नाउ इन दिस मॉलिक्यूल यू आर गोइंग टू गेट r o माइ ए प् तो जो लिथियम प्लस होती है वो इधर आ जाती है ठीक है एंड यू हैव हाइड्रोजन और ये जो इनकमिंग हाइड्रोजन है दैट विल बी योर हेयर ठीक है सो आफ्टर दैट सेकंड स्टेप में हम लोग य पर क्या ऐड करेंगे सेकंड स्टेप में हम लोग ऐड करेंगे योर वर्क अप वर्क अप एसिडिक वर्क अप और बेसिक वर्क मेनली हम लोग वीक डाइल्यूट एसिड से हम लोग करते डाइल्यूट एल का सलूशन ऐड कर देते एंड नाउ वी आ गोइंग टू गेट दिस प्रोडक्ट सो दिस इज योर प्राइमरी अल्कोहल ओके नाउ व्हाट यू नीड टू अंडरस्टैंड दैट ये जो रेट होती है ना ये जो एल्डिहाइड के केस में जो रेट होती है न्यूक्लि फाइल का जो रेट होती है वो काफी फास्ट होती है स्टेरिक क्राउडिंग का भी एक रीजन होती है और ये ज्यादा रिएक्टिव होती है क्योंकि हाइड्रोजन एंड आर है तो यहां पे ये अटैक और भी फीजिबल हो जाती है बट इन केस ऑफ कीटोन कीटोन के केस में य ये यहां पे आर है यहां पे भी आर है तो कीटोन के केस में इसका रेट थोड़ी बहुत स्लो हो जाती है सो यू नीड टू अंडरस्टैंड दैट ठीक है यू नीड टू अंडरस्टैंड दैट कीटोन और एल्डिहाइड की अगर बात करें तो एल्ट में जो रेट होती है वह काफी फास्ट होती है ट्राई टू अंडरस्टैंड जो एल्ट का जो रेट होती है वह काफी फास्ट होती है सो व्हाट यू कैन राइट डाउन हियर व्हाट यू कैन राइट डाउन हियर दैट एल्डिहाइड के केस में रेट काफी फास्ट होती है कीटोन के केस में रेट काफी कम होती है थोड़ी देर में हम लोग डिस्कस भी करेंगे नाउ नेक्स्ट पार्ट इज दैट एक अगर रियल एग्जांपल की अगर बात करें सो वी हैव अ दिस ये जो मैकेनिज्म है यहां पे और एक चीज को मेंशन किया गया है इन योर क्लेटन बुक जिसको हम लोग नॉर्मली करते नहीं है बट स्टिल वी आर गोइंग टू डिस्कस हियर दैट यू नो यह जो स्टेप है यह जो स्टेप है ठीक है लेट्स से यू हैव दिस यू हैव दिस r ठीक है c डबल ब o h सो दिस इज योर एल्डिहाइड तो एल्डिहाइड में ना इसको सही से समझ लेना ठीक है एल्डिहाइड में इफ यू ऐड योर सोडियम बोरो हाइड्राइड दैट मींस b ए bh4 nabh4 तो ये एस h माइनस को अटक करेगी और यह कार्बोन ओपन हो जाएगी क्लियर हो गया सो वी हैव आर सी ठीक है ए ए ठीक है एंड वी हैव अच्छा इसको ना ऐसे करके लिख देते ओ माइनस क्लियर हो गया वही ओ माइनस में अभी क्या हो जाएगी देखो य एक हाइड निकल जाएगी सो नाउ य पे bh3 तो यहां पे ये आ जाएगी bh3 o माइ जो क्रिएट हुआ है ना वो और एक bs3 को जाके अटैक कर देगी अच्छा यहां पे जब अटैक करेगी देखो एक-एक स्टेप को मैं कर देता हूं फर्स्ट ऑफ ऑल यू आर गोइंग टू गेट o माइन तो अगर हाइड्राइड निकल जाएगी तो यहां से क्या रहेगी यू हैव अ bh3 मॉलिक्यूल राइट तो bh3 इज अ लुईस एस तो o माइनस जाके बोन में जाके अटैक कर देगी इमीडिएट सो ऐसे करके मैकेनिज्म को प्रपोज किया गया है सो नाउ वी हैव हाइड्रोजन और o और य हैव ए बोर अच्छा बोरनन में और एक बॉन्ड अगर बन जाएगी तो बोरनन नेगेटिव हो जाती है ठीक है सो य है हाइड्रोजन हाइड्रोजन हाइड्रोजन एंड दे आर सेइंग दैट ये जो हाइड्राइट है दैट विल बी अनदर यू नो हाइड्राइड सोर्स जो और एक और एक कार्बोनियम से एल्डिहाइड में जाके अटैक कर सकती है ठीक है और एक एल्डिहाइड में जाके ये अटैक कर सकते हैं तो ऐसे करके हम लोग कर सते बट द से सेक ऑफ सिंपलीसिटी इफ यू ऐड सम एसिडिक वर्क अप इफ यू डू द एसिडिक वर्क अप देन इमीडिएट यूर गो टू गे द अल्कोहल सो दैट इज द जनरल मैकेनिज्म हम लोग रियल एग्जांपल के ऊपर बात करते हैं अगर हम लोग रियल एग्जांपल की बात करेंगे लेट्स से यू हैव अ मिथक्स इड एल्डिहाइड ठीक है तो इस एल्डिहाइड में इफ यू ऐड सोडियम बोरो हाइड्रेट मैं एक-एक स्टेप में कर रहा हूं सोडियम बोरोहाइट्राइड फर्स्ट स्टेप में अगर रिएक्शन कंप्लीट हो जाएगी देन यू कैन डू द वर्क अप दोनों एक साइड मत करना दोनों एक साथ करोगे तो ये जो दो रिएजेंट देख रहे हो ना सोडियम बोरोहाइट्राइड हमम बोरो हाइड्रेट फिर भी सही है लिथियम एलुमिनियम हाइड्रेट में अगर इफ यू ऐड अ वाटर तो ये आग लग सकती है सो दैट इज वेरी डेंजरस रिएजेंट बी केयरफुल तो जब भी हम लोग डिस्कस करें ट्राई टू अंडरस्टैंड ट्राई टू अंडरस्टैंड द केमिस्ट्री लिथियम एल्युमिनियम हाइड्राइड में लिथियम मेटल भी है काफी रिएक्टिव होती है ना अल्कली मेटल काफी रिएक्टिव होती है वाटर के बिजनेस में यू नो हाइड्रोजन गैस रिमूव करके आग लगा देती है राइट डेंजरस भी होती है ओके काफी सारे इ इंसिडेंट बेस्ड ऑन दिस केमिस्ट्री के ऊपर होती है सो बी केयरफुल सोडियम बोरो हाइड्राइड को पहले रिएक्शन में ऐड करो रिएक्शन जब कंप्लीट हो जाएगी तो अंडर सम कंडीशन ठीक है आइस के प्रेजेंस में लो टेंपरेचर में यू कैन ऐड वाटर स्लोली ठीक है ये धीरे-धीरे सो फॉर सेक ऑफ़ सिंपलीसिटी यू कैन ऐड अमोनियम क्लोराइड सम बेसिक यू नो बेसिक सॉल्ट वाटर मिक्सचर भी ऐड किया जाता है ठीक है तो ऐसे करके ऐड किया जाता है नाउ वी हैव दिस एओ एंड नाउ वी आर गोइंग टू गेट करेस्पॉन्डिंग्ली तो ये सिंपल रिएक्शन है तो बेसिक ऑक्सीडेशन ऑफ योर बेसिक ऑक्सी बेसिक सॉरी बेसिक रिडक्शन ऑफ़ योर योर एल्डिहाइड क्लियर हो गया सो दिस इज़ बेसिक रिडक्शन ऑलवेज रिमेंबर जो सोडियम अ जो सोडियम अ बोरो हाइड्राइट होती है सोडियम बोरो हाइड्राइड इज अ वेरी वीक रिड्यूस एजेंट दिस इज योर वीक रिड्यूस एजेंट रिड्यूस एजेंट का मतलब क्या है ये बेसिकली ये सोडियम बोरो हाइड्राइड का जो bh4 माइन यूनिट है येसे जो हाइड्राइड माइनस का अटैक होती है ना दैट इज नॉट दैट मच स्ट्रांग दैट मींस इट कैन रिड्यूस इट कैन रिड्यूस इट कैन इट कैन रिड्यूस एल्डिहाइड एल्डिहाइड बट इसका रेट फास्ट होती है एंड कीटोन तो यह जो है ना इट कैन रिड्यूस र एल्डिहाइड एंड कीटोन बट एल्डिहाइड के केस में इसका रेट फास्ट होती है इसको याद रखना ठीक है बाब ओके उसके बाद देखो व्ट व डिस्कसिंग अबाउट दिस रिडक्शन केमिस्ट्री नाउ इफ यू टेक एनी पी ठीक है सो इन दिस मॉलिक्यूल यू कैन ड सोडियम बोरो हाइड्राइड इन दिस मॉलिक्यूल यू कैन ड सोडियम बोरो हाइड्राइड इन मिनल सॉल्वेंट ठीक है ओ हाइड्रोजन ठीक है एंड नाउ आफ्टर दैट यू ग टू गेट दिस प्रोडक्ट ठीक है सिमिलरली यू कैन टेक सम कीटोन साइक्लिक कीटोन राइट इन दिस मॉलिक्यूल यू कैन ड सोडियम बोरो हाइड्राइड इन प्रस ऑफ अल्कोहल सॉल्वेंट आइसो प्रोपनल यू आर गोइंग टू गट दिस प्रोडक्ट क्लियर हो गया ट्स वई आई टोल्ड यू दैट ये जो रिडक्शन होती है सोडियम बोरो हाइड्राइड इट कैन रिड्यूस योर एल्डिहाइड एंड कीटोन ट्राई टू अंडरस्टैंड इट कैन रिड्यूस योर एल्डिहाइड एंड कीटोन बट एल्डिहाइड की रेट ज्यादा फास्ट होती हैन कीटोन बिकॉज एल्डिहाइड इज मोर रिएक्टिव एल्डिहाइड का जो इलेक्ट्रोफीलिसिटी है वो ज्यादा होती है कौन सी सेंटर की कार्बन की तो इसको भी याद रखना ठीक है नाउ अगर कोई ऐसा कंपाउंड है जहां पे एल्डिहाइड ग्रुप भी है एंड ए स्टार है ए स्टार का रिएक्टिविटी बहुत कम होती है ठीक है सो लेट्स से यू हैव एल्डिहाइड ओवर देयर एंड देन यू हैव अ ए स्टार सीओटी ए स्टार नाउ इन दिस मॉलिक्यूल इन दिस मॉलिक्यूल इफ यू ऐड सोडियम सोडि बोइड एंड देन य हैव नी अल्कोहल कंटेनिंग सॉल्वेंट इनल सोडियम बोरोहाइट्राइड सली एल्डिहाइड में जाके अटैक करेगी ए स्टार में जाके अटैक नहीं करेगी सो यू हैव अनदर ग्रुप ओवर हियर दिस इ योर ए स्टार क्लियर हो गया बाबू एंड दिस इज योर कीटोन साइड सॉरी एल्डिहाइड तो एल्डिहाइड एस्टर में व्हाट इज मोर रिएक्टिव सेंटर एल्डिहाइड में जाएग अच्छा सोडियम बोरो हाइड्राइड कैन नॉट रिड्यूस योर एस्टर कार्बोन ग्रुप ठीक है यहां से ये जाके यही कार्बोन में जाके अटैक नहीं करेगी नेवर एवर पॉसिबल ही नहीं है ठीक है बिकॉज ऑफ द रिएक्टिविटी रीजन ठीक है भी यनो एस्ट का रिडक्शन नहीं कर पाएगी सोडियम बोरो हाइड ठीक है सो नाउ वी आर गोइंग टू गेट अल्कोहल ओवर देयर क्लियर हो गया दैट यू नीड टू अंडरस्टैंड सिमिलरली अगर हम लोग कोई नाइट्रो ग्रुप डाल देते नाइट्रो ग्रुप इज मोर नाइट्रो ग्रुप अगर रहेगी ना रिंग में तो वो और भी डिफिशिएंट हो जाती है इलेक्ट्रॉनिकली काफी डिफिशिएंट हो जाती है सो वी हैव नाइट्रो ग्रुप समर हियर नाउ इन दिस मॉलिक्यूल इन दिस मॉलिक्यूल इफ यू ड सोडियम बोरो हाइड्राइड ठीक है सोडियम बोरो हाइड्राइड फॉलो बाय योर मिनल सॉल्वेंट वाटर देन ये जो एल्डिहाइड सेंटर है इसका इजली रिडक्शन हो जाती है बट इसका रेट काफी फास्ट होती है ट्राई टू अंडरस्टैंड बिकॉज दिस इज मोर इलेक्ट्रॉन डेफिसिट ड्यू टू दिस इलेक्ट्रॉन ड ड्राइंग ग्रुप ch2oh क्लियर हो गया तो ऐसे करके हम लोग इजली इसका रिडक्शन भी कर सकते हैं ठीक है इसको सही से प्रैक्टिस कर लेना ठीक है काफी इंपोर्टेंट है सिमिलरली सिमिलरली इफ आई टेक सिमिलरली इफ आई टेक दिस मॉलिक्यूल सोडियम बोरो हाइड्राइट सो यहां पे यू आर गोइंग टू गेट यू आर गोइंग टू गेट बर एंड देन हाइड्रोजन क्लियर हो गया दैट यू नीड टू अंडरस्टैंड नाउ उसके बाद हम लोग कोई और न्यूक्लि फाइल के ऊपर डिस्कस करेंगे सो मेनली हम लोग अभी डिस्कस करेंगे योर अनदर काइंड ऑफ न्यूक्लि फाइल दैट कैन बी योर ch3 माइन ठीक है तो ये स3 माइन कहां से आ सकती है सो लेट्स से यू हैव अ ए ए आई ठीक है ए एई ठीक है और एक चीज य मेंशन कर देता हूं जस्ट फॉरगेट टू मेंशन सो यू हैव यू हैव बी ए4 राइट ये जो बी फ देख रहे हो य बी फ ट इ आइसो इलेक्ट्रॉनिक विथ योर ch4 बी ए4 एंड ch4 यह सब आइसो इलेक्ट्रॉनिक है दोनों में अगर इलेक्ट्रॉन काउंट करोगे ना टोटल इलेक्ट्रॉन काउंट करोगे तो सेम मिलेगी य कैन कॉल इट आइसो इलेक्ट्रॉनिक सो वी कैन राइट डाउन एज ए आइसो इलेक्ट्रॉनिक ठीक है सो दिस इज योर आइसो इलेक्ट्रॉनिक bh4 माइनस एंड ch4 बट एटक कहां से होती है b4 से हाइड्राइड स हो कार्बन से नहीं जाती है मिथेन से ये मीथेन से कभी भी हाइड्राइड नहीं जाएगी इंपॉसिबल राइट ओके नाउ क्लियर हो गया अभी हम लोग ए एआई के ऊपर डिस्कस करेंगे सेम केमिस्ट्री हम लोग मिथाइल यू नो ए एआई राइट एमी एआई के ऊपर डिस्कस करेंगे राइट नाउ बिफोर गोइंग टू डिस्कस दैट लिथियम रिएजेंट अ यू हैव अ लिथियम एंड मैग्नीशियम दे आर वेरी इलेक्ट्रो पॉजिटिव मेटल राइट लिथियम के अगर ग्रुप की बात करें हाइड्रोजन तो छोड़ो लिथियम सोडियम पोटेशियम रूबियन सीसियम मैग्नीशियम की बात करें वेलियम मैग्नीशियम कैल्शियम स्टोस वेडियम ऐसे होती है ना पेरी टेबल में सो दे आर वेरी यू नो यू नो रिएक्टिव मेटल और यू कैन से काफी पावरफुल येय जो इलेक्ट्रो पॉजिटिव होती है ना दैट इज द राइट टर्म इलेक्ट्रो पॉजिटिव टर्म दैट मींस यू कैन राइट डाउन लिथियम प्लस एंड मैग्नीशियम टू प्लस दैट मींस इट कैन गिव यू इले न किसको इलेक्ट्रॉन देगी दैट मींस अगर कोई भी ए एई होगी तो लिथियम इजली इलेक्ट्रॉन दे देगी तो ए के ऊपर माइनस हो जाएगी मैग्निशियम की अगर बात करें तो me2 अच्छा सॉरी सॉरी ये नहीं होती है क हम लोग कुछ कॉपर की सोच रहे थ सो ए एज ब तो यहां पे मैग्नीशियम तो टू प्लस होती है इलेक्ट्रॉन दे देगी तो यहां पे ये माइनस हो जाती है एंड ब्रोमीन भी माइनस है तो इसको बैलेंस किया जाता है सो दिस इज योर लिथियम रिएजेंट एंड दैट इज योर ग्रिक रिएजेंट राइट तो इसके ऊपर अभी हम हम लोग डिस्कस करेंगे तो ये ऑर्गेनो मैग्नीशियम ऑर्गेनो लिथियम रिएजेंट ऑर्गेनो लिथियम ऑर्गेनो मैग्नीशियम क्लियर हो गया नाउ यहां पे देखो सही से ये जो ए माइनस होती है दैट विल बी योर न्यूक्लि फाइल हियर एंड मिथल मैग्नीशियम ब्रोमाइड में ए इज आल्सो योर न्यूक्लि फाइड सो आई कैन राइट डाउन r स ए सो दिस इज योर अ एल्डिहाइड नाउ इन दिस एल्डिहाइड फर्स्ट स्टेप में अगर हम लोग ए ए आई ऐड करेंगे और एक चीज मैं मेंशन कर देता हूं यह दोनों जो मेटल है काफी दोनों जो लिथियम और मैग्नीशियम रिएजेंट है ये और भी सेंसिटिव होती है टुवर्ड योर मॉइश्चर मॉइश्चर मींस यर यू कैन नॉट ऐड वाटर इन दिस मेडियम ठीक है वो डिस्ट्रॉय हो जाएगी तो यह मॉइश्चर सेंसिटिव होती है राइट वेपर सेंसिटिव वाटर वे परर सेंसिटिव होती है तो इस रिएक्शन को इनर्ट कंडीशन में किया जाता है कौन सी कंडीशन इनर्ट कंडीशन इनर्ट सॉल्वेंट इनर्ट सॉल्वेंट कंडीशन आल्सो यू कैन नॉट ऐड प्रोटोनेटेड सॉल्वेंट क्लियर हो गया प्रोटोनेटेड सॉल्वेंट मींस यू कैन नॉट ऐड मेथनल यू कैन नॉट ऐड टर्स आइसो प्रोपनल सॉरी आइसो प्रोपनल स ऐसे हम लोग प्रोटोनेटेड सॉल्वेंट हम लोग ऐड नहीं कर सकते द रीजन इज दैट इफ यू ऐड दिस नो प्रोटोनेटेड सॉल्वेंट तो ए एआई जो है वो इजली इस हाइड को लेके वो डिस्ट्रॉय हो जाएगी सो व्हाट इज द यू नो पर्पस हम लोग एम एआई क्यों ऐड कर रहे क्योंकि एम एआई जाके इस कार्बोन में जाके अटैक करेगी एंड इस कार्बोन को ओपन कर देगी सो दैट्ची सॉल्वेंट देखो एप्रोटो यू कैन टेक टीए टीए टीटीएफ का स्ट्रक्चर क्या है सो नाउ यू कैन आस्क मी सर यहां पे भी तो स ब है हाइड्रोजन तो इस सॉल्वेंट में है राइट यहां पे हाइड्रोजन नहीं है ch2 बट दैट इज नॉट एसिडिक वी आर टॉकिंग अबाउट प्रोटिक एसिडिक सॉल्वेंट जैसे अगर अल्कोहल की बात करेंगे तो अल्कोहल में मेथेनॉल है ऑक्सीजन के साथ है हाइड्रोजन सो दिस हाइड्रोजन इज़ यू नो एसिडिक इन नेचर सो दैट h विल बी योर h+ राइट सो वी कैन नॉट यूज़ दिस सॉल्वेंट बट इंस्टेड ऑफ़ दैट वी कैन यूज़ टीएचएफ ड्राई टीएचएफ वी राइट यहां पे हम लोग ड्राई टीएचएफ लिख देते हैं ड्राई मतलब यहां पे कोई भी मॉइश्चर नहीं होना चाहिए मीनिंग समझना चाहिए ठीक है यू कैन नॉट ऐड एनी अ यू कैन नॉट ऐड नॉर्मल सॉल्वेंट यू टू यूज द ड्राई सॉल्वेंट ठीक है एंड नाउ वी आर गोइंग टू गेट स ए o - l+ ठीक है सेकंड स्टेप में ट्राई टू अंडरस्टैंड सेकंड स्टेप में दिस इज द फर्स्ट स्टेप सेकंड स्टेप में हम लोग ऐड करेंगे अ यू नो वर्क अप सो य कैन कॉल इट एज अ वर्क अप फर्स्ट स्टेप में हम लोग ऐड नहीं करेंगे तो ब्लास्ट हो जाएगी आफ्टर दैट वी हैव अ स ए एंड वी हैव अ o तो ये जो माइनस है o माइनस वो यहां से मीडियम से फोटोन लेगी और एक चीज़ यहां पे मेंशन रखो मेंशन करना कि ये जो अटैक हम लोग कर रहे हैं ना बाबू यहां पे ठीक है यहां पे जब हम लोग इसका अटैक कर रहे ना तो यह काफी लो टेंपरेचर में किया जाता है हाई टेंपरेचर में नहीं किया जाता है सो अनदर पार्ट दैट यू हैव टू रिमेंबर लो टेंपरेचर कंडीशन जो ऑर्गेनो लिथियम और गन रिजन होते है वो हम लोग आइस कंडीशन और भी लोअर टेंपरेचर में परफॉर्म करते हैं ठीक है सो नाउ लेट्स राइट डाउन सम अदर रियल प्रॉब्लम रियल मॉलिक्यूल सो एक एमी होगी सो सिमिलरली इफ यू टेक एनी कीटोन सिमिलरली इफ यू टेक सम कीटोन एंड ड सम अनदर ऑर्गेनो लिथियम रिएजेंट दैट कैन बी एन बूली काफी फेमस रिएजेंट है यू नो वेरी फेमस रिएजेंट मे बी देयर विल बी अ फुल टू थ्री आवर्स लेक्चर ऑन दिस ब्यूटे लिथियम रीजन व्हाट आर द यू नो एप्लीकेशन हम लोग किसको कैसे इसको यूज कर सकते हैं ठीक है वो भी हम लोग अलग सीरीज करेंगे बाद में सो देन दिस माइनस ये जाके अप अटक कर देगी एंड नाउ वी आर गोइंग टू गेट दिस ओपन हो जाएगी राइट सो नाउ वी आर गोइंग टू गेट आर आर अच्छा दिस इज द फर्स्ट स्टेप ठीक है दिस इज ब्यूटे लिथियम ठीक है दिस इज योर ब्यूटे लिथियम सेकंड स्टेप में हम लोग हाइड्रोलिक से वर्क अप एसिडिक वर्क अप नॉर्मल वर्क अप भी कर सकते हैं वर्क तो यहां पे बन जाएगी ओ टूथ 4 सो दिस इज योर टरी अल्कोहल द नेक्स्ट पार्ट दैट वी टू डिस्कस इन दिस वीडियो दैट विल बी अनदर इंपोर्टेंट पॉइंट दैट यू हैव टू अंडरस्टैंड जैसे हम लोग न्यूक्लि फाइल एडिशन की बात कर रहे थे ना हम लोग जब न्यूक्लि फाइल एडिशन की बात कर रहे थे सो फार वी हैव डिस्कस अबाउट द हाइड्राइड न्यूक्लि फाइल वी हैव डिस्कस अबाउट द नाइट्राइल न्यूक्लि फाइल एमी न्यूक्लि फाइल एज वेल नाउ और एक चीज यहां पे मैं मेंशन कर देता हूं थोड़ा एनएमआर का हम लोग हेल्प भी लेंगे एनएमआर सबको पता है यू हैव फॉर्मल डिहाइड इन रियलिटी यह जो कार्बोन ग्रुप देख रहे हो ना इट इ हाईली रिएक्टिव टू मच रिएक्टिव तो ये ना पॉलीमराइज हो जाती है रीजन ट य पॉलीमराइज हो जाएगी इमीडिएट य पॉलीमराइज हो जाती है तो यह काफी रिएक्टिव होती है ठीक है ओके नाउ यह जो कार्बोन का जो कार्बन देख रहे हो इसका अगर हम लोग 13 सी की 13 सी जो एनएमआर होती है इसका जो वैल्यू आना चाहिए केमिकल शिफ्ट का जो वैल्यू आना चाहिए ट इज योर एक्सपेक्टेड वैल्यू जो आना चाहिए वो बेसिकली अराउंड 160 अराउंड 200 पीपीएम इसके बीच आना चाहिए इन बिटवीन ट वैल्यू के अंदर आना चाहिए हम लोग अजूम कर सकते बट इन दिस मॉलिक्यूल इफ यू ड वाटर तो यह रिएक्शन काफी फास्ट होती है इमीडिएट वाटर जाके ना वाटर का जो लोन पेर वो जा न्यूक्लि फाइल के र अटक करगी एंड यू आर गोइंग टू गेट ए एज ओ ओ ए तो य से बन ओपन हो जाएगी क्लियर हो गया सो नाउ वी आर गोइंग टू गेट हाइड्रेट हाइड्रेटेड पार्ट और यू कैन से ये एक नंबर कार्बन है राइट तो वी कैन राइट न वव डाइल इन रियलिटी ये जो कार्बन का सिग्नल आना चाहिए प्रीवियसली दैट वाज योर sp2 हाइब्रिडाइज्ड इज कार्बन नाउ इसका जो वैल्यू आता है वो अराउंड 83 टू 85 पीपीएम इस रेंज में आती है ठीक है तो इस चीज को ना सही से याद रखना ठीक है एंड आल्सो आई टोल्ड यू टू दैट कि एल्डिहाइड का जो रेट होती है वो काफी फास्ट होती है राइट न्यूक्लियोफिल के एडिशन टुवर्ड एल्डिहाइड का जो रेट होती है के इक्विलियम वैल्यू काफी फास्ट होती है इन कंपेरिजन टू योर कीटोन तो इफ आई कैन राइट डाउन अ जेनरल स्ट्रक्चर आर आ डैश इन दिस मॉलिक्यूल इफ यू ऐड वाटर जो वाटर जो बेसिकली यहां पे न्यूक्लि फाइल के तौर पे काम करही है तो मैं यहां पे मेंशन कर देता हूं न्यूक्लि फाइल तो वही न्यूक्लि फाइल जब ये एल्डिहाइड में और यू नो r स r ड में जब अटैक करेगी ना तो इसको हम लोग कैसे लिख सकते हैं सही से देखो बाबू वी हैव अ के इक्विलियम के इक्वि म का जो मेन डेफिनेशन मैंने थोड़ी देर पहले ही मेंशन किया था k इक्विलियम इज बेसिकली kf2 एंड इक्विलियम कंडीशन फॉरवर्ड बाय बैकवर्ड एट कांस्टेंट राइट नाउ वी वी आर गोइंग टू गेट r आड वी हैव अ हाइड्रेटेड सॉल्ट नाउ ट्राई टू अंडरस्टैंड सम बेसिक एग्जांपल जैसे अगर हम लोग रियल मॉलिक्यूल की बात करेंगे एसीटोन एसीटोन एसीटोन में दैट इज द स्ट्रक्चर r ch3 स तो इसका जो इक्विलियम कांस्टेंट का वैल्यू होता है k इक्विलियम इसका जो वैल्यू होती है वो अराउंड 0.001 होती है ठीक है सिमिलरली इफ यू टेक एसिट एल्डिहाइड इफ यू टेक एसिट एल्डिहाइड इफ यू टेक एसिड एल्डिहाइड बिकॉज़ दिस इज एल्डिहाइड इसका रेट और भी फास्ट हो जाती है तो k इक्विलियम का वैल्यू ए इन प्रेजेंस ऑफ़ वट इन वाटर ए नफा इसका जो प्रोडक्ट बनेगी ठीक है ट विल बी इसका जो रेट होती कास्ट और भी हाई हो जाती है ठीक है सिमिलरली इफ यू टेक द ओनली फॉर्मल डिहाइड य जो मॉलिक्यूल है ना फॉर्मल डिड य टेक ओनली द फॉर्मल डिड बिकॉज य हैव हाइड्रोजन हाइड्रोजन ओनली इसका रेट देखो काफी हाई होती 280 काफी हाई होती है ना सिमिलरली इंस्ट ऑट इ यू टेक ओनली इसको सही से देख लो cf3 सडल ब cf3 समझ गए हो हेक्सा फ्लोरो एसीटोन ठीक है तो इस केस में जो रेट होती है वो और भी हाई हो जाती है अराउंड क्योंकि ये ट्राई फ्लोरो ग्रुप है ना इसलिए वो कार्बोन का रिएक्टिविटी बढ़ जाती है ट्राई फ्लोरो फ्लोरिन है ना सो वन मोर इलेक्ट्रोनेगेटिव ग्रुप है तो इसका रेट देखो कितनी ज्यादा हो जाती है सो ट्राई टू अंडरस्टैंड दिस केमिस्ट्री वेरी ब्यूटीफुल केमिस्ट्री है ठीक है ऐसे करके हम लोग के इक्विलियम को डिफाइन कर सकते हैं अगर के इक्विलियम का वैल्यू ज्यादा है इसका मतलब क्या है फॉर्मेशन और कंस प्रोडक्ट यह जो प्रोडक्ट का कंसंट्रेशन है व ज्यादा है यह जो प्रोडक्ट का कंसंट्रेशन है वो और भी ज्यादा है दैट मींस रेट भी काफी फास्ट होती है ठीक है सिमिलरली वी कैन डिस्कस सम अदर केमिस्ट्री जैसे सो फार वी हैव डिस्कस द नॉर्मल कार्बोनेल कंपाउंड नॉर्मल कार्बोनेल कंपाउंड के मतलब वी हैव डिस्कस द एसाइक्लिक कंपाउंड एसाइक्लिक मीन समझ रहे हो ना साइक्लिक एसाइक्लिक तो हम लोग साइक्लिक कंपाउंड के ऊपर भी डिस्कस कर सकते हैं साइक्लिक कंपाउंड की अगर बात करें तो वी कैन टेक सम बेसिक एग्जांपल बेसिक एग्जांपल अगर बात करें लेट से व कैन टेक अबाउट सम साइक्लिक कंपाउंड अगर हम लोग साइक्लिक कंपाउंड की बात करें सो कौन सी साइक्लिक कंपाउंड ले सकते देखो सबसे सिंपलेस्ट स्मले साइक्लिक कंपाउंड और साइक्लिक यू नो व टॉकिंग अबाउट द कार्बोल कंपाउंड सो साइक्लिक र दिस दिस कंपाउंड साइक्लो प्रोनो ठीक है नाउ इन दिस कंपाउंड व्ट इज द बन एंगल व्ट इज द बन एंगल दिस 0 डिग्री बन एंगल ठीक है व्ट इ द बन एंगल 120 डिग्री सो व्ट इज द हाइब्रिड दिस इ बेसिकली र sp2 हाइब्रिड 120 डिग्री बोन एंगल क्लियर हो गया उसके बाद अगर हम लोग य प वाटर ऐड कर देंगे इ य ड वाटर जसे कार्ब अटक करगी य ओपन हो जाएगी क्लियर हो गया तो य पर क्या बनेगी देखो ओ य प ब एंगल कितना है स्टल अच्छा य मैंने गलती कर दिया एक मैंने गलती कि य 60 डिग्री है दोनों केस में 6 है नाउ हाइब्रिड कांसेप्ट बिकॉज ये जो कार्बन हैट इ बेसिकली sp3 हाइब्रिड नाउ व्ट इ य बन एंगल कितने है मैं लिख देता देखो बोन एंगल इन दिस मॉलिक्यूल इज योर 60 डिग्री बिकॉज यह जो मॉलिक्यूल है स्टार्टिंग पार्ट में जो प्रोडक्ट बना है इसमें बन एंगल कितना है यव 60 डिग्री नाउ बेस्ड ऑन यर हाइब्रिड बेस्ड ऑन योर हाइब्रिडाइज्ड एंगल यहां पे होना चाहिए बिकॉज sp2 तो 120 डिग्री होना चाहिए था एंड इसमें इसका एंगल कितना होना चाहिए था 109 डिग्री होना चाहिए था राइट सो व्हाट इज द डिफरेंस व्हाट इज द डिफरेंस ठीक है व्हाट इज द डिफरेंस यहां पे देखो 60° का डिफरेंस है यहां पे कितना डिफरेंस है करके देखो 109 - 60° का डिफरेंस है 109 - 60° अराउंड 49 डि क्लियर हो गया तो ज्यादा स्टेबल पार्ट कौन सी है यही है ज्यादा स्टेबल क्योंकि यहां पे डिफरेंस बहुत कम है डिफरेंस बहुत ज्यादा कहां पे है यू नो कार्बोन में दैट्ची ये देखो इसका एरो जो है ये गलत हो गया ये एरो जो है वो फॉरवर्ड डायरेक्शन में ज्यादा है बैकवर्ड में बहुत कम है क्लियर हो गया तो ऐसे करके हम लोग साइक्लिक कंपाउंड में भी इसका एक्सप्लेनेशन कर सकते हैं एंड आल्सो जब भी हम लोग कार्बोन कंपाउंड की बात करते हैं देयर इज अ वेरी ब्यूटीफुल केमिस्ट्री वी कॉल इट एज ए एसिट एंड हेमी एसिट कंपाउंड जैसे यू हैव अ एल्डिहाइड सही से देखो यू हैव अ एल्डिहाइड एंड यू हैव अ कीटोन कंपाउंड ठीक है यू हैव एल्डिहाइड एंड यू हैव अ कीटोन कंपाउंड नाउ ये जो एल्डिहाइड एंड न कंपाउंड है ना यहां पे एक तो होती है हेमी एटल और एक होती है एटल ठीक है नाउ ये जो हेमी एटल होती है एल्डिहाइड में सो वी कैन राइट डाउन आर सी ए र एंड योर ओ क्लियर हो गया इसको हम लोग क्या बोलते हैं एल्डिहाइड में एक हो सकता है हेमी एटल हेमी एटल का मतलब हम लोग इसको ऐसे करके समझते हैं लेट्स से एक जनरल एग्जांपल ले लेते हैं एल्डिहाइड इज योर नॉर्मल एसिड एल्डिहाइड तो यहां पे अगर कोई भी अल्कोहल ऐड कर देंगे जैसे ओ एंड देन यू हैव o ठीक है इसको हम लोग क्या बोलते हैं एटल एटल का जो बेसिक सॉरी दिस इज यो हेमी एटल नाउ यहां पे एटल को हम लोग कैसे लिखेंगे देखो यहां पे एडल को हम लोग ऐसे लिखेंगे कि यू हैव ए r यू हैव अ हाइड्रोजन मान लेते हैं कोई भी अल्कोहल आया है ड अनदर अल्कोहल माइटी इज देयर डड ठीक है तो इसको हम लोग बोलते हैं एटल सो यहां पर देखो एक ओ है एक ओ है इन हेमी एटल अच्छा एटल में क्या होती है ch3 हाइड्रोजन ओ एंड ओ तो एटल में य हैव ओ ग्रुप दोनों ओ है बट हेमी एटल में एक र एक ओ है सोट इ द डिफरेंस य हैव टू रिमेंबर ठीक है नाउ इन केस ऑफ कीटोन कीटोन को हमलोग कैसे लिख सकते जैसे r1 आ2 दोनों केस में r1 आ2 मैं लिख देता हूं एंड हेमी एटल में दैट विल बी ड इन एटल दैट विल बी र ड एंड देन र डबल ड क्लियर हो गया सो दैट थिंग यू नीड टू अंडरस्टैंड प्रॉपर्ली व्हाट इज योर हेमी एटल व्हाट इज एटल नाउ नेक्स्ट पार्ट इज दैट कैसे हम लोग इस एटल को फॉर्म करते कैसे हम लोग हेमी एटल को फॉर्म करते हैं रिमेंबर द जनरल फार्मूला क्योंकि काफी टाइम ना जब हम लोग एग्जांपल में डिस्कस करेंगे तो हम लोग बहुत जल्दी बोले दिस इज़ योर एटल इ दिस इज योर हेमी एटल दैट टाइम आई एम नॉट गोइंग टू एक्सप्लेन एटल का जनरल स्ट्रक्चर क्या है और हेमी एटल का जनरल स्ट्रक्चर क्या है सो यू हैव टू रिमेंबर ठीक है सो नाउ कैसे हम लोग इसको फॉर्म कर सकते हैं लेट्स से यू हैव दिस एल्डिहाइड ग्रुप सो नाउ वी हैव आर ए ओ अच्छा मैं एक एक स्टेप को मैं य पे ये कर रहा हूं तो मैं प्रोडक्ट को मैं पहले लिख देता हूं उसके बाद मैं एकएक करके एक्सप्लेन करूंगा कैसे इसका मैकेनिज्म होती है तो देखो इस मैकेनिज्म को हम लोग नीचे करते ओ सॉरी सॉरी देखो व्हाट इज द मैकेनिज्म लेट्स से वी कैन टेक दिस इथेनॉल तो यहां पे इथेनॉल जो है ऑक्सीजन का लोन प्र न्यूक्लि फाइल के ौर प काम कर य करेगी ये ओपन हो जाएगी ठीक है नाउ वी आर गोइंग टू गेट r हाइड्रोजन o माइ o हाइड्रोजन प्लस ओ माइ क्लियर हो गया ठीक है उसके बाद क्या करना है बाबू देखो यहां पे उसके बाद क्या करना है यू कैन टेक सम अनदर इथेनॉल मॉलिक्यूल एंड यही ओ जो है यह बेसिकली य से हाइड को ले सकती है और इसको ऐसे करके न्यूट्रलाइज कर देगी ठीक है नाउ ट गेट आर हाइड्रोजन ओटी ठीक है एंड नेक्स्ट स्टेप में क्या हो जाएगी यू हैव हाइड्रोजन ओटी क्लियर हो गया दैट वे यू कैन गेट दिस हेमी एटल कंपाउंड ऑफ योर एल्डिहाइड क्लियर दैट इ जनरल मैकेनिज्म आई होप सबको समझ में आया है बट व्हेन एवर यू हैव अ साइक्लिक कंपाउंड तो साइक्लिक कंपाउंड में ये हेमी एटल फॉर्मेशन जो है वो फैसा होती है ड्यू टू सम यू नो बन एंगल रीजन सो आई कैन राइट डाउन य वि दिस हाइड्रो कंपाउंड हाइड्रोक्सी टूथ 4 एंड देन यू हैव एल्डिहाइड ठीक है तो इसको हम लोग ऐसे लिख सकते हैं कि ये जो मॉलिक्यूल ऑक्सी इंट्रा मॉलिक्यूलर ये य पट कर देगी सो देखो यहां पे 1 2 3 4 फ सो य फाइव मेंबर रिंग यहां पे फॉर्म हो जाएगी ठीक है सो नाउ वी कैन राइट डाउन o योर हाइड्रोजन टूथ ठीक है वी हैव o माइनस हियर h तो नंबरिंग को मैं य पे डाल देती हूं सो वी हैव अ वनटू 3 फर एंड फ तो यहां पे न्यू बॉन्ड कहां पे बनी है देखो बाबू यहां पे न्यू बन कौन सी है तो ये जो ग्रीन कलर की बन है दैट इज योर न्यू ब ठीक है नाउ नेक्स्ट स्टेप में क्या हो जाएगी यह जो नेगेटिव चार्ज है यह पॉजिटिव हो जाएगी ना ऑक्सीजन का लोन प्र इधर अटैक करेगी तो ये इसको ले लेगी एंड ये न्यूट्रलाइज कर देगी सो नाउ वी हैव दिस साइक्लिक कंपाउंड देखने में कैसा लग रहा है ठीक है ना ओके सो य o एंड देन यू हैव हाइड्रोजन ट्राई टू अंडरस्टैंड दिस इज योर दिस इज योर हेमी एसिट एज वेल क्योंकि ओ है oc3 यहां पे एक कार्बन में एक हाइड्रोजन है यहां पे ch2 है यहां पे o3 है समझ रहे हो सो दिस इज द पार्ट ऑफ हेमी एटल देखो यहां पे oc2 है यहां पे एक ch2 है एक हाइड्रोजन है दिस इज द एल्डिहाइड पार्ट एंड यू हैव और तो और एंड o2 बट दिस इज साइक्लिक बट दैट इज योर हेमी एटल मॉलिक्यूल आई होप सबको समझ में आया राइट सो दिस इज़ योर हेमी एटल साइक्लिक हेमी एटल सो वी कैन राइट डाउन साइक्लिक हेमी एसिट ट्राई टू अंडरस्टैंड द केमिस्ट्री इज वेरी वेरी इंपोर्टेंट सो द सेम केमिस्ट्री वी कैन टू द ग्लूकोज टाइप ऑफ मॉलिक्यूल कार्बोहाइड्रेट मॉलिक्यूल लेट्स आई एम गोइंग टू ड्रॉ दिस इसको सही से प्रैक्टिस कर लेना ठीक है बाबू सही से प्रैक्टिस कर ले ना एंड आई होप दिस विल बी रियली रियली हेल्पफुल फॉर योर एग्जाम पॉइंट ऑफ व्यू इफ यू कैन अंडरस्टैंड दिस केमिस्ट्री व्हाट एवर वी हैव डिस्कस सो फार ठीक है एंड ये जो एसिट और हेमी एसिट फॉर्मेशन होती है ना ये एसिडिक कंडीशन में फीजिबल होती है और फैसा इल होती है रेट फास्ट हो जाती है इन एसिडिक एज वेल एज बेसिक कंडीशन वो भी हम लोग इसके बाद डिस्कस करेंगे सो फर्स्ट ऑफ ऑल इफ यू हैव दिस ब्यूटीफुल मॉलिक्यूल तो सेम मॉलिक्यूल को हम कैसे ड्र कर सकते नंबरिंग कर देते नट 3 4 5 6 ठीक है ऐसे नहीं ड्र कर सकते ना ऐसे कर सकते न टू ्र फोर फ स नहीं सिक्स इधर हो जाएगी सोस क्योंकि यहां प जो हम लोग अटैक करेंगे ना यही अल्कोहल को अटैक करेंगे और वी गोइंग टू गेट द सिक्स मेंबर इन हियर ठीक है तो देखो यहां पे अभी देखो इसका स्टी केमिस्ट्री दैट इज वेरी वेरी इंपोर्टेंट इसको स से प्रैक्टिस कर लेना ये ठीक है य इधर आएगी ठीक है देन ये इधर आ जाएगी ये इधर आ जाएगी तो वो हम लोग ग्लूकोज में डिस्कस करेंगे यहां प हम लोग डिटेल्स में य नहीं करेंगे तो यहां पे जो ओ है वही जाके अटक कर और ये ऐसे करके ओपन हो जाएगी राइट तो यहां प भी देखो हम लोग ना जो फॉर्मेशन कर रहे दिस इज आल्सो योर हेमी एसिट साइक्लिक ग्लूकोज फॉर्मेशन राइट हेमी एसिट हम लोग इसको बोलते हैं राइट दिस इज आल्सो योर हेमी एसिट कंपाउंड राइट सो वी कैन राइट डाउन र ओ ठीक है सो वी हैव दिस नाउ ये जो हम लोग कार्बोन से अल्कोहल में कन्वर्ट कर रहे हैं ना यह दो तरीके से हम लिख सकते हैं ये एक ओज नीचे हो सकती है और एक ओज ना ऊपर भी हो सकती है ऊपर को हम लोग बीटा बोलती है जब नीचे ओज आ जाएगी उसको हम लोग अल्फा बोलेंगे ठीक है इसलिए ना दोनों का जो मिक्सचर होती है ना उसको हम लोग इस बॉन्ड के थ्रू लिख देते हैं बिकॉज दिस इज अ रेसमिक कन ऊपर बीटा और अल्फा दोनों हो सकती है ठीक है नाउ य गइ टू गेट दिस बाकी जो स्टी केमिस्ट है वो रिमन सेम रहेगी क्लियर हो गया सो दैट इज वेरी वेरी इंपोर्टेंट पार्ट सो दिस इज योर साइक्लिक ग्लूकोज ठीक है दिस इज योर साइक्लिक ग्लूकोज सो वी कैन राइट डाउन साइक्लिक ग्लूकोज मॉलिक्यूल ठीक है क्लियर हो गया एंड देन यू हैव नेक्स्ट पार्ट वी कैन राइट डाउन दिस ओ अगर ये फाइ मेंबर हम लोग हेमी एटल की बात करेंगे सो य कैन टेक अनदर मॉलिक्यूल यर ये जो ऑक्सीजन का इसको हम लोग कैसे ड्र कर सकते बाबू देखो वी कैन ड्र इट लाइक दिस वे टूथ टूथ अच्छा फोर फ फोर फ तो हम लोग इस फ अक कर ठीक है ंडर फ मेंबर कट करेंगे तो फ मेंबर नंबर का जो ऑक्सीजन है वक करेंगे ठीक सो मोटिव टो में फ मेंबर नहीं गलत हो गया इसको सही से प्रैक्टिस कर लेना य काफी इंपोर्टेंट है सोस र वन नंबर दिस ू नंबर नंबर फोर नंबर सो टू नंबर में जो अल्कोहल होगी ओ होगी ट विल बी दिस तीन नंबर में जो होगी ट विस चार नंबर में छोड़ो पांच नंबर स2 ठीक है नाउ ये जो ऑक्सीजन का लोन प जा पटक करेगी और इसको ओपन कर देगी क्लियर हो गया तो ऐसे करके हम लोग इजली एक साइक्लिक इसको हम लोग राइबोस भी बोल सकते हैं साइक्लिक इस कार्बन सेंटर को हमने क्रिएट किया है राइट दिस इज योर हेमी एसिट सो आई कैन राइट डाउन र राइबोस मॉलिक्यूल सॉरी एंड आल्सो ये जो हेमी एटल फॉर्मेशन होते हैं वो एसिडिक एज वेल एज बेसिक कंडीशन में फीजिबल होते हैं तो वो पार्ट हम लोग नेक्स्ट वीडियो में डिस्कस करेंगे नेक्स्ट वीडियो में हम लोग कंजू पार्ट को भी इनकॉरपोरेट कर लेंगे |
9994 | https://fixyourenglish.quora.com/What-is-the-difference-between-affluence-and-opulence | What is the difference between affluence and opulence? - Fix your English - Quora
Something went wrong. Wait a moment and try again.
Try again
Skip to content
Skip to search
Sign In
Fix your English
A space for English learners, fix your grammar, vocabulary and spoken English
Follow
·
219.6K 219.6K
Rajagopalan J
Lives in Gurugram, Haryana, India (2010–present) · 5y ·
Rajagopalan J
Writing to improve English and Grammar on Quora. · 7y ·
What is the difference between affluence and opulence?
What is the difference between affluence and opulence?
Both words are synonymous, indicating wealth and richness.
Affluence is more appropriately used to denote individual family wealth or richness in the society, e.g. The Rockfeller family is a very affluent one, or The majority population of so and so region is affluent.
Opulence goes well to indicate the richness or grandeur of something like a mansion or a palace or monument with regard to the architecture or lavish art work, etc, like The Durbar Hall of Rashtrapathi Bhawan looks opulent and beautiful.
4.1K views
·
15 upvotes
·
2 shares
·
7 comments
315 views ·
View upvotes
·
Upvote ·
9 1
About the Author
Rajagopalan J
Happily Retired from Service
Worked at Government of India 1974–2010
Studied at Kumbakonam, Tamil Nadu, India
Lives in Gurugram, Haryana, India 2010–present
8M content views 44.7K this month
Active in 18 Spaces
Knows Tamil
View more in Fix your English
About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press ·
© Quora, Inc. 2025 |
9995 | https://artofproblemsolving.com/wiki/index.php/Spiral_similarity?srsltid=AfmBOooqNdTt70z3LQmLUM7uk4ne_cuaVrcb9_sawaywjxg56DZfF_o8 | Art of Problem Solving
Spiral similarity - AoPS Wiki
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical CampusVisit the Virtual Campus
Sign In
Register
online school
Class ScheduleRecommendationsOlympiad CoursesFree Sessions
books tore
AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates
community
ForumsContestsSearchHelp
resources
math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten
contests on aopsPractice Math ContestsUSABO
newsAoPS BlogWebinars
view all 0
Sign In
Register
AoPS Wiki
ResourcesAops Wiki Spiral similarity
Page
ArticleDiscussionView sourceHistory
Toolbox
Recent changesRandom pageHelpWhat links hereSpecial pages
Search
Spiral similarity
Page made by vladimir.shelomovskii@gmail.com, vvsss
Contents
1 Definition
2 Simple problems
2.1 Explicit spiral similarity
2.2 Hidden spiral symilarity
2.3 Linearity of the spiral symilarity
2.4 Construction of a similar triangle
2.5 Center of the spiral symilarity for similar triangles
2.6 Spiral similarity in rectangle
2.7 Common point for 6 circles
2.8 Three spiral similarities
2.9 Superposition of two spiral similarities
2.10 Spiral similarity for circles
2.11 Remarkable point for spiral similarity
2.12 Remarkable point for pair of similar triangles
2.13 Remarkable point’s problems
2.13.1 Problem 1
2.13.2 Problem 2
2.13.3 Problem 3
2.13.4 Problem 4
2.13.5 Solutions
2.14 Japan Mathematical Olympiad Finals 2018 Q2
Definition
A spiral similarity is a plane transformation composed of a rotation of the plane and a dilation of the plane having the common center. The order in which the composition is taken is not important.
Any two directly similar figures are related either by a translation or by a spiral similarity (directly similar figures are similar and have the same orientation).
The transformation is linear and transforms any given object into an object homothetic to given.
On the complex plane, any spiral similarity can be expressed in the form where is a complex number. The magnitude is the dilation factor of the spiral similarity, and the argument is the angle of rotation.
The spiral similarity is uniquely defined by the images of two distinct points. It is easy to show using the complex plane.
Let with corresponding complex numbers and so For any points and the center of the spiral similarity taking to point is also the center of a spiral similarity taking to This fact explain existance of Miquel point.
Case 1 Any line segment can be mapped into any other using the spiral similarity. Notation is shown on the diagram.
is circle is circle
is any point of is circle is the image under spiral symilarity centered at
is the dilation factor,
is the angle of rotation.
Case 2 Any line segment can be mapped into any other using the spiral similarity. Notation is shown on the diagram. is circle (so circle is tangent to is circle tangent to is any point of is circle is the image under spiral symilarity centered at is the dilation factor,
is the angle of rotation.
Simple problems
Explicit spiral similarity
Given two similar right triangles and Find and
Solution
The spiral similarity centered at with coefficient and the angle of rotation maps point to point and point to point
Therefore this similarity maps to
Hidden spiral symilarity
Let be an isosceles right triangle Let be a point on a circle with diameter The line is symmetrical to with respect to and intersects at Prove that
Proof
Denote Let cross perpendicular to in point at point
Then
Points and are simmetric with respect so
The spiral symilarity centered at with coefficient and the angle of rotation maps to and to point such that
Therefore
Linearity of the spiral symilarity
Points are outside
Prove that the centroids of triangles and are coinsite.
Proof
Let where be the spiral similarity with the rotation angle and
A vector has two parameters, modulo and direction. It is not tied to a center of the spiral similarity. Therefore We use the property of linearity and get Let be the centroid of so
is the centroid of the
Construction of a similar triangle
Let triangle and point on sideline be given. Construct where lies on sideline and lies on sideline
Solution
Let be the spiral symilarity centered at with the dilation factor and rotation angle
so image of any point lies on The spiral symilarity centered at with the dilation factor and rotation angle maps into and therefore the found triangle is the desired one.
Center of the spiral symilarity for similar triangles
Let triangle and point on sideline be given. where lies on sideline and lies on sideline The spiral symilarity maps into Prove
a)
b) Center of is the First Brocard point of triangles and
Proof
a) Let be the spiral symilarity centered at with the dilation factor and rotation angle
Denote Similarly
b) It is well known that the three circumcircles and have the common point (it is in the diagram).
Therefore is cyclic and
Similarly,
Similarly,
Therefore, is the First Brocard point of
is cyclic Similarly,
Therefore is the First Brocard point of and
Therefore the spiral symilarity maps into has the center the angle of the rotation
Spiral similarity in rectangle
Let rectangle be given. Let point
Let points and be the midpoints of segments and respectively.
Prove that
Proof
Let be the midpoint
is a parallelogram
and are corresponding medians of and
There is a spiral similarity centered at with rotation angle that maps to Therefore
Common point for 6 circles
Let and point on sideline be given. where lies on sideline and lies on sideline
Denote
Prove that circumcircles of triangles have the common point.
Proof
so there is the spiral symilarity taking to Denote the center of the center of is the secont crosspoint of circumcircles of and but this center is point so these circles contain point . Similarly for another circles.
Three spiral similarities
Let triangle be given. The triangle is constructed using a spiral similarity of with center , angle of rotation and coefficient
A point is centrally symmetrical to a point with respect to
Prove that the spiral similarity with center , angle of rotation and coefficient taking to
Proof
Corollary
Three spiral similarities centered on the images of the vertices of the given triangle and with rotation angles equal to the angles of take to centrally symmetric to with respect to
Superposition of two spiral similarities
Let be the spiral similarity centered at with angle and coefficient
Let be spiral similarity centered at with angle and coefficient
Let
Prove: a) is the crosspoint of bisectors and
b)
Algebraic proof
We use the complex plane Let Then Geometric proof
Denote
Then
Let be the midpoint be the point on bisector such that be the point on bisector such that Then
is the crosspoint of bisectors and
Corollary
There is another pair of the spiral similarities centered at and with angle coefficients and
In this case
Spiral similarity for circles
Let circle cross circle at points and Point lies on
Spiral similarity centered at maps into
Prove that points and are collinear.
Proof
Arcs
Corollary
Let points and be collinear.
Then exist the spiral similarity centered at such that
Let circle cross circle at points and
Points and lie on
Let be the tangent to be the tangent to
Prove that angle between tangents is equal angle between lines and
Proof
There is the spiral similarity centered at such that Therefore angles between these lines are the same.
Remarkable point for spiral similarity
Circles and centered at points and respectively intersect at points and Points and are collinear. Point is symmetrical to with respect to the midpoint point Prove:
a) b)
Proof
a) cross in midpoint
b) is parallelogram
Denote
Corollary
Let points and be collinear. Then
Therefore is the crosspoint of the bisectors and
Remarkable point for pair of similar triangles
Let
Let the points and be the circumcenters of and
Let point be the midpoint of
The point is symmetric to with respect point
Prove:
a) point be the crosspoint of the bisectors and
b)
Proof
is parallelogram Denote
Similarly,
The statement that was proved in the previous section.
Remarkable point’s problems
Problem 1
Let a convex quadrilateral be given,
Let and be the midpoints of and respectively.
Circumcircles and intersect a second time at point
Prove that points and are concyclic.
Problem 2
Let triangle be given.
Let point lies on sideline
Denote the circumcircle of the as , the circumcircle of the as .
Let be the circumcenter of
Let circle cross sideline at point
Let the circumcircle of the cross at point
Prove that
Problem 3
The circles and are crossed at points and points
Let be the tangent to be the tangent to
Point is symmetric with respect to
Prove that points and are concyclic.
Problem 4
The circles and are crossed at points and points
Let be the tangent to be the tangent to
Points and lye on bisector of the angle
Points and lye on external bisector of the angle
Prove that and bisector are tangent to the circle
bisector is tangent to the circle
Solutions
Solutions are clear from diagrams.
In each case we use remarcable point as the point of bisectors crossing.
Solution 1
We use bisectors and . The points and are concyclic.
Solution 2 We use bisectors of and
Solution 3
We use bisectors of and .
is the circumcenter of
Circle is symmetric with respect diameter
Point is symmetric to with respect diameter
Therefore
Solution 4
Let
Let be midpoint be midpoint .
We need prove that and
Denote
The angle between a chord and a tangent is half the arc belonging to the chord.
is tangent to
is diameter Similarly, is diameter is tangent to
Let be the spiral similarity centered at
Points and are collinear Points and are collinear
Therefore is tangent to
Similarly, is tangent to
Japan Mathematical Olympiad Finals 2018 Q2
Given a scalene let and be points on lines and respectively, so that
Let be the circumcircle of and the reflection of across
Lines and meet again at and respectively.
Prove that and intersect on
Proof
Let be the orthocenter of
Point is symmetrical to point with respect to height
Point is symmetrical to point with respect to height
is centered at
is symmetrical with respect to heightline
is symmetrical to point with respect to height
is symmetrical to point with respect to height
The isosceles triangles
a) are concyclic.
b) is the spiral center that maps to maps to
Therefore are concyclic and are concyclic.
This article is a stub. Help us out by expanding it.
Retrieved from "
Category:
Stubs
Art of Problem Solving is an
ACS WASC Accredited School
aops programs
AoPS Online
Beast Academy
AoPS Academy
About
About AoPS
Our Team
Our History
Jobs
AoPS Blog
Site Info
Terms
Privacy
Contact Us
follow us
Subscribe for news and updates
© 2025 AoPS Incorporated
© 2025 Art of Problem Solving
About Us•Contact Us•Terms•Privacy
Copyright © 2025 Art of Problem Solving
Something appears to not have loaded correctly.
Click to refresh. |
9996 | https://www.simplechemconcepts.com/mole-concept-and-chemical-calculations-difference-between-relative-atomic-mass-relative-molecular-mass-relative-formula-mass-and-molar-mass/ | Skip to primary navigation
Skip to main content
Skip to primary sidebar
O Level Chemistry & IP Chemistry Notes by 10 Year Series Author - Chemistry Specialist
O Level Chemistry Made Easy
Mole Concept and Chemical Calculations: Difference between Relative Atomic Mass, Relative Molecular Mass, Relative Formula Mass and Molar Mass
Many students are confused when it comes to the difference between the following terms commonly used in Mole Concept & Chemical Calculations, namely:
Relative Atomic Mass
Relative Molecular Mass
Relative Formula Mass
Molar Mass
In order to differentiate them (as well as see how they are connected to each other), we have to know their definitions first.
Let’s take a look at them one at a time…..
A) Relative Atomic Mass, Ar
The relative atomic mass of an atom is the average mass of one atom of that element compared to 1/12 of the mass of one carbon-12 atom.
Basically, it is not practical for scientists to use actual masses of atoms in scientific calculations since atoms have very small masses.
As such, scientists compare masses of different atoms with reference to the carbon-12 atom (which is an isotope of carbon).
To be more precise, the masses of all elements (in terms of atom) listed in The Periodic Table are always compared to 1/12 of the mass of one atom of carbon-12.
The symbol for relative atomic mass is Ar.
Relative atomic mass is a ratio and therefore has no unit.
If you refer to the Periodic Table, you will notice that the relative atomic masses of some elements are not whole numbers. These elements exist as a mixture of isotopes (which have different mass numbers).
The relative atomic masses that you see in the Periodic Table are calculated based on the relative percentage abundance of the isotopes. So, relative atomic mass is basically an average value after considering all the different isotopes of the element.
For GCE O-Level Pure Chemistry students (syllabus code 6092) in Singapore, the Periodic Table is not as precise as the one used by the GCE A-Level H2 Chemistry students (syllabus code 9729).
The following Periodic Table is used by GCE O-Level Chemistry students.
Notice that chlorine (atomic number 17) has a relative atomic mass, Ar of 35.5 which is not a whole number. In fact, most elements in the Periodic Table exist as isotopes and thus their relative atomic mass is not a whole number.
If you look at the Periodic Table used by more advanced Chemistry syllabus, such as GCE A-Level H2 Chemistry, most of the elements are not whole numbers. They are presented with 1 decimal place.
Let’s take a look at a classic example on how relative atomic mass is being calculated.
Example:
Chlorine exists as chlorine-35 (% abundance of 75%) and chlorine-37 (% abundance of 25%).
Hence, the relative atomic mass of chlorine = (75/100 x 35) +(25/100 x 37) = 35.5 (as shown in the Periodic Table)
Recap on the definition of Isotopes:
Atoms of the same element
Same number of protons and electrons
Different number of neutrons
A) Relative Molecular Mass, Mr
The relative molecular mass of a molecule is the average mass of one molecule of that element or compound compared to 1/12 of the mass of one carbon-12 atom.
Many elements and compounds exist as covalent molecules.
The mass of a molecule (element or compound) is measured in terms of its relative molecular mass.
The symbol for relative molecular mass is Mr.
Relative molecular mass is also a ratio and therefore has no unit.
To calculate the relative molecular mass of a molecule, you need to add together all the relative atomic masses of all the atoms in its chemical formula.
Due to isotopic effects, actual molecular masses could be different when the atoms of each element present are isotopes.
Let me give you an example.
Carbon is known to exist in two isotopes: carbon-12 and carbon 13.
Oxygen is known to exist in three isotopes: oxygen-16, oxygen-17 and oxygen-18.
As such, we can have carbon dioxide, CO2 molecules with different molecular masses.
Note that for mole calculations, we tend to just use the relative atomic masses (average values) of the elements (as seen in the Periodic Table) and calculate the relative molecular masses (average values).
C) Relative Formula Mass, Mr
The relative formula mass of an ionic compound is the average mass of one unit of that ionic compound compared to 1/12 of the mass of one carbon-12 atom.
Ionic compounds such as sodium chloride and magnesium oxide do no exist as molecules.
As such, it is confusing to use the term relative molecular mass for ionic compounds.
Instead, we use the term relative formula mass.
Relative formula mass is given the same symbol of Mr and have no unit.
It is calculated in the same way as relative molecular mass for molecules, by looking at the chemical formula and adding up the relative atomic masses of atoms.
D) Molar Mass
Molar mass refers to the mass of one mole of a substance (which could be an element or a compound).
The molar mass of an element (in terms of atom) is equal to its relative atomic mass (Ar) in grams.
The molar mass of a molecular substance is equal to its relative molecular mass (Mr) in grams.
The molar mass of an ionic compound is equal to its relative formula mass (Mr) in grams.
Example:
The relative atomic mass of magnesium is 24. This means that one mole of magnesium atoms has a mass of 24 g. The molar mass of magnesium will thus be 24 g/mol.
The relative molecular mass of carbon dioxide is 44. This means that one mole of carbon dioxide molecules has a mass of 44 g. The molar mass of carbon dioxide will thus be 44 g/mol.
In terms of Mole Concepts and Chemical Calculations, do note that there is really “no difference” between Molar Mass and Relative Atomic Mass (Ar) or Relative Molecular/Formula Mass (Mr). When it comes to calculation, you will end up with the same numerical answer.
And for this reason, i have seen educators (teachers and professional tutors) use the two terms interchangeably and causally.
Let me give you an example….
Chemists count particles by using a unit known as the Mole (SI unit is “mol”), which are used to determine chemical formulae of substances when they perform calculations.
One mole of any substance contains 6 x 1023 particles. This is known as Avogadro’s constant or Avogadro’s number.
You will probably come across two similar looking formulae involving Mole and Mass when you are googling and searching online.
A) Mole of Element (mol) = Mass of Element (g) / Relative Atomic Mass of Element
Example: Say we have 24 g of magnesism
Answer: Mole of Mg = (24 g) / (24) = 1 mol
B) Mole of Element (mol) = Mass of Element (g) / Molar Mass of Element (g/mol)
Example: Say we have 24 g of magnesism
Answer: Mole of Mg = (24 g) / (24 g/mol) = 1 mol
As you can see, based on SI unit, Formula B is the more precise one because the units will cancel off nicely to give you “mol”.
However, you can also see that both formulae will give you the same answers of “1 mol”.
It also sometimes depends on the syllabus content given by their country’s Ministry of Education. School teachers tend to follow very closely. As a professional Chemistry specialist tutor in Singapore, i also follow it closely, so that i don’t have to argue with students who are attending my GCE O-Level and IP Pure Chemistry Weekly Tuition classes, whenever they say that the formula they saw in their school is slightly different.
Let me show you what i mean.
“Chemistry Matters” Textbook by Marshall Cavendish Education publisher is one of the two approved Pure Chemistry textbooks by Ministry of Education Singapore (MOE) which all Sec 3 and 4 Chemistry students have to get as a reference guide.
The formula to calculate the number of moles of substances is slightly different in 1st Edition (before year 2013) and in 2nd/3rd Edition (2013 onwards).
Formula A is printed in the 1st Edition (for use before Year 2013).
Formula B is printed in the 2nd Edition (for use from 2013-2017). The newest 3rd edition (2014 onwards) also follow this version.
So for all my customised Chemistry notes, YouTube videos and chemistry blogs before 2013, i have been using Formula A. I started to use Formula B from 2013 onwards.
You can see that i was using Formula A (use of relative molecular mass, Mr instead of Molar Mass) in one of my Chemistry YouTube Videos produced in April 2009 and has about 15,000 views. Wow, time really flies, this video is already 11 years old!
Hope the above is clear and you know the differences as well as the connection between the four common terms used in Mole Concept and Chemical Calculations.
Enjoy learning Chemistry with understanding! Get it right the 1st time!
If you have any questions, leave me a comment below. Feel free to share this blog post with your friends.
Subscribe to my blog to receive 2 updates per month sent to your email!
PS: Under related articles below, there are several blog post discussions and questions related to Mole Concept and Chemical Calculations. You can also do a keyword search using the search box at the top right hand corner.
Related Articles:
O Level Chemistry – Isotopes Part 2
O Level Chemistry – Atomic Structure
O Level Chemistry: Difference between Atom, Molecule & Particle
O Level Chemistry: Mole Concepts / Chemical Calculations
O Level Chemistry: Mole Concepts / Chemical Calculations
Share:
FacebookWhatsAppEmailLinkedIn
Reader Interactions
Comments
Sunita shah says
Thank you? it is really helpful to me.
2. John Schatz says
I might pick your brain from time to time.
I’m taking Chemistry 11. I’m from Vancouver, BC, Canada.
Leave a Reply |
9997 | https://www.isinj.com/mt-usamo/Applied%20Combinatorics%20(6th%20Edition)%20by%20Alan%20Tucker%20Wiley%20(2012).pdf | This page is intentionally left blank www.itpub.net APPLIED COMBINATORICS This page is intentionally left blank www.itpub.net APPLIED COMBINATORICS ALAN TUCKER SUNY Stony Brook John Wiley & Sons, Inc.
VP AND PUBLISHER Laurie Rosatone PROJECT EDITOR Shannon Corliss MARKETING MANAGER Debi Doyle MARKETING ASSISTANT Patrick Flatley PRODUCTION MANAGER Janis Soo ASSISTANT PRODUCTION EDITOR Elaine S. Chew COVER ILLUSTRATOR & DESIGNER Seng Ping Ngieng This book was set in Times by Aptara, Inc. and printed and bound by Courier Westford, Inc. The cover was printed by Courier Westford, Inc.
This book is printed on acid free paper. ∞ ⃝ Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship.
Copyright C ⃝2012, 2007, 2002, 1995 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com.
Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201)748-6008, website
Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return mailing label are available at www.wiley.com/go/returnlabel. If you have chosen to adopt this textbook for use in your course, please accept this book as your complimentary desk copy. Outside of the United States, please contact your local sales representative.
Library of Congress Cataloging-in-Publication Data Tucker, Alan, 1943 July 6-Applied combinatorics / Alan Tucker. — 6th ed.
p. cm.
Includes bibliographical references and index.
ISBN 978-0-470-45838-9 (acid free paper) 1. Combinatorial analysis. 2. Graph theory. I. Title.
QA164.T83 2012 511′.6—dc23 2011044318 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 www.itpub.net PREFACE Combinatorial reasoning underlies all analysis of computer systems. It plays a similar role in discrete operations research problems and in finite probability. Two of the most basic mathematical aspects of computer science concern the speed and logical structure of a computer program. Speed involves enumeration of the number of times each step in a program can be performed. Logical structure involves flow charts, a form of graphs. Analysis of the speed and logical structure of operations research algorithms to optimize efficient manufacturing or garbage collection entails similar combinatorial mathematics. Determining the probability that one of a certain subset of equally likely outcomes occurs requires counting the size of the subset.
Such combinatorial probability is the basis of many nonparametric statistical tests.
Thus,enumerationandgraphtheoryareusedpervasivelythroughoutthemathematical sciences.
This book teaches students how to reason and model combinatorially. It seeks to develop proficiency in basic discrete math problem solving in the way that a calculus textbook develops proficiency in basic analysis problem solving.
The three principal aspects of combinatorial reasoning emphasized in this book are the systematic analysis of different possibilities, the exploration of the logical structure of a problem (e.g., finding manageable subpieces or first solving the problem with three objects instead of n), and ingenuity. Although important uses of combina-torics in computer science, operations research, and finite probability are mentioned, these applications are often used solely for motivation. Numerical examples involving the same concepts use more interesting settings such as poker probabilities or logical games.
Theory is always first motivated by examples, and proofs are given only when their reasoning is needed to solve applied problems. Elsewhere, results are stated without proof, such as the form of solutions to various recurrence relations, and then applied in problem solving. Occasionally, a few theorems are stated simply to give students a flavor of what the theory in certain areas is like.
For decades, collegiate curriculum recommendations from the Mathematical Association of America have included combinatorial problem solving as an impor-tant component of training in the mathematical sciences. Combinatorial problem solving underlies a wide spectrum of important subjects in the computer science curriculum. Indeed, it is expected that most students in a course using this book will be computer science majors. For both mathematics majors and computer science majors, v vi Preface this author believes that general reasoning skills stressed here are more important than mastering a variety of definitions and techniques.
This book is designed for use by students with a wide range of ability and maturity (sophomores through beginning graduate students). The stronger the students, the harder the exercises that can be assigned. The book can be used for a one-quarter, two-quarter, or one-semester course depending on how much material is used. It may also be used for a one-quarter course in applied graph theory or a one-semester or one-quarter course in enumerative combinatorics (starting from Chapter 5). A typical one-semester undergraduate discrete methods course should cover most of Chapters 1 to 3 and 5 to 8, with selected topics from other chapters if time permits.
Instructors are strongly encouraged to obtain a copy of the instructor’s guide accompanying this book. The guide has an extensive discussion of common student misconceptions about particular topics, hints about successful teaching styles for this course, and sample course outlines (weekly assignments, tests, etc.).
The sixth edition of this book draws upon features from all the earlier editions.
For example, the game of Mastermind that appeared at the beginning of the first edition has been brought back, and a closing Postlude about cryptanalysis has been added. The suggested solutions to selected enumeration exercises from the second and third editions have returned. Of course, there are also new exercises. Also, the numbers were changed in many of the old exercises in the counting chapters (to guard against student groups accumulating old solution sets).
Many people gave useful comments about early drafts and the first edition of this text; Jim Frauenthal and Doug West were especially helpful. The idea for this book is traceable to a combinatorics course taught by George Dantzig and George Polya at Stanford in 1969, a course for which I was the grader. Many instructors who have used earlier editions of this book have supplied me with valuable feedback and suggestions that have, I hope, made this edition better. I gratefully acknowledge my debt to them. Ultimately, my interest in combinatorial mathematics and in its effective teaching rests squarely on the shoulders of my father, A. W. Tucker, who had long sought to give finite mathematics a greater role in mathematics as well as in the undergraduate mathematics curriculum. Finally, special thanks go to former students of my combinatorial mathematics courses at Stony Brook. It was they who taught me how to teach this subject.
Alan Tucker Stony Brook, New York www.itpub.net CONTENTS PRELUDE xi PART ONE GRAPH THEORY 1 CHAPTER 1 ELEMENTS OF GRAPH THEORY 3 1.1 Graph Models 3 1.2 Isomorphism 14 1.3 Edge Counting 24 1.4 Planar Graphs 31 1.5 Summary and References 44 Supplementary Exercises 45 CHAPTER 2 COVERING CIRCUITS AND GRAPH COLORING 49 2.1 Euler Cycles 49 2.2 Hamilton Circuits 56 2.3 Graph Coloring 68 2.4 Coloring Theorems 77 2.5 Summary and References 86 Supplement: Graph Model for Instant Insanity 87 Supplement Exercises 92 CHAPTER 3 TREES AND SEARCHING 93 3.1 Properties of Trees 93 3.2 Search Trees and Spanning Trees 103 3.3 The Traveling Salesperson Problem 113 vii viii Contents 3.4 Tree Analysis of Sorting Algorithms 121 3.5 Summary and References 125 CHAPTER 4 NETWORK ALGORITHMS 127 4.1 Shortest Paths 127 4.2 Minimum Spanning Trees 131 4.3 Network Flows 135 4.4 Algorithmic Matching 153 4.5 The Transportation Problem 164 4.6 Summary and References 174 PART TWO ENUMERATION 177 CHAPTER 5 GENERAL COUNTING METHODS FOR ARRANGEMENTS AND SELECTIONS 179 5.1 Two Basic Counting Principles 179 5.2 Simple Arrangements and Selections 189 5.3 Arrangements and Selections with Repetitions 206 5.4 Distributions 214 5.5 Binomial Identities 226 5.6 Summary and References 236 Supplement: Selected Solutions to Problems in Chapter 5 237 CHAPTER 6 GENERATING FUNCTIONS 249 6.1 Generating Function Models 249 6.2 Calculating Coefficients of Generating Functions 256 6.3 Partitions 266 6.4 Exponential Generating Functions 271 6.5 A Summation Method 277 6.6 Summary and References 281 CHAPTER 7 RECURRENCE RELATIONS 283 7.1 Recurrence Relation Models 283 7.2 Divide-and-Conquer Relations 296 7.3 Solution of Linear Recurrence Relations 300 7.4 Solution of Inhomogeneous Recurrence Relations 304 www.itpub.net Contents ix 7.5 Solutions with Generating Functions 308 7.6 Summary and References 316 CHAPTER 8 INCLUSION–EXCLUSION 319 8.1 Counting with Venn Diagrams 319 8.2 Inclusion–Exclusion Formula 328 8.3 Restricted Positions and Rook Polynomials 340 8.4 Summary and Reference 351 PART THREE ADDITIONAL TOPICS 353 CHAPTER 9 POLYA’S ENUMERATION FORMULA 355 9.1 Equivalence and Symmetry Groups 355 9.2 Burnside’s Theorem 363 9.3 The Cycle Index 369 9.4 Polya’s Formula 375 9.5 Summary and References 382 CHAPTER 10 GAMES WITH GRAPHS 385 10.1 Progressively Finite Games 385 10.2 Nim-Type Games 393 10.3 Summary and References 400 POSTLUDE 401 APPENDIX 415 A.1 Set Theory 415 A.2 Mathematical Induction 420 A.3 A Little Probability 423 A.4 The Pigeonhole Principle 427 A.5 Computational Complexity and NP-Completeness 430 GLOSSARY OF COUNTING AND GRAPH THEORY TERMS 435 BIBLIOGRAPHY 439 SOLUTIONS TO ODD-NUMBERED PROBLEMS 441 INDEX 475 This page is intentionally left blank www.itpub.net PRELUDE This book seeks to develop facility at combinatorial reasoning, which is the basis for analyzing a wide range of problems in computer science and discrete applied mathematics. As a warm-up exercise for such reasoning, this Prelude presents the game of Mastermind. Mastermind was introduced in the 1970s and attained such popularity in England that in 1975 a British National Mastermind Championship was held with overflow crowds. Masterminduses the same type of combinatorialreasoning that underlies the mathematics in this book but uses it in a recreational setting.
The objective of the game is to guess a secret code consisting of colored pegs.
The secret code is a row of four pegs that may be chosen (with repeats) from the colors red (R), white (W), yellow (Y), green (G), blue (Bu), and black (Bk). Each guess of a possible secret code is scored to give some information about how close the guess is to the real secret code. Specifically, the player who chose the secret code indicates (1) how many of the code pegs in the guess are both of the right color and in the right position in the row, and (2) how many of the code pegs are of the right color (occur somewhere in the secret code) but in the wrong position. These two pieces of information are recorded in the form of black keys for (1) and white keys for (2). The game ends when the secret code has been correctly guessed, that is, a score of four black keys is given to a guess. The guesses in Example 1 indicate how this scoring procedure works.
Example 1: Mastermind Scoring Secret Code R Bu Y Y Scoring Comments Guess 1 Bu W G Y •o A peg can receive at most one key; so the Y earns one black key.
Guess 2 Y G G Y •o Guess 3 Bk G G W A null score is actually very helpful; it eliminates all three colors.
Guess 4 Bu Bu R R •o There is only one blue peg in the secret code, and so only one blue peg earns a key; it gets the best key possible, black.
xi xii Prelude This game has many good associated counting problems. How many secret codes are possible, how many different sets of black and white scoring keys are possible, how many secret codes are possible given a particular first guess and its scoring? This game is conceptually similar to several important problems in information classification and retrieval. In computer recognition of human speech, chemical compounds, or other complex data sets, a number of cleverly planned queries must be made about the data.
Compilers perform a variety of sequential tests, usually in the form of binary search trees (discussed in Chapter 3), to identify commands in the text they are compiling.
Many of the theoretical analyses associated with efficient recognition require exactly the same reasoning and techniques as the Mastermind counting problems.
Although we justify our discussion of Mastermind in terms of related counting problems, the game itself provides enjoyable recreation and we encourage readers to play it with a friend. In the absence of a playing companion, the exercises in this section present games in which enough guesses have been made and scored to enable one to determine the secret code. The following example illustrates how these exercises can be analyzed.
Example 2: Find a Secret Code The following guesses and scoring have been occurred. Readers should try to determine the secret code for themselves before reading the analysis below.
Scoring Guess 1 W G Bu R •o Guess 2 Bk Y R Bu •o Guess 3 R R G Y ooo Guess 4 Y R R W •• Consider color red. Guess 3 with two reds and three scoring keys indicates that there must be at least one red peg in the secret code; but since the keys are white, the red(s) must be in position 3 or 4. However, guess 4 has only black keys and the red pegs are in positions 2 and 3. Since guess 3 indicates that there cannot be a red peg in position 2, we conclude that the secret code has exactly one red peg in position 3 and no other red peg (or else the red peg in position 2 in Guess 1 would earn a white key). Note that the black key in guess 2 must also be due to the red in position 3.
Guess 3 now implies that the green and yellow pegs must have earned two of the three white keys (since one of the reds received no key). Since yellow is somewhere in the secret code, then guess 4 tells us that yellow must be in position 1. So the secret code is of the form Y R . Moreover, yellow cannot also occur in position 2 or 4, because it appears there in guesses 2 and 3, respectively, without earning a black key. So there is only one Y, as well as just one R, in the secret code. Since the keys www.itpub.net Prelude xiii in guesses 2 and 4 are now known to be for yellow and red, the other colors in these two guesses—black, blue, and white—cannot be in the secret code. Then the only remaining color that can appear in positions 2 and 4 is green, and the secret code is Y G R G.
More information about strategies for playing Mastermind can be found in The Official Mastermind Handbook by Ault .
EXERCISES 1. Determine the secret code for G Y R Bu oooo G Bu R Y ••oo 2. Determine the two possible secret codes (one has no repeated colors) for Bk Bu Y W oo R W Bu G •oo Bu G R G oo Y R W Bu •oo 3. Determine the secret code for Bu Bu Bk W ••o Y W Bk W •• W Bu R G •o Y Bu Bk R •• 4. Determine the two possible secret codes for R G Y W oo Y Bk W Bu •o G R R G •o Bk Bu G R •o W R Bk R o 5. Find a fourth guess whose scoring will allow you to determine the secret code for G Y Bk R •• Y Bu G W •o Bu W Y Y o 6. Find a fourth guess whose scoring will allow you to determine the secret code for G R Bu W • G Y Bu G • Y Bk Bk Y o xiv Prelude 7. A seventh color, orange (O), has been added in this game. Determine the secret code for O R Bu G oo Y O W Y o R Bk Bu G ooo O R R Y •o Bk Bu O R oo 8. A seventh color, orange (O), has been added. Determine the two possible secret codes for O R G Bu oo Y Y W O •o R Bu Bu Bk •o Bk O Y R •o Bu Bk O Y •oo 9. A seventh color, orange (O), has been added. Determine the seven possible secret codes for O Y W Bk o W Bk O R •o G O R R • Y R Bu O oo 10. There are now five positions and two extra colors: orange (O) and pink (P).
Determine the secret code for Bk G G O O ••o W O Bk G R ••o O Bk P Y Bu •o G P Y R W oo 11. There are now five positions and three extra colors: orange (O) pink (P), and violet (V). Determine the secret code for R V Bk G O oo V R V R Y •o Bk W W Bu Bk • P O P O G oo G Y Bu W P •oo W P V Bu Y •oo Y Bu R V Bk ooo 12. Find the probability that your initial guess in a Mastermind game is correct.
13. (a) What sets of four or fewer black and white keys can never occur in Master-mind scores?
(b) How many different sets of black and white keys can occur in Mastermind scores?
www.itpub.net Prelude xv 14. We call two guesses in Mastermind similar if one can be obtained from the other by a permutation of positions and/or a permutation of colors. How many different (nonsimilar) guesses are there?
15. Suppose your first guess uses four different colors and the score is four white keys. How many different secret codes are possible? (Note that all four-color guesses are similar; see Exercise 14.) 16. Suppose your first guess uses three different colors (one color is repeated) and its score is one black and three whites. How many different secret codes are possible? (Note that all three-color guesses are similar; see Exercise 14.) 17. Suppose your first guess uses at least two colors and its score is no keys. What is the minimum number of secret codes that are eliminated by this guess?
18. Consider the simplified Mastermind game in which there are four pegs, each of a different color, and the secret code consists of some arrangement of these four pegs. Develop a complete strategy for playing this game so that you can determine the secret code after at most three guesses (by the fourth guess, you get a score of four black keys).
19. Consider the simplified Mastermind game in which there are three positions and three colors of pegs. Find an optimal first guess for this game. A guess is optimal if for some number k, all possible scores of the guess leave at most k possible secret codes and some possible score of any other guess leaves at least k possible secret codes.
20. Consider the simplified Mastermind game in which there are four positions but only two colors of pegs. Find an optimal first guess for this game (see Exercise 19).
21. Show that no matter what the scores of these four guesses, any secret code can be correctly guessed (using the scores of these first four guesses) in at most two more guesses.
R R W W R Bk R Bk Bu Bu G G Bu Y Bu Y 22. Write a computer program to make up secret codes and score guesses.
23. Write a program to play Mastermind.
REFERENCE 1. L. Ault, The Official Mastermind Handbook, Signet Press, New York, 1976.
This page is intentionally left blank www.itpub.net PART ONE GRAPH THEORY This page is intentionally left blank www.itpub.net CHAPTER 1 ELEMENTS OF GRAPH THEORY 1.1 GRAPH MODELS The first four chapters of this book deal with graphs and their applications. A graph G = (V, E) consists of a finite set V of vertices and a set E of edges joining dif-ferent pairs of distinct vertices.∗Figure 1.1a shows a depiction of a graph with V = {a, b, c, d} and E = {(a, b), (a, c), (a, d), (b, d), (c, d)}. We represent vertices with points and edges, and lines joining the prescribed pairs of vertices. This defini-tion of a graph does not allow two edges to join the same two vertices. Also, an edge cannot “loop” so that both ends terminate at the same vertex—an edge’s end vertices must be distinct. The two ends of an undirected edge can be written in either order, (b, c) or (c, b). We say that vertices a and b are adjacent when there is an edge (a, b).
Sometimes the edges are ordered pairs of vertices, called directed edges. In a directed graph, all edges are directed. See the directed graph in Figure 1.1b. We write (b⃗ , c) to denote a directed edge from b to c. In a directed graph, we allow one edge in each direction between a pair of vertices. See edges (a⃗ , c) and (c⃗ , a) in Figure 1.1b.
The combinatorial reasoning required in graph theory, and later in the enumer-ation part of this book, involves different types of analysis than are used in calculus and high school mathematics. There are few general rules or formulas for solving these problems. Instead, each question usually requires its own particular analysis.
This analysis sometimes calls for clever model-building or creative thinking, but more often consists of breaking the problem into many cases (and subcases) that are easy enough to solve using simple logic or basic counting rules. A related line of reasoning is to solve a special case of the given problem and then to find ways to extend that a c b d a b d c (a) (b) Figure 1.1 ∗What this book calls a graph is referred to in many graph theory books as a simple graph. In general, graph theory terminology varies a little from book to book.
3 4 Chapter 1 Elements of Graph Theory reasoning to all the other cases that may arise. The underlying theme here is summa-rized by a famous quote from the great problem-solver George Polya: “The challeng-ing part is asking the right questions. Then the answers are easy.” In graph theory, combinatorial arguments are made a little easier by the use of pictures of the graphs. For example, a case-by-case argument is much easier to construct when one can draw a graphical depiction of each case.
Graphs have proven to be an extremely useful tool for analyzing situations in-volving a set of elements in which various pairs of elements are related by some property. The most obvious examples of graphs are sets with physical links, such as electrical networks, where electrical components (transistors) are the vertices and connecting wires are the edges; or telephone communication systems, where tele-phones and switching centers are the vertices and telephone lines are the edges. Road maps, oil pipelines, and subway systems are other examples.
Another natural form of graphs is sets with logical or hierarchical sequenc-ing, such as computer flowcharts, where the instructions are the vertices and the logical flow from one instruction to possible successor instruction(s) defines the edges; or an organizational chart, where the people are the vertices and if person A is the immediate superior of person B, there is an edge (A⃗ , B). Computer data struc-tures, evolutionary trees in biology, and the scheduling of tasks in a complex project are other examples.
The emphasis in this book will be on problem solving, with problems about general graphs and applied graph models. Observe that we will usually not have any numbers to work with, only some vertices and edges. At first, this may seem to be highly nonmathematical. It is certainly very different from the mathematics that one learns in high school or in calculus courses. However, disciplines such as computer science and operations research contain as much graph theory as they do standard numerical mathematics.
This section consists of a collection of illustrative examples about graphs. We will solve each problem from scratch with a little logic and systematic analysis. Many of these examples will be revisited in greater depth in subsequent chapters.
The following three graph theory terms are used in the coming examples. A path P is a sequence of distinct vertices, written P = x1–x2– · · · –xn, with each pair of consecutive vertices in P joined by an edge. If in addition there is an edge (xn, x1), the sequence is called a circuit, written x1–x2– · · · –xn–x1. For example, in Figure 1.1a, b-d-a-c forms a path, while a-b-d-c-a forms a circuit. A graph is connected if there is a path between every pair of vertices. The removal of certain edges or vertices from a connected graph G is said to disconnect the graph if the resulting graph is no longer connected—that is, if at least one pair of vertices is no longer joined by a path. The graph in Figure 1.1a is connected, but the removal of edges (a, b) and (b, d) will disconnect it.
Example 1: Matching Suppose that we have five people A, B, C, D, E and five jobs a, b, c, d, e, and that various people are qualified for various jobs. The problem is to find a feasible one-to-one matching of people to jobs, or to show that no such matching can exist. We www.itpub.net 1.1 Graph Models 5 A B C D E a b c d e Figure 1.2 can represent this situation by a graph with vertices for each person and for each job, with edges joining people with jobs for which they are qualified. Does there exist a feasible matching of people to jobs for the graph in Figure 1.2?
The answer is no. The reason can be found by considering people A, B, and D.
These three people as a set are collectively qualified for only two jobs, c and d. Hence there is no feasible matching possible for these three people, much less all five people.
An algorithm for finding a feasible matching, if any exists, will be presented in Chapter 4. Such matching graphs in which all the edges go horizontally between two sets of vertices are called bipartite. Bipartite graphs are discussed further in Section 1.3.
Example 2: Spelling Checker A spelling checker looks at each word X (represented in a computer as a binary number) in a document and tries to match X with some word in its dictionary, which typically contains close to 100,000 words. To understand how this checking works, we consider the simplified problem of matching an unknown letter X with one of the 26 letters in the English alphabet. In the spirit of the strategy humans use to home in on the page in a dictionary where a given word appears, the computer search procedure would first compare the unknown letter X with M, to determine whether X ≤M or X > M. The answer to this comparison locates X in the first 13 letters of the alphabet or the second 13 letters, thus cutting the number of possible letters for X in half.
This strategy of cutting the possible matches in half can be continued with as many comparisons as needed to home in on X’s letter. For example, if X ≤M, then we could test whether or not X ≤G; if X > M, we could test whether X ≤S.
This testing procedure is naturally represented by a directed graph called a tree.
Figure 1.3 shows the first three rounds of comparisons for the letter-matching pro-cedure. The vertices represent the different letters used in the comparisons. The left descending edge from a vertex Q points to the letter for the next comparison if X ≤Q, and the right descending edge from Q points to the next letter if X > Q.
For our original spelling-checker problem, a word processor would use a similar, but larger, tree of comparisons. With just 12 rounds of comparisons, it could reduce M G S W P J D Figure 1.3 6 Chapter 1 Elements of Graph Theory the number of possible matches for an unknown word X from 100,000 down to 25, about the number of words in a column of a page in a dictionary. (Once a list was reduced to about 25 possibilities, a computer search for X would usually run linearly down that list, just as a human would.) Chapter 3 examines trees and their use in various search problems. Trees can be characterized as graphs that are connected and that have a unique path between any pair of vertices (ignoring the directions of directed edges). The next example uses trees in a very different way.
Example 3: Network Reliability Suppose the graph in Figure 1.4 represents a network of telephone lines (or electrical transmission lines). We are interested in the network’s vulnerability to accidental disruption. We want to identify sets of those lines and switching centers that must stay in service to avoid disconnecting the network.
There is no telephone line (edge) whose removal will disconnect the telephone network (graph). Similarly, there is no vertex whose removal disconnects the graph.
Is there any pair of edges whose removal disconnects the graph? There are several such pairs. For example, we see that if the two edges incident to a are removed, vertex a is isolated from the rest of the network. A more interesting disconnecting pair of edges is (b, c), (j, k). It is left to the reader as an exercise to find all disconnecting sets consisting of two edges for the graph in Figure 1.4.
Let us take a different tack. Suppose we want to find a minimal set of edges needed to link together the 11 vertices in Figure 1.4. There are several possible minimal connecting sets of edges. By inspection, we find the following one: (a, b), (b, c), (c, d), (d, h), (h, g), (h, k), (k, j), ( j, f ), ( j, i), (i, e); the edges in this minimal connecting set are darkened in Figure 1.4. A minimal connecting set will always be a tree. One interesting general result about these sets is that if the graph G has n vertices, then a minimal connecting set for G (if any exists) always has n −1 edges.
The number of edges incident to a vertex is called the degree of the vertex.
Example 4: Street Surveillance Now suppose the graph in Figure 1.4 represents a section of a city’s street map. We want to position police officers at corners (vertices) so that they can keep every block (edge) under surveillance—that is, every edge should have police officers at (at least) one of its end vertices. What is the smallest number of police officers that can do this job?
Let us try to get a lower bound on the number of police officers needed. The map has 14 blocks (edges). Corners b, c, e, f, h, and j each have degree 3, and corners a, d, g, i, and k each have degree 2. Since four vertices can be incident to at most 4 × 3 = 12 edges but there are 14 edges in all, we will need at least five police officers. We shall now try to find a set of five vertices incident to all the edges. If we can find such a set, we know that it is the best (smallest) solution possible.
If all five police officers were positioned at degree-3 vertices, then 5 × 3 = 15 edges are watched by the five police officers. Since there are only 14 edges, some www.itpub.net 1.1 Graph Models 7 a e i f b j g c d h k Figure 1.4 edge would be covered by police officers at both end vertices. If four police officers are at degree-3 vertices and one at a degree-2 vertex, then exactly 14 edges are watched—and no edge need be covered at both ends. (If fewer than four of the five police officers are at degree-3 vertices, we could not watch all 14 edges). With these general observations, we are ready for a systematic analysis to try to find five vertices that are collectively adjacent to all 14 edges.
Consider edge (c, d). Suppose it is watched by an officer at vertex d. Then vertex c (the other end vertex of edge (c, d)) cannot also have an officer, since we noted above that if we use a degree-2 vertex, such as d, then no edge can be watched from both end vertices. However, if vertex c cannot be used, then edge (c, g) must be watched from its other end vertex g. But now we are using two degree-2 vertices, d and g. We noted above that at most one degree-2 vertex can be used. We got into this trouble by assuming that edge (c, d) is watched from vertex d.
Now assume no officer is at vertex d. Then we must watch edge (c, d) with an officer at vertex c. Similarly, edge (d, h) can be watched only by placing an officer at vertex h. Next look at edge (h, k). It is already watched by vertex h. Then we assert that (h, k) cannot also be watched by an officer at vertex k, since k has degree-2 and we noted above that if we use a degree-2 vertex, no edge can be watched from both ends. We conclude that there cannot be an officer at vertex k. Then edge (k, j) can be watched only by placing an officer at vertex j. We now have officers required to be at vertices c, h, and j.
Similar reasoning shows that with an officer at vertex j, there cannot be an officer at vertex i; then there must be an officer at vertex e; then there cannot be an officer at vertex a; and then there must be an officer at vertex b. In sum, we have shown that we should place police officers at vertices c, h, j, e, and b. A check confirms that these five vertices do indeed watch all 14 edges. A smallest number of police officers for this surveillance problem has been found. Note that since our reasoning forced us to use exactly these five vertices, no other set of five vertices can work.
At the beginning of this example, we showed that at least five corners were needed to keep all the blocks (edges) under surveillance. Now we have produced a set of five corners that achieve such surveillance. It then follows that five is the minimum number of corners.
We conclude this example by noting that in this surveillance situation, one can also consider watching the vertices rather than the edges: How few officers are needed to watch, that is, be at or adjacent to, all the vertices? We use the same type of argument as in the block surveillance problem to get a lower bound on the number of corners needed for corner surveillance. An officer at vertex x is considered to be watching vertex x and all vertices adjacent to x. There are 11 vertices, and six of these 8 Chapter 1 Elements of Graph Theory vertices watch four vertices (themselves and three adjacent vertices). Thus three is the theoretical minimum. This minimum can be achieved. Details are left as an exercise.
A set C of vertices in a graph G with the property that every edge of G is incident to at least one vertex in C is called an edge cover. The previous example was asking for an edge cover of minimal size in Figure 1.4. The reasoning in Example 4 illustrates the kind of systematic case-by-case analysis that is common in graph theory.
The analysis in the previous example also illustrates a principle that is used over and over again in graph theory and other combinatorial settings. Namely, to show a graph has some property—in this case, the existence of a five-vertex edge cover—we assume that the property exists and deduce useful consequences of this assumption.
The key consequence for the graph in Figure 1.4 was as follows: (∗) if an edge (x, y) links a 3-degree vertex x with a 2-degree vertex y then at most one of x and y can be used in a five-vertex edge cover A subsequent consequence of () concerning the pair x, y is that if we want to use vertex x (and not y) in a minimal edge cover to cover (x, y), then to cover the other edge at y—call it (y, z)—vertex z would also have to be in the minimal edge cover.
We give the mnemonic name Assumptions generate helpful Consequences— the AC Principle, for short—to this strategy of assuming that a graph has a desired property in order to deduce useful consequences, consequences we use to help us show that the graph indeed has this property. The AC Principle can also be used to show that a graph does not have some property: to do so, we deduce consequences under the assumption that the graph does have the property, and then show that these consequences lead to a contradiction.
Example 5: Scheduling Meetings Consider the following scheduling problem. A state legislature has many committees that meet for one hour each week. One wants a schedule of committee meeting times that minimizes the total number of hours of meetings—but such that two committees with overlapping membership cannot meet at the same time.
This situation can be modeled with a graph in which we create a vertex for each committee and join two vertices by an edge if they represent committees with overlapping membership. Suppose that the graph in Figure 1.4 now represents the membership overlap of 11 legislative committees. For example, vertex c’s edges to vertices b, d, and g in Figure 1.4 indicate that committee c has overlapping members with committees b, d, and g.
A set of committees can all meet at the same time if there are no edges between the corresponding set of vertices. A set of vertices without an edge between any two is called an independent set of vertices. Our scheduling problem can now be restated as seeking a minimum number of independent sets that collectively include all vertices.
This problem is discussed in depth in Section 2.3.
How many committees can meet at one time? We are asking the following graph question: What is the largest independent set of the graph? It is very hard in general www.itpub.net 1.1 Graph Models 9 a e i f b j g c d h k Figure 1.4 to find the largest independent set in a graph. For the graph in Figure 1.4, a little examination shows that there is one independent set of size 6, a, d, f, g, i, k. All other independent sets have five or fewer vertices.
One goal of graph theory is to find useful relationships between seemingly unre-lated graph concepts that arise from different settings. Now we show that independent sets are closely related to edge covers. If V is the set of vertices in a graph G, then I will be an independent set of vertices if and only if V −I is an edge cover! Why?
Because if there are no edges between two vertices in I, then every edge involves (at least) one vertex not in I—that is, a vertex in V −I. Conversely, if C is an edge cover so that all edges have at least one end vertex in C, then there is no edge joining two vertices in V −C. So V −C is an independent set. Check that in Figure 1.4, the vertices not in the independent set a, d, f, g, i, k form edge cover b, c, e, h, j.
A consequence of this relationship is that if I is an independent set of largest possible size in a graph, then V −I will be an edge cover of smallest possible size.
So finding a maximal independent set is equivalent to finding a minimal edge cover.
We next give an example involving directed graphs.
Example 6: Influence Model Suppose psychological studies of a group of people determine which members of the group can influence the thinking of others in the group. We can make a graph with a vertex for each person and a directed edge (p1⃗ , p2) whenever person p1 influences p2. Let the graph in Figure 1.5a represent a set of such influences. Now let us ask for a minimal subset of people who can spread an idea through to the whole group, either directly or by influencing someone who will influence someone else, and so forth. In graph-theoretic terms, we want a minimal subset of vertices with directed paths to all other vertices (a directed path from p1 to pk is an edge b a c d e f g b a d c e f g (a) (b) Figure 1.5 10 Chapter 1 Elements of Graph Theory sequence (p1⃗ , p2), (p2⃗ , p3) . . . (pk−1⃗ , pk)). Such a subset of influential vertices is called a vertex basis.
To aid us, we can build a directed-path graph for the original graph with the same vertex set and with a directed edge (pi⃗ , p j) if there is a directed path from pi to p j in the original graph. Figure 1.5b shows the directed-path graph for the graph in Figure 1.5a. Now our original problem can be restated as follows: Find a minimal subset of vertices in the new graph with edges directed to all other vertices. This is just a directed-graph version of the vertex-covering problem mentioned at the end of Example 4. Observe that any vertex in Figure 1.5b with no incoming edges must be in this minimal subset (since no other vertices have edges to it); vertex b is such a vertex. Since b has edges to a, c, and d, then e, f, and g are all that remain to be “influenced.” Either e, f, or g “influence” these three vertices. Then b, e, or b, f, or b, g are the desired minimal subsets of vertices.
1.1 EXERCISES S u m m a r y o f E x e r c i s e s Thefirstsixexercisesinvolvesimplegraph models. Exercises 7–24 present examples and extensions of the models presented in the examples in this section.
1. Suppose interstate highways join the six towns A, B, C, D, E, F as follows: I-77 goes from B through A to E; I-82 goes from C through D, then through B to F; I-85 goes from D through A to F; I-90 goes from C through E to F; and I-91 goes from D to E.
(a) Draw a graph of the network with vertices for towns and edges for segments of interstates linking neighboring towns.
(b) What is the minimum number of edges whose removal prevents travel be-tween some pair of towns?
(c) Is it possible to take a trip starting from town C that goes to every town without using any interstate highway for more than one edge (the trip need not return to C)?
2. (a) Suppose four teams, the Aces, the Birds, the Cats, and the Dogs, play each other once. The Aces beat all three opponents except the Birds. The Birds lost to all opponents except the Aces. The Dogs beat the Cats. Represent the results of these games with a directed graph.
(b) A dominance order is a listing of teams such that the ith team in the order beats the (i + 1) st team. Find all dominance orders for part (a).
3. (a) A schedule is to be made with five football teams. Each team is to play two other teams. Explain how to make a graph model of this problem.
(b) Show that except for interchanging names of teams, there is only one possible graph in part (a).
4. Suppose there are six people—John, Mary, Rose, Steve, Ted, and Wendy—who pass rumors among themselves. Each day John talks with Mary and Wendy; Mary talks with John, Rose, and Steve; Rose talks with Mary, Steve, and Ted; Steve talks with Mary, Rose, Ted, and Wendy; Ted talks with Rose, Steve, and Wendy; www.itpub.net 1.1 Graph Models 11 and Wendy talks with John, Steve, and Ted. Whatever people hear one day they pass on to others the next day.
(a) Model this rumor-passing situation with a graph.
(b) How many days does it take to pass a rumor from John to Steve? Who will tell it to Steve?
(c) Is there any way that if two people stopped talking to each other, it would take three days to pass a rumor from one person to all the others?
5. (a) Give a direction to each edge in Figure 1.4 so that there are directed routes from any vertex to any other vertex.
(b) Do part (a) so as to minimize the length of the longest directed path between any pair of vertices. Explain why a smaller minimum is not possible.
6. (a) What is the length of the longest possible path (with the most vertices) in the graph in Figure 1.3, ignoring directions of edges?
(b) What is the length of the longest possible circuit (with the most vertices) in the graph in Figure 1.4?
7. Find a matching, or explain why none exists for the following graphs: (a) (b) A B C D a b c d A B C D a b c d 8. Give another reason why Figure 1.2 has no matching by considering the appro-priate subset of jobs (showing that they cannot all be filled).
9. We generalize the idea of matching in Example 1 to arbitrary graphs by defining a matching to be a pairing off of adjacent vertices in a graph. For example, one possible matching in Figure 1.1a is a-b, c-d. Which of the following graphs have a matching? If none exists, explain why.
(a) Figure 1.4 (b) a b c d e f g h 10. (a) Suppose a dictionary in a computer has a “start” from which one can branch to any of the 26 letters: at any letter one can go to the preceding and succeeding letters. Model this data structure with a graph.
(b) Suppose additionally that one can return to “start” from letters c or k or t.
Now what is the longest directed path between any two letters?
11. Build the complete testing tree in Example 2 to identify one of the 26 letters of the alphabet.
12. Repeat Example 2 using three-way comparisons (less than, greater than, or equal to) to identify one of the 26 letters.
12 Chapter 1 Elements of Graph Theory 13. Suppose eight current varieties of chipmunk evolved from a common ancestral strain through an evolutionary process in which at various stages one ancestral variety split into two varieties (none of the ancestral varieties survive when they split into two new varieties).
(a) Explain how one might model this evolutionary process with a graph.
(b) What is the total number of splits that must have occurred?
14. In Example 3, find a minimal connecting set of edges containing neither (a, b) nor (b, c ).
15. (a) What are the other sets of two edges whose removal disconnects the graph in Figure 1.4 besides (a, b), (a, e) and (c, d), (d, h)? Either produce others or give an argument why no others exist.
(b) Find all sets of two vertices whose removal disconnects the remaining graph in Figure 1.4.
16. (a) For the following graph, find all sets of two vertices whose removal discon-nects the graph of remaining vertices.
(b) Find all sets of two edges whose removal would disconnect the graph.
a b c d e f g 17. Find a minimal edge cover and a minimal set of vertices adjacent to all other vertices for the graph in Figure 1.2.
18. In Figure 1.4, find all sets of three vertices that are adjacent to all the other vertices. Give a careful logical analysis to justify your answer.
19. Repeat Example 4 for minimal block and corner surveillance when the network in Figure 1.4 is altered by adding edges ( f, g), (g, j) and deleting (b, f ).
20. Repeat Example 4 for the edge cover and minimal corner surveillance when the network is formed by a regular array of north–south and east–west streets of size: (a) 3 streets by 3 streets (b) 4 streets by 4 streets (c) 5 streets by 5 streets 21. (a) A queen dominates any square on a chessboard in the same row, column, or diagonal as the queen. How few queens can dominate all squares on an 8 by 8 chessboard?
(b) Repeat this problem for bishops, which dominate only diagonals.
22. Solve the committee scheduling problem for the committee overlap graph in Figure 1.4. That is, what is the minimum number of independent sets needed to cover all vertices?
23. (a) Find a maximum independent set in the following graphs: (i) Figure 1.1a (ii) Figure 1.2 www.itpub.net 1.1 Graph Models 13 (b) Use your result in part (a) to produce a minimal edge cover in these graphs.
24. What is the largest independent set in a circuit of length 7? Of length n?
25. (a) What is the largest independent set possible in a connected seven-vertex graph? Draw the graph.
(b) What is the largest independent set possible in a seven-vertex graph (need not be connected)? Draw the graph.
26. Find a vertex basis in the following directed graphs: (a) Figure 1.1b (b) Figure 1.3 (c) Figure 1.4 with edges directed by alphabetical order [e.g., edge (a, e) is directed from a to e] 27. Show that the vertex basis in a directed graph is unique if there is no sequence of directed edges that forms a circuit in the graph.
28. A game for two players starts with an empty pile. Players take turns putting one, two, or three pennies in the pile. The winner is the player who brings the value of the pile up to 16c /.
(a) Make a directed graph modeling this game.
(b) Show that the second player has a winning strategy by finding a set of four “good” pile values, including 16c /, such that the second player can always move to one of the “good” piles (when the second player moves to one of the good piles, the next move of the first player must be to a non-good pile, and from this position the second player has a move to a good pile, etc.).
29. The parsing of a sentence can be represented by a directed graph, with a vertex S (for the whole sentence) having edges to vertices Su (subject) and P (predicate), then Su and P having edges to the parts into which they are decomposed into pieces, and so on.
Consider the abstract grammar with decomposition rules: S →AB, S →BA, A →ABA, B →BAS, and B →S. For example, BAABA can be “parsed” as shown below.
S B A B S B A A A Find a parsing graph for each of the following (or explain why no parsing exists): (a) BABABABA (b) BBABAABA 14 Chapter 1 Elements of Graph Theory 1.2 ISOMORPHISM In this section we investigate some of the basic structure of graphs. We are interested in properties that distinguish one vertex in a graph from another vertex and, more generally, that distinguish one graph from another graph. We motivate this discussion with the question: how can we tell if two graphs are really the same graph, but drawn differently and with different names for the vertices? For example, are the two five-vertex graphs in Figure 1.6 different versions of the same graph?
Agraphcanbedrawnonasheetofpaperinmanydifferentways.Thus,itisusually possible to draw a graph in two ways that would lead a casual viewer to consider the drawings to be “different” graphs. This motivates the following definition.
Two graphs G and G′ are called isomorphic if there exists a one-to-one corre-spondence between the vertices in G and the vertices in G′ such that a pair of vertices are adjacent in G if and only if the corresponding pair of vertices are adjacent in G′.
Such a one-to-one correspondence of vertices that preserves adjacency is called an isomorphism. A useful way to think of isomorphic graphs is as follows: the first graph can be redrawn on a transparency that can be exactly superimposed over a drawing of the second graph.
To be isomorphic, two graphs must have the same number of vertices and the same number of edges. The two graphs in Figure 1.6 pass this initial test. Both graphs have one vertex, e and 5, respectively, at the end of just one edge. Then any isomorphism of these two graphs must match e with 5. Also, the vertices at the other ends of the edge from e and 5 must be matched; that is, d matches with 4. (Think of superimposing one graph over the other.) The remaining three vertices in each graph are mutually adjacent (forming a triangle) and also are all adjacent to d or 4, respectively. Thus the matching a with 1, b with 2, and c with 3 (or any other matching of these two subsets of three vertices) will preserve the required adjacencies. The correspondence a −1, b −2, c −3, d −4, e −5 is then an isomorphism, and the two graphs are isomorphic.
To visualize how they can be made to look the same, think of moving vertices 4 and 5 in the right graph upward and to the right [past edge (1,3)], so that 1, 2, 3, 4 form a quadrilateral with crossing diagonals.
a d b c 2 3 5 4 1 e Figure 1.6 www.itpub.net 1.2 Isomorphism 15 Recall that the degree deg(x) of a vertex is the number of edges incident to the vertex. Degrees are preserved under isomorphism—that is, two matched vertices must have the same degree. Then in Figure 1.6, e has to be matched with 5 and d matched with 4 because they are the unique vertices of degree 1 and 4 in their respective graphs. Further, two isomorphic graphs must have the same number of vertices of a given degree. For example, if they are to be isomorphic, the two graphs in Figure 1.6 must both have the same number of vertices of degree 3—they do; both have three vertices of degree 3.
A subgraph G′ of a graph G is a graph formed by a subset of vertices and edges of G. If two graphs are isomorphic, then subgraphs formed by corresponding vertices and edges must be isomorphic. In Figure 1.6, removal of vertices e and 5 (and their incident edges) leaves two isomorphic subgraphs consisting of four mutually adjacent vertices. Once this subgraph isomorphism is noted, isomorphism of the whole graphs is easily demonstrated.
Subgraphs can be used to test for isomorphism in the following way. If a graph G has a set of six vertices forming a chordless circuit of length 6 (chordless means there are no other edges between these six vertices except the six edges forming the circuit), then any graph isomorphic to G must also have a set of six vertices forming such a chordless 6-circuit.
A graph with n vertices in which each vertex is adjacent to all the other vertices is called a complete graph on n vertices, denoted Kn. A complete graph on two vertices, K2, is just an edge. Complete subgraphs are in a sense the building blocks of all larger graphs. For example, both graphs in Figure 1.6 consist of a K4 and a K2 joined at a common vertex. Conversely, every graph on n vertices is a subgraph of Kn.
Before examining other pairs of graphs for isomorphism, let us mention the practical importance of determining whether two graphs are isomorphic. Researchers working with organic compounds build up large dictionaries of compounds that they have previously analyzed. When a new compound is found, they want to know if it is already in the dictionary. Large dictionaries can have many compounds with the same molecular formula but differing in their structure as graphs (and possibly in other ways). Then one must test the new compound to see if its graph-theoretic structure is the same as the structure of one of the known compounds with the same formula (and the same in other ways)—that is, whether the new compound is graph-theoretically isomorphic to one of a set of known compounds. A similar problem arises in designing efficient integrated circuitry for a computer. If the design problem has already been solved for an isomorphic circuit (or if a piece of the new network is isomorphic to a previously designed circuit), then valuable savings in time and money are possible.
Example 1: Simple Isomorphism Are the two graphs in Figure 1.7 isomorphic?
Both graphs have eight vertices and 10 edges. Let us examine the degrees of the different vertices. We see that b, d, f, h and 3, 4, 7, 8 have degree 2, while the other verticeshavedegree3.Thenthetwographshavethesamenumberofverticesofdegree 2 and the same number of degree 3. The respective subgraphs of the four vertices 16 Chapter 1 Elements of Graph Theory a b c d e h g f 1 2 3 4 5 8 7 6 Figure 1.7 of degree 2 (and the edges between these degree-2 vertices) in each graph must be isomorphic if the whole graphs are isomorphic. However, there are no edges between any pair of b, d, f, h, while the other subgraph of degree-2 vertices has two edges: (4, 3) and (8, 7). So the subgraphs of degree-2 vertices are not isomorphic, and hence the two full graphs are not isomorphic. The reader can also check that the two subgraphs of degree-3 vertices in each graph are not isomorphic.
The vertices of degree 2 in the left graph in Figure 1.7 form a subgraph of mutually nonadjacent vertices. Such a subgraph is called a set of isolated vertices.
Let us review the reasoning used in Example 1. It is a contrapositive version of the AC Principle, Assumptions generate helpful Consequences, introduced after Example 4 in Section I.1. The contrapositive statement is that if a consequence is false, then the assumption must be false. In this case, we assume that two graphs G and G′ are isomorphic. A consequence of this assumption is that G2 and G′ 2 must also be isomorphic, where G2 (G′ 2) is the subgraph of G (G′) generated by its vertices of degree 2. For the graphs in Example 1, the contrapositive statement is that if G2 and G′ 2 are not isomorphic, then the assumption that G and G′ are isomorphic must be false.
Example 2: Isomorphism in Symmetric Graphs Are the two graphs in Figure 1.8 isomorphic?
The two graphs both have seven vertices and 14 edges. Every vertex in both graphs has degree 4. Further, both graphs exhibit all the symmetries of a regular 7-gon. With no distinctions possible among vertices within the same graph, our only option is to try to construct an isomorphism. To do this, we assume that there is an isomorphism and use the AC Principle to deduce properties of an isomorphism for these two graphs that can guide us to construct such an isomorphism. If at some point a 1 2 3 4 5 6 7 b c d e f g Figure 1.8 www.itpub.net 1.2 Isomorphism 17 f c b g 7 2 4 5 Figure 1.9 in our construction a contradiction arises, then we know our assumption was false and there is no isomorphism.
Start with vertex a in the left graph. By rotational symmetry, we can match a to any vertex in the right graph (that is, if the two graphs are isomorphic, there will exist an isomorphism with a matched to any vertex in the right graph). Let us use the match a −1.
The set of neighbors of a (vertices adjacent to a) must be matched with the set of neighbors of 1. Let us look at the subgraphs formed by these neighbors of a and 1.
See Figure 1.9. Both subgraphs are paths: one is f to g to b to c, and the other is 7 to 4 to 5 to 2. The isomorphism must make these path subgraphs isomorphic. Thus, f and c must be matched with 7 and 2 (matching ends of the two paths). By the left-right symmetry of the graphs, it makes no difference which way f and c are matched—say f – 7 and c – 2. Then to complete the isomorphism of neighbors of a and 1, we must match g with 4 and b with 5. Now there remain only two unmatched vertices in each graph: d, e and 3, 6. Vertex g is adjacent to e but not d, and its matched vertex 4 is adjacent to 3 but not to 6. Thus we must match e with 3 and d with 6.
In sum, allowing for symmetries to match a with 1 and f with 7, we conclude that if the graphs are isomorphic, one isomorphism must be a −1, b −5, c −2, d −6, e −3, f −7, g −4. Checking edges, we see that the graphs are indeed isomorphic with this matching (if this matching were found not to be an isomorphism, then the two graphs would not be isomorphic, since the matches we made were all forced except for the symmetries involving the matches of a and f ).
Given a graph G = (V, E), its complement is a graph G = (V, E) with the same set of vertices but now with edges between exactly those pairs of vertices not linked in G. The union of the edges in G and G forms a complete graph. Two graphs G1 and G2 will be isomorphic if and only if G1 and G2 are isomorphic. The isomorphism problem in Example 2 is easy to answer using complements. Figure 1.10 shows the a 1 2 3 4 5 6 7 b c d e f g Figure 1.10 18 Chapter 1 Elements of Graph Theory a b c d e f g h 1 3 7 8 5 6 4 2 Figure 1.11 complements of the two graphs in Figure 1.8. Clearly, both these complementary graphs are just a (twisted) circuit of length 7 and hence are isomorphic.
In general, if a graph has more pairs of vertices joined by edges than pairs not joined by edges, then its complement will have fewer edges and thus will probably be simpler to analyze.
Example 3: Isomorphism of Directed Graphs Are the two directed graphs in Figure 1.11 isomorphic?
Each graph has eight vertices and 12 edges, and each vertex has degree 3. If we break the degree of a vertex into two parts, the in-degree (number of edges pointed in toward the vertex) and out-degree (number of edges pointed out), we see that each graph has four vertices of in-degree 2 and out-degree 1, and each graph has four vertices of in-degree 1 and out-degree 2. We could try to build an isomorphism as in the previous example by starting with a match (by a symmetry argument) between a and 1 and then matching their neighbors (with edge directions also matched), and so forth.
However, there is a basic difference in the directed path structure of the two graphs. We will exploit this difference to prove nonisomorphism. In the left graph we can draw a directed path from any given vertex to any other vertex by going clockwise around the circle of vertices: the outer edges form a directed circuit through all the vertices in the left graph. But in the right graph, all edges between the vertex subsets V1 = {1, 2, 3, 4} and V2 = {5, 6, 7, 8} are directed from V1 to V2, and so there can be no directed paths from any vertex in V2 to any vertex in V1 (nor is there a directed circuit through all the vertices). Thus, the two graphs are not isomorphic.
1.2 EXERCISES 1. List all nonisomorphic undirected graphs with four vertices.
2. List all nonisomorphic directed graphs with three vertices.
3. Draw two nonisomorphic graphs with (a) Six vertices and 10 edges (b) Nine vertices and 13 edges www.itpub.net 1.2 Isomorphism 19 4. If directions are ignored, are the two graphs in Figure 1.11 isomorphic?
5. Which of the following pairs of graphs are isomorphic? Explain carefully.
(a) a e f g h b 1 2 3 4 5 6 7 8 d c (b) 1 b c d e f 2 3 4 5 6 a (c) a b c d e f g h i 1 2 3 4 5 6 7 8 9 (d) 1 6 5 4 3 2 a b c d e f 20 Chapter 1 Elements of Graph Theory (e) a b c d e f 1 2 3 4 5 6 (f ) a b c d e f g 1 2 3 4 5 6 7 (g) 5 4 3 2 1 6 7 b d c f e a g (h) 1 2 3 4 5 6 7 a b c d e f g www.itpub.net 1.2 Isomorphism 21 (i) a b c d 1 2 3 4 5 6 e f (j) 1 2 3 4 5 6 a b c d e f (k) 1 2 3 4 5 11 10 9 8 7 6 a d b c h j g k f e i (l) a b c d e f g h i j 1 2 3 4 5 6 7 8 9 10 22 Chapter 1 Elements of Graph Theory 6. Which of the following pairs of graphs are isomorphic? Explain carefully.
(a) (b) 5 d c b a f e 6 1 4 3 2 4 e d b a f c 6 1 3 5 2 (c) (d) 5 d e f g h a b c 6 7 8 1 2 3 4 h 8 7 6 4 1 2 3 5 f c e g a b d (e) (f) 2 6 5 7 8 4 1 3 d c b a h g f e 7 g 8 9 1 5 6 2 3 4 e f d c b a i h (g) (h) a b c d e f g h i j 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 a b c d e f g h www.itpub.net 1.2 Isomorphism 23 7. Which pairs of graphs in this set are isomorphic?
6 5 4 3 2 1 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 8. Which pairs of graphs in this set are isomorphic?
1 2 3 4 5 6 7 8 9 10 11 12 16 15 14 13 17 18 19 20 21 22 23 24 25 2627 28 29 30 31 32 33 34 35 36 37 38 39 40 9. Suppose each edge in the graphs in Figure 1.8 is directed from the smaller (numerically or alphabetically) end vertex to the larger end vertex. Are the two resulting directed graphs isomorphic?
10. Are the following pairs of directed graphs isomorphic?
a b c d e 1 2 3 4 5 1 2 3 6 5 4 a b c d e f 11. Show that all 5-vertex graphs with each vertex of degree 2 are isomorphic.
12. Are there any 6-vertex graphs with three edges incident to each vertex that are not isomorphic to one of the graphs in Exercise 7?
13. What are the sizes of the largest complete subgraphs in the two graphs in Exercise 6(g)?
24 Chapter 1 Elements of Graph Theory 14. Build 6-vertex graphs with the following degrees of vertices, if possible. If not possible, explain why not.
(a) Three vertices of degree 3 and three vertices of degree 1 (b) Vertices of degrees 1, 2, 2, 3, 4, 5 (c) Vertices of degrees 2, 2, 4, 4, 4, 4 1.3 EDGE COUNTING There is very little in the way of general assertions that can be made about all graphs.
There is one useful general theorem, a formula for counting edges.
Theorem 1 In any graph, the sum of the degrees of all vertices is equal to twice the number of edges.
Proof Summing the degrees of all vertices counts all instances of some edge being incident at some vertex. But each edge is incident with two vertices, and so the total number of such edge–vertex incidences is simply twice the number of edges. The theorem is now proved. ◆ As an illustration of Theorem 1, consider the graph in Figure 1.12 with six vertices, three of degree 4, two of degree 3, and one of degree 2. The sum of the degrees is 4 + 4 + 4 + 3 + 3 + 2 = 20. This sum must equal twice the number of edges. The reader can check that the number of edges in this graph is 10.
For the sum of degrees to be an even integer, there must be an even number of odd integers in the sum. Thus we obtain the following Corollary In any graph, the number of vertices of odd degree is even.
Let us now look at uses of this theorem and corollary.
a b c d e f Figure 1.12 www.itpub.net 1.3 Edge Counting 25 Example 1: Use of Theorem 1 Suppose we want to construct a graph with 20 edges and have every vertex of degree 4.
How many vertices must the graph have?
Let v denote the number of vertices. The sum of the degrees of the vertices will be 4v, and by the theorem this sum must be twice the number of edges: 4v = 2 × 20 = 40.
Hence v = 10.
Example 2: Edges in a Complete Graph How many edges are there in Kn, a complete graph on n vertices?
Recall that Kn has an edge between all possible pairs of vertices. At any given vertex, there will be edges going to each of the n −1 other vertices in Kn, and so each vertex has degree n −1. The sum of the degrees of all n vertices in Kn will be n(n −1). Since this sum equals twice the number of edges, the number of edges is n(n −1)/2.
Example 3: Impossible Graph Is it possible to have a group of seven people such that each person knows exactly three other people in the group?
If we model this problem using a graph with a vertex for each person and an edge between each pair of people who know each other, then we would have a graph with seven vertices all of degree 3. But this is impossible by the Corollary—the number of vertices of odd degree must be even—and so no such set of seven people can exist.
Recall that a graph G is connected if every pair of vertices in G is joined by a path in G. If G is not connected, its vertices can be partitioned into connected pieces, called components. Formally, a component H is a connected subgraph of G such that there is no path between any vertex in H and any vertex of G not in H. The component of G containing a particular vertex x consists of x and all vertices that may be reached from x by a path in G. Because each component of G is a graph in its own right, this section’s Corollary applies to each component as well as to G. We next present a puzzle that seems to have no relation to graphs.
Example 4: Mountain Climbers Puzzle Two people start at locations A and Z at the same elevation on opposite sides of a mountain range whose summit is labeled M. See Figure 1.13a. We pose the following puzzle: is it possible for the people to move along the range in Figure 1.13a to meet at M in a fashion so that they are always at the same altitude every moment? We shall show this is possible for any mountain range like Figure 1.13a. The one assumption we make is that there is no point lower than A (or Z) and no point higher than M.
We make a range graph whose vertices are pairs of points (PL, PR) at the same altitude with PL on the left side of the summit and PR on the right side, such that one of the two points is a local peak or valley (the other point might also be a peak or 26 Chapter 1 Elements of Graph Theory A B C D E M T U V W X Y Z (A, Z) (C, X) (E, W) (M, M) (C, T) (C, V) (B, U) (D, U) (D, Y) (a) (b) Figure 1.13 valley). The vertices for the range in Figure 1.13a are shown in the graph in Figure 1.13b. We make an edge joining vertices (PL, PR) and (P′ L, P′ R) if the two people can move constantly in the same direction (both going up or both going down) from point PL to point P′ L and from PR to P′ R, respectively. Our question now: is there a path in the range graph from the starting vertex (A, Z) to the summit vertex (M, M)? For the graph in Figure 1.13b, the answer is obviously yes.
We claim that vertices (A, Z) and (M, M) in any range graph have degree 1, whereas every other vertex in the range graph has degree 2 or 4. (A, Z) has degree 1 because when both people start climbing up the range from their respective sides, they have no choice initially but to climb upward until one arrives at a peak. In Figure 1.13a, the first peak encountered is C on the left, and so the one edge from (A, Z) goes to (C, X). A similar argument applies at (M, M). Next consider a vertex (PL, PR) where one point is a peak and the other point is neither peak nor valley, such as (E, W). From the peak we can go down in either direction: at W, we can go down toward Z or toward U. In either direction, the people go until one (or both) reaches a valley. At (E, W), the two edges go to (D, Y) and (D, U). Thus such a vertex has degree 2. A similar argument applies if one point (but not both) is a valley. It is left as an exercise for the reader to show that if a vertex (PL, PR) consists of two peaks or two valleys, such as (D, U), it will have degree 4. (A vertex consisting of a valley and a peak will have degree 0—why?) Suppose there were no path from (A, Z) to (M, M) in the range graph. Thus, these two vertices are in different components of the range graph. We use the fact that starting vertex (A, Z) and summit vertex (M, M ) are the only vertices of odd degree to obtain a contradiction. The component of the range graph consisting of (A, Z) and all the vertices that can be reached from (A, Z) would form a graph with just one vertex of odd degree, namely, (A, Z). This contradicts the Corollary, and so any range graph must have a path from (A, Z) to (M, M).
Many interesting properties in graph theory are dependent on certain sets of edges having even size. Euler cycles, discussed in Section 2.1, arise when all vertices have even degree.
In Example 2 of Section 1.1, we considered a matching problem involving the graph showninFigure1.14.Theverticesontheleftrepresentedpeopleandthevertices on the right represented jobs. An edge links a left vertex to a right vertex to indicate www.itpub.net 1.3 Edge Counting 27 A B C D E a b c d e Figure 1.14 that a certain person can perform a certain job. There can never be an edge between two vertices on the left or between two vertices on the right. Such a graph is called a bipartite graph. Formally, a graph G is bipartite if its vertices can be partitioned into two sets V1 and V2 and every edge joins a vertex in V1 with a vertex in V2.
Bipartite graphs can be characterized by all circuits in such graphs having even length (if there are no circuits, the graph is also bipartite), where the length of a circuit or path is the number of edges in it.
Theorem 2 A graph G is bipartite if and only if every circuit in G has even length.
Proof Note that it is sufficient to prove this theorem for connected bipartite graphs. We claim that if the theorem is true for each connected component of a disconnected bipartite graph G, then it is true for G (components were formally defined just above Example 4). This claim follows from G’s being bipartite if and only if each of its components is bipartite, and any circuit in G’s having even length if and only if any circuit in each of its components has even length.
First we show that if G is bipartite, then any circuit has an even length. If G is bipartite so that it can be drawn with all edges connecting a left vertex with a right vertex, then any circuit x1–x2–x3 · · · –xn–x1 has alternately a left vertex, then a right vertex, then a left vertex, and so on, assuming the first vertex x1 is on the left. Odd-subscripted vertices are on the left, and even-subscripted vertices on the right. See Figure 1.15a. Since xn is adjacent to x1, xn must be on the right, so its subscript is even. That is, there are an even number of vertices in the circuit. Any circuit has the same number of edges as vertices, and thus this circuit has even length.
x1 x3 x5 xn – 1 x2 x4 x6 xn a b c Q' Q (a) (b) Figure 1.15 28 Chapter 1 Elements of Graph Theory Suppose next that any circuit in G (there may be no circuits) has even length. We show how to construct a bipartite arrangement of G. We use the AC Principle and assume that a bipartition exists, and use properties of that bipartition to construct it.
Take any given vertex, call it a, and put it on the left. Put all vertices adjacent to a on the right. Next put all vertices that are two edges away from a, that is, at the end of some path of length 2 from a, on the left. In general, if there is a path of odd length between a and a vertex x, put x on the right. If there is a path of even length between a and x, put x on the left.
There cannot be distinct paths P and P′ between a and x of odd and of even lengths, respectively, since taking P from a to x and then returning to a on P′ yields an odd-length circuit. This is impossible, since all circuits have even length. (If P and P′ have a vertex q in common besides a and x, then a further argument is needed to show that there is a circuit of odd length. See Exercise 15 for details.) Similarly, we argue that there cannot be an edge between two vertices, say,b and c, both on the left. There must exist even-length paths Q, Q′ joining a with b and c, respectively (since b and c are on the left). See Figure 1.15b, in which Q is dashed and Q′ is dotted. Observe that Q′ followed by the edge (c, b) yields an odd-length path from a to b. This is impossible, since we just proved that there cannot be both an even-length path (Q) and an odd-length path (Q′ plus (a, b)) from a to any other vertex in G. By similar reasoning, two vertices on the right cannot be adjacent. Thus, we have a bipartite arrangement of G. ◆ Example 5: Testing for a Bipartite Graph Is the graph in Figure 1.16a bipartite?
Pick any vertex, say a, and put it on the left. We follow the approach in the second half of the proof of Theorem 2. Put vertices joined to a by an even-length path on the left and vertices joined to a by an odd-length path on the right. If all the circuits in this graph are even-length, then the reasoning in the above proof guarantees that our placement of vertices will yield a bipartite arrangement. If we end up with two vertices on the left (or on the right) being adjacent, then the graph cannot be bipartite.
In this case, the construction succeeds, as shown in Figure 1.16b.
a b c d e f g h i j k l m n o p a b c e h l j n m o f g d i k p (a) (b) Figure 1.16 www.itpub.net 1.3 Edge Counting 29 1.3 EXERCISES 1. How many vertices will the following graphs have if they contain: (a) 12 edges and all vertices of degree 3.
(b) 21 edges, three vertices of degree 4, and the other vertices of degree 3.
(c) 24 edges and all vertices of the same degree.
2. For each of the following questions, describe a graph model and then answer the question.
(a) Must the number of people at a party who do not know an odd number of other people be even?
(b) Must the number of people ever born who had (have) an odd number of brothers and sisters be even?
(c) Must the number of families in Alaska with an odd number of children be even?
(d) For each vertex x in the following graph, let s(x) denote the number of vertices (including x) adjacent to at least one of x’s neighbors. Must the number of vertices with s(x) odd be even? Is this true in general?
a b c d e f g 3. What is the largest possible number of vertices in a graph with 19 edges and all vertices of degree at least 3?
4. Is any subgraph of a bipartite always bipartite? Prove, or give a counterexample.
5. What constraint must be placed on a bipartite graph G to guarantee that G’s complement will also be bipartite?
6. If a graph G has n vertices, all of which but one have odd degree, how many vertices of odd degree are there in G, the complement of G?
7. Show that a complete graph with m edges has (1 + 8m)/2 vertices.
8. Let G be an n-vertex graph that is isomorphic to its complement G. How many edges does G have? (Hint: Use Exercise 5.) 9. Suppose all vertices of a graph G have degree p, where p is an odd number. Show that the number of edges in G is a multiple of p.
10. There used to be 26 football teams in the National Football League (NFL) with 13 teams in each of two conferences (each conference was divided into divisions, but that is irrelevant here). An NFL guideline said that each team’s 14-game schedule should include exactly 11 games against teams in its own conference and three games against teams in the other conference. By considering the right 30 Chapter 1 Elements of Graph Theory part of a graph model of this scheduling problem, show that this guideline could not be satisfied!
11. Prove a directed version of Theorem 1: The sum of the in-degrees of vertices in a directed graph equals the sum of the out-degrees of vertices, and further, each sum equals the number of edges.
12. Build the range graph for each of the following mountain ranges and use the graph to find a solution to the problem in Example 4.
(a) (b) A B C D E F G H I M U V X W Y Z A B C D M R S T U V W X Y Z 13. Prove in a range graph that if a vertex (PL, PR) consists of two peaks or two valleys, it will have degree 4.
14. Prove in a range graph that if a vertex (PL, PR) consists of a valley and a peak, it will have degree 0.
15. Determine whether the following graphs are bipartite. If so, give the partition into left and right vertices as in Figure 1.16b.
(a) Figure 1.4 (b) Figure 1.7 (left graph) (c) Figure 1.12 16. Determine whether the following graphs are bipartite. If so, give the partition into left and right vertices as in Figure 1.16b.
a f c h d g b e d k m a e f g b h c i j n 17. Suppose x and y are the only two vertices of odd degree in graph G, and x and y are not adjacent to each other. Show that G is connected if and only if the graph obtained from G by adding edge (x, y) is connected.
