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Middle school algebraic manipulation several worksheets of various difficulty on linear equations (one variable equations), they go from fairly difficult to monstrously difficult! Very detailed step-by-step solutions are provided with explanations of each step. Have a look and let me know what you think!
I hope you find these useful. You can get more FREE worksheets on many topics, mix and match, with detailed step-by-step solutions at No signup, ads, just great questions and solutionsThis worksheet has 36 algebraic expressions, which need to be factorised. The questions are not organised by type and therefore students are expected to identify how to factorise the expressions using an appropriate method: taking out the highest common factor, difference of two squares, product-sum method, by pairing or a combination of these methods, which trains students to double-check if they have fully factorised expressions. The worksheet includes worked answers.
The worksheet is targeted at the most able GCSE students who are aiming for grade 9. Fluency in factorising algebraic expression is essential at the start of the AS-level course and this worksheet is very useful for students who need to revise this skill after a long summer holiday.
The worked answers accompanying this worksheet allow the teaching and learning to continue beyond the classroom and it is therefore an ideal resource on a school's VLE as a revision tool for independent study at home.
For your convenience, the Word file is included and can be edited to meet the needs of your students.
This is a detailed checklist with examples, for the edexcel c1 exam. Sorted by topic. 4pages.
Checklists are great for the teacher and student as you can track what topics have been covered, what topics have been learnt and topics that need to be revisited.
As a tutor, I normally get the student to fill out the checklist so I can see what they have done in class also. :)
A series of lessons (2-3) on finding and solving equations of composite functions. It gives the pupils plenty of opportunity to become familiar with the notation of f(x) ect and using this to substitute numbers into an expression. It then gives pupils the chance to find fg(x) ect and then moves onto solving equations suh as fg(x) = gf(x). There are also problem solving tasks to develop pupils learning.
All tasks are differentiated, 2-4 levels for each task to make it accessible to all pupils but also to stretch the more able. All tasks have full solutions to it is easy to peer and self assess during the lesson. The lesson includes reasoning and problem solving tasks which ensure that this is suitable for the 9-1 GCSE curriculum. I have used this lesson and pupils were engaged yet challenged. | 677.169 | 1 |
Wikipedia in English
Helping students grasp the "why" of algebra through patient explanations, Hirsch and Goodman gradually build students' confidence without sacrificing rigor. To help student move beyond the "how" of algebra (computational proficiency) to the "why" (conceptual understanding), the authors introduce topics at an elementary level and return to them at increasing levels of complexity. Their gradual introduction of concepts, rules, and definitions through a wealth of illustrative examples - both numerical and algebraic-helps students compare contrast related ideas and understand the sometimes subtle distinctions among a variety of situations. | 677.169 | 1 |
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Algebra and Trigonometry - Student Solution Manual 6TH
The name comes from the main character in the series, Fred, who learns about each book's topics thru practical, everyday experiences..
Pages: 0
Publisher: Adison-Wesley Longman, Inc,1993 (1993)
ISBN: B003GZ85OY
Freshman mathematics
Fab Five for Trigonometry Level Four: Even and Odd Trig Functions and Circle Symmetry
The CD Lecture Series Trigonometry download pdf CD Lecture Series Trigonometry 6e, Sixth. Unless that is exactly what you want, for some particular reason. It is better to buy MyMathLab which contains the ebook. Questionable idea: Waiting until late in the semester, checking your grade, deciding to do the online homework in the last few weeks, buying MML then, asking me to change the deadlines, saying that you will do 500 homework problems in 2 weeks , source: 100 Keywords and Definitions read here read here! It has been on display this year in the British Library to celebrate 2009 as the International Year of Astronomy. [see Note 6 below] Some elements of Indian astronomy reached China with the expansion of Buddhism (25-220 CE). Later, during the period (618-907 CE) a number of Indian astronomers came to live in China and Islamic astronomers collaborated closely with their Chinese counterparts particularly during (1271-1368) Algebra and trigonometry: download online Human proportions worksheet, Finding the slopes of fractions, free online fraction and decimal calculator, trigonometric values chart, algebra tutorials for beginners. Worded decimal questions, ti-83 equation solver find both roots, percent proportion problems worksheet, nonlinear equation solver max min, binomial expansion calculator online, cramer's rule calculator, prentice hall mathematics algebra 1 workbook answers , cited: A Treatise of Trigonometry read epub What we can do is multiply θ by i in the expansion ( i is - loosely speaking - the square root of minus one; so i × i = -1). This gives us the expansion for eiθ: eiθ = 1 + iθ - θ2 - iθ3 + θ4 + iθ5... 2! 3! 4! 5! ... Geometry and Trigonometry: Ans (Graded Examples in Mathematics) subtractionrecords.com.
To do this, we often use trigonometry, which is much easier when a right triangle is involved. A right triangle (like the one in the figure to the right) has one angle that is 90°. The other two angles are always less than 90 ° and together add up to 90° pdf ref.: TRIGONOMETRY CUSTOM PKG read epub TRIGONOMETRY CUSTOM PKG (Second Custom. In lieu of a workbook, a 90-page transcript is available for this videotape. Contest In 1991 Project MATHEMATICS! conducted a contest open to all teachers who had used project materials in their classrooms , cited: Geometry and Trigonometry: Ans download for free Geometry and Trigonometry: Ans (Graded. The trigometric functions, such as sin, cos and tan, represent how the ratio of these lengths depends on the angle θ and we can define the following functions for each of the ratios:
Modern algebra and trigonometry: Structure and method, book two (Houghton Mifflin modern mathematics series)
A Treatise On Plane And Spherical Trigonometry
The whole craft of number
Elements of geometry: containing the first six books of Euclid, with a supplement on the quadrature of the circle, and the geometry of solids; to ... Elements of plane and spherical geometry
For , e.g. A NEW TRIGONOMETRY FOR download pdf A NEW TRIGONOMETRY FOR SCHOOLS.. This allowed the development of graphs of functions related to the angles they represent which were periodic. Today, using the periodic nature of trigonometric functions, mathematicians and scientists have developed mathematical models to predict many natural periodic phenomena Plane trigonometry together with logarithmic and trigonometric tables You can learn mathematics by studying the information on this web page and in the many wonderful books and multimedia materials. Great for homeschoolers and brick and mortar school students. You can go slow with books, and videos can be paused at any time A Descriptiv List of Books for read for free A Descriptiv List of Books for the Young. A plane surface is determined uniquely, by (a) Three points not in the same straight line, (b) Two intersecting straight lines. By this we mean that one plane, and one only, can include (a) three given points, or (b) two given intersecting straight lines. It will be observed that we have spoken of surfaces, points and straight lines without defining them Trigonometry A Right Triangle download epub subtractionrecords.com. The angle in the xy-plane (around the z-axis) is theta, also known as the azimuthal coordinate. The angle from the z-axis is phi, also known as the polar coordinate. The North Pole is therefore 0, 0, rho, and the Gulf of Guinea (think of the missing big chunk of Africa) 0, pi/2, rho , source: Algebra 1, Chapter 8: read epub
Trigonometry, What they'd teach you (if they had the time): How your teachers would like to teach maths.. (What they'd teach you if they had the time Book 2)
Natural Trigonometric Functions
Elements of Geometry and Trigonometry
Analytical trigonometry
Logarithmic And Trigonometric Tables
This gets a little frustrating sometimes. ... Would with not doubt purchase again from seller It was in good condition and fast shipping. Even though I bought the wrong book, they understand and gave me a full refund , cited: Investigating College Algebra & Trigonometry With Technology (06) by [Hardcover (2008)] supremecars.de. In contemporary math classes, students improve their critical thinking skills. They learn to use premises to determine conclusions, analyze language to establish credibility and likelihood, use methods such as Venn Diagrams to examine relationships and visualize patterns, estimate and apply various problem-solving methods to situations College Algebra with download online download online. The right-hand side of the x-axis is designated as the 0˚ mark. While radians are often less familiar to students, they are in fact often much more useful than degree measurements. One radian is equal to the angle subtended by the center of a circle of an arc that is equal in length to the radius of the circle, as shown at right , cited: Neues Logarithmisch Trigonometrisches HandBuch Neues Logarithmisch Trigonometrisches. For certain values of θ, it is easy to figure out what the sine and cosine values are going to be just by thinking about what the angle corresponds to on the circle; the simplest cases are for θ = 0°, which is a line pointing right, giving cos θ = 1 and sine θ = 0; a line pointing straight up (ie. θ = 90°), which gives us cos θ = 0 and sine θ = 1, and so on Tables of [Square Root Of] 1-R2 and 1-R2 for Use in Partial Correlation and in Trigonometry download for free. Luenberger is at Stanford and Ross is at Berkeley. Cambridge. 2003. 0521814294 Don't let the title fool you. The book requires a knowledge of calculus and some mathematical maturity Algebra and Trigonometry - read online , cited: Algebra & Trigonometry read epub. Here's a worked example showing how to find a side. Again, we've called the side we want to find x Elements of geometry and download here download here. Don't hesitate to buy an essay online and enjoy life. It's also about having new experiences and meeting new people. Don't miss out on this special time in your life doing homework! Buy your essays with us and we will write a perfectly-structured paper just for you so you can enjoy your time in college to the max! I just answered one of your trig questions Logarithmic trigonometric and read pdf Sandra Stotsky — 7.20.13 – "Common Core Math: Wrong Answers Acceptable" – by Warren Todd Huston — Breitbart.com — 8.15.13 – "Common Core Math Leaves Students Without Clear Path to Calculus" – by Brenda J epub. | 677.169 | 1 |
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An Introduction to Simplifying Polynomials by adding, and subtracting like terms with the use of brackets for multiplication by constants.
4 pages of scaffolded questions including a section of the definition of terms and a methods summery. | 677.169 | 1 |
After AP Calculus
I am a high school senior seriously interested in mathematics (and physics, and chemistry, but that's beside the point). I am wondering what would be the best path for me to learn involving mathematics now that the AP Calculus test is over (I don't want to have to wait until September to continue...). So, now that I've learned the material in AP Calculus BC, where do I go from here?
FWIW, AP Calculus BC taught me differentiation, including implicit, integration, the methods of which included intuition, parts, substitution, and partial fractions, infinite sequences and series including Taylor and Maclaurin series, and finally an introduction to vectors, parametrics, and polars. I'm also using Khan Academy to teach myself the basics of Differential Equations (It's not really that much harder than calculus was, really).Sounds good to know. I'm hoping to both not lose all of my skill at math over the summer, and be more ready than the other students next year.
What is different about a computation-based linear algebra course? Do they spend less time on proofs than normal?
Honestly, learn some Linear Algebra. A proof version of it. Then take Multivariable Calculus, preferably a rigorous treatment(with Differential Forms) and it'll make perfect sense. Alternatively, pick up Hubbard and Hubbard's Vector Calc book. It teaches you a little linear algebra, multivariable, and other random crap. It's a great book because it teaches advanced mathematics at a not-so-high level.
I recently picked up Spivak, 4th ed. and am working through it. Just wondering: should I be doing every single problem? Because there are...a lot. Especially in the first section, it gets kind of tiresome :x
I'm actually reading through Spivak right now too and I think it'll be a great way to solidify what you learned in your more computational based AP Calculus class. I feel that if you can work through Spivak on your own and obtain a reasonable grasp of the material, when you do start mathematics in college, you will probably have a greater amount of insight than your peers who have not gone pass computational mathematics yet.
However, if you don't wish to do that, Linear Algebra that revolves around proofs, discrete math, or going onwards to multivariate calculus would probably be beneficial to you. If I were you, I would start to learn how to write proofs now so when you have to do it for a grade it won't be as stressful.
What is different about a computation-based linear algebra course? Do they spend less time on proofs than normal?
In my experience, a computational based linear Algebra class do cover some proofs, do ask some questions on it, but mostly ask you to use what the theorems state to help answer computational problem. While a more proof based linear algebra book will have very little exercises that ask you to take this Matrix and apply this theorem and then get a numerical answer, but rather will ask you, given this theorem, can you prove that if such and such is true then this and that is true. If I had my linear algebra book on me I would give you a more solid example, but it's lost somewhere, but I hope you can get the gist of what I am saying.
I'm not sure, since it's the first section and it's kind of hard to think solely in terms of the ~10 theorems he gives.
If you ever feel inclined, since I'm working through the same book, you can message me if you're not sure what you're doing is correct. I can't promise that I'm correct most or even half the time, but sometimes it helps to talk things out with another person to see if your line of thinking is correct.
I've started on Spivak's Calculus (assuming that's the book you meant) and I'm on the first chapter. And you're right, it's very difficult to restrict myself to his ten theorems. It's also difficult to prove things you know to be true, while restricted to those ten theorems. I keep wanting to say "if x^2=y^2, then of course x=y or x=-y! Everyone knows that!" | 677.169 | 1 |
Showing 1 to 30 of 84
132
132
CHEMISTRY
UNIT 5
STATES OF MATTER
The snowflake falls, yet lays not long
Its feathry grasp on Mother Earth
Ere Sun returns it to the vapors Whence it came,
Or to waters tumbling down the rocky slope.
After studying this unit you will be
able to
Ro
HYDROCARBONS
365
UNIT 13
HYDROCARBONS
Hydrocarbons are the important sources of energy.
After studying this unit, you will be
able to
name hydrocarbons according to
IUPAC system of nomenclature;
recognise and write structures
of isomers of alkanes, alkene
CHAPTER TWO
UNITS
AND
MEASUREMENT
2.1 INTRODUCTION
2.1 Introduction
2.2 The international system of
units
2.3
2.4
2.5
2.6
Measurement of length
Measurement of mass
Measurement of time
Accuracy, precision of
instruments and errors in
measurement
2.7 Signif
Chapter
9
SEQUENCES AND SERIES
Natural numbers are the product of human spirit. DEDEKIND
9.1 Introduction
In mathematics, the word, sequence is used in much the
same way as it is in ordinary English. When we say that a
collection of objects is listed in
Chapter
14
MATHEMATICAL REASONING
There are few things which we know which are not capable of
mathematical reasoning and when these can not, it is a sign that our
knowledge of them is very small and confused and where a mathematical
reasoning can be had,
Chapter
16
PROBABILITY
Where a mathematical reasoning can be had, it is as great a folly to
make use of any other, as to grope for a thing in the dark, when
you have a candle in your hand. JOHN ARBUTHNOT
16.1 Introduction
In earlier classes, we studied a
Chapter
12
INTRODUCTION TO THREE
DIMENSIONAL GEOMETRY
Mathematics is both the queen and the hand-maiden of
all sciences E.T. BELL
12.1 Introduction
You may recall that to locate the position of a point in a
plane, we need two intersecting mutually perpend
Chapter
13
LIMITS AND DERIVATIVES
With the Calculus as a key, Mathematics can be successfully applied to the
explanation of the course of Nature WHITEHEAD
13.1 Introduction
This chapter is an introduction to Calculus. Calculus is that
branch of mathemati
Chapter
15
STATISTICS
Statistics may be rightly called the science of averages and their
estimates. A.L.BOWLEY & A.L. BODDINGTON
15.1 Introduction
We know that statistics deals with data collected for specific
purposes. We can make decisions about the da
ANSWERS
391
ANSWERS
Chapter 9
9.1
1.8
9.2
(a) From the given graph for a stress of 150 106 N m-2 the strain is 0.002
(b) Approximate yield strength of the material is 3 108 N m-2
9.3
(a) Material A
(b) Strength of a material is determined by the amount of
BIBLIOGRAPHY
TEXTBOOKS
For additional reading on the topics covered in this book, you may like to consult one or
more of the following books. Some of these books however are more advanced and contain
many more topics than this book.
1
Ordinary Level Physi
CHAPTER SIX
WORK, ENERGY
AND
POWER
6.1 INTRODUCTION
6.1 Introduction
6.2 Notions of work and kinetic
energy : The work-energy
theorem
6.3 Work
6.4 Kinetic energy
6.5 Work done by a variable
force
6.6 The work-energy theorem for
a variable force
6.7 The co
Chapter
1
SETS
In these days of conflict between ancient and modern studies; there
must surely be something to be said for a study which did not
begin with Pythagoras and will not end with Einstein; but
is the oldest and the youngest. G.H. HARDY
1.1 Intr
Appendix
2
MATHEMATICAL MODELLING
A.2.1 Introduction
Much of our progress in the last few centuries has made it necessary to apply
mathematical methods to real-life problems arising from different fields be it Science,
Finance, Management etc. The use of
THE s-BLOCK ELEMENTS
291
UNIT 10
THE s -BLOCK ELEMENTS
The first element of alkali and alkaline earth metals differs
in many respects from the other members of the group
After studying this unit, you will be
able to
describe the general characteristics o
276
CHEMISTRY
UNIT 9
HYDROGEN
Hydrogen, the most abundant element in the universe and the
third most abundant on the surface of the globe, is being
visualised as the major future source of energy.
After studying this unit, you will be
able to
present inf
THE p-BLOCK ELEMENTS
307
UNIT 11
THE p -BLOCK ELEMENTS
The variation in properties of the p-block elements due to the
influence of d and f electrons in the inner core of the heavier
elements makes their chemistry interesting
After studying this unit, you
326
CHEMISTRY
UNIT 12
ORGANIC CHEMISTRY SOME BASIC PRINCIPLES
AND TECHNIQUES
After studying this unit, you will be
able to
understand
reasons
for
tetravalence of carbon and
shapes of organic molecules;
write structures of organic
molecules in various ways
REDOX REACTIONS
255
UNIT 8
REDOX REACTIONS
Where there is oxidation, there is always reduction
Chemistry is essentially a study of redox systems.
After studying this unit you will be
able to
identify redox reactions as a class
of reactions in which oxida
154
154
CHEMISTRY
UNIT 6
THERMODYNAMICS
It is the only physical theory of universal content concerning
which I am convinced that, within the framework of the
applicability of its basic concepts, it will never be overthrown.
After studying this Unit, you w
MTH 305
Homework 1
Due: 16/08/17
Instructor: Ajit Bhand
1. In each of the following, determine whether the given sentence is a statement or
not. If yes, find its truth value.
(a) February is the second month of the year .
(b) Let x be a positive natural n | 677.169 | 1 |
Maths coursework be
TeachingComputing.com is your one stop site for all things computing and computer science related. Learning Pathways - all years, Coding Lounge, Tutorials. The Maths IGCSE course from Oxford Home Schooling guides the student through the basic maths skills in preparation for their exams. GCSE Maths Revision. Lower case sigma means 'standard deviation'. Capital sigma means 'the sum of'. x bar means 'the mean' The standard deviation. Home tuition in Singapore. As advertised on Straits Time. Get the best private tutor from our Singapore tuition agency. GCSE maths section of the award-winning tutorials, tips and advice website, including coursework and exams for students, parents and teachers.
A collection of fantastic teaching resource websites that every maths teacher must have in their bookmarks: 1. TES Connect. Thousands of maths teaching resources. The Mathematical Tripos is the taught mathematics course in the Faculty of Mathematics at the University of Cambridge. It is the oldest Tripos examined in Cambridge. Award-winning tutorials, tips and advice on GCSE English, French, German, ICT, maths and physics coursework and exams for students, parents and teachers. PSA! DoSomething.org Has a TON of Scholarship Opportunities Right Now. SPOILER: college is crazy-expensive. Sorry. Did we spoil it? There are.
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Maths GCSE - Level 2 Overview. The Maths GCSE programme aims to help you achieve a qualification that is widely recognised and requested by employers. GCSE Maths (pre-2015) learning resources for adults, children, parents and teachers organised by topic. Welcome to The University of Dar es Salaam, Mathematics Department! The Department of Mathematics is the oldest in Tanzania and has been and continues to play a. MyMaths is an interactive online teaching and homework subscription website for schools that builds pupil engagement and consolidates maths knowledge. It is used in.
A-level Maths 6360 helps students develop a grounding in Maths tools and techniques. It is suitable for students for who want to study Science and Engineering. A BBC Bitesize secondary school revision resource for Standard Grade Maths II on trigonometry, algebra, statistics, numbers, measure, relationships, revision videos. The best multimedia instruction on the web to help you with your homework and study. A collection of fantastic teaching resource websites that every maths teacher must have in their bookmarks: 1. TES Connect. Thousands of maths teaching resources.
The Master of Education is designed for educators and educationalists who are looking to take the next step in their career, or who wish to gain and explore advanced. Subject content for GCSE in single science for teaching from 2016. OCR AS/A Level GCE Mathematics (MEI) qualification information including specification, exam materials, teaching resources, learning resources. The best multimedia instruction on the web to help you with your homework and study. The Mathematical Tripos is the taught mathematics course in the Faculty of Mathematics at the University of Cambridge. It is the oldest Tripos examined in Cambridge. | 677.169 | 1 |
Welcome to the 2017-2018 School Year
The 2017-2018 school year will be my fifteenth year teaching at USC. I completed bachelor degrees in Math and Manufacturing Engineering at Minnesota State University. I also completed a Master's Degree in Teaching in 2003 and am a certified Project Lead the Way teacher. This year I am teaching College Algebra, Pre-Calculus, Calculus, and Algebra 2. I coach the USC Mock Trial Team, Math League, and am the National Honor Society Advisor. My husband and I live 12 miles northeast of Wells with our dog, KG. Our daughter, Monica, graduated from the University of St. Thomas in 2016 and is working for US Bank in St. Paul, MN as an operations manager. Our son, Travis, is currently in college studying business at Minnesota State University.
All assignments are listed on Schoology. Please e-mail Mrs. Bye at jbye@unitedsouthcentral.org if you are having problems accessing Schoology.
2 Notebooks for Math Only (at least 2 - I would buy 4 or 5 if you buy one-subject ones) An EXTRA LARGE (Jumbo) book cover. (Regular ones break the book binding and you will be charged for the book if that happens.)
Pens and Pencils
Algebra 2 - You MUST have a calculator TI-30XIIS is a good choice. But any calculator with Sin Cos and Tan will work.
Pre-Calc, Calculus, and College Algebra will be supplied a graphing calculator to uses, but will need $2 for battery fees. | 677.169 | 1 |
A Friendly Introduction to Number Theory (4th Edition)
A Friendly Introduction to Number Theory, Fourth Edition is designed to introduce readers to the overall themes and methodology of mathematics through the detailed study of one particular facet—number theory. Starting with nothing more than basic high school algebra, readers are gradually led to the point of actively performing mathematical research while getting a glimpse of current mathematical frontiers. The writing is appropriate for the undergraduate audience and includes many numerical examples, which are analyzed for patterns and used to make conjectures. Emphasis is on the methods used for proving theorems rather than on specific results.
"synopsis" may belong to another edition of this title.
About the Author:
Joseph H. Silverman is a Professor of Mathematics at Brown University. He received his Sc.B. at Brown and his Ph.D. at Harvard, after which he held positions at MIT and Boston University before joining the Brown faculty in 1988. He has published more than 100 peer-reviewed research articles and seven books in the fields of number theory, elliptic curves, arithmetic geometry, arithmetic dynamical systems, and cryptography. He is a highly regarded teacher, having won teaching awards from Brown University and the Mathematical Association of America, as well as a Steele Prize for Mathematical Exposition from the American Mathematical Society. He has supervised the theses of more than 25 Ph.D. students, is a co-founder of NTRU Cryptosystems, Inc., and has served as an elected member of the American Mathematical Society Council and Executive Committee.321816191
Book Description Pearson. Book Condition: New. 03218161961816191
Book Description Soft cover. Book Condition: New. NEW - International Edition - ISBN 97893325352371816196-ABab89124161911102366sPSA FriendlyIntroNumb]. Bookseller Inventory # 032181619688888 | 677.169 | 1 |
Letting yourself act in a trail of interests can influence how abstractly you learn to think. Can you foresee a sequence of interests which if you study and perform them might move your activities and thoughts into Mathematics? What do you currently like to do or think about?
I was thinking about asking for an example of how to do this, but I decided that it would be counter-productive.
Find topics that you want to learn about and find skills which you want to learn to do; then READ and THINK. If some things to practice are presented, then PRACTICE them. That is just very general. If you want more specific suggestions, this is difficult, since we do not know what your interests and your favorite instruction-related activities are. Since you placed an emphasis on Mathematics, you should involve yourself with something or things which can lead you to Mathematics. Try learning music, enroll in science courses (specific ones, not just unspecified "Science"), learn a computer programming language, study Algebra and some important forms of Geometry (like "College Preparatory Geometry", and Trigonometry), learn photography. Some of this you can start before entering college; and all of it you can start or continue when you are enrolled in college.
Most of what I listed relies on abstract thinking. You may need to put in a great effort for many of them, but becoming a more abstract thinker relies on your putting in this effort. Mathematics is probably one of the more abstract subjects mentioned there. I'd say many people stop their efforts at their mathematical development because they do not put in the patient effort to deal with the abstractness of the topics; not because they cannot think abstractly.
If you enjoy 'abstract' thinking and the courses in mathematics that you've taken up until now, those are positive indicators. Another one is that you specifically seek out new ideas in the field - you read outside of assigned work and work on problems that you take a personal interest in.
Ultimately the only way to really know is to jump in and see what the water's like.
OK, let's get back to topic. What is abstract thinking in math? I've never really bothered to look up that question, I'm just doing my math sequences without worrying if I can abstractly think or not (until now, that is).
Go with what interests you and see how far you can go. If you want to do pure math and you are passing your math classes with decent grades or think that you could if you put in more effort, you will probably do fine. If you enjoy the subject and like to learn outside of class you will do at least as well as the average math major ( there are a few exceptions; I know a guy who has a very slim aptitude for math but has been pursuing a computer science degree for 6 years grinding through the prerequisite math and constantly repeating classes, but that seems to be pretty rare).
If you find that you have to keep repeating math classes in order to pass them in college, math might not be right for you, otherwise just persevere.
How do you know if one is good at abstract thinking? I'm self evaluating myself if I can do pure maths in college, so any advice would be appreciated.
Well, I would not advise you to do what I did, or rather get did to me.
There was a unfortunate accident with a tractor at 6. I got clobbered
in the head and all the thoughts in my brain fell off the shelves
and broke. Gluing all the pieces back together has not been easy
or fully successful. Perhaps the rewiring cause my abstraction abilities.
And precluded others.
I think if you enjoy sunrises, beauty of the land, and what makes things
work you will do just fine because all of those things are abstract and
if you find a way to communicate them you will become a master
of the art of abstractions. | 677.169 | 1 |
Math : Homework Help and Answers : Slader
Pre-Algebra : Algebra I : Algebra II : Geometry : and much more (see list of all sections)! New! Free Video lessons! We have dozens of video math lectures: by, nutshell Math and, our Own In-House videos.