18. In the second part of the proof of Theorem 2, one can encounter the situation in which there exist paths P and P′ between a and x, P of odd length and P′ of even length, and these two paths have one or more vertices in common. One must show that a subset of the edges on these two paths forms an odd-length circuit.
Let q be the first vertex on P, starting from a, that also lies on P′. Show that either www.itpub.net 1.4 Planar Graphs 31 the circuit from a along P to q and then back on P′ to a, or the edge sequence from q along P to x and then back on P′ to q, has odd length. In the latter case, if the edge sequence is not a circuit, then it has a vertex q′ on both P and P′. Repeat the same reasoning considering the circuit on P from q to q′ and then back on P′ to q or the edge sequence from q′ along P to x and back along P′ to q′.
1.4 PLANAR GRAPHS The most natural examples of graphs are street maps and telephone networks. The graphs that arise from such physical networks usually have the property that they can be depicted on a piece of paper without edges crossing (different edges meet only at vertices). We say that a graph is planar if it can be drawn on a plane without edges crossing. We use the term plane graph to refer to a planar depiction of a planar graph.
The two graphs in Figure 1.17 are both planar. The graph in Figure 1.17b is a plane graph. The graph in Figure 1.17a is planar, since it can be redrawn in the form of the graph in Figure 1.17b.
Our principal focus in this section is determining whether a graph is planar. We take two approaches, both based on the AC Principle. The first approach involves a systematic method for trying to draw a graph edge-by-edge with no crossing edges, in the same spirit as when we tried to determine if two graphs are isomorphic. The second approach develops some theory with a goal of finding useful properties of planar graphs. If a graph does not satisfy one or more of these properties, then we know that it cannot be planar.
Remember that if a graph G has been drawn with edges crossing, this does not mean the graph is nonplanar. There may be another way to draw the graph without edges crossing, as illustrated by the graph in Figure 1.17a, which can be redrawn to be the plane graph in Figure 1.17b.
Probably the most important need today for testing whether a graph is planar arises in designing electronic circuits. Complex integrated circuits are nonplanar and require several layers of (planar) circuit connections in their wiring. But the number of layers is limited and so a major problem in integrated circuit design is decomposing a large circuit into a minimal number of subcircuits that are known to be planar.
a b f c d e d f b c e a (a) (b) Figure 1.17 32 Chapter 1 Elements of Graph Theory A more mundane, but still important, use of planarity testing arises in checking data-entry errors in planar networks. When a large planar graph such as a city’s street network is entered on data terminals for computerized analysis, it is a common error-checking technique to test first whether the graph as typed in is indeed planar (most data-entry errors would make the graph nonplanar).
Planar graphs were first studied extensively by mathematicians over 100 years ago in connection with a map-coloring problem.
Example 1: Map Coloring One of the most famous problems in mathematics concerns map coloring. The question is how many colors are needed to color countries on some map so that any pair of countries with a common border are given different colors. A map of countries is a planar graph with edges as borders and vertices where borders meet. See Figure 1.18a. However, a closely related planar graph called a dual graph of the map graph is more useful. The dual graph is obtained by making a vertex for each country and an edge between vertices corresponding to two countries with a common border.
See Figure 1.18b. Normally, a vertex is also included for the unbounded region surrounding the map.
The question in the dual graph now is how many colors are needed to “color” the vertices such that adjacent vertices have different colors. In Figure 1.18b, vertices A, B, C, D form a complete subgraph on four vertices and so each requires a different color, four colors in all. With four colors, we can also properly color the remaining vertices.
One of the most famous unsolved problems in all of mathematics during the last century was the conjecture that all planar graphs can be properly colored with only four colors. In trying to resolve this conjecture, mathematicians developed a large theory about planar graphs. In 1976 the four-color conjecture was proven true by Appel and Haken using an immense computer-generated, case-by-case, exhaustive analysis (there were 1955 classes of graphical configurations to be considered, each involving numerous subcases). We will take a closer look at graph coloring in the next chapter.
We now use the AC Principle to try to find a planar drawing of a graph. We assume it to be planar. As with isomorphism between two graphs, we want to be A B C D E F G A B C D E F G (a) (b) Figure 1.18 www.itpub.net 1.4 Planar Graphs 33 able to conclude that a graph is not planar if our construction fails. We shall call our approach the circle–chord method. It starts by finding a circuit that contains all the vertices of our graph (though such circuits do not exist for all graphs, they are common in the types of graphs we will be considering in this section). We draw this circuit as a large circle. The remaining noncircuit edges, which we will call chords, must be drawn either inside the circle or outside the circle in a planar drawing.
We choose a first chord and draw it, say, outside the circle. If properly chosen, this chord will force certain other chords to be drawn inside the circle (if also placed outside the circle, they would have to cross the first chord). These inside chords will force still other chords to be drawn outside, and so on. After the first chord is drawn, the choice of placing subsequent chords inside or outside is forced. Thus, if we reach a point where a new chord will have to cross some previous chord, whether the new chord is drawn inside or outside, we can claim that the graph must be nonplanar. That is, our construction based on the assumption that the graph had a planar depiction led to a contradiction. If all the chords can be added without crossing other chords, then the graph is planar.
A critical decision is whether the first chord drawn should go inside or outside the circle. We claim that it makes no difference, because of the following inside–outside symmetry of a circle. Consider two maps of the earth, the first with the North Pole at the center of the map, the second with the South Pole at the center. Suppose each map has a path drawn in the Northern Hemisphere linking two cities on the equator.
In the first map (with the Northern Hemisphere inside the equator) the path is inside the equator’s circle, whereas in the second map the path is outside the equator. Think of the circle formed by the circuit in the circle–chord method as the equator and the first chord as the path between two cities. Whether the chord is inside or outside the circle is just a matter of which “map” of the earth one uses.
We note that very efficient algorithms exist to test whether a graph is planar, whereas there is no efficient algorithm known to test whether two graphs are isomor-phic. The planarity testing algorithms are fairly complicated and beyond the scope of this text.
Example 2: Circle–Chord Method Use the circle–chord method to determine whether the graph in Figure 1.19a is planar.
Let us look for a circuit with all eight vertices. One possibility is a-f-c-h-d-g-b-e-a.
a b c d e f g h a f c h e b g d (a) (b) Figure 1.19 34 Chapter 1 Elements of Graph Theory 1 2 3 4 5 6 1 2 3 4 5 6 (a) (b) Figure 1.20 Now try to add the other four edges (a, h), (b, f ), (c, g), (d, e). By inside–outside symmetry, we can start by drawing (a, h) outside. See Figure 1.19b. Then (b, f ) and (c, g) must go inside. Then (d, e) must go outside. So the graph is planar, as shown in Figure 1.19b.
Example 3: Showing K 3,3 Is Nonplanar Show that K3,3, the graph in Figure 1.20a, is nonplanar. The notation K3,3 indicates that this graph is a complete bipartite graph consisting of two sets of three vertices with each vertex in one set adjacent to all vertices in the other set. Applying the circle–chord method, we form a circuit containing all six vertices in K3,3, and then try to add the remaining edges (not in the circuit) as inside and outside chords.
There are several choices for a 6-vertex circuit. Suppose we use the circuit 1– 4 – 2–5–3–6–1anddrawitinacircleasshowninFigure1.20b.Nexttheedges(1,5),(2,6), and (3, 4) must be added. First draw chord (1, 5). By the inside–outside symmetry of a circle discussed above, we can assume that (1, 5) is drawn inside the circuit, as in Figure 1.20b. Then (2, 6) must be drawn outside the circuit to avoid crossing chord (1, 5). Finally, we must draw (3, 4): if drawn outside the circuit, (3, 4) would have to cross chord (2, 6); if drawn inside the circuit, (3, 4) would have to cross chord (1, 5).
Thus K3,3 cannot be drawn in a planar depiction. Hence K3,3 is nonplanar.
Using a mixture of theory and careful, case-by-case analysis, it is possible to prove that any nonplanar graph always contains a K3,3 or a K5 (the complete graph on five vertices shown in Figure 1.21a) as a subgraph or a slight modification of these two graphs. It is left as an exercise to show that the circle–chord method (used in 1 2 3 4 5 6 1 2 3 4 5 (a) (b) Figure 1.21 www.itpub.net 1.4 Planar Graphs 35 Example 3) can be used to show that K5 is nonplanar. Thus, these two graphs are the “reason” that a graph cannot be drawn in a planar fashion.
We have to allow a slight variation in K3,3 and K5 in nonplanarity analysis. Figure 1.21b shows a K3,3, graph that has been subdivided by adding vertices in the middle of some of its edges. The resulting graph is no longer a K3,3 and does not contain K3,3 as a subgraph, yet it is still nonplanar (repeatedly adding a vertex in the middle of an edge cannot make a nonplanar graph planar). We say that a subgraph is a K 3,3 configuration if it can be obtained from a K3,3 by adding vertices in the middle of some edges. A K 5 configuration is defined similarly. The following planar graph characterization theorem was first proved by the Polish mathematician Kuratowski.
Theorem 1 (Kuratowski, 1930) A graph is planar if and only if it does not contain a subgraph that is a K5 or K3,3 configuration.
If the circle–chord method shows that a graph is nonplanar, then by Theorem 1 this graph has a subgraph that is a K5 or K3,3 configuration. Finding such a configu-ration can sometimes be tricky. However, the following observation is helpful: Most small nonplanar graphs contain a K3,3 configuration. All but one of the nonplanar graphs in the exercises have K3,3 configurations. Note also that the depiction of a K3,3 in Figure 1.20b as a 6-vertex circle with three chords joining pairs of opposite vertices is the way that a K3,3 configuration normally arises in a nonplanar graph.
Example 4: Finding a K 3,3 Use the circle–chord method to determine whether the graph in Figure 1.22a is planar.
If it is nonplanar, find a subgraph that is a K3,3 configuration.
First we seek a circuit that visits all vertices. Many such circuits exist. Choose the circuit a-b-c-d-e-f-g-h-a, shown in Figure 1.22b. Say we pick (a, d) as our first chord to draw. By inside–outside symmetry, it makes no difference whether we draw (a, d) inside or outside the circuit. Put it inside. Then (b, e) must go outside to avoid intersecting (a, d). Next look for another chord reaching across the circle. Observe a b c d e f g h a b d e f g h c a b c d e g (a) (b) (c) Figure 1.22 36 Chapter 1 Elements of Graph Theory that chord (c, g) cannot be drawn inside without crossing (a, d), nor drawn outside without crossing (b, e). So the graph is nonplanar.
Next we look for a K3,3 configuration. We can simplify the problem by restricting our attention to the nonplanar subgraph in Figure 1.22b, whose crossing edges proved that the full graph had to be nonplanar. A K3,3 configuration has six vertices of degree 3 (corresponding to the vertices of a K3,3) plus some number of vertices of degree 2 that subdivide the edges of a K3,3. The way to find a K3,3 configuration in a subgraph is to eliminate edge subdivisions in the graph (remove each vertex of degree 2 and combine the two edges at each such vertex into a single edge). Figure 1.22c shows the subgraph in Figure 1.22b with subdivisions removed. The graph in Figure 1.22c looks just like the depiction of a K3,3 in Figure 1.20b. Thus, the subgraph in Figure 1.22b was a K3,3 configuration.
Finding a K3,3 configuration in the graph in Figure 1.22a (without using the subgraph in Figure 1.22b) would be difficult. The challenging problem in finding a K3,3 configuration in a general nonplanar graph is the following. Let z be some vertex of degree 3 in the original graph and suppose that just two of the z’s edges, say (z, r) and (z, q), are part of a K3,3 configuration. Then z corresponds to a subdivision vertex in this K3,3 configuration, and these two edges of z need to be fused into a single edge (r, q) to find the underlying K3,3. That is, z disappears and a new edge (r, q) is created. Using the subgraph produced by the circle–chord method makes it much easier to identify vertices of degree 2 in a K3,3 configuration whose two edges should be fused together.
There are many different plane graph depictions that can be drawn for a planar graph. For example, we can redraw the plane graph in Figure 1.23a by making the region bounded by the triangle (d, e, f ) very large and bringing vertex a to the right side, as in Figure 1.23b. Now flip the part of the graph above and to the right of the triangle inside the triangle, obtaining the plane graph in Figure 1.23c. The triangle (d, e, f ) has become the outside boundary of the whole graph. The boundary of any region can be converted to the outside boundary of the whole graph by a similar process.
Despite this variability in plane graph depictions of a planar graph, one important property of the plane depictions does not change. The number of regions is always d b e a c g f a c e b d f g g c e f d b a (a) (b) (c) Figure 1.23 www.itpub.net 1.4 Planar Graphs 37 the same. For simplicity, assume that G is a connected planar graph. (Recall that connected means having paths between every pair of vertices.) If v and e denote the number of vertices and edges, respectively, in G, then a plane graph depiction of G will always have a number of regions r given by the formula, r = e −v + 2. Remember that the unbounded region outside the graph is counted as a region. This remarkable formula for r was discovered by Euler 360 years ago.
Theorem 2 Euler’s Formula (1752) If G is a connected planar graph, then any plane graph depiction of G has r = e −v + 2 regions.
Proof Let us draw a plane graph depiction of G edge by edge. We choose successive edges so that at every stage we have a connected subgraph. Let Gn denote the connected plane graph obtained after n edges have been added, and let vn, en, and rn denote the number of vertices, edges, and regions in Gn, respectively. Initially we have G1, which consists of one edge, its two end vertices, and the one (unbounded) region. Then e1 = 1, v1 = 2, r1 = 1, and so Euler’s formula is valid for G1, since r1 = e1 −v1 + 2: 1 = 1 −2 + 2. We obtain G2 from G1 by adding an edge at one of the vertices in G1.
In general, Gn is obtained from Gn−1 by adding an nth edge at one of the vertices of Gn−1. The new edge might link two vertices already in Gn−1. If it does not, the other end vertex of the nth edge is a new vertex that must be added to Gn.
We will now use the method of induction (see Appendix A.2) to complete the proof. We have shown that the theorem is true for G1. Next we assume that it is true for Gn−1 for any n > 1, and prove that it is true for Gn. Let (x, y) be the nth edge that is added to Gn−1 to get Gn. There are two cases to consider.
In the first case, x and y are both in Gn−1. Then they are on the boundary of a common region K of Gn−1, possibly the unbounded region [if x and y were not on a common region, edge (x, y) could not be drawn in a planar fashion, as required]. See Figure 1.24a. Edge (x, y) splits K into two regions. Then rn = rn−1 + 1, en = en−1 + 1, vn = vn−1. So each side of Euler’s formula grows by 1. Hence, if the formula was true for Gn−1, it will also be true for Gn.
In the second case, one of the vertices x, y is not in Gn−1—say it is x. See Figure 1.24b. Then adding (x, y) implies that x is also added, but no new regions are formed (i.e., no existing regions are split). Thus rn = rn−1, en = en−1 + 1, vn = vn−1 + 1, and y x K y x (a) (b) Figure 1.24 38 Chapter 1 Elements of Graph Theory the value on each side of Euler’s formula is unchanged. The validity of Euler’s formula for Gn−1 implies its validity for Gn.
So each increase in r is balanced in Euler’s formula by an increase in e or v. By induction, the formula is true for all Gn’s and hence true for the full graph G. ◆ Example 5: Using Euler’s Formula How many regions would there be in a plane graph with 10 vertices each of degree 3?
By Theorem 1 in Section 1.3, the sum of the degrees, 10 × 3, equals 2e, and so e = 15. By Euler’s formula, the number of regions r is r = e −v + 2 = 15 −10 + 2 = 7 Theorem 2 has the following corollary that can often be used to show quickly that a graph is nonplanar.
Corollary If G is a connected planar graph with e > 1, then e ≤3v −6.
Proof Let us define the degree of a region analogously to the degree of a vertex to be the number of edges incident to a region—that is, the number of edges on its boundary.
If an edge occurs twice along a boundary, as does (x, y) in region K in Figure 1.25a, the edge is counted twice in region K’s degree; for example, region K has degree 10 and region L has degree 3 in Figure 1.25a. Observe that each region in a plane graph must have degree ≥3, for a region of degree 2 would be bounded by two edges joining the same pair of vertices and a region of degree 1 would be bounded by a loop edge (see Figure 1.25b), but parallel edges and loops are not allowed in graphs.
Since each region in a plane graph has degree ≥3, the sum of the degrees of all regions will be at least 3r. But this sum of degrees of all regions must equal 2e, since this sum counts each edge twice, that is, each of an edge’s two sides is part of some boundary (this is the same type of argument as used to show that the sum of the K L y x (a) (b) Figure 1.25 www.itpub.net 1.4 Planar Graphs 39 vertices’ degrees equals 2e). Thus, 2e = (sum of regions’ degrees) ≥3r, or 2 3e ≥r.
Combining this inequality with Euler’s formula (Theorem 2), we have 2 3e ≥r = e −v + 2 or 0 ≥1 3e −v + 2 Solving for e, we obtain e ≤3 v −6. ◆ Example 6: Showing That K5 Is Nonplanar Use the Corollary to prove that K5, the complete graph on 5 vertices, is nonplanar.
The graph K5 has v = 5 and e = 10 (see Example 2 in Section 1.3 for how to find the number of edges in a complete graph). Then 3v −6 = 3 × 5 −6 = 9. But the Corollary says that e ≤3v −6 must be true in a connected planar graph, and so assuming K5 is a planar graph leads to a contradiction—namely, that the Corollary is not true. So K5 cannot be planar.
The Corollary should not be misinterpreted to mean that if e ≤3v −6, then a connected graph is planar. Many nonplanar graphs also satisfy this inequality. For example, K3,3 with v = 6 and e = 9 satisfies it.
Our two theorems and corollary have laid the foundation for a mathematical theory of planar graphs. In the process, we have acquired a practical aid for showing that a graph is nonplanar. One way to extend this theory is to make the inequality in the Corollary “stronger”—that is, to get a smaller upper bound on e. Recall that the key step in proving the Corollary was the observation that every region has degree at least 3. This led to the inequality 2e ≥3r. Suppose that a certain connected graph G (with at least two edges) is known to be bipartite. By Theorem 2 in Section 1.3, all the circuits in a bipartite graph have even length. Then no region in this graph can have degree 3 (since this would imply a boundary circuit of length 3). Then every region in a bipartite planar graph must have degree ≥4. Summing the degrees of all regions, we now obtain the inequality 2e = (sum of degrees of regions) ≥4r. Reworking the Corollary with the inequality 2e ≥4r, we have 2 4e ≥r = e −v + 2, and hence e ≤2v −4 Every connected planar graph that is bipartite must satisfy this inequality. Con-sider our “favorite” bipartite graph K3,3. K3,3 has v = 6 and e = 9 and, as noted above, satisfies the corollary inequality e ≤3v −6. But since it is bipartite, K3,3 would also have to satisfy the new inequality e ≤2v −4 if it were planar. It does not: 9 ̸≤2 × 6 −4.
1.4 EXERCISES S u m m a r y o f E x e r c i s e s The first six exercises involve determining whether various graphs are planar and drawing planar graphs in different ways. Exer-cise 10 involves duality. Exercises 15–26 build on Euler’s formula and the corollary e ≤3v −6. The other exercises introduce new concepts.
40 Chapter 1 Elements of Graph Theory 1. Draw a dual graph of the planar graph (include a vertex for the unbounded region outside the graph) in (a) Figure 1.17b (b) Figure 1.18b 2. Show that K5 is nonplanar by the method in Example 2.
3. Which of the following graphs are planar? Find K3,3 or K5 configurations in the nonplanar graphs (almost all are K3,3).
(a) (b) (c) (d) c b a d f e h g f e d c b j i a h g f e d c b a i c b a g f e d (e) (f) (g) (h) a a i h g f e d c b g b c f e a b c e d c b a f d d e f (i) (j) (k) (l) c d e a b g h j i f a b c d e f g h b d c a e h g f c b a f e g d 4. Redraw the graph in Figure 1.19b so that the infinite region is bounded by the circuit a-f-c-h-a.
5. (a) For what values of n is Kn planar?
(b) For what values of r and s is the complete bipartite graph Kr,s planar? (Kr,s is a bipartite graph with r vertices on the left side and s vertices on the right side and edges between all pairs of left and right vertices.) www.itpub.net 1.4 Planar Graphs 41 6. A complete tripartite Kr,s,t is a generalization of a complete bipartite graph (see part (b) of the previous exercise). There are three subsets of vertices: r in the first subset, s in the second subset, and t in the third subset. Every vertex in one particular subset is adjacent to every vertex in the other two subsets; that is, a vertex is adjacent to all vertices except those in its own subset. Determine all the triples r, s, t for which Kr,s,t is planar.
7. In each case, give the values of r, e, or v (whichever is not given) assuming that the graph is planar. Then either draw a connected, planar graph with the property, if possible, or explain why no such planar graph can exist.
(a) Six vertices and seven regions (b) Eight vertices and 13 edges (c) Six vertices and 14 edges (d) 14 edges and nine regions (e) Six vertices all of degree 4 (f) Five regions and 10 edges (g) Six regions all with four boundary edges (h) Seven vertices all of degree 3 (i) 12 vertices and every region has four boundary edges (j) 17 regions and every vertex has degree 5 8. If a connected planar graph with n vertices all of degree 4 has 10 regions, deter-mine n.
9. A line graph L(G) of a graph G has a vertex of L(G) for each edge of G and an edge of L(G) joining each pair of vertices corresponding to two edges in G with a common end vertex.
(a) Show that L(K5) is nonplanar.
(b) Find a planar graph whose line graph is nonplanar.
10. The construction of a dual D(G) can be applied to any plane graph G: draw a vertex of D(G) in the middle of each region of G and draw an edge e∗of D(G) perpendicular to each edge e of G; e∗connects the vertices of D(G) representing the regions on either side of e.
(a) A dual need not be a graph. It might have two edges between the same pair of vertices or a self-loop edge (from a vertex to itself). Find two planar graphs with duals that are not graphs because they contain these two forbidden situations.
(b) Show that the duals of the two different plane depictions of the graph in Figures 1.23a and 1.23c are isomorphic.
(c) Show that the degree of a vertex in the dual graph D(G) equals the number of boundary edges of the corresponding region in the planar graph G.
(d) Find a planar graph that is isomorphic to its own dual.
(e) Show for any plane depiction of a graph G that the vertices of G correspond to regions in D(G).
11. (a) Show that if a circuit in a planar graph encloses exactly two regions, each of which has an even number of boundary edges, then the circuit has even length.
42 Chapter 1 Elements of Graph Theory (b) Show that if a circuit in a planar graph encloses a collection of regions, each of which has an even number of boundary edges, then the circuit has even length.
12. The crossing number c(G) of a graph G is the minimum number of pairs of crossing edges in a depiction of G. For example, if G is planar, then c(G) = 0.
Determine c(G) for the following graphs: (a) K3,3 (b) K5 (c) Figure 1.22a 13. A graph G is critical nonplanar if G is nonplanar but any subgraph obtained by removing a vertex is planar.
(a) Which of the following graphs are critical nonplanar?
(i) K3,3 (ii) K5 (iii) Figure 1.22a (b) Show that critical nonplanar graphs must be connected and cannot have a vertex whose removal disconnects the graph.
14. A graph G is called maximal planar if G is planar but the addition of another edge between two nonadjacent vertices will make the graph nonplanar.
(a) Show that every region of a connected, maximal planar graph will be trian-gular.
(b) If a connected, maximal planar graph has n vertices, how many regions and edges does it have?
15. Suppose G is a planar graph that is not necessarily connected, as required in Euler’s formula (Theorem 2). Recall that a component H is a connected subgraph of G with the property that there is no path between any vertex in H and any vertex of G not in H.
(a) Find the appropriate modification of Euler’s formula for a planar graph with c components.
(b) Show that the corollary is valid for unconnected planar graphs.
16. Prove that if a graph G has 11 vertices, then either G or its complement G must be nonplanar. (Hint: Determine the total number N11 of edges in a complete graph on 11 vertices; if the result were false and G and its complement were each planar, how many of the N11 edges could be in each of these two graphs?) 17. Mimic the argument in the corollary to prove that e ≤3r −6 in a planar graph with each vertex of degree ≥3.
18. (a) Prove that every connected planar graph has a vertex of degree at most 5.
[Hint: Assume that every vertex has degree at least 6 and obtain a contradic-tion as follows. Get a lower bound, involving v, on the sum of the degrees of vertices (similar to the lower bound on the sum of the degrees of regions obtained in the proof of the corollary). Since this sum of degrees equals 2e, the corollary’s bound on e can be used to get an upper bound, also involving v, on this sum of degrees. Combining these two bounds yields the desired contradiction.] www.itpub.net 1.4 Planar Graphs 43 (b) Show that part (a) immediately generalizes to any (unconnected) planar graph.
19. Prove that every connected planar graph with less than 12 vertices has a vertex of degree at most 4. [Hint: Assume that every vertex has degree at least 5 to obtain a lower bound on e (together with the upper bound on e in the corollary) that implies v ≥12.] 20. If G is a connected planar graph with all circuits of length at least k, show that the inequality e ≤3v −6 can be strengthened to e ≤ k k−2(v −2). (Hint: The degree of a region will be at least k.) 21. (a) Show that every circuit in the graph in Exercise 3(l) has at least five edges.
(b) Use part (a) and the result of Exercise 20 to show that this graph is nonplanar.
22. (a) Give an example of a graph with regions consisting solely of squares (regions bounded by four edges) and hexagons, and with vertices of degree at least 3.
(b) Write an expression for the sum of the degrees of the vertices (=2e) in such a graph in terms of v and s, the number of squares. Then use Exercise 17 to get an upper bound on 2e. Deduce that any graph of the sort defined in part (a) has at least six squares.
(c) If each vertex has degree 3, show that any graph of the sort defined in part (a) has exactly six squares.
23. If G is a connected planar graph where e = 3v −6, show that every region is triangular (has three boundary edges).
24. A Platonic graph is a planar graph in which all vertices have the same degree d1 and all regions have the same number of bounding edges d2, where d1 ≥3 and d2 ≥3. A Platonic graph is the “skeleton” of a Platonic solid, for example, an octahedron.
(a) If G is a Platonic graph with vertex and face degrees d1 and d2, respectively, then show that e = 1 2d1v and r = (d1/d2)v.
(b) Using part (a) and Euler’s formula, show that v(2d1 + 2d2 −d1d2) = 4d2.
(c) Since v and 4d2 are positive integers, we conclude from part (b) that 2d1 + 2d2 −d1d2 > 0. Use this inequality to prove that (d1 −2)(d2 −2) < 4.
(d) From part (c), find the five possible pairs of positive (integral) values of d1, d2.
25. Suppose that l lines are drawn through a circle and these lines form p points of intersection (involving exactly two lines at each intersection). How many regions r are formed inside the circle by these lines? Assume that the lines end at the edge of the circle at 2l distinct points.
26. Consider an overlapping set of four circles A, B, C, D. One would like to position the circles so that every possible subset of the circles forms a region, e.g., four regions each contained in just one (different) circle, six regions formed by the intersection of two circles (AB, AC, AD, BC, BD, CD), four regions formed by the 44 Chapter 1 Elements of Graph Theory intersection of three of the four circles, and one region formed by the intersection of all four circles. Prove that it is not possible to have such a set of 15 bounded regions.
27. Show that the following graphs can be drawn on the surface of a doughnut (torus) without crossing edges: (a) K3,3 (b) K5 (c) K6 1.5 SUMMARY AND REFERENCES This chapter introduced graphs, their applications, and some of their basic structures.
This text takes a user-oriented approach to graph theory. Readers interested in a more formal graph theory text presenting the subject as an interesting area of pure mathematics should see the books by Bondy and Murty , West , or Wilson .
Section 1.1 introduced a set of illustrative graph models. The basic structure of graphs was explored in Section 1.2 under the guise of determining what makes two graphs different. Section 1.3 presented some useful edge-counting results. The final section introduced the important class of planar graphs. It surveyed ad hoc and theoretical approaches for determining whether a graph is planar.
The history of graph theory begins with the work of L. Euler in 1736 on Euler cycles(discussedinSection2.1).Euler’sformulaforplanargraphs,originallystatedin terms of polyhedra, was proved in 1752. Bits of graph theory appeared in papers about topology and geometric games, but it was not until around 1850 that formal studies of graphs began to appear. One was A. Cayley’s 1857 paper counting the number of trees (discussed in Chapter 3). Another was G. Kirchhoff’s 1847 paper presenting an algebra of circuits and introducing graphs in the study of electrical circuits. This same paper contains Kirchhoff’s famous current and voltage laws (Kirchhoff was 21 when he wrote this historic paper). The term graph was first used by J. Sylvester in 1877. The first book on graph theory, by D. Konig, did not appear until 1936. An excellent sourcebook on the history of graph theory is Graph Theory 1736–1936 by Biggs, Lloyd, and Wilson .
See the General References (at the end of the book) for a list of other introductory texts on graph theory.
1.
J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, American Elsevier, New York, 1976.
2.
N. Biggs, E. Lloyd, and R. Wilson, Graph Theory 1736–1936, Cambridge University Press, Cambridge, 1999.
3.
D. West, Introduction to Graph Theory, 2nd ed., Prentice-Hall, Saddle River, N. J., 2001 4.
R. Wilson, Introduction to Graph Theory, 4th ed., John Wiley and Sons, New York, 1997.
www.itpub.net Supplementary Exercises 45 SUPPLEMENTARY EXERCISES S u m m a r y o f E x e r c i s e s Graph theory is a field famous for its in-teresting problems. Several exercises introduce new graph concepts, such as strong connectedness (Exercise 14) and cut-set (Exercise 25). Exercise 30 is a very famous problem in Ramsey theory (see Appendix A.4). For more problems in graph the-ory, see any of the graph theory texts listed in the References at the end of this text.
1. Suppose that there are seven committees with each pair of committees having a common member and each person being on two committees. How many people are there?
2. Show that the complement of Kn, a complete graph on n vertices, is a set of n isolated vertices.
3. A graph is regular if all vertices have the same degree. If a graph with n vertices is regular of degree 3 and has 18 edges, determine n.
4. If a graph has 50 edges, what is the least number of vertices it can have?
5. Show that at least two vertices have the same degree in any graph with at least two vertices. (Hint: Be careful about vertices of degree 0.) 6. Show that an undirected graph with all vertices of degree ≥2 must contain a circuit (edges cannot be repeated in a circuit).
7. If every vertex in the graph G has degree 2, does every vertex lie on a circuit?
Prove, or give a counterexample.
8. If every vertex in a graph G has degree ≥d, then show that G must contain a circuit of length at least d + 1.
9. If every vertex in a directed graph G has positive out-degree (at least one out-wardly directed edge), (a) Must G contain a directed circuit?
(b) Must every vertex of G be on a directed circuit?
10. If G is a connected graph that is not a complete graph, show that some vertex, call it x, has two neighbors, call them y, z, that are not adjacent to each other [that is, there are edges (x, y) and (x, z) but not edge (y, z)].
11. Show that if a graph is not connected, then its complement must be connected.
12. Show that removal of some vertex x disconnects the connected undirected graph G if and only if there are two vertices a and b in G such that all paths in G from a to b pass through x.
13. Let G be a connected graph such that G-x is not connected for all but two vertices x of G. Show that G is a path.
14. A directed graph G is called strongly connected if there is a directed path from x to y for any two vertices x, y in G. Direct the edges in the following graphs to 46 Chapter 1 Elements of Graph Theory make the graphs strongly connected. If not possible, explain why. (An application of this problem is making streets in a city one-way.) (a) Figure 1.4 (b) Figure 1.2 15. Prove that G is strongly connected (see Exercise 14) if and only if G’s vertices cannot be partitioned into two sets V1, V2 such that there are no edges from a vertex in V1 to a vertex in V2.
16. A bridge is an edge in a connected graph whose removal disconnects the graph.
Show that if a graph G contains a bridge, then it cannot have a circuit that contains all vertices of G.
17. Show that if every edge in a connected graph G lies on a circuit, then G cannot have a bridge (a bridge is defined in Exercise 16).
18. Prove that the edges of a connected undirected graph G can be directed to create a strongly connected graph (see Exercise 14) if and only if there is no bridge in G (see Exercise 16).
19. Draw planar graphs with the following types of vertices, if possible: (a) Six vertices of degree 3 (b) i vertices of degree i, i = 1, 2, 3, 4, 5 (c) Two vertices of degree 3 and four vertices of degree 5 20. Let I be a set of independent vertices in a graph G and let C be a set of vertices that forms a complete subgraph. Show that I and C have at most one vertex in common.
21. Is it possible to have a connected simple graph containing six vertices with degrees: 5, 2, 2, 2, 2, 1? If so, draw such a graph.
22. Mr. Megabucks invites three married couples to his penthouse for dinner. Upon arrival, Mr. Megabucks and the six guests shake the hands of some of the other people (none of the guests shakes hands with his or her own spouse). Suppose each of the six guests shakes a different number of hands (possibly one person shakes no hands). By building a graph model of this situation (with seven ver-tices for the six guests and Mr. Megabucks), determine exactly how many hands Mr. Megabucks must have shaken. (Hint: First determine the six different num-bers of hands shaken by the different guests.) 23. A round-robin tournament can be represented by a complete directed graph with vertices for competitors and an edge (a⃗ , b) if a beats b. Each competitor plays every other competitor once. A vertex’s (competitor’s) score will be its out-degree (number of victories). Show that if vertex x has a maximum score among vertices in a round-robin tournament, then for any other vertex y, either there is an edge (x⃗ , y) or for some w there are edges (x⃗ , w) and (w⃗ , y).
24. Suppose the round-robin tournament graph (see Exercise 23) has no directed circuits. We define a ranking of vertices (competitors) as follows. A vertex with no outward edge has a rank of 0. In general, a vertex has rank k if it has an edge directed to a rank k −1 vertex and all other edges directed to lower ranks www.itpub.net Supplementary Exercises 47 (≤k −1). Show that a directed complete graph with no directed circuits always has such a ranking, and that each vertex will have a different rank.
25. A trail is a sequence of vertices with consecutive vertices joined by a distinct edge (no edge can be repeated). Unlike a path, a vertex can be visited any number of times in a trail. A cycle is a trail that starts and ends at the same vertex. A cycle that repeats no vertices is a circuit.
(a) Show that a subset of the vertices on a trail from x to y can be used to make a path from x to y.
(b) Prove, or give a counterexample: If x and y lie on a cycle, then they must lie on a circuit.
(c) Show that the edges in a cycle can be partitioned into a collection of circuits.
(d) Show that if C is an odd-length cycle, then a subset of C’s edges forms an odd-length circuit.
26. Consider a collection of circles (of varying sizes) in the plane. Make a circle graph with a vertex for each circle and an edge between two vertices when they correspond to two circles that cross (if one circle properly contains another, there would be no edge).
(a) Draw a family of circles whose circle graph is isomorphic to K4.
(b) Draw a family of circles whose circle graph is the graph in Figure 1.3 (ignoring edge directions).
(c) Draw a family of circles whose circle graph is isomorphic to K3,3.
27. A cut-set S is a set of edges in a connected undirected graph G whose removal disconnects G, but such that no proper subset of S can disconnect G.
(a) Find a cut-set of minimal size in Figure 1.22a.
(b) Show that every cut-set has an even number of edges in common with any circuit (remember that 0 is an even number).
28. A graph with n vertices and n + 2 edges must contain two edge-disjoint circuits.
Prove or give a counterexample.
29. Show that if an n-vertex graph has more than 1 2(n −1)(n −2) edges, then it must be connected. (Hint: The most edges possible in a disconnected graph will occur when there are two components, each complete subgraphs.) 30. Show that an n-vertex graph cannot be a bipartite graph if it has more than 1 4n2 edges.
31. Suppose that G is a connected graph containing no triangles and that every pair of two non-adjacent vertices in G has exactly two neighbors in common. Show that every vertex of G must have the same degree. (Hint: Show that any pair of adjacent vertices must have the same degree.) 32. (Famous Ramsey theory problem) Let each edge of a complete graph on six vertices be painted red or white. Show that there must always be either a red triangle of three edges or a white triangle of three edges.
48 Chapter 1 Elements of Graph Theory 33. (a) Find a graph that is isomorphic to its own complement.
(b) Show that any self-complementary graph [as in part (a)] must have either 4k or 4k + 1 vertices, for some integer k. (Hint: Use the fact that G and G both must have the same number of edges.) 34. Suppose that each path in a certain 7-vertex planar graph contains an even number of edges (zero edge or two edges or four edges, etc.). Draw the graph.
(Hint: This is a “trick” problem.) 35. Show that a directed graph has no directed circuits if and only if its vertices can be indexed x1, x2, . . . , xn, so that all edges are of the form (xi⃗ , x j), i < j.
36. Let Km,n be a complete bipartite graph, with m > n. What is the size of the smallest edge cover of Km,n? What is the size of the largest independent set?
37. A line graph L(G) of a graph G has a vertex of L(G) for each edge in G and an edge between two vertices in L(G) corresponding to two edges of G with a common end vertex.
(a) Draw a line graph of the left graph in Figure 1.6.
(b) Show that each vertex in L(Kn) has degree 2(n −2).
(c) Find all graphs that are isomorphic to their own line graph.
38. Show that if a graph H is the line graph (see Exercise 37) of some graph, then the edges of H can be partitioned into a collection of complete subgraphs such that each vertex of H is in exactly two such complete subgraphs.
39. An automorphism of a graph is an isomorphism (1 −1 mapping preserving adjacency) of the vertices of a graph with themselves. Find an automorphism of the graph in (a) Figure 1.1a (b) Figure 1.4 (c) Figure 1.13 40. (a) Show that there is no way to pair off the 14 vertices in the graph below with seven edges.
(b) Generalize part (a) to the problem of trying to use 31 dominoes to cover the 62 squares of an 8 × 8 chessboard with its two opposite corner squares removed.
41. Suppose circuits C1 and C2 have common edges (but C1 ̸= C2). Show that the edges in (C1, ∪C2) −(C1, ∩C2) form a circuit (or collection of circuits).
42. If the graph G has 2n vertices and no triangles, then show that G cannot have more than n2 edges.
www.itpub.net CHAPTER 2 COVERING CIRCUITS AND GRAPH COLORING 2.1 EULER CYCLES In this chapter we examine two graph-theoretic concepts that have many important applications. One is edge sequences, which visit either all the edges in a graph once or all the vertices once. The other is graph coloring, a concept introduced in Example 1 of Section 1.4.
In some applications, such as a highway network linking a group of cities, it is natural to permit several edges to join the same pair of vertices. Such generalized graphs are called multigraphs; loop edges, of the form (a, a), are also allowed in multigraphs. Figure 2.1b displays a multigraph.
At the beginning of Chapter 1, we defined a path P = x1–x2–· · ·–xn to be a sequence of distinct vertices with each pair of consecutive vertices in P joined by an edge. When there is also an edge (xn, x1), the sequence is called a circuit, written x1–x2–· · ·–xn–x1. Just as we generalized our notion of a graph to a multigraph, now we generalize our definitions of path and circuit in order to allow repeated vertices. A trail T = x1–x2–· · ·–xn is a sequence of vertices (not necessarily distinct) in which, like a path, consecutive vertices are joined by an edge. However, no edge can be repeated in a trail. In Figure 2.1b, A–B–D–B–C is a trail (we assume that segments B–D and D–B on this trail are using the two different edges joining B and D). When there is also an edge (xn, x1), the sequence of vertices is called a cycle, written x1–x2–· · ·–xn–x1.
Example 1: Konigsberg Bridges The old Prussian city of Konigsberg was located on the banks of the Pregel River.
Part of the city was on two islands that were joined to the banks and each other by seven bridges, as shown in Figure 2.1a. The townspeople liked to take walks, or Spaziergangen, about the town across the bridges. Several people were apparently bothered by the fact that no one could figure out a walk that crossed each bridge just once, for they brought this problem to the attention of the famous mathemati-cian Leonhard Euler. He solved the Spaziergangen problem, thereby giving birth to graph theory and immortalizing the Seven Bridges of Konigsberg in mathematics texts.
49 50 Chapter 2 Covering Circuits and Graph Coloring A B C D A D C B (a) (b) Figure 2.1 We can model this walk problem with a multigraph having a vertex for each body of land and an edge for each bridge. See Figure 2.1b. The desired type of walk corresponds to what we now call an Euler cycle. An Euler cycle is a cycle that contains all the edges in a graph (and visits each vertex at least once). Readers should convince themselves that no Euler cycle exists for this multigraph.
A multigraph possessing an Euler cycle will have to have an even degree at each vertex, since each time the cycle passes through a vertex it uses two edges. A second obvious property is that the multigraph must be connected. Euler showed that these two properties are also sufficient to guarantee the existence of an Euler cycle. Before proving this theorem, let us first build an Euler cycle by ad hoc methods in a graph that is connected and has even-degree vertices. Then we extend our construction to a general proof of the Euler cycle theorem.
Observe first that the even-degree condition is so powerful that it creates an apparent paradox. Suppose we start at some vertex a in a graph G with all even-degree vertices and start building an Euler cycle by tracing out an arbitrary sequence of edges (without repeating any edges). The assumption of all even degrees implies not only that there will be an even number of edges the first time we come to any vertex, but also that there will be even number of unused edges after we leave that vertex, since entering and leaving a vertex uses two edges. This in turn implies that every time we come to that vertex again, it will have an even number of unused edges.
We can only be forced to stop if we enter a vertex with just one unused edge (the edge by which we enter the vertex), but we just showed that this can never happen. So it appears that we are able to continue our sequence forever, even though any graph has only a finite number of edges. The paradox is resolved by noting that when we started off from a (leaving it without first having entered it), a became an odd-degree vertex and so our sequence of edges must eventually end at a.
Example 2: Building an Euler Cycle Build an Euler cycle for the graph in Figure 2.2a that is connected and has even-degree vertices.
Let us start by blindly tracing out a trail from vertex a. Recall that a trail is like a path, except that vertices may be repeated. Suppose we go a-d-j-n-o-k-l-h-f-e-b-a.
www.itpub.net 2.1 Euler Cycles 51 c a b d e f g m l h o k n j i c d e i j k g m h (a) (b) Figure 2.2 Now we are back to a. Note that the even degree property means that we can always leave any vertex we enter except a. There are no vertices of degree 1 or that are reduced to degree 1 after being visited several times. Thus, any trail we trace from a will never be forced to end at another vertex and so must eventually come back to a, forming a cycle; in this case the cycle is a circuit (no vertex is visited twice).
Next consider the graph of remaining edges, shown in Figure 2.2b. It is no longer connected, but all vertices still have even degree (removing the cycle reduced each degree by an even amount). Each connected piece of the remaining graph has an Euler cycle: d-c-i-j-k-e-d and h-g-m-h. These two cycles can be inserted into the original cycle at vertices d and h, respectively, yielding the Euler cycle a-d-c-i-j-k-e-d-j-n-o-k-l-h-g-m-h-f-e-b-a.
Theorem 1 (Euler, 1736) A multigraph has an Euler cycle if and only if it is connected and has all vertices of even degree.
Proof The proof generalizes the construction in Example 2. As noted above, when an Euler cycle exists, the multigraph must be connected and have all vertices of even degree.
Suppose a multigraph G satisfies these two conditions. Then pick any vertex a and trace out a trail. As in Example 2, the even-degree condition means that we are never forced to stop at some other vertex, and so eventually the trail must terminate at a (possibly it passes through a several times before finally being forced to stop at a).
Recall that upon leaving a at the start, a now has odd degree. This is why we are eventually forced to stop at a. Let C be the cycle thus generated and let G′ be the multigraph consisting of the remaining edges of G–C. As in Example 2, G′ may not be connected but it must have all vertices of even degree. Since the original graph was connected, C and G′ must have a common vertex or else there is no path from vertices in C to vertices in G′. Let a′ be such a common vertex. Now build a cycle C′ tracing through G′ from a′ just as C was traced in G from a. Incorporate C′ into the cycle C at a′, as done in Example 2, to obtain a new larger cycle C′. Repeat this 52 Chapter 2 Covering Circuits and Graph Coloring process by tracing a cycle in the graph G′′ of still remaining edges and incorporate the cycle into C′ to obtain C′′. Continue until there are no remaining edges. ◆ Let us now consider an application of Euler cycles to an urban systems problem.
Example 3: Routing Street Sweepers Suppose the graph of solid edges in Figure 2.3 represents a collection of blocks to be swept by a street sweeper in a certain district of some city from 10 A.M. to 11 A.M.
(whenparkingontheseblocksisforbidden).Theloopingedgeatkrepresentsacircular street. We want a tour that sweeps each solid edge once. That is, we want an Euler cycle of the solid edges. Unfortunately, the graph of edges to be swept in such applica-tions rarely has all vertices of even degree. Frequently the graph is also not connected.
In such cases we must traverse extra edges, called deadheading edges, to obtain the desired tour. This creates a new problem: minimizing the number of deadheading edges.
The desired tour will be an Euler cycle of the multigraph of sweeping and dead-heading edges—a multigraph because some edges may be repeated in deadheading.
By the theorem, this multigraph must be connected and have all vertices of even degree. Thus to minimize deadheading, the deadheading edges should be a minimal set of extra edges with the property that when added to the original graph all vertices have even degree and the new multigraph is connected. Suppose the dashed edges in Figure 2.3 are such a minimal set of extra edges.
There is one more problem to be faced once we have a multigraph possessing an Euler cycle. We want to build an Euler cycle that minimizes turns at corners, especially U-turns. Even right-hand turns can tie up traffic in busy cities. We can successively look at each corner (vertex) and pair off the edges at the corner so as to minimize disruption on the tour’s passes through that corner. For the multigraph in Figure 2.3, we would go straight through corners d, e, h, i on visits to these corners; we would go straight through j once and turn once [going between ( f , j) and (k, j)]; we must turn both times at k; and at all other corners (of degree 2) we would have forced turns.
The short lines near each vertex in Figure 2.3 indicate these edge pairings. Of course, such edge pairings are very unlikely to produce a single Euler cycle. In Figure 2.3, we get two cycles from these pairings: a-b-e-i-m-l-h-d-a and c-d-e-f-j-k-k-j-i-h-g-c. Now a b d c g h l m j i e f k Figure 2.3 www.itpub.net 2.1 Euler Cycles 53 we break the two cycles at a common vertex and fuse them together. For example, at d, change the edge pairings to d The result is an Euler cycle with optimal pairings at all but one corner.
A large street network with 200 edges typically forms only three or four cycles when minimal-disruption edge pairing is performed at each corner, and so only a few pairings need to be changed. In practice, it is also necessary to use directed graphs since street sweepers cannot move against the traffic on their side (curb) of the street.
It is interesting to note that the street sweeping problem gives rise to an alternative proof of the Euler cycle theorem as follows.
Alternative Proof of Theorem The even-degree condition means that we can pair off (and link together) the edges at each vertex. These linked edges form a set of cycles. The connectedness condition means that these cycles have common vertices where they can be joined to form a single cycle—an Euler cycle. ◆ This proof is not as intuitive or pictorial as our original path-tracing method, but it is both simpler and more applicable.
We conclude this section by extending the concept of an Euler cycle to an Euler trail. An Euler trail is a trail that contains all the edges in a graph (and visits each vertex at least once).
Corollary A multigraph has an Euler trail, but not an Euler cycle, if and only if it is connected and has exactly two vertices of odd degree.
Proof Suppose a multigraph G has an Euler trail but not an Euler cycle. Call it T. Then the starting and ending vertices of T must have odd degree while all other vertices have even degree (by the same reasoning that showed all vertices have even degree in an Euler cycle). Also the graph must be connected.
On the other hand, suppose that a multigraph G is connected and has exactly two vertices, p and q, of odd degree. Add a supplementary edge (p, q) to G to obtain the graph G′. G′ is connected and has all vertices of even degree. Hence by the Euler cycle theorem, G′ has an Euler cycle, call it C. Now remove the edge (p, q) from C.
This removal reduces the Euler cycle to an Euler trail that includes all edges of G. ◆ 54 Chapter 2 Covering Circuits and Graph Coloring 2.1 EXERCISES S u m m a r y o f E x e r c i s e s The first four exercises involve trying to build Euler cycles. Exercises 5–8 and 11–14 present extensions and other questions related to the Euler cycle theorem. The remaining questions involve some modeling and further theory, and the last two exercises ask for computer programs.
1. (a) Build an Euler cycle for the right graph in Figure 1.8.
(b) Build an Euler trail for the graph in Figure 1.22a with edge (d, g) removed.
2. (a) For which values of n does Kn, the complete graph on n vertices, have an Euler cycle?
(b) Are there any Kn that have Euler trails but not Euler cycles?
(c) For which values of r and s does the complete bipartite graph Kr,s have an Euler cycle?
3. Find a graph G with seven vertices such that G and its complement both have an Euler cycle.
4. What is the minimum number of times one must raise one’s pencil in order to draw the graph in Figure 1.4?
5. (a) Can a graph with an Euler cycle have a bridge (an edge whose removal disconnects the graph)? Prove or give a counterexample.
(b) Give an example of a 10-edge graph with an Euler trail that has a bridge.
6. Give an argument for and an argument against the statement that a 1-vertex graph (with no edges) has an Euler cycle.
7. SupposethatinthedefinitionofanEulercycle,wedroptheseeminglysuperfluous requirement that the Euler cycle visit every vertex and require only that the cycle include every edge. Show that now the theorem is false. Draw a graph that illustrates why the theorem is now false.
8. Prove that if a connected graph has a 2k vertices of odd degree, then there are k disjoint trails that contain all the edges.
9. The matrix below marks with a 1 each pair of the set of racers A, B, C, D, E, F who are to have a drag race together. It is most efficient if a racer can run in two races in a row (but not in three in a row). Is it possible to design a sequence of races such that one of the racers in each race (except the last race) also runs in the following race (but not three in a row)? If possible, give the sequences of races (pairs of racers); if not, explain why not.
A B C D E F A — 1 0 1 1 0 B 1 — 1 1 0 1 C 0 1 — 0 1 0 D 1 1 0 — 1 1 E 1 0 1 1 — 1 F 0 1 0 1 1 — www.itpub.net 2.1 Euler Cycles 55 10. Is it possible for a knight to move around an 8 × 8 chessboard so that it makes every possible move exactly once (consider a move between two squares con-nected by a knight to be completed when the move is made in either direction)?
11. Show that in Example 3 the minimum set of deadheading edges in any sweeping problem will be a collection of edges forming paths between pairs of different odd-degree vertices.
12. Prove the directed version of the Euler cycle theorem: a directed multigraph has a directed Euler cycle if and only if the multigraph is connected (when directions are ignored) and the in-degree equals the out-degree at each vertex.
(a) Model your proof after the argument in the proof of the theorem.
(b) Model your proof after the argument in the alternative proof of the theorem.
(c) Build a directed Euler cycle for the graph below.
a b c d e j i h g f 13. State and prove a directed multigraph version of the corollary.
14. A directed graph is called strongly connected if there is a directed path from any given vertex to any other vertex. Show that if a directed graph possesses a directed Euler cycle, then it must be strongly connected.
15. Try to find a minimal set of edges in the graph below whose removal produces an Euler cycle. (Hint: Tricky.) e a f b g h c d i j 16. The line graph L(G) of a graph G has a vertex for each edge of G, and two of these vertices are adjacent if and only if the corresponding edges in G have a common end vertex.
(a) Show that L(G) has an Euler cycle if G has an Euler cycle.
(b) Find a graph G that has no Euler cycle but for which L(G) has an Euler cycle.
17. Consider the following algorithm due to Fleury for building an Euler cycle, when one exists, in a single pass through a graph (without later adding side cycles as in the proof of the theorem). Starting at a chosen vertex a, build a cycle and erase edges after they are used (also erase vertices when they become isolated points). The one rule to follow in the cycle building is that one never 56 Chapter 2 Covering Circuits and Graph Coloring chooses an edge whose erasure will disconnect the resulting graph of remaining edges.
(a) Apply this algorithm to build Euler cycles for the graph in (i) Figure 2.2a (ii) Figure 2.3 (including deadheading edges) (b) Prove that this algorithm works.
(c) Does this algorithm work for Euler trails? Explain.
18. Suppose we are given an undirected connected graph representing a network of two-way streets.
(a) Show that there always exists a tour of the network in which a person drives along each side of every street once.
(b) Show that the tour in part (a) can be generated by the following rule: at any intersection, do not leave by the street first used to reach this intersection unless all other streets from the intersection have been used.
19. A set of eight binary digits (0 or 1) are equally spaced about the edge of a disk.
We want to choose the digits so that they form a circular sequence in which every subsequence of length three is different. Model this problem with a graph with four vertices, one for each different subsequence of two binary digits. Make a directed edge for each subsequence of three digits whose origin is the vertex with the first two digits of the edge’s subsequence and whose terminus is the vertex with the last two digits of the edge’s subsequence.
(a) Build this graph.
(b) Show how an Euler cycle (which exists for this graph) will correspond to the desired 8-digit circular sequence.
(c) Find such an 8-digit circular sequence with this graph model.
(d) Repeat the problem for 4-digit binary sequences.
20. Write computer programs for finding an Euler cycle, when one exists, in a multi-graph: (a) Use the method in the proof of the theorem.
(b) Repeat part (a) for a directed graph.
(c) Use the method in the alternative proof of the theorem.
21. Write a program to implement the algorithm in Exercise 17.
2.2 HAMILTON CIRCUITS In this section we explore Hamilton circuits and paths, circuits and paths that visit each vertex in a graph exactly once. Hamilton circuits arise in operations research problems such as routing a delivery truck that must visit a set of stores. In these www.itpub.net 2.2 Hamilton Circuits 57 applications, the most efficient solution will be obtained by finding a minimal-cost Hamilton circuit (this problem is discussed in Section 3.4).
The problem of determining whether a graph has an Euler cycle was answered by the Euler cycle theorem, which tells us that a graph has an Euler cycle if and only if all vertices have even degree and the graph is connected. Such a nice, sim-ple answer is very unusual in graph theory (which is probably why the Euler cycle theorem was the first result proved in the field of graph theory). In this sec-tion, we return to graph theory normalcy: There is no simple way to determine whether or not an arbitrary graph has a Hamilton circuit or a Hamilton path. In-deed, finding a Hamilton circuit or path in a graph is an NP-complete problem (see Appendix A.5).
Finding Hamilton circuits by inspection, when they exist, is usually not too hard in moderate-sized graphs, but proving that no Hamilton circuit exists in a given graph can be very difficult. Such a proof typically involves the same type of reasoning with the AC Principle needed to show that two graphs are not isomorphic or that a graph is nonplanar. At the end of this section we present a sampling of the theory that has been developed about the existence of Hamilton circuits. These theorems give various special conditions on a graph that guarantee the existence of a Hamilton circuit. Most graphs satisfy none of these conditions.
Our focus will be proving that a Hamilton circuit does not exist in particular graphs. This nonexistence problem requires the type of systematic logical analysis that is the essence of most applied graph theory. To prove nonexistence, we must begin building parts of a Hamilton circuit and then show that the construction must always fail. This is similar to the way in Section 1.4 that we showed a graph to be nonplanar by building a circuit and then adding chords in a structured fashion that forced two edges to cross.
Our analysis uses three simple rules that must be satisfied by the set of edges used to build a Hamilton circuit. The first step in applying the AC Principle (Assumptions generate helpful Consequences) is that if a graph has a Hamilton circuit, then any such Hamilton circuit must contain exactly two edges incident to each vertex. From this consequence three further consequences follow.
Rule 1.
If a vertex x has degree 2, both of the edges incident to x must be part of any Hamilton circuit.
Rule 2.
No proper subcircuit—that is, a circuit not containing all vertices—can be formed when building a Hamilton circuit.
Rule 3.
Once the Hamilton circuit is required to use two edges at a vertex x, all other (unused) edges incident at x must be removed from consideration.
Example 1: Nonexistence of Hamilton Circuit I Show that the graph in Figure 2.4 has no Hamilton circuit.
We can apply Rule 1 at vertices a, b, d, and e. To indicate that the two edges at each of these vertices must be used, we draw a little line segment from one edge to 58 Chapter 2 Covering Circuits and Graph Coloring a b e d c Figure 2.4 the other, as shown in Figure 2.4. There are two types of contradictions that have now arisen. First, using Rule 1 for vertices a and d forces the Hamilton circuit to use edges (a, c), (a, d), (d, c) in forming a triangle; this violates Rule 2. Second, using Rule 1 for all four vertices of degree 2 forces the Hamilton circuit to contain more than two edges incident at c. Either of these difficulties proves that this graph has no Hamilton circuit.
In the next two examples, our three rules will be used in the following sort of sequential reasoning. When Rule 3 requires us to delete edges at some vertex q, one of q’s deleted edges may go to a vertex r of degree 3. After the deletion of (q, r), r has degree 2, requiring by Rule 1 that its remaining two edges be used. As certain edges are forced to be deleted, other edges are forced to be used, which in turn forces still other edges to be deleted, and so on.
Example 2: Nonexistence of Hamilton Circuit II Show that the graph in Figure 2.5 has no Hamilton circuit.
We can apply Rule 1 at vertices a and g, so that the subpaths b-a-c and e-g-i must be part of any Hamilton circuit. Next consider vertex i. We already know that (g, i) must be part of the circuit. Since the graph is symmetric with respect to edges (i, j) and (i, k), it does not matter which of these two edges we choose as the other edge incident to i on the Hamilton circuit. Suppose we pick (i, j). If we obtain a contradiction using (i, j), then by symmetry we would also obtain a contradiction with (i, k). The situation can be visualized as follows: if we marked (i, j) and then held a page with the graph in front of us (facing away from us) and viewed it in a mirror, it would appear that we had chosen (i, k); every subsequent move starting with (i, j) would have a mirror image move starting from (i, k).
Having chosen to use (i, j) as the second edge at i, we delete the other edge at i, (i, k), by Rule 3. See Figure 2.5. Each time we apply a rule, we need to check how a b c d e f g h i j k Figure 2.5 www.itpub.net 2.2 Hamilton Circuits 59 other edges and vertices are affected. Does any vertex now have only two remaining edges at it, allowing Rule 1 to be applied? Or are two edges now required to be used at some vertex, allowing Rule 3 to be applied? Also, is there some edge that, if used, would complete a subcircuit, allowing Rule 2 to be used to delete it?
Deleting (i, k) reduces the degree of k to 2, and so Rule 1 requires that we use both remaining edges incident at k, ( j, k) and (h, k). Edge ( j, k) is the second edge used that is incident to j. Then by Rule 3, j’s other edge ( f , j) must be deleted. This deletion reduces the degree of f to 2. See Figure 2.5. So we must use the two remaining edges at f, (b, f ) and ( f , e). Edge (b, f ) is the second edge used at b, and so edge (b, d) at b must be deleted. Similarly, edge ( f , e) is the second edge used at e, and so e’s other edges, (e, d) and (e, h), must be deleted. (Note that after having determined the two edges at j and k, we have the path e-g-i-j-k-h, and the edge e-h must again be deleted by Rule 2 to avoid a subcircuit.) Deleting (e, h) forces the use at h of (c, h). Using (c, h) forces the deletion at c of (c, d). However, now we have deleted all the edges incident to d from consideration on a Hamilton circuit. This contradiction implies that G cannot have a Hamilton circuit.
Note that this graph does have Hamilton paths—for example, a-b-f-e-g-i-j-k-h-c-d.
It should be emphasized that the following is not a useful line of reasoning to show that a graph has no Hamilton circuit: start from some vertex and construct a route visiting successive vertices and show that all attempts to find a circuit through all vertices fail. Even in a simple graph like the one in Figure 2.5, there are hundreds of possible beginning subpaths for a Hamilton circuit that would all have to be checked.
The approach presented here requires much less work than a rigorous trial-and-error effort.
Example 3: Nonexistence of Hamilton Circuit III Show that the graph in Figure 2.6 has no Hamilton circuit.
Note that this graph has vertical and horizontal symmetry (although vertex n is off to one side, its adjacencies have a square-like symmetry). It sometimes takes a j a b c k n l g h f e d m i Figure 2.6 60 Chapter 2 Covering Circuits and Graph Coloring bit of trial-and-error experimenting to find a good vertex at which to start trying to build a Hamilton circuit when there are no vertices of degree 2. We seek a vertex with the property that once two edges are chosen at the vertex, then the use of Rules 1 and 3 will force the successive deletion and inclusion of many edges. Vertex e is such a vertex. We can either use two edges incident at e from opposite sides (180◦apart) or use two edges incident at e that form a 90◦angle. We must examine both cases to show that no Hamilton circuit can exist. (Hamilton circuits frequently have many such subcases that must all be checked out.) Case I Suppose we use two edges incident at e from opposite sides. By symmetry, they can either be edges from d and f or from b and h. Suppose we choose (d, e) and (e, f ). Then by Rule 3, we can delete edges (e, b) and (e, h). Then at b and at h we must use both remaining edges, getting subpaths a-b-c and g-h-i. Now at d we can use either edge (d, a) or edge (d, g). The two cases are symmetrical with respect to the edges chosen for the circuit thus far. So, without loss of generality, we can choose edge (d, a).
At f, we cannot use ( f , c) or else subcircuit a-b-c-f-e-d-a results. So we must use ( f , i). See Figure 2.6. Since we have used two edges at vertices a and i, the other edges at these vertices can be deleted by Rule 3. We now obtain several inconsistencies.
Vertices j, k, l, and m each now have degree 2. But each is incident to n, and so using the two remaining edges at each causes four edges to be used at n. We conclude that there is no Hamilton circuit in Case I.
Case II Suppose we use two edges incident at e that form a 90◦angle. By symme-try it does not matter which of the four 90◦angle pairs of edges we choose. Suppose we choose (b, e) and (d, e). See Figure 2.7. Then by Rule 3 we can delete edges (e, f ) and (e, h). Then at f and h we must use both remaining edges getting sub-paths c-f-i and g-h-i. See Figure 2.7. By Rule 3 at i, we can delete edges (i, k) and (i, l). By Rule 1, we must use the remaining edges at k and l, forming the subcircuit i-f-c-k-n-l-g-h-i.
Having obtained contradictions in both cases, we have proved that the graph has no Hamilton circuit.
j a b c k n l g h f e d m i Figure 2.7 www.itpub.net 2.2 Hamilton Circuits 61 We now present a few of the theoretical results about the existence of Hamilton circuits and paths. (See for proofs of Theorems 1, 2, and 3.) Theorem 1 (Dirac, 1952) A graph with n vertices, n > 2, has a Hamilton circuit if the degree of each vertex is at least n/2.
Theorem 2 (Chvatal, 1972) Let G be a connected graph with n vertices, and let the vertices be indexed x1, x2, . . . , xn, so that deg(xi) ≤deg(xi+1). If for each k ≤n/2 either deg(xk) > k or deg(xn−k) ≥n −k, then G has a Hamilton circuit.
Theorem 3 (Grinberg, 1968) Suppose a planar graph G has a Hamilton circuit H. Let G be drawn with any planar depiction, and let ri denote the number of regions inside the Hamilton circuit bounded by i edges in this depiction. Letr′ i be the number of regions outside the circuit bounded by i edges. Then the numbers ri and r′ i satisfy the equation i (i −2)(ri −r′ i) = 0 (∗) Just as the inequality e ≤3v −6 for planar graphs in the corollary in Section 1.4 could be used to prove that some graphs are not planar, Theorem 3 can be used to show that some planar graphs cannot have Hamilton circuits.
Example 4: Application of Theorem 3 Show that the planar graph in Figure 2.8 has no Hamilton circuit.
We have indicated that the number of bounding edges inside each of the regions in the planar depiction of the graph in Figure 2.8. There are three regions with four edges and six regions with six edges. Thus, no matter where a Hamilton circuit is drawn (if it exists), we know that r4 +r′ 4 = 3 and r6 +r′ 6 = 6. Observe that for this graph, (∗) reduces to 2(r4 −r′ 4) + 4(r6 −r′ 6) = 0 c d e f g h i j k l m a 4 6 6 6 6 6 q n o p b 6 4 4 Figure 2.8 62 Chapter 2 Covering Circuits and Graph Coloring We cannot have r6 −r′ 6 = 0, that is, r6 = r′ 6 = 3, for then (∗) would require r4 −r′ 4 = 0 or r4 = r′ 4—which is impossible, since r4 + r′ 4 = 3. If r6 −r′ 6 ̸= 0, then |r6 −r′ 6| ≥2, and so |4(r6 −r′ 6)| ≥8. Now it is impossible to satisfy (∗), since even if r4 = 3, r′ 4 = 0 (or r4 = 0, r′ 4 = 3), |2(r4 −r′ 4)| ≤6. Thus, it is impossible for (∗) to be valid for this graph, and so no Hamilton circuit can exist.
We next present a theorem involving directed graphs and Hamilton paths.
A tournament is a directed graph obtained from a complete (undirected) graph by giving a direction to each edge.
Theorem 4 Every tournament has a Hamilton path.
Proof The proof is byinduction.Fora2-vertextournament, adirectedHamiltonpathtrivially exists. Next assume by induction that any tournament with n −1 vertices, for n ≥3, has a directed Hamilton path and let us prove that a n-vertex tournament G has a directed Hamilton path.
Remove a vertex z from G, leaving a tournament G′ with n −1 vertices. By the induction assumption, G′ has a Hamilton path H = x1–x2–x3–· · ·–xn−1. If the edge between z and x1 is (z⃗ , x1), then z can be added to the front of H to ob-tain a Hamilton path for G. Similarly, if the edge between z and xn−1 is (xn−1⃗ , z), then z can be added to the end of H to obtain a Hamilton path for G. So assume that the edge from the first vertex x1 of H points toward z and that the edge from the last vertex xn−1 of H points from z. Then for some consecutive pair on H, xi−1, xi, the edge direction must change—that is, we have edges (xi−1⃗ , z) and (z⃗ , xi). We can insert z between xi−1 and xi in H to obtain a Hamilton path x1–x2–· · ·–xi−1–z–xi– · · ·–xn−1. ◆ We conclude this section with an application of Hamilton paths to a problem in coding theory.
Example 5: Gray Code When a spacecraft is sent to distant planets and transmits pictures back to earth, these pictures are transmitted as a long sequence of numbers, each number being a darkness value for one of the dots in the picture. For simplicity, assume the darkness numbers range between 1 and 8. These numbers are actually sent as a sequence of 0s and 1s. A straightforward encoding scheme would be to express each number in its binary representation, that is, 1 as 001, 2 as 010, 3 as 011, and so on, ending with 8 as 000.
However, a better scheme, called a Gray code, uses an encoding with the property that two consecutive numbers are encoded by binary sequences that are almost the www.itpub.net 2.2 Hamilton Circuits 63 same, differing in just one position. For example, a fragment of a Gray code might be 4 as 010, 5 as 011, 6 as 001. The advantage of such an encoding is that if an error from “cosmic static” causes one binary digit in a sequence to be misread at a receiving station on earth, then the mistaken sequence will often be interpreted as a darkness number that is almost the same as the true darkness number. For example, in the preceding fragment of a Gray code, if 011 (5) were transmitted and an error in the last position caused 010 (4) to be received, the resulting small change in darkness would not seriously affect the picture. (Of course, some errors will cause substantial inaccuracies.) With this background, we now translate the problem of finding a Gray code for the 8 darkness numbers into the problem of finding a Hamilton circuit in a graph. We define the graph as follows: Each vertex corresponds to a 3-digit binary sequence, and two vertices are adjacent if their binary sequences differ in just one place. The graph is shown in Figure 2.9a. Observe that it is a cube (the binary sequences can be thought of as the coordinates of a cube drawn in three dimensions). A similar graph can be drawn for the 16 4-digit binary sequences, or for any given n, the 2n n-digit binary sequences.
We claim that the order in which vertices (binary sequences) occur along a Hamil-ton path in this graph produces a Gray code. That is, 1 is encoded as the first binary sequence (vertex) in a given Hamilton path, 2 as the second binary sequence, and so on. This process yields a Gray code because consecutive vertices in the Hamil-ton path, which encode consecutive darkness numbers, will correspond to binary sequences that differ in just one position. Figure 2.9b illustrates how a Hamilton path in the graph produces a Gray code.
The graph with one vertex for each n-digit binary sequence and an edge joining vertices that correspond to sequences that differ in just one position is called an n-dimensional cube, or hypercube. The graph in Figure 2.9a is a 3-dimensional cube (or standard cube). An n-dimensional cube has 2n vertices, each of degree n. It has the property that the longest path between any two vertices has length n.
A generalization of a Hamilton circuit is used to solve the Instant Insanity puzzle in the supplement to this chapter.
5 = 011 6 = 111 3 = 110 2 = 100 1 = 000 4 = 010 8 = 001 7 = 101 011 111 110 100 000 010 001 101 (a) (b) Figure 2.9 64 Chapter 2 Covering Circuits and Graph Coloring 2.2 EXERCISES S u m m a r y o f E x e r c i s e s The first 10 exercises involve the exis-tence or nonexistence of Hamilton paths and circuits. Exercises 9 and 10 introduce other potential aids for proving nonexistence. Exercises 15–19 involve applications of Hamilton circuits. The last five exercises involve theory.
1. (a) Draw a graph with a Hamilton circuit but no Euler cycle.
(b) Draw a graph with an Euler cycle but no Hamilton circuit.
2. Find a Hamilton circuit in each of the following graphs.
(a) (b) (c) 3. Find a Hamilton circuit in the following graph.
a b c d e f g h i j k 4. In each of the following graphs, prove that no Hamilton circuit exists: a b c d e f g h i a b c d e f g h (a) (b) www.itpub.net 2.2 Hamilton Circuits 65 5 4 6 1 2 3 7 e f g a b c d i h a b c e d g f h i n m j k l (c) (d) (e) a d g h e f b c i ( f ) a b c d e f l g h i j k m (h) a f b g c j k i e h d (g) a b c d e f g i j k l m a b c d e f g h i j h ( j) ( i) c k b i l m d n f e a g h o j c n b m g i d j e a f k h (l) (k) 66 Chapter 2 Covering Circuits and Graph Coloring a c f k q p n o j e d h b m g l i a b c d g f h i e (n) (m) a h k d e f b c i j g a b c d e j f g h i ( p) (o) 5. Find a Hamilton path in Figure 2.8 and show that no Hamilton circuit exists (using the reasoning in Examples 1, 2, 3).
6. Show that there can be no Hamilton circuit in the following graph using both edges (a, f ) and (c, h).
e a f b g h c i j d 7. Prove that the following graphs have no Hamilton circuits.
a b c d e f g h i j a b c d e f g h i j k l m n o p (a) (b) www.itpub.net 2.2 Hamilton Circuits 67 8. Use Theorem 3 to show that the following planar graphs have no Hamilton circuit: (a) Exercise 4(a) (b) Exercise 4(b) (c) Exercise 4(p) 9. Recall from Example 1 in Section 1.1 that a graph is bipartite if the vertices can be divided into two sets; for convenience call them blue vertices and red vertices, such that every edge connects a blue and a red vertex.
(a) Show that if a connected bipartite graph has a Hamilton circuit, then the numbers of red and blue vertices must be equal. Further, show that if a bipartite graph has an odd number of vertices, then it has no Hamilton circuit.
(b) Show that if a connected bipartite graph has a Hamilton path, then the num-bers of reds and blues can differ by at most one.
(c) Use part (a) to show that the following graphs have no Hamilton circuit: (i) Figure 2.8 (ii) Exercise 4(m) (iii) Exercise 7(b) 10. Suppose a set I of k vertices in a graph G is an independent set—that is, no pair of vertices in I are adjacent. Then for each x in I, deg(x) −2 of the edges incident to x will not be used in a Hamilton circuit. Summing over all vertices in I, we have e′ = x∈I(deg(x) −2) = { x∈I(deg(x)} −2k edges that cannot be used in a Hamilton circuit.
(a) Let v and e be the numbers of vertices and edges in G, respectively. Show that if e −e′ < v, then G can have no Hamilton circuit.
(b) Why is the claim in part (a) valid only when I is a set of nonadjacent vertices?
(c) With a suitably chosen set I, use part (a) to show that the following graphs have no Hamilton circuits: (i) Figure 2.6 (ii) Exercise 4(p) (iii) Exercise 7(b) 11. (a) Draw a 4-dimensional hypercube graph.
(b) Use the graph in part (a) to find a Gray code for encoding the numbers 1 through 16 as 4-bit binary sequences.
12. Let the distance between two vertices in a connected graph be defined as the number of edges in the shortest path connecting those two vertices. Then the diameter of a graph is defined to be greatest distance between any two vertices in the graph. Show that a 4-dimensional hypercube has diameter 4. In general, show that a k-dimensional hypercube has diameter k. Note: No graph with 2k vertices, all of degree k, has a smaller diameter than k. This minimum is achieved by a k-dimensional hypercube.
13. Find a connected, cubic graph (all vertices have degree 3) with no Hamilton circuit.
14. Show without citing any theorems stated in this section that any 6-vertex, undi-rected graph with all vertices of degree 3 has a Hamilton circuit.
15. Find a path of knight’s moves visiting all squares exactly once on an 8 × 8 chessboard. (Hint: Very tricky—get help on the Internet.) 68 Chapter 2 Covering Circuits and Graph Coloring 16. Suppose a classroom has 25 students seated in desks in a square 5 × 5 array. The teacher wants to alter the seating by having every student move to an adjacent seat (just ahead, just behind, on the left, or on the right). Show that such a move is impossible.
17. (a) Describe how to construct a circuit including all squares of an n × n chess-board, n even, using a rook. Then do the same using a king.
(b) Repeat part (a) for n odd.
18. Consider 27 little cubes arranged in a 3 × 3 × 3 array (as in Rubik’s Cube).
Form an associated graph with 27 vertices, one for each little cube, and with two vertices adjacent if they have touching faces (not just edges). Does this graph have a Hamilton path starting at the vertex corresponding to the middle inside cube and ending at one of the vertices corresponding to a corner cube?
19. (a) How many different Hamilton circuits are there in Kn, a complete graph on n vertices?
(b) Show that Kn, n prime ≥3, can have its edges partitioned into 1 2(n −1) disjoint Hamilton circuits.
(c) If 17 professors dine together at a circular table during a conference, and if each night each professor sits next to a pair of different professors, how many days can the conference last?
20. (a) If a graph G has an Euler cycle, show that L(G), the line graph of G (see Exer-cise 16 of Section 2.1 for the definition of a line graph), has a Hamilton circuit.
(b) If G has a Hamilton circuit, show that L(G) has a Hamilton circuit.
(c) Show that the converses of parts (a) and (b) are false by finding counter-examples.
21. Show that if G is not a complete graph, then it is possible to direct the edges of G so that there is no directed Hamilton path.
22. Show that in a tournament (defined preceding Theorem 4) it is always possible to rank the contestants so that the person ranked ith beats the person ranked (i + 1)st. (Hint: Use Theorem 4.) 23. Show that Theorem 1 is false if the requirement of degree ≥1 2n is relaxed to just ≥1 2(n −1).
24. (a) Prove for n ≥3 that an undirected graph with n vertices and at least n−1 2 + 2 edges must have a Hamilton circuit.
(b) Show that part (a) is false if there are only n−1 2 + 1 edges.
2.3 GRAPH COLORING In Example 1 of Section 1.4 we introduced the problem of map coloring—coloring the countries of a map so that two countries with a common border are assigned dif-ferent colors. The problem of showing that any map can be 4-colored tantalized www.itpub.net 2.3 Graph Coloring 69 mathematicians for 100 years until a computer-assisted proof was obtained by Appel and Haken in 1976. More recently, graph coloring has been applied to a variety of problems in computer science, operations research, and experiment de-sign. Recall that coloring countries in a map is equivalent to coloring vertices, with adjacent vertices getting different colors, in the graph obtained by making a ver-tex for each country and an edge between vertices representing countries with a common border; see Example 1 in Section 1.4. In general, a coloring of a graph G assigns colors to the vertices of G so that adjacent vertices are given different colors.
In this section we show how to determine the minimal number of colors required to color a given graph. This minimal number of colors is called the chromatic number of a graph. We also give some applications of graph coloring. In the next section we will present some theorems about graph coloring. In a coloring of a graph, the vertices that have a common color will be mutually nonadjacent (no pair is joined by an edge).
In Example 5 of Section 1.1 we introduced the term independent set to refer to such a set of mutually nonadjacent vertices. We shall revisit that example later in this section.
For graphs with 15 or fewer vertices, it is usually not difficult to guess a graph’s chromatic number. To verify rigorously that the chromatic number of a graph is a number k, we must also show that the graph cannot be properly colored with k −1 colors. Proving that a graph cannot be (k −1)-colored is, like proving that a graph has no Hamilton circuit, an NP-complete problem (see Appendix A.5). In this case, the goal is to show that any (k −1)-coloring we might construct for the graph must force two adjacent vertices to have the same color.
Example 1: Simple Graph Coloring Find the chromatic number of the graph in Figure 2.10.
Looking at the inner square with crossing diagonals, we see that vertices a, c, e, g are mutually adjacent—that is, they form a complete subgraph. They each require a different color in a proper coloring: four colors in all. Once four colors are available, it is easy to properly color the remaining vertices b, d, f , h. Each of them is adjacent to only two other vertices, and so at most two out of the four colors need ever be avoided with these vertices. Let us use the numbers 1, 2, 3, 4 as the “names” of our colors. Then one possible 4-coloring of the graph is shown in Figure 2.10.
b, 3 a, 1 h, 3 d, 1 f, 1 e, 3 g, 4 c, 2 Figure 2.10 70 Chapter 2 Covering Circuits and Graph Coloring a, 1 b, 2 e, 1 d, 2 c, 1 f, 3 Figure 2.11 In this problem it is immediately clear that the graph cannot be 3-colored, since some adjacent pair of vertices in the complete subgraph formed by a, c, e, g would have the same color in a 3-coloring. So the chromatic number of this graph is 4.
Example 1 points out two important rules. First, a complete subgraph on k vertices requires k colors [cannot be (k −1)-colored]. Second, when building a k-coloring of some graph, we can ignore all vertices of degree 0.
14. Determine the chromatic polynomial Pk(G) for the following graphs: (a) (b) (d) (e) (c) 15. Show that a graph is k-colorable if and only if its edges can be directed so that there is no directed circuit and its longest path has length k −1.
16. Use the fact that every planar graph with fewer than 12 vertices has a vertex of degree ≤4 (Exercise 19 in Section 1.4) to prove that every planar graph with less than 12 vertices can be 4-colored.
17. Show that a graph with at most two odd-length circuits can be 3-colored.
18. Prove that for any positive integer k, there exists a triangle-free graph G with χ(G) = k.
(a) The proof should be by induction on k, the chromatic number. Initially for k = 3, we use the graph G3 consisting of a 5-circuit.
(b) Assuming one can construct Gk, a triangle-free graph with χ(Gk) = k, one constructs Gk+1 by making k copies of Gk and then adding (nk)k vertices, 86 Chapter 2 Covering Circuits and Graph Coloring where nk is the number of vertices in Gk. Each new vertex has as its set of neighbors a different k-tuple consisting of one vertex from each copy of Gk.
Confirm that this new graph is the desired Gk+1.
2.5 SUMMARY AND REFERENCES This chapter presented two important graph-theoretic concepts: covering cycles or circuits and coloring. Section 2.1 discussed Euler cycles—cycles that traverse every edge exactly once. Section 2.2 discussed Hamilton circuits—circuits that visit every vertex exactly once. Both types of covering edge sets arise naturally in operations research routing problems. Despite the similarity in the definitions of Euler cycles and Hamilton circuits, determining the existence of Euler cycles and Hamilton circuits in a graph are as different as graph-theoretic problems can be. Euler’s theorem allows one to quickly decide whether an Euler cycle exists. On the other hand, except in special cases, the existence or nonexistence of a Hamilton circuit can be determined only by a laborious systematic search to try all possible ways of constructing a Hamilton circuit.
Section 2.3 introduced graph coloring with ad hoc coloring schemes and some applications of coloring. Section 2.4 gave a sampling of coloring theory, highlighted by a proof of the fact that any planar graph can be 5-colored. The stronger theorem proved in 1976 by Appel and Haken , that planar graphs are 4-colorable, was the motivation of much of the research in graph theory over the previous 100 years. The search for a proof of the Four Color Theorem led to reformulations of this theorem in terms of Hamilton circuits and other graph concepts whose properties were then examined.
Euler’s 1736 analysis of Euler cycles was the first paper on graph theory. Euler’s paper (translated in Biggs, Lloyd, and Wilson ) makes very interesting reading. It is instructive to see how awkward Euler’s writing was when he lacked the modern terminology of graph theory. The first use of the concept of a Hamilton circuit occurred in a 1771 paper by A. Vandermonde that presented a sequence of moves by which a knight could tour all positions of a chessboard (without repeating a position). The name “Hamilton” refers to W. Hamilton, whose algebraic research led him to consider special types of circuits and paths on the edges of a dodecahedron [the graph in Exercise 2(c) in Section 2.2]. Hamilton even had a game marketed that involved finding a Hamilton circuit on a dodecahedron (Hamilton’s instructions for this game are reprinted in ). See Barnette for a good history of the Four Color Problem, its restatements, and final solution by Appel and Haken.
1. K. Appel and W. Haken, “Every planar map is 4-colorable,” Bull. Am. Math. Soc.
82 (1976), 711–712.
2. D. Barnette, Map Coloring and the Four Color Problem, Mathematical Associ-ation of America, Washington, DC, 1984.
www.itpub.net Supplement: Graph Model for Instant Insanity 87 3. N. Biggs, E. Lloyd, and R. Wilson, Graph Theory 1736–1936, Cambridge Uni-versity, Cambridge, 1999.
4. J. Bondy and U. Murty, Graph Theory with Applications, American Elsevier, New York, 1976.
5. J. O’Rourke, Art Gallery Theorems and Algorithms, Oxford University Press, New York, 1987.
SUPPLEMENT: GRAPH MODEL FOR INSTANT INSANITY This supplement presents a clever solution to the Instant Insanity puzzle devised by Blanche Decartes (an alleged pseudonym for the famous graph theorist W. Tutte). The solution uses a generalized form of Hamilton circuit in which all vertices are covered by a collection of vertex-disjoint circuits rather than a single circuit.
The Instant Insanity puzzle consists of four cubes whose faces are colored with one of the four colors: red (R), white (W), blue (B), and green (G). The six faces on the ith cube are denoted: fi—front face, li—left face, bi—back face, ri—right face, ti—top face, and ui—under face. In Figure 2.20, the first cube is colored: l1 = B,r1 = W, f1 = R, b1 = R, t1 = G, u1 = B. The objective of this puzzle is to place the four cubes in a pile (cube 1 on top of cube 2 on top of . . . etc.) so that each side of the pile has one face of each color. For example, Figure 2.21 shows one solution for the pile of cubes given in Figure 2.20.
We shall work with the four cubes shown in Figure 2.20. Determining how to arrange these four cubes in a pile that is an Instant Insanity solution is a very difficult task. Observe that there are 24 symmetries of a cube, and thus 244 = 331,776 different piles that can be built. An enumeration tree search will be immense, although symmetries of the face colors and the constraint of no repeated colors on a side will eliminate many possibilities. A computer program to do this search is easily written.
Fortunately, we can model this puzzle with a 4-vertex graph in such a fashion that the graph-theoretic restatement of the puzzle can be solved by inspection in a few minutes.
Before presenting the graph model, we need to discuss a simple decomposition principle for this puzzle. Arranging the cubes in a pile so that left and right sides of the pile have one face of each color is “independent” of arranging the front and back sides with one face of each color. By “independent” we mean that once cube Cube 1 Cube 2 Cube 3 Cube 4 Figure 2.20 88 Chapter 2 Covering Circuits and Graph Coloring Figure 2.21 i is arranged so that a given pair of opposite faces are on the left and right side of the pile, then any remaining pair of opposite faces on cube i can be on the front and back sides of the pile.
Let l∗ i ,r∗ i , f ∗ i , b∗ i denote the colors of the four respective faces of cube i visible on the four sides of the pile when cube i is reoriented to obtain an Instant Insanity solution. Suppose that l∗ i = t1 = G and r∗ 1 = u1 = B, that is, the top face t1 of cube 1 in Figure 2.20, which is green, is reoriented to be the left side l∗ 1, and the under face u1, which is blue, is reoriented to be the right side r∗ 1. Then by rotating cube 1 about the centers of these two faces, we can get f ∗ 1 = B, b∗ 1 = W, or f ∗ 1 = R, b∗ 1 = R, or f ∗ 1 = W, b∗ 1 = B—all possible remaining choices for front and back sides. Since this same left–right and front–back “independence” holds for the other cubes, we see that the puzzle can be broken into two disjoint problems.
Decomposition Principle 1.
Pick one pair of opposite faces on each cube for the left and right sides of the pile so that these two sides of the pile will have one face of each color.
2.
Pick a different pair of opposite faces on each cube for the front and back sides of the pile so that these two sides will have one face of each.
Now we are ready to present our graph model (due to F. de Carteblanche).
Actually we use a multigraph (in which multiple edges and loops are allowed). Make one vertex for each of the four colors. For each pair of opposite faces on cube i, create an edge with label i joining the two vertices representing the colors of these two opposite faces. For opposite faces l1 = B,r1 = W on cube 1, we draw an edge labeled 1 between vertex B and vertex W; for f1 = R, b1 = R, we draw a loop labeled 1 at vertex R; and for t1 = G, u1 = B, we draw an edge labeled 1 between vertices G and B. The edges for the other cubes are drawn similarly. Figure 2.22 shows this multigraph for the cubes in Figure 2.20.
www.itpub.net Supplement: Graph Model for Instant Insanity 89 G R 4 1 1 3 2 G 2 4 3 W B 1 3 4 2 Figure 2.22 By the decomposition principle, we can break the puzzle into a left–right part and a front–back part. We initially consider just the left–right part—that is, find one pair of opposite faces on each cube so that the left and right sides of the pile have one face of each color.
Let us simplify this left–right problem slightly by asking only for a set of four opposite-face pairs, one pair from each cube, such that among this total set of eight faces each color appears twice. Later we will show how to ensure that each color appears once on the left side and once on the right side. Since a color corresponds to a vertex, a cube to an edge number, and a pair of opposite faces to an edge, this simplified left–right problem has the following graph-theoretic restatement: find four edges, one with each number, such that the family of eight end vertices of these four edges contains each vertex twice.
This condition on the end vertices of the four edges is equivalent to requiring that the subgraph formed by these four edges has each vertex of degree 2 (a self-loop counts as degree 2 at its vertex). Note that a subgraph with all vertices of degree 2 is just a circuit or collection of disjoint circuits (a self-loop is a circuit of length 1). In an n-vertex multigraph, a set of n edges forming disjoint simple circuits is called a factor. Observe that a factor is a natural generalization of a Hamilton circuit. In our Instant Insanity model, let the term labeled factor denote a factor in which each edge number appears once. In Figure 2.23 we show three possible labeled factors for the multigraph in Figure 2.22.
The simplified left–right problem in our Instant Insanity graph model reduces to: find a labeled factor.
G R 4 1 2 G W B 3 W 2 R 3 G 1 G 4 B G 1 G W B 2 4 R 3 (a) (b) (c) Figure 2.23 90 Chapter 2 Covering Circuits and Graph Coloring Cube 1 Cube 2 Cube 3 Cube 4 Figure 2.24 We next show how a labeled factor can be transformed into an arrangement of the cubes in which the left and right sides of the pile have one face of each color.
We direct the edges in each circuit in a consistent direction—say, in the clockwise direction (see Figure 2.23). Consider the labeled factor in Figure 2.23a. As we go around this circuit, we arrange each cube with the color of the tail vertex of an edge as the color on the left side of the cube, and the color of the head vertex on the right side. One can start with any edge on the circuit; we pick the edge labeled 1.
This 1-edge in Figure 2.23a goes from B to G, and so we arrange cube 1 so that l∗ 1 (the left face of cube 1 in the pile) = B and r∗ 1 = G (see Figure 2.24). Following the 1-edge on the circuit (in the clockwise orientation of the circuit), we next en-counter a 4-edge from G to R. Accordingly we arrange cube 4 so that l∗ 4 = G and r∗ 4 = R. Next comes a 2-edge from R to W followed by a 3-edge from W to B. So we arrange cubes 2 and 3 with l∗ 2 = R,r∗ 2 = W and l∗ 3 = W,r∗ 3 = B. Figure 2.24 shows the left and right sides (only) of the pile made by the four cubes arranged as just described.
This process assures that each color appears once on each side, since each vertex (color) is at the head of one edge and at the tail of one edge. If we had chosen the labeled factor in Figure 2.23b, we would have used the clockwise traversal procedure for all three circuits, yielding l∗ 1 = B,r∗ 1 = G,l∗ 4 = G,r∗ 4 = B for one circuit, and l∗ 3 = r∗ 3 = R and l∗ 2 = r∗ 2 = W for the two self-loops.
G 3 G W B 2 4 R 1 Figure 2.25 www.itpub.net Supplement: Graph Model for Instant Insanity 91 We are now ready to solve the Instant Insanity puzzle. Restating the decomposi-tion principle in terms of labeled factors: Graph-Theoretic Formulation of Instant Insanity 1.
Find two edge-disjoint labeled factors in the graph of the Instant Insanity puzzle, one for left–right sides and one for front–back sides.
2.
Use the clockwise traversal procedure to determine the left–right and front–back arrangements of each cube.
It is not hard to find two disjoint labeled factors by inspection. The three labeled factors in Figure 2.23 all use the same 1-edge (between B and G). So no two of these three factors are disjoint. The easiest approach is to find one labeled factor, delete its edges, and look for a second labeled factor. If none is found, start with a different labeled factor. For example, suppose we use the labeled factor in Figure 2.23a as our first factor. After deleting its edges from the graph in Figure 2.22, we easily find a second labeled factor. Such a second labeled factor is shown in Figure 2.25 (the reader should be able to find a second factor).
We use the factor from Figure 2.23a to arrange the left and right sides, as shown in Figure 2.24. Next we rotate each cube about the centers of its left and right faces to arrange the front and back faces according to a clockwise traversal of the circuits in Figure 2.25. Starting with the 2-edge, we make f ∗ 2 = W, b∗ 2 = B, f ∗ 3 = B, b∗ 3 = G, f ∗ 4 = G, b∗ 4 = W and f ∗ 1 = b∗ 1 = R. Figure 2.26 shows the resulting solution of the Instant Insanity puzzle. Now go buy or borrow a set of Instant Insanity cubes and show your friends that you learned something really useful from this book!
Figure 2.26 92 Chapter 2 Covering Circuits and Graph Coloring SUPPLEMENT EXERCISES 1.
Find all other labeled factors of the multigraph in Figure 2.22 besides the ones in Figure 2.23 and Figure 2.25. Give an argument in the process to show that there can be no other labeled factors.
2.
Find all Instant Insanity solutions (a disjoint pair of labeled factors) to the game with associated graph (a).
G B R W B G G B R W W R 2 4 1 2 2 3 4 3 3 4 1 1 2 1 4 2 3 4 3 4 1 3 2 4 3 1 2 3 2 3 4 1 4 1 2 1 (a) (b) (c) 3.
Find all Instant Insanity solutions to the game with the associated graph (b) shown above.
4.
Show carefully that Instant Insanity graph (c) shown above does not possess a solution, that is, a pair of disjoint labeled factors.
5.
(a) If a graph G has a Hamiltonian circuit, must it also have a factor? Prove true or give a counterexample.
(b) Repeat part (a) for the case of G having an Euler cycle.
6.
Find a factor in the following graphs, if possible: a b c d h g e f a b c g h i d e f i k h g e f j c b a d l (a) (b) (c) 7.
Show that the definition of a factor in a graph implies that each vertex is incident to exactly two edges (or one self-loop) in a factor.
www.itpub.net CHAPTER 3 TREES AND SEARCHING 3.1 PROPERTIES OF TREES The most widely used special type of graph is a tree. There are two ways to define trees. In undirected graphs, a tree is a connected graph with no circuits. Alternatively, one can define a tree as a graph with a designated vertex called a root such that there is a unique path from the root to any other vertex in the tree. This second definition applies to directed as well as undirected graphs. Theorem 1 proves the equivalence of these two definitions.
Intuitively, a tree looks like a tree. See the examples of trees in Figure 3.1. The vertex labeled a is a root for each of the trees in Figure 3.1. The equivalence of the above two definitions is proved in Exercise 5. To illustrate the equivalence, observe that in the tree in Figure 3.1a, the addition of an edge (g⃗ , h) would create a circuit (when edge directions are ignored) and would simultaneously create a second path from root a to h via g.
Trees are a remarkably powerful tool for organizing information and search pro-cedures. In this chapter, we survey some of the diverse settings in which trees can be used to represent and analyze search procedures. These settings include solv-ing puzzles (Section 3.2), solving the “traveling salesperson” problem (Section 3.3), and sorting lists (Section 3.4). In the next chapter, trees play a critical role in sev-eral of the network algorithms. In this first section, we present some basic proper-ties of trees and introduce some convenient terminology for working with trees. We prove several useful counting formulas about trees and illustrate some uses of these formulas.
Observe that if a tree is an undirected graph (with no directed edges), then any vertex can be the root. For example, the tree in Figure 3.1b is drawn so that a appears to be a root, but the tree in Figure 3.1c, which is a redrawing of Figure 3.1b, has no single vertex that is a natural root—that is, any vertex can be the root.
In most of this chapter we will be using trees with directed edges. Following common terminology, we call a directed tree a rooted tree. A rooted tree T has a unique root, for if vertices a and b were both roots of T, then there would be paths from a to b and from b to a forming a circuit. An undirected tree is unrooted in the sense that it has no one particular root. An undirected tree can be made into a rooted tree by choosing one vertex as the root and then directing all edges away from the root. For example, to root the undirected tree in Figure 3.1b at vertex a, we would simply direct all the edges from left to right.
93 94 Chapter 3 Trees and Searching a b d e f h i l j k g c a b c d e f g h i j k l m j k e i h d a b c g f l m (a) (b) (c) Figure 3.1 The standard way to draw a rooted tree T is to place the root a at the top of the figure. Then the vertices adjacent from a are placed one level below a, and so on, as in Figure 3.1a. We say that the root a is at level 0, vertices b and c in that tree are at level 1, vertices d, e, f, and g in that tree are at level 2, and so forth. The level number of a vertex x in T is the length of the (unique) path from the root a to x.
For any vertex x in a rooted tree T, except the root, the parent of x is the vertex y with an edge (y⃗ , x) into x (the unique edge directed into x). The children of x are vertices z with an edge directed from x to z. Children have level numbers 1 greater than x. Two vertices with the same parent are siblings. The parent–child relationship extends to ancestors and descendants of a vertex. In Figure 3.1a, vertex e has b as its parent, h and i as its children, d and f as its siblings, a as its other ancestor, and l as its other descendant. Observe that each vertex x in a tree T is the root of the subtree of x and its descendants. For easy reference, there is a glossary of tree-related terminology at the end of this text.
We now show that the two definitions of an (undirected) tree presented at the start of the chapter, and two additional characterizations of trees, are equivalent.
Theorem 1 Let T be a connected graph. Then the following statements are equivalent: (a) T has no circuits.
(b) Let a be any vertex in T . Then for any other vertex x in T , there is a unique path Px between a and x.
(c) There is a unique path between any pair of distinct vertices x, y in T .
(d) T is minimally connected, in the sense that the removal of any edge of T will disconnect T .
Proof First note that (b) and (c) are immediately equivalent; let vertex a in (b) be vertex y in (c). We shall show that (a) implies (d), (d) implies (c), and (c) implies (a).
(a) Î(d): Suppose that the removal of some edge (a, b) does not disconnect T . Then the graph T −(a, b) contains a path P between a and b. However, the edges of P plus edge (a, b) form a circuit, contradicting (a).
www.itpub.net 3.1 Properties of Trees 95 (d) Î (c): Suppose there are two different paths P1, P2 between two vertices a, b in T . Let e = (u, v) be the first edge on P1, starting from a, that is not on P2 (possibly, u = a or v = b). T −e is disconnected by (d), so u and v must be in different components. But following P2 from u to b and then coming back along P1 from b to v creates a path Q connecting u and v. Note: To make sure Q does not repeat any edges, we actually should follow P2 until the first vertex w also on P1 is encountered (such a w might occur before we get to b) and then go back on P1 to v.
(c) Î (a): T cannot contain a circuit, because the edges of a circuit provide two different paths joining any two vertices that lie on the circuit. ◆ As minimally connected graphs, trees have as few edges as possible. The follow-ing theorem gives a simple formula for the number of edges in an n-vertex tree.
Theorem 2 A tree with n vertices has n – 1 edges.
Proof Assume that the tree is rooted; if undirected, make it rooted as described above. We can pair off a vertex x with the unique incoming edge (y⃗ , x) from its parent y. Since each vertex except the root has such a unique incoming edge, there are n −1 nonroot vertices and hence n −1 edges. ◆ Vertices of T with no children are called leaves of T. Vertices with children are called internal vertices of T. If every internal vertex of a rooted tree has m children, we call T an m-ary tree. If m = 2, T is a binary tree.
Theorem 3 Let T be an m-ary tree with n vertices, of which i vertices are internal. Then, n = mi + 1.
Proof Each vertex in a tree, other than the root, is the child of a unique vertex (its parent).
Each of the i internal vertices has m children, and so there are a total of mi children.
Adding the one nonchild vertex, the root, we have n = mi + 1. ◆ Corollary Let T be an m-ary tree with n vertices, consisting of i internal vertices and l leaves.
If we know one of n, i, or l, then the other two parameters are given by the following formulas: (a) Given i, then l = (m −1)i + 1 and n = mi + 1.
(b) Given l, then i = (l −1)/(m −1) and n = (ml −1)/(m −1).
(c) Given n, then i = (n −1)/m and l = [(m −1)n + 1]/m.
96 Chapter 3 Trees and Searching Winner . . .
. . .
. . .
. . .
Figure 3.2 The proof of the corollary’s formulas follow directly from n = mi + 1 (Theorem 3) and the fact that l + i = n. Details are left to Exercise 8.
Example 1 If 56 people sign up for a tennis tournament, how many matches will be played in the tournament?
The tournament proceeds in a binary-tree-like fashion. The entrants are leaves and the matches are the internal vertices. See Figure 3.2. Given l = 56 and m = 2, we de-termine i from part (b) of the corollary: i = (l −1)/(m −1) = (56 −1)/(2 −1) = 55 matches.
Example 2 Suppose a telephone chain is set up among 100 parents to warn of a school closing.
It is activated by a designated parent who calls a chosen set of three parents. Each of these three parents calls given sets of three other parents, and so on. How many parents will have to make calls? Repeat the problem for a telephone tree of 200 parents.
Such a telephone chain is a rooted tree with 100 vertices. An edge corresponds to a call and an internal vertex corresponds to a parent who makes a call. Since we know n = 100 and that the tree is ternary (3-ary), part (c) of the corollary can be used to determine i, the number of callers: i = (n −1)/m = (100 −1)/3 = 33.
When we repeat the computation for an organization of 200 people, we get i = (200 −1)/3 = 66 1 3 internal vertices. By Theorem 3, a ternary tree must have a number of vertices n equal to 3i + 1, for some i—that is, n ≡1 (mod 3). But 200 ≡2 (mod 3), and so we do not have a true ternary tree. Either 199 or 202 parents would give a ternary tree. As a practical matter, with 200 people there will be 66 parents who each make 3 calls and one parent who makes just one call (one internal vertex with one child).
The height of a rooted tree is the length of the longest path from the root or, equivalently, the largest level number of any vertex. A rooted tree of height h is called balanced if all leaves are at levels h and h −1. Balanced trees are “good” trees. The telephone chain tree in Example 2 should be balanced to get the message to everyone as quickly as possible. A tennis tournament’s tree should be balanced to be fair; otherwise some players could reach the finals by playing several fewer matches than other players. Making an m-ary tree balanced will minimize its height (Exercise 12).
The tree in Figure 3.1b, with a as root, is a balanced binary tree of height 3.
www.itpub.net 3.1 Properties of Trees 97 Theorem 4 Let T be an m-ary tree of height h with l leaves. Then, (a) l ≤mh, and if all leaves are at height h,l = mh.
(b) h ≥⌈logml⌉, and if the tree is balanced, h = ⌈logml⌉.
Proof The expression ⌈r⌉denotes that the smallest integer ≥r; that is, ⌈r⌉rounds r up to the next integer.
(a) An m-ary tree of height 1 has m1 = m leaves (children of the root). Now we use induction on h to show that an m-ary tree of height h has at most mh leaves, with l = mh if all leaves are at level h. An m-ary tree of height h can be broken into m subtrees rooted at the m children of the root. See Figure 3.3. These m subtrees have height of at most h −1. By induction, each of these subtrees has at most mh−1 leaves, and if all leaves are at height h −1 in the subtrees, each has exactly mh−1. The m subtrees combined have at most m × mh−1 = mh leaves, and if all leaves are at height h, there are exactly mh leaves.
(b) Taking the logarithm base m on both sides of the inequality l ≤mh yields logml ≤ h. Since h is an integer, we have ⌈logml⌉≤h. If the tree is balanced with height h, then the largest possible value for l is l = mh (if all leaves are at level h), and the smallest possible value is l = (mh−1 −1) + m (with m leaves at level h and the rest at level h −1). So mh−1 < l ≤mh. Taking logarithms on both sides yields h −1 < logm l ≤h, or h = ⌈logm l⌉. ◆ The most common use of trees is in searching. The following two sequential testing examples, one a basic computer science problem and the other a logical puzzle, illustrate the use of trees in searching.
Example 3 Let us reexamine the dictionary look-up problem discussed in Example 2 of Section 1.1. We want to identify an unknown word (number) X by comparing it to words in a set (dictionary) to which X belongs. This time our comparison test will be a three-way Figure 3.3 98 Chapter 3 Trees and Searching H D B F J N A C E G I K M L Figure 3.4 branch (less than, equal to, greater than). The test procedure can be represented by a binary, or almost binary, tree. If X were one of the first 14 letters of the alphabet, then Figure 3.4 is such a binary search tree. Each vertex is labeled with the letter tested at that stage in the procedure. The procedure starts by testing X against H. The left edge from a vertex is taken when X is less than the letter and the right edge when X is greater. Such a tree may have one internal vertex with just one child if the number of vertices is even (as is the case for vertex N in Figure 3.4). This tree is built by making the middle letter in the list (in this case, G or H) the root. The left child of the root is the middle letter in the left subtree (in this case, D), and so on.
To minimize the number of tests needed to recognize any X, that is, the height of the search tree, we should make the tree balanced. Suppose that X were known to belong to a set of n “words.” What is the maximum number of tests that would be needed to recognize X? By the corollary to Theorem 3, a binary search tree with n vertices has l = (2 −1)n + 1 2 = n + 1 2 leaves. Then the maximum number of tests needed to recognize X is the height of a balanced 1 2(n + 1)-leaf search tree. By Theorem 4, h = ⌈log2[(n + 1)/2]⌉= ⌈log2(n + 1)⌉−1.
Example 4 A well-known logical puzzle has n coins, one of which is counterfeit—too light or too heavy—and a balance to compare the weight of any two sets of coins (the balance can tip to the right, to the left, or be even). For a given value of n, we seek a procedure for finding the counterfeit coin in a minimum number of weighings. Sometimes one is told whether the counterfeit coin is too light or too heavy. If we are told that the fake coin is too light, how many weighings are needed for n coins?
Our testing procedure will form a tree in which the root is the first test, the other internal vertices are the other tests, and the leaves are the solutions—that is, which coin is counterfeit. See the testing procedure in Figure 3.5 for eight coins. The coins are numbered 1 through 8, and the left edge is followed when the left set of coins in a test is lighter, the middle edge when both sets have the same weight, and the right edge when the right set of coins is lighter. Note that when weighing 4 and 5 (and already knowing that 1, 2, 3, 6, 7, 8 are not the light coin), the balance cannot be even.
www.itpub.net 3.1 Properties of Trees 99 123⊥678 4⊥5 1⊥3 6⊥8 1 2 3 4 5 6 7 8 Figure 3.5 The test tree is ternary, and with n coins there will be n leaves—that is, n different possibilities of which coin is counterfeit. Theorem 4 tells us that the test tree must have height at least ⌈log3n⌉to contain n leaves. For the light counterfeit coin problem, this bound can be achieved by successively dividing the current subset known to have the fake coin into three almost equal piles and comparing two of the piles of equal size, as in Figure 3.5.
If the counterfeit coin could be either too light or too heavy, then the problem is harder (see Exercise 27). A particular coin will appear at 2 leaves in the test tree, once when the coin is determined to be too light and another time when too heavy.
Determining the number of leaves and height of more complex search trees is a major concern in the field of computer science called analysis of algorithms. In Section 3.4, we determine the number of leaves and height of search trees that arise when sorting a list of n items. Recurrence relations for the number of leaves and height of some other search trees are discussed in Section 7.2 in the enumeration part of this book. We conclude this section with a formula for the number of different undirected trees on n labeled vertices. Let the labels be the numbers 1 through n. For example, there are three different labeled trees on three labels. Each 3-label tree is a path of two edges with the difference being which of the three labels is the one middle vertex: 1–2–3, 1–3–2, and 2–1–3 (switching the position of the two leaves does not produce a different tree). While the formula was first proved by Cayley, we present a simpler proof due to Prufer.
Theorem 5 (Cayley, 1889) There are nn−2 different undirected trees on n labels.
Proof Observe that nn−2 is the number of sequences of the n labels of length n −2. We now construct a one-to-one correspondence between trees on n labels and (n −2)-length sequences of the n labels. Recall that for simplicity, we let the labels be the numbers 1, 2, . . . , n.
For any tree on n numbers, we form a sequence (s1, s2, . . . , sn−2) of length n −2 as follows. Let l1 be the leaf in the tree with the smallest number and let s1 be the number of the one vertex adjacent to it. For the tree in Figure 3.6, the leaf with the smallest number is 1 and the number of its neighboring vertex is 6. So s1 = 6.
We delete leaf l1 from the tree and repeat this process. For the tree in Figure 3.6, l2, 100 Chapter 3 Trees and Searching 7 1 6 3 8 4 2 5 Figure 3.6 the smallest numbered leaf in the tree after 1 is deleted is 4, and its neighbor is 2.
So s2 = 2. Continuing we have l3 = 5 and s3 = 2, then l4 = 2 and s4 = 3, then l5 = 6 and s5 = 3, and then l6 = 7 and s6 = 3. We stop when the remaining tree has been reduced to two leaves joined by an edge. The 6-label sequence for the 8-label tree in Figure 3.6 is thus (6, 2, 2, 3, 3, 3). Such sequences are called Prufer sequences.
Next we show that any such (n −2)-length sequence of n items defines a unique n-item tree. We simply reverse the procedure in the preceding paragraph used to build the sequence. Observe that leaves (vertices of degree 1) will never appear in the sequence. The first number of the sequence is the neighbor of the smallest numbered leaf. From what we just observed, this smallest numbered leaf is the smallest number that does not appear in the sequence. For the sequence (6, 2, 2, 3, 3, 3), 1 is the smallest number not in the sequence, so 1 is the leaf with 6 as its neighbor.
Nowwesetthesmallestleaf(label1)aside(itspositioninthetree—aleafadjacent to the first item in the sequence—is determined) and we consider the first item (item 6) as a leaf that will be adjacent to some item in the remaining sequence. We then repeat the process of identifying the smallest leaf in the remaining (n −1)-label tree specified by the remaining (n −3)-label sequence. For the remaining sequence (2, 2, 3, 3, 3), label 4 is the smallest of the remaining numbers (label 1 has been deleted) not in the sequence, so item 4 is a leaf and is adjacent to label 2, the first number of the remaining sequence. Continuing in the reduced sequence (2, 3, 3, 3), label 5 is the smallest leaf of the remaining numbers and is adjacent to label 2. Now label 2 becomes a potential “leaf” with respect to the remaining sequence (3, 3, 3). Note that the available leaves are currently labels 2, 6, 7, 8. So label 2, the smallest available leaf, is adjacent to item 3. In the sequence (3,3), label 6 is adjacent to label 3. Finally, label 7 is adjacent to label 3. There remain only labels 3 and 8, which must be adjacent to each other.
The preceding construction of a labeled tree from the given Prufer sequence can be applied to any Prufer sequence. Thus the correspondence between n-label trees and sequences of length n −2 is one-to-one and the theorem is proved. ◆ 3.1 EXERCISES S u m m a r y o f E x e r c i s e s Exercises 3–18 present theory about trees.
Exercises 19–29 involve various modeling problems with trees.
1. Draw all nonisomorphic trees with (a) Four vertices (b) Five vertices (c) Six vertices 2. Suppose a connected graph has 20 edges. What is the maximum possible number of vertices?
www.itpub.net 3.1 Properties of Trees 101 3. Show that all trees are 2-colorable.
4. Show that all trees are planar.
5. Show that an undirected connected graph G is a tree (i.e., has a unique path from root to each vertex) if any one of the following conditions holds: (a) G has n vertices and n −1 edges.
(b) G has fewer edges than vertices.
(c) Removal of any edge disconnects G.
6. Reprove Theorem 2 by using the fact that trees are planar (Exercise 4) and Euler’s formula (Theorem 2 in Section 1.4).
7. Show that any tree with more than one vertex has at least two vertices of degree 1.
8. Prove the following parts of the corollary to Theorem 3: (a) Part (a) (b) Part (b) (c) Part (c).
9. Reprove that l ≤mh in an m-ary tree of height h by counting the maximum possible number of choices at each internal vertex when building a path from the root to a leaf.
10. What is the maximum number of vertices (internal and leaves) in an m-ary tree of height h?
11. Show that the fraction of internal vertices among all vertices in an m-ary tree is about 1/m.
12. For a given h, show that the height of an m-ary tree with k leaves is minimized when the tree is balanced.
13. What is the size of the largest and smallest numbers of vertices of degree 1 possible in an n-vertex tree, for n > 2?
14. Show that a graph, not necessarily connected, is a tree if it has no circuits but the addition of any edge (between two existing vertices) always creates a circuit.
15. Show that the size of the largest independent set (defined in Example 5 of Section 1.1) in an n-vertex tree is at least n/2.
16. A forest is an unconnected graph that is a disjoint union of trees. If G is an n-vertex forest of t trees, how many edges does it have?
17. Show that the sum of the level numbers of all l leaves in a binary tree is at least l⌈log2 l⌉, and hence the average leaf level is at least ⌈log2 l⌉.
18. Let T be an undirected tree. If the choice of vertex x to be the root yields a rooted tree of minimal height, then x is call a center of T. Show that any undirected tree has at most two centers.
19. Consider the problem of summing n numbers by adding together various pairs of numbers and/or partial sums, for example, {[(3 + 1) + (2 + 5)] + 9}.
(a) Represent this addition process with a tree. What will internal vertices rep-resent?
(b) What is the smallest possible height of an “addition tree” for summing 100 numbers?
102 Chapter 3 Trees and Searching 20. If T is a balanced 5-ary tree with 80 internal vertices, (a) How many leaves does T have?
(b) How many edges does T have?
(c) What is the height of T ?
21. Every year, the NCAA basketball championship tournament features 64 teams that must compete for the national title. There are 16 teams from the East, 16 teams from the South, 16 teams from the West and 16 teams from the Midwest.
In each division (East, South, West, and Midwest), the teams compete in a single elimination tournament to determine the best team from their respective division.
Then, the top East team plays the top South team and the top West team plays the top Midwest team. The winners of these games go on to play a final match to determine the champion. How many games are played in total?
22. What type of search procedure is represented by the search tree below?
A B C 23. A tree can be used to represent a binary code; a left branch is a 0 and a right branch a 1. The path to a letter (vertex) is its binary code. To avoid confusion, one sometimes requires that the initial digits of one letter’s code cannot be the code of another letter (e.g., if K is encoded as 0101, then no letter can be encoded as 0 or 01 or 010). Under this requirement, which vertices in a tree represent letters?
How many letters can be encoded using n-digit binary sequences?
24. Suppose that each player in a tennis tournament (like the binary-tree tournament in Example 1) brings a new can of tennis balls. One can is used in each match and the other can is taken by the match’s winner along to the next round. Use this fact to show that a tennis tournament with n entrants has n −1 matches.
25. Consider a tennis tournament T (with the tree structure illustrated in Figure 3.2) with 32 entrants.
(a) How many players are eliminated (lose) in the first two rounds of matches?
(b) Suppose that the losers in the first two rounds of the tournament qualify for a losers’ tournament T ′. How many players are eliminated in the first two rounds of matches in T ′? Note that all tennis tournaments are balanced trees; the number of people playing matches in each round after the first round is a power of 2.
(c) Suppose that the people who lose in the first two rounds of T ′ qualify for another losers’ tournament T ′′. How many players are eliminated in the first two rounds of matches in T ′′?
(d) Suppose that the people who lose in the first two rounds of T ′′ qualify for another losers’ tournament and so on until finally there is just one grand www.itpub.net 3.2 Search Trees and Spanning Trees 103 loser (the last tournament has two people). How many losers’ tournaments are required to determine this grand loser?
26. Suppose that a chain letter is started by someone in the first week of the year.
Each recipient of the chain letter mails copies on to five other people in the next week. After six weeks, how much money in postage (40c / a letter) has been spent on these chain letters?
27. (a) Repeat Example 3 assuming now that only a two-way branch (less than, greater than or equal to) is available. Draw a balanced search tree for the first 13 letters and determine the height of an n-letter search.
(b) Suppose a two-way branching search tree for letters A, B, C, D, E is to take advantage of the following letter frequencies: A 20%, B 20%, C 30%, D 10%, E 20%. Build a two-way tree that minimizes the average number of tests required to identify a letter.
28. Repeat Example 4 for 20 coins with at most one too light.
29. Suppose we have four coins and possibly one coin is either too light or too heavy (all four might be true).
(a) Show how to determine which of the nine possible situations holds with just two weighings, if given one additional coin known to be true.
(b) Show that two weighings are not sufficient without the extra true coin.
30. Let T be an m-ary tree with n vertices, consisting of i internal vertices and l leaves. Suppose that m is an even number. Show that n always has to be an odd number. Give two (small) examples for the same value of m illustrating that i can be either even or odd.
31. In the proof of Theorem 5, we showed that a Prufer sequence (s1, s2, . . . , sn−2) uniquely described a tree on n items. Construct the trees with the following Prufer sequences.
(a) (4, 5, 6, 2) (b) (2, 8, 8, 3, 5, 4) (c) (3, 3, 3, 3, 3, 3) 3.2 SEARCH TREES AND SPANNING TREES Trees provide a natural framework for finding solutions to problems that involve a sequence of choices, whether hunting through a graph for a special vertex or finding one’s way out of a maze or searching for the cheapest solution to a vehicle rout-ing problem. Most of the problems in the two preceding chapters—isomorphism, Hamilton circuits, minimal colorings, and placing police on street corners—require tree-based searching for computerized solutions. By letting the sequential choices be internal vertices in a rooted tree and the solutions and “dead ends” be the leaves, we can organize our search for possible solutions. Whether searching in a graph or in a 104 Chapter 3 Trees and Searching maze or through all solutions to an optimization problem, the foremost concern in the search procedure is that it be exhaustive—that is, guaranteed to check all possibilities.
In this section we present applications of tree searches that involve games rather than operations research applications. Most operations research tree enumeration problems involve very large trees and use special tree “pruning” algorithms. As an example of such applications, we solve a small Traveling Salesperson problem in Section 3.3. The next chapter, “Network Algorithms,” discusses four important opti-mization algorithms that implicitly use trees to search through graphs.
We start with searching in a graph. In many applications, graph algorithms are needed to test whether a graph has a certain property, such as connectedness or planarity, or to count all occurrences of a given structure, such as circuits or complete subgraphs.Thealgorithmsusuallyemployaspanningtreeinsearchingamongvertices and edges of a graph to check for these properties or structures. A spanning tree of a graph G is a subgraph of G that is a tree containing all vertices of G. Spanning trees can be constructed either by depth-first (backtrack) search or by breadth-first search.
To build a depth-first spanning tree, we pick some vertex as the root and begin building a path from the root composed of edges of the graph. The path continues until it cannot go any further without repeating a vertex already in the tree. The vertex where this path must stop is a leaf. We now backtrack to the parent of this leaf and try to build a path from the parent in another direction. When all possible paths from this parent y and its other children have been built, we backtrack to the parent of y, and so on, until we come back to the root and have checked all other possible paths from the root.
To build a breadth-first spanning tree, we pick some vertex x as the root and put all edges leaving x (along with the vertices at the ends of these edges) in the tree.
Then we successively add to the tree the edges leaving the vertices adjacent from x, unless such an edge goes to a vertex already in the tree. We continue this process in a level-by-level fashion.
It is important to note that if the graph is not connected, then no spanning tree exists. We thus have the following algorithm as a result.
Algorithm to Test Whether an Undirected Graph Is Connected Use a depth-first or breadth-first search to try to construct a spanning tree. If all vertices of the graph are reached in the search, a spanning tree is obtained and the graph is connected. If the search does not reach all vertices, the graph is not connected.
A formal proof that depth-first (breadth-first) searching does search all vertices in a connected graph is left as an exercise (Exercise 9).
An adjacency matrix of an (undirected) graph is a (0,1)-matrix with a 1 in entry (i, j) if vertex xi and vertex x j are adjacent; entry (i, j) is 0 otherwise.
Example 1: Testing for Connectedness Is the undirected graph G whose adjacency matrix is given in Figure 3.7a connected?
Let us perform a depth-first search of G starting with x1 as the root. At each successive vertex, we pick the next edge on the tree to be the edge going to the lowest numbered vertex not already in the tree. So from x1 we go to x2.
www.itpub.net 3.2 Search Trees and Spanning Trees 105 x1 x2 x3 x4 x5 x6 x7 x8 x1 0 1 1 1 0 1 0 0 x2 1 0 1 0 1 0 1 0 x3 1 1 0 0 0 0 0 0 x4 1 0 0 0 0 0 1 0 x5 0 1 0 0 0 0 1 0 x6 1 0 0 0 0 0 0 0 x7 0 1 0 1 1 0 0 1 x8 0 0 0 0 0 0 1 0 (a) x5 x8 x7 x2 x3 x4 x6 x1 x1 x2 x3 x4 x5 x8 x7 x6 (b) (c) Figure 3.7 From x2 we go to x3. Since x3 is not adjacent to any other vertex besides x1 and x2 (which are already in the tree), we backtrack from x3 to x2 and continue the search from x2, going to x5, then to x7, and then to x4. At x4 we backtrack to x7 and go to x8. From x8, we must backtrack all the way back to x1. From x1, we go to x6.
This finishes the search—all vertices have been visited. The spanning tree obtained is shown in Figure 3.7b.
The result of a breadth-first search is shown in Figure 3.7c.
The computation time required to make a depth-first search of a graph is propor-tional to the number of edges in the graph (each edge in the spanning tree is traversed twice, and edges that cannot be used are tried just twice, once from each end vertex).
If an undirected graph is not connected, then we can find its components (connected pieces) by applying a depth-first (or breadth-first) search at any vertex to find one component, then apply this search starting at a vertex not on the previous tree to find another component, and continue finding additional components until no unused vertices are left.
We note one important property of a breadth-first search. A breadth-first spanning tree consists of shortest paths from the root to every other vertex in the graph. A proof of this claim is left to Exercise 8. If we only wanted to find a shortest path from the root to a particular vertex x in a graph, then the breadth-first search could stop as soon as x was reached (without trying to construct a full spanning tree).
Now we apply the techniques of depth-first and breadth-first search to games. In a maze, the vertices will be the intersection points of paths. In puzzles, the vertices will be the different configurations of the puzzle and the edges will be the possible moves.
106 Chapter 3 Trees and Searching S E w Figure 3.8 Example 2: Traversing a Maze Consider the maze in Figure 3.8. We start at the location marked with an S and seek to reach the end marked with an E.
We use a depth-first search. For mazes, there is a convenient rule of thumb (whose verification is left to Exercise 17) for constructing a depth-first search: stick to the right wall in the maze. When we come to a dead end, we follow the right wall to the end wall, along the end wall, and then backtrack along the left wall (now the right wall as we leave the dead end). When we come to a previously visited corner, we put an artificial (wiggly) dead-end wall to stop us from actually reaching that corner (as at S in Figure 3.8). In the maze in Figure 3.8 we use solid lines to indicate the (forward) pathbuilding and dashed lines for backtracking. Because the maze is easily searched directly, we have not drawn the search tree for this problem (in which S would be the root, other corners internal vertices, and dead ends and E the leaves).
If the tree of possible paths is large, then the breadth-first method quickly becomes unwieldy. The depth-first method that traces only one path at a time is much easier to use by hand or to program. Further, in cases where we need to find only one of the possible solutions, it pays to go searching all the way down a path for a solution rather than to take a long time building a large number of partial paths, only one of which in the end will actually be used. On the other hand, when we want a solution involving a shortest path or when there may be very long dead-end paths (while solution paths tend to be relatively short), then the breadth-first method is better. All the network search algorithms in the next chapter use breadth-first searches.
Example 3: Pitcher-Pouring Puzzle Suppose we are given three pitchers of water, of sizes 10 quarts, 7 quarts, and 4 quarts.
Initially the 10-quart pitcher is full and the other two empty. We can pour water from one pitcher into another, pouring until the receiving pitcher is full or the pouring pitcher is empty. Is there a way to pour among pitchers to obtain exactly 2 quarts in the 7- or 4-quart pitcher? If so, find a minimal sequence of pourings to get 2 quarts in the 7-quart or 4-quart pitcher.
www.itpub.net 3.2 Search Trees and Spanning Trees 107 (6, 4, 0) (0, 6, 4) (0, 7, 3) (3, 3, 4) (3, 7, 0) (6, 0, 4) (10, 0, 0) 0 1 2 3 4 5 6 7 0 1 2 3 4 (0, 4) (2, 4) (3, 4) (4, 4) (6, 4) (7, 3) (7, 1) (7, 0) (6, 0) (4, 0) (3, 0) (0, 3) (a) (b) Figure 3.9 Thepositions,orvertices,inthisenumerationproblemareorderedtriples(a,b,c), the amounts in the 10-, 7-, and 4-quart pitchers, respectively. Actually, it suffices to record only (b, c), the 7- and 4-quart pitcher amounts, since a = 10 −b −c. A directed edge corresponds to pouring water from one pitcher to another. Let us draw the tree on a b, c-coordinate grid as shown in Figure 3.9a. The grid is bounded by b = 7, c = 4, b + c = 10. Pouring between the 10- and 7-quart pitchers will be a horizontal edge, between 10- and 4-quart pitchers a vertical edge, and between 7- and 4-quart pitchers a diagonal edge with slope −1. The beginning of the same tree is shown in the standard form in Figure 3.9b.
The root of this search tree is (0, 0). Since we want a minimal sequence of pourings (a shortest path in the search tree), we will use a breadth-first search. From the root, we can get to positions (7, 0) and (0, 4). From (7, 0), we can get to new positions (7, 3) and (3, 4), and from (0, 4), we can get to new positions (6, 4) and (4, 0). The tree built thus far is shown in Figure 3.9b.
From (7, 3), the only new position is (0, 3), and from (3, 4), the only new position is (3, 0). From (6, 4), the only new position is (6, 0), and from (4, 0), the only new position is (4, 4). We have now checked all paths of length 3. The only new moves now are from (4, 4) to (7, 1) and from (6, 0) to (2, 4). But (2, 4) has 2 quarts in the 7-quart pitcher. So (0, 0) to (0, 4) to (6, 4) to (6, 0) to (2, 4) is a sequence of pourings to obtain 2 quarts.
Example 4: Missonaries—Cannibals Puzzle Suppose three missionaries and three cannibals must cross a river in a two-person boat. Show how this can be done so that at no time do the cannibals outnumber the missionaries on either side (unless there are no missionaries on a shore).
We use vertices to denote the states of the puzzle when the boat is on one of the shores. We label each vertex with an ordered pair (Nm, Nc), where Nm is the number of missionaries and Nc the number of cannibals on the near shore. When the boat is on the near shore, we put an asterisk after the label. We start at vertex (3,3)∗. Each edge is labeled with an arrow showing the direction of the boat along with who is in the boat: one m for each missionary in the boat and one c for each cannibal in the boat. The only legal initial moves from vertex (3,3)∗are for two cannibals or else a missionary 108 Chapter 3 Trees and Searching cc → c ← cc → c ← mm → mc ← mm → c ← cc → c ← cc → (3, 3) (3, 1) (3, 2) (3, 0) (3, 1) (1, 1) (2, 2) (0, 2) (0, 3) (0, 1) (0, 2) (0, 0) Figure 3.10 and a cannibal to cross the river with a cannibal left on the far shore and the other rower returning with the boat to the near shore. The readers should try to solve this puzzle for themselves. We show a solution to the game in Figure 3.10, which starts with two cannibals in the first crossing. The key step occurs on the sixth move when two people (a missionary and a cannibal) cross from the far shore to the near shore.
Suppose we have built a tree to provide the framework for searching or organizing information. Searching for a particular vertex or processing information in the tree normally involves a depth-first type of traversal of the spanning tree. However, there are several times during a traversal when internal vertices can be checked. A preorder traversal of a tree is a depth-first search that examines an internal vertex when the vertex is first encountered in the search. A postorder traversal examines an internal vertex when last encountered (before the search backtracks away from the vertex and its subtree). If a tree is binary, we can define an inorder traversal that checks an internal vertex in between the traversal of its left and right subtrees. Figures 3.11a 1 2 3 4 8 5 6 7 8 6 1 5 7 4 2 3 (a) (b) Figure 3.11 www.itpub.net 3.2 Search Trees and Spanning Trees 109 d + ÷ – ÷ e q × ÷ + b a c f g Figure 3.12 and 3.11b display numberings of the vertices of the tree in Figure 3.11a according to preorder and postorder traversals, respectively. (We assume the left child is visited before the right child at each internal vertex.) An important property of preorder and postorder traversals is that in a preorder traversal a vertex precedes its children and all its other descendants, whereas in a postorder traversal a vertex follows its children and all its other descendants.
We now give examples of each type of traversal. In the binary search tree example of a dictionary look-up (Example 2 in Section 3.1), the alphabetical order of the vertices corresponds to inorder traversal. In searching through a graph for a vertex with a specified label, we should test each new vertex as it is encountered on a depth-first search to see if it has the desired label. There is no advantage to postponing such testing. Thus, in searching for a special vertex, vertices should be examined according to a preorder traversal.
We demonstrate the need for postorder traversal with an arithmetic tree. The arithmetic expression ((a + b) ÷ c) × (((q ÷ (d + e)) −f ) ÷ g) can be decomposed into a binary tree as shown in Figure 3.12. The internal vertices are arithmetic operations and the leaves variables. To evaluate this arithmetic expression, we need to execute the operations specified by internal vertices according to a pos-torder traversal of the arithmetic tree—that is, we cannot perform an operation until the subexpressions represented by the operation vertex’s two subtrees are evaluated.
3.2 EXERCISES S u m m a r y o f E x e r c i s e s The first six exercises involve building spanning trees and the next five involve properties of spanning trees. Exercises 12–25 are based on the puzzle examples in this section. Exercises 26–27 involve traversals.
1. Find depth-first spanning trees for each of these graphs; (a) K8 (a complete graph on eight vertices) (b) The graph in Figure 2.5 in Section 2.2 110 Chapter 3 Trees and Searching (c) The graph in Figure 2.10 in Section 2.3 (d) K3,3 (a complete bipartite graph) 2. Find breadth-first spanning trees of each of the graphs in Exercise 1.
3. Find all spanning trees (up to isomorphism) in the following graphs: (a) Figure 2.4 (b) K4 (c) a b c d e (d) K3,3 (a complete bipartite graph) 4. Test the graph whose adjacency matrix is given below to see if it is connected.
x1 x2 x3 x4 x5 x6 x7 x8 x1 0 0 1 0 0 1 0 1 x2 0 0 1 0 1 0 1 0 x3 1 1 0 1 0 0 0 0 x4 0 0 1 0 1 1 1 0 x5 0 1 0 1 0 0 0 1 x6 1 0 0 1 0 0 0 0 x7 0 1 0 1 0 0 0 1 x8 1 0 0 0 1 0 1 0 5. Consider an undirected graph with 25 vertices x2, x3, . . ., x26 with edges (xi, x j) if and only if integers i and j have a common divisor. How many components does this graph have? Find a spanning tree for each component.
6. Show that in an n-vertex graph, a set of n −1 edges that form no circuits is a spanning tree.
7. Show that a connected undirected graph with just one spanning tree is a tree.
8. Show that a breadth-first spanning tree contains shortest paths from the root to every other vertex.
9. (a) Prove that a depth-first search reaches all vertices in an undirected connected graph.
(b) Repeat part (a) for breadth-first search.
10. (a) Show that the height of any depth-first spanning tree of a graph G starting from a given root a must be at least as large as the height of a breadth-first spanning tree of G with root a.
(b) By starting with different roots, it might be possible for a depth-first spanning tree of G to have a smaller height than a breadth-first spanning tree of G.
Find a graph in which this can happen.
11. A cut-set is a set S of edges in a connected graph G whose removal disconnects G, but no proper subset of S disconnects G. Show that any cut-set of G has at least one edge in common with any spanning tree of G.
www.itpub.net 3.2 Search Trees and Spanning Trees 111 12. (a) Use a depth-first search to find your way through this maze from S to E.
(b) How many ways through this maze (without cycles or backing up) are there?
S E 13. Repeat Example 3 using a depth-first search.
14. Perform a breadth-first search of the graph in Figure 3.8 to find the shortest path from S to E.
15. Find another (actually shorter) way in Example 3 to get 2 quarts in one pitcher.
16. (a) Repeat Example 3 with pitchers of size 8, 5, and 3 with an objective of 4 quarts in one pitcher.
(b) Repeat Example 3 with pitchers of sizes 12, 8, and 5, and an objective of 2 quarts in one pitcher.
17. Show that the stick-to-the-right-hand-wall rule will always get one out of a two-dimensional maze.
18. Use a depth-first search in Example 3 to show that any amount between 0 and 10 quarts can be obtained in one of the pitchers.
19. Suppose that a dog, a goat, and a bag of tin cans are to be transported across a river in a ferry that can carry only one of these three items at once (along with a ferry driver). If the dog and goat cannot be left alone on a shore when the driver is not present, nor can the goat and tin cans, find a scheme for getting all across the river.
20. Three jealous wives and their husbands come to a river. The party must cross the river (from near shore to far shore) in a boat that can hold at most two people. Find a sequence of boat trips that will get the six people across the river without ever letting any husband be alone (without his wife) in the presence of another wife.
21. RepeatExercise20thistimewithfourjealouswivesandtheirhusbands,ifpossible.
22. Do Exercise 20, this time with four jealous wives and husbands and a three-person boat.
23. Repeat Example 4 with the additional condition that only one of the cannibals can row (all the missionaries can row).
24. The following is a variation of the pitcher-pouring problem. An hourglass is a timepiece in which a specified period of time will elapse as sand in the top half empties through the narrow opening to the bottom half. If an hourglass is designed to take 10 minutes for the sand to flow down and after, say, 7 minutes the glass is turned upside down, then the sand flow will be reversed and it will take 7 more minutes for the sand to empty. Suppose you have two hourglasses, a 4-minute hourglass and a 7-minute hourglass. Describe a scheme for starting them and turning them upside down so as to measure out a 9-minute period of time. (Hint: In the beginning, you start one or both hourglasses; then a decision 112 Chapter 3 Trees and Searching must be made whenever an hourglass empties whether to restart it and whether to start or invert the other hourglass.) 25. A group of four people have to cross a bridge that will accommodate only two people at a time. It is night and there is one flashlight. Any party who crosses, either one or two people, must have the flashlight with it. (The flashlight must be walked back and forth; it cannot be thrown.) Find a way for all four to get to the other side of the bridge in 17 minutes, given that person A takes 1 minute to cross the bridge, B takes 2 minutes, C takes 5 minutes, and D takes 10 minutes.