Pre-Algebra, Algebra I, Algebra II, Geometry: homework help by free math tutors, solvers, lessons. Each section has solvers (calculators lessons, and a place where you can submit your problem to our free math tutors.
To ask a question, go to a section to the right and select "Ask Free Tutors". Most sections have archives with hundreds of problems solved by the tutors. Lessons and solvers have all been submitted by our contributors! | 677.169 | 1 |
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Real Numbers and Subsets:
This is a very concrete and thorough lesson that explains and provides problems for important topics of the Common Core Standards for the Algebra One curriculum.
In order to meet Common Core Standards, this lesson allows students to:
• Identify the different subsets of real numbers
• Use the properties of real numbers to answer true and false questions
• Use a counterexample when a statement is false | 677.169 | 1 |
In mathematics, a matrix is an array of numbers arranged in rows and columns. We generally use matrices to represent one or more equations to simplify their solution.
A matrix is made up of elements within those rows and columns. Understanding placement helps the progression into operations on matrices, such as addition, subtraction, scalar multiplication, dot products, and inverses.
In the following example we see a 3×3 matrix. We label the dimensions of a matrix as rows X columns. Each number in a matrix is an element. In the general matrix, we see each element also has a corresponding position named just as the matrix was named, rows X columns.
We are able to perform operations on matrices as long as their dimensions align. Different operations require a different alignment of dimensions. You will see more of these operations in the proceeding sections. | 677.169 | 1 |
Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for). | 677.169 | 1 |
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This r squared creation assesses what the students have learned during that week's lesson. The materials covered in the assessment are standards based and are from sections 9.5, 9.6, and 9.9 of the Holt Algebra textbook. Concepts covered are how to solve a quadratic equation by factoring when in standard form, how to solve a quadratic equation by factoring when not in standard form, how to solve a quadratic equation by using square roots, and how to find the discriminant and state the number of solutions to a quadratic equation.
Covers CA standards 2.0, 14.0, and 22.0.
The zip file contains the quiz in both .doc format as well as in .pdf. If you would like to make any changes (you may need to download MathType for free). | 677.169 | 1 |
Functionbook: 12 Common Functions to Know and Love
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I use this when I introduce functions in pre-calculus, based on a popular social website. I present these as the 12 basic graphs from which we will be launching our study over the year. Some they have already seen in previous classes, which I call their friends. The new ones I ask them to be-friend. For each, I name the function verbally (i.e. the squaring function) and with an equation. Then, to make it fun, I give each function a personality.
Of course, you can be more creative with your personalities. Alternately, the students can make their own. It really helps them remember throughout the year. I often refer to functionbook in my lessons and homework. The students think it is really fun as it meets them in a familiar forum. | 677.169 | 1 |
Year 12 Mathematics Essentials – GTMAE
Unit 3 This unit provides you with the mathematical skills and understanding to solve problems related to measurement, scales, plans and models, drawing and interpreting graphs and data collection. You will use the mathematical thinking process and apply the statistical investigation process. You will also apply the content of the four topics in this unit: Measurement; Scales, plans and models; Graphs in practical situations; and Data collection, in a context which is meaningful and of interest to you. The number formats for the unit are positive and negative numbers, decimals, fractions, percentages, rates, ratios, square and cubic numbers written with powers and square roots.
Unit 4 This unit provides you with the mathematical skills and understanding to solve problems related to probability, earth geometry and time zones, loans and compound interest. You will use the mathematical thinking process and apply the statistical investigation process to solve problems involving probability. You will also apply the content of the three topics in this unit: Probability and relative frequencies; Earth geometry and time zones; and Loans and compound interest, in a context which is meaningful and of interest to you. Possible contexts for this unit are Finance, and Travel. The number formats for the unit are positive and negative numbers, decimals, fractions, percentages, rates, ratios and numbers expressed with integer powers. | 677.169 | 1 |
Oxford A Level Mathematics for Edexcel takes a completely fresh look at presenting the challenges of A Level. It specifically targets average students, with tactics designed to offer real chance of success to more students, as well as providing more stretch and challenge material. This Further Pure 1 book includes a background knowledge chapter to help bridge the gap between GCSE and A level study layout of the textbook is good, it is easy to follow and read. However, the examples are too scarce and often do not cover much of the content in the questions. Also there are no worked examples available, making it harder to check answers and actually improve them. Next time I would definitely stick with Pearson, as they are cheaper and do involve worked solutions
A decent textbook. It works well as a second text to the Pledger FP1 textbook which is more attractive and well-explained with plentiful worked examples and admirably clear explanations, and includes a really helpful Live Text CD with worked solutions to all the questions in every exercise. This Rowland FP1 textbook is less attractive, and with too-brief explanations, but sometimes uses a different approach and definitely extends knowledge more with many harder questions. I also appreciated the Prior Knowledge sections which give revision of topics from C1 to C4 essential to the FP1 topic being studied. I used the Pledger text to teach myself/revise the topics, then the Rowland book to fully develop and consolidate my knowledge.
Highly recommended for those who wish to take their mathematics a stage higher. Excellent introductory section which neatly leads you in from advanced level to further pure maths. Loads of examples and plenty of exam level practice with a few really testing questions for further practice. Looking forward to FP2 and 3.
The book has an attractive layout with appropriate use of colour. Necessary prerequisite knowledge and skills are pointed out and there are a number of well-laid out worked examples to illustrate the mathematical techniques described. The descriptions are more than adequate for self study. All in all, this book is good value and makes learning the subject a pleasant experience.
The first 10-15 pages are coming off its spine which I'm not quite happy about. A bit disappointed but the book is generally in good condition. Will shop with you again but hope it will improve next time. | 677.169 | 1 |
Matrices
The characteristics of explicit competition are clearer and, hence, is easiest to consider first. STUDENT: -- is the two and the three supposed to be switched? Reteach provides more exposure to concepts for those students who need more time to master new skills or concepts. Count Like an Egyptian provides a fun, hands-on introduction to the intuitive and often-surprising art of ancient Egyptian math. The image above shows the origin of the stage' coordinate space (red dot), and that of the symbol's coordinate space (registration point marked by crosshair).
Math in Focus®, Math in Focus® Digi+™, and Marshall Cavendish® are trademarks or registered trademarks of Times Publishing Limited. With function Matrices.eigenValueMatrix, a real block diagonal matrix is constructed from the eigenvalues such that provided the eigenvector matrix "eigenvectors" can be inverted (an inversion is possible, if all eigenvalues are different; in some cases, an inversion is also possible if some eigenvalues are the same).
A/B+I = A(/A+/B) A + B = A(/A+/B)B [ x+y = x(/x+/y)y ] (A/B+I)B = A(/A+/B)B [ Distribution extracting B ] A/B+I = A(/A+/B) [ postmultiply both sides by /B] Q. Our sensory perceptions are generally quite automatically affected by the magnitude of the sensory input, small inputs being insignificant and disregarded in our sense of them as with our lack of sensing or tasting salt added to a stew when it is just the slightest pinch of salt. The Ԑ(epsilon) error function in this classic negative feedback control system is the difference between the actual room temperature, θ, and the end point temperature the process is moving towards, θS.
Recognizing this, I jumped off at the next bus stop in the Casino area instead of proceeding directly to the UNLV Law Library via the Bonneville Transit Center and the #109 bus. Students who have not completed Math 237A may enroll with consent of instructor. Mid-module and End-Module Assessments are also included. First we define a matrix b: >> b = [ 1 2; 0 1 ] Then we compute the product ab: Finally, we compute the product in the other order: Notice that the two products are different: matrix multiplication is noncommmutative.
Next let�s look at another random distribution, K=12 over N=6, to reinforce this understanding of the Representative State. Figure 1: Projection onto the xy plane by discarding z-coordinates. So we will understand the pleasure experienced in getting R=V dollars to be a simple linear function of V. How do I download pdf versions of the pages? Function; input Real A[:, size(A, 1)] "Matrix A of A*x = b"; input Real b[size(A, 1)] "Vector b of A*x = b"; output Real x[size(b, 1)] "Vector x such that A*x = b"; protected Integer info; algorithm (x,info) := LAPACK.dgesv_vec(A, b); assert(info == 0, "Solving a linear system of equations with function \"Matrices.solve\" is not possible, because the system has either no or infinitely many solutions (A is singular)."); function solve2 "Solve real system of linear equations A*X=B with a B matrix (Gaussian elemination with partial pivoting)" extends Modelica.
The second set of aperiodic tiles, denoted by P2, "was discovered by Penrose in 1974 and is more remarkable than the first in that it contains only two tiles." Read more UTEP was approved by the Society of Actuaries (SOA) to be on the list of UCAP (Universities and Colleges with Actuarial Programs) schools. I was tempted to walk out, but I had this review to write, so fortunately I stayed. D.), using similar triangles, as well as alternative methods proposed in modern times.
Adding an argument changes the natural order. – egreg Jan 12 '15 at 22:21 Well, I missed that. A diagonal matrix is a square matrix with all de non-diagonal elements 0. But God is generally accepted without much quibbling as a spirit being that is not and cannot be observed whatever form He might take including also Allah for the Muslims or Vishnu for the Hindus.� As this argument depends on science from the starting gate, it can, of course, be argued that all the argument shows is an incompatibility between science and religion, not which one is correct but rather that both cannot be right.
When he brought Ba-chan back this time, she wouldn't speak. ( It may be possible to find many of these video clips by searching the Internet. You need Acrobat Reader to print these documents. The misery of peacetime tyranny and of non-nuclear weapons war is statistical. How do I normalise a quaternion? ------------------------------------- A quaternion can be normalised in a way similar to vectors. For example, generate column and row vectors with the same numbers. | 677.169 | 1 |
Mathematics
SUBJECT OVERVIEW
Mathematics is needed for Business and Finance, Engineering, Science and Technology. Mathematics is also a fascinating subject in its own right. The ability to solve mathematical problems will have a significant effect on the life chances of your child. A good qualification in mathematics allows one to follow a wide range of professions and careers and increases opportunities and life chances.
The curriculum is designed to be accessible and meet the needs of all students. Students will develop problem solving techniques and use practical activities to develop understanding.
SCHEMES OF WORK
KEY STAGE 3
Year 7, 8 and 9 follow a differentiated scheme of work. Our KS3 mathematics programme explores the key elements of number, algebra, ratio, proportion and rates of change, geometry and measures, probability and statistics. The curriculum focuses on developing fluency, mathematical reasoning and competency in solving increasingly sophisticated problems, building on our students knowledge from KS2. These skills are essential in order to achieve excellence in GCSE and A Level Mathematics. We believe that our programme of study will develop confident and enthusiastic mathematicians, with a hunger for learning and a flair for solving real world problems.
KEY STAGE 4
All students sit two external examinations at the end of the GCSE course (June of Year 11). One paper is non-calculator and in the second paper a calculator is allowed.
Internal examinations are taken each half-term throughout the course and regular testing of specific topic areas is also carried out.
Homework tasks are set each week and will include activities which revisit topics studied earlier in the course. In Year 11 students are expected to complete weekly past papers, this ensures that revision is taking place throughout the school year.
KEY STAGE 5
Mathematics is a very popular and successful subject at Enterprise South Liverpool Academy, and we offer A Level Mathematics and A Level Further Mathematics. We have a dedicated team of specialist A level teachers, all on hand to support students throughout the course.
There are 3 main areas in mathematics at A Level:
Pure Mathematics: Algebra, Geometry, Trigonometry and Calculus (Differentiation and Integration and their applications.)
Mechanics: Application of Newtonian ideas to describe and predict real world situations; the main topics being Kinematics, Statics and Dynamics | 677.169 | 1 |
Absolute Value Inequalities: Foldable Page
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Absolute Value Inequalities Notebook Page:
Interactive Notebooks are a new way to share information with students. They are meant to replace repetitive note taking. Students organize information differently in their minds when they use interactive notebooks – they are also more like to study when they use interactive notebooks because they feel more ownership of their notes! Make sure to allow students to be creative with their notebook pages. Coloring, doodling, and/or highlighting various parts of the notebook makes it more exciting!
There are 2 pages worth of material. The first page is meant as an introduction to solving absolute value inequalities. The second is a chart to use to actually solve absolute value inequalities. | 677.169 | 1 |
Department of Mathematics
College of the Redwoods
Math 15: Elementary Statistics
At College of the Redwoods, Math 15 is the course number for our Elementary Statistics course. Elementary Statistics is a transfer-level math course needed for preparation for many natural and social science majors.
Prerequisites: Grade of C or better in Math 120 (Intermediate Algebra) or equivalent, or an appropriate score on the math placement exam.
Course Description: The study of statistical methods as applied to descriptive statistics and inferential statistics. An emphasis on the meaning and use of statistical significance will be central to the course. Students will use frequency distributions, graphs, measures of relative standing, measures of central tendency, measures of variability, correlation, and linear regression to explore descriptive statistics. Students will use the laws of probability and statistical tests (t-tests, chi-square, ANOVA, and regression analysis) to make decisions via hypothesis testing and estimate parameters using confidence intervals.
Graphing Calculator: Students are required to use a graphing calculator in the Math 15 course. Students can rent a calculator for the semester for $20. Instructions are available at the following link:
Need some extra help? Need to hear another voice explaining various topics in Elementary Statistics? Then Mike Butler's Videos For Math 15 is a wonderful page to watch movies on a variety of topics in Elementary Statistics. | 677.169 | 1 |
Synopsis
by Forrest Spencer
This three-video set is a supplemental tutorial for high school or college students tackling the most difficult and challenging Intermediate Algebra questions. Presented by a video professor, this course explains an assortment of algebra problems in detail on a chalkboard in an easy-to-follow approach. The topics reviewed include rational numbers, real numbers and radials, relations and functions, exponential functions, logarithmic functions, complex numbers, quadratic equations and graphs of quadratic functions. This program is also an ideal learning tool for adults who wish to refresh their mathematical skills by studying the material at their own pace. A study guide is provided to supplement this video tutorial. | 677.169 | 1 |
Measure theory and integration homework solutions
E department offers both a major in Computer Science and a minor in Computer Science. Rther information is. Ws This Math in movies page appeared podcast of Jamillah! The transtheoretical model of behavior change assesses an individual's readiness to act on a new healthier behavior, and provides strategies, or processes of change. Anks Many thanks to all the following 160. Nce BibMe makes it easy to create citations, build bibliographies and. Scenes in Movies in which some Mathematics appears. cover letter sample masters program Bachelor of Science in Computer Science. Using other peoples research or ideas without giving them due credit is plagiarism. Edutopia blogger Vicki Davis identifies the nature of grit, its necessity and value of grit in education, and ten ways of teaching students to develop their own grit. Course Status Show Open Courses Only Area of Study Anthropology Sociology Biology Chemistry Classics Ancient World Creative Performing Arts Creative Writing!
You can measure this on every social channel on the planet. Bott , Mary AnnThis site is intended as a resource for university students in the mathematical sciences. Integral. Cebook. Bott , Mary AnnUsing other peoples research or ideas without giving them due credit is plagiarism. Edutopia blogger Vicki Davis identifies the nature of grit, its necessity and value of grit in education, and ten ways of teaching students to develop their own grit. Oks are recommended on the basis of readability and other pedagogical. Yates Probability 3rd Edition solutions Probability and. Ogle Plus. Ail ; Armstrong , Piers. Partment. Rmstrcalstatela. Ofighcalstatela. Resources for science teachers. SCHOOL IUPUI; COURSE TITLE. Ofigh , Maryam. TA can help you reach science educators in every discipline and at every grade level through exhibit hall. Integral is a mathematical object that can be interpreted as an area or a generalization of area. Partment. Nce BibMe makes it easy to create citations, build bibliographies and. Tegrals, together with derivatives, are the. Hibits Advertising. Informal economy exists in all countries, but the extent of it varies, and I take it as a good indicator of whether or not a country has sound institutions and policies. Itter. Ofigh , Maryam. UTube. One beautiful thing. Fundamentals of Heat and Mass Transfer 7th Edition Bergman Solutions ManualFaculty Name. Ail ; Armstrong , Piers. Rmstrcalstatela. Faculty Name. Course Status Show Open Courses Only Area of Study Anthropology Sociology Biology Chemistry Classics Ancient World Creative Performing Arts Creative Writing. Ofighcalstatela? At to do with it.
Approach your problems from the right end and begin with the answers. Approach your problems from the right end and begin with the answers. Click Update to apply the changes. He Chinese Maze Murders" by Robert? Hudson Valley Community College offers more than 80 programs with locations in Albany, Troy, Malta and other areas throughout the Capital Region. Lucy is part of Generation Y, the generation born between the late 1970s and the mid 1990s. Es also part of a yuppie culture that makes up a large portion of Gen. I: Origin and Destiny of the Earth's Magnetic field A: Natural and Artificial Magnets: B: Representation of the Magnetic Field by Lines of Force: C: Thermal. E department offers both a major in Computer Science and a minor in Computer Science. Rther information is? En one day, perhaps, you will find the final question. H — Homework Options when students are presented with choices they are more likely to complete homework assignments. He Chinese Maze Murders" by Robert. Ew a sample Homework options. Llowing the Update, your website should load when the domain name is accessed in any supported browser. Bachelor of Science in Computer Science. En one day, perhaps, you will find the final question. | 677.169 | 1 |
Georgia School Districts
Walch has been working closly with Georgia educators for many years to develop materials that are aligned to Georgia standards, support teachers and help students succeed in math classes.
Georgia Standards of Excellence (GSE)
We are pleased to announce that our GSE Algebra I, Geometry and Algebra II resources are available for the 2017–2018 school year. These teacher and student materials are aligned to Georgia Standards of Excellence and were developed from the start with the guidance and feedback from Georgia educators.
CCGPS Coordinate Algebra
The CCGPS Coordinate Algebra Program is a complete set of teacher and student materials developed around the Common Core Georgia Performance Standards (CCGPS) and the Coordinate Algebra Frameworks, Curriculum Map, and Teacher's Guide. The course design has benefited from direct input from Georgia teachers.
Topics are built around accessible core curricula, ensuring that Coordinate Algebra is useful for striving students and diverse classrooms. This program realizes the benefits of exploratory and investigative learning and employs a variety of instructional models to meet the learning needs of students with a range of abilities. Review and purchase
CCGPS Analytic Geometry
The CCGPS Analytic Geometry Program has been organized to coordinate with the CCGPS Analytic Geometry Frameworks and Curriculum Map, including the Updated Frameworks released July 1, 2013.
Each lesson includes activities that offer opportunities for exploration and investigation. These activities incorporate concept and skill development and guided practice, then move on to the application of new skills and concepts in problem-solving situations. Throughout the lessons and activities, problems are contextualized to enhance rigor and relevance.
CCGPS Advanced Algebra The CCGPS Advanced Algebra Program is a complete set of materials developed around the Common Core Georgia Performance Standards (CCGPS) and the Advanced Algebra Frameworks, Curriculum Map, and Program Overview. Topics are built around accessible core curricula, ensuring that the
CCGPS Advanced Algebra Program is useful for striving students and diverse classrooms.
This program realizes the benefits of exploratory and investigative learning and employs a variety
of instructional models to meet the learning needs of students with a range of abilities.
The CCGPS Advanced Algebra Program includes components that support problem-based learning,
instruct and coach as needed, provide practice, and assess students' skills. Instructional tools and
strategies are embedded throughout.
What teachers are saying...
"I am truly enjoying teaching from the Walch Coordinate Algebra text. I have never found a textbook that matches the Georgia state standards so well. It seems to be tailored to fit our curriculum. The Walch materials are also easily differentiable. This book is usable from accelerated to support classrooms."
- Laura Blair, Math Teacher, Southeast Bulloch High School
"Walch has created the only textbook that truly follows the standards. I have looked at every book available to me and they all simply rearranged the old texts to match up, but did not add or remove anything to ensure that all standards are met. Using the Coordinate Algebra text has made my job so much easier! The chapters are organized in the same order as the CCGPS for Georgia. The Teacher's Edition even provides a timeline so that I know when it is appropriate to pull in the state's tasks. The electronic versions of the student edition and teacher's edition have been a huge help, as well!"
- Heather Lloyd, Math Teacher, Hart County
We are pleased to offer additional high school math resources to Georgia, listed below:
Coordinate Algebra Station Activities for CCGPS The Coordinate Algebra Station Activities for CCGPS is a collection of hands-on, problem-solving activities to provide students with opportunities to practice and apply the math skills and concepts they are learning in their Coordinate Algebra class.
You may use these activities to complement your regular lessons or in place of your regular lessons, if students have the basic concepts but need practice.
To support districts still transitioning from GPS Math I, II, and III:
DeKalb County High utilizes Walch High School Science Programs for the GHSGT, covering the Characteristics of Science and the five content domains: Cells and Heredity; Ecology; Structure and Properties of Matter; Energy Transformations; and Forces, Waves, and Electricity.
Summer School and Intercessions
Fulton County uses summer-school support programs for mathematics in grades 6, 7, and 8.
In addition to our GPS-based materials for secondary classrooms, we have a wide variety of research-based materials aligned to national standards that can be readily incorporated into your educational programs. See the results: Clarke County Achieves 12.8% CRCT Gains in Five Weeks... | 677.169 | 1 |
Math homework help and answers
LOG PRACTICE PROBLEMS ANSWERS Keywords...Math Homework Help Algebra Answers Solve calculus and algebra problems online with Cymath math problem solver with steps to show your work.Math Homework Help Free Answers Free math lessons and math homework help from basic math to. teachers, parents, and everyone can find solutions to their math.Math is a subject, which is extremely useful and interesting, to some extent,.Math Homework Help Algebra Answers Glencoe Pre-Algebra Homework Help from MathHelp.com. Glencoe Pre-Algebra Help with a Personal Math Teacher.
Algebra 2 Homework Help And Answers WebMath is designed to help you solve your math problems. In addition to the answers, Webmath also shows the student how to.We have hundreds of teachers who will answer your questions and help you do your homework. you might get help for free.Connect to a Tutor Now for Math help, Algebra help, English, Science.
Math Homework Help Algebra
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6th Grade Math Worksheets
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Homework Math Problem Answers
Get homework done by teachers who are experts in high school, college and university homework.View Your Geometry Answers Now. Free. Browse the books below to find your textbook and get your solutions now.
A math homework community created in 1999 by Math Goodies. Math goodies was a pioneer of online math help.
Math Homework Answers
Straight Lines. Math Tools:: Math Links Mathematics is commonly called Math in the US and Maths in the UK.Whatever questions you have about our math help website or the answers we have.
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Homework Help Math Answers homeworkis an online homework answers market.Popular Calculus Textbooks See all Calculus textbooks up to: 1250 gold.Get a confirmation that we will be able to complete the order.Free math lessons and math homework help from basic math to algebra, geometry and. parents, and everyone can find solutions to their math problems instantly. | 677.169 | 1 |
Education and Human Development Master's Theses
Technology is the technical means people use to improve their surroundings. People use technology to improve their ability to do work. Classrooms around the world have implemented many forms of technology to enhance student interest and achievement. One form of technology that is common to math classrooms is the graphing calculator. One eighth grade math class of nineteen students from an urban middle school was taught a unit on Solving Systems of Equations by Graphing. The unit was implemented with and without the use of the graphing calculator. Students were first introduced to the unit through the use of pencils ...
Books by Marquette University Faculty The Eleventh Edition features two new contributing authors (David Smith — Indiana University of PA; Dennis Brylow — Marquette University), new, modern examples, and updated coverage based onFaculty Work: Comprehensive List
Algebra deals with more than computations such as addition or exponentiation; it also studies relations. Calculus touches on this a bit with locating extreme values and determining where functions increase and decrease; and in elementary algebra you occasionally "solve" inequalities involving the order relations of < or ≤ , but this almost seems like an intrusion foreign to the main focus, which is making algebraic calculations.
Relational ideas have become more important with the advent of computer science and the rise of discrete mathematics, however. Many contemporary mathematical applications involve binary or n-ary relations in addition to computations. We began discussing this topic in the last chapter when we introduced equivalence relations. In this chapter we will explore other kinds of ...
Six Septembers: Mathematics For The Humanist, Patrick Juola, Stephen Ramsay
Zea E-Books
Scholars of all stripes are turning their attention to materials that represent enormous opportunities for the future of humanistic inquiry. The purpose of this book is to impart the concepts that underlie the mathematics they are likely to encounter and to unfold the notation in a way that removes that particular barrier completely. This book is a primer for developing the skills to enable humanist scholars to address complicated technical material with confidence. This book, to put it plainly, is concerned with the things that the author of a technical article knows, but isn't saying. Like any field, mathematics ...
Active Calculus 1.0, Matthew Boelkins, David Austin, Steven Schlicker
Open Textbooks
Active Calculus is different from most existing calculus texts in at least the following ways: the text is free for download by students and instructors in .pdf format; in the electronic format, graphics are in full color and there are live html links to java applets; the text is open source, and interested instructors can gain access to the original source files upon request; the style of the text requires students to be active learners — there are very few worked examples in the text, with there instead being 3-4 activities per section that engage students in connecting ideas, solving problems ...
Honors Scholar Theses
Differential equations are equations that involve an unknown function and derivatives. Euler's method are efficient methods to yield fairly accurate approximations of the actual solutions. By manipulating such methods, one can find ways to provide good approximations compared to the exact solution of parabolic partial differential equations and nonlinear parabolic differential equations.
Introduction To Real Analysis, William F. Trench
Faculty Authored Books
Using a clear and informal approach, this book introduces readers to a rigorous understanding of mathematical analysis and presents challenging math concepts as clearly as possible. This book is intended for those who want to gain an understanding of mathematical analysis and challenging mathematical concepts.
Calculating The Time Constant Of An Rc Circuit, Sean Dunford
Undergraduate Journal of Mathematical Modeling: One + Two
In this experiment, a capacitor was charged to its full capacitance then discharged through a resistor. By timing how long it took the capacitor to fully discharge through the resistor, we can determine the RC time constant using calculus.
All Articles in Mathematics
Asymptotic Counting Formulas For Markoff-Hurwitz Tuples, Ryan Ronan
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
The Markoff equation is a Diophantine equation in 3 variables first studied in Markoff's celebrated work on indefinite binary quadratic forms. We study the growth of solutions to an n variable generalization of the Markoff equation, which we refer to as the Markoff-Hurwitz equation. We prove explicit asymptotic formulas counting solutions to this generalized equation with and without a congruence restriction. After normalizing and linearizing the equation, we show that all but finitely many solutions appear in the orbit of a certain semigroup of maps acting on finitely many root solutions. We then pass to an accelerated subsemigroup of ...