When two people cross, they must go at the speed of the slower person.
26. List the vertices in order of a preorder traversal and a postorder traversal of (a) Figure 3.1a (b) Figure 3.1b (c) Figure 3.4 27. Generalize the arithmetic tree in Figure 3.12 to include unary operations such as inverses or sin( ). Give the tree for the following expression: sin(((a + ((b × c)−1 + ((a + d) × e))) −(c + e)) + (a −b)−1) 28. Let T be a spanning tree in a connected undirected graph G. Show that when any non-tree edge is added to T, a unique circuit results.
29. Prove that any spanning tree T ′ of a graph G can be converted into any other spanning tree T ′′ of G by a sequence of spanning trees T1, T2, . . . , Tm where T ′′ = T1, T ′′ = Tm, and Tk is obtained from Tk−1 by removing one edge of Tk−1 that was in T ′ and adding one edge in T ′ (m is the number of edges by which T ′ and T ′′ differ). (Hint: Use Exercise 28 and induction on m.) 30. Suppose that during a preorder traversal of a binary tree T, we write down a 1 for each internal vertex and a 0 for each leaf in the traversal, building a sequence of 1s and 0s. If T has n leaves, the sequence will have n 0s and n −1 1s. We call this sequence the characteristic sequence of T. (Such a sequence determines a unique tree.) (a) Find the binary tree with the characteristic sequence 110100110100100.
(b) Prove that the last two digits in any characteristic sequence are 0s (assuming n ≥2).
(c) Prove that a binary sequence with n 0s and n −1 1s, for some n, is a characteristic sequence of some binary tree if and only if the first k digits of the sequence contain at least as many 1s as 0s, for 1 ≤k ≤2n −2.
31. (a) Write a program for building a depth-first spanning tree of a graph whose adjacency matrix is given.
(b) Repeat part (a) for a breadth-first spanning tree.
32. Use a breadth-first inverted spanning tree T (edges directed from children to par-ents) to build an Euler cycle in a directed graph possessing an Euler circuit as fol-lows. Starting at the root of T, trace out a path such that at any vertex x we choose onlytheedgeinTtotheparentofx(inT)iftherearenootherunusededgesleaving x at this stage. Verify the correctness of the algorithm. Program this algorithm.
www.itpub.net 3.3 The Traveling Salesperson Problem 113 3.3 THE TRAVELING SALESPERSON PROBLEM In this section, we illustrate the use of trees in graph optimization problems of oper-ations research. The traveling salesperson problem seeks to minimize the cost of the route for a salesperson to visit a set of cities and return to home. One seeks a minimal-cost Hamilton circuit in a complete graph having an associated cost matrix C. Entry cij in C is the cost of using the edge from the ith vertex to the jth vertex. The traveling salesperson problem arises in many different guises in operations research, and is famously difficult (it is an NP-complete problem; see Appendix A.5). One example is planning the movement of an automated drill press making holes at specified locations on printed circuit boards.
We present two approaches to solving the traveling salesperson problem. Both approaches use trees, but in two very different ways. The first approach is to system-atically consider all possible ways to build Hamilton circuits in search of the cheapest one. It uses a “branch and bound” method to limit the number of different vertices in the search tree that must be inspected in search of a minimal solution. Each vertex in the tree will represent a partial Hamilton circuit, and each leaf a complete Hamilton circuit. Note that a complete graph on n vertices has (n −1)! different Hamilton cir-cuits; for example, a 50-city (vertex) problem has 49! ≈6 × 1062 possible circuits.
The “branch and bound” method reduces substantially the number of circuits that must be checked, although not enough to make it practical to solve large traveling salesperson problems.
This leads to another strategy, our second approach, which is to construct a near-minimal solution using an algorithm that is quite fast. The huge amount of time required to find exact solutions to problems that involve enumeration of very large trees, such as the traveling salesperson problem, has led researchers to concentrate on heuristic, near-optimal algorithms for such problems.
Note that the numbers in the cost matrix may themselves only be estimates—e.g., travel times by truck—and so any solution with these numbers is only an estimate.
Finding a minimal coloring of any arbitrary graph or testing for isomorphism are other problems we have seen that have underlying search trees with similarly huge numbers of possible solutions. Near-minimal coloring algorithms have received much attention, but unfortunately, a “near-isomorphism” is usually of no value.
We consider a small traveling salesperson problem with four vertices x1, x2, x3, x4. Let the cost matrix for this problem be the matrix in Figure 3.13. Entry cij is the cost of going from vertex (city) i to vertex j. (Note that we do not require cij = cji.) The infinite costs on the main diagonal indicate that we cannot use these entries. A Hamilton circuit will use four entries in the cost matrix C, one in each row and in Figure 3.13 114 Chapter 3 Trees and Searching Figure 3.14 each column, such that no proper subset of entries (edges) forms a subcircuit. This latter constraint means that if we choose, say, entry c23, then we cannot also use c32, for these two entries form a subcircuit of length 2. Similarly, if entries c23 and c31 are used, then c12 cannot be used.
We first show how to obtain a lower bound for the cost of this traveling salesperson problem. Since every solution must contain an entry in the first row, the edges of a minimal tour will not change if we subtract a constant value from each entry in the first row of the cost matrix (of course, the cost of a minimal tour will change by this constant). We subtract as large a number as possible from the first row without making any entry in the row negative; that is, we subtract the value of the smallest entry in row 1, namely 3. Do this for the other three rows also. We display the altered cost matrix in Figure 3.14.
All rows in Figure 3.14 now contain a 0 entry. After subtracting a total of 3 + 3 + 5 + 4 = 15 from the different rows, a minimal tour using the cost data in Figure 3.15 will cost 15 less than a minimal tour using original cost data in Figure 3.14. Still, the edges of a minimal tour for the altered problem are the same edges that form a minimal tour for the original problem.
In a similar fashion, we can subtract a constant from any column without changing the set of edges of a minimal tour. Since we will want to avoid making any entries negative, we consider subtracting a constant only from columns with all current entries positive. The only such column in Figure 3.14 is column 4, whose smallest value is 1. So we subtract 1 from the last column in Figure 3.14 to get the matrix in Figure 3.15. Every row and column in Figure 3.15 now contains a 0 entry.
The cost of a minimal tour using Figure 3.15 has been reduced by a total of 15 + 1 = 16 from the original cost using Figure 3.13. We can use this reduction of cost to obtain a lower bound on the cost of a minimal tour: A minimal tour using the costs in Figure 3.15 must trivially cost at least 0, and hence a minimal tour using Figure 3.13 must cost at least 16. In general, the lower bound for the traveling salesperson problem equals the sum of the constants subtracted from the rows and columns of Figure 3.15 www.itpub.net 3.3 The Traveling Salesperson Problem 115 Figure 3.16 the original cost matrix to obtain a new cost matrix with a 0 entry in each row and column.
Now we are ready for the branching part of the “branch and bound” method. We look at an entry in Figure 3.15 that is equal to 0. Say c12. Either we use c12 or we do not use c12. We “branch” on this choice. In the case that we do not use c12, we represent the no-c12 choice by setting c12 = ∞. The smallest value in row 1 of the altered Figure 3.15 is now c14 = 3, and so we can subtract this amount from row 1. Similarly, we can subtract 1 from column 2. If we do not use entry c12, we obtain the new cost matrix in Figure 3.16. Hence, any tour for the original problem (Figure 3.13) that does not use c12 must cost at least 16 + (3 + 1) = 20.
If we use c12 to build a tour for Figure 3.15, then the rest of the tour cannot use another entry in row 1 or column 2; also entry c21 must be set equal to ∞(to avoid the subcircuit x1 −x2 −x1). The new smallest value of the reduced matrix in row 2 is 1, and so we subtract 1 from row 2 to obtain the cost matrix in Figure 3.17. The lower bound for a tour in Figure 3.17 using c12 is now 16 + 1 = 17.
Since our lower bound of 17 is less than the lower bound of 20 when we do not use c12, we continue our consideration of tours using c12 by determining bounds associated with whether or not to use a second entry with value 0. We will continue to extend these partial tours using c12 as long as the lower bound is ≤20. If the lower bound were to exceed 20, then we would have to go back and consider partial tours not using c12. Our tree of choices for this problem will be a binary tree whose internal vertices represent choices of the form: use entry ci j or do not use ci j (see Figure 3.19).
At any stage, as long as the lower bounds for partial tours using ci j are less than the lower bound for tours not using ci j, we do not need to look at the subtree of possible tours not using ci j.
We extend a tour using c12 by considering an entry with value 0 in Figure 3.17.
This next 0 entry need not connect with c12—that is, need not be of the form ci1 or c2 j—but for simplicity we shall pick an entry in row 2. The only 0 entry in row 2 of Figure 3.17 is c24.
Again we have the choice of using c24 or not using c24. Not using c24 will increase the lower bound by 2 (the smallest entry in row 2 of Figure 3.17 after we set c24 = ∞; Figure 3.17 116 Chapter 3 Trees and Searching Figure 3.18 column 4 still has a 0). Using c24 will not increase the lower bound, and so we further extend the partial tour using c24 along with c12. Again we delete row 2 and column 4 in Figure 3.17 and set c41 = ∞(to block the subcircuit x1−x2−x4−x1). Figure 3.18 shows the new remaining cost matrix (all rows and columns still have a 0 entry).
Now the way to finish the tour is clear: use entries c43 and c31 for a complete tour x1−x2−x4−x3−x1. We actually have no choice since not using either of c43 or c31 forces us to use an ∞(which represents a forbidden edge). Since c43 = c31 = 0, this tour has a cost equal to the lower bound of 17 in Figure 3.18. Further, this tour must be minimalsinceitscostequalsourlowerbound.Recheckingthetour’scostwiththeorig-inal cost matrix in Figure 3.13, we have c12 + c24 + c43 + c31 = 3 + 5 + 4 + 5 = 17.
We summarize the preceding reasoning with the decision tree in Figure 3.19 summarizing choices we made and their lower bounds (L.B.).
One general point should be made about how to take best advantage of the branch-and-bound technique. At each stage, we should pick as the next entry on which to branch (use or do not use the entry), the 0 entry whose removal maximizes the increase in the lower bound. In Figure 3.15, a check of all 0 entries reveals that not using entry c43 will raise the lower bound by 3 + 3 = 6 (3 is the new smallest value in row 4 and in column 3). So c43 would theoretically have been a better entry than c12 to use for the first branching, since the greater lower bound for the subtree of tours not Solutions without c12 All solutions Solutions with c12 Solutions without c24 Solutions with c24 Solutions without c43 Solutions with c43 Solution with c31 L.B. = 16 L.B. = 20 L.B. = 17 L.B. = 19 L.B. = 17 No solution L.B. = 17 Figure 3.19 www.itpub.net 3.3 The Traveling Salesperson Problem 117 Figure 3.20 using c43 makes it less likely that we would ever have to check possible tours in that subtree.
Now consider the 6 × 6 cost matrix in Figure 3.20. We present the first few branchings in Figure 3.21 for this cost matrix. We circle the 0 entry on which we branch at each stage.
The branch-and-bound method is used not only to solve optimization problems.
It is also a central tool in artificial intelligence. Computers that play chess make a tree of possible future moves and then assign some sort of “value” to the resulting chessboard situations. A move that leads to a bad situation gets a very high value, and future moves from this situation are not pursued. At more promising situations, all reasonable next moves are considered and their resulting situations evaluated.
The more powerful the computer, the more moves into the future it can examine, although the procedure for evaluating positions is just as important as the machine’s speed.
Next we present a quicker algorithm for obtaining good (near-minimal) tours, when the costs are symmetric (i.e., ci j = c ji) and the costs satisfy the triangle inequality—cik ≤ci j + c jk. These two assumptions are satisfied in most traveling salesperson problems. After describing the algorithm and giving an example of its use, we will prove that at worst the approximate algorithm’s tour is always less than twice the cost of a true minimal tour. A bound of twice the true minimum may sound bad, but in many cases where a ballpark figure is needed (and the exact minimal tour can be computed later if needed), such a bound is quite acceptable. Furthermore, in practice our approximate algorithm finds a tour that is close to the true minimum.
This algorithm uses a successive nearest-neighbor strategy.
Approximate Traveling Salesperson Tour Construction 1.
Pick any vertex as a starting circuit C1 consisting of 1 vertex.
2.
Given the k-vertex circuit Ck, k ≥1, find the vertex zk not on Ck that is closest to a vertex, call it yk, on Ck.
3.
Let Ck+1 be the k + 1-vertex circuit obtained by inserting zk immediately in front of yk in Ck.
4.
Repeat steps 2 and 3 until a Hamilton circuit (containing all vertices) is formed.
118 Chapter 3 Trees and Searching Figure 3.21 Tree of traveling salesperson partial solutions www.itpub.net 3.3 The Traveling Salesperson Problem 119 Figure 3.22 Example 1: Approximate Traveling Salesperson Tour Let us apply the preceding algorithm to the 6-vertex traveling salesperson problem whose cost matrix is given in Figure 3.22. Name the vertices x1, x2, x3, x4, x5, x6.
We will start with x1 as C1. Vertex x4 is closest to x1, so C2 is x1−x4−x1. Ver-tex x3 is the vertex not in C2 closest to a vertex in C2—namely, closest to x4; thus C3 = x1−x3−x4−x1. There are now two vertices, x2 and x6, 3 units from vertices in C3. Suppose we pick x2. It is inserted before x3 to obtain C4 = x1−x2−x3−x4−x1.
Vertex x6 is still 3 units from x1, so we insert x6 before x1, obtaining C5 = x1−x2−x3−x4−x6−x1. Finally, x5 is within 4 units of x3 and x6. Inserting x5 be-fore x6, we obtain our near-minimal tour C6 = x1−x2−x3−x4−x5−x6−x1, whose cost we compute to be 19.
In this case, the length of this tour obtained with the Approximate Tour Construc-tion algorithm is quite close to the minimum (which happens to be 18). The length of the approximate tour typically depends on the starting vertex. If we suspected that our approximate tour was not that close to an optimal length, then by trying other vertices as the starting vertex (that forms C1) and applying the algorithm, we will get other near-minimal tours. Taking the shortest of this set of tours generated by the Approximate Tour Construction would give us an improved estimate for the true minimal tour.
Theorem The cost of the tour generated by the approximate tour construction is less than twice the cost of the minimal traveling salesperson tour.
Proof Optional Suppose we are successively building the k-vertex circuits Ck according to the ap-proximate tour construction. Let Sk be a subset of edges in the true minimal tour C∗ that connect the vertices in Ck to the other vertices in G(Sk is described precisely below). When k = 1, we let S1 consist of C∗−e∗, where e∗is the costliest edge in C∗.
Since C1 is a single vertex and C∗−e∗is a Hamilton path, S1 = C∗−e∗will connect C1 with the rest of G.
For concreteness, see the 8-vertex graph in Figure 3.23a, where C1 is the vertex x6 and S1 = C∗−e∗is the Hamilton path of solid edges. If z1 = x3 and so C2 is 120 Chapter 3 Trees and Searching x1 x2 x3 x4 x5 x6 x7 x8 x1 x2 x3 x4 x5 x6 x7 x8 x1 x2 x3 x4 x5 x6 x7 x8 C2 C3 (c) (b) (a) Figure 3.23 x6−x3−x6, then we set S2 = S1−(x5, x6) (C2 is shown by dashed edges and S2 by solid edges in Figure 3.23b). In general, we will obtain Sk+1 from Sk by removing the first edge on the path in Sk from Ck to the new vertex zk being added to Ck. In Figure 3.23c, suppose that z2 = x8 and y2 = x6. So x8 is inserted into C2 between x3 and x6 to obtain C3 = x6−x8−x3 −x6. Edge (x6, x7) is removed from S2 to get S3, since (x6, x7) is the first edge on the path in S2 from C2 (specifically x6) to x8.
By the shortest edge rule for picking z2 = x8 and y2 = x6, we know that c68 [the cost of edge (x6, x8)] is the smallest cost among all edges between C2(= x6−x3−x6) and the rest of G. Thus c68 ≤c67 [where (x6, x7) was the edge removed from S2 to get S3]. Using this fact and the triangle inequality, we shall now prove that in-serting x8 into C2 to get C3 has a net increase in cost ≤2c67. The increase of C3 over C2 is c68 + c38 −c36, since edges (x6, x8) and (x3, x8) replace (x3, x6).
But by the triangle inequality, c38 ≤c36 + c68, or equivalently c38−c36 ≤c68. Thus c68 + (c38−c36) ≤c68 + c68 ≤2c67, as claimed.
This same argument can be applied when we insert each successive zk into Ck to prove that the increase of Ck+1 over Ck is at most twice the cost of the edge dropped from Sk to get Sk+1. Starting from C1 and repeating this bound on the insertion cost, we see that Cn, the final near-optimal Hamilton circuit, is bounded by twice the sum of the costs of the edges in S1 = C∗−e∗, which is less than twice the cost of the minimal traveling salesperson tour C∗. ◆ 3.3 EXERCISES 1. Subtract the value of each nondiagonal entry in Figure 3.13 from 10. Now solve the traveling salesperson problem for this new cost matrix.
2. Solve the traveling salesperson problem for the cost matrix in Figure 3.20.
3. Write a computer program to solve a six-city traveling salesperson problem and use it for the cost matrix in Figure 3.22.
4. Use ad hoc arguments to show that the cost of a minimal tour for the cost matrix in Figure 3.22 is 18.
5. Every month a plastics plant must make batches of five different types of plastic toys. There is a conversion cost cij in switching from the production of toy i to www.itpub.net 3.4 Tree Analysis of Sorting Algorithms 121 toy j, shown in the matrix. Find a sequence of toy production (to be followed for many months) that minimizes the sum of the monthly conversion costs.
To T1 T2 T3 T4 T5 From T1 — 3 2 4 3 T2 4 — 4 5 5 T3 5 3 — 4 4 T4 3 5 1 — 6 T5 5 4 2 3 — 6. The assignment problem is a matching problem with n people and n jobs and a cost matrix with entry cij representing the “cost” of assigning person i to job j.
The goal is a one-to-one matching of people to jobs that minimizes the sum of the costs. Set all diagonal entries in Figure 3.13 equal to 5 and solve this 4 × 4 assignment problem by a branch-and-bound approach. How does an assignment problem differ from a traveling salesperson problem?
7. Use the approximate algorithm to find approximate traveling salesperson tours for the cost matrices in the following (use just entries above the main diagonal): (a) Figure 3.13 (b) Figure 3.20 (c) Exercise 5 8. Consider the following rule for building approximate traveling salesperson tours.
Starting with a single-vertex tour T1, successively add the vertex whose insertion into Tk to form Tk+1 minimizes the increase in cost—that is, if x is inserted between xk and xk+1, then ckr + cr(k+1) −ck(k+1) should be minimal over all choices of xr and xk.
(a) Can you prove an upper bound on this method similar to the one found for the approximate algorithm (making the same assumptions)?
(b) Apply this method to the cost matrix in (i) Figure 3.20 (using just entries above the main diagonal) and (ii) Figure 3.22.
9. Find a 3 × 3 cost matrix for which two different initial lower bounds can be ob-tained (with different sets of 0 entries) by subtracting from the rows and columns in different orders.
10. Make up a 5 × 5 cost matrix for which the Approximate Tour Construction finds the following: (a) An optimal tour (b) A fairly costly tour (at least 50% over the true minimum).
3.4 TREE ANALYSIS OF SORTING ALGORITHMS One of the basic combinatorial procedures in computer science is sorting a set of items. Items are sorted according to their own numerical value or the value of some 122 Chapter 3 Trees and Searching associated variable. To simplify our discussion, we will assume that all items have distinct numerical values.
Some of the sorting procedures we mention will use binary trees explicitly. All the procedures use binary testing trees implicitly—that is, the procedures use a sequence of tests that compare various pairs of items. A binary testing tree model of sorting has one important consequence. The tree must have at least n! leaves if there are n items, since there are n! possible permutations, or rearrangements, of the set. By Theorem 4 in Section 3.1, the height of a binary testing tree must be at least ⌈log2n!⌉, which is approximately n log2n. Thus we obtain the theorem.
Theorem In the worst case, the number of binary comparisons required to sort n items is at least O(n log2n).
The theorem refers to the worst case, because certain sorting algorithms will sort some sets very quickly (corresponding to short paths in the binary testing tree). One can also show that the average number of binary comparisons required to sort n items is at best of order O(n log2 n) (see Exercise 3). A function g(n) is said to be of order O( f (n)) if, for large values of n, g(n) ≤cf (n), for some constant c.
The best known sorting algorithm is called bubble sort, so named because small items move up the list the way bubbles rise in a liquid. It is compactly written as FOR m ←2 TO n DO FOR i←n STEP −1 TO m DO IF Ai≤Ai −1, THEN interchange items Ai and Ai −1 This procedure will always require (n −1) + (n −2) + · · · + 1 = 1 2n(n −1) binary comparisons. Thus this procedure requires O(n2) comparisons, as opposed to the theoretical bound of O(n log2 n) comparisons. To see how much faster O(n2) grows than O(n log2 n), observe that for n = 50, n2 = 2500, and n log2 n ≈ 300; and for n = 500, n2 = 250,000, and n log2 n ≈4500.
The simplest-to-state sorting procedure that achieves the O(n log2 n) bound is merge sort. It recursively subdivides the original list and successive sublists in half (or as close to half as possible) until each sublist consists of one item. Then it succes-sively merges the sublists in order. The subdivision process is naturally represented as a balanced binary tree and the merging as a reflected image of the tree.
Example 1: Merge Sort Sort the list 5, 4, 0, 9, 2, 6, 7, 1, 3, 8 using a merge sort.
The subdivision tree and subsequent ordered merges are shown in Figure 3.24.
To analyze the number of binary comparisons in a merge sort, we make the simplifying assumption that n = 2r, for some integer r. Then the subdivision tree will have sublists of size 2r−1 at the level-1 vertices. In general, there will be 2r−k items in the sublists at level k. At level r, there are leaves each with one item.
www.itpub.net 3.4 Tree Analysis of Sorting Algorithms 123 4, 5 5 4 6 7 0 9 2 6, 7 1 3 8 3, 8 1, 6, 7 2, 9 0, 4, 5 0, 2, 4, 5, 9 1, 3, 6, 7, 8 0,1, 2, 3, 4, 5, 6, 7, 8, 9 5⎜ 4 0 9 2 1 3 8 7 6 4 5 6⎜ 7 5, 4⎜ 0 9⎜ 2 6, 7⎜ 1 3⎜ 8 6, 7, 1⎜ 3, 8 5, 4, 0⎜ 9, 2 5, 4, 0, 9, 2⎜ 6, 7, 1, 3, 8 Figure 3.24 In the merging tree, first the pairs of leaves are ordered at each vertex on level r −1; this requires one binary comparison (to see which leaf item goes first). See Figure 3.24. In general, at each vertex on level k we merge the two ordered sublists (of 2r−k−1 items) of the two children into an ordered sublist of 2r−k items; this merging will require 2r−k −1 binary comparisons (verification of this number is left as an exercise). Finally, the two ordered sublists of 2r−1 items at level 1 are merged at the root. The number of binary comparisons at all the vertices on level k is 2k(2r−k −1) since there are 2k different vertices on level k. Summing over all levels, we compute the total number of binary comparisons r−1 k=0 2k(2r−k −1) = r−1 k=0 (2r −2k) = r−1 k=0 2r − r−1 k=0 2k = r2r −(2r −1) = (log2 n)n −(n −1) since n = 2r and hence r = log2 n. Thus the number of binary comparisons in a merge sort is O(n log2 n). However, extra computer time is required to implement the initial subdivision process. This extra work also requires only O(n log2 n) steps.
The O(n log1 n) bound is also achieved by the tree-based sorting procedure called heap sort. A heap is a binary or almost-binary (some internal vertices may have just one child) tree with each internal vertex’s value being numerically greater than the values of its children. Figure 3.25a shows a heap involving the numbers 0 through 9.
The root of the heap will have the largest value in the set. If we remove the root of 124 Chapter 3 Trees and Searching 9 8 6 3 4 7 1 2 0 5 8 7 6 3 4 2 1 0 5 (a) (b) Figure 3.25 the heap (and place it at the end of the ordered list we are building), then we can reestablish a heap by making the larger child of the root the new root and recursively replacing each internal vertex moved up with the larger of its two children. The arrows in Figure 3.25a show how vertices move up when the root 9 is removed. The new heap is shown in Figure 3.25b. Now we again remove the root of the heap (and make it the next-to-last item in the sorted list). Repeat this procedure until the heap is emptied and the sorted list complete. The one important missing step is creating the initial heap. This problem and further discussion of heap sort are left for the exercises.
3.4 EXERCISES S u m m a r y o f E x e r c i s e s The first exercises are based on the sorting methods presented in this section. The remaining exercises present other sorting schemes.
1. Apply merge sort to the following sequences: (a) 6, 9, 5, 0, 3, 1, 8, 4, 2, 7 (b) 15, 27, 4, −7, 9, 13, 8, 28, 12, 20, −80 2. Show that at most n −1 comparisons are needed to merge two sorted sublists into a single sorted list of n items.
3. Use the result of Exercise 17 in Section 3.1 to show that the average number of binary comparisons required to sort n items is at least O(n log2 n).
4. (a) Describe how to build an initial heap from an unordered list of n items (the initial heap should be a balanced tree).
(b) Use your method in part (a) to make a heap for the lists in Exercise 1.
(c) Apply heap sort to the heaps in part (b).
5. Showthatheapsortrequires O(n log2 n)comparisonstosortnitems(thisincludes the initial construction of a heap).
6. Modify the following sorting methods to allow for repeated (two or more equal) items: (a) Bubble sort (b) Merge sort (c) Heap sort www.itpub.net 3.5 Summary and References 125 7. Write a computer implementation of (use a recursive language such as PASCAL): (a) Merge sort (b) Heap sort 8. Consider the following sorting scheme, which we call tree sort. We build a “dictionary look-up” binary tree recursively. The first item in the list to be sorted is the root. The second item is a left or right child of the root, depending on whether it numerically precedes or follows the root. We continue to add each successive item as a leaf to this growing tree. For example, if a list begins 8, 4, 11, 6,. . ., the tree after four items would be 8 4 11 6 When all items have been incorporated into the tree, the sorted order is obtained by an inorder traversal of the tree (inorder traversals were defined in Section 3.2).
Build the tree for the list in Exercise 1(a).
9. Consider the following sorting scheme for the list A1, A2, . . . , An(n = 2r). First do a sort (one comparison) of Ai and Ai+n/2; call this the ith ordered pair, for i = 1, 2, . . . , n/2. Next do a merge sort of the ith and the (i + n/4)th ordered pairs; call this the ith 4-tuple. Next do a merge sort of the ith and the (i + n/8)th ordered 4-tuples. Continue until there is just one sorted list of all n items.
(a) Apply this sorting method to the list 10, 12, 7, 0, 5, 8, 11, 15, 1, 6, 3, 9, 13, 4, 2, 14.
(b) How many comparisons does this sorting method require for a list of n numbers?
3.5 SUMMARY AND REFERENCES This chapter examined a variety of search and data organization problems. The com-mon graph-theoretic tool for all these problems was trees. Section 3.1 presented basic properties and terminology of trees. Section 3.2 introduced spanning trees and demonstrated the uses of depth-first and breadth-first search. Section 3.3 gave tree-based branch-and-bound and heuristic approaches to a famous optimization problem, the traveling salesperson problem. Section 3.4 looked at the decision trees underlying sorting algorithms.
The first paper implicitly using trees was Kirchhoff’s 1847 fundamental paper about electrical networks. Cayley was the first person to use the term tree in an 1857 formula for counting ordered trees (Theorem 5 in Section 3.1). Searching methods 126 Chapter 3 Trees and Searching have been around for years (see Lucas ), but a systematic development of this subject came only in recent years with the advent of digital computers. There are several good computer science books about searching, sorting, and graph algorithms.
See Kruse or Sedgewick . For a good survey of the Traveling Salesperson Problem, see Lawler et al. .
1.
R. Kruse, Data Structures and Program Design in C++, Prentice-Hall, Upper Saddle River, NJ, 1998.
2.
E. Lawler, J. Lenstra, A. Rinnooy Kan, and D. Shmoys, The Traveling Salesman Problem, John Wiley, New York, 1985.
3.
E. Lucas, Recreations Mathematiques, Gauthier-Villars, Paris, 1891.
4.
R. Sedgewick, Algorithms in C, 3rd. ed., Addison-Wesley, Reading, MA, 1997.
www.itpub.net CHAPTER 4 NETWORK ALGORITHMS 4.1 SHORTEST PATHS In this chapter we present algorithms for the solution of four important network optimization problems: shortest paths, minimum spanning trees, maximum flows, and the transportation problem. By a network we mean a graph with a non-negative integer k(e) assigned to each edge e. This integer will typically represent the “length” or “cost” of an edge, in units such as miles or dollars, or represent “capacity” of an edge, in units such as megawatts or gallons per minute. The optimization problems we shall discuss arise in hundreds of different guises in management science and system analysis settings. Thus good systematic procedures for their solution are essential. In the case of network flows, we shall see that the flow optimization algorithm can also be used to prove several well-known combinatorial theorems.
We begin with an algorithm for a relatively simple problem, finding a shortest path in a network from point a to point z. We do not say the shortest path because, in general, there may be more than one shortest path from a to z. For the rest of this section, let us assume that all networks are undirected and connected.
Let us immediately eliminate one possible shortest path algorithm: Determine the lengths of all paths from a to z, and choose a shortest one. The computer is fast, but not that fast—such enumeration is already infeasible for most networks with 100 vertices. So when we find a shortest path, we must be able to prove it is shortest without explicitly comparing it with all other a–z paths. Although the problem is now starting to sound difficult, there is a straightforward algorithmic solution.
The algorithm we present is due to Dijkstra. This algorithm gives shortest paths from a given vertex a to all other vertices. Let k(e) denote the length of edge e. Let the variable m be a “distance counter.” For increasing values of m, the algorithm labels vertices whose minimum distance from vertex a is m. The first label of a vertex x will be the previous vertex on the shortest path from a to x. The second label of x will be the length of the shortest path from a to x.
Shortest Path Algorithm 1.
Set m = 1 and label vertex a with (−, 0) (the “−” represents a blank).
2.
Check each edge e = (p, q) from some labeled vertex p to some unlabeled vertex q. Suppose p’s labels are [r, d(p)]. If d(p) + k(e) = m, label q with (p, m).
127 128 Chapter 4 Network Algorithms 3.
If all vertices are not yet labeled, increment m by 1 and go to Step 2. Otherwise go to Step 4. If we are only interested in a shortest path to z, then we go to Step 4 when z is labeled.
4.
For any vertex y, a shortest path from a to y has length d(y), the second label of y. Such a path may be found by backtracking from y (using the first labels) as described below.
Observe that instead of concentrating on the distances to specific vertices, this algorithm solves the questions: how far can we get in 1 unit, how far in 2 units, in 3 units, . . . , in m units, . . . ? Formal verification of this algorithm requires an induction proof (based on the number of labeled vertices). The edges that make up the shortest paths from a to the other vertices can be shown to form a spanning tree (Exercise 11).
The key idea is that to find a shortest path from a to any other vertex we must first find shortest paths from a to the “intervening” vertices. If Pk = (s1, s2, . . . , sk) is a shortest path from s1 = a to sk, then Pk = Pk−1 + (sk−1, sk), where Pk−1 = (s1, s2, . . . , sk−1) is a shortest path to sk−1. Similarly, Pk−1 = Pk−2 + (sk−2, sk−1), and so on.
To record a shortest path to sk all we need to store (as the first part of a label in the above algorithm) is the name of the next-to-last vertex on Pk—namely, sk−1.
Preceding sk−1 on the shortest path from a is sk−2, the next-to-last vertex on Pk−1. By continuing this backtracking process, we can recover all of Pk.
The algorithm given above has two significant inefficiencies. First, if all sums d(p) + k(e) in Step 2 have values of at least m′ > m, then the distance counter m should be increased immediately to m′. Second, one does not need to repeat the computation of the expressions d(p) + k(e) in Step 2 every time m increases. Instead, each unlabeled vertex q can be given a “temporary” label (p, d∗(q)), where d∗(q) equals the minimum of the d(p) + k(e), for those labeled vertices p with an edge e = (p, q) to q. Thus, d∗(q) represents the current shortest path length to q using labeled vertices. Now q’s temporary label needs to be updated only when a newly labeled vertex p is adjacent to q. Details of this and other improvements in the form of Dijkstra’s algorithm given here can be found in Ahuja et al. .
Example 1: Shortest Path A newly married couple, upon finding that they are incompatible, want to find a shortest path from point N (Niagara Falls) to point R (Reno) in the road network shown in Figure 4.1. We apply the shortest path algorithm. First N is labeled (−, 0).
For m = 1, no new labeling can be done [we check edges (N, b), (N, d), and (N, f)]. For m = 2, d(N) + k(N, b) = 0 + 2 = 2, and we label b with (N, 2). For m = 3, 4, no new labeling can be done. For m = 5, d(b) + k(b, c) = 2 + 3 = 5. So we label c with (b, 5). We continue to obtain the labeling shown in Figure 4.1. Backtracking from R, we find the shortest path to be N–b–c–d–h–k–j–m–R with length 24.
If we want simultaneously to find the shortest distances between all pairs of vertices (without directly finding all the associated shortest paths), we can use the www.itpub.net 4.1 Shortest Paths 129 h (d, 9) c (b, 5) b (N, 2) N (—, 0) d (c, 7) e (d, 11) g (h, 13) f (N, 10) i (f, 14) m (j, 19) k (h, 14) j (k, 17) R (m, 24) 5 2 6 6 6 4 8 10 2 3 2 6 2 4 3 4 5 5 3 12 20 4 Figure 4.1 following simple algorithm due to Floyd. Let matrix D have entry dij = ∞(or a very large number) if there is no edge from the ith vertex to the jth vertex; otherwise dij is the length of the edge from xi, to x j. Then Floyd’s algorithm is most easily stated with the following deceptively simple computer program: FOR k←1 TO n DO FOR i ←1 TO n DO FOR j ←1 TO n DO IF di k+ dkj< di j THEN di j←di k+ dkj; When finished, dij will be the shortest distance from the ith vertex to the jth vertex.
4.1 EXERCISES S u m m a r y o f E x e r c i s e s The first six exercises involve shortest path calculations. The remaining exercises discuss associated theory (fairly easy theory), and the last exercise asks for a program.
1. Use the shortest path algorithm to find the shortest path between vertex c and vertex m in Figure 4.1.
2. Find the shortest path between the following pairs of vertices in the network in Figure 4.2 in Section 4.2: (a) a and y (b) d and r (c) e and g 3. The network below shows the paths to success from L (log cabin) to W (White House). The first number on an edge is the time (number of years) it takes to traverse the edge; the second is the number of enemies you make in taking that edge. Use the shortest path algorithm to answer the following: (a) Find the quickest path to success (from L to W).
(b) Find the quickest path to success that avoids points c, g, and k.
(c) Find the path to success (from L to W) that minimizes the total number of enemies you make.
130 Chapter 4 Network Algorithms L d c e g i l W m j h b f k a 7, 2 8, 4 8, 5 8, 4 12, 4 8, 3 6, 2 5, 5 6, 3 7, 2 8, 3 5, 2 5, 1 4, 1 3, 1 5, 1 7, 2 6, 4 4, 2 4, 1 3, 2 4, 2 5, 1 6, 3 4, 2 5, 3 4, 4 3, 3 (d) Alter the order in which edges are checked in Step 2 of the shortest path algorithm to get another minimum-enemies path. How many such minimum-enemies paths are there?
4. With reference to Exercise 3, let the roughness index R of a path to success be R = T + 2E, where T is the time to get from L to W and E is the total number of enemies made.
(a) Find the smoothest (least rough) path to success.
(b) Find the smoothest path to success that includes edge ( f, i); this edge can be traversed in either direction.
5. Ignore the numbers on the edges in Exercise 3 and use the shortest path algorithm to find the following shortest (fewest edges) paths: (a) Shortest path from L to W (b) Shortest path from L to W including vertex d (c) Shortest path from L to W including both vertex e and vertex m 6. Suppose that the edges in Exercise 3 are directed according to the alphabetical order of the endpoints, where L precedes a and W follows m [so edge (r⃗, s) goes from r to s if r precedes s in the alphabet]. Find the quickest path from L to W in this directed network.
7. Prove by induction on m that Dijkstra’s shortest path algorithm finds the shortest path from a to every other vertex in the network.
8. Prove that Floyd’s algorithm finds the shortest path between all pairs of vertices.
9. Make up an example to show that Dijkstra’s algorithm fails if negative edge lengths are allowed.
10. (a) Show that if the edges are properly ordered (and the edges are checked in this order in Step 2), the shortest path algorithm will produce any given shortest path from a to z when more than one shortest path exists.
(b) Order the edges of the figure in Exercise 3 so that in Exercise 3(c), the “shortest” path found will be (L, a, c, d, f, g, k, m, W).
11. Show that in the shortest path algorithm, the edges used in Step 2 to label new vertices form a spanning tree.
12. The transitive closure of a directed graph G is obtained by adding to G an edge (xi⃗, x j) for each nonadjacent pair xi, x j with a directed path from to xi to x j. Let www.itpub.net 4.2 Minimum Spanning Trees 131 dij = 1 if (xi⃗, x j) is an edge in G and = 0 otherwise. Replace the IF statement in Floyd’s algorithm with IF d i k· dkj > d i j THEN di j←1 Show that this revised Floyd’s algorithm finds the transitive closure of G.
13. Write a computer program implementing Dijkstra’s shortest path algorithm.
4.2 MINIMUM SPANNING TREES A minimum spanning tree in a network is a spanning tree whose sum of edge lengths k(e) is as small as possible. Minimum spanning trees arise in a variety of important commercial settings, such as finding a minimum-length fiber-optic network to link a given set of sites. This problem appears to be harder than finding a shortest path between two given vertices, but with the proper algorithm it is actually easier to solve by hand than the shortest path problem. The reason for the simple solution is that there are straightforward “greedy” algorithms that can build a minimum spanning tree by successively picking a shortest available edge.
We present two greedy algorithms for building a minimum spanning tree. Let n denote the number of vertices in the network.
Kruskal’s Algorithm Repeat the following step until the set T has n −1 edges (initially T is empty): add to T the shortest edge that does not form a circuit with edges already in T.
Prim’s Algorithm Repeat the following step until tree T has n −1 edges: add to T the shortest edge between a vertex in T and a vertex not in T (initially pick any edge of shortest length).
In both algorithms, when there is a tie for the shortest edge to be added, any of the tied edges may be chosen. A proof of the validity of Kruskal’s algorithm is given in Exercise 12. We prove the validity of Prim’s algorithm shortly.
Example 1: Minimum Spanning Tree We seek a minimum spanning tree for the network in Figure 4.2. Both algorithms start with a shortest edge. There are three edges of length 1: (a, f ), (l, q), and (r, w).
Suppose we pick (a, f ). If we follow Prim’s algorithm, the next edge we would add is (a, b) of length 2, then ( f, g) of length 4, (g, l), then (l, q), then (l, m), and so forth.
The next-to-last addition would be either (m, n) or (o, t) both of length 5 [suppose we choose (m, n)], and either one would be followed by (n, o). The final tree is indicated with darkened lines in Figure 4.2.
On the other hand, if we follow Kruskal’s algorithm, we first include all three edges of length 1: (a, f ), (l, q), (r, w). Next we would add all the edges of length 2: (a, b), (e, j), (g, l), (h, i), (l, m), (p, u), (s, x), (x, y). Next we would add almost all the edges of length 3: (c, h), (d, e), (k, l), (k, p), (q, v), (r, s), (v, w), but not (w, x) unless 132 Chapter 4 Network Algorithms e j o t y x w v u p k f a b c d g l q h m r i n s 1 7 3 2 5 2 1 3 3 4 10 1 7 8 8 2 2 9 5 6 3 5 4 4 2 4 2 5 3 3 8 6 2 7 3 2 4 3 5 7 Figure 4.2 (r, s) were omitted [if both were present we would get a circuit containing these two edges together with edges (r, w) and (s, x)]. Next we would add all the edges of length 4 and finally either (m, n), or (o, t) to obtain the same minimum spanning tree(s) produced by Prim’s algorithm. This similarity is no coincidence (see Exercise 14).
The difficult part in the minimum spanning tree problem is proving the minimality of the trees produced by the two algorithms.
Theorem Prim’s algorithm yields a minimum spanning tree.
Proof For simplicity, assume that the edges all have different lengths. Let T ′ be a minimum spanning tree chosen to have as many edges as possible in common with the tree T ∗ constructed by Prim’s algorithm.
If T ∗̸= T ′, let ek = (a, b) be the first edge chosen by Prim’s algorithm that is not in T ′. This means that the subtree Tk−1, composed of the first k −1 edges chosen by Prim’s algorithm, is part of the minimum tree T ′. In Figure 4.3, edges of Tk−1 are in bold, ek’s edge is dashed, and the other edges of T ′ are drawn normally. Since ek is not in the minimum spanning tree T ′, there is a path, call it P, in T ′ that connects a to b. At least one of the edges of P must not be in Tk−1, for otherwise P ∪ek would form a circuit in the tree Tk formed by the first k edges in Prim’s algorithm. Let e∗be the first edge along P (starting from a) that is not in Tk−1 (see Figure 4.3). Note that one end vertex of e∗is in Tk−1.
If e∗is shorter than ek, then on the kth iteration, Prim’s algorithm would have incorporated e∗, not ek. If e∗is greater than ek, we remove e∗from T ′ and replace it e b ek a Figure 4.3 www.itpub.net 4.2 Minimum Spanning Trees 133 with ek. The new T ′ is still a spanning tree (see Exercise 10 for details), but its length is shorter—this contradicts the minimality of the original T ′. ◆ 4.2 EXERCISES S u m m a r y o f E x e r c i s e s The first five exercises involve building minimum spanning trees. Exercises 6–15 are variations and theory questions about the two minimum spanning tree algorithms.
1. The network below shows the paths to success from L (log cabin) to W (White House). The first number is the cost of building a freeway along the edge and the second is the number of trees (in thousands) that would have to be cut down.
L d c e g i l W m j h b f k a 7, 2 8, 4 8, 5 8, 4 12, 4 8, 3 6, 2 5, 5 6, 3 7, 2 8, 3 5, 2 5, 1 4, 1 3, 1 5, 1 7, 2 6, 4 4, 2 4, 1 3, 2 4, 2 5, 1 6, 3 4, 2 5, 3 4, 4 3, 3 (a) Use Kruskal’s algorithm to find a minimum-cost set of freeways connecting all the vertices together.
(b) Court action by conservationists rules out use of edges (c, e), (d, f ), and (k, W). Now find the minimum-cost set of freeways.
(c) Find two nonadjacent vertices such that the tree in part (a) does not contain the cheapest path between them.
(d) Use Prim’s algorithm to find a set of connecting freeways that minimizes the number of trees cut down.
2. With reference to Exercise 1, suppose that the set of freeways must include city c or city d (but not necessarily both) and all the other cities. Find the minimum-cost set of freeways.
3. Find a minimum spanning tree for the network in Figure 4.1 using (a) Prim’s algorithm (b) Kruskal’s algorithm 4. With reference to Exercise 1, suppose that the governor’s summer home is along edge ( f, i). Find a minimum-cost set of freeways such that ( f, i) is in that set.
5. Find a maximum spanning tree (whose sum of edge lengths is maximum) for the network in Figure 4.2.
6. If each edge has a different cost, show that the minimum spanning tree is unique.
7. Modify Kruskal’s algorithm so that it finds a minimum spanning tree that contains a prescribed edge. Prove that your modification works.
8. Modify Prim’s algorithm so that it finds a maximum spanning tree.
134 Chapter 4 Network Algorithms 9. In the proof of the Theorem, show that (a) ek = (a, b) has one end vertex (a or b) in the tree Tk−1.
(b) e∗has one of its end vertices in the tree Tk−1.
10. Show that the new T ′ (mentioned in the last sentence of the proof of the Theorem) with e∗replaced by ek is a tree—that is, that T ′ is connected and circuit-free.
11. Let T be the spanning tree found by Prim’s algorithm in an undirected, connected network N.
(a) Prove that T contains all edges of shortest length in N unless such edges include a circuit.
(b) Prove that if e∗= (a, b) is any edge of N not in T and if P is the unique path in T from a to b, then for each edge e in P, k(e) ≤k(e∗).
(c) Prove that part (b) characterizes a minimum spanning tree—that is, that any spanning tree T (not just those formed by Prim’s algorithm) is a minimum spanning tree if and only if part (b) is always true for T.
12. Prove that Kruskal’s algorithm gives a minimum spanning tree.
13. Construct an undirected, connected network with eight vertices and 15 edges that has a minimum spanning tree containing the shortest path between every pair of vertices.
14. Suppose that the edges of the undirected, connected network N are ordered and that in both Prim’s and Kruskal’s algorithms, when there is a tie for the next edge to be added, the smaller indexed edge is chosen.
(a) Prove that the edges can be ordered so that Prim’s algorithm will yield any given minimum spanning tree.
(b) Prove that the edges can be ordered so that Kruskal’s algorithm will yield any given minimum spanning tree.
(c) Prove that with ordered edges, both algorithms give the same tree.
15. Given an undirected, connected n-vertex graph G with lengths assigned to each edge, we form a graph G N whose vertices correspond to minimum spanning trees of G with two vertices v1, v2 adjacent if the corresponding minimum spanning trees T1, T2 differ by one edge—that is, T1 = T2 −e′ + e′′ (for some e′, e′′).
(a) Produce an 8-vertex network G such that G N is a chordless 4-circuit.
(b) Prove that if T1 and T2 are minimum spanning trees which differ by k edges, that is |T1 ∩T2| = n −k, then in G N there is a path of length k between the corresponding vertices.
16. Write a computer program to implement (as efficiently as possible) the following: (a) Kruskal’s algorithm (b) Prim’s algorithm www.itpub.net 4.3 Network Flows 135 4.3 NETWORK FLOWS In this section we interpret the integer k(e) associated with edge e in a network as a capacity. We seek to maximize a “flow” from vertex a to vertex z such that the flow in each edge does not exceed that edge’s capacity. Many transport problems are of this general form—for example, maximizing the flow of oil from Houston to New York through a large pipeline network (here the capacity of an edge represents the capacity in barrels per minute of a section of pipeline), or maximizing the number of telephone calls possible between New York and Los Angeles through the cables in a telephone network. It is convenient to assume initially that all networks are directed.
We define an a–z flow f (e) in a directed network N to be an integer-valued function f (e) defined on each edge e— f (e) is the flow in e— together with a source vertex a and a sink vertex z satisfying the following conditions. In(x) and Out(x) denote the sets of edges directed into and out from vertex x, respectively.
(a) 0 ≤f (e) ≤k(e) (b) For x ̸= a or z, e∈In(x) f (e) = e∈Out(x) f (e).
To simplify our analysis, we assume that the flow goes from a to z, never in the reverse direction: (c) f (e) = 0 if e ∈In(a) or e ∈Out(z).
A sample flow is shown in Figure 4.5; the capacity and flow in each edge e are written beside the edge: k(e), f(e). The assumption of integer capacities and flows is not restrictive—that is, the units we count could be thousandths of an ounce rather than barrels. The requirement that there be a single supply vertex, the source a, and a single demand vertex, the sink z, also turns out not to be a restriction. The reason is that we can recast network problems with multiple sources and multiple sinks into a single-source, single-sink network, as illustrated in the following example.
Example 1: Flow Networks with Supplies and Demands Consider the network of solid edges in Figure 4.4 with supplies and demands. Vertex b can supply up to 60 units of flow, and vertices c and d can each supply 40 units.
Vertices h, i, and j have flow demands of 50, 40, and 40 units, respectively. Can we a b e h z j g d c f i 60 30 20 30 30 30 30 30 30 20 30 40 40 50 40 40 40 50 40 20 50 30 Figure 4.4 136 Chapter 4 Network Algorithms meet all the demands? The sources could be oil refineries and the sinks oil-truck distribution centers, or the sources factories and the sinks warehouses.
We model this network problem with a standard one-source, one-sink network as follows. Make b, c, and d regular nonsource vertices. Add a new source vertex a and edges (a⃗, b), (a⃗, c), and (a⃗, d) having capacities 60, 40, and 40, respectively.
This construction simulates the role of the capacitated sources b, c, d. Next make h, i, and j regular nonsink vertices and add a new sink vertex z and edges (h⃗, z), (i⃗, z), and ( j⃗, z), having capacities 50, 40, and 40, respectively.
A flow satisfying the original demands is equivalent to a flow in the new network that saturates the edges coming into z, that is, a flow of value 130.
Let (P, P) denote the set of all edges (x⃗, y) with vertex x ∈P and y ∈P) (where P denotes the complement of P). We call such a set (P, P) a cut. The cut ({a, b, c}, {d, e, z}) in Figure 4.5 consists of the edges (b⃗, d), (b⃗, e), (c⃗, e); the edge (d⃗, c) is not in the cut because it goes from P to P. This cut is represented in Figure 4.5 by a dashed line that separates the vertices in P = {a, b, c} from the vertices in P; the edges in the cut are the edges crossing the dashed line from left to right. We call (P, P) an a–z cut if a ∈P and z ∈P. The a–z cuts in a network are important, because all flow from a to z must cross each a–z cut. The combined capacity of the edges in any a–z cut is thus an upper bound on how much flow can get from a to z.
Let f (e) be an a–z flow in the network N and let P be a subset of vertices in N not containing a or z. Summing together the conservation-of-flow equations in condition (b) for each x ∈P, we obtain x∈P e∈In(x) f (e) = x∈P e∈Out(x) f (e) Certain f (e) terms occur on both sides of the preceding equality: namely, edges from one vertex in P to another vertex in P. After eliminating such f (e) from both sides, the left side becomes e∈(P,P) f (e) and the right side e∈(P,P) f (e). Thus we have (b′) For each vertex subset P not containing a or z, e∈( ¯ P,P) f (e) = e∈(P,P) f (e) That is, the flow into P equals the flow out of P. The following intuitive result is readily verified with (b′).
a b d z e c 7, 4 5, 5 4, 0 4, 3 8, 8 5, 5 8, 6 7, 6 4, 1 3, 2 Figure 4.5 www.itpub.net 4.3 Network Flows 137 Theorem 1 For any a–z flow f (e) in a network N, the flow out of a equals the flow into z.
Proof Assume temporarily that N contains no edge (a⃗, z). Let P be all vertices in N except a and z. So P = {a, z}.
Remember that the only flow into P from {a, z} must be from a, since condition (c) forbids flow from z. Similarly by condition (c), all flow out of P must go to z. Then (P, P) consists of edges going into z (from P) and (P, P) consists of edges going out of a (to P). Thus flow out of a = e∈(P,P) f (e) by(b′) = e∈(P,P) f (e) = flow into z The flow equality still holds if there is flow in an edge (a⃗, z). ◆ The value of the a–z flow f (e), denoted | f |, equals the sum of the flow in edges coming out of a, or equivalently by Theorem 1, the flow into z. Let us consider the question of how large | f | can be. One upper bound is the sum of the capacities of the edges leaving a, since | f | = e∈Out(a) f (e) ≤ e∈Out(a) k(e) Similarly, the sum of the capacities of the edges entering z is an upper bound for | f |. Intuitively, | f | is bounded by the sum of the capacities of any set of edges that cut all flow from a to z—that is, the edges in an a–z cut.
We define the capacity k(P, P) of the cut (P, P) to be k(P, P) = e∈(P,P) k(e) The capacity of the a–z cut (P, P), where P = {a, b, c}, in Figure 4.5 is 13. This tells us that no a–z flow in the network in Figure 4.5 can have a value greater than 13. This motivates the following theorem.
Theorem 2 For any a–z flow f and any a–z cut (P, P) in a network N, | f | ≤k(P, P).
Proof Informally, | f | should equal the total flow from P to P [which is bounded by k(P, P)], butwecannotproveanequalityofthistypeusingcondition(b′)withoutfirstmodifying N. We cannot use (b′) in N because condition (b′) requires that P not contain the source a.
Expand the network N by adding a new source vertex a′ with an edge e′ = (a′⃗, a) of immense capacity. Assign a flow value of | f | to e′, yielding a valid a′–z flow in the expanded network (see Figure 4.6). In effect, the old source a now gets its flow 138 Chapter 4 Network Algorithms a’ a P P z Figure 4.6 from the “super source” a′. Note that a′ will be part of P. Observe that new source a′ is playing a role similar to the source a we added to the multiple-source network in Example 1.
In the new network, we can apply condition (b′) to P. It says that the flow into P, which is at least | f | (other flow could come into P along edges from P), equals the flow out of P. Thus | f | ≤ e∈(P,P) f (e) by(b′) = e∈(P,P) f (e) ≤ e∈(P,P) k(e) = k(P, P) ◆ (∗) Since Theorem 2 says that the value of an a-z flow can never exceed the capacity of any a-z cut, it follows that if the value of some a-z flow f ∗equals the capacity of some a-z cut, then f ∗has to be a flow of the maximum possible value. We shall show that for any flow network, we can always construct an a-z flow whose value equals the capacity of some a-z cut. As just noted, such an a-z flow will be guaranteed to be a flow of maximum value.
To see what properties a flow must have to make the inequality in Theorem 2 an equality, we need to look carefully at the two inequalities in expression () that was used to prove Theorem 2.
Corollary 2a For any a–z flow f and any a–z cut (P, P) in a network N, | f | = k(P, P) if and only if (i) For each edge e ∈(P, P), f (e) = 0.
(ii) For each edge e ∈(P, P), f (e) = k(e).
Further, when | f | = k(P, P), f is a maximum flow and (P, P) is an a–z cut of minimum capacity.
Proof Consider the two inequalities in () of the preceding proof. The first inequality is an equality—the flow from P into P (in the expanded network) equals | f |—if condition (i) holds; otherwise the flow into P is greater than | f |. The second inequality is an equality—the flow out of P equals k(P, P)—if condition (ii) holds; otherwise it is less. Thus equality holds in () if and only if conditions (i) and (ii) are both true. The last sentence in the corollary follows directly from Theorem 2. ◆ www.itpub.net 4.3 Network Flows 139 We have developed all the concepts needed to present our flow maximizing algorithm. We first, discuss an intuitive but faulty technique that can sometimes be used as a shortcut in place of the correct algorithm. After the fault in the shortcut is exposed, the correct algorithm can be more easily understood and appreciated.
All normal flows can be decomposed into a sum of unit-flow paths from a to z, for short, a–z unit flows (abnormal flows that cannot be so decomposed are discussed in Exercise 23). For example, in a telephone network, the flow from New York to Los An-geles can be decomposed into paths of individual telephone calls. Similarly, flow of oil in a pipeline network can be decomposed into the paths of each individual petroleum molecule. Formally, an a–z unit flow fL along a–z path L is defined as fL(e) = 1 if e is in L and = 0 if e is not in L.
This suggests a way to build a maximum flow. We build up the flow as much as possible by successively adding a–z unit flows together, always being sure not to exceed any edge’s capacity. An additional unit flow can use only unsaturated edges, edges where the present flow does not equal the capacity. So we must build paths consisting of unsaturated edges. We define the slack s(e) of edge e in flow f by s(e) = k(e) −f (e) If s is the minimum slack among edges in the a–z unit flow fL, then we can put an additional flow along L of s fL = fL + fL + fL + · · · + fL (s times).
If f1, f2, . . . , fm are a–z unit flows, then f = f1 + f2 + · · · + fm will satisfy conditions (b) and (c) in the definition of a flow (given at the beginning of this section) since f1, f2, . . . , fm satisfy these conditions. If in addition, ϕ satisfies condition (a)— f (e) ≤k(e), for all e—then f is a valid flow.
Example 2: Building a Flow with Flow Paths Let us use the method just outlined to build a maximum a–z flow for the network in Figure 4.7a. Note, as an upper bound, that the value of a flow cannot exceed 10, the sum of the capacities of edges going out of a.
We start with no flow—that is, f (e) = 0, for all e. Now we find some path from a to z, for example, the a–z path L1 = a–b–d–z. The minimum slack on L1 is 3 (at the start, the slack of each edge is just its capacity). So to our initial zero flow, we add the flow 3 fL1. All edges except (b⃗, d) are still unsaturated.
Suppose that we next find the a–z path L2 = a–c–e–z, also with minimum slack 3. Our current flow is 3 fL1 + 3 fL2, as shown in Figure 4.7b. The path L3 = a–b–e–z with minimum slack 2 can be used to get the augmenting flow 2 fL3. Figure 4.7c shows the remaining unsaturated edges with their slacks.
The only a–z path in Figure 4.7c is L4 = a–c–d–z with minimum slack 1. After adding the flow fL4 (see Figure 4.7d), we can get no further than from a to c. Observe that we have saturated the edges in the a–z cut (P, P), where P = {a, c}. The value of the final flow, 9, equals the capacity k(P, P) of this cut, and so by Corollary 2a the flow must be maximum.
140 Chapter 4 Network Algorithms a z e c d b 5 3 1 6 6 6 3 5 a z e c d b 5, 3 3, 3 1, 0 6, 0 6, 3 6, 3 3, 3 5, 3 a z e c d b 2 1 4 1 3 a z e c d b 5, 4 3, 3 1, 1 6, 2 6, 5 6, 4 3, 3 5, 5 e b 6, 5 (a) (b) (c) (d) (Only capacities are shown.) 3 fL1 + 3 fL2 Slack in unsaturated edges for 3f L1 +3f L2 + + 2f L3 3f L1 +3f L2 + 2f L3 fL4 0 0 0 0 a z d c 5, 2 1, 1 3, 1 3, 0 6, 1 6, 6 5, 5 (e) ( f ) f and saturated cut P P , ) with P = ( ( a, c, e) _ 0 0 P P , ) ( _ Redrawn networking highlighting cut a c e z d b 5, 5 3, 0 6, 1 6, 6 5, 2 6, 5 1, 1 3, 1 a z d c e b 5, 4 1, 1 3, 3 3, 2 6, 3 6, 6 6, 3 5, 5 ( g) Corrected maximum flow Figure 4.7 www.itpub.net 4.3 Network Flows 141 Example 3: Faulty Flow Building Suppose the network in Figure 4.7a is redrawn as in Figure 4.7e, with the positions of d and e switched.
Let us again choose augmenting a–z flow paths across the top and bottom of the network, now having sizes 5 and 1, respectively. Since edges (a⃗, b) and (c⃗, d), are saturated by these flow paths, the only possible a–z path along unsaturated edges is L5 = a–c–e–z with minimum slack 1 [the minimum occurring in edge (e, z)]. After adding the a–z unit flow fL5, we get the flow f0 shown in Figure 4.7e; the network is redrawn in Figure 4.7f to make the cut (P0, P0) clearer. The cut (P0, P0), where P0 = {a, c, e}, is saturated, and so no more augmenting a–z unit flows exist. Yet | f0| = 7 and k(P0, P0) = 12. Remember that a flow of size 9 was obtained for this same network in Example 2! What has happened?
We now see that an arbitrary sequence of augmenting a–z unit flows need not inevitably yield a maximum flow. We are also faced with a flow f0 and a saturated a–z cut (P0, P0) such that | f0| < k(P0, P0). Corollary 2a implies that there must be some flow in an edge e ∈(P0, P0). Looking at Figure 4.7f, we see that the flow path L′ = a–b–e–z crosses the a–z cut (P0, P0) backward (from P to P0) on the edge (b⃗, e); or, equivalently, L′ crosses the cut forward twice, on edge (a⃗, b) and again on (e⃗, z). Thus the 5 units of flow along L′ use up 10 units of capacity in the cut, whence k(P0, P0) is 5 units greater than | f0|.
The reason that the sequence of augmenting flow paths in this example did not lead to a maximum flow can be explained intuitively as follows. By sending 5 units of flow along L′ (see Figure 4.7e), we have routed all the flow passing through b on to e and none of it to d. Then only 1 unit of flow passing through c can be routed on to e and then along edge (e⃗, z). But much of the flow through c must go to e, since the capacity of (c⃗, d) is only 1. In sum, the initial 5-unit flow along a–b–e–z was a “mistake” because some of the capacity in edge (e⃗, z) should have been “reserved” for flow from c.
How can we avoid or correct such mistakes? If we understood where the mistakes were made, we could change some of the flow paths and try a new sequence of augmenting flow paths. However, we are likely to make other mistakes in subsequent constructions. Indeed, there may be certain networks in which it is impossible not to make such a mistake, no matter what sequence of flow paths is used. In terms of cuts, we may always end with a saturated a–z cut that one of our flow paths crosses twice.
Fortunately,thereisaproceduretocorrect“mistakes”andtherebyfurtherincrease the flow. The method will not look for edges that must be “reserved” for later flow paths, as suggested above (that is too hard a problem). Rather, it looks for flow that is going the wrong way (backward) across an a–z cut. Then it finds a way to decrease the backward flow without changing the forward flow across the cut, the result being more total flow across the cut (and through the network). The following example presents the idea behind this procedure.
Example 3: (continued) The edge (b⃗, e) contains 5 units of flow going backward across the saturated cut (P0, P0), where P0 = {a, c, e}. By how much can the flow in (b⃗, e) be reduced?
142 Chapter 4 Network Algorithms Condition (b) of a flow—flow in equals flow out, at each vertex—requires that a reduction of the flow into e from b must be compensated by an increase to e from elsewhere in P0 (if the compensating flow comes from P0 we would have a new backward flow). Such an increase must in the end come from a. Thus, we need an a–e flow path in P0. The only such path is K1 = a–c–e. Similarly, a reduction of the flow out of b to e must be compensated by an increase in flow out of b to somewhere else in P0. So we need a b–z flow path in P0. The only such path is K2 = b–d–z.
The minimum slack along K1 is 2 and the minimum slack along K2 is 3. Then we can decrease the flow in (b⃗, e) by 2 while increasing the flow in K1 and K2 by 2.
Figure 4.7g shows the resulting maximum flow. Note that this new maximum flow is different from the maximum flow in Figure 4.7d, although both saturate the a–z cut ({a, c}, {b, d, e, z}).
A chain in a directed graph is a sequence of edges that forms a path when the direction of the edges is ignored. A unit-flow chain from a to z along the a–z chain K is a “flow” fK with a value of 1 in each edge of K forwardly directed, a value of −1 in each edge of K backwardly directed, and a value of 0 elsewhere. Note that fK is not really a flow because it can assume a negative value on some edges. However, if f is a flow that already has positive values in each backwardly directed edge of K and has slack in each forwardly directed edge of K, then f + fK is a valid flow. The flow correction made in the continuation of Example 3 consisted of a (2-unit) a–z flow chain along chain K = a–c–e–b–d–z.
When no backwardly directed edges occur in a flow chain, then it is simply a flow path. We shall see that any sequence of augmenting a–z flow chains can be extended to a maximum flow. Flow chains are the appropriate generalization of flow paths that both build additional flow and simultaneously correct possible “mistakes.” We now present a flow chain algorithm to increase the value of a flow. The algorithm is designed so that if it fails to obtain an augmenting a–z flow chain, it will produce a saturated a–z cut whose capacity equals the value | f | of the current flow f.
Thus by Corollary 2a, f would be maximum.
The algorithm recursively tries to build augmenting flow chains from a to all vertices in a manner reminiscent of the shortest path algorithm in Section 4.1. The algorithm assigns two labels to a vertex q: (p±, (q)), where p is the previous vertex on a flow chain from a to q, and (q) is the amount of additional flow that can be sent from a to q. That is, (q) is the minimum slack among the edges of the chain from a to q. On a backwardly directed edge e, the slack is the amount of flow that can be removed, namely f(e). A + superscript on p means flow is being added to edge (p⃗, q); a −superscript means flow is being subtracted from edge (q⃗, p).
Augmenting Flow Algorithm 1.
Give vertex a the labels (−, ∞).
2.
Call the vertex being scanned p with second label (p). Initially, p = a.
(a) Check each incoming edge e = (q⃗, p). If f (e) > 0 and q is unlabeled, then label q with [p−, (q)], where (q) = min[(p), f (e)].
www.itpub.net 4.3 Network Flows 143 (b) Check each outgoing edge e = (p⃗, q). If s(e) = k(e) −f (e) > 0 and q is unlabeled, then label q with [p+, (q)], where (q) = min[(p), s(e)].
3.
If z has been labeled, go to Step 4. Otherwise choose another labeled vertex to be scanned (which was not previously scanned) and go to Step 2. If there are no more labeled vertices to scan, let P be the set of labeled vertices, and now (P, P) is a saturated a–z cut. Moreover, | f | = k(P, P), and thus f is maximum.
4.
Find an a–z chain K of slack edges by backtracking from z as in the shortest path algorithm. Then an a–z flow chain fK along K of (z) units is the desired augmenting flow. Increase the flow in the edges of K by (z) units (decrease flow if edge is backward directed in K).
Like the shortest path algorithm, the flow algorithm extends partial-flow chains from currently labeled vertices to adjacent unlabeled vertices and the edges used to label vertices form a spanning tree. Before we prove that repeated application of our algorithm always leads to a maximum flow, let us give some examples.
Example 3: (continued) Let us apply our augmenting flow algorithm to the flow in Figure 4.7e.
Vertex a is labeled (−, ∞) by Step 1 of the algorithm. Next we apply Step 2b at a (Step 2a does not apply at a since the definition of a flow does not allow flow into the source a). The edge (a⃗, b) is saturated, but (a⃗, c) has slack 5 −2 = 3. So we label c (a+, 3). At c, Step 2a finds no incoming flow from an unlabeled vertex, but Step 2b finds slack in edge (c⃗, e) going to unlabeled vertex e. We label e (c+, 2) [2 is the minimum of (c), the extra flow we can get to c, and the slack in edge (c⃗, e)].
At e, Step 2a finds a positive flow entering on edge (b, −e) from unlabeled vertex b.
We label b (e, −2). From b, we label d (b+, 2), and from d we label z (d+, 2). Sink z is now labeled and so Step 4 tells us we can get (z) = 2 more units of flow from a to z.
Backtracking with the labels, the flow chain K (in backwards order) is z–d–b–e–c–a.
The new flow f0 + 2 fK is shown in Figure 4.7f.
Example 4: Using Augmenting Flow Algorithm Consider the network shown in Figure 4.8a. We seek a maximum flow from a to z.
If a maximum flow were being found by a computer, it would have to start with a zero flow. When solving a flow problem by hand, we can speed the process by starting with a (nonzero) flow obtained by inspection (in small networks we can often obtain a maximum flow by inspection). Let the initial flow be f = 4 fK1 + 4 fK2 + 5 fK3, where K1 = a–b–e–z, K2 = a–c–d–f–z, and K3 = a–d–f–z. See Figure 4.8a. The reader may see that we sent too much flow from c to d. Some of it should have gone directly from c to f.
We now apply the labeling algorithm to the flow f. Label vertex a (−, ∞).
Scanning edges at a, there cannot be incoming edges to the source with flow [by part (b) of the definition of a flow], but there are three outgoing edges: edge (a⃗, b) has slack s(a⃗, b) = 2 > 0 and b is unlabeled, so we label b (a+, 2), where 2 is the 144 Chapter 4 Network Algorithms a (—, •) b (a+, 2) 6, 4 7, 4 4, 4 12, 9 3, 0 5, 4 4, 4 5, 0 4, 0 4, 0 7, 5 9, 9 e (b+, 2) z (f +, 2) f (c+, 2) d (a+, 2) a (—, •) b (a+, 2) 6, 4 7, 4 4, 4 12, 11 3, 2 5, 2 4, 4 5, 0 4, 0 4, 0 7, 7 9, 9 e (b+, 2) z (f +, 1) f (c+, 1) d (c+, 2) a (—, •) b (a+, 1) 6, 5 7, 5 4, 4 12, 12 3, 3 5, 2 4, 4 5, 0 4, 0 4, 1 7, 7 9, 9 e (b+, 1) z f d (c+, 1) c (d–, 2) c (e+, 2) c (e+, 1) (a) (b) (c) Figure 4.8 minimum of (a) (= ∞) and s(a⃗, b); edge (a⃗, c) has no slack; edge (a⃗, d) has slack s(a⃗, d) = 2 and d is unlabeled, so we label it (a+, 2).
Next we scan b (we will scan vertices in the order that they were labeled). There are two incoming edges at b: Edge (a⃗, b) has f (a⃗, b) = 4 > 0 but a is already labeled; edge (c⃗, b) has no flow. There is one outgoing edge at b: edge (b⃗, e) has slack s(b⃗, e) = 3 and e is unlabeled, so we label e (b+, 2), where 2 is min[(b), s(b⃗, e)].
Next we scan at d. There are two incoming edges at d: edge (a⃗, d) comes from a labeled vertex; edge (c⃗, d) has f(c⃗, d) = 4 and c is unlabeled, so using Step 2a we label c (d , −2), where 2 = min[(d), f(c⃗, d)]. There is one outgoing edge at d: edge (d⃗, f ) is saturated.
No labeling can be done from e. At c we label f with (c+, 2). At f we label z with ( f +, 2).
Since z is now labeled, the labeling procedure terminates. We can send 2 [= (z)] units in the augmenting a–z flow chain fK4, where K4 = a–d–c– f –z, found by the backtracking procedure. Recall that since edge (c⃗, d) is backwardly directed in K4, the flow chain fK4 subtracts 2 units from the current flow in edge (c⃗, d). The new flow f ′ = f + 2 fK4 is shown in Figure 4.8b.
We eliminate all the current labels and restart the labeling algorithm from scratch.
Label vertex a(−, ∞). From a we label b with (a+, 2). At b we label e with (b+, 2).
At e we label c with (e+, 2). At c we label d with (c+, 2) and f with (c+, 1). At d we can make no new labels. At f we label z with (f +, 1). The augmenting a–z flow chain is fK5, with K5 = a–b–e–c– f –z, using forward edge (e⃗, c). Our new flow is f ′′ = f + 2 fK4 + fK5, shown in Figure 4.8c.
www.itpub.net 4.3 Network Flows 145 The reader may have observed that the flow f ′′ is maximum. The incoming edges at z are now saturated. Let us again apply the augmenting flow algorithm to flow f ′′.
At a we label b; at b we label e; at e we label c; and at c we label d. No more vertices can be labeled. Let P be the set of labeled vertices. Then (P, P) is the saturated a–z cut specified by the algorithm with | f ′′| = 16 = k(P, P).
In applying our algorithm, we must always check incoming edges in Step 2a for the possibility of minus labeling of vertices, even though this labeling is very infrequent. Recall that the minus labeling corresponds to correcting a mistaken flow assignment. A permissible shortcut would be to use only positive labeling (as in the faulty procedure discussed earlier) until no new flow paths can be found, and then apply the full algorithm to hunt for “mistakes.” The use of this shortcut serves to increase the importance of having a rigorous proof that, when repeatedly applied to any given flow, our algorithm will yield a maximum flow.
Theorem 3 For any given a–z flow f, a finite number of applications of the augmenting flow algo-rithm yields a maximum flow. Moreover, if P is the set of vertices labeled during the final (unsuccessful) application of the algorithm, then (P, P) is a minimum a–z cut.
Proof There are two main parts to the proof. First, if f is the current flow and fK is the augmenting a–z unit flow chain (along chain K) found by the algorithm with m = (z), then we must show that the new flow f + m fK is indeed a legal flow in the network. Both f and fK satisfy flow conditions (b) and (c) and are integer-valued.
Hence f + m fK also satisfies (b) and (c) and is integer-valued. The labeling algorithm is designed so that m = (z) is the minimum slack (of the appropriate kind) along chain K and hence f + m fK satisfies flow condition (a): 0 ≤f (e) + m fK (e) ≤k(e).
Since m is a positive integer, each new flow is larger by an integral amount. The capacities and the number of edges are finite, and so the algorithm must eventually halt—fail to label z. Let P be the set of labeled vertices when the algorithm halts.
Clearly (P, P) is an a–z cut, since a is labeled and z is not. Observe that there cannot be an unsaturated edge from a labeled vertex p to an unlabeled vertex q, or else at p we could have labeled q in Step 2b. Similarly, there cannot be a flow in an edge from an unlabeled vertex q to a labeled vertex p, or else again at p we could have labeled q in Step 2a. Thus, both conditions of Corollary 2a are satisfied. Hence the value of the final flow equals k(P, P) and is maximum. Also (P, P) is a minimum a–z cut. ◆ Corollary 3a Max Flow–Min Cut Theorem In any directed flow network, the value of a maximum a–z flow is equal to the capacity of a minimum a–z cut.
Let us now indicate how flows in a directed network can be used to model a large variety of extensions in directed and undirected networks. In Example 1, we showed 146 Chapter 4 Network Algorithms a b e z g c d f 30 10 10 10 10 20 20 20 30 30 10 5 5 10 20 5 Figure 4.9 how to model a problem with several supplies and demands by a network with one source and one sink.
Example 5: Undirected Networks Suppose the undirected network in Figure 4.9 represents a network of telephone lines (the capacity of an edge is the number of calls the line can handle). We wish to know the maximum number of calls that the network can simultaneously carry between locations a and z. That is, we seek the value of a maximum flow in this network.
To make the network directed, we can replace each undirected edge (x, y) by the two edges (x⃗, y) and (y⃗, x), each with the same capacity as (x, y). An equivalent approach is to allow a directed flow in undirected edges. If e = (x, y), f (e) would be a number with an “arrow” indicating whether the flow goes from x to y or from y to x.
Step 2 of the flow algorithm is modified as follows: when checking edges at a labeled vertex p, edges with a flow directed inward are treated like incoming edges and edges with no flow or flow away from p are treated like outgoing edges.
Example 6: Edge-Disjoint Paths in a Graph We are going to send messengers from a to z in the graph shown in Figure 4.10.
Because certain edges (roads) may be blocked, we require each messenger to use different edges. How many messengers can be sent? That is, we want to know the number of edge-disjoint paths.
We convert this path problem into a network flow problem by assigning unit capacities to each edge. One could think of the flow as “flow messengers,” and the unit capacities mean that at most one messenger can use any edge. The number of edge-disjoint paths (number of messengers) is thus equal to the value of a maximum flow in this undirected network. (See Example 5 for flows in undirected networks.) Observe that we have implicitly shown that a maximum a–z flow problem for a unit-capacity network is equivalent to finding the maximum number of edge-disjoint a–z paths in the associated graph (where edge capacities are ignored). This equiva-lence can be extended using multigraphs to all networks by replacing each k-capacity a z Figure 4.10 www.itpub.net 4.3 Network Flows 147 a d c b z 400, 2 600, 1 400, 2 300, 2 300, 2 600, 1 600, 1 600, 1 a0 a1 a2 a3 a4 b4 b3 b2 b1 b0 c0 c1 c2 c3 c4 d4 d3 d2 d1 d0 z0 z1 z2 z3 z4 (b) (a) (edges directed downward and to right) Figure 4.11 edge (x, y) with k multiple unit-capacity edges and now proceeding with the above conversion.
Example 7: Dynamic Network Flows We want to know how many autos can be shipped from location a to location z in four days through the network in Figure 4.11a. We assume that each edge (x⃗, y) is the route of a train that leaves location x daily for a nonstop run to location y. The first number associated with an edge is the capacity of the trains (number of autos). The second number is the number of days the trip takes. Autos may be left temporarily at any location in the network.
We turn this dynamic problem into a static a–z flow problem by adding the dimension of time to our network: each vertex x is replaced by five vertices x0, x1, x2, x3, x4, where the subscript refers to the ith day in the four-day shipping time (the starting day is day 0). For each original edge (x⃗, y), which takes k days to traverse, we make edges (x0⃗, yk), (x1⃗, yk+1), . . . , (x4−k⃗, y4), each with the same capacity as (x⃗, y). For each vertex x, we make four edges of the form (xi⃗, xi+1) with very large (in effect, infinite) capacity; these edges correspond to the provision that permits autos to be left temporarily at any vertex from one day to the next. See Figure 4.11b.
A maximum flow in the new network gives the maximum dynamic flow in the original network. Note that for vertices other than a and z, the range of the subscripts of usable vertices will be at most 1 to 3. Nondaily trains could easily be incorporated into the model.
148 Chapter 4 Network Algorithms It is hoped that the preceding examples have impressed the reader with the ver-satility of our basic static a–z flow model. More examples are to be found in the exercises.
4.3 EXERCISES S u m m a r y o f E x e r c i s e s Exercises 1–17 mimic or extend the flow computations and models in Examples 3–7. Exercises 18–38 develop the theory of network flows (later exercises are very challenging). Exercises 39–42 present pro-gramming projects.
In the following problems, unless directed otherwise, the reader should route most of the flow by inspection and then, when a near-optimal flow is obtained, use the augmenting flow algorithm to get further flow and afterwards a minimum cut.
1. Apply the augmenting flow algorithm to the flow in Figure 4.5.
2. Find a maximum a–z flow and minimum capacity a–z cut in the following networks: a b d z e c 12 10 5 4 18 12 10 8 a b e f c d z 10 5 10 5 10 10 15 7 7 15 7 7 a 8 2 4 4 8 6 6 4 4 2 2 2 4 b e z g d c f (b) (a) (c) 3. Find a maximum flow from a to z in the network of Figure 4.9 (using the associated directed network). Also give a minimum capacity a–z cut.
4. Treat the first number assigned to each edge in Exercise 3 of Section 4.1 as a capacity and let the edges be directed by the alphabetical order of their endpoints with L preceding a and W following n [e.g., edge (f, i) goes from f to i].
(a) Find a maximum L–W flow and a minimum L–W cut in this network.
(b) By inspection, find a different maximum L–W flow.
(c) Make a misrouted flow that yields a saturated L–W cut whose capacity is greater than the flow (similar to the situation in Figure 4.7e). Now apply the algorithm.
www.itpub.net 4.3 Network Flows 149 (d) In addition, let the second number of each edge be a lower bound on the flow in that edge. Try to find an L–W flow satisfying both constraints.
5. Delete vertices p through y in Figure 4.2 in Section 4.2 and treat the remaining edgenumbersascapacities(edgesarestillundirected).Usetheundirectedversion of the flow algorithm suggested in Example 5.
(a) Find a maximum flow from b to j and a minimum b–j cut.
(b) Build a b–j flow that saturates edge (k, l) (in either direction) and then apply the algorithm.
(c) Treat the edge numbers as lower bounds and modify the algorithm to find a maximum b–j flow, using the whole network (a through y). Remember that there is no flow into b or out of j.
6. Is there a flow meeting the demands in Figure 4.4?
7. Suppose vertices b, c, d in Figure 4.4 have unlimited supplies. How much flow can be sent to the set {h, i, j}? Explain your model.
8. Vertices b, c, d have supplies 30, 20, 10, respectively, and vertices j, k have demands of 30 and 25 in this network.
d i k j g e b c h f 10 10 10 10 10 10 40 30 15 15 15 10 5 20 20 20 15 20 10 10 25 30 20 (a) Find a flow satisfying the demands, if possible.
(b) Reverse the direction of edge (h, g) and repeat part (a).
9. Solve the messenger problem in Example 6.
10. Suppose that up to three messengers can use each edge in Example 6. Now how many messengers can be sent? Is the answer with such a modification always just three times the answer to the original problem?
11. (a) Ignoring the numbers of the edges in Exercise 3 of Section 4.1, what is the size of the largest set of edge-disjoint paths from L to W?
(b) What is the size of the largest set of paths from L to W such that no edge is used by more than five paths?
150 Chapter 4 Network Algorithms 12. Suppose that no more than five units of flow can go through each intermediate vertex b, c, d, in Figure 4.8a. Now find a maximum flow in this revised network and associated minimum a–z cut.
13. Suppose that no more than 20 units of flow can go through each intermediate vertex b, c, d, e, f, g in Figure 4.4. Now find the maximum flow in this revised network.
14. What is the size of a largest set of vertex-disjoint paths from a to z in Figure 4.10?
15. Solve the dynamic flow problem in Example 7.
16. In Example 7, suppose that the trains do not run every day. Let trains depart from a and c on Monday, Wednesday, and Friday, and from b and d on Tuesday, Thursday, and Saturday. In one week, Monday through Sunday, how many cars can be sent from a to z in that network?
17. Suppose that it takes one day to traverse each edge in Figure 4.5. How many units can be moved from a to z in five days in this network?
18. In the proof of Theorem 3, show that f 0 = m fK satisfies the second part of condition (b).
19. (a) Prove that if a directed network contains edges (x⃗, y) and (y⃗, x) for some x, y, then the augmenting flow algorithm would never make assignments that would have flow occurring simultaneously in both edges.
(b) Could edges (x⃗, y) and (y⃗, x) both get flow if the augmenting flow algorithm in Step 2 checked outgoing edges before incoming edges?
20. (a) Restate the augmenting flow algorithm so that the labeling starts at z and “works back” to a.
(b) Restate the augmenting flow algorithm for undirected networks (see Example 5). Sketch a proof of this algorithm.
21. Show that the set of edges used to label vertices in Steps 2a and 2b of the augmenting flow algorithm form a tree rooted at a.
22. (a) Give a weakened replacement for condition (b) in the definition of an a–z flow. The new condition should still ensure that the net flow is from a to z.
(b) Suppose condition (b) is eliminated. Can there exist maximum flows that violate condition (b)? Prove or give a counterexample.
23. Build a flow in the network in Figure 4.8a with the prescribed properties: (a) Its value | f | is 0, but not all edges have 0 flow.
(b) Its value is 2, but it cannot be decomposed into a sum of a–z flow paths.
24. (a) Show that for a flow f in a (directed or undirected) network, if f is circuit-free—that is, there is no set of edges with flow that form a (directed) circuit— then f can be decomposed into a sum of a–z flow paths. (Hint: Prove by induction on the value of f.) (b) Use the proof to get a decomposition algorithm for any such f.
www.itpub.net 4.3 Network Flows 151 (c) Conclude that any flow f can be decomposed into | f | a–z flow paths plus a set of circuits.
25. (a) Prove that starting from a circuit-free flow (see Exercise 24), perhaps a zero flow, the maximum flow generated by the augmenting flow algorithm is circuit-free and hence [by Exercise 24(b)] can be decomposed into a sum of a–z flow paths.
(b) Use part (a) and Exercise 24(b) to find the routings of a maximum set of phone calls in Example 5.
26. How many times is an edge checked in Step 2 to perform one complete it-eration of the augmenting flow algorithm (resulting in increased flow or a min-cut)?
27. Show that if a flow is decomposed into unit-flow paths and if each unit-flow path crosses a given saturated cut once, then the flow is maximum.
28. A cut-set in an undirected graph G is a set S of edges whose removal disconnects G such that no proper subset of S disconnects G. Prove that in an undirected flow network, every cut-set that separates a and z is an a–z cut and every minimum a–z cut is a cut-set.
29. Let G be a connected, undirected graph and a, b be any two vertices in G.
(a) Show that there are k edge-disjoint paths between a and b if and only if every a–b cut has at least k edges.
(b) Show that there are k vertex-disjoint paths between a and b if and only if every set of vertices disconnecting a from b has at least k vertices.
30. As mentioned in Example 6, we can model a flow network N (directed or undi-rected) by another multigraph flow network N ′ in which each edge has unit capacity; N ′ has the same vertices as N and for each edge e in N there are k(e) edges in N ′ paralleling e. Since a flow in N ′ takes 0 or 1 values on the edges, we can drop the capacities in N ′ to get a multigraph G′. A flow in N ′ is just a subset of edges (with flow) in G′.
(a) Characterize the subsets of edges in G′ that correspond to an a–z flow in N ′.
(b) Restate the augmenting flow algorithm in terms of G′.
(c) Using the multigraph model, prove that any a–z flow f in N ′ contains | f | a–z flow paths.
(d) Using the multigraph model and assuming the result in Exercise 29(a), prove Corollary 3a (max flow–min cut theorem).
31. Suppose the numbers on the edges in a directed network represent lower bounds for the flow. State a decreasing flow algorithm. Sketch a proof of this algorithm and deduce the counterpart of Corollary 3a.
32. (a) Explain how the algorithm in Exercise 31 can be applied to a bipartite graph to find a minimum set of edges incident with every vertex in the graph.
(b) Find such an edge set for the bipartite graph in Figure 4.12 of Section 4.4.
152 Chapter 4 Network Algorithms 33. Suppose that we have upper and lower bounds k1(e) and k2(e), respectively, on the flow in each edge e in a directed network, and that we are given a feasible flow (satisfying these constraints).
(a) Modify the augmenting flow algorithm so that it can be used to construct a maximum flow from a given feasible flow.
(b) Prove that the maximum flow has value equal to the minimum of k1 (P, P) − k2(P, P), among all a–z cuts (P, P), where k1(S, S) and k2(S, S) are the sums of the upper and lower bounds of edges in a cut (S, S).
34. Consider a directed network with supplies and demands as in Example 1. Let z(P) be the total demand of vertices in set P and a(P) be the total supply of vertices in P.
(a) Prove that the demands can be met if and only if for all sets P, z(P) − a(P) ≤k(P, P). (Hint: Generalize the reasoning in Example 1.) (b) Prove that the supplies can all be used if and only if for all P, a(P) − z(P) ≤k(P, P).
35. Suppose the edges in Figure 4.11a in Example 7 were undirected. How would we construct a static flow model to simulate this dynamic flow problem so that a maximum static flow in the new network would correspond to a maximum dynamic flow? Are there any difficulties?
36. Suppose an undirected flow network is a planar graph and a (at the left side) and z (at the right side) are both on the unbounded region surrounding the network. Draw edges extending infinitely to the left from a and to the right from z; give them infinite capacity. This divides the unbounded region into two unbounded regions, an upper and a lower unbounded region. Now form the dual network (see Section 1.4) of this planar network with each dual edge assigned the capacity of the original edge it crosses.
(a) Show that a path from the upper unbounded region’s vertex to the lower unbounded region’s vertex in the dual network corresponds to an a–z cut in the original network. Thus, a shortest such path in the dual network is a minimum a–z cut in the original network.
(b) Draw the dual network for the network in Figure 4.9 and find a shortest path (using the algorithm in Section 4.1) corresponding to a minimum cut in the original network.
37. Suppose an undirected flow network N is a planar graph and a (at the left side) and z (at the right side) are both on the unbounded region surrounding the network. Consider the following flow-path building heuristic. Starting from a, build an a–z flow path by choosing at each vertex x the first unsaturated edge in clockwise order starting from the edge used to enter x.
(a) Apply this heuristic to the network in Figure 4.9 to find a maximum flow.
(b) Show that repeated use of this heuristic yields a maximum flow in N.
www.itpub.net 4.4 Algorithmic Matching 153 38. Prove that repeated use of the augmenting flow algorithm yields a maximum flow in a finite number of applications for networks with irrational capacities provided that vertices are ordered (indexed) and that the next vertex scanned in Step 2 of the algorithm is the labeled vertex with lowest index.
39. Write a program to find maximum flows in directed networks (the networks are input data).
40. Write a program to find the maximum number of paths in an undirected graph between two given vertices such that the following are true: (a) The paths are edge disjoint.
(b) The paths are vertex disjoint.
41. Write a program that when given a network with an a–z flow of value k, will extract from the flow k unit-flow paths from a to z.
42. Write a program to find, for a given pair of vertices a and b in a given connected graph, a minimum set of vertices whose removal disconnects a from b (see Exercise 29).
4.4 ALGORITHMIC MATCHING In this section, we apply network flows to the theory of matchings. Recall that a bipartite graph G = (X, Y, E) is an undirected graph with two specified vertex sets X and Y and with all edges of the form (x, y), x ∈X, y ∈Y. See Figure 4.12. Bipartite graphs are a natural model for matching problems. We let X and Y be the two sets to be matched and edges (x, y) represent pairs of elements that may be matched together.
A matching in a bipartite graph is a set of independent edges (with no common endpoints). The thicker edges in Figure 4.12 constitute a matching. An X-matching is a matching involving all vertices in X. A maximum matching is a matching of the largest possible size. As with network flows, we cannot always obtain a maximum matching in a bipartite graph by simply adding more edges to a non-maximum matching. The matching indicated in Figure 4.12 cannot be so increased, even though it is not maximum.
A typical matching problem involves pairing off compatible boys and girls at a dance or the one-to-one assignment of workers to jobs for which they are trained.
b c d e f g h i j k m Figure 4.12 154 Chapter 4 Network Algorithms A closely related problem is to find a set of distinct representatives for a collection of subsets. We need to pick one element from each subset without using any element twice. In the bipartite graph model, we make one X-vertex for each subset, one Y-vertex for each element, and an edge (x, y) whenever element y is in subset x. Now an X-matching picks a distinct representative element for each subset. Conversely, any matching problem can be modeled as a set-of-distinct-representatives problem.
We employ a modification of the trick in Example 1 of Section 4.3 to turn a matching problem into a network flow problem. Associate a supply of 1 at each X-vertex and a demand of 1 at each Y-vertex. The capacities of the edges from X to Y can be any large positive integers, but it is convenient to assume that these capacities are∞.WealsoassumethatedgesaredirectedfromXtoY.Nowweapplythetechnique in Example 1 of Section 4.3 with source a connected by a unit-capacity edge to each X-vertex and sink z connected by a unit capacity edge from each Y-vertex.
We call such a network a matching network. See Figure 4.13. The X-Y edges used in an a–z flow constitute a matching. A maximum flow is a maximum matching.
A flow saturating all edges from source a corresponds to an X-matching.
Example 1: Maximum Matching Suppose the bipartite graph in Figure 4.12 represents possible pairings of boys b, c, d, e, f with girls g, h, i, j, k, m. A tentative matching indicated by thicker edges in Figure 4.12 was made. Although this matching cannot be extended, we still wonder whether a complete X-matching is possible.
As has happened before in flow problems, we have made a “mistake” in this matching and now need to make some reassignments. We convert this matching into the corresponding flow in the associated matching network: darkened edges in Figure 4.13 have a flow of 1, other edges have a flow of 0. Now we apply the augmenting flow algorithm. See Figure 4.13.
From a, the only outgoing edge that is not saturated is (a⃗, c), since c is the only X-vertex not involved in the initial matching. We label c (a+, 1). From c we can label the two Y-vertices, g and i, adjacent to c with the label (c+, 1). At g the one outgoing edge—to z—is saturated. Then there must be an incoming edge to g with positive flow, namely (b⃗, g), coming from the unlabeled X-vertex b. We label b (g−, 1). By similar reasoning, from i we label f (i−, 1). At b and f we can label Y-vertices not currently matched to b and f . We label j (b+, 1) and m ( f +, 1).
a (–, ∞) (g–, 1) b g (c+, 1) (a+, 1) c (m–, 1) d (j–, 1) e (i–, 1) f h (d+, 1) i (c+, 1) j (b+, 1) k (e+, 1) m (f +, 1) z (k+, 1) Figure 4.13 www.itpub.net 4.4 Algorithmic Matching 155 Summarizing our labeling thus far, we are trying to find a match for c (the only X-vertex currently unmatched). We have possible matches of c with g and i, but to make either match, we must sever the current match of g with b or of i with f and rematch b or f with other Y-vertices—in this network, b can be rematched with j or f rematched with m. Continuing this reasoning, to rematch b with j requires us to delete j’s current match with e and find a new match for e. Or to rematch f with m requires us to delete m’s current match with d and find a new match for d. Now either d can be matched with h (currently unmatched) or e can be matched with k (also currently unmatched). The fact that h and k are unmatched is reflected in the fact that from either h or k one can label z. See the labels in Figure 4.13.
If we label z (k+, 1), the augmenting flow chain prescribed by the algorithm is a–c–g–b–j–e–k–z. Our augmenting flow chain specifies that we add edge (c⃗, g) to our matching, remove edge (b⃗, g), add edge (b⃗, j), remove edge (e⃗, j), and add edge (e⃗, k). The new flow corresponds to the matching b–j, c–g, d–m, e–k, f–i. Another application of the algorithm would label only a. The set of edges leaving a now forms a (saturated) minimum a–z cut—a sign that we have an X-matching.
Observe that the action of the augmenting flow algorithm in matching problems can be described as follows: starting from an unmatched vertex x1 in X, we go on a nonmatching edge (an edge not currently used in the matching) to a matched Y-vertex y1, then we move back from y1 along a matching edge to a matched X-vertex x2, then we go on a nonmatching edge from x2 to a matched vertex y2, and so on until a nonmatching edge (xk, yk) goes from a matched X-vertex xk to an unmatched Y-vertex yk. In short, starting from an unmatched X-vertex, we create an odd-length alternating path L of nonmatching and matching edges in search of an unmatched Y-vertex. Given such a path, we get a new, larger set of matching edges by interchanging the roles of matching and nonmatching edges on L.
We have seen that in bipartite graphs, a matching is analogous to a flow in a network. What then is the bipartite graph counterpart to an a–z cut? Actually it is convenient to restrict our attention to finite-capacity a–z cuts. The corresponding concept is an edge cover, a set S of vertices such that every edge has a vertex of S as an endpoint. Recall that edge covers were encountered in a street surveillance problem in Example 4 of Section 1.1.
Lemma LetG=(X,Y,E)beabipartitegraphandletNbethematchingnetworkassociatedwith G. For any subsets (possibly empty) A ⊆X and B ⊆Y, S = A ∪B is an edge cover if and only if (P, P)) is a finite capacity a–z cut in N, where P = a ∪(X −A) ∪B.
In terms of P, S = (P ∩X) ∪(P ∩Y). Further, |S| = k(P, P).
Proof A finite capacity a–z cut cannot contain an edge between X and Y (whose capacity is ∞). Then a finite-capacity cut (P, P) must consist of edges of the form (a⃗, x), x ∈A and (y⃗, z), y ∈B for some sets A ⊆X and B ⊆Y. These edges block all flow (and thus are an a–z cut) if and only if any X–Y edge starts at some x ∈A or 156 Chapter 4 Network Algorithms a P P P P X Y • z P P 1 1 ∩ ∩ P P X Y ∩ ∩ Figure 4.14 ends at some y ∈B—that is, if and only if S = A ∪B is an edge cover. In terms of A and B, P = a ∪(X – A) ∪B, and in terms of P, A = P ∩X and B = P ∩Y.
See Figure 4.14. Also, |S| = k(P, P), since the edges in (P, P) all have unit capacity. ◆ We now prove two famous theorems about matching in bipartite graphs.
Theorem 1 (K¨ onig-Egervary) In a bipartite graph G = (X, Y, E), the size of a maximum matching equals the size of a minimum edge cover.
Matchings correspond to flows in the associated matching network and, by the lemma, edge covers correspond to (finite capacity) a–z cuts. Theorem 1 is simply a bipartite-graph restatement of the max flow–min cut theorem (Corollary 3a in Section 4.2).
In the next theorem, the range R(A) of A denotes the set of vertices adjacent to a vertex in A.
Theorem 2 (Hall’s Marriage Theorem) A bipartite graph G = (X, Y, E) has an X-matching if and only if for each subset A ⊆X, |R(A)| ≥|A|.
Proof The condition is necessary, for if |R(A)| < |A|, a matching of the vertices in A is clearly impossible, and so an X-matching is impossible.
Next we show that if |R(A)| ≤|A|, for all A ⊆X, then G has a X-matching. By Theorem 1, an X-matching exists if and only if each edge cover S has size |S| ≥|X|.
If S contains only X-vertices—that is, S = X—then the result is immediate. We must show that S cannot be made smaller by dropping some X-vertices and in their place using a smaller number of Y-vertices. If A is the set of X-vertices not in S, then S ∩Y must contain the vertices in R(A) to cover the edges between A and R(A). Then we have |S| = |S ∩X| + |S ∩Y| = |X| −|A| + |S ∩Y| ≥|X| −|A| + |R(A)| www.itpub.net 4.4 Algorithmic Matching 157 Since |R(A)| ≥|A|, then |X| −|A| + |R(A)| ≥|X|, and so |S| ≥|X|, proving that there is a matching of size X. ◆ Theorems 1 and 2 are the starting point for a large family of matching theorems (for example, see Exercises 18 and 23).
Example 2 Country A would like to spy on all meetings in its territory between diplomats from Country X and from Country Y. We know which pairs of diplomats are likely to meet.
We can make a bipartite graph expressing this likely-to-meet relationship. Country A cannot afford to assign spies to every X diplomat or to every Y diplomat. Instead it wants to find a minimum set S of diplomats such that every possible meeting would involve a diplomat of S.
Country A hires a graduate of this course, who immediately sees that this min-imum spying problem is really a minimum edge cover problem in disguise. The problem can thus be solved by using the augmenting flow algorithm to find a min-imum a–z cut in the associate matching network and, from the a–z cut, obtain a minimum edge cover using the lemma.
Example 3 Suppose there are n people and n jobs, each person is qualified for k jobs, and for each job there are k qualified people. Is it possible to assign each person to a (different) job he or she can do?
We model this problem using a bipartite graph G = (X, Y, E) with X-vertices for people and Y-vertices for jobs. Note that each vertex will have degree k. The question is now: is there an X-matching? From Theorem 2, we can deduce a yes answer as follows: Let A be any subset of X. Since each vertex has degree k, there will be k|A| edges leaving A. Since at most k of these edges can go to any one vertex in Y, it follows that R(A) has at least |A| vertices. Then, by Theorem 2, there is an X-matching.
We close this section with an application of matching networks to the sports problem of determining late in a season whether a particular team still has a math-ematical chance of being conference champion. The network model and associated analysis presented here are due to Schwartz .
Example 4: Elimination from Contention Suppose we know how many games each of the teams in a sports conference has won to date and how many games remain to be played between each pair of teams. We want to know if there exists a scenario under which a specified team could finish the season with the most wins in the conference.
Let us consider a concrete example with four teams, the Bears, the Lions, the Tigers, and the Vampires, with the records of wins and games to play in Figure 4.15.
Suppose we wonder if it is possible for the Bears (currently with fewest wins) to end 158 Chapter 4 Network Algorithms Wins Games with with with with Team to date to play Bears Lions Tigers Vampires Bears (t1) 16 7 — 2 2 3 Lions (t2) 22 7 2 — 3 2 Tigers (t3) 20 8 2 3 — 3 Vampires (t4) 19 8 3 2 3 — Figure 4.15 the season with the most wins. The problem is that even assuming the Bears win the rest of their games to finish with 23 wins, the outcomes of the games between the other teams may always result in one of those teams finishing with more wins than the Bears. Thus, the challenge is to find an answer to the following question: Does there exist a way to assign the wins in the games among other teams so that each of those teams finishes with at most 23 wins?
We describe how to construct a matching network model to answer this question.
Figure 4.16 gives the network for the data in Figure 4.15. The X-vertices will represent the other teams besides the Bears. In Figure 4.16 the X-vertices will be L (for Lions), T (for Tigers), and V (for Vampires). The Y-vertices will represent all different pairs of other teams. In this case, LT, LV, and TV.
The size of the flow in the edge from source a to X-vertex ti will stand for the number of games won by the i-th team ti in the rest of the season. The capacity of the edge (a⃗, ti) will be 23 −wi, where wi denotes the number of wins to date by ti.
Observethatifteamti haswonwi gamestodate,andifitwinsatmost23−wi gamesin the rest of the season, then ti cannot finish with more than 23 wins, enabling the Bears to be the champion or co-champion (assuming the Bears win all remaining games).
The flow in the X–Y edge (t⃗, (ti, t j)) will represent the number of wins of ti over t j in their remaining games. We can let the capacity of all X–Y edges be ∞. The flow in the edge from the Y-vertex (ti, t j) to the sink z represents the number of games that will be played between teams ti and t j. The capacity of ((ti, t j)⃗, z) is the number of scheduled remaining games between ti and t j. Conservation of flow at vertex ti requires that flow in (the number of games won in the rest of the season) = flow out (the number of future wins over various other teams). Conservation of flow at the vertex (ti, t j) requires that flow in (the wins of ti over t j and of t j over ti) = flow out (the number of games these two teams will play).
a L LT TV z V T LV 1, 1 0 1 0 3 1 3, 3 4, 4 3, 3 3, 3 2, 2 3 (middle edges have capacity •) Figure 4.16 www.itpub.net 4.4 Algorithmic Matching 159 To have a flow that represents a possible real scenario, we require that the edges from Y to zallbesaturated—thenumberofgamesplayedbetweenti andt j istheactual numberofgamesscheduledbetweenthem.Ifaflowexistsinthisnetworkthatsaturates the Y–z edges, it produces a scenario in which the Bears end the season with the most wins. Figure 4.16 presents a flow saturating the Y–z edges showing how the Bears could end up with the most wins (actually, all four teams would tie for first place).
Optional We now extend the preceding sports network model to give a necessary and sufficient condition for a particular team t ∗to be able to be conference champion.
For a subset S of teams, let w(A) denote the total number of wins to date by all teams in S and let r(S) be the total number of games remaining to be played between all pairs of teams in S. Observe that for any subset S of teams, the total number of wins at the end of the season by teams in S will be at least w(S) + r(S) (because when two teams in S play each other, one of them must win). Let W∗be the final number of wins by t ∗if t∗wins all remaining games. One constraint that guarantees that t∗ cannot finish with the most wins is if the average number of wins by any subset S not containing t∗exceeds W∗—that is, for any subset S not containing t∗, {w(S) +r(S)}/|S| > W ∗ (∗) We shall show that constraint (∗) is sufficient as well as necessary to characterize when team t∗is eliminated. If the min-cut is the set of Y–z edges, then there is a flow saturating all Y–z edges and specifying a scenario in which t∗has the most wins (as described in the example above). Suppose there is an a–z cut (P, P) of smaller capacity. We now show that (∗) is satisfied by some subset S.
(P, P) cannot contain any of the infinite-capacity edges from X to Y, and so it consists solely of a–X edges and Y–z edges. Recall that edge (a⃗, ti) has capacity W∗– wi. Then if A = P ∩X, the a–X edges in (P, P) have capacity |A|W∗−w(A).
One can show that if ti and t j are in P, then (ti, t j) must be in P (details are left as an exercise). Then the Y–z edges in (P, P) have capacity (ti,t j)̸⊂Arij, where rij denotes the capacity of edge ((ti, t j)⃗, z) (= the number of games remaining to be played between ti and t j) and we sum over all (ti, t j) pairs for which at least one of ti or t j is not in A. So k(P, P) = [|A|W ∗−w(A)] + (ti,t j)̸⊂Arij.
Assuming k(P, P) is less than ri j, the capacity of all the edges from Y to z, then |A| W∗−w(A) + (ti,t j)̸⊂A rij < rij ⇒|A| W∗−w(A) < rij − (ti,t j)̸⊂A rij = (ti,t j)⊂A ri j = r(A) ⇒|A|W ∗< w(A) +r(A) ⇒W ∗< {w(A) +r(A)}/|A| This last inequality is (∗), as claimed, with the subset being A. The teams in A average more wins than t∗, and hence t∗cannot be conference champion.
160 Chapter 4 Network Algorithms 4.4 EXERCISES S u m m a r y o f E x e r c i s e s Exercises 1–14 involve matching net-works. Exercises 15–24 develop theory.
1. Bill is liked by Ann, Diana, and Lolita; Fred is liked by Bobbie, Carol, and Lolita; George is liked by Ann, Bobbie, and Lolita; John is liked by Carol and Lolita; and Larry is liked by Diana and Lolita. We want to pair each girl with a boy she likes. (Make girls the vertices on the left side.) (a) Set up the associated matching network and maximize its flow to solve this problem.
(b) Make a flow corresponding to a partial pairing that has Bill with Diana and Fred with Carol along with two other matches (chosen by the reader). Now apply the flow augmenting algorithm to increase the matching to a complete matching.
2. Suppose there are five committees: committee A’s members are a, c, e; committee B’s members are b, c; committee C’s members are a, b, d; committee D’s members are d, e, f; and committee E’s members are e, f. We wish to have each committee send a different representative to a convention.
(a) Set up the associated matching network and maximize its flow to solve this problem.
(b) Make a flow corresponding to the partial assignment: A sends e, B sends b, C sends a, and D sends f. Now apply the flow-augmenting algorithm to increase the matching.
3. Let us repeat Exercise 1, but this time Bill gets a total of five dates, Fred four dates, George three, John five, and Larry three, while Ann gets four, Bobbie three, Diana five, Carol four, and Lolita four. Compatible pairs may have any number of dates together. Model with a network to find a possible set of pairings.
4. Let us repeat Exercise 1, but this time we want to pair each boy twice (with two different girls) and each girl twice. Find the pairings using an appropriate network flow model.
5. Suppose that there are six universities and each will produce five mathematics Ph.D.s this year, and there are five colleges that will be hiring seven, seven, six, six, five math Ph.D.s, respectively. No college will hire more than one Ph.D. from any given university. Will all the Ph.D.s get a job? Explain.
6. In Example 4, is it possible for the Vampires to be champions (or co-champions) if they win all remaining games? Build the appropriate network model.
7. (a) In the following table of remaining games, is it possible for the Bears to be co-champions, or sole champions, if they win all remaining games? Build the appropriate network model and show the required flow, if possible. Answer the question with a feasible flow in the network you created or with an explanation of why one is not possible. If there are co-champions with the Bears, who are they?
www.itpub.net 4.4 Algorithmic Matching 161 (b) Is it possible for the Bear to be sole champions?
Wins Games with with with with Team to date to play Bears Lions Tigers Vampires Bears 21 7 — 1 2 4 Lions 26 6 1 — 3 2 Tigers 26 7 2 3 — 2 Vampires 22 8 4 2 2 — 8. In the following table of remaining games, is it possible for the Vikings to be co-champions, or sole champions, if they win all remaining games? Build the appropriate network model and flow. Answer the question with a feasible flow in the network you created or with an explanation of why one is not possible. If there are co-champions with the Bears, who are they?
Wins Games with with with with Team to date to play Vikings Huns Romans Mongols Vikings 22 6 — 2 2 2 Huns 27 6 2 — 2 2 Romans 26 6 2 2 — 2 Mongols 25 6 2 2 2 — 9. In the following table of remaining games, is it possible for the Bears to be co-champions, or sole champions, if they win all remaining games? Build the appropriate network model. Answer the question with a feasible flow in the network you created or with an explanation of why one is not possible.
If there are co-champions with the Bears, who are they?
Wins Games with with with with Team to date to play Bears Lions Tigers Vampires Bears 18 8 — 1 3 4 Lions 25 6 1 — 2 3 Tigers 24 7 3 2 — 2 Vampires 22 9 4 3 2 — 10. In the following table of remaining games, is it possible for the Bears to be co-champions, or sole champions, if they win all remaining games? Build the appropriate network model. Answer the question with a feasible flow in the network you created or with an explanation of why one is not possible.
If there are co-champions with the Bears, who are they?
162 Chapter 4 Network Algorithms Wins Games with with with with Team to date to play Bears Lions Tigers Vampires Bears 20 6 — 1 2 3 Lions 25 6 1 — 3 2 Tigers 25 6 2 3 — 1 Vampires 22 6 3 2 1 — 11. There are n boys and n girls in a computer dating service. The computer has made nm pairings so that each boy dates m different girls and each girl dates m different boys (m < n).
(a) Show that it is always possible to schedule the nm dates over m nights—that is—that the pairings may be partitioned into m sets of complete pairings.
(b) Show that in part (a), no matter how the first k complete pairings are selected (0 < k < m), the partition can always be completed.
12. We want to construct an n × m matrix whose entries will be nonnegative integers such that the sum of the entries in row i is ri, and the sum of the entries in column j is c j. Clearly the sum of the ris must equal the sum of the c js.
(a) What other constraints (if any) should be imposed on the ris and c js to ensure such a matrix exists?
(b) Construct such a 5 × 6 matrix with row sums 20, 40, 10, 13, 25 and column sums all equal to 18.
13. We have a group of people and each is a member of a subset of committees. In addition, each person graduated from (exactly) one of three different universities.
In this extension of the “distinct representatives” problem, we seek a unique person to represent each committee with the added constraint that a third of the representatives must have graduated from each of the three universities. Describe how build a network to model this problem. (Assume that m, the number of committees, is a multiple of 3.) 14. (Due to Bacharach) We are given an n by n matrix with numerical entries that all have one digit to the right of the decimal point (e.g., 13.3). We want to round the entries to whole numbers in a fashion so that the sum of the rounded entries in each column (and row) is a rounded value of the original column (row) sum; e.g., if the first column has entries 2.5, 6.4, 5.7 summing to 16.6, then the sum of the rounded values of these three entries would have to be 16 or 17.
(a) Describe how to build a matching-type network flow model of this problem.
Let X-vertices represent the columns and Y-vertices represent the rows. Note that this model will have lower as well as upper bounds for each edge.
(b) Find the required rounding, if possible, using the network model developed in part (a) for the following matrix: 4.5 7.5 2.5 6.8 4.3 5.7 3.6 1.6 4.3 www.itpub.net 4.4 Algorithmic Matching 163 15. Give necessary and sufficient conditions for the existence of a circuit or a set of vertex-disjoint circuits that pass through each vertex once in a directed graph G = (V, E ). [Hint: Make a bipartite graph G′ = (X, Y, E) with X and Y copies of V, and for each edge (v1⃗, v2) in G, G′ has an edge (x1, y2); restate the problem.] 16. Prove that a bipartite graph G = (X, Y, E) has a matching of size t if and only if for all A in X, |R(A)| ≥|A| + t −|X| = t −|X −A|. (Hint: Add |X| −t new vertices to Y and join each new Y-vertex to each X-vertex.) 17. Show that every bipartite-graph matching problem can be modeled as a set-of-distinct-representatives problem.
18. In the analysis following Example 4, show that if ti and t j are in P, then vertex (ti, t j) is in P.
19. Prove Theorem 2 using Exercise 34 in Section 4.3, which applies directly to the bipartite network, not the augmented a–z matching network.
20. Prove Theorem 2 for complete Y-matchings (without simply interchanging the roles of X and Y). By symmetry, the same condition is required, but the set A in the reproof is chosen differently from the A in the text’s proof.
21. Let δ(G) = maxA⊆X(|A| −|R(A)|. δ(G) is called the deficiency of the bipartite graph G = (X, Y, E) and gives the worst violation of the condition in Theorem 2.
Note that δ(G) ≥0 because A = Ø is considered a subset of X.
(a) Use Exercise 16 to prove that a maximum matching of G has size |X| −δ(G).
(b) Given a maximum matching of size t = |X| −δ(G) [assume δ(G) > 0], describe how the associated minimum edge covering of Theorem 1 can be used to find an A such that |A| −|R(A)| = δ(G).
22. (a) Show that the size of the largest independent set of vertices (mutually non-adjacent vertices) in G = (X, Y, E) is equal to |Y| + δ(G) (see Exercise 21).
Describe how to find such an independent set.
(b) Use part (a) to find such an independent set in Figure 1.3.
23. (Due to J. Hopcroft) Suppose each vertex of a bipartite graph G has degree 2r, for some r. Partition the edges of G into circuits, and delete every other edge in each circuit. Repeat this process on the new graph (where each vertex now has degree 2r−1), and continue repeating until a graph is obtained with each vertex of degree 1. The edges in this final graph constitute a matching of the vertices of G. Show that such a partition of edges into circuits exists in each successive graph and can be found in a number of steps proportional to the number of edges in the current graph.
24. Suppose we are given a partial matching to an arbitrary graph (such a matching is a set of edges with no common endpoints).
(a) Prove that a generalization of the interchange method along a path alternating between nonmatching and matching edges (described following Example 1) can be used to increase the size of the partial matching until it is a maximum matching.
164 Chapter 4 Network Algorithms (b) Randomly pick a partial matching for the graph in Figure 4.9 and use this method to get a maximum matching.
25. Write a computer program to find a maximum matching and a minimum edge cover in a bipartite graph (the graph is input data).
4.5 THE TRANSPORTATION PROBLEM In this section we apply our knowledge of spanning trees, network flows, and match-ings developed in the previous sections to study network flows in which there is a cost charged to flow along an edge. We will consider networks whose underlying graphs are bipartite, as in the matching networks in Section 4.4. Similar to the matching problems, the edges here have unlimited capacity and the vertices have supplies and demands—now integers larger than 1. We refer to the supply vertices as warehouses and the demand vertices as stores. The bipartite graph G = (X, Y, E) is assumed to be complete, with an edge (i, j) from each warehouse Wi to each store Sj. What is new is that there is a cost ci j charged for shipping an item on edge (i, j). The goal is to find a routing of all the items from warehouses to stores that minimizes the transportation costs. See the sample problem in Figure 4.17. This optimization problem is appropriately called the Transportation Problem. It was one of the first optimization problems studied in operations research.
Warehouse i has a supply of size si, and store j has a demand of size d j. We assume that the total supplies, summed over all warehouses, are adequate to meet the demand at all the stores. A solution to a transportation problem specifies the amount xi j to ship on each edge (i, j) so that the total shipments leaving each warehouse do not exceed the warehouse’s supply and the total shipments arriving at each store equal the demand of that store. The goal is to find, among all solutions, a minimum-cost STORES Demand W A R E H O U S E S W A R E H O U S E S S T O R E S Supply 1 2 2 6 7 30 30 6 20 60 50 6 4 1 2 3 40 60 30 30 2 4 6 6 6 1 2 1 2 3 7 Figure 4.17 www.itpub.net 4.5 The Transportation Problem 165 solution. A mathematical statement of the transportation problem is minimize i, j ci jxi j such that j xi j ≤si for each i, i xi j = d j for each j, and xi j ≥0 While spanning trees do not appear anywhere in the statement of the problem, it turns out that they are central to solving this problem: The set of edges used for shipments in the optimal solution form a spanning tree. Our strategy to find an optimal solution will start with an initial (nonoptimal) solution that is a spanning tree. We will repeatedly find a cheaper spanning tree solution by dropping one edge from the current spanning tree and replacing it with another edge to form a cheaper spanning tree solution.
The data describing a Transportation Problem are usually presented in a table called a transportation tableau. See the tableau and associated bipartite graph for a sample transportation problem in Figure 4.17. (The graph representation is so clut-tered that it is only drawn for small problems.) Our analysis of how to solve the transportation problem will be developed in terms of this example. The supplies si of the warehouses appear on the right side of the tableau, and the demands d j of the stores appear at the bottom. The shipping cost ci j for edge (i, j) appears in the upper right half of entry (i, j) in the tableau. The number xi j in the lower left half of entry (i, j) tells how much is shipped from warehouse i to store j.
[Note: we refer interchangeably to (i, j) as an edge or an entry in the tableau.] The xi j’s in Figure 4.17 are a (nonoptimal) solution for this transportation prob-lem. In the graph in Figure 4.17, we have thickened the edges used in the solution.
The reader can check that the sum of the xi j’s in each row i is ≤si and that the sum of the xi j’s in each column j exactly equals d j. The cost of the sample solu-tion in Figure 4.17’s tableau, which uses the four edges (1,1), (2,1), (2,2), (3,2) is 30 × $4 + 30 × $6 + 30 × $6 + 20 × $6 = $600.
To simplify the problem, we will assume that the sum of the supplies exactly equals the sum of the demands. If the supplies exceed the demands, we can create a “dummy” store that takes in all excess supply. In Figure 4.17, the total of the sup-plies is 130 and the total of the demands is 110. In this case, we would create a dummy store with a demand of 130 −110 = 20, as shown in the tableau and graph in Figure 4.18. The solution in Figure 4.17 has been appropriately modified in Figure 4.18 so that the sum of the xi j’s in row i now exactly equals si. The cost of ship-ping a unit along any edge to this dummy store will be 0, since no real costs are involved in this gimmick to balance supply and demand. With 0 costs for shipping to the dummy store, the total transportation cost does not change from Figure 4.17 to Figure 4.18.
We will now show that it suffices to limit ourselves to solutions involving a set of edges that does not contain any circuits. Recall that a set of edges that contains no circuit forms a tree or collection of trees.
166 Chapter 4 Network Algorithms STORES Demand W A R E H O U S E S W A R E H O U S E S S T O R E S Supply Dummy 1 2 2 6 7 30 30 6 20 60 50 20 6 0 0 0 4 1 2 3 40 60 30 30 10 10 2 0 4 0 6 6 6 0 1 2 1 2 Dummy 3 7 Figure 4.18 Lemma 1 Let S be a solution to a transportation problem involving a set of edges E(S) that contains a circuit. Then there is another solution S∗that costs the same or less than S, where E(S∗) is a subset of E(S) containing no circuits.
Proof We will not give a formal proof of this Lemma, but rather illustrate it with an exam-ple and explain how this example generalizes to all solutions for all transportation problems.
Consider the solution S displayed in Figure 4.18. Observe that the edges used for shipments in S, (1,1), (2,1), (2,2), (3,2), (3,3), (1,3), form a circuit; see the darkened edges in the graph in Figure 4.18. We can modify S by in-creasing the shipments by 1 on the first, third, and fifth edges of this circuit, what we call the odd edges of this circuit starting at (1,1), and reducing the shipments by 1 on the even edges (the other edges) of this circuit. These balancing changes keep the row and column sums of the modified xi j’s the same, and so the modified shipments are again a valid solution.
Let us check how the original cost of the solution changes with this modification of S. One more unit along (1,1), (2,2), (3,3) increases the cost by 4 + 6 + 0 (the costs for those three edges) = 10. One less unit along (2,1), (3,2), and (1,3) reduces the cost by 6 + 6 + 0 = 12. The net change is 10 −12 = −2. Thus, this modification reduces the cost of the solution by 2. To get the largest reduction in cost possible by this modification, let us increase the shipments in the odd edges of this circuit as much as possible, with corresponding decreases on the even edges of the circuit. The critical constraint in our modification is that we cannot decrease the shipment in any even edge of the circuit below 0. So we examine the shipment levels in the even edges in S. The smallest number is 10 on edge (1,3). So we can increase shipment levels by 10 on the odd edges and reduce them by 10 on the even edges. Note that by driving www.itpub.net 4.5 The Transportation Problem 167 STORES Demand W A R E H O U S E S Supply Dummy 1 2 2 6 7 20 40 6 10 60 50 20 6 0 0 0 4 1 2 3 40 60 30 40 20 Figure 4.19 the level of shipments down to 0 on edge (1,3), we have dropped this edge from the subset of edges used in the new solution. Thus, the new solution is circuit-free. The new tableau is given in Figure 4.19. The reader should check that the new solution in Figure 4.19 has a cost of $580, which, as expected, is $20 less than the cost of the solution in Figure 4.18.
If it had turned out that our modification of increasing shipments by 1 on odd edges and decreasing shipments by 1 on even edges had increased the cost, then we would reverse the strategy and decrease shipments on odd edges as much as possible—until one of the odd edges has a shipment of zero—and increase shipments on even edges correspondingly. Also, if our modification resulted in no change in the cost, we would still follow the original strategy of increasing odd-edge shipments and decreasing even-edge shipments as much as possible in order to get a new solution of the same cost that was circuit-free.
Because the graph for a transportation problem is bipartite, every circuit in the graph will be even-length, by Theorem 2 in Section 1.3. So any circuit’s edges can be divided into odd and even edges, as done in the preceding example. Then, whenever a solution contains a circuit, we can apply the modification of increasing shipments in odd edges and decreasing shipments in even edges as much as possible, or the reverse strategy. This will produce a new solution that is less costly, or possibly the same cost, and which does not contain that circuit. If the original solution contained several circuits, the modification would have to be applied repeatedly until all circuits were broken. ◆ From Lemma 1, it follows that we only need to consider solutions which contain no circuits—that is, solutions which are spanning trees or spanning forests; a spanning forest is a disconnected collection of trees incident to all vertices. For the moment, let us assume that all circuit-free solutions are spanning trees. In the following discussion, we shall explain how to add one or more edges to a spanning forest solution, if one arises, to convert it into a spanning tree solution.
168 Chapter 4 Network Algorithms STORES Demand W A R E H O U S E S Supply Dummy 1 2 6 7 6 20 50 20 0 0 6 2 3 60 30 Figure 4.20 Our solution strategy will have two phases. First, find some initial spanning tree solution by an ad hoc method. Second, repeatedly modify the current spanning tree solution to find a cheaper solution, until a cheapest solution is found. The simplest way to find such an initial solution is by an ad hoc method called the Northwest Corner Rule. In Exercise 8, we present another way of obtaining an initial solution that typically is closer to the optimal solution.
Phase I: Find an Initial Solution by the Northwest Corner Rule We will use our sample transportation problem. It turns out that the solution in Figure 4.19 is the one obtained by the Northwest Corner Rule. This rule is applied as follows.
Start at the northwest corner of the tableau—that is, at entry (1,1). Make x11 as large as possible. The value will be the minimum of s1 and d1. For our problem, x11 = min (40, 60) = 40. We have totally used up the supply at warehouse 1, and so we delete row one from the tableau and reduce the demand at store 1 to 60 −40 = 20.
The resulting modified tableau is shown in Figure 4.20.
To meet the rest of the demand at store 1, we turn to the northwest corner in the remaining tableau shown in Figure 4.20—that is, entry (2,1), and make this entry as large as possible. So x21 = min (40, 20) = 20. Now we have satisfied the demand at store 1, and so we delete column one from the tableau and reduce the supply at warehouse 2 to 40. See Figure 4.21.
STORES Demand W A R E H O U S E S Supply Dummy 2 0 6 0 50 20 6 2 3 40 30 Figure 4.21 www.itpub.net 4.5 The Transportation Problem 169 STORES Demand W A R E H O U S E S Supply Dummy 1 2 2 6 7 20 40 6 10 60 50 20 6 0 0 0 4 1 2 3 40 60 30 40 20 Figure 4.22 We continue the procedure of repeatedly making the entry in the northwest corner of the remaining tableau as large as possible. The complete Northwest Corner Rule solution is shown in Figure 4.19, which for convenience is displayed in Figure 4.22.
Note that if the value we assign the current northwest entry in the current tableau uses up the supplies at the first (remaining) warehouse and also satisfies the demand of the first remaining store, then we would have to delete both the first row and the first column of the current tableau. This will lead to a disconnected set of edges in the solution—that is, a spanning forest. To avoid this outcome, we arbitrarily keep either the first row or first column, although its supplies or demand is 0.
Next we explain the second phase of how to improve the current solution. There are three steps to determining the improvement.
Phase II: Finding a Better Solution Step II.A. Determining Selling Prices at Warehouses and Stores At this point, the initial solution is the Northwest Corner Rule solution in Figure 4.22. We now introduce “selling prices” for the commodity at the warehouses and stores, based on the transportation costs in the initial solution. To start the pricing process, we need to pick an arbitrary price u1 for the commodity at warehouse 1, say u1 = $10. We use edge (1,1) in our initial solution to transport the commodity from warehouse 1 to store 1 at a cost of c11 = $4 per unit. Thus, if we buy the commodity at warehouse 1 for $10 and ship it for $4 to store 1, the commodity should sell for v1 = u1 + c11 = $10 + $4 = $14 at store 1.
Store 1 also receives shipments from warehouse 2 with a transportation cost of c21 = $6 per unit. Now we reverse the reasoning used to determine the price at warehouse 1 from the price at store 1. Given that the commodity’s price at store 1 is $14 and it costs $6 to ship from warehouse 2 to store 1, the price at warehouse 2 should be u2 = v1 −c21 = $14 −$6 = $8. Warehouse 2 also ships to store 2 at a cost of c22 = $6 so the price at store 2 should be v2 = u2 + c22 = $8 + $6 = $14. Using the shipping costs on the edges in our current spanning tree solution, we can continue 170 Chapter 4 Network Algorithms this process to calculate selling prices u3, v3 for the commodity at warehouse 3 and at store 3, respectively.
The complete set of selling prices is warehouse 1: u1 = $10 store 1: v1 = $14 warehouse 2: u2 = $8 store 2: v2 = $14 warehouse 3: u3 = $8 dummy: v3 = $8 (1) Summarizing, we set u1 equal to any arbitrary value. Then we determine successive ui’s and vj’s, as in the example, using the conditions that vj −ui = ci j, for each edge (i, j) in the current solution. Note that the calculation of prices in this fashion is only possible if the edges used in the solution do not form a circuit; for details of this claim, see Exercise 9.
Lemma 2 Let S be a spanning tree solution of a transportation problem using edge set E(S), and let ui and vj be prices at warehouse i and store j, respectively, based on solution S according to Step II.A. The transportation cost of this solution, i, j ci jxi j, summed over edges (i, j) in E(S), equals the income from the sale of the stores’ demands at the stores’ prices minus the cost of the warehouses’ supplies at the warehouses’ prices. That is, i, j ci jxi j = j vjd j − i uisi (2) Proof For each edge (i, j) in E(S), the prices are determined by the condition that ci j = vj −ui. Then ci jxi j = (vj −ui)xi j (3) When one sums (3) over all edges in E(S), the total of the shipments xi j out of warehouse i is si and the total of the shipments into store j is d j. Thus, the right-hand side of the sum of the (3)s over all edges in E(S) is equal to the right-hand side of (2). The left-hand side of this sum of (3)s is clearly the left-hand side of (2). ◆ To illustrate Lemma 2, for the solution in Figure 4.22 with the associated prices in (1), the right side of (2) is ($14 × 60 + $14 × 50 + $8 × 20) −($10 × 40 + $8 × 60 + $8 × 30) = $1700 −$1120 = $580 As noted earlier, $580 is the sum of the transportation costs for this solution.
Step II.B. Determining Which Edge to Add to the Current Solution We now look at the edges that are not used in the current solution. Consider edge (1,2) with cost c12 = $2. Warehouse 1’s selling price is $10 and Store 2’s selling price is $14. However, if we buy the commodity at warehouse 1 for $10 and ship it on edge (1,2) for $2, we can sell it at store 2 for $10 + $2 = $12, reducing the selling price at store 2 by $2 per item. A reduced price at store 2 will cause a reduced profit—store www.itpub.net 4.5 The Transportation Problem 171 sales minus warehouse purchases—which by Lemma 2 equals a reduced transporta-tion cost. This means that a cheaper solution can be obtained by incorporating edge (1,2) into the solution. [To maintain a spanning tree solution when edge (1,2) is added, some edge in the current solution would have to be dropped—a choice made in Step II.C.] Before using edge (1,2), we check the other edges not in the current solution to see how much each of them could reduce the profit. The results are summarized below.
edge (1 , 2): c12 = $2 < v2 −u1 = $14 −$10 = $4 decrease of $2 edge (1 , 3): c13 = $0 > v3 −u1 = $8 −$10 = −$2 increase of $2 edge (2, 3): c23 = $0 = v3 −u2 = $8 −$8 = $0 no change edge (3 , 1): c31 = $7 > v1 −u3 = $14 −$8 = $6 increase of $1 (4) We see that only edge (1,2) yields a reduction. So we add edge (1,2) to the current solution. If there were no edge that reduces the cost of the solution, then the current solution is optimal and we are finished.
Step II.C. Determining a Cheaper Spanning Tree Solution In Step II.B, an edge is chosen to be added to the current solution. For the example in Figure 4.22, the choice is edge (1,2). When this edge is added to the set of edges in the current solution, we have a unique circuit, by Exercise 28 in Section 3.2. For the solution in Figure 4.22, the circuit is (1,2), (2,2), (2,1), (1,1). As noted in the proof of Lemma 1 above, we get a cheaper solution by increasing the flow in the odd edges, (1,2) and (2,1), as much as possible while decreasing the flow in the even edges, (2,2) and (1,1), correspondingly. The net reduction is c12 + c21 −c11 −c22 = 2 + 6 −4 −6 = −2. Note that this calculation confirms our previous analysis that we will save $2 for each unit shipped on edge (1,2).
The limiting constraint is when the shipment in an even edge decreases to 0. In our example, since the current shipment in both x11 and x22 is 40, we can increase the shipments in (2,1) and (1,2) by 40 and reduce the shipments in (1,1) and (2,2) by 40.
The new solution, cheaper by 40 × $2 = $80 than the previous solution, is shown in Figure 4.23.
STORES Demand W A R E H O U S E S Supply Dummy 1 2 2 6 7 60 0 6 10 60 50 20 6 0 0 0 4 1 2 3 40 60 30 40 20 Figure 4.23 172 Chapter 4 Network Algorithms Note that although the flow in edges (1,1) and (2,2) both decreased to 0, we cannot drop both edges. If we did, the new solution would not be a spanning tree. We arbitrarily pick one of (1,1) and (2,2) to stay in the new solution but with a shipment level of 0.
In Figure 4.23, we picked (2,2).
Now we repeat the three parts of Phase II to find a still better solution.
In Step II. A, using the solution in Figure 4.23, we compute new prices at warehouses and stores. Starting with u1 = $10 and using edge (1,2), we find v2 = u1 + c12 = $10 + $2 = $12. Continuing as before, we obtain the following set of prices: warehouse 1: u1 = $10 store 1: v1 = $12 warehouse 2: u2 = $6 store 2: v2 = $12 warehouse 3: u3 = $6 dummy: v3 = $6 (5) In Step II.B, we see if any edges not in the current solution produce a reduction in the transportation costs. The calculations yield edge (1,1): c11 = $4 > v1 −u1 = $12 −$10 = $2 increase of $2 edge (1,3): c13 = $0 > v3 −u1 = $6 −$10 = −$4 increase of $4 edge (2,3): c23 = $0 = v3 −u2 = $6 −$6 = $0 no change edge (3,1): c31 = $7 > v1 −u3 = $12 −$6 = $6 increase of $1 (6) Since none of these edges reduces the cost of the current solution, the current solution is optimal, and we are finished.
EXERCISES Exercises 1–7 ask for the solution of transportation problems. Exercise 8 presents a better starting rule. Exercise 9 examines why prices cannot be uniquely determined if the solution contains a circuit.
1.
Solve the following transportation problems in which warehouse 1 has 30 units and warehouse 2 has 30 units and in which store 1 needs 20 units and store 2 needs 40 units. The tables below give the transportation costs.
(a) Store 1 2 (b) Store 1 2 Warehouse 1 $6 $2 Warehouse 1 $4 $3 2 $4 $3 2 $6 $4 2.
Solve the following transportation problems in which warehouse 1 has 50 units and warehouse 2 has 20 units and in which store 1 needs 20 units and store 2 needs 30 units. The tables below give the transportation costs. Note that a dummy store will be needed for the excess supplies.
(a) Store 1 2 (b) Store 1 2 Warehouse 1 $8 $4 Warehouse 1 $4 $4 2 $3 $6 2 $5 $7 www.itpub.net 4.5 The Transportation Problem 173 3.
Solve the following transportation problems in which warehouse 1 has 30 units, warehouse 2 has 30 units, and warehouse 3 has 30 units and in which store 1 needs 40 units and store 3 needs 50 units. The tables below give the transportation costs.
(a) Store 1 2 (b) Store 1 2 Warehouse 1 $4 $6 Warehouse 1 $5 $4 2 $5 $3 2 $6 $5 3 $6 $8 3 $3 $7 4.
Solve the following transportation problems in which warehouse 1 has 30 units, warehouse 2 has 30 units, and warehouse 3 has 30 units and in which store 1 needs 40 units and store 2 needs 40 units. The tables below give the transportation costs. Note that a dummy store will be needed for the excess supplies.
(a) Store 1 2 (b) Store 1 2 Warehouse 1 $5 $2 Warehouse 1 $8 $3 2 $9 $5 2 $4 $5 3 $4 $8 3 $4 $9 5.
Solve the following transportation problems in which warehouse 1 has 30 units, warehouse 2 has 30 units and warehouse 3 has 30 units and in which store 1 needs 20 units, store 2 needs 20 units, and store 3 needs 50 units. The tables below give the transportation costs.