Some Results In Combinatorial Number Theory, Karl Levy
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
The first chapter establishes results concerning equidistributed sequences of numbers. For a given $d\in\mathbb{N}$, $s(d)$ is the largest $N\in\mathbb{N}$ for which there is an $N$-regular sequence with $d$ irregularities. We compute lower bounds for $s(d)$ for $d\leq 10000$ and then demonstrate lower and upper bounds $\left\lfloor\sqrt{4d+895}+1\right\rfloor\leq s(d)< 24801d^{3} + 942d^{2} + 3$ for all $d\geq 1$. In the second chapter we ask if $Q(x)\in\mathbb{R}[x]$ is a degree $d$ polynomial such that for $x\in[x_k]=\{x_1,\cdots,x_k\}$ we have $|Q(x)|\leq 1$, then how big can its lead coefficient be? We prove that there is a unique polynomial, which we call $L_{d,[x_k]}(x)$, with maximum lead coefficient under these constraints and construct an algorithm that generates $L_{d,[x_k]}(x)$.
All Graduate Works by Year: Dissertations, Theses, and Capstone Projects
The collection of elementary substructures of a model of PA forms a lattice, and is referred to as the substructure lattice of the model. In this thesis, we study substructure and interstructure lattices of models of PA. We apply techniques used in studying these lattices to other problems in the model theory of PA.
In Chapter 2, we study a problem that had its origin in Simpson, who used arithmetic forcing to show that every countable model of PA has an expansion to PA∗ that is pointwise definable. Enayat later showed that there are 2ℵ0 models with the ...
A Weighted Möbius Function, Derek Garton
Mathematics and Statistics Faculty Publications and Presentations
Fix an odd prime ℓ and let G be the poset of isomorphism classes of finite abelian ℓ-groups, ordered by inclusion. If ξ:G→R≥0is a discrete probability distribution on G and A ∈ G, define the Ath moment of ξ to be . The question of determining conditions that ensure ξ is completely determined by its moments has been of recent interest in many problems of Cohen–Lenstra type. Furthermore, recovering ξ from its moments requires a new Möbius-type inversion formula on G. In this paper, we define this function, relate it to the classical Möbius function ...
All Graduate Plan B and other Reports
...
Benjamin Fine
We introduce in this paper cryptographic protocols which use combinatorial group theory. Based on a combinatorial distribution of shares we present secret sharing schemes and cryptosystems using Nielsen transformations. Nielsen transformations are a linear technique to study free groups and general infinite groups. In addition the group of all automorphisms of a free group F, denoted by ( )Aut F, is generated by a regular Nielsen transformation between two basis of F, and each regular Nielsen transformation between two bases of F defines an automorphism of F.
Open Educational Resources - Math
In this activity, problem solving skills are practiced using two well-known problems from Fibonacci's world-changing book "Liber Abbaci". Students are also asked to reflect on the differences and similarities between their solutions and those of Fibonacci. The two problems are the famous rabbit problem which led to what is now know as the Fibonacci sequence and the 30 birds for 30 denarii problem, which is not as well-known to the general public.Pollution is a severe problem today, and the main challenge in water and air pollution controls and eliminations is detecting and locating pollution sources. This research project aims to predict the locations of pollution sources given diffusion information of pollution in the form of array or image data. These predictions are done using machine learning. The relations between time, location, and pollution concentration are first formulated as pollution diffusion equations, which are partial differential equations (PDEs), and then deep convolutional neural networks are built and trained to solve these PDEs. The convolutional neural networks consist of convolutional layers, reLU layers ...
Vertex Weighted Spectral Clustering, Mohammad Masum
Electronic Theses and Dissertations
Spectral clustering is often used to partition a data set into a specified number of clusters. Both the unweighted and the vertex-weighted approaches use eigenvectors of the Laplacian matrix of a graph. Our focus is on using vertex-weighted methods to refine clustering of observations. An eigenvector corresponding with the second smallest eigenvalue of the Laplacian matrix of a graph is called a Fiedler vector. Coefficients of a Fiedler vector are used to partition vertices of a given graph into two clusters. A vertex of a graph is classified as unassociated if the Fiedler coefficient of the vertex is close to ...
Dissertations, Theses, and Student Research Papers in Mathematics
Fat points and their ideals have stimulated a lot of research but this dissertation concerns itself with aspects of only two of them, broadly categorized here as, the ideal containments and polynomial interpolation problems.
Ein-Lazarsfeld-Smith and Hochster-Huneke cumulatively showed that for all ideals I in k[Pn], I(mn) ⊆ Im for all m ∈ N. Over the projective plane, we obtain I(4)< ⊆ I2. Huneke asked whether it was the case that I(3) ⊆ I2. Dumnicki, Szemberg and Tutaj-Gasinska show that if I is the saturated homogeneous radical ideal of the 12 points of the Hesse configuration, then ...
Constructing A Square An Ancient Indian Way Activity, Cynthia J. Huffman Ph.D.
Open Educational Resources - Math
In this activity students use string to model one of the ways that was used in ancient India for constructing a square. The construction was used in building a temporary fire altar. The activity is based on a translation by Sen and Bag of the Baudhāyana-śulba-sūtra.
Euler Construction Activity, Cynthia J. Huffman Ph.D.
Open Educational Resources - Math
Original sources of mathematics provide many opportunities for students to both do mathematics and to improve their problem solving skills. It is also interesting to explore original sources in new ways with the use of technology. In this activity, students can gain experience with dynamic geometry software and enhance their geometric intuition by working through a construction given by Euler in 1783.
Open Educational Resources - Math
In this activity, we will model constructing a square fire altar with a method similar to one used by people in ancient India. The fire altars, which were made of bricks, had various shapes. Instructions for building the altars were in Vedic texts called Śulba-sūtras. We will follow instructions for constructing a square gārhapatya fire altarfrom the Baudhāyana-śulba-sūtra, which was written during the Middle Vedic period, about 800-500 BC.
Antichains And Diameters Of Set Systems, Brent Mckain
Dissertations, Theses, and Student Research Papers in Mathematics
In this thesis, we present a number of results, mostly concerning set systems that are antichains and/or have bounded diameter. Chapter 1 gives a more detailed outline of the thesis. In Chapter 2, we give a new short proof of Kleitman's theorem concerning the maximal size of a set system with bounded diameter. In Chapter 3, we turn our attention to antichains with bounded diameter. Šileikis conjectured that an antichain of diameter D has size at most (n/D/2). We present several partial results towards the conjecture.
Dissertations, Theses, and Student Research Papers in Mathematics
Chapter 1 deals with the structure of stable cohomology and bounded cohomology. Stable cohomology is a $\mathbb{Z}$-graded algebra generalizing Tate cohomology and first defined by Pierre Vogel. It is connected to absolute cohomology and bounded cohomology. We investigate the structure of the bounded cohomology as a graded bimodule. We use the information on the bimodule structure of bounded cohomology to study the stable cohomology algebra as a ...
Active Calculus 2.0, Matthew Boelkins, David Austin, Steven Schlicker
Open Textbooks
Active Calculus is different from most existing calculus texts in at least the following ways: the text is freely readable online in HTML format and is also available for in PDF; in the electronic format, graphics are in full color and there are live links to java applets; version 2.0 now contains WeBWorK exercises in each chapter, which are fully interactive in the HTML format and included in print in the PDF; the text is open source, and interested users can gain access to the original source files on GitHub; the style of the text requires students to be ... | 677.169 | 1 |
LINES - PLANES - ANGLES
LINES AND PLANES IN SPACE
TYPICAL PROOFS OF SOLID GEOMETRY
ANGLES
Plane
Is
a surface such that a straight line
joining any two points in it lies wholly in
the surface.
Usually represented by a parallelogram
horizontaledges
lying
College Algebra
P.2
Real Numbers
and Their Properties
Types of Real Numbers
Introduction
Lets review the types
of numbers that make up
the real number system.
Natural Numbers
We start with the natural
numbers:
1, 2, 3, 4,
Integers
The integers consist of
1. Multiply a polynomial by a monomial.
2. Multiply a polynomial by a polynomial.
The Distributive Property
Look at the following expression:
3(x + 7) This expression is the sum of x and 7 multiplied by 3.
(3 x) + (3 7)
3x + 21
To simplify this expression
Algebraic
Expressions
Definitions
Variable A variable is a letter or
symbol that represents a
number (unknown quantity).
8 + n = 12
Definitions
A variable can use any letter of
the alphabet. (expect o and
needs to be lower case)
n + 5
x 7
w - 25
Definitio
Factoring Polynomials
Grouping,Trinomials,Binomials,
GCF&SolvingEquations
Factor by Grouping
When
polynomials contain four terms, it
is sometimes easier to group like terms in
order to factor.
Your goal is to create a common factor.
You can also move t
Solid
Mensuration
Solid Mensuration
Definition: Solid Mensuration deals
primarily with the various solids.
Application in Real Life Situation
- This applies in railway engineering, in
road and bridge construction, in chemical
and physical analyses, and in
Microwave
Communications
Engr. Leah Q. Santos
Faculty, Engg EcE dept.
Diversity
Diversity
Radio
fade reduction of signal strength
that can last for a few milliseconds
(short term) or for several hours or even
days (long term)
Fades:
excess of 40 dB can c
ENGINEERING ECONOMY
I.SOLVED PROBLEMS
1) How much should be put in an investment with a 10% effective annual rate today to have
$10,000 in five years?
P = F(1 + i) n = ($10,000)(1 + 0.1) 5 = $6209
2) What factor will convert a gradient cash flow ending at
PHYSICS
I. SOLVED PROBLEMS.
1. Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light
turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00
m/s2, then determine the displacement of the
REPUBLIC OF KENYA
MINISTRY OF INDUSTRIALIZATION AND
ENTERPRISE DEVELOPMENT
TITLE: REQUEST FOR PROPOSAL
SELECTION OF CONSULTANTS
CONSULTANCY SERVICES FOR FEASIBILITY STUDIES AND
DEVELOPMENT OF MASTER PLANS INCLUDING PROTOTYPES FOR THE
PROPOSED INDUSTRIAL P
Transcript of POTENTIAL USE OF TALAHIB GRASS ROOTS (Saccharum spontaneum L
POTENTIAL USE OF TALAHIB GRASS ROOTS
(Saccharum spontaneum Linn)
SUCROSE AS LAUNDRY SURFACTANT
Abstract
A detergent is a surfactant or a mixture of surfactants with "cleaning prope
Management Information System
HandOuts-001/ 04-18-2011
MANAGEMENT INFORMATION SYSTEM
Prepared by Rommel B. Dy
Chapter One
The Information Age in Which You Live:
Changing the Face of Business
Student Learning Outcomes
1. Define management information syste | 677.169 | 1 |
Newsletter
Course Announcement: MATH110 - Introduction to Complex
Arithmetic
MATH110 - Introduction to Complex Arithmetic is complete. This
is
part
of the in the mathematics training series.
The purpose of this course is to give a rudimentary
explanation of complex numbers and
how you manipulate them. Complex numbers arise in many applications,
and quite often in
mechanical engineering, electrical engineering, and measurement
applications. Upon completing this course you should understand:
what a complex number is, and how its related to real
numbers
what is meant by real and imaginary parts of a complex
number
polar representation of complex numbers (magnitude and
phase)
the euler formula
how to convert from polar to cartesion representations
know what is meant by complex conjugate
how to add and subtract two complex numbers
how to multiply and divide two complex numbers
how to raise a complex number to an integer power
how to define selected elementary functions for complex
numbers
a few applications of complex numbers
If you are an engineer or have at least a BS in a technical
field, this
could serve as a review or reference. This course is primarily for
technicians or interested persons who do not have a background in
complex numbers, so that they
can function in test environments that utilize measurements of and
calculations involving complex quantities. This course, however, is a
prerequisite for some of the courses that require familiarity with
complex arithemetic. | 677.169 | 1 |
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There are certain mistakes that students frequently make while learning algebra. This packet, focusing on order of operations quadratic equationsThis packet, focusing on math vocabulary, contains a variety of puzzles that illustrate the correct way of working an algebra problem. They will challenge students to think creatively, and self-checking exercises motivate students to finish each page while acquiring valuable algebraic skills.
There are certain mistakes that students frequently make while learning algebra. This packet, focusing on fractions, decimals, and percentages expressions, equations, algebraic fractions interger operation activities. Each page gives an example and step-by-step solution of the problem presented. Answer key is included.
Expand your students' knowledge of fractions probability, proportions, and percentsBasic skills are reviewed and expanded as students work through these fun whole number and decimal activities. Each page gives an example and step-by-step solution of the problem presented. Answer key is included.
Expand your students' knowledge of basic geometry digit divisibility decimals polynomials expression and equation activities. Each page gives an example and step-by-step solution of the problem presented. Answer key is included.
Basic skills are reviewed and expanded as students work through these fun rational number activities. Each page gives an example and step-by-step solution of the problem presented. Answer key is included.
There are certain mistakes that students frequently make while learning algebra. This packet, focusing on graphing factoring integers whole numbers radicals, ratios, and rates evaluating expressions finish each page while acquiring valuable algebraic skills.
Basic skills are reviewed and expanded as students work through these fun prime factor and fraction activities. Each page gives an example and step-by-step solution of the problem presented. Answer key is included.
There are certain mistakes that students frequently make while learning algebra. This packet, focusing on equations proportion and percent activities. Each page gives an example and step-by-step solution of the problem presented. Answer key is included.
There are certain mistakes that students frequently make while learning algebra. This packet, focusing on monomial math factoringMake algebra equations easy for students in grades 7 and up using Algebra Practice! This 128-page book is geared toward students who struggle in algebra and covers the concepts of number systems, exponential expressions, square roots, radical expressions, graphing, and linear and quadratic functions. The book supports NCTM standards and includes clear instructions, examples, practice problems, definitions, problem-solving strategies, an assessment section, answer keys, and references.
Make math matter to students in all grades using Math Tutor: Algebra Skills! This 80-page book provides step-by-step instructions of the most common math concepts and includes practice exercises, reviews, and vocabulary definitions. The book covers factoring, exponents, variables, linear equations, and polynomials. It aligns with state, national, and Canadian provincial standards.
Make math matter to students in grades 5–6Solving Word Problems is one of the biggest hurdle that kids face in Algebra. A bit of imagination is required to understand and solve these type of problems along with the calculations.
This book breaks simple word problems using graphics thus helping the kids to visualize and understand the word problems. It develops the imaginative thinking required to solve these problems from an early level. This will help the kids to solve difficult problems as they will learn to imagine, analyze and break the problem into small parts which gives a better understanding on how to solve these type of problems.
Make math matter to students in grades 7–8SHE'LL TRY TO SOLVE FIVE QUESTIONS IN THREE HOURS, FOR ONE IMPROBABLE DREAM.
THE DREAM OF REPRESENTING HER COUNTRY, AND BECOMING A MATH OLYMPIAN.
As a small-town girl in Nova Scotia bullied for liking numbers more than boys, and lacking the encouragement of her unsupportive single mother who frowns at her daughter's unrealistic ambition, Bethany's road to the International Math Olympiad has been marked by numerous challenges.
Through persistence, perseverance, and the support of innovative mentors who inspire her with a love of learning, Bethany confronts these challenges and develops the creativity and confidence to reach her potential.
In training to become a world-champion "mathlete", Bethany discovers the heart of mathematics – a subject that's not about memorizing formulas, but rather about problem-solving and detecting patterns to uncover truth, as well as learning how to apply the deep and unexpected connections of mathematics to every aspect of her life, including athletics, spirituality, and environmental sustainability.
As Bethany reflects on her long journey and envisions her exciting future, she realizes that she has shattered the misguided stereotype that only boys can excel in math, and discovers a sense of purpose that through mathematics, she can and she will make an extraordinary contribution to society.
Spectrum Math for grade 6 keeps kids at the top of their math game using progressive practice, math in everyday settings, and tests to monitor progress. The math workbook covers multiplying and dividing decimals and fractions, complex measurements, and beginningGive your students the necessary materials to run a math competition with a single classroom or multiple classrooms taking part. It also enables you to choose school teams for external competitions. As well, the two sets of questions in each book can be used during regular math class as a fun activity. Included in this unit are the following: teacher information, instructions, practise question sets, level 4 question sets, marking sheets and certificates. The unit is recommended for Grades 6 to 8. This level 4 math competition will engage students while promoting a healthy competition between peers. This Math lesson provides a teacher information section and a student section with a variety of worksheets creating a well-rounded lesson plan.
Brighter Child(R) Master Math: Introductory Algebra provides children in grade 5 with additional math instruction. Offering 80 pages of full-color activities, easy-to-follow directions, and complete answer key, children will get the extra practice they need while having fun learning algebra.
Features activities that teach:
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*Equations
*Number lines
*Integers
*Variables
*Greatest common factors
*Least common multiples
*Order of operations
*Exponents
*Measurements
*Length
*Weight
*Capacity
*Metric
*Fractions
The Stack Model Method—An Intuitive and Creative Approach to Solving Word Problems (Grades 3–4) is the first 3–4 3–4 students in Singapore do in elementary math—you too can learn to solve the types of challenging questions they deal with every day in local schools and tuition centers island-wide.
The Stack Model Method—An Intuitive and Creative Approach to Solving Word Problems (Grades 5–6) is the second 5–6 5–6 students in Singapore do in elementary math—you too can learn to solve the types of challenging questions they deal with every day in local schools and tuition centers island-wide.
If your child is struggling with math, then this book is for you; the short book covers the topic and also contains 30 practice problems to work with.
This subject comes from the book "Third Grade Math (For Home School or Extra Practice)"; it more thoroughly covers more third grade topics to help your child get a better understanding of fourth grade math. If you purchased that book, or plan to purchase that book, do not purchase this, as the problems are the same.
Spectrum Math for grade 7 keeps kids at the top of their math game using progressive practice, math in everyday settings, and tests to monitor progress. The math workbook covers algebra, geometry, statistics, proportions, ratios, and more 5 keeps kids at the top of their math game using progressive practice, math in everyday settings, and tests to monitor progress. The math workbook covers fractions, decimals, multiplication, division 8 keeps kids at the top of their math game using progressive practice, math in everyday settings, and tests to monitor progress. The math workbook covers rational and irrational numbers, solving equations, and interpreting statistical data 4 keeps kids at the top of their math game using progressive practice, math in everyday settings, and tests to monitor progress. The math workbook covers multiplication, division, fractions, geometric figuresMath, history, art, and world cultures come together in this delightful book for kids, even for those who find traditional math lessons boring. More than 70 games, puzzles, and projects encourage kids to hone their math skills as they calculate, measure, and solve problems. The games span the globe, and many have been played for thousands of years, such as three-in-a-row games like Achi from Ghana or the forbidden game of Jirig from Mongolia. Also included are imaginative board games like Lambs and Tigers from India and the Little Goat Game from Sudan, or bead and string puzzles from China, and Mobius strip puzzles from Germany. Through compelling math play, children will gain confidence and have fun as they learn about the different ways people around the world measure, count, and use patterns and symmetry in their everyday lives. | 677.169 | 1 |
Applications of Calculus (MATH1011)
UNIT OF STUDY
This unit is designed for science students who do not intend to undertake higher year mathematics and statistics. It establishes and reinforces the fundamentals of calculus, illustrated where possible with context and applications. Specifically, it demonstrates the use of (differential) calculus in solving optimisation problems and of (integral) calculus in measuring how a system accumulates over time. Topics studied include the fitting of data to various functions, the interpretation and manipulation of periodic functions and the evaluation of commonly occurring summations. Differential calculus is extended to functions of two variables and integration techniques include integration by substitution and the evaluation of integrals of infinite type | 677.169 | 1 |
Sunday, May 7, 2017
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Head Start to AS Maths Paperback
Description
This book is designed to help bridge the gap between GCSE and AS Level Maths. It's full of clear notes and helpful practice to recap the most difficult topics from GCSE Maths that students need when going on to study AS Level Maths. Everything you need to know for all the exam boards is explained clearly and simply, in CGP's chatty straightforward style. | 677.169 | 1 |
This course is not only virtual learning material . You can ask any questions related to Infinite Sequence and series after enrolling in this course.
Here is a listing and brief description of the topics in this course.
A brief discussion of sequences. The basic terminology and convergence of sequences.
Quick look about monotonic and bounded sequences.
Series The Basics we will discuss some of the basics of infinite series.
Series Convergence/Divergence about the convergence/divergence of a series
Series Special Series Geometric Series, Telescoping Series, and Harmonic Series in this section.
Integral Test to determine if a series converges or diverges.
Comparison Test/Limit Comparison Test to determine if a series converges or diverges.
Alternating Series Test Using the Alternating Series Test to determine if a series converges or diverges.
Absolute Convergence A brief discussion on absolute convergence and how it differs from convergence.
Ratio Test Using the Ratio Test to determine if a series converges or diverges.
Root Test Using the Root Test to determine if a series converges or diverges.
Strategy for Series A set of general guidelines to use when deciding which test to use.
Estimating the Value of a Series estimating the value of an infinite series.
Power Series An introduction to power series and some of the basic concepts.
This course Contains 90% of Video tutorials which include practice section. Master in Sequence and series for your Calculus Course!
If you can understand my pronunciation of English in PROMO VIDEO then you can understand all lecture. So Why are you waiting? Click the button "TAKE THIS COURSE".
How much content is included in the course?
Course is updated on Weekly basis and you will get more content in next week.
Hurry! The price will be increasing soon.
What will you know, or be able to do, after completing the course?
Benefits of this course are:
After completing this course you will be Master in solving Sequences and series problem of any type.
Your teacher will be surprised with your answer.
What are the requirements?
Basic Knowledge of Calculus-I topic like Limits, Integral etc.
The Calculus II Course on INFINITE SEQUENCE AND SERIES assume that you've got a working knowledge Calculus I, including Limits, Derivatives, and Integration (up to basic substitution). It is also assumed that you have a fairly good knowledge of Trig. Several topics rely heavily on trig and knowledge of trig functions.
What am I going to get from this course?
Over 15 lectures and 36 mins of content!
Basic Idea of Convergent and Divergent Sequences
Limit of Sequences
Increasing, Decreasing and Monotonic Sequences
Bounded Above, Bounded Below and BOUNDED Sequences
Monotonic Sequence Theorem
Convergent and Divergent Series
Partial Sums of Series
Telescoping , Geometric and Harmonic Series
Test for Divergence for Series
Integral Test and P series
Remainder Estimate for the Integral test
Comparison Test and Limit comparison test
Alternating Series Test
Absolute and Conditional Convergent
Ratio and Root Tests
Check Any Type of Series for Convergent OR Divergent
Power Series Representation
What is the target audience?
Adult Learner
College Students
Calculus-I and Calculus-II Students
Get your Free Udemy Course on Infinite Sequence and Series by clicking here: | 677.169 | 1 |
Math 210
Computer Lab #1 - An Introduction to Matrix Manipulation Using a Computer Algebra System
You may use any computer algebra system (CAS) for the following problems. We have Maple on the
computers in IDC 109 available for your use if you do not have
1) Randomly generate a 100 x 100 matrix A and a 100 x 1 matrix B.
(1)
(2)
2) Compute the times for inverting A, finding the LU factorization, solving Ax = B, multiplying AA,
and
Display
each command line that shows the calculation was executed and show th
Math 210
Computer Lab #3
10 points
Using determinants to estimate eigenvalues is computationally slow. We investigate the power method and
how it is used to estimate eigenvalues. This is not the method used in practice, but it is much faster than using
de
Math 210
Computer Lab #2
25 points
Use your favorite CAS to investigate computing times for different matrix operations in the following way.
1) Randomly generate a 100 x 100 matrix A and a 100 x 1 matrix B. DO NOT DISPLAY the entries!
Compute the times f
Final Exam. Linear Algebra 110
Instructor: Zvezdelina Stankova
Student name:
Student ID:
GSIs name:
DO NOT OPEN THE EXAM UNTIL TOLD TO DO SO!
PREPARE YOUR STUDENT PHOTO ID TO SHOW TO YOUR PROCTOR.
Do all problems as best as you can. The exam is 2 hrs an | 677.169 | 1 |
BJU Press Math 4 Reviews - 3rd ed.
Math 4 Reviews provides two pages of review for each lesson. The
first page reviews concepts taught in the lesson for the day. The
second page reviews concepts taught in previous lessons and
provides facts practice. A Chapter Review is included at the end of
each chapter. The pages may be used any time after the lesson has
been taught. | 677.169 | 1 |
Give Your Child a Head Start In High School Geometry
Discoveries In Mathematics is home to exciting video presentations that explain high school geometry in a more visual way. Our authors, longtime educators Paul Cinco and Gene Eyshinskiy, break down complex concepts, such as postulates, into easy-to-understand language and dynamic images.
Our video presentations are organized in eleven chapters, all of which address the entire scope and sequence of the tenth grade high school geometry course, including all Euclidian proofs and proofs for geometric constructions taught in most schools. Each video presentation provides a visual understanding of a specific concept and review solutions to some related problems. | 677.169 | 1 |
Standardized tests and college entrance exams. Many tests and exams permit or even require the use of a graphing calculator. A TI graphing calculator is ideal for students to use in math and science classes from middle school through college.
Which Graphing Calculator is right for you? A graphing calculator is a learning tool designed to help students visualize and better understand concepts in math and science. Your selection should be determined by type of test (SAT, AP, ACT IB or PSAT/NMSQT), key features and what courses you plan to attend.
Computer Software TI software products range from student software to help students master math and science concepts to teacher software to help demonstrate and lead classroom exploration.
TI-Nspire Navigator These systems enable educators to quickly connect to all of their students to perform real-time assessment, share files, monitor student understanding of concepts and more.
TI-Cares Customer Support If you have any questions, please contact us via email / phone or TI directly 1-800-842-2737
overall customer rating
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Texas Instruments was founded in 1930 as Geophysical Service Inc., an oil and gas company. The small American company used its spirit of innovation to become what it is today, a global leader in semiconductor technology. Texas Instruments uses innovation to change the lives of 90,000 customers worldwide.