(a) Store 1 2 3 (b) Store 1 2 3 Warehouse 1 $4 $6 $5 Warehouse 1 $8 $2 $4 2 $2 $4 $6 2 $4 $6 $4 3 $5 $3 $7 3 $5 $4 $5 6.
Solve the following transportation problems in which warehouse 1 has 40 units, warehouse 2 has 30 units, and warehouse 3 has 50 units and in which store 1 needs 50 units, store 2 needs 10 units, and store 3 needs 40 units. The tables below give the transportation costs. A dummy store will be needed.
(a) Store 1 2 3 (b) Store 1 2 3 Warehouse 1 $7 $2 $5 Warehouse 1 $8 $3 $2 2 $3 $5 $4 2 $3 $5 $7 3 $4 $6 $3 3 $6 $4 $5 7.
Solve the following transportation problems in which warehouse 1 has 30 units, warehouse 2 has 30 units, warehouse 3 has 30 units, and warehouse 4 has 30 174 Chapter 4 Network Algorithms units, and in which store 1 needs 40 units, store 2 needs 30 units, and store 3 needs 50 units. The tables below give the transportation costs.
(a) Store 1 2 3 (b) Store 1 2 3 Warehouse 1 $4 $3 $6 Warehouse 1 $7 $5 $3 2 $8 $7 $4 2 $2 $4 $6 3 $6 $3 $7 3 $5 $6 $5 4 $7 $3 $8 4 $7 $3 $3 8.
This exercise presents a better initial spanning tree solution, called the Minimum-Cost Rule. Instead of picking the northwest corner entry at each stage, this method picks the minimum-cost entry in the current tableau. For the tableau in Figure 4.18, the minimum cost entries are (1,3), (2,3), and (3,3), all of cost $0. In the case of a tie, we can pick any entry. Suppose we pick (1,3).
The Minimum-Cost Rule ships as much as possible along edge (1,3)—that is, x13 = min (40, 20) = 20. Now we delete the column for the dummy store, since its demand has been met, and reduce the supplies at warehouse 1 to 60 −20 = 40.
The minimum cost entry in the reduced tableau is (1,2) with a cost of $2. So we ship as much as possible along edge (1,2). Then x12 = min(40, 50) = 40. Now we delete the row for the warehouse 1, since its supplies have been used, and reduce the demand at Store 2 to 50 −40 = 10. We continue in this fashion.
Solve the following transportation problems using the Minimum-Cost Rule: (a) Exercise 4(a) (b) Exercise 5(a) (c) Exercise 6(a) 9.
Apply Step II.A to the tableau in Figure 4.18 to try to determine selling prices at warehouses and stores. Since the solution in this tableau uses edges forming a circuit, show that there are two routes which can be used to determine the selling price at store 3. Are the selling prices at store 3 determined from the two different routes the same?
4.6 SUMMARY AND REFERENCES Inthischapterwepresentedalgorithmsforthreebasicnetworkoptimizationproblems: shortest path, minimum spanning trees, maximum flow, and the transportation prob-lem. Principal emphasis was placed on a thorough discussion of maximum flows. We showed how these flows could be applied to a wide variety of other network problems.
In Section 4.4 we used flow models to develop a combinatorial theory of matching.
All the material about flows in this chapter is discussed in greater detail in the pioneer-ing work Flows in Networks by Ford and Fulkerson and in Network Flow Theory by Ahuja et al. . We have omitted discussion of the speed of these algorithms.
The interested reader is referred to Ahuja et al. for efficient implementations of www.itpub.net 4.6 Summary and References 175 the shortest path and maximum flow algorithms. Modern maximum flow algorithms require less than O(n3) operations for an n-vertex network.
It is often natural in network flow problems to have costs associated with edges so that when many possible maximum flows exist, one can ask for a least-cost maxi-mum flow. Such problems are called trans-shipment and transportation problems.
Similarly, in a matching problem with many solutions (X-matchings), one can ask for a least-cost matching. Such a problem is called an assignment problem. Efficient al-gorithms exist for all these minimization problems (see , , or ). Furthermore, any flow optimization problem, with or without the abovementioned minimization, is a problem of optimizing a linear function of the edge flows subject to linear equalities and inequalities, such as the flow constraints (a), (b), and (c) of Section 4.3. Such a constrained linear optimization problem is called a linear program. Linear program-ming is a principal tool of operations research, and good algorithms exist for solving linear programs. However, it is much more efficient to solve network problems with the network-specific algorithms presented in this chapter.
1.
R. Ahuja, T. Magnanti, and J. Orlin, Network Flow Theory, Algorithms and Applications, Prentice-Hall, Englewood Cliffs, NJ, 1993.
2.
L. Ford and D. Fulkerson, Flows in Networks, Princeton University Press, Princeton, NJ, 1962.
3.
F. Hillier and G. Lieberman, Introduction to Operations Research, Holden Day, San Francisco, 1988.
4.
B. Schwartz, “Possible winners in partially completed tournaments,” SIAM Review 8 (1966), 302–308.
This page is intentionally left blank www.itpub.net PART TWO ENUMERATION This page is intentionally left blank www.itpub.net CHAPTER 5 GENERAL COUNTING METHODS FOR ARRANGEMENTS AND SELECTIONS 5.1 TWO BASIC COUNTING PRINCIPLES This is the most important chapter in this book. It develops the fundamental skills of combinatorial reasoning. We present a few basic formulas, involving permutations and combinations, most of which the reader has seen before. Then we examine a variety of word problems involving counting and show how they can be broken down into sums and products of simple numerical factors. Having read through the examples as passive readers, students next must assume the active role of devising solutions on their own to the exercises. The first exercises at the end of each section are similar to the examples discussed in the section. The later exercises, however, have little in common with the examples except that they require the same general types of logical reasoning, clever insights, and mathematical modeling. Facility with these three basic skills in problem solving, as much as an inventory of special techniques, is the key to success in most combinatorial applications. In sum, for many students, this will be the most challenging and most valuable chapter in this book.
Subsequent chapters in the enumeration part of this book develop specialized theory and techniques that considerably simplify the solution of specific classes of counting problems. The usefulness of this theory is particularly evident after one has grappled with counting problems in this chapter using only first principles. Unfortu-nately, there is no such theory to assist in the solution of most real-world counting problems.
In order to motivate our problem solving, we mention applications to probability and statistics, computer science, operations research, and other disciplines. However, the details of the counting problems arising from such applications often appear tedious if one is not actively working in the area of application. So instead, we will base many of the worked-out examples (and exercises) on recreational problems, such as poker probabilities. The solution of these problems requires the same mathematical skills used in more substantive applications.
179 180 Chapter 5 General Counting Methods for Arrangements and Selections Theproblem-solvingskillsmentionedabovearebuildingblocksinamoregeneral strategy: asking the right questions about how to solve a problem. These questions range over a number of issues, such as what techniques to use, how to break the problem into manageable pieces, what unseen challenges are lying beneath the surface in the statement of the problem, what the first step is in solving the problem, and many more. For example, a critical question we repeatedly must ask at the start of each counting problem is: are the outcomes to be enumerated in this problem unordered sets (combinations), or ordered sets (sequences)? It is typically much harder to ask the right questions than it is to find the answers to these questions.
When an instructor or textbook presents the solution to a new type of counting problem, the steps in the solution may not be too difficult to follow, but the hard part of the solution—knowing what the right steps are—is frequently never discussed. This text will try to give proper attention to asking the right questions as well as answering these questions.
This section starts with two elementary but fundamental counting principles whose simplicity masks both their power and the ease with which they can be misused.
The Addition Principle If there are r1 different objects in the first set, r2 different objects in the second set, . . . , and rm different objects in the mth set, and if the different sets are disjoint, then the number of ways to select an object from one of the m sets is r1 +r2 + · · · +rm.
The Multiplication Principle Suppose a procedure can be broken into m successive (ordered) stages, with r1 different outcomes in the first stage, r2 different outcomes in the second stage,. . . ,andrm differentoutcomesinthemthstage.Ifthenumberofoutcomes at each stage is independent of the choices in previous stages and if the com-posite outcomes are all distinct, then the total procedure has r1 ×r2 × · · · ×rm different composite outcomes.
Remember that the addition principle requires disjoint sets of objects and the multiplication principle requires that the procedure break into ordered stages and that the composite outcomes be distinct. The validity of these two principles follows directly from the definitions of addition and multiplication of integers. That is, the sum a + b is the number of items resulting when a set of a items is added to a set of b items; and the product a × b is the number of sequences (A,B), when A can be any of a items and B can be any of b items. The two principles are standard mth-order extensions of these two binary operations.
www.itpub.net 5.1 Two Basic Counting Principles 181 Example 1: Rolling Dice Two dice are rolled, one green and one red. Each die has faces numbered 1 through 6.
(a) How many different outcomes of this procedure are there?
(b) What is the probability that there are no doubles (not the same value on both dice)?
(a) The question one needs to ask to get started is: what is an outcome in this problem? An outcome is actually a composite outcome of two smaller outcomes—namely, the number appearing on the green die when it is rolled and the number appearing on the red die.
To count the number of outcomes for a problem, it is essential to have a way to list outcomes on a piece of paper. In this case, a natural way to express the outcomes is ordered pairs of numbers, where the first number is the red die’s value and the second number the green die’s value. Then the collection of outcomes will be all 2-number sequences with each number between 1 and 6 (inclusive).
Thus, there are 6 × 6 outcomes.
(b) We calculate the probability of no doubles using the probability formula for an event: the size of the subset of outcomes producing the desired event divided by the number of all possible outcomes (see Appendix A.3 for details).
The denominator in the probability fraction is 36, the total number of outcomes determined in part (a). To get the numerator—the number of outcomes with no doubles—we again use the multiplication principle. The constraint “no doubles” must be recast as “the value of the green die is different from the value of the red die.” Once the red die is rolled, then there are just five permissible other values for the green die, independent of the particular value of the red die. So there are 6 × 5 = 30 outcomes with no doubles, and the probability of no doubles is 30/36 = 5/6.
The number of no-doubles in Example 1(b) could also be answered by subtracting the six outcomes of doubles from all 36 outcomes, yielding 36 −6 = 30 outcomes with no doubles.
Example 2: Arranging Books There are five different Spanish books, six different French books, and eight different Transylvanian books. How many ways are there to pick an (unordered) pair of two books not both in the same language?
The outcomes are unordered pairs of books. “Unordered” means that there is not a first book in the pair, and so we cannot break the outcomes into a first stage (first book) and a second stage—that is, the multiplication principle does not apply. So we face the question of how we recast or decompose this problem in such a way that the multiplication or additional principle can be used. Typically when we face this challenge, we need to break the problem into smaller parts to which the principles apply. This raises the question of what the parts are into which we should break the 182 Chapter 5 General Counting Methods for Arrangements and Selections problem. Once we get the right parts, the solution to each part is usually easy to find. If one recognizes a similarity between the analysis of the current problem and a previously solved problem, then the right parts might be found by analogy. Without such help, the best strategy is to pick a special case of the problem that is easy to solve. Then solve another special case and look for a pattern for solving the other cases.
One possible case in this problem is when the two books consist of a Spanish and a French book. This case has two stages: first, pick a Spanish book in five ways, and then pick a French book in six ways. The multiplication principle applies giving an answer of5 × 6.Itisnothardtoseethattheothertwocases—(i)aSpanishandaTransylvanian book and (ii) a French and a Transylvanian book—will work out similarly with answers of 5 × 8 and 6 × 8, respectively. The outcomes in three cases are clearly disjoint, and so the addition principle can be used to add the numbers of outcomes in the three cases to get the total number of pairs of books: 5 × 6 + 5 × 8 + 6 × 8 = 30 + 40 + 48 = 118.
The preceding example typifies a basic way of thinking in combinatorial problem solving: find a small enough special case for one to solve, then use the same solution method on the other cases, and add up the answers to all the special cases. There may be cleverer ways to solve the problem, but if we can reduce the original problem to a set of subproblems with which we are familiar, then we are less likely to make a mistake.
Example 3: Sequences of Letters How many ways are there to form a three-letter sequence using the letters a, b, c, d, e, f (a) with repetition of letters allowed? (b) without repetition of any letter?
(c) without repetition and containing the letter e? (d) with repetition and containing e?
(a) Withrepetition, we have six choicesfor each successive letter in thesequence.
So by the multiplication principle there are 6 × 6 × 6 = 216 three-letter sequences with repetition.
(b) Without repetition, there are six choices for the first letter. For the second letter,there arefivechoices,correspondingtothefiveremainingletters(what-ever the first choice was). Similarly for the third letter, there are four choices.
Thus there are 6 × 5 × 4 = 120 three-letter sequences without repetition.
(c) It is often helpful to make a diagram displaying the positions in a sequence, even a sequence with just three letters: −−− Such a diagram helps focus on choices involving the positions. Now we face a hard question in starting to solve this problem. How do we recast the constraint of at least one e in the sequence into a form that can be analyzed in terms of the addition and multiplication principles? There are typically www.itpub.net 5.1 Two Basic Counting Principles 183 several different ways to analyze a new constraint like this. Here we employ the case-by-case approach that we used in Example 2. One appealing version of this approach is to consider the different places where there must be an e.
As the following diagrams show, the e could be in the first position or the second position or the third position: e e e −−− −−− −−− In each diagram, there are five choices for which of the other five let-ters (excluding e) goes in the first remaining position and four choices for which of the remaining four letters goes in the other position. Thus there are 5 × 4 + 5 × 4 + 5 × 4 = 60 three-letter sequences with e. Another approach to solving this problem is given in Exercise 19.
(d) Let us try the approach used in part (c) when repetition is allowed. As before, there are three choices for e’s position. For any of these choices for e’s position, there are 6 × 6 = 36 choices for the other two positions, since e and the other letters can appear more than once. But the answer of 36 + 36 + 36 = 108 is not correct.
The addition principle has been violated because the outcomes in the three cases of where the e must be are not distinct. Consider the sequence e c e −−− It was generated two cases: once when e was put in the first position followed by c e as one of the 36 choices for the latter two positions, and a second time when e was put in the last position with e c in the first two positions.
We must use an approach for breaking the problem into parts that ensures distinct outcomes. Let us decompose the problem into disjoint cases based on where the first e in the sequence occurs. First suppose the first e is in the first position: e −−− Then there are six choices (including e) for the second and for the third positions—6 × 6 ways.
Next suppose the first e is in the second position: e −−− no e Then there are five choices for the first position (cannot be e) and six choices for the last position—5 × 6 ways. Finally, let the first (and only) e be in the last position: e −−− no e no e There are five choices each for the first two positions—5 × 5 ways. The correct answer is thus (6 × 6) + (5 × 6) + (5 × 5) = 91.
184 Chapter 5 General Counting Methods for Arrangements and Selections The hardest part about solving most counting problems is finding a structure in the problem that allows it to be broken into subcases or stages. In other words, the difficulty is in “getting started.” At the same time, one must be sure that the decomposition into cases or stages generates outcomes that are all distinct.
Many counting problems require their own special insights. For such problems, knowing the solution to problem A is typically of little help in solving problem B. This skill cannot be acquired in reading textbook examples. It is only gained by working many problems.
The following example illustrates this type of special insight in combinatorial problem solving.
Example 4: Nonempty Collections How many nonempty different collections can be formed from five (identical) apples and eight (identical) oranges?
Readers with some experience in combinatorial problem solving may want to break the problem into subcases based on the number of objects in the collections.
Any one of these subcases can be counted quite easily, but there are 13 possible subcases; that is, collections can have 1 or 2 or . . . up to 13 pieces of fruit.
In counting different possibilities, we must concentrate on what makes one col-lection different from another collection. The answer is, the number of apples and/or the number of oranges will be different in different collections. Then we can charac-terize any collection by a pair of integers (a, o), where a is the number of apples and o is the number of oranges.
Now the number of collections is easy to count. There are six possible values for a (including 0)—0, 1, 2, 3, 4, 5—and nine possible values for o. Together there are 6 × 9 = 54 different collections. (Note that we multiply 6 and 9, not add them, because a collection combines any number of apples and any number of oranges; we add if we want to count the ways to get some amount of apples or some amount of oranges, but not both.) Since the problem asked for nonempty collections and one of the possibilities allowed was (0,0), the desired answer is 54 −1 = 53.
Here is one piece of advice to consider when one is stuck and cannot get started with a problem. Try writing down in a systematic fashion some (a dozen or so) of the possible outcomes. In listing outcomes, one should start to see a pattern emerge. Think of the list as being part of one particular subcase. Then ask: how many outcomes would the list need to include to complete that subcase? Next ask what other (hopefully similar) subcases need to be counted. Once a first subcase has been successfully analyzed, other subcases are usually easier.
We summarize the two key facets we have encountered in combinatorial problem-solving. First, we must find a way to recast the constraints in a problem so that some combination of the addition and multiplication principles can be applied. Second, to use these principles we must find a way to break the problem into pieces or stages, and www.itpub.net 5.1 Two Basic Counting Principles 185 be sure that the outcomes in the different pieces are distinct. As noted at the outset, these concerns are instances of the general challenge of asking the right questions in problem-solving.
5.1 EXERCISES S u m m a r y o f E x e r c i s e s The early exercises are straightforward.
Then the exercises become more challenging and require analysis that will be different for each problem. For the harder problems, readers must devise their own method of solution rather than mimic a method used in one of the text’s examples. All exercises should be read carefully two times to avoid misinterpretation.
Some problems include possible answers for comment. Try to infer from each expression the reasoning that would have generated such an answer. If the answer is wrong, point out the mistake in the reasoning.
The word “between” is always used in the inclusive sense; that is, “integers between 0 and 50” means 0, 1, 2, . . . , 49, 50. The probability of a particular outcome is the number of such particular outcomes divided by the number of all outcomes.
See Appendix A.3 for more about probability.
1. (a) How many ways are there to pick a sequence of two different letters of the alphabet that appear in the word CRAB? In STATISTICS?
(b) How many ways are there to pick first a vowel and then a consonant from CRAB? From STATISTICS?
2. (a) How many integers are there between 0 and 50 (inclusive)?
(b) How many of these integers are divisible by 2?
(c) How many (unordered) pairs of these integers are there whose difference is 5?
3. A store carries eight styles of pants. For each style, there are 12 different possible waist sizes, five pants lengths, and four color choices. How many different types of pants could the store have?
4. How many different sequences of heads and tails are possible if a coin is flipped 100 times? Using the fact that 210 = 1024 ≈1000 = 103, give your answer in terms of an (approximate) power of 10.
5. How many six-letter “words” (sequence of any six letters with repetition) are there? How many with no repeated letters?
6. How many ways are there to pick a man and a woman who are not husband and wife from a group of n married couples?
7. Given 10 different English books, six different French books, and four different German books, (a) How many ways are there to select one book?
(b) How many ways are there to select three books, one of each language?
186 Chapter 5 General Counting Methods for Arrangements and Selections (c) How many ways are there to make a row of three books in which exactly one language is missing (the order of the three books makes a difference)?
8. There are seven different roads between town A and town B, four different roads between town B and town C, and two different roads between town A and town C.
(a) How many different routes are there from A to C altogether?
(b) How many different routes are there from A to C and back (any road can be used once in each direction)?
(c) How many different routes are there from A to C and back in part (b) that visit B at least once?
(d) How many different routes are there from A to C and back in part (b) that do not use any road twice?
9. How many ways are there to pick two different cards from a standard 52-card deck such that (a) The first card is an Ace and the second card is not a Queen?
(b) The first card is a spade and the second card is not a Queen? (Hint: Watch out for the Queen of spades.) 10. How many nonempty collections of letters can be formed from four As and eight Bs?
11. How many ways are there to roll two distinct dice to yield a sum evenly divisible by 3?
12. How many six-letter “words” (sequences of letters with repetition) are there in which the first and last letter are vowels? In which vowels appear only (if at all) as the first and last letter?
Comment on these possible answers to the second part: (a) 52214, (b) 52264, (c) 212264, and (d) 266 −212264.
13. (a) How many different six-digit numbers are there (leading zeros, e.g., 00174, not allowed)?
(b) How many even six-digit numbers are there?
(c) How many six-digit numbers are there with exactly one 3?
(d) How many six-digit palindromic numbers (numbers that are the same when the order of their digits is inverted, e.g., 137731) are there?
14. How many different numbers can be formed by various arrangements of the six digits 1, 1, 1, 1, 2, 3?
15. What is the probability that the top two cards in a shuffled deck do not form a pair?
16. (a) How many different outcomes are possible when a pair of dice, one red and one white, are rolled two successive times?
(b) What is the probability that each die shows the same value on the second roll as on the first roll?
www.itpub.net 5.1 Two Basic Counting Principles 187 (c) What is the probability that the sum of the two dice is the same on both rolls?
(d) What is the probability that the sum of the two dice is greater on the second roll?
17. A rumor is spread randomly among a group of 10 people by successively having one person call someone, who calls someone, and so on. A person can pass the rumor on to anyone except the individual who just called.
(a) By how many different paths can a rumor travel through the group in three calls? In n calls?
(b) What is the probability that if A starts the rumor, A receives the third calls?
(c) What is the probability that if A does not start the rumor, A receives the third call?
18. (a) How many different license plates involving three letters and two digits are there if the three letters appear together either at the beginning or end of the license?
(b) How many license plates involving one, two, or three letters and one, two, or three digits are there if the letters must appear in a consecutive grouping?
19. Re-solve the problem in Example 3 of counting the number of three-letter se-quences without repetition using a, b, c, d, e, f that have an e by first counting the number with no e.
20. What is the probability that the sum of two randomly chosen integers between 20 and 40 inclusive is even (the possibility of the two integers being equal is allowed)?
Comment on the answers (a) 1/2, (b) 11/21, and (c) 112/212.
21. How many three-letter sequences without repeated letters can be made using a, b, c, d, e, f in which either e or f (or both) is used?
Comment on the answers (a) 3 × 2 × 4 × 3, (b) 3 × 2 × 5 × 4, (c) 3 × 2 × 4 × 4 −3 × 2 × 4, and (d) 6 × 5 × 4 −4 × 3 × 2.
22. How many ternary (0, 1, 2) sequences of length 10 are there without any pair of consecutive digits the same?
23. How many integers between 1,000 and 10,000 are there with (make sure to avoid sequences of digits with leading 0s): (a) Distinct digits?
(b) Repetition of digits allowed but with no 2 or 4?
(c) Distinct digits and at least one of 2 and 4 must appear?
24. How many 12-digit decimal sequences are there that start and end with a sequence of at least two 3s?
25. How many sequences of length 5 can be formed using the digits 0, 1, 2, . . . , 9 with the property that exactly two of the 10 digits appear, e.g., 05550?
188 Chapter 5 General Counting Methods for Arrangements and Selections 26. What is the probability that an integer between 0 and 9,999 has exactly one 8 and one 9?
Comment on the answers (a) 4 × 3/104, (b) 4 × 3 × 8 × 7/104, and (c) 2 × 4 × 3 × 82/104.
27. How many different five-letter sequences can be made using the letters A, B, C, D with repetition such that the sequence does not include the word BAD—that is, sequences such as ABADD are excluded.
28. (a) How many election outcomes are possible with 20 people each voting for one of seven candidates (the outcome includes not just the totals but also who voted for each candidate)?
(b) How many election outcomes are possible if only one person votes for can-didate A and only one person votes for candidate D?
29. There are 15 different apples and 10 different pears. How many ways are there for Jack to pick an apple or a pear and then for Jill to pick an apple and a pear?
30. How many times is the digit 5 written when listing all numbers from 1 to 100,000?
Comment on the answers (a) 4, (b) 5 × 104, and (c) 1 + 10 + 100 + 1000.
31. How many times is “25” written when listing all numbers from 1 to 100,000?
(This is an extension of the previous exercise.) 32. What is the probability that if one letter is chosen at random from the word RECURRENCE and one letter is chosen from RELATION, the two letters are the same?
33. How many four-digit numbers are there formed from the digits 1, 2, 3, 4, 5 (with possible repetition) that are evenly divisible by 4?
34. How many nonempty collections of letters can be formed from n As, n Bs, n Cs and n Ds?
35. There are 50 cards numbered from 1 to 50. Two different cards are chosen at random. What is the probability that one number is twice the other number?
36. If two different integers between 1 and 100 inclusive are chosen at random, what is the probability that the difference of the two numbers is 15?
37. If three distinct dice are rolled, what is the probability that the highest value is twice the smallest value?
38. How many different numbers can be formed by the product of two or more of the numbers 3, 4, 4, 5, 5, 6, 7, 7, 7?
39. There are 10 different people at a party. How many ways are there to pair them off into a collection of five pairings?
40. A chain letter is sent to five people in the first week of the year. The next week each person who received a letter sends letters to five new people, and so on.
How many people have received letters in the first five weeks?
41. How many ways are there to place two identical rooks in a common row or column of an 8 × 8 chessboard? an n × m chessboard?
www.itpub.net 5.2 Simple Arrangements and Selections 189 42. How many ways are there to place two identical kings on an 8 × 8 chessboard so that the kings are not in adjacent squares? on an n × m chessboard?
43. How many ways are there to place two identical queens on an 8 × 8 chessboard so that the queens are not in a common row, column, or diagonal?
44. How many different positive integers can be obtained as a sum of two or more of the numbers 1, 3, 5, 10, 20, 50, 82?
45. How many ways are there for a man to invite some (nonempty) subset of his 10 friends to dinner?
46. How many different rectangles can be drawn on an 8 × 8 chessboard (the rectan-gles could have sides of length 1 through 8; two rectangles are different if they contain different subsets of individual squares)?
47. How many ways are there to place a red checker and a black checker on two black squares of a checkerboard so that the red checker can jump over the black checker? (A checker jumps on the diagonal from in front to behind.) 48. Use induction to verify the following formally: (a) The addition principle (b) The multiplication principle 49. On the real line, place n white pegs at positions 1, 2, . . . , n and n blue pegs at positions −1, −2, . . . , −n (0 is open). Whites move only to the left, blues to the right. When beside an open position, a peg may move one unit to occupy that position (provided it is in the required direction). If a peg of one color is in front of a peg of the other color that is followed by an open position (in the required direction), a peg may jump two units to the open position (the jumped peg is not removed). By a sequence of these two types of moves (not necessarily alternating between white and blue pegs), one seeks to get the positions of the white and blue pegs interchanged. (See the article on this game in Mathematics Teacher, January 1982.) (a) Play this game for n = 3 and n = 4.
(b) Use a combinatorial argument to show that in general, n2 + 2n moves (unit steps and jumps) are required to complete the game.
5.2 SIMPLE ARRANGEMENTS AND SELECTIONS A permutation of n distinct objects is an arrangement, or ordering, of the n objects.
An r-permutation of n distinct objects is an arrangement using r of the n objects. An r-combination of n distinct objects is an unordered selection, or subset, of r out of the n objects. We use P(n,r) and C(n,r) to denote the number of r-permutations and 190 Chapter 5 General Counting Methods for Arrangements and Selections r-combinations, respectively, of a set of n objects. From the multiplication principle we obtain P(n, 2) = n(n −1), P(n, 3) = n(n −1)(n −2), P(n, n) = n(n −1)(n −2) × · · · × 3 × 2 × 1 In enumerating all permutations of n objects, we have n choices for the first position in the arrangement, n −1 choices (the n −1 remaining objects) for the second position, . . . , and finally one choice for the last position. Using the notation n! = n(n −1)(n −2) · · · × 3 × 2 × 1 (n! is said “n factorial”), we have the formulas P(n, n) = n!
and P(n,r) = n(n −1)(n −2) × · · · × [n −(r −1)] = n!
(n −r)!
Our formula for P(n, r) can be used to derive a formula for C(n, r). All r-permutations of n objects can be generated by first picking any r-combination of the n objects and then arranging these r objects in any order. Thus P(n, r) = C(n, r) × P(r, r), and solving for C(n, r) we have C(n,r) = P(n,r) P(r,r) = n!/(n −r)!
r!
= n!
r!(n −r)!
The numbers C(n, r) are frequently called binomial coefficients because of their role in the binomial expansion (x + y)n. We study the binomial expansion and identities involving binomial coefficients in Section 5.5. It is common practice to write the expression C(n, r) as n r and to say “n choose r” [we usually write C(n, r) in this book]. Note that C(n, r) = C(n, n – r), since the number of ways to pick a subset of r out of n distinct objects equals the number of ways to throw away a subset of n – r.
The rest of this section is a continuation of the combinatorial problem-solving examples in the previous section. Along with the introduction of permutations and combinations, these examples involve more complicated situations. With many new constraints to keep straight, readers may find themselves becoming confused about even basic issues such as when to add and when to multiply. One must think very carefully about each step in the analysis of a counting problem.
There are two ways to analyze arrangement problems. We have the option of either picking which item goes in the first position, then which item goes in the second position, and so on, or picking which position to choose for the first item, which position to choose for the second item, and so on.
Example 1: Ranking Wizards How many ways are there to rank n candidates for the job of chief wizard? If the ranking is made at random (each ranking is equally likely), what is the probability that the fifth candidate, Gandalf, is in second place?
www.itpub.net 5.2 Simple Arrangements and Selections 191 A ranking is simply an arrangement, or permutation, of the n candidates. So there are n! rankings. Intuitively, a random ranking should randomly position Gandalf, and so we would expect that he has probability 1/n of being second, or being in any given position. Equivalently, each of the n candidates should have the same probability of being in second place. Formally, we calculate the probability of Gandalf being second using the probability formula for an event: the subset of outcomes producing the desired event divided by all possible outcomes (see Appendix A.3 for details).
Prob(Gandalf second) = no. of rankings with Gandalf second total no. of rankings Now comes the hard question. How do we count outcomes where Gandalf is ranked second? The answer lies in the observation preceding this example about the two ways to analyze arrangement problems: (i) picking an item for the first position, then the second position, and so on; or (ii) picking a position for the first item, then for the second item, and so on. We need to use the second approach and make Gandalf the first wizard to be given a position.
We must put Gandalf in second place—1 way—and then put the remaining wizards in the remaining n −1 places—(n −1)! ways. So Gandalf is second in (n −1)! rankings, and Prob(Gandalf second) = (n −1)!/n! = 1/n, as expected.
To count arrangements where a particular position is fixed—in the preceding example, that Gandalf is in second place—we do not count the ways to make Gandalf second (there is only one way to put Gandalf in second place) but rather count all the ways to arrange the remaining wizards in the remaining positions. We count what has not been constrained.
Example 2: Arrangements with Repeated Letters How many ways are there to arrange the seven letters in the word SYSTEMS? In how many of these arrangements do the three Ss appear consecutively?
The first question we must ask is: how do we modify the formula for arrangements of distinct items to incorporate the situation where one item may be repeated. In the previous example, we handled the new constraint, Gandalf ranked second, first and then arranged other wizards. In this example, we take the opposite approach and delay arranging the multiple Ss until the end.
We first pick the positions that the other four letters E, M, T, Y will occupy in the seven-letter arrangement, and then the three Ss will fill the remaining three positions in one way. There are seven possible positions for E, six for M, five for T, and four for Y. Thus there are P(7, 4) = 7!/3! = 840 arrangements. (A general formula for counting arrangements with repeated objects is given in the next section.) Here again we counted arrangements in terms of where successive items should be placed rather than what should go in successive positions.
Next we consider the case where the three Ss appear consecutively, that is, the three Ss are all side-by-side. The “trick” for handling this new consecutivity constraint 192 Chapter 5 General Counting Methods for Arrangements and Selections is to realize that when the three Ss are grouped together they now become a single composite letter. So the problem reduces to arranging the five distinct letters, Y, T, E, M and SSS (treated as a single letter), which can be done in 5! = 120 ways.
Another way to look at this problem is to think of temporarily setting aside two of the Ss, arranging the five remaining letters, Y, T, E, M, S, in 5! ways, and then in each resulting arrangement inserting beside the S the other two Ss.
Example 3: Binary Sequences How many different 8-digit binary sequences are there with six 1s and two 0s?
A good starting question is: what distinguishes one 8-digit binary sequence with six 1s from another such sequence? The answer is the positions of the six 1s (or the two 0s). Though this problem initially reads as an arrangement problem, what must be counted is the different possible placements of the six 1s, that is, different possible subsets of six of the eight positions in the binary sequence. (A sequence with 1s in positions 1, 2, 3, 4, 5, 6 is the same as a sequence with 1s in positions 6, 5, 2, 3, 4, 1—the order does not matter, just the collection of positions involved.) So the answer is C(8, 6) = 28.
We could alternatively have focused on picking a subset of two of the eight positions for 0s.
Example 4: Poker Probabilities (a) How many 5-card hands (subsets) can be formed from a standard 52-card deck?
(b) If a 5-card hand is chosen at random, what is the probability of obtaining a flush (all five cards in the hand are in the same suit)?
(c) What is the probability of obtaining three, but not four, Aces?
(a) A 5-card hand is a subset of five cards chosen from the 52 cards in a deck, and so there are C(52, 5) = 52!/(47!5!) = 2,598,960 different 5-card hands.
(b) To find the probability of a flush, we need to find the number of 5-card subsets with all cards of the same suit. There are four suits, and a subset of five cards from the 13 cards in a given suit can be chosen in C(13, 5) = 13!/(5!8!) = 1287 ways. So there are 4 × 1287 = 5148 flushes, and Prob(5-card hand is a flush) = 5148 2,598,960 = 0.00198 (≈0.2%) (c) To count the number of hands with exactly three Aces, we must pick three of the four Aces—done in C(4, 3) = 4 ways—and then fill out the hand with two cards chosen from the 48 non-Ace cards—done in C(48, 2) = 1128 ways. So there are 4 × 1128 = 4512 hands with exactly three Aces, and Prob(5-card hand has exactly three Aces) = 4512 2,598,960 = .00174 www.itpub.net 5.2 Simple Arrangements and Selections 193 Note that to count hands with three Aces in Example 4, we implicitly used the multiplication principle to multiply the ways to pick three Aces times the ways to fill out the hand with two non-Ace cards. However, a hand is an unordered collection, and by ordering it into two parts, we might generate two outcomes that are really the same set, violating the distinctness condition of the multiplication principle. In this problem, hands could safely be decomposed into an Aces part and a non-Aces part because the types of cards in the two parts were different. The next example illustrates how this disjointness condition can be violated.
Example 5: Forming Committees A committee of k people is to be chosen from a set of seven women and four men.
How many ways are there to form the committee if (a) The committee consists of three women and two men?
(b) The committee can be any positive size but must have equal numbers of women and men?
(c) The committee has four people and one of them must be Mr. Baggins?
(d) The committee has four people and at least two are women?
(e) The committee has four people, two of each sex, and Mr. and Mrs. Baggins cannot both be on the committee?
(a) Applying the multiplication principle to disjoint collections, we can count all committees of three women and two men by composing all subsets of three women with all subsets of two men, C(7, 3) × C(4, 2) = 35 × 6 = 210 ways.
(b) To count the possible subsets of women and men of equal size on the committee, we must know definite sizes of the subsets. That is, we must break the problem into the four disjoint subcases: one woman and one man, two each, three each, and four each (there are only four men available). So the total num-ber is the sum of the possibilities for these four subcases, [C(7, 1) × C(4, 1)] + [C(7, 2) × C(4, 2)] + [C(7, 3) × C(4, 3)] + [C(7, 4) × C(4, 4)] = 7 × 4 + 21× 6 + 35 × 4 + 35 × 1 = 329.
(c) If Mr. Baggins must be on the committee, this simply means that the problem reduces to picking three other people on the 4-person committee from the remaining 10 people (seven women and three men). Observe the similarity of this constraint with the Gandalf-second constraint in Example 1. In both cases, we need to focus on the ways of arranging or selecting the other people. So far we have made all selections from the set of men or the set of women. Now the other three people must be chosen from the set of all remaining people, and so the answer is C(10, 3) = 120. (It is easy to get in a mindset where one uses the same sets in the current problem that one used in the previous problem, here the set of women and the set of men—problem solvers must always be alert to this trap.) (d) One approach is to pick two women first, C (7, 2) = 21 ways, and then pick any two of the remaining set of nine people (five women and four men). However, counting all committees in this fashion counts some outcomes more than once, 194 Chapter 5 General Counting Methods for Arrangements and Selections since any woman in one of these committees could be chosen as either one of the first two women or one of the two remaining people. For example, if Wi denotes theithwomanand Mi theithman,then(W1,W3)composedwiththetworemaining people (W2, M3) yields the same set as (W1, W2) composed with (W3, M3).
A correct solution to this problem must use a subcase approach, as in part (b). That is, break the problem into three subcases that specify exactly how many women and how many men are on the committee: two women and two men, three women and one man, and four women. The answer is thus [C(7, 2) × C(4, 2)] + [C(7, 3) × C(4, 1)] + C(7, 4) = 21 × 6 + 35 × 4 + 35 = 301.
(e) We need to recast the condition “Mr. and Mrs. Baggins cannot both be on the committee” into several simpler subcases in which we know exactly which of Mr. and Mrs. Baggins is on the committee. Note that the possibility of neither Mr. nor Mrs. Baggins’ being on the committee satisfies the given condition.
There are three subcases to consider.
The first subcase is that Mrs. Baggins is on the committee and Mr. Baggins is not. Then one more woman must be chosen from the remaining six women and two more men must be chosen from the remaining three men (Mr. Baggins is ex-cluded). This can be done in C(6, 1) × C(3, 2) = 6 × 3 = 18 ways. The other two cases are: Mrs. Baggins is off and Mr. Baggins is on, which by a similar argument yields C(6, 2) × C(3, 1) = 15 × 3 = 45 ways; and neither is on the committee, C(6, 2) × C(3, 2) = 15 × 3 = 45 ways. The total answer is 18 + 45 + 45 = 108.
An easier solution to this problem can be obtained by taking a complementary approach. We can consider all C(7, 2) × C(4, 2) 2-women–2-men committees and then subtract the forbidden committees that contain both Bagginses. The forbidden committees are formed by picking one more woman and one more man to join Mr. and Mrs. Baggins—done in C(6, 1) × C(3, 1) ways. This approach yields a simpler formula for the same answer: 21 × 6 −6 × 3 = 108.
Parts (b), (d), and (e) of Example 5 illustrate an important point. When counting the ways to pick elements in a given subset, as a part of a more complex problem, one needs to specify the number of elements in the subset. If the size of the subset can vary, then one must break the problem into subcases so that the size of the subset is a fixed number in each subcase.
The mistake of counting the same outcome twice, which arose in part (d) of Example 5, arises in many guises. The following principle should help the reader avoid this problem.
The Set Composition Principle Supposeasetofdistinctobjectsisbeingenumeratedusingthemultiplicationprinciple, multiplying the number of ways to form some first part of the set by the number of ways to form a second part (for a given first part). The possible members of the first and second parts must be disjoint. Another way to express this condition is: given any set S thus constructed, one must be able to tell uniquely which elements of S are in the first part of S and which elements are in the second part.
www.itpub.net 5.2 Simple Arrangements and Selections 195 In Example 5(a), the part W of three women and the part M of two men are disjoint, and so by the set composition principle the number of committees with three women and two men is the size of W times the size of M. In the “committee with at least two women” problem in Example 5(d), the method of choosing two of the seven women first and then picking any remaining two people (women or men) violates the Set Com-position Principle because the members of the two parts are not disjoint—any woman could be in either the first or second part. For example, in the committee {W1, W2, W3, M3} it is impossible to say which two women are chosen in the first part. The scheme of picking two women and then any two remaining people generates the committee {W1, W2, W3, M3} in C(3, 2) different ways—namely, all compositions of (i) two of the three women W1, W2, W3; with (ii) M3 plus the remaining woman in W1, W2, W3 not chosen in (i).
In Example 2 about arrangements of the letters of SYSTEMS, we dealt with the constraint that certain letters had to be grouped together in the arrangement. In Example 2, it was three consecutive Ss. In the following continuation of that example, we consider the constraint of requiring a particular letter to appear somewhere before another letter in the arrangement. We also show how to grapple with the two constraints simultaneously.
Example 2: (continued) Arrangements with Repetitions How many arrangements of the seven letters in the word SYSTEMS have the E occurring somewhere before the M? How many arrangements have the E somewhere before the M and the three Ss grouped consecutively?
The key to the constraint of E being somewhere before the M (not necessarily immediately before the M) is to focus on the pair of positions where E and M will go.
Thus, we start by picking which of the two out of the seven positions in an arrangement are where the E and M will go—C(7, 2) = 21 ways—and then we put E in the first of this pair of positions and M in the second one. Now we fill in the five other positions in the arrangement by picking a position for the Y and the T—P(5, 2) = 5 × 4 = 20 ways—and then putting the three Ss in the three remaining positions. The answer is thus 21 × 20 = 420.
While it may sound scary to deal with the two constraints at once, it often turns out to be less hard than expected if one handles the constraints in the right order. If we first pick the pair of the positions for the E and the M, things get messy for the consecutivity constraint because the different placements of the E and M will impact differently the positions in the arrangement where there is enough room to place three consecutive Ss.
SowetrydealingwiththeconsecutiveSsfirst.Thissimplyrequiresthatwe“glue” the three Ss together into one composite letter and consider the reduced problem of arranging five letters, Y, T, E, M, (SSS). Now we turn to the other constraint and pick the pair of positions, out of the five new positions, for the E and the M—C(5, 2) = 10 ways—with the E going in the first of the two chosen positions. Then we arrange the Y, T and (SSS) in the remaining three positions in 3! = 6 ways. So the answer is 10 × 6 = 60.
196 Chapter 5 General Counting Methods for Arrangements and Selections The following type of counting problem arises frequently in quality-control problems.
Example 6: Counting Defective Products A manufacturing plant produces ovens. At the last stage, an inspector marks the ovens A (acceptable) or U (unacceptable). How many different sequences of 15 As and Us are possible in which the third U appears as the twelfth letter in the sequence?
This problem is a binary sequence problem similar to Example 5 except now the elements are A and U, rather than 0 and 1. If the third U appears at the twelfth letter in the sequence, then the subsequence composed of the first 11 letters must have exactly two Us (and nine As). Following the reasoning in Example 5, there are C(11, 2) = 55 possible sequences for the first 11 letters. There is one way to pick the twelfth letter— it is specified to be U. The remaining three letters in the sequence can be either A or U—23 = 8 possibilities. All together, there are 55 × 1 × 8 = 440 sequences.
Example 7: Probability of Repeated Digits What is the probability that a 4-digit campus telephone number has one or more repeated digits?
There are 104 = 10,000 different 4-digit phone numbers. We break the problem of counting 4-digit phone numbers with repeated digits into four different cases of repetitions: (a) All four digits are the same.
(b) three digits are the same, the other is different.
(c) two digits are the same, the other two digits are also the same (e.g., 2828).
(d) two digits are the same, the other two digits are each different (e.g., 5105).
(a) There are 10 numbers formed by repeating one of the 10 digits four times.
(b) One way to decompose the process of generating these case-(b) numbers is as follows (several other decompositions are possible). First pick which digit appears once—10 choices—then where it occurs in the 4-digit number—four choices—and finally which other digit appears in the other three positions—nine choices. This yields 10 × 4 × 9 = 360 numbers.
(c) First pick which two digits are each to appear twice—C(10, 2) = 45 choices—and then how to arrange these four digits: pick which two positions are used by, say, the smaller of the digits—C(4, 2) = six ways. This yields 45 × 6 = 270 numbers.
(d) First pick which pair of digits appear once—C(10, 2) = 45 choices—then pick a position for, say, the smaller of these two digits and a position for the larger digit—4 × 3 = 12 choices—and finally pick which other digit appears in the remainingtwopositions—eightchoices.Thisyields45×12×8=4320numbers.
In sum, there are 10 + 360 + 270 + 4320 = 4960 4-digit phone numbers with a repeated digit. The probability of a repeated digit is thus 4960/10,000 = 0.496 ≈0.5.
www.itpub.net 5.2 Simple Arrangements and Selections 197 We note that there happens to be a simpler way, using the complementary set, to count phone numbers with repeated digits: count numbers with no repeated dig-its. These are just the P(10, 4) = 5040 four-permutations of the 10 digits. So the remaining repeated-digit numbers amount to 10,000 −5040 = 4960.
One point of caution about cases (c) and (d) where two different digits both occur once or both occur twice. In case (d), we pick the two digits occurring once as an unordered pair in C(10, 2) ways and arrange those digits (and then pick the digit to go in the remaining two positions) rather than pick a first digit, position it, then pick a second digit, position it (and then pick the digit to go in the remaining two positions)—10 × 4 × 9 × 3 × 8 ways. In this latter (wrong) approach, we cannot tell for a telephone number such as 2529 whether the 5 was chosen first and put in the second position and then the 9 chosen next and put in the fourth position, or whether the 9 was chosen first and put in the fourth position and then the 5 chosen next and put in the second position. The disjointness requirement of the multiplication principle is being violated and each outcome in case (d) would be counted twice.
Example 8: Voter Power [Optional] We consider a way of measuring the influence of different players in weighted voting.
Suppose that in a 5-person regional council there are three representatives from small towns, call them a, b, c, who each cast one vote, and there are two representatives from large towns, call them D, E, who each cast two votes. With a total of seven votes cast, it takes four votes (a majority of votes) in favor of legislation to enact it. Suppose that in forming a coalition to vote for some legislation, the people join the coalition in order (an arrangement of the people). The pivotal person in a coalition arrangement is the person whose vote brings the number of votes in the coalition up to four. For example, in the coalition arrangement bDcaE, the pivotal person is c. A measure of the “power” of a person p in the council is the fraction of coalition arrangements in which p is the pivotal person. This measure of power is called the Shapley–Shubik index.
Determine the Shapley–Shubik index of person a and person D in this council— that is, determine the fraction of all coalition arrangements in which a and D, respec-tively, are pivotal. (By symmetry, the other 1-vote people b and c will have the same index as a, and similarly E will have the same index as D.) If a is pivotal in a coalition arrangement, then people with (exactly) three votes must precede a in the arrangement and people with three votes must follow a. Since there are only two other 1-vote people, the three votes preceding a must come from one 1-vote person—b or c—and one 2-vote person—D or E. Then the beginning of the coalition can be formed in 2 (choice of 1-vote person) × 2 (choice of 2-vote person) × 2 (whether 1-vote or 2-vote person goes first) = 8 ways. The remain-ing 1-vote and remaining 2-vote person will follow a, with two ways to arrange them. In total there are 8 × 2 = 16 coalition arrangements in which a is pivotal.
There are 5! = 120 arrangements in all, and so the Shapley–Shubik index of a is 16/120 = 4/30.
If D is pivotal in a coalition arrangement, there can be people with two or three votes preceding D. Suppose there are two votes before D and three votes after D.
198 Chapter 5 General Counting Methods for Arrangements and Selections Either the arrangement starts with two of the three 1-vote people—arranged 3 × 2 ways—then D, followed by the other 1-vote person and E in either order—two ways— or the arrangement starts with E, then D, followed by an arrangement of the three 1-vote people—3! ways. In total, there are (3 × 2 × 2) + 3! = 18 arrangements with two votes before D and three votes after D. By interchanging the people before D with the people after D in these arrangements, we obtain the arrangements with of three votes before D and two votes after D. So there are 18 of the latter arrangements. In total, there are 18 + 18 = 36 arrangements in which D is pivotal, and so D’s Shapley– Shubik index is 36/120 = 9/30.
Observe that a 2-vote person has an index 2 1 4 times the size of a 1-vote person.
We close this section by noting that for large values of n, n! can be approximated by the number sn = √ 2πn(n/e)n, where e is Euler’s constant (e = 2.718 . . . ). This approximation is due to Stirling and its derivation is given in most advanced calculus texts (see Buck ). The error |n! −sn| increases as n increases, but the relative error |n! −sn|/sn is always less than 1/11n. The following table gives some sample values of n!, sn, and |n! −sn|/sn.
n n!
sn |n! −sn|/sn 1 1 .922 .085 2 2 1.919 .042 5 120 118.02 .017 10 3,628,800 3,598,600 .008 20 2.433 × 1018 2.423 × 1018 .004 100 9.333 × 10157 9.328 × 10157 .0005 5.2 EXERCISES S u m m a r y o f E x e r c i s e s As in the previous section, most of these exercises require individual analysis, different for each problem. Remember to read problems carefully to avoid misinterpretation. Pay special attention to whether a problem involves arrangements or subsets. Note that the problems assume that people are distinct objects (no identical people).
1. How many ways are there to arrange the cards in a 52-card deck?
2. How many different 5-letter “words” (sequences) are there with no repeated letters formed from the 26-letter alphabet?
3. How many ways are there to distribute six different books among 13 children if no child gets more than one book?
Comment on the answers (a) C(13, 6), (b) 136, and (c) C(13, 6)6!
4. Howmanywaysaretheretoseatsixdifferentboysandsixdifferentgirlsalongone sideofalongtablewith12seats?Howmanywaysifboysandgirlsalternateseats?
www.itpub.net 5.2 Simple Arrangements and Selections 199 5. How many arrangements are there of the seven letters in the word UNUSUAL?
6. How many ways are there to pick a subset of four different letters from the 26-letter alphabet?
7. How many ways are there to pick a 5-person basketball team from 14 possible players? How many teams if the weakest player and the strongest player must be on the team?
8. There are nine white balls and four red balls in an urn. How many different ways are there to select a subset of six balls, assuming the 13 balls are different? What is the fraction of selections with four whites and two reds?
9. If a fair coin is flipped 11 times, what is the probability of nine or more heads?
10. What is the probability that an arrangement of a, b, c, e, f, g begins and ends with a vowel?
11. Given six distinct pairs of gloves, 12 distinct gloves in all, how many ways are there to distribute two gloves to each of six sisters (a) If the two gloves someone receives might both be for the left hand or right hand?
(b) If each sister gets one left-hand glove and one right-hand glove?
12. How many ways are there to partition 12 people into: (a) Three groups of sizes two, four, and six?
(b) Two (unordered) groups of size six?
13. How many 7-letter sequences (formed from the 26 letters in the alphabet, with repetition allowed) contain exactly one A and exactly two Bs?
14. (a) On a 10-question test, how many ways are there to answer exactly eight questions correctly?
(b) Repeat part (a) with the requirement that the first or second question, but not both, are answered correctly.
(c) Repeat part (a) in the case that three of the first five questions are answered correctly.
15. How many n-digit ternary (0, 1, 2) sequences with exactly nine 0s are there?
Comment on the answers (a) 3n−9, (b) 2n−9, and (c) C(n, 8)3n−9.
16. What is the probability that a five-card poker hand has the following?
(a) Four Aces (b) Four of a kind (c) Two pairs (not four of a kind or a full house) (d) A full house (three of a kind and a pair) (e) A straight (a set of five consecutive values) (f) No pairs (possibly a straight or flush) 17. What is the probability that a randomly generated n-digit ternary sequence has exactly k 0s?
200 Chapter 5 General Counting Methods for Arrangements and Selections 18. What is the probability that a randomly chosen 10-card hand has exactly three three-of-a-kinds (and no four-of-a-kinds)?
19. How many 10-letter sequences (formed from the 26 letters in the alphabet, with repetition allowed) contain exactly m Rs and exactly n Ts?
20. Of a company’s personnel, seven work in design, 14 in manufacturing, four in testing, five in sales, two in accounting, and three in marketing. A committee of six people is to be formed to meet with upper management.
(a) In how many ways can the committee be formed if there must be exactly two members from the manufacturing department?
(b) Lucy works in the design department and her husband Ricky works in marketing. In how many ways can the committee be formed if they cannot both be on the committee together?
(c) In how many ways can the committee be formed if there must be at least two members from the manufacturing department?
21. How many ways can a committee be formed from four men and six women with (a) At least two men and at least twice as many women as men?
(b) Between three and five people, and Ms. Wonder is excluded?
(c) Five people, and not all of the three O’Hara sisters can be on the committee?
(d) Four members, at least two of whom are women, and Mr. and Mrs. Baggins cannot both be chosen?
22. Suppose a class of 50 students has 20 males and 30 females. The instructor will pick one (different) student to compete in three different national competitions— in mathematics, chemistry, and English.
(a) What is the probability that there is exactly one female student selected?
(b) What is the probability that there are at least two male students selected?
23. Suppose that campus telephone numbers consist of any four digits (repetition allowed).
(a) What is the probability that the digit 6 appears at least twice in a campus telephone number?
(b) What is the probability that a campus telephone number contains exactly two different digits (e.g., 2444)?
(c) What is the probability that a campus telephone number consists of four distinct digits in ascending order (e.g., 2578)?
24. There are eight applicants for the job of dog catcher and three different judges who each rank the applicants. Applicants are chosen if and only if they appear in the top three in all three rankings.
(a) How many ways can the three judges produce their three rankings?
(b) What is the probability of Mr. Dickens, one of the applicants, being chosen in a random set of three rankings?
www.itpub.net 5.2 Simple Arrangements and Selections 201 25. There are six different French books, eight different Russian books, and five different Spanish books. How many ways are there to arrange the books in a row on a shelf with all books of the same language grouped together?
26. If 13 players are each dealt four cards from a 52-card deck, what is the probability that each player gets one card of each suit?
27. How many 10-letter (sequences) are there using five different vowels and five different consonants (chosen from the 21 possible consonants)? What is the probability that one of these words has no consecutive pair of consonants?
Comment on the answers (a) 3(n−1)!, (b) (3!(n−3)!)3, and (c) 3((n−1)!)3.
28. What is the probability that a random 9-digit Social Security number has at least one repeated digit?
29. What is the probability that an arrangement of a, b, c, d, e, f has (a) a and b side-by-side?
(b) a occurring somewhere before b?
30. How many ways are there to pair off 10 women at a dance with 10 out of 20 available men?
31. How many triangles are formed by (assuming no three lines cross at a point) (a) Pieces of n nonparallel lines; for example, the four lines below form four triangles: acd, abf, efd, and ebc?
(b) Pieces of n lines, m of which are parallel and the others mutually nonparallel?
a b c d f e 32. There are three women and five men who will split up into two four-person teams. How many ways are there to do this so that there is (at least) one woman on each team?
33. A man has n friends and invites a different subset of four of them to his house every night for a year (365 nights). How large must n be?
34. How many arrangements of JUPITER are there with the vowels occurring in alphabetic order?
Comment on the answers (a) 3!4!, (b) 7!/3!, and (c) C(7, 3)4!
35. Determine the Shapley–Shubik index of a 1-vote person and a 2-vote person in the councils with the following make up: (a) Three 1-vote people and one 2-vote person.
(b) Two 1-vote people and two 2-vote people (majority is four).
202 Chapter 5 General Counting Methods for Arrangements and Selections (c) Four 1-vote people and three 2-vote people (majority is six).
(d) Four 1-vote people and four 2-vote people (majority is seven).
36. How many 7-card hands dealt are there with three pairs (each of a different kind, plus a seventh card of a different kind)?
37. How many arrangements of the 26 letters of the alphabet in which (a) a occurs before b?
(b) a occurs before b and c occurs before d?
(c) the five vowels appear in alphabetical order?
38. A student must answer five out of 10 questions on a test, including at least two of the first five questions. How many subsets of five questions can be answered?
39. How many 6-letter sequences are there with at least three vowels (A, E, I, O, U)? No repetitions are allowed.
40. How many arrangements of 1, 1, 1, 1, 2, 3, 3 are there with the 2 not beside either 3?
41. How many arrangements of INSTRUCTOR are there in which there are exactly two consonants between successive pairs of vowels?
42. How many “words” can be formed by rearranging INQUISITIVE so that U does not immediately follow Q?
43. Suppose a subset of 60 different days of the year are selected at random in a lottery. What is the probability that there are five days from each month in the subset? (For simplicity, assume a year has 12 months with 30 days each.) 44. Suppose a subset of eight different days of the year is selected at random.
What is the probability that each day is from a different month? (For simplicity, assume a year has 12 months with 30 days each).
45. What is the probability of randomly choosing a permutation of the 10 digits 0, 1, 2, . . . , 9 in which (a) An odd digit is in the first position and 1, 2, 3, 4, or 5 is in the last position?
(b) 5 is not in the first position and 9 is not in the last position?
46. (a) What is the probability that a 5-card hand has at least one card of each suit?
(b) Repeat for a 6-card hand.
47. What is the probability that a 5-card hand has (a) At least one of each of the four values: Ace, King, Queen, and Jack?
(b) The same number of hearts and spades?
48. How many n-digit decimal (0, 1, 2) sequences are there with k 1s?
49. What fraction of all arrangements of INSTRUCTOR have: (a) Three consecutive vowels?
(b) Two consecutive vowels?
www.itpub.net 5.2 Simple Arrangements and Selections 203 50. If one quarter of all 3-subsets of the integers 1, 2, . . . , m contain the integer 5, determine m.
51. How many ways are there to form an (unordered) collection of four pairs of two people chosen from a group of 30 people?
52. What fraction of all arrangements of GRACEFUL have: (a) F and G appearing side by side?
(b) No pair of consecutive vowels?
53. How many arrangements of SYSTEMATIC are there in which each S is followed by a vowel (this includes Y)?
54. How many arrangements of MATHEMATICS are there in which each consonant is adjacent to a vowel?
55. (a) What is the probability that k is the smallest integer in a subset of four different numbers chosen from 1 through 20 (1 ≤k ≤17)?
(b) What is the probability that k is the second smallest?
56. What is the probability that the difference between the largest and the smallest numbers is k in a subset of four different numbers chosen from 1 through 20 (3 ≤k ≤19)?
57. How many ways are there to place eight identical black pieces and eight identical white pieces on an 8 × 8 chessboard?
58. How many 8-letter arrangements can be formed from the 26 letters of the alphabet (without repetition) that include at most three of the five vowels and in which the vowels appear in alphabetical order? (Hint: Break into cases.) 59. What is the probability that two (or more) people in a random group of 25 people have a common birthday? (This is the famous Birthday Paradox Problem.) 60. How many ways are there to arrange n (distinct) people so that (i) Mr. and Mrs.
Smith are side by side; and (ii) Mrs. Tucker is k positions away from the Smiths (i.e., k −1 people between Mrs. Tucker and the Smiths)?
61. A basketball team has five players, three in ‘‘forward” positions (this includes the ‘‘center”) and two in ‘‘guard” positions. How many ways are there to pick a team if there are six forwards, four guards, and two people who can play forward or guard?
62. A family has two boys and three girls to send to private schools. There are five boys’ schools, eight girls’ schools, and three coed schools. If each child goes to a different school, how many different subsets of five schools can the family choose for their children?
63. Given a collection of 2n objects, n identical and the other n all distinct, how many different subcollections of n objects are there?
64. A batch of 50 different automatic typewriters contains exactly 10 defective machines. What is the probability of finding (a) At least one defective machine in a random group of five machines?
(b) At least two defective machines in a random group of 10 machines?
204 Chapter 5 General Counting Methods for Arrangements and Selections (c) The first defective machine to be the kth machine taken apart for inspection in a random sequence of machines?
(d) The last defective machine to be the kth machine taken apart?
65. Ten fish are caught in a lake, marked, and then returned to the lake. Two days later 20 fish are again caught, two of which have been marked.
(a) Find the probability of two of the 20 fish being marked if the lake has k fish (assuming the fish are caught at random).
(b) What value of k maximizes the probability?
66. Professor Grinch’s telephone number is 6328363. Mickey remembers the collection of digits but not their order, except that he knows the first 6 is before the first 3. How many arrangements of these digits with this constraint are there?
67. How many arrangements of the letters in PREPOSTEROUS are there in which the first vowel to appear is an E?
68. For the given map of roads between city A and city B, A B (a) How many routes are there from A to B that do not repeat any road (a road is a line segment between two intersections)?
(b) How many ways are there for two people to go from A to B without both ever traversing the same road in the same direction?
69. What is the probability that a random five-card hand has (a) Exactly one pair (no three of a kind or two pairs)?
Comment on the answers (a) 13 × C(4, 2) × C(48, 3)/C(52, 5), (b) 13 × C(4, 2) × 48 × 44 × 40/C(52, 5), and (c) (52 × 3 × 48 × 44 × 40)/ (52 × 51 × 50 × 49 × 48).
(b) One pair or more (three of a kind, two pairs, four of a kind, full house)?
(c) The cards dealt in order of decreasing value?
(d) At least one spade, at least one heart, no diamonds or clubs, and values of all spades greater than the values of all hearts?
70. How many ways are there to pick a group of n people from 100 people (each of a different height) and then pick a second group of m other people such that all people in the first group are taller than the people in the second group?
71. How many subsets of three different integers between 1 and 90 inclusive are there whose sum is (a) An even number?
(b) Divisible by 3?
(c) Divisible by 4?
www.itpub.net 5.2 Simple Arrangements and Selections 205 72. How many arrangements are there of the letters in REPETITIVENESS such that the Ss are consecutive AND the first I comes (somewhere) before the first E ?
73. How many arrangements are there of the letters in STATISTICIANS such that the Is are consecutive and the first S comes (somewhere) before the first T?
74. How many arrangements of the letters in MATHEMATICS are there such that the last vowel in the arrangement is an I.
75. In a class of 20 students, one student is chosen president, another one is chosen secretary, and another is chosen treasurer. Such elections are held each week for six weeks. A student can be elected to one of the positions more than once. How many sequences of six election outcomes are possible?
76. Sweaters are made in a design with three bands of colors: a top color, a middle color, and a bottom color. Eight colors are available and a color can appear in only one level. How many subsets of 12 different sweaters are there?
77. Each Saturday for 12 weeks, a different pair is chosen from eight senior citizens to play a chess match. How many sequences of 12 different pairs are possible?
78. How many arrangements of PEPPERMILL are there in which MP appear con-secutively or LP appear consecutively but not both MP and LP are consecutive?
79. How many arrangements of UNUSUALLY are there in which SU appear con-secutively or LU appear consecutively but not both SU and LU are consecutive?
80. How many ways are there for a woman to invite different subsets of three of her five friends on three successive days? How many ways if she has n friends?
81. How many triangles can be formed by joining different sets of three corners of an octagon? How many triangles if no pair of adjacent corners is permitted?
82. How many arrangements of five 0s and ten 1s are there with no pair of consecutive 0s?
83. (a) How many points of intersection are formed by the chords of an n-gon (assuming no three of these lines cross at one point)?
(b) Into how many line segments are the lines in part (a) cut by the intersection points?
(c) Use Euler’s formula r = e −v + 2 and parts (a) and (b) to determine the number of regions formed by the chords of an n-gon.
84. Given 14 positive integers, 12 of which are even, and 16 negative integers, 11 of which are even, how many ways are there to pick 12 numbers from this collection of 30 integers such that six of the 12 numbers are positive and six of the 12 numbers are even?
85. How many triangles are formed by (assuming no three lines cross at a point): (a) Pieces of three chords of a convex 10-gon such that the triangles are wholly within the 10-gon (a corner of the 10-gon cannot be a corner of any of these triangles)?
(b) Pieces of three chords or outside edges of a convex n-gon?
206 Chapter 5 General Counting Methods for Arrangements and Selections 86. How many arrangements of the integers 1, 2, . . . , n are there such that each integer differs by 1 (except the first integer) from some integer to the left of it in the arrangement?
87. A man has seven friends. How many ways are there to invite a different subset of three of these friends for a dinner on seven successive nights such that each pair of friends are together at just one dinner?
5.3 ARRANGEMENTS AND SELECTIONS WITH REPETITIONS In this section we discuss arrangements and selections with repetition—arrangements ofacollectionwithrepeatedobjects,suchasthecollectionb,a,n,a,n,a,andselections when an object can be chosen more than once, such as ordering six hot dogs chosen from three varieties. We motivate the formulas for these counting problems with the two examples just mentioned.
Example 1: Arrangements of banana How many arrangements are there of the six letters b, a, n, a, n, a?
Consider a possible arrangement: n a b n a a −−−−−− This problem is solved by an extension of reasoning used to solve the problem of counting 8-digit binary sequences with six 1s (Example 3 in Section 5.2). The key is to focus on the subset of positions where the as go and the subset of positions where the ns go. For example, the above arrangement is characterized by having as in positions 2, 5, 6, ns in positions 1, 4, and b in position 3. We count the arrangements by first choosing the subset of three positions in the arrangement where the as will go—C(6, 3) = 20 ways—then the subset of two positions (out of the remaining three) where the ns will go—C(3, 2) = three ways—and finally the last remaining position gets the b—1 way. Thus there are C(6, 3) × C(3, 2) × C(1, 1) = 20 × 3 × 1 = 60 arrangements.
Theorem 1 If there are n objects, with r1 of type 1, r2 of type 2, . . . , and rm of type m, where r1 +r2 + · · · +rm = n, then the number of arrangements of these n objects, denoted P(n; r1, r2, . . . , rm), is P(n;r1,r2, . . . ,rm) = n r1 n −r1 r2 n −r1 −r2 r3 · · · n −r1 −r2 · · · −rm−1 rm = n!
r1!r2! . . .rm!
() www.itpub.net 5.3 Arrangements and Selections with Repetitions 207 Proof 1 First pick r1 positions for the first types, then r2 of the remaining positions for the second types, and so on. A mathematically precise proof of the “etc.” part requires induction (see Exercise 35). The second line of () is just a simplification that results from canceling factorials in the binomial coefficients; for example, P(6; 3, 2, 1) = 6 3 3 2 1 1 = 6!
3!3! × 3!
2!1! × 1!
1! = 6!
3!2!1!
◆ Proof 2 This proof is similar to our derivation of C(n, r) through the equation P(n, r) = C(n, r) × P(r, r). Suppose that for each type, the ri objects of type i are given subscripts numbered 1, 2, . . . , ri to make each object distinct. Then there are n! arrangements of the n distinct objects. Let us enumerate these n! arrangements of distinct objects by enumerating all P(n; r1, r2, . . . , rm) patterns (without subscripts) of the objects, and then for each pattern placing the subscripts in all possible ways. For example, the pattern baanna can have subscripts on as placed in the 3! ways: ba1a2nna3 ba2a1nna3 ba3a1nna2 ba3a2nna1 ba1a3nna2 ba2a3nna1 For each of these 3! ways to subscript the as, there are 2! ways to subscript the ns. Thus, in general a pattern will have r1! ways to subscript the r1 objects of type 1, r2! ways for type 2, and rm! ways for type m. Then n! = P(n;r1,r2, . . . ,rm)r1!r2! . . .rm!
or P(n;r1,r2, . . . ,rm) = n!
r1!r2! . . .rm!
◆ Example 2: Ordering Hot Dogs How many different ways are there to select six hot dogs from three varieties of hot dog?
To solve such a selection-with-repetition problem, we recast it as an arrangement problems as follows. Suppose the three varieties are regular dog, chili dog, and super dog. Let a selection be written down on an order form (when a person places this order) in the following fashion: Regular Chili Super x xxxx x Each x represents a hot dog. The request shown on the form above is one regular, four chili, and one super. Since the hot dog chef knows that the sequence of dogs on 208 Chapter 5 General Counting Methods for Arrangements and Selections the form is regular, chili, super, the request can simply be written as x|xxxx|x without column headings.
Any selection of r hot dogs will consist of some sequence of r xs and two |s.
Conversely, any sequence of r xs and two |s represents a unique selection: the xs before the first | count the number of regular dogs; the xs between the two |s count chilis; and the final xs count supers. So there is a one-to-one correspondence between orders and such sequences. Counting the number of sequences of six xs and two |s is an arrangement-with-repetition problem whose answer is P(8; 6, 2) = 8 6 2 2 = 8!
6!2!.
As discussed in Example 3 of Section 5.2, counting such sequences of xs and |s is simply a matter of picking the subset of positions where the xs (or the |s) go—again, 8 6 ways.
Theorem 2 The number of selections with repetition of r objects chosen from n types of objects is C(r + n −1,r).
Proof We make an “order form” for a selection just as in Example 2, with an x for each object selected. As before, the xs before the first | count the number of the first type of object, the xs between the first and second |s count the number of the second type, . . . , and the xs after the (n −1)-st | count the number of the nth type (n −1 slashes are needed to separate n types). The number of sequences with r xs and n −1 |s is C(r + (n −1),r). ◆ Example 3: Grouping Classes Nine students, three from Ms. A’s class, three from Mr. B’s class, and three from Ms. C’s class, have bought a block of nine seats for their school’s homecoming game.
If three seats are randomly selected for each class from the nine seats in a row, what is the probability that the three A students, three B students, and three C students will each get a block of three consecutive seats?
In probability problems, we seek the number of favorable outcomes divided by all outcomes. The first question is, what is the set of all outcomes in this problem?
They are the P(9; 3, 3, 3) = 9!/3!3!3! = 1680 ways to arrange three As, three Bs and three Cs in the row of nine seats.
If the three students of each class are to sit together, then we want to count outcomes that are arrangements of the three blocks, AAA, BBB, and CCC. Thus, instead of nine letters, we really are working now with three composite letters. There are 3! = 6 ways to arrange these three composite letters. So the probability that each class sits together is 6/1680.
Example 4: Sequencing Genes The genetic code of organisms is stored in DNA molecules as a long string of four nucleotides: A (adenine), C (cytosine), G (guanine), and T (thymine). Short strings of www.itpub.net 5.3 Arrangements and Selections with Repetitions 209 DNA can be “sequenced”—the sequence of letters determined—by various modern biotech methods. Although the DNA sequence for a single gene typically has hundreds or thousands of letters,thereexistspecialenzymesthatwillsplitalongstringintoshort fragments (which can be sequenced) by breaking the string immediately following each appearance of a particular letter.
Suppose a C-enzyme (which splits after each appearance of C) breaks a 20-letter string into eight fragments, which are identified to be: AC, AC, AAATC, C, C, C, TATA, TGGC. Note that each fragment, except the last one on the string, must end with a C. How many different strings could have given rise to this set of fragments?
Since the fragment TATA does not end with a C, it must go at the end of the string. The other seven fragments can occur in any order. These fragments consist of three Cs, two ACs, one AAATC, and one TGGC. Treat each fragment as a letter, similar to the reasoning in Example 3. Then we must arrange seven letters, three of one type, two of a second type, and one each of two other types. There are thus P(7; 3, 2, 1, 1) = 420 possible arrangements of the fragments to form a 20-letter string.
If we use an A-enzyme to break the same string into fragments and look at all the possible arrangements of these fragments and then do the same with a G-enzyme and a T-enzyme, there will normally be only one string that appears in all four sets of arrangements. This will be the true original string. More sophisticated variations on this approach are used to determine the DNA sequence of entire genes.
Example 5: Sequences with Varying Repetitions How many ways are there to form a sequence of 10 letters from four as, four bs, four cs, and four ds if each letter must appear at least twice?
To apply Theorem 1 we need to know exactly how many as, bs, cs, and ds will be in the arrangement. Thus we have to break this problem into a set of subproblems that each involves sequences with given numbers of as, bs, cs, and ds. There are two categories of letter frequencies that sum to 10 with each letter appearing two or more times. The first category is four appearances of one letter and two appearances of each other letter. The second is three appearances of two letters and two appearances of the other two letters.
In the first category, there are four cases for choosing which letter occurs four times and P(10; 4, 2, 2, 2) = 18,900 ways to arrange four of one letter and two of the three others. In the second category, there are C(4, 2) = 6 cases for choosing which two of the four letters occur three times and P(10; 3, 3, 2, 2) = 25,200 ways to arrange three of two letters and two of the two others. The final answer is 4 × 18,900 + 6 × 25,200 = 226,800 ways.
Example 6: Selecting Doughnuts How many ways are there to fill a box with a dozen doughnuts chosen from five different varieties with the requirement that at least one doughnut of each variety is picked?
210 Chapter 5 General Counting Methods for Arrangements and Selections The starting question to solve this problem is: how do we recast selection with at least one of each kind in terms of an unconstrained selection with repetition or some other counting problem we know how to solve? Handling this constraint involves an insight that is hard to find but follows immediately from the way people make such a doughnut selection. They would first pick one doughnut of each variety and then pick the remaining seven doughnuts any way they pleased. There is no choice (only one way) in picking one doughnut of each type. The choice occurs in picking the remaining seven doughnuts from the five types. Think of placing one doughnut of each type in a box and then covering them with a sheet of waxed paper, and then choosing the remaining seven doughnuts. The variety in outcomes involves only the seven doughnuts chosen to go on top of the waxed paper. So the answer by Theorem 2 is C(7 + 5 −1, 7) = 330. (Note that the set composition principle does not apply here, because the objects are not all distinct.) Note that in the selection with repetition, we are only concerned with counting how many of each type we have. With this objective, it is easy to handle lower bounds for the types (as in Example 6), since this constraint recasts the problem into counting how many more than the lower bound we have of each type. However, in arrangement with repetition, lower bound constraints are much more complex. For each outcome in the selection with repetition problem—that is, when we consider specific amounts of each type, we have a subproblem of counting how many ways there are to arrange the items in this selection.
Example 7: Selections with Lower and Upper Bounds How many ways are there to pick a collection of exactly 10 balls from a pile of red balls, blue balls, and purple balls if there must be at least five red balls? If at most five red balls?
ThefirstproblemissimilartoExample6.Weputfiveredballsinabox,coverthem with waxed paper, and then select five more balls arbitrarily (possibly including more red balls). The five balls above the waxed paper can be chosen in C(5 + 3 −1, 5) = 21 ways.
There is no direct way to count selections when there is an upper bound of the number of repetitions of some object. To handle the constraint of at most five red balls, we count the complementary set. Of all C(10 + 3 −1, 10) = 66 ways to pick 10 balls from the three colors without restriction, there are C(4 + 3 −1, 4) = 15 ways to choose a collection with at least six red balls (put six red balls in a box, cover with waxed paper, and then arbitrarily choose four more balls). So there are 66 −15 = 51 ways to choose 10 balls without more than five red balls.
Example 8: Arrangements with Restricted Positions We return to Example 1 about arrangements of the letters in banana, but now with some constraints of the sort encountered in Section 5.2. How many arrangements are there of the letters b, a, n, a, n, a such that: www.itpub.net 5.3 Arrangements and Selections with Repetitions 211 (a) The b is followed (immediately) by an a: We use the method for counting ar-rangements with consecutive letters introduced in Example 2 of Section 5.2; that is, we glue the b and one of the as together to form the multiletter ba. Now we want to count all arrangements of the five “letters”: ba, a, a, n, n. By Theorem 1, there are 5!/1!2!2! = 30 arrangements.
(b) The pattern bnn never occurs: We solve this problem by counting all arrangements of b, a, n, a, n, a without constraint and then subtracting off the arrangements with the pattern bnn. Arrangements with this pattern are handled the same way as the pattern ba in part (a). We treat bnn as a single multiletter and now count arrangements of the four letters bnn, a, a, a—4!/1!3! = 4 ways. Subtracting the four forbidden arrangements from all 6!/1!3!2! = 60 arrangements yields the answer, 60 −4 = 56.
(c) The b occurs before any of the as (not necessarily immediately before an a): in other words, the relative order of the b and 3 as is b–a–a–a. We use the method of handling relative order introduced in the continuation of Example 2 in Section 5.2. We pick a subset of four positions (for these four letters) from the six positions in an arrangement—C(6, 4) = 15 ways. We put the b, a, a, a in these four positions in the required relative order—one way. Next we fill the two remaining positions in the arrangement with the two ns—one way. So the final answer is 15 × 1 × 1 = 15.
5.3 EXERCISES S u m m a r y o f E x e r c i s e s Most of these problems are not-too-difficult variations on the section’s examples. Be careful to distinguish between arrangements and subsets problems.
1. How many ways are there to roll a die seven times and obtain a sequence of outcomes with three 1s, two 5s, and two 6s?
2. How many ways are there to arrange the letters in STATISTICAL?
3. (a) How many 8-digit numbers can be formed with the digits 3, 5, and 7?
(b) What fraction of the numbers in part (a) have three 3s, two 5s, and three 7s?
4. How many ways are there to invite one of four different friends over for dinner on five successive nights such that no friend is invited more than three times?
5. How many ways are there to pick a collection of nine coins from piles of pennies, nickels, dimes, and quarters?
6. If four identical dice are rolled, how many different outcomes can be recorded?
7. How many ways are there to select a committee of 17 politicians chosen from a room full of indistinguishable Democrats, indistinguishable Republicans, and indistinguishable Independents if every party must have at least two members on the committee? If, in addition, no group may have a majority of the committee members?
212 Chapter 5 General Counting Methods for Arrangements and Selections 8. Ten different people walk into a delicatessen to buy a sandwich. Four always order tuna fish, two always order chicken, two always order roast beef, and two order any of the three types of sandwich.
(a) How many different sequences of sandwiches are possible?
(b) How many different (unordered) collections of sandwiches are possible?
9. How many ways are there to pick a selection of coins from $1 worth of identical pennies, $1 worth of identical nickels, and $1 worth of identical dimes if (a) You select a total of 9 coins?
(b) You select a total of 16 coins?
Comment on the answers: (a) 16 + 3 −1 16 , (b) 10 k=0 16 k (16 −k) + 2 −1 (16 −k) , and (c) 16 + 3 −1 16 − 16 k=11 16 k (16 −k) + 2 −1 (16 −k) 10. (a) How many ways are there to distribute seven identical apples and six identical pears to three distinct people such that each person has at least one pear?
(b) How many ways are there to distribute seven distinct applies and six distinct pears to three distinct people such that each person has at least one pear?
11. How many ways are there to have a collection of eight fruits from a large pile of identical oranges, apples, bananas, peaches, and pears if the collection should include exactly two different kinds of fruits?
12. How many ways are there to pick nine balls from large piles of (identical) red, white, and blue balls plus one pink ball, one lavender ball, and one tan ball?
13. How many numbers greater than 3,000,000 can be formed by arrangements of 1, 2, 2, 4, 6, 6, 6?
14. How many different rth-order partial derivatives does f (x1, x2, . . . , xn) have?
15. How many 8-digit sequences are there involving exactly six different digits?
16. How many 9-digit numbers are there with twice as many different odd digits involved as different even digits (e.g., 945222123 with 9, 3, 5, 1 odd and 2, 4 even).
17. How many arrangements are possible with five letters chosen from MISSISSIPPI?
18. How many ways are there to select an unordered group of eight numbers between 1 and 25 inclusive with repetition? In what fraction of these ways is the sum of these numbers even?
19. How many arrangements of letters in REPETITION are there with the first E occurring before the first T?
www.itpub.net 5.3 Arrangements and Selections with Repetitions 213 20. In an international track competition, there are five United States athletes, four Russian athletes, three French athletes, and one German athlete. How many rankings of the 13 athletes are there when (a) Only nationality is counted?
(b) Only nationality is counted and all the U.S. athletes finish ahead of all the Russian athletes?
21. How many arrangements of the letters in MATHEMATICS are there in which TH appear together but the TH is not immediately followed by an E (not THE)?
22. How many arrangements of the letters in PEPPERMILL are there with (a) The M appearing to the left of all the vowels?
(b) The first P appearing before the first L?
23. How many arrangements of the letters in MISSISSIPPI in which (a) The M is followed (immediately) by an I?
(b) The M is beside an I—that is, an I is just before or just after the M. Possibly there is an I both just before and just after the M—special care is required to make sure you count arrangements with the pattern IMI correctly.
24. How many arrangements of PREPOSTEROUS are there in which the five vowels are consecutive?
25. How many ways to select a subset of eight doughnuts from three types of dough-nuts if at most two doughnuts of the first type can be chosen?
26. How many sequences of outcomes are possible if one rolls two identical dice 10 successive times?
27. How many ways are there to split a group of 2n αs, 2n βs, and 2n γ s in half (into two groups of 3n letters)? (Note: The halves are unordered; there is no first half.) 28. How many ways are there to place nine different rings on the four fingers of your right hand (excluding the thumb) if (a) The order of rings on a finger does not matter?
(b) The order of rings on a finger is considered? (Hint: Tricky.) 29. Show that P(10; k1, k2, k3) = 310, where k1, k2, k3 are nonnegative integers ranging over all possible triples such that k1 + k2 + k3 = 10.
30. How many arrangements of six 0s, five 1s, and four 2s are there in which (a) The first 0 precedes the first 1?
(b) The first 0 precedes the first 1, which precedes the first 2?
31. How many arrangements are there of 4n letters, four of each of n types of letters, in which each letter is beside a similar letter?
32. How many ways are there for 10 people to have five simultaneous telephone conversations?
214 Chapter 5 General Counting Methods for Arrangements and Selections 33. When a coin is flipped n times, what is the probability that (a) The first head comes after exactly m tails?
(b) The ith head comes after exactly m tails?
34. How many arrangements are there of TINKERER with two but not three consecutive vowels?
35. How many arrangements are there of seven as, eight bs, three cs, and six ds with no occurrence of the consecutive pairs ca or cc?
36. How many arrangements of five αs, five βs, and five γ s are there with at least one β and at least one γ between each successive pair of αs?
37. (a) Use induction to give a rigorous proof of Theorem 1 (Proof 1).
(b) Use induction to prove that n r1 n −r1 r2 n −r1 −r2 r3 · · · · · n −r1 −r2 · · · · · −rm−1 rm = n!
r1!r2! · · · .rm!
5.4 DISTRIBUTIONS Generally a distribution problem is equivalent to an arrangement or selection problem with repetition. Specialized distribution problems must be broken up into subcases that can be counted in terms of simple permutations and combinations (with and without repetition). General guidelines for modeling distribution problems are Distributions of distinct objects are equivalent to arrangements and Distributions of identical objects are equivalent to selections Basic Models for Distributions Distinct Objects The process of distributing r distinct objects into n different boxes is equivalent to putting the distinct objects in a row and stamping one of the n different box names on each object. The resulting sequence of box names is an arrangement of length r formed from n items (box names) with repeti-tion. Thus there are n × n × · · ·× n (r ns) = nr distributions of the r distinct www.itpub.net 5.4 Distributions 215 objects. If ri objects must go in box i, 1 ≤i ≤n, then there are P(r; r1, r2, . . . , rn) distributions.
Identical Objects The process of distributing r identical objects into n differ-ent boxes is equivalent to choosing an (unordered) subset of r box names with repetition from among the n choices of boxes. Thus there are C(r + n −1,r) = (r + n −1)!/r!(n −1)! distributions of the r identical objects.
Example 1: Assigning Diplomats Howmanywaysaretheretoassign100differentdiplomatstofivedifferentcontinents?
How many ways if 20 diplomats must be assigned to each continent?
Accordingtothemodelfordistributionsofdistinctobjects,thisassignmentequals the number of sequences of length 100 involving the five continental destinations— 5100 such sequences (think of the diplomats lined up in a row holding attach´ e cases stamped with their destination). The constraint that 20 diplomats go to each continent means that each continent name should appear 20 times in the sequence. This can be done P(100; 20, 20, 20, 20, 20) = 100!/(20!)5 ways.
Example 2: Bridge Hands In bridge, the 52 cards of a standard card deck are randomly dealt 13 apiece to players North, East, South, and West. What is the probability that West has all 13 spades?
That each player has one Ace?
There are P(52; 13, 13, 13, 13) distributions of the 52 cards into four different 13-card hands, using the same reasoning as in the preceding example. Distributions in which West gets all the spades may be counted as the ways to distribute to West all the spades—one way—times the ways to distribute the 39 non-spade cards among the three other hands—P(39; 13, 13, 13) ways. So the probability that West has all the spades is 39!
(13!)3 52!
(13!)4 = 1 52!
13!39! = 1 52 13 The simple form of this answer can be directly obtained by considering West’s possible hands alone (ignoring the other three hands). A random deal gives West one of the C(52, 13) possible 13-card hands. Thus, the unique hand of 13 spades has probability 1/C(52, 13).
To count the ways in which each player gets one Ace, we divide the distribution up into an Ace part—four! ways to arrange the four Aces among the four players—and a non-Ace part—P(48; 12, 12, 12, 12) ways to distribute the remaining 48 non-Ace cards, 12 to each player. So the probability that each player gets an Ace is 4!48!
(12!)4 52!
(13!)4 = 13!4 12!4 × 4!48!
52! = 134 52 4 = 0.105 216 Chapter 5 General Counting Methods for Arrangements and Selections Example 3: Distributing Candy How many ways are there to distribute 20 (identical) sticks of red licorice and 15 (identical) sticks of black licorice among five children?
Using the identical objects model for distributions, we see that the ways to dis-tribute 20 identical sticks of red licorice among five children is equal to the ways to select a collection of 20 names (destinations) from a set of five different names with repetition. This can be done C(20 + 5 −1,20) = 10,626 ways. By the same modeling argument, the 15 identical sticks of black licorice can be distributed in C(15 + 5 −1,15) = 3876 ways. The distributions of red and of black licorice are disjoint procedures. The multiplication principle applies and so the number of ways to distribute red and black licorice is 10,626 × 3876 = 41,186,376.
Example 4: Distributing a Combination of Identical and Distinct Objects How many ways are there to distribute four identical oranges and six distinct apples (each a different variety) into five distinct boxes? In what fraction of these distributions does each box get exactly two objects?
There are C(4 + 5 −1, 4) = 70 ways to distribute four identical oranges into five distinct boxes and 56 = 15,625 ways to put six distinct apples in five distinct boxes.
These two processes are disjoint, and so there are 70 × 15,625 = 1,093,750 ways to distribute the four identical and six distinct fruits.
The additional constraint of two objects in each box substantially complicates matters. To count constrained distributions of distinct objects, we must know exactly how many of the distinct objects should go in each box. But in this problem, the number of distinct objects that can go in a box depends on how many identical objects are in the box. So we deal with the (identical) oranges first. There are three possible categories of distributions of the four oranges (without exceeding two in any box).
Case 1 Two (identical) oranges in each of two boxes and no oranges in the other three boxes. The two boxes to get the pair of oranges can be chosen in C(5, 2) = 10 ways, and the six (distinct) apples can then be distributed in those three other boxes, two to a box, in P(6; 2, 2, 2) = 90 ways. So Case 1 has 10 × 90 = 900 possible distributions.
Case 2 Two (identical) oranges in one box, one orange in each of two other boxes, and the remaining two boxes empty. The one box with two oranges can be chosen C(5, 1) = 5 ways, and the two boxes with one orange can be chosen in C(4, 2) = 6 ways [or combining these two steps, we can think of arranging the num-bers 2, 1, 1, 0, 0, among the five boxes in P(5; 1, 2, 2) = 30 ways]. Then the two boxes still empty will each get two apples and the two boxes with one orange will get one apple. Thus, six distinct apples can then be distributed in P(6; 2, 2, 1, 1) = 180 ways, for a total of 5 × 6 × 180 = 5400 ways.
www.itpub.net 5.4 Distributions 217 Case 3 One orange each in four of the five boxes. This can be done in C(5, 4) = five ways, and then the apples can be distributed in P(6; 2, 1, 1, 1, 1) = 360 ways, for a total of 5 × 360 = 1800 ways.
Summing these cases, there are 900 + 5400 + 1800 = 8100 distributions with two objects in each box. The fraction of such distributions among all ways to distribute the four oranges and six apples is 8100/1,093,750 = .0074—a lot smaller than one might guess. (If all 10 fruits were distinct, the fraction with two in each box would be about 50 percent larger. Why?) Example 5: Distributing Balls Show that the number of ways to distribute r identical balls into n distinct boxes with at least one ball in each box is C(r −1, n −1). With at least r1 balls in the first box, at least r2 balls in the second box, . . . , and at least rn balls in the nth box, the number is C(r −r1 −r2 −· · · −rn + n −1, n −1).
The requirement of at least one ball in each box can be incorporated into an equivalent selection-with-repetition model (as in the doughnut selection problem in Section 5.3). One could form such a collection of box destinations by putting one label for each of the n boxes on a tray, covering these labels with a piece of waxed paper, and then picking the remaining r −n labels in all possible ways.
Alternatively, we can handle this constraint directly in terms of the boxes as follows. First put one ball in each box and then put a false bottom in the boxes to conceal the ball in each box. Now it remains to count the ways to distribute without restriction the remaining r −n balls into the n boxes.
With either approach, the answer is C((r −n) + n −1, (r −n)) = [(r −n) + n −1]!
(r −n)!(n −1)! = C(r −1, n −1) ways In the case where at least ri balls must be in the ith box, we first put ri balls in the ith box, and then distribute the remaining r −r1 −r2 −· · · −rn balls in any way into the n boxes. This can be done in C((r −r1 −r2 −· · · −rn) + n −1, (r −r1 −r2 −· · · −rn)) = C((r −r1 −r2 −· · · −rn) + n −1, n −1) ways The next counting problem will play a critical role in building generating func-tions in the next chapter.
Example 6: Integer Solutions How many integer solutions are there to the equation x1 + x2 + x3 + x4 = 12, with xi ≥0? How many solutions with xi ≥1? How many solutions with x1 ≥2, x2 ≥ 2, x3 ≥4, x4 ≥0?
218 Chapter 5 General Counting Methods for Arrangements and Selections By an integer solution to this equation, we mean an ordered set of integer values for the xis summing to 12, such as x1 = 2, x2 = 3, x3 = 3, x4 = 4. We can model this problem as a distribution-of-identical-objects problem or as a selection-with-repetition problem. Let xi represent the number of (identical) objects in box i or the number of objects of type i chosen. The integer on the right side of the equation is the number of objects to be distributed or selected. Using either of these models, we see that the number of integer solutions is C(12 + 4 −1, 12) = 455.
Solutions with xi ≥1 correspond in these models to putting at least one object in each box or choosing at least one object of each type. Solutions with x1 ≥2, x2 ≥2, x3 ≥4, x4 ≥0 correspond to putting at least two objects in the first box, at least two in the second, at least four in the third, and any number in the fourth (or equivalently in the selection-with-repetition model). Formulas for these two types of distribution problems were given in Example 5. The respective answers are C(12 −1, 4 −1) = 165 and C((12 −2 −2 −4) + 4 −1, 4 −1) = C(7, 3) = 35.
Equations with integer-valued variables are called diophantine equations. They are named after the Greek mathematician Diophantus, who studied them 2,250 years ago.
We have now encountered three equivalent forms for selection with repetition problems.
Equivalent Forms for Selection with Repetition 1.
The number of ways to select r objects with repetition from n different types of objects.
2.
The number of ways to distribute r identical objects into n distinct boxes.
3.
The number of nonnegative integer solutions to x1 + x2 + · · · + xn = r.
It is important that the reader be able to restate a problem given in one of the foregoing settings in the other two. Many students find version 2 the most convenient way to look at such problems because a distribution is easiest to picture on paper (or in one’s head). Indeed, the original argument with order forms for hot dogs used to derive our formula for selection with repetition was really a distribution model.
Version 3 is the most general (abstract) form of the problem. It is the form needed in Chapter 6 to build generating functions.
We next present three problems that we solve by recasting as problems involving distributions of identical or distinct objects.
Example 7: Ingredients for a Witch’s Brew A warlock goes to a store with $5 to buy ingredients for his wife’s Witch’s Brew.
The store sells bat tails for 25c / apiece, lizard claws for 25c / apiece, newt eyes for www.itpub.net 5.4 Distributions 219 25c / apiece, and calf blood for $1 a pint bottle. How many different purchases (subsets) of ingredients will $5 buy?
The first step is to make our unit of money 25c / (a quarter). So the warlock has 20 units to spend, with blood costing 4 units and the other three items each costing 1 unit. An integer-solutions-of-equation model of this problem is T + S + E + 4B = 20, T, S, E, B ≥0 The simplest way to handle the fact that B has a coefficient of 4 is to break into cases by specifying exactly how much blood is bought. If 1 pint is bought, B = 1, then we have T + S + E = 16,anequationwithC(16 + 3 −1, 16) = 153 nonnegative integer solutions. In general, if B = i, then we have T + S + E = 20 −4i, an equation with C((20 −4i) + 3 −1, 20 −4i) = C(22 −4i, 2) solutions, for i = 0, 1, 2, 3, 4, 5.
Summing these possibilities, we obtain 22 2 + 18 2 + 14 2 + 10 2 + 6 2 + 2 2 = 536 different purchases Example 8: Binary Patterns What fraction of binary sequences of length 10 consists of a (positive) number of 1s, followed by a number of 0s, followed by a number of 1s, followed by a number of 0s? An example of such a sequence is 1110111000.
There are 210= 1024 10-bit binary sequences. We model the problem of counting this special type of 10-bit binary sequences as a distribution problem as follows. We create four distinct boxes, the first box for the initial set of 1s, the second box for the following set of 0s, and so on. We have 10 identical markers (call them xs) to distribute into the four boxes. Each box must have at least one marker, since each subsequence of 0s or of 1s must be nonempty. By Example 5 the number of ways to distribute 10 xs into four boxes with no box empty is C(10 −1, 4 −1) = 84. So there are 84 such binary sequences and thus 84/1024 ≈8% of all 10-bit binary sequences are of this special type.
There are three constraints in arrangement problems that are handled by “tricks.” One is requiring certain elements to be consecutive; this is handled by gluing the elements together. The second is requiring one element to occur before another, or more generally, specifying the relative order of a subset of elements; this is handled by picking the subset of positions in the arrangement where the ordered elements will appear. The next problem presents the third, and most complex, “trick” constraint— namely, requiring certain elements to be nonconsecutive (i.e., never side by side).
Example 9: Nonconsecutive Vowels How many arrangements of the letters a, e, i, o, u, x, x, x, x, x, x, x, x (eight xs) are there if no two vowels can be consecutive?
220 Chapter 5 General Counting Methods for Arrangements and Selections The right question to start with is: in which positions in the arrangement will the xs appear to ensure the vs (v = vowel) are nonconsecutive. Observe that once we know where the vs goes, it is easy to count the ways to order the five vowels in those positions. In a fashion similar to the preceding binary patterns example, the placement of xs can be modeled by creating boxes before, between, and after the vs and distributing the xs into the six resulting boxes with at least one x in the middle four boxes (to ensure that no two vowels are consecutive). The following pattern illustrates one possible distribution of the xs: box 1 v x x box 2 v x box 3 v x box 4 v xxx box 5 v x box 6 Initially we put one x in boxes 2, 3, 4, 5. Then we distribute the remaining four xs into the six boxes without constraint—C(4 + 6 −1, 4) = 126 ways. This counts all possible patterns of xs and vs (the positions of the vs are forced by the way we choose the xs).
Next in each pattern, we arrange a, e, i, o, u in 5! = 120 ways among the five positions with vs. In total, there are 120 × 126 = 15,120 arrangements.
We close this section with a table summarizing the different basic types of count-ing problems we have encountered in this chapter.
Ways to Arrange, Select, or Distribute r Objects from n Items or into n Boxes Arrangement Combination (Ordered Outcome) (Unordered Outcome) or or Distribution of Distribution of Distinct Objects Identical Objects No repetition P(n, r) C(n, r) Unlimited repetition nr C(n +r −1,r) Restricted repetition P(n;r1,r2, . . .rm) — 5.4 EXERCISES S u m m a r y o f E x e r c i s e s Be careful to distinguish whether a prob-lem involves distinct or identical objects. Exercises 31–34 involve problem restate-ment (no actual numerical answers are sought). Exercise 51 presents an important alternative approach for analyzing nonconsecutivity problems.
1. How many ways are there to distribute 27 identical jelly beans among three children: (a) Without restrictions?
(b) With each child getting nine beans?
(c) With each child getting at least one bean?
www.itpub.net 5.4 Distributions 221 2. How many ways are there to distribute 18 different toys among four children?
(a) Without restrictions?
(b) If two children get seven toys and two children get two toys?
3. In a bridge deal, what is the probability that: (a) West has four spades, two hearts, four diamonds, and three clubs?
(b) North and South have four spades, West has three spades, and East has two spades?
(c) One player has all the Aces?
(d) All players have a (4, 3, 3, 3) division of suits?
4. How many ways are there to distribute eight (identical) apples, six oranges, and seven pears among three different people (a) Without restriction?
(b) With each person getting at least one pear?
5. How many ways are there to distribute 18 chocolate doughnuts, 12 cinnamon doughnuts, and 14 powdered sugar doughnuts among four school principals if each principal demands at least two doughnuts of each kind?
6. How many distributions of 18 different objects into three different boxes are there with twice as many objects in one box as in the other two combined?
7. How many ways are there to arrange the letters in VISITING with no pair of consecutive Is?
8. How many ways are there to arrange 12 identical apples and five different oranges in a row so that no two oranges will appear side by side?
9. (a) How many arrangements of the letters in COMBINATORICS have no con-secutive vowels?
(b) In how many of the arrangements in part (a) do the vowels appear in alpha-betical order?
10. How many ways are there to arrange the 26 letters of the alphabet so that no pair of vowels appear consecutively (Y is considered a consonant)?
11. If you flip a coin 18 times and get 14 heads and four tails, what is the probability that there is no pair of consecutive tails?
12. How many integer solutions are there to x1 + x2 + x3 + x4 + x5 = 31 with (a) xi ≥0 (b) xi > 0 (c) xi ≥i(i = 1, 2, 3, 4, 5) 13. How many integer solutions are there to x1 + x2 + x3 = 0 with xi ≥−5?
14. How many positive integer solutions are there to x1 + x2 + x3 + x4 < 50?
15. How many ways can a deck of 52 cards be broken up into a collection of unordered piles of sizes (a) Four piles of 13 cards?
(b) Three piles of eight cards and four piles of seven cards?
222 Chapter 5 General Counting Methods for Arrangements and Selections 16. Consider the problem of distributing 10 distinct books among three different people with each person getting at least one book. Explain why the following solution strategy is wrong: first select a book to give to the first person in 10 ways; then select a book to give to the second person in nine ways; then select a book to give to the third person in eight ways; and finally distribute the remaining seven books in 73 ways.
17. Each of 10 employees brings one (distinct) present to an office party. Each present is given to a randomly selected employee by Santa (an employee can get more than one present). What is the probability that at least two employees receive no presents?
18. How many ways are there to distribute k balls into n distinct boxes (k < n) with at most one ball in any box if (a) The balls are distinct?
(b) The balls are identical?
19. How many ways are there to distribute three different teddy bears and nine identical lollipops to four children (a) Without restriction?
(b) With no child getting two or more teddy bears?
(c) With each child getting three “goodies”?
20. Suppose that 30 different computer games and 20 different toys are to be distributed among three different bags of Christmas presents. The first bag is to have 20 of the computer games. The second bag is to have 15 toys. The third bag is to have 15 presents, any mixture of games and toys. How many ways are there to distribute these 50 presents among the three bags?
21. Suppose a coin is tossed 12 times and there are three heads and nine tails. How many such sequences are there in which there are at least five tails in a row?
22. How many binary sequences of length 20 are there that (a) Start with a run of 0s—that is, a consecutive sequence of (at least) one 0— then a run of 1s, then a run of 0s, then a run of 1s, and finally finish with a run of 0s?
(b) Repeat part (a) with the constraint that each run is of length at least 2.
23. How many binary sequences of length 18 are there that start with a run of 1s—that is, a consecutive sequence of (at least) one 1—then a run of 0s, then a run of 1s, and then a run of 0s, such that one run of 1s has length at least 8?
24. What fraction of all arrangements of EFFLORESCENCE has consecutive Cs and consecutive Fs but no consecutive Es?
25. How many arrangements of MISSISSIPPI are there with no pair of consecu-tive Ss?
26. How many ways are there to distribute 15 identical objects into four different boxes if the number of objects in box 4 must be a multiple of 3?
www.itpub.net 5.4 Distributions 223 27. If n distinct objects are distributed randomly into n distinct boxes, what is the probability that (a) No box is empty?
(b) Exactly one box is empty?
(c) Exactly two boxes are empty?
28. How many ways are there to distribute eight balls into six boxes with the first two boxes collectively having at most four balls if (a) The balls are identical?
(b) The balls are distinct?
29. In Example 4, if all 10 pieces of fruit were distinct, what would be the fraction of outcomes with two pieces of fruit in each box? Why is this fraction greater when the pieces of fruit are distinct?
30. (a) How many ways are there to make 35 cents change in 1952 pennies, 1959 pennies, and 1964 nickels?
(b) In 1952 pennies, 1959 pennies, 1964 nickels, and 1971 quarters?
31. State an equivalent distribution version of each of the following arrangement problems: (a) Arrangements of eight letters chosen from piles of as, bs, and cs (b) Arrangements of two as, three bs, four cs (c) Arrangements of 10 letters chosen from piles of as, bs, cs, and ds with the same number of as and bs (d) Arrangements of four letters chosen from two as, three bs, and four cs 32. State an equivalent arrangement version of each of the following distribution problems: (a) Distributions of n distinct objects into n distinct boxes (b) Distributions of 15 distinct objects into five distinct boxes with three objects in each box (c) Distributions of 15 distinct objects into five boxes with i objects in the ith box, i = 1, 2, 3, 4, 5 (d) Distributions of 12 distinct objects into three distinct boxes with at most three objects in box 1 and at most five objects in box 2 33. State an equivalent distribution version and an equivalent integer-solution-of-an-equation version of the following selection problems: (a) Selections of six ice cream cones from 31 flavors (b) Selections of five marbles from a group of five reds, four blues, and two pinks (c) Selections of 12 apples from four types of apples with at least two apples of each type 224 Chapter 5 General Counting Methods for Arrangements and Selections (d) Selections of 20 jelly beans from four different types with an even number of each type and not more than eight of any one type 34. State an equivalent selection version and an equivalent integer-solution-of-an-equation version of the following distribution problems: (a) Distributions of 30 black chips into five distinct boxes (b) Distributions of 18 red balls into six distinct boxes with at least two balls in each box (c) Distributions of 20 markers into four distinct boxes with the same number of markers in the first and second boxes 35. How many election outcomes are possible (numbers of votes for different candidates) if there are three candidates and 30 voters? If, in addition, some candidate receives a majority of the votes?
36. How many election outcomes in the race for class president are there if there are five candidates and 40 students in the class and (a) Every candidate receives at least two votes?
(b) One candidate receives at most one vote and all the others receive at least two votes?
(c) No candidate receives a majority of the votes?
(d) Exactly three candidates tie for the most votes?
37. How many numbers between 0 and 10,000 have a sum of digits (a) Equal to 7?
(b) Less than or equal to 7?
(c) Equal to 13?
38. How many integer solutions are there to the equation x1 + x2 + x3 + x4 ≤15 with xi ≥−10?
39. How many nonnegative integer solutions are there to the equation 2x1 + 2x2+ x3 + x4 = 12?
40. How many nonnegative integer solutions are there to the pair of equations x1 + x2 + · · · + x6 = 20 and x1 + x2 + x3 = 7?
41. How many nonnegative integer solutions are there to the inequalities x1 + x2 + · · · + x6 ≤20 and x1 + x2 + x3 ≤7?
42. How many nonnegative integer solutions are there to x1 + x2 + · · · · + x5 = 20 (a) With xi ≤10?
(b) With xi ≤8?
(c) With x1 = 2x2?
43. How many ways are there to split four red, five blue, and seven black balls among (a) Two boxes without restriction?
(b) Two boxes with no box empty?
44. How many ways are there to arrange the letters in UNIVERSALLY so that the four vowels appear in two clusters of two consecutive letters with at least two consonants between the two clusters? An example of such an arrangement is LNUILYSVAER.
www.itpub.net 5.4 Distributions 225 45. How many ways are arrange the letters in UNIVERSALLY so that no two vowels occur consecutively and also the consonants appear in alphabetical order?
46. How many 8-letter arrangements can be formed from the 26 letters of the alphabet (without repetition) that include at most three of the five vowels and in which the vowels are nonconsecutive?
47. How many arrangements of letters in INSTITUTIONAL have all of the following properties simultaneously?
(a) No consecutive Ts (b) The 2 Ns are consecutive (c) Vowels in alphabetical order 48. How many arrangements of the letters in INSTRUCTOR have all of the following properties simultaneously?
(a) The vowels appearing in alphabetical order (b) At least 2 consonants between each vowel (c) Begin or end with the 2 Ts (the Ts are consecutive) 49. How many arrangements of the letters in STATISTICS have all of the following properties simultaneously?
(a) No consecutive Ss (b) Vowels in alphabetical order (c) The 3 Ts are consecutive (appear as 3 Ts in a row) 50. (a) How many arrangements are there of REVISITED with vowels not in increasing order—that is, an I before one (or both) of the Es?
(b) How many arrangements with no consecutive Es and no consecutive Is?
(c) How many arrangements with vowels not in increasing order and no consecutive Es and no consecutive Is?
51. This exercise presents an alternative approach to counting arrangements with a certain type of object not occurring consecutively. Take Exercise 7, which seeks arrangements of VISITING with no consecutive Is. Arrange the five consonants in 5! ways and then pick a subset of three positions (with no repetition) for Is from amongthesixlocationsbefore,between,andafterthefiveconsonants—donein C (6, 3) ways.
(a) Rework Example 9 using this approach.
(b) Use this approach to solve Exercise 10.
52. Among all arrangements of WISCONSIN without any pair of consecutive vowels, what fraction have W adjacent to an I?
53. How many bridge deals are there in which North and South get all the spades?
54. What is the probability in a bridge deal that each player gets at least three honors (an honor is an Ace, or King, or Queen, or Jack)?
226 Chapter 5 General Counting Methods for Arrangements and Selections 55. How many ways are there to distribute 15 distinct oranges into three different boxes with at most eight oranges in a box?
56. How many ways are there to distribute 20 toys to m children such that the first two children get the same number of toys if (a) The toys are identical?
(b) The toys are distinct?
57. How many ways are there to distribute 25 different presents to four people (including the boss) at an office party so that the boss receives exactly twice as many presents as the second most popular person?
58. How many ways are there to distribute r identical balls into n distinct boxes with exactly m boxes empty?
59. How many subsets of six integers chosen (without repetition) from 1, 2, . . . , 20 are there with no consecutive integers (e.g., if 5 is in the subset, then 4 and 6 cannot be in it)?
60. How many arrangements are there of eight αs, six βs, and seven γ s in which each α is beside (on at least one side) another α? (Hint: Watch out for two clusters of αs occurring consecutively.) 61. How many arrangements are there with n 0s and m 1s, with k runs of 0s? [A run is a consecutive set (1 or more) of the same digit; e.g., 0001110100 has three (underlined) runs of 0s.] 62. What is the probability that a random arrangement of a deck of 52 cards has exactly k runs of hearts (see Exercise 61 for a definition of a run)?
63. How many binary sequences of length n are there that contain exactly m occurrences of the pattern 01?
64. Howmanyways aretheretodistribute20distinctflagsonto12distinctflagpolesif (a) In arranging flags on a flagpole, the order of flags from the ground up makes a difference?
(b) No flagpole is empty and the order on each flagpole is counted?
65. How many ways are there to distribute r identical balls into n distinct boxes with the first m boxes collectively containing at least s balls?
66. How many ways are there to deal four cards to each of 13 different players so that exactly 11 players have a card of each suit?
5.5 BINOMIAL IDENTITIES In this section we show why the numbers C(n, r) are called binomial coefficients.
Then we present some identities involving binomial coefficients. We use three techniques to verify the identities: combinatorial selection models, “block walking,” and the binomial expansion. The study of binomial identities is itself a major subfield www.itpub.net 5.5 Binomial Identities 227 of combinatorial mathematics and we will necessarily just scratch the surface of this topic.
Consider the polynomial expression (a + x)3. Instead of multiplying (a + x) by (a + x) and the product by (a + x) again, let us formally multiply the three factors term by term: (a + x)(a + x)(a + x) = aaa + aax + axa + axx + xaa + xax + xxa + xxx Collecting similar terms, we reduce the right-hand side of this expansion to a3 + 3a2x + 3ax2 + x3 (1) The formal expansion of (a + x)3 was obtained by systematically forming all products of a term in the first factor, a or x, times a term in the second factor times a term in the third factor. There are two choices for each term in such a product, and so there are the 2 × 2 × 2 = 8 formal products obtained above. If we were expanding (a + x)10, we would obtain 210 = 1024 different formal products.
Now we ask the question, how many of the formal products in the expansion of (a + x)3 contain k xs and (3 −k) as? This question is equivalent to asking for the coefficient of a3−kxk in (1). Since all possible formal products of as and xs are formed, and since formal products are just three-letter sequences of as and xs, we are simply asking for the number of all three-letter sequences with k xs and (3 −k) as. The answer is C(3, k) and so the reduced expansion for (a + x)3 can be written as 3 0 a3 + 3 1 a2x + 3 2 ax2 + 3 3 x3 By the same argument, we see that the coefficient of an−kxk in (a + x)n will be equal to the number of n-letter sequences formed by k xs and (n −k) as, that is, C(n, k). If we set a = 1, we have the following theorem.
Binomial Theorem (1 + x)n = n 0 + n 1 x + n 2 x2 + · · · + n k xk + · · · + n n xn Another proof of this expansion, using induction, is given in the Exercises. Just as important as the binomial theorem is the equivalence we have established between the number of k-subsets of n objects and the coefficient of xk in (1 + x)n. We exploit this equivalence extensively in the next chapter.
Let us now consider some basic properties of binomial coefficients. The most important identity for binomial coefficients is the symmetry identity n k = n!
k!(n −k)! = n n −k (2) 228 Chapter 5 General Counting Methods for Arrangements and Selections In words, this identity says that the number of ways to select a subset of k objects out of a set of n objects is equal to the number of ways to select a group of n −k of the objects to set aside (the objects not in the subset).
The other fundamental identity is n k = n −1 k + n −1 k −1 (3) This identity can be verified algebraically. We will give a combinatorial argument instead. We classify the C(n, k) committees of k people chosen from a set of n people into two categories, depending on whether or not the committee contains a given person P. If P is not part of the committee, there are C(n −1, k) ways to form the committee from the other n −1 people. On the other hand, if P is on the committee, the problem reduces to choosing the k −1 remaining members of the committee from the other n −1 people. This can be done C(n −1, k −1) ways. Thus C(n, k) = C(n −1, k) + C(n −1, k −1).
The following example presents another binomial identity that can be verified algebraically or by a combinatorial argument.
Example 1 Show that n k k m = n m n −m k −m (4) The left-hand side of (4) counts the ways to select a group of k people chosen from a set of n people and then to select a subset of m leaders within the group of k people.
Equivalently, as counted on the right side, we could first select the subset of m leaders from the set of n people and then select the remaining k −m members of the group from the remaining n −m people. Note the special form of (4) when m = 1: k n k = n n −1 k −1 or n k = n k n −1 k −1 (5) For a fixed integer n, the values of the binomial coefficients C(n, k) increase as k increases as long as k ≤n/2. Then the values of C(n, k) decrease as k increases for k ≥n/2 + 1. To verify this assertion, we observe that the binomial coefficients are increasing if and only if the ratio C(n, k)/C(n, k −1) is greater than 1.
n k n k −1 = n!
k!(n −k)!
n!
(k −1)! (n −k + 1)!
= n −k + 1 k So the C(n, k)s are increasing when (n −k + 1)/k > 1 or equivalently when n −k + 1 > k. Solving for k in terms of n, we have k < (n + 1)/2. This is the same bound on integer values of k as k ≤n/2.
www.itpub.net 5.5 Binomial Identities 229 Using (3) and the fact that C(n, 0) = C(n, n) = 1 for all nonnegative n, we can recursivelybuildsuccessiverowsinthefollowingtableofbinomialcoefficients,called Pascal’s triangle. Each number in this table, except the first and last numbers in a row, is the sum of the two neighboring numbers in the preceding row.
Table of binomial coefficients: kth number in row n is n k 1 4 6 4 1 1 3 3 1 1 2 1 1 1 1 k = 0 n = 4 n = 3 n = 2 n = 1 n = 0 k = 1 k = 2 k = 3 k = 4 Pascal’s triangle has the following nice combinatorial interpretation, due to George Polya. Consider the ways a person can traverse the blocks in the map of streets shown in Figure 5.1. The person begins at the top of the network, at the spot marked (0, 0), and moves down the network (down the page) making a choice at each intersection to go right or left (for simplicity, let “right” be your right as you look at this page,nottherightofthepersonmovingdownthenetwork).Welabeleachstreetcorner in the network with a pair of numbers (n, k), where n indicates the number of blocks traversed from (0, 0) and k the number of times the person chose the right branch at intersections. Figure 5.1 shows a possible route from the start (0, 0) to the corner (6, 3).
0, 0 1, 0 2, 0 3, 1 4, 2 5, 3 6, 3 o Start Figure 5.1 230 Chapter 5 General Counting Methods for Arrangements and Selections Any route to corner (n, k) can be written as a list of the branches (left or right) chosen at the successive corners on the path from (0, 0) to (n, k). Such a list is just a sequence of k Rs (right branches) and (n −k) Ls (left branches). To go to corner (6, 3) following the route shown in Figure 5.1, we have the sequence of turns LLRRRL.
Let s(n, k) be the number of possible routes from the start (0, 0) to corner (n, k) (moving down in the network). This is the number of sequences of k Rs and (n −k) Ls, and so s(n, k) = C(n, k). Another useful interpretation of binomial coefficients is the committee selection model, used above to verify Eqs. (3) and (4).
Let us show how our “block-walking” model for binomial coefficients can be used to get an alternate proof of identity (3). At the end of a route from the start to corner (n, k), a block walker arrives at (n, k) from either corner (n −1, k) or corner (n −1, k −1). For example, to get to corner (6, 3) in Figure 5.1, the person either goes to corner (5, 3) and branches left to (6, 3), or goes to corner (5, 2) and branches right to (6, 3). Thus, the number of routes from (0, 0) to corner (n, k) equals the number of routes from (0, 0) to (n −1, k) plus the number of routes from (0, 0) to (n−1, k−1).
We have verified identity (3): s(n, k) = s(n −1, k) + s(n −1, k −1).
We now list seven well-known binomial identities and verify three of them (the others are left as exercises). These binomial identities are of much practical interest. Expressions involving sums and products of binomial coefficients arise fre-quently in complicated real-world counting problems. These identities can be used to simplify such expressions.
n 0 + n 1 + n 2 + · · · + n n = 2n (6) n 0 + n + 1 1 + n + 2 2 + · · · + n +r r = n +r + 1 r (7) r r + r + 1 r + r + 2 r + · · · + n r = n + 1 r + 1 (8) n 0 2 + n 1 2 + n 2 2 + · · · + n n 2 = 2n n (9) r k = 0 m k n r −k = m + n r (10) m k = 0 m k n r + k = m + n m +r (11) m−r k = n−s m −k r n + k s = m + n + 1 r + s + 1 (12) Here C(n, r) = 0 if 0 ≤n < r. These identities can be explained by the “committee” type of combinatorial argument used for Eqs. (2), (3), and (4) or by www.itpub.net 5.5 Binomial Identities 231 “block-walking arguments.” We give a committee argument for Eq. (6). Consider the following two ways of counting all subsets (of any size) of a set of n people: (a) summing the number of subsets of size 0, of size 1, of size 2, and so on—this yields the left-hand side of Eq. (6)—and (b) counting all subsets by whether or not the first person is in the subset, whether or not the second person is in the subset, and so on—this yields 2 × 2 × 2 . . . × 2 (n times) = 2n, the right-hand side of (6). A simpler proof of (6) is given at the end of this section.
In general, a combinatorial argument for proving such identities consists of spec-ifying a selection counted by the term on the right side, partitioning the selection into subcases, and showing that the terms in the summation count the subcases.
The advantage of proofs involving block walking is that we can draw pictures of the “proof.” The picture shows that all the routes to a certain corner—this amount is the right-hand side of the identity—can be decomposed in terms of all routes to certain intermediate corners (or blocks)—the sum on the left-hand side of the identity.
Example 2 Verify identity (8) by block-walking and committee-selection arguments.
As an example of this identity, we consider the case where r = 2 and n = 6. The corners (k, 2), k = 2, 3, 4, 5, 6, are marked with a in Figure 5.1 and corner (7, 3) is marked with an o. Observe that the right branches at each starred corner are the locations of last possible right branches on routes from the start (0, 0) to corner (7, 3).
After traversing one of these right branches, there is just one way to continue on to corner (7, 3), by making all remaining branches left branches. In general, if we break all routes from (0, 0) to (n + 1, r + 1) into subcases based on the corner where the last right branch is taken, we obtain identity (8).
We restate the block-walking model as a committee selection: If the kth turn is right, this corresponds to selecting the kth person to be on the committee; if the kth turn is left, the kth person is not chosen. We break the ways to pick r + 1 members of a committee from n + 1 people into cases depending on who is the last person chosen: the (r + 1)st, the (r + 2)nd, . . . , the (n + 1)st. If the (r + k + 1)st person is the last chosen, then there are C(r + k,r) ways to pick the first r members of the committee.
Identity (8) now follows.
Example 3 Verify identity (9) by a block-walking argument.
The number of routes from (n, k) to (2n, n) is equal to the number of routes from (0, 0) to (n, n −k), since both trips go a total of n blocks with n −k to the right (and k to the left). So the number of ways to go from (0, 0) to (n, k) and then on to (2n, n) is C(n, k) × C(n, n −k). By (2), C(n, n −k) = C(n, k), and thus the number of routes from (0, 0) to (2n, n) via (n, k) is C(n, k)2. Summing over all k—that is, over all intermediate corners n blocks from the start—we count all routes from (0, 0) to (2n, n). So this sum equals C(2n, n), and identity (9) follows.
232 Chapter 5 General Counting Methods for Arrangements and Selections Now we show how binomial identities can be used to evaluate sums whose terms are closely related to binomial coefficients.
Example 4 Evaluate the sum 1 × 2 × 3 + 2 × 3 × 4 + · · · + (n −2)(n −1)n.
The general term in this sum (k −2)(k −1)k is equal to P(k, 3) = k!/(k −3)!.
Recall that the numbers of r-permutations and of r-selections differ by a factor of r!
That is, C(k, 3) = k!/(k −3)!3! = P(k, 3)/3!, or P(k, 3) = 3!C(k, 3). So the given sum can be rewritten as 3!
3 3 + 3!
4 3 + · · · + 3!
n 3 = 3!
3 3 + 4 3 + · · · + n 3 By identity (8), this sum equals 3!
n + 1 4 .
Example 5 Evaluate the sum 12 + 22 + 32 + · · · + n2.
A strategy for problems whose general term is not a multiple of C(n, k) or P(n, k) is to decompose the term algebraically into a sum of P(n, k)-type terms. In this case, the general term k2 can be written as k2 = k(k −1) + k. So the given sum can be rewritten as [(1 × 0) + 1] + [(2 × 1) + 2] + [(3 × 2) + 3] + · · · + [n(n −1) + n] = [(2 × 1) + (3 × 2) + · · · + n(n −1)] + (1 + 2 + 3 + · · · + n) = 2 2 2 + 2 3 2 + · · · + 2 n 2 + 1 1 + 2 1 + · · · + n 1 = 2 n + 1 3 + n + 1 2 by identity (8).
Note that as part of Example 5, we showed that 1 + 2 + 3 + · · · + n = C(n + 1, 2) = 1 2n(n + 1), a result we verify by induction in Example 1 of Appendix A.2.
There is another simple way to verify certain identities with binomial coefficients.
We start with the binomial expansion in the binomial theorem. Then we substitute appropriate values for x. The following identities can be obtained from the bino-mial expansion: (6) by setting x = 1, (13), below by setting x = −1, and (14) by differentiating both sides of the binomial expansion and setting x = 1.
(1 + 1)n = 2n = n 0 + n 1 + n 2 + · · · + n n (6) (1 −1)n = 0 = n 0 − n 1 + n 2 −· · · + (−1)n n n (13) www.itpub.net 5.5 Binomial Identities 233 or n 0 + n 2 + · · · = n 1 + n 3 + · · · = 2n−1 (13′) (since the sum of both sides is 2n) n(1 + x)n−1 = 1 n 1 + 2 n 2 x + 3 n 3 x2 + · · · + n n n xn−1 (14) and so n(1 + 1)n−1 = n2n−1 = 1 n 1 + 2 n 2 + 3 n 3 + · · · + n n n (14′) 5.5 EXERCISES S u m m a r y o f E x e r c i s e s Combinatorial identities are a well-developed field whose surface was barely scratched in this section. The later problems in this exercise set go well beyond the level of examples worked in this section.
1. (a) Verify identity (3) algebraically (writing out the binomial coefficients in factorials).
(b) Verify identity (4) algebraically.
(c) Verify that C(n, k)/C(n, k −1) = (n −k + 1)/k.
2. Verify the following identities by block-walking: (a) (6) (b) (7) (c) (10) (d) (11) 3. Verify the following identities by a committee-selection model: (a) (7) (b) (9) (c) (10) (d) (11) (e) (13) 4. Verify the following identities by mathematical induction. [Hint: Use (3)].
(a) (3) (b) (5) (c) (6) (d) (7) (e) (13′) 5. Prove the Binomial theorem by mathematical induction.
6. Show that identity (7) can be obtained as a special case of (11).
7. Show that C(2n, n) + C(2n, n −1) = 1 2C(2n + 2, n + 1).
8. If C(n, 3) + C(n + 3 −1, 3) = P(n, 3), find n.
9. Show by a combinatorial argument that (a) 2n 2 = 2 n 2 + n2 (b) (n −r) n +r −1 r n r = n n +r −1 2r 2r r 10. Show that C(k + m + n, k)C(m + n, m) = (k + m + n)!/(k!m!n!).
234 Chapter 5 General Counting Methods for Arrangements and Selections 11. (a) Show that n 1 + 6 n 2 + 6 n 3 = n3 (b) Evaluate 13 + 23 + 33 + · · · + n3.
12. (a) Evaluate n k=0 12(k + 1)k(k −1). (Hint: Use Example 4.) (b) Evaluate n k=0 (2 + 3k)2.
(c) Evaluate n k=0 k(n −k).
13. (a) Evaluate the sum 1 + 2 n 1 + · · · (k + 1) n k + · · · + (n + 1) n n by breaking this sum into two sums, each of which is an identity in this section.
(b) Evaluate the sum n 0 + 2 n 1 + n 2 + 2 n 3 + · · · 14. By setting x equal to the appropriate values in the binomial expansion (or one of its derivatives, etc.), evaluate (a) n k=0 (−1)k n k (b) n k=0 k(k −1) n k (c) n k=0 2k n k (d) n k=1 k3k n k (e) n k=1 (−1)kk n k (f) n k=0 1 k + 1 n k (g) n k=0 (2k + 1) n k 15. Show that n k=m k r = n + 1 r + 1 − m r + 1 .
16. Show that n k=1 m + k −1 k = m k=1 n + k −1 k .
17. Show that n−1 k=0 P(m + k, m) = P(m + n, m + 1) (m + 1) .
18. (a) Consider a sequence of 2n distinct people in a line at a cashier. Suppose n of the people owe $1 and n of the people are due a $1 payment. Show that the number of arrangements in which the cashier never goes into debt www.itpub.net 5.5 Binomial Identities 235 (i.e., at every stage at least as many people have paid in $1 as were paid out $1) is equal to 2n n − 2n n+1 . [Hint: Use a symmetry argument in a block-walking type of model to demonstrate a one-to-one correspondence between sequences where at some stage the cashier goes (at least) $1 into debt and all sequences of 2n people in which n + 1 of the people are owed $1.] (b) Repeat part (a) with the following constraint: if the first person pays $1, how many arrangements of the people are possible in which the cashier always has a positive amount of money until the last person in line (who is owed $1)?
19. Show that n 0 + n 1 + n 2 + · · · + n n 2 = 2n k=0 2n k .
20. Find the value of k that maximizes (a) n k (b) 2n + k n 2n −k n .
21. Evaluate n k=1 n k n k −1 .
22. (a) Prove that n 1 + 3 n 3 + 5 n 5 + · · · = 2 n 2 + 4 n 4 + 6 n 6 + · · · (b) What is the value of the sum on each side?
23. Evaluate n 0 −2 n 1 + 3 n 2 + · · · + (−1)n(n + 1) n n .
24. Show that n−1 k=0 (−1)k n k + 1 = 1. (Hint: Rewrite as a binomial coefficient with k alone in the bottom.) 25. Show that n k=0 (2n)!
k!2(n −k)!2 = 2n n 2 .
26. Give a combinatorial argument to evaluate n k=0 n k m k , n ≤m.
27. (a) For any given k, show that an integer n can be represented as n = m1 j + m2 j + 1 + · · · + mk k where 0 < m1 < m2 < · · · < mk, k ≥1.
(b) Prove that for a given k, the representation in part (a) is unique. [Hint: Use the fact that n k −1 = n −1 k + n −2 k −1 + · · · + n −k 1 . 28. Show that k j=0 n + k −j −1 k −j m + j −1 j = n + m + k −1 k .
236 Chapter 5 General Counting Methods for Arrangements and Selections 29. Show that m k=0 P(m, k) P(n, k) = 1 n m m k=0 n −k n −m = n + 1 n −m + 1, m ≤n.
30. Show that m k=0 m!(n −k)!
n!(m −k)! = n + 1 n −m + 1, m ≤n.
31. If C2n = 1 n + 1 2n n , show that C2m = m k=1 C2k−2C2m−2k.
32. Show that m k=0 n k n −k m −k = 2m n m , m < n.
33. Consider the problem of three-dimensional block walking. Show by a combina-torial argument that P(n; i −1, j, k) + P(n; i, j −1, k) + P(n; i, j, k −1) = P(n + 1; i, j, k).
34. Give a combinatorial argument to show that (x1 + x2 + · · · + xk)n = P(n; i1, i2, . . . , ik)xi1 1 xi2 2 · · · xik k where the sum is over all i1 + i2 + · · · + ik = n, ij ≥0.
35. Show for sums over all i1 + i2 + · · · + ik = n, ij ≥0, that (a) P(n; i1, i2, . . . , ik) = kn (b) i1i2 · · · ik P(n; i1, i2, . . . , ik) = P(n, k)kn−k 5.6 SUMMARY AND REFERENCES This chapter introduced the basic formulas and logical reasoning used in arrangement and selection problems. There were four formulas for arrangements and selections with and without repetition, and there were three principles for composing subprob-lems. But these formulas and principles were just the building blocks for constructing answers to dozens of examples and hundreds of exercises. These problems required a thorough, logical analysis before one could begin to decompose them into tractable subproblems. Such logical analysis of possibilities arises often in computer science, probability, and operations research. It is the basic methodology of discrete mathe-matics and the most important skill for a student to develop in this course. Note that such logical reasoning is the very essence of mathematical model building.
The n! formula for arrangements was known at least 2,500 years ago (see David for more details of the history of combinatorial mathematics). Problems involv-ing binomial coefficients and the binomial expansion were mentioned in a primitive way in Chinese, Hindu, and Arab works 800 years ago. Pascal’s triangle appears in a fourteenth-century work of Shih-Chieh (see text cover). The first appearance of the triangle in the West was 200 years later. However, it was not until the end of the seventeenth century that Jacob Bernoulli gave a careful proof of the binomial the-orem. The examination of probabilities related to gambling by Pascal and Fermat www.itpub.net Supplement: Selected Solutions to Problems in Chapter 5 237 around 1650 was the beginning of modern combinatorial mathematics. The formula for selection with repetition was discovered soon afterward in the following context: the probability when flipping a coin that the nth head appears after exactly r tails is 1 2 r+nC(n +r −1,r) (there are n “boxes” before and between the occurrences of the n heads for placing the r tails). The formula for arrangements with repetition was discovered around 1700 by Leibnitz in connection with the multinomial theorem (see Exercise 34 of Section 5.5). Jacques Bernoulli’s Ars Conjectandi (1713) was the first book presenting basic combinatorial methods. The first comprehensive textbook on permutation and combination problems was written by Whitworth in 1901.
See General References (at end of text) for a list of other introductory texts on enumeration.
1. R. Buck, Advanced Calculus, 3rd ed., McGraw-Hill, New York, 1978.
2. F. David, Games, Gods, and Gambling: A History of Probability and Statistical Ideas, Dover Press, New York, 1998.
3. W. Whitworth, Choice and Chance, 5th ed. (1901), Hafner Press, New York, 1901.
SUPPLEMENT: SELECTED SOLUTIONS TO PROBLEMS IN CHAPTER 5 Section 5.1 Exercise 13 (a) How many different 6-digit numbers are there (leading zeros, e.g., 00174, not allowed)?
(b) How many even 6-digit numbers are there?
(c) How many 6-digit numbers are there with exactly one 3?
(d) How many 6-digit palindromic numbers (numbers that are the same when the order of their digits is inverted, e.g., 137731) are there?
Answer (a) All numbers from 100,000 to 999,999 form the 6-digit numbers. Answer is 9 × 10 × 10 × 10 × 10 × 10.
(b) For a number to be even, the rightmost position must have an even digit (0 or 2 or 4 or 6 or 8). Answer: 9 × 10 × 10 × 10 × 10 × 5.
(c) If leftmost digit is 3, there are five numbers. If any other digit is 3, 8 × 94 numbers.
Total is 95 + 5 × (8 × 94).
(d) A palindromic number must have the same digit in the leftmost and rightmost position and similarly for the second-leftmost and second-rightmost positions.
Then the problem becomes one of counting the ways to pick values for three leftmost digits (the three right digits are then forced by palindromic symmetry)— 9 × 10 × 10 ways.
238 Chapter 5 General Counting Methods for Arrangements and Selections Exercise 30 How many times is the digit 5 written when listing all numbers from 1 to 100,000?
Answer It is simpler to answer this question for numbers between 0 and 99,999 (adding 0 and omitting 100,000 makes no difference since neither contains a 5). We thus ask how many times 5 appears in writing 5-digit sequences (5-digit numbers with leading 0s allowed). Consider the following problem: How many times does a 5 occur in the third (middle) position in these 5-digit sequences—that is, how many 5-digit sequences are there with a 5 in the third position? The answer is 104. For all five positions in the sequence, the answer is 5 × 104.
Exercise 33 How many four-digit numbers are there formed from the digits 1, 2, 3, 4, 5 (with possible repetition) that are evenly divisible by 4?
Answer A number is divisible by 4 if and only if the number formed by its two rightmost digits is divisible by 4. Also to be divisible by 4, the 1s digit (the rightmost digit) must be even—in this problem, 2 or 4. Observe that composing any given 10s digit with the two even 1s digits, such as 32, 34, we get two consecutive even numbers. Exactly one of every two consecutive even numbers is divisible by 4. So to generate a number divisible by 4 using digits 1, 2, 3, 4, 5 there can be any of these digits in the 1000s, the 100s and 10s positions followed by one choice for the 1s position—53 ways.
Supplement: Selected Solutions to Problems in Chapter 5 Section 5.2 Exercise 49 What fraction of all arrangements of INSTRUCTOR has (a) Three consecutive vowels?
(b) Two consecutive vowels?
Answer Total number of arrangements is C(10, 2) × C(8, 2) × 6! and so the outcomes in parts (a) and (b) will be divided by this amount to get the probabilities of these events.
(a) Replace the sequence of three consecutive vowels by a special symbol V. Now we arrange the eight letters, C, N, R, R, S, T, T, V. We pick a pair (subset of size 2) of positions for the two Rs, then pick a pair of (remaining) positions for the Ts, and then arrange the four distinct letters—C(8, 2) × C(6, 2) × 4! arrangements.
Next there are 3! arrangements of the three vowels to put in place of V. Answer: 3! × C(8, 2) × C(6, 2) × 5!
www.itpub.net Supplement: Selected Solutions to Problems in Chapter 5 239 (b) Initially proceed as in part (a), now with V as a double vowel and V ′ as a single vowel. There are C(9, 2) × C(7, 2) × 5! arrangements of C, N, R, R, S, T, T, V, V ′.
There are three choices for the vowel to be V ′ and two arrangements of the remain-ing vowels for V. In total, 3 × 2 × C(9, 2) × C(7, 2) × 5! arrangements. However, if V is followed (immediately) by V ′ or if V ′ is followed by V, we obtain three vowels in a row. Every arrangement with three vowels in a row will be generated in two different ways—e.g., TIOUCRRTNS arises from T(IO)(U)CRRTNS and T(I)(OU)CRRTNS. Subtracting arrangements with three vowels in a row [part (a)], we have {3 × 2 × C(9, 2) × C(7, 2) × 5!} −{3! × C(8, 2) × C(6, 2) × 4!}.