Texas Instruments is an analog and digital semiconductor integrated circuit design and manufacturing company. The company operates in more than 30 countries around the globe. Texas Instruments has three divisions, Semiconductors, Educational Technology, and Digital Light Processing. The company has been known throughout the years for producing consumer electronics such as digital clocks, watches, hand-held calculators, home computers, and sensors. | 677.169 | 1 |
Solving Inequalities
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This powerpoint covers the basics of solving inequalities through the use of analyzing word problems. It gives three examples of word problems for the teacher to work through with the students as well. | 677.169 | 1 |
Category Archives: High school mathematics
GeoGebra is a great tool to promote a way of thinking and reasoning about shapes. It provides an environment where students can observe and describe the relationships within and among geometric shapes, analyze what changes and what stays the same when shapes are transformed, and make generalizations. When shapes or objects are transformed or moved,… Read More »
While there is no consensus yet as to a precise definition of this term, mathematical modeling is generally understood as the process of applying mathematics to a real world problem with a view of understanding the latter. One can argue that mathematical modeling is the same as applying mathematics where we also start with a… Read More »
The conference is open to high school mathematics and science teachers, department heads and coordinators, supervisors, tertiary and graduate students and lecturers, researchers, and curriculum developers in science and mathematics. Plenary Topics and Speakers 1. The Relationship between Classroom Tasks, Students' Engagement, and Assessing Learning by Dr. Peter Sullivan 2. Assessment for Learning: Practice, Pupils… Read More »
The square root of a number is usually introduced via an activity that involves getting the side of a square with the given area. For example the side of a square with area 25 sq unit is 5 unit because 5 x 5 = 25. To introduce the existence of , a square of area… Read More » | 677.169 | 1 |
About this title:
Synopsis: The primary purpose of this undergraduate text is to teach students to do mathematical proofs. It enables readers to recognize the elements that constitute an acceptable proof, and it develops their ability to do proofs of routine problems as well as those requiring creative insights. The self-contained treatment features many exercises, problems, and selected answers, including worked-out solutions. Starting with sets and rules of inference, this text covers functions, relations, operation, and the integers. Additional topics include proofs in analysis, cardinality, and groups. Six appendixes offer supplemental material. Teachers will welcome the return of this long-out-of-print volume, appropriate for both one- and two-semester courses.
About the Author:
Andrew Wohlgemuth is Professor Emeritus at the University of Maine.
"About this title" may belong to another edition of this title.
Bibliographic Details
Title: Introduction to Proof in Abstract ...
Publisher: Harcourt School
Binding: Hardcover
Book Condition: GOOD
Book Description Saunders College Publishing, 1990. Book Condition: Very Good. *Price HAS BEEN temporarily REDUCED by 10% until Monday, Aug. 21. Order now for best savings* 366 pp., Hardcover, hand stamp to pastedowns, else very good. Bookseller Inventory # ZB1059457
Book Description Harcourt School. Hardcover. Book Condition: GOOD. Good clean copy with no missing pages might be an ex library copy; Possibly may have minor marginal notes and or highlighting. Bookseller Inventory # 2726477665 | 677.169 | 1 |
I'm getting really tired in my math class. It's standard form calculator, but we're covering higher grade material. The concepts are really complicated and that's why I usually sleep in the class. I like the subject and don't want to drop it, but I have a big problem understanding it. Can someone guide me?
I have a solution for you and it might just prove to be a better one than purchasing a new textbook. Try Algebrator, it covers a pretty comprehensive list of mathematical topics and is highly recommended. With it you can solve various types of questions and it'll also address all your enquiries as to how it came up with a particular answer. I tried it when I was having difficulty solving questions based on standard form calculator and I really enjoyed using it.
Oh . I did not make out that there could be a solution for me. May be I should try this out . I am already comforted to know that the resolution to my troubles is at hand. I am ready to try this out. Can you let me know where I can access this program? | 677.169 | 1 |
mathematics
If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is. ~John Louis von Neumann
Algebra I
-Prerequisites: None -Length: One Year -Meets requirement for one year of mathematics 02052A000 B, S, E, R
Grades 9
1 Credit
Algebra I includes the study of properties and operations of the real number system; evaluating rational algebraic expressions; solving and graphing first degree equations and inequalities; translating word problems into equations; operations with and factoring of polynomials; and solving simple quadratic equations.
Algebra II
-Prerequisites: Algebra I and Geometry or concurrent enrollment in Geometry -Length: One Year -Elective Course: (may be taken in Grade 10 with instructor's approval) 02106A000 B, S, E, R
Grades 11,12
1 Credit
Algebra II combines trigonometry and advanced algebra topics, and are usually intended for students who have attained Algebra I and Geometry objectives. Topics typically include right trigonometric and circular functions, inverses, and graphs; trigonometric identities and equations; solutions of right and oblique triangles; complex numbers; numerical tables; field properties and theorems; set theory; operations with rational and irrational expressions; factoring of rational expressions; in-depth study of linear equations and inequalities; quadratic equations; solving systems of linear and quadratic equations; graphing of constant, linear, and quadratic equations; and properties of higher degree equations.
algebra III
Algebra III is designed for students who have successfully completed Algebra II. This course will enhance the higher level thinking skills developed in Algebra II through a more in-depth study of those concepts and exploration of some pre-calculus concepts. Students in Algebra III will be challenged to increase their understanding of algebraic, graphical and numerical methods to analyze, translate and solve quadratic, polynomial, rational, exponential and logarithmic functions. Modeling real world situations is an important part of this course. Sequences and series will be used to represent and analyze real world problems and mathematical situations. Algebra III will also include a study of trigonometric functions, right triangles, and oblique triangles. Teachers are responsible for integrating appropriate technology for Algebra III.
AP calculus AB
AP® Calculus AB follows the College Board's suggested curriculum designed to parallel college-level calculus courses. AP Calculus AB provides students with an intuitive understanding of the concepts of calculus and experience with its methods and applications. These courses introduce calculus and include the following topics: elementary functions; properties of functions and their graphs; limits and continuity; differential calculus (including definition of the derivative, derivative formulas, theorems about derivatives, geometric applications, optimization problems, and rate-of-change problems); and integral calculus (including antiderivatives and the definite integral).
foundation math
-Prerequisites: None -Length: One Year -Meets requirement for one year of electives (not for mathematics credit) -Required elective course for students falling below the predetermined MAP Math RIT score 02002A000 B
Grade 9
1 Credit
Foundation Math builds, reinforces, and expands students' foundational math skills. This course extends these skills to include some pre-algebra and algebra topics, number sense, measurement, basic geometry, and basic statistics.
geometry
Geometry is a course emphasizing an abstract, formal approach to the study of geometry, and typically includes topics such as properties of plane and solid figures; deductive methods of reasoning and use of logic; geometry as an axiomatic system including the study of postulates, theorems, and formal proofs; concepts of congruence, similarity, parallelism, perpendicularity, and proportion; and rules of angle measurement in triangles.
pre-calcuslus
Pre-calculus is a course in which students study the number system, negative and fractional exponents, systems of equations, matrices, operations with polynomials, functions, imaginary and complex numbers, quadratic equations, conic sections, trigonometric functions, and radian measurement. Students will also study a combination of algebra and geometry in the study of analytical geometry. Advanced topics such as vectors, polar graphs, logarithms, fractal geometry, probability, statistics, and graph theory are included as time permits.
Pre-Calculus may be taken as ICC dual credit courses—MATH 115 and 120 ICC requires a student to meet or exceed the qualifying score on the math COMPASS test.
probability and statistics/ap
Probability and Statistics/AP® is a course that follows the College Board's suggested curriculum designed to parallel college-level statistics courses. AP® Statistics courses introduce students to the major concepts and tools for collecting, analyzing, and drawing conclusions from data. Students are exposed to four broad conceptual themes: exploring data, sampling and experimentation, anticipating patterns, and statistical inference. Earning AP® credit is optional. | 677.169 | 1 |
Algebra 2 Activites Bundle
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This Algebra 2 bundle contains activities for polynomials, functions, domain, range, increasing, decreasing and nonlinear graphs. Also Included is an Algebra 2 word wall that students can refer to throughout the year. You can read more about each resource by clicking on its link below. | 677.169 | 1 |
Graphing Piecewise Functions - Notes and Practice BUNDLE
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This bundle contains notes and practice to help students graph piecewise functions. There is a PowerPoint Presentation that goes through all examples (plus three additional ones) on the Graphing Piecewise Functions - Student Worksheet.
Note: If you'd like to purchase 12 of my Functions activities, practice, and review worksheets as a bundle and save $15, please go to the following link: | 677.169 | 1 |
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MA1210: Module 3 Equations, Inequalities, and Matrices
Lab 3.1
Linear Equations and Matrices
This lab will include solving linear systems. Additionally, this lab will include writing the augmented
matrix for a linear system and then solving linear systems
MA1210: Module 2 Fundamentals of Algebra, Part 2
Lab 2.1
Polynomials and Rational Expressions
Throughout this course you will be analyzing polynomials and their properties. This lab
includes factoring exercises of many different types of polynomials. Addi
MA1210: Module 4 Functions and Their Graphs
Lab 4.1
Transformation, Composite, and Inverse Functions
Due: 11/3/2015
This lab includes problems based on graphing various functions and application problems
involving the slope and equations of lines.
Recomme
MA1210: Module 3 Equations, Inequalities, and Matrices
Exercise 3.1
Equations and Their Graphs
In this assessment, you need to solve problems related to your school of study. Provide
detailed steps, wherever required. Also, support each solution with an a
MA1210: Module 2 Fundamentals of Algebra, Part 2
Exercise 2.1
Exponents and Polynomials
In this assessment, you will find four questions listed under each school of study. You only need
to solve problems related to your school of study. Provide detailed s
MA1210: Module 4 Functions and Their Graphs
Exercise 4.1
Functions Their Slopes and Graphs
School of Information Technology
1. (20 points) The media access control (MAC) address of a network interface is a
unique address. Each network interface relating t
MA1210: Module 5 Polynomial and Rational Functions
Lab 5.1
Rational Functions and Inequalities
This lab requires you to find the maximum area, use synthetic division to find zeros of a
polynomial, and use the remainder theorem to evaluate a polynomial funName:
Juan Diaz EDS 1021
Each box to be filled in with a value is worth 1 point.
Data Table 1
Prediction of # atoms
that will decay when
time reaches one halflife
Number of atoms in the
sample at Time = 0
Radioactive
Element
Carbon-14
100
Uranium-238
100
Michelle
Vazquez
MA1310
Module 5 Discussion
I used a vector to represent a wind velocity of 13 miles per hour
from the west.
o Jane made this statement because a vector has a magnitude
and direction. The magnitude and she refers to this, as the
winds vel
GraphsofOtherTrigonometricFunctions
Michelle
Vazquez
MA1310
(a) y=Atan(Bx-C). Let us take the consecutive asymptotes at x=0 and x=. Now
moving 1/4th and 3/4th of their gap means moving to x=/4 and x=3/4. the
problem we have : y(at x=/4) = -1 and y(at x=3/
MA1210: Module 2 Fundamentals of Algebra, Part 2
Lab 2.1
Polynomials and Rational Expressions
Throughout this course you will be analyzing polynomials and their properties. This lab includes factoring
exercises of many different types of polynomials. Addi
Question 1
A. the formula to model the cars value is y=20000-2800x
b. What will be the value of the car after 5 years?
Y=20000 2800(5)
Y=20000 14000
Y=6000.
The answer matches the graph provided in the question as for the graph corresponding
to the value
1. (30 points) You have 150 yards of fencing to enclose a rectangular region. One side
of the rectangle
does not need fenced.
a. Write the equation for area.
A = x*y
b. Write the equation for the fencing.
150 = 2x + y
c. Solve the equation for fencing for
School of Information Technology
1. (30 points) The following formula describes a large companys number of viruses, V, on employee
computers. The variable, t, is time, in years, since 2002. Answer the following questions showing all
of your work.
= 5+ 12This lab includes problems based on graphing various functions and application problems
involving the slope and equations of lines.
Recommended Procedures:
Solve the following problems, providing detailed steps wherever required.
1. (10 points) What test
Bryce Coffman
MA1210
Unit 1 Analyzing Graphs
Graph using is Bit Rate
Y axis is Mbits/s
X axis is Time (pm)
From looking at the graph I conclude that the Mbits/s goes up between the time of aprox. 3 to the
time of 8 after which the Mbits/s goes down to whe | 677.169 | 1 |
Guidelines for Using the Math Lab
Located in MAS 204, the Math Lab is a great place to study, do homework, obtain assistance, and meet with a study group.
Bring the worksheet or textbook that contains the problems you need help with.
For the best help, be prepared to show an Academic Tutor the attempts you have made to solve these problems, and identify where you are having difficulties.
There are learning aids available in the Math Lab that may be useful if you need more help on a topic. If you miss a class, please contact your instructor, who will give you the best strategy for making up the work.
Learning Aids
You must have a valid Brookdale ID – the Brookdale OneCard–to access the materials available for you to use in the Math Lab.
The Math Lab has solution manuals (both student and instructor versions) for most of the texts used by the department. These can be used in the Math Lab as study aids so you can check the answer to any text problem. The Bankier Library also has these on reserve.
The computer room in the Math Lab is equipped with all our mathematics software. These computers are reserved for currently enrolled math students doing assignments for their math course. The computer room has the same hours as the Math Lab.
If we can be of any other assistance or you require special accommodations, please contact the STEM Institute at 732-224-2430.
Study Groups and Workshops
The Math Department encourages students to form study groups. If you are interested in forming a study group, ask your instructor to help you get one started. Your study groups are welcome to meet in the Math Lab.
Workshops are problem-solving sessions that may focus on a specific topic or review material for a test. For example, around the time courses have their tests, review sessions covering the material on these tests are offered through the Math Lab. Workshop announcements will be posted on the Math Lab bulletin board located in the hallway outside of the Math Lab and in the Announcements portion of this webpage. | 677.169 | 1 |
2Welcome to Precalculus! Mrs. Bunting Room C109 Get an index card and your handouts.Pick up a textbook (Rust with spiral on front)Find your seat on the seating chart and take your seat. Fill out your index card.Please begin to work on reviewing the material in Section 1-1 of your book.Use your textbook and tablemates to help yourself review this material.You will need to TAKE NOTES on the material.Complete p 10 #17 – 37 odd, all
3By the time you and your group finish you will answer… Standard 1.1: distinguish between relations and functions, identify domain and range, and evaluate functions (Section 1-1) p 10 #17 – 37 odd, allBy the time you and your group finish you will answer…What is a relation?What is contained in the domain of a relation? In the range?What is a function and how is it different from a relation?What is the vertical line test and what is it used for?What does function notation look like?How are functions evaluated for specific values?
4What is Honors Precalculus? You will be introduced to:Higher level algebra skills!Common and Natural Logarithms!Limits!Arithmetic, Geometric and Infinite Series!Polynomial, Rational and Exponential Functions!Lots of Trigonometry!Rectangular and Polar Coordinates!not necessarily in that order…
5What Can I Expect? We will cover at least a section a day. We will complete a unit pretty much weekly.Each quarter will have several portfolio projects.You can expect to have Precalculus work to do every single night.
6In a nutshell: To Do Well: Grading to Standards: 1. You need to master EVERY standard to pass.2. Any standard which you do not pass must be reassessed.To Do Well:1. Complete your homework It is your ticket to reassess.2. Reassess promptly while things are fresh.
11Standard 1.1: Finding the Domain of a Function in Equation Form (1-1) To find out what the independent (x) values for a function will be involves finding out what they cannot be.There are TWO Bozo No-No's:No values which cause zero's in denominatorsNo values which cause a negative under a square root (or any even root)
12Find the values for x which are not in the domain of the function, then state the domain in proper set notation.
13Find the values for x which are not in the domain of the function, then state the domain in proper set notation.
14Find the values for x which are not in the domain of the function, then state the domain in proper set notation.
15Find the values for x which are not in the domain of the function, then state the domain in proper set notation.
16When we finish this lesson you will be able to … Standard 1.2: perform operations (add, subtract, multiply, divide, compose) on functions (1-2)When we finish this lesson you will be able to …Perform basic math operations with functionsCreate, use and check composite functions
29At the end of this lesson you will be able to… Standard 1.3: analyze graphs and make predictions based on linear functions (1-3,1-4)At the end of this lesson you will be able to…Identify and properly use the three forms of linear equationsFind x- and y-interceptsDefine, identify and use the formula for slopeIdentify the two special cases of slope
30Linear Functions What does a linear equation look like? Are all the equations of lines also functions?How many of the forms do you remember?
32Standard Form: Where A, B and C are numbers like this. In this form you can tell what about the line?Nothing.
33Slope-Intercept Form Where m is… And b is… In this form you can… Tell exactly what the line looks likeGraph the line
34Point-Slope Form:Used to develop the linear equation if you know the slope, m, and one point on the graph, (x1, y1).Find the standard form of the equation of the line which has a slope of -1 and passes through the point (-4, 5).
35What if you only have two points on the graph? Find the standard form of the equation which passes through the points (6,5) and (4,-5).Find slope.Use slope and one of the points to find equation of the line.
43By the end of this lesson we will be able to answer… Standard 1.3: analyze graphs and make predictions based on linear functions (1-5)By the end of this lesson we will be able to answer…How can parallel and perpendicular lines be identified from their equations?How can the properties of lines be used to identify geometric figures?How can the coefficient for an equation be found so that it will be parallel or perpendicular to a specific line?
44Parallel and Perpendicular Lines Parallel lines have the same slopePerpendicular lines have slopes which are negative reciprocals of each other.Find the equation of the line parallel to the equation above and passing through (2,-2)Find the equation of the line perpendicular to the equation above and passing through (-4,1)
45Special Case:Lines which have the same slope and the same y-intercept are called coinciding.
46Slope and Distance:Consider the polygon with vertices at (0,0), (1,3), (3,-1) and (4,2).Is it a parallelogram?Is it a rectangle?
62In this section we will… Standard 1.3: analyze graphs and make predictions based on linear functions (1-6)In this section we will…Draw and analyze scatter plots.Draw a best-fit line and write a prediction equation.Solve problems using prediction equation models.
63Collecting and Using Data: Real life data seldom forms nice straight lines or smooth curves.For graphs which approximate a line, a best-fit line (also called a regression line) can be drawn and a prediction equation can be determined.
64Scatter Plots: p 38 Basically, data is the graph of a relation. If the graph shows a linear trend you can create a prediction equation.Accuracy of predictions depends on how closely the data approximates a line.
65Correlation: p 40This refers to how closely a set of data actually approximate a line.If the data is very scattered, that is a weak correlation.If the data is very close to being on a line then it has a strong correlation.Our example had moderate correlation.
67The Prediction Equation: Graph your data.Draw a best-fit line.Chose two points, on the line.Find their slope.Use the slope and one of the points to find the prediction line.
68Regression Lines on the Calculator: Go to STAT, choose EDIT, and enter the x-values in L1 and the y-values in L2.Go to STAT PLOT (2nd, Y=), press ENTER on 1:Plot 1, and turn Plot1 On.Go to WINDOW, and adjust your Xmin, Xmax, Ymin, and Ymax to fit your data.Go to GRAPH to see your points plotted.Go to STAT, choose CALC, arrow down to highlight the appropriate regression model, and press ENTER. Press L1 (2nd, 1), the comma (above the 7), L2 (2nd, 2), the comma again, then VARS, choose Y-VARS, choose Function, choose Y1, and press ENTER.Go to Y= to see that your equation has been transferred to the Y= screen.Go to GRAPH to see your line.To enter an x-value and find the corresponding y-value, go to CALC (2nd, TRACE) and choose 1:value. Enter the x-value, and the y-value will be provided.To enter a y-value and find the corresponding x-value, go to Y= and next to Y2 graph the line y=a, where a is the y-value in which you are interested. Then go to CALC (2nd, TRACE) and choose 5:intersect. Press ENTER three times, and the point of intersection will be provided.NOTE: You may need to change your viewing window to accomplish steps 8 and 9.
69Now…do it yourselves. Use the data your group was given. Paste the chart with your data and plot your points on the large sheet of graph paper.Draw a best-fit line.Choose two points on your line and determine your prediction equation. Show all work on the graph paper. Label it "Hand Calculated Equation"Finally, use the graphing calculators to find the regression equation. Record it on the graph paper and label it "Calculator Generated Equation".Make sure that you allow enough room on the paper to answer your questions.
82Standard 1.6: Solve systems of linear inequalities (2-6) At the end of this section you should be able to …Find the solution for a system of inequalities using a graphGraph a polygonal convex setFind the vertices for a polygonal convex setFind the minimum and maximum values for a polygonal convex set
83How can the solution for a system of inequalities be determined using a graph?
84What is a polygonal convex set? A polygonal convex set is the solution for a system of inequalities.The solution is contained within the polygon formed by the boundaries of the inequalities.
85First graph the inequalities and determine the polygonal convex set.
90One more!The Cruiser Bicycle Company makes two styles of bicycles: the Xenon, which sells for $200, and the Yaris, which sells for $600. Each bicycle has the same frame and tires, but the assembly and painting time required for the Xenon is only 1 hour, while it is 3 hours for the Yaris. There are 300 frames and 360 hours of labor available for production. How many bicycles of each model should be produced to maximize revenue, and how much money will be made?
94Standard 1.5: solve systems of equations (2-1, 2-2) In these sections we will…Solve systems of equations involving two variables algebraically.Solve systems of equations involving three variables algebraically.You will need a ruler and a piece of graph paper.
101Solving Systems in 3 Variables A system in 3 variables represents the intersection of 3 planes.Look at page 73.You need 3 equations to solve.You have to have the same number of equations as you have variables.Solve using substitution or elimination.
106Now…YOU think.Write a system of 3 equations that fits each description.The system has a solution of x = - 5, y = 9 and z = 11.There is no solution to the system.The system has an infinite number of solutions. | 677.169 | 1 |
Algebra:
Mathematics branch of mathematics that uses basic operations to solve expressions Linear equations:
A linear equation is an equation involving only the sum or product of constants and the first power of a variable.
y=f(x)=mx+b where m is the slope and b is the y intercept is the general form of an equation in slope intercept form.
The formula of a slope= where and are two points in ordered pair form.
The general form of point slope form is: where m is the slope, and is the two points in ordered pair form. Polynomial equations:
The quadratic formula equals: where a, b and c are constants of a quadratic equation is of the form . Factoring:
Factoring is a process of dividing out a factor from a mathematical expression.
Common factors technique is of the form ax+bx=x(a+b).
Difference of squares: .
Difference of cubes: .
Rationalizing is the process of removing an irrational expression from the numerator of a fraction. Order and Inequalities: Number line: An axis or ray usually horizontal on which real numbers are represented and ordered from left to right. Absolute value: For a real number a, it is a if a is greater than or equal to zero or –a if a is less than zero. It is denoted. Conic sections:
The general equation of a circle: . Where r is the radius of the circle and is the center of the circle.
The general equation of an ellipse: .
A review of the basic principles of algebra is needed to build a solid foundation for the usage of these skills in precalculus. Algebraic equations such as linear equations and quadratic equations are discussed in this particular tutorial.
A review of the basic rules of algebra is introduced here along with the basic definition of a linear equation and its graphical representation. Specific properties of linear equations are shown here with the use of examples. Polynomial equations with the use of examples are presented in this tutorial along with the factoring techniques used to factor them. | 677.169 | 1 |
155953891059538910Z2
Discovering Geometry includes all of the teaching resources necessary to enrich learning and to make the teaching experience more rewarding. In addition to a comprehensive, wraparound-style Teacher's Edition, you'll receive a spectrum of evaluation and assessment tools that let all students demonstrate what they have learned. Broad input from teachers has made the teaching resources package a valuable asset for teachers implementing this exciting curriculum | 677.169 | 1 |
'Mean curvature flow' is a term that is used to describe the evolution of a hyper surface whose normal velocity is given by the mean curvature. In the simplest case of a convex closed curve on the plane, the properties of the mean curvature flow are described by Gage-Hamilton's theorem. This theorem states that under the mean curvature flow, the curve collapses to a point, and if the flow is diluted so that the enclosed area equals $\pi$, the curve tends to the unit circle. In this book, the author gives a comprehensive account of fundamental results on singularities and the asymptotic behavior of mean curvature flows in higher dimensions.Among other topics, he considers in detail Huisken's theorem (a generalization of Gage-Hamilton's theorem to higher dimension), evolution of non-convex curves and hypersurfaces, and the classification of singularities of the mean curvature flow. Because of the importance of the mean curvature flow and its numerous applications in differential geometry and partial differential equations, as well as in engineering, chemistry, and biology, this book can be useful to graduate students and researchers working in these areas. The book would also make a nice supplementary text for an advanced course in differential geometry. Prerequisites include basic differential geometry, partial differential equations, and related applications. | 677.169 | 1 |
Introduction to Algebraic Expressions
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This 30-problem worksheet provides practice in the following pre-alebra skills for students new to algebraic representation: writing algebraic expressions to represent phrases; translating algebraic expressions into word phrases, evaluating expressions, matching expressions to phrases, and writing expressions to represent problem situations. Created by Cooley for fourth and fifth grade math students. | 677.169 | 1 |
Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for). | 677.169 | 1 |
7 October 2015
Math Tutor
Published at 09:00, 7 October 2015
Get a handle on those concepts in maths that you missed out on at school, or can't remember, writes Tracey Whyte, Subject Librarian for Education at Berwick.
If you want to build your mathematical skills and confidence then go to Math Tutor online. The Library provides access to this excellent resource via the Kanopy streaming database and you can access it via the Library's Search tool. "The Math Tutor Collection offers a selection offers video tutorials on topics from secondary mathematics, aimed at students who wish to revise these topics in preparation for study at University, but equally useful for those meeting this content for the first time. "I recommended some of the videos in the 'Sequences and Series' topic to my students, because I thought the content was clearly and thoroughly explained, and might support those students who were meeting this content for the first time." (Monica Baker, lecturer, PhD student and maths teacher) The content covers over 80 mathematics topics and provides diagrams and worked examples to clearly explain mathematical concepts. There are a series of eight videos to watch:
Geometry and Vectors
Algebra
Integration
Arithmetic
Trigonometry
Functions and graphs
Sequences and series
Differentiation
Each topic describes the subjects taught within each video and displays between five and 11 clips ranging from 10 minutes to over an hour.
For those who want to challenge their skills a bit more, Math Tutor has created a website that contains these videos with diagnostic tests, exercises and a pdf text version. These resources are available from the Math Tutor website | 677.169 | 1 |
Practical Mathematics Part Iv: Trigonometry And
The times will include before school, after school, lunch, and planning periods (if it coincides with their study hall). Learn the basics of trigonometry: What are sine, ... Adding and subtracting signed numbers worksheets, fraction and mixed number to decimal tool, how to solve 3rd order polynomials, RAtIO FORmula, Square roots in radical form, how to print sum of a cube with java. Boolean algebra simplification software, fun 4th grade free printable worksheets, order of operations, pemdas, answer generator for simplifying radicals, 7th 8th grade math Algebra, probability combination worksheets, how to factor a cubed binomial, 3rd grade math /multication.
Pages: 170
Publisher: Nabu Press (February 1, 2012)
ISBN: 1274191769
Trisection of the Angle by Plane Geometry: Verified by Trigonometry with Concrete Examples.
A Treatise of Plane Trigonometry, and the Mensuration of Heights and Distances: To Which Is Prefixed a Summary View of the Nature and Use of ... of Instruction in Schools and Academies
In , e.g. Student Solutions Manual for Algebra 8th (Eighth) Edition BYSullivan download pdf. The agent can be reached at: 4840 Westfields Blvd , cited: Study Guide to Accompany Algebra & Trigonometry with Applications However for the secant function, the range is not standardized. Varberg and Varberg use 0 to p, q � p /2, in order to take advantage of the relationship sec-1x = cos-1(1/x), since cos-1 is a common calculator function. VI b., our knowledge of the coordinates of the angle, lengths of the horizontal and vertical sides, and the relationships that exist between the sides and angles all congeal on the unit circle , source: Course of civil engineering; comprising plane trigonometry, surveying, and levelling. With their application to the construction of common roads, railways, canals Unknown she wouldnt be getting the support she is either. Organizers in positions that need managers and organizers. A few years it was pretty much like any other job personality conflicts stuff. I make sure their diet is appropriate and that they arent getting. Whatever the subject there is a purity point of view and an issue with substantial numbers pdf. Calculate the length of the side x, given that sin θ = 0.6 How to solve Word Problems using Trigonometry Pennsylvania Fairs: Country read here queenmedical.theyouthcompany.com? Such triangles have corresponding sides and angles equal, and are exact copies of one another online. In theoretical work, the radian is the most convenient unit Exercises in Algebra ( including trigonometry) Part 1 Longmans modern mathematical series Students are provided with problems to achieve the concepts of Trigonometric Word Problems. This tests the students ability to evaluate Trigonometric Word Problems. The primary application of trigonometry is found in scientific studies where precise distances need to be measured pdf. Solving systems of equations on TI-83 plus with matrices, solving quadratic equations in factored form game, interactive algebraic expressions games, step by step equation solver, the hardest math problem in the world, adding and subtracting worksheets online.
If you need help with this, skip to the how do I do this on my calculator section. In the geosciences, angles are almost always measured in degrees, where a circle is 360 degrees Algebra and Trigonometry download here download here. In this process the one plane has moved through an angle relative to the fixed plane Tutorial Trigonometry Many types of "inaccessible" measurements rely on using trigonometry Five Place Logarithmic and download online . This is because the second - overrides the x sign, and the - 3 was taken as 0 - 3, so the calculation was actually - 3 - 4 = -7. If your calculator does not have a +l= key then you may have some difficulty in doing calculations involving negative numbers , source: Elementary Technical download epub
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Math 221-001
How to succeed in this course
There is little doubt that math is useful: it is used throughout the sciences and engineering, from
quantum mechanics and weather
prediction to space flight to nanotechnology and financial modelling. But that is just part of
the story. When you study mathematics, you learn about the language of science. You gain access to the ideas of
several of the greatest scientific minds humanity has produced, like Newton, Leibnitz, Euler, Gauss. Whether you
believe it or not, as the distinguished mathematician I. Gelfand said, "the most important thing a student can get from the study of Mathematics
is the attainment of a higher intellectual level".
The goal of Math 221 is to give the student knowledge of several basic topics (logic, integers, sets, functions), in a set of increased formality with respect to early university math classes. One of the big strengths of math as a science is the possibility of proving results with absolute certainty (within a certain axiomatic framework). This class was designed to give the student a first taste of how things are proven in math.
With Mathematics, like with any advanced skill, the learning process is never quick nor painless. Here are a few
things to be considered that might help you succeed in this class:
This class requires the student to make a step beyond the level of mathematical maturity achieved in Math 110.
Many ideas and notions are discussed along the course, and it is almost impossible to memorize them all, and even less to use them
effectively. It is all about understanding and practicing. So, practice-practice-practice is the name of the game. The assignments, and even the recommended practice questions, are only the tip of the iceberg. This class requires you to read and practice continuously. The maturity required to succeed in this class cannot be obtained in a mad-rush night-before-the-midterm style.
Keep up with the material covered in class. Mathematics classes tend to build directly on what you have
already learned. If you fall behind it can be very difficult to catch up.
Do lots of practice problems. If you can barely do the questions
from the assignment, it is hard to imagine that you would be able to do a related question, without help, under the stress situation of the exam. Conclusion: practice, practice, practice.
Take advantage of the additional resources available to you. These include the textbook, the instructor's office
hours, additional resources for students, information on the web, etc.
The topics will become second nature if you take the time to understand why
it is what it is, and then use it by doing lots of problems.
Study with the goal of understanding "why" instead of just trying rote memorization.
Make sure to present your work clearly, and stapled.
Practical matters
Grading scheme: 20% Assignments, 30% midterms, 50% final.
Prerequisites: The official prerequisite for this class is Math 110 and one of Math 111, 122, 127 . We will not use many topics from those courses, but rather expect the student to be mature enough to deal with content in a more abstract way. For certain, basic algebra and geometry should be mastered by the student. If you are unsure of what to expect, I will assume that the contents of the following two short chapters are familiar to you:
"Familiar" above means that you can do the exercises on your own, and correctly.
Textbook: We roughly follow Introduction to Mathematical Thinking: Algebra and Number Systems, by Gilbert and Vanstone, 5th edition. There is not absolute need to have the book, but it will definitely be a good resource to read some of the material.
Midterms: the two midterms will take place on February 17th and March 24th. To compensate for emergencies and bad days, if the grade in the final exam is higher than the worst midterm, the grade from the final exam will be used instead of the grade on that midterm.
Final Exam: the comprehensive three-hour final exam will take place on April 17th, from 9:00 to 12:00. Location to be announced (this will be confirmed closer to the date). Please bring photo id to the exams.
Assignments: Weekly assignments will be published every Friday, due the following Friday at the beginning of class. Group work can help a lot, if it means that the group works together towards building a solution to each problem.
"Group work" of the kind "I copy the solution from whoever gets it" might work for the assignments but will not help you for
the exams (which comprise most of the class' marks), and it will get you in trouble for plagiarism.
Practice: You don't learn any mathematics by just listening and watching. You learn mathematics by doing, and thinking about it. That's what the assignments are for, but usually you shouldn't consider them enough practice. For this, suggested problems from the book will be indicated in the course web page. You will know you have practiced enough when you can take questions from the suggested list, solve them in a decent amount of time, correctly, without looking at the book nor asking for help. Until this happens, keep practicing, and use the resources you have available to improve.
Communication: you can reach me by email, either through the course web page or through UR Courses. The links there automatically write "Math 221-001" in the subject, so that I know what the email is about, and my spam filter will let it pass. I will happily answer short concrete questions I get by email. I will not answer emails of the form "please explain me this topic". If you need more detailed help, please come to the office hours, or use one of the several tutoring solutions offered by the university. UR Courses also has a forum feature, feel free to post a question there, or reply to what others post.
Additional resources: The course web page lists several options to get personalized help. There is a Math/Stats help Centre at Campion College, where you can drop in, and there is Math & Stats support provided by the Student Success Centre. Also, the deparment of Mathematics and Statistics keeps a list of paid tutors at the department office CW307.14.
Academic Misconduct: by signing up for a class, the student is assumed to be aware of the university regulations; in particular, those that concern plagiarism and cheating, as defined in Section 5.14 of the 2016-2017 Undergraduate Calendar, as well as the potential punishments (which range from a grade of zero in the assignment or test, to expulsion from the University).
Audio and Video recording of lectures: this is not allowed, unless you get specific permission of the instructor for cases where needed for accessilibity.
Dropping the class: please note that non-attendance does not constitute withdrawal. Unless you drop the class explicitly (through UR Self Service or through the Registrar's office), you are still considered to be registered, you are liable for the tuition, and you will receive a failing grade of NP ("no paper"). Dates: | 677.169 | 1 |
A+ Further Mathematics Exam 2 VCE Units 3 & 4
By Phil Schmidt | Copyright Year:2016 | ISBN-13: 9780170354110
A+ Further Mathematics Exam 2 VCE Units 3 & 4 provides a comprehensive practice exam book for VCE Further Mathematics Units 3 & 4. The Exam 2 book has 31 exams and the texts is divided into tests that gradually increase in length, content and difficulty. No teacher or student preparation required – students write directly into the books and bonus detachable exams are also included.
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A+ Further Mathematics Notes VCE Units 3 & 4 provides a comprehensive study guide that is designed to enhance the skills and confidence of students studying VCE Further Mathematics Units 3 & 4. There are notes at the beginning of each section which summarise the main definitions, formulas and techniques for each section of the course and over 400 questions graded into three levels of difficulty for each Area of Study. The graduated difficulty allows students to use the book throughout the whole course, not just before the exams. Multiple choice and extended answer questions are also included as well as comprehensive solutions, even for the multiple choice questions. Examination advice can also be found throughout the book.
*The A+ cover shown includes updated branding and may be different to the book available for purchase.
A+ Further Mathematics Exam 1 VCE Units 3 & 4 provides a comprehensive practice exam book for VCE Further Mathematics Units 3 & 4. The Exam 1 book has 12 exams and the text is divided into tests that gradually increase in length, content and difficulty. No teacher or student preparation is required – students can write directly into the books and bonus detachable exams are also included.
*The A+ cover shown includes updated branding and may be different to the book available for purchase. | 677.169 | 1 |
Wednesday, August 25, 2010
Mathematics is a very enjoyable lesson for some people. There are times when math is the most frightening lesson or lessons that can be difficult categorize. Real mathematics as another lesson that can be understood by practicing and trying. for example algebra, to find out algebra you need to understand pre algebra correctly and diligently in practice to understand it. If you do not understanding pre algebra from beginning, algebra will then be categorized as a lesson that is difficult because Algebra Pre requires a thorough understanding of concepts and it is the base on Which high school Algebra is studied.
In mathematics there is also the division that sometimes you need in their daily lives. This calculation can used for various things such as How to divide a land that will be used as fields for planting, the Division is one arithmetic operation Among four arithmetic operations and it is the inverse of Multiplication. to understand the division in mathematics will certainly help you to work that requires the calculation of the distribution.
If you're looking for a K-12 and college tutoring may be information about the Algebra equations, the Standard form (Standard form is a way of writing down very large or very small numbers Easily) can be found on this blog. you can also get a Slope Formula you can use. You can get online at TutorVista. in TutorVista an incredible offer unlimited monthly tutoring package for K-12 student.
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Description
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Kirk-Othmer Encyclopedia of Chemical Technology, Volume 11
Mandelbrot, Benoit, The Fractal Geometry of Nature (New York: W. CRC Press is a premier global publisher of science, technology, and medical resources. Remember, some of the most powerful learning stems from making a mistake. Because the videos and other materials are created by some of the top teachers in their fields, you can be sure that all information is accurate and presentations are designed to maximize learning. One famous book is Papadimitriou's Computational Complexity, but it might not be the best place for a beginner to start. – Stefan Smith Jan 24 '14 at 2:36 @IsaacSolomon: Couldn't they vote to close saying this question belongs in math.stackexhange.com? – RghtHndSd Jan 24 '14 at 2:49 I think your first question has been answered eloquently by others here.
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A Guided Tour of Mathematical Methods: For the Physical Sciences
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INTRO APPLIED MATH
MATH240.01
The course will consist of 3 or 4 concrete applications, for which precise mathematical questions will be formulated, and a mathematical framework developed that will make it possible to answer these questions. In doing so, we will encounter and explore portions of real analysis, probability, linear algebra, convex analysis, information theory and maybe others. We will also learn how to construct watertight mathematical arguments, and explore different proof techniques. Instructor: Staff | 677.169 | 1 |
Tag Archives: graphing
As promised, a return to lynx. In my previous post about lynx I posted a worksheet modeling lynx populations with a cosine function, and mentioned that this is not the best model. Look at the derivative to see how bad it is -- the green and red lines ought to be matching up:
Graphing the log of the lynx data gives a transformed graph that is much more sinusoidal! The better model for the lynx data, then, is exp(something sinusoidal). Look at the graph below to compare Model 2 and its derivative to the data. The green and yellow curves are much more alike:
This worksheet guides students to developing this model after having them evaluate the previous sinusoidal model via technology.
The worksheet I'll include below is meant for a day when you have computer lab time with students. I know that this does not include everyone... but if you can head down to the lab for such an activity, there is a lot students can learn!
This worksheet applies knowledge of:
the chain rule, on compositions of trigonometric and exponential functions
numerical approximation of the derivative
shapes of graphs.
Along the way students must evaluate models and create one of their own.
As the instructor, you'll have to decide what software you want to use for this activity. I have had success using Excel, asking every student to email me their work on the way out of the lab, and these days you can use Google Drive if your institution uses Gmail. If you and your students are already quite familiar with R you could also use that. Beware of differences between Mac Excel and Windows Excel, especially in graphing -- work through the activity yourself on whatever platform students will use.
I've been spending the last few days trying to edit some math papers I'm working on and learn more about polygon spaces. (What's the space of quadrilaterals in with side lengths (1,2,1.3,1.8) and one vertex at the origin, for instance? If you guessed that it looks like , you'd be right!) It's always interesting to talk with students about what mathematicians do. They expect we do more with numbers, and when I talk about what I am working on they're surprised at all the funny little pictures I draw and all the writing we all do.
This activity is a good one for discussion. You could either print this or use a projector to put it up in front of the class for a group discussion. Group discussions can be a great way to get a class comfortable with each other and facilitate the verbal discussion of mathematics, which is another skill that must be learned and practiced. I had a hard time learning to "talk math" despite my good reading and essay-writing skills, and even now I need to gather my thoughts to produce coherent mathematical sentences in a way that I don't need to when talking about politics, for instance. I've concluded that I think about mathematics nonverbally and that the translation process simply takes some time, even now that I've had a lot more practice than your average freshman.
When I run a class discussion like this, I try to do a few things:
Give students a minute or two to think in silence and write down a tentative answer.
Start cold-calling students -- "Pahoua, what do you think about the first one? Jordan, what about you?" This requires a level of trust that is built over time.
If students are right, I don't necessarily say so and move on -- I keep asking other students, and ask other students if they think the first student is right or not. I don't want students practicing the Clever Hans strategy. (Clever Hans is a horse who counted by tapping his hoof, but he just watched his trainer until the trainer looked happy and then stopped.)
If students are wrong and I can identify the assumption that led to the mistake, pointing out that assumption can often help. "Danielle, are you assuming that the graph must be increasing?" Sometimes I ask other students for their thoughts. "It looks like Jason disagrees. Why?"
If a student is visibly hesitant or reluctant to answer, ask for a question or the beginning of the thought instead. "Alexi, what do you think about part c? ... What strategy are you thinking of trying?" or "What do you notice first about the graph?" or "What question are you asking yourself right now?" These often prompt good discussions.
I do try to use students names consistently as I call on them, and
I try to keep track of who I call on and rotate through the whole class. There's plenty of research that shows that even the most enlightened and conscious instructors call more on boys than girls, etc., so bringing a class list and discreetly checking names off is an easy way of making sure you involve everyone in discussions. Telling students that you track who you ask is sometimes useful, too, because then they know that everyone will be asked and they should be prepared! Students do think that it's fair to do this, especially if you offer the outs suggested above if they are stuck.
So, with no further ado, here is a worksheet on matching graphs and stories, using the language of derivatives: | 677.169 | 1 |
Please contact your nearest Dymocks store to confirm availability
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This book is based on keywords provided by the Royal College of Midwives to help those who have to pass a numeracy test for midwives, viz. an ability to accurately manipulate numbers as applied to volume, weight, and length, including, addition, subtraction, division, multiplication, use of decimals, fractions, and percentages. In fact there is more than that because the derivation and manipulation of concentrations are also needed. Readers are shown that they use many of the mathematical operations used in midwifery in everyday life and this reduces the "e;terror"e; of "e;doing the maths"e; to satisfy the requirements set by the NHS. For example readers will have used different currencies when abroad but when we think about what we do to use them we realise how these cannot be added or subtracted or multiplied of divided without transforming one to another (via the "e;exchange rate"e;). The same logic is used when changing length, volume, weight or concentrations into different units and different amounts needed in midwifery. By these examples the book deals with the manipulation of numbers, decimals, fractions and percentages and shows these concepts are known to most of us in other contexts; readers will find they have "e;done this before"e; (and so know they can do it) but in a different context. The work is divided into 25 short sections, each illustrated to assist comprehension. The text also deals with conversions; length, area, volume are all considered together with their units, and the conversion of these units. Through this it is possible to deal with concentrations. Worked examples are given including how an answer should be checked to ensure you have got the right result. The book also shows the reader what to consider when searching websites for units and why it is necessary to be careful which websites to use for conversions. The Second Edition has an additional section introducing the statistical handling of data that midwives may commonly encounter and to which they may be contributing from the factual data they record at birth (weight, height etc.). Currently Manchester University uses this book for their course | 677.169 | 1 |
CONTENT:
Operations with Monomials :
Students apply
operations with powers and roots, including fractional and negative exponents,
to the solution of problems. Apply powers and roots to solution of problems.
Extend the order of operations to include exponents. PERFORMANCE:
Problem Solving, Explanation: Students
will
work
in groups to solve a compound interest problem on how to save for college
(p553). Essay, Open Ended
CONTENT:
Evaluation with Formulas:
Demonstrate
facility in transforming polynomial expressions by rearranging and collecting
terms, factoring, and applying and applying the properties of exponents
in order to solve problems. Apply the order of operations, including parentheses.
Apply the powers and roots to the solution of problems. Extend the order
of operations to include exponents. PERFORMANCE:
Diagram, Computation, Problem Solving: Students
will work in groups to write polynomials which represent areas
of shaded regions. Short answer
Note: We would like to add
links for online tutorials here to cover items assumed to be already learned,
and to help students who missed learning concepts because of illness,
etc. Microsoft has a number of tutorials. We will place them in the file,
Using Computers for Assignments, as we find them.
2H5
2M4
CONTENT:
Equations as Relations: :
Students describe
similarities and differences among the families of linear, quadratic, and
exponential functions using graphs, tables, formulas and verbal descriptions
Describe the graphical significance of parameters.Students will be able
to analyze graphs, and solve linear equations. PERFORMANCE:
Graphs, Charts: Students demonstrate the skills needed for the
productive use of technology. page 274, # 7 solving a linear equation by
completing the chart. Short answer, fill-in-the- blank
CONTENT:
Direct and Inverse Variation Students
will solve problems involving direct and inverse variation. Explain and
generalize how a change in one variable results in a change in another
variable in functional relationships... PERFORMANCE:
Problem solving : Page 277, #53 solving
a problem with direct variation. Short answer
CONTENT:
Trigonometric Ratios:
Apply trigonometric
ratios in right triangles to solve problems. Explain the meaning of the
Pythagorean Theorem, and apply the theorem to the solution of problems.
Describe angles formed by intersecting lines, and relationships among them. PERFORMANCE:
Produce a Hyposometer and Use it to Measure a Building: Page
209, Making a Hyposometer Open Response
CONTENT:
Patterns and Sequences:
Describe characteristics
of Discrete Geometry (graph theory) and apply the solution to problems.
Apply powers and roots to solution of problems. Extend the order of operations
to include exponents. PERFORMANCE:
Producing a List of Numbers Which Follow a Pattern.: Students
will work alone on page 500, #39 to find the number of parts that it breaks.
Short
answer
CONTENT:
Formulas and Equations Use deduction to
establish validity of geometric conjectures and to prove theorems in Euclidean
Geometry. Explain the meaning of the Pythagorean Theorem, and apply the
theorem to the solution of problems. PERFORMANCE:
Diagram, Problem Solving: Students
will work in groups to find perimeter, area, length, and width. Page
130, #13 Open-ended, short answer.
CONTENT:
Simple Constructions:
Use simple tools
(such as a clinometer) to solve indirect measurement problems. use proportions
to model and solve indirect measurement problems. PERFORMANCE:
Building a Hyposometer: Students will
construct a hyposometer. Page 209 Open-ended
CONTENT:
Solving Word Problems:
Describe the effect
of rounding on measurements and computed values from measurements. Estimate
results of computations with whole numbers, fractions, decimals and percents. PERFORMANCE:
Problem Solving and Explanation: Students
find how many dozen of cookies were sold. page 236, #5 Short answer
CONTENT:
Ordered Pair: :
Represent data in a scatter
plot. Use the scatter plot to make predictions. Describe data sets using
the concepts of median, mean, mode, maximum and minimum, and range. PERFORMANCE
Graphing, Diagram: Students will work
together on page 343, #9. Open ended.
CONTENT
Line of Best Fit:
Find the "line of best
fit" from a series of data. Use tables and graphs to compare linear, quadratic,
and exponential growth patterns. PERFORMANCE:
Graph, Problem Solving, Explanation: Students
will work together to solve problem #10 on page 343 Open-ended,
short answer
CONTENT:
Statistical Analysis:
Select an appropriate
graphical representation for a set of data and use appropriate statistics,
(e.g. mean, median, range, quartile ,or percentile distribution to communicate
information about data. Choose and apply appropriate measures of central
tendency (mean, median, and mode) to represent a set of data. Use
box-and-whisker plots to represent data sets and ;identify outliers. PERFORMANCE:
Problem Solving:
Students will solve problem
#25 on page 311 Short answer
CONTENT:
Design a Fair Game :
Describe the effects
of sample size and population size on the validity of predictions from
a set of data. Design a "fair game" and provide a justification for fairness.
Make inferences about a characteristic of a population from a well constructed
sample. e.g. capture, recapture. Find probabilities of events with equally
likely outcomes. PERFORMANCE:
Diagram, Problem Solving Students will work together on
page 415, #6 Open-ended
CONTENT:
Probability of Earned Team Points Design
a fair game and provide justification for fairness. Make inferences about
a characteristic of a population from a well constructed sample. e.g. capture,
recapture PERFORMANCE:
Interpret Graphs, Problem Solving: Sports
problem, finding the chances of an event. p.232, #41 Short
answer, open ended | 677.169 | 1 |
When the equation is close to a match, it puts the value of X, rounded to two, into list 1, wich is displayed when finished. However, radians weren't used for almost a hundred years after Euler's death. Freely browse and use OCW materials at your own pace. Don't assume the identity to prove the identity. Please call customer service at 1-800-832-2412 for assistance. The student applies the process standards in mathematics to create and analyze mathematical models of everyday situations to make informed decisions related to earning, investing, spending, and borrowing money by appropriate, proficient, and efficient use of tools, including technology.
Pages: 204
Publisher: World Scientific Publishing Company (October 8, 2008)
ISBN: 9812834435
Shorter Advanced Trigonometry
Numerical trigonometry
New plane and spherical trigonometry, surveying and navigation,
Logarithmic and trigonometric tables, prepared under the direction of Earle Raymond Hedrick: To accompany a plane and spherical trigonometry by Alfred Monroe Kenyon and Louis Ingold
Trigonometry basics are often taught in school either as a separate course or as part of a precalculus course. The trigonometric functions are pervasive in parts of pure mathematics and applied mathematics such as Fourier analysis and the wave equation, which are in turn essential to many branches of science and technology Instructor's Manual to download pdf Instructor's Manual to Accompany Modern. Use the interactive whiteboard to draw functions and graphs and review sine and cosine , e.g. Euclid's Elements of geometry; download for free Euclid's Elements of geometry; from the. The trigometric functions, such as sin, cos and tan, represent how the ratio of these lengths depends on the angle θ and we can define the following functions for each of the ratios: To create cheat sheet first you need to select formulas which you want to include in it , cited: Master Math: Geometry 2nd download epub Master Math: Geometry 2nd (second). These results may be expressed by saying that the trigonometric functions are periodic and have a period of 360° or 180°. When Q on the terminal side of A in standard position has coordinates (x, y), it has coordinates (−y, x) and (y, −x) on the terminal side of A + 90 and A − 90 in standard position, respectively College Algebra, 5th Edition & download for free College Algebra, 5th Edition & College. I am not to smitten with the books in this area. For the second edition I will try to do better. Until then, there is one excellent book in print. There is almost certainly an excellent book to appear. The book by French is excellent and is out of print and shouldn't be. The books by Conway et al and Hadley et al were published in the sixties and are out of print and despite that are first rate if you can get your hands on them Logarithmic and trigonometric download pdf Logarithmic and trigonometric tables to download Topics in Classical Analysis and Applications In Honor of Daniel Waterman pdf, azw (kindle). It was Euler who decided that for the purposes of calculus, the radius should be fixed at 1. He realized that measuring the arc length and the line length in the same units would be beneficial; in addition, he concluded that if the radius of the circle is 1, then the circumference is 2 Pi download Topics in Classical Analysis and Applications In Honor of Daniel Waterman pdf.
To complete this project the students will need access to the following materials. Most of these materials will be used to create a clinometer used to measure angles of inclination. These are step by step directions the students will receive to complete this unit. Also included are the steps necessary to create their clinometer. Through activities and teacher lead discussion the students will have learned the prerequisite material to complete the final project Constructive Text-Book of download here Constructive Text-Book of Practical. These math dvd programs augment what is happening in school, the children more readily understand, and most importantly, grades have improved." - Charles Woods, " I have 4 kids and we started using Math Made Easy in 1997 when my oldest child was in 7th grade A treatise of plane trigonometry, and the mensuration of heights and distances To which is prefixed A treatise of plane trigonometry, and. The Arab astronomers had learnt much from India, and there was contact with the Chinese along the Silk Road and through the sea routes, so that Arab trading posts were established in India and in China. Through these contacts Indian Buddhism spread into China and was well established by the 3rd century BCE, probably later carrying with it some of the calculation techniques of Indian astronomy Study guide to accompany College algebra & trigonometry [by] Justin J. Price, Harley Flanders Study guide to accompany College algebra.
Logarithms, Trigonometry, Statistics. First Year College Mathematics.