Exercise 55 (a) What is the probability that k is the smallest integer in a subset of four different numbers chosen from 1 through 20 (1 ≤k ≤17)?
(b) What is the probability that k is the second smallest?
Answer (a) As in most counting problems, the key here is to focus on the numbers that remain to be chosen, not on what one is told must be in the subset. To be concrete, initially assume k = 8. For 8 to be the smallest number in the sub-set, the other three numbers in the subset must be larger than 8—chosen in C (20 −8, 3) ways. For a general k, the probability is C(20 −k, 3)/C(20, 4).
(b) If 8 is the second smallest number in the subset, there is one smaller number— chosen in C(8 −1, 1) ways—and two larger numbers—chosen in C(20 −8, 2) ways. In general, the probability is C(k −1, 1) × C(20 −k, 2)/C(20, 4).
Exercise 59 What is the probability that two (or more) people in a random group of 25 people have a common birthday?
Answer The “trick” in this problem is that it is easier to count the probability that no one has a common birthday and subtract this probability from 1. We want the fraction of possible birthday dates for 25 people in which everyone has a different birthday.
The denominator, all possibilities of various birthdays for the 25 (different) people, is 36525. The numerator, the possibilities where everyone has a different birthday, is P(365, 25). The desired probability equals 1 −P(365, 25)/36525. (With a calculator or computer, one determines this probability to be about .57.) Exercise 63 Given a collection of 2n objects, n identical and the other n all distinct, how many different subcollections of n objects are there?
240 Chapter 5 General Counting Methods for Arrangements and Selections Answer Thetrickistodecidewhichofthendistinctobjectswillbeinthecollection.Anysubset of distinct objects can be chosen in C(n, 0) + C(n, 1) + C(n, 2) + · · · + C(n, n) ways, with the remaining elements made up of identical objects—done in one way (since the objects are identical). An alternative approach is to say that we have the choice to use or not use each distinct object—2n outcomes.
Exercise 69 What is the probability that a random 5-card hand has (a) Exactly one pair (no three of a kind or two pairs)?
(b) One pair or more (three of a kind, two pairs, four of a kind, full house)?
(c) The cards dealt in order of decreasing value?
(d) At least one spade, at least one heart, no diamonds or clubs, and the values of the spades are all greater than the values of the hearts?
Answer The denominator is C(52, 5).
(a) There are C(13, 1) × C(4, 2) ways to pick a pair (two cards of the same kind). To fill out the rest of the hand, pick one card of a second kind—48 ways—then one card of a third kind—44 ways—and finally one card of a fourth kind—40 ways.
The problem is that this rest-of-hand count is ordered—that is, the sequence of choices 7♠, Q♥, 5♦yields the same rest-of-hand as 5♦, Q♥, 7♠. Divide by 3!
to “un-order” this rest-of-hand, yielding the answer: {C(13, 1) × C(4, 2) × (48 × 44 × 40/3!)}/C(52, 5).
Another approach for the rest-of-hand is to pick a subset of three other kinds that will appear in the rest-of-hand—C(12, 3) ways—and pick a card of each of these kinds—43 ways, yielding {C(13, 1) × C(4, 2) × C(12, 3) × 43}/C(52, 5).
(b) Determine the probability of each of the five cards being of a different kind and subtract this probability from 1 : 1 −C(13, 5) × 45/C(52, 5).
(c) Any 5-card hand has a 1/5! chance of being dealt in a particular order (increasing or otherwise).
(d) Pick a subset of the five kinds (values) that will appear in the hand—C(13, 5) ways. The lowest k (1 ≤k ≤4) kinds must be hearts and the other kinds spades.
It remains only to pick the value of k—four choices: 4 × C(13, 5). (Wasn’t that sneaky!) Exercise 83 (a) How may points of intersection are formed by the chords of an n-gon (assuming no three of these lines cross at one point)?
(b) Into how many line segments are the lines in part (a) cut by the intersection points?
www.itpub.net Supplement: Selected Solutions to Problems in Chapter 5 241 (c) Use Euler’s formula r = e −v + 2 and parts (a) and (b) to determine the number of regions formed by the chords of an n-gon.
Answer (a) Every subset of four vertices of the n-gon forms a unique intersection point generated by the intersection of the two chords joining opposite vertices in the subset of four vertices. Thus the answer is C(n, 4).
(b) The number of chords is C(n, 2) −n (all pairs of vertices are connected by chords except for the n edges of the polygon). Each intersection point splits two line segments, increasing the number of line segments by 2. Then the answer to part (b) is C(n, 2) −n + 2C(n, 4).
(c) The total number of edges e equals the number of line segments—C(n, 2) −n + 2C(n, 4) [from part (b)]—plus the number of edges of the polygon—n edges— yielding e = C(n, 2) + 2C(n, 4). The number of vertices v is the number of polygon vertices plus intersection points, v = n + C(n, 4). By Euler’s formula, the number r of regions (excluding the infinite region) is r = e −v + 1 = [C(n, 2) + 2C(n, 4)] −[n + C(n, 4)] + 1 = C(n, 2) + C(n, 4) −n + 1.
Exercise 87 A man has seven friends. How many ways are there to invite a different subset of three of these friends for a dinner on seven successive nights such that each pair of friends are together at just one dinner?
Answer We need to find all possible unordered collections of seven 3-subsets such that each pair of the seven friends appears in exactly one subset. Consider the following table of one such collection of 3-subsets (an X means that the element of that row is in the subset of that column): Subsets 1 2 3 4 5 6 7 F A X X X r B X X X i C X X X e D X X X n E X X X d F X X X s G X X X Let the first three 3-subsets be the ones involving friend A. There are 6 2 × 4 2 × 2 2 3! = 15 collections of 3-subsets involving A—we count all ways to pair off the other six friends into 3-subsets containing A, and divide by 3! so that the pairings are not ordered. Let the first subset involve B as well as A. A general collection of 242 Chapter 5 General Counting Methods for Arrangements and Selections three 3-subsets with A has the form (A, B, C′), (A, D′, E′), (A, F′, G′) where C′ is the other friend in the subset with A and B, D′ is the (alphabetically) first of the friends not in the first subset, E′ is the other friend in the subset with A and D′, and F′ is alphabetically earlier than G′. Replacing C, D, E, F, and G by C′, D′, E′, F′, and G′ in the above table, the only remaining choice is whether B or C′ forms 3-subsets with D′, F′ and E′, G′—two choices. So there are 15 × 2 collections of seven 3-subsets of the seven friends such that each pair appears in one 3-subset.
Each such collection can be arranged over the seven successive nights in 7! ways.
So the answer is 15 × 2 × 7!.
Section 5.3 Exercise 27 How many ways are there to split a group of 2n αs, 2n βs, and 2n γ s in half (into two groups of 3n letters)? (Note: The halves are unordered; there is no first half.) Answer Count the ways to select 3n letters from the three types of letters and then subtract outcomes with 2n + 1 or more of one letter—C(3n + 3 −1, 3n) −3 × C((n −1) + 3 −1, n −1). Since each split forms two groups of 3n letters, it appears we should divide this count of 3n-letter groups by 2. However, the split in which each group consists of n letters of each type is not double counted. So the answer is 1 2[C(3n + 3 −1, 3n) −3 × C((n −1) + 3 −1, n −1) −1] + 1.
Exercise 30 How many arrangements of six 0s, five 1s, and four 2s are there in which (a) The first 0 precedes the first 1?
(b) The first 0 precedes the first 1, which precedes the first 2?
Answer (a) Position the four 2s among the 15 positions—C(15, 4) ways—then put a 0 in the first of the remaining positions—one way—and then pick five other positions for the remaining 0s—C(10, 5) ways. Answer: C(15, 4) × C(10, 5).
(b) Put a 0 in the first position—one way—then pick five other positions for the remaining 0s—C(14, 5) ways—then put a 1 in the first of the remaining positions—1 way—and then pick four other positions for the remaining 1s— C(8, 4) ways: C(14, 5) × C(8, 4).
Exercise 31 How many arrangements are there of 4n letters, four of each of n types of letters, in which each letter is beside a similar letter?
www.itpub.net Supplement: Selected Solutions to Problems in Chapter 5 243 Answer There cannot be exactly three consecutive letters the same, because that would leave the fourth letter of that type alone. So there are either two in a row or four in a row. But four in a row is the same as two consecutive two-in-a-rows. So our problem reduces to counting arrangements of 2n objects (two-in-a-rows) of n types with 2 of each type.
There are (2n)!/(2!)n such arrangements.
Exercise 33 When a coin is flipped n times, what is the probability that (a) The first head comes after exactly m tails?
(b) The ith head comes after exactly m tails?
Answer (a) There are 2n outcomes in all. The sequence of flips begins with m successive tails followed by a head. The sequence can be completed in 2n−(m+1) ways. Probability: 2n−(m+1)/2n = 2−(m+1).
(b) If the ith head comes after exactly m tails, then the first m + (i −1) flips con-tain m tails and i −1 heads—C(m + (i −1), m) ways. The remaining flips are unrestricted—2n−(m+i) ways. Probability: C(m + (i −1), m)2n−(m + i)/2n = C(m + i −1, m)2−(m+i).
Exercise 36 How many arrangements of five αs, five βs and five γ s are there with at least one β and at least one γ between each successive pair of αs?
Answer There are three cases: 1.
Exactly one β and exactly one γ between each pair of αs: between each of the four pairs of αs, the β or the γ can be first—24 ways. The fifth β and fifth γ along with the sequence of the rest of the letters can be considered as three objects to be arranged—3! ways. Altogether, 24 × 3! = 96 ways.
2.
Exactly one β between each pair of αs and two γ s between some pair of αs (or two βs between some pair of αs and exactly one γ between each pair of αs): there are four choices for between which pair of αs the two γ s go and three ways to arrange the two γ s and one β there. There are 23 choices for whether the β or the γ goes first between the other three pairs of αs and two choices for at which end of the arrangement the fifth β goes. Multiplying by 2 for the case of two βs between some pair of αs, we obtain 2 × (4 × 3 × 23 × 2) = 384 ways.
3.
Two βs between some pair of αs and two γ s between some pair of αs: There are two subcases. If the two βs and two γ s are between the same pair of αs, there are four choices for which pair of αs, C(4, 2) ways to arrange them between this pair of αs, and 23 choices for whether the β or the γ goes first between the other three 244 Chapter 5 General Counting Methods for Arrangements and Selections pairs of αs. If two βs and two γ s are between the different pairs of αs, there are 4 × 3 ways to pick between which αs the two βs and then between which αs the two γ s go, 32 ways to arrange the two γ s and one β and to arrange the one γ and two βs, and 22 choices for whether the β or the γ goes first between the other two pairs of αs. Together, 4 × C(4, 2) × 23 + 4 × 3 × 32 × 22 = 1056 ways.
All together, the three cases give us a total of 96 + 384 + 1056 = 1536 arrangements.
Section 5.4 Exercise 27 If n distinct objects are distributed randomly into n distinct boxes, what is the proba-bility that (a) No box is empty?
(b) Exactly one box is empty?
(c) Exactly two boxes are empty?
Answer (a) n!/nn.
(b) Pick which box is empty—n choices—then which other box gets two objects— n −1 choices—then which two objects go into this box—C(n, 2) choices—and then distribute the remaining n −2 objects into the remaining n −2 boxes, one per box—(n −2)! ways. Probability: n × (n −1) × C(n, 2) × (n −2)!/nn.
(c) Pick which two boxes are empty—C(n, 2) choices. Two cases: Case (i). One box has three objects: pick the box with three objects—n −two choices— then pick which three objects go in this box—C(n, 3) choices—and then distribute the remaining n −three objects into the remaining n −three boxes—(n −3)! ways. Case (ii). Two boxes which each have two ob-jects: pick the two boxes with two objects—C(n −2, 2) choices—then pick which two objects go into the first 2-object box and which two ob-jects go into the second 2-object box—C(n, 2) × C(n −2, 2) choices— and then distribute the remaining n −four objects into the remaining n −4 boxes—(n −4)! ways. The probability is C(n, 2) × {(n −2) × C(n, 3) × (n −3)! + C(n −2, 2) × C(n, 2) × C(n −2, 2) × (n −4)!}/nn.
Exercise 28 How many ways are there to distribute eight balls into six boxes with the first two boxes collectively having at most four balls if (a) The balls are identical?
(b) The balls are distinct?
www.itpub.net Supplement: Selected Solutions to Problems in Chapter 5 245 Answer Break into five cases of zero or one or two or three or four balls in first two boxes.
(a) 4 k=0 C(k + 2 −1, k) × C(8 −k + 4 −1, 8 −k).
(b) 4 k=0 C(8, k) × 2k × 48−k: if k distinct balls are in the first two boxes, pick which k balls—C(8, k) ways—then decide for each ball into which of the first two boxes it goes—2k ways—and then distribute remaining 8 −k distinct balls into the other four boxes—48−k ways.
Exercise 47 How many arrangements of the letters in INSTITUTIONAL have all of the following properties simultaneously?
(a) No consecutive Ts (b) The 2 Ns are consecutive (c) Vowels in alphabetical order Answer First, we handle the constraint of consecutive N’s by gluing the two Ns together into one letter. Next there are C((9−2) + 4 −1, (9−2)) = C(10, 7) patterns of three Ts and nine non-Ts (including the double N as a single letter) with no consecutive Ts. There remains the subproblem of ordering the nine non-Ts in the nine positions chosen for them. There are several approaches possible. We use the following strategy. First place the double N in the ordering—nine choices—then place the L—eight choices—then place the S—seven choices. Now there is just one way to put the six vowels in the remaining six places in alphabetical order. The total answer is C(10, 7)× 9 × 8 × 7.
Exercise 52 Among all arrangements of WISCONSIN without any pair of consecutive vowels, what fraction have W adjacent to an I?
Answer Arrangements of WISCONSIN without any pair of consecutive vowels: C[(6 −2) + 4 −1, (6 −2)] = C(7, 4) patterns of vowels and consonants with-out consecutive vowels; three ways to distribute I, I, O among three vowel positions; 6!/2!2! ways to distribute consonants among consonant positions. Denominator in fraction is then C(7, 4) × 3 × 6!/2!2!
For the numerator, we look at three cases: 1.
W is beside an I that is the first vowel. If W is just before the I (glue W and I together), there are C((5 −2) + 4 −1, (5 −2)) = C(6, 3) patterns for distributing 246 Chapter 5 General Counting Methods for Arrangements and Selections the other consonants to assure no consecutive vowels, whereas if W is just after I, there are C((5 −1) + 4 −1, (5 −1)) = C(7, 4) patterns. In either case, there are two ways to order the other vowels and 5!/2!2! ways to arrange the other consonants.
2.
If W is beside an I that is the last vowel, the number of outcomes is the same as in case (a).
3.
If W isbeside anIthatisthemiddlevowel, thereare C((5 −1) + 4 −1, (5 −1)) = C(7, 4) patterns for distributing the other consonants to assure no consecutive vowels whether W is just before or just after the I—two choices for W. Again the other vowels and other consonants can be placed in 2 × 5!/2!2! ways. We must subtract the arrangements with the subsequence IWI (that is, W is the only consonant between two Is): C((5 −1) + 3 −1, (5 −1)) × 2 × 5!/2!2! In total, the numerator is [2 × C(6, 3) + 4 × C(7, 4) −C(6, 4)] × 2 × 5!/2!2!
Exercise 59 How many subsets of six integers chosen (without repetition) from 1, 2, . . . , 20 are there with no consecutive integers (e.g., if 5 is in the subset, then 4 and 6 cannot be in it)?
Answer Form a binary sequence of length 20 with six 1s and 14 0s to represent which integers are in the subset (a 1 in the ith position means that i is in the subset). In this form, we seek all 20-digit binary sequences with six nonconsecutive 1s. C((14 −5) + 7 −1, (14 −5)).
Exercise 61 How many arrangements are there of n 0s and m 1s with k runs of 0s? A run is a consecutive set (1 or more) of the same digit; e.g., 0001110100 has three (underlined) runs of 0s.
Answer Represent a run no matter what its length as an R. Then we arrange m 1s and k Rs with no consecutive Rs. Using the reasoning in Exercise 59, there are C(m + 1, k) arrangements. Next pick how many 0s there are in each run. We select distribute the n 0s into k boxes (runs) with at least one 0 in each box—C(n −1, k −1) ways: C(m + 1, k) × C(n −1, k −1).
Exercise 64 How many ways are there to distribute 20 distinct flags onto 12 distinct flagpoles if (a) In arranging flags on a flagpole, the order of flags from the ground up makes a difference?
(b) No flagpole is empty and the order on each flagpole is counted?
www.itpub.net Supplement: Selected Solutions to Problems in Chapter 5 247 Answer (a) Lay the flagpoles on the ground end-to-end with a slash (|) between flag-poles. Then we need to count all arrangements of the 20 distinct flags and (12 −1)|s—31!/11! ways.
(b) First we distribute 20 identical flags into 12 flagpoles with no flagpole empty in C(20 −1, 12 −1) ways. Now as in part (a), lay the flagpoles end to end. Then arrange the 20 distinct flags among the 20 places where there are (identical) flags in 20! ways: C(20 −1, 12 −1) × 20!
This page is intentionally left blank www.itpub.net CHAPTER 6 GENERATING FUNCTIONS 6.1 GENERATING FUNCTION MODELS In this chapter, we introduce the concept of a generating function. Generating func-tions are developed in this chapter to handle special constraints in selection and arrangement problems with repetition. They are used in Chapters 7, 8, and 9 to solve other combinatorial problems. Generating functions are one of the most abstract problem-solving techniques introduced in this text. But once understood, they are also the easiest way to solve a broad spectrum of combinatorial problems.
Suppose ar is the number of ways to select r objects in a certain procedure. Then g(x) is a generating function for ar if g(x) has the polynomial expansion g(x) = a0 + a1x + a2x2 + · · · + arxr + · · · + anxn If the function has an infinite number of terms, it is called a power series.
In Section 5.5 we verified the well-known binomial expansion (1 + x)n = 1 + n 1 x + n 2 x2 + · · · + n r xr + · · · + n n xn Then g(x) = (1 + x)n is the generating function for ar = C(n,r), the number of ways to select an r-subset from an n-set. Recall that we derived the expansion of (1 + x)n by first considering the formal multiplication of (a + x)3: (a + x)(a + x)(a + x) = aaa + aax + axa + axx + xaa + xax + xxa + xxx When a = 1, we obtained (1 + x)(1 + x)(1 + x) = 111 + 11x + 1x1 + 1xx + x11 + x1x + xx1 + xxx (1) Such a formal expansion lists all ways of multiplying a term in the first factor times a term in the second factor times a term in the third factor. The problem of determining the coefficient of xr in (1 + x)3, and more generally in (1 + x)n, reduces to the problem of counting the number of different formal products with r xs and (n −r) 1s. Such 249 250 Chapter 6 Generating Functions formal products are all sequences of r xs and (n −r) 1s. So the coefficient of xr in (1 + x)3 is C(3,r), and in (1 + x)n is C(n,r).
It is very important that multiplication in a product of several polynomial factors be viewed as generating the collection of all formal products obtained by multiplying together a term from each polynomial factor. If the ith polynomial factor contains ri different terms and there are n factors, then there will be r1 ×r2 ×r3 × · · · ×rn differ-ent formal products. For example, there will be 2n formal products in the expansion of (1 + x)n.
In the expansion of (1 + x + x2)4, the set of all formal products will be sequences of the form ⎧ ⎨ ⎩ 1 x x2 ⎫ ⎬ ⎭· ⎧ ⎨ ⎩ 1 x x2 ⎫ ⎬ ⎭· ⎧ ⎨ ⎩ 1 x x2 ⎫ ⎬ ⎭· ⎧ ⎨ ⎩ 1 x x2 ⎫ ⎬ ⎭ (2) that is, a 1 or an x or an x2 in each of the entries in the product, such as x1x2x.
In this chapter we are primarily concerned with multiplying polynomial factors in which the powers of x in each factor have coefficient 1, factors such as (1 + x + x2 + x3) or (1 + x2 + x4 + x6 + · · ·). These factors are completely specified by the set of different exponents of x. Note that 1 = x0. Thus expansion (1) can be rewritten (x0 + x1)(x0 + x1)(x0 + x1) = x0x0x0 + x0x0x1 + x0x1x0 + x0x1x1 + x1x0x0 + x1x0x1 + x1x1x0 + x1x1x1 And the formal products in expansion (2) can be written as ⎧ ⎨ ⎩ x0 x1 x2 ⎫ ⎬ ⎭· ⎧ ⎨ ⎩ x0 x1 x2 ⎫ ⎬ ⎭· ⎧ ⎨ ⎩ x0 x1 x2 ⎫ ⎬ ⎭· ⎧ ⎨ ⎩ x0 x1 x2 ⎫ ⎬ ⎭ or xe1xe2xe3xe4 0 ≤ei ≤2 (3) The problem of determining the coefficient of xr when we multiply several such polynomial factors together can be restated in terms of exponents. Consider the coefficient of x5 in the expansion of (1 + x + x2)4. It is the number of different formal products, such as x2x0x2x1, formed in expansion (3) whose sum of exponents is 5. Determining the coefficient of x5 in (1 + x + x2)4 can be modeled as an integer-solution-to-an-equation problem (Example 6 of Section 5.4). The number of formal products xe1xe2xe3xe4 0 ≤ei ≤2 equaling x5 is the same as the number of integer solutions among the exponents to e1 + e2 + e3 + e4 = 5 0 ≤ei ≤2 According to the equivalent forms of selection-with-repetition discussed in Sec-tion 5.4, the preceding integer-solution-to-an-equation problem is equivalent to the problem of selecting five objects from a collection of four types, with at most two objects of each type. It is also equivalent to the problem of distributing five identical objects into four distinct boxes with at most two objects in each box.
www.itpub.net 6.1 Generating Function Models 251 More generally, the coefficient of xr in (1 + x + x2)4—that is, the number of formal products xe1xe2xe3xe4 0 ≤ei ≤2 equaling xr—will be the number of integer solutions among the exponents to e1 + e2 + e3 + e4 = r 0 ≤ei ≤2 This problem in turn equals the number of ways of selecting r objects from four types with at most two of each type (or of distributing r identical objects into four boxes with at most two objects in any box). Thus (1 + x + x2)4 is the generating function for ar, the number of ways to select r objects from four types with at most two of each type (or to perform the equivalent distribution).
At this stage, we are concerned only with how to build generating function models for counting problems. In the next section we will see how various algebraic manipulations of generating functions permit us to evaluate desired coefficients.
We have shown how the coefficients of (1 + x + x2)4 can be interpreted as the so-lutions to a certain selection-with-repetition or distribution-of-identical-objects prob-lem. This line of reasoning can be reversed: given a certain selection-with-repetition or distribution problem, we can build a generating function whose coefficients are the answers to this problem. We now give some examples of how to build such generating functions. We use the intermediate model of an integer-solution-to-an-equation to aid in the construction of generating function models.
Example 1 Find the generating function for ar, the number of ways to select r balls from three green, three white, three blue, and three gold balls.
This selection problem can be modeled as the number of integer solutions to e1 + e2 + e3 + e4 = r 0 ≤ei ≤3 Here e1 represents the number of green balls chosen, e2 the number of white, e3 blue, and e4 gold. To be more concrete, suppose r = 6. Then the integer equation model is e1 + e2 + e3 + e4 = 6 0 ≤ei ≤3 We want to construct a product of polynomial factors such that when multiplied out formally (as described above), we obtain all products of the form xe1xe2xe3xe4, with each exponent ei between 0 and 3. In the case r = 6 a possible formal product would be x2x0x1x3. Then we need four factors, and each factor should consist of an “inventory” of the powers of x from which ei is chosen. That is, each factor should be (x0 + x1 + x2 + x3). The desired generating function is thus (x0 + x1 + x2 + x3)4 or (1 + x + x2 + x3)4.
Example 2 Find a generating function for the number of ways to select r doughnuts from five chocolate, five strawberry, three lemon, and three cherry doughnuts. Repeat with the additional constraint that there must be at least one of each type.
252 Chapter 6 Generating Functions The initial selection problem can be modeled as the number of integer solutions to e1 + e2 + e3 + e4 = r, 0 ≤e1, e2 ≤5, 0 ≤e3, e4 ≤3 Here e1 represents the number of chocolate doughnuts selected, e2 the number of strawberry doughnuts, e3 the number of lemon doughnuts, and e4 the number of cherry doughnuts. We want to construct a product of polynomial factors such that when multiplied out formally, we obtain all products of the form xe1xe2xe3xe4 with each ei bounded as in the integer solutions model. For e1 and e2 the factor is (x0 + x1 + x2 + x3 + x4 + x5). For e3 and e4, the factor is (x0 + x1 + x2 + x3). Then the required generating function is (x0 + x1 + x2 + x3 + x4 + x5)2(x0 + x1 + x2 + x3)2.
When the additional constraint is added of at least one doughnut of each type, the integer solution model becomes e1 + e2 + e3 + e4 = r, 1 ≤e1, e2 ≤5, 1 ≤e3, e4 ≤3 Then the polynomial factor for e1 and e2 becomes (x1 + x2 + x3 + x4 + x5) and for e3 and e4 becomes (x1 + x2 + x3). The required generating function is (x1 + x2 + x3 + x4 + x5)2(x1 + x2 + x3)2.
Example 3 Use a generating function to model the problem of counting all selections of six objects chosen from three types of objects with repetition of up to four objects of each type. Also model the problem with unlimited repetition.
This selection problem can be modeled as the number of integer solutions to e1 + e2 + e3 = 6 0 ≤ei ≤4 This problem does not ask for the general solution of the ways to select r objects. How-ever, in building a generating function, we automatically model all values of r, not just r = 6. Wanting a solution to the problem for six objects means that we are interested only in the coefficient of x6—that is, the ways xe1xe2xe3 can equal x6. We build a generating function with a factor of (1 + x + x2 + x3 + x4) for each xei. The desired generating function is (1 + x + x2 + x3 + x4)3, and we want the coefficient of x6 in it.
Permitting unlimited repetition means that any number can be chosen of each type and so any exponent value is possible (although in this particular problem no exponent can exceed 6). From this point of view, the answer is the coefficient of x6 in (1 + x + x2 + x3 + · · ·)3, where “ + · · ·” means that the factor is an infinite series.
In the unlimited repetition case above, we could have used the generating function (1 + x + x2 + x3 + x4 + x5 + x6)3, since we wanted only the coefficient of x6 and thus greater powers of x could not be used. However, we shall see in the next section that it is easier to use infinite series in generating functions. In selection-with-repetition problems, when we ask for six objects from three types with unlimited repetition, www.itpub.net 6.1 Generating Function Models 253 we do not add the constraint of at most six of any type—it is implicit in the problem and there is no need to make it explicit. The same situation applies with generating functions.
Example 4 Find a generating function for ar, the number of ways to distribute r identical objects into five distinct boxes with an even number of objects not exceeding 10 in the first two boxes and between three and five in the other boxes.
While the constraints may be a bit contrived in this example, it is still easy to model the problem as an integer-solution-to-an-equation problem with appropriate constraints. We want to count all integer solutions to e1 + e2 + e3 + e4 + e5 = r e1, e2 even 0 ≤e1, e2 ≤10, 3 ≤e3, e4, e5 ≤5 To generate all formal products of the form xe1xe2xe3xe4xe5, with the given constraints on the eis, we need a product of five factors, each containing the inventory of the powers of x permitted for its xei. For example, the inventory for xe1 is (1 + x2 + x4 + x6 + x8 + x10). The required generating function is g(x) = (1 + x2 + x4 + x6 + x8 + x10)2(x3 + x4 + x5)3.
With a little practice, generating function models become simple to build. How-ever, the concept behind generating functions is far from simple. In the previous chapter, we solved similar selection and distribution problems with explicit formu-las involving binomial coefficients. Now we are modeling these problems as some coefficient, which represents the number of formal products in the multiplication of certain polynomial factors. All we write down is the polynomial factors.
A generating function cannot be designed to model a selection or distribution problem for just one amount—say, six objects. It must model the problem for all possible numbers of objects. A generating function model stores all the necessary information about these subproblems in one function.
In the next section, determining the value of a specified coefficient of a generating function will be reduced to a series of rote algebraic operations. The fact that the function models a complex selection or distribution problem will be irrelevant. The combinatorial reasoning arises solely in the construction of generating functions.
6.1 EXERCISES S u m m a r y o f E x e r c i s e s The first 21 exercises involve simple gen-erating function modeling. The remaining problems involve more challenging mod-eling; Exercises 25–29 use multinomial generating functions.
254 Chapter 6 Generating Functions 1. For each of the following expressions, list the set of all formal products in which the exponents sum to 4.
(a) (1 + x + x3)2(1 + x)2 (b) (1 + x + x2 + x3 + x4)2 (c) (1 + x3 + x4)2(1 + x + x2)2 (d) (1 + x + x2 + x3 + · · ·)3 2. Buildageneratingfunctionforar,thenumberofintegersolutionstothefollowing equations: (a) e1 + e2 + e3 + e4 + e5 = r, 0 ≤ei ≤5 (b) e1 + e2 + e3 = r, 0 < ei < 6 (c) e1 + e2 + e3 + e4 = r, 2 ≤ei ≤7 e1 even, e2 odd (d) e1 + e2 + e3 + e4 = r, 0 ≤ei (e) e1 + e2 + e3 + e4 = r, 0 < ei, e2, e4 odd, e4 ≤3 3. Build a generating function for ar, the number of r selections from (a) Five red, five black, and four white balls (b) Five jelly beans, five licorice sticks, eight lollipops with at least one of each type of candy (c) Unlimited amounts of pennies, nickels, dimes, and quarters (d) Six types of lightbulbs with an odd number of the first and second types 4. Build a generating function for ar, the number of distributions of r identical objects into (a) Five different boxes with at most three objects in each box (b) Three different boxes with between three and six objects in each box (c) Six different boxes with at least one object in each box (d) Three different boxes with at most five objects in the first box 5. Use a generating function for modeling the number of 5-combinations of the letters M, A, T, H in which M and A can appear any number of times but T and H appear at most once. Which coefficient in this generating function do we want?
6. Use a generating function for modeling the number of different selections of r hot dogs when there are four types of hot dogs.
7. Use a generating function for modeling the number of distributions of 16 choco-late bunny rabbits into four Easter baskets with at least three rabbits in each basket. Which coefficient do we want?
8. (a) Use a generating function for modeling the number of different election outcomes in an election for class president if 25 students are voting among four candidates. Which coefficient do we want?
www.itpub.net 6.1 Generating Function Models 255 (b) Suppose each student who is a candidate votes for herself or himself. Now what is the generating function and the required coefficient?
(c) Suppose no candidate receives a majority of the vote. Repeat part (a).
9. Find a generating function for ar, the number of r-combinations of an n-set with repetition.
10. Given one each of u types of candy, two each of ν types of candy, and three each of w types of candy, find a generating function for the number of ways to select r candies.
11. Find a generating function for ak, the number of k-combinations of n types of objects with an even number of the first type, an odd number of the second type, and any amount of the other types.
12. Find a generating function for ar, the number of ways to distribute r identical objects into q distinct boxes with an odd number between r1 and s1 in the first box, an even number between r2 and s2 in the second box, and at most three in the other boxes. Note that r1 and s1 are assumed to be odd numbers; r2 and s2 are assumed to be even numbers.
13. Find a generating function for ar, the number of ways n distinct dice can show a sum of r.
14. Find a generating function for ar, the number of ways a roll of six distinct dice can show a sum of r if (a) The first three dice are odd and the second three even (b) The ith die does not show a value of i 15. Build a generating function for ar, the number of integer solutions to e1 + e2 + e3 + e4 = r, −3 ≤e j ≤3.
16. Find a generating function for the number of integers between 0 and 999,999 whose sum of digits is r.
17. Find a generating function for the number of selections of r sticks of chewing gum chosen from eight flavors if each flavor comes in packets of five sticks.
18. (a) Useageneratingfunctionformodelingthenumberofdistributionsof20iden-tical balls into five distinct boxes if each box has between two and seven balls.
(b) Factor out an x2 from each polynomial factor in part (a). Interpret this revised generating function combinatorially.
19. Use a generating function for modeling the number of ways to select five integers from 1, 2, . . . , n, no two of which are consecutive. Which coefficient do we want for n = 20? For a general n?
20. Explain why (1 + x + x2 + · · · + xr)4 is not a proper generating function for ar, the number of ways to select r objects from four types with repetition. What is the correct generating function?
21. Explain why (1 + x + x2 + x3 + x4)r is not a proper generating function for the number of ways to distribute r jelly beans among r children with no child getting more than four jelly beans.
256 Chapter 6 Generating Functions 22. Show that the generating function for the number of integer solutions to e1 + e2 + e3 + e4 = r, 0 ≤e1 ≤e2 ≤e3 ≤e4, is (1 + x + x2 + · · ·)(1 + x2 + x4 + · · ·) (1 + x3 + x6 + · · ·)(1 + x4 + x8 + · · ·) 23. Find a generating function for the number of ways to make r cents change in pennies, nickels, and dimes.
24. A national singing contest has five distinct entrants from each state. Use a gen-erating function for modeling the number of ways to pick 20 semifinalists if (a) There is at most one person from each state.
(b) There are at most three people from each state.
25. Find a generating function g(x, y) whose coefficient of xr ys is the number of ways to distribute r chocolate bars and s lollipops among five children such that no child gets more than three lollipops.
26. (a) Find a generating function g(x, y, z) whose coefficient xr yszt is the number of ways to distribute r red balls, s blue balls, and t green balls to n people with between three and six balls of each type to each person.
(b) Suppose also that each person gets at least as many red balls as blues.
(c) Suppose also that the first three people get equal numbers of reds and blues.
(d) Suppose also that no one gets the same number of green and red balls.
27. Find a generating function g(x, y, z) whose coefficient of xr yszt is the number of ways eight people can each pick two different fruits from a bowl of apples, oranges, and bananas for a total of r apples, s oranges, and t bananas.
28. Find a generating function (x1, x2, . . . , xm) whose coefficient of xr1 1 xr2 2 . . . xrm m is the number of ways n people can pick a total of r1 chairs of type 1, r2 chairs of type 2, . . .rm chairs of type m if (a) Each person picks one chair (b) Each person picks either two chairs of one type or no chairs at all (c) Person i picks up to i chairs of exactly one type 29. If g(x1, x2, . . . , x p) = (x1 + x2 + · · · + x p)n, p > n, how many terms of g(x1, . . . , x p)’s expansion have no exponent of any xi greater than 1? What is the coefficient of one of these terms?
6.2 CALCULATING COEFFICIENTS OF GENERATING FUNCTIONS We now develop algebraic techniques for calculating the coefficients of generating functions. All these methods seek to reduce a complex generating function to a simple binomial-type generating function or a product of binomial-type generating functions.
www.itpub.net 6.2 Calculating Coefficients of Generating Functions 257 For easy reference, we list in Table 6.1 all the polynomial identities and expansions to be used in this section.
The rule for multiplication of generating functions in (6) is simply the stan-dard formula for polynomial multiplication. Identity (1) can be verified by poly-nomial “long division.” We restate it, multiplying both sides of (1) by (1 −x), as (1 −xm+1) = (1 −x)(1 + x + x2 + · · · + xm). We verify that the product of the right-hand side is 1 −xm+1 by “long multiplication.” 1 + x + x2 + · · · + xm 1 −x 1 + x + x2 + · · · + xm () −x −x2 −x3 · · · −xm −x m+1 1 −xm+1 If m is made infinitely large, so that 1 + x + x2 + · · · + xm becomes the infinite series 1 + x + x2 + · · ·, then the multiplication process (∗) will yield a power series in which the coefficient of each xk, k > 0, is zero [the reader can confirm this in (∗)]. We conclude that (1 −x)(1 + x + x2 + · · ·) = 1. [Analytically, this equation is valid for |x| < 1; the “remainder” term xm+1 in (∗) goes to 0 as m becomes infinite.] Dividing both sides of this equation by (1 −x) yields identity (2).
Expansion (3), the binomial expansion, was explained at the start of Section 6.1.
Expansion (4) is obtained from (3) by expanding (1 + y)n, when y = −xm: [1 + (−xm)]n = 1 + n 1 (−xm) + n 2 (−xm)2 + · · · + n k (−xm)k + · · · + n n (−xm)n Table 6.1 Polynomial Expansions (1) 1 −xm+1 1 −x = 1 + x + x2 + · · · + xm (2) 1 1 −x = 1 + x + x2 + · · · (3) (1 + x)n = 1 + n 1 x + n 2 x2 + · · · + n r xr + · · · + n n xn (4) (1 −xm)n = 1 − n 1 xm + n 2 x2m + · · · + (−1)k n k xkm + · · · + (−1)n n n xnm (5) 1 (1 −x)n = 1 + 1 + n −1 1 x + 2 + n −1 2 x2 + · · · + r + n −1 r xr + · · · (6) If h(x) = f (x)g(x), where f (x) = a0 + a1x + a2x2 + · · · and g(x) = b0 + b1x + b2x2 + · · ·, then h(x) = a0b0 + (a1b0 + a0b1)x + (a2b0 + a1b1 + a0b2)x2 + · · · + (arb0 + ar−1b1 + ar−2b2 + · · · + a0br)xr + · · · 258 Chapter 6 Generating Functions By identity (2), (1 −x)−n, or equivalently 1 1−x n, equals (1 + x + x2 + x3 + · · ·)n (7) Let us determine the coefficient of xr in (7) by counting the number of formal products whose sum of exponents is r. If ei represents the exponent of the ith term in a formal product, then the number of formal products xe1xe2xe3 . . . xen whose exponents sum to r is the same as the number of integer solutions to the equation e1 + e2 + e3 + · · · + en = r ei ≥0 In Example 5 of Section 5.4, we showed that the number of nonnegative integer solutions to this equation is C(r + n −1,r). Thus the coefficient of xr in (7) is C(r + n −1,r). This verifies expansion (5).
With formulas (1) to (6) we can determine the coefficients of a variety of gener-ating functions: first, perform algebraic manipulations to reduce a given generating function to one of the forms (1 + x)n, (1 −xm)n, or (1 −x)−n, or a product of two such expansions; then use expansions (3) to (5) and the product rule (6) to obtain any desired coefficient. We illustrate some common reduction methods in the following examples.
Example 1 Find the coefficient of x16 in (x2 + x3 + x4 + · · ·)5. What is the coefficient of xr?
To simplify the expression, we extract x2 from each polynomial factor and then apply identity (2).
(x2 + x3 + x4 + · · ·)5 = [x2(1 + x + x2 + · · ·)]5 = x10(1 + x + x2 + · · ·)5 = x10 1 (1 −x)5 Thus the coefficient of x16 in (x2 + x3 + x4 + · · ·)5 is the coefficient of x16 in x10 (1 −x)−5. But the coefficient of x16 in this latter expression will be the coefficient of x6 in (1 −x)−5 [i.e., the x6 term in (1 −x)−5 is multiplied by x10 to become the x16 term in x10(1 −x)−5]. From expansion (5), we see that the coefficient of x6 in (1 −x)−5 is C(6 + 5 −1, 6).
More generally, the coefficient of xr in x10(1 −x)−5 equals the coefficient of xr−10 in (1 −x)−5—namely, C((r −10) + 5 −1, (r −10)).
Observe that (x2 + x3 + x4 + · · ·)5 is the generating function ar, for the number of ways to select r objects with repetition from five types with at least two of each type. In the last chapter, we solved such a problem by first picking two objects in each type—one way—and then counting the ways to select the remaining r −10 objects—C((r −10) + 5 −1, (r −10)) ways. In the generating function analysis in Example 1, we algebraically picked out an x2 from each factor for a total of x10 and then found the coefficient of xr−10 in (1 + x + x2 + · · ·)5, the generating function for selection with unrestricted repetition of r −10 from five types.
www.itpub.net 6.2 Calculating Coefficients of Generating Functions 259 The standard algebraic technique of extracting the highest common power of x from each factor corresponds to the “trick” used to solve the associated selection problem. Such correspondences are a major reason for using generating functions: the algebraic techniques automatically do the combinatorial reasoning for us.
Example 2 Use generating functions to find the number of ways to collect $15 from 20 distinct people if each of the first 19 people can give a dollar (or nothing) and the twentieth person can give either $1 or $5 (or nothing).
This collection problem is equivalent to finding the number of integer solutions to x1 + x2 + · · · + x19 + x20 = 20 when xi = 0 or 1, i = 1, 2, . . . 19, and x20 = 0 or 1 or 5. The generating function for this integer-solution-of-an-equation problem is (1 + x)19(1 + x + x5). We want the coefficient of x15. The first part of this generating function has the binomial expansion (1 + x)19 = 1 + 19 1 x + 19 2 x2 + · · · + 19 r xr + · · · + 19 19 x19 If we let f (x) be this first polynomial and let g(x) = 1 + x + x5, then we can use (6) to calculate the coefficient of x15 in h(x) = f (x)g(x). Let ar be the coefficient of xr in f (x) and br the coefficient of xr in g(x). We know that ar = 19 r and that b0 = b1 = b5 = 1 (other bis are zero).
Then the coefficient of x15 in h(x) = f (x)g(x) is, by (6), a15b0 + a14b1 + a13b2 + · · · + a0b15 which reduces to a15b0 + a14b1 + a10b5 since b0, b1, b5 are the only nonzero coefficients in g(x). Substituting the values of the various as and bs in (8), we have 19 15 × 1 + 19 14 × 1 + 19 10 × 1 = 19 15 + 19 14 + 19 10 The answer in Example 2 could be obtained directly by breaking the collection problem into three cases depending on how much the twentieth person gives: $0 or $1 or $5. In each case, the subproblem is counting the ways to pick a subset of the other 19 people to obtain the rest of the $15. The generating function approach automatically breaks the problem into three cases and solves each, doing all the combinatorial reasoning for us.
Example 3 How many ways are there to distribute 25 identical balls into seven distinct boxes if the first box can have no more than 10 balls but any number can go into each of the other six boxes?
260 Chapter 6 Generating Functions The generating function for the number of ways to distribute r balls into seven boxes with at most 10 balls in the first box is (1 + x + x2 + · · · + x10)(1 + x + x2 + · · ·)6 = 1 −x11 1 −x 1 1 −x 6 = (1 −x11) 1 1 −x 7 usingidentities(1)and(2).Let f (x) = 1 −x11 and g(x) = (1 −x)−7.Usingexpansion (5), we have g(x) = (1 −x)−7 = 1 + 1 + 7 −1 1 x + 2 + 7 −1 2 x2 + · · · + r + 7 −1 r xr + · · · We want the coefficient of x25 (25 balls distributed) in h(x) = f (x)g(x). As in Example 2, we need to consider only the terms in the product of the two polynomials (1 −x11) and 1 1−x 7 that yield an x25 term. The only nonzero coefficients in f (x) = (1 −x11) are a0 = 1 and a11 = −1. So the coefficient of x25 in f (x)g(x) is a0b25 + a11b14 = 1 × 25 + 7 −1 25 + (−1) × 14 + 7 −1 14 The combinatorial interpretation of the answer in Example 3 is that we count all the ways to distribute without restriction the 25 balls into the seven boxes, C(25 + 7 − 1, 25) ways, and then subtract the distributions that violate the first box constraint, that is, distributions with at least 11 balls in the first box, C((25 −11) + 7 −1, (25 −11)) (first put 11 balls in the first box and then distribute the remaining balls arbitrarily). Again, generating functions automatically performed this combinatorial reasoning.
The next example employs all the techniques used in the first three examples to solve a problem that cannot be solved by the combinatorial methods of the previous chapter.
Example 4 How many ways are there to select 25 toys from seven types of toys with between two and six of each type?
The generating function for ar, the number of ways to select r toys from seven types with between two and six of each type, is (x2 + x3 + x4 + x5 + x6)7 We want the coefficient of x25. As in Example 1, we extract x2 from each factor to get [x2(1 + x + x2 + x3 + x4)]7 = x14(1 + x + x2 + x3 + x4)7 www.itpub.net 6.2 Calculating Coefficients of Generating Functions 261 Now we reduce our problem to finding the coefficient of x25−14 = x11 in (1 + x + x2 + x3 + x4)7. Using identity (1), we can rewrite this generating function as (1 + x + x2 + x3 + x4)7 = 1 −x5 1 −x 7 = (1 −x5)7 1 1 −x 7 Let f (x) = (1 −x5)7 and g(x) = (1 −x)−7. By expansions (4) and (5), respectively, we have f (x) = (1 −x5)7 = 1 − 7 1 x5 + 7 2 x10 − 7 3 x15 + · · · g(x) = 1 1 −x 7 = 1 + 1 + 7 −1 1 x + 2 + 7 −1 2 x2 + · · · + r + 7 −1 r xr + · · · To find the coefficient of x11, we need to consider only the terms in the product of the two polynomials (1 −x5)7 and 1 1−x 7 that yield x11. The only nonzero coefficients in f (x) = (1 −x5)7 with a subscript ≤11 (larger subscripts can be ignored) are a0, a5, and a10 [see the expansion of f (x) above]. The products involving these three coefficients that yield x11 terms are a0b11 + a5b6 + a10b1 = 1 × 11 + 7 −1 11 + − 7 1 × 6 + 7 −1 6 + 7 2 × 1 + 7 −1 1 The following combinatorial interpretation can be given to the final answer in Ex-ample 4. The first term, C(11 + 7 −1, 11), counts the number of ways to select 11 toys from seven types of toys with no restriction, where 11 is the number of additional toys to select after we first pick two toys of each type. The next term, −7C(6 + 7 −1, 6), subtracts seven cases of a violation where we pick at least five additional toys of some type (we pick five of some type and then pick 11 −5 = 6 more toys from the 7 types). The final term, C(7, 2)C(1 + 7 −1, 1) adds back all C(7, 2) cases where some pair of violations occurred—that is, with at least five chosen from a pair of types [we pick five of two types and then pick 11 −(2 × 5) = 1 more toy from the seven types]. The logic behind this combinatorial approach will be developed in Chapter 8.
We close this section by showing how generating functions can be used to verify binomial identities. We express the right side of the identity as a particular coefficient in some generating function h(x) and the left side as the same coefficient in a product of generating functions f (x) and g(x), where h(x) = f (x)g(x).
Example 5: Binomial Identity Verify the binomial identity n 0 2 + n 1 2 + · · · + n n 2 = 2n n 262 Chapter 6 Generating Functions The right-hand side of the identity is the coefficient of xn in (1 + x)2n. The left-hand side terms involves coefficients in (1 + x)n. The product of generating functions we want is (1 + x)n(1 + x)n. That is, let f (x) = g(x) = (1 + x)n so that f (x)g(x) = h(x) = (1 + x)2n. Then ar = br = C(n,r). By the product rule (6), the coefficient of xn in f (x)g(x) is a0bn + a1bn−1 + · · · + anb0 = n 0 n n + n 1 n n −1 + · · · + n 0 n n = n 0 2 + n 1 2 + · · · + n n 2 since n r = n n −r Equating coefficients of xn in (1 + x)n(1 + x)n = (1 + x)2n, we obtain the required identity.
The reader is encouraged to compare the generating function proof of the pre-ceding combinatorial identity with the combinatorial proof of the same identity given in Example 3 of Section 5.5.
6.2 EXERCISES S u m m a r y o f E x e r c i s e s The first 29 exercises are similar to the examples of coefficient calculation in this section. Exercises 31–36 involve binomial identities (several of these problems are quite tricky). Exercises 38–42 and 44 intro-duce the topic of probability generating functions (some background in probability is helpful).
1. Find the coefficient of x10 in (1 + x + x2 + x3 + · · ·)n.
2. Find the coefficient of xr in (x5 + x6 + x7 + · · ·)8.
3. Find the coefficient of x7 in (1 + x2 + x4)(1 + x)m.
4. Find the coefficient of x16 in (x + x2 + x3 + x4 + x5)(x2 + x3 + x4 + · · ·)5.
5. Find the coefficient of x17 in (x2 + x3 + x4 + x5 + x6 + x7)3.
6. Find the coefficient of x47 in (x10 + x11 + · · · + x25)(x + x2 + · · · + x15) (x20 + · · · + x45).
7. Find the coefficient of x32 in (x3 + x4 + x5 + x6 + x7)7.
8. Find the coefficient of x24 in (x + x2 + x3 + x4 + x5)8.
9. Find the coefficient of x16 in (x + x2 + x3 + x4 + x5 + x6 + x7)4.
10. Find the coefficient of x36 in (x2 + x3 + x4 + x5 + x6 + x7 + x8)5.
www.itpub.net 6.2 Calculating Coefficients of Generating Functions 263 11. Find the coefficient of x11 in (a) x2(1 −x)−10 (d) x + 3 1 −2x + x2 (b) x2 −3x (1 −x)4 (e) bmxm (1 −bx)m+1 (c) (1 −x2)5 (1 −x)5 12. Give a formula similar to (1) for (a) 1 + x4 + x8 + · · · + x24 (b) x20 + x40 + · · · + x180 13. Find the coefficient of x8 in (x2 + x3 + x4 + x5)5.
14. Find the coefficient of x18 in (1 + x3 + x6 + x9 + · · · )6.
15. Find the coefficient of x12 in (a) (1 −x)8 (b) (1 + x)−1 (c) (1 + x)−8 (d) (1 −4x)−5 (e) (1 + x3)−4 16. Find the coefficient of x25 in (1 + x3 + x8)10.
17. Use generating functions to find the number of ways to select 10 balls from a large pile of red, white, and blue balls if (a) The selection has at least two balls of each color (b) The selection has at most two red balls (c) The selection has an even number of blue balls 18. Use generating functions to find the number of ways to distribute r jelly beans among eight children if (a) Each child gets at least one jelly bean (b) Each child gets an even number of beans 19. How many ways are there to place an order for 12 chocolate sundaes if there are five types of sundaes, and at most four sundaes of one type are allowed?
20. How many ways are there to paint the 10 identical rooms in a hotel with five colors if at most three rooms can be painted green, at most three painted blue, at most three red, and no constraint is laid on the other two colors, black and white?
21. How many ways are there to distribute 20 cents to n children and one parent if the parent receives either a nickel or a dime and (a) The children receive any amounts?
(b) Each child receives at most 1c /?
22. How many ways are there to get a sum of 25 when 10 distinct dice are rolled?
264 Chapter 6 Generating Functions 23. How many ways are there to select 300 chocolate candies from seven types of candy if each type comes in boxes of 20 and if at least one but not more than five boxes of each type are chosen? (Hint: Solve in terms of boxes of chocolate.) 24. How many different committees of 40 senators can be formed if the two senators from the same state (50 states in all) are considered identical?
25. How many ways are there to split six copies of one book, seven copies of a second book, and 11 copies of a third book between two teachers if each teacher gets 12 books and each teacher gets at least two copies of each book?
26. How many ways are there to divide five pears, five apples, five doughnuts, five lollipops, five chocolate cats, and five candy rocks into two (unordered) piles of 15 objects each?
27. How many ways are there to collect $24 from four children and six adults if each person gives at least $1, but each child can give at most $4 and each adult at most $7?
28. If a coin is flipped 25 times with eight tails occurring, what is the probability that no run of six (or more) consecutive heads occurs?
29. If 10 steaks and 15 lobsters are distributed among four people, how many ways are there to give each person at most five steaks and at most five lobsters?
30. Show that (1 −x −x2 −x3 −x4 −x5 −x6)−1 is the generating function for the number of ways a sum of r can occur if a die is rolled any number of times.
31. Use generating functions to show that r k=0 m k n r −k = m + n r 32. Use the equation (1 −x2)n (1 −x)n = (1 + x)n to show that m/2 k=0 (−1)k n k n + m −2k −1 n −1 = n m m ≤n and m even 33. Use binomial expansions to evaluate (a) m k=0 m k n r + k (b) r k=0 (−1)k n k n r −k (c) n k=0 2k n k 34. (a) Evaluate n2 k=n1 k + 5 k .
(b) Evaluate m k=0 n −k m −k .
35. (a) Show that ∞ k=0 1 2 k n + k −1 k = 2n.
(b) Evaluate ∞ k=0 1 2 k k.
www.itpub.net 6.2 Calculating Coefficients of Generating Functions 265 36. Why can’t you set x = −1 in formula (5) to “prove” the following?
1 2 n = ∞ k=0 (−1)k n + k −1 k 37. If g(x) is the generating function for ak, then show that g(k)(0)/k! = ak.
38. A probability generating function PX(t) for a discrete random variable X has a polynomial expansion in which pr, the coefficient of tr, is equal to the probability that X = r.
(a) If X is the number of heads that occur when a fair coin is flipped n times, show that PX(t) = ( 1 2)n(1 + t)n.
(b) If X is the number of heads that occur when a biased coin is flipped n times with probability p of heads (and q = 1 −p), show that PX(t) = (q + pt)n.
(c) If X is the number of times a fair coin is flipped until the fifth head occurs, find PX(t).
(d) Repeat part (c) until the mth head occurs and probability of a head is p.
39. (a) The expected value E(X) of a discrete random variable X is defined to be prr. Show that E(X) = P′ X(1)—that is, (d/dt)PX(t), with t set equal to 1.
(b) Find E(X) for the random variables X in Exercise 38.
40. (a) The second moment E2(X) of a discrete random variable X is defined to be prr2. Show that E2(X) = P′ X(1) + P′′ X(1).
(b) Find E2(X) for the random variables X in Exercises 38(b) and 38(c).
41. Suppose a fair coin is flipped until the mth head occurs and suppose that no more than s tails in a row occur. If X is the number of flips, find PX(t).
42. Experiments A′ and A′′ have probabilities p′ and p′′ of success in each trial and are performed n′ and n′′ times, respectively. Let X′ and X′′ be the number of successes in the respective experiments, and let X be the total number of successes on both experiments. Verify the following.
(a) PX(t) = P′ X(t)P′′ X(t) (b) E(X) = E(X′) + E(X′′) (c) E2(X) = E2(X′) + E2(X′′) + E(X′X′′) 43. Suppose a red die is rolled once and then a green die is rolled as many times as the value on the red die. If ar is the number of ways that the (variable length) sequence of rolls of the green die can sum to r, show that the generating function for ar is f ( f (x)), where f (x) = (x + x2 + x3 + x4 + x5 + x6).
44. Suppose X is the random variable of the number of minutes it takes to serve a person at a fast-food stand. Suppose Y is the random variable of the number of people who line up to be served at the stand in one minute. Let Z be the number of people who line up while a person is being served. Show that PZ(t) = PX(PY(t)).
266 Chapter 6 Generating Functions 6.3 PARTITIONS In this section we discuss partitions and their generating functions. Unfortunately, there is no easy way to calculate the coefficients of most of these generating functions.
A partition of a group of r identical objects divides the group into a collection of (unordered) subsets of various sizes. Analogously, we define a partition of the integer r to be a collection of positive integers whose sum is r. Normally we write this collection as a sum and list the integers of the partition in increasing order. For example, the seven partitions of the integer 5 are 1 + 1 + 1 + 1 + 1 1 + 1 + 1 + 2 1 + 1 + 3 1 + 2 + 2 1 + 4 2 + 3 5 Note that 5 is a “trivial” partition of itself.
Let us construct a generating function for ar, the number of partitions of the integer r. A partition of an integer is described by specifying how many 1s, how many 2s, and so on, are in the sum. Let ek denote the number of ks in a partition. Then 1e1 + 2e2 + 3e3 + · · · + kek + · · · +rer = r Intuitively, we can think of picking r objects from an unlimited number of piles where the first pile contains single objects, the second pile contains objects stuck together in pairs, the third pile contains objects stuck together in triples, and so on. To model this integer-solution-to-an-equation problem with a generating func-tion, we need polynomial factors whose formal multiplication yields products of the form ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ x0 x1 x2 .
.
.
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ (x2)0 (x2)1 (x2)2 .
.
.
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ (x3)0 (x3)1 (x3)2 .
.
.
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ · · · ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ (xk)0 (xk)1 (xk)2 .
.
.
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ · · · The generating function g(x) must be g(x) = (1 + x + x2 + x3 + · · · + xn + · · ·) · (1 + x2 + x4 + x6 + · · · + x2n + · · ·) · (1 + x3 + x6 + x9 + · · · + x3n + · · ·) .
.
.
· (1 + xk + x2k + x3k + · · · + xkn + · · ·) .
.
.
If we set y = x2, then the second factor in g(x) becomes 1 + y + y2 + · · · = (1 −y)−1.
Thus 1 + x2 + x4 + x6 + · · · + x2n + · · · = (1 −x2)−1 www.itpub.net 6.3 Partitions 267 Similarly, the kth factor can be written as (1 −xk)−1. For partitions up to r = m, we need the first m polynomial factors. But for arbitrary values of r, we need an infinite number of polynomial factors. Then g(x) = 1 (1 −x)(1 −x2)(1 −x3) · · · (1 −xk) · · · We now consider some more specialized partition problems.
Example 1 Find the generating function for ar, the number of ways to express r as a sum of distinct integers.
We must constrain the standard partition problem not to allow any repetition of an integer. For example, there are three ways to write 5 as a sum of distinct integers 1 + 4, 2 + 3, and 5. The appropriate modification of the generating function for unrestricted partitions is g(x) = (1 + x)(1 + x2)(1 + x3)(1 + x4) · · · (1 + xk) · · · Example 2 Find a generating function for ar, the number of ways that we can choose 2c /, 3c /, and 5c / stamps adding to a net value of rc /.
The problem is equivalent to the number of integer solutions to 2e2 + 3e3 + 5e5 = r 0 ≤e2, e3, e5 The appropriate generating function is (1 + x2 + x4 + x6 + · · ·)(1 + x3 + x6 + x9 + · · ·) · (1 + x5 + x10 + x15 + · · ·) Example 3 Show with generating functions that every positive integer can be written as a unique sum of distinct powers of 2.
The generating function for ar, the number of ways to write an integer r as a sum of distinct powers of 2, will be similar to the generating function for sums of distinct integers in Example 1, except that now only integers that are powers of 2 are used.
The generating function is g∗(x) = (1 + x)(1 + x2)(1 + x4)(1 + x8) · · · (1 + x2k) · · · To show that every integer can be written as a unique sum of distinct powers of 2, we must show that the coefficient of every power of x in g∗(x) is 1. That is, show that g∗(x) = 1 + x + x2 + x3 + · · · = 1 1 −x 268 Chapter 6 Generating Functions or equivalently (1 −x)g∗(x) = 1 We prove this identity by repeatedly using the factorization (1 −xk)(1 + xk) = 1 −x2k.
(1 −x)g∗(x) = (1 −x)(1 + x)(1 + x2) (1 + x4)(1 + x8) · · · = (1 −x2) (1 + x4)(1 + x8) · · · = [ (1 −x4) ] (1 + x4)(1 + x8) · · · .
.
.
= 1 By successively making the replacement (1 −xk)(1 + xk) = 1 −x2k, we even-tually eliminate all factors in (1 −x)g∗(x). Formally, the coefficient of any specific xk in (1 −x)g∗(x) must be 0. So (1 −x)g∗(x) = 1 and g∗(x) = 1 + x + x2 + · · ·, as required.
A convenient tool for studying partitions is a diagram known as Ferrers diagram.
A Ferrers diagram displays a partition of r dots in a set of rows listed in order of decreasing size. The partition 1 + 2 + 2 + 3 + 7 of 15 is shown in the Ferrers diagram in Figure 6.1a. If we transpose the rows and columns of a Ferrers diagram of a partition of r, we get a Ferrers diagram of another partition of r. This diagram is called the conjugate of the original Ferrers diagram. For example, Figure 6.1b shows the conjugate of the Ferrers diagram in Figure 6.1a. This new Ferrers diagram represents the partition of 15, 1 + 1 + 1 + 1 + 2 + 4 + 5. Clearly, transposing is unique: two Ferrers diagrams have equal conjugates if and only if they are equal.
Example 4 Show that the number of partitions of an integer r as a sum of m positive integers is equal to the number of partions of r as a sum of positive integers, the largest of which is m.
(a) (b) Figure 6.1 www.itpub.net 6.3 Partitions 269 If we draw a Ferrers diagram of a partition of r into m parts, then the Ferrers diagram will have m rows. The transposition of such a diagram will have m columns, that is, the largest row will have m dots. Thus there is a one-to-one correspondence between these two classes of partitions.
6.3 EXERCISES S u m m a r y o f E x e r c i s e s Exercises 1–10 involve generating func-tion models for partitions. The next seven exercises use Ferrers diagrams to verify partition identities. The next five exercises are more difficult partition problems.
1. List all partitions of the integer: (a) 4, (b) 6.
2. Find a generating function for ar, the number of partitions of r into (a) Even integers (b) Distinct odd integers 3. Find a generating function for the number of ways to write the integer r as a sum of positive integers in which no integer appears more than three times.
4. Find a generating function for the number of integer solutions of 2x + 3y + 7z = r with (a) x, y, z ≥0 (b) 0 ≤z ≤2 ≤y ≤8 ≤x 5. Find a generating function for the number of ways to make r cents’ change in pennies, nickels, dimes, and quarters.
6. Find a generating function for the number of ways to distribute r identical objects into (a) Three indistinguishable boxes (b) n indistinguishable boxes (n ≤r) 7. (a) Show that the number of partitions of 10 into distinct parts (integers) is equal to the number of partitions of 10 into odd parts by listing all partitions of these two types.
(b) Show algebraically that the generating function for partitions of r into distinct parts equals the generating function for partitions of r into odd parts, and hence the numbers of these two types of partitions are equal.
8. Show with generating functions that every positive integer has a unique decimal representation.
9. (a) Prove the result in Example 3 by induction.
(b) Prove the result of Example 3 directly by recursively substituting (1 + xk) = (1 −x2k)/(1 −xk) in g∗(x).
10. The equation in Example 3, g∗(x)(1 −x) = 1, can be rewritten 1 −x = 1/g∗(x).
Use this latter equation to prove that among all partitions of an integer r,r ≥2, into powers of 2, there are as many such partitions with an odd number of parts as with an even number of parts.
270 Chapter 6 Generating Functions 11. Let R(r, k) denote the number of partitions of the integer r into k parts.
(a) Show that R(r, k) = R(r −1, k −1) + R(r −k, k).
(b) Show that r k=1 R(n −r, k) = R(n,r).
12. Use a Ferrers diagram to show that the number of partitions of an integer into parts of even size is equal to the number of partitions into parts such that each part occurs an even number of times.
13. Interpret the integer multiplication mn, “m times n,” to be the sum of m ns. Prove that mn = nm.
14. Show that the number of partitions of the integer n into three parts equals the number of partitions of 2n into three parts of size < n.
15. Show that the number of partitions of n is equal to the number of partitions of 2n into n parts.
16. Show that any number of partitions of 2r + k into r + k parts is the same for any k.
17. Show that the number of partitions of r + k into k parts is equal to (a) The number of partitions of r + k+1 2 into k distinct parts (b) The number of partitions of r into parts of size ≤k 18. Show that the number of ordered partitions of n is 2n−1. For example, the ordered partitions of 4 are: 4, 1 + 3, 2 + 2, 3 + 1, 1 + 1 + 2, 1 + 2 + 1, 2 + 1 + 1, 1 + 1 + 1 + 1. (Hint: Write n 1s in a row and determine all the ways to partition this sequence into clusters of 1s).
19. (a) Find a generating function for an, the number of partitions that add up to at most n.
(b) Find a generating function for an, the number of partitions of n into three parts in which no part is larger than the sum of the other two.
(c) Find a generating function for an, the number of different (incongruent) triangles with integral sides and perimeter n.
20. Show that 2(1 −x)−3[(1 −x)−3 + (1 + x)−3] is the generating function for the number of ways to toss r identical dice and obtain an even sum.
21. (a) A partition of an integer r is self-conjugate if the Ferrers diagram of the partition is equal to its own transpose. Find a one-to-one correspondence between the self-conjugate partitions of r and the partitions of r into distinct odd parts.
(b) The largest square of dots in the upper left-hand corner of a Ferrers diagram is called the Durfee square of the Ferrers diagram. Find a generating function for the number of self-conjugate partitions of r whose Durfee square is size k (a k × k array of dots). (Hint: Use a 1 −1 correspondence between these and the partitions of r −k2 into even parts of size at most 2k.) www.itpub.net 6.4 Exponential Generating Functions 271 (c) Show that (1 + x)(1 + x3)(1 + x5) . . .
= 1 + ∞ k=1 xk2 (1 −x2)(1 −x4)(1 −x6) . . . (1 −x2k) 22. Let n k denote the number of partitions of n distinct objects into k nonempty subsets. Show that n+1 k = k n k + n k−1 .
23. Write a computer program to determine the number of all partitions of an inte-ger r (a) Into k parts (b) Into any number of parts (c) Into distinct parts 6.4 EXPONENTIAL GENERATING FUNCTIONS In this section we discuss exponential generating functions. They are used to model and solve problems involving arrangements with repetition. The generating functions used in the previous sections to model selection with repetition problems are called ordinary generating functions. Exponential generating functions involve a more complicated modeling process than do ordinary generating functions. Consider the problem of finding the number of different words (arrangements) of four letters when the letters are chosen from an unlimited supply of as, bs and cs, and the word must contain at least two as. The possible selections of four letters to form the word are {a, a, a, a}, {a, a, a, b}, {a, a, a, c}, {a, a, b, b}, {a, a, b, c}, and {a, a, c, c}. The number of arrangements possible with each of these six selections is 4!
4!0!0!
4!
3!1!0!
4!
3!0!1!
4!
2!2!0!
4!
2!1!1!
4!
2!0!2!
respectively. So the total number of words will be the sum of these six terms.
The selections of letters used in such a word range over all sets of four letters chosen with repetition from as, bs, and cs with at least two as. Equivalently, the number of such selections equals the number of integer solutions to the equation e1 + e2 + e3 = 4 2 ≤e1, 0 ≤e2, e3 (1) At first sight, the four-letter words problem seems similar to previous problems that could be modeled by (ordinary) generating functions. However, in this case we do not want each integer solution of (1) to contribute 1 to the count of the num-ber of possible words. Instead it must contribute 4!/(e1!e2!e3!) words. In terms of 272 Chapter 6 Generating Functions generating functions, the coefficient of x4 will count all formal products with associ-ated coefficients (e1 + e2 + e3)!
e1!e2!e3!
xe1xe2xe3 2 ≤e1, 0 ≤e2, e3 whose exponents sum to 4 (a much harder problem). Fortunately, exponential gener-ating functions yield formal products of exactly this form.
An exponential generating function g(x) for ar, the number of arrangements with r objects, is a function with the power series expansion g(x) = a0 + a1x + a2 x2 2! + a3 x3 3! + · · · + ar xr r! + · · · We build exponential generating functions in the same way that we build ordinary generating functions: one polynomial factor for each type of object; each factor has a collection of powers of x that are an inventory of the choices for the number of objects of that type. However, now each power xr is divided by r!.
As an example, let us consider the four-letter word problem with at least two as.
We claim that the exponential generating function for the number of r-letter words formed from an unlimited number of as, bs, and cs containing at least two as is g(x) = x2 2! + x3 3! + x4 4! + · · · 1 + x + x2 2! + x3 3! + x4 4! + · · · 2 (2) The coefficient of xr in (2) will be the sum of all products (xe1/e1!) (xe2/e2!) (xe3/e3!), where e1 + e2 + e3 = r, 2 ≤e1, 0 ≤e2, e3. If we divide xr by r! and com-pensate by multiplying its coefficient by r!, then the xr term in g(x) becomes e1+e2+e3=r r!
e1!e2!e3!
xr r!
2 ≤e1, 0 ≤e2, e3 This coefficient of xr/r! is just what we wanted. So the exponential generating func-tion for the number of such r-letter words is indeed the expression (2). What makes exponential generating functions work is the “trick” of dividing xr by r! and multi-plying the coefficient of xr by r!.
Before we show how to calculate coefficients of an exponential generating func-tion, let us give a few other examples of exponential generating function models.
Example 1 Find the exponential generating function for ar, the number of r arrangements without repetition of n objects.
We know that the answer is P(n,r). Since there is no repetition, the exponential generating function is (1 + x)n—our old binomial friend! The coefficient of xr in www.itpub.net 6.4 Exponential Generating Functions 273 (1 + x)n is n r . However, now we want the coefficient of xr r! in (1 + x)n: n r xr = n!
(n −r)!r!xr = n!
(n −r)!
xr r!
Thus ar = n!/(n −r)! = P(n,r), as expected.
Example 2 Find the exponential generating function for ar, the number of different arrangements of r objects chosen from four different types of objects with each type of object appearing at least two and no more than five times.
The number of ways to arrange the r objects with ei objects of the ith type is r!/e1!e2!e3!e4! To sum up all such terms with the given constraints, we want the coefficient of xr/r! in (x2/2! + x3/3! + x4/4! + x5/5!)4.
Example 3 Find the exponential generating function for the number of ways to place r (distinct) people into three different rooms with at least one person in each room. Repeat with an even number of people in each room.
Recall that distributions of distinct objects are equivalent to arrangements with repetition; see Section 5.4. The required exponential generating functions are x + x2 2! + x3 3! + x4 4! + · · · 3 and 1 + x2 2! + x4 4! + · · · 3 There are few identities or expansions to use in conjunction with exponential generating functions. As a result, there are only a limited number of exponential generating functions whose coefficients can be easily evaluated. The fundamental expansion for exponential generating functions is ex = 1 + x + x2 2! + x3 3! + · · · + xr r! + · · · (3) Replacing x by nx in (3), we obtain enx = 1 + nx + n2x2 2!
+ n3x3 3!
+ · · · + nrxr r!
+ · · · (4) The power series in (3) is the Taylor series for ex, which is derived in all calculus texts.
This expansion and the companion expansion for enx are valid for all values of x. There is no way to factor out a common power of x in exponential generating functions, since the power of x must be matched with r! in the denominator. The best that can be done is to subtract the missing (lowest) powers of x from ex. For example, the exponential factor representing two or more of a certain type of objects can be written x2 2! + x3 3! + x4 4! + · · · = ex −1 −x 274 Chapter 6 Generating Functions Two useful expansions derived from (3) are 1 2(ex + e−x) = 1 + x2 2! + x4 4! + x6 6! + · · · (5) 1 2(ex −e−x) = x + x3 3! + x5 5! + x7 7! + · · · (6) Let us work some examples using (3) to (6).
Example 4 Find the number of different r arrangements of objects chosen from unlimited supplies of n types of objects.
In Chapter 5 we would have solved this problem by arguing that there are n choices for the type of object in each of the r positions, for a total of nr different arrangements. Now let us solve this problem with exponential generating functions.
The exponential generating function for this problem is 1 + x + x2 2! + x3 3! + · · · n = (ex)n = enx By (4), the coefficient of xr/r! in this generating function is nr.
Example 5 Find the number of ways to place 25 people into three rooms with at least one person in each room.
In Example 3, we found the exponential generating function for this problem x + x2 2! + x3 3! + · · · 3 = (ex −1)3 To find the coefficient of xr/r! in (ex −1)3, we first must expand this binomial expression in ex (ex −1)3 = e3x −3e2x + 3ex −1 From (4), we get e3x −3e2x + 3ex −1 = ∞ r=0 3r xr r! −3 ∞ r=0 2r xr r! + 3 ∞ r=0 xr r! −1 So the coefficient of x25/25! is 325 −(3 × 225) + 3.
Example 6 Find the number of r-digit quaternary sequences (whose digits are 0, 1, 2, and 3) with an even number of 0s and an odd number of 1s.
www.itpub.net 6.4 Exponential Generating Functions 275 The exponential generating function for this problem is 1 + x2 2! + x4 4! + x6 6! + · · · x + x3 3! + x5 5! + x7 7! + · · · 1 + x + x2 2! + x3 3! + · · · 2 Using identities (5) and (6), we can write this expression as 1 2(ex + e−x) × 1 2(ex −e−x)exex = 1 4(e2x −e0 + e0 −e−2x)exex = 1 4(e2x −e−2x)e2x = 1 4(e4x −1)= 1 4 ∞ r=0 4r xr r! −1 Then for r > 0, the coefficient of xr/r! is 1 44r = 4r−1. The simple form of this answer suggests that there should be some combinatorial argument for obtaining this short answer directly.
6.4 EXERCISES S u m m a r y o f E x e r c i s e s Most of the first 15 exercises are similar to the examples in this section. Exercises 21 and 22 are a continuation of the probability generating function exercises in Section 6.2.
1. Find the exponential generating function for the number of arrangements or r objects chosen from five different types with at most five of each type.
2. Find the exponential generating function for the number of ways to distribute r people into six different rooms with between two and four in each room.
3. Find an exponential generating function for ar, the number of r-letter words with no vowel used more than once (consonants can be repeated).
4. (a) Find the exponential generating function for sn,r, the number of ways to distribute r distinct objects into n distinct boxes with no empty box. Consider n a fixed constant.
(b) Determine sn,r. The number sn,r/n! is called a Stirling number of the second kind.
5. Find the exponential generating function, and identify the appropriate coefficient, for the number of ways to deal a sequence of 13 cards (from a standard 52-card deck) if the suits are ignored and only the values of the cards are noted.
6. How many ways are there to distribute eight different toys among four children if the first child gets at least two toys?
7. How many r-digit ternary sequences are there with (a) An even number of 0s?
(b) An even number of 0s and even number of 1s?
(c) At least one 0 and at least one 1?
276 Chapter 6 Generating Functions 8. How many ways are there to make an r-arrangement of pennies, nickels, dimes, and quarters with at least one penny and an odd number of quarters? (Coins of the same denomination are identical.) 9. How many 10-letter words are there in which each of the letters e, n, r, s occur (a) At most once?
(b) At least once?
10. How many r-digit ternary sequences are there in which (a) No digit occurs exactly twice?
(b) 0 and 1 each appear a positive even number of times?
11. How many r-digit quaternary sequences are there in which the total number of 0s and 1s is even?
12. Find the exponential generating function for the number of ways to distribute r distinct objects into five different boxes when b1 < b2 ≤4, where b1, b2 are the numbers of objects in boxes 1 and 2, respectively.
13. (a) Find the exponential generating functions for pr, the probability that the first two boxes each have at least one object when r distinct objects are randomly distributed into n distinct boxes.
(b) Determine pr.
14. Find an exponential generating function for the number of distributions of r distinct objects into n different boxes with exactly m nonempty boxes.
15. Find an exponential generating function with (a) ar = 1/(r + 1) (b) ar = r!
16. Show that if g(x) is the exponential generating function for ar, then g(k)(0) = ak, where g(k)(x) is the kth derivative of g(x).
17. If f (x) = ∞ r=0 ar xr r!
g(x) = ∞ r=0 br xr r!
and h(x) = f (x)g(x) = ∞ r=0 cr xr r!
then show that cr = ∞ k=0 r k akbr−k 18. Show that exey = ex+y by formally multiplying the expansions of ex and ey together.
19. Find a combinatorial argument to show why the answer in Example 6 is 4r−1.
20. Show that ex/(1 −x)n is the exponential generating function for the number of ways to choose some subset (possibly empty) of r distinct objects and distribute them into n different boxes with the order in each box counted.
www.itpub.net 6.5 A Summation Method 277 21. APoissonrandomvariableXhas pr = Prob(x = r) = μr r! e−μ.Findtheprobability generating function for X (see Exercise 34 in Section 6.2).
22. Let P(x) = ∞ k=0 pkxk be the probability generating function for the discrete random variable X—that is, pk is the probability that X = k.
(a) Show that the exponential generating function for mk, the kth moment of X, mk = ∞ j=0 jk p j is P(ex).
(b) The kth factorial moment of X, m∗ k, is defined to be equal to ∞ j=k j!/( j −k)!p j. Show that the exponential generating function for m∗ k is P(x + 1).
(c) If X is the number of heads when n coins are flipped, find m1, m2, m∗ 2. (Hint: Use Exercise 16.) (d) If X is Poisson (see Exercise 21), find m1 and m∗ 1.
6.5 A SUMMATION METHOD In this section we show how to construct an ordinary generating function h(x) whose coefficient of xr is some specified function p(r) of r, such as r2 or C(r, 3). Then we use h(x) to calculate the sums p(0) + p(1) + · · · + p(n), for each positive n. The following four simple rules for constructing a new generating function from an old one will be used repeatedly in this section. Assume that A(x) = anxn, B(x) = bnxn, and C(x) = cnxn.
Rule 1. If bn = dan, then B(x) = dA(x), for any constant d.
Rule 2. If cn = an + bn, then C(x) = A(x) + B(x).
Rule 3. If cn = n i=0 aibn−i, then C(x) = A(x)B(x).
Rule 4. If bn = an−k, except bi = 0 for i < k, then B(x) = xk A(x).
Rule (3) is simply expression (6) from Section 6.2. The other rules are immediate.
The other basic operation for our coefficient construction is to multiply each coefficient ar in a generating function g(x) by r. We claim that the new generating function g∗(x) with a∗ r = rar is obtained by differentiating g(x) and then multiplying by x—that is, g∗(x) = x d dx g(x) . If g(x) = a0 + a1x + a2x2 + a3x3 + · · · + arxr + · · · (1) then differentiation of g(x) yields d dx g(x) = a1 + 2a2x + 3a3x2 + · · · +rarxr−1 + · · · (2) 278 Chapter 6 Generating Functions and now multiplying by x (Rule 4) gives g∗(x) = x d dx g(x) = a1x + 2a2x2 + 3a3x3 + · · · +rarxr + · · · (3) Note that the a0 term in g(x) disappears in (2) because 0a0 = 0.
Combining this operation with Rules (1) and (2), we can repeatedly multiply the coefficient of xr by r or a constant and can add such coefficients together. This permits us to form polynomials in r.
The natural question now is: To what coefficients do we apply these operations?
The natural answer is: When in doubt start with the unit coefficients ar = 1 of the generating function 1 1 −x = 1 + x + x2 + x3 + · · · + xr + · · · Example 1 Build a generating function h(x) with ar = 2r2.
Starting with 1/(1 −x), we multiply its coefficients first by r using the generating function operations shown in Eqs. (2) and (3) to obtain x d dx 1 1 −x = x 1 (1 −x)2 = 1x + 2x2 + 3x3 + · · · +rxr + · · · Now we repeat these operations on x/(1 −x)2 to obtain x d dx x (1 −x)2 = x(1 + x) (1 −x)3 = 12x + 22x2 + 32x3 + · · · +r2xr + · · · Finally we multiply by 2 to obtain the required generating function h(x) = 2x(1 + x) (1 −x)3 = (2 × 12)x + (2 × 22)x2 + (2 × 32)x3 + · · · + (2 ×r2)xr + · · · Example 2 Build a generating function h(x) with ar = (r + 1)r(r −1).
We could multiply (r + 1)r(r −1) out getting ar = r3 −r, obtain generating functions for r3 and r as an Example 1, and then subtract one generating function from the other. It is easier, however, to start with 3!(1 −x)−4, whose coefficient ar equals ar = 3!
r + 4 −1 r = 3!(r + 3)!
r!3!
= (r + 3)!
r!
= (r + 3)(r + 2)(r + 1) www.itpub.net 6.5 A Summation Method 279 Then the power series expansion of 3!(1 −x)−4 is 3!
(1 −x)4 = (3 × 2 × 1) + (4 × 3 × 2)x + (5 × 4 × 3)x2 + · · · + (r + 3)(r + 2)(r + 1)xr + · · · (4) Compare (4) with the desired generating function h(x) = (3 × 2 × 1)x2 + (4 × 3 × 2)x3 + (5 × 4 × 3)x4 + · · · + (r + 1)r(r −1)xr + · · · The generating function h(x) that we seek is just the series in (4) multiplied by x2.
So h(x) = 3!x2(1 −x)−4.
Generalizing the construction in Example 2, we see that (n −1)!(1 −x)−n has a coefficient ar = (n −1)!C(r + n −1,r) = [r + (n −1)][r + (n −2)] · · · (r + 1) Any coefficient involving a product of decreasing terms, such as (r + 1)r(r −1), can be built from (n −1)!(1 −x)−n as in Example 2, where n −1 is the number of terms in the product.
Thus far the construction of an h(x) with specified coefficients has been just an exercise in algebraic manipulations of generating functions. The following easily verified theorem gives some purpose to this exercise.
Theorem If h(x) is a generating function where ar is the coefficient of xr, then h∗(x) = h(x)/(1 −x) is a generating function of the sums of the ars. That is, h∗(x) = a0 + (a0 + a1)x + (a0 + a1 + a2)x2 + · · · + r i=0 ai xr + · · · This theorem follows from Rule (3) for the coefficients of the product h∗(x) = f (x)h(x), where f (x) = 1/(1 −x). Now we return to the previous examples.
Example 1: (continued) Evaluate the sum 2 × 12 + 2 × 22 + 2 × 32 + · · · + 2n2.
The generating function h(x) for ar = 2r2 was found in Example 1 to be 2x(1 + x)/(1 −x)3. Then by the theorem, the desired sum a1 + a2 + · · · + an is the coefficient of xn in h∗(x) = h(x)/(1 −x) = 2x(1 + x)/(1 −x)4 = 2x(1 −x)−4 + 2x2(1 −x)−4 280 Chapter 6 Generating Functions The coefficient of xn in 2x(1 −x)−4 is the coefficient of xn−1 in 2(1 −x)−4, and the coefficient of xn in 2x2(1 −x)−4 is the coefficient of xn−2 in 2(1 −x)−4. Thus, the given sum equals 2 (n −1) + 4 −1 (n −1) + 2 (n −2) + 4 −1 (n −2) = 2 n + 2 3 + 2 n + 1 3 Example 2: (continued) Evaluate the sum 3 × 2 × 1 + 4 × 3 × 2 + · · · + (n + 1)n(n −1).
The generating function h(x) for ar = (r + 1)r(r −1) was found in Example 2 to be h(x) = 6x2(1 −x)−4. By the theorem the desired sum is the coefficient of xn in h∗(x) = h(x)/(1 −x) = 6x2(1 −x)−5. This coefficient is the xn−2 term in 6(1 −x)−5—namely, 6C((n −2) + 5 −1, n −2) = 6C(n + 2, 4).
Note that these two summation problems can also be solved with the summation method in Section 5.5 based on binomial identities. For comparison, see the analysis of the sum in Example 2 given in Example 4 of Section 5.5 [note that the general term of this sum in Section 5.5 is (n −2)(n −1)n instead of (n + 1)n(n −1)]. Both methods have their advantages.
6.5 EXERCISES 1.
Find ordinary generating functions whose coefficient ar equals (a) r (b) 13 (c) 3r2 (d) 3r + 7 (e) r(r −1)(r −2)(r −3).
2.
Evaluate the following sums (using generating functions): (a) 0 + 1 + 2 + · · · + n (b) 13 + 13 + · · · + 13 (c) 0 + 3 + 12 + · · · + 3n2 (d) 7 + 10 + 13 + · · · + (3n + 7) (e) 4 × 3 × 2 × 1 + 5 × 4 × 3 × 2 + · · · + n(n −1)(n −2)(n −3) 3.
Find a generating function with ar = r(r + 2) (do not add together generating functions for r2 and for 2r).
4.
(a) Show how r2 and r3 can be written as linear combinations of P(r, 3), P(r, 2), and P(r, 1).
(b) Use part (a) to find a generating function for 3r3 −5r2 + 4r.
5.
Find a generating function for (a) ar = (r −1)2 (b) ar = 1/r 6.
If h(x) is the ordinary generating function for ar, what is the coefficient of xr in h(x)(1 −x) (give your answer in terms of the ars)?
7.
Verify Theorem 1 in this section.
www.itpub.net 6.6 Summary and References 281 8.
If h(x) is the ordinary generating function for ar, find the generating function for sr = ∞ k=r+1 ak, assuming all srs are finite and ar →0 as r →∞.
6.6 SUMMARY AND REFERENCES Generating functions are at once a simple-minded and a sophisticated mathematical model for counting problems; simple-minded because polynomial multiplication is a familiar, seemingly well understood part of high school algebra, and sophisticated be-cause with standard algebraic manipulations on generating functions, one can solve complicated counting problems. These algebraic manipulations automatically per-form the correct combinatorial reasoning for us! Note that generating functions are an elementary example of the algebraic approach that pervades the research frontiers of contemporary mathematics. Letting algebraic expressions, whether they model combinatorial, geometric, or functional information, do the work for us is what much of modern mathematics is all about.
In this chapter, generating functions were used to model selection and arrange-ment problems with constrained repetition. Partition problems were also modeled (but were not solved). Finally, we showed how to construct generating functions whose coefficient for xr was a given function of r and used these generating functions to evaluate related sums. In the next three chapters, generating functions will be used to model and solve other combinatorial problems. Exercise 38 of Section 6.2 introduces one type of generating function used in probability theory (for more, see Feller ).
Laplace and Fourier transforms in analysis are also generating functions [the Fourier transform of a function f (t) can be viewed as a generating function for the Fourier coefficients of f (t)].
The first use of combinatorial generating functions was by DeMoivre around 1720. He used them to derive a formula for Fibonacci numbers (this derivation is given in Section 7.5). In 1748, Euler used generating functions in his work on par-tition problems. The theory of combinatorial generating functions, developed in the late eighteenth century, was primarily motivated by parallel work on probability gen-erating functions (see Exercises 38 to 42 in Section 6.2 and Exercises 21 and 22 in Section 6.4). Laplace made many contributions to both theories and presented the first complete treatment of both in his 1812 classic Th´ eorie Analytique des Probabilit´ es.
For a full discussion of the use of generating functions in combinatorial math-ematics, see MacMahon . For a nice presentation of partition problems, also see Cameron .
1. P. Cameron, Combinatorics: Topics, Techniques, Algorithms, Cambridge Univer-sity Press, Cambridge, 1994.
2. W. Feller, An Introduction to Probability Theory and Its Applications, vol. I, 2nd ed., John Wiley & Sons, New York, 1957.
3. P. MacMahon, Combinatory Analysis, vols. I and II (1915), reprinted in one vol-ume, Chelsea Publishing, New York, 1960.
This page is intentionally left blank www.itpub.net CHAPTER 7 RECURRENCE RELATIONS 7.1 RECURRENCE RELATION MODELS In this chapter we show how a variety of counting problems can be modeled with recurrence relations. We then discuss methods of solving several common types of recurrence relations.
A recurrence relation is a recursive formula that counts the number of ways to do a procedure involving n objects in terms of the number of ways to do it with fewer objects. That is, if ak is the number of ways to do the procedure with k objects, for k = 0, 1, 2, . . . , then a recurrence relation is an equation that expresses an as some function of preceding aks, k < n. The simplest recurrence relation is an equation such as an = 2an−1. The following equations display some of the forms of recurrence relations that we will build to model counting problems in the chapter: an = c1an−1 + c2an−2 + · · · + cran−r where ci are constants an = can−1 + f (n) where f (n) is some function of n an = a0an−1 + a1an−2 + · · · + an−1a0 an,m = an−1,m + an−1,m−1 Just as mathematical induction is a proof technique that verifies a formula or assertion by inductively checking its validity for increasing values of n, so a recurrence relation is a counting technique that solves an enumeration problem by recursively computing the answer for successively larger values of n.
The observant reader will remember that mathematical induction also involves an initial step of verifying the formula or assertion for some starting (smallest) value of n. The same is true for a recurrence relation. We cannot recursively compute the next an unless some initial values are given. If the right-hand side of a recurrence relation involves the r preceding aks, then we need to be given the first r values, a0, a1, . . . , ar−1. For example, in the relation an = an−1 + an−2, knowing only that a0 = 2 is insufficient. If given also that a1 = 3, then we use the relation to obtain a2 = a1 + a0 = 3 + 2 = 5; a3 = a2 + a1 = 5 + 3 = 8; a4 = 8 + 5 = 13; and so on. The information about starting values needed to compute with a recurrence relation is called the initial conditions.
283 284 Chapter 7 Recurrence Relations If we can devise a recurrence relation to model a counting problem we are studying and also determine the initial conditions, then it is usually possible to solve the problem for moderate sizes of n, such as n = 20, quickly by recursively computing successive values of an up to the desired n. For larger values of n, a programmable calculator or computer is needed.
For many common types of recurrence relations, there are explicit formulas for an. Sections 7.2–7.4 discuss some of these solutions. However, it is frequently easier to determine recursively the value of a0, a1, . . . up to the desired value, say, a12 than to compute a12 from a complicated general formula for an.
Example 1: Arrangements Find a recurrence relation for the number of ways to arrange n distinct objects in a row. Find the number of arrangements of eight objects.
Letan denotethenumberofarrangementsofndistinctobjects.Therearenchoices for the first object in the row. This choice can be followed by any arrangement of the remaining n −1 objects; that is, by the an−1 arrangements of the remaining n −1 objects. Thus an = nan−1. Substituting recursively in this relation, we see that an = nan−1 = n[(n −1)an−2] = · · · = n(n −1)(n −2) · · · × 2 × 1 = n!
In particular, a8 = 8!
Of course, we already know that the number of arrangements of n objects is n!
It is often useful to try out a new technique on an old problem to see how it works before using it on new problems. Now for some new problems.
Example 2: Climbing Stairs An elf has a staircase of n stairs to climb. Each step it takes can cover either one stair or two stairs. Find a recurrence relation for an, the number of different ways for the elf to ascend the n-stair staircase.
It is easy to check with a figure that a1 = 1, a2 = 2 (two one-steps or one two-steps) and a3 = 3. For n = 4, Figure 7.1a depicts one way of climbing the stairs, taking successive steps of sizes 1, 2, 1. Other possibilities are one two-stair step either preceded by or followed by or in between two one-stair steps, or two two-stair steps, or four one-stair steps. In all, we count five ways to climb the four stairs; so a4 = 5.
1 2 3 4 1 n . . .
an – 1 ways 1 n . . .
an – 2 ways 2 (b) (c) (a) Figure 7.1 www.itpub.net 7.1 Recurrence Relation Models 285 Is there some systematic way to enumerate the ways to climb four stairs that breaks the problem into parts involving the ways to climb three or fewer stairs?
Clearly, once the first step is taken there are three or fewer stairs remaining to climb.
Thus we see that after a first step of one stair, there are a3 ways to continue the climb up the remaining three stairs. If the first step covers two stairs, then there are a2 ways to continue up the remaining two stairs. So a4 = a3 + a2. We confirm that the values for a4, a3, a2 satisfy this relation: 5 = 3 + 2. This argument applies to the first step when climbing any number of stairs, as is shown in Figures 7.1b and 7.1c. Thus an = an−1 + an−2.
In Section 7.3 we obtain an explicit solution to this recurrence relation. The rela-tion an = an−1 + an−2 is called the Fibonacci relation. The numbers an generated by the Fibonacci relation with the initial conditions a0 = a1 = 1 are called the Fibonacci numbers. They begin 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89. Fibonacci numbers arise natu-rally in many areas of combinatorial mathematics. There is even a journal, Fibonacci Quarterly, devoted solely to research involving the Fibonacci relation and Fibonacci numbers. Fibonacci numbers have been applied to other fields of mathematics, such as numerical analysis. They occur in the natural world—for example, the arrangements of petals in some flowers. For more information about the occurrences of Fibonacci numbers in nature, see .
Example 3: Dividing the Plane Suppose we draw n straight lines on a piece of paper so that every pair of lines intersect (but no three lines intersect at a common point). Into how many regions do these n lines divide the plane?
Again we approach the problem initially by examining the situation for small values of n. With one line, the paper is divided into two regions. With two lines, we get four regions—that is, a2 = 4. See Figure 7.2a. From Figure 7.2b, we see that a3 = 7. The skeptical reader may ask: how do we know that three intersecting lines will always create seven regions? Let us go back one step, then.
Clearly two intersecting lines will always yield four regions, as shown in Figure 7.2a. Now let us examine the effect of drawing the third line (labeled “3” in Figure 7.2b). It must cross each of the other two lines (at different points). Before, between, and after these two intersection points, the third line cuts through three of the regions formed by the first two lines (this action of the third line does not depend on how it is drawn, just that it intersects the other two lines). So in severing three regions, the third line must form three new regions, actually creating six new regions out of three old regions. Thus a3 = a2 + 3 = 4 + 3 = 7, independently of how the third line is drawn.
2 3 4 1 2 3 1 2 1 (b) (c) (a) Figure 7.2 286 Chapter 7 Recurrence Relations Similarly, the fourth line severs four regions before, between, and after its three intersection points with the first three lines (see Figure 7.2c), so that a4 = a3 + 4 = 7 + 4 = 11. In general, the nth line must sever n regions before, between, and after its n −1 intersection points with the first n −1 lines. So an = an−1 + n.
Example 4: Tower of Hanoi The Tower of Hanoi is a game consisting of n circular rings of varying size and three pegs on which the rings fit. Initially all the rings are placed on the first (leftmost) peg with the largest ring at the bottom covered by successively smaller rings. See Figure 7.3a. By transferring the rings among the pegs, one seeks to achieve a similarly tapered pile on the third (rightmost) peg. The complication is that each time a ring is transferred to a new peg, the transferred ring must be smaller than any of the rings already piled on this new peg; equivalently, at every stage in the game there must be a tapered pile (or no pile) on each peg.
Find a recurrence relation for an, the minimum number of moves required to play the Tower of Hanoi with n rings. How many moves are needed to play the six-ring game? Try playing this game with a dime, penny, nickel, and quarter (four rings) before reading our solution.
The key observation is that if, say, the six smallest rings are on peg A and we want to move them to peg C, we must first “play the five-ring Tower-of-Hanoi game” to get the five smallest rings from peg A to peg B, then move the sixth smallest ring from peg A to peg C, and then again “play the five-ring game” from peg B to peg C.
See Figure 7.3. (Of course, to move the five smallest rings we must move the fifth smallest from A to B, which means playing four-ring games from A to C and from C to B, etc.) Thus to move the n rings from A to C, the n −1 smallest rings must first be moved from A to B, then the largest (nth) ring moved from A to C, and then the n −1 smallest rings moved from B to C.
If an is the number of moves needed to transfer a tapered pile of n rings from one peg to another peg, then the previous sentence yields the following recurrence relation: an = an−1 + 1 + an−1 = 2an−1 + 1. The initial condition is a1 = 1, and so a2 = 2a1 + 1 = 3; a3 = 2a2 + 1 = 7; a4 = 2a3 + 1 = 15; a5 = 2a4 + 1 = 31; and a6 = 2a5 + 1 = 63. So the six-ring game requires 63 moves. Note that the ans thus far fit the formula an = 2n −1.
A B C A B C (b) (a) Figure 7.3 Tower of Hanoi www.itpub.net 7.1 Recurrence Relation Models 287 Example 5: Money Growing in a Savings Account A bank pays 4 percent interest each year on money in savings accounts. Find re-currence relations for the amounts of money a gnome would have after n years if it follows the investment strategies of (a) Investing $1,000 and leaving it in the bank for n years (b) Investing $100 at the end of each year If an account has x dollars at the start of a year, then at the end of the year (i.e., start of the next year) it will have x dollars plus the interest on the x dollars, provided no money was added or removed during the year. Then for part (a), the recurrence relation is an = an−1 + .04an−1 = 1.04an−1. The initial condition is a0 = 1000. For part (b), the relation must reflect the $100 added (which earns no interest since it comes at the end of the year). So an = 1.04an−1 + 100, with a0 = 0.
A wide range of common financial activities can be modeled by recurrence relations, including college saving plans and mortgages. A mortgage is similar to part (b) of Example 5, except now the value of the account is what you owe, not what you own. One starts with a large initial debt—e.g., a0 might be 300,000—and each period, typically one month, interest adds to the debt while monthly payments are large enough to offset the interest and slowly reduce the debt. Closed-form solutions to recurrence relations lead to the formulas that bankers use to determine the monthly payments for a mortgage of a given size, interest rate, and term.
Example 6: Making Change Find a recurrence relation for the number of different ways to hand out a piece of chewing gum (worth 1c /) or a candy bar (worth 10c /) or a doughnut (worth 20c /) on successive days until nc / worth of food has been given away.
This problem can be treated similarly to the stair-climbing problem in Example 2.
That is, if on the first day we hand out 1c / worth of chewing gum, we are left with (n −1)c / worth of food to give away on following days; if the first day we hand out 10c / worth of candy, we have (n −10)c / worth to dispense the next days; and if 20c / then (n −20)c / the next days. So an = an−1 + an−10 + an−20 with a0 = 1 (one way to give nothing—by giving 0 pieces of each item), and implicitly, ak = 0 for k < 0.
Example 7: A Forbidden Subsequence Find a recurrence relation for an, the number of n-digit ternary sequences without any occurrence of the subsequence “012.” Recall that a ternary sequence is a sequence composed of 0s, 1s, and 2s. We start with the same analysis used in the elf stair-climbing problem. If the first digit in an n-digit ternary sequence is 1, then there are an−1(n −1)-digit ternary sequences without the pattern 012 that can follow that initial 1. Similarly if the first digit is 2.
However, there is a problem if the first digit is a 0. Among the (n −1)-digit ternary sequences without the pattern 012 that might follow the initial 0 are sequences that 288 Chapter 7 Recurrence Relations start with 12. While such (n −1)-digit sequences do not contain the pattern 012, the n-digit sequences of 0 followed by such (n −1)-digit sequences do start with the pattern 012. We correct this mistake by subtracting off all n-digit sequences starting with 012 (but with 012 not appearing later in the sequence). Such sequences are formed by 012 followed by any (n −3)-digit ternary sequence with no 012 pattern— there are an−3 such sequences. Thus the desired recurrence relation is an = an−1 + an−1 + (an−1 −an−3) = 3an−1 −an−3.
The preceding examples developed recurrence relation models either by breaking a problem into a first step followed by the same problem for a smaller set (Examples 1, 2, 6 and 7) or by observing the change in going from the case of n −1 to the case of n (Examples 3 and 5). Example 4 was a modified form of the former procedure in which the move of the nth ring (largest ring) was seen to be preceded and followed by an (n −1)-ring game.
There are two simple methods for solving some of the relations seen thus far. The first is recursive backward substitution: wherever an−1 occurs in the relation for an, we replace an−1 by the relation’s formula for an−1 (involving an−2) and then replace an−2, and so on. In Example 3, the relation an = an−1 + n becomes an = (an−2 + n −1) + n = · · · = 1 + 2 + 3 + · · · + n −1 + n. We used this method in Example 1 to obtain an = n(n −1)(n −2) × · · · × 2 × 1.
The second method is to guess the solution to the relation and then to verify it by mathematical induction. In Example 4, we noted that an = 2n −1 for the first six values of the recurrence relation an = 2an−1 + 1, a1 = 1. We prove an = 2n −1 by induction as follows. It is seen to be true for n = 1. Assuming an−1 = 2n−1 −1, we then have an = 2an−1 + 1 = 2(2n−1 −1) + 1 = 2n −2 + 1 = 2n −1.
We now consider more complex recurrence relations involving two variable re-lations and simultaneous relations.
Example 8: Selection Without Repetition Let an,k denote the number of ways to select a subset of k objects from a set of n distinct objects. Find a recurrence relation for an,k.
Observe that an,k is simply n k . We break the problem into two subcases based on whether or not the first object is used. There are an−1,k k-subsets that do not use the first object, and there are an−1,k−1 k-subsets that do use the first object. So we have an,k = an−1,k + an−1,k−1. This is the Pascal’s triangle identity [identity (3) from Section 5.5].
The initial conditions are an,0 = an,n = 1 for all n ≥0 (and an,k = 0, k > n).
Example 9: Distributions Find a recurrence relation for the ways to distribute n identical balls into k distinct boxes with between two and four balls in each box. Repeat the problem with balls of three colors.
In the spirit of the previous example, we consider how many balls go into the first box. If we put two in the first box (one way to do this), then there are an−2,k−1 ways to put the remaining n −2 identical balls in the remaining k −1 boxes. Continuing www.itpub.net 7.1 Recurrence Relation Models 289 this line of reasoning, we see that an,k = an−2,k−1 + an−3,k−1 + an−4,k−1. The initial conditions are a2,1 = a3,1 = a4,1 = 1 and an,1 = 0.
Ifnowthreecolorsareallowed,thereareC(2 + 3 −1, 2) = 6waystopickasubset of two balls for the first box from three types of colors with repetition. (Review the start of Section 5.3 for a discussion of selection with repetition.) Similarly, there are C(3 + 3 −1, 3) = 10waystopickthreeballsfromthreetypesandC(4 + 3 −1, 4) = 15ways to pick four balls from three types. Then an,k = 6an−2,k−1 + 10an−3,k−1 + 15an−4,k−1 with initial conditions a2,1 = 6, a3,1 = 10, a4,1 = 15, and an,1 = 0.
The problem with all balls identical can be solved by generating functions (see Example 3 of Section 6.1), but recurrence relations are the only practical approach with the extra constraint of different types of balls.
Example 10: Placing Parentheses Find a recurrence relation for an, the number of ways to place parentheses to multiply the n numbers k1 × k2 × k3 × k4 × · · · × kn on a calculator.
To clarify the problem, observe that there is one way to multiply (k1 × k2), so a2 = 1. There are two ways to multiply k1 × k2 × k3, namely, [(k1 × k2) × k3] and [k1 × (k2 × k3)]; so a3 = 2. It is not clear what a0 and a1 should be, but to make the eventual recurrence relation have a simple form, we let a0 = 0 and a1 = 1. To find a recurrence relation for an, we look at the last multiplication (the outermost parenthesis) in the product of the n numbers. This last multiplication involves the products of two multiplication subproblems: (k1 × k2 × · · · × ki) × (ki+1 × ki+2 × · · · × kn) where i can range from 1 to n −1. The numbers of ways to parenthesize the two respective subproblems are ai and an−i, and so there are aian−i ways to parenthesize both subproblems. Summing over all i, we obtain the recurrence relation (for n ≥2) an = a1an−1 + a2an−2 + . . . + an−1a1.
Example 11: Systems of Recurrence Relations Find recurrence relations for (a) The number of n-digit ternary sequences with an even number of 0s (b) The number of n-digit ternary sequences with an even number of 0s and an even number of 1s (a) We use the first-step analysis of the stair-climbing model to find a recurrence relation for an, the number of n-digit ternary sequences with an even number of 0s.
If an n-digit ternary sequence starts with a 1, then we require an even number of 0s in the remaining (n −1)-digit sequence—an−1 such sequences. Similarly, if an n-digit ternary sequence starts with a 2, there are an−1 (n −1)-digit sequences with even 0s. If the n-digit sequence starts with a 0, we require an odd number of 0s in the remaining n −1 digits—3n−1 −an−1 such sequences, since all 3n−1(n −1)-digit 290 Chapter 7 Recurrence Relations sequences minus the even-0s (n −1)-digit sequences yields the odd-0s (n −1)-digit sequences. In sum, an = 2an−1 + (3n−1 −an−1) = an + 3n−1.
(b) We will need simultaneous recurrence relations for an, the number of n-digit ternary sequences with even 0s and even 1s; bn, the number of n-digit ternary sequences with even 0s and odd 1s; and cn, the number of n-digit se-quences with odd 0s and even 1s. Observe that 3n −an −bn −cn is the number of n-digit ternary sequences with odd 0s and odd 1s. An n-digit ternary sequence with even 0s and even 1s is obtained either by having a 1 for the first digit fol-lowed by an (n −1)-digit sequence with even 0s and odd 1s, or a 0 followed by an (n −1)-digit sequence with odd 0s and even 1s, or a 2 followed by an (n −1)-digit sequence with even 0s and even 1s. Thus an = bn−1 + cn−1 + an−1. Similar anal-yses yield bn = an−1 + (3n−1 −an−1 −bn−1 −cn−1) + bn−1 = 3n−1 −cn−1 and cn = an−1 + (3n−1 −an−1 −bn−1 −cn−1) + cn−1 = 3n−1 −bn−1. The initial conditions are a1 = b1 = c1 = 1. To recursively compute values for an, we must simultaneously compute bn and cn.
We close this section with a few words about difference equations. The first (backward) difference an of the sequence (a0, a1, a2, . . .) is defined to be an = an −an−1. The second difference is 2an = an −an−1 = an −2an−1 + an−2, and so on. A difference equation is an equation involving an and its differences, such as 22an −3an + an = 0. Observe that an−1 = an −(an −an−1) = an −an an−2 = an−1 −an−1 = (an −an) −(an −an) = an −2an + 2an(∗) Similar equations can express an−k in terms of an, an, . . . , kan. Thus any re-currence relation can be rewritten as a difference equation, by expressing the an−ks on the right-hand side of (∗) in terms of an and its differences. Conversely, by writing an as an −an−1, and so on, any difference equation can be written as a recurrence relation.
Difference equations are commonly used to approximate differential equations when solving differential equations on a computer. Difference equations have wide use in their own right as models for dynamical systems for which differential equations (which require continuous functions) are inappropriate. They are used in economics in models for predicting the gross national product in successive years. They are used in ecology to model the numbers of various species in successive years. As noted above, any difference equation model can also be formulated as a recurrence relation; however, the behavior of a dynamical system is easier to analyze and explain in terms of differences. Refer to Sandefur for further information about difference equations and their applications.
Example 12: Two-Animal Population Model Assume that if undisturbed by foxes, the number of rabbits increases each year by an amount αrn, where rn is the number of rabbits, but when foxes are present, each rabbit has probability β fn of being eaten by a fox ( fn is the number of foxes). Foxes alone decrease by an amount γ fn each year, but when rabbits are present, each fox has probability δrn of feeding and raising up a new young fox (death of foxes is included www.itpub.net 7.1 Recurrence Relation Models 291 in the γ fn term). Give a pair of simultaneous difference equations describing the number of rabbits and foxes in successive years.
The information given about yearly changes in the two populations yields the following difference equations: rn = αrn −βrn fn fn = −γ fn + δrn fn 7.1 EXERCISES S u m m a r y o f E x e r c i s e s The first 39 exercises call for recurrence relation modeling similar to that in the examples, with multiple indices and equa-tions required in Exercises 28–40 and difference equations in Exercises 41–43. The remaining exercises are more advanced problems.
1. Find a recurrence relation for the number of ways to distribute n distinct objects into five boxes. What is the initial condition?
2. (a) Find a recurrence relation for the number or ways the elf in Example 2 can climb n stairs if each step covers either one or two or three stairs?
(b) How many ways are there for the elf to climb four stairs?
3. Find a recurrence relation for the number of ways to arrange cars in a row with n spaces if we can use Cadillacs or Hummers or Fords. A Hummer requires two spaces, whereas a Cadillac or a Ford requires just one space.
4. (a) Find a recurrence relation for the number of ways to go n miles by foot walking at 2 miles per hour or jogging at 4 miles per hour or running at 8 miles per hour; at the end of each hour a choice is made of how to go the next hour.
(b) How many ways are there to go 12 miles?
5. Find a recurrence relation for the number of ways to distribute a total of n cents on successive days using 1971 pennies, 1951 nickels, 1967 nickels, 1959 dimes, and 1975 quarters.
6. (a) Find a recurrence relation for the number of n-digit binary sequences with no pair of consecutive 1s.
(b) Repeat for n-digit ternary sequences.
(c) Repeat for n-digit ternary sequences with no consecutive 1s or consecutive 2s.
7. Find a recurrence relation for the number of pairs of rabbits after n months if (1) initially there is one pair of rabbits who were just born, and (2) every month each pair of rabbits that are over one month old have a pair of offspring (a male and a female).
8. Show that the binomial sum sn = n + 1 0 + n 1 + n −1 2 + · · · satisfies the Fibonacci relation.
292 Chapter 7 Recurrence Relations 9. Find a recurrence relation for the number of ways to arrange n dominoes to fill a 2-×-n checkerboard.
10. Find a recurrence relation for an for the number of bees in the nth previous generation of a male bee, if a male bee is born asexually from a single female and a female bee has the normal male and female parents. The ancestral chart at the right shows that a1 = 1, a2 = 2, a3 = 3.
f m m f f f m 11. Find a recurrence relation for an for the number of ways for an image to be reflected n times by internal faces of two adjacent panes of glass. The diagram below shows that a0 = 1, a1 = 2, and a2 = 3.
0 reflections 1 reflection 2 reflections 12. Find a recurrence relation for the number of regions created by n mutually inter-secting circles on a piece of paper (no three circles have a common intersection point).
13. (a) Find a recurrence relation for the number of regions created by n lines on a piece of paper if k of the lines are parallel and the other n −k lines intersect all other lines (no three lines intersect at one point).
(b) If n = 9 and k = 3, find the number of regions.
14. Show that each of the following rules for playing the Tower of Hanoi works.
(a) On odd-numbered moves, move the smallest ring clockwise one peg (think of the pegs being at the corners of a triangle), and on even-numbered moves, make the only legal move not using the smallest ring.
(b) Number the rings from 1 to n in order of increasing size. Never move the same ring twice in a row. Always put even-numbered rings on top of odd-numbered rings (or on an empty peg) and put odd-numbered rings on top of even-numbered rings (or on an empty peg).
15. Find a recurrence relation for the amount of money in a savings account after n years if the interest rate is 6 percent and $50 is added at the start of each year.
www.itpub.net 7.1 Recurrence Relation Models 293 16. (a) Find a recurrence relation for the amount of money outstanding on a $30,000 mortgage after n years if the interest rate is 8 percent and the yearly payment (paid at the end of each year after interest is computed) is $3,000.
(b) Use a calculator or computer to determine how many years it will take to pay off the mortgage.
17. Each day Angela eats lunch at a deli, ordering one of the following: chicken salad, a tuna sandwich, or a turkey wrap. Find a recurrence relation for the number of ways for her to order lunch for the n days if she never orders chicken salad three days in a row.
18. Find a recurrence relation for the number of n-letter sequences using the letters A, B, C such that any A not in the last position of the sequence is always followed by a B.
19. (a) Find a recurrence relation for the number of sequences of 1s, 3s, and 5s whose terms sum to n.
(b) Repeat part (a) with the added condition that no 5 can be followed by a 1.
(c) Repeat part (a) with the condition of no subsequence of 135.
20. (a) Find a recurrence relation for the number of ways to arrange three types of flags on a flagpole n feet high: red flags (1 foot high), gold flags (1 foot high), and green flags (2 feet high).
(b) Repeat part (a) with the added condition that there may not be three 1-foot flags (red or gold) in a row.
(c) Repeat part (a) with the condition of no red above gold above green (in a row).
21. Find a recurrence relation for an, the number of ways to give away $1 or $2 or $3 for n days with the constraint that there is an even number of days when $1 is given away.
22. Find a recurrence relation to count the number of n-digit binary sequences with at least one instance of consecutive 0s.
23. Find a recurrence relation for the number of n-digit quaternary (0, 1, 2, 3) se-quences with at least one 1 and the first 1 occurring before the first 0 (possibly no 0s).
24. Find a recurrence relation for the number of n-digit ternary (0, 1, 2) sequence in which no 1 appears anywhere to the right of any 2.
25. Find a recurrence relation for the number of n-digit ternary sequences that have the pattern “012” occurring for the first time at the end of the sequence.
26. A switching game has n switches, all initially in the OFF position. In order to be able to flip the ith switch, the (i −1)st switch must be ON and all earlier switches OFF. The first switch can always be flipped. Find a recurrence relation for the total number of times the n switches must be flipped to get the nth switch ON and all others OFF.
294 Chapter 7 Recurrence Relations 27. Find a recurrence relation for the number of ways to pair off 2n people for tennis matches.
28. Find a recurrence relation for the number of ways to pick k objects with repetition from n types.
29. Find a recurrence relation for the number of ways to select n objects from k types with at most three of any one type.
30. Find a recurrence relation for an,k, the number of ways to order n doughnuts from k different types of doughnuts if two or four or six doughnuts must be chosen of each type.
31. Find a recurrence relation for the number of partitions of the integer n into k parts.
32. Find a recurrence relation for the number an,m,k of distributions of n identical objects into k distinct boxes with at most four objects in a box and with exactly m boxes having four objects.
33. Find a system of recurrence relations for computing the number of n-digit binary sequences with an even number of 0s and an even number of 1s.
34. Find a system of recurrence relations for computing the number of n-digit qua-ternary sequences with (a) An even number of 0s (b) An even total number of 0s and 1s (c) An even number of 0s and an even number of 1s 35. Find a system of recurrence relations for computing the number of n-digit binary sequences with exactly one pair of consecutive 0s.
36. Find a system of recurrence relations for the number of n-digit binary sequences with k adjacent pairs of 1s and no adjacent pairs of 0s.
37. Find a system of recurrence relations for computing the number of ways to hand out a penny or a nickel or a dime on successive days until n cents are given such that the same amount of money is not handed out on two consecutive days.
38. Find a recurrence relation for the number of ways to pair off 2n points on a circle with nonintersecting chords. (Hint: The recurrence involves products of aks as in Example 11.) 39. Find a recurrence relation for the number of ways to divide an n-gon into triangles with noncrossing diagonals.
40. Find a recurrence relation for the number of binary trees with n labeled leaves.
41. Find an and 2an if an equals (a) 3n + 2 (b) n2 (c) n3 42. Let fn be the amount of food that can be bought with n dollars. Let pn be the “perceived” value of the $n of food. Suppose the increase in perceived value with www.itpub.net 7.1 Recurrence Relation Models 295 $1 more of food equals the relative, or percentage, increase in the actual amount of food. Find a difference equation relating pn and fn.
43. (a) Findarecurrencerelationforthenumberofpermutationsofthefirstnintegers such that each integer differs by one (except for the first integer) from some integer to the left of it in the permutation. What is the initial condition?
(b) Solve the relation in part (a) by guessing and verifying the guess by induction.
(c) Give a direct combinatorial answer to this problem.
44. (a) Find a recurrence relation for f (n, k), the number of k subsets of the integers 1 through n with no pair of consecutive integers.
(b) Show that n/2 k=0 f (n, k) = Fn+1, the (n + 1)st Fibonacci number (n even), where F0 = F1 = 1.
45. Find a system of recurrence relations for computing an, the number of (un-ordered) collections of (identical) pennies, (identical) nickels, (identical) dimes, and (identical) quarters whose value is n cents.
46. Find a recurrence relation for the number of ways a coin can be flipped 2n times with n heads and n tails and (a) The number of heads at any time never be less than the number of tails (b) The number of heads equal the number of tails only after all 2n flips 47. Find a recurrence relation for the number of incongruent integral-sided triangles whose perimeter is n (the relation is different for n odd and n even).
48. Find a system of recurrence relations for computing the number of spanning trees in the “ladder” graph with 2n vertices.
. . .
49. Verify the following identities for Fibonacci numbers (Fi is the ith Fibonacci number) by induction or combinatorial argument. Here F0 = F1 = 1.
(a) n i=0 Fi = Fn+2 −1 (b) n i=0 F2 i = Fn Fn+1 (c) n k=0 F2k = F2n+1 (d) Fn Fn+2 = F2 n+1 + (−1)n (e) F1 −F2 + F3 −· · · −F2n = −F2n−1 50. (a) Show that Fn+m = Fm Fn + Fm−1Fn−1.
(b) From part (a) conclude that Fn−1 divides Fkn−1.
296 Chapter 7 Recurrence Relations 7.2 DIVIDE-AND-CONQUER RELATIONS In this section we present a special class of recurrence relations that arise frequently in the analysis of recursive computer algorithms. These are algorithms that use a “divide-and-conquer” approach to recursively split a problem into two subproblems of half the size. The dictionary search in Example 3 of Section 3.1 is such an algorithm. A binary tree is explicitly or implicitly associated with most “divide-and-conquer” algorithms.
Conversely, many counting problems involving trees (the subject of Chapter 3) are most easily solved with divide-and-conquer recurrence relations.
The total number of steps an required by a divide-and-conquer algorithm to process an n-element problem frequently satisfies a recurrence relation of the form an = can/2 + f (n) (1) The following table indicates the form of the solution of (1) for some common values of c and f (n) (⌈r⌉denotes the smallest integer m with m ≥r): c f(n) an c = 1 d d⌈log2 n⌉+ A c = 2 d An −d c > 2 dn Anlog2 c + 2d 2 −c n c = 2 dn dn(⌈log2 n⌉+ A) The constant A is to be chosen to fit the initial condition.
If a problem is recursively split into k parts instead of two parts, then one should replace 2 by k everywhere in the foregoing table, except the solution for c = k and f (n) = d becomes An −d/(k −1). For example, the recurrence relation an = can/k + dn c ̸= k has the solution an = Anlogk c + kd k −c n (2) The solutions for an given in the foregoing table are easily verified by substitution.
Consider the case an = can/2 + dn, c > 2. Substituting the table’s solution of an = Anlog2 c + [2d/(2 −c)]n into can/2 + dn, we have can/2 + dn = c A n 2 log2 c + 2d 2 −c n 2 + dn = cAnlog2 c 2log2 c + cdn 2 −c + (2 −c)dn 2 −c = cAnlog2 c c + cdn + (2 −c)dn 2 −c = Anlog2 c + 2d 2 −c n = an www.itpub.net 7.2 Divide-and-Conquer Relations 297 The following examples illustrate such “divide-and-conquer” recurrences and their solution.
Example 1: Rounds in a Tournament In a tennis tournament, each entrant plays a match in the first round. Next, all winners from the first round play a second-round match. Winners continue to move on to the next round, until finally only one player is left—the tournament winner. Assuming that tournaments always involve n = 2k players, for some k, find and solve a recurrence relation for the number of rounds in a tournament of n players.
In terms of binary trees, an is the height of a balanced binary tree with n = 2k leaves. Since half the players are eliminated in each round, the number of rounds increases by 1 when the number of players doubles. The recurrence relation for an, the number of rounds, is thus an = an/2 + 1 Fromtheforegoingtable,thesolutionofthisrecurrencerelationis an = log2 n + A.
To determine A, we observe that 0 = a1 = log21 + A = 0 + A, and so A = 0.
Example 2: Finding the Largest and Smallest Numbers in a Set Build a recurrence relation model to count the number of comparisons that must be made in the following algorithm for finding the largest number l and the smallest number s in a set S of n distinct integers. Then solve this recurrence relation.
Initially suppose that n is an even number. Assume that we have already found l1 and s1, the largest and smallest numbers, respectively, in the first half of S (the first n/2 numbers) and have found l2 and s2, the largest and smallest numbers in the second half of S. Then make two comparisons, one between l1 and l2 and the other between s1 and s2, to find the largest number l and smallest number s in S.
The associated recurrence relation for the number of comparisons in this proce-dure is an = 2an/2 + 2, for n ≥4 and even. If n is odd and we split S almost equally, the relation would be an = a(n+1)/2 + a(n−1)/2 + 2. Observe that a1 = 0, since the one number is both largest and smallest. And a2 = 1, since we can determine the larger and smaller number in a two-element set with one comparison. With these two rela-tions along with a1 and a2, we can recursively determine the number of comparisons needed for any n.
Next we solve the recurrence relation an = 2an/2 + 2, n ≥4, with a2 = 1. The foregoing table tells us that the solution will be of the form an = An −2. We confirm this by substituting an = An −2 into both sides of the relation an = 2an/2 + 2: An −2 = an = 2an/2 + 2 = 2(An 2 −2) + 2 = An −4 + 2 = An −2 We can use the initial condition a2 = 1 to determine A: 1 = a2 = A1(2) −2 or A1 = 3 2 298 Chapter 7 Recurrence Relations So an = 3 2n −2 is the number of comparisons needed to find the largest and smallest number according to the procedure given previously. (It is possible to prove that one cannot do better than 3 2n −2 comparisons.) Example 3: Efficient Multidigit Multiplication Normally one must do n2 digit-times-digit multiplications to multiply two n-digit numbers. Use a divide-and-conquer approach to develop a faster algorithm.
Let us initially assume that n is a power of 2. Let the two n-digit numbers be g and h. We split each of these numbers into two n/2-digit parts: g = g110n/2 + g2 h = h110n/2 + h2 Then g × h = (g1 × h1)10n + (g1 × h2 + g2 × h1)10n/2 + g2 × h2 (3) Observe that g1 × h2 + g2 × h1 = (g1 + g2) × (h1 + h2) −g1 × h1 −g2 × h2 and so we need to make only three n/2-digit multiplications, g1 × h1, g2 × h2, and (g1 + g2) × (h1 + h2) to determine g × h in (3) (actually (g1 + g2) or (h1 + h2) may be (n/2 + 1)-digit numbers, but this slight variation does not affect the general magnitude ofoursolution).Ifan representsthenumberofdigit-times-digitmultiplicationsneeded to multiply two n-digit numbers by the foregoing procedure, then the procedure yields the recurrence an = 3an/2.
By the table at the start of this section (where d = 0), an is proportional to nlog2 3 = n1.6, a substantial improvement over n2.
In some settings, we are not so interested in an exact formula for an as we are in the general rate of growth for an. The following theorem from Cormen, Leiserson, and Rivest gives bounds on such growth.
Theorem Letan = can/k + f (n)bearecurrencerelationwithpositiveconstant candthepositive function f (n).
(a) If for large n, f (n) grows proportional to nlogk c [that is, there are positive constants p and p′ such that pnlogk r ≤f (n) ≤p′nlogk c], then an grows proportional to nlogk c log2 c.
(b) If for large n, f (n) ≤pnq, where p is a positive constant and q < logk c, then an grows at most at a rate proportional to nlogk c.
www.itpub.net 7.2 Divide-and-Conquer Relations 299 7.2 EXERCISES 1.
Solve the following recurrence relations assuming that n is a power of 2 (leaving a constant A to be determined): (a) an = 2an/2 + 5 (b) an = 2an/4 + n (c) an = an/2 + 2n −1 (d) an = 3an/3 + 4 (e) an = 16an/2 + 5n (f) an = 4an/2 + 3n 2.
Find and solve a recurrence relation for the number of matches played in a tournament with n players, where n is a power of 2.
3.
In a large corporation with n salespeople, every 10 salespeople report to a local manager, every 10 local managers report to a district manager, and so forth until finally 10 vice-presidents report to the firm’s president. If the firm has n salespeople, where n is a power of 10, find and solve recurrence relations for (a) The number of different managerial levels in the firm (b) The number of managers (up through president) in the firm 4.
In a tennis tournament, each player wins k hundreds of dollars, where k is the number of people in the subtournament won by the player (the subsection of the tournament including the player, the player’s victims, and their victims, and so forth; a player who loses in the first round gets $100). If the tournament has n contestants, where n is a power of 2, find and solve a recurrence relation for the total prize money in the tournament.
5.
Consider the following method for rearranging the n distinct numbers x1, x2, . . . , xn in order of increasing size (n is a power of 2). Pair the integers off {x1, x2}, {x3, x4}, and so forth. Compare each pair and put the smaller number first. Next pair off the pairs into sets of four numbers and merge the ordered pairs to get ordered 4-tuples. Continue this process until the whole set is ordered. Find and solve a recurrence relation for the total number of comparisons required to rearrange n distinct numbers. (Hint: First find the number of comparisons needed to merge two ordered k-tuples into an ordered 2k-tuple).
6.
In a standard elimination tournament, a player wins $100k when he/she wins a match in the kth round (e.g., first round win earns $100, second round win $200).
Develop and solve a recurrence relation for an, the total amount of money given away in a tournament with n entrants, where n is assumed to be a power of 2.
7.
Verify by substitution the form of solution given in the text to the following recurrence relations: (a) an = an/2 + d 300 Chapter 7 Recurrence Relations (b) an = 2an/2 + d (c) an = 2an/2 + dn 8.
(a) Use a divide-and-conquer approach to devise a procedure to find the largest and next-to-largest numbers in a set of n distinct integers.
(b) Give a recurrence relation for the number of comparisons performed by your procedure.
(c) Solve the recurrence relation obtained in part (b).
9.
(a) Use a divide-and-conquer approach to devise a procedure to find the largest number in a set of n distinct integers.
(b) Give a recurrence relation for the number of comparisons performed by your procedure.
(c) Solve the recurrence relation obtained in part (b).
7.3 SOLUTION OF LINEAR RECURRENCE RELATIONS In this section we show how to solve recurrence relations of the form an = c1an−1 + c2an−2 + · · · + cran−r (1) where the cis are given constants. There is a simple technique for solving such rela-tions. Readers whohavestudiedlineardifferentialequationswithconstantcoefficients will see a great similarity between their solution and the form of solutions we discuss here. The general solution to (1) will involve a sum of individual solutions of the form an = αn. To determine what α is, we simply substitute αk for ak in (1), yielding αn = c1αn−1 + c2αn−2 + · · · + crαn−r (2) We can reduce the power of α in all terms in (2) by dividing both sides by αn−r: αr = c1αr−1 + c2αr−2 + · · · + cr (3) or, equivalently, αr −c1αr−1 −c2αr−2 −· · · −cr = 0 (4) Equation (4) is called the characteristic equation of the recurrence relation (1).
It has r roots, some of which may be complex (but we shall initially assume that there are no multiple roots).
If α1, α2, . . . , αr are the r roots of (4), then, for any i, 0 ≤i ≤r, an = αn i is a solution to the recurrence relation (1). It is easy to check that any linear combination of such solutions is also a solution (Exercise 8). That is, an = A1αn 1 + A2αn 2 + · · · + Arαn r (5) is a solution to (1), for any choice of constants Ai, 1 ≤i ≤r.
www.itpub.net 7.3 Solution of Linear Recurrence Relations 301 Recall that for a recurrence relation involving an−1, an−2, . . . , an−r on the right side, we need to be given the initial conditions of the first r values a0, a1, a2, . . . , ar−1.
Let us denote such a set of initial values by a′ 0, a′ 1, a′ 2, . . . , a′ r−1. Then the Ais must be chosen to satisfy the r constraints: A1αk 1 + A2αk 2 + · · · + Arαk r = a′ k 0 ≤k ≤r −1 (6) The r linear equations in (6) can be solved by Gaussian elimination to determine the r constants Ai (remember that at this stage the αis are known). With the Ai’s determined, we will have the desired solution for an, a solution that satisfies (4), and hence the recurrence relation (1), and satisfies the initial conditions a0 = a′ 0, a1 = a′ 1, . . . , ar−1 = a′ r−1. If the solution of the characteristic equation (4) has a root α∗of multiplicity m, then αn ∗, nαn ∗, n2αn ∗, . . . , n(m−1)αn ∗can be shown to be the m associated individual solutions to be used in (5) and (6).
Example 1: Doubling Rabbit Population Every year Dr. Finch’s rabbit population doubles. He started with six rabbits. How many rabbits does he have after eight years? After n years?
Ifan isthenumberofrabbitsaftern years,thenan satisfiestherelationan = 2an−1.
We are also given a0 = 6. Substituting an = αn, we obtain αn = 2αn−1 or, dividing by αn−1, α = 2. So an = 2n is the one individual solution, and an = A2n is the general solution. The initial condition is 6 = a0 = A20, or A = 6. The desired solution is then an = 6 × 2n. After eight years, we have a8 = 6 × 28 = 6 × 256 = 1536 rabbits.
Example 2: Second-Order Linear Recurrence Relation Solve the recurrence relation an = 2an−1 + 3an−2 with a0 = a1 = 1.
Setting an = αn, we get the characteristic equation αn = 2αn−1 + 3αn−2 which yields α2 = 2α + 3 This may be written α2 −2α −3 = 0 or (α −3)(α + 1) = 0. That is, the roots are +3 and −1. So the basic solutions of the recurrence relation are an = 3n and an = (−1)n, and the general solution is an = A13n + A2(−1)n Now we determine A1 and A2 by using the initial conditions 1 = a0 = A130 + A2(−1)0 = A1 + A2 1 = a1 = A131 + A2(−1)1 = 3A1 −A2 We solve these two simultaneous equations to obtain A1 = 1 2, A2 = 1 2 (add the two equations together to eliminate A2 yielding 2 = 4A1 or A1 = 1 2, and then determine 302 Chapter 7 Recurrence Relations A2). The required solution to the recurrence relation with the given initial conditions is an = 1 2 × 3n + 1 2 × (−1)n Example 3: Solution of Fibonacci Relation Find a formula for the number of ways for the elf in Example 2 of Section 7.1 to climb the n stairs.
The recurrence relation obtained in Example 2 of Section 7.1 was the Fibonacci relation an = an−1 + an−2, with the initial conditions a1 = 1, a2 = 2, or equivalently, a0 = a1 = 1. The associated characteristic equation is obtained by setting an = αn: αn = αn−1 + αn−2 which reduces to α2 = α + 1 or α2 −α −1 = 0 Using the quadratic formula, we get α = 1 2(1)[−(−1) ± (−1)2 −4(1)(−1)] = 1 2(1 ± √ 5) That is, we have roots 1 2 + 1 2 √ 5 and 1 2 −1 2 √ 5, and the general solution of the problem is an = A1 1 2 + 1 2 √ 5 n + A2 1 2 −1 2 √ 5 n It is left as an exericse to show that A1 = 1 √ 5 1+ √ 5 2 and A2 = −1 √ 5 1− √ 5 2 . We note the surprising fact that to generate the Fibonacci sequence of integers 1, 1, 2, 3, 5, 8, 13, . . . , we need powers of 1 2 + 1 2 √ 5 and 1 2 −1 2 √ 5 .
Optional: The following example illustrates a solution of a recurrence relation that has complex and multiple roots.
Example 4: Complex and Multiple Roots Find a formula for an satisfying the relation an = −2an−2 −an−4 with a0 = 0, a1 = 1, a2 = 2, and a3 = 3.
Substituting an = αn, we obtain αn = −2αn−2 −αn−4, which yields the charac-teristic equation α4 + 2α2 + 1 = (α2 + 1)2 = 0. The roots of this equation are α = +i and α = −i (where i = √−1) and each root has multiplicity 2. Recall that when α is a double root, the associated recurrence relation solutions are an = αn and an = nαn.
So the general solution is an = A1in + A2nin + A3(−i)n + A4n(−1)n www.itpub.net 7.3 Solution of Linear Recurrence Relations 303 The initial conditions yield the equations 0 = a0 = A1i0 + A20i0 + A3(−i)0 + A40(−i)0 = A1 + 0 + A3 + 0 1 = a1 = A1i1 + A21i1 + A3(−i)1 + A41(−i)1 = i(A1 + A2 −A3 −A4) 2 = a2 = A1i2 + A22i2 + A3(−i)2 + A42(−i)2 = −A1 −2A2 −A3 −2A4 3 = a3 = A1i3 + A23i3 + A3(−i)3 + A43(−i)3 = i(−A1 −3A2 + A3 + 3A4) Solving these four simultaneous equations in four unknown Ai’s, we obtain A1 = −3 2i A2 = −1 2 + i A3 = 3 2i A4 = −1 2 −i Then the solution of the recurrence relation is an = −3 2in+1 + −1 2 + i nin + 3 2i(−i)n + −1 2 −i n(−i)n We remind the reader that for specific values of n, such as n = 12, it is eas-ier to determine a12 in the two preceding examples by recursively calculating a3, a4, a5 up to a12 from the recurrence relation than to solve the initial-condition equations.
7.3 EXERCISES 1. If $500 is invested in a savings account earning 8 percent a year, give a formula for the amount of money in the account after n years.
2. Find and solve a recurrence relation for the number of n-digit ternary sequences with no consecutive digits being equal.
3. Solve the following recurrence relations: (a) an = 3an−1 + 4an−2, a0 = a1 = 1 (b) an = an−2, a0 = a1 = 1 (c) an = 2an−1 −an−2, a0 = a1 = 2 (d) an = 3an−1 −3an−2 + an−3, a0 = a1 = 1, a2 = 2 4. Determine the constants A1 and A2 in Example 3. First show that the initial conditions a1 = 1, a2 = 2 are equivalent to the initial conditions a0 = 1, a1 = 1.