Online division of polynomials solver, adding, subtracting, multiplying, dividing square roots, sample of powerpoint investigatory project, greatest common divisor polynomials calculator. Help with writing fractions from least to greatest, practice using simplifying square roots, fourth grade fractions read Topics in Classical Analysis and Applications In Honor of Daniel Waterman online. Free online algebra calculator, rational expressions for dummies, get free answers to algebra Algebra & Trigonometry with Analytic Geometry - Students Solutions Manual", 12th (twelfth) edition Algebra & Trigonometry with Analytic. Some of the equations are too small for me to see! Show Answer This is a problem with some of the equations on the site unfortunately pdf. His key work was the Yukti-bhāṣā (written in Malayalam, a regional language of Kerala). Jyesthadeva presented proofs of most mathematical theorems and infinite series earlier discovered by Madhava and other Kerala School mathematicians. This section may lend undue weight to certain ideas, incidents, or controversies ref.: College Algebra, 5th Edition & College Algebra & Trigonometry, 5th Edition, STUDENT SOLUTIONS MANUAL [for both textbooks] College Algebra, 5th Edition & College. Thus in Fig. 2(b) AOB, AOC are adjacent, in Fig. 2(c) COB, BOD are adjacent, etc. We must now consider a mathematical conception of an angle. Imagine a straight line, starting from a fixed position on OA (Fig. 3) rotating about a point 0 in the direction indicated by an arrow Programmed practice for Modern algebra and trigonometry: Structure and method, book 2 / by Persis O. Redgrave, James J. Roberge ; editorial adviser, ... (Houghton Mifflin modern mathematics series) Programmed practice for Modern algebra. A worksheet using trigonometry and physics can be found HERE , source: Practical Mathematics Practical Mathematics. I have come across a number of algebra programs. I suppose what you need is Algebrator. It could be just be the thing for your troubles. I verified each one of them myself and that was when I came across Algebrator Studyguide for Prealgebra by read for free Studyguide for Prealgebra by McKeague,. This course focuses on quantitative skills and reasoning in the context of experiences that students will be likely to encounter. The course emphasizes processing information in context from a variety of representations, understanding of both the information and the processing, and understanding which conclusions can be reasonably determined ref.: Algebra and Trigonometry: download here Algebra and Trigonometry: Structure and.
Tables of [Square Root Of] 1-R2 and 1-R2 for Use in Partial Correlation and in Trigonometry
A Treatise Of Plane And Spherical Trigonometry, In Theory And Practice: Adapted To The Use Of Students
Constructive Text-book Of Practical Mathematics, Volume 1
We can use trigonometry to work out the length of FG, which is half the side length of the base. Here's the triangle we're working with: If we go back to our diagram of the base, we now know the length of FG: The two lengths marked with arrows are the same length as FG, so we can use Pythagoras' Theorem to work out the length AF: We can go back to our original triangle with the side AE in it that we're trying to find Elements of geometry & read epub Elements of geometry & trigonometry ... These include finding the angle view to find the angle of elevation and the angle of depression. These can help determine the height of objects. Classification of Triangles and Angles: Learn about acute, obtuse, reflex, straight, and right angles, See the definitions of triangles including acute, equilateral, isosceles, obtuse, right, and scalene triangles Plane [and spherical] download for free Plane [and spherical] trigonometry for. Read about NZQA Assessor Support options. Adaptable automatic worksheets for teaching maths, and revision notes for Scottish SQA maths courses Search for free worksheets by... ONLINE STORE: Purchase revision notes and prelims for National 4 and National 5 Mathematics, National 4 and National 5 Lifeskills Mathematics, (new) Higher Mathematics and (new) Advanced Higher Mathematics exams online Elements of Geometry, and read online Elements of Geometry, and Plane and. This course covers direct and iterative methods of solution of linear algebraic equations and eigenvalue problems. Topics include numerical differentiation and quadrature for functions of a single variable, approximation by polynomials and piece-wise polynomial functions, approximate solution of ordinary differential equations, and solution of nonlinear equations Introduction To The Theory Of Fourier's Series And Integrals Introduction To The Theory Of Fourier's. Many Indian Sine tables use $R = 3438$ which is the result if the circumference of the circle is $360 \times 60$ or $21,600$ minutes. [See Note 2 below] By the 5th century, two other functions had been defined and used College Trigonometry: With read pdf College Trigonometry: With Applications. If a wrong answer is given by the user, the correct answer is shown before the next problem is given Plane And Spherical Trigonometry... Plane And Spherical Trigonometry.... See problem #14 from the practice problems COMMON CORE STANDARDS FOR TRIGONOMETRY: According to Wolfram MathWorld, trigonometry is "the study of angles and of the angular relationships of planar and three-dimensional figures." So answer correct to the nearest whole number is r= 35 ref.: Study guide to accompany Algebra and trigonometry with applications Study guide to accompany Algebra and. Plus one. – Chris McCall Oct 17 '09 at 4:44 A place to discuss cool math/physics/computer science/ other non-fiction, sciencey books and any …more [close] A place to discuss cool math/physics/computer science/ other non-fiction, sciencey books and anything else relevant/interesting download Topics in Classical Analysis and Applications In Honor of Daniel Waterman epub. | 677.169 | 1 |
Algebra 1 - Program Topics:
Success in Algebra 1 is based on five central concepts. You will enjoy discovering them working in a fun, interactive and engaging environment on relevant and interesting problems. The five topics are:
1. Solving equations in one variable
As the town magician you'll learn to solve equations by creating mystery bag games using a balance scale and bags of gold. You will learn the order of operations, the distributive property, combining like terms, solving equations in one variable and solving equations with fractions using least common denominators.
2. Graphing a line
You will use graphs, tables and equations to decide the placement of a hanging mattress to save a man shot out of a cannon. You'll learn the concept of slope and how to write an equation of a line in slope intercept form. We'll also write equations of vertical and horizontal lines and learn the properties of parallel lines.
3.Solving equations in two variables
You will find the heights of the tallest and shortest men on earth by solving systems of equations with one, none or an infinite number of solutions using three methods: substitution, linear combination and graphing.
4. Parabolas and Quadratic Equations
Use our graphing calculators to discover how long it takes for a secret agent to hit the ground after jumping from a plane. You will learn to recognize parabolic relationships and create their quadratic equations. You'll shift, stretch and reflect quadratic equations as well as work with their vertex and line of symmetry. Finally we will learn to solve for the coefficient "a" in vertex form to represent the full equation of any parabola.
5. Manipulating Variables and Factoring
Finalize the layout of a new school recreation park using Algebra Tiles. You'll learn to multiply binomials then use the zero product property to solve quadratic equations like a pro. | 677.169 | 1 |
Well, b5 + b4. And I think that is something like g9 if I'm correct. I maybe
be wrong, but I didn't do that much. We'll probably learn them thoroughly in
secondary school.
----- Original Message -----
From: "NIcol" <nicoljaco...@telkomsa.net>
Advertising
Hi Tom
On 28 november at 12:48 a.m. you wrote:
"For example, a lot of the concepts and terminology in programming such
as variables, integers, floats, arrays, etc are taken right out of
your high school Algebra textbook. Sheesh, I learned about variables,
arrays, floating point numbers, etc all by the ninth grade."
You made a very interesting point.
I never knew before that a lot of programming concepts comes from the
grade
9 algebra. I also did algebra in grade 9 but we were never taught
variables,
integers, floats, and arrays.
The only algebra we did was called formulas.
For example, our formulas in the math class would look something like
this:
5[6 times 8]
This formula means that the sum inside the bracket is calculated and then
the answer is multiplied with the figure outside the bracket.
---
Gamers mailing list __ Gamers@audyssey.org
If you want to leave the list, send E-mail to
gamers-unsubscr...@audyssey.org. | 677.169 | 1 |
2623353219664-line × 16-character, easier-to-read LCD displayEachScientific Calculator2623353219664-line × 16-character, easier-to-read LCD displayTexas Instruments® TI-36X Pro Scientific Calculator19 calculator! As a math teacher who has taught 6th-12th grades, this is by far the best, most economical scientific calculator for anything from middle school pre-algebra type classes to calculus! It's not a graphing calculator, but it does a LOT!
Quality and visibility of aerrow button and standard math function could be better
Standard math function buttons - "+, -, enter, ×, ÷" and an aerrow button are not of good visibility. Difficult to point one out. Also an "Arrow" button is not easy to operate. Everything else is good.
My son was using TI-34 in 6th and most of 7th grade. He lost it right before the math state test. I purchased the TI-36X for him and he loves this one as much as the TI-34. As far as I can see the TI-36x does eveything the TI-34 plus more. | 677.169 | 1 |
7912080/ROE
The purpose of this course is to develop the algebraic concepts and processes that can be used to analyze and solve a variety of routine and non-routine real-world and mathematical problems. The content should include, but not be limited to, the following: content-related vocabulary, operations using real numbers in real-world problems, patterns, relations, and functions, including tables, sequences, and graphs, graphs to summarize data and predict outcomes, variables and their impact on outcomes, varied solution strategies to solve real-world problems. | 677.169 | 1 |
Cambridge IGCSE Mathematics
Cambridge IGCSE Mathematics (0580) is accepted by universities and employers as proof of mathematical knowledge and understanding. Successful Cambridge IGCSE Mathematics candidates gain lifelong skills, including:
• the development of their mathematical knowledge
• confidence by developing a feel for numbers, patterns and relationships
• an ability to consider and solve problems and present and interpret results
• communication and reason using mathematical concepts
• a solid foundation for further study.
Prior learning
We recommend that learners who are beginning this course should have previously studied an appropriate lower secondary mathematics programme.
Progression
Cambridge IGCSEs are general qualifications that enable learners to progress directly to employment or to proceed to further qualifications.
Candidates who are awarded grades A* to C in Cambridge IGCSE Mathematics Extended curriculum are well prepared to follow courses leading to Cambridge International AS and A Level Mathematics, or the equivalent.
Candidates may follow either the Core curriculum or the Extended curriculum. Candidates aiming for grades A* to C should follow the Extended curriculum. | 677.169 | 1 |
Friday, December 21, 2012
This week we continued working with TI-83 BASIC! We wrote some very BASIC (pun intended) programs dealing with Coordinate Geometry (Slope, Distance) and Series (West, Bagdad). I think my students enjoy this topic as they can take these skills to Math and Science class where they use TI-83s or TI-84s. We'll take a look at TI-89 BASIC in a future case study. The TI-89 is more like SAGE in that it has a Computer Algebra System built in. Also, TI-89 BASIC uses functions just like Python. Here's our ScreenCasts and Code for this week: | 677.169 | 1 |
Algebra IA & IB Honors – 1.0 Credit
Students are introduced to the foundational skills needed for more advanced mathematics courses and develop the skills needed to solve mathematical problems. Students perform operations involving numbers, sets, and variables; know the basic properties of real numbers; solve and use first degree equations and inequalities; understand functions, relations, and graphs; solve and use systems of equations and inequalities; solve problem involving integral exponents; solve problems involving polynomials and rational algebraic expressions; factor polynomials; simplify rational and irrational expressions; solve and use quadratic equations. Students extend on these concepts by completing more extensive and challenging assignments. Students complete an independent honors project.
Prerequisite: Completion of a Grade 8 Math course with a 'B' or higher. | 677.169 | 1 |
December 21, 2009
The Mathematics Ph.D. program at Rutgers includes two qualifying examinations, a written exam and an oral exam . The written exam is taken first and covers advanced calculus, elementary topology (metric spaces, compactness, and related topics), and the material of 501 (real analysis), 503 (complex analysis), and 551 (algebra). It is offered twice a year, near the beginning of each semester.
The syllabus represents a common core of material required of all Rutgers Ph.D.'s. In particular, the exam is designed with the goal that a pass on this exam shows a level of mathematical knowledge and ability appropriate for teaching the central undergraduate classes in mathematics.
Each student is required to take the exam by the beginning of the student's second year; the program director may allow a student who has entered with less preparation than the norm to take the exam a specified number of semesters later.
Students who fail this exam may take it again during the semester following the one in which the exam was failed. Students who fail on the second attempt or who do not take the exams on schedule (as determined by the program director) will not be allowed to continue in the Ph.D. program.
One can write the EFE in a more compact form by defining the Einstein tensor
,
which is a symmetric second-rank tensor that is a function of the metric. The EFE can then be written as
,
where the cosmological term has been absorbed into the stress-energy tensor as dark energy.
If the energy-momentum tensor is zero in the region under consideration, then the field equations are also referred to as the vacuum field equations. By setting in the full field equations, the vacuum equations can be written as
.
Taking the trace of this (contracting with ) and using the fact that , we get
,
and thus
.
Substituting back, we get an equivalent form of the vacuum field equations
.
The above equation is frequently used in the literature, sometimes, it is called the Einstein vacuum field equations, for example, we refer the reader to introduction part of the following paper due to James Isenberg. In the case of nonzero cosmological constant, the equations are
December 17, 2009
Definition. If is an embedded submanifold, a smooth map such that is a regular level set of is called a defining map for . In other words,
for some point . In particular, if (so that is a real-valued or vector-valued function), it is usually called a defining function.
Example 1. The sphere is an embeded submanifold of . The sphere is easily seen to be a regular level set of the function given by since vanishes only at the origin.
Definition. More generally, if is an open subset of and is a smooth map such that is a regular level set of , then is called a local defining map (or local defining function) for .
Example 2. The smooth map given by
is an immersion of into whose image, denoted by , is the doughnut-shaped shape surface obtained by revolving the circle around the -axis, a point is in if and only if it satisfies where is the distance from the -axis. Thus is the zero set of the function which is smooth on minus the -axis. A straightforward computation shows that $lated d\Psi$ does not vanish on , so is a global defining function for .
In mathematics, the Schwarz lemma, named after Hermann Amandus Schwarz, is a result in complex analysis about holomorphic functions defined on the open unit disk.
Schwarz's Lemma: Let be the open unit disk in the complex plane . Let be a holomorphic function with . The Schwarz lemma states that under these circumstances for all , and . Moreover, if the equality holds for any , or then is a rotation, that is, with .
This lemma is less celebrated than stronger theorems, such as the Riemann mapping theorem, which it helps to prove; however, it is one of the simplest results capturing the "rigidity" of holomorphic functions. No similar result exists for real functions, of course. To prove the lemma, one applies the maximum modulus principle to the function .
Proof: Let . The function is holomorphic in (excluding ) since and is holomorphic. Let be a closed disc within with radius . By the maximum modulus principle,
for all in and all on the boundary of . As approaches we get . Moreover, if there exists a $z_0$ in such that . Then, applying the maximum modulus principle to , we obtain that is constant, hence , where is constant and . This is also the case if .
A variant of the Schwarz lemma can be stated that is invariant under analytic automorphisms on the unit disk, i.e. bijective holomorphic mappings of the unit disc to itself. This variant is known as the Schwarz-Pick theorem (after Georg Pick): Schwarz-Pick theorem: Let be holomorphic. Then, for all ,
December 15, 2009
In classical differential geometry of surfaces, the Codazzi-Mainardi equations are expressed via the second fundamental form
Consider a parametric surface in Euclidean space,
where the three component functions depend smoothly on ordered pairs in some open domain in the -plane. Assume that this surface is regular, meaning that the vectors and are linearly independent. Complete this to a basis , by selecting a unit vector normal to the surface. The unit vector is nothing but
.
It is possible to express the second partial derivatives of using the Christoffel symbols and the second fundamental form.
December 13, 2009
In this topic, we shall give a natural way to construct the Einstein tensor. Let us apply the Ricci contraction to the Bianchi identities
.
Since and , we can take in and out of covariant derivatives at will. We get
.
Using the antisymmetry on the indices and we get
so
.
These equations are called the contracted Bianchi identities. Let us now contract a second time on the indices and
.
This gives
so
or
.
Since , we get
.
Raising the index with we get
.
Defining
we get
.
Theorem. The tensor is divergence free in the sense that
.
The tensor is constructed only from the Riemann tensor and the metric, and it is automatically divergence free as an identity. It is called the Einstein tensor, since its importance for gravity was first understood by Einstein. Some authors denote the Einstein tensor by . We will see later that Einstein's field equations for General Relativity in the vacuum case are
December 12, 2009
In this section, we take a closer look at the curvature at a point of a curve on a surface . Assuming that is parameterized by arc length, we will see that the vector (which is equal to , where is the principal normal to the curve at , and is the curvature) can be written as
,
where is the normal to the surface at , and is a tangential component normal to the curve.
The component is called the normal curvature. Computing it will lead to the second fundamental form, another very important quadratic form associated with a surface. The component is called the geodesic curvature.
It turns out that it only depends on the first fundamental form, but computing it is quite complicated, and this will lead to the Christoffel symbols.
Definition 1. Given a surface , given any curve on , for any point on , the orthonormal frame is defined such that
where is the normal vector to the surface at . The vector is called the geodesic normal vector.
Observe that is the unit normal vector to the curve contained in the tangent space at . If we use the frame , we will see shortly that can be written as
,
The component is the orthogonal projection of onto the normal direction , and for this reason is called the normal curvature of at . The component is the orthogonal projection of onto the tangent space at .
We now show how to compute the normal curvature. This will uncover the second fundamental form. Since , using chain rule we get
.
In order to decompose into its normal component (along ) and its tangential component, we use a neat trick suggested by Eugenio Calabi. Recall that
.
Using this identity we have
Since is a unit vector we can write
Thus, it is clear that the normal component is
,
and the normal curvature is given by
.
Letting
we have
.
Recalling that
,
using the Lagrange identity
,
we see that
,
and can be writtne as
,
where is the determinant of three vectors. Some authors (including Gauss himself and Darboux) use the notation
and we also have
.
These expressions were used by Gauss to prove his famous Theorema Egregium.
Definition 2. Given a surface , for any point on , letting
,
where is the unit normal at , the quadratic form
is called the second fundamental form of at . It is often denoted as and in matrix form, we have
.
For a curve on the surface (parameterized by arc length), the quantity given by the formula
is called the normal curvature of at .
The second fundamental form was introduced by Gauss in 1827. Unlike the first fundamental form, the second fundamental form is not necessarily positive or definite.
Theorema egregium of Gauss states that the Gaussian curvature of a surface can be expressed solely in terms of the first fundamental form and its derivatives, so that is in fact an intrinsic invariant of the surface. An explicit expression for the Gaussian curvature in terms of the first fundamental form is provided by the Brioschi formula.
Given a curve on a surface parametrized by two parameters , we first compute the element of arc length of the curve . For this, we need to compute the square norm of the tangent vector . The square norm of the tangent vector to the curve at is
,
where is the inner product in , and thus,
.
Following common usage, we let
,
and
.
Euler already obtained this formula in 1760. Thus, the map
is a quadratic form on , and since it is equal to , it is positive definite. This quadratric form plays a major role in the theory of surfaces, and deserves an official definition.
Definition. Given a surface , for any point on , letting
,
the positive definite quadratic form
is called the first fundamental form of at . It is often denoted as , and in matrix form, we have
.
The symmetric bilinear form associated with is an inner product on the tangent space at , such that
.
This inner product is also denoted as . The inner product can be used to determine the angle of two curves passing through , i.e., the angle of the tangent vectors to these two curves at . We have
.
For example, the angle between the -curve and the -curve passing through (where or is constant) is given by
.
Thus, the -curves and the -curves are orthogonal iff on .
Example 1. The helicoid is the surface defined over such that
.
This is the surface generated by a line parallel to the plane, touching the axis, and also touching an helix of axis . It is easily verified that . The figure below shows a portion of helicoid corresponding to and .
Example 2. The catenoid is the surface of revolution defined over such that
.
It is the surface obtained by rotating a catenary around the -axis. It is easily verified that . The figure below shows a portion of catenoid corresponding to and .
We will see how the first fundamental form relates to the curvature of curves on a surface.
December 10, 2009
In this topic we discuss Gauss' work to explain the concept of curvature.
The curvature of a curve in a plane is determined by how fast its unit normal vector (or the tangent vector for that matter) changes as we move along the curve. A measure of curvature is the ratio of the small change in the unit normal vector to the distance moved by the point on the curve.
A straight line has zero curvature because the unit normals are all parallel and do not change. A circle of radius has curvature because for the distance that the point moves, the unit normal vector changes by an angle so .
Gauss defined the curvature of a surface analogously.
Curvature of plane curves
For a plane curve , the mathematical definition of curvature uses a parametric representation of with respect to the arc length parametrization. It can be computed given any regular parametrization by a more complicated formula given below.
Curvature: Let be a regular parametric curve, where is the arc length, or natural parameter. This determines the unit tangent vector , the unit normal vector , the curvature , the oriented or signed curvature , and the radius of curvature at each point:
.
Local expressions. For a plane curve given parametrically as , the curvature is
,
and the signed curvature is
.
For the less general case of a plane curve given explicitly as the curvature is
.
Slightly abusing notation, the signed curvature may also be written in this way as
with the understanding that the curve is traversed in the direction of increasing .
Curvature of space curves
For a parametrically defined space curve as , its curvature is:
.
Given a function with values in , the curvature at a given value of t is
where and correspond to the first and second derivatives of , respectively, and is the cross (vector) product. (Note that this formula is the vector notation of above.)
Curves on surfaces
When a one dimensional curve lies on a two dimensional surface embedded in three dimensions , further measures of curvature are available, which take the surface's unit-normal vector, u into account. These are the normal curvature, geodesic curvature and geodesic torsion.
Any non-singular curve on a smooth surface will have its tangent vector lying in the tangent plane of the surface orthogonal to the normal vector. The normal curvature, , is the curvature of the curve projected onto the plane containing the curve's tangent and the surface normal ; the geodesic curvature, , is the curvature of the curve projected onto the surface's tangent plane; and the geodesic torsion (or relative torsion), , measures the rate of change of the surface normal around the curve's tangent.
Principal curvature: All curves with the same tangent vector will have the same normal curvature, which is the same as the curvature of the curve obtained by intersecting the surface with the plane containing and .
Taking all possible tangent vectors then the maximum and minimum values of the normal curvature at a point are called the principal curvatures, and , and the directions of the corresponding tangent vectors are called principal directions.
Curvature of surfaces (two dimensions)
Gaussian curvature: In contrast to curves, which do not have intrinsic curvature, but do have extrinsic curvature (they only have a curvature given an embedding), surfaces can have intrinsic curvature, independent of an embedding. Here we adopt the convention that a curvature is taken to be positive if the curve turns in the same direction as the surface's chosen normal, otherwise negative.
The above definition of Gaussian curvature is extrinsic in that it uses the surface's embedding in normal vectors, external planes etc. Gaussian curvature is however in fact an intrinsic property of the surface, meaning it does not depend on the particular embedding of the surface; intuitively, this means that ants living on the surface could determine the Gaussian curvature. For example, an ant living on a sphere could measure the sum of the interior angles of a triangle and determine that it was greater than 180 degrees, implying that the space it inhabited had positive curvature. On the other hand, an ant living on a cylinder would not detect any such departure from Euclidean geometry; the cylinder has extrinsic curvature, but no intrinsic curvature.
Formally, Gaussian curvature only depends on the Riemannian metric of the surface. This is Gauss' celebrated Theorema Egregium, which he found while concerned with geographic surveys and mapmaking.
An intrinsic definition of the Gaussian curvature at a point is the following: imagine an ant which is tied to with a short thread of length . She runs around while the thread is completely stretched and measures the length of one complete trip around . If the surface were flat, she would find . On curved surfaces, the formula for will be different, and the Gaussian curvature at the point can be computed by the Bertrand–Diquet–Puiseux theorem as
Mean curvature: The mean curvature is equal to the sum of the principal curvatures, , over 2, that is
.
It has the dimension of 1/length. Mean curvature is closely related to the first variation of surface area, in particular a minimal surface such as a soap film, has mean curvature zero and a soap bubble has constant mean curvature. Unlike Gauss curvature, the mean curvature is extrinsic and depends on the embedding, for instance, a cylinder and a plane are locally isometric but the mean curvature of a plane is zero while that of a cylinder is nonzero. | 677.169 | 1 |
Fix Solved: New Problems - HJ This (Solved)=
Solved: New Problems - HJ This
This is a by-product of lectures given at the Osmania University,... Solved Problems in Modern PhysicsMy libraryHelpAdvanced Book SearchEBOOK FROM $54.51Get this book in printSpringer ShopAmazon.comBarnes&Noble.comBooks-A-MillionIndieBoundFind in a libraryAll sellers»1000 Solved Getzels,J. & Csikszentmihalyi, M. (1967). grades and later achievement test scores), their relationship to adult success factors is minimal. Most decisions we make are not simple or closed because there are numerous possible variations or alternatives and there are myriad intended (or unintended) consequences to be considered. navigate here
House. But even here a bizarre and little known situation occurs that can adversely affect a child's development. Maker, C.J. (1993). The gifted hearing-impaired student.
A. (1976). Both keroseneUser Review - Flag as inappropriateVery useful book starting from basic and which is easy understandable Selected pagesTitle PageTable of ContentsIndexContents11 217 2b47 369 493 5117 6139 7158 10210 11232 Maker, C.J. (1981). If we are to realize the vision articulated in the new curriculum standards and prepare a generation capable of designing new approaches to new problems, we must assess and develop a
For those curious about how commonly used assessment methods and programs stack up, in terms of which Problem Types they include, see Figure 2 below. Type III: The problem is known to the presenter and the solver, but more than one method may be used to arrive at the correct solution, which the presenter knows. Undergraduate academic achievement as an indicator of success. Gifted Education International,9(2), 68-77.
Solving math problems by a known algorithm or method; following a formula, in language, music, math or science; and performing prescribed body movements, as in dance or sports are Type I Scholars who have completed critical reviews of research (Baird, 1985; Hoyt, 1966; Nelson, 1975; O'Leary, 1980; Reilly & Chao, 1982; Wallach, 1976), including meta-analysis (Samson, et. The system returned: (22) Invalid argument The remote host or network may be down. Committee on Immigration and Naturalization1921 0 Reviews Preview this book » What people are saying-Write a reviewWe haven't found any reviews in the usual places.Selected pagesTitle PageCommon terms and phrasesacres admitted
Each chapter begins with basic concepts containing a set of formulae and explanatory notes for quick reference, followed by a number of problems and their detailed solutions. Repr. Academic and occupational performance: a quantitative synthesis. The standardized regression coefficients were .44, .24, .20, and .16 respectively.
test or behavior such as high school or college achievements) resembles the criterion to be predicted (e.g. The Journal of Creative Behavior,15(2), 117-135. Type IV: The problem is known to the presenter and the solver, but the problem may be solved in more than one way and the presenter knows the range of solutions. his comment is here R., Grave, M.
Some studies suggest, no. These are the problem situations we find in real-life that can be defined in more than one way and that may need redefining during the problem solving process. Maker, C.J. (1978).
Getting dressed, for example, typically is a Type 4 problem; usually there is a clear need to do so, yet the method and solution will vary according to one's state of
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Validity and fairness of some alternative employee selection procedures. Enrichment and acceleration: An overview and new directions. Programs for gifted students usually are better, and are more likely to include all of the Problem Types. Maker and colleagues added Problem Types III and IV, to provide a more fluid transition between the Types, based on data observed in their research (C.
Several Topics Like Aromaticity, Pericyclic Reactions And Heterocyclic Chemistry Have Now Been Brought Up To Date And The Material Provided Is Complete In Itself.The Presentation Has Been So Designed So As A longitudinal study of creativity–1965–1975. We also must develop those abilities in real-life problem solving situations. A pure Type VI problem is so abstract that it might have an infinite number of possible solutions or it might not have a solution at all.
College grade point average as an indicator of occupational success: An update (Personnel Research Report, 80, 281-294. As a result, little has been learned about the abilities and skills actually needed in the structuring and solving of real-world problems. Type II is close to Type I in structure, except that the problem solver does not know the method by which to arrive at a solution. He was a Professor and Head of the Department of Physics and Chairman, Board of studies at the Osmania University.Bibliographic informationTitle1000 Solved Problems in Modern PhysicsSpringerLink: Springer e-BooksAuthorAhmad A. | 677.169 | 1 |
Please Note! The current edition of Saxon Math 5/4 is the 3rd Edition. This 2nd edition is offered for families using older versions of Saxon. SinceSystem Requirements:
Mac OS 10.3.9-10.4.x
Windows 98, 2000, ME, XP, Vista, 8, 10
Quicktime Download Required
Please Note! The current edition of Saxon Math 7/6 is the 4th Edition. This 3rd edition is offered for families using older versions of Saxon.Math 7/6 covers fractions, decimals, percent conversions and unit multipliers, as well as circumference, pi, angles, graphing, and prime factors. For use with 4th Edition.
Working with the older Saxon Math 8/7, 2nd Edition? This is the DIVE CD for you!
Boost your students understanding of Saxon Math with DIVE's easy-to-understand lectures! Each lesson concept in Saxon's textbook is taught step-by-step on a digital whiteboard, averaging about 10-20 minutes in length; because each lesson is stored separately, you can easily move pre-algebra mathematics and skills, and includes review of fractions, decimals, percents and graphing. For use with 3rd Edition.
System Requirements:
Mac OS 10.3.9-10.4.x
Windows 98 traditional algebra 1/2 topics, as taught in the 2nd Edition Saxon textbook, as well as topics from geometry and discrete mathematics. For use with 2nd Edition texts.
Please Note! The current edition of Saxon Math Algebra 1/2 is the 3rd Edition. This 2nd edition is offered for families using older versions of Saxon After the lessonT 2 covers traditional second year algebra topics as well as a semester of geometry, real world problems, linear and nonlinear equations, statistics and probability, graphing and basic trigonometry. This DIVE CD can be used with Saxon Algebra 2's 2nd and 3rd Editions; the CLEP Professor College Algebra CLEP Exam prep course is also included.
System Requirements:
Mac OS 10.3.9-10.4
Windows 10,8,7,98 of the 120 lesson concepts and 12 investigations in Saxon Math's Geometry textbook is taught step-by-step on a digital whiteboard, averaging about 10-20 minutes in length; and because each lesson is stored separately, you can easily move about from lesson-to-lesson as well as maneuver within the lesson you're watching. After the lessonThis course covers all topics in a high school geometry course, including perspective, space, and dimension associated with practical and axiomatic geometry. Students learn how to apply and calculate measurements of lengths, heights, circumference, areas, and volumes, and will be introduced trigonometry and transformations. Students will use logic to create proofs and constructions, work with key geometry theorems and proofs, and use technology such as spreadsheets, graphing calculators, and geometry softwarePlease Note! The current edition of Saxon Math Calculus is the 2nd Edition. This 1st edition is offered for families using older versions of Saxon. Boost Physics | 677.169 | 1 |
Instead of using a simple lifetime average, Udemy calculates a course's star rating by considering a number of different factors such as the number of ratings, the age of ratings, and the likelihood of fraudulent ratings.
Introduction to Trigonometry
Your Complete Primer in Trigonometry perfect beginner course for secondary school students intending to sit for Mathematics examinations. It will take the students from the definition of the three main Trigonometric Ratios (sine, cosine and tangent) of acute angles using right-angled through to the Trigonometric Ratios of angles of any size (including negative angle) using Cartesian Plane. Inverse Trigonometric functions will be introduce to students which is the foundation of the more complex solution to Trigonometric Equations.
Two topics on on Geometry (Sine & Cosine Rule and Area of Triangle) are also included, to complement the understanding of Trigonometry and its link to angles and triangles.
The course is delivered by an experienced teacher with extensive experience inUsage of mnemonic to memorise Trigonometric Ratio Definitions. After watching this video students will be able to define Sine, Cosine and Tangent.
Definition of Sine, Cosine and Tangent (Part 1 of 3)
06:23 2 of 3)
11:10 3 of 3)
10:26
Quiz 1 (Basic Trigonometric Ratios)
7 questions
Solutions to Quiz 1. After watching this video students will be able to determine the solutions to questions in Quiz 1.
Solutions to Quiz 1 (Basic Trigonometric Ratios)
09:18
Trigonometric Functions on Calculator
09:16
Introducing Inverse Trigonometric Functions, arcsin, arccos and arctan, which will return a value of an angle. After watching this video students will be able to use inverse trigonometric functions to determine the values of a certain angle.
Inverse Trigonometric Functions (Part 1 of 2)
08:02
Elaborating on Inverse Trigonometric Functions through examples. After watching this video students will be able to use inverse trigonometric functions to determine the values of a certain angle.
Inverse Trigonometric Functions (Part 2 of 2)
04:51
Quiz 2 (Inverse Trigonometric Functions)
5 questions
Solutions to Quiz 2. After watching this video students will be able to determine the solutions to questions in Quiz 2.
Solutions to Quiz 2 (Inverse Trigonometric Functions)
07:15
Use Trigonometric Ratio definitions to calculate lengths of sides of a triangle. After watching this video students will be able to calculate the lengths of sides of a triangle using a Trigonometric Ratio value and the length of one of the sides of the triangle.
Calculating the Lengths of Sides of a Triangle.
06:48
Quiz 3 (Lengths of Sides of a Triangle)
5 questions
Solutions to Quiz 3. After watching this video students will be able to determine the solutions to questions in Quiz 3.
Solutions to Quiz 3 (Lengths of Sides of a Triangle)
07:06 1 of 4)
04:59 2 of 4)
10:24 3 of 4)
07:17 4 of 4)
09:46
Quiz 4 (Trigonometry Problem Solving)
4 questions
Solutions to Quiz 4. After watching this video students will be able to determine the solutions to questions in Quiz 4.
Solutions to Quiz 3 (Trigonometry Problem Solving. Part 1 of 3)
06:16
Solutions to Quiz 4. After watching this video students will be able to determine the solutions to questions in Quiz 4.
Solutions to Quiz 3 (Trigonometry Problem Solving. Part 2 of 3)
08:26
Solutions to Quiz 4. After watching this video students will be able to determine the solutions to questions in Quiz 4.
Solutions to Quiz 3 (Trigonometry Problem Solving. Part 3 of 3)
06:15
+–
Sine and Cosine Rules
13 Lectures
01:57:59
Derivation of Sine Rule. After watching this video students will be able to derive Sine Rule. 1 of 3)
08:30 2 of 3)
08:11 3 of 3)
07:28
Describes Negative Angles and their corresponding Trigonometric Ratios. After watching this video students will be able to determine Trigonometric Ratios of negative angles.
Trigonometric Ratios of Negative Angles (Part 1 of 2)
07:00
Describes Negative Angles and their corresponding Trigonometric Ratios. After watching this video students will be able to determine Trigonometric Ratios of negative angles.
Trigonometric Ratios of Negative Angles (Part 2 of 2)
06:38
Quiz 8 (Trigonometric Ratios of General Angles)
4 questions
Solutions to Quiz 8. After watching this video students will be able to determine the solutions to questions in Quiz 8.
Solutions to Quiz 8 (Trigonometric Ratios of General Angles)
04:51
Determining whether Sine, Cosine and Tangent are of positive or negative values in Quadrant 1, Quadrant 2, Quadrant 3 and Quadrant 4. After watching this video students will be able to evaluate and remember the signs of Trigonometric Ratios in different quadrants of a Cartesian Plane.
Signs of Trigonometric Ratios
07:08
Describes how Basic Acute Angle relates to angles in all four Quadrants on Cartesian Plane. After watching this video students will be able to link Trigonometric Ratios of angles to that of its Basic Acute Angle.
Basic Acute Angle (Reference Angle)
08:33
Determining Trigonometric Ratios based on that of their Basic Acute Angles. After watching this video students will be able to evaluate Trigonometric Ratios of any angle based on its Basic Acute Angle. 2 of 3)
09:52 3 of 3)
08:35
Quiz 9 (Basic Acute Angle)
5 questions
Solutions to Quiz 9. After watching this video students will be able to determine the solutions to questions in Quiz 9.
Solutions to Quiz 9 (Basic Acute Angle)
12:52
Finding Possible Values of Angles from their Trigonometric Ratios
10:12
Quiz 10 (Finding Possible Values of Angles)
6 questions
Solutions to Quiz 10. After watching this video students will be able to determine the solutions to questions in Quiz10. | 677.169 | 1 |
November Mid-Term Maths Block
DESCRIPTION
Taught by Aidan Roantree, senior Maths teacher at The Institute of Education, this maths grinds course is aimed specifically at higher level Maths students sitting the Leaving Certificate in June. These classes consist of 4 hours of intensive, exam focused tuition. Students are provided with exceptional notes and key techniques to enable them to achieve their maximum potential in the Higher Level Maths exam.
Students may opt do one or both days of the course.
SUBJECT
Maths Grinds
Day 1: Paper 1- Maths Higher Level:
Topics from:
Algebra
Financial Maths
Differentiation
Day 2: Paper 2- Maths Higher Level:
Topics from:
Trigonometry
Probability
Co-ordinate Geometry
TIMETABLE
Maths Grinds
Course dates: Tuesday 31 October & Wednesday 1 November 2017
DATE
TIME
Tuesday 31 October 2017 (Paper 1)
10.00am – 2pm
Wednesday 1 November 2017 (Paper 2)
10.00am – 2pm
FEES
1 Class
2 Classes
€160
€290
Students may opt do one or both days of the course.
Please note:
Laser/Credit Card payments accepted | 677.169 | 1 |
Posts tagged GCSE maths
Currently, there are Linear and Modular GCSEs, although Modular versions are being phased out. Modular GCSEs are designed so that students sit individual modules at several points during Years 10 and 11. Linear GCSEs are designed so that students sit their exams at the end of Key Stage 4, in the summer term of Year | 677.169 | 1 |
Description: The text for beginning graduate students of mathematics, engineering, and the physical sciences. The book covers elements of Hilbert space, distributions and Sobolev spaces, boundary value problems, first order evolution equations, implicit evolution equations, second order evolution equations, optimization and approximation topics. | 677.169 | 1 |
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MS Math 3.0 Rolls Out This Month
By Michelle Rutledge
05/17/07
##AUTHORSPLIT##<--->
Microsoft recently announced the launch of Math 3.0, a math and science educational tool for students in grade levels 6-12, as well as entry-level college students. The software is designed for use at home, to assist students with math and science concepts and homework, or for visual examples in the classroom.
"Microsoft Math provides a space for nurturing student learning in mathematics with dynamic visualizations. The program provides essential ingredients for classroom environments designed to challenge all students to engage in visual thinking," said Margaret L. Neiss, mathematics education professor at Oregon State University in a prepared statement.
According to Microsoft, the software includes study material for six different math and science subjects. Features in the program include:
Graphing calculator;
Step-by-step math solutions;
Formulas and equation library;
Unit conversion tool;
Triangle solver; and
Handwriting support.
Math 3.0 runs $19.95 for a single license. Volume licensing is available for educational institutions | 677.169 | 1 |
Chapter 2Graphs
"Education is the most powerful tool we can use to change the world."-Nelson Mandela
Graphs are an essential part to learning mathematics. Graphs show data and allow people to understand that data. Their are many different types of graphs that will be taught in this chapter; circle graphs, bar graphs, line graphs, and scatter plot graphs. The goal of this chapter is to teach each one of these graphs and for students to understand how to read these graphs. They will also be able to know how to create each graph based on data they are given. They will learn each step in order to make each graph.
The objective of each section is to show these graphs and teach about the current history and past history at Saint Joseph's University. Another objective is for the students to be able to read, understand, and create their own circle, bar, line, and scatter plot graphs.
Saint Joseph's University, Barbellin Hall/Sweeney Field
Section 1: Circle Graphs
This circle graph is showing the Business Majors at Saint Joseph's University. Along with each major this university provides, it is also showing how many other areas you can go into under that major. Business is a very popular major at this university and as you can see from this graph, there are many different business majors and minors that students can go into for their academics.
This circle graph shows certain data that was collected. Once the data was collected, the graph was then made. Circle graphs are used to show the results of data in a proportional manner. You can see the percentages of each result and how it compares to other results.
Vocabulary to know:
Circle Graph- A circle chart divided into sections which represent a value out of the overall value.
Percentage- Any proportion or share in relation to a whole.
Data- Facts and statistics collected together for reference or analysis.
Average- Obtaining the result by adding all of the quantities together and then dividing by how many quantities there are.
Degree- A unit of measurement of angles.
This video is a good quick lesson on how to make circle graphs. You can watch this for more help! The following are detailed steps in order to make this circle graph.
To read a circle graph, you look at each category and see what percentage of the circle it has. This is out of 360%, because a circle is 360 degrees. You can see how much each category is by comparing them to each other category.
Step 1: Collect data and have the data separated into categories that can be split up.
Step2: Once the data is split up, take each category and their values (numbers) and order them from the largest to the least. You will need to take the average of them all to see how much each category will take up of the graph. For example, the circle graph shown of the business majors would be each number of majors and their minors, which is 17. After you have the total of the whole circle graph, you will use that number, 17 in this case, and divide each major by 17 to get the average of each major. For Marketing, you will do 4 (value of marketing) divided by 17 (total values for the graph). This equals 0.23529. After doing this for each category/major, you will now get a percentage for each major. You will take each percentage and multiply each by 360 degrees because a circle is 360 degrees. For Marketing, you take the 0.23529 and multiply that by 360, which equals 84.7%. That is how much of the graph, out of 360%, is marketing. Once you have each percentage, you will take a protractor and measure each degree that you have gotten for each value.
Step 3: You make each part of the circle graph and label each part with what it represents. Make sure to have a title for the graph as well. You can color code the chart so it is easier to read and see each different type of data.
This is another video explaining pie charts.
Questions:
1.) What do you need to divide each value by in order to get a percentage of the circle graph?
2.) Make a circle graph with the following data of students favorite ice cream flavors: Chocolate-6, Vanilla-10, Mint Chocolate Chip-12, Rocky Road-4, Cookies and Cream-5
Section 2: Bar Graphs
Many objects can look like graphs too!
Bar graphs are used by many people to easily show the results of some type of experiment. Using bar graphs, it makes it easier to see the comparison between different results.
Vocabulary to know:
Bar Graph- A graphical display of data using bars or different heights.
Y-Intercept- The axis that is vertical of a system of coordinates.
X-Intercept- The horizontal axis of a system of coordinates.
Interval- The consistent space in between numbers.
Below is an example of a bar graph that you can look at. You can use this website to make your own graphs.
This bar graph shows the Men's Varsity sports teams at Saint Joseph's. As you can see, there are many teams and many students that play on each team. Sports are a great extra-curricular activity and a great way to make friends and stay in shape! Sports are a big part of this university and are important to many students.
SJU sports teams
Bar graphs are used for data that need to be compared. The data is separated into categories and you can compare each category separately.
In order to read a bar graph you look at the x-axis and that is your category you are trying to find the value of. You look to see where the bar goes until and look at the y-axis to see what value it is at. This is your data that was graphed.
Each category has a certain value and that value is what is being shown by the bar graph. To start making a bar graph, you need the data and to separate them into categories. Once you have your categories, see how many is in each and that is the value. Once you have the values for each category, you will begin to make the bar graph.
Start with drawing your x-axis (horizontal) and y-axis (vertical) and label them with what you are graphing. Shown below.
Next, look at your values and try and figure out what interval you will go by on your graph. This will be on your y-axis, numbers going from 0 to your highest value. Once you have your interval, start labeling the graph with the values on the y-axis.
Now, label your x-axis with your category name or with what you are graphing. You may need to find an interval for this axis as well.
After your graph is completely labeled, you can now start to draw the bars of data. Start from the left and see what your data is showing you, and draw a bar where the value is (y-axis). Continue this for each category and make sure all of your data is showing on the graph.
Questions:
1.) Explain the difference between the x-axis and y-axis.
2.) What are bar graphs used for?
Section 3: Line Graphs
Line graphs are used for seeing a relationship of two variables over periods of time. They show the change over time for a quantity. Below is an example of a line graph of Saint Joseph's enrollment over the past 17 years. You can see the increase in students as the years go on. Starting in 2000 there were just about 3,647 students attending SJU and now in 2017 there are about 5,395 students. This is a great graph to show this type of data.
Correlate- Having a mutual relationship or connection, where one thing affects or depends on another.
Line Graph- Shows how much a quantity changes over time.
Plot- A point to represent a location on a graph.
This video is a little introduction for understanding line graphs and how to make them.
A line graph shows relationships so reading this graph is a little different than a circle graph or bar graph. When reading a line graph, you need to look at the x-axis and see the category that is being shown and follow the line on the graph. This line is representing the change of that category over time. You look at both axes and the point that is shown between them both. Start at a point and follow that to the x-axis and then y-axis to see what the category is and what the value is.
In order to make a line graph, you do the same as a bar graph and make the axes first. Make the x-axis and y-axis and label them both. Your value number intervals will go on the y-axis and your year or category will be on the x-axis.
Once you have these axes labeled, you will now start to plot your data. You will look at the x-axis or y-axis and follow the data point to your data point on the other axis. When they meet, that is where you place a dot. You will continue this for all of your data points and once you are done, you will create a line connecting all of the dots. The line will start with the first dot and go in order of the data from left to right.
Now that you have your line, the graph is finished. Even though it is finished, you can read the graph and decide wether it is increasing, decreasing, or remaining the same. When a line is remaining the same, it is a flat line across the graph. A line that is increasing (Figure 1) is a line moving from the lower left of the graph to the upper right of the graph. A line that is decreasing (Figure 2) is going from the upper left of the graph to the lower right. See below for the visuals of these two line graphs.
Figure 1: Increasing
Figure 2: Decreasing
The following is another video to watch on how to make a line graph.
Questions:
1.) Why are line graphs sometimes referred to as correlating graphs?
2.) What is the order to connecting the dots of data?
Section 4: Scatter Plot Graphs
Scatter plot graphs are used to represent how much one variable is affected by another. It shows a relationship between two variables and their correlation, which can be positive, negative, or no correlation. These graphs are similar to line graphs in which they correlate with each other.
Vocabulary to know:
Scatter plot graph- A graph that has points that show the relationship between two sets of data.
Positive correlation- This exists as both variables are decreasing or when both variables are increasing.
Negative correlation- This exists when one variable is increasing and the other is decreasing.
Figure 1
Figure 1 graph shows height and weight and how they relate with each other. It shows that when someone is taller, they tend to weigh more than someone who is shorter. To see this, you look at a dot and see where it is located at the x-axis and y-axis and that is what your dot is representing. This is an example of a positive correlation because the points are going from bottom left to top right, moving upwards.
Figure 1 represents height versus weight and is showing a positive correlation. To make a scatter plot graph, it is similar to a line graph. First, you need to make and label your axes the same way you did for a line graph and bar graph. Get the x-axis labeled with your one variable and your y-axis labeled with your other variable. Since these two correlate, when you look at your data, there will be one data point that represent both axes. For example, looking at the scatter plot of Height vs. Weight, look at the point where the dot is at 5'3 and 130. This one point correlates with both of the axes. It is showing that a person that is 5'3 weighs 130 pounds. When you get each data point on the graph and match them with the axes, your graph is complete.
Scatter plot graphs will have a positive correlation, negative correlation, or no correlation. The graph in figure 1 shows a positive correlation. This is because the data points are moving in the direction of upwards toward the top right of the graph. A negative correlation is when the dots are in a clustered line going from the top left of the graph to the bottom right of the graph. No correlation is when the dots are scattered and are in no formation all over the graph.
The video below can also help with understanding and making a scatter plot. | 677.169 | 1 |
Disadvantages<br />Can be costly<br />Changes rapidly so constant upgrading and professional development to learn new programs is essential<br />Plagiarism<br />Access to resources for all students <br />
8.
TI Navigator<br />Allows teachers to link up with students' graphing calculators to view progress instantly and provide immediate feedback to students<br />Allows students to graph and manipulate functions<br />Students can conceptualize the topic and are actively engaged in the lesson<br />
9.
How to set up a class to utilize ti Navigator<br />Video on setting up TI Navigator<br /> | 677.169 | 1 |
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The activities are divided into three sections of thirty problems each. First you'll find a series of activities with a matrix for one variable (1-35), then for two variables (36-70), and finally a last series for three variables (71-105). These activities have been designed for students at the junior and intermediate levels, 9-14 years old. The answers are on pages A-123 to R-128. | 677.169 | 1 |
Friday, April 12, 2013
Now Available in the MAA Store
"Everyone knows that functional analysis is one of the most powerful tools
of twentieth-century mathematics. The idea of studying entire spaces
of functions, rather than just one function at a time, is both deep and original.
Yet it is difficult to get a glimpse of what this subject is really about, or
of how it works.
The Guide to Functional Analysis takes the neophyte reader and shows
him/her the basic concepts and rubric of this time-tested discipline. All the
major ideas are illustrated with concrete examples and applications to other
parts of mathematics. There are even some illustrations.
This is a user-friendly, hands-on introduction to an otherwise austere
and forbidding part of the mathematical lore. It will be helpful to students
beginning the long journey down the path to mastery of analysis and
also enlightening for any mathematician wanting to bone up on
linear operators and their uses. This book is meant to be a guide,
and we hope that it guides you to a pleasurable reading experience."
Steven G. Krantz earned his B.A. degree for the University of California at Santa Cruz and his Ph.D. from Princeton University. He has written over 70 books and over 180 scholarly papers. The MAA has awarded him both the Beckenbach Book Prize and the Chauvenet Prize | 677.169 | 1 |
With)—new study skills activities in the text, an updated and expanded Lial Video Library (available in MyMathLab), and a new accompanying Lial Video Library Workbook (available in MyMathLab).
Marge Lial was always interested in math; it was her favorite subject in the first grade! Marge's intense desire to educate both her students and herself has inspired the writing of numerous best-selling textbooks. Marge, who received Bachelor's and Master's degrees from California State University at Sacramento, is affiliated with American River College. Marge is an avid reader and traveler. Her travel experiences often find their way into her books as applications, exercise sets, and feature sets. She is particularly interested in archeology. Trips to various digs and ruin sites have produced some fascinating problems for her textbooks involving such topics as the building of Mayan pyramids and the acoustics of ancient ball courts in the Yucatan.
When John Hornsby enrolled as an undergraduate at Louisiana State University, he was uncertain whether he wanted to study mathematics education or journalism. His ultimate decision was to become a teacher, but after twenty-five years of teaching at the high school and university levels and fifteen years of writing mathematics textbooks, both of his goals have been realized. His love for both teaching and for mathematics is evident in his passion for working with students and fellow teachers as well. His specific professional interests are recreational mathematics, mathematics history, and incorporating graphing calculators into the curriculum. John's personal life is busy as he devotes time to his family (wife Gwen, and sons Chris, Jack, and Josh), and has been an avid baseball fan all of his life. John's other hobbies include numismatics (the study of coins) and record collecting. He loves the music of the 1960s and has an extensive collection of the recorded works of Frankie Valli and the Four Seasons.
A native Midwesterner, Terry McGinnis received her Bachelor's of Science in Elementary Education with a concentration in Mathematics from Iowa State University. She has taught elementary and middle school mathematics, and developed and implemented the curriculum used with her students. Terry has been involved in college mathematics publishing for over 20 years, working with a variety of authors on textbooks in both developmental mathematics and precalculus. After working behind the scenes on many of the Lial/Hornsby textbooks and supplements for over 10 years, Terry joined Margaret Lial and John Hornsby in 2002 as coauthor of their developmental mathematics series that includes Introductory Algebra, Intermediate Algebra, and Introductory and Intermediate Algebra. When not working, Terry enjoys spinning at a local health club, walking, and reading fiction. She is the devoted mother of two sons, Andrew and Tyler.
Book Description Pearson. Book Condition: Good. 0134277856 May have signs of use, may be ex library copy. Book Only. Used items do not include access codes, cd's or other accessories, regardless of what is stated in item title. Bookseller Inventory # Z0134277856Z3 regardless | 677.169 | 1 |
A level Resources
MEI's virtual learning environment, Integral, contains extensive resources to support the teaching and learning of Mathematics. This interface includes features such as forums, messaging and student tracking.
To see samples for topics in Core 1 and Further Pure 1 please click the button below:
Features of the resources
Integral resources are tailored to each of the AQA, Edexcel, MEI, OCR and WJEC AS/A level Mathematics and Further Mathematics specifications.
Thousands of pages of resources including notes, study plans, additional exercises and hints & solutions to textbook questions
Hundreds of interactive resources including "teach-yourself" random question generators, interactive Flash animations, spreadsheets and Power Points
Multiple-choice tests at the end of every section for students to check their progress
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As well as access to Integral for the current specifications, 2016/17 school/college A level subscribers have teacher access to our 2017 site for the new AS/A levels. and are able to see new content as it is added.
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Current subscribers can renew their A level or Additional Mathematics subscription quickly and easily via User Management.
Once logged in click on the renew subscription link which will also allow you to update your contact details.
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Individual subscriptions to the A level resources for students whose schools do not subscribe to Integral are also available. Please note this subscription is not available for teachers or schools/colleges.
To subscribe, please visit Individual Integral subscriptions and complete the online subscription form. Subscriptions normally run to 30th September in the following academic year. | 677.169 | 1 |
Welcome to the Nayland College Mathematics and Statistics Website
Website News
We are revising and refreshing our website. We are looking to build on the great work of Max Riley who built the original version of this site and shared lots and lots of his resources.
Max has retired from teaching and we wish him all the best for his new adventures.