5. Find and solve a recurrence relation for the number of ways to arrange flags on an n-foot flagpole using three types of flags: red flags 2 feet high, yellow flags 1 foot high, and blue flags 1 foot high.
304 Chapter 7 Recurrence Relations 6. Find and solve a recurrence relation for the number of ways to make a pile of n chips using red, white, and blue chips and such that no two red chips are together.
7. Find and solve a recurrence relation for pn, the value of a stock market indicator that obeys the rule that the change this year (from the previous year) equals twice last year’s change. Suppose p0 = 1, p1 = 4.
8. Show that any linear combination of solutions to (1) is itself a solution to (1).
9. Show that if the characteristic equation (4) has a root α∗of multiplicity 3, then n jαn ∗, for j = 0, 1, 2, are solutions of (1).
10. Show that if Fn is the nth Fibonacci number in the Fibonacci sequence starting F0 = F1 = 1, then lim n→∞ Fn+1 Fn = 1 + √ 5 2 11. If the recurrence relation an = c1an−1 + c2an−2 has a general solution an = A13n + A26n, find c1 and c2.
7.4 SOLUTION OF INHOMOGENEOUS RECURRENCE RELATIONS A recurrence relation is called homogeneous if all the terms of the relation involve some ak. In the preceding section, we presented a method for solving any homoge-neous linear recurrence relation an = c1an−1 + c2an−2 + · · · + cran−r. When an addi-tional term involving a function of n (or a constant) appears in a recurrence relation, such as an = can−1 + f (n) (1) then the recurrence relation is said to be inhomogeneous.
In this section, we discuss methods for solving inhomogeneous recurrence re-lations of the form of (1). The key idea in solving these relations is that a general solution for an inhomogeneous relation is made up of a general solution to the as-sociated homogeneous relation [obtained by deleting the f (n) term] plus any one particular solution to the inhomogeneous relation.
For (1), the homogeneous relation is an = can−1, whose general solution is an = Acn. Suppose a∗ n is some particular solution of (1)—that is, a∗ n = ca∗ n−1 + f (n).
Then we see that an = Acn + a∗ n satisfies (1): an = Acn + a∗ n = Acn + [ca∗ n−1 + f (n)] = c(Acn−1 + a∗ n−1) + f (n) = can−1 + f (n) www.itpub.net 7.4 Solution of Inhomogeneous Recurrence Relations 305 The constant A in the general solution is chosen to satisfy the initial condition, as in the previous section (A cannot be determined until a∗ n is found).
There is one special case of (1) that we can restate as an enumeration problem treated in previous chapters. If c = 1, then (1) becomes an = an−1 + f (n) (2) We can iterate (2) to get a1 = a0 + f (1) a2 = a1 + f (2) = [a0 + f (1)] + f (2) a3 = a2 + f (3) = [a0 + f (1) + f (2)] + f (3) .
.
.
an = a0 + f (1) + f (2) + f (3) + · · · + f (n) = a0 + n k=1 f (k) So an is just the sum of the f (k)s plus a0. In Sections 5.5 and 6.5 we pre-sented methods for summing functions of n. Either method can be used to solve (2).
Example 1: Summation Recurrence Solve the recurrence relation an = an−1 + n with initial condition a1 = 2, obtained in Example 3 of Section 7.1, for the number of regions created by n mutually intersecting lines.
The initial condition of a1 = 2 can be replaced by the initial condition a0 = 1 (no lines means that the plane is one big region). By the foregoing discussion, we see that an = 1 + (1 + 2 + 3 + · · · + n). The expression to be summed can be written as 1 1 + 2 1 + 3 1 + · · · + n 1 (3) By identity (7) of Section 5.5, this sum equals n+1 2 = 1 2n(n + 1). Then an = 1 + 1 2n(n + 1).
When c ̸= 1 in (1), there are known solutions to (1) to use for specific functions f (n), similar to the situation for “divide-and conquer” recurrences presented in Section 7.2. We present a table for the simplest f (n)s. These solutions can be derived by generating function methods introduced in the next section.
f (n) Particular solution p(n) d, a constant B dn B1n + B0 dn2 B2n2 + B1n + B0 edn Bdn 306 Chapter 7 Recurrence Relations The Bs are constants to be determined. If f (n) were a sum of several different terms, we would separately solve the relation for each separate f (n) term and then add these solutions together to get a particular solution for the composite f (n).
There is one case in which the particular solution for f (n) = edn will not work.
This involves the relation an = dan−1 + edn—note here that the homogeneous so-lution an = Adn has the same form as f (n). Then one must try a∗ n = Bndn as the particular solution.
Example 2: Solving the Tower of Hanoi Puzzle Solve the recurrence relation an = 2an−1 + 1 with a1 = 1 obtained in Example 4 of Section 7.1 for the number of moves required to play the n-ring Tower of Hanoi puzzle. The general solution to the homogeneous equation an = 2an−1 is an = A2n.
We find a particular solution to the inhomogeneous relation by setting a∗ n = B [this is the form of a particular solution given in the foregoing table when f (n) is a constant]. Substituting in the relation, we have B = a∗ n = 2a∗ n−1 + 1 = 2B + 1 or B = −1 So a∗ n = −1 is the particular solution, and the general inhomogeneous solution is an = A2n + a∗ n = A2n −1. We now can determine A from the initial condition: 1 = a1 = A21 −1, or 1 = 2A −1. Hence A = 1, and the desired solution is an = 2n −1.
Example 3: Compound Inhomogeneous Term Solve the recurrence relation an = 3an−1 −4n + 3 × 2n to find its general solution.
Also find the solution when a1 = 8.
The general solution to the homogeneous equation an = 3an−1 is an = A3n. We solve for a particular solution of the relation separately for each inhomogeneous term.
For an = 3an−1 −4n, we try the form a∗ n = B1n + B0, obtaining B1n + B0 = a∗ n = 3a∗ n−1 −4n = 3[B1(n −1) + B0] −4n.
(4) We now equate the constant terms and the coefficients of n on each side of (4): Constant terms: B0 = −3B1 + 3B0 (5) n terms: B1n = 3B1n −4n or B1 = 3B1 −4 (6) Solving for B1 in (6), we obtain B1 = 2. Substituting B1 = 2 in (5), we obtain B0 = 3.
So a∗ n = 2n + 3 is a particular solution of an = 3an−1 −4n.
Next for an = 3an−1 + 3 × 2n, we try a+ n = B2n, obtaining B2n = a+ n = 3a+ n−1 + 3 × 2n = 3(B2n−1) + 3 × 2n (7) Dividing both sides of (7) by 2n−1, we get 2B = 3B + 6, or B = −6. So a+ n = −6 × 2n is a particular solution of an = 3an−1 + 3 × 2n. Combining our particular solutions with the general homogeneous solution, we obtain the general inhomogeneous solution an = A3n + 2n + 3 −6 × 2n www.itpub.net 7.4 Solution of Inhomogeneous Recurrence Relations 307 When a1 = 8, we can determine A: 8 = a1 = A31 + 2(1) + 3 −6 × 21 = 3A −7 Thus A = 5, and the solution is 5 × 3n + 2n + 3 −6 × 2n.
7.4 EXERCISES 1. Solve the following recurrence relations: (a) an = an−1 + 3(n −1), a0 = 1 (b) an = an−1 + n(n −1), a0 = 3 (c) an = an−1 + 3n2, a0 = 10 2. Find and solve a recurrence relation for the number of infinite regions formed by n infinite lines drawn in the plane so that each pair of lines intersects at a different point.
3. Find and solve a recurrence relation for the number of different square subboards of any size that can be drawn on an n × n chessboard.
4. Find and solve a recurrence relation for the number of different regions formed when n mutually intersecting planes are drawn in three-dimensional space such that no four planes intersect at a common point and no two planes have parallel intersection lines in a third plane. [Hint: Reduce to a two-dimensional problem (Example 1).] 5. Find and solve a recurrence relation for the number of regions into which a convex n-gon is divided by all its diagonals, assuming no three diagonals intersect at a common point. (Hint: Sum the inhomogeneous term using a special case of an identity from Section 5.5.) 6. If the average of two successive years’ production 1 2(an + an−1) is 2n + 5 and a0 = 3, find an.
7. Solve the recurrence relation an = 1.04an−1 + 100, a0 = 0, from part (b) of Example 5 in Section 7.1.
8. Suppose a savings account earns 5 percent a year. Initially there is $1,000 in the account, and in year k, $10k are withdrawn. How much money is in the account at the end of n years if (a) Annual withdrawal is at year’s end?
(b) Withdrawal is at start of year?
9. Solve the following recurrence relations: (a) an = 3an−1 −2, a0 = 0 (b) an = 2an−1 + (−1)n, a0 = 2 (c) an = 2an−1 + n, a0 = 1 (d) an = 2an−1 + 2n2, a0 = 3 308 Chapter 7 Recurrence Relations 10. Solve the recurrence relation an = 3an−1 + n2 −3, with a0 = 1.
11. Solve the recurrence relation an = 3an−1 −2an−2 + 3, a0 = a1 = 1.
12. Find and solve a recurrence relation for the number of n-digit ternary sequences in which no 1 appears to the right of any 2.
13. Find and solve a recurrence relation for the earnings of a company when the rate of increase of earnings increases by $10 × 2k in the kth year from the previous year, where a0 = 20 and a1 = 1020.
14. Show that the general solution to any inhomogeneous linear recurrence relation is the general solution to the associated homogeneous relation plus one particular inhomogenous solution.
15. Show that the form of the particular solution of (1) given in the table in this section is correct for (a) f (n) = d (b) f (n) = dn (c) f (n) = dn2 (d) f (n) = dn 16. Show that if f (n) in (1) is the sum of several different terms, a particular solution for this f (n) may be obtained by summing particular solutions for the individual terms.
17. Find a general solution to an −5an−1 + 6an−2 = 2 + 3n.
18. If the recurrence relation an −c1an−1 + c2an−2 = c3n + c4 has a general solution an = A12n + A25n + 3n −5, find c1, c2, c3, c4.
19. Solve the following recurrence relations when a0 = 1 (a) a2 n = 2a2 n−1 + 1 (Hint: Let bn = a2 n.) (b) an = −nan−1 + n! [Hint: Define an appropriate bn as in part (a).] 7.5 SOLUTIONS WITH GENERATING FUNCTIONS Most recurrence relations for an can be converted into an equation involving the generating function g(x) = a0 + a1x + · · · + anxn + · · ·. This associated functional equation for g(x) can often be solved algebraically and the resulting expression for g(x) expanded in a power series to obtain an as the coefficient of xn. Some of the algebraic manipulations of the functional equations may be new to the reader.
We will treat g(x) as if it were a standard single variable, such as y, and treat other functions of x as constants. For example, the functional equation g(x) = x2g(x) −2x can be solved by rewriting the equation as g(x)(1 −x2) = −2x and hence g(x) = −2x(1 −x2)−1. Similarly, the functional equation (1 −x2)[g(x)]2 −4xg(x) + 4x2 = 0 (1) can be solved by the quadratic formula that we normally apply to equations such as ay2 + by + c = 0. Now a = (1 −x2), b = −4x, and c = 4x2. Intuitively, for each www.itpub.net 7.5 Solutions with Generating Functions 309 particular value of x, g(x) is the solution of (1). Thus, by the quadratic formula, the solution to (1) is g(x) = 1 2(1 −x2)[4x ± 16x2 −16x2(1 −x2)] = 1 2(1 −x2)(4x ± 4x2) So g(x) = 2(x + x2)/(1 −x2) or 2(x −x2)/(1 −x2). If there are two (or more) possible solutions, only one will normally make sense as a generating function for an (e.g., have a power series expansion with the correct value for the initial condition a0).
Now let us show by example how some of the recurrence relations obtained in Section 7.1 can be converted into functional equations for an associated generating function.
Example 1: Summation Recurrence Find a functional equation for g(x) = a0 + a1x + · · · + anxn + · · · where an satisfies the recurrence relation an = an−1 + n, when n ≥1, obtained in Example 3 of Section 7.1. The initial condition was a0 = 1. Solve the functional equation and expand g(x) to find an.
Using this recurrence relation for every term in g(x) except a0, we have anxn = an−1xn + nxn, n ≥1. Summing the terms, we can write g(x) −a0 = ∞ n=1 anxn = ∞ n=1 (an−1xn + nxn) (2) = x ∞ n=1 an−1xn−1 + ∞ n=1 nxn (3) = x ∞ m=0 amxm + ∞ n=0 n 1 xn (4) = xg(x) + x (1 −x)2 (5) Line (3) is obtained from line (2) by breaking up the sum of the two xn terms into sums of each term, and by rewriting an−1xn as xan−1xn−1 (in order to make the power of x correspond with the subscript of an−1). Line (4) is obtained from line (3) by reindexing the first sum with m = n −1, and by adding the “phantom” (zero) term 0x0 to the second sum and rewriting n as C(n, 1). The first series is the generating function g(x) multiplied by x. The second series has a generating function obtained by the construction presented in Section 6.5. Equating line (5) with g(x) −a0 [the left side of line (2)] and setting a0 = 1, we have the required functional equation for g(x): g(x) −1 = xg(x) + x (1 −x)2 (6) 310 Chapter 7 Recurrence Relations Solving for g(x), we rewrite (6) as g(x) −xg(x) = 1 + x (1 −x)2 or g(x)(1 −x) = 1 + x (1 −x)2 Thus g(x) = 1 (1 −x) + x (1 −x)3 The coefficient of xn in (1 −x)−1 is just 1 and in x(1 −x)−3 is C((n −1) + 3 − 1, n −1) = C(n + 1, n −1) = C(n + 1, 2). Then g(x) = 1 (1 −x) + x (1 −x)3 = ∞ n=0 xn + ∞ n=0 n + 1 2 xn = ∞ n=0 1 + n + 1 2 xn and so an = 1 + C(n + 1, 2)—the same answer as obtained for this recurrence relation in Example 1 of Section 7.4.
Example 2: Fibonacci Relation Use generating functions to solve the recurrence relation an = an−1 + an−2, with a1 = 1, a2 = 2 obtained in Example 2 of Section 7.1.
The initial conditions, a1 = 1, a2 = 2, are equivalent to a0 = 1, a1 = 1. Then using the same power series summation approach as in the previous example, we obtain g(x) −a0 −a1x = ∞ n=2 anxn = ∞ n=2 (an−1xn + an−2xn) = x ∞ n=2 an−1xn−1 + x2 ∞ n=2 an−2xn−2 = x ∞ m=1 amxm + x2 ∞ k=0 akxk = x[g(x) −a0] + x2g(x) Setting a0 = 1 and a1 = 1, we have the functional equation g(x) −1 −x = x[g(x) −1] + x2g(x) or g(x) −xg(x) −x2g(x) −1 = 0 and so g(x)(1 −x −x2) = 1 or g(x) = 1/(1 −x −x2) Observe that this denominator is closely related to the characteristic equation of this recurrence relation, given in Example 3 of Section 7.3. In general, g(x) will have a denominator 1 + c1x + c2x2 + · · · + crxr if and only if xr + cr−1xr−1 + c2xr−2 + · · · + cr is the characteristic equation of the associated recurrence relation, and so 1 −αx is a factor of the denominator of g(x) if and only if α is a root of the characteristic equation. The numerator will depend on the initial conditions.
www.itpub.net 7.5 Solutions with Generating Functions 311 By the quadratic formula, we can factor 1 −x −x2 = 1 −1 2(1 + √ 5)x][1 −1 2(1 − √ 5)x For simplicity, let us define α1 = 1 2(1 + √ 5) and α2 = 1 2(1 − √ 5). Then we have g(x) = 1 (1 −α1x)(1 −α2x) = α1/ √ 5 1 −α1x −α2/ √ 5 1 −α2x (7) The decomposition of g(x) into two fractions in (7) is obtained by the method of partial fractions, the reverse process of combining two fractions into a common fraction. Setting y = α1x in the first fraction on the right side in (7), we have α1/ √ 5 1 −α1x = α1 √ 5 1 1 −y = α1 √ 5 ∞ n=0 yn = α1 √ 5 ∞ n=0 αn 1xn The same type of expansion exists for y = α2x. Then an, the coefficient of xn in the power series expansion of g(x), is an = 1 √ 5 αn+1 1 −1 √ 5 αn+1 2 = 1 √ 5 1 + √ 5 2 n+1 −1 √ 5 1 − √ 5 2 n+1 Observe that it is much easier to determine a10 by recursively computing a2, a3, . . . up to a10 using the Fibonacci relation than by setting n = 10 in the foregoing formula for an.
Example 3: Selection without Repetition Let gn(x) be a family of generating functions gn(x) = an,0 + an,1x + · · · + an,kxk + · · · + an,nxn satisfying the relation an,k = an−1,k + an−1,k−1, with an,0 = an,n = 1 (and an,k = 0, k > n) obtained in Example 8 of Section 7.1 for an,k, the number of k-subsets of an n-set. Find a functional relation among the gn(x)s and solve it to obtain a formula for an,k. Using the power series summation method, we obtain gn(x) −1 = n k=1 an,kxk = n k=1 (an−1,kxk + an−1,k−1xk) = n k=1 an−1,kxk + x n−1 h=0 an−1,hxh = gn−1(x) −1 + xgn−1(x) Thus gn(x) = gn−1(x) + xgn−1(x) = (1 + x)gn−1(x) 312 Chapter 7 Recurrence Relations The resulting recurrence gn(x) = (1 + x)gn−1(x) is solved just like the recurrence an = can−1. The solution is gn(x) = (1 + x)ng0(x) = (1 + x)n since g0(x) = a0,0 = 1. Now, by the binomial theorem, we have an,k = C(n, k).
Optional: The rest of this section involves more complicated computations. The reader may skip this material. First we consider a nonlinear recurrence relation and solve it using generating functions.
Example 4: Placing Parentheses Solve the recurrence relation an = a1an−1 + a2an−2 + · · · + aian−i + · · · + an−1a1 obtained in Example 10 of Section 7.1 for the number of ways to place parenthe-ses when multiplying n numbers.
Observe that if g(x) = a0 + a1x + · · · + anxn + · · ·, then the right-hand side of this equation is simply the coefficient of xn, for n ≥2, in the product g(x)g(x) = (0 + a1x + · · · + anxn + · · ·)2. Using the power series summation method, we have (recall that a0 = 0 and a1 = 1) g(x) −1x = ∞ n=2 anxn = ∞ n=2 (a1an−1 + a2an−2 + · · · + an−1a1)xn = [g(x)]2 Solving this quadratic equation in g(x) as described at the start of this section, we obtain g(x) = 1 2(1 ± √ 1 −4x). To make a0 = 0 [i.e., g(0) = 0], we want the solution g(x) = 1 2 −1 2 √ 1 −4x.
This g(x) requires a new type of generating function expansion called thegeneral-ized binomial theorem. The power series expansion (1 + y)q = q 0 + q 1 y + q 2 y2 + · · · + q n yn + · · · , where q is any real number, has a coefficient q n of yn defined as q n = q(q −1)(q −2) × · · · × [q −(n −1)] n!
(8) [The formula for this generalized coefficient q n arises from the Taylor series for (1 + y)q; see any standard calculus text.] Using (8), the coefficient of xn in √1 −4x is [we think of √1 −4x as (1 + y)1/2, where y = −4x] 1 2 n (−4)n = 1 2(−1 2)(−3 2) × · · · × [−1 2(2n −3)] n!
(−4)n = −1 × 1 × 3 × 5 × · · · × (2n −3) n!
2n = −2 n 2n −2 n −1 (9) The last step in (9) is obtained by multiplying certain numbers in the numerator by appropriately selected powers of 2 (details are left to Exercise 6). Multiplying the www.itpub.net 7.5 Solutions with Generating Functions 313 final expression in Eq. (9) by −1 2, we obtain the coefficient of xn in −1 2 √ 1 −4x, an = 1 n 2n −2 n −1 n ≥1 The expression 1 2C(2n −2, n −1) arises in various combinatorial settings and is called the nth Catalan number. We note, as an aside, that while where parentheses are placed makes no real difference when multiplying numbers, if we were working with a complex product of matrices then the placement of parentheses has an impor-tant impact on the amount of computation required. Next let us consider generating functions for simultaneous recurrence relations.
Example 5: Simultaneous Recurrence Relations Use generating functions to solve the set of simultaneous recurrence relations obtained in Example 11 of Section 7.1: an = an−1 + bn−1 + cn−1, bn = 3n−1 −cn−1, cn = 3n−1 −bn−1, a1 = b1 = c1 = 1 Let A(x), B(x), and C(x) be the generating functions for an, bn, and cn, respectively.
We use the power series summation method to obtain A(x) −a0 = ∞ n=1 anxn = ∞ n=1 an−1xn + ∞ n=1 bn−1xn + ∞ n=1 cn−1xn = x ∞ m=0 amxm + x ∞ m=0 bmxm + x ∞ m=0 cmxm = x A(x) + x B(x) + xC(x) B(x) −b0 = ∞ n=1 bnxn = ∞ n=1 3n−1xn − ∞ n=1 cn−1xn = x(1 −3x)−1 −xC(x) C(x) −c0 = ∞ n=1 cnxn = ∞ n=1 3n−1xn − ∞ n=1 bn−1xn = x(1 −3x)−1 −x B(x) It is always desirable to state initial conditions in terms of a0, b0, c0. Solving our three recurrence relations for a0, b0, c0 given a1 = b1 = c1 = 1, we get 1 = b1 = 30 −c0 or c0 = 0. Similarly, we find that b0 = 0 and a0 = 1. Then our functional equations are A(x) −1 = x A(x) + x B(x) + xC(x) or A(x) = 1 1 −x [x B(x) + xC(x) + 1] (10) 314 Chapter 7 Recurrence Relations and B(x) = x 1 −3x −xC(x) (11) C(x) = x 1 −3x −x B(x) (12) We can solve (11) and (12) for B(x) and C(x) simultaneously. Multiplying (12) by x and using this expression for xC(x) in (11), we have B(x) = x 1 −3x −xC(x) = x 1 −3x −x x 1 −3x −x B(x) ⇒B(x)(1 −x2) = x −x2 1 −3x ⇒B(x) = (1 −x)x (1 −x2)(1 −3x) = x (1 + x)(1 −3x) = 1 4 1 −3x − 1 4 1 + x The last step is again a partial fraction decomposition. The coefficient of xn in 1 4(1 −3x)−1 is 1 43n and in −1 4(1 + x)−1 is −1 4(−1)n. So bn, the coefficient of xn in B(x), is 1 4[3n −(−1)n]. Equations (11) and (12) are symmetric with respect to B(x) and C(x), and so C(x) = B(x) and cn = bn = 1 4[3n −(−1)n].
Next we solve for A(x) in (10): A(x) = 1 1 −x [x B(x) + xC(x) + 1] = 2x 1 −x B(x) + 1 1 −x [since B(x) = C(x)] = 2x 1 −x 1 4 1 −3x − 1 4 1 + x + 1 1 −x = 1 2x (1 −x)(1 −3x) − 1 2x 1 −x2 + 1 1 −x = 1 4 1 −3x − 1 4 1 −x − 1 2x 1 −x2 + 1 1 −x The coefficient of xn in 1 4(1 −3x)−1 is 1 43n, in −1 4(1 −x)−1 is −1 4, in −x 2(1 −x2)−1 is −1 2, n odd, or 0, n even, and in (1 −x)−1 is 1. Collecting these terms, we get an = 1 4(3n + 3), n even, and = 1 4(3n + 1), n odd.
7.5 EXERCISES 1. Find functional equations for the generating functions whose coefficients satisfy the following relations: (a) an = an−1 + 2, a0 = 1 (b) an = 3an−1 −2an−2 + 2, a0 = a1 = 1 www.itpub.net 7.5 Solutions with Generating Functions 315 (c) an = an−1 + n(n −1), a0 = 1 (d) an = 2an−1 + 2n, a0 = 1 2. Solve the recurrence relations in Exercise 1 using generating functions.
3. Find functional equations for the generating functions whose coefficients satisfy the following relations: (a) an = n−1 i=0 aian−1−i (n ≥1), a0 = 1 (b) an = n−2 i=2 aian−i (n ≥3), a0 = a1 = a2 = 1 (c) an = n−1 i=1 2ian−i (n ≥2), a0 = a1 = 1 4. Find a functional equation and solve it for the sequence of generating functions Fn(x) = n k=0 an,kxk whose coefficients satisfy the following: (a) an,k = an,k−1 −2an−1,k−1, an,0 = 0 (b) an,k = 2an−1,k −3an,k−1, an,0 = 1 5. Verify the form of particular solutions to inhomogeneous recurrence relations in the table in Section 7.4.
6. Verify in (9) that −1 × 1 × 3 × 5 × · · · × (2n −3) n!
2n = −2 n 2n −2 n −1 7. Find a recurrence relation and solve it with generating functions for the number of ways to divide an n-gon into triangles with noncrossing diagonals. (Hint: Use reasoning similar to Example 4.) 8. Find a recurrence relation and associated generating function for the number of n-digit ternary sequences that have the pattern “012” occurring for the first time at the end of the sequence.
9. Find a recurrence relation and associated generating function for the number of different binary trees with n leaves.
10. Find a recurrence relation for an,k, the number of k-subsets of an n set with repetition. Find an equation for Fn(x) = ∞ k=0 an,kxk, and solve for Fn(x) and an,k.
11. Let an,k be the probability that k successes occur in an experiment with n trials if each trial has probability p of success. Find a recurrence relation for an,k. Use this relation to find and solve an equation for Fn(x) = n k=0 an,kxk.
12. (a) Find a recurrence relation for an,k, the number of k-permutations of n elements.
316 Chapter 7 Recurrence Relations (b) Show that Fn(x) = n k=0 an,kxk satisfies the differential equation [F′ n−1(x) denotes the derivative of Fn−1(x)] Fn(x) = (1 + x)Fn−1(x) + x2F′ n−1(x) (c) Find a functional equation for Gn(x) = n k=0 an,kxk/k!
13. Find and solve a system of recurrence relations allowing one to determine the number of n-digit quaternary sequences with an odd number of 1s and an odd number of 2s.
14. Find and solve simultaneous recurrence relations for determining the number of n-digit ternary sequences whose sum of digits is a multiple of 3.
15. (a) Find a recurrence relation for an, the number of ways to partition n distinct objects among n indistinguishable boxes (some boxes may be empty).
(b) Let g(x) = ∞ n=0 anxn/n! where a0 = 1. Show that g(x) satisfies the differ-ential equation g′(x) = g(x)ex. Solve this equation for g(x).
16. (a) Define sn,r as numbers such that n r=0 sn,rxr = x(x −1)(x −2) · · · (x −n + 1). Find a recurrence relation for sn,r.
(b) Find a differential equation for Fn(x) = n r=0 sn,rxr/r!
7.6 SUMMARY AND REFERENCES In this chapter we saw that recurrence relations are one of the simplest ways to solve counting problems. Without fully understanding the combinatorial process, as was required in Chapter 5, we now need only express a given problem for n objects in terms of the problem posed for fewer numbers of objects. Once a recurrence relation has been found, then starting with a1 (the solution for one object), we can successively compute the solutions for 2, 3, . . . up to any (moderate) value of n. Or we can try one of the techniques in the later sections of this chapter to solve the recurrence relation explicitly.
The first recurrence relation in mathematical writings was the Fibonacci relation.
In his work Liber abaci, published in 1220, Leonardo di Pisa, known also as Fibonacci, posed a counting problem about the growth of a rabbit population (Exercise 7 in Section 7.1). The number an of rabbits after n months was shown to satisfy the Fibonacci relation an = an−1 + an−2. As mentioned in Section 6.6, DeMoivre gave the first solution of this relation 500 years later in 1730 using the generation function derivation given in Example 2 of Section 7.5. The Fibonacci relation and numbers have proven to be amazingly ubiquitous. For example, it has been shown that the ratios of Fibonacci numbers provide an optimal way (in a certain sense) to divide up an interval when searching for the minimum of a function in this interval (see Kiefer ). The appearance of Fibonacci numbers in rings of leaves around flowers is discussed in Adler .
www.itpub.net 7.6 Summary and References 317 The methods for solving recurrence relations appeared originally in the devel-opment of the theory of difference equations, cousins of differential equations. For a good presentation of the methods and applications of difference equations, see Sandefur .
1. I. Adler, “The consequence of constant pressure in phyllotaxis,” J. Theoretical Biology 65 (1977), 29–77.
2. T. Cormen, C. Leiserson, and R. Rivest, An Introduction to Algorithms, 3rd ed., MIT Press, Cambridge, MA, 2009.
3. J. Kiefer, “Sequential minimax search for a maximum,” Proceedings of American Math. Society 4 (1953), 502–506.
4. J. Sandefur, Discrete Dynamical Modeling, Oxford University Press, New York, 1993.
This page is intentionally left blank www.itpub.net CHAPTER 8 INCLUSION–EXCLUSION 8.1 COUNTING WITH VENN DIAGRAMS In this chapter we develop a set-theoretic formula for counting problems involving several interacting properties in which either all properties must hold or none must hold. An example is counting all five-card hands with at least one card in each suit, or equivalently, all five-card hands with no void in any suit. In the process of solving such problems, we have to count the subsets of outcomes in which various combinations of the properties hold. We use Venn diagrams to depict these different combinations. See Appendix A.1 for a review of the essentials of sets and Venn diagrams.
Let us begin with a one-property Venn diagram, and then progress to two- and three-property problems. In Figure 8.1 we show a set A within a universe U. The complementary set A consists of all elements of U not in A. Let N(S) denote the number of elements in set S. We define N = N(U). Then N(A) = N −N(A), or N(A) = N −N(A). This is similar to the situation in probability where the probability of an event E is 1 minus the probability of the complementary event E.
Suppose, for example, that there is a “universe” of 100 students in a math course and there are 30 students who are not math majors in the course—that is, N(A) = 30, where A is the set of math majors. Then the number N(A) of math majors in the course is N(A) = N −N(A) = 100 −30 = 70.
Next consider a problem with two sets. Let the universe U be all students in a school, let F be the set of students taking French, and let L be the set of students taking Latin. See Figure 8.2. We want formulas for the number of students taking French or Latin N(F ∪L) and the number taking neither language N(F ∩L) in terms of N, N(F), N(L), and N(F ∩L). Note that N(F ∪L) is not simply N(F) + N(L), because N(F) + N(L) counts each student taking both languages two times. Thus, we must know how many students take both languages, the number N(F ∩L). Subtracting N(F ∩L) from N(F) + N(L) corrects the double counting of students taking two languages. That is, N(F ∪L) = N(F) + N(L) −N(F ∩L) (1) By de Morgan’s law (Equation BA3 of Appendix A.1), F ∩L = F ∪L, and so N(F ∩L) = N(F ∪L) = N −N(F ∪L) 319 320 Chapter 8 Inclusion–Exclusion _ A A Figure 8.1 Combining this equation with (1), we have N(F ∩L) = N −N(F ∪L) = N −N(F) −N(L) + N(F ∩L) (2) Formula (2) is the 2-set version of the general n-set inclusion–exclusion formula that we present in the next section. It is called an inclusion–exclusion formula because first we include the whole set, N; then we exclude (subtract) the single sets F and L; and then we include (add) the 2-set intersection F ∩L. With more sets, this alternating inclusion and exclusion process will continue several rounds.
Example 1: Students Taking Neither Language If a school has 100 students with 50 students taking French, 40 students taking Latin, and 20 students taking both languages, how many students take no language?
In this problem, N = 100, N(F) = 50, N(L) = 40, and N(F ∩L) = 20. We need to determine N(F ∩L). By Eq. (2), we have N(F ∩L) = N −N(F) −N(L) + N(F ∩L) = 100 −50 −40 + 20 = 30 The next example applies this set-theoretic formula to a counting problem that has no obvious set-theoretic structure.
Example 2: Restricted Arrangements How many arrangements of the digits 0, 1, 2,. . . , 9 are there in which the first digit is greater than 1 and the last digit is less than 8?
Before using formula (2) to solve this problem, let us consider a more direct approach and see why it fails. For the first digit, there are eight choices. Then for the last digit, there are . . . the number of choices depends on whether or not an 8 or 9 was chosen for the first digit. If an 8 or 9 were chosen for the first digit, there will be eight choices for the last digit, while if neither 8 nor 9 were chosen for the first digit, F L F L ⊃ Figure 8.2 www.itpub.net 8.1 Counting with Venn Diagrams 321 there will be seven choices for the last digit. Because of this difficulty, we shall solve the problem with formula (2).
Let U be all arrangements of 0, 1, 2,. . . , 9. Let F be the set of all arrangements with a 0 or 1 in the first digit, and let L be the set of all arrangements with an 8 or 9 in the last digit. Then the number of arrangements with first digit greater than 1 and the last digit less than 8 is N(F ∩L).
We have N = 10!, N(F) = 2 × 9! (two choices for the first digit followed by any arrangement for the remaining 9 digits), and N(L) = 2 × 9! Similarly, N(F ∩L) = 2 × 2 × 8!. Then by (2), N(F ∩L) = 10! −(2 × 9!) −(2 × 9!) + (2 × 2 × 8!) The strategy used to solve Example 2 and most other examples in this chapter is as follows: when faced with multiple constraints that are difficult to enumerate, we try to solve the problem by counting the sizes of the complementary sets for these constraints, as well as the intersections of these complementary sets. The inclusion– exclusion formula tells us how to put together the answers to the complementary-set calculations solve the original problem.
There is one important notational convention to watch out for. The sets in the inclusion–exclusion formula are always defined so that the final answer (the left side of (2)) is the number of items that are in none of the sets. Thus, we need to define sets that represent the complements of the original constraints we are given. For example, in Example 2, the set of arrangements where the first digit is not greater than 1—the complement of the given constraint on the first digit—is defined to be F, so that the original constraint of the first digit being greater than 1 is now F.
Consider next a problem with three sets. We extend Figure 8.2 with the additional set G of students taking German, as shown in Figure 8.3. We want a formula for N(F ∩L ∩G). A first guess might be N(F ∩L ∩G) ?
= N −N(F) −N(L) −N(G) As in Figure 8.2, this formula double-counts (that is, subtracts twice) the students in two of the sets, F, L, and G. We can correct this first formula by adding the number of students taking two languages. Thus we propose the formula N(F ∩L ∩G) ?
= N −N(F) −N(L) −N(G) +N(F ∩L) + N(L ∩G) + N(F ∩G) (3) F L G Figure 8.3 322 Chapter 8 Inclusion–Exclusion Number of Languages + [N(F ∩L) Taken by −[N(F) + N(L) + N(L ∩G) Student N + N(G)] + N(F ∩G)] −N(F ∩L ∩G) 0 +1 0 0 0 1 +1 −1 0 0 2 +1 −2 +1 0 3 +1 −3 +3 −1 Figure 8.4 Figure 8.4 shows how many times a student will be added and subtracted in parts of this formula. A student taking no language is counted once (by the term N)—such students are exactly the ones we want to count. The challenge is to make sure that all other students are counted a net of 0 times. The students taking one language are counted once by N and subtracted once by the term −[N(F) + N(L) + N(G)], for a net count of 0. The students taking two languages are counted once by N, subtracted twice by −[N(F) + N(L) + N(G)] (since they are in exactly two of the three sets), and then added once by the term +[N(F ∩L) + N(L ∩G) + N(F ∩G)] (since they are in exactly one of the three pairwise intersections), for a net count of 0.
Finally, we consider the students taking all three languages. They are counted once by N, then subtracted three times by the sum of the three sets (since they are in all three sets), then added three times by the pairwise intersections (since they are in all three of these subsets). This yields a net count of 1 −3 + 3 = 1. Then we must correct formula (3) by subtracting N(F ∩L ∩G) to make the net count of students with all three languages 0: N(F ∩L ∩G) = N −[N(F) + N(L) + N(G)] + [N(F ∩L) + N(L ∩G) + N(F ∩G)] −N(F ∩L ∩G) (4) For general sets A1, A2, A3, we rewrite (4) as N(A1 ∩A2 ∩A3) = N − i N(Ai) + i j N(Ai ∩A j) −N(A1 ∩A2 ∩A3) (5) where the sums are understood to run over all possible i and all i,j pairs, respectively.
Example 3: Students Taking None of Three Languages If a school has 100 students with 40 taking French, 40 taking Latin, and 40 taking German, 20 students are taking any given pair of languages, and 10 students are taking all three languages, then how many students are taking no language?
Here N = 100, N(F) = N(L) = N(G) = 40, N(F ∩L) = N(L ∩G) = N(F ∩G) = 20, and N(F ∩L ∩G) = 10. Then by (4), the number of students taking no language is N(F ∩L ∩G) = 100 −(40 + 40 + 40) + (20 + 20 + 20) − 10 = 30.
www.itpub.net 8.1 Counting with Venn Diagrams 323 Next we apply (5) to two counting problems that cannot be solved by methods developed in the three previous chapters.
Example 4: Relatively Prime Numbers How many positive integers ≤70 are relatively prime to 70?
Let U be the set of integers between 1 and 70. The phrase “relatively prime to 70” means “have no common divisors with 70.” The prime divisors of 70 are 2, 5, and 7. Then we want to count the number of integers ≤70 that do not have 2 or 5 or 7 as divisors. Let A1 be the set of integers in U that are evenly divisible by 2, or equivalently, integers in U that are multiples of 2; A2 be integers evenly divisible by 5; and A3 be integers evenly divisible by 7. Then the number of positive integers ≤70 that are relatively prime to 70 equals N(A1 ∩A2 ∩A3). We find N = 70 N(A1) = 70/2 = 35 N(A2) = 70/5 = 14 N(A3) = 70/7 = 10 The integers evenly divisible by both 2 and 5 are simply the integers evenly divis-ible by 10. Thus N(A1 ∩A2) = 70/10 = 7. By similar reasoning, N(A2 ∩A3) = 70/ (5×7) = 2, N(A1 ∩A3) = 70/(2×7) = 5, and N(A1 ∩A2 ∩A3) = 70/(2×5×7) = 1.
So by Eq. (5): N(A1 ∩A2 ∩A3) = 70 −(35 + 14 + 10) + (7 + 2 + 5) −1 = 24 Example 5: Ternary Sequences with No Voids How many n-digit ternary (0, 1, 2) sequences are there with at least one 0, at least one 1, and at least one 2? How many n-digit ternary sequences with at least one void (missing digit)?
Let U be the set of all n-digit ternary sequences. Formula (5) counts outcomes in which none of a set of properties holds. So we must formulate the first part of this problem in terms of outcomes for which none of a set of properties holds. The solution is to define Ai to be the number of n-digit ternary sequences with no is, for i = 0, 1, 2.
(Note that instead of numbering the sets A1, A2, A3, we are using A0, A1, A2.) Then the number of sequences with at least one of each digit will be N(A0 ∩A1 ∩A2).
The number of n-digit ternary sequences is N = 3n. The number of n-digit ternary sequences with no 0s is simply the number of n-digit sequences of 1s and 2s. Thus, N(A0) = 2n. Similarly, N(A1) = N(A2) = 2n. The only n-digit sequence with no 0s and no 1s is the sequence of all 2s. Then N(A0 ∩A1) = 1; also N(A1 ∩A2) = N(A0 ∩A2) = 1. Finally, there is no ternary sequence with no 0s and no 1s and no 2s. Then by (5), N(A0 ∩A1 ∩A2) = 3n −(2n + 2n + 2n) + (1 + 1 + 1) −0 = 3n −3 × 2n + 3 Now we turn to the second part of this problem involving n-digit ternary se-quences with at least one void. The phrase “at least” is used in a very different way here than it was used in the first part. At least one void means a void of the digit 0 or 324 Chapter 8 Inclusion–Exclusion a void of the digit 1 or a void of digit 2. In terms of the Ai defined above, we want to count the union of the Ai’s—namely, N(A1 ∪A2 ∪A3). Thus, the sequences with at least one void are exactly the complement of the sequences with no voids that were counted in the first part. So the answer to the second part is N(A1 ∪A2 ∪A3) = N −N(A1 ∩A2 ∩A3) = 3n −(3n −3 × 2n + 3) = 3 × 2n −3 The reader must be constantly alert for union problems when seeing the phrase “at least” in inclusion–exclusion problems. In one case (the more common case), we are counting outcomes with no voids of any type—an intersection problem—which means at least one outcome of the first type and at least one outcome of the second type and etc. In the other case, we are counting outcomes with at least one of a set of properties—a union problem—which means an outcome with the first property or an outcome with the second property or etc. A general formula for union problems will be presented in the next section.
Observe that whereas polynomial algebra was used in Chapter 6 to model count-ing problems and recurrence relations were used in Chapter 7, now we are using a set-theoretic model. This approach does not eliminate combinatorial enumeration as the other models did. We still must solve the subproblems of finding N(Ai), N(Ai ∩A j), and so forth, but these are much easier problems.
Sometimes non-standard input data for a Venn diagram are given. The following example shows how to determine the sizes of all possible subsets in a Venn diagram in such circumstances.
Example 6: Nonstandard Constraints Suppose there are 100 students in a school and there are 40 students taking each language, French, Latin, and German. Twenty students are taking only French, 20 only Latin, and 15 only German. In addition, 10 students are taking French and Latin.
How many students are taking all three languages? No language?
We draw the Venn diagram for this problem and number each region as shown in Figure 8.5. Let Ni denote the number of students in region i, for i = 1, 2, . . . , 8.
Students taking only French are the subset F ∩L ∩G, region 1; so N1 = 20. Similarly, the other information given us says N5 = 20 and N7 = 15. Students taking both French and Latin are the subset F ∩L = (F ∩L ∩G) ∪(F ∩L ∩G), regions 2 and 3. So F L G 1 3 4 6 7 5 2 8 Figure 8.5 www.itpub.net 8.1 Counting with Venn Diagrams 325 N2 + N3 = 10. The set of students taking French is F, which consists of regions 1, 2, 3, and 4. So 40 = N(F) = N1 + (N2 + N3) + N4 = 20 + (10) + N4, or N4 = 10 Similarly, set L consists of regions 2, 3, 5, and 6, and so 40 = N(L) = (N2 + N3) + N5 + N6 = (10) + 20 + N6, implying N6 = 10.
Since G consists of regions 3, 4, 6, and 7, and since we were given N7 = 15 and have just found that N4 = N6 = 10, then 40 = N(G) = N3 + N4 + N6 + N7 = N3 + 10 + 10 + 15 or N3 = 5 But region 3 is the subset F ∩L ∩G of students taking all three languages. Thus, there are five trilingual students.
The general line of attack in these problems is to break the Venn diagram into the eight regions shown in Figure 8.5. For each subset whose size is given, write that number as a sum of Ni’s of regions i in that subset. By combining these equations (and sometimes solving them simultaneously), one can eventually determine the number of elements in each region. Then the size of any subset is readily found as the sum of the sizes of the regions in that subset. For example, in the preceding paragraph we determined all Nis except N2 and N8. But N2 + N3 = 10 and N3 was found to be 5; thus N2 = 5. N8 = N(F ∩L ∩G) is the number of students taking no language.
Since all regions total to N, then N8 = N − 7 i=1 Ni = 100 −(20 + 5 + 5 + 10 + 20 + 10 + 15) = 100 −85 = 15 Note that because there are eight underlying variables (the sizes of the eight basic regions), we need to be given eight pieces of information to be able to solve such a problem.
8.1 EXERCISES S u m m a r y o f E x e r c i s e s The first 30 exercises are similar to the examples in this section. The last six exercises involve more complicated Venn diagram arguments; see the last paragraph in Example 6 for the strategy for solving these problems.
1. How many 8-letter “words” using the 26-letter alphabet (letters can be repeated) either begin or end with a vowel?
2. How many 9-digit Social Security numbers are there with repeated digits?
3. How many n-digit ternary sequences are there in which at least one pair of consecutive digits are the same?
326 Chapter 8 Inclusion–Exclusion 4. What is the probability that at least two heads (not necessarily consecutive) will appear when a coin is flipped eight times?
5. What is the probability that a 7-card hand has at least one pair (possibly two pairs, three of a kind, full house, or four of a kind)?
6. If n people of different heights are lined up in a queue, what is the probability that at least one person is just behind a taller person?
7. How many 5-digit numbers (including leading 0s) are there with exactly one 8 and no digit appearing exactly three times?
8. Suppose 40% of all families own a dishwasher, 30% own a trash compacter, and 20% own both. What percentage of all families own neither of these two appliances?
9. Among 700 families, 150 families have no children, 180 have only boys, and 200 have only girls. How many families have boy(s) and girl(s)?
10. How many ways are there to pick five people for a committee if there are six (different) men and eight (different) women and the selection must include at least one man and one woman?
11. Suppose a bookcase has 300 books, 70 in French and 100 about mathematics.
How many non-French books not about mathematics are there if (a) There are 40 French mathematics books?
(b) There are 60 French nonmathematics books?
12. How many arrangements of the 26 different letters are there that (a) Contain either the sequence “the” or the sequence “aid”?
(b) Contain neither the sequence “the” nor the sequence “math”?
13. If you pick an integer between 1 and 1,000, what is the probability that it is either divisible by 7 or even (or both)?
14. How many arrangements of the letters in INVITING are there in which the three Is are consecutive or the two Ns are consecutive (or both)?
15. A school has 200 students with 85 students taking each of the three subjects: trigonometry, probability, and basket-weaving. There are 30 students taking any given pair of these subjects, and 15 students taking all three subjects.
(a) How many students are taking none of these three subjects?
(b) How many students are taking only probability?
16. Suppose 60% of all college professors like tennis, 65% like bridge, and 50% like chess; 45% like any given pair of recreations.
(a) Should you be suspicious if told 20% like all three recreations?
(b) What is the smallest percentage who could like all three recreations?
17. How many numbers between 1 and 30 are relatively prime to 30?
18. How many numbers between 1 and 280 are relatively prime to 280?
www.itpub.net 8.1 Counting with Venn Diagrams 327 19. How many ways are there to assign 20 different people to three different rooms with at least one person in each room?
20. How many n-digit numbers are there with at least one of the digits 1 or 2 or 3 absent? [Hint: This is a union problem; find N(A1 ∪A2 ∪A3).] 21. How many arrangements are there of MURMUR with no pair of consecutive letters the same?
22. If three couples are seated around a circular table, what is the probability that no wife and husband are beside one another?
23. How many ways are there to form a committee of 10 mathematical scientists from a group of 15 mathematicians, 12 statisticians, and 10 operations researchers with at least one person of each different profession on the committee?
24. Find the number of ways to arrange the six numbers 1, 2, 3, 4, 5, 6 such that either in the arrangement 1 is immediately followed by 2, or 3 is immediately followed by 4, or 5 is immediately followed by 6. [Hint: This is a union problem.] 25. How many ways are there to deal a 6-card hand that contains at least one Jack, at least one 8, and at least one 2?
26. How many ways are there to arrange the letters in the word MISSISSIPPI so that either all the Is are consecutive or all the Ss are consecutive or all the Ps are consecutive?
27. How many arrangements are there of TAMELY with either T before A, or A before M, or M before E? By “before,” we mean anywhere before, not just immediately before. [Hint: This is a union problem.] 28. How many arrangements are there of MATHEMATICS with both Ts before both As, or both As before both Ms, or both Ms before the E? Note that “before” is used as in Exercise 27.
29. TheBernsteins,Hendersons,andSmithseachhavefivechildren.Ifthe15children of these three families camp out in five different tents, where each tent holds three children, and the 15 children are randomly assigned to the five tents, what is the probability that every family has at least two of its children in the same tent?
30. Suppose 45% of all newspaper readers like wine, 60% like orange juice, and 55% like tea. Suppose 35% like any given pair of these beverages and 25% like all three beverages.
(a) What percentage of the readers likes only wine?
(b) What percentage of the readers likes exactly two of these three beverages?
31. Suppose that among 40 toy robots, 28 have a broken wheel or are rusted but not both, six are not defective, and the number with a broken wheel equals the number with rust. How many robots are rusted?
32. Suppose a school with n students offers two languages, PASCAL and BASIC. If 30 students take no language, 70 students do not take just PASCAL (i.e., either they do not take PASCAL or they take both languages), 80 students do not take just BASIC, and 20 students take both languages, determine n.
328 Chapter 8 Inclusion–Exclusion 33. Suppose a school with 120 students offers yoga and karate. If the number of students taking yoga alone is twice the number taking karate (possibly, karate and yoga), if 25 more students study neither skill than study both skills, and if 75 students take at least one skill, then how many students study yoga?
34. A survey of 150 college students reveals that 83 own cars, 97 own bikes, 28 own motorcycles, 53 own a car and a bike, 14 own a car and a motorcycle, seven own a bike and a motorcycle, and two own all three.
(a) How many students own just a bike?
(b) How many students own a car and a motorcycle but not a bike?
35. Suppose that among 150 people on a picnic, 90 bring salads or sandwiches, 80 bring sandwiches or cheese, 100 bring salads or cheese, 50 bring cheese and either salad or sandwiches (possibly both), 60 bring at least two foods, and 20 bring all three foods. How many people bring just salads? How many people bring just sandwiches?
36. In a class of 30 children, 20 take Latin, 14 take Greek, and 10 take Hebrew. If no child takes all three languages and eight children take no language, how many children take Greek and Hebrew? [Hint: Use formula (5) to determine the value of the expression N(L ∩G) + N(L ∩H) + N(G ∩H).] 8.2 INCLUSION–EXCLUSION FORMULA In this section we generalize the inclusion–exclusion formula for counting N(A1 ∩ A2 ∩A3) to n sets A1, A2, . . . , An. To simplify notation, we will omit the intersection symbol “∩” in expressions and write intersected sets as a product. For example, A1 ∩A2 ∩A3 would be written A1A2A3. Using this new notation, the number of elements in none of the sets A1, A2, . . . , An will be written N(A1A2 · · · An). Recall that an inclusion–exclusion formula is so called because of the way it successively includes (adds) and excludes (subtracts) the various k-tuple intersections of sets.
Theorem 1 Inclusion–Exclusion Formula Let A1, A2, · · · An, be n sets in a universe U of N elements. Let Sk denote the sum of the sizes of all k-tuple intersections of the Ais. Then N(A1A2 · · · An) = N −S1 + S2 −S3 + · · · + (−1)kSk + · · · + (−1)nSn (1) Proof To clarify the definition of the Sks, S1 = i N(Ai), S2 = i j N(Ai A j), Sk is the sum of the N(A j1 A j2 · · · A jk)s for all sets of k A j’s, and finally Sn = N(A1A2 · · · An). We www.itpub.net 8.2 Inclusion–Exclusion Formula 329 prove this formula by the same method used for N(F L G) in the previous section: we shall show that the net effect of (1) is to count any element in none of the sets A j once and to count elements in one or more A j’s a net of 0 times.
If an element is in none of the A j’s—that is, is in A1A2 · · · An, then it is counted once in the right-hand side of (1) by the term N and is not counted in any of the Sks.
So the count is 1 for each element in A1A2 · · · An, as required. An element in exactly one A j is counted once by N, is subtracted once by S1 (since it is in one of the A j’s), and is counted in none of the other Sks—for a count of 1 −1 = 0, as required. Now more generally let us show that an element x that is in exactly m of the A j’s has a net count of 0 in (1). Element x is counted once by N, is counted m times by S1 (since x is in m Ai’s), is counted C(m, 2) times by S2 [since x is in the intersection Ai A j for the C(m, 2) pairwise intersections involving two of the m sets containing x],. . ., and, in general, is counted C(m, k) times by Sk, k ≤m. It is not counted in Sh when h > m. So the net count of x in (1) is 1 − m 1 + m 2 − m 3 + · · · + (−1)k m k + · · · + (−1)m m m (2) This alternating sum of binomial coefficients can be evaluated from the binomial expansion (1 + x)m = 1 + m 1 x + m 2 x2 + · · · + m k xk (3) + · · · + m m xm If we set x = −1 in (3), the right side of (3) is now the expression in (2), whereas the left side of (3) becomes [1 + (−1)]m = 0m = 0. Thus, (2) equals 0, as required. ◆ Corollary Let A1, A2 · · · An be sets in the universe U. Then N(A1 ∪A2 ∪· · · ∪An) = S1 −S2 + S3 −· · · + (−1)k−1Sk + · · · + (−1)n−1Sn (4) Proof We write formula (1) as N(A1A2 · · · An) = N −[S1 −S2 + S3 −· · · + (−1)n−1Sn] (5) Next we observe that the number of elements in none of the sets equals the total number of elements minus the number of elements in one or more sets—that is, N(A1A2 · · · An) = N −N(A1 ∪A2 ∪· · · ∪An) (6) Comparing (5) and (6), we see that the expression in brackets on the right side of (5) is N(A1 ∪A2 ∪· · · ∪An). ◆ 330 Chapter 8 Inclusion–Exclusion Before giving examples of the inclusion–exclusion formula, we want to empha-size an important logical point about applying this formula. To use this formula in a counting problem, one must select a universe U and a collection of sets Ai in that universe such that the outcomes to be counted are the subset of elements in U that are in none of the Ais. That is, the Ais represent properties not satisfied by the outcomes being counted.
Example 1: Card Hands with No Suit Voids How many ways are there to select a 6-card hand from a regular 52-card deck such that the hand contains at least one card in each suit? How many 6-card hands with a void in at least one suit?
The universe U should be the set of all 6-card hands. We need to define the sets Ai such that hands with at least one card in each suit are in none of the Ais. With a moment’s thought, we see that at least one card in a suit is equivalent to no void in the suit. Thus, we let A1 be the set of 6-card hands with a void in spades; A2 a void in hearts; A3 a void in diamonds; and A4 a void in clubs. Now the first part of the problem asks for N(A1A2A3A4), and we can use the inclusion–exclusion formula.
We must next calculate N, S1, S2, S3, and S4. As noted in Chapter 5, a 6-card hand is simply a subset of six cards, and so N = C(52, 6). The size of A1, the set of hands with a void in spades, is simply the number of 6-card hands chosen from the 52 −13 = 39 nonspade cards. So N(A1) = C(39, 6). Likewise, N(Ai) = (39, 6), i = 2, 3, 4, and so S1 = 4 × C(39, 6). The hands in A1A2 are hands chosen from the 26 non-(spades or hearts) cards, and so N(A1A2) = C(26, 6). There are C(4, 2) = 6 different intersections of two out of the four sets, and so S2 = 6 × C(26, 6). There are C(4, 3) = 4 different triple intersections of the sets and each has C(13, 6) hands.
So S3 = 4 × C(13, 6). Finally, a hand cannot be void in all four suits, and so S4 = 0.
Then by (1), N(A1A2A3A4) = 52 6 −4 39 6 + 6 26 6 −4 13 6 + 0 The second part of the problem, counting 6-card hands with a void in at least one suit, asks for N(A1 ∪A2 ∪A3 ∪A4), which by formula (4) in the Corollary, equals S1 −S2 + S3 −S4. So the answer for the second part is 4 39 6 −6 26 6 + 4 13 6 −0 We note that, in general, Sk is a sum of C(n, k) different k-tuple intersections of the n Ai’s.
Example 2: Distributions with an Empty Box How many ways are there to distribute r distinct objects into five (distinct) boxes with at least one empty box?
www.itpub.net 8.2 Inclusion–Exclusion Formula 331 We do not need to determine N(A1A2A3A4A5) in this problem, because it does not concern outcomes where some property does not hold for all boxes. Rather, this is a union problem, using the corollary’s formula.
Let U be all distributions of r distinct objects into five boxes. Let Ai be the set of distributions with a void in box i. Then the required number of distributions with at least one void is N(A1 ∪A2 ∪· · · A5). We have N = 5r, N(Ai) = 4r (distributions with each object going into one of the other four boxes), N(Ai A j) = 3r, and so forth.
As just noted, there are C(5, k) subsets in Sk, k = 1, 2, . . . , 5. Thus by (4), N(A1 ∪A2 ∪· · · A5) = S1 −S2 + S3 −S4 + S5 = 5 1 4r − 5 2 3r + 5 3 2r − 5 4 1r + 0 It is easy to mistake union problems, which use phrases such as “with at least one empty box,” with standard inclusion–exclusion problems, which use phrases such as “at least one object in every box.” The former ask for at least one of a set of properties to hold, whereas the latter ask for every property to hold. Moreover, the latter problems must be solved by using the complementary properties, such as “box i is empty,” and determining all ways for none of these complementary properties to hold. The reader has to reason carefully through a counting problem and determine whether to frame the solution as counting outcomes where none of a set of properties hold or outcomes where one or more properties hold.
Another common source of confusion is the subscripts of the Sis and the sub-scripts of the Ai’s. In example 1, students sometimes define A1 to be all hands with a void in one suit, A2 to be all hands with a void in two suits, and so on. This is wrong.
The subscript of the As is an ordinal (ordering) number. A1 denotes the first set (in Example 1, the set of all hands with a void in spades), A2 the second set, A3 the third set, . . . . The subscript of the Ss is a cardinal (magnitude) number. S1 is the sum of the sizes of all single sets, S2 the sum of the sizes of all pairwise intersections of sets, S3 the sum of the sizes of all 3-way intersections of sets, . . . .
Example 3: Upper Bounds on Integer Solutions How many different integer solutions are there to the equation x1 + x2 + x3 + x4 + x5 + x6 = 20 0 ≤xi ≤8 This type of problem was solved with generating functions in Section 6.2. Now we solve it with an inclusion–exclusion approach. Let U be all integer solutions with xi ≥0, and let Ai be the set of integer solutions with xi > 8, or equivalently xi ≥9.
Then the number of solutions with 0 ≤xi ≤8 will be N(A1A2 · · · A6).
Recalling the formulas for integer solutions of equations from Section 5.5 (Example 5) we have N = 20 + 6 −1 20 = 25 20 332 Chapter 8 Inclusion–Exclusion To count N(A1), the outcomes with x1 ≥9, we consider the xi’s to be amounts of objects chosen when a total of 20 objects are chosen from six types. The outcomes with at least nine of the first type can be generated by first picking nine of the first type and then picking the remaining 20 −9 objects without restriction from the six types. This reasoning applies to all N(Ai): N(Ai) = (20 −9) + 6 −1 (20 −9) = 16 11 By a similar reasoning, now first choosing nine from two types and then choosing the remaining 20 −9 −9 objects without restriction, we have N(Ai A j) = (20 −9 −9) + 6 −1 (20 −9 −9) = 7 2 For a solution to be in three or more Ai’s, the sum of the respective xi’s would exceed 20—impossible. So Sj = 0 for j ≥3, and N(A1A2 · · · A6) = N −S1 + S2 = 25 20 − 6 1 16 11 + 6 2 7 2 Recall that to use generating functions to solve the preceding problem, we would seek the coefficient of x20 in (1 + x + · · · + x8)6 = [(1 −x9)/(1 −x)]6 = (1 −x9)6(1 −x)−6 The coefficient of x20 in this product is a0b20 + a9b11 + a18b2, where ak is the coeffi-cient of xk in (1 −x)−6 and bk is the coefficient of xk in (1 −x9)6. This coefficient of x20 turns out to be exactly the foregoing expression for N(A1A2 · · · A6). The factor (1 −x9)6 = 1 −C(6, 1)x9 + C(6, 2)x18 . . . does the inclusion–exclusion task of sub-tracting cases where one xi is at least 9 and adding back cases where two xi’s are at least 9. By using generating functions to solve this problem, we did not need to know anything about the inclusion–exclusion complexities of this problem. Generat-ing functions automatically performed the required inclusion–exclusion calculations!
Example 4: Retrieving Hats What is the probability that if n people randomly reach into a dark closet to retrieve their hats, no person will pick his own hat?
The probability will be the fraction of outcomes in which no person gets her own hat. Our universe U will be all ways for the n people to successively select a different www.itpub.net 8.2 Inclusion–Exclusion Formula 333 hat. So N = n! If Ai is the set of outcomes in which person i gets her own hat, then N(A1A2 · · · An) counts the required outcomes where no one gets her own hat.
Then N(Ai) = (n −1)!, since given that person i gets her hat, the number of possible outcomes is all ways for the other n −1 people to select hats. Similarly N(Ai A j) = (n −2)!, and generally N(A j1 A j2 · · · A jk) = (n −k)! for k-way intersec-tions. Since Sk is a sum of C(n, k) k-way terms, we obtain by Eq. (1) N(A1A2 · · · An) = N −S1 + S2 + · · · + (−1)kSk + · · · + (−1)nSn = n! − n 1 (n −1)! + n 2 (n −2)! + · · · +(−1)k n k (n −k)! + · · · + (−1)n n n 0!
= n k=0 (−1)k n k (n −k)!
Recalling that n k = n!
k!(n −k)!, we see that n k (n −k)! = n!
k!· So N(A1A2 · · · An) = n k=0 (−1)kn!
k!
= n!
n k=0 (−1)k k!
and now the probability that no person gets her own hat is N(A1A2 · · · An) N = n!
n k=0 (−1)k k!
n! = 1 −1 + 1 2! −1 3! + · · · + (−1)n n!
(7) This alternating series is the first n + 1 terms of ∞ k=0(−1)k/k!, which is the power series for ex when x = −1. The series converges very fast. The difference between e−1 and (7) is always less than 1/n! For example, e−1 = 0.367879 . . . and for n = 8, the series in (7) equals 0.367888 (even for n = 5, it is 0.366). Thus, for all but very small n, the desired probability is essentially e−1. The answer is independent of n.
The problem treated in Example 4 is equivalent to asking for all permutations of the sequence 1, 2, . . . , n such that no number is left fixed, that is, no number i is still in the ith position. Such rearrangements of a sequence are called derangements.
The symbol Dn is used to denote the number of derangements of n integers. From Example 4, we have Dn = n!
n k=0 (−1)k k!
≈n!e−1 334 Chapter 8 Inclusion–Exclusion e1 x1 e4 x4 e2 x3 e5 x2 e3 Figure 8.6 For exact values of Dn, it is usually easier to use the following simple recurrence relation: Dn = nDn−1 + (−1)n n ≥2 See Exercise 36 for details on deriving this recurrence.
Example 5: Graph Coloring How many ways are there to color the four vertices in the graph shown in Figure 8.6 with n colors (such that vertices with a common edge must be different colors)?
We label the edges e1, e2, e3, e4, e5, as shown in Figure 8.6. The universe should be the set of all ways to color the vertices. So N = n4 (n choices for each vertex). The situation we must avoid is coloring adjacent vertices with the same color. For each edge ei, we define the set Ai to be all colorings of the graph in which the vertices at either end of edge ei have the same color. Then N(A1A2A3A4A5) will be the desired number of proper colorings with n colors.
Then N(Ai) = n3, since the two vertices connected by ei get one color and the two other vertices each get any color, and S1 = 5 × n3. Similarly, N(Ai A j) = n2 and S2 = C(5, 2) × n2. A little care must be used with three-way intersections. Specifi-cally, edges e1, e2, e3 interconnect three, not four, vertices, and so N(A1A2A3) = n2 (one color for vertices x1, x2, x3 and one color for vertex x4). Likewise, N(A3A4A5) = n2. Any other set of three edges interconnect to all four vertices, and so the associ-ated N(Ai A j Ak) equals n. Then S3 = 2 × n2 + [C(5, 3) −2] × n. Four or five edges always interconnect all four vertices, and so S4 = C(5, 4) × n and S5 = n . Then the answer to our problem is N(A1A2A3A4A5) = n4 − 5 1 n3 + 5 2 n2 − 2n2 + 5 3 −2 × n + 5 4 n −n = n4 −5n3 + 8n2 −4n Note that the preceding expression is the chromatic polynomial of this graph (see Section 2.4).
We conclude this section with two generalizations of the inclusion–exclusion formula. Many readers may want to skip this material.
www.itpub.net 8.2 Inclusion–Exclusion Formula 335 Theorem 2 If A1, A2, . . . , An are n sets in a universe U of N elements, then the number Nm of elements in exactly m sets and the number N ∗ m of elements in at least m sets are given by Nm = Sm − m + 1 m Sm+1 + m + 2 m Sm+2 + · · · + (−1)k−m k m Sk + · · · + (−1)n−m n m Sn (8) N ∗ m = Sm − m m −1 Sm+1 + m + 1 m −1 Sm+2 + · · · + (−1)k−m k −1 m −1 Sk + · · · + (−1)n−m n −1 m −1 Sn (9) Proof The formula for Nm can be proved in a fashion similar to the proof of the inclusion– exclusion formula. All elements in exactly m sets will be counted once in Sm, and not counted in any other term in (8), as required. We must show that elements in more than m sets are counted a net of 0 times in (8). In this formula the count is slightly trickier to sum. Readers who dislike such technicalities should skip this proof.
The count for an element in r sets, m ≤r ≤n, is C(r, k) in Sk, and so the net count of this element in (8) is r m − m + 1 m r m + 1 + · · · + (−1)k−m k m r k (10) + · · · + (−1)r−m r m r r Remember that the element is not counted in Sk for k > r. Recall from Example 1 of Section 5.5 that the number of ways to pick k objects from r and then pick m special objects from those k is equal to the number of ways to pick the m special objects from the r first and then pick k −m more from the remaining r −m. So k m r k = r m r −m k −m Using this substitution, (10) now becomes r m − r m r −m 1 + · · · + (−1)k−m r m r −m k −m + · · · + (−1)r−m r m r −m r −m = r m 1 − r −m 1 + · · · + (−1)k−m r −m k −m + · · · + (−1)r−m r −m r −m 336 Chapter 8 Inclusion–Exclusion As in the proof of Theorem 1, the expression in brackets here is just the expansion of (1 + x)r−m with x = −1 So (10) sums to 0, as required.
The formula for N ∗ m can be verified with induction by showing that formula (9) for N ∗ m satisfies N ∗ m = Nm + N ∗ m+1: “In at least m sets” means “in exactly m sets” or “in at least m + 1 sets.” ◆ Example 6 Find the number of 4-digit ternary sequences with exactly two 1s. Also find the number with at least two 1s.
Let U be all 4-digit ternary sequences. If Ai is the set of 4-digit ternary sequences with a 1 in position i, then N2 and N ∗ 2 are the numbers of 4-digit ternary sequences with exactly two 1s and at least two 1s, respectively. Then N = 34 and S1 = C(4, 1)33, S2 = C(4, 2)32, S3 = C(4, 3)31, and S4 = C(4, 4)30 = 1. Now by formulas (8) and (9), the desired numbers are N2 = S2 − 3 4 S3 + 3 2 S4 = 4 2 32 − 3 2 4 3 31 + 4 2 1 = 24 N ∗ 2 = S2 − 2 1 S3 + 3 1 S4 = 4 2 32 − 2 1 4 3 31 + 3 1 1 = 33 The observant reader may have noted that N2 can be computed directly by a simple combinatorial argument.
8.2 EXERCISES S u m m a r y o f E x e r c i s e s These exercises require a combination of inclusion–exclusion modeling and Chapter 5 type of enumeration skills (to solve the subproblems, some of which are tricky). Exercises 37–47 use Theorem 2.
1. How many m-digit decimal sequences (using digits 0, 1, 2, . . . , 9) are there in which digits 1, 2, 3 all appear?
2. How many ways are there to roll eight distinct dice so that all six faces appear?
3. What is the probability that a 9-card hand has at least one 4 of a kind?
4. How many positive integers ≤630 are relatively prime to 630?
5. What is the probability that a 13-card bridge hand has (a) At least one card in each suit?
(b) At least one void in a suit?
(c) At least one of each type of face card (face cards are Aces, Kings, Queens, and Jacks)?
6. Given six pairs of non-identical twins, how many ways are there for six teachers to each choose two children with no one getting a pair of twins?
www.itpub.net 8.2 Inclusion–Exclusion Formula 337 7. How many arrangements are there of a, a, a, b, b, b, c, c, c, without three consecutive letters the same?
8. Given 2n letters, two of each of n types, how many arrangements are there with no pair of consecutive letters the same?
9. How many permutations of the 26 letters are there that contain none of the sequences MATH, RUNS, FROM, or JOE?
10. How many integer solutions of x1 + x2 + x3 + x4 = 28 are there with (a) 0 ≤xi ≤12?
(b) −10 ≤xi ≤20?
(c) 0 ≤xi, x1 ≤6, x2 ≤10, x3 ≤15, x4 ≤21?
11. How many ways are there to distribute 26 identical balls into six distinct boxes with at most six balls in any of the first three boxes?
12. How many 5-digit numbers (including leading 0s) are there with no digit appear-ing exactly two times?
13. How many ways are there for a child to take 10 pieces of candy with four types of candy if the child does not take exactly two pieces of any type of candy?
14. Santa Claus has five toy airplanes of each of n plane models. How many ways are there to put one airplane in each of r(r ≥n) identical stockings such that all models of planes are used?
15. A wizard has five friends. During a long wizards’ conference, it met any given friend at dinner 10 times, any given pair of friends five times, any given threesome of friends three times, any given foursome two times, and all five friends together once. If in addition it ate alone six times, determine how many days the wizards’ conference lasted.
16. How many secret codes can be made by assigning each letter of the alphabet a (unique) different letter? Give an approximate answer using Euler’s constant e.
17. How many ways are there to distribute 10 books to 10 children (one to a child) and then collect the books and redistribute them with each child getting a new one?
18. How many arrangements of 1, 2, . . . , n are there in which only the odd integers must be deranged (even integers may be in their own positions)?
19. How many ways are there to assign each of five professors in a math department to two courses in the fall semester (i.e., 10 different math courses in all) and then assign each professor two courses in the spring semester such that no professor teaches the same two courses both semesters?
20. Consider the following game with a pile of n cards numbered 1 through n.
Successively pick a different (random) number between 1 and n and remove all cards in the pile down to, and including, the card with this number until the pile is empty. If the chosen card number has already been removed, pick another number.
What is the approximate probability that at some stage the number chosen is the card at the top of the pile? (Hint: Approach as an arrangement with repetition.) 338 Chapter 8 Inclusion–Exclusion 21. There are 15 students, three (distinct) students each from five different high schools. There are five admissions officers, one from each of five colleges. Each of the officers successively picks three of the students to go to their college.
How many ways are there to do this so that no officer picks three students from the same high school?
22. The rooms in the circular house plan shown below are to be painted using eight colors such that rooms with a common doorway must be different colors. In how many ways can this be done?
23. How many ways are there to color the vertices with n colors in the following graphs such that adjacent vertices get different colors?
(a) (b) (c) 24. The four walls and ceiling of a room are to be painted with five colors available.
How many ways can this be done if bordering sides of the room must have different colors?
25. (a) How many arrangements of the integers 1 through n are there in which i is never immediately followed by i + 1, for i = 1, 2, . . . , n −1?
(b) Show that your answer equals Dn + Dn−1.
26. Repeat Exercise 25(a) now considering n to be followed by 1.
27. If two identical dice are rolled n successive times, how many sequences of outcomes contain all doubles (a pair of 1s, of 2s, etc.)?
28. How many ways are there to seat n couples around a circular table such that no couple sits next to each other?
29. If there are n families with five members each, how many ways are there to seat all 5n people at a circular table so that each person sits beside another member of his/her family? (Hint: This is not a standard inclusion–exclusion problem but uses a type of inclusion–exclusion argument.) 30. How many arrangements of a, a, a, b, b, b, c, c, c have no adjacent letters the same? (Hint: This is tricky—not a normal inclusion–exclusion problem.) 31. Suppose that a person with seven friends invites a different subset of three friends to dinner every night for one week (seven days). How many ways can this be done so that all friends are included at least once?
www.itpub.net 8.2 Inclusion–Exclusion Formula 339 32. There are six tennis players and each week for a month (four weeks), a different pair of the six play a tennis match. How many ways are there to form the sequence of 4 matches so that every player plays at least once?
33. There are 10 different people. Each person orders three donuts, chosen from five types of donuts. How many ways are there to do this such that (i) at least one person chooses all three donuts of the first type, (ii) at least one person chooses all three donuts of the second type, . . ., (v) at least one person chooses all three donuts of the fifth type?
(a) Two (or more) people may choose the same collection of three donuts.
(b) Each person must choose a different collection of three donuts.
34. There are eight Broadway musicals and they offer a special three-night package (Friday, Saturday, Sunday nights) where one can order one ticket that is good for three different musicals on successive nights (a sequence of three different musicals). A travel agent plans to order 30 of these tickets for a tour group of 30 people. How many ways are there to order a subset of 30 such tickets with the constraint that each of the eight musicals appears on at least one ticket?
35. A company produces eight different designs for sweaters. Each sweater is made from three different pieces of cloth (top piece, middle piece, bottom piece).
There are six different colors available for each piece, and the three pieces in a sweater must each be a different color. How many collections (subsets) of eight designs are possible if each color appears in at least one of the designs?
36. Find the number of ways to give each of six different people seated in a circle one of m different types of entrees if adjacent people must get different entrees.
37. How many ways are there to distribute r distinct objects into n indistinguishable boxes with no box empty?
38. (a) Show that Dn satisfies the recurrence Dn = (n −1)(Dn−1 + Dn−2).
(b) Rewriting the recurrence in part (a) as Dn −nDn−1 = −[Dn−1 −(n −1) Dn−2], iterate backwards to obtain the recurrence Dn = nDn−1 + (−1)n.
(c) Use part (b) to make a list of Dn values up to n = 10.
(d) Use part (a) to show that D(x) = e−x(1 −x)−1 is the exponential generating function for Dn.
39. How many ways are there to distribute r distinct objects into n distinct boxes with exactly three empty boxes? With at least three empty boxes?
40. How many ways are there to deal a six-card hand with at most one void in a suit?
41. How many ways are there to arrange the letters in INTELLIGENT with at least two consecutive pairs of identical letters?
42. If n balls labeled 1, 2, . . . , n are successively removed from an urn, a rencontre is said to occur if the ith ball removed is numbered i. If the n balls are removed in random order, what is the probability that exactly k rencontres occur? Show that this probability is about e−1/k!.
340 Chapter 8 Inclusion–Exclusion 43. Show that the number of ways to place r different balls in n different cells with m cells having exactly k balls is (−1)nn!r!
m!
n j=m (−1) j (n −j)r−jk ( j −m)!(n −j)!(r −jk)!(k!) j 44. (a) If g(x) is the (ordinary) generating function for Nm (see Theorem 2), show that g(x) = n k=0 Sk(x −1)k. This g(x) is called the hit polynomial.
(b) Show that 2[g(1) + g(−1)] is the number of elements in an even number of Ais.
(c) Usepart(b)todeterminethenumberofn-digitternarysequenceswithaneven number of 0s. Simplify your answer with a binomial expansion summation.
45. Show that Sm = n k=m k m Nm 46. Useacombinatorialargument(withinclusion–exclusion)toprovethefollowing: (a) m k=0 (−1)k m k n −k k = n −m n −r , m ≤r ≤n (b) m k=0 (−1)k n k n −k m −k = 0 (c) n k=m (−1)k−m n k = n −1 m −1 (d) n k=0 (−1)k n k n −k +r −1 r = r −1 n −1 47. Use Theorem 2 to show that n m = n k=m (−1)k−m k m n k 2n−k 48. Prove formula (9) in Theorem 2.
49. In Example 4, let the random variable X = number of people who get their own hat. Find E(X).
8.3 RESTRICTED POSITIONS AND ROOK POLYNOMIALS In this section we consider the special problem of counting arrangements of n objects when particular objects can appear only in certain positions. We will solve one such problem involving five objects in careful detail. In the process we will indicate how to generalize our analysis to any restricted-positions problem.
www.itpub.net 8.3 Restricted Positions and Rook Polynomials 341 a b c d e Letters 1 2 3 4 5 Positions Figure 8.7 Consider the problem of finding all arrangements of a, b, c, d, e with the restric-tions indicated in Figure 8.7. That is, a may not be put in position 1 or 5; b may not be put in 2 or 3; c not in 3 or 4; and e not in 5. There is no restriction on d. A permissible arrangement can be represented by picking five unmarked squares in Figure 8.7, with one square in each row and each column. For example, the permissible arrangement badec corresponds to picking squares (a, 2), (b, 1), (c, 5), (d, 3), (e, 4).
When viewed in terms of the 5 × 5 array of squares, the arrangement problem can be thought of as a matching problem, matching letters with positions. In Section 4.4 we also considered matching problems. There we wanted to determine whether any complete matching existed. Here we want to count how many complete matchings there are.
We use the inclusion–exclusion formula, expression (1) of the previous section, to count the number of permissible arrangements for Figure 8.7. Let U be the set of all arrangements of the five letters without restrictions. So N = 5!. Let Ai be the set of arrangements with a forbidden letter in position i (note that we could equally well define the properties in terms of the ith letter being in a forbidden position). The numberofpermissiblearrangementswillthenbe N(A1A2A3A4A5).IntermsofFigure 8.7, Ai is the set of all collections of five squares, each in a different row and column such that the square in column i is a darkened square. We obtain N(Ai) by counting the ways to put a forbidden letter in position i times the 4! ways to arrange the remaining four letters in the other four positions (we do not worry about forbidden positions for these letters). Then N(A1) = 1 × 4!, N(A2) = 1 × 4!, N(A3) = 2 × 4!, N(A4) = 1 × 4!, and N(A5) = 2 × 4!. Collecting terms, we obtain S1 = 5 i=1 N(Ai) = 1 × 4! + 1 × 4! + 2 × 4! + 1 × 4! + 2 × 4!
= (1 + 1 + 2 + 1 + 2)4! = 7 × 4!
Observe that (1 + 1 + 2 + 1 + 2) = 7 is just the number of the darkened squares in Figure 8.7. Since each choice of a darkened square (i.e., some letter in some forbidden position) leads to 4! possibilities, then S1 = (number of darkened squares) × 4!
for any restricted-positions problem with a 5 × 5 family of darkened squares similar to Figure 8.7.
342 Chapter 8 Inclusion–Exclusion Next, N(Ai A j) will be the number of ways to put (different) forbidden letters in positions i and j times the 3! ways to arrange the remaining three letters. Or equivalently, the ways to pick two darkened squares, one in column i and one in column j (and in different rows), times 3!. The reader can check that N(A1A2) = 1 × 3!
N(A1A3) = 2 × 3!
N(A1A4) = 1 × 3!
N(A1A5) = 1 × 3!
N(A2A3) = 1 × 3!
N(A2A4) = 1 × 3!
N(A2A5) = 2 × 3!
N(A3A4) = 1 × 3!
N(A3A5) = 4 × 3!
N(A4A5) = 2 × 3!
Collecting terms, we obtain S2 = i j N(Ai A j) = (1 + 2 + 1 + 1 + 1 + 1 + 2 + 1 + 4 + 2)3! = 16 × 3!
The number 16 counts the ways to select two darkened squares, each in a different row and column. Generalizing, we will have Sk = number of ways to pick k darkened squares each in a different row and column × (5 −k)!
(1) Since letter d’s row in Figure 8.7 has no darkened squares, there is no way to pick five darkened squares, each in a different row and column. Thus S5 = 0. On the other hand, tedious case-by-case counting apparently awaits us for S3 and S4. Instead, let us try to develop a theory for determining the number of ways to pick k darkened squares, each in a different row and column.
This darkened squares selection problem can be restated in terms of a recreational mathematics question about a chess-like game. A chess piece called a rook can capture any opponent’s piece on the chessboard in the same row or column as the rook (provided there are no intervening pieces). Instead of using a normal 8 × 8 chessboard, we “play chess” on the “board” consisting solely of the darkened squares in Figure 8.7.
Countingthenumberofwaystoplacekmutuallynoncapturingrooksonthisboard of darkened squares is equivalent to our original subproblem of counting the number of ways to pick k darkened squares in Figure 8.7, each in a different row and column.
The phrase “k mutually noncapturing rooks” is simpler to say and more pictorial.
A common technique in combinatorial analysis is to break a big messy problem into smaller manageable subproblems. We will develop two breaking-up operations to help us count noncapturing rooks on a given board B.
The first operation applies to a board B that can be decomposed into disjoint sub-boards B1 and B2,—that is, subboards involving different sets of rows and columns.
Often a board has to be properly rearranged before the disjoint nature of the two subboards can be seen.
When the rows and columns of Figure 8.7 are rearranged as shown in Figure 8.8, it is obvious that the three darkened squares in rows a and e and columns 1 and 5 are disjoint from the four darkened squares in rows b and c and columns 2, 3, and 4. Let B be the board of darkened squares in Figure 8.8, let B1 be the three darkened squares in rows a and e, and let B2 be the four darkened squares in rows b and c.
www.itpub.net 8.3 Restricted Positions and Rook Polynomials 343 Letters 1 5 2 3 4 Positions B1 B2 a e d b c Figure 8.8 Define rk(B) to be the number of ways to place k noncapturing rooks on board B,rk(B1) the number of ways to place k noncapturing rooks on subboard B1, and rk(B2) the number of ways to place k noncapturing rooks on subboard B2. There are three ways to place one rook on subboard B1 in Figure 8.8, since B1 has three squares, and similarly four ways to place one rook on subboard B2. Then r1(B1) = 3 and r1(B2) = 4. A little thought shows that there is only one way to place two rooks on subboard B1 and three ways to place two rooks on subboard B2, so that r2(B1) = 1 and r2(B2) = 3. Note that rk(B1) = rk(B2) = 0 for k ≥3, since each subboard has only two rows. It will be convenient to define r0 = 1 for all boards.
Observe next that since B1 and B2 are disjoint, placing, say, two noncapturing rooks on the whole board B can be broken into three cases: placing two noncapturing rooks on B1 (and none on B2), placing one rook on each subboard, or placing two noncapturing rooks on B2. Thus we see that r2(B) = r2(B1) +r1(B1)r1(B2) +r2(B2) or, using that fact that r0(B2) = r0(B1) = 1, r2(B) = r2(B1)r0(B2) +r1(B1)r1(B2) +r0(B1)r2(B2) = 1 × 1 + 3 × 4 + 1 × 3 = 16 (2) Recall that 16 is the number obtained earlier when summing all N(Ai A j) to count all ways to pick two darkened squares each in a different row and column.
The reasoning leading to (2) applies to rk(B) for any k and for any board B that decomposes into two disjoint subboards B1 and B2.
Lemma If B is a board of darkened squares that decomposes into the two disjoint subboards B1 and B2, then rk(B) = rk(B1)r0(B2) +rk−1(B1)r1(B2) + · · · +r0(B1)rk(B2) (3) The observant reader may notice that (3) is very similar to formula (6) in Sec-tion 6.2 for the coefficient of a product of two generating functions. That is, if f (x) = arxr and g(x) = brxr, then the coefficient of xk in h(x) = f (x)g(x) is akb0 + ak−1b1 + · · · + a0bk. We will now exploit this similarity.
We define the rook polynomial R(x, B) of the board B of darkened squares to be R(x, B) = r0(B) +r1(B)x +r2(B)x2 + · · · 344 Chapter 8 Inclusion–Exclusion Remember that r0(B) = 1 for all B. Note that the rook polynomial depends only on the darkened squares, not on the size of the original assignment diagram. Then for B1 and B2 as defined above, we found that R(x, B1) = 1 + 3x + 1x2 and R(x, B2) = 1 + 4x + 3x2 Moreover, by the correspondence between (3) and the formula for the product of two generating functions, we see that rk(B), the coefficient of xk in the rook poly-nomial R(x, B) of the full board, is simply the coefficient of xk in the product R(x, B1)R(x, B2). That is, R(x, B) = R(x, B1)R(x, B2) = (1 + 3x + 1x2)(1 + 4x + 3x2) = 1 + [(3 × 1) + (1 × 4)]x + [(1 × 1) + (3 × 4) + (1 × 3)]x2 + [(1 × 4) + (3 × 3)]x3 + (1 × 3)x4 = 1 + 7x + 16x2 + 13x3 + 3x4 This product relation is true for any such B, B1, and B2.
Theorem 1 If B is a board of darkened squares that decomposes into the two disjoint subboards B1 and B2 then R(x, B) = R(x, B1)R(x, B2) Without meaning to belittle the role of generating functions in Theorem 1, we shouldobservethatthegeneratingfunctionswereusedheresolelybecausepolynomial multiplication corresponds to the subboard composition rule given in (3). That is, plugging the rk(Bi)s into two polynomials and multiplying them together is a more familiar way to organize the computation required by (3). Shortly we will decompose a board B into nondisjoint subboards, and rook polynomials will then play a truly essential role.
Now that R(x, B) has been determined for the board of darkened squares in Figure 8.8 and hence in Figure 8.7, we can solve our original problem about the permissible arrangements of a, b, c, d, e. Expression (1) for Sk can now be rewritten Sk = rk(B)(5 −k)!
By the inclusion–exclusion formula, the number of permissible arrangements is N(A1A2A3A4A5) = N −S1 + S2 −S3 + S4 −S5 = 5! −r1(B)4! +r2(B)3! −r3(B)2! +r4(B)1! −r5(B)0!
= 5! −7 × 4! + 16 × 3! −13 × 2! + 3 × 1! −0 × 0!
= 120 −168 + 96 −26 + 3 −0 = 25 The values for the rk(B)s came from the rook polynomial R(x, B) for Figure 8.8 computed above. This rook-polynomial-based variation on the inclusion–exclusion formula is valid for any arrangement problem with restricted positions.
www.itpub.net 8.3 Restricted Positions and Rook Polynomials 345 Theorem 2 The number of ways to arrange n distinct objects when there are restricted positions is equal to n! −r1(B)(n −1)! +r2(B)(n −2)! + · · · + (−1)krk(B)(n −k)!
+ · · · + (−1)nrn(B)0!
(4) where the rk(B)s are the coefficients of the rook polynomial R(x, B) for the board B of forbidden positions.
Let us summarize the little theory we have developed.
1.
Given a problem of counting arrangements or matchings with restricted positions, display the constraints in an array with darkened squares for forbidden positions, as in Figure 8.7.
2.
Try to rearrange the array so that the board B of darkened squares can be decom-posed into disjoint subboards B1 and B2.
3.
By inspection, determine the rk(Bi)s, the number of ways to place k noncapturing rooks on subboard Bi.
4.
Use therk(Bi)s to form the rook polynomials R(x, B1) and R(x, B2), and multiply R(x, B1)R(x, B2) to obtain R(x, B).
5.
Insert the coefficient rk(B) of R(x, B) in formula (4).
Example 1: Sending Birthday Cards How many ways are there to send six different birthday cards, denoted C1, C2, C3, C4, C5, C6, to three aunts and three uncles, denoted A1, A2, A3, U1, U2, U3, if aunt A1 would not like cards C2 and C4; if A2 would not like C1 or C5; if A3 likes all cards; if U1 would not like C1 or C5; if U2 would not like C4; and if U3 would not like C6?
The forbidden positions information is displayed in Figure 8.9. We rearrange the board by putting together rows with a darkened square in the same column, and putting together columns with a darkened square in the same row. For example, rows C1 and C5 both have darkened squares in columns A2 and U1, so we put rows C1 and C5 beside one another and columns A2 and U1 beside one another; similarly for C1 C2 C3 C4 C5 A1 A2 A3 U1 U2 U3 C6 Figure 8.9 346 Chapter 8 Inclusion–Exclusion C1 C5 C3 C2 C4 A2 U1 A3 A1 U2 U3 C6 Figure 8.10 rows C2 and C4 and columns A1 and U2. We get the rearrangement shown in Figure 8.10. Thus the original board B of darkened squares decomposes into the two disjoint subboards, B1 in rows C1 and C5, and B2 in rows C2, C4, and C6. Actually B2 itself decomposes into two disjoint subboards B′ 2 and B′′ 2, where B′′ 2 is the single square (C6, U3). By inspection, we see that R(x, B1) = 1 + 4x + 2x2 R(x, B2) = R(x, B′ 2)R(x, B′′ 2) = (1 + 3x + x2)(1 + x) So R(x, B) = R(x, B1)R(x, B2) = (1 + 4x + 2x2)(1 + 3x + x2)(1 + x) = 1 + 8x + 22x2 + 25x3 + 12x4 + 2x5 Then the answer to the card-mailing problem is 6 k=0 (−1)krk(B)(6 −k)!
= 6! −8 × 5! + 22 × 4! −25 × 3! + 12 × 2! −2 × 1! + 0 × 0!
= 720 −960 + 528 −150 + 24 −2 + 0 = 160 Now let us consider the problem of determining the coefficients of R(x, B) when the board B does not decompose into two disjoint subboards. Consider the board B shown in Figure 8.11. Let us break the problem of determining rk(B) into two cases, depending on whether or not a certain square s is one of the squares chosen for the k noncapturing rooks. Let Bs be the board obtained from B by deleting square s, and let B∗ s be the board obtained from B by deleting square s plus all squares in the same row or column as s. If s is the square indicated in Figure 8.11, then Bs and B∗ s are as s Figure 8.11 www.itpub.net 8.3 Restricted Positions and Rook Polynomials 347 s s Bs Bs Figure 8.12 shown in Figure 8.12. If square s is not used, we must place k noncapturing rooks on Bs. If square s is used, then we must place k −1 noncapturing rooks on B∗ s . Hence we conclude that rk(B) = rk(Bs) +rk−1(B∗ s ) (5) Using the generating function methods introduced in Section 7.5 for turning a recurrence relation into a generating function, we obtain from (5) R(x, B) = k rk(B)xk = k rk(Bs)xk + k rk−1(B∗ s )xk = k rk(Bs)xk + x h rh(B∗ s )xh = R(x, Bs) + x R(x, B∗ s ) Multiplying the rook polynomial R(x, B∗ s ) by x reflects the fact that one rook is placed in square s in combination to any placement of rooks on board B∗ s . Observe that Bs and B∗ s both break into disjoint subboards whose rook polynomials are easily determined by inspection: R(x, Bs) = (1 + 3x)(1 + 2x) = 1 + 5x + 6x2 R(x, B∗ s ) = (1 + 2x)(1 + x) = 1 + 3x + 2x2 R(x, B) = R(x, Bs) + x R(x, B∗ s ) = (1 + 5x + 6x2) + x(1 + 3x + 2x2) = 1 + 6x + 9x2 + 2x3 These results apply to any board B and any square s in B.
Theorem 3 Let B be any board of darkened squares. Let s be one of the squares of B, and let Bs and B∗ s be as defined above. Then R(x, B) = R(x, Bs) + x R(B∗ s ) Theorem 3 provides a way to simplify any board’s rook polynomial. If the boards Bs and B∗ s do not break up into disjoint subboards, we can reapply Theorem 3 to Bs and B∗ s . It is important that the square s be chosen to split up B as much as possible.
Example 2: Nondecomposable Constraints In Example 1, suppose that the tastes of uncle U1 change and now he would not like card C2 but would like C1. The new board of forbidden positions is shown in 348 Chapter 8 Inclusion–Exclusion A2 U1 A1 U2 U3 C1 C5 C2 C4 C6 s Figure 8.13 Figure 8.13. The square in the bottom right corner, call it t, is still disjoint from the other squares, call them board B1. Board B1 cannot be decomposed into disjoint subboards, and so we must use Theorem 3. The square s that breaks up B1 most evenly is (C2, U1), indicated in Figure 8.13.
Boards Bs and B∗ s shown in Figure 8.14 both decompose into simple disjoint subboards: R(x, Bs) = (1 + 3x + x2)(1 + 3x + x2) = 1 + 6x + 11x2 + 6x3 + x4 R(x, B∗ s ) = (1 + 2x)(1 + 2x) = 1 + 4x + 4x2 Then R(x, B1) = R(x, Bs) + x R(x, B∗ s ) = (1 + 6x + 11x2 + 6x3 + x4) +x(1 + 4x + 4x2) = 1 + 7x + 15x2 + 10x3 + x4 and R(x, B) = R(x, B1)R(x, t) = (1 + 7x + 15x2 + 10x3 + x4)(1 + x) = 1 + 8x + 22x2 + 25x3 + 11x4 + x5 Now by Theorem 2, the number of ways to send birthday cards is 6! −8 × 5! + 22 × 4! −25 × 3! + 11 × 2! −1 × 1! + 0 × 0! = 159 Note that uncle U1’s change in tastes changed the final rook polynomial only slightly: r4(B) changed from 12 to 11, r5(B) changed from 2 to 1, and the final answer changed from 160 to 159.
s s Bs Bs Figure 8.14 www.itpub.net 8.3 Restricted Positions and Rook Polynomials 349 8.3 EXERCISES S u m m a r y o f E x e r c i s e s The first nine exercises are similar to the examples in this section; the next seven exercises develop theory about rook polyno-mials and combinatorial theory based on rook polynomials.
1. Describe the associated chessboard of darkened squares for finding all derange-ments of 1, 2, 3, 4, 5.
2. Find the rook polynomial for the following boards: (a) (d) (e) (b) (c) 3. Find the number of matchings of five men with five women given the constraints in the figure below on the left, where the rows represent the men and the columns represent the women.
4. Find the number of matchings of five men with five women given the constraints in the figure above on the right, where the rows represent the men and the columns represent the women.
5. Seven dwarfs D1, D2, D3, D4, D5, D6, D7 each must be assigned to one of seven jobs in a mine, J1, J2, J3, J4, J5, J6, J7. D1 cannot do jobs J1 or J3; D2 cannot do J1 or J5; D4 cannot do J3 or J6; D5 cannot do J2 or J7; D7 cannot do J4. D3 and D6 can do all jobs. How many ways are there to assign the dwarfs to different jobs?
350 Chapter 8 Inclusion–Exclusion 6. A pair of two distinct dice are rolled six times. Suppose none of the ordered pairs of values (1, 5), (2, 6), (3, 4), (5, 5), (5, 3), (6, 1), (6, 2) occur. What is the probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice?
7. A computer dating service wants to match four women each with one of five men.
If woman 1 is incompatible with men 3 and 5; woman 2 is incompatible with men 1 and 2; woman 3 is incompatible with man 4; and woman 4 is incompatible with men 2 and 4, how many matches of the four women are there?
8. Suppose five officials O1, O2, O3, O4, O5 are to be assigned five different city cars: an Escort, a Lexus, a Nissan, a Taurus, and a Volvo. O1 will not drive an Escort or a Nissan; O2 will not drive a Taurus; O3 will not drive a Lexus or a Volvo; O4 will not drive a Lexus; and O5 will not drive an Escort or a Nissan. If a feasible assignment of cars is chosen randomly, what is the probability that (a) O1 gets the Volvo?
(b) O2 or O5 get the Volvo? (Hint: Model this constraint with an altered board.) 9. Calculate the number of words that can be formed by rearranging the letters EERRIE so that no letters appears at one of its original positions—for example, no E as the first, second, or sixth letter.
10. Find the rook polynomial for a full n × n board.
11. An ascent in a permutation is a consecutive pair of the form i, i + 1. The ascents in the following permutation are underlined: 12534.
(a) Design a chessboard to represent a permutation of 1, 2, 3, 4, 5 so that a check in entry (i, j) means that i is followed immediately by j in the permutation.
Darken entries so as to exclude all ascents.
(b) Find the rook polynomial for the darkened squares in part (a).
(c) Find the number of permutations of 1, 2, 3, 4, 5 with k ascents. (Hint: See Theorem 2 in Section 8.2.) 12. Let Rn,m(x) be the rook polynomial for an n × m chessboard (n rows, m columns, all squares may have rooks).
(a) Show that Rn,m(x) = Rn−1,m(x) + mx Rn−1,m−1(x).
(b) Show that d dx Rn,m(x) = nmRn−1,m−1(x). [Hint: Use identity (5) in Section 5.5.] 13. Find two different chessboards (not row or column rearrangements of one another) that have the same rook polynomial.
14. Consider all permutations of 1, 2, . . . , n in which i appears in neither position i nor i + 1 (n not in n or 1). Such a permutation is called a menage. Let Mn(x) be the rook polynomial for the forbidden squares in a menage. Let M∗ n(x) be the rook polynomial when n may appear in position 1, and let M0 n(x) be the rook polynomial when both 1 and n may appear in position 1.
(a) Show that M∗ n(x) = x M∗ n(x) + M0 n(x), M0 n(x) = x M0 n−1(x) + M∗ n(x), and Mn(x) = M∗ n(x) + x M∗ n−1(x).
www.itpub.net 8.4 Summary and Reference 351 (b) Using the initial conditions M∗ 1(x) = 1 + x, M0 1(x) = 1, show by induction that M∗ n(x) = n k=0 2n −k k xk and M0 n(x) = n−1 k=0 2n −k −1 k xk (c) Find Mn(x).
15. Given an n × m chessboard Cn,m (see Exercise 12) and a board C of darkened squares in Cn,m, the complement C′ of C in Cn,m is the board of nondarkened squares.
(a) Show that rk(C′) = k j=0 (−1) j n −j k −j m −j k −j (k −j)!rk(C) (b) Show that R(x, C′) = xn R(1/x, C).
16. Use Theorem 2 in Section 8.2 to derive a formula for counting arrangements when exactly k elements appear in forbidden positions.
8.4 SUMMARY AND REFERENCE Frequently the combinatorial complexities in the counting problems in Chapter 5 arose from simultaneous constraints such as “with at least one card in each suit.” In this chapter, we formed a simple set-theoretic model for such problems and solved once and for all the combinatorial logic of this model. The resulting formula, the inclusion–exclusion formula, was then applied to various counting problems. This formula requires the proper set-theoretic restatement of a problem and the solution of some fairly straightforward subproblems, but in return the formula eliminates all the worry about logical decomposition (as well as worry about counting some outcomes twice). After having no help in their problem-solving in Chapter 5, readers should find it easy to appreciate fully the power of a formula that does much of the reasoning for them. This is what mathematics is all about!
The last section on rook polynomials provides a nice mini-theory about organiz-ing the inclusion–exclusion computations in arrangements with restricted positions.
The inclusion–exclusion formula was obtained by J. Sylvester about 100 years ago, although in the early eighteenth century the number of derangements had been cal-culated by Montmort and a (noncounting) set union and intersection version of the formula was published by De Moivre. Rook polynomials were not invented until the mid-twentieth century (see Riordan ).
1. J. Riordan, An Introduction to Combinatorial Analysis, John Wiley & Sons, New York, 1958.
This page is intentionally left blank www.itpub.net PART THREE ADDITIONAL TOPICS This page is intentionally left blank www.itpub.net CHAPTER 9 POLYA’S ENUMERATION FORMULA 9.1 EQUIVALENCE AND SYMMETRY GROUPS In this chapter we examine a special class of counting problems. Consider the ways of coloring the corners of a square black, white, or red: There are three color choices at each of the four corners, giving 3 × 3 × 3 × 3 = 81 different colorings. Suppose, however, that the square is not in a fixed position but is unoriented, like a square molecule in a liquid. Now how many different corner colorings are there? The unori-ented figure being colored could be a n-gon or a cube, and the edges or faces could be colored instead of the corners. The floating square problem is equivalent to finding the number of different (unoriented) necklaces of four beads colored black, white, or red. Polya’s motivation in developing his formula for counting distinct colorings of unoriented figures came from a problem in chemistry, the enumeration of isomers.
The difficulty in these problems comes from the geometric symmetries of the figure being colored. We develop a special formula, based on this set of symmetries, to count all distinct colorings of a figure. With a little more work, we also obtain a generating function that gives a pattern inventory of the distinct colorings. For example, the pattern inventory of black–white colorings of the corners of a cube with all geometric symmetries allowed is b8 + b7w + 3b6w2 + 3b5w3 + 7b4w4 + 3b3w5 + 3b2w6 + bw7 + w8 where the coefficient of biwi is the number of nonequivalent colorings with i black corners and j white corners.
The presentation in this chapter avoids the notational complexities and abstract framework of traditional presentations of Polya’s Enumeration Formula. Instead, we develop the theory for Polya’s Formula by means of a simple concrete example— namely, black–white colorings of the corners of an unoriented (floating) square. Once one understands the needed concepts and analysis for 2-coloring an unoriented square, it will be easy to state the general formula and apply it to other geometric structures.
While students may be used to seeing a general formula developed and then applied in subsequently examples, many important mathematical results were first obtained by working special cases and then generalizing.
It is impossible to draw an unoriented object; any picture shows it in a fixed position. Thus, we start our analysis with the 24 = 16 black–white colorings of the 355 356 Chapter 9 Polya’s Enumeration Formula C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 Figure 9.1 fixed square. See Figure 9.1. We can partition these 16 colorings into subsets of col-orings that are equivalent when the square is floating. There are six such subsets (see the groupings of colored squares shown in Figure 9.1), and so there are six different 2-colorings of the floating square. The set of fixed colorings would be too large if a harder sample problem were used, such as 2-colorings of a cube or 3-colorings of the square.
Weseekatheoryandformulatoexplainwhytherearesixsuchdistinct2-colorings of the square. Note that the six subsets of equivalent colorings vary in size.
To define the partition of a set into subsets of equivalent elements, we first define the general concept of the equivalence of two elements a and b. We write this equivalence as a ∼b. The fundamental properties of an equivalence relation are (i) Transitivity: a ∼b, b ∼c ⇒a ∼c (ii) Reflexivity: a ∼a (iii) Symmetry: a ∼b ⇒b ∼a Allotherpropertiesofequivalencecanbederivedfromthesethree.Anybinaryrelation with these three properties is called an equivalence relation. Such a relation defines a partition into subsets of mutually equivalent elements called equivalence classes.
Example 1: Equivalence Relations (a) For a set of people, being the same weight is an equivalence relation; all people of a given weight form an equivalence class.
(b) For a set of numbers, differing by an even number is an equivalence relation; the even numbers form one equivalence class and the odd numbers the other class.
(c) For a set of figures, having the same number of corners is an equivalence relation; for each n, an equivalence class consists of all n-corner figures.
Next we turn our attention to the motions that map the square onto itself (see Figure 9.2). These motions, or symmetries, are what make the foregoing colorings equivalent to another one. Before we develop a theory about symmetries and their www.itpub.net 9.1 Equivalence and Symmetry Groups 357 a b d c 2 π a b d c 6 π a b d c 5 π a b d c 1 π a b d c 4 π a b d c 3 π a b d c 7 π a b d c 8 π Figure 9.2 relation to coloring equivalence, let us take a closer geometric look at the symmetries of a square and of some other figures.
Example 2: Symmetries of Even n-gons Figure 9.2 displays the set of motions that map the square onto itself, the symmetries of the square. How was this set obtained? More generally, what is the set of symmetries of an n-gon, for n even?
The symmetries of the square divide into two classes: rotations—circular motions in the plane; and reflections (flips)—motions using the third dimension. The rotations, about the center of the square, are easy to find. Each rotation is an integral multiple of the smallest (nonzero) rotation. The rotations are π2 = 90◦rotation, π3 = 180◦ rotation, π4 = 270◦rotation, and π1 = 360◦(or 0◦) rotation.
Reflections are a little harder to visualize since they are motions in three dimen-sions. The reflections are π5 = reflection about the vertical axis, π6 = reflection about the horizontal axis, π7 = reflection about opposite corners a and c, and π8 = reflection about opposite corners b and d.
In a regular n-gon, the smallest rotation is (360/n)◦. Any multiple of this (360/n)◦ rotation is again a rotation, and so there are n rotations in all. There are two types of reflections for a regular even n-gon: flipping about the middles of two opposite sides and flipping about two opposite corners. Since there are n/2 pairs of opposite sides and n/2 pairs of opposite corners, a regular even n-gon will have n/2 + n/2 = n reflections. Summing rotations and reflections, we find that a regular even n-gon has 2n symmetries.
We leave it as an exercise to show that these 2n symmetries just described are distinct. To show that there are at most 2n symmetries of an even n-gon, consider a particular corner, call it x, and the edge e incident to x on the clockwise side. The position of x and the relative position of e after the action of a symmetry totally determine the symmetry (readers should convince themselves of this). A symmetry could map x to any of the n corners of the n-gon—n choices—and e could be mapped 358 Chapter 9 Polya’s Enumeration Formula Figure 9.3 Symmetric reflection of a pentagon to either side of x—2 choices. In total, there are 2n possible different symmetries of the even n-gon.
Example 3: Symmetries of Odd n-gons Describe the symmetries of a pentagon, and more generally, of an n-gon for odd n.
AsnotedinExample2,anyregularn-gonhasnrotationalsymmetries.Apentagon will have five rotational symmetries of 0◦, 72◦, 144◦, 216◦, and 288◦. However, the reflections discussed in Example 2 about opposite sides or opposite corners do not exist in the pentagon. Instead, we reflect about an axis of symmetry running from one corner to the middle of an opposite side (see Figure 9.3). There are five such reflections, for a total of 10 symmetries.
In a regular odd n-gon, there are n such flips, along with n rotations. Summing rotations and reflections, we find that a regular odd n-gon also has 2n symmetries.
We leave it as an exercise for the reader to show that these 2n symmetries are all distinct. The same argument used in Example 2 shows that there are only these 2n symmetries.
Example 4: Symmetries of a Tetrahedron Describe the symmetries of a tetrahedron.
A tetrahedron consists of four equilateral triangles that meet at six edges and four corners (see Figure 9.4). Besides the motion leaving all corners fixed—call it the 0◦motion—we can revolve 120◦or 240◦about a corner and the center of the opposite face (see Figure 9.4a), or we can revolve 180◦about the middle of opposite edges (see Figure 9.4b). Since there are four pairs of a corner and opposite face and three pairs of opposite edges, we have a total of 1 (0◦motion) + 4 × 2 + 3 = 12 symmetries. It is left as an exercise to check that these 12 symmetries are distinct and that no other symmetries exist.
(a) (b) Figure 9.4 Symmetric revolutions of a tetrahedron www.itpub.net 9.1 Equivalence and Symmetry Groups 359 The symmetries of a square are naturally characterized by the way they permute the corners of the square. Thus, the 180◦rotation π3 (see Figure 9.2) can be described as the corner permutation: a →c, b →d, c →a, d →b; in tabular form, we write a b c d c d a b . The 90◦rotation π2 can be described as a →b, b →c, c →d, d →a, or a →b →c →d →a.
A permutation of the form x1 →x2 →x3 · · · →xn →x1 is called a cyclic per-mutation or cycle. Thus, π2 is a cycle of length 4. Cycles are usually written in the form (x1x2x3 . . . xnx1). So π2 = (abcd). Any permutation can be expressed as a product of disjoint cycles (proof of this claim is an exercise). For example, π3 = (ac)(bd), π4 = (adcb), and π7 = (a)(bd)(c). The reader should find cycle de-compositions for the other πis. The depiction of a motion as in Figure 9.2, with arrows indicating the mapping at each corner, will make it easier to trace out cycles in later calculations.
In permuting the corners, the symmetries create permutations of the colorings of the corners. For example, if Ci is the ith square in Figure 9.1, then π3 is the following permutation of colorings: π3 = C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C1 C4 C5 C2 C3 C8 C9 C6 C7 C10 C11 C14 C15 C12 C13 C16 (1) The point is that while a symmetry πi is easily visualized by how it moves the corners of the square, what we are really interested in is the way πi takes one coloring into another (making them equivalent). Thus, we formally define our coloring equivalence as follows: Colorings C and C′ are equivalent, C ∼C′, if there exists a symmetry πi such that πi(C) = C′ (2) The properties of the set G of symmetries that interest us are the ones that make the relation C ∼C′ in Eq. (2) an equivalence relation. These properties of G are (here πi · π j means applying motion πi followed by motion π j): 1.
Closure: If πi, π j ∈G, then πi · π j ∈G; for example, in Figure 9.2, π2 · π5 = π7.
2.
Identity: G contains an identity motion πI such that πI · πi = πi and πi · πI = πi; in Figure 9.2, πI is π1.
3.
Inverses: For each πi ∈G, there exists an inverse in G, denoted π−1 i , such that π−1 i · πi = πI and πi · π−1 i = πI; for example, in Figure 9.2, π−1 2 = π4.
Observe that closure makes our coloring relation ∼satisfy transitivity [property (i) of an equivalence relation]. For suppose C ∼C′ and C′ ∼C′′. Since C ∼C′, there mustexistπi ∈G suchthatπi(C) = C′.Similarly,thereisa π j ∈G suchthatπ j(C′) = C′′. Then by closure, there exists πk = πi · π j ∈G with πk(C) = (πi · π j)(C) = C′′.
Thus C ∼C′′. Similarly, properties (2) and (3) of the symmetries imply that our col-oring relation satisfies properties (ii) and (iii) of an equivalence relation, respectively (see Exercise 16).
360 Chapter 9 Polya’s Enumeration Formula Acollection G ofmathematicalobjectswithabinaryoperationiscalledagroupif it satisfies properties 1, 2, and 3 along with the associativity property—(πi · π j) · πk = πi · (π j · πk). Thus we have the following theorem.
Theorem Let G be a group of permutations of the set S (corners of a square) and T be any collection of colorings of S (2-colorings of the corners). Then G induces a parti-tion of T into equivalence classes with the relation C ∼C′ ⇔some π ∈G takes C to C′.
Note that S could be any set of objects and T could be any possible collection of colorings. The following simple lemma about groups lies at the heart of the counting formula developed in the next section.
Lemma For any two permutations πi, π j in a group G, there exists a unique permutation πk = π−1 i · π j in G such that πi · πk = π j.
Proof First we show that πi · πk = π j. Since πk = π−1 i · π j, πi · πk = πi · π−1 i · π j = πi · π−1 i · π j (by associativity) = πI · π j = π j as claimed. Next we show that πk is unique. Suppose there also exists a permutation π′ k such that πi · π′ k = π j. Then πi · πk = πi · π′ k. Multiplying the equation by π−1 i , we have π−1 i · (πi · πk) = π−1 i · (πi · π′ k) ⇒(π−1 i · πi) · πk = π−1 i · πi · π′ k ⇒πI · πk = πI · π′ k ⇒πk = π′ k ◆ 9.1 EXERCISES S u m m a r y o f E x e r c i s e s The first 13 exercises continue the exam-ples of equivalence and symmetries given in this section. The remaining exercises develop basic aspects of group theory associated with symmetries. Prior experience with modern algebra is needed for most of these latter problems.
1. Which of the following relations are equivalence relations? State your reasons.
(a) ≤(less than or equal to), for a set of numbers (b) = (equal to), for a set of numbers (c) “Difference is odd,” for a set of numbers (d) “Being blood relations,” for a group of people (e) “Having a common friend,” for a group of people www.itpub.net 9.1 Equivalence and Symmetry Groups 361 2. Which of the following collections with given operations are groups? For those collections that are groups, which elements are the identities?
(a) The nonnegative integers 0, 1, 2, . . . with addition (b) The integers 0, 1, 2, . . ., n −1 with addition modulo n (c) All polynomials (with integer coefficients) with polynomial addition (d) All nonzero fractions with regular multiplication (e) All invertible 2 × 2 real-valued matrices with matrix multiplication 3. Find all symmetries of the following figures (indicate with arrows where the corners move in each symmetry, as in Figure 9.2): a c b a b c d e f a b c d (a) (b) (c) 4. Write the following symmetries or permutations as a product of cyclic permuta-tions: (a) π2 (b) π3 (c) π6 (d) πI (e) π4 · π7 (f) π7 · π4 (g) 1 2 3 4 5 6 7 3 5 4 6 7 1 2 5. For the symmetries of the square listed in Figure 9.2, give the associated permu-tation of 2-colorings, as in (1), for (a) π1 (b) π2 (c) π5 (d) π7 6. (a) List all 2-colorings of the three corners of a triangle.
(b) For the following symmetries of a triangle, give the associated permutation of 2-colorings, as in (1), for (i) π = 120◦rotation (ii) π = flip about vertical axis 7. Show that the eight symmetries of a square listed in Figure 9.2 are all distinct.
8. Show that the 2n symmetries of an n-gon (even or odd) mentioned in Examples 2 and 3 are all distinct.
9. Show that the 12 symmetries of a tetrahedron listed in Example 4 are all distinct.
Show that there are exactly 12 symmetries of a tetrahedron.
10. Find the symmetry of the square equal to the following products (remember that πi · π j means applying motion πi followed by motion π j): (a) π2 · π4 (b) π2 · π5 (c) π7 · (π2 · π8) (d) (π7 · π6) · π3 11. (a) Write out the 6 × 6 multiplication table for the product of all pairs of sym-metries of a triangle.
(b) Repeat part (a) for integers 1, 2, 3, 4 with multiplication modulo 5.
362 Chapter 9 Polya’s Enumeration Formula (c) Repeat part (a) for the following group of permutations of 1, 2, 3, 4: 1 2 3 4 1 2 3 4 1 2 3 4 2 1 4 3 1 2 3 4 3 4 1 2 1 2 3 4 4 3 2 1 12. Find two symmetries of the square πi, π j such that πi · π j ̸= π j · πi (this means the group of symmetries of a square is noncommutative).
13. Many organic compounds consist of a basic structure formed by carbon atoms (carbon atoms are the corners of a floating figure), plus submolecular groups called radicals that are attached to each carbon atom (the carbons are like cor-ners and the radicals like colors). Suppose such an organic molecule has six carbon atoms, with four radicals of type A and two radicals of type B. Suppose this molecule has three isomers—that is, three different ways that the two types of radicals can be distributed. Which of the following two hypothetical carbon structures would have three isomers with four As and two Bs?
14. Prove that an equivalence relation partitions a set into disjoint subsets of mutually equivalent elements.
15. Give a procedure for decomposing any permutation into a product of cycles.
16. Show that properties (2) and (3) of a group G of permutations imply properties (ii) and (iii), respectively, of the associated equivalence relation defined in the theorem.
17. Let S be a set and G a group of permutations of S. For any two subsets S1, S2 of S, define S1 ∼S2 to mean that for some π ∈G, S1 = π(S2) (= {π(s) | s ∈S2}).
Show that ∼is an equivalence relation.
18. Prove that the set of permutations of the 2-colorings of a square [see (1)] forms a group.
19. Prove that for any prime p, the integers 1, 2, . . ., p −1 with multiplication modulo p form a group.
20. A subset of elements in a group G is said to generate the group if all elements in G can be obtained as (repeated) products of elements in the subset.
(a) Which of the following subsets generate the group of symmetries of a square?
(i) π1, π2, π3 (ii) π2, π5 (iii) π3, π6 (iv) π6, π7 (b) Show that the group of symmetries of a regular n-gon can be generated by a subset of two elements.
21. If a subset G′ of elements in a group G is itself a group, then G′ is called a subgroup of G.
www.itpub.net 9.2 Burnside’s Theorem 363 (a) Show that G′ = {π1, π2, π3, π4} and G′′ = {π1, π7} are subgroups of the group G of symmetries of a square.
(b) Find another 4-element subgroup of G containing π3.
(c) Find all subgroups of the group G of symmetries of a square.
22. (a) How many different binary relations on n elements are possible?
(b) How many symmetric binary relations are possible?
23. Show that πi ∼π j if there exists π ∈G such that πi = π−1π jπ is an equivalence relation.
24. A transposition is a cycle of size 2—that is, a permutation that interchanges the positions of just two elements (and leaves all other elements fixed). Show by induction that any permutation of n elements can be written as a composition of transpositions.
25. For a given group G of n elements, define the function fπ on G as follows: for each π′ ∈G, fπ(π′) = π · π′.
(a) Show that fπ is a one-to-one mapping for any π ∈G.
(b) If we define fπ1 ∗fπ2 = fπ1·π2, show that the set { fπ} forms a group.
9.2 BURNSIDE’S THEOREM We now develop a theory for counting the number of different (nonequivalent) 2-colorings of the square. More generally, in a set T of colorings of the corners (or edges or faces) of some figure, we seek the number N of equivalence classes of T induced by a group G of symmetries of this figure.
Suppose there is a group of s symmetries acting on the c colorings in T. Let EC be the equivalence class consisting of C and all colorings C′ equivalent to C—that is, all C′ such that for some π ∈G, π(C) = C′. If each of the s πs takes C to a different coloring π(C), then EC would have s colorings. Note that the set of π(C)s includes C since πI(C) = C (πI is the identity symmetry). If every equivalence class is like this with s colorings, then sN = c: (number of symmetries) × (number of equivalence classes) = (total number of colorings) Solving for N, we have N = c/s.
Consider, for example, the c = n! oriented seatings of n people around a round table. There are s = n cyclic rotations of the seatings, and each equivalence class consists of n seatings. Thus, the number of equivalence classes (cyclicly nonequivalent seatings) is N = n!/n = (n −1)!
On the other hand, suppose we have a small round table with three positions for chairs (each 120◦apart), and white and black chairs are available. There are 23 = 8 ways to place a white or black chair in each position. See Figure 9.5. There are three 364 Chapter 9 Polya’s Enumeration Formula 1 1 1 1 1 3 Multiplicity 3 Multiplicity 1 Figure 9.5 cyclicrotationsofthetablepossible,0◦,120◦,and240◦.Wehavec = 8“colorings”and s = 3 symmetries, but the number of equivalence classes cannot be N = 8 3, a fraction!
It is true that the three arrangements of one black and two white chairs (or vice versa) form an equivalence class, since 0◦, 120◦, and 240◦rotations move the one black chair to different positions. However, an arrangement of three black chairs (or three white chairs) forms an equivalence class by itself. See Figure 9.5. Any rotation maps this arrangement of three black chairs into itself, that is, leaves it fixed.
We need to correct the numerator in the formula N = c/s by adding the multiplic-ities of an arrangement—that is, when one or more symmetries map the arrangement to itself instead of to other arrangements. In this way, when multiplicities are counted, every equivalence class will have s members. Since two symmetries, along with the 0◦ symmetry, leave the all-black-chair arrangement fixed and similarly for the all-white, then counting arrangements in Figure 9.5 with multiplicities we have the correct answer N = (3 + 1 + 1 + 1 + 1 + 1 + 1 + 3)/3 = 12/3 = 4 The “multiplicity” correction is even more complicated for 2-colorings of our square. Here the size of an equivalence class of colorings may be 1 or 2 or 4, but never 8 (= the number of symmetries of the square). The first problem is that several πs, besides the identity symmetry π1, may leave a coloring Ci fixed—that is, π(Ci) = Ci.
The other problem is that if Ck is another coloring in Ci’s equivalence class, there may be several πs all taking Ci to Ck. For example, the coloring C10 (see Figure 9.1) is fixed by symmetries π1, π3, π7, π8, and is mapped to C11 by symmetries π2, π4, π5, π6.
By the lemma in Section 9.1, the symmetries π2, π4, π5, π6 taking C10 to C11 can be written in the form π = π2 · π′, where π′ is a symmetry that leaves C11 fixed, or else π2 followed by π′ would not take C10 to C11. For example, π5 = π2 · π8: a b c d b a d c = a b c d b c d a · a b c d c b a d Similarly π2 = π2 · π1, π4 = π2 · π3, π6 = π2 · π7. Conversely, given any π∗that leaves C11 fixed, π2 · π∗takes C10 to C11 and so π2 · π∗must be one of π2, π4, π5, π6.
Thus there is a 1 −1 correspondence between the πs that take C10 to C11 and the πs that leave C11 fixed. Therefore, to count the colorings in an equivalence class E with appropriate multiplicities (i.e., coloring C11 has multiplicity 4 since four different πs www.itpub.net 9.2 Burnside’s Theorem 365 take C10 to C11), it suffices to sum over the colorings in E the number of πs that leave each coloring fixed.
In the case of the equivalence class consisting of C10 and C11, each of C10 and C11 have multiplicity 4, so that the size of their equivalence class including multiplicities is 4 + 4 = 8 (= s, the number of symmetries), as required.
In general, when multiplicities are counted, each equivalence class E will have s elements. If φ(x) denotes the number of πs that leave the coloring x fixed, then x∈E φ(x) = s.
Formal proof that x∈E φ(x) = s: Let x1, x2, . . . , xm be the colorings in equivalence class E and let π1, π2, . . . , πs be the group of symmetries. These πs can be divided into m groups, R1, R2, . . . , Rm, where Ri is the set of πs that map x1 to xi. As shown above for x1 = C10 and xi = C11, the number of πs mapping x1 to xi equals φ(xi), the number of πs that leave xi fixed. Thus, x∈E φ(x) sums the number of πs in R1, in R2, . . . , and in Rm. But the sum of the Rs is just the total number of symmetries, s.
Summing over all equivalence classes, we obtain the following theorem, first proved by Burnside more than 100 years ago.
Theorem (Burnside, 1897) Let G be a group of permutations of the set S (corners of a square). Let T be any collection of colorings of S (2-colorings of the corners) that is closed under G. Then the number N of equivalence classes is N = 1 |G| x∈T φ(x) or N = 1 |G| π∈G (π) (∗) where |G| is the number of permutations and (π) is the number of colorings in T left fixed by π.
By “closed under G,” we mean that for all π ∈G and x ∈T, π(x) ∈T . This closure property is automatic when T is the set of all corner 2-colorings of the square.
In Section 9.4 we need to apply formula (∗) to special subsets of S, such as the set of colorings with two corners black and two corners white, that are closed under G.
The two sums in the theorem both count all instances of some coloring being left fixed by some π, the first sums over the different colorings, the second sums over different πs. Formula (∗) will turn out to be more useful in later computations.
We informally summarize the spirit behind formula (∗) as follows. The total number c of all colorings of our square is equal to (π1), since the identity symmetry π1 leaves all colorings fixed. If each of the eight symmetries mapped a coloring C into eight different colorings, then we would have eight colorings in each equivalence class and hence a total c/8 = (π1)/|G| equivalence classes. However, for any coloring C, the colorings in the collection of π(C)s as π takes on all the different symmetries 366 Chapter 9 Polya’s Enumeration Formula 1 2 1 2 3 1 2 3 4 (a) (b) (c) Figure 9.6 in G are never distinct. This means that each equivalence class does not contain 8 elements. The terms (πi) in (∗) add the “multiplicities” of repeated colorings so that each equivalence class has eight colorings in the sum in (∗).
Example 1: 2-Colored Batons A baton is painted with equal-sized cylindrical bands. Each band can be painted black or white. If the baton is unoriented as when spun in the air, how many different 2-colorings of the baton are possible if the baton has (a) 2 bands? (b) 3 bands?
(c) 4 bands?
The batons with two bands, three bands, and four bands are pictured in Figure 9.6.
Irrespective of the number of bands, there are two symmetries of a baton: π1 is a 0◦ revolution of the baton—π1 is the identity symmetry—and π2 is a 180◦revolution of the baton.
(a) For the 2-band baton, the set of 2-colorings left fixed by π1 is all 2-colorings of the baton. There are 22 = 4 2-colorings, and so (π1) = 4. The set of 2-colorings left fixed by π2 consists of the all-black and all-white coloring, and so (π2) = 2.
By Burnside’s theorem, the number of different colorings is 1 2 [(π1) + (π2)] = 1 2(4 + 2) = 3.
(b) For the 3-band baton, all 23 2-colorings are left fixed by π1, and so (π1) = 23 = 8. The set of 2-colorings left fixed by π2 can have any color in the middle band (band 2) and a common color in the two end bands, and so (π2) = 2 × 2 = 4. The number of different colorings is 1 2 [(π1) + (π2)] = 1 2(8 + 4) = 6.
(c) For the 4-band baton, all 24 2-colorings are left fixed by π1, and so (π1) = 24 = 16. The set of 2-colorings left fixed by π2 have a common color for the end bands and a common color for the inner bands, so (π2) = 2 × 2 = 4. The number of different colorings is 1 2 [(π1) + (π2)] = 1 2(16 + 4) = 10.
Example 2: 3-Colored Batons How many different 3-colorings of the bands of an n-band baton are there if the baton is unoriented (as in Example 1)?
As in Example 1, the symmetries of the baton are a 0◦revolution and a 180◦ revolution. We apply formula (∗) for the group of the 0◦and 180◦revolution acting on an n-band baton with three colors. There are 3n colorings of the fixed baton and so (0◦) = 3n. The number of colorings left fixed by a 180◦spin depends on whether n is even or odd.
If n is even, each of the n/2 bands on one half of the baton can be any color—3n/2 choices—andthenforthecoloringtobefixedbya180◦spin,eachofthesymmetrically opposite bands must be the corresponding color. So (180◦) = 3n/2 and we have from formula (∗): N = 1 2(3n + 3n/2).
www.itpub.net 9.2 Burnside’s Theorem 367 To enumerate batons left fixed by a 180◦spin when n is odd, we can use any color for the “odd” band in the middle of the baton—three choices. Each of the (n −1)/2 bands on one side of the middle band can be any color—3(n−1)/2 choices— and again the other (n −1)/2 bands must be colored symmetrically. So ((180◦) = 3 × 3(n−1)/2 = 3(n+1)/2 and N = 1 2(3n + 3(n+1)/2).
Example 3: 3-Colored Necklaces Suppose a necklace can be made from beads of three colors—black, white, and red.
How many different necklaces with n beads are there?
When the n beads are positioned symmetrically about the circle, the beads occupy the positions of the corners of a regular n-gon. Thus, our question asks for the number of corner colorings of an n-gon using three colors. The answer depends on what is meant by “different.” If the beads are not allowed to move about the necklace—that is, the n-gon is fixed—the answer is 3 × 3 × · · · × 3 = 3n (three color choices at each of n corners). A more realistic interpretation of our problem would allow the beads to move freely about the circle—that is, the n-gon rotates freely (but in this case flips will not be allowed). We employ formula (∗) to count the number N of equivalence classes of these 3-colorings induced by the rotational symmetries of an n-gon. We shall now do the calculations for n = 3. A more general technique for larger n is developed in the next section.
There are 33 = 27 3-colorings of a 3-bead necklace, and three rotations of 0◦, 120◦, 240◦. The 0◦rotation leaves all colorings fixed, and so (0◦) = 27. The 120◦ rotation cannot fix colorings in which some color occurs at only one corner. It follows that the 120◦rotation fixes just the monochromatic colorings. Thus, (120◦) = 3. The 240◦rotation is a reverse 120◦rotation, and so (240◦) = 3. By formula (∗), we have N = 1 3(27 + 3 + 3) = 11 9.2 EXERCISES S u m m a r y o f E x e r c i s e s The exercises continue the application of Burnside’s theorem to count colorings introduced in Examples 1, 2, and 3.
1. How many different n-bead necklaces are there using beads of red, white, blue, and green (assume necklaces can rotate but cannot flip over)?
(a) n = 3 (b) n = 4 2. How many different ways are there to place a diamond, sapphire, or ruby at each of the four vertices of this pin?
368 Chapter 9 Polya’s Enumeration Formula 3. Fifteen balls are put in a triangular array as shown. How many different arrays can be made using balls of three colors if the array is free to rotate?
4. How many different ways are there to 2-color the 64 squares of an 8 × 8 chess-board that rotates freely?
5. A merry-go-round can be built with three different styles of horses. How many five-horse merry-go-rounds are there?
6. A domino is a thin rectangular piece of wood with two adjacent squares on one side (the other side is black). Each square is either blank or has 1, 2, 3, 4, 5, or 6 dots.
(a) How many different dominoes are there?
(b) Check your answer by modeling a domino as an (unordered) subset of two numbers chosen with repetition for 0, 1, 2, 3, 4, 5, 6.
7. Two n-digit decimal sequences consisting of digits 0, 1, 6, 8, 9 are vertically equivalent if reading one upside-down produces the other—that is, 0068 and 8900. Howmanydifferent(verticallyinequivalent)n-digit(0,1,6,8,9)sequences are there?
8. Howmanydifferentwaysaretheretocolorthefivefacesofanunorientedpyramid (with a square base) using red, white, blue, and yellow?
9. (a) Find the group of all possible permutations of three objects.
(b) Find the number of ways to distribute 12 identical balls in three indistin-guishable boxes. [Hint: First let boxes be distinct, and then use part (a).] 10. How many ways are there to 3-color the n bands of a baton if adjacent bands must have different colors?
11. How many ways are there to 3-color the corners of a square with rotations and reflections allowed if adjacent corners must have different colors?
12. (a) How many ways are there to distribute 12 red jelly beans to four children, a pair of identical female twins and a pair of identical male twins?
(b) Repeat part (a) with the requirement that each child must have at least one jelly bean.
13. (a) Show that for any given i, the subset Gi of symmetries of the square that leave coloring Ci (of Figure 9.1) fixed is a group (a subgroup of all symmetries).
(b) Find G4.
(c) Find G7.
www.itpub.net 9.3 The Cycle Index 369 9.3 THE CYCLE INDEX Without further theory, we would find it very difficult to apply Burnside’s theorem to counting different colorings of an unoriented figure. Recall that Burnside’s theorem in Section 9.2 says that the number N of equivalence classes in a set T of colorings of a figure with respect to a group of symmetries G is N = 1 |G| π∈G (π) where (π) is the number of colorings in T left fixed by π. If the set T were all 3-colorings of the corners of a 10-gon or a cube, it would seem close to impossible to determine (π)s, the number of colorings left fixed by various symmetries π of the figure. However, we shall show that (π) can be determined easily from the structure of π. We develop the theory for this simplified calculation of (π) in terms of 2-colorings of a square.
Let us apply Burnside’s theorem to the 2-colorings of the square. Initially we determine the number of 2-colorings left fixed by motion πi by inspection (using Figures 9.1 and 9.2). As we count (πi) for each πi, we look for a pattern that would enable us to predict mathematically which colorings must be left fixed by each πi.
It is helpful to make a table of the πis and the colorings that they leave fixed; see columns (i) and (ii) in Figure 9.7 [column (iii) is developed later].
The 0◦rotation π1 leaves each corner fixed and hence it leaves all colorings fixed.
So (π1) = 16. The 90◦rotation π2 cyclicly permutes corners a, b, c, d. Being left fixed by the 90◦rotation means that each corner in a coloring has the same color after the 90◦rotation as it did before. Since π2 takes a to b, then a coloring left fixed by π2 must have the same color at a as at b. Similarly, such a coloring must have the same color at b as at c, the same color at c as at d, and the same at d as at a. Taken together, these conditions imply that only the colorings of all white or all black corners, C1 (i) (ii) (iii) Motion Colorings Left Cycle Structure πi Fixed by πi Representation π1 16—all colorings x4 1 π2 2—C1, C16 x4 π3 2—C1, C10, C11, C16 x2 2 π4 2—C1, C16 x4 π5 2—C1, C6, C8, C16 x2 2 π6 2—C1, C7, C9C16 x2 2 π7 8—C1, C2, C4, C10 x2 1x2 C11, C12, C14, C16 π8 8—C1, C3, C5, C10 x2 1x2 C11, C13, C15, C16 Figure 9.7 370 Chapter 9 Polya’s Enumeration Formula and C16, are left fixed. Thus (π2) = 2. In general, a coloring C will be left fixed by π if and only if for each corner ν, the color of C at ν is the same as the color at π(ν) so that the symmetry leaves the color at π(ν) unchanged.
Next we consider the 180◦rotation π3. Looking at the depiction of π3 in Figure 9.2, we see that π3 causes corners a and c to interchange and corners b and d to interchange. It follows that a coloring left fixed by π3 must have the same color at cor-ners a and c and the same color at b and d (no further conditions are needed). With two color choices for a, c and with two color choices for b, d, we can construct 2 × 2 = 4 colorings that will be left fixed—namely, C1, C10, C11, C16. Hence (π3) = 4.
Symmetry π4 is a rotation of 270◦or −90◦, and so the symmetry is similar to π2.
Hence (π4) = (π2) = 2. The horizontal rotation π5 interchanges corners a, b and interchanges corners c, d. Like the 180◦rotation π3, symmetry π5 will leave a coloring fixed if corners a and b havethesamecolor—twochoices,whiteorblack—andcorners c and d have the same color—two choices. Then like π3, we have (π5) = 2 × 2 = 4 colorings fixed, namely, C1, C6, C8, C16.
A pattern is becoming clear. The reader should now be able quickly to pre-dict that π6 will also leave 2 × 2 = 4 2-colorings of the square fixed, for again this symmetry interchanges two pairs of corners. Formally, an interchange is a cyclic per-mutation on two elements. All our enumeration of fixed colorings has been based on the fact that if πi cyclicly permutes a subset of corners (that is, the corners form a cycle of πi), then those corners must all be the same color in any coloring left fixed by πi.
As mentioned in Section 9.1, any πi can be represented as a product of disjoint cycles. For example, π3 = (ac)(bd) and π4 = (adcb). For each symmetry, we need to get such a cyclic representation and then count the number of ways to assign a color to each cycle of corners. For future use, let us also classify the cycles by their length.
It will prove convenient to encode a symmetry’s cycle information in the form of a product containing one x1 for each cycle of 1 corner in πi, one x2 for each cycle of size 2, and so forth. This expression is called the cycle structure representation of a symmetry.
The cycle structure representation of π2 and π4 is x4, since each consists of one 4-cycle: π2 = (abcd) and π4 = (adcb). The cycle structure representation for π3, π5, and π6 is x2x2 or x2 2, since each consists of two 2-cycles. Column (iii) in Figure 9.7 gives the cycle structure representation of each symmetry.
What about π1? Previously, it sufficed to say that π1 leaves all colorings fixed.
Now it is time to point out that a corner left fixed by a permutation is classified as a 1-cycle. Thus π1 consists of four 1-cycles. Its cycle structure representation is then π4 1 .
We “predict” that π1 leaves 24 = 16 colorings fixed—that is, π1 leaves all 2-colorings fixed.
For any symmetry π of any figure, the number of colorings left fixed will be given by setting each x j equal to 2 (or, in general, the number of colors available) in the cycle structure representation of π, that is, (π) = 2number of cycles in π www.itpub.net 9.3 The Cycle Index 371 Let us apply this theory to π7 and π8. For each of these reflections, the cycle structure representation is seen to be x2 1x2 (follow the arrows in Figure 9.2), and thus for each we can find 22 × 2 = 8 colorings that are left fixed. Note that for 180◦flips around opposite corners and midpoints of opposite edges, all corners will be in cycles of size 2, unless a corner is left fixed by the flip.
To obtain the number of different 2-colorings of the floating square with Burnside’s Theorem, we sum the numbers in column (ii) of Figure 9.7 and divide by 8: 1 |G| π∈G (π) = 1 8(16 + 2 + 4 + 2 + 4 + 4 + 8 + 8) = 1 8(48) = 6 There is a slightly simpler way to get this result. First, algebraically sum the cycle structure representations of each symmetry, collecting like terms together, and then divide by 8. From column (iii) of Figure 9.7, we obtain 1 8(x4 1 + 2x4 + 3x2 2 + 2x2 1x2).
This expression is called the cycle index PG(x1, x2, . . . , xk) for a group G of sym-metries. By setting each xi = 2 in this cycle index—that is, PG(2, 2, . . . , 2)—we get the same answer. (Before, the steps were reversed: we first set xi = 2 in each cycle structure representation and then added.) Suppose that instead of two colors, we had three colors. Then the same reasoning applies, but now there are three choices for the color of the corners in each cycle. If a symmetry has k cycles, then it will leave 3k 3-colorings of the square fixed, and the number of different 3-colorings will be PG(3, 3, . . . , 3). More generally, for any m, PG(m, m, . . . , m) will be the number of nonequivalent m-colorings of an unoriented square. The argument used to derive this coloring counting formula with the cycle index of a square is valid for colorings of any set with associated symmetries.
Theorem Let S be a nonempty set of elements and G be a group of symmetries of S that acts to induce an equivalence relation on the set of m-colorings of S. Then the number of nonequivalent m-colorings of S is given by PG(m, m, . . . , m).
Example 1: Coloring Necklaces Use this theorem to re-solve the problem of Example 3 in Section 9.2 of counting n-bead necklaces with black, white, and red beads.
Recall that beads on the necklace can rotate freely around the circle, but that reflections are not permitted, and that the number of different 3-colored strings of n beads is equal to the number of 3-colorings of a cyclicly unoriented n-gon. For n = 3, the rotations are of 0◦, 120◦, and 240◦with cycle structure representations of x3 1, x3, and x3, respectively. Thus, PG = 1 3(x3 1 + 2x3). The number of 3-colored strings of three beads is PG(3, 3, 3) = 1 3(33 + 2 × 3) = 11. More generally, the number of m-colored necklaces of three beads is PG(m, m, m) = 1 3(m3 + 2m).
Let us try a more complicated case: n = 8 (see Figure 9.8). The rotations are of 0◦, 45◦, 90◦, 135◦, 180◦, 225◦, 270◦, and 315◦. The 0◦rotation consists of eight 372 Chapter 9 Polya’s Enumeration Formula a b c d e f g h Figure 9.8 1-cycles. The 45◦rotation is the cyclic permutation (abcdefgh). The 90◦rotation has the cycle decomposition (aceg) (bdfh). The 135◦rotation is the cyclic permutation (adgbehcf ). The 180◦rotation has the cyclic decomposition (ae) (bf ) (cg) (dh). The cycle structure representations are thus 0◦rotation, x8 1, 45◦rotation, x8, 90◦rotation, x2 4; 135◦rotation, x8; and 180◦rotation, x4 2. The 225◦, 270◦, and 315◦rotations are reverse rotations of 135◦, 90◦, 45◦, respectively, and have the corresponding cycle structure representations. Collecting terms, we obtain PG = 1 8 x8 1 + 4x8 + 2x2 4 + x4 2 The number of different m-colored necklaces of eight beads is 1 8(m8 + 4m + 2m2 + m4) For m = 3, we have 1 8(38 + 4 × 3 + 2 × 32 + 34) = 1 8(6561 + 12 + 18 + 81) = 834 There are two helpful observations about rotations of an n-gon. First, for a given rotation, all its cycles will be of the same size, since it does not matter at which corner you start the rotation’s cyclic permutation of the corners. It follows that the size of any rotation’s cycles must divide n, the number of corners. This tells us what the possible sizes of cycles can be for a rotation. For example, for a 9-gon, the only possible cycle size for any rotation is 1, 3, or 9 (the divisors of 9).
Second, for any rotation, the cycle size will be the minimum number of times we must iterate this rotation to generate a cumulative rotation that is a multiple of 360◦, since to map a corner back to its original position (completing a cycle), a rotation must move a corner a multiple of 360◦. If a rotation of an n-gon moves each corner k places clockwise around the n-gon, then the length of the rotation’s cycle will be the minimum times we must iterate this rotation so that the repeated shifts of k places equal a multiple of n. In particular, if k has no common divisors with n, the rotation will form a single cycle of length n. For example, the 135◦degree rotation of the 8-gon in Example 1 moved each corner three places clockwise. Since the smallest number of 3s that sum to a multiple of 8 is eight 3s (3 × 8 = 24), the 135◦rotation creates a single cycle of length 8.
www.itpub.net 9.3 The Cycle Index 373 Example 2: Coloring Corners of a Tetrahedron Use the theorem to determine the number of 3-colorings of the four corners of a floating tetrahedron. In Example 4 in Section 9.1 we listed the 12 symmetries of the tetrahedron (see Figure 9.4 in Section 9.1): the 0◦revolution, the eight revolutions of 120◦and 240◦about a corner and the middle of the opposite face, and the three revolutions of 180◦about the middle of opposite edges.
The 0◦revolution has the cycle structure representation x4 1. The 120◦revolution about corner a and the middle of face bcd has the cyclic decomposition (a)(bcd ) and its cycle structure representation is x1x3. By symmetry, the other 120◦and 240◦ revolutions have this same cycle structure representation. The 180◦revolution about the middle of edges ab and cd has the cyclic decomposition (ab)(cd ) and its cycle structure representation is x2 2. By symmetry, the other 180◦revolutions have the same cycle structure representation. Thus we have PG = 1 12 x4 1 + 8x1x3 + 3x2 2 The number of different corner 3-colorings is PG(3, 3, 3, 3) = 1 12(34 + 8 × 3 × 3 + 3 × 32) = 1 12(81 + 72 + 27) = 15 9.3 EXERCISES S u m m a r y o f E x e r c i s e s The first 11 exercises count distinct color-ings of unoriented figures. The remaining problems involve associated theory. Note that “floating” means that all possible rotations and reflection are allowed.
1. How many ways are there to color the corners of a floating square using four different colors?
2. How many ways are there to 4-color the corners of a pentagon that is (a) Distinct with respect to rotations only?
(b) Distinct with respect to rotations and reflections?
3. How many ways are there to 3-color the corners of a hexagon that is (a) Distinct with respect to rotations only?
(b) Distinct with respect to rotations and reflections?
(c) Findtwo3-coloringsofthehexagonthataredifferentinpart(a)butequivalent in part (b).
374 Chapter 9 Polya’s Enumeration Formula 4. How many different n-bead necklaces (cyclicly distinct) can be made from three colors of beads when (a) n = 7 (b) n = 9 (c) n = 10 (d) n = 11 5. Find the number of different m-colorings of the vertices of the following floating figures.
a h b g c f d e f a b c d e g (d) (e) (f) a e d b c a c b d a h g e d b i j k l f c (a) (b) (c) 6. (a) Find the cycle index for the group of symmetries of a square in terms of permutations of edges, not corners.
(b) How many ways are there to 3-color the edges of a floating square?
(c) How many ways are there to 3-color both edges and corners of a floating square?
(d) Why is the following not the case? (number of floating 3-colorings of corners) (number of floating 3-colorings of edges) = (number of floating 3-colorings of edges and corners) Explain.
7. How many ways are there to m-color the edges of the floating figures in Exercise 5? (Hint: The cycle index now is for symmetries of the edges.) 8. (a) Find the number of different n-bead 3-colored necklaces (cyclicly distinct) in which each color appears at least once when (i) n = 3, (ii) n = 4, (iii) n = 7.
(b) Repeat part (a) when necklaces may reflect as well as rotate.
9. Find the number of different 2-sided dominoes (two squares of 1 to 6 dots or a blank on each side of the domino).
10. (a) Let G be the group of all 4! permutations of 1, 2, 3, 4. Find PG.
(b) Usepart(a)tofindthenumberofwaystopaintfouridenticalmarbleseachone of three colors (check your answer by modeling this problem as a selection-with-repetition problem).
(c) Use part (a) to find the number of ways to put 12 balls chosen from three colors into four indistinguishable boxes with three balls in each box.
www.itpub.net 9.4 Polya’s Formula 375 11. (a) Find the number of 2 × 4 chessboards distinct under rotation whose squares are colored red or black.
(b) Suppose that two chessboards are also considered equivalent (aside from rotational symmetry) if one can be obtained from the other by complementing red and black colors. How many different 2 × 4 chessboards are there?
12. Show that if xk1 1 xk2 2 · · · xkm m is the cycle index for a symmetry π of an n-gon (expressed in terms of a permutation of the corners), then 1k1 + 2k2 + · · · + mkm = n.
13. In solving for the number of corner 2-colorings of some unoriented figure, sup-pose we are given the cycle index PG∗of the group G∗of induced permutations of the 2-colorings [as in (1) in Section 9.1], instead of the usual cycle index PG of the group of symmetries of the figure. What integer values should be substituted now for each xi in PG∗to get the number of 2-colorings, or will no substitution work? Explain.
14. Let S be some set of n objects and G a group of permutations of S. For subsets S1, S2 of S, define the equivalence relation S1 ∼S2 if for some π ∈G, S1 = π(S2) (= {π(s) | s ∈S2}). Show that the number of equivalence classes equals PG(2, 2, . . .). (There are 2n subsets of S in all.) 15. Find the number of m-colorings of the corners of a p-gon, where p is a prime (p > 2), if (a) Only rotations are allowed (b) Rotations and reflections are allowed 9.4 POLYA’S FORMULA We are now ready to address our ultimate goal of a formula for the pattern inven-tory. Recall that the pattern inventory is a generating function that tells how many colorings of an unoriented figure there are using different possible collections of colors. For black–white colorings of the unoriented square, the pattern inventory is 1b4 + 1b3w + 2b2w2 + 1bw3 + 1w4. For example, the term 1b3w tells us that there is one nonequivalent coloring with three black (b) corners and one white (w) corner. The coefficients in the pattern inventory can be viewed as the results of several Burnside theorem–type counting problems. Recall that Burnside’s theorem says that the num-ber N of equivalence classes in T, a collection of colorings of S (where S is the corners of a square or the faces of a tetrahedron, etc.), caused by a group of symmetries of S is N = 1 |G| π∈G (π) () 376 Chapter 9 Polya’s Enumeration Formula In the case of 2-colorings of the floating square, we divide the set of all colorings in Figure 9.1 into sets based on the numbers of black and white corners: T0 = {C1} T1 = {C2, C3, C4, C5} T2 = {C6, C7, C8, C9, C10, C11} T3 = {C12, C13, C14, C15} T4 = {C16} The coefficient of b3w can be obtained from () if we let the group G of symmetries of the square act on just the set T1. Recall that formula () from Burnside’s theorem applies to any collection of colorings that is closed under G. The term “closed under G” means that for any π ∈G, and any coloring C ∈T, π(C) equals another coloring in T. T1 is closed under G, since any symmetry acting on a coloring with three black corners and one white corner yields another coloring with three black corners and one white corner. The same is true for the other Tks. Thus, the coefficient of b4−kwk in the pattern inventory is the result of () when Tk is the set on which G acts.
Let us try to solve these five subproblems simultaneously. In Figure 9.9 we have duplicated the table in Figure 9.7 and added a new column (iv). In the first row of column (iv), we write a polynomial whose coefficients give the numbers of 2-colorings in each Tk left fixed by π1, then in the second row of the table we write a polynomial for the numbers of 2-colorings in each Tk left fixed by π2, then by π3, and so forth.
Then we total up the b4 term in each row (the number of 2-colorings with four blacks) and divide by 8 to get the coefficient of b4 in the pattern inventory, total up the b3w term in each row and divide by 8 to get the coefficient of b3w, and so forth.
Since the action of π1 leaves all Cs fixed, the first row’s coefficients are 1, 4, 6, 4, 1. We write b4 + 4b3w + 6b2w2 + 4bw3 + w4; this is an inventory of fixed colorings.
For π1, the inventory of fixed colorings is an inventory of all colorings. Observe that this inventory is simply (b + w)4 = (b + w)(b + w)(b + w)(b + w), one (b + w) for each corner. For π2, the inventory is b4 + w4. For π3, we find by observation that the inventory is b4 + 2b2w2 + w4. This expression factors into (b2 + w2)2.
(i) (ii) (iii) (iv) Motion Colorings Left Cycle Structure Inventory of Colorings πi Fixed by πi Representation Left Fixed by πi π1 16—all colorings x4 1 (b + w)4 = 1b4 + 4b3w + 6b2w2 + 4bw3 + 1w4 π2 2—C1, C16 x4 (b4 + w4) = 1b4 + 1w4 π3 2—C1, C10, C11, C16 x2 2 (b2 + w2)2 = 1b4 + 2b2w2 + 1w4 π4 2—C1, C16 x4 (b4 + w4) = 1b4 + 1w4 π5 2—C1, C6, C8, C16 x2 2 (b2 + w2)2 = 1b4 + 2b2w2 + 1w4 π6 2—C1, C7, C9, C16 x2 2 (b2 + w2)2 = 1b4 + 2b2w2 + 1w4 π7 8—C1, C2, C4, C10 x2 1 x2 (b + w)2(b2 + w2) = 1b4 + 2b3w + 2b2w2 + 2bw3 + 1w4 C11, C12, C14, C16 π8 8—C1, C3, C5, C10 x2 1 x2 (b + w)2(b2 + w2) = 1b4 + 2b3w + 2b2w2 + 2bw3 + 1w4 C11, C13, C15, C16 Figure 9.9 www.itpub.net 9.4 Polya’s Formula 377 Just as we did before when counting the total number of colorings fixed by the action of some π, let us look for a pattern in the inventories of fixed colorings. Again the key to the pattern is the fact that in a coloring left fixed by π, all corners in a cycle of π must have the same color. Since π2 has one cycle involving all four corners, the possibilities are thus all corners black or all corners white; hence the inventory is b4 + w4. The motion π3 has two 2-cycles (ac) and (bd). Each 2-cycle uses two blacks or two whites in a fixed coloring. Hence the inventory of a cycle of size two is b2 + w2.
The possibilities with two such cycles have the inventory (b2 + w2)(b2 + w2).
The inventory of fixed colorings for πi will be a product of factors (b j + w j), one factor for each j-cycle of the πi. So we need to know the number of cycles in πi of each size. But this is exactly the information encoded in the cycle structure representation.
Indeed, setting x j = (b j + w j) in the representation yields precisely the inventory of fixed colorings for πi. By this method we compute the rest of the inventories of fixed colorings. See Figure 9.9. For π7 especially, the inventory should be checked against the list of colorings in column (ii). The pattern inventory is obtained by adding together the inventories of fixed colorings, collecting like-power terms, and dividing by 8.
As before in Section 9.3, we get a more compact formula and save some com-putation by first adding together the cycle structure representations and dividing by 8, and then setting each x j = (b j + w j) and doing the polynomial algebra all at once.
Again the first step in this approach yields the cycle index PG(x1, x2, . . . , xk). Thus by setting x j = (b j + w j) in PG, we obtain the pattern inventory.
If three colors, black, white, and red, were permitted, each cycle of size j would have an inventory of (b j + w j +r j) in a fixed coloring. So we would set x j = (b j + w j +r j) in PG. The preceding argument applies for any number of colors and any figure. In greater generality we have the following theorem.
Theorem (Polya’s Enumeration Formula) Let S be a set of elements and G be a group of permutations of S that acts to induce an equivalence relation on the colorings of S. The inventory of nonequivalent colorings of S using two colors is given by the generating function PG((b + w), (b2 + w2), (b3 + w3), . . . , (bk + wk)). The inventory using colors c1, c2, . . . , cm is PG m j=1 c j, m j=1 c2 j, . . . , m j=1 ck j For a moment, let us return to the problem of counting the total number of nonequivalent 2-colorings. This number is simply the sum of the coefficients in the pattern inventory. To sum coefficients, we set the indeterminants, b and w (and hence their powers), equal to 1, or, equivalently, set x j = 2 in PG. If m colors were allowed, we would set x j = m in PG, obtaining the same formula as in the theorem in Section 9.3.
As in many other generating function problems, the actual expansion of the generating function for a pattern inventory can be quite tedious. We are expanding expressions of the form (ci 1 + ci 2 + · · · + Ci m)r. When m and r get large, it is time to turn to computer algebra software.
378 Chapter 9 Polya’s Enumeration Formula Example 1: Pattern Inventory for 3-Bead Necklaces Determine the pattern inventory for 3-bead necklaces distinct under rotations using black and white beads. Repeat using black, white, and red beads.
From Example 1 of Section 9.3, we know PG = 1 3(x3 1 + 2x3). Substituting x j = (b j + w j), we get 1 3[(b + w)3 + 2(b3 + w3)] = 1 3[(b3 + 3b2w + 3bw2 + w3) + (2b3 + 2w3)] = 1 3(3b3 + 3b2w + 3bw2 + 3w3) = b3 + b2w + bw2 + w3 This result could be obtained empirically. There is only one way to color all beads black or all white. If one bead is white and the others black, then by rotation the white bead can occur anywhere; thus there is only one necklace with one white and two blacks. The same is true by symmetry for a necklace with one black and two whites.
Now consider 3-bead necklaces using three colors. We substitute x j = (b j + w j +r j), in PG to obtain 1 3[(b + w +r)3 + 2(b3 + w3 +r3)]. Instead of expanding the polynomials in this expression, we use indirect means. Perhaps again each inventory term has coefficient 1. There is a general test for whether all inventory coefficients are 1: Compare N, the number of terms in the pattern inventory, with N, the total number of patterns [i.e., PG(m, m, . . . , m)]. Since N equals the sum of the coefficients in the inventory, then N ∗= N if and only if each term has coefficient 1. The number N of terms in the pattern inventory when n elements (corner, beads, etc.) are colored with m colors is C(n + m −1, n) (see Exercise 19). For the case at hand, m = 3, n = 3, and so N ∗= 3+3−1 3 = 10. From Example 1 of Section 9.3, we know that N = 11.
Since N ∗̸= N, we know that all terms do not have coefficient 1.
On the other hand, the only way for N ∗= N + 1 is that nine terms in the inventory must have coefficient 1 and one term coefficient 2. But as argued above, there is only one necklace with all beads the same color and only one necklace with one bead of color A and the other two beads of color B. The only other possibility for a 3-bead necklace using three colors is to have one bead of each color. Thus, there must be two necklaces with one bead of each color (the necklaces are any cyclic order of the three colors and the reverse cyclic order), and the pattern inventory for black, white and red 3-bead necklaces is b3 + w3 +r3 + b2w + b2r + w2b + w2r +r2b +r2w + 2bwr Example 2: Pattern Inventory for 7-Bead Necklaces Find the number of 7-bead necklaces distinct under rotations using three black and four white beads.
We need to determine the coefficient of b3w4 in the pattern inventory. Each rotation, except the 0◦rotation, is a cyclic permutation when the number of beads is a prime [see Exercise 13(a)], so PG = 1 7(x7 1 + 6x7). The pattern inventory is 1 7[(b + w)7 + 6(b7 + w7)].
www.itpub.net 9.4 Polya’s Formula 379 Sincethefactor6(b7 + w7)inthepatterninventorycontributesnothingtotheb3w4 term, we can neglect it. Thus the number of 3-black, 4-white necklaces is simply 1 7[(coefficient of b3w4 in (b + w)7] = 1 7 7 3 Example 3: Pattern Inventory for Edge 2-Colorings of a Tetrahedron Find the pattern inventory of black–white edge colorings of a tetrahedron.
Although we calculated the cycle index for corner symmetries of the tetrahedron in Example 2 of Section 9.3, we need a different cycle index for edge symmetries.
Since the set of objects to be colored is the six edges, we need to consider the symmetries of the tetrahedron as permutations of the edges.
The 0◦revolution clearly leaves all edges fixed and thus has cycle structure representation x6 1. The 120◦(or 240◦) revolution about a corner and the middle of the opposite face cyclicly permutes the edges incident to that corner and cyclicly permutes the edges bounding the opposite face (see Figure 9.4 in Section 9.1). Thus, the 120◦revolution has cycle structure representation x2 3.
The 180◦revolution about opposite edges leaves those two edges fixed (see Figure 9.4). Since two applications of a 180◦revolution return the tetrahedron to its original position, the other four edges not left fixed must be in 2-cycles. Thus, the 180◦ revolution has cycle structure representation x2 1x2 2. Then PG = 1 12(x6 1 + 8x2 3 + 3x2 1x2 2).
Substituting x j = (b j + w j), we get 1 12[(b + w)6 + 8(b3 + w3)2 + (b + w)2(b2 + w2)2] = 1 12[(b6 + 6b5w + 15b4w2 + 20b3w3 + 15b2w4 + 6bw5 + w6) + (8b6 + 16b3w3 + 8w6) + (3b6 + 6b5w + 9b4w2 + 12b3w3 + 9b2w4 + 6bw5 + 3w6)] = 1 12(12b6 + 12b5w + 24b4w2 + 48b3w3 + 24b2w4 + 12bw5 + 12w6) = b6 + b5w + 2b4w2 + 4b3w3 + 2b2w4 + bw5 + w6 Example 4: Pattern Inventory for Corner 2-Colorings of a Cube Find the pattern inventory for corner 2-colorings of a floating cube.
The symmetries of the cube involve revolutions about opposite faces, about oppo-site edges, and about opposite corners. First, of course, there is the identity symmetry, with cycle structure representation x8 1.
(a) Opposite faces: As a concrete example, we revolve about the center point of the pair of opposite faces abcd and efgh; see Figure 9.10a. A 90◦revolution yields the permutation (abcd) (efgh) with cycle structure representation x2 4. A 270◦revolution has the same structure. A 180◦revolution yields the permutation (ac)(bd)(eg)(fh) with cycle structure representation x4 2. There are three pairs of 380 Chapter 9 Polya’s Enumeration Formula a d e f g c b h a d f e g c b h a d f g c b h e (a) (b) (c) Figure 9.10 Revolutions of the cube. (a) Revolution about opposits faces.
(b) Revolution about opposite edges. (c) Revolution about opposite corners.
opposite faces and so the total contribution to the cycle index of opposite-face revolutions is 6x2 4 + 3x4 2.
(b) Opposite edges: As a concrete example, we revolve about the middle of oppo-site edges ad and fg; see Figure 9.10b. A 180◦revolution yields the permuta-tion (ad)(bh)(ce)(fg) with cycle structure representation x4 2. There are six pairs of opposite edges, and so the total contribution of opposite edge revolutions is 6x4 2.
(c) Oppositecorners:Asaconcereteexample,werevolveabouttheoppositecornersa and g; see Figure 9.10c. A 120◦revolution yields the permutation (a)(bde)(chf )(g) with cycle structure representation x2 1x2 3. One way to see what this permutation does is by noting that the three corners b, d, e adjacent to a must be cyclically permuted in any motion that leaves a fixed (and similarly for the corners adjacent to g). A 240◦revolution has the same structure. There are four pairs of opposite corners and so the contribution of opposite-corner revolutions is 8x2 1x2 3.
We leave it to the reader (see Exercise 17) to verify that we have enumerated all symmetries of the cube and that these symmetries are all distinct.
Collecting terms, we find PG = 1 24(x8 1 + 6x2 4 + 9x4 2 + 8x2 1x2 3). The pattern inven-tory for corner colorings of the cube using black and white is thus 1 24 (b + w)8 + 6(b4 + w4)2 + 9(b2 + w2)4 + 8(b + w)2(b3 + w3)2 As in previous examples, the coefficients of the terms b8, b7w, bw7, and w8 are readily seen to be 1. The b6w2, b5w3, and b4w4 terms in the four factors of the generating function are (· · · + 28b6w2 + 56b5w3 + 70b4w4 + · · ·), 6(· · · + 2b4w4 + · · ·), 9(· · · + 4b6w2 + 6b4w4 + · · ·), and 8(· · · + b6w2 + 2b5w3 + 4b4w4 + · · ·), re-spectively. Summing and dividing by 24 we have 1 24(· · · + 72b6w2 + 72b5w3 + 168b4w4 + · · ·) = 3b6w2 + 3b5w3 + 7b4w4. (It is easy to detect errors in these cal-culations, since most errors will result in a noninteger coefficient.) By symmetry, we fill out the pattern inventory to obtain b8 + b7w + 3b6w2 + 3b5w3 + 7b4w4 + 3b3w5 + 3b2w6 + bw7 + w8 www.itpub.net 9.4 Polya’s Formula 381 9.4 EXERCISES S u m m a r y o f E x e r c i s e s The first 16 exercises use Polya’s enu-meration formula, and the remaining problems involve associated theory. Note that “floating” means that all possible rotations and reflections are allowed.
1. Find the pattern inventory for black and white corner colorings of a floating pentagon.
2. Find an expression for the pattern inventory for black–white, n-bead necklaces (rotations only) and find the number of necklaces with three white beads and the rest black: (a) n = 6 (b) n = 9 (c) n = 10 (d) n = 11 3. Find the pattern inventory for black, white, and red corner colorings of a floating square.
4. Find an expression for the pattern inventory for the 2-colorings (rotations only) of the 16 squares in a 4 × 4 chessboard.
5. Find an expression for the pattern inventory for black–white corner colorings of the floating figures in Exercise 5 of Section 9.3.
6. (a) Find the pattern inventory for corner 2-colorings of a floating pyramid (with a square base).
(b) Repeat part (a) for edge colorings.
(c) Repeat part (a) for face colorings.
7. Find an expression for the pattern inventory for edge 2-colorings of the floating figures in Exercise 5 in Section 9.3.
8. Find an expression for the pattern inventory for edge 2-colorings of a floating cube and find the number of edge 2-colorings with three white and nine black edges.
9. Find an expression for the pattern inventory for face 2-colorings of (a) A floating tetrahedron (b) A floating cube 10. (a) Find the pattern inventory for the corner 3-colorings of a floating pentagon with adjacent corners different colors.
(b) Repeat part (a) for a floating tetrahedron.
(c) Repeat part (a) for a floating cube.
11. Find an expression for the pattern inventory for black–white colorings of four indistinguishable balls (see Exercise 10 in Section 9.3).
12. Give an empirical argument (without use of the cycle index) to show that there are ⌊n 2⌋different n-bead necklaces with 2 white beads and n −2 black beads (⌊r⌋ is the largest integer ≤r).
13. Given that p is a prime and p-bead necklaces are made of black and white beads, (a) Show that each rotation except 0◦is a cyclic permutation of the corners (b) What is the number of such necklaces with exactly k white beads?
382 Chapter 9 Polya’s Enumeration Formula 14. (a) Suppose that instead of coloring the corners of a floating square, we attach 0 or 1 or 2 identical jelly beans at a corner. Find a generating function, and expand it, for the number of different squares with a total of k beans at its corners. (Hint: In the inventory of patterns left fixed, the exponents within a factor will vary.) (b) Repeat part (a) with each corner getting a red jelly bean or a white jelly bean or a red and a white jelly bean. Now find a generating function for the number of squares with j whites and k reds.
15. Suppose n batons with small holes drilled through their midpoints are strung along a piece of wire and each end of each baton is painted one of three possible colors (red, white, blue). The wire is fixed, but the batons revolve about their centers. How many indistinguishable configurations are there for (a) n = 2 (b) n = 3 16. How many distinct (nonisomorphic) unlabeled graphs are there with four ver-tices? (Hint: Each possible edge has two possible colors: “edge present” and “edge absent.”) 17. Show that the 24 symmetries of a cube listed in Example 4 are all distinct and include all symmetries of a cube.
18. Characterize the geometric figures that when floating have only one coloring with one black corner and the others white.
19. Show that there are C(n + m −1, n) different terms in the pattern inventory for m-colorings of corners of an n-gon.
20. Let ak,n denote the minimum number of coefficients in the pattern inventory of all k-colorings of the corners of an n-gon needed so that using symmetry all other coefficients in the pattern inventory are known. Find a generating function gk(x) for ak,n.
9.5 SUMMARY AND REFERENCES Polya’s enumeration formula is important, practically because it solves important problems, mathematically because it is an elegant marriage of group theory and generating functions, and pedagogically because it is the only truly powerful combi-natorial formula we will see in this book (where a difficult problem can be solved by “plugging” the right numbers into a specialized formula). Of equal importance is the manner in which Polya’s formula has been developed: Sections 9.3 and 9.4 could be read as a case study in the experimental derivation of a mathematical theory. (This approach follows the pedagogical style Polya used when he taught this material.) Stu-dents interested in a more rigorous development of this theory and its extensions are referred to , which includes an English translation of Polya’s original 1937 paper, www.itpub.net 9.5 Summary and References 383 “Kombinatorische Anzahlbestimmungen f¨ ur Gruppen, Graphen und chemische Verbindungen.” In a more formal development, one defines a coloring as a function f from a set S (of corners) to a set R (of colors). The function f6, corresponding to C6 (in Figure 9.1), would be written in tabular form as a b c d w w b b . We define the composition π · f of a motion and a color function to be a new color function f ′; for example, π2 · f6 maps a →b →w, b →c →b, c →d →b, d →a →w, or π2 · f6 = a b c d w b w b = f7.
The coloring permutation induced by a symmetry π maps each f to π · f. Although tedious, a formal calculation of the coloring permutation is thus possible (before we did it by inspection). We define coloring equivalence by f ∼f ′ if and only if there exists π such that π · f = f ′. Our theory can readily be restated in terms of this new definition.
Polya’s enumeration formula has an important application to another field of combinatorial mathematics. It is used to enumerate families of graphs (see Exercise 16 in Section 9.4). This application was pioneered by F. Harary; see Harary and Palmer .
1. F. Harary and E. Palmer, Graphical Enumeration, Academic Press, New York, 1973.
2. G. Polya and R. C. Reade, Combinatorial Enumeration of Groups, Graphs, and Chemical Compounds, Springer-Verlag, New York, 1987.
This page is intentionally left blank www.itpub.net CHAPTER 10 GAMES WITH GRAPHS 10.1 PROGRESSIVELY FINITE GAMES In this chapter we develop the theory of progressively finite games and apply it to Nim-type games. While clearly not an important use of graphs, our graph-theoretic analysis of these games illustrates some very abstract strategies for attacking very practical problems that seem far removed from such abstraction. Further, it is essential to study how graph models can be used to solve a variety of real-world problems, but there is a more personal reward in learning how graphs can permit one to win at certain games.
A game in which two players take turns making a move until one player wins (no ties are allowed) is called progressively finite if (1) there are a finite number of different positions in the game and (2) the play of the game must end after a finite number of moves.
Our objective in this section is winning: how to determine winning strategies for progressively finite games. We can model a progressively finite game by a directed graph with a vertex for each position that can occur in the play of the game and a directed edge for each possible move from one position to another.
Observe that the graph of a progressively finite game cannot contain any directed circuits, since players could move around and around a circuit of positions forever (violating the constraint on finite play). Thus, no positions can ever be repeated in the play of a game. Rather, the game moves inexorably toward some final position that is a win for one of the players. Games such as checkers and chess that permit ties and repetition of positions are not progressively finite. Most progressively finite games are “takeaway”-type games of the sort illustrated by the games in Examples 1 and 2.
Example 1: Restricted Takeaway Game A set of 16 objects is placed on a table. Two players take turns removing 1, 2, 3, or 4 objects. The winner is the player who removes the last object. The graph of this game is shown in Figure 10.1 (all edges are directed from left to right).
Example 2: Inverted Takeaway Game Starting with an empty pile, two players add 1 penny or 2 pennies or 1 nickel to the pile until the value of the pile is the square of a positive integer ≥2 or until the value 385 386 Chapter 10 Games with Graphs 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Figure 10.1 exceeds 40. The player whose addition brings the value of the pile to one of these critical amounts takes all the money in the pile. The graph of this game is shown in Figure 10.2 (all edges are directed from left to right; arrows point to the game-winning positions). This game is an inverted form of a “takeaway” game in which 1, 2, or 5 objects are withdrawn from a pile until the pile is reduced to certain critical sizes.
The reader is encouraged to try playing these games with a friend. A winning strategy for the first player (who makes the first move) in the game in Example 1 can be found with a little thought. The game in Example 2 is harder. The details of this second game only serve to confuse systematic attempts to find a winning strategy.
It is easier to develop a general theory of winning strategies in progressively finite games and then apply the theory to Example 2.
A winning position in a progressively finite game is a position at which play stops and the player who moved to this position is declared the winner.
In the graph of a progressively finite game, vertices with 0 out-degree must repre-sent winning positions, for if no edges leave a vertex, the game must stop at that vertex.
We call such vertices winning vertices. The graph of the game in Example 1 has just one winning vertex, numbered 0, while the graph of the game in Example 2 has six winning vertices, numbered 4, 9, 16, 25, 36, and “over 40.” Every progressively finite game must have at least one winning position or else play would go on without end.
A winning strategy for a player is a rule that tells the player which move to make at each stage in a game so as to ensure that the player will eventually win, that is, finally move to a winning position. Obviously only one player can have a winning strategy.
Any vertex adjacent to a winning vertex is a “losing” vertex in the sense that if one player moves to a “losing” vertex, the other player can then move to a winning vertex (and win the game). The vertices numbered 1, 2, 3, and 4 are losing vertices in the game in Example 1. The vertices numbered 2, 3, 7, 8, 11, 14, 15, 20, 23, 24, 31, 34, 35, 37, 38, 39, and 40 are losing vertices in the game in Example 2.
Stepping back one more move from the end of the game, we see that if all edges from a vertex x go to losing vertices, then x is a “prewinning” vertex. Whenever player A moves to such a prewinning vertex, then player B’s next move must be to a losing vertex and now player A can move to a winning vertex. Vertex 5 in the game . . .
. . .
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 24 25 26 36 37 38 39 40 Over 40 Figure 10.2 www.itpub.net 10.1 Progressively Finite Games 387 in Example 1 is a prewinning vertex, since all edges from 5 go to losing vertices (1, 2, 3, and 4). Vertices 6 and 33 are the prewinning vertices in the game in Example 2; all other vertices adjacent to losing vertices are also adjacent to nonlosing vertices.
The general theory of winning strategies in progressively finite games is based on a recursive extension of the preceding reasoning. We seek good vertices, such as winning and prewinning vertices, that win or lead only to bad vertices. When our opponent is forced to move to a bad vertex, such as losing vertices, then we will always be able to move to another good vertex. A winning strategy for the first player will tell the player how to find good vertices to move to from any bad vertex. This pattern of play, with the first player moving to successive good vertices and the second player forced to move to bad vertices, will continue until finally the first player reaches a good vertex that is a winning vertex. If the game starts at a good vertex, and so the first player’s initial move must be to a bad vertex, then roles are reversed and it is the second player who now has a winning strategy using the good vertices.
Example 1: (continued) The winning strategy for the first player in this game is to move to a vertex whose number is a multiple of 5. These are the good vertices. Thus, the move from the starting vertex 16 is to vertex 15 (i.e., the first player removes one object). From 15, the second player must move to one of the vertices 11, 12, 13, or 14. From any of these bad vertices the first player now moves to vertex 10. Whatever the second player’s next move is, the first player will always be able to move to the prewinning vertex 5, and one round later the first player will win. Note that if the game started with only 15 objects, then the roles would be reversed and the second player would be able to use this winning strategy.
We formalize the concept of good vertices with the following definition. A kernel in a directed graph is a set of vertices such that both of the following are true: 1.
There is no edge joining any two vertices in the kernel.
2.
There is an edge from every nonkernel vertex to some kernel vertex.
The kernel of the graph in Example 1 is the set of vertices numbered 0, 5, 10, and 15.
Theorem 1 If the graph of a progressively finite game has a kernel K, then a winning strategy for the first player is to move to a kernel vertex on every turn. However, if the starting vertex is in the kernel, then the second player can use this winning strategy.
Proof First we show that all winning vertices must be in K. The reason is that vertices not in a kernel must have an edge directed to a vertex in the kernel. But winning vertices have 0 out-degree. Thus, all winning vertices must be in any kernel.
388 Chapter 10 Games with Graphs a b c d e Figure 10.3 Next we show that moving to a kernel vertex on every turn is a winning strategy for the first player. By property (1) of a kernel, when the first player moves to a kernel vertex, the second player must then move to a nonkernel vertex. Then by property (2), the first player can always move from the nonkernel vertex to a kernel vertex. The play proceeds in this fashion with the first player always moving to a kernel vertex and the second player always moving to a nonkernel vertex. Since the game is progressively finite, the play must eventually end. Since all winning vertices are in the kernel, the first player must win.
If the starting vertex is in the kernel, then the roles of the first and second players are reversed and the second player can always move to a kernel vertex for an eventual win. ◆ The preceding proof does not show explicitly how successively moving to kernel vertices leads to a winning vertex. The proof is existential. It shows only that by moving to kernel vertices the first player must eventually arrive at a winning vertex.
A more immediate problem is to show that the graph of every progressively finite game has a kernel and then to find this kernel. Not all graphs have kernels.
For example, the graph in Figure 10.3 has no kernel. To show that this graph has no kernel, we argue thus. If there were a kernel, we could assume by the symmetry in this graph that vertex a is in the kernel. Then vertex e, which has an edge to a, cannot be in the kernel. Vertex d’s only outward edge goes to nonkernel vertex e, and so d must be in the kernel. Similarly, c cannot be in the kernel and then b must be in the kernel. But now the edge (a, ⃗b) joins two kernel vertices—a contradiction.
Fortunately, graphs of progressively finite games do have kernels. To demon-strate the existence of kernels in such graphs, we need to organize the vertices in a progressively finite graph into levels based on the “distance” of the vertices from a winning vertex. We recursively define the level l(x) of vertex x in a directed graph and the sets Lk of vertices at level ≤k as follows. Let s(x) = {y | x has an edge to y} be the set of successors of x. Then l(x) = 0 ⇔s(x) is empty and L0 = {x |l(x) = 0} l(x) = 1 ⇔x / ∈L0 and s(x) ⊆L0 and L1 = L0 ∪{x |l(x) = 1} and, in general, l(x) = k ⇔x / ∈Lk−1 and s(x) ⊆Lk−1 and Lk = Lk−1 ∪{x |l(x) = k} Observe that Lk −Lk−1 is the set of vertices at level k.
www.itpub.net 10.1 Progressively Finite Games 389 It can be shown that l(x) is the length of the longest path in the directed graph that starts at x (see Exercise 12). Since all paths in a progressively finite graph have finite length, the longest path from a vertex x has finite length. Further, the longest path from x must end at a vertex of out-degree 0; otherwise the path could be extended.
Then starting from vertices of out-degree 0, the assignment of level numbers by this recursive definition will eventually reach all vertices.
It follows from this definition of level that every vertex at level 0 is a winning vertex and that every vertex at level 1 is a losing vertex. A vertex at level 2 is also a losing vertex if it is adjacent to a vertex at level 0; it is a prewinning vertex only if all its successors are at level 1. In general, every vertex at level k(k > 0) must be adjacent to a vertex at level k −1 and possibly other vertices at lower levels, but cannot be adjacent to any other vertex at level k (or greater). Now we can prove the fundamental theorem for progressively finite games.
Theorem 2 Every progressively finite game has a unique winning strategy. That is, the graph of every progressively finite game has a unique kernel.
Proof The proof is by induction on the levels or, more precisely, on the sets Lk. Let Kk be the set of kernel vertices in Lk. First consider the set L0. L0 consists of vertices with 0 out-degree. These are the winning vertices. As noted in the proof of Theorem 1, all winning vertices must be in the kernel. Thus K0 = L0.
Next let us inductively assume for n ≥1 that Kn−1 is the unique, well-defined set of kernel vertices in Ln−1. We show that we can find a unique set of level-n vertices that must be added to Kn−1 to form the kernel Kn for Ln. By the way that level numbers were defined, l(x) = n means that s(x) ⊆Ln−1. If a level-n vertex x is adjacent to no kernel vertex of Kn−1, this x must be in Kn (since by the definition of a kernel, any vertex with no successors in the kernel must itself be in the kernel). On the other hand, if x is adjacent to a kernel vertex of Kn−1, x cannot be in the kernel.
Hence Kn = Kn−1 ∪{x |l(x) = n and s(x) ∩Kn−1 = Ø} is the unique, well-defined set of kernel vertices in Ln. It follows by mathematical induction that the graph has a unique kernel. By Theorem 1, this kernel is the unique winning strategy. ◆ The proof of Theorem 2 tells us how to build a kernel. First put the winning verticesinthekernel.Thenrecursivelyaddtheverticesatincreasinglevelsnotadjacent to the current set of kernel vertices. We implement this procedure for finding kernels using a labeling rule called a Grundy function.
Definition of Grundy Function g(x) For each vertex x in a directed graph, g(x) is the smallest nonnegative integer not assigned to any of x’s successors.
390 Chapter 10 Games with Graphs We shall prove shortly that vertices with Grundy number 0 are exactly the set of kernel vertices. In the next section, Grundy numbers will be seen to play a fundamental role in more complex games.
In the graph of a progressively finite game, Grundy values can be easily deter-mined using a level-by-level approach. The vertices on level 0 have no successors and so their Grundy number will be 0 (the smallest nonnegative integer). Next we determine g(x) for vertices x at level 1, then vertices at level 2, and so forth. In this way, all of a vertex x’s successors are assigned Grundy numbers before it is time to determine the Grundy number of x, since x’s successors are at lower levels. When we come to x, we can check the Grundy function values of s(x), x’s successors, and set g(x) equal to the smallest nonnegative integer not assigned to any vertex in s(x).
Actually we do not need to proceed in a totally level-by-level fashion. We can use any method that does not try to assign a vertex its Grundy number until all the vertex’s successors have Grundy numbers.
No Grundy function can be defined for the graph in Figure 10.3. Each vertex x in Figure 10.3 has a successor whose Grundy number must be defined before g(x) can be determined. Even if we try to invent simultaneously Grundy numbers for all vertices in Figure 10.3 at once, no Grundy function exists (details are left to the reader).
The Grundy numbers for the vertices in Figures 10.1 and 10.2 are given in the following tables. Note that the vertices in Figure 10.1 that are in the kernel—namely, 0, 5, 10, 15—all have Grundy number 0. This is no accident.
Table of Grundy Numbers for Figure 10.1 Vertex x 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 g(x) 1 0 4 3 2 1 0 4 3 2 1 0 4 3 2 1 0 Table of Grundy Numbers for Figure 10.2 Vertex x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 g(x) 0 3 1 2 0 1 0 2 1 0 3 1 0 3 1 2 0 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 0 2 1 0 3 2 1 0 2 0 1 2 0 31 32 33 34 35 36 37 38 39 40 over 40 1 2 0 1 2 0 1 3 2 1 0 Theorem 3 The graph of a progressively finite game has a unique Grundy function. Further, the vertices with Grundy number 0 are the vertices in the kernel.
Proof The recursive level-by-level construction of a Grundy function for the graph of a progressively finite game gives each vertex a unique Grundy number, as described www.itpub.net 10.1 Progressively Finite Games 391 above. By the definition of a Grundy function, a vertex x with Grundy number 0 cannot have a successor with Grundy number 0 (or else x would have to have a different number). Similarly any vertex y with Grundy number g(y) = k > 0 must have an edge to some vertex with Grundy number 0 (or else y’s number would be 0).
Thus, the set of vertices with Grundy number 0 satisfies the two defining properties of a kernel. ◆ Example 2: (continued) By Theorem 3 and the foregoing table of Grundy numbers for Figure 10.2, we see that the kernel is the set 0, 4, 6, 9, 12, 16, 18, 21, 25, 27, 30, 33, 36, “over 40.” Since the starting vertex is in the kernel, the second player has the winning strategy in this game. A play of the game might proceed as follows (let player A be the first player and B be the second player): first A moves to 1, then B moves to kernel vertex 6, then A must move to 11 (a move to 7 or 8 lets B win at 9), then B moves to kernel vertex 12, then A must move to 17, then B moves to kernel vertex 18, then A moves to 20, and then B moves to the winning vertex 25 (and collects the 25 cents).
10.1 EXERCISES S u m m a r y o f E x e r c i s e s The first 10 exercises involve finding ker-nels and Grundy functions in various graphs of progressively finite games. Exercises 11–17 involve proofs of properties of kernels, Grundy functions, and level numbers.
Exercises 18 and 19 present two more complicated progressively finite games.
1. Find a kernel in the following graphs or show why none can exist: a b c d e a c i g h d f b e b c d a f (b) (c) (a) 2. Suppose that there are 25 sticks and two players take turns removing up to five sticks. The winner is the player who removes the last stick. Which player has a winning strategy? Describe this strategy.
3. Repeat Example 2 with 2, 3, or 7 cents added each time. Find the set of positions in the kernel.
4. Show that in Example 2 the second player can always win by the second move.
5. Suppose in Example 2 that the first player A knows he/she will lose and wants to minimize his/her loss. What is the smallest winning amount A can force B 392 Chapter 10 Games with Graphs (the second player) to accept (such that if B tried to make the game go longer, then A could get into the kernel and win)?
6. (a) Suppose we have a pile of seven red sticks and 10 blue sticks. A player can remove any number of red sticks or any number of blue sticks or an equal number of red and blue sticks. The winner is the player who removes the last stick. Find the set of positions in the kernel.
(b) Repeat part (a) with a limit of removing at most five sticks of one color (or both colors) on a move.
7. Repeat Example 1—but now the player to remove the last object loses. Describe the winning strategy for this game.
8. Find the Grundy function for graphs or games in (a) Exercise 1(a) (b) Exercise 1(b) (c) Exercise 3 9. Show that there is no Grundy function for the graph in Figure 10.3.
10. Find a directed graph possessing a kernel but no Grundy function.
11. If W(S) is the set of vertices without an edge directed to any vertex in the set S, show that a set S is a kernel if and only if S = W(S).
12. (a) Show that if l(x) = k for a vertex x in the progressively finite graph G, then k is the length of the longest path starting at x in G.
(b) Show that if g(x) = k, there is a path of length k starting at x in G.
13. Show that for any vertex x in a progressively finite graph, g(x) ≤l(x).
14. Show that if a directed graph G and every subgraph of G (obtained by deleting various vertices) have kernels, then G has a Grundy function.
15. Show that both the level numbers and the Grundy function in a progressively finite graph G constitute proper colorings of G.
16. Show that no matter how the edges are directed in a bipartite graph, it will always have a Grundy function.
17. Show that the graph of a progressively finite game can have only a finite number of vertices. [Hint: Show that if there are an infinite number of vertices, then there must be an infinite path (infinite play).] 18. Consider the following graph game. Player A tries to make a path with a set of vertices from the left to the right side of this graph. Player B tries to make a path from top to bottom. The players take turns picking vertices until one player gets the desired path.
www.itpub.net 10.2 Nim-Type Games 393 (a) Show that this is a progressively finite game.
(b) Find a winning strategy for Player A.
19. The game of kayles has a row of n equally spaced stones. Two players alternate turns of removing one or two consecutive stones (with no intervening spaces).
The player to remove the last stone wins.
(a) Draw the graph and find its Grundy function for a four-stone game.
(b) Repeat part (a) for a six-stone game.
10.2 NIM-TYPE GAMES In this section we extend the theory of progressively finite games to takeaway games involving several piles of objects. The simplest game of this form is called Nim, a game in which two players take turns removing any number they wish from one of the piles. The winner is the player who removes the last object from the last remaining (nonempty) pile. While the positions in the two games in Examples 1 and 2 in the previous section could be described with a single nonnegative integer representing the size or monetary value of the single pile, the position in a Nim game requires a vector of nonnegative integers (p1, p2, . . . , pm), the kth number pk representing the current size of the kth pile.
Example 1: Game of Nim Consider the Nim game with four piles of sticks: one stick in the first pile, two sticks in the second pile, three in the third, and four in the fourth. See Figure 10.4. We represent the initial position of this game with the vector (1, 2, 3, 4).
Let the first and second players be named A and B, respectively. A sample play of the game might go as follows. First A removes all four sticks from the fourth pile.
The new position is (1, 2, 3, 0). Next B removes one stick from the third pile to produce position (1, 2, 2, 0). Now A removes the one stick in the first pile to produce position (0, 2, 2, 0). A has been playing a winning strategy—that is, moving into kernel positions—and is now about to win. If B removes all of the second or third pile, A will remove the other pile; or if B removes just one stick, A will remove one stick from the other pile and A will win on the next round. The reader is encouraged to play this Nim game with a friend.
Figure 10.4 394 Chapter 10 Games with Graphs 0 1 2 3 4 Figure 10.5 The game in Example 1 has 2 × 3 × 4 × 5 = 120 different positions (the ith pile has i + 1 possible sizes, 0, 1, . . . , i). Thus, it would be very cumbersome to draw the graph of this game and compute its kernel with a Grundy function, as in the previous section. In general, a Nim game with m piles and ni objects in the ith pile will have (n1 + 1)(n2 + 1) · · · (nm + 1) different positions. The only way that we could ever play winning Nim without a computer is if we could determine the Grundy number of a position (vertex) directly from the position vector (p1, p2, . . . , pm). Fortunately, such direct computation is possible.
First we need to examine the Grundy function for a single-pile Nim game. Con-sider the Nim game with one pile of four sticks. See Figure 10.5. The graph of this game has edges (i, ⃗j), for all 0 ≤j ≤i ≤4. The Grundy number of vertex 0, the winning vertex, is 0; the Grundy number of vertex 1 is 1; and so on. In any one-pile Nim game, vertex i will have a Grundy number of i, since vertex i has edges to all lower-numbered vertices. Of course, the strategy of any one-pile Nim game is trivial: the first player removes the whole pile and wins. The nice form of this single-pile Grundy function will simplify the computation of Grundy functions for multi-pile Nim games.
Although the graph of a multi-pile Nim game is too complex to draw, we can still describe it symbolically. We have a vertex for each position (p1, p2, . . . , pm), where 0 ≤pi ≤ni (ni is the initial size of the ith pile). Since the only permissible moves are removing some amount from one pile, the as-sociated graph has edges from vertex (p1, p2, . . . , pm) to vertex (q1, q2, . . . , qm) for each pair of vertices such that for one j, q j, < p j, and for all i ̸= j, qi = pi. This graph is in some sense a “composition” of the graphs for each pile, for if we fix all pi except one, say p3, then the subgraph of vertices (p1, p2, 0, p4, . . . , pm), (p1, p2, 1, p4, . . . , pm), . . . , (p1, p2, n3, p4, . . . , pm) is ex-actly the graph for pile 3 alone.
This type of composition of graphs can be formalized as follows. The direct sum H = H1 + H2 + · · · + Hm of graphs H1, H2, . . . , Hm with vertex sets X1, X2, . . . , Xm, respectively, has vertex set X = {(x1, x2, . . . , xm) | xi ∈Xi, 1 ≤i ≤m} and edges defined by the successor sets s((x1, x2, . . . , xm)) = {(y, x2, x3, . . . , xm) | y ∈s(x1)} ∪{(x1, y, x3, . . . , xm1) | y ∈x(x2)} .
.
.
∪{(x1, x2, . . . , xm−1, y) | y ∈s(xm)} It follows from this definition that the graph G of an m-pile Nim game is the direct sum of the graphs Gi of the ith pile: G = G1 + G2 + · · · + Gm.
www.itpub.net 10.2 Nim-Type Games 395 (1, 2) 3 (0, 2) 2 (2, 2) 0 (0, 0) 0 (0, 1) 1 (2, 1) 3 (1, 0) 1 (2, 0) 2 (1, 1) 0 0 1 2 (a) (b) Figure 10.6 Example 2: Graph of 2 × 2 Nim Consider the simple Nim game with two piles of two objects each. Figure 10.6a shows the graph G1 = G2 of one pile alone. Figure 10.6b shows the graph G = G1 + G2 of the two-pile game. Find the Grundy function and kernel of this graph.
The Grundy numbers for the vertices are shown in Figure 10.6b. The vertices with Grundy number 0 are in the kernel. Since g((2, 2)) = 0, the second player has the winning strategy, according to Theorems 1 and 3 in the previous section.
Next we show how to compute the Grundy number of a vertex (x1, x2, . . . , xm) in a direct sum of graphs from the Grundy numbers g(xi) of the vertices xi in the component graphs Gi. The computation to be performed on these Grundy numbers is called a digital sum.
The digital sum c of nonnegative integers c1, c2, . . . , cm, written c = c1 ˙ + c2 ˙ + c3 ˙ + · · · ˙ + cm is computed in the following manner. Let c(k) be the kth binary digit in a binary expansion of c, that is, c = c(0) + c(1)2 + c(2)22 + · · · , and similarly let c(k) i be the kth digit in a binary expansion of ci. Then c(k) ≡c(k) 1 + c(k) 2 + · · · + c(k) m (modulo 2). That is, the kth binary digit of c is 1 if the sum of the kth binary digits of the c js is odd, and the kth binary digit of c is 0 if the sum of the kth binary digits of the c js is even.
Example 3: Digital Sum Compute the digital sum 2 ˙ + 12 ˙ + 15 ˙ + 8. We write the numbers 2, 12, 15, and 8 in binary form and determine the sum (modulo 2) of the digits in each column, as illustrated in Figure 10.7. Translating the value of the binary sum back into an integer, we see that this digital sum equals 9.
We now present a remarkable theorem. In essence, it says that digital sums are the only way to win at Nim.
Theorem If the graphs G1, G2, . . . , Gn possess Grundy functions g(x), then the direct sum G = G1 + G2 + · · · + Gm possesses a Grundy function g(x), where x = (x1, x2, . . . , xm), 396 Chapter 10 Games with Graphs Figure 10.7 defined by g(x) = g((x1, x2, . . . , xm)) = g1(x1) ˙ + g2(x2) ˙ + · · · ˙ + gm(xm) Proof We must show that g(x) is the smallest nonnegative integer that is not equal to any number in the set {g(y) | y ∈s(x)}. This definition of a Grundy function can be broken into two parts: (1) if y ∈s(x), then g(x) ̸= g(y), and (2) for each nonnegative integer b < g(x), there exists some y ∈s(x) with g(y) = b.
Part (1): Show that if y ∈s(x), then g(x) ̸= g(y). Since y = (y1, y2, . . . , ym) ∈ s((x1, x2, . . . , xm)), then by the definition of direct sum, for some j, y j ∈s(x j), and for all i ̸= j, yi = xi. If ck = gk(xk) and dk = gk(yk), then c j ̸= d j and ci = di. So g(x) = c1 ˙ + c2 ˙ + · · · ˙ + c j ˙ + · · · ˙ + cm = c′ ˙ + c j g(y) = d1 ˙ + d2 ˙ + · · · ˙ + d j ˙ + · · · ˙ + dm = c′ ˙ + d j where c′ is the digital sum of all cs except c j. It is not hard to show that c′ ˙ + c j = c′ ˙ + d j if and only if c j = d j (see Exercise 9). Since c j ̸= di, we conclude that g(x) ̸= g(y), as required.
Part (2): Show that for each nonnegative integer b < g(x), there exists some y ∈s(x) with g(y) = b. A general proof of this part is fairly technical (see Berge , p. 25, for details). The practical side of this proof, discussed below, is finding a y with g(y) = 0 (a kernel vertex) and moving to it. A formal proof of part (2) is a generalization of the discussion below. ◆ Corollary The vertex (p1, p2, . . . , pm) in the graph of an m-pile Nim game has the Grundy number p1 + p2 ˙ + · · · ˙ + pm. Thus (p1, p2, . . . , pm) is a kernel vertex if and only if p1 ˙ + p2 ˙ + · · · ˙ + pm = 0.
The reader should go back to the Nim game of two piles of two objects each in Example 2 and check that the Grundy numbers obtained for the vertices of the graph of that game are the digital sums of the pile sizes.
www.itpub.net 10.2 Nim-Type Games 397 Figure 10.8 Example 1: (continued) The Grundy number of the starting vertex of the Nim game in Figure 10.4 is the digital sum 1 ˙ + 2 ˙ + 3 ˙ + 4 = 4, as computed in Figure 10.8. The first player A wants to decrease the size of one of the piles so that the new digital sum is 0 (a kernel position). That is, A should alter the binary digits in one row of Figure 10.8 so that the sum (bottom) row is 0 0 0. For the sum row to become all 0s, every column now having an odd number of 1s should have one of its digits changed (either a 1 to a 0 or a 0 to a 1) to make the number of 1s in that column even.
Since the sum row has a 1 in just the 22 column, we can change the (single) 1 in that column to a 0. That is, pile 4’s binary expansion should be changed from 1 0 0 to 0 0 0, and so player A should remove all sticks in the fourth pile. This new position (1, 2, 3, 0) has a Grundy number of 0. Note that to make the number of 1s even in the 22 column, we could not change any 0 to a 1, for the new binary expansion in the altered row would then be a larger number—an impossible move in Nim.
We now generalize the method of finding a vertex with Grundy number 0 given in the preceding example. Form a digital sum table as in Figure 10.8 for the current game position. Pick a row e having a 1 in the leftmost column that has an odd number of 1s—that is, row e should have a 1 in the leftmost column with a 1 in the sum row. In row e, change the digit in every column having a 1 in the sum row. After this change of digits, every column will have an even number of 1s, and so the sum row will be all 0s. Thus, this new position will be a kernel vertex. Note that since the leftmost digit that is changed in row e is a 1 (this is how row e was chosen), changing digits in row e will yield a smaller number, call it h. The first player should thus decrease the size of the eth pile to a size of h objects.
Example 4: Another Game of Nim Consider the Nim game of four piles with two, three, four, and six sticks shown in Figure 10.9a. The digital sum table for the initial position is shown in Figure 10.9b.
The 21 column is the leftmost column with a 1 in the sum row, and so we must change a row with a 1 in the 21 column. We can use the first, second, or fourth row. Suppose that we choose the first row. Then since the 21 and 20 columns in the sum row have 1s, we change the digits in these columns in the first row. The new first row is 0 0 1. Thus, the 398 Chapter 10 Games with Graphs Figure 10.9 first player should reduce the size of the first pile to 1. The reader should continue the play of this Nim game with a friend (or against oneself) to practice this kernel-finding rule.
10.2 EXERCISES 1. Find the Grundy number of the initial position and make the first move in a winning strategy for the following Nim games: (a) (b) (c) (d) 2. Suppose that no more than two sticks can be removed at a time from any pile.
Repeat the games in Exercise 1 with this additional condition.
3. Suppose that no more than i sticks can be removed at a time from the ith pile (piles are numbered from top to bottom). Repeat Exercise 1 with this additional condition.
4. Suppose that only one or two or five sticks can be removed from each pile. Repeat the games in Exercise 1 with this constraint.
5. Suppose that only one or four sticks can be removed from each pile. Repeat the games in Exercise 1 with this constraint.
6. For the Nim game in Exercise 1(c), find moves that yield positions with Grundy number equal to (a) 1 (b) 2 (c) 3 www.itpub.net 10.2 Nim-Type Games 399 7. Suppose three copies of the game in Example 2 of Section 11.1 are played simultaneously. Players stop adding money to a pile when the value of the pile is a square or exceeds 40. The player who adds the last amount to the last pile wins the money in all three piles. Note that there is a table of Grundy numbers for this game near the end of the previous section.
(a) If initially there are 2c / in two piles and 1c / in the third pile, what is the Grundy number of this position and what is a correct winning move?
(b) Find all kernel positions in which the sum of the money in the first two piles is less than 10c / and the third pile is empty.
(c) Using the table of Grundy numbers for the one-pile game near the end of Section 11.1, write a computer program to compute the next winning move in this three-pile game.
8. Draw the direct sums involving the graphs shown below.
(a) G1 + G1 (b) G1 + G2 + G3 x1 x2 x4 x3 G1 y1 z1 y2 z2 G2 G3 9. Prove the assertion in the proof of the theorem that c′ ˙ + c j = c′ ˙ + d j if and only if c j = d j.
10. Generalize the argument for finding kernel vertices in Nim (preceding Example 4) to obtain part (2) of the proof of the theorem.
11. Consider the variation of Nim in which the last player to move loses. Show that the winning strategy is to play the regular last-player-wins Nim strategy but when only one pile has more than one stick now decrease that pile’s size to 1 instead of 0 (or to 0 instead of 1).
12. Write a computer program to play winning Nim.
13. Consider the following variation on Nim. Place C(n + 1, 2) identical balls in a triangle, like bowling pins. Two players take turns removing any number of balls in a set that all lie on a straight line. The player to remove the last ball wins.
n = 4 (a) Find a winning first move for this game when n = 3.
(b) Repeat for n = 4.
400 Chapter 10 Games with Graphs 10.3 SUMMARY AND REFERENCES The first published analysis of Nim by C. Bouton appeared in 1902. The Grundy function for progressively finite games was presented by P. Grundy in 1939. It was independently discovered a few years earlier by the German mathematician Sprague.
This chapter has only scratched the surface of the theory of finitely progressive games. Interested readers should turn to On Winning Ways by Berlekamp, Conway, and Guy . The reader should also see the classic book about mathematics and games, Mathematical Recreations and Essays by Coxeter and Ball .
1. E. Berlekamp, J. Conway, and R. Guy, On Winning Ways, 2 volumes, Academic Press, New York, 1982.
2. C. Berge, The Theory of Graphs, Methuen-Wiley, New York, 1962.
3. C. L. Bouton, “Nim, a game with a complete mathematical theory,” Ann. Math. 3 (1902), 35–39.
4. H. Coxeter and W. Ball, Mathematical Recreations and Essays, University of Toronto Press, 1972.
5. P. M. Grundy, “Mathematics and games,” Eureka, January, 1939.
www.itpub.net POSTLUDE A Prelude opened this textbook with the puzzle Mastermind that introduced com-binatorial reasoning in a recreational setting. This Postlude looks at a cryptanalysis problem that is again of a recreational nature but also illustrates the less structured side of combinatorial reasoning as it often occurs in real-world problems. In particular, we will look at a simple cryptographic scheme in which the analysis of underlying combinatorial problems is complicated by the somewhat random pattern of letters in English text.
P.1 Letter Frequencies There have been many tables produced of the relative frequencies of letters in English writing, starting with Samuel Morse (of Morse code fame). We use fre-quencies averaged over several tables.
Most Common Letters in English Text Vowels Consonants E 12% T 9% A, I, O 8% N, R, S 6–8% D, L 4% The least frequent letters, all consonants, are: J, Q, X, Z below .05% As we shall see, it is also useful to know which pairs of consecutive letters, called digraphs, are most frequent. The eight most common digraphs are Most frequent: TH Second most frequent: HE Next six most frequent: ER, RE and AN, EN, IN, ON N is unique among the very frequent letters in that close to 90% of its occurrences are preceded by a vowel; other frequent letters have a much wider range of other letters preceding them. Some other frequent digraphs that can be helpful are ES, SE ED, DE ST TE, TI, TO OF 401 402 Postlude Frequent consonants tend to appear beside vowels but vowels do not occur side by side often and similarly frequent consonants do not appear side by side often except for TH and ST. So there is a quasi-bipartite graph-like relationship with vowels as one set of vertices and frequent consonants as the other set of vertices in the bipartition), and the frequent digraphs are the edges.
There is one triple of consecutive letters, called a trigraph, that stands out, namely, THE. THE is four times as common as any other trigraph in English text. It is frequent both as a three-letter word and as a part of other words.
If we are given an encoded message, we could count the frequency of each letter in the message to determine single-letter frequencies. To get information of how often various letters occur before and after other letters, we build what is called a trigraphic frequency table. For each occurrence of a letter, we record the letter just preceding and the letter following this letter. To illustrate, consider the following cryptogram.
Here spaces have been removed between words, but for readability letters are written in groups of five.
Cryptogram FJYHP KKYRH YKYRF HYVYK PRQYI SFIFP RNAVP PUDQC CAYJY COQRF JYRYD TQYCO JPMIY FJQIN YSVTP VFJYT QVIFF QKYQR FJYES IFIFM OYRFI JSWYP TFYRK QIAYJ QWYOQ RSVPD ONRPQ INTSI JQPRP MFIQO YQFQI JPEYO FJSFF JYQRO PPVIY FFQRB DQCCA VQRBP MFFJY AYIFQ RFJQI NYSVI BVSOM SFQRB HCSII FJYVY DQCCA YRPOY CQWYV QYIPT OPKQR PIEQX XSSCC PDYOQ RIQOY The trigraphic frequency table for the cryptogram is given in Table P.1. Letter frequencies appear at the top of each column in Table P.1. When a trigraph is repeated in some letter’s column, the trigraph is underlined. To illustrate how the table is constructed, consider the beginning of the cryptogram: FJYHP KKY . . . For each occurrence of a letter, we enter the letter just before it and the letter just after it. For the first letter F, we enter .J in F’s column (the “.” indicates that since F is the first letter of the message, no letter precedes it). For the second letter J in the message, we enter FY in J’s column. For the third letter Y, we enter JH in Y’s column. For the fourth letter H, we enter YP in H’s column; for the fifth letter P, we enter HK in P’s column. For the sixth letter K, we enter PK in K’s column. Since the seventh letter is also K, next in K’s column, we enter KY.
For the frequent letters, their columns of trigraph data can be overwhelming, and so it is helpful to make a digraph table for each frequent letter, such as F, listing the frequency of letters that occur two or more times Before F and After F. See Table P.2.
We shall be referring to the data in Tables P.1 and P.2 repeatedly through this Postlude.
Finally, we list the sequences of three or more letters that are repeated several times in the message.
seven times: FJY three times: DQCCA, RFJ, QRB (also DQCCAY two times) www.itpub.net Postlude 403 Table P.1 Trigraph table for cryptogram 6 4 12 6 3 27 0 4 22 16 7 0 5 5 13 21 31 20 13 6 1 12 3 2 37 0 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z NV RD QC UQ YS .J YP YS FY PK PI RA CQ HK RY YH IF DQ PD YY SY QX JH CY RP CA YT PY RH RY FF YY KY FO IY CJ KR DC YF YV VP AP QY XS KR IY IV YO PO IQ SI FY MY FY YY PF OR MY FR OR PQ EI YQ ST QY HK CV RH YO BQ IP BC QN OP YP PF IT YQ VP TY PN JW PF PF KR YY HS QC YQ RJ VF FQ QY OS IY DN PU JI QF RV NS QI HV CY CA PY YJ SF FY RQ QY JM QN YY TI PO SP VK HS VJ FF FY PQ YF TV TV QF JF PI QI QC IF FJ IS RP YT FK YF YV AQ AJ CA FQ QA YQ SM VD YR YK VO SI JC YQ RJ QN IQ PY RQ KI QS MF BS JR SC II SJ IP TP QR JW NP CI YY RD CP IM FQ FS YQ RM OR PP XS YQ QC RI QJ FY QY JE PI QO SC IF TY VF FY OP JP QB NS MI YF FQ PV IO QB JT QQ QN FY BM YF QF KQ OJ VB RO FI QB JE SF SI IT YR YP OR FJ IF OK FR QP WP YF YP RI DC QI FR FQ PE CD VR AJ MF RQ FR WO FJ JI OQ IQ FR EO RJ DC JQ SQ CW IF IJ VY JA KR AI EX NS OR JV IQ VD AR OC WV QI DO O.
Note that longer repeats, such as DQCCA and DQCCAY, can be found by looking at trigraph (3-letter) repeats and concatenating these repeats together. That is, DQCCA is built from the repeated trigraphs of DQC, QCC, and CCA. Observe that this 5-letter sequence is probably a word, since the chances are extremely low that a repeated 5-letter sequence would be formed by a common ending of three different words followed by a common start of three other different words.
Now we start the decoding process. We typically begin with the letters in the English word THE. The very frequent trigraph FJY is a perfect fit to be the encoding of THE, since (i) THE is the most frequent trigraph in English text and FJY is the most frequent trigraph in the cryptogram (occurring 7 times), (ii) E is the most frequent letter in English text and Y is the most frequent letter in our cryptogram, (iii) TH is the most frequent digraph in English text and FJ is the most frequent digraph in 404 Postlude Table P.2 Digraph repetitions for frequent letters in table P.1 12 27 21 16 13 22 31 20 14 12 36 C F I J O P Q R S V Y Bef Bef Bef Bef Bef Bef Bef Bef Bef Bef Bef Aft C4 F4 F4 F10 C2 J2 D3 P3 J2 A2 A4 3C Q3 I6 Q5 I3 Q2 G2 F4 Q10 S2 P2 I2 2D Y3 M2 S3 Y2 Y3 P2 I3 Y6 Y2 S3 J8 2F R5 V3 R4 J4 Y3 K3 3I Aft S3 Y3 Aft Aft V2 K2 Aft Aft N2 2J 3A Y2 2P 2P O2 3B 3F Aft O4 2K 4C Aft 4Q 2Q Aft T2 4F 3I 3I Q3 30 2O Aft 6F 3S 4Y 2D V2 4P 3V 2P V2 3Q 4F 3J 8Y 2M Y3 2Q W3 6R 4I 3N 2P 2Y 2S 10J 2Q 2R Aft 3V 5Q 2Y 2T 3C 2V 4I 2O 10R 3Y 2W the cryptogram (occurring 10 times), and (iv) T is one of the most frequent letters in English text and F is one of the most frequent letters in our cryptogram. We write the information about the encoding of THE as TP = FC, HP = JC, EP = YC, where the P subscript stands for Plain text and the C subscript stands for Code text.
Once we know that EP = YC, we can look for frequent letters that are rarely beside Y, keeping in mind that vowels do not occur side-by-side often. Looking through the digraph information in Figure P.2, we see that YC has no repeated occurrences before or after PC (YC appears before PC once and not at all after PC). So PC is extremely likely to be a vowel. To find another vowel, we can look at frequent code letters that have few occurrences of YC and PC before and after them. An excellent candidate vowel is SC which has no PC’s before or after it, no YC’s after it, but two YC’s before it. No other frequent letter has only two YC’s beside it (before or after).
Finally, recall that in English text the letter N is distinctive because it occurs almost exclusively with the frequent vowels (A, E, I, O) before it. One code letter has exactly this characteristic—namely, RC. Nineteen of the 20 occurrences of RC are preceded by PC (three times), QC (10 times), and YC (six times). We already identified PC and YC as vowels. So NP = RC, and QC is another vowel. While QC has YC(= EP) occurring three times before and three times after it, QC has no PC before or after it and has no SC before or after it; recall that SC was identified as a likely vowel in the preceding paragraph. Thus, YC(= EP), PC, QC and SC are extremely likely to be the four frequent vowels AP, EP, IP, and OP, although we do not know which vowel corresponds to each of these code letters other than that YC = EP. Among the four most frequent English consonants NP, RP, SP, and TP, we have identified that www.itpub.net Postlude 405 TP = FC and NP = RC. In sum, we have made a very good start at breaking this cryptogram by only considering aggregated data about the frequency of single letters, digraphs, and one trigraph.
P.2 Keyword Transpose Encoding In this Postlude we shall consider the following scheme for encoding text called keyword transpose encoding. Interested readers can learn more about the basics of cryptography and some of the associated mathematics from the references at the end of this Postlude or from web sources. We use a given keyword, suppose it is MORNING, to create a one-to-one mapping of plain-text letter to code letters. The first step is to build an array of letters in which the first row consists of the letters in MORNING, with repeated letters omitted (that is, we drop the second N), and the remaining rows are constructed by listing the rest of the 26 letters (those not in the keyword) in alphabetical order row-by-row, as follows: M O R N I G A B C D E F H J K L P Q S T U V W X Y Z The number of columns in the array, which we denote by ncol, is the number of distinct letters in the keyword. MORNING has six distinct letters in it, and so ncol = 6 for this array. The number of rows in the array, denoted nrow, equals ⌈26/ncol⌉, where ⌈⌉is used to indicate that we round the quotient up to the next whole number.
With the keyword MORNING, nrow = ⌈26/ncol⌉= ⌈26/6⌉= 5.
Now we create the encoding sequence by listing code letters in the array taken column-by-column. We place this sequence of code letters underneath the plain-text alphabet: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z M A H S Y O B J T Z R C K U N D L V I E P W G F Q X That is, the plain-text letters AP, BP, CP, DP, EP are encoded with the code letters that appear going down the first column of the array: MC, AC, HC, SC, YC. This encoding is called a transpose scheme because we build the array by listing the code letters in some sort of row-by-row scheme and we then transpose the array by taking the code letters column-by-column to get the encoding sequence. This process gives a fairly good scrambling of plain-text letters into code letters, and can be generated from the memory of a simple keyword.
Now we consider how we can use the structure of a keyword transpose scheme to help us decipher the message. The first step is to determine ncol, the number of columns in the array or equivalently the length of the keyword (not counting repeated letters). Let us consider how the plain-text letters are assigned to locations in the 406 Postlude keyword transpose array for keywords of length 6, length 7, and length 8. We do not know the code letters in the array, so we leave the code positions blank. However, we do know where the plain-text letters go. In each of these arrays, we insert the encoding of the letters in the English word THE that we determined in the previous section; namely, TP = FC, HP = JC, EP = YC, Table P.3: Possible Keyword Transpose Arrays Table P.3a: Array for ncol = 6 Plain text: A F K O S W Code text: Plain text: B G L P T X Code text: F Plain text: C H M Q U Y Code text: J Plain text: D I N R V Z Code text: Plain text: E J Code text: Y Table P.3b: Array for ncol = 7 Plain text: A E I M Q U X Code text: Y Plain text: B F J N R V Y Code text: Plain text: C G K O S W Z Code text: Plain text: D H L P T Code text: J F Table P.3c: Array for ncol = 8 Plain text: A E I L O R U X Code text: Y Plain text: B F J M P S V Y Code text: Plain text: D G K N Q T W Z Code text: F Plain text: A H Code text: J www.itpub.net Postlude 407 Remember that the code letters in the first row of the array spell the keyword (with repeated letters omitted) and then the succeeding rows list the non-keyword code letters in alphabetical order.
Consider the encoding TP = FC. Since F is about a quarter of the way through the alphabet, if code letter FC is not in the keyword (appearing in the first row of the array), then FC will appear about a quarter of the way through the alphabetical listing of the other code letters in the remaining rows of the array. We look at where FC is placed in the three different encoding arrays given above when we make the assignment TP = FC. For ncol = 6, TP is located towards the end of the second row— i.e., corresponding to a code letter not in the keyword that is about a quarter of the way through the alphabet. This is exactly where FC should appear if FC is not in the keyword. In the second scheme where ncol = 7, TP is located at the very end of the array and would be encoded as the last letter not in the keyword. Thus TP = FC is impossible for this array. The situation is no better when ncol = 8, for again TP is located near the end of the listing of the letters not in the keyword. We conclude, based on the encoding TP = FC, that only the keyword transpose encoding array with ncol = 6 is possible among these three arrays.
Next we perform the same analysis for the encoding of HP and of EP. JC (=HP) is about 40% of the way through the alphabet, and, since J is a very infrequent letter, JC is unlikely to appear in a keyword. Then the encoding HP = JC requires that HP appear in a position that corresponds to a code letter that is 40% of the way through the listing of non-keyword code letters. For ncol = 6, HP is well placed (see the 6-column array). For ncol = 7 and ncol = 8, HP has a location that corresponds to a letter near the end of the alphabet—impossible. So again, only for ncol = 6, does HP = JC fit the encoding array.
Finally we look at the encoding EP = YC. Y is a relatively infrequent letter that is the next-to-last letter in the alphabet. The code letters YC and ZC are both unlikely to be in the keyword, so the code letter YC will normally appear as the next to the last letter in the non-keyword code letters. For ncol = 6, EP is positioned at the next-to-letter non-keyword code letter. Exactly the right place for EP = YC. For ncol = 7 and ncol = 8, EP is assigned to the second letter in the keyword. While there are words in which Y is the second letter, such as CYPHER, the odds of Y not being in the keyword are vastly greater than Y being the second letter in the keyword.
When we look at how the encodings of each of TP = FC, HP = JC, and EP = YC fit the three different arrays, we see that the array with ncol = 6 is the only one that makes sense. We note that if we had looked at other arrays with greater or fewer columns than the above three arrays, we would have found that TP or HP would have to correspond to a code letter not in the keyword that was near the end of the alphabet—impossible given TP = FC and HP = JC. So we can be very confident that ncol = 6.
As an aside, note that if YC is the most frequent code letter in a cryptogram, we can be quite certain that EP = YC and ncol = 6. Also if a 7-letter keyword is used, then TP = ZC (see Table P.3b); furthermore, this is the only array where ZC can be the encoding of a very frequent English letter. Thus, if ZC is a very frequent code letter in a cryptogram, then TP = ZC and ncol = 7.
408 Postlude The underlying logical argument here was used many times in the graph theory part of this text, where it was called the AC Principle: Assumptions generate helpful Consequences. An example of the AC Principle is that if a graph is planar, then there are a number of properties that such a graph must have, such as e ≤3v – 6. If one of these properties does not hold, then the graph cannot be planar. Here, we tried assuming different lengths of the keyword, and then checked for each length whether the known encodings for the plain-text letters T, H, and E were consistent with the required position of their code letters in the different arrays. When the encodings did not fit an array, we could eliminate the associated keyword lengths.
Let us continue with our efforts to determine the encoding. In the previous section, we used trigraphic analysis to show that NP = RC and that PC, QC, and SC were very likely to be the three other frequent vowels AP, IP, and OP. We can now conclude that JP = ZC since JP must be encoded as the last of the non-keyword letters (since we know EP = YC). We add this information to the 6-letter keyword transpose array, obtaining Table P.3a(i).
Table P. 3a (i) Plain text: A F K O S W Code text: {P,Q,S} {P,Q,S} Plain text: B G L P T X Code text: F Plain text: C H M Q U Y Code text: J Plain text: D I N R V Z Code text: {P,Q,S} R Plain text: E J Code text: Y Z Determining the size of the keyword transpose array places tremendous con-straints on encoding possibilities. While it can be hard to determine the encoding for plain-text letters that appear in the keyword row of the array (its first row), the encoding of several frequent plain-text letters in other rows of the array is usually possible. For example, consider the plain-text vowel IP, which is the other frequent vowel (besides EP) that does not appear in the first row. Ignore for a moment that IP is a vowel and concentrate solely on its position in the array. IP appears just before NP (= RC) in the fourth row of our encoding array, so IP corresponds to a code letter that appears before, or almost before, RC in the list of non-keyword code letters. The letter just before R in the alphabet is Q and since Q is an extremely infrequent letter and unlikely to be in the keyword, it is virtually certain that QC is the code letter just before RC, in the listing of non-keyword code letters—that is, IP = QC. We confirm this by noting that QC is known to be a frequent vowel and hence is one of our three possibilities for the encoding of IP.
www.itpub.net Postlude 409 Next look at RP, the other frequent consonant (besides TP, and NP) that does not appear in the first row. RP appears just after NP = RC in the fourth row. The letter following R in the alphabet is S, but SC is known to be a vowel. The next two code letters TC and UC have too low a frequency (see Table P.1) to be RP. Next is VC, the only remaining (in alphabetical order) high-frequency letter except for YC, which is known to encode EP. So based on frequency alone, VC is the only possible encoding of RP. Moreover, VC has repeated occurrences of YC (= EP) both before and after it, matching the fact that ER and RE are both frequent English digraphs. This solidifies the assignment RP = VC.
Now note that there are two (undetermined) code-letter positions at the end of the fourth row following VC (= RP); also we know that the fifth row starts with YC (= EP). Since there are just two letters, W and X, in the alphabet between V and Y, we conclude that the two code letters at the end of the fourth row are W followed by Z; that is VP = WC and ZP = XC. So we now have determined five of the six code letters in the fourth row Fourth row of array: Plain text: D I N R V Z Code text: Q R V W X The other two frequent code letters that are known to be vowels, PC and SC, must correspond to AP and OP, which are in the keyword. Also the fact that TC and UC are missing in the fourth row means that these code letters are also in the keyword.
Low-frequency letters can also be useful in finding entries in this array. For example, QP is a very infrequent English letter and appears in the third row of our array. QP will be encoded by an infrequent code letter that occurs soon after JC, the encoding of HP, the second code letter in the third row. Looking in Table P.1 at the frequencies of letters after JC, LC stands out with zero frequency; all other code letters near LC are at least moderately frequent. Thus it is extremely likely that QP = LC. Since there is only one letter between JC and LC in the alphabet, the position in-between JC and LC in the third row must be occupied by KC. That is, MP = KC.
WenowrewritethearrayinTableP.3a(ii)withthecodewordswehavedetermined so far.
Table P. 3a (ii) Plain text: A F K O S W Code text: {P,S} {P,S} (T and U also known to be in keyword) Plain text: B G L P T X Code text: F Plain text: C H M Q U Y Code text: J K L Plain text: D I N R V Z Code text: Q R V W X Plain text: E J Code text: Y Z 410 Postlude We have determined the encodings of many plain-text letters. As shown in Table P.3a(ii), we know four of the six letters in the keyword, and so there are only two additional code letters that belong in the keyword row. Let us identify where in the remaining rows these two remaining code letters in the keyword must occur.
There are three sequences of missing code letters: (i) There are four unfilled code letter positions in the second row preceding FC, and there are five letters of the alphabet preceding F. Thus one of the code letters AC, BC, CC, DC, EC must be in the keyword.
(ii) There are two unfilled code letter positions between FC and JC, and there are three letters of the alphabet between F and J. Thus one of the code letters GC, HC, I must be in the keyword.
(iii) There are three unfilled code positions between LC and QC, and there are four letters of the alphabet between L and Q. However, we know that one of these letters, PC, must be in the keyword. Then the other three letters must occupy the three unfilled positions. That is, UP = MC, YP = NC, and DP = OC, Let us turn now to the two plain-text vowels in the first row of the array, AP and OP, that we know correspond collectively to the code letters, PC and SC. One of the ways to try to identify the encoding of AP is to look in the message for common short words that consist an A in combination with other letters that are already decoded.
Such a word is THAT, since the TP and HP are known. Another consecutive pair of common words is HAVE BEEN (currently we know the encoding of all these letters except AP and BP). In the fourth row of our message starting with the eleventh letter we find a possible (THAT)P, namely the sequence (FJSF)C. Moreover, these four code letters are followed by (FJY)C, which we know is the encoding of (THE)P. The odds of the phrase “THAT THE” in our plain-text message—i.e. AP = SC—are vastly greater than that of a word ending in “THIT” is followed by “THE”—i.e., AP = PC.
So we conclude with high confidence that AP = SC. It then follows that the other plain-text vowel in the keyword row OP must be encoded as PC. As an aside, we note there is no pair of consecutive YC’s (= EP) in this cryptogram, and hence no point in trying to find an encoding of HAVE BEEN.
We have found the encodings of the four frequent plain-text vowels and all the most frequent plain-text consonants except SP. There is just one code letter occurring more than 12 times that we have not identified—namely, IC, which occurs 21 times. It is natural to look for digraph information to confirm that SP = IC. The list of frequent English digraphs near the beginning of this Postlude contained SE, ES, and ST. And indeed in the digraph information for IC we find that YC (= EP) both precedes and follows IC several times and that FC (= TP) follows IC six times. Based on the high frequency of IC and these digraphs with IC, it is extremely likely that IC = SP. Observe also that for the two missing non-keyword code letters between FCand JC in the array, there are now only two choices of code letters: GC and HC, since IC (= SP) is in the keyword row. Thus XP = GC and CP = HC.
www.itpub.net Postlude 411 The only moderately frequent plain-text letter we have not identified now is LP, and the only unidentified code letter with moderate frequency is CC, which occurs 12 times. The location in the array is good for the match CC = LP, since this is the assignment that would result if AC, BC, or CC were all non-keyword letters, so that CC would be the third code letter in the second row of the array. Further, no other unidentified code letter occurs more than six times. However, there are no frequent English digraphs with L to look for. Based on the frequency of CC and its position in the array, the match CC = LP looks good.
There is one additional piece of information to use. Recall that we found the repeated sequence (DQCCAY)C in our message. If CC = LP, then we are forced by the missing letters in the beginning of the second row of the array to assign AC = BP and BC = CP. Then we decode (DQCCAY)C as ( ILLBE)P To help us, we note that (FJYVY)C = (THERE)P immediately precedes one of the occurrencesof(DQCCAY)C yieldingthedecodedsequenceoflettersTHERE ILLBE.
We recognize the phrase THERE WILL BE, confirming that CC = LP. We have additionally determined that DC = WP. Back in the second row of the array, the final missing code letter can be assigned (now that we know DC is in the keyword). Thus EC = PP.
We write the array with all the encodings we have identified: Table P. 3 (iii) Plain text: A F K O S W Code text: S P I D (T and U are also in keyword) Plain text: B G L P T X Code text: A B C E F G Plain text: C H M Q U Y Code text: H J K L M N Plain text: D I N R V Z Code text: O Q R V W X Plain text: E J Code text: Y Z We note that we have been able to decode all but two of the code letters without determining the keyword. We accomplished this by using the structure of the encod-ing array, frequency data for code letters and digraphs, and the repeated sequence (DQCCAY)C. The only time we looked at the cryptogram to guess a particular word involved the encoding THAT when we knew the encoding of T and H. Finishing up the decoding, we now easily deduce that the keyword is STUPID.
One does not always get a near-complete decoding of the code letters in a cryp-togram by applying the above methods, as occurred here. However, it is virtually 412 Postlude always possible to determine the encoding of THE and N. Then the number of columns in the keyword transpose array can be determined, and from the array structure, some other frequent code letters can be decoded. If further progress is difficult, one can substitute into the cryptogram the plain-text decodings of all the known code letters.
From all the resulting word fragments, the plain-text equivalents for other code letters can be deduced and then confirmed by checking that these decodings fit properly into the keyword transpose array. Alternating between using the array structure and using word fragments formed by deciphered code letters, one can normally decode any message. In the very worst case, one might have to guess the plain-text equivalent for one or two frequent code letters and then apply these proposed decodings in the cryptogram to see if they produce words and word fragments that validate the guesses.
Readers will hopefully recognize in the preceding reasoning the same sorts of step-by-step analysis, aided by insights unique to the particular problem at hand, that arose repeatedly in the graph theory and counting chapters of this text. The important difference here is that we face the additional challenge of randomness, a complication that characterizes many of the applications of combinatorial reasoning in real-world problems. Preparing readers to analyze the combinatorial situations they will face in future employment is the ultimate goal of this text.
POSTLUDE EXERCISES: The following cryptograms are all encoded using the keyword transpose method discussed above.
1. Decode this cryptogram: TDHQP HKFHZ PIYHF KNOYI ZHFAH EDLAI EXZTZ OEZQE XXAHT AXHZN YHFBN FOTXX ZWHPJ PZHOP YFNLF TOOED LTDVT YYXEK TZEND PYFNL FTOOE DLBNF TXXNZ WHFKN OYIZH FPEZE PHPZE OTZHV ZWTZE DZQNJ HTFPZ WHFHQ EXXND XJAHN DHZWN IPTDV CNAPQ NFXVQ EVHBN FKNOY IZHFY FNLFT OHFD 2. Decode this cryptogram (note: the sequence JPENFYV occurs five times): FKYVY STDDB YQYLP UHFVG FTPUH GFFKY JPLEN FYVJY UFYVU YIFSY YAFPE VPFYH FFKYV YHFVT JFTPU HPUYU VPDDL YUFHT UJPLE NFYVH JTYUJ YJPNV HYHTU FKYRG DDFKY JPLEN FYVJY UFYVS TDDBY JDPHY QQNVT UCFKY EVPFY HFHGU QJNVV YUFJP LENFY VHJTY UJYHF NQYUF HFPDQ FPVNU FKYTV EVPCV GLHGF JPLEN FYVHG FPFKY VDPUC THDGU QHJKP PDH 3. Decode this cryptogram: PCFIQ BZWOR ZEVOF ZRMFZ WMRKG CYWZW OQGIJ QEGRO SOGOF QZIOZ OTYOG MOFJO VMFRQ XHMFK JGIYZ QKGCP RSWOF YGQBO RRQGZ EJNOG CRRMK FOVZW MRWQP OSQGN YGQAX OPZWO GOSOG OLEMZ OCBOS RZEVO FZRSW QWCHO AOOFR ZEPYO VAIZW MRNOI SQGVZ GCFRY QROOF JQVMF KRJWO POKQQ VXEJN www.itpub.net Postlude 413 4. Decode this cryptogram: RPQPO LTPUG KYDVP PEPEG KYBVH DKGKY PVQVY FOCGG KHGGK YTOMA YVPEY RBYFS THDCH THVBV HDKSF CYFFG KHTPV YNOHC GPGKV YYGSM YFGKY TOMAY VPEWY VGSJY FSTGK YBVHD KMSTO FFSIS EGKYB VHDKS FASDH VGSGY GKYTO MAYVP EYRBY FSFAP OTRYR AQGUP GSMYF GKYTO MAYVP EWYVG SJYFM STOFE POV 5. Decode this cryptogram: GKXUX SLTIG FNINX TGGID ILXSA QIPYF TGGID XFUTK IYGKX RXJIR STCIA GKXJI PULXJ UQEGI CUFNL SLLPE EILXR GIBXR ITXQI PASUL GTXXR GIBPS DRGKX GUSCU FEKGF BDXFT RGKXT DIIMA IUGKX YIURG KXTXH GGUQG IASTR GKXVI YXDLB QDIIM STCAI UAUXO PXTGJ IRXDX GGXUL GKFGR ITIGF EEXFU AUXOP XTGDQ BXLSR XGKXJ IRXDX GGXUG KFGLG FTRLA IUGKX EDFST GXHGD XGGXU XTXHG GUQGI ASTRY KSJKJ IRXDX GGXUS LTCII RDPJM 6. Decode this cryptogram: JDBXB CXBXY UWXIJ DCJTX WRBII WXYJQ EBXKS VVYIB FBKQX GTJWM XCUIS FJDBM XCTDJ DBWXG JCEBD WUBRS FCVJD CJKSV VABNB XGDCX OJWOB QWOBD WKBNB XIJYO BKJIK DWEFW KDSUK BVVAB VSBNB JDCJJ DBQXG TJWMX CUIJD SIGBC XKSVV ABJDB ICUBI SVVGU BIICM BIJDC JDBDC IYIBO SFTCI JGBCX I 7. Decode this cryptogram: JUFYR QYPXI JNSIX XAMPE VNENV AHYEN QIVND ZFUEA REKNE ZPLNN YUXXU XJPWP XIJAN GNEUK NQPRZ PBZWN JPFFN DJNFN DZNMN EJIQN QZWIQ HNUEI DZWNY UQZQZ RVNDZ QWUGN ANNDG NEHAP IOZNE PRQUD VPBZN DIDZN EERYZ NUJPF FNDJN FNDZQ YNULN EQQYN JIUXN XNJZE PDIJV NGIJN QZPVN ZNJZW IVVND JUDQP BANNE SIXXA NRQNV 8. Decode this cryptogram: FKXUX JITFS TOXFI BXUXE IUFLF KHFFK XOTSF XRLFH FXLHF FIUTX QCXTX UHDYS DDHTT IOTJX FKHFF KXMIL FYHTF XRJUS MSTHD LSTFK XOTSF XRLFH FXLHU XTIYH DDAIU MXUIA ASJXU LIAAS THTJS HDSTL FSFOF SITLF KXKOT FAIUF KXLXE XIEDX YSDDB XSTFX UTHFS ITHDL STJXM HTQIA FKXMK HVXXL JHEXR AUIMF KXJIO TFUQ 9. Decode this cryptogram: ZWEUE BRFRZ RNFNB VLGVY WUEYE FUONB ZWEZE SZANN MJNCV GEGEV URFLZ WEGEV GEOEH EGVXU RBBEG EFZUE BRFRZ RNFOZ WEEFL RFEEG OCOEN FEOEZ NBZEG TOZWE TVZWE TVZRK RVFOC OEVFN ZWEGO EZVFU ZWEYW JORKR OZOCO EVFNZ WEGOE ZZWEU EBRFR ZRNFJ NCOWN CXUCO EROZW ENFEZ WVZRO EVORE OZBNG JNCGV YYXRK VZRNF 414 Postlude 10. Decode this short cryptogram (warning: E is not the most frequent letter): MYTKI JIRUL AZOAH MIJAC UYGII JIUJA CHETR JMRUY MJFAG RMRPJ FTMEX ALAZU YMRQM OAZEX OAZRU ARTRI TGELG IJHAR JITJU AVYMO YVTLV IFMPY UXIOM XIUAP A REFERENCES 1. H. Gaines, Cryptanalysis, Dover, New York, 1956.
2. R. Lewand, Cryptological Mathematics, Mathematical Association of America, Washington, D.C., 2000.
3. A. Sinkov and T. Feil, Elementary Cryptanalysis, 2nd ed., Mathematical Associ-ation of America, Washington, D.C., 2009.
www.itpub.net APPENDIX A.1 SET THEORY A set is a collection of distinct objects. In contrast to a sequence of objects, a set is unordered. Usually we refer to the objects in a set as the elements, or members, of the set. These elements may themselves be sets, as in the set of all 5-card hands; each hand is a set of five cards. A family is a collection in which multiple appearances of objects is allowed; for example, {a, a, a, b, b, c}.
A set S is a subset of set T if every element of S is also an element of T. In the caseofa5-cardhand,weareworkingwitha5-cardsubsetofthesetofall52cards.Any set is trivially a subset of itself. A proper subset is a nonempty subset with fewer ele-ments than the whole set. Two sets with no common members are called disjoint sets.
Thistextusescapitalletterstodenotesetsandlowercaseletterstodenoteelements (unless the elements are themselves sets). The number of elements in a set S is denoted by N(S) or |S|. The symbol ∈represents set membership, for example, x ∈S means that x is an element of S; and x / ∈S means that x is not a member of S. The symbol ⊆ represents subset containment—for example, T ⊆S means that T is a subset of S.
There are three ways to define a set with formal mathematical notation: 1.
By listing its elements, as in S = {x1, x2, x3, x4} or T = {{1, −1}, {2, −2}, {3, −3}, {4, −4}, {5, −5}, . . .}—implicitly, T is the set of all pairs of a positive integer and its negative 2.
By a defining property, as in P = {p | p is a person taking this course} or R = {r| there exist integers s and t with t ̸= 0 and r = s/t}—R is the set of all rational numbers 3.
As the result of some operation(s) on other sets (see below) There are two special sets we often use: the empty, or null, set, written Ø; and the universal set of all objects currently under consideration, written U.
It is important to bear in mind that in many real-world problems, sets cannot be precisely defined or enumerated. For example, the set of all ways in which a large computer program can fail is ill defined. A census of the population of the United States involves substantial error for several important subcategories of the populace, yet the official United States population was given in 1990 as 248,709,873. Beware of any calculations based on imprecise sets!
415 416 Appendix The three basic operations on sets that we will use are 1.
The intersection of S and T, S ∩T = {x ∈U | x ∈S and x ∈T } 2.
The union of S and T, S ∪T = {x ∈U | x ∈S or x ∈T } 3.
The complement of S, S = {x ∈U | x / ∈S} A fourth operation that is sometimes useful is 4.
The difference of S of minus T, S −T = {x ∈U |x ∈S and x / ∈T } Observe that S −T can be expressed in terms of the preceding operations as S −T = S ∩T .
Example 1: Selecting Calculators A sample of eight different brands of calculators consists of four machines that have rechargeable batteries and four that do not. Also, four out of the eight have memory.
We want to select four calculators to be taken apart for thorough analysis. Half of this set of four should be rechargeable and half should have memory. How many ways can such a set of four machines be chosen from the eight brands?
To answer this question, we must know how many machines are rechargeable and have memory, how many are rechargeable and have no memory, and so on.
That is, if R is the set of rechargeable machines and M the set of machines with memory, then we need information about the sizes of the intersections N(R ∩M), N(R ∩M), N(R ∩M), and N(R ∩M). Assume that there are two machines in each of these four subsets. See Figure A1.1. Then one strategy to get the desired mix of four machines would be to pick one machine from each category in Figure A1.1 (2 × 2 × 2 × 2 = 16 choices). There are two other strategies for getting the four ma-chines from Figure A1.1 (see Exercise 4 at the end of this section).
The study of set expressions involving the foregoing set operations and associated laws is called Boolean algebra. Three of the most important laws of Boolean algebra are BA1. S = S BA2. S ∩T = S ∪T BA3. S ∪T = S ∩T To visualize set expressions, we use a picture called a Venn diagram. Figure A1.2a shows a Venn diagram for the sets S and T. The whole rectangle represents Rechargeable Not rechargeable 2 2 2 2 Memory No memory Figure A1.1 www.itpub.net A.1 Set Theory 417 S T S T Figure A1.2 the universe U of all elements under consideration. The circles represent the sets S and T. In Figure A1.2b, the darkened region represents S ∩T and the striped area (S ∪T ) −(S ∩T ). Note that in particular problems, certain regions in a Venn diagram may be empty sets.
The next example illustrates how laws BA2 and BA3 can simplify a counting problem.
Example 2: Counting with Boolean Algebra Consider the universe U of all 522 outcomes obtained by picking one card from a deck of 52 playing cards, replacing that card, and then again picking a second card (possibly the same card). Suppose that we want to compute the size of the set Q of outcomes with at least one spade or at least one heart. Let S be the set of outcomes with a void in spades—that is, in which no spade is chosen on either pick. Let H be the set of outcomes with a void in hearts. Write the set Q as an expression in terms of S and H, and calculate N(Q).
The set Q equals S ∪H, the union of the set of outcomes with one or more spades (not a void in spades) and the set of outcomes with one or more hearts. The set S ∪H is the shaded area in Figure A1.3. By BA3, S ∪H = S ∩H, where S ∩H is the set of outcomes with no spades and no hearts (S ∩H is the unshaded region in Figure A1.3). That is, S ∩H is the set of outcomes where each pick is one of the 26 diamonds or clubs. Thus N(S ∩H) = 262. By Figure A1.2, N(S ∩H = N(U) −N(S ∩H) = 522 −262 = 2704 −676 = 2028, and so N(S ∪H) = N(S ∩H) = 2028.
Set theory, and, more generally, nonnumerical mathematics, were studied little until the nineteenth century. G. Peacock’s Treatise on Algebra, published in 1830, first suggested that the symbols for objects in algebra could represent nonnumeric entities. A. De Morgan discussed a similar generalization for algebraic operations S H Figure A1.3 418 Appendix a few years later. G. Boole’s Investigation of the Laws of Thought (1854) extended and formalized Peacock’s and De Morgan’s work to present a formal algebra of sets and logic. The great philosopher-mathematician Bertrand Russell has written, “Pure Mathematics was discovered by Boole.” Subsequently it was found that nu-merical mathematics also needs to be defined in terms of set theory in order to have a proper mathematical foundation (see Halmos’s Naive Set Theory). See Chap-ter 26 of Boyer’s History of Mathematics for more details about the history of set theory.
EXERCISES 1. Let A be the set of all positive integers less than 30. Let B be the set of all positive integers that end in 7 or 2. Let C be the set of all multiples of 3. List the numbers in the following sets: (a) A ∩(B ∩C) (b) A ∩(B ∪C) (c) A ∩(B ∪C) (d) A −(B ∩C) 2. If A = {1, 2, 4, 7, 8}, B = {1, 5, 7, 9}, and C = {3, 7, 8, 9} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then find set expressions (if possible) equal to (a) {7} (c) {1} (e) {4, 8, 10} (b) {6, 10} (d) {2, 7, 9} (f) {3, 5, 6, 7, 9, 10} 3. Suppose that in Example 1 we were given only the additional information that two calculators have memory but are not rechargeable. Show that we can now deduce how many machines are in each of the other three boxes in Figure A1.1.
4. What are the other two strategies in Example 1 to pick a subset of four calculators, two of which are rechargeable and two of which have memory, given the numbers in Figure A1.1? How many choices are there for each of these ways?
5. Suppose we are given a group of 20 people, 13 of whom are women. Suppose 8 of the women are married. In each of the following cases, tell whether the additional piece of information is sufficient to determine the number of married men. If it is, give this number.
(a) There are 12 people who are either married or male (or both).
(b) There are 8 people who are unmarried.
(c) There are 15 people who are female or unmarried.
6. Label each region of the Venn diagram in Figure A1.2a with the set expression it represents (e.g., the region where both sets intersect would be labeled A ∩B).
7. Draw Venn diagrams and shade the area representing the following sets: (a) A −B (b) A ∪B (c) (A ∪B) ∩(A ∩B) (d) A −(B −A) 8. Write as simple a set expression as you can for the shaded areas in the following Venn diagrams.
www.itpub.net A.1 Set Theory 419 A A B B (a) (b) 9. Create a Venn diagram for representing sets A, B, and C. The diagram should have regions for all possible intersections of A, B, and C. How many different regions are there? Label each region with the set expression it represents (e.g., the region where all three sets intersect would be labeled A ∩B ∩C).
10. Draw Venn diagrams for sets A, B, and C and shade the areas representing the sets in Exercise 1.
11. Verify with Venn diagrams the following laws of Boolean algebra: (a) (A ∪B) ∩A = A (b) (A ∩B) ∪A = A (c) A ∪A = U (d) (A ∪B) ∩C = (A ∩C) ∪(B ∩C) (e) (A ∩B) ∪C = (A ∪C) ∩(B ∪C) 12. Use Venn diagrams to determine which pairs of the following set expressions are equivalent.
(a) (A ∩B) ∪(A ∩B) (b) A −(B −(A −B)) (c) (A −B) ∩(A ∪B) (d) (A ∩(B −A)) 13. Use Law BA2 to prove that A ∩B ∩C = A ∪B ∪C [note that A ∩B ∩C = (A ∩B) ∩C = A ∩(B ∩C)].
14. Use Law BA3 to prove that A ∪B ∪C = A ∩B ∩C [note that A ∪B ∪C = (A ∪B) ∪C = A ∪(B ∪C)].
15. Use Laws BA1, BA2, and BA3 plus the identities A ∩A = A, A ∪A = A, A ∩B = B ∩A, and A ∪B = B ∪A to simplify the following set expressions: (a) (A ∪B) ∩(A ∩B) (b) A −(B ∪A) (c) (A ∪C) ∩(A ∪B) ∩(B ∪C) (d) (A ∪B) ∩[(C ∪B) −(A ∩B)] 16. Suppose two dice are rolled. How many outcomes have a 1 or a 2 showing on at least one of the dice?
17. Suppose that a card is drawn from a deck of 52 cards (and replaced before the next draw) three times. Let S be the set of outcomes where the first card is a spade, H be the set where the second card is a heart, and C be the set where the third card is a club. For each of the following sets E, write E as a set expression in terms of S, H, C, and determine N(E).
(a) Let E be the set of outcomes contained in at least one of the sets S, H, C.
420 Appendix (b) Let E be the set of outcomes contained in at least one but not all three of the sets S, H, C.
(c) Let E be the set of outcomes contained in exactly two of the sets S, H, C.
A.2 MATHEMATICAL INDUCTION The most useful, and simplest, proof technique in combinatorial mathematics and computer science is mathematical induction. Let pn denote a statement involving n objects. Then an induction proof that pn is true for all n ≥0 requires two steps: 1.
Initial step: Verify that p0 is true.
2.
Induction step: Show that if p0, p1, p2, . . . , pn−1 are true, then pn must be true.
Sometimes pn will be true only for n ≥k. Then the initial step is to verify that pk is true, and in the induction step we assume only that pk, pk + 1, . . . , pn −1 are true.
Example 1: Summation Formula Let sn denote the sum of the integers 1 through n—that is, sn = 1 + 2 + 3 + · · · + n.
Show that sn = 1 2n(n + 1).
We use mathematical induction to verify this formula. Since s0 is not defined, the initial step is to verify the formula for s1. The formula says s1 = 1 21(1 + 1) = 1— obviously correct. For the induction step, we assume that the formula is true for s1, s2, . . . , sn−1. In this problem (as in most induction problems), we need only to assume that the formula is true for sn−1.
sn−1 = 1 + 2 + 3 + · · · + (n −1) = 1 2(n −1)[(n −1) + 1] = 1 2(n −1)n We now use this expression for sn−1 to prove that the formula is true for sn.
sn = [1 + 2 + · · · + (n −1)] + n = sn−1 + n = 1 2(n −1)n + n = 1 2[(n −1)n + 2n] = 1 2n(n + 1) This completes the induction proof that sn = 1 2n(n + 1) for all positive n.
Example 2: Population Growth Model Suppose that the population of a colony of ants doubles in each successive year. A colony is established with an initial population of 10 ants. How many ants will this colony have after n years?
Let an denote the number of ants in the colony after n years. We are given that a0 = 10. Since the colony’s population doubles annually, then a1 = 20, a2 = 40, and a3 = 80. Looking at the first few values of an, we are led to conjecture that www.itpub.net A.2 Mathematical Induction 421 an = 2n × 10. For the initial step, we check that a0 = 10 = 20 × 10, as required. For the induction step, we assume that an−1 = 2n−1 × 10. Now we use the annual doubling property of the colony: an = 2an−1 = 2(2n−1 × 10) = 2n × 10 Example 3: Prime Factorization A prime number is an integer p > 1 that is divisible by no other positive integer besides 1 and itself. Show that any integer n > 1 can be written as a product of prime numbers.
We prove this fact by mathematical induction. Since the assertion concerns in-tegers n > 1, the initial step is to verify that 2 can be written as a product of prime numbers. But 2 is itself a prime. Thus 2 is trivially the product of primes (i.e., actually of a single prime), itself. Next assume that the numbers 2, 3, . . . , n −1 can be written as a product of primes, and use this assumption to prove that n can also be written as a product of primes. If n is itself a prime, then as in the case of 2, there is nothing more to do. Suppose n is not a prime, and so there is an integer m that divides n and for some integer k, n = km. Since k and m must be less than n, they can each be written as a product of primes. Multiplying these two prime products for k and m together yields the desired representation of n as a product of primes.
Although the Greeks used certain iterative arguments in geometric calculations and the principle of reductio ad absurdum, mathematical induction was first used explicitly by Maurolycus around 1550. Pascal used an induction argument in 1654 to verify the additive property of binomial coefficients in the array now known as Pascal’s triangle [this property is identity (3) in Section 5.5]. The actual term induction was coined by De Morgan 200 years later. See Bussey, “Origin of mathematical induction,” American Math. Monthly, 1917, for a fuller discussion of the history of mathematical induction.
EXERCISES 1. Prove by induction that 1 + 3 + · · · + (2n + 1) = (n + 1)2.
2. Prove by induction that 12 + 22 + · · · + n2 = n(n + 1)(2n + 1) 6 .
3. Prove by induction that −12 + 22 −32 + · · · + (−1)nn2 = (−1)n n(n + 1) 2 .
4. Prove by induction that (1 × 2) + (2 × 3) + · · · + n(n + 1) = n(n + 1)(n + 2) 3 .
5. Prove by induction that 13 + 23 + · · · + n3 = n2(n + 1)2 4 .
422 Appendix 6. Prove by induction that (1 + 2 + · · · + n)2 = 13 + 23 + · · · + n3 (assume the result in Example 1).
7. Prove by induction that 1 1 × 2 + 1 2 × 3 + · · · + 1 n × (n + 1) = n n + 1.
8. Prove by induction that 1 1 × 4 + 1 4 × 7 + 1 7 × 10 + · · · + 1 (3n −2) × (3n + 1) = n 3n + 1.
9. Prove by induction that (1 × 1!) + (2 × 2!) + · · · + (n × n!) = (n + 1)! −1.
10. Prove by induction that for a ̸= 1, 1 −an+1 1 −a = 1 + a + a2 + · · · + an.
11. Prove by induction that 12 + 32 + 52 + · · · + (2n −1)2 = n(2n −1)(2n + 1) 3 .
12. If the number an of calf births at Dr. Smith’s farm after n years obeys the law an = 3an−1 −2an−2 and in the first two years a1 = 3 and a2 = 7, then prove by induction that an = 2n+1 −1.
13. Prove that any positive integer has a unique factorization into primes. You may assume the result of Example 3; you need to use the fact that if a prime p divides a product of positive integers, then p divides one of the integers.
14. Write a computer program to find and print the prime factorizations of the first 50 integers.
15. Prove by induction that for any integer m > 0, m × n = n × m. By r × s, we mean the sum of r copies of s.
16. Prove by induction that for integers n ≥5, 2n > n2.
17. Prove that the number of different subsets (including the null set and full set) of a set of n objects is 2n.
18. Prove by induction that A1 ∩A2 ∩. . . ∩An = A1 ∪A2 ∪ ∪An (see Exercise 13 in Section A.1).
19. Prove by induction that if n distinct dice are rolled, the number of outcomes where the sum of the faces is an even integer equals the number of outcomes with an odd sum.
20. Prove by induction that the number of n-digit binary sequences with an even num-ber of 1s equals the number of n-digit binary sequences with an odd number of 1s.
21. Prove by induction that the sum of the cubes of three successive positive integers is divisible by 9.
22. The cat population in a dormitory has the property that the number of cats in one year is equal to the sum of the number of cats in the two previous years. If in the first year there was one cat and if in the second year there were two cats, then prove by induction that the number of cats in the nth year is equal to 1 √ 5 1 2 + 1 2 √ 5 n+1 − 1 2 −1 2 √ 5 n+1 www.itpub.net A.3 A Little Probability 423 23. Why cannot one prove by induction that the number of binary sequences of all finite lengths is finite?
24. What is wrong with the following induction proof that all elements x1, . . . , xn in a set of n elements are equal?
(a) Initial step (n = 1): The set has one element x1 equal to itself.
(b) Induction step: Assume x1 = x2 = · · · = xn−1. Since also xn−1 = xn by induction assumption (when xn−1, xn are considered alone as a two-element set), then x1 = x2 = · · · = xn−1 = xn.
25. What is wrong with the following induction proof that for a ̸= 0, an = 1?
(a) Initial step (n = 0): a0 = 1—always true.
(b) Induction step: Assume an−1 = 1 and now an = an−1an−1 an−2 = 1.
A.3 A LITTLE PROBABILITY Historically, counting problems have been closely associated with probability. Indeed, any problem of the form “how many hobbits are there who . . . ” has the closely related form “what fraction of all hobbits . . . ,” which in turn can be posed probabilistically as “what is the probability that a randomly chosen hobbit . . . ?” The famous Pascal’s tri-angle for binomial coefficients was developed by Pascal around 1650 in the process of analyzing some gambling probabilities. The probability of getting at least seven heads on 10 flips of a fair coin, the probability of being dealt a five-card poker hand (from a well-shuffled deck) with a pair or better, and the probability of finding a faulty calcu-lator in a sample of 20 machines if 5 percent of the machines from which the sample was drawn are faulty—all these probabilities are essentially counting problems.
Two hundred years ago, the French mathematician Laplace first defined proba-bility as follows: Probability = number of favorable cases total number of cases This definition corresponds to the “person in the street’s” intuitive idea of what prob-ability is. In this text, we treat probability problems only where Laplace’s definition of probability applies. Implicit in this definition is the assumption that each case is equally likely. If the total number of cases is infinite, for example, all real numbers between 0 and 1, then we would not be able to use Laplace’s definition. A more subtle difficulty is that in some probability problems, “cases” have to be carefully defined or else they may not all be equally likely, as, for example, the possible numbers of heads observed when a coin is flipped 10 times. To clarify this difficulty, we need to introduce a little of the basic terminology of probability theory.
An experiment is a clearly defined procedure that produces one of a given set of outcomes. These outcomes are called elementary events and the set of all 424 Appendix elementary events is called the sample space of the experiment. We are interested only in experiments where the elementary events are equally likely. An event that is a subset of several elementary events is called a compound event.
For example, when a single die is rolled, then obtaining a specific number, such as 5, is an elementary event, whereas obtaining an even number is a compound event.
If S is the sample space of the experiment and E is an event in S, then Laplace’s definition of probability says that the probability of event E, prob(E), is prob(E) = N(E) N(S) In most instances, the size of S is easily determined, and so the problem of deter-mining prob(E) reduces to counting the number of outcomes (elementary events) in eventE.Returningtothedieroll,thesamplespaceofoutcomesis S = {1, 2, 3, 4, 5, 6}.
If the event E is obtaining an even number, then E = {2, 4, 6}, and prob(E) = 3 6 = 1 2.
Many experiments we discuss involve a repeated (or simultaneous) series of sim-ple experiments. Each round of the simple experiment is called a trial. For example, the experiment of flipping a coin three times involves three successive trials of the simple experiment of flipping a coin. Rolling two dice and recording the sum of the two values rolled involves the two simultaneous simple experiments of rolling a single die. In any experiment involving multiple trials, the elementary events are the sequences of outcomes of the simple experiments. If the simple experiments are performed simultaneously, we number the simple experiments and list their outcomes in order of the experiments’ numbers.
The sample space of elementary events for the experiment of flipping a coin three times is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT } (1) The sample space for the experiment of rolling two dice is S = {(i, j) | 1 ≤i ≤6, 1 ≤j ≤6} Let us consider more closely the experiment of rolling two dice and recording the sum of the two values on each die (the sample space is the set S listed above).
In terms of our formal definition of events, the sum of the two values equaling k is not an elementary event but rather a compound event. The event, the sum of the two dice equals 7, is the subset of elementary events S1 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} Thus, prob(sum = 7) = N(S7)/N(S) = 6/36. Similarly, S2 = {(1, 1)} and so prob (sum = 2) = N(S2)/N(S) = 1/36. Observe that a sum of 7 is six times as likely as a sum of 2. If we had considered the possible sums of the two dice as the elemen-tary events, we would have had a sample space S∗= {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} and each sum would have had probability 1/N(S∗) = 1/11—clearly a mistake. The same type of error would have been made if we were to regard the number of heads as the elementary events in the experiment of flipping three coins [see the correct sample space listed above in (1)].
www.itpub.net A.3 A Little Probability 425 Suppose we have a box containing five identical red balls and 20 identical black balls. Our experiment is to draw a ball. What is the sample space of elementary events?
All we can record as an outcome is the color of the ball, leading to the sample space S = {red, black}. But then by Laplace’s definition of probability, picking a red ball will have probability 1 2. Although this experiment consists of just a single trial, we are clearly making the same sort of mistake that occurs when the sums of dice are treated as elementary events. The reader is encouraged to pause a moment and try to supply his or her own explanation for resolving this red–black ball paradox before reading the next paragraph.
The logical solution is to consider not what color ball we withdraw from the box but where in the box we direct our hand to grasp a ball. There are 25 different spatial positions where balls are located in the box. So, in this sense, there are 25 different events. We resolve this complication about “identical” objects with the following rule.
Identical Objects Rule In probability problems, there are no collections of identical objects; all objects are distinguishable.
In complex experiments one should always take a moment to be sure that the elementary events are properly identified.
The original motivation for studying probability was the same as used in this book, games of chance—rolling dice, flipping coins, and card games. Cardano cal-culated the odds for different sums of two dice in the 1500s. Around 1600, Galileo calculated similar odds for three dice. A series of letters about gambling probabilities between Pascal and Fermat, written around 1650, constitute the real beginning of probability theory. Jacques Bernoulli’s Ars Conjectandi (1713) was the first system-atic treatment of probability (and associated combinatorial methods). One hundred years later, Laplace published his epic treatise Th´ eorie Analytique des Probabiliti´ es, a work containing both the definition of probability in a finite sample space used in this book and also advanced calculus–based derivations of modern probability theory.
See F. N. David, Games, Gods, and Gaming: A History of Probability (Dover, 1998) for an excellent history of probability theory.
EXERCISES 1. Anintegerbetween5and12inclusiveischosenatrandom.Whatistheprobability that it is even?
2. In the experiment of flipping a coin three times, let Ek be the compound event that the number of heads equals k. Determine prob(Ek), for k = 0, 1, 2, 3.
3. In the experiment of rolling two distinct dice, find the probability of the following events: (a) Both dice show the same value.
426 Appendix (b) The sum of the dice is even.
(c) The sum of the dice is the square of one of the die’s value.
4. A die is rolled three times. What is the probability of having a 5 appear at least two times?
5. Find the probability that in a randomly chosen arrangement of the letters h, a, t, the following occurs: (a) The letters are in alphabetical order.
(b) The letter h occurs somewhere after the letter t in the arrangement.
6. Find the probability that in a randomly chosen (unordered) subset of two numbers from the set 1, 2, 3, 4, 5 the following occurs: (a) The subset is {1, 2}.
(b) 1 is not in the subset.
7. Anurncontainssixredballsandthreeblackballs.Ifaballischosen,thenreturned, and a second ball chosen, what is the probability that one of the following is true?
(a) Both balls are black.
(b) One ball is black and one is red.
8. An urn contains two black balls and three red balls. If two different balls are successively removed, what is the probability that both balls are of the same color?
9. Two boys and two girls are lined up randomly in a row. What is the probability that the girls and boys alternate?
10. Five distinct dice are rolled. What is the probability of getting at least one 6?
11. If a school has 100 students of whom 50 take French, 40 take Spanish, and 20 take French and Spanish, then what is the probability that a randomly chosen student takes no language?
12. What is the probability that an integer between 1 and 50 inclusive is divisible by 3 or 4?
13. There are three urns each with two balls: One urn has two black balls, the second has two red balls, and the third a black and a red ball. If an urn is randomly chosen and a randomly chosen ball in that urn is red, what is the probability that the other ball in this urn is also red? (Hint: Let the sample space be the set of all experimental outcomes where a red ball is chosen first.) 14. What should the sample space be for the following problem: An urn has 10 black balls and five red balls; four balls are removed but not seen (our eyes are shut) and then a fifth is removed and observed; now what is the probability that the fifth ball is red?
15. What should the elementary events be in each of the following experiments so that they are equally likely?
(a) A fair coin is tossed until two heads occur or until the coin is flipped 10 times.
www.itpub.net A.4 The Pigeonhole Principle 427 (b) A positive integer is chosen at random.
(c) Two positive integers are chosen at random and their difference is recorded.
(d) A ball is randomly chosen from an urn with two red and two black balls and this ball is replaced and a ball chosen again, if necessary, until a red ball is obtained.
A.4 THE PIGEONHOLE PRINCIPLE The pigeonhole principle is one of the most simple-minded ideas imaginable, and yet its generalizations involve some of the most profound and difficult results in all of combinatorial theory. This topic is included in these appendices because it does not fit naturally into any chapter of this book. More mathematical texts on combinatorics devote a whole chapter to Ramsey theory—for example, see Brualdi’s Introductory Combinatorics, 4th ed. (Prentice Hall, 2004). For a thorough treatment of Ramsey theory, see Graham, Rothschild, and Spencer, Ramsey Theory (John Wiley, 1990).
Pigeonhole Principle If there are more pigeons than pigeonholes, then some pigeonhole must contain two or more pigeons. More generally, if there are more than k times as many pigeons as pigeonholes, then some pigeonhole must contain at least k + 1 pigeons.
This principle is also called the Dirichlet drawer principle. An application of it is the observation that two people in New York City must have the same number of hairs on their heads. New York City has over 7,300,000 people, and the average scalp contains 100,000 hairs. Indeed, the pigeonhole principle allows us to assert that it is theoretically possible to fill a subway car with about 73 New Yorkers, all of whom have the same number of hairs on their head.
The following three problems suggest some of the diverse generalizations of the pigeonhole principle.
1.
How large a set of distinct numbers between 1 and 200 is needed to assure that two numbers in the set have a common divisor?
2.
How large a set of distinct numbers between 1 and n is needed to assure that the set contains a subset of five equally spaced numbers a1, a2, a3, a4, a5—that is, a2 −a1 = a3 −a2 = a4 −a3 = a5 −a4?
3.
Given a positive integer k, how large a group of people is needed to ensure that either there exists a subset of k people in the group who all know each other or there exists a subset of k people none of whom know each other?
In Problem 3, it might be possible that for large values of k, there are groups of arbitrarily large size that do not meet the k mutual friends or k mutual strangers property. Such is not the case. One of the theorems of Ramsey theory is that there 428 Appendix exists a finite number rk such that any group with at least rk people must satisfy this property. The theorem does not say what number rk is. Actually, rk has been determined for only a few small values of k.
Most generalizations of the pigeonhole principle require special, research-level skills rather than the combinatorial problem-solving logic and techniques that this text seeks to develop. However, we include the basic pigeonhole principle in this text because it can occasionally be very helpful. The following examples illustrate an obvious and two not-so-obvious uses of the principle.
Example 1: Pooling Responses There are 20 small towns in a region of west Texas. We want to get three people from one of these towns to help us with a survey of their town. If we go to any particular town and advertise for helpers, we know from past experience that the chances of getting three respondents are poor. Instead, we advertise in a regional newspaper that reaches all 20 towns. How many responses to our ad do we need to assure that the set of respondents will contain three people from the same town?
By the pigeonhole principle, we need more than 2 × 20 = 40 responses.
Example 2: Connecting Computers with Printers We have 15 minicomputers and 10 printers. Every five minutes, some subset of the computers requests printers. How many different connections between various computers and printers are necessary to guarantee that if 10 (or fewer) computers want a printer, there will always be connections to permit each of these computers to use a different printer?
Using a variant of the pigeonhole principle, we see that there must be at least 60 connections. Otherwise one of the 10 printers, call it printer A, would be connected to five (or fewer) computers and if none of the five computers connected to A were in the subset of 10 computers seeking printers, then printer A could not be used by any of these 10 computers. It happens that exactly 60 connections, if properly made, will solve the connection problem (see Exercise 15).
Example 3: Subsets Summing to 4 Show that any collection of 8 positive integers whose sum is 20 has a subset summing to 4.
We show that the collection must have one of the following four subsets with a sum of 4: (a) four 1s; (b) two 2s; (c) two 1s and a 2; (d) a 1 and a 3. That one of these possibilities must hold is a pigeonhole-type argument.
If S contained no 1 or 2, then its sum would be at least 3 × 8 = 24. So by the pigeonhole principle, S must contain a 1 or a 2.
Suppose S contains a 2. We are finished if there is a second 2 or if there are also two 1s. Possibly there is one 1, but all other integers are at least 3. If there is no 1, the other 7 integers, each at least 3, must sum to 20 −2 = 18—impossible by www.itpub.net A.4 The Pigeonhole Principle 429 a pigeonhole principle–type argument. If there is one 1, the other 6 integers, each at least 3, must sum to 20 −2 −1 = 17—again impossible.
Suppose S contains at least one 1 but no 2. If there is a 3 in S, case (d) applies. If there are four 1s in S, case (a) applies. The alternative is no 2s and no 3s in S and at most three 1s. Then there are at least five other integers in S of size at least 4 whose sum must be less than 20—impossible.
EXERCISES 1. Given a group of n women and their husbands, how many people must be chosen from this group of 2n people to guarantee the set contains a married couple?
2. Show that at a party of 20 people, there are two people who have the same number of friends.
3. In a round-robin tournament, show that there must be two players with the same number of wins if no player loses all matches.
4. Given 10 French books, 20 Spanish books, 8 German books, 15 Russian books, and 25 Italian books, how many books must be chosen to guarantee that there are 12 books of the same language?
5. If there are 48 different pairs of people who know each other at a party of 20 people, then show that some person has four or fewer acquaintances.
6. A professor tells three jokes in her ethics course each year. How large a set of jokes does the professor need in order never to repeat the exact same triple of jokes over a period of 12 years?
7. Show that given any set of seven distinct integers, there must exist two integers in this set whose sum or difference is a multiple of 10.
8. Show that if n + 1 distinct numbers are chosen from 1, 2, . . . , 2n, then two of the numbers must always be consecutive integers.
9. Suppose the numbers 1 through 10 are randomly positioned around a circle. Show that the sum of some set of three consecutive numbers must be at least 17.
10. A computer is used for 99 hours over a period of 12 days, an integral number of hours each day. Show that on some pair of two consecutive days, the computer was used for at least 17 hours.
11. Show that any subset of eight distinct integers between 1 and 14 contains a pair of integers k, l such that k divides l.
12. Show that in any set of n integers, n ≥3, there always exists a pair of integers whose difference is divisible by n −1.
13. Show that any subset of n + 1 distinct integers between 2 and 2n (n ≥2) always contains a pair of integers with no common divisor.
14. Show that any set of 16 positive integers (not all distinct) summing to 30 has a subset summing to n, for n = 1, 2, . . . , 29.
430 Appendix 15. In Example 2, find the required set of 60 computer–printer connections.
16. There used to be six computers and 10 printers in a large computing center. Each computer was connected to some subset of printers. Now the 10 old printers are being replaced by six more reliable printers, but temporarily the computers will still be allowed to believe that there are 10 printers. Four dummy printers will pass on computer requests to the six other real printers. How many connections between the four dummy printers and the six real printers are needed to handle any set of six printing requests?
17. Show that for any set S of 10 distinct numbers between 1 and 60, there always exist two disjoint subsets of S (not necessarily using all the numbers in S) both of whose numbers have the same sum.
18. Two circular disks each have 10 0s and 10 1s spaced equally around their edges in different orders. Show that the disks can always be superimposed on top of each other so that at least 10 positions have the same digit.
19. A student will study basketweaving for at least an hour a day for seven weeks, but never more than 11 hours in any one week. Show that there is some period of successive days during which the student studies a total of exactly 20 hours.
20. If G is an n-vertex graph in which each vertex has degree ≥(n −1)/2, show that G is connected, i.e., there exists a path joining every pair of vertices in G.
21. Show that any sequence of n2 + 1 distinct numbers contains an increasing sub-sequence of n + 1 numbers or a decreasing subsequence of n + 1 numbers.
22. Show that at any party with at least six people, there exists either a set of three mutual friends or a set of three mutual strangers.
A.5 COMPUTATIONAL COMPLEXITY AND NP-COMPLETENESS This section gives a summary of the basic ideas of computational complexity. Not surprisingly, a major concern of computer science is the speed of a computation, whether it be a numerical algorithm to plot the trajectory of a spacecraft or an opera-tions research procedure to schedule airline crews or an information retrieval search in a database. For some computing problems, such as sorting (in alphabetical order) a list of n words, fast procedures are known. For other problems, such as the Traveling Salesperson Problem (determining the shortest Hamilton circuit in an n-vertex graph where each edge has a length), the calculation can take an incredibly long time for moderate sizes of n, e. g., n = 100.
The standard way to describe the computational complexity, or speed, of a procedure is with a function f (n) whose value is the maximum number of computer steps (e.g., additions, multiplications, comparisons) required to execute the procedure when n is the “size” of the input data. The size could be the number of vertices in a graph or the number of items to be sorted. About the best possible function would be www.itpub.net A.5 Computational Complexity and NP-completeness 431 a linear function. As the size of the input grows, the computational time would grow at a proportional rate.
An even faster function is the logarithmic function. With the proper search tree, one can identify an unknown word in a dictionary of n words in log2n steps (compar-isons), as discussed in Section 3.1. The key is that one repeatedly compares whether the unknown word is before or after a sequence of test words, so that each comparison cuts the possible answers in half. The possible answers are leaves in a binary tree, and the internal vertices are the test words. The complexity of the search is the height of the tree. A balanced binary tree with n leaves has height log2n.
A slightly slower function than linear would be of the form Anlog2n, for some constant A. This was the number of comparisons of the Heap Sort algorithm for sorting n items, presented in Section 3.4. There are n! possible orderings of a list of n items. To choose among n! possible answers in only Anlog2n comparisons requires an algorithm, like the log2n recognition algorithm for an unknown word, that repeatedly splits the set of possible orderings in half. Note that log2(n!) is bounded by n log2n.
Less efficient algorithms have a computational complexity that is a polynomial function of n. Recall that a quadratic function grows by a factor of 4 when n doubles, a cubic grows by a factor of 8, and an kth order function grows by a factor of 2k when n doubles. With the speed of modern computers, especially parallel supercomputers, algorithms with complexity n4 or n5 can solve moderately large problems—e.g., n = 500. Recall that computational complexity functions look at the worst-case behavior. Algorithms that have a worst-case complexity of n4 might be much faster for typical problems.
A much worse type of computational complexity is an exponential function such as 2n or n!. For a complexity of n!, simply increasing n by 1, say, from 49 to 50, increases the computation effort by a factor of 50. All known algorithms for the Traveling Salesperson Problem have exponential complexity. Computer science the-oreticians try to determine lower bounds on the computational complexity of solving a certain problem. Of particular interest is the question of whether polynomial algo-rithms can exist for very hard problems such as the Traveling Salesperson Problem.
Cook developed a useful framework for discussing hard problems. He took as his starting hard problem the satisfiability problem, namely, determining whether a log-ical proposition in n variables is valid. No faster algorithm is known than evaluating a logical expression for all possible 2n values of True or False for each variable (see Section A.1 for background on logical propositions). Then Cook looked at modeling the satisfiability problem in terms of other hard problems, and conversely modeling other hard problems as a satisfiability problem.
A problem is called NP-complete if the problem can model the satisfiability problem and can be modeled by it. The modeling effort is required to take polynomial time. These models are usually very complicated. A comparatively simple example of this modeling process is presented in Example 1, below which shows how to model the problem of deciding whether a graph is vertex 3-colorable as a satisfiability problem. For a more complete treatment of NP-completeness models, see any analysis of algorithms text. We note that the formal definition of NP-completeness is more complicated, but our definition here is equivalent.
432 Appendix To be NP-complete, a problem has to be posed as a question with a yes or no answer. For example, the question of whether an arbitrary n-vertex graph can be colored with three colors is NP-complete, as is the question of whether a graph can be colored with any specific number of colors greater than two. To find the chromatic number of an n-vertex graph (the minimum number of colors), one does a binary search by asking whether the graph is n/2-colorable; if so, whether it is n/4 colorable; if not, whether it is 3n/8 colorable, etc.—homing in on the smallest number of colors that can color the graph. This strategy requires log2n coloring questions.
Observe that any NP-complete problem can model any other NP-complete prob-lem, via the intermediate model as a satisfiability problem. The consequence of this equivalence is that if an algorithm of polynomial complexity is found for solving any one of these NP-complete problems, then it can be used to provide a polynomial algorithm for all NP-complete problems.
It is generally believed that no polynomial algorithm will ever be found for an NP-complete problem. Thousands of very hard problems have been shown to be NP-complete. Among the graph problems known to be NP-complete are the existence of a Hamilton circuit or Hamilton path, graph colorability, and the size of the largest com-plete subgraph. It is still not known if determining whether two graphs are isomorphic is an NP-complete problem.
Example 1: Modeling Graph 3-Coloring as a Satisfiability Problem Reformulate the problem of deciding whether an n-vertex graph G can be 3-colored as a satisfiability problem. We will create a collection of simple disjunctions (proposi-tions using just logical “or”). The satisfiability problem will be determining whether there is some choice of the values (True or False) of the logical variables that makes all the disjunctions true. The number of values needs to be a polynomial function of n. In this case, we will use 3n variables.
The choices we have in graph coloring are which color the ith vertex should be as follows: is it color 1 or color 2 or color 3? We define three logical variables for each of these possibilities: xi,1, xi,2, xi,3 with xi, j = True if the ith vertex has color j and = False otherwise.
First we develop disjunctions to represent the constraint that each vertex has exactly one color, that is, exactly one of xi,1, xi,2, xi,3 is true. The following compound propositions express this constraint.
(xi,1 ∧¯ xi,2 ∧¯ xi,3) ∨(¯ xi,1 ∧xi,2 ∧¯ xi,3) ∨(¯ xi,1 ∧¯ xi,2 ∧xi,3), i = 1, 2, . . . , n By repeatedly applying the distributive law for disjunction and eliminating redundant terms, we obtain the following a set of disjunctions that are equivalent to the previous compound propositions.
xi,1 ∨xi,2 ∨xi,3, ¯ xi,1 ∨¯ xi,2, ¯ xi,1 ∨¯ xi,3, ¯ xi,2 ∨¯ xi,3, i = 1, 2, . . . , n (1) www.itpub.net A.5 Computational Complexity and NP-completeness 433 Next we need to express the coloring constraint that for each edge (νi, ν j), of G, νi and ν j must not have the same color—that is, at least one of νi and ν j does not have color k.
¯ xi,k ∨¯ x j,k, k = 1, 2, 3, (νi, ν j) is an edge in G, (2) Satisfying the collection of simple disjunctive propositions in (1) and (2) is the desired propositional formulation of the question of whether G can be 3-colored.
This page is intentionally left blank www.itpub.net GLOSSARY OF COUNTING AND GRAPH THEORY TERMS Arrangement: An arrangement is a sequence or ordered list of objects. An r-arrangement is an arrangement with r objects. An arrangement may or may not allow repetition of objects. There are P(n,r)r-arrangements without rep-etition of r objects chosen from n objects. There are nr r-arrangements with repetition of r objects chosen from n types of objects.
Binomial coefficient: A binomial coefficient is a coefficient in the polynomial expansion of a binomial expression such as (a + x)n. The coefficient of xr in (1 + x)n is written C(n,r) or n r . This coefficient equals the number of distinct r-subsets of an n-set.
Bipartite graph: A bipartite graph G = (X, Y, E) is a graph whose vertices are partitioned into two vertex sets, X and Y, and every edge in G joins a vertex in X with a vertex in Y.
Chromatic number: The chromatic number of a graph is the smallest number of colors that can be used in a coloring of a graph. See Coloring a graph.
Chromatic polynomial: A polynomial Pk(G) that tells how many ways there are to k-color the vertices of a graph G.
Circuit: A circuit is a sequence of vertices (x1, x2, x3, . . . , xn) where x1 = xn, and xi is adjacent to xi+1. A vertex may not appear more than once in a circuit (except for the same vertex as the starting and ending vertex).
Coloring a graph: A coloring of a graph G assigns a color to each vertex so that adjacent vertices have different colors. One can also color edges so that edges with a common end/vertex have different colors.
Combination: A combination is a subset of objects or, equivalently, an unordered collection of objects. Objects may or may not be repeated in a combination. There are C(n,r) different combinations without repetition of r objects chosen from n objects. There are C(n +r −1,r) different combinations with repetition of r objects chosen from n types of objects.
Complete graph Kn and complete bipartite graph Km,n: Kn is a graph on n vertices with an edge joining every pair of vertices. K3 is a triangle. K2 is an edge. Km,n is a bipartite graph with m and n vertices in its two vertex sets and all possible edges between vertices in the two sets.
435 436 Glossary of Counting and Graph Theory Terms Complementary graph or complement: The complementary graph G = (V, E) of a graph G = (V, E) has the same vertex set V as G does. A pair of vertices are joined by an edge in G if and only if they are not joined by an edge in G.
Component: An unconnected graph G consists of a collection of components or “connected pieces.” A connected graph consists of a single component. Formally, a component of G consists of some particular vertex x and all vertices reachable from x by a path in G.
Connected graph: A graph is connected if there is a path joining any given pair of vertices. A directed graph is connected if it is connected when treated as an undirected graph (with all edge directions ignored).
Cycle: A cycle is a sequence of consecutively linked edges whose starting vertex is the ending vertex and in which no edge can appear more than once. Unlike a circuit, a vertex can be visited any number of times in a cycle.
Derangement: A derangement of a given arrangement of distinct objects is a re-arrangement such that no object is in the same position it had in the original arrangement.
Directed graph, directed edge: A graph is a directed graph if each edge (a, ⃗b) is directed, going from a to b. A directed graph may contain two oppositely directed edges joining two vertices, such as (a, ⃗b) and (b, ⃗a).
Distribution: A distribution is an assignment of a given set of objects, which may be identical or distinct, to a set of distinct destinations. Unless explicitly prohibited, more than one object may go to the same destination.
Edge cover: An edge cover is a set S of vertices such that every edge in any graph is incident to one vertex in S.
Euler cycle, Euler trail: An Euler cycle (trail) is a cycle (trail) that contains all the edges in a graph. Furthermore, it must visit each vertex at least once.
Generating function: A generation function g(x) for an, the number of ways to do some procedure with n objects, is a polynomial or power series with the expansion g(x) = a0 + a1x + a2x2 + · · · + anxn + · · ·. Such a function is also called an ordinary generating function in contrast to an exponential generating function g(x), which has the form g(x) = a0 + a1 x 1! + a2 x2 2! + · · · + an xn n! + · · ·.
Graph: A graph G = (V, E) consists of a finite set V of vertices and a finite set E of edges. Each edge e = (a, b) joins two different vertices a, b (a ̸= b). Also, two edges cannot join the same pair of vertices. Unless G is a directed graph (see Directed graph), (a, b) and (b, a) are the same edge (order does not matter).
Hamilton circuit, Hamilton path: A Hamilton circuit (path) is a circuit (path) that contains every vertex of a graph.
Independent set: A set of vertices in a graph is independent if no pair of them are adjacent.
Isomorphism, isomorphic graphs: Two graphs G1 and G2 are isomorphic if there exists a matching, called an isomorphism, of vertices in G1 with the vertices in www.itpub.net Glossary of Counting and Graph Theory Terms 437 G2 so that two vertices in G1 are adjacent if and only if the corresponding two vertices in G2 are adjacent. Informally, two graphs are isomorphic if they are “the same graph” except that their vertices have different names.
Map coloring: A map of countries is properly colored by assigning a color to each country so that countries with a common border get different colors.
Matching: A matching in a bipartite graph G = (X, Y, E) is a subset of edges that pair off, in a one-to-one fashion, some vertices in X with some vertices in Y.
Multigraph: A multigraph is a generalized graph in which (1) multiple edges are allowed—two or more edges can join the same two vertices; and (2) loops are allowed—edges of the form (a, a).
Network: A network is a graph, usually a directed graph, with a positive integer k(e) assigned to each edge e of the graph.
Network flow: A flow is a function on the edges of a network that satisfies certain constraints listed at the start of Section 4.3.
Partition: A partition of a collection of identical objects divides the objects into a collection of groups of various sizes. One can also speak of a partition of an integer n as a collection of positive integers that sum to n.
Path: A path is a sequence of vertices (x1, x2, x3, . . . , xn) such that xi is adjacent to xi+1. A vertex may not appear more than once in a path, except possibly x1 = xn.
Permutation: A permutation of a set or sequence of objects is an arrangement of the set or sequence, normally with no repetition allowed. An r-permutation is an r-arrangement of r objects chosen from the set or sequence.
Planar graph: A graph is planar if there exists a way to draw it on a sheet of paper so that no edges cross. A plane graph is a planar graph drawn so that no edges cross.
Recurrence relation: A recurrence relation is an equation such as an = 2an−1 + 3an−2, in which an, the number of ways to do some procedure with k objects, is expressed in terms of other ak’s, where k < n.
Selection: A selection is an unordered collection of objects. Objects may or may not be repeated in a selection. There are C(n, k) different selections without repetition of k objects chosen from n objects. There are C(n + k −1, k) different selections with repetition of k objects chosen from n types of objects.
Subgraph: A subgraph is a graph that is contained in another graph. If G′ = (V ′, E′) is a subgraph of G = (V, E), then V ′ ⊆V and E′ ⊆E.
Trail: A trail is a sequence of consecutively linked edges in which no edge can appear more than once. Unlike a path, a vertex can be visited any number of times in a trail.
Tree: A tree is a graph with a special vertex called a root and for each vertex x, other than the root, there is a unique path from the root to x. An undirected tree is characterized as a connected graph with no circuits. For tree-related terms, see the following Subglossary of Tree Terminology.
438 Glossary of Counting and Graph Theory Terms Subglossary of Tree Terminology Ancestors: Ancestors of vertex x are the set of vertices on the path from the root to x.
Backtracking search: See Depth-first search.
Balanced tree: A tree with all leaves at level h and h −1, where h is the height of the tree.
Binary tree: A tree in which each internal vertex has exactly two children.
Breath-first search: From the root, find all vertices z adjacent to the root, then all vertices adjacent to one of the zs, and so on.
Child: Children of vertex x are vertices y with an edge (x, ⃗y) from x to y.
Depth-first search: Also known as backtracking search, builds a path from the root as far as possible; one backtracks from the current vertex when the path cannot be extended to a previously unvisited vertex and backs up the current path until a vertex is found at which a side path may be constructed to a new vertex.
Descendant: Descendants of vertex x are the set of vertices z whose path from the root to z contains x.
Height: The largest level number in a tree.
Inorder traversal: In a binary tree, starting from the root, this traversal recursively lists the vertices of the left subtree of a particular vertex x, then x, and then the vertices in the right subtree of x.
Internal vertex: A vertex with children; internal vertex and parent are equivalent terms.
Leaf: A vertex with no children; a leaf’s only incident edge comes in from its parent.
Level or level number: The length of the path from the root to a given vertex—e.g., level 2 of a tree consists of all vertices whose path from the root has length 2.
m-ary tree: A tree in which all internal vertices have m children.
Parent: The parent of vertex x is the unique vertex z with an edge (z, ⃗x) from z to x.
Postorder traversal: Lists vertices in the order they are last encountered in a depth-first search of a tree.
Preorder traversal: Lists vertices in the order they are first encountered in a depth-first search of a tree.
Root: The special vertex in a tree with a unique path to any other specified vertex; in a directed or rooted tree, the root is the unique vertex at level 0 that has no parent.
Rooted tree: A directed tree with all edges directed away from the root.
Sibling: Siblings of vertex x are those vertices with the same parent as x.
Spanning tree: A tree that is a subgraph of a connected graph and that contains all vertices of the graph.
Subtree: A connected subgraph of a tree.
www.itpub.net BIBLIOGRAPHY GRAPH THEORY AND ENUMERATION COMBINED V. K. Balakrishnan, Theory and Problems of Combinatorics, Schaum’s Outline Series, McGraw-Hill, New York, 1995.
K. Bogart, Introductory Combinatorics, 3rd ed., Harcourt Brace Jovanovich, San Diego, 2000.
M. Bona, A Walk Through Combinatorics: An Introduction to Enumeration and Graph Theory, 2nd ed., World Scientific Publishing, Singapore, 2006.
R. Brualdi, Introductory Combinatorics, 5th ed., Prentice Hall, Upper Saddle River, NJ, 2009.
P. Cameron, Combinatorics: Topics, Techniques, Algorithms, Cambridge University Press, Cambridge, 1994.
M. Erickson, Introduction to Combinatorics, John Wiley & Sons, New York, 1996.
R. Graham, D. Knuth, and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, 2nd ed., Addison-Wesley, Reading, MA, 1994.
J. Harris, Combinatorics and Graph Theory, Springer, New York, 2010.
D. Mazur, Combinatorics: A Guided Tour, Mathematical Association of America, Washington, DC, 2009.
R. Merris, Combinatorics, 2nd ed., John Wiley & Sons, New York, 2003.
F. Roberts and B. Tesman, Applied Combinatorics, 3rd ed., Prentice Hall, Upper Saddle River, NJ, 2009.
D. Stanton, R. Stanton, and D. White, Constructive Combinatorics, Springer, New York, 1986.
J. VanLint and R. Wilson, A Course in Combinatorics, 2nd ed., Cambridge University Press, Cambridge, 2001.
W. Wallis and J. George, Introduction to Combinatorics, CRC Press, Boca Raton, 2011.
GRAPH THEORY J. Aldous, R. Wilson, and S. Best, Graphs and Applications: An Introductory Approach, Springer, New York, 2000.
G. Agnarsson and R. Greenlaw, Graph Theory: Modeling, Applications, and Algo-rithms, Pearson, 2008.
439 440 Bibliography V. Balakrishnan, Schaum’s Outline of Graph Theory, McGraw-Hill, New York, 1997.
C. Berge, Graphs, 2nd ed., Elsevier Science, New York, 1985.
N. Biggs, E. Lloyd, and R. Wilson, Graph Theory 1736–1936, Cambridge University Press, Cambridge, 1999.
A. Bondy and U. Murty, Graph Theory, Springer, New York, 2008.
M. Capobianco and J. Molluzzo, Examples and Counterexamples in Graph Theory, North Holland, New York, 1978.
G. Chartrand and L. Lesniak, Graphs and Digraphs, 5th ed., Chapman and Hall, New York, 2004.
J. Gross and J. Yellen, Graph Theory and Its Applications, CRC Press, New York, 2005.
N. Hartsfield and G. Ringel, Pearls in Graph Theory: A Comprehensive Introduction, Dover, New York, 2003.
D. Marcus, Graph Theory: A Problem-Oriented Approach, Mathematical Association of America, Washington, DC, 2008.
O. Ore, Graphs and Their Uses, Mathematics Association of America, Washington, DC, 1990.
R. Trudeau, Introduction to Graph Theory, Dover, New York, 1994.
D. West, Introduction to Graph Theory, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 2001.
R. Wilson, Introduction to Graph Theory, 5th ed., Longman, Reading, MA, 2010.
ENUMERATION R. Allenby and A. Slomson, How to Count: An Introduction to Combinatorics, CRC Press, New York, 2010.
T. Andreescu and Z. Feng, A Path to Combinatorics for Undergraduates: Counting Strategies, Birkhauser, Boston, 2004.
M. Hall, Combinatorial Theory, 2nd ed., Blaisdell, Boston, 1986.
D. Marcus, Combinatorics: A Problem-Oriented Approach, Mathematical Associa-tion of America, Washington, DC, 1999.
G. Martin, Counting: The Art of Enumerative Combinatorics, Springer, New York, 2010.
G. Polya, R. Tarjan and D. Woods, Notes on Introductory Combinatorics, Birkhauser, New York, 1983.
J. Riordan, An Introduction to Combinatorial Analysis, John Wiley & Sons, New York, 1958.
H. Ryser, Combinatorial Mathematics, Mathematics Association of America, Washington, DC, 1963.
R. Stanley, Enumerative Combinatorics, Wadsworth, Belmont, CA, 2001.
W. Whitworth, Choice and Chance, Hafner Press, New York, 1901.
www.itpub.net SOLUTIONS TO ODD-NUMBERED PROBLEMS PRELUDE SOLUTIONS 1. R Bu G Y.
3. Y Bu Bu W.
5. Bu Bk Bk Bu or Bk Bu Bu Bk.
7. G Y R Bk.
9. W W R Bu, W Bu R Bu, W Bu R W, Bk Bk R Bu, Bu Bk R Bu, Bu Bu Bk R.
11. O R Y Bu P.
13. (a) Three black and one white. (b) 14.
15. 9.
17. 1040.
19. Two of one color and one of a second color.
CHAPTER ONE SOLUTIONS Section 1.1 1. (a) F A E C D B 82 90 90 82 82 77 77 91 85 85 (b) 2(C, D), (C, E), (c) Yes, several routes.
3. (a) A 5-circuit.
5. (a) Many possibilities, (b) min = 4, several possibilities.
7. (a) {A-a, B-b, C-d, D-c}.
(b) A and C each can fill only job b.
441 442 Solutions to Odd-Numbered Problems 9. (a) No possible (odd number of vertices), (b) a, c, e collectively must be matched with just b and e.
13. (a) Vertex = variety of chipmunk, if A splits into B and C, then make edges (A, ⃗B) and (A, ⃗C), (b) 7 splits.
15. (a) 4 other pairs.
(b) {b, j}, {c, h} and 6 other pairs.
17. C, E, c, d (answer for both parts of the question).
19. Minimal block surveillance—6, minimal corner surveillance—3.
21. (a) 5: squares (2, 4), (3, 4), (4, 4), (5, 4), (8, 4), (b) 8.
23. (a) (i) {b, c}, (ii) {A, a, B, b, D, e}, (b) (i) {a, d}, (ii) {C, c, d, E}.
25. (a) 6, K1,6 (b) 7, isolated vertices.
29. (a) (b) not B A A B A A B A A B A S Section 1.2 1.
Plus complements of first 4 graphs 5. (a) No, right graph only has circuits of length 5, (b) Yes, a-6, b-1, c-3, d-5, e-2, f -4, (c) Yes, a-3, b-4, c-7, d-1, e-6, f -8, g-2, h-5, i-9, (d) No, degree-2 vertex is part of triangle only in right graph, (e) No, right graph has one more edge, (f) No, subgraphs of vertices of degree 3 do not match, (g) Yes, a-7, b-3, c-5, d-4, e-1, f -2, g-6, (h) No, degree-3 vertices are adjacent only in left graph, (i) Yes, a-1, b-3, c-5, d-2, e-4, f -6, (j) No, degree-2 vertices are adjacent only in left graph, (k) Yes, a-1, b-8, c-4, d-7, e-6, f -9, g-11, h-10, i-2, j-3, k-5, (l) Yes, a-1, b-2, c-3, etc.
7. Graphs 1-6, 13-18, 31-36, 37-42 mutually isomorphic, 7-12, 19-24, 25-30 isomorphic.
9. No, building on the isomorphism in Example 2, b →c but 5 →2.
13. 3 in each.
www.itpub.net Solutions to Odd-Numbered Problems 443 Section 1.3 1. (a) 12, (b) 9, (c) 8 or 10 or 20 or 40.
3. 12.
5. Solve for n in terms of m in the formula m = n(n −1)/2.
7. If v vertices, then e = 1 2 vp edges (where 1 2v is an integer since p is odd).
9. Sum of in-degrees (or out-degrees) = number of edges, since sum counts each edge once.
13. (a) No, (b) Yes, (c) No.
Section 1.4 1. (a) (b) d a e b f c A B C D E F G 3. (a) No, delete (b, c), (e, f ): {a, e, f }, {b, c, d}, (b) No, delete (b, j), (e, g): {a, d, h}, {c, f, i}, (c) No, many possible K3,3, (d) No, delete (a, b), (b, c), (d, e), ( f, g): {a, d, e}, {c, f, g}, (e) Yes, (f) No, delete a and (b, c), (g) Yes (h) No, delete (e) : {a, b, c, d, f } (i) No, delete (d, e): {a, c, e}, {b, f, g}, (j) Delete ( f, g): {a, c, e}, {b, d, h}, 5. (a) n ≤4, (b) r or s ≤2.
7. (a) Possible, e = 11, (b) Possible, r = 7, (c) Not possible, (d) Possible, v = 7, (e) Possible, e = 12, r = 8, (f) Possible, v = 7, (g) Possible, e = 12, v = 8, (h) Not possible (parity violation), (i) Possible, e = 20, r = 10, (j) Not possible.
9. (a) Degree of vertices in K5 is 4 ⇒degree of vertices in L(K5) is 2 × (4 −1) = 6; L(K5) has v = 10 and e = 1 2 deg = 3v = 30 ⇒e ≤ \ 3v −6; 444 Solutions to Odd-Numbered Problems (b) 11. (a) Circuit length = sum of number of boundary edges of R1 and R2 −2; (b) Circuit length = sum of number of boundary edges of each enclosed region minus 2 × (number of edges interior to circuit).
13. (a) K3,3 and K5 are critical nonplanar.
15. (a) r = e −v + c + 1, (b) Using (a), corollary becomes e ≤3v −3c −3 (≤3v −6).
19. If false, deg ≥5 and so 5v ≤sum of degrees = 2e ≤2(3v −6) = 6v −12, that is, 5v ≤6v −12 or 12 ≤v—impossible.
25. v = p + 2l, e = 1 2 deg = 2p + 3l, answer = r −1 = e −v + 1 = p +l + 1.
Supplement 1. Vertex = committee, edge = committee overlap (person). Graph is K7 with e = 1 2 deg = 1 2(6v) = 21.
3. n = 12.
7. Yes, each component of G is a circuit.
9. (a) Yes,traceoutanysequenceofedgesandeventuallyavertex z willberepeated, between first and second visit to z a circuit is formed, (b) No—e.g., 11. Vertices in different components of G are directly adjacent in G; vertices in same component are joined in G by a path of length 2 via any vertex in other component of G.
15. If: suppose not strongly connected with no path from a to b—let V1 consist of a and all vertices that can be reached by a directed path from a, V2 is other vertices, only if: obvious.
17. A bridge edge cannot lie on a circuit.
19. (a) Yes, see Exer. 7 in Section 1.2, (b) No, odd number of vertices of odd degree, (c) No; e = 1 2 deg = 13 ⇒e ≤ \ 3v −6.
21. Possible, a K1,5 with two extra edges.
23. Consider a vertex y that beat x, [i.e., (y, ⃗x)]; if y beat every competitor that beat x, then y would have a greater score that x—not possible—and so for some w, (x, ⃗w) and (w, ⃗y).
25. (a) Repeatedlyremovesidecircuitsuntiltrailfrom x to y hasnorepeatedvertices.
(b) x y www.itpub.net Solutions to Odd-Numbered Problems 445 (c) Similar argument to part (a).
(d) If odd number of edges of cycle are partitioned into circuits [part (d)], then some circuit must have odd number of edges.
27. (a) (h, a), (h, g).
29. Two-component, n-vertex graph with the most edges is a Kn−1 plus an isolated vertex; it has 1 2 (n −1)(n −2) edges.
33. (a) 5-circuit or 3-edge path.
(b) By hint, if G has n vertices, number of edges in G = 1 2(number of edges in Kn) = 1 2[ 1 2n(n −1)] = 1 4n(n −1). Since n and n −1 are not each divisible by 2, then either n or n −1 must be divisible by 4, i.e., n = 4k or n = 4k + 1.
35. If: obvious, only if: let xn be vertex with 0 out-degree (if no such vertex, there is a directed circuit—Exer. 9(a), move xn from graph and let xn−1 be vertex with 0 out-degree in remaining graph, continue indexing in this fashion.
37. (a) ad ab ac bd cd de bc (b) Each edge of Kn is incident to n −2 other edges at each end vertex, 2(n −2) incidences in all, (c) Can be no vertices of degree 0 or 1 and so all degrees ≥2; since e = 1 2 deg, then to have e = v (so that G and L(G) have same number of vertices), all deg = 2 ⇒G is any circuit.
39. (a) b ↔c, a and d fixed, (b) None, (c) a ↔c, e ↔f , b and d fixed.
41. Sketch of proof: show that each vertex in (C1 ∪C2) −(C1 ∩C2) has even degree and then repeatedly trace a path (without repeating any edge) until a vertex x is revisited and remove the circuit formed between the first and second visit to x.
CHAPTER TWO SOLUTIONS Section 2.1 1. (a) Many possibilities, (b) Many possibilities with b and f as end vertices.
3. One example is a graph consisting of a circuit of 7 edges.
5. (a) No, once bridge crossed there is no way back to starting vertex, (b) Many possibilities, e.g., a 10-edge path.
7. An isolated vertex added to a connected graph with even degrees now has a Euler cycle but is not connected.
446 Solutions to Odd-Numbered Problems 9. Build a graph with a vertex for each racer and an edge for each race; a Euler trail corresponds to a sequence of races in which each racer is in two consecutive races; this graph has the desired Euler trail because only A and F have odd degree.
11. A set of deadheading edges must have one edge at each odd-degree vertex; joining these edges at odd-degree vertices together (without changing the parity of the degree of other vertices) requires a set of paths.
13. A directed graph has a Euler trail if and only if at all but two vertices indegree = outdegree and at those two, indegree and outdegree differ by 1. Proof: add an extra edge so that indegree = outdegree at two unbalanced vertices and resulting graph has Euler cycle; remove added edge yielding desired Euler trail.
15. No such Euler cycle containing all vertices.
17. (a) Many possibilities, (b) If at every stage graph of remaining edges is connected, then just before using last edge E and being forced to stop at starting point, E is only edge remaining and once taken there are no remaining edges, (c) Applies to Euler trails.
19. (a) 01 001 011 11 111 110 10 100 000 00 010 101 (b) Concatenating the first digits of each node on an Euler cycle produces desired sequence, (c) Many possibilities, e.g., cycle 00-00-01-11-11-10-01 (-00) produces 00011101.
Section 2.2 1. Many possibilities—e.g., (a) , (b) 3. a-g-c-b-f-e-i-k-h-d-j-a.
5. Path: m-b-c-d-e-q-a-l-k-j-o-p-n-f-g-h-i, no circuit: by symmetry at p, delete any edge, choose p-m, at m and c rule 1 forces subcircuit m-b-c-d-m.
7. (a) Rule 1 at f , h, j forces subcircuit e-f-g-h-i-j-e, (b) By symmetry at p, use m-p-n and delete p-o forcing k-o-i, at m and n, if both use edge going up (to g) then subcircuit p-m-g-n-p, if both m, n use edge down then subcircuit p-m-k-o-i-n-p, so by symmetry use m-g and n-i, deleted i-h, i-c, i-j forcing g-h-b-c-d-j-k, plus k-o-i-n-p-m-g, yielding a subcircuit.
www.itpub.net Solutions to Odd-Numbered Problems 447 9. (a) Hamilton circuit alternates between red and blue vertices and so must have equal numbers of each, (b) Follows from part (a), (c) (i), (ii) Have odd number of vertices, (iii) Has 9 vertices in one part and 7 vertices in other part of bipartition.
11. (a) 0110 0111 0011 0001 0000 0010 0101 0100 1110 1111 1011 1001 1000 1101 1100 1010 Edges between vertices at corresponding positions of the two cubes (b) Many possibilities: 1-0000, 2-0001, 3-0011, 4-0010, 5-0110, 6-0111, 7-0101, 8-0100, 9-1100, 10-1101, 11-1111, 12-1110, 13-1010, 14-1011, 15-1001, 16-1000 13.
15. Many solutions but very tedious; the following heuristic works: at each stage, look at all possible squares (not yet visited) a knight’s move from current square and move to a square with the minimum number of possible squares for the next move.
17. (a) Rook: starting at upper left corner, go down first column, square by square, at bottom of first column move right to bottom of second column, move up second column, square by square, continue going down and up successive columns finishing at top square of right column from which one moves back to top square of left column, king: similar to rook, except in columns 2, 3, . . ., n −1, avoid top square, when top square of right column reached, move left along top squares of each column to return to top left square, (b) Rook: not possible (variation on reasoning in Exer. 14), king: (1, 1)-(1, 2)-(2, 2)-(1, 3) . . . (2, n), now go left and right covering rows 3 through n avoiding left square in each row except row n, finishing at (1, n) and now move up first column to starting square.
19. (a) n!, (b) Form the Hamilton circuits by placing vertices in a circle: first circuit formed by joining consecutive vertices, second circuit formed by joining vertices 2 positions apart (one intervening vertex), third circuit formed by joining vertices 3 positions apart, and so on (no subcircuits because n prime); 448 Solutions to Odd-Numbered Problems (c) Form complete graph with professors as vertices, answer: 8 days, using part (b) to form 8 Hamilton circuits out of the edges of K17.
21. If x, y nonadjacent, direct all edges inward at both x and y, now x and y can only be on a Hamilton path if they are the last vertex on the path—cannot be two last vertices.
23. Many possibilities—e.g., Section 2.3 1. (a) 3, has K3, (b) 4, an attempt to 3-color outer 6 vertices gives b, d, f different colors so that c requires fourth color, (c) 3, not bipartite, (d) 4, a, b, e, g form a K4, (e) 3, odd circuit, (f) 2, (g) 4, isomorphic to graph in 1(b), (h) 4, a, c, g, i form a K4, (i) 5, a, b, c, d, f form a K5, (j) 3, (k) 2, (l) 2, (m) 4, an attempt to 3-color vertices clockwise around the circle starting at a forces j to have same color as a, (n) 4, graph contains a 5-wheel (see Example 2), (o) 4, graph contains a 5-wheel (see Example 2), (p) 3, outer pentagon (odd-length circuit) cannot be 2-colored, (q) 4, an attempt to 3 color the sequence of vertices a, e, b, f , c, g, d forces d to have same color as a, (r) 4, a 3-coloring forces a and h to have different colors, similarly for i and p so that a, h, i, p act like a K4.
3. (a) Many possibilities, (b) Many possibilities, 5. (b), (g), (p), (q).
7. (a) {a, b, c}, {d, e, f }, {g, h}, (b) {a, b, e}, {c, d}, { f , g}, (c) {a, b, c}, {d, e, f }, {g, h, i}, { j, k}.
9. No, Nevada (or Kentucky or West Virginia) and its neighboring states have duals that form odd-length wheels.
11. 3.
13. Vertices = classes, edges = two classes with a common student, colors = class times.
15. Vertices = ships, edges = overlapping visit, colors = piers.
www.itpub.net Solutions to Odd-Numbered Problems 449 17. (a) Yes, (b) Yes.
Section 2.4 1. Proceed as in Theorem 5 using induction and the fact that there is a vertex x of degree 5, but since 6 colors are available, x is immediately colored with a color different from those used by its 5 neighbors.
3. The chromatic number of G is the maximum of the chromatic numbers of its components.
5. If the edge chromatic number is 2, the maximum degree must be 2. If the vertex chromatic number is 2, then the graph is bipartite. Combining these two facts, we see that the graph must be a path or an even-length circuit.
7. (a) 3-color one triangle and then extend by successively coloring a vertex that lies on a triangle with two previously colored vertices, (b) Process in part (a) Yields unique coloring.
9. (a) If not connected, then k colors are needed to color one of G’s components and removing a vertex from another component will not reduce χ(G), (b) If deg(x) ≤k −2, then if G −x could be k −1 colored, also G could be k −1 colored by giving x one of the k −1 colors not used by one of x’s k −2 (or less) neighbors, (c) If x disconnects so that G −x has components G′ and G′′ then one of G′ ∪x or G′′ ∪x requires k colors and removing a vertex from the other component will not reduce χ(G).
11. (a) For n = 1, χ(G) + χ(G) = 2; assume for n −1 and consider an n-vertex graph G; by induction we can color G −x and G −x, for any given vertex x, with a total of n colors for both graphs (possibly fewer); x has a total of n −1 edges in G and G and so is not adjacent to one of the n color classes in one of G or G and can be added to that color class, although in the other graph x may require an additional color—for a total of n + 1, as required, (b) χ(G)) ≥size of largest complete subgraph in G = q, size of largest indepen-dent set in G, and by Exer. 8(a), χ(G)q ≥n, (c) Square both sides of inequality, new inequality follows immediately from (b) and the fact that a2 + b2 ≥2ab.
13. (k2 −6k + 8) = 0 when k = 4 and so Pk(G) = 0 for k = 4, but the Four Color Theorem says that any planar graph can be 4-colored (i.e., has a positive number of 4-colorings).
15. If: label with numbers that are the length of the longest path starting at the ver-tex, adjacent vertices must have different-length longest paths since edge (x, ⃗y) implies that x’s longest path will be at least one greater that y’s longest path length, only if: let colors be numbers 0, 1, . . . , k −1 and direct edges from larger to smaller numbers.
17. Pick two nonadjacent vertices, each on one of the odd-length circuits, and give them a third color. The remaining graph has no odd-length circuits and hence 450 Solutions to Odd-Numbered Problems is bipartite (2-colorable). Some details are required to show that the two such non-adjacent vertices can be found.
Supplement 1.
G R W B W R G B 4 2 1 3 1 3 4 2 3. One labeled factor must be 1: W-B, 2: G-R, 3: B-G, 4: R-W, second labeled factor can be 1 and 2: G-B, 3: R-R, 4: W-W or 1: B-G, 2: W-B, 3: R-W, 4: G-R (or interchange 1 and 2).
5. (a) Yes, a Hamilton circuit is a factor, (b) Having a Euler circuit has no relation to having a factor.
CHAPTER THREE SOLUTIONS Section 3.1 1.
(a) (c) (b) 3. No odd circuits ⇒bipartite (Theorem 2 of Sect. 1.3) ⇒2-colorable.
5. (a) See answer to part (b), (b) Since G is connected, a subset of edges can be chosen to form a tree con-taining all vertices of G, but this subgraph has one fewer edges that vertices (by Theorem 1) and if G contained other edges, besides those in this tree, it would have as many vertices as edges, (c) If G has a circuit, removal of an edge an circuit does not disconnect G ⇒G has no circuits; now use part (a).
7. Start at any vertex and trace a trail (no repeated edges); since no vertex is ever visited twice (if so, circuit would result) and graph is finite, trail must end at a vertex x of degree 1; now start trail-building again at x to get a second vertex of degree 1.
www.itpub.net Solutions to Odd-Numbered Problems 451 9. Height h ⇒each path to a leaf passes through at most h internal vertices with a choice of m children to go to at each internal vertex, for a total of at most mh paths to a leaf.
11. i/n = [by Corollary part (a)] i/(mi + 1) ≈i/mi = 1/m.
13. Largest n −1, smallest 2.
15. A tree is bipartite, one vertex class in the bipartition must have at least n/2 vertices.
17. Unbalancing a tree: makes sum of level numbers larger and so smallest sum occurs when binary tree is balanced, in which case each level number is ⌊log2 l⌋or ⌊log2 l⌋+ 1, and sum of level numbers is at least l⌊log2 l⌋.
19. (a) Internal vertices are +s 9 5 2 1 3 + + + + (b) ⌈log2 100⌉= 7.
21. 63.
23. Leaves = letters, each leaf = n-digit binary sequence ⇒2n letters.
25. (a) 24, (b) 16, (c) 12, (d) 9 tournaments (original and 8 losers’ tournaments).
27. (a) log2 n + 1 2 = ⌊log2(n + 1)⌋−1, (b) E C B A D 29. (a) 1L 3H 2L 1 2 4L All fair 4H 4 T 2H 3L 1H 1 2 12 3T (b) First weighing is either one coin on either side or two coins on either side, in either case some outcome has more than 3 possibilities that must be dis-tinguished in the one additional weighing (impossible, for example with one on each side, there are 5 possibilities if the scales balance).
452 Solutions to Odd-Numbered Problems Section 3.2 1. (a) Any 8-vertex path, (b) Many possibilities, e.g., a-b-d-c-h-k-i-g-e-f-j, (c) Many possibilities, e.g., a-b-c-d-e-f-g-h; many possibilities, e.g., any 6-vertex path.
3. All trees on 5 vertices [see solution to Exer. 1(b) in Section 3.1], (b) All trees on 4 vertices [see solution to Exer. 1(a) in Section 3.1], (c) a a d b c d c , e b e a d b c e (d) A path of length 5 on a tree with 3 vertices of level 1 and 2 vertices at level 2, places of d and c can be interchanged in last tree.
5. 4 components, x17, x19, x23 isolated vertices, depth-first spanning tree for other component has path x2-x4-x6-x3-x9-x12-x8-x10-x5-x15-x20-x14-x16-x18-x22-x24-x26-x13, plus edges from x20 to x25, from x14 to x7 to x21, and from x22 to x11.
7. If the connected graph is not a tree, it has a circuit, which contains an edge e not on the one spanning tree; a second spanning tree must exist that contains e.
9. (a) If not all vertices reached, some reached vertex would have an edge to an unreached vertex, but a depth-first search would use such edge, (b) Immediate.
11. If C has no edge of a spanning tree T , removal of C could not disconnect graph (spanning tree’s edges connected graph).
13.
4 3 5 6 7 2 1 0 1 2 3 4 (6, 4) (7, 3) (7, 0) (6, 0) (0, 4) (2, 4) (0, 3) 15. (0, 0)-(0, 4)-(4, 0)-(4, 4)—now 2 quarts in 10-quart pitcher.
17. Right-hand-wall rule is same as depth-first search, which takes leftmost branch at every intersection (corner).
19. The ferryman takes goat across and returns alone; next he takes the dog across and brings the goat back; next he takes the tin cans across and returns alone, and finally he takes the goat across.
21. Not possible.
25. A and B cross (2 min.) and A returns (1 min.); next C and D cross (10 min.) and B returns (2 min.); finally A and B cross again (2 min.).
www.itpub.net Solutions to Odd-Numbered Problems 453 27.
e c – a b a Sin x –1 c b x e + d a + + – + + –1 Section 3.3 1. Cost is 11: 1-3-2-4-1.
5. Cost is 14: T1-T5-T4-T3-T2-T1.
7. (a) 1-4-3-2, (b) 1-4-5-3-2-6-1, (c) 2-3-5-1-4-2.
9. 2 1 2 2 2 2 2 1 2 Section 3.4 1. Outcomes are ordered lists.
3. A binary comparison tree has n! outcomes or leaves (as noted at the beginning of this section); so average number of comparisons = average leaf level = (by Exer. 17 in Section 3.1) log2n! = n log2n.
5. If the initial heap is balanced [as decribed in Exer. 4 (a)], then largest level number is log2n and the number of comparisons to adjust the heap each time the root is removed will equal the largest level number—i.e., log2n; for n iterations of removing the root and readjusting the heap, there will be nlog2n comparison; constructing initial heap is similar.
9. (b) O(nlog2n).
CHAPTER FOUR SOLUTIONS Section 4.1 1. Length 14: c-d-h-k- j-m.
3. (a) 31: L-c-d- f -g-k-W, (b) 32: L-b-h- j-m-W, (c) 13: L-a-c-d- f -g-k-W, (d) L-c-d- f -g-k-W, 4 paths.
454 Solutions to Odd-Numbered Problems 5. (a) 5: L-b-h- j-m-W, (b) 6: L-c-d- f -g-k-W, (c) 6: L-c-e-g- j-m-W.
7. If algorithm has found shortest paths to all vertices ≤m unit from a, then a vertex x will be distance m + 1 from a if and only if the length k(y, x) from a vertex y (y is closer to a than x) plus the distance d(y) of y from a equals m + 1—this is exactly the test that the algorithm performs.
9. Many possible examples—e.g., a directed circuit from a to b to c to a with two edges of length 1 and the third edge of length−3.
11. Define a tree by letting the first label of a vertex be its parent.
Section 4.2 1. (a) 59: path L-a-c-d-h- f -g-i-k-W plus edges (b, d), (d, e), (g, j), (k,l), (k, m), (b) 60: replace (k, W) by (m, W), (c) L, j (other possibilities), (d) Modify part (b) by deleting (a, c), (g, i), and adding (a, b), (g, k).
3. (a, b) 39: path N-b-c-d-e-g- j-m-R plus edges (d, h), (e, f ), ( f, i), ( j, k).
5.
7. Modification: let initial edge be prescribed edge.
11. (a) If all edges of shortest length do not form circuit and they are not all in T ′, then add an omitted one and remove a longer edge in the resulting tree (as in proof of Prim’s algorithm) to obtain a shorter minimal spanning tree—impossible.
(b) Same reasoning as in part (a), (c) If: part (b) is property of minimal spanning tree used to prove validity of Prim’s algorithm; only if: verified in part (b).
Section 4.3 1. max flow = 13, P = {a, b, c}.
3. max flow = 50, P = { f, z}.
5. (a) max flow = 13, P = {a, b, c}, www.itpub.net Solutions to Odd-Numbered Problems 455 (c) 2 7 3 8 11 10 14 7 12 4 10 1 10 10 8 7 6 13 9 3 2 8 4 3 8 3 4 6 a b c d e f g h i j k l m n o p q r s t v w x y u j 5 4 7 4 14 7 12 8 16 5 4 5 7. Set capacities of edges out of a and into z equal to 100 (equivalent to unlimited flow), max flow = 150, P = {a, b, c, d}.
9. 5, build paths by choosing leftmost unused edge leaving each vertex.
11. (a) 3 (3 edges leaving L), (b) 15, quintuple flow in part (a).
13. See Exer. 12 for modeling the vertex capacity constraint; max flow = 40, P = { fi, fo, go, z}.
15. max flow = 2100, route 400 on a0-d2-z3-z4, 300 on a0-c2-z4, 400 on a0-b1-z3-z4, 200 on a0-b1-c2-d3-z4, 400 on a0, a1-d3-z4, 400 on a0-a1-b2-z4.
17. max flow = 36.
19. (a) Algorithm tries to reduce flow in incoming edge before adding flow to out-going edge, (b) Yes.
21. Define tree with a vertex’s parent being its first label.
23. (a) Put flow of 1 in the two edges between c and e; other edges no flow, (b) Take flow in (a); and add a 2-unit flow path a-d-f-z.
27. If each flow path crosses a cut once, then the capacity of the cut = sum of values of such flow paths = value of flow.
29. (a) This is max flow–min cut theorem for model in Example 6, (b) Make edge capacities 1 and build vertex constraints of 1 (see Exer. 12), result is max flow–min cut theorem for this network.
31. One needs to build an initial feasible flow (requires at least one edge entering and one edge leaving each vertex, except a and z), then use same algorithm to reduce, instead of increase flow along a flow path.
33. (a) In step 2a of augmenting flow algorithm, incoming flow cannot be reduced below its lower bound value; otherwise use same algorithm.
456 Solutions to Odd-Numbered Problems Section 4.4 1. (a) Several possibilities, see (b), (b) Using pairings A-G, Lo-J, first labels are (except for a, all second labels are 1): Bo-a+, F-Bo+, G-Bo+, C-F−, A-G−, J-C+, Bi-A+, Lo-J −, D-Bi−, La-Lo+, z-La+, new matching A-G, D-Bi, Lo-La, Bo-F, C-J.
3. Give edges from a and into z the appropriate supplies and demands. Middle edges still ∞, one solution: A-Bi (3 dates), D-Bi (2), Bo-F (1), Lo-F (2), C-F (1), A-G (1), Bo-G (2), C-J (3), Lo-J (2), D-La (3).
5. No, schools with demand of 7 Ph.D.s can hire at most 6 (one from each university).
7. (a) Can be co-champions with either Lions or Tigers, (b) Not possible; Lions and Tigers must play each other 3 times, but each team can only win once.
9. Only possible to be co-champions, jointly with the other three teams.
11. (a, b) Make complete bipartite graph, each vertex of degree n; by Example 3 there is a pairing for the first night; remove these edges; now again by Example 3, there is a pairing for the second night; continue in this fashion.
13. Start with a standard set-of-distinct-representatives matching network; replace the source a with 3 sources (one for each university), each with capacity m/3 and unit-capacity edges to each university’s graduates.
15. Necessary and sufficient condition: for any set S of vertices, |S| ≤|s(S)|, where s(S) is the set of successors of vertices in S (vertices with an edge coming in from a vertex in S); this condition guarantees A complete matching in hinted bipartite graph, which corresponds to a set of edges in original graph with one edge out of and one edge into each vertex.
17. For each left-side vertex, x, form a set of x’s (right-side) neighbors.
Section 4.5 1. (a) x12 = 30, x21 = 20, x22 = 10, (b) x11 = 20, x12 = 10, x22 = 30.
3. (a) x11 = 30, x22 = 30, x31 = 10, x32 = 20, (b) x12 = 30, x21 = 10, x22 = 20, x31 = 30.
5. (a) x13 = 30, x21 = 20, x23 = 10, x32 = 20, x33 = 10, (b) x12 = 20, x13 = 10, x21 = 20, x23 = 10, x33 = 30.
7. (a) x11 = 30, x23 = 30, x32 = 10, x33 = 20, x42 = 30, (b) x13 = 30, x21 = 30, x31 = 10, x33 = 20, x42 = 30.
CHAPTER FIVE SOLUTIONS Section 5.1 1. (a) 12, 20, (b) 3, 6.
3. 8 × 12 × 5 × 4.
5. 266, 26 × 25 × 24 × 23 × 22 × 21.
www.itpub.net Solutions to Odd-Numbered Problems 457 7. (a) 10 + 6 + 4, (b) 10 × 6 × 4, (c) 3 × {10 × 9 × (6 + 4) + 6 × 5 × (10 + 4) + 4 × 3 × (10 + 6)}.
9. (a) 4 × 47 = 188, (b) (1 × 48) + (12 × 47).
11. 12.
13. See Supplement at end of Chapter 5.
15. 52 × 48/52 × 51.
17. (a) 10 × 9 × 8 × 8, 10 × 9 × 8n−2, (b) 9 × 8 × 1/9 × 8 × 8, (c) 9 × 8 × 7 × 1/9 × 9 × 8 × 8.
19. (6 × 5 × 4) −(5 × 4 × 3).
21. (6 × 5 × 4) −(4 × 3 × 2) or (2 × 3 × 4 × 3) + (3 × 2 × 4).
23. (a) 9 × 9 × 8 × 7, (b) 7 × 8 × 8 × 8, (c) 9 × 9 × 8 × 7 −7 × 7 × 6 × 5.
25. 10 × 9 × (25 −2)/2 27. 45 −3 × 42 29. 15 × 10(14 + 9).
31. 4 × 103.
33. See Supplement at end of Chapter 5.
35. 2 × 25/50 × 49.
37. [(3 × 3 × 2) + (3 × 3!)]/63—pick the possible smallest value (3 choices), count ways of picking middle value and then arranging them (3 × 2 possibilities when middle value same as largest or smallest).
39. 9 × 7 × 5 × 3.
41. 16 × 8 × 7/2, n × m × (m + n −2)/2.
43. {[28 × (64 −22) (Queen on edge of board)] + [20 × (64 −24) (Queen one away from edge of board)] + [12 × (64 −26)] + [4 × (64 −28)]}/2, 45. 210 −1—each friend is or is not the subset (subtract 1 for the case of no one invited).
47. 2 × 3 × 6.
Section 5.2 1. 52!
3. P(13, 6).
5. P(7, 4).
7. C(14, 5), C(12, 3).
9. {C(11, 9) + C(11, 10) + C(11, 11)}/211 11. (a) C(12, 2) × C(10, 2) × C(8, 2) × C(6, 2) × C(4, 2), (b) 6! × 6!.
13. 7 × C(6, 2) × 244.
15. C(n, 8)2n−8.
17. C(n, k) × 2n−k/3n.
19. C(10, m) × C(10 −m, n) × 2410−m−n.
458 Solutions to Odd-Numbered Problems 21. (a) C(4, 2) × {C(6, 4) + C(6, 5) + C(6, 6)} + [C(4, 3) × C(6, 6)], (b) C(9, 3) + C(9, 4) + C(9, 5), (c) C(10, 5) −C(7, 2), (d) {[C(4, 2) × C(6, 2)] −(3 × 5)} + {[C(4, 1) × C(6, 3)] −C(5, 2)} + C(6, 4).
23. (a) {C(4, 2) × 92 + C(4, 3) × 9 + C(4, 4)}/104, (b) C(10, 2) × (24 −2)/104, (c) C(10, 4)/104.
25. 3! × 6! × 8! × 5!.
27. C(21, 5) × 10!; 2 × 5!2/10!.
29. (a) 2 × 5!/6! = 1/3, (b) C(6, 2) × 4!/6!
31. (a) C(n, 3), (b) C(n −m, 3) + [m × C(n −m, 2)] 33. 12.
35. (a) 1-vote 1/6, 2-vote 1/2, (b) 1-vote 1/6, 2-vote 1/3, (c) 1-vote 468/7!, 2-vote 1056/7!, (d) 1-vote 396/7!, 2-vote 864/7!.
37. (a) 26!/2, (b) 26!/22, (c) C(26, 5) × 21!.
39. C(5, 3) × P(6, 3) × P(21, 3) + C(5, 4) × P(6, 4) × P(21, 2) + 21 × 6!.
41. 4 × 3! × C(7, 2) × C(5, 2) × 3!
43. C(30, 5)12/C(360, 60).
45. (a) {[(3 × 4) + (2 × 5)] × 8!}/10!, (b) {(1 × 9 × 8!) + (8 × 8 × 8!)}/10!.
47. (a) {[C(4, 1)4 × C(36, 1)] + [4 × C(4, 2) × C(4, 1)3]}/C(52, 5), (b) {C(26, 5) + C(13, 1)2C(26, 3) + C(13, 2)2 × C(26, 1)}/C(52, 5).
49. See Supplement at end of Chapter 5.
51. {C(30, 2) × C(28, 2) × C(26, 2) × C(24, 2)}.
53. 4 × 3 × 8!/2!/2!.
55. (a) See Supplement at end of Chapter 5.
57. C(64, 8) × C(56, 8).
59. See Supplement at end of Chapter 5.
61. C(6, 3) × C(4, 2) + 2 × [C(6, 2) × C(4, 2) + C(6, 3) × C(4, 1)] + [6 × C(4, 2) + C(6, 2) × C(4, 1) + C(6, 3)] 63. See Supplement at end of Chapter 5.
65. (a) C(10, 2) × C(k −10, 18)/C(k, 20), (b) k = 100.
67. C(12, 5) × C(4, 2) × 2 × C(7, 2) × C(5, 2) × C(3, 2).
69. See Supplement at end of Chapter 5.
71. (a) C(45, 3) + C(45, 2) × C(45, 1), (b) [3 × C(30, 3) (three integers with same value mod 3)] + C(30, 1)3 (each integer with different value mod 3), (c) as in (b), break into cases based on the value mod 4: C(22, 3) + [C(23, 2) × C(22, 1)] + [2 × C(23, 2) × C(45, 1)] + [C(23, 1) × C(22, 2)] + 222 × 23.
73. C(11, 6) × C(5, 2) × 5!/2!.
75. P(20, 3)6, 77. P(C(8, 2), 12), 79. 8!/2!2! + (8!/2! −7!/2!) −(7! −6!/2!) www.itpub.net Solutions to Odd-Numbered Problems 459 81. C(8, 3), C(8, 3) −(8 × 5).
83. See Supplement at end of Chapter 5.
85. (a) C(10, 6), (b) C(10, 6) + [5 × C(10, 5)] + [4 × C(10, 4)] + C(10, 3).
87. See Supplement at end of Chapter 5.
Section 5.3 1. 7!/3!2!2!.
3. (a) 38, (b) (8!/3!2!3!)/38.
5. C(9 + 4 −1, 9).
7. C(11 + 3 −1, 11), subtract cases where one party has a majority (9 or more) C(11 + 3 −1, 11) −[3 × C(4 + 3 −1, 4)].
9. (a) C(9 + 3 −1, 9), (b) C(16 + 3 −1, 16) −C(5 + 3 −1, 5).
11. C(5, 2) × C(6 + 2 −1, 6).
13. 6!/3!2!1! + 6!/2!2!1!1!.
15. C(10, 6) × {[C(6, 2) × 8!/2!2!1!4] + [6 × 8!/3!1!5]}.
17. Considerthecasesof:(i)4ofoneletterand1ofanotherletter,(ii)3ofonekindand 2 of another letter, (iii) 3 of one letter and 1 of two others, (iv) 2 of two letters and 1 of another, and (v) 2 of one letter and 1 of three others—(2 × 3 × 5!/4!1!) + (2 × 2 × 5!/3!2!) + [2 × C(3, 2) × 5!/3!1!1!] + [C(3, 2) × 2 × 5!/2!2!1!] + (3 × 5!/ 2!1!3).
19. 3 × 10!/2!4!.
21. 10!/2!2! −9!/2!2!.
23. (a) 10!/4!3!2!, (b) 2 × 10!/4!3!2! −9!/4!2!2!.
25. C(8 + 3 −1, 8) −C(5 + 3 −1, 5).
27. See Supplement at end of Chapter 5.
29. Counts all (1, 2, 3)-sequences of length 10 two ways: left side looks at all cases of k1 1s, k2 2s, and k2 3s, while right side counts all (unrestricted) 10-digit sequences of 1s, 2s, 3s.
31. See Supplement at end of Chapter 5.
33. See Supplement at end of Chapter 5.
35. Must have b or d after each c (except possibly last c); sum is over number of bs following cs (first sum when last c followed by b or d, second sum when last c is at end of sequence: C(3, k)21!/7!3!(8 −k)![6 −(3 −k)]! + C(2, k)20!/7!2!
(8 −k)![6 −(2 −k)]!.
Section 5.4 1. (a) C(27 + 3 −1, 27), (b) 1, (c) C(24 + 3 −1, 24).
3. (a) 13!/4!4!3!2!/C(52, 13), (b) [13!/4!4!2!3! × 39!/9!9!11!10!]/(52!/13!4), 460 Solutions to Odd-Numbered Problems (c) 4 × C(48, 9)/C(52, 13), (d) 4! × (13!/4!3!3)4/(52!/13!)4.
5. C(10 + 4 −1, 10) × C(4 + 4 −1, 4) × C(6 + 4 −1, 6).
7. C[(5 −2) + 4 −1, (5 −2)] × 5!.
9. (a) C(9, 4) × 8!/2! × 5!/2!2!, (b) C(9, 4) × 8!/2!.
11. C[(14 −3) + (5 −1), (14 −3)]/C(18, 14).
13. C(15 + 3 −1, 15).
15. (a) [52!/13!4]/4!, (b) [52!/8!37!4]/3! × 4!.
17. 1 −(10! −10 × 9 × 10!/2!)/1010.
19. (a) 43 × C(9 + 4 −1, 9), (b) P(4, 3) × C(9 + 4 −1, 9), (c) Distribute teddies and fill out each child with lollipops: 43.
21. 4 × C(4 + 4 −1, 4).
23. 2 × C(7 + 4 −1, 7) −1.
25. C[(7 −3) + 5 −1, (7 −3)] × 7!/4!2!1!.
27. See Supplement of Selected Solutions at end of Chapter 5.
29. [10!/2!5]/510.
31. (a) Distributions of 8 distinct items into 3 boxes, (b) Distributions of 9 distinct items into 3 boxes with two items in the first box, etc.
33. (a) Distributions of 6 identical objects into 31 boxes, 31 i=1 xi = 6, (b) Distributions of 5 identical objects into 3 boxes with at most 4 objects in first box, etc., 3 i=1 xi = 5, x1 ≤5, x2 ≤4, x3 ≤2.
35. C(30 + 3 −1, 30), 3 × C[(30 −16) + 3 −1, (30 −16)].
37. (a) C(7 + 4 −1, 7), (b) C(7 + 5 −1, 7) + 1 (c) [C(13 + 4 −1, 13)] −[4 × C(3 + 4 −1)].
39. 6 k=0 C(k + 2 −1, k) × C[(12 −2k) + 2 −1, (12 −2k)].
41. 7 k=0 C(7 + 3 −1, 7) × C[(20 −k) + 4 −1, (20 −k)].
43. (a) 5 × 6 × 8, (b) (5 × 6 × 8) −2.
45. 4! × C(4 + 5 −1, 4).
47. See Supplement at end of Chapter 5.
49. C(3 + 4 −1, 3) × 5!/3!
51. (a) 5! × C(9, 5), (b) 21! × 5! × C(22, 5).
53. 13 k=0 C(13, k) × C(39, 13 −k) × C(26 + k, k) × C(26, 13).
55. 15!/a!b!c! summing over all a, b, c 3-tuples where a + b + c = 15 with no letter greater than 7.
57. (3! × 25!/16!8!1!) + (3 × 25!/14!7!4! × 24) + {3 × [25!/12!6!7! × (27 −2) − 25!/6!6!1!]} + 25!/10!5!3.
59. See Supplement at end of Chapter 5.
61. See Supplement at end of Chapter 5.
63. C[(n −2m) + (2m + 2) −1, (n −2m)]{simplifies to C(n + 1, 2m + 1).
65. 5 k=0 C(k + m −1, k) × C[(r −k) + (n −m) −1, (r −k)].
www.itpub.net Solutions to Odd-Numbered Problems 461 Section 5.5 11. (b) [C(n + 1, 2)] + [6 × C(n + 1, 3)] + [6 × C(n + 1, 4)].
13. (a) 2n + (n × 2n−1), (b) 1.5 × 2n.
21.
1 2 × C(2n + 2, n + 1) −C(2n, n).
23. 0.
CHAPTER SIX SOLUTIONS Section 6.1 1. (a) 7 products—xxxx, x311x, x31x1, x3x11, 1x31x, 1x3x1, xx311, (b) 5 products—1x4, xx3, x2x2, x3x, x41, (c) 7 products—x4111, 1x411, x311x, x31x1, 1x31x, 1x3x1, 11x2x2, (d) 15 products—x411, x3x1, x31x, x2x21, x2xx, x21x2, xx31, xx2x, xxx2, x1x3, 1x41, 1x3x, 1x2x2, 1xx3, 11x4.
3. (a) (1 + x + x2 + x3 + x4 + x5)2(1 + x + x2 + x3 + x4), (b) (x + x2 + x3 + x4 + x5)2(x + x2 + · · · + x7 + x8), (c) (1 + x + x2 + · · ·)4, (d) (x + x3 + x5 + · · ·)2(1 + x + x2 + · · ·)4.
5. (1 + x + x2 + · · ·)2(1 + x)2, coef. of x5.
7. (x3 + x4 + x5 + · · ·)4, coefficient of x16.
9. (1 + x + x2 + · · ·)n.
11. (1 + x2 + x4 + · · ·)(x + x3 + x5 + · · ·)(1 + x + x2 + · · ·)n−2.
13. (x + x2 + x3 + x4 + x5 + x6)n.
15. (x−3 + x−2 + x−1 + 1 + x + x2 + x3)4.
17. (1 + x5 + x10 + · · ·)8.
19. (1 + x + x2 + · · ·)(x + x2 + x3 + · · ·)4(1 + x + x2 + · · ·), coefficient of x15, gen-erally xn−5.
21. Cannot have a variable number of factors.
23. (1 + x + x2 + · · ·)(1 + x5 + x10 + · · ·)(1 + x10 + x20 + · · ·).
25. (1 + x + x2 + · · ·)5(1 + y + y2 + y3)5.
27. (xy + xz + yz)8.
29. C(p, n), n!.
Section 6.2 1. C(10 + n −1, 10).
3. C(m, 7) + C(m, 5) + C(m, 3).
5. C(11 + 3 −1, 11) −C(3, 1) × C(5 + 3 −1, 5).
7. C(11 + 7 −1, 11) −7 × C(6 + 7 −1, 6) + C(7, 2) × C(1 + 7 −1, 1).
9. C(12 + 4 −1, 12) −4 × C(5 + 4 −1, 5).
11. (a) C(9 + 10 −1, 9), (b) C(9 + 4 −1, 9) −3 × C(10 + 4 −1, 10), 462 Solutions to Odd-Numbered Problems (c) 0, (d) C(10 + 2 −1, 10) + 3 × C(11 + 2 −1, 11), (e) bm × C(11, m).
13. 0.
15. (a) 0, (b) 1, (c) C(12 + 8 −1, 12), (d) 412 × C(12 + 5 −1, 12), (e) C(4 + 4 −1, 4).
17. (a) (x2 + x3 + x4 + . . .)3, C((10 −6) + 3 −1, (10 −6)), (b) (1 + x + x2)(1 + x + x2 + . . .)2, C(10 + 3 −1, 10) −C(7 + 3 −1, 7), (c) (1 + x2 + x4 + . . .)(1 + x + x2 + . . .)2, C((10 −2k) + 2 −1, (10 −2k)).
19. C(12 + 5 −1, 12) −5 × C(7 + 5 −1, 7) + C(5, 2) × C(2 + 5 −1, 2).
21. (a) C(15 + n −1, 15) + C(10 + n −1, 10), (b) C(n, 15) + C(n, 10).
23. C(8 + 7 −1, 8) −7 × C(3 + 7 −1, 3).
25. C(6 + 3 −1, 6) −C(3 + 3 −1, 3) −C(2 + 3 −1, 2).
27. C(14 + 10 −1, 14) −4 × C(10 + 10 −1, 10) −6 × C(7 + 10 −1, 7) + C(4, 2) × C(6 + 10 −1, 6) + 4 × 6 × C(3 + 10 −1, 3) −4 × C(2 + 10 −1, 2)+ C(6, 2).
29. [C(10 + 4 −1, 10) −4 × C(4 + 4 −1, 4)] × [C(15 + 4 −1, 15) −4 × C(9 + 4 −1, 9)} + C(4, 2) × C(3 + 4 −1, 3)].
33. (a) C(m + n, m +r), (b) 0, if r odd, C(n,r/2), r even, (c) 3n.
35. (b) 2.
39. (b) n/ 2, pn, 10, m/p.
41. Px(t) = 1 2 m 1 −(t/2)s+1 1 −(t/2) m .
Section 6.3 1. (a) 5 partitions—4, 3 + 1, 2 + 1 + 1, 2 + 2, 1 + 1 + 1 + 1, (b) 11 partitions—6, 5 + 1, 4 + 2, 4 + 1 + 1, 3 + 3, 3 + 2 + 1, 3 + 1 + 1 + 1, 2 + 2 + 2, 2 + 2 + 1 + 1, 2 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1 + 1.
3. (1 + x + x2 + x3)(1 + x2 + x4 + x6)(1 + x3 + x6 + x9) . . ..
5. (1 −x)−1(1 −x5)−1(1 −x10)−1(1 −x25)−1.
7. (b) (1 + x)(1 + x2)(1 + x3) . . . = 1 −x2 1 −x 1 −x4 1 −x2 1 −x6 1 −x3 1 −x8 1 −x4 . . .
= (after canceling) 1 (1 −x)(1 −x3)(1 −x5) . . ..
19. (a) Multiply the partition generating function (given just before Example 1) times 1 1 −x ; (b) (x3 + x6)/(1 −x3)(1 −x4)(1 −x6), (c) Same as (b).
www.itpub.net Solutions to Odd-Numbered Problems 463 21. (a) Let the number of dots forming the first row and first column (= 2k −1 if first row and column have length k) in a self-conjugate Ferrers diagram be the length of the first row in the distinct, odd-parts diagram; delete the first row and column of the self-conjugate diagram and use the number of dots in the reduced self-conjugate diagram to define the second row of the distinct, odd-parts diagram, etc.
Section 6.4 1.
1 + x + x2 2! + x3 3! + x4 4! + x5 5!
5 .
3. (1 + x)5e21x.
5.
1 + x + x2 2! + x3 3! + x4 4!
13 , coefficient of x13/13!.
7. (a) 1/2(3r + 1), (b) 1/4[3r + 2 + (−1)r], (c) 3r −2 × 2r + 1.
9. (a) 2210 + 4 × P(10, 1) × 229 + 6 × P(10, 2) × 228 + 4 × P(10, 3) × 227 + P(10, 4) × 226, (b) 2610 −4 × 2510 + 6 × 2410 −4 × 2310 + 2210.
11.
1 24r.
13. (a) ex −2 × e(n−1)x/n + e(n−2)x/n, (b) [nr −2(n −1)r + (n −2)r]/nr.
15. (a) (ex −1)/x, (b) (1 −x)−1.
21. eμ(x−1).
Section 6.5 1. (a) x/(1 −x)2, (b) 13/(1 −x), (c) 3x(1 + x)/(1 −x)3, (d) [3x/(1 −x)2] + [7/(1 −x)], (e) 4!x4/(1 −x)5.
3. x(3 −x)/(1 −x)3.
5. (a) (4x2 −3x + 1)/(1 −x)3, (b) log e(1 −x).
CHAPTER SEVEN SOLUTIONS Section 7.1 1. an = 5an−1, a1 = 5.
3. an = 2an−1 + an−2.
5. an = an−1 + 2an−5 + an−10 + an−25.
7. an = an−1 + an−2.
464 Solutions to Odd-Numbered Problems 9. an = an−1 + an−2.
11. an = an−1 + an−2.
13. (a) an = an−1 + n, n > k, initial condition: ak = k + 1, (b) 43.
15. an = 1.06(an−1 + 50).
17. an = 2an−1 + 2an−2 + 2an−3.
19. (a) an = an−1 + an−3 + an−5, (b) an = an−1 + an−3 + an−5 −an−6, (c) an = an−1 + an−3 + an−5 −an−9.
21. an = an−1 + 3n−1.
23. an = 2an−1 + 4n−1.
25. an = 3an−1 −an−3.
27. an = (2n −1)an−1.
29. an,k = an,k−1 + an−1,k−1 + an−2,k−1 + an−3,k−1.
31. an,k = an−1,k−1 + an−k,k.
33. an = bn−1 + cn−1, bn = cn = 2n−1 −bn−1 −cn−1.
35. an = such sequences starting with a 0, bn = such sequences starting with a 1, cn = n-digit binary sequences with no consecutive 0s, an = bn−1 + cn−3, bn = an−1 + bn−1, cn = cn−1 + cn−2.
37. pn = ways to hand out a penny, nickel, or dime on successive days with a penny on the first day, nn and dn are defined similary, pn = nn−1 + dn−1, nn = pn−5 + dn−5, dn = pn−10 + nn−10.
39. an = an−1a3 + an−2a4 + · · · + a3an−1.
41. (a) 3, 0, (b) 2n −1, 2, (c) 3n2 −3n + 1, 6n −6.
43. (a) an = 2an−1, (b) an = 2n−1, (c) The first k integers in the sequence will form a set of consecutive integers, the (k + 1)-st integer can be the next larger or the next smaller number to extend this consecutive set.
45. an = an−1, n not a multiple of 5, an = an−1 + 1, n a multiple of 5 but not 10 or 25, an = an−1 + 2, n a multiple of 10 or 25.
47. an = an−3, n even, an = an−3 + n + 1 4 .
Section 7.2 1. (a) An −5, (b) An1/2 + 2n, (c) A + 4n −⌊log2(n)⌋, (d) An −2, (e) an = An4 −5n/7, (f) an = An2 −3n.
3. (a) an = an/10 + 1, an = log10 n, (b) an = 10an/10 + 1, an = 1 9n −1 9.
5. an = 2an/2 + n −1, an = n log2 n −n + 1.
9. (a) Pick largest from first half and largest from second half and compare, www.itpub.net Solutions to Odd-Numbered Problems 465 (b) an = 2an/2 + 1, (c) an = n −1.
Section 7.3 1. (1.08)n × 1000.
3. (a) an = 2 54n + 3 5(−1)n, (b) an = 1, (c) an = 2, (d) an = 1 2n2 −1 2n + 1.
5. an = 2an−1 + an−2, an = 2 + √ 2 4 (1 + √ 2)n + 2 − √ 2 4 (1 − √ 2)n.
7. pn −pn−1 = 2(pn−1 −pn−2), pn = (3 × 2n) −2.
11. c1 = 9, c2 = −18.
Section 7.4 1. (a) an = 3C(n, 2) + 1, (b) an = 2C(n + 1, 3) + 3, (c) an = 6C(n + 2, 3) −3C(n + 1, 2) + 10.
3. an = an−1 + 2C(n, 2) + n, an = 2C(n + 1, 3) + C(n + 1, 2).
5. an = an−1 + n−3 k=1{k × (n −2 −k) + 1}, an = C(n, 4) + C(n, 2) −n + 1.
7. an = 1250(1.04)n −1250.
9. (a) −3n + 1, (b) 5 32n + 1 3(−1)n, (c) (3 × 2n) −n −2, (d) (15 × 2n) −2n2 −8n −12.
11. 3 × 2n −3n −2.
13. an = 2an−1 −an−2 + (10 × 2k), an = 960n −20 + (40 × 2n).
17. A × 2n + B × 2n + 3 2n + 25 4 .
19. (a) an = √ 2n+1 −1, (b) an = n!, n even, an = 0, n odd.
Section 7.5 1. (a) g(x) −1 = xg(x) + 2x 1 −x , (b) g(x) −x −1 = 3xg(x) −3 −2x2g(x) + 2x2 1 −x , (c) g(x) −1 = xg(x) + 2x2 (1 −x)3 , (d) g(x) −1 = 2xg(x) + 2x 1 −2x .
7. an = a2an−1 + a3an −2 + · · · + an−1a2, n ≥3, a2 = 1, an = 1 n −1 C(2n −4, n −2).
9. an = a1an−1 + a2an−2 + · · · + an−1a1, n ≥2, a1 = 1, an = 1 n C(2n −2, n −1).
11. an,k = pan−1,k−1 + qan−1,k, Fn(x) = (q + px)n.
466 Solutions to Odd-Numbered Problems 13. Similar to recurrence relations in Example 5 except with 4n−1 replacing 3n−1 and a1 = 0 instead of a1 = 1 (still b1 = c1 = 1), an = 3 15(4n −1), n even, = 2 15(4n −4), n odd.
15. (a) an = C(n −1, k −1)an−k, (b) g(x) = eex−1.
CHAPTER EIGHT SOLUTIONS Section 8.1 1. 268 −212 × 266.
3. 3n −3 × 2n−1.
5. {C(52, 7) −C(13, 7) × C(4, 1)7}/C(52, 7).
7. 5 × (94 −9 × 8 × C(4, 3).
9. 700 −200 −180 −150.
11. (a) 300 −70 −100 + 40.
(b) 300 −100 −60.
13. (142 + 500 −71)/1000.
15. (a) 200 −3 × 85 + 3 × 30 −15, (b) 85 −2 × 30 + 15.
17. 30 −15 −10 −6 + 5 + 3 + 2 −1.
19. 320 −3 × 220 + 3 × 1.
21. 6!/2!3 −3 × 5!/2!2 + 3 × 4!/2! −3!.
23. C(37, 10) −[C(27, 10) + C(25, 10) + C(22, 10)] + [C(15, 10) + C(12, 10) + 1].
25. C(52, 6) −3 × C(48, 6) + 3 × C(44, 6) −C(40, 6).
27. 3 × C(6, 2) × 4! −{2 × C(6, 3) × 3! + 6!/2!2} + C(6, 4) × 2!.
29. (15!/3!5 −3 × 5! × 10!/2!5 + 3 × 5!3 −5!3)/(15!/3!5).
31. 20.
33. N(Y) = N(Y −K) + N(Y ∩K) = 50 + 20.
35. 25, 5.
Section 8.2 1. 10m −3 × 9m + 3 × 8m −7m.
3. 13 × C(48, 5) −C(13, 2) × 44.
5. (a) {C(52, 13) −C(4, 1) × C(39, 13) + C(4, 2) × C(26, 13) −C(4, 3) × C(13, 13)}/C(52, 13), (b) {C(4, 1) × C(39, 13) −C(4, 2) × C(26, 13) + C(4, 3) × C(13, 13)}/C(52, 13), (c) {C(52, 13) −C(4, 1) × C(48, 13) + C(4, 2) × C(44, 13) −C(4, 3) × C(40, 13) + C(36, 13)}/C(52, 13).
7. 9!/3!3 −3 × 7!/3!2 + 3 × 5!/3! −3!.
9. 26! −{3 × 23! + 24!} + {2 × 20! + 2 × 21!} −18!.
www.itpub.net Solutions to Odd-Numbered Problems 467 11. C(26 + 6 −1, 26) −3 × C(19 + 6 −1, 19) + 3 × C(12 + 6 −1, 12) −C(5 + 6 −1, 5).
13. C(10 + 4 −1, 10) −C(4, 1) × C(8 + 3 −1, 8) + C(4, 2) × 7 −C(4, 3) × 1.
15. 27.
17. ≈10!2/e.
19. 10!/2!5 × {10!/2!5 −C(5, 1) × 8!/2!4 + C(5, 2) × 6!/2!3 −C(5, 3) × 4!/2!2 + C(5, 4) −C(5, 5)}.
21. 15!/3!5 −5 × 5 × 12!/3!4 + C(5, 2) × 5 × 4 × 9!/3!3 −C(5, 3) × 5 × 4 × 3 × 6!/3!2 + 5 × 5! −5!.
23. (a) n5 −C(5, 1) × n4 + C(5, 2) × n3 −C(5, 3) × n2 + {C(5, 4) −C(5, 5)} × n; (b) n5 −C(5, 1) × n4 + C(5, 2) × n3 −{(C(5, 3) −1) × n2 + n3} + {(C(5, 4) −2) × n + 2 × n2} −n.
(c) n5 −C(7, 1) × n4 + C(7, 2) × n3 −{C(7, 3) −3) × n2 + 3 × n3} + {(C(7, 4) −14) × n + 14 × n2} −{(C(7, 5) −2) × n + 2 × n2} + {C(7, 6) −C(7, 7)} × n.
27. If 21 = C(2 + 6 −1, 2), then (−1)k × C(6, k) × (21 −k)n.
29. 5!n × {(2n −1)! − C(n, k) × (2n −1 −k)!}.
31. P(C(7, 3), 7) −7 × P(C(6, 3), 7) + C(7, 2) × P(C(5, 3), 7).
33. C(P(6, 3), 8) −6 × C(P(5, 3), 8) + C(6, 2) × C(P(4, 3), 8).
35.
(−1)k × C(n, k) × (n −k)r /n!.
37. n−1 k=3(−1)k+1 × C(k, 3) × C(n, k) × (n −k)r, n−1 k=3(−1)k+1 × C(k −1, 2) × C(n, k) × (n −k)r.
39. [C(5, 2) × 9!/2!3] −[2 × C(5, 3) × 8!/2!2] + [3 × C(5, 4) × 7!/2!] −[4 × 6!].
47. n k=0 k × n j=k(−1) j−k × C( j, k) × n!/j!
.
Section 8.3 1. 5 × 5 board with darkened squares on main diagonal.
3. 5! −8 × 4! + 20 × 3! −16 × 2! + 4 × 1!.
5. 7! −(9 × 6!) + (30 × 5!) −(46 × 4!) + (32 × 3!) −(8 × 2!).
7. 5! −(7 × 4!) + (16 × 3!) −(13 × 2!) + (2 × 1!).
9. 3.
11. (a) 4 × 5boardwithdarkenedsquaresin4positionsjusttorightofmaindiagonal, (b) (x + 1)4, (c) 5 j=k(−1)k+ j × C( j, k) × C(n −1, j) × (n −j)!.
13. 2 × 2 array of darkened squares and “L” (a column of 3 squares beside a single square both have 1 + 4x + 2x2).
CHAPTER NINE SOLUTIONS Section 9.1 1. (a) Not symmetric, (b) Yes, (c) Not transitive, 468 Solutions to Odd-Numbered Problems (d) Not transitive, (e) Not transitive.
3. (a) 6 symmetries (as in Example 3), (b) 4 symmetries, (c) 1 symmetry.
5. (a) All C j left fixed, (b) C1C2C3C4C5C6C7C8C9C10C11C12C13C14C15C16 C1C3C4C5C2C7C8C9C6C11C10C13C14C15C12C16 (c) C1C2C3C4C5C6C7C8C9C10C11C12C13C14C15C16 C1C3C2C5C4C6C9C8C7C11C10C15C14C13C12C16 (d) C1C2C3C4C5C6C7C8C9C10C11C12C13C14C15C16 C1C2C5C4C3C9C8C7C6C10C11C14C13C12C15C16 11. (a) a, b, c are rotations or 0◦, 120◦, and 240◦, respectively d, e, f are flips around vertical axis, axis 30◦clockwise of vertical, and axis 30◦counterclockwise of vertical; row is first symmetry, column second symmetry: a b c d e f a a b c d e f b b c a f d e c c a b e f d d d e f a b c e e f d c a b f f d e b c a (b) Straightforward.
(c) Let a = 1 2 3 4 1 2 3 4 , b = 1 2 3 4 2 1 4 3 , c = 1 2 3 4 3 4 1 2 , d = 1 2 3 4 4 3 2 1 , a b c d a a b c d b b a d c c c d a b d d c b a 13. Only left structure (right structure has 6 isomers).
21. (b) {π1, π3, π5, π6} or {π1, π3, π7, π8}, (c) In addition to subgroups in (b) and G, G′, G′′, other subgroups are {π1πi}, for i = 3, 5, 6, 8.
Section 9.2 1. (a) 24, (b) 70.
3.
1 3[315 + (2 × 35)].
5. 51.
www.itpub.net Solutions to Odd-Numbered Problems 469 7.
1 2(5n + 5n/2) n even, and 1 2[5n + (3 × 5(n−1)/2)] n even.
9. (a) In cycle form: π1 = (1)(2)(3), π2 = (12)(3), π3 = (13)(2), π4 = (1)(23), π5 = (123), π6 = (132), (b) (π1) = C(12 + 3 −1, 12) = 91, (π2) = (π3) = (π4) = 7, (π5) = (π6) = 1, answer: 1 6 (91 + 7 + 7 + 7 + 1 + 1) = 19.
11. (π1) = 18, (π3) = 6, (π7) = (π8) = 12, and other (πi)′ = 0, answer 1 8(18 + 6 + 12 + 12) = 6; 13. (b) {π1, π7}, (c) {π1, π6}.
Section 9.3 1. 55.
3. (a) 130, (b) 92, (c) Cyclic color sequence on hexagon of R-W-B-R-W-W and R-B-W-R-W-W.
5. (a) 1 6(m4 + 2m2 + 3m3), (b) 1 8(m12 + 2m3 + 3m6 + 2m7), (c) 1 2(m5 + m3), (d) 1 4(m8 + 3m4), (e) 1 12(m7 + 2m2 + 2m3 + 4m4 + 3m5), (f) Same as (b).
7. (a) 1 6(m6 + 2m2 + 3m4), (b) 1 8(m12 + 2m3 + 3m6 + 2m7), (c) 1 2(m6 + m4), (d) 1 4(m10 + m5 + m6 + m7), (e) 1 12(m12 + 2m2 + 2m4 + m6 + 6m7), (f) 1 8(m16 + 2m4 + m8 + 4m9).
9.
1 4[74 + (3 × 72)] = 637.
11. (a) 1 2(28 + 24) = 136, (b) 1 4(28 + 24 + 0 + 24) = 72.
13. (πi) = number of cycles of length 1.
15. (a) 1 p[m p + (p −1) × m], (b) 1 2p{m p + [(p −1) × m] + (p × m(p+1)/2)}.
Section 9.4 1. b5 + b4w + 2b3w2 + 2b2w2 + bw4 + w5.
3. b4 + w4 +r4 + b3w + b3r + bw3 + w3r + br3 + wr3 + 2b2w2 + 2b2r2 + 2w2r2 + 2b2wr + 2bw2r + 2bwr2.
5. (a) 1 6{(b + w)4 + 2(b3 + w3)(b + w) + 3(b2 + w2)(b + w)2}, (b) 1 8{(b + w)12 + 2(b4 + w4)3 + 3(b2 + w2)6 + 2(b2 + w2)5(b + w)2}, (c) 1 2{(b + w)5 + (b + w)(b2 + w2)2}, 470 Solutions to Odd-Numbered Problems (d) 1 4{(b + w)8 + 3(b2 + w2)4}, (e) 1 12{(b + w)7 + 2(b6 + w6)(b + w) + 2(b3 + w3)2(b + w) + 4(b2 + w2)3 (b + w) + 3(b2 + w2)2(b + w)3}, (f) Same as (b).
7. (a) 1 6{(b + w)6 + 2(b3 + w3)2 + 3(b2 + w2)2(b + w)2}, (b) 1 8{(b + w)12 + 2(b4 + w4)3 + 3(b2 + w2)6 + 2(b2 + w2)5(b + w)2}, (c) 1 2{(b + w)6 + (b + w)2(b2 + w2)2}, (d) 1 4{(b + w)10 + (b2 + w2)5 + (b + w)2(b2 + w2)4 + (b + w)4(b2 + w2)3}, (e) 1 12{(b + w)12 + 2(b6 + w6)2 + 2(b3 + w3)4 + (b2 + w2)6 + 6(b2 + w2)5 (b + w)2}, (f) 1 8{(b + w)16 + 2(b4 + w4)4 + (b2 + w2)8 + 4(b2 + w2)7(b + w)2}.
9. (a) 1 12{(b + w)4 + 8(b3 + w3)(b + w) + 3(b2 + w2)2}, (b) 1 24{(b + w)6 + 6(b4 + w4)(b + w)2 + 3(b2 + w2)2(b + w)2 + 6(b2 + w2)3 + 8(b3 + w3)2}.
11.
1 24{(b + w)4 + 6(b4 + b4) + 8(b3 + w3)(b + w) + 3(b2 + w2)2 + 6(b2 + w2) (b + w)2}.
13. (a) If not a cyclic rotation of all corners, the length of the cycle would have to divide p—impossible, (b) C(p, k)/p.
15. (a) 36, (b) 216.
CHAPTER TEN SOLUTIONS Section 10.1 1. (a) {a, c} or {b, d}, (b) f , (c) No kernel, consider directed 5-circuit b, a, d, g, h, b, a—if a is K (kernel), then d not in K, then g in K, then h not in K, then b in K—impossible since a in K; similar sort of argument (also involving c, f, e) if a not in K.
3. {3, 4, 9, 11, 12, 16, 17, 21, 25, 26, 27, 31, 32, 36, over 40}.
5. A goes to 2, B must go to 4 or else A will win.
7. Move to multiples of 5k + 1 (initial position is win for second player).
9. By symmetry assume g(a) = 0, then g(e) = 1, then g(d) = 0, then g(c) = 1, then g(b) = 0, but now two kernel vertices are adjacent.
11. S is a kernel if and only if all vertices not in S have an edge to a vertex in S while no vertex in S has an edge to a vertex in S—that is, if and only if W(S) = S.
13. Follows from parts (a) and (b) of Exercise 12 since g(x) = k means there is a path of length k starting at x while l(x) is length of longest path starting at x; longest path is at least length k (maybe there is another longer path starting at x).
www.itpub.net Solutions to Odd-Numbered Problems 471 15. Suppose x and y are adjacent because there is an edge from x to y; then x must have a larger level number than y and a different Grundy value from its successor y.
17. If there were an infinite number of vertices, then one of the finite number of starting vertices, call it x1, must have an infinite number of vertices reachable from it, and one of the finite number of successors of x1, call it x2, must have an infinite number of vertices reachable from it, and one of the finite number of successors of x2, call it x3, must have an infinite number of vertices reachable from it, and so on, without end.
19. (a) Let a = 0000, b = 000, c = 0−00, d = 00−0, e = 00, f = 0−0, g = 0, h =−(win); s(a) = {b, c, d, e, f }, s(b) = {e, f, g}, s(c) = s(d) = {e, f, g}, s(e) = {g, h}, s( f ) = {g}, s(g) = h, s(h) = Ø; g( f ) = g(h) = 0, g(a) = g(g) = 1, g(e) = 2, g(b) = g(c) = g(d) = 3.
Section 10.2 1. (a) 7, remove 3 from 4th pile, (b) 0, (c) 4, remove 4 from 2nd, 3rd, or 4th pile, (d) 0.
3. (a) 3, remove 3 from 3rd pile or 2 from 4th pile, (b) 2, remove 2 from 3rd or 4th pile, (c) 0, (d) 0.
5. (a) 0, (b) 0, (c) 1, remove 1 from 4th pile, (d) 3, remove 1 from 3rd pile.
7. (a) 3, add nickel to 3rd pile, (b) (0, 0), (0, 4), (0, 6), (0, 9), (1, 1), (2, 2), (2, 5), (2, 8), (3, 3), (3, 7), (4, 4), (4, 6), (5, 5).
9. If c j = d j, then trivially c′ ˙ + c j = c′ ˙ + d j; if c′ + c j = c′ ˙ + d j, then c j and d j must have 1s in the same positions in their binary representations—that is, they must be equal.
11. Immediately the proposed strategy works.
13. (a) Remove three balls along one of the three lines formed by ball on one of the three sides of the arrangement.
POSTLUDE SOLUTIONS 1. THEORIES 3. COMPLETE 5. FAMILY 7. UNIFORM 9. VERTICES 472 Solutions to Odd-Numbered Problems APPENDIX SOLUTIONS Section A.1 1. (a) 12, 27 (b) 2, 3, 6, 7, 9, 12, 15, 17, 18, 21, 22, 24, 27, (c) 1, 4, 5, 8, 10, 11, 13, 14, 16, 19, 20, 23, 25, 26, 28, 29, (d) all 1 ≤k ≤29 except 12, 27.
3. We are given N(R ∩M) = 2 as well as that N(M) = N(R) = N(M) = N(R) = 4; then N(R ∩M) = N(M) −N(R ∩M) = 4 −2 = 2, N(R ∩M) = N(R) −N(R ∩M) = 4 −2 = 2, and clearly N(R ∩M) = 2.
5. (a) Impossible, (b) Yes, 20 −8 −8 = 4, (c) 20 −15 = 5.
7.
A B A B (a) (b) (c) (A ∪B) ∩(A ∩B) = (A ∩B), see Figure A1.3, (d) A −(B −A) = A; here all expressions involve ∩, so that ABC = A ∩B ∩C.
9.
ABC ABC ABC ABC ABC _ _ _ _ ABC _ _ _ _ ABC _ _ _ _ ABC _ _ _ __ 17. (a) E = S ∪H ∪C, 523 −393, (b) E = (S ∪H ∪C) ∩(S ∩H ∩C), 523 −393 −133, (c) E = (S ∩H ∩C) ∪(S ∩H ∩C) ∪(S ∩H ∩C), 3 × 132 × 39.
Section A.2 23. One can prove by induction only a property that is a function of n—for example, one can prove that there are a finite number of binary sequences of length ≤n.
www.itpub.net Solutions to Odd-Numbered Problems 473 25. The initial step only assumes that n ≥1, not n ≥2, but for n = 1, an−2 is unde-fined.
Section A.3 1. 1/2.
3. (a) 1/6, (b) 18/36 = 1/2, (c) 3/36 = 1/12.
5. (a) 1/6, (b) 1/2.
7. (a) 1/3 × 1/3 = 1/9, (b) 2 × 1/3 × 2/3 = 4/9.
9. (2 × 2 × 2)/4! = 1/3.
11. 1 −[(50 + 40 −20)/100] = 3/10.
13. 2/3.
15. (a) All sequences with k tails, 0 ≤k ≤8, and one head followed by a head, (b) All positive integers, (c) All ordered pairs of positive integers, (d) All sequences of k black balls, k ≥0, followed by a red ball.
Section A.4 1. n + 1.
15. Printer i is connected to computers i, i + 1, i + 2, i + 3, i + 4, i + 5.
This page is intentionally left blank www.itpub.net INDEX A AC Principle, 16 Addition Principle, 180 Adjacency matrix, 103 Adjacent vertices, 3 Adler, I., 316, 317 Agnarsson, G., 439 Ahuja, R., 128, 174, 175 Aldous, J., 439 Allenby, R., 440 Ancestors in a tree, 94, 438 Andreescu, T., 440 Appel, K., 32, 69, 80, 86 Arrangement(s), 190, 435 with repetition, 206 Assignment problem, 121, 175 Art gallery problem, 79 Augmenting flow algorithm, 142 Ault, L., xiii, xv Automorphism of a graph, 48 B Backtracking in a graph, 104, 438 Balakrishnan, V., 439, 440 Balanced tree, 96, 438 Ball, W., 400 Barnette, D., 86 Berge, C., 400, 440 Berlekamp, E., 400 Bernoulli, Jacob, 237 Bernoulli, Jacques, 237, 425 Best, S., 439 Biggs, N., 44, 86, 440 Binomial coefficient, 190, 228, 435 Binomial identities, 230, 261, 291, 329 Binomial Theorem, 227 Bipartite graph, 5, 27, 67, 153, 435 deficiency, 163 Birthday paradox, 203 Blockwalking, 230 Boat crossing puzzles, 108 Bogart, K., 439 Bondy, J., 44, 87 Bono, M., 439 Boole, G., 418 Boolean algebra, 416 Bouton, C., 400 Boyer, C., 418 Branch-and-bound search, 113 Breadth-first search, 104, 438 Bridge in a graph, 46 Bridge probabilities, 215 Brook’s Theorem, 80 Brualdi, R., 439 Bubble sort, 122 Buck, R., 198, 237 Burnside’s Lemma, 365 Bussey, W., 421 C C(n,r), 190 Cameron, P., 281, 439 Capacity of a cut, 137 of an edge, 135 Capobianco, M., 440 Cardano, B., 425 Catalan number, 313 Cayley, A., 44, 99, 1125 Center of a tree, 101 Chain in a network, 142 Characteristic equation, 300, 310 Characteristic sequence of a tree, 112 Chartrand, G., 440 475 476 Index Chessboard, generalized, 342 Children in a tree, 94, 438 Chromatic number, 69, 80, 435 Chromatic polynomial, 81, 435 Chvatal’s theorem, V., 61 Circle graph, 51 Circle-chord method, 343 Circuit in a graph, 4, 435 length of a circuit, 27 Closure in a group, 359 Code text, 404 Coin balancing, 98 Color critical graph, 79, 85 Coloring a graph, 69, 334, 433 Combination(s), 190, 435 Complement: of a chessboard, 351 of a graph, 17, 85, 435 of a set, 416 Complete graph, 15, 25, 435 Component of a graph, 25, 436 Computational complexity, 430 Configuration in a graph, 35 Conjugate diagram, 268 Connected graph, 4, 435 Strongly connected, 55 Test for connectedness, 104 Conway, J., 400 Cook, S. 431 Cormen, T., 298, 317 Coxeter, H., 400 Crossing number, 42 Cryptogram, 401 Cube, symmetries of, 379 Cut in network, 136 a-z cut, 136 Cut-set, 47, 151 Cycle, 49, 435 Cycle in a permutation, 359 Cycle index, 371 Cycle structure representation, 370 D David, F., 236, 237, 425 Deadheading edge, 52 Decomposition principle for Instant Insanity, 88 Deficiency of a bipartite graph, 163 Degree of a vertex, 6, 24, in-degree and out-degree, 18 Degree of a region, 38 De Carteblanche, F., 88 De Moivre, A., 281, 316, 351 De Morgan, A., 389, 417, 421 Depth-first search, 104, 438 Derangement, 333, 436 Descendants in a tree, 94, 438 Diameter of a graph, 67 Dictionary search, 5, 97 Difference equation, 290 Difference of sets, 416 Digraph, 401 Digital sum, 395 Dijkstra’s algorithm for shortest paths, 127 Dirac’s theorem, 61 Direct sum of graphs, 394 Directed graph, 3, 436 Dirichlet drawer principle, 427 Distribution, 215, 289, 436 Divide-and-conquer relations, 296 Dual graph, 32 Durfee square, 270 E Edge, 3, 24 directed edge, 3, 436 Edge chromatic number, 80 Edge coloring, 73 Edge cover, 8, 155, 436 Edge coloring, 73 Elements of a set, 415 Eliminating a team from contention, 157 Equivalence relation, 356 equivalence class, 356 Erickson, M., 439 Euler, L., 37, 44, 49, 51, 86, 281 Euler’s constant e, 198, 333 Euler cycle, 50, 112, 436 Euler’s formula for graphs, 37 www.itpub.net Index 477 Euler trail, 53, 436 Event of outcomes, 423 compound event, 424 elementary event, 423 Expected value of a random variable, 265 Experiment, 423 F Factor in a graph, 89 labeled factor, 89 Family of elements, 415 Feil, T., 414 Feller, W., 281 Feng, Z., 440 Fermat, P., 236, 425 Ferrers diagram, 268 Fibonacci, 316 Fibonacci numbers, 281, 285, 302, 310, 316 Fibonacci relation, 285 Fisk’s theorem, 79 Fleury’s algorithm for Euler cycles, 55 Flow in network, 157 dynamic flow, 147 value of a flow, 137 Floyd’s algorithm for shortest paths, 129 Ford, L., 174, 175 Forest of trees, 101 Four-color problem, 32, 69, 80, 86 Fourier, J., 281 Fourier transform, 281 Fulkerson, D., 174, 175 G Gaines, H., 414 Galileo, 425 Garbage collection, 72 Generating functions, 249, 308, 436 exponential, 272 Generators of a group, 362 George, J., 440 Graham, R., 427, 439 Graph, 3, 436 Gray code, 62 Greenlaw, R., 440 Grinberg’s theorem, 61 Gross, J., 440 Group of symmetries, 360 Grundy, P., 390, 400 Grundy function, 390, 395 Guy, R., 400 H Haken, W., 32, 69, 80, 86 Hall, M., 440 Hall’s Marriage Theorem, 156 Halmos, P., 418 Hamilton, W., 86 Hamiltonian circuit and path, 56, 86, 119, 436 Harary, F., 383, 440 Harris, J., 440 Hartsfield, N., 440 Heap, 123 Heap sort, 123 Height of a tree, 96, 438 Hillier, F., 175 Hopcroft, J., 163 Hypercube, 63 I Identical Objects Rule, 425 Identity in a group, 359 Inclusion-exclusion formula, 328 Inclusion-exclusion principle, 320 Independent edges, 153 Independent set, 8, 69, 436 Induction, 420 Initial conditions, 283 Instant Insanity puzzle, 87 Integer solutions of an equation, 217, 250, 332 Interest problems, 287 Internal vertex in a tree, 95, 438 Intersection of sets, 416 Inverse of a symmetry, 359 Isolated vertices, 16 Isomers of organic compounds, 362 Isomorphism of graphs, 14, 436 478 Index K Kn (complete graph on n vertices), 15 Kan, A. R., 126 Kayles, 393 Kernel of game, 387 Keys in Mastermind, xi Keyword, 405 Keyword transpose encoding, 405 Kiefer, J., 316, 317 Kirchhoff, G., 44, 125 Knight’s tour, 55, 67 Knuth, D., 440 Konig, D., 44 Konig-Egevary theorem, 156 Konigsberg bridges, 49 Kruse, R., 126 Kruskal’s algorithm for minimal spanning trees, 131 Kuratowski’s theorem, 35 L Laplace, S., 281, 423, 425 Laplace transform, 281 Lawler, E., 126 Leaves of a tree, 95, 438 Leibnitz, G., 237 Leiserson, C., 298, 317 Lenstra, J., 126 Lesniak, L., 440 Level in a game, 388 Level numbers in a tree, 94, 438 Letter frequencies, 401 Lewand, R., 414 Lieberman, G., 175 Line graph, 41, 48, 55, 68 Linear program, 175 Lloyd, E., 44, 86, 440 Lucas, E., 126 M MacMahon, P., 281 Magnanti, T., 175 Map coloring, 32, 437 Marcus, D., 440 Martin, G., 440 Mastermind, xi Matching in a graph, 4, 153, 437 maximal, 153 X-matching, 153 Matching network, 154 Maurolycus, 421 Max flow-min cut theorem, 145 Maximal planar graph, 42 Maze searching, 106 Mazur, D.,439 Member of a set, 415 Menage, 350 Merge sort, 122 Meriss, R., 440 Minimal spanning tree, 131 Minimum cost rule, 174 Missionaries-cannibals puzzle, 108 Molluzzo, J., 440 Moments of a random variable, 265 k-th moment, 277 Montmort, P., 351 Mountain climber’s puzzle, 25 Multigraph, 49, 437 Multiplication, fast, 298 Multiplication Principle, 180 Murty, U., 44, 87, 440 N Network, 127, 437 Network flow, 135, 437 Nim game, 393, 400 Northwest corner rule, 168 NP-completeness, 57, 69, 113, 431 Null set, 415 O Ore, O., 440 Orlin, J., 175 O’Rourke, J., 79, 87 P P(n,r), 190 Palmer, E., 383 Parent in a tree, 94, 438 Parenthesization, 289, 312 www.itpub.net Index 479 Partitions, 266, 437 of an integer, 266 self-conjugate, 270 Pascal, B., 236, 421, 423, 425 Pascal’s triangle, 229, 236 Patashnik, O., 440 Path in a graph, 4, 437 directed path, 10 length of a path, 27 Pattern inventory, 353 Peacock, G.,417 Permutation, 190, 437 r-permutation, 190 Pigeonhole Principle, 427 Pisa, L., 316 Pitcher pouring puzzle, 106 Plain text, 404 Planar graph, 31, 80, 152, 437 Plane graph, 31 Platonic graph, 43 Poker probabilities, 192, 331 Polya, G., 229, 377, 382, 383, 440 Polya’s enumeration formula, 377 Polygon, 78 Power series, 249 Prim’s algorithm for minimal spanning tree, 131 Probability of an event, 423 Probability generating function, 265, 277 Progressively finite game, 385 Prufer sequence, 100 R Ramsey theory, 47, 427 Random variable, 265 Range graph, 25 Range in a matching, 156 Recurrence relation, 283, 437 homogeneous relation, 300 inhomogeneous relation, 304 systems of recurrence relations, 289 Reade, R., 383 Reflexivity in a relation, 356 Region, 37 Regular graph, 45 Rencontre, 339 Ringel, G., 440 Riordan, J., 351, 440 Rivest, R., 298, 317 Roberts, F., 439 Rook, non-capturing, 342 Rook polynomial, 343 Root of a tree, 93, 438 Rotation of a figure, 356 Rothschild, B., 427, 439 Round-robin tournament, 46, 73 Rubik’s cube, 68 Run in a sequence, 222 Ryser, H., 440 S Sample space, 424 Sandefur, J., 290, 317 Satisfiability problem, 431 Saturated and unsaturated edges, 139 Scheduling problems, 8, 29, 71, 72 Schwartz, B., 157, 175 Secret code. xi Sedgewick, R., 126 Selection, 190, 437 Selection with repetition, 207 equivalent forms, 218, 223 Self-complementary graph, 48 Set, 415 Set Composition Principle, 194 Set of distinct representatives, 154 Shapley-Shubik index, 197 Shih-Chieh, C., 237 Shmoys, D., 126 Shortest path algorithms, 127 Sibling in a tree, 94, 438 Sink in a network, 135 Sinkov, A., 414 Slack in an edge, 139 Slomson, A., 440 Sorting algorithms, 121 Source in a network, 135 Spanning tree, 104, 131, 167 438 Spaziergangen problem, 49 480 Index Spencer, J., 427 Stanley, R., 440 Stanton, D. and R., 440 Stirling number, 275 Stirling’s approximation for n!, 198 Street surveillance, 6 Street sweeping, 52 Strongly connected graph, 45 Subdivided edge, 35 Subgraph, 15, 437 Subgroup, 362 Subset, 415 proper subset, 415 Subtree, 94, 438 Successor of a vertex, 388 Summation methods: using binomial identities, 230 using generating functions, 277 using recurrence relations, 305 Sylvester, J., 44, 351 Symmetry: of a geometric figure, 356 in a relation, 356 T Takeaway games, 385 Tetrahedron, symmetries of, 358 Tournaments, 44, 62, 73, 102, 297 Tower of Hanoi game, 286 Trail, 47, 49, 437 Transitive closure, 130 Transitivity in a relation, 356 Transportation problem, 164 Transportation tableau, 165 Transposition in a permutation, 363 Traveling salesperson problem, 113, 125, 431 Traversal of tree, 118 inorder, postorder, preorder, 118, 438 Tree, 93, 437 binary, 95, 438 m-ary, 95, 438 rooted, 93, 438 Tree sort, 125 Trial of experiment, 424 Triangulation of a graph, 78 Trigraph, 402 Trigraph frequency table, 402 Tripartite graph, 41 Tutte, W., 87 U Union of sets, 416 Unit-flow chain, 142 Unit-flow path, 139 Universal set, 415 V Value of a flow, 137 Vandermonde, A., 86 VanLint, J., 440 Venn diagram, 320, 417 Vertex, 3 Vertex basis, 10 Vizing’s theorem, 80 Voter power, 197 W Wallis, W., 439 West, D., 44, 440 Wheel graph, 70 White, D., 440 Whitworth, W., 237 Wilson, R., 44, 86, 440 Winning position in a game, 386 Winning strategy in a game, 386 Winning vertex, 386 Woods, J., 440 Y Yellen, J., 440 www.itpub.net |
9998 | https://www.teacherspayteachers.com/store/math-shop | Math Shop | Teachers Pay Teachers
Log InSign Up
Cart is empty
Total:
$0.00
View Wish ListView Cart
Grade
Elementary
Preschool
Kindergarten
1st grade
2nd grade
3rd grade
4th grade
5th grade
Middle school
6th grade
7th grade
8th grade
High school
9th grade
10th grade
11th grade
12th grade
Adult education
Resource type
Student practice
Independent work packet
Worksheets
Assessment
Graphic organizers
Task cards
Flash cards
Teacher tools
Classroom management
Teacher manuals
Outlines
Rubrics
Syllabi
Unit plans
Lessons
Activities
Games
Centers
Projects
Laboratory
Songs
Clip art
Classroom decor
Bulletin board ideas
Posters
Word walls
Printables
Seasonal
Holiday
Black History Month
Christmas-Chanukah-Kwanzaa
Earth Day
Easter
Halloween
Hispanic Heritage Month
Martin Luther King Day
Presidents' Day
St. Patrick's Day
Thanksgiving
New Year
Valentine's Day
Women's History Month
Seasonal
Autumn
Winter
Spring
Summer
Back to school
End of year
ELA
ELA by grade
PreK ELA
Kindergarten ELA
1st grade ELA
2nd grade ELA
3rd grade ELA
4th grade ELA
5th grade ELA
6th grade ELA
7th grade ELA
8th grade ELA
High school ELA
Elementary ELA
Reading
Writing
Phonics
Vocabulary
Grammar
Spelling
Poetry
ELA test prep
Middle school ELA
Literature
Informational text
Writing
Creative writing
Writing-essays
ELA test prep
High school ELA
Literature
Informational text
Writing
Creative writing
Writing-essays
ELA test prep
Math
Math by grade
PreK math
Kindergarten math
1st grade math
2nd grade math
3rd grade math
4th grade math
5th grade math
6th grade math
7th grade math
8th grade math
High school math
Elementary math
Basic operations
Numbers
Geometry
Measurement
Mental math
Place value
Arithmetic
Fractions
Decimals
Math test prep
Middle school math
Algebra
Basic operations
Decimals
Fractions
Geometry
Math test prep
High school math
Algebra
Algebra 2
Geometry
Math test prep
Statistics
Precalculus
Calculus
Science
Science by grade
PreK science
Kindergarten science
1st grade science
2nd grade science
3rd grade science
4th grade science
5th grade science
6th grade science
7th grade science
8th grade science
High school science
By topic
Astronomy
Biology
Chemistry
Earth sciences
Physics
Physical science
Social studies
Social studies by grade
PreK social studies
Kindergarten social studies
1st grade social studies
2nd grade social studies
3rd grade social studies
4th grade social studies
5th grade social studies
6th grade social studies
7th grade social studies
8th grade social studies
High school social studies
Social studies by topic
Ancient history
Economics
European history
Government
Geography
Native Americans
Middle ages
Psychology
U.S. History
World history
Languages
Languages
American sign language
Arabic
Chinese
French
German
Italian
Japanese
Latin
Portuguese
Spanish
Arts
Arts
Art history
Graphic arts
Visual arts
Other (arts)
Performing arts
Dance
Drama
Instrumental music
Music
Music composition
Vocal music
Special education
Speech therapy
Social emotional
Social emotional
Character education
Classroom community
School counseling
School psychology
Social emotional learning
Specialty
Specialty
Career and technical education
Child care
Coaching
Cooking
Health
Life skills
Occupational therapy
Physical education
Physical therapy
Professional development
Service learning
Vocational education
Other (specialty)
Math Shop
Rated 4.73 out of 5, based on 252 reviews
4.7(252)
412 Followers
Hertfordshire, United Kingdom
Follow
About the store
Hi and welcome to my store! I have been teaching math for over 16 years now. As well as whole classes, I also run small group interventions and tutor special needs students. After school I teach online math lessons.
Read more
Featured resources
Easter Math Distributive Property 7th Grade Math
$3.00 Original Price $3.00
Easter Math Coloring Combining Like Terms
$3.00 Original Price $3.00
April Fools Day Math Coloring Activities Algebra Combining Like Terms
$3.00 Original Price $3.00
April Fool’s Day Math
$3.00 Original Price $3.00
All resources
Search resources
Search
Sort by: Most Recent
Best Sellers
Rating
Price (Ascending)
Price (Descending)
Most Recent
Grade
Subject
Price
Format
All filters
Filters
Grade
Elementary
Preschool
Kindergarten
1st grade
2nd grade
3rd grade
4th grade
5th grade
Middle school
6th grade
7th grade
8th grade
High school
9th grade
10th grade
11th grade
12th grade
Not grade specific
More
Subject
Art
Visual arts
Math
Algebra
Algebra 2
Applied math
Arithmetic
Basic operations
Decimals
Fractions
Geometry
Math test prep
Measurement
Mental math
Numbers
Order of operations
Place value
Statistics
Other (math)
Science
Earth sciences
Environment
Physical science
Physics
Social studies
Geography
For all subjects
More
Price
Free
Under $5
$5-10
$10 and up
Format
Image
PDF
More
Resource type
Classroom decor
Posters
Printables
Clip art
Hands-on activities
Activities
Games
WebQuests
Instruction
Handouts
More
Student assessment
Assessment
Student practice
Independent work packet
Worksheets
Homework
Task cards
Standard
Theme
Seasonal
Autumn
Back to school
End of year
Spring
Summer
Winter
Holiday
Christmas-Chanukah-Kwanzaa
Earth Day
Easter
Halloween
Hispanic heritage month
New Year
Presidents' Day
St. Patrick's Day
Thanksgiving
Valentine's Day
More
Audience
TPT sellers
Homeschool
More
Language
English (UK)
More
All Resources 570
August/September Math 2
Back to School Math 18
Math Coloring Activities 168
Christmas Math 48
Add Subtract Multiply Divide 44
Algebra 6-8 202
Algebra 9-12 23
Clip Art 4
Code Breakers 12
Color by Number 22
December January 29
Decimals 16
Earth Day Math 22
Easter Math 20
Fall Autumn 7
Fathers Day Math 5
February 31
Fractions 3-7 24
Geography 16
Geometry 55
Groundhog Day Math 9
Growing Bundles 3
Halloween Math 56
June July 13
March April May 69
Math Logic Puzzles 47
Math Posters 2
Math Recipes 29
Math Test Prep 13
Mental Math 1
Metric Measures 41
Mothers Day Math 20
New Year Math 29
October November 15
Patriotic Math 12
Percents 19
Proportions and Ratios 41
Real World Math 21
Sports Day Math 19
Spring Math 24
St Patrick's Day Math 22
Summer Math 6
Thanksgiving Math 45
Valentines Day Math 36
Winter Math 8
Valentines Day Mardi Gras February Math Activities
Created by
Math Shop
Valentines Day Math and Mardi Gras Fat Tuesday 6th Grade Math Activities This is a fun pack of February Math resources. Printable and ready to go. All answers are included. ☞ Visit Math Shop to see more of my resources.
5 th - 8 th
Algebra, Math
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
Mardi Gras Math Activities Carnival Themed Algebra Fat Tuesday
Created by
Math Shop
Mardi Gras Math Activities Carnival Algebra Fat Tuesday Math Activities. This is a fun pack of February Math resources. Printable and ready to go. All answers are included. ☞ Visit Math Shop to see more of my resources.
5 th - 8 th
Algebra, Math
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
Mardi Gras Math Coloring Activities Combining Like Terms Fun Math
Created by
Math Shop
Mardi Gras Carnival Themed Math Activities for Fat Tuesday. Combining Like Terms Math Coloring Collecting Like Terms 6th Grade Algebra 1 Coloring Activity. Simplifying algebraic expressions including positive and negative numbers. This is great for Mardi Gras fun! Students will combine like terms to generate equivalent expressions. For example, simplifying -6r + 5d – r – 8d to produce the equivalent simplified expression. ✦ Notes:There are 26 questions. Each answer occurs twice, once with a
5 th - 8 th
Algebra, Math
CCSS, TEKS, VA SOL
6.EE.A.3
, 6.EE.A.4
, 7.EE.A.1
+3
Also included in:Mardi Gras Math Activities Carnival Themed Algebra Fat Tuesday
$3.00
Original Price $3.00
Add to cart
Wish List
Combining Like Terms Simplifying Algebraic Expressions 6th Grade Math
Created by
Math Shop
Combining Like Terms Simplifying Algebraic Expressions 6th Grade Math This is a fun pack of math resources. Just print and go! All answers are included.Students will simplify expressions by combining like terms including positive and negative numbers. For example, simplifying -7b + 2c + 5b – 3c ☞ Visit Math Shop to see more of my resources.
5 th - 8 th
Algebra, Basic Operations, Math
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
Combining Like Terms Simplifying Algebraic Expressions 6th Grade Math Coloring
Created by
Math Shop
Combining Like Terms Simplifying Algebraic Expressions Cat Themed 6th Grade Math Coloring Activity. Students will simplify expressions by combining like terms including positive and negative numbers. For example, simplifying -7b + 2c + 5b – 3cAll answers are included. ✦ Notes:There are 28 questions. Each answer occurs twice, once with a number and once with a color. The matching color can then be used to color the picture in a fun color by number activity. This activity should take around a
5 th - 8 th
Algebra, Basic Operations, Math
CCSS, TEKS, VA SOL
6.EE.A.3
, 6.EE.A.4
, 7.EE.A.1
+3
Also included in:Combining Like Terms Simplifying Algebraic Expressions 6th Grade Math
$3.00
Original Price $3.00
Add to cart
Wish List
Thanksgiving Math Coloring Distributive Property Combining Like Terms Algebra
Created by
Math Shop
Thanksgiving Math Coloring Distributive Property Combining Like Terms Algebra. This is a discounted algebra bundle. All answers are included.Please click on the individual products listed in the bundle to see previews and descriptions of each one. Click here for Algebra grades 6 to 8 resources.Click here for Algebra grades 9 to 12 resources. ☞ Visit Math Shop to see more of my resources.
6 th - 8 th
Algebra, Math
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
Thanksgiving Math Coloring Distributive Property Combining Like Terms Algebra
Created by
Math Shop
Thanksgiving Math Themed Algebra Distributive Property Equivalent Expressions and Combining Like Terms 6th Grade Math Coloring Activities. This is great for Thanksgiving fun! The resource is printed in black and white for the students to color in. ✦ Notes:Students will apply the properties of operations to generate equivalent expressions. For example, applying the distributive property to the expression 3(2t - g) + 3(4g - 3t) y to produce the equivalent simplified expression. Each answer occur
6 th - 8 th
Algebra, Math
CCSS, TEKS, VA SOL
6.EE.A.3
, 6.EE.A.4
, 6.EE.B.6
+5
Also included in:Thanksgiving Math Coloring Distributive Property Combining Like Terms Algebra
$3.00
Original Price $3.00
Rated 5 out of 5, based on 1 reviews
5.0(1)
Add to cart
Wish List
Christmas Math Algebra Combining Like Terms Simplifying Expressions
Created by
Math Shop
Christmas Math Algebra Combining Like Terms Simplifying Expressions 6th Grade Algebraic Expressions Bundle. This is a discounted bundle. All answers are included.Please click on the individual products listed in the bundle to see previews and descriptions of each one. ✦✦ Click here for Algebra grades 6 to 8 resources. ✦✦ ✦✦ Click here for Algebra grades 9 to 12 resources. ✦✦ ✦ Notes:All answers are included, so you can easily mark students work. ♥ You may also like: Middle School Math Workshe
6 th - 8 th
Algebra, Math
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
Happy Holidays Math Coloring Activity Distributive Property Algebra 1
Created by
Math Shop
Happy Holidays Math Winter Snowman Themed Algebra 1 Distributive Property Equivalent Expressions and Combining Like Terms 6th Grade Math Coloring Activities. This is great for Christmas Holiday and Winter fun! This resource is printed in black and white for the students to color in. ✦ Notes:Students will apply the properties of operations to generate equivalent expressions. For example, applying the distributive property to the expression 2(6d + 9y) - 8(d + 3y) to produce the equivalent simpli
6 th - 8 th
Algebra, Math
CCSS, TEKS, VA SOL
6.EE.A.3
, 6.EE.A.4
, 6.EE.B.6
+5
Also included in:Christmas Math Algebra Combining Like Terms Simplifying Expressions
$3.00
Original Price $3.00
Rated 5 out of 5, based on 1 reviews
5.0(1)
Add to cart
Wish List
Math Worksheets Composite Shapes Geometry Area and Perimeter
Created by
Math Shop
Math Worksheets Composite Shapes Geometry Area and Perimeter Bundle. This is a discounted bundle. All answers are included. Measurements are in inches.✦✦ Click here for more Geometry resources. ✦✦ Please click on the individual products listed in the bundle to see previews and descriptions of each one. ☞ Visit Math Shop to see more of my resources.
4 th - 7 th
Geometry, Math, Measurement
Bundle (2 products)
$4.00 Original Price $4.00
$3.20
Price $3.20
Rated 4 out of 5, based on 1 reviews
4.0(1)
Add to cart
Wish List
Area and Perimeter of Composite Shapes Area of Composite Figures Decomposing
Created by
Math Shop
This is a fun resource of composite shapes piled up to make a skyscraper challenge. Students find the area and perimeter of each composite shape by finding missing sides and splitting into rectangles. These worksheets are in inches. Click here for centimetres.There are 2 versions so that you can print in color or black and white. ✦ Objectives:To practice: finding the area and perimeter of rectangles.finding the area and perimeter of composite shapes by decomposing into rectangles.finding the ar
4 th - 7 th
Applied Math, Geometry, Math
CCSS, TEKS, VA SOL
3.MD.C.5
, 3.MD.C.7
, 3.MD.D.8
+8
Also included in:Math Worksheets Composite Shapes Geometry Area and Perimeter
$1.00
Original Price $1.00
Add to cart
Wish List
Math Worksheets Composite Shapes Geometry Area and Perimeter
Created by
Math Shop
Math Worksheets Composite Shapes Geometry Area and Perimeter Bundle. This is a discounted bundle. All answers are included. Measurements are in centimetres.✦✦ Click here for more Geometry resources. ✦✦ Please click on the individual products listed in the bundle to see previews and descriptions of each one. ☞ Visit Math Shop to see more of my resources.
4 th - 7 th
Geometry, Math, Measurement
Bundle (2 products)
$4.00 Original Price $4.00
$3.20
Price $3.20
Rated 4 out of 5, based on 2 reviews
4.0(2)
Add to cart
Wish List
Area and Perimeter of Composite Shapes Area of Composite Figures Decomposing
Created by
Math Shop
This is a fun resource of composite shapes piled up to make a skyscraper challenge. Students find the area and perimeter of each composite shape by finding missing sides and splitting into rectangles. These worksheets are in centimetres. Click here for inches.There are 2 versions so that you can print in color or black and white. ✦ Objectives:To practice: finding the area and perimeter of rectangles.finding the area and perimeter of composite shapes by decomposing into rectangles.finding the ar
4 th - 7 th
Applied Math, Geometry, Math
CCSS, TEKS, VA SOL
3.MD.C.5
, 3.MD.C.7
, 3.MD.D.8
+8
Also included in:Math Worksheets Composite Shapes Geometry Area and Perimeter
$1.00
Original Price $1.00
Add to cart
Wish List
Math Worksheets Composite Shapes Geometry Area Perimeter Volume
Created by
Math Shop
Math Worksheets Composite Shapes Geometry Area Perimeter Volume Bundle. This is a discounted bundle. All answers are included. Measurements are in inches.✦✦ Click here for more Geometry resources. ✦✦ Please click on the individual products listed in the bundle to see previews and descriptions of each one. ☞ Visit Math Shop to see more of my resources.
4 th - 7 th
Geometry, Math, Measurement
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
Math Worksheets Composite Shapes Geometry Area Perimeter Volume Metric System
Created by
Math Shop
Math Worksheets Composite Shapes Geometry Area Perimeter Volume Metric System Bundle. This is a discounted bundle. All answers are included. Measurements are in centimetres.✦✦ Click here for more Geometry resources. ✦✦ Please click on the individual products listed in the bundle to see previews and descriptions of each one. ☞ Visit Math Shop to see more of my resources.
4 th - 7 th
Geometry, Math, Measurement
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
Math Composite Shapes Geometry Worksheets Area and Perimeter
Created by
Math Shop
Math Composite Shapes Geometry Worksheets Area and Perimeter Geometry Bundle. This is a discounted bundle. All answers are included. Measurements are in centimetres.✦✦ Click here for more geometry resources. ✦✦ Please click on the individual products listed in the bundle to see previews and descriptions of each one. ☞ Visit Math Shop to see more of my resources.
4 th - 7 th
Geometry, Math, Measurement
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
Summer and Winter Math Logic Puzzles
Created by
Math Shop
Are you looking for some seasonal math activities? Keep your students engaged with these summer and winter themed logic puzzles, printable and ready to go! These puzzles are perfect for introducing algebraic methods. All answers are included.✎ Contents:This bundle includes: Summer Math Logic PuzzlesWinter Math Logic Puzzles✦ Objectives:These resources have been designed to practice: Operations with numbers.Solving word problems.Developing logic skills.Developing problem solving skills. Order
5 th - 8 th
Math
CCSS, TEKS, VA SOL
6.EE.A.2
, 6.EE.A.2c
, 6.EE.B.5
+18
Bundle (2 products)
$2.00
Original Price $2.00
Add to cart
Wish List
Winter Math Logic Puzzles FREEBIE
Created by
Math Shop
Get ready for Winter with these FREE Math Logic Puzzles. These fun math worksheets are ready to print and go, no prep needed. The problems can be solved by trial and error or can be used to introduce algebraic methods. There are 2 versions included so that you can print in black and white or color. All answers are included. Enjoy this FREEBIE and check out my store for more fun math puzzles and activities. ✦✦✦ See more Math Logic Puzzles here. ✦✦✦ ✦✦✦ See Math Coloring Worksheets here.✦✦✦ ✎
5 th - 8 th
Algebra, Math
CCSS, TEKS, VA SOL
6.EE.A.2
, 6.EE.B.6
, 7.EE.B.4
+4
Also included in:Fun Friday Math Activities and Logic Puzzles
FREE
Rated 4.8 out of 5, based on 10 reviews
4.8(10)
Log in to Download
Wish List
Fathers Day Math Coloring Worksheets for Algebra 1
Created by
Math Shop
Fathers Day or Grandparents Day Math Coloring Algebra 1. These are algebra worksheets with coloring answer sheets. They are perfect to color in and glue onto the front of a card for Fathers Day or Grandparents Day. ✦✦✦ Click here for more Algebra grades 6 to 8 resources. ✦✦✦ ☞ Visit Math Shop to see more of my resources.
5 th - 8 th
Algebra, Math, Order of Operations
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
Mothers Day and Fathers Day Math Coloring Algebra 1
Created by
Math Shop
Fathers Day Mothers Day Grandparents Day Math Coloring Algebra 1. These are algebra worksheets with coloring answer sheets. They are perfect to color in and glue onto the front of a card for Fathers Day, Mothers Day or Grandparents Day. ✦✦✦ Click here for more Algebra grades 6 to 8 resources. ✦✦✦ ☞ Visit Math Shop to see more of my resources.
5 th - 8 th
Algebra, Basic Operations, Math
Bundle (2 products)
$6.00 Original Price $6.00
$4.50
Price $4.50
Add to cart
Wish List
1
2
3
4
5
Showing 1-20 of 570 results
About the store
Experience
Hi and welcome to my store! I have been teaching math for over 16 years now. As well as whole classes, I also run small group interventions and tutor special needs students. After school I teach online math lessons.
Teaching style
I want my students to come out of my lessons and feel that they have learnt something new! I love it when they gain confidence in themselves and math.
My own education history
I have a Math Degree and a Computer Science Masters Degree.
Additional biographical information
I started my store in 2018 to share the resources I love to create for my students. When I am not busy working, I enjoy spending time with my 2 grown up sons and my rescue dog :-)
Back to top
TPT is the largest marketplace for PreK-12 resources, powered by a community of educators.
Facebook
Instagram
Pinterest
Twitter
About
Who we are
We're hiring
Press
Blog
Gift Cards
Support
Help & FAQ
Security
Privacy policy
Student privacy
Terms of service
Tell us what you think
Updates
Get our weekly newsletter with free resources, updates, and special offers.
Get newsletter
IXL family of brands
IXL Comprehensive K-12 personalized learning
Rosetta Stone Immersive learning for 25 languages
Wyzant Trusted tutors for 300 subjects
Education.com 35,000 worksheets, games, and lesson plans
Vocabulary.com Adaptive learning for English vocabulary
Emmersion Fast and accurate language certification
Thesaurus.com Essential reference for synonyms and antonyms
Dictionary.com Comprehensive resource for word definitions and usage
SpanishDictionary.com Spanish-English dictionary, translator, and learning
FrenchDictionary.com French-English dictionary, translator, and learning
Ingles.com Diccionario inglés-español, traductor y sitio de aprendizaje
ABCya Fun educational games for kids
© 2025 by IXL Learning
|Protected by reCAPTCHAPrivacy•Terms |
9999 | https://www.aft.com/support/product-tips/fittings-losses-k-factor-in-aft | Fittings K Factors in AFT Fathom - AFT Blog
This website stores cookies on your computer. These cookies are used to collect information about how you interact with our website and allow us to remember you. We use this information in order to improve and customize your browsing experience and for analytics and metrics about our visitors both on this website and other media. To find out more about the cookies we use, see our Privacy Policy.
Accept
Contact Us
Help Site
Products
Software Overview
Fathom
Arrow
Impulse
xStream
CHEMCAD
Chempak Database
Educational Use
How to Order
Consulting
Nuclear Validation (NVV)
Support
Support
Request Support
Installation
Support Upgrade & Maintenance
Services
Problems We Solve
Consulting
Nuclear Validation (NVV)
Downloads
Current Versions
Fathom
Arrow
Impulse
xStream
Chempak
Previous Versions
Free Demos
Learning Center
Training Seminar
Case Studies
Technical Papers
Tips and Tricks
Application Topics
Video Tutorials
Webinars
Channel Partners
About
Contact Us
Mission and History
Industries We Serve
Events Calendar
News
Blog
Customer Testimonials
Careers
Directions and Lodging
AFT Privacy Policy
Fathom Student Latest Release
Home
Support
Product Tips
Michel Monnet
Don't throw a fit over your Fitting K Factors - Part 1: Fittings & Losses in AFT Fathom
AFT Blog
Welcome to the Applied Flow Technology Blog where you will find the latest news and training on how to use AFT Fathom, AFT Arrow, AFT Impulse, AFT xStream and other AFT software products.
Home
Categories
Tags
Authors
Categories: All CategoriesSearch...
Suggested keywords
Search
x
Search
Font size:+–
Print
7 minutes reading time(1386 words)
Don't throw a fit over your Fitting K Factors - Part 1: Fittings & Losses in AFT Fathom
Tips and Tricks
Michel Monnet
Friday, 15 July 2022
6169 Hits
0 Comments
AFT Fathom, AFT Arrow, AFT Impulse, and AFT xStream all have a tab in the Pipe Properties window called "Fittings & Losses" which allows users to quickly specify K factors for components within a pipeline such as valves, elbows, and area changes. The Fittings & Losses allow you to simplify your model and avoid workspace clutter by integrating the K factor data into the pipe properties instead of drawing each junction on the workspace. However, this feature should be implemented differently depending on which AFT application you are using. In Part 1 of this two part blog series, we will discuss how to use Fittings & Losses in AFT Fathom. Special considerations for Fittings & Losses in AFT Arrow, Impulse, and xStream will be covered in Part 2 at a later date.
When should I use the Fittings & Losses feature?
Generally, the Fittings & Losses feature should be used when you want to simplify a model, such that instead of using a junction icon on the workspace to represent a K factor, you would incorporate the loss into the pipe properties. For example, say you have a pipe in your system that snakes its way around multiple rooms/floors in a building and has 10 different 90 degree bends. Instead of drawing the 10 bend junctions and 12 connecting pipes you could combine the loss information into the Fittings & Losses of a single pipe whose length equals the total length of all 12 pipes. Or maybe you have an expansion joint at the end of your pipe because it is connecting to a valve or component that has a larger nominal size compared to the pipe. You can incorporate this expansion into the Fittings & Losses instead of drawing a separate area change junction and additional pipe.
Something to consider is that Fittings & Losses won't work for alternate loss models like resistance curves or Cv values. If you want to use a loss model other than a K factor you should use a dedicated junction. Furthermore, Fittings & Losses should only be used on components that do not change state during the simulation or across different scenarios in your model.
Does using the Fittings & Losses feature instead of a junction affect results?
In AFT Fathom the results such as flowrate, velocity, and total pressure drop will be identical when using Fittings & Losses versus drawing the junctions individually on the workspace. When performing hydraulic calculations, the solver will create a total resistance in a unique flow path that includes friction and losses from junctions. It does not matter if the K factors are specified inside a pipe using Fittings & Losses, or separate junctions in the flow path – so long as the value of the K factors are identical. For AFT Arrow, Impulse, and xStream it gets more complicated, and as such will be discussed in Part 2. Furthermore, if you are interested in knowing the pressure at the inlet or outlet of a component, you will need to use a junction because the Fittings & Losses distribute the pressure loss over the entire pipe instead of applying it at a single point.
Consider the example below where one scenario has a valve junction and two pipes versus another scenario that has one pipe with a valve loss added to the Fittings & Losses. Valve J2 is a 100% open Ball valve from the Crane handbook source. The boundary conditions are 40 psig on the upstream side, and 0 psig on the downstream side, with 1 inch Steel – ANSI pipes carrying water at 60 degF. The results for the first scenario with a valve junction show that the flowrate is 19.12 gpm, velocity is 7.09 ft/s, and total head loss is 46.15 ft (P1+P2 head loss + valve head loss; 23.052 + 0.05 ft). The results for the second scenario with Fittings & Losses are equivalent for flowrate and velocity, and within rounding error for the head loss of 46.14 feet.
Results for scenario with valve junction (left):
Results for scenario with valve loss added in the Fittings & Losses (right):
How do I add a K factor to the Fittings & Losses?
A pipe material/size must be defined before being able to use Fittings & Losses. When inside the Pipe Properties window of a defined pipe, click on the Fittings & Losses tab. From within the Fittings & Losses tab, you can click Specify Fittings & Losses from the lower right corner. Then, navigate the tabs on the top of the Pipe Fittings & Losses window to find the type of component you wish to specify. The components will be organized into sections for different valve types, bend angles, etc. Simply enter a quantity for the components of interest and the total K factor will be summed up automatically.
Note:if you have the equivalent lengths method enabled, the list of components will look different, and the data is sourced from the NFPA 13 standard.
By default, our applications contain library data for different types of components that have been collected from sources such as Crane, Idelchik, Miller, and Darby. You can tell which source the K factor is from because the name will end with parentheses enclosing the first letter of the author's name – (C), (I), (M), or (D).
Pro Tip:you can double click on the picture graphic of the component in the bottom left of the window to see the full reference information including the page number that the data came from.
Most of the handbook data is based on nominal pipe size, usually for a common material such as commercial steel with turbulent flow of water. It is always preferable to use a K factor from a vendor datasheet because it matches your exact diameter and material, but if this information is unavailable or you don't have a specific component selected, the default Fittings & Losses can provide an initial estimate. You can add your own Fittings & Losses components which may be useful if you find yourself reusing the same vendor component multiple times in a system or between projects/models.
To add your own components, go to the Library menu and click Edit Fittings & Losses (versions prior to Fathom 12 used 'Databases' instead of 'Libraries'). Then click Create New Loss Item and define your component type and K factor. You can also use the Import button to import this data if you have a .csv file with the correct formatting. This can help keep your vendor information organized in separate Excel files and easily share between colleagues. See our Help File page on the Fittings & Losses Library for more information.
User defined Fittings & Losses will display a "(U)" at the end of the name in the lists. You can filter the lists to see only your user defined components from your local user library if desired.
Pipes that have Fittings & Losses specified will display a red ampersand symbol on the workspace.
Potential Sources of Error
If you are using custom component losses that you added with the Library Manager and change your pipe material/size, the K factors will not be updated automatically. Don't forget to update your Fittings & Losses with new vendor data for the new nominal size components.
Furthermore, when area changes are modeled with Fittings & Losses, Fathom does not account for the fact that the velocity will be changed due to the different diameter. The hydraulic calculations use the velocity that corresponds with the pipe inner diameter, and the area change merely represents a K factor.
Summary
In conclusion, Fittings & Losses can be a useful way to simplify your model by incorporating simple K factors into a pipe's properties. Fathom's default libraries contain many component losses gathered from handbook data by Crane, Idelchik, Miller, and Darby. Furthermore, you can easily add or import your own vendor component data using the Library Manager. For more information, visit our Help Site page on Fittings & Losses.
Share
Tweet
Tags:
AFT FathomFittings & LossesK FactorPipe Properties
Steam Piping in Nuclear Power Plants Is Not As Saf...
Space Debris, Forever Chemicals and Zombie Movies ...
About the author
Michel Monnet
Michel Monnet
Michel Monnet is an Applications Engineer with Applied Flow Technology (AFT) where he provides technical support to individual clients using AFT products. Michel is a graduate from the University of Colorado Boulder with his Bachelor of Science in Mechanical Engineering.
Author's recent posts
More posts from author
Tuesday, 12 December 2023 Awesome Annotation Tool Additions
Monday, 09 October 2023 Compressor Maps
Friday, 13 January 2023 Restore Your Sanity, and Your Corrupted Model Files
Related Posts
The Opposite of Good Vibrations - Evaluating FIV, AIV, FIP Guidelines from AFT Fathom and AFT Arrow
Tips and Tricks
Know Your Pump & System Curves - Part 3
Tips and Tricks
Know Your Pump & System Curves – Part 2B
Tips and Tricks
When a Lowly Engineering Student Ends Up Working With One of His Exalted Professors
President's Perspective
Turning Negative Experiences Into New Directions
President's Perspective
Comments
No comments made yet. Be the first to submit a comment
Tuesday, 01 July 2025
Submit Your Comment
Log in
© 1996 - 2025 Applied Flow Technology
Mobile Nav
Products
AFT Software Overview
AFT Fathom
AFT Arrow
AFT Impulse
AFT xStream
Chempak Database
AFT Flow Expert Package
Educational Use
Pricing & How to Order
Support
Services
Problems We Solve
Consulting
Software Development
Nuclear Validation (NVV)
Support
Installation Instructions
License Information
USB Key Support
Support Upgrade & Maintenance (SUM)
Downloads
Current Versions
AFT Fathom
AFT Arrow
AFT Impulse
AFT xStream
Chempak
Previous Versions
Free Demos
Free Viewers
Learning Center
Training Seminar
Case Studies
Technical Papers
White Papers
Tips and Tricks
Application Topics
Video Tutorials
Webinars
Upcoming Webinars
Webinar Library
Channel Partners
Channel Partner Locations
About
Contact Us
Vision & Mission
Events Calendar
Platinum Pipe Award
News
Blog
Customer Testimonials
Careers
History
Directions and Lodging
AFT Privacy Policy
Login to your account
Don't have an account yet? Register now!
Username
Password
[x] Remember Me
Forgot your password?
Forgot your username?
Create an account
Fields marked with an asterisk () are required.
Name
Username
Password
Verify password
Email
Verify email
Captcha
Reload Captcha
Register |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.