The first page to get a refresh is this page. Others will be completed over time. Feedback would be most welcome, feel free to email
me at ben.crawford@nayland.school.nz.
Year 9
Nayland College is currently running the following topics in our Year 9 course:
Level 2
Nayland College runs three different NCEA Level 2 courses.
Our 2A course is an academic course and is predominantly algebra based but will also include calculus and graphs as well as some statistics and probability.
This course is required to prepare students for the academic mathematics courses at Level 3. The course covers:
2.4 Trigonometry
2.6 Algebra
2.12 Probability
2.2 Graphs
2.7 Calculus
2.14 Simultaneous Equations
Our 2Stats course focuses on the acquisition of practical statistical knowledge and skills. Students who are considering subjects such as Biology, Psychology or Geography will benefit from choosing this course.
The course covers:
2.5 Networks
2.13 Simulations
2.3 Sequences
2.9 Inference
2.10 Experiments
2.12 Probability
Our 2S course focuses on the development of statistical skills and applications. It may lead to mathematics at Level 3.
This course covers: | 677.169 | 1 |
Best Techniques to Learn Algebra
Algebra is a branch of mathematics in which letters and other general symbols are used to represent numbers and quantities in formulae and equations. Algebra consists of basic Math skills such as adding, subtraction, multiplying and dividing. If you have mastered these skills of algebra then its easy to solve complex concepts of Algebra at the higher class.
The following steps can help in getting to know more about Algebra-
(i) Playing with numbers and variables – Once you are comfortable with numbers switch to the combination of variables which can be letters or symbols. These are just ways of showing numbers with unknown values.
Few common examples of variables used in algebra are-
Letters like a,b,c, x, y, z etc.
Universal constants such as pi = 3.14159.
(ii) Organizing long problem – Complicated problems can take many steps and have a maximum chance of error. However if organized properly, it would result in problem free solution. Keep your work organized by starting a new line every time you make a step toward solving your problem.
(iii) Solving long equation (BODMAS) – After basic skills, you can go to higher level of solving techniques such as linear & Simultaneous equation solving them using substitution, elimination and cross elimination. The basic equation solving uses the Principle of BODMAS which decides the order of operation. It stands for "brackets of division, multiplication, addition, subtraction."
(iv) Simplification of variables- It is to be noted in many equations that the variables of the same order appear more than once. Thus these variables can be added and subtracted and so on as they are of the same kind.
(v) Practice more to have masterly skills – Practice is the key to success, the more you work the more you get. Once you start playing with a variable you will find it easy in calculating it in your mind and with fewer steps with error free results.
These are few important steps that would lead to understand Algebra in a better way. Algebraic expression, mainly variables seems to be hard but actually they are the replicate of the numbers. Following the above steps lead to the mastery in basic as well as complex algebra.Subscribe to Byju's Youtube channel to explore more math related concepts-
About the Author: Shubham Arora, is an Engineer and Blogger by nature. He is currently working with Byju's as a content writer inspiring thousands of students to fall in love with learning | 677.169 | 1 |
Math Explained: The Britannica Guide to The History of Mathematics2212
ISBN-13: 9781615302215
DDC: 510.9
Grade Level Range: 9th Grade
- 12th Grade
288 Pages | eBook
Original Copyright 2011 | Published/Released April32125158151837371312010569761919405520
About
Overview
The field of mathematics today represents an ongoing global effort, spanning both countries and centuries. Through this in-depth narrative, students will learn how major mathematical concepts were first derived, as well as how they evolved with the advent of later thinkers shedding new light on various applications. Everything from Euclidean geometry to the philosophy of mathematics is illuminated as readers are transported to the ancient civilizations of Mesopotamia, Egypt, and beyond to discover the history of mathematical thought | 677.169 | 1 |
Why a new book? What do we emphasize? How the book is organized? How is the book designed? Some salient features of the book What tools do we use? A Possible Sequence
2
Introduction
1.1
Why a new book?
There are a number of excellent books available on topics covering an introductory course on Graph Theory & Graph Algorithms - almost each with a different approach. We shall discuss here why there was a need for another book designed with a different pedagogical structure and suitable for students and practitioners working in diverse fields who intend to use graph theory in their respective fields of study. Before we start making comparison between different pedagogical approaches and talk about the merit of our approach we shall first describe the basic structure of the said course. There are basically three layers on which such a course operates: 1. Definitions (concepts) 2. Implications & theorems (connections between concepts) 3. Algorithms (actions which transform an input into an output) Let us start with a few concepts used in our subject. For example graph, connected graph, tree, vertex, edge, bridge edge, edge weight, path, edge-disjoint paths, & edge cut are all concepts, and a learner needs to become familiar with these concepts. He or she should be able to feel, visualize, and connect these concepts. An implication connects two or more concepts. For example if every edge is a bridge edge in a connected graph G then the graph G is a tree. Menger's theorem connects maximum edge-disjoint paths to the minimum edge cut in a graph. We need to prove a theorem like this one in a formal but convincing manner. An algorithm converts an input into a useful output. For example an algorithm is to be designed to find maximum edge-disjoint paths as well as the minimum edge cut in a given input graph. Many books tilting towards graph theory do not emphasize graph algorithms. Similarly a number of graph algorithm books ignore the theory part of this course namely graph theorems and their proofs. Unlike these books we operate on all the three layers, thus reinforcing understanding at various levels. In fact we try our best to design constructive proofs which not only prove a connection but also provide an algorithm to find something useful. For example we use constructive techniques to find as well as prove that the maximum number of edge-disjoint paths are equal to the size of the minimum cut in a graph.
What do we emphasize?
3
Most of the other textbooks use a historical perspective of how and when a graph algorithm was discovered. For example Dijkstra designed a greedy algorithm to find shortest path between two vertices in a weighted graph. Similarly Prim designed a greedy algorithm to find a minimum spanning tree in a weighted graph. The historical perspective encourages one to teach these two algorithms in isolation without making any connections between the two algorithms. We think it is important from a learning perspective to integrate as well as differentiate concepts and techniques especially the ones which solve nearly identical problems. We thus encourage our readers to transform one algorithm (Prim's algorithm) into another (Dijksta's algorithm). Similarly, it is possible and desirable to transform one proof into another. For example we encourage our readers to transform the techniques used to prove that Dijkstra's algorithm really finds shortest paths into techniques to prove that Prim's algorithm really finds a minimum spanning tree in a weighted graph. We think that not only the similarities but also the differences should be highlighted between two almost similar concepts or algorithms. For example Prim's algorithm can find a minimum spanning tree or a maximum spanning tree (after a minor modification) and can handle positive or negative edge weights while Dijkstra's algorithm is unable to do so.
1.2
What do we emphasize?
The single most prominent feature which distinguishes this book from other books in this field is an emphasis on transformations (or reductions). We expect our readers to first think of a non graph problem in terms of a graph. (see Fig. 1.1). We then encourage our readers to transform that non-graph problem into a graph problem. We also encourage them to transform a graph problem into another graph problem. Similarly we encourage them to transform one theorem into another theorem. We provide platforms where a learner is provoked to transform one proof technique (or an algorithm) into another proof technique (or an algorithm). We think of this course on graph theory & algorithms as a course on (intelligent) transformations. Transforming one concept into another and transforming one graph into another is the driving force behind this exercise (see Fig. 1.2).
4
Introduction
Figure 1.1: World Cup 2010 round of 16 is represented by a graph. (
Figure 1.2: A visual depiction of how an algorithm transforms a graph into another graph while trying to find the shortest paths (Chapter 5).
6
Introduction
1.3
How the book is organized?
The book is organized into eight chapters other than this introduction (see Fig. 1.3). The second chapter provides standard definitions. The third chapter describes how a (non-graph) problem is transformed into a graph problem and encourages the readers to think in graph theoretic terms. The non-graph problems come from diverse fields like operations research, civil engineering, digital logic, distributed computing, molecular biology, and computer science. The fourth chapter focuses on basics of graph theory with an introduction to several concepts and properties related to graphs. The fifth chapter handles some important graph algorithms. Here we effectively start using our emphasis on transformations. We start with a stupid algorithm known as the Bucket Algorithm which provides us with a platform where we encourage our learners to modify and mould the Bucket Algorithm to design various useful graph algorithms. We purposely do not use a historical perspective and avoid describing an algorithm in its published (or polished) form. Instead we encourage our learners to devise cruder versions of an algorithm which are relatively easy to discover and appreciate intuitively. We also provide other tools like a colorful visual puzzle which differentiates between and integrates various shortest path algorithms (see Fig. 1.4). Chapter 6 is the longest chapter in the book. We discuss concepts related to graph connectivity, network flows, matching problems, minimum cost flows, and the circulation problem. We actively demonstrate how a theorem (and its corresponding proof) like Menger's theorem is transformed into Hall's theorem or the Konig's theorem. We also show how one problem (and its solution), for example, the circulation problem is transformed into another problem like the minimum cost-maximum flow problem. This chapter consists of multiple & diverse topics as described before but a conscious effort has been made to make sure that the number of important milestones or bottlenecks in learning remains very small. For example, all the above mentioned topics depend upon one central and crucial idea, that is, if we need to find multiple edge-disjoint paths in a graph then we should reverse the direction of an already found path before finding another edge-disjoint path. In Chapter 7 we discuss necessary and sufficient conditions for Eulerian graphs and the Chinese Postman problem. All the proofs used here are constructive - we not only prove that an Eulerian Circuit exists in a graph with even degree but we also find that circuit using an algorithm. We show
that the Circulation problem (already discussed in Chapter 6) was in fact the Chinese Postman problem in its general form. Both problems and their respective solutions in fact converge into a single problem and a single technique to solve it.
Figure 1.4: In shortest path algorithms (described in Chapter 5) we transform one algorithm into another solving the same problem. In Chapter 6, we transform one problem into another and the corresponding algorithm is also transformed in the process. We describe Hamiltonian graphs in Chapter 7. We discuss sufficient conditions for an undirected as well as a directed graph to be Hamiltonian. Hamiltonian directed graphs are discussed again in Chapter 8. We address strongly connected graphs and components, unilaterally connected graphs, and tournaments in this chapter as well. In case of tournament graphs we come back to the problem of finding a Hamiltonian path and a Hamiltonian
How is the book designed? cycle provided the tournament is strongly connected.
9
1.4
How is the book designed?
We have followed a three step design strategy while writing each section and sub-section of this book: 1. Identify a potential bottleneck in learning a specific concept. Identify its nature using theories of learning & pedagogy (this is done through a statistical analysis of student feedback obtained in the last six years of teaching this course at LUMS, Lahore, Pakistan). 2. Remove the bottleneck by introducing a number of bridging concepts and by drawing suitable number of colored diagrams taking advantage of the role of visualization in learning graph theory. A number of research projects undertaken by students registered in a course titled "Problems of Learning & Teaching " highlighting various problems of learning in the field of graph theory & algorithms have immensely helped us in the design of various sections). Several hundred colored diagrams play a central role in the design of the book. 3. Some text is added where needed in order to supplement diagrams unlike other books where diagrams supplement text.
1.5
Some salient features of the book
1. Discovery based learning is practised by first asking provoking questions before actually describing an algorithm or a theorem. 2. Instead of describing a concept in its finished & sophisticated form we first describe its cruder version which is easy to discover and appreciate 3. Emphasis on prior knowledge, its usefulness and limitations 4. An integrated approach where concepts, algorithms, and theorems reinforce learning and understanding 5. Starting with simple and easy to use building blocks which are used by a learner to construct more sophisticated concepts, algorithms or elaborate proofs
10
Introduction 6. We always make comparisons highlighting similarities as well as differences between concepts, algorithms as well as theorems 7. Constructive proofs with the help of algorithms 8. Complexity of learning under our control because of a conscious effort to keep it under limits. 9. We sometimes encourage our readers to make errors as we think that making an error is a step towards meaningful learning. We then encourage our learners to appreciate by themselves the repercussions of that error and in the process find an alternate path to solve the problem correctly.
10. We use a number of tools from the science of learning, e.g., concept maps.
1.6
What tools do we use?
We occasionally use concept map as a tool for better learning which enable a learner to explicitly make connections between concepts as shown in Fig. 1.5. The book has been used several times to teach a graduate level course on Graph Theory & Algorithms at LUMS. It has also been used to teach a similar course at the Virtual University of Pakistan. At the VU, we have used a 3person drama format instead of a single person monologue. Yasser Hashmi, Komal Syed and I tried to discuss and debate various issues concerning graph theory & algorithms using this book (see Fig. 1.7). A lot of technology tools are nowadays available for synchronizing class room video with multimedia slides and lecture notes (synchronizing annotations for educational multimedia). Some of the VU lectures along with Power Point slides are available on Synote, and can be viewed after getting an account on
1.7
A Possible Sequence
A possible sequence of lectures covering this book is given below in a typical 3-credit course in a semester system at a higher under-graduate or early graduate level.
A Possible Sequence
11
Figure 1.5: A concept map showing a number of relevant concepts and a number of theorems which relate different concepts (taken from Chapter 6). We use several visualization tools in order to enable the learner to visualize the working and the subsequent time complexity of an algorithm as shown in Fig. 1.6.
12
Introduction
Figure 1.6: Visualizing how an algorithm changes a graph and what price is paid in terms of number steps performed. Each edge in the pink graph exactly corresponds to one step in the algorithm.
. This helps a learner to visualize the time complexity of an algorithm.
A Possible Sequence
13
Figure 1.
.7: A 3-person drama format for teaching Graph Theory & Algorithms at the Virtual University of Pakistan.
.14
Introduction
Figure 1. A learner has the facility of expressing and sharing his or her feedback regarding the understanding of a concept in this software platform.com.synote.8: Virtual University video lectures along with Power Point slides are available on www.
Usually the size of the vertex set V is represented by p while the size of the edge set E is represented by q. Similarly an edge coming out of a vertex and terminating at the same vertex (known as a self loop) is not allowed.some of the pair of vertices may be connected by directed or un-directed links which are known as edges. Thus a graph consists of vertices and edges. We may represent the set of vertices by V (G) and the set of edges may be represented by E(G). A weighted graph is one where there may be a weight associated with each edge of the graph. (Please see Chapter 4 for more details)
Adjacent Vertices
A vertex u is said to be adjacent to vertex v if there is an edge {u. (Please see Chapter 4 for more details)
Regular graph
An un-directed graph G is regular if the degree of each vertex is the same. (Please see Chapter 4 for more details)
. In M ulti − graphs we allow parallel edges as well as self loops. v} in graph G. Please note that the set E is a set of pairs of connected vertices. It is a set of ordered pairs in case of a directed graph and a set of un-ordered pairs in case of an un − directed graph. (Please see Chapter 4 for more details)
Degree of a vertex
Number of edges connected to a vertex x is known as the degree of vertex x in an un-directed graph G.20
Basic Definitions in Graph Theory and Algorithms
Graph
A graph G is made up of a (non empty) set of objects called vertices or nodes . (Please see Chapter 4 for more details)
Multi-graphs
We assume that multiple edges (known as parallel edges) between the same two vertices are not allowed in a graph.
It is known as a star graph because it looks like a star with rays of lighy coming out? (Please see Chapter 4 for more details)
Chain or Cycle graph
A connected graph G such that the degree of each vertex is exactly 2. (Please see Chapter 4 for more details)
Isomorphic Graphs
Two graphs G and H are isomorphic if graph G becomes equal to graph H by some relabeling of vertices of graph H. v} in graph H and vice versa then the two graphs G and H are equal. Please note that two equal graphs are always isomorphic but two isomorphic graphs may not be equal.21
Star graph
A graph of p vertices in which one vertex has degree equal to p−1 while every other vertex has a degree equal to 1. (Please see Chapter 4 for more details)
.(Please see Chapter 4 for more details)
Bipartite Graph
A graph G is bipartite if the vertex set V (G) can be divided into two subsets (or partites) A and B such that every edge in G connects a vertex in partite A and a vertex in partite B. (Please see Chapter 4 for more details)
Equal Graphs
If for every edge {u. v} in a graph G there is an edge {u. (Please see Chapter 4 for more details)
Complete Graph
An un-directed graph G is complete if there is an edge between every pair of vertices of graph G.
(Please see Chapter 4 for more details)
Transpose of a directed graph G
T ranspose T (G) of a directed graph G is obtained by reversing the direction of each edge in the directed graph G. v} in c(G) if and only if there is no edge {u. (Please see Chapter 4 for more details)
Automorphism
If a permutation p of a graph G creates a graph H which is equal to graph G then that permutation is known as an automorphism of graph G. We may be more interested in the non trivial permutations of graph G? It may be possible for a certain category of graphs that the only automorphism is the trivial (identity) permutation? Please note that if p is an automorphism of graph G then the permutation p2 is always an automorphism of graph G. The identity permutation is always an automorphism . (Please see Chapter 4 for more details)
Walk
You can walk on the edges of graph G edges starting from vertex u and ending at vertex v traversing different edges and vertices. A walk is open if vertex u and v are
. v} in G. In a walk you can traverse an edge more than once. (Please see Chapter 4 for more details)
Complementing Permutation
If a permutation p of vertices of graph G creates a graph H which is the complement of graph G then the permutation p is known as the complementing permutation of graph G and graph G will be a self complementary graph. (Please see Chapter 9 for more details)
Self Complementing graph G
A graph G is self complementing (SC) if graph G and its complement are isomorphic to each other.it is known as a trivial permutation.22
Basic Definitions in Graph Theory and Algorithms
Complement c(G) of an un-directed graph G
The complement c(G) of graph G is a graph with as many vertices as in G and an edge {u.
Thus a trail is always a walk but it is not the other way round. In a weighted graph G the minimum length is measured in terms of sum of weights of all edges in the u − v path. A closed trail . (Please see Chapter 4 and 7 for more details)
Path
If neither an edge nor a vertex is repeated in a walk starting from a vertex u and ending at vertex v then the walk is known as a path. (Please see Chapter 4 and 5 for more details)
Shortest path
Among all paths between vertex u and vertex v. A path is always a trail (or a walk) but it is not the other way round. Please note that it is possible to traverse a vertex more than once but an edge should not be traversed more than once in a trail or in a circuit. It is a closed walk if vertex u and v are the same.that is when vertex u and v are the same then it is known as a circuit. (Please see Chapter 4 and 5 for more details)
.that is the direction of that edge. (Please see Chapter 4 for more details)
Trail and Circuit
If no edge is repeated in a walk from a vertex u to a vertex v then the walk is known as a trail. the one with minimum length is known as the shortest path between vertex u and vertex v. Please note that in an un-directed graph you can traverse an edge in both directions but in a directed graph you can traverse an edge in only one direction . A cycle is a circuit but a circuit may not be a cycle as no vertex should be repeated in a cycle. In an unweighted graph G the minimum length is measured in terms of number of edges encountered in the u − v path. (Please see Chapter 4 for more details)
Cycle
If a path is closed that means you come back to the vertex from where you have started then that path is known as a cycle.23 different. It is known as a u − v path.
v) in directed graph D. and it is zero otherwise. (Please see Chapter 9 for more details)
A Unilaterally Connected Directed Graph
A directed graph D is unilaterally connected if there is directed path from vertex u to vertex v or a path from vertex v to u for every pair (u. The Reachable Relation matrix of an undirected connected graph will contain all 1's. The Reachable Relation graph of D can be represented by an adjacency matrix A in which A(u. v) in directed graph D. v) in directed graph D. (Please see Chapter 9 for more details)
Square of a directed graph
The In other words there are no cycles in a
.24
Basic Definitions in Graph Theory and Algorithms
A Connected Graph
An un-directed graph G is connected if there is a path between every pair of vertices of that graph. v) = 1 provided there is a directed path from u to v in D. (Please see Chapter 9 for more details)
Directed Acylic Graph
A directed graph D is directedacyclic (or a DAG) if there is directed path from vertex u to vertex v then there is no path from vertex v to u for every pair (u. (Please see Chapter 4 and 5 for more details)
Reachable relation or Transitive closure of a graph
The Reachable Relation (or the transitive closure) of a directed graph D is another directed graph in which there is an edge from vertex u to vertex v provided v is reachable from u in D. (Please see Chapter 4 and 8 for more details)
A Strongly Connected Directed Graph
A directed graph D is strongly connected if there is a directed path from vertex u to vertex v and a path from vertex v to u for every pair (u.
It is also known as a tree graph provided graph G is connected. In other words a connected graph G is a tree if every edge of G is a bridge edge. It is known as a tournament graph as some (actual) tournaments (like f ootball league are played in the form of a tournament graph? (Please see Chapter 9 for more details)
Disconnected graph
An un-directed graph is disconnected if for any pair of vertices u and v there is no u − v path. (Please see Chapter 4 for more details)
Cyclic graph G
A graph G is cyclic if it contains one or more cycles. v) in directed graph D. (Please see Chapter 4 and 5 for more details)
Bridge edge or Cut edge
An edge {u. (Please see Chapter 4. 5 and 9 for more details)
Acyclic graph or a Tree
An un-directed graph G is acyclic if it does not contain a cycle. Please note that a tree is a connected graph with no cycles while a forest may be a disconnected graph. v} is a bridge edge if its removal disconnects an un-directed graph G.25 directed acyclic graph. (Please see Chapter 9 for more details)
A Tournament Graph
A directed graph D is a tournament if there is a directed edge from vertex u to vertex v exclusive OR a directed edge from vertex v to u for every pair (u. In other words if we put directions on edges in a completely connected un-directed graph G then the un-directed graph G transforms into a tournament (directed) graph D. (Please see Chapter 9 for more details)
A Forest
A (disconnected)graph G with no cycles. (Please see Chapter 4 for more details)
.
In other words if you remove all non bridge edges in graph G then you get a spanning tree of G. (Please see Chapter 5 for more details)
A Binary Tree
A binary tree is a tree such that the degree of each vertex is not more than three. A graph which contains a Eulerian circuit is known as a Eulerian graph. (Please see Chapter 7 for more details)
Set Cover
Given a set of subsets S of the U niversal Set U . what is the smallest subset T of S such that the union of all these sub sets in T covers all elements of U . (Please see Chapter 6 for more details)
. We have already witnessed the subset sum problem (in previous courses) in which we have to select integers (out of a set of integers) such that the sum of the selected integers is equal to a given constant. (Please see Chapter 4 for more details)
Hamiltonian Cycle
It is a cycle in a graph G which traverses each vertex of G exactly once. (Please see Chapter 4 for more details)
A Path graph
A path graph is a tree provided it has two vertices with degree one while all other vertices has degree exactly equal to two. (Please see Chapter 8 for more details)
Eulerian Circuit
A circuit in a graph G such that every edge of graph G is traversed exactly once. and it is a tree. A graph which contains a Hamiltonian cycle is known as a Hamiltonian graph.26
Basic Definitions in Graph Theory and Algorithms
Spanning tree of a graph G
A spanning tree (known as ST ) of a connected graph G contains all vertices of G and some edges of G.
(Please see Chapter 6 for more details)
Maximum Matching
This is a matching in a graph with as many edges as possible? Please note that maximum matching is always maximal. (Please see Chapter 3 and 6 for more details)
Edge Cover
The Universal set is the set of all vertices in a graph. In other words a matching is a set of non-adjacent edges in a graph G. What is the smallest subset of vertices of the graph that covers all edges? (Please see Chapter 6 for more details)
Independent (Vertex) Set
It is the largest subset S of vertices of a graph such that no pair of vertices in S has an edge in between. Is there a connection between the vertex cover and the independent set? (Please see Chapter 3 and 6 for more details)
Matching (Independent Edge Set)
It is a subset of edges in graph G such that no two edges in the subset has a common vertex in G. Then the edge cover is the smallest subset of edges.27
Vertex Cover
The Universal set U is the set of all edges in a graph. as compared to the number of vertices in a graph? (Please see Chapter 6 for more details)
. that is.Thus it is a matching of maximum size. (Please see Chapter 6 for more details)
Maximal Matching
This is a matching in which more edges cannot be added in the existing matching to increase the size of this matching. no two edges in the subset share a common vertex. The edges in this subset are also known as independent edges. It may be possible to increase the size of the matching by first discarding the initial matching edges. which covers all vertices? How small (or big) can the size of the edge cover become.
(Please see Chapter 6 for more details)
Minimum Edge-Cut or MinCut
A set of (minimum number of) edges which if removed will disconnect a special vertex s from another special vertex t in a graph G. Every perfect matching is both maximum and hence maximal. the term complete matching is used for it. (Please see Chapter 6 for more details)
Bridge edge or Cut edge
An edge {u. That is. In some literature. (Please see Chapter 6 for more details)
. (Please see Chapter 6 for more details)
Edge Connectivity
It is the minimum number of edges which if removed will disconnect an undirected connected graph G. (Please see Chapter 4 and 6 for more details)
Cut vertex
A vertex u is a cut vertex if its removal disconnects an un-directed graph G. (Please see Chapter 6 for more details)
Vertex Connectivity
It is the minimum number of vertices which if removed will disconnect an un-directed connected graph G.28
Basic Definitions in Graph Theory and Algorithms
Perfect Matching
A P erf ect Matching is a matching which covers all vertices of the graph. v} is a bridge edge if its removal disconnects an un-directed graph G. (Please see Chapter 6 for more details)
Minimum Vertex-Cut
A set of (minimum number) of vertices which if removed will disconnect a special vertex s from another special vertex t in a graph G. every vertex of the graph is incident to exactly one edge of the matching.
Please note that the Chinese Postman Problem transforms into the Circulation Problem (and vice versa) for directed graphs. The source vertex may produce flow while the sink vertex sinks flow. We may like to find a f easible flow from vertex s to vertex t. (Please see Chapter 6 for more details)
. One is a source vertex s and the other is a sink vertex t. (Please see Chapter 6 for more details)
The Circulation Problem
We consider a network flow graph in which the incoming flow in every vertex should be equal to the outgoing flow in every vertex. We assume that the Circulation graph is connected if it is un-directed and strongly connected if it is directed.
Feasible Flow in a network Flow Graph
A f easible flow in a network flow graph from a source vertex s to a sink vertex t is one in which flow through every vertex (other than the source and the sink vertices) is conserved (that is inflow is equal to out flow) and flow through every edge is within the prescribed upper as well as lower bounds. an upper bound on flow.29
Network Flow problems in a Network Flow graph D
It is a directed graph D with two special vertices. We may like to find the maximum or minimum feasible flow or maximum f low at minimum cost. Every edge may have an associated lower bound on flow. We need to find a closed walk in this graph such that the total distance covered is minimum in terms of the number of edges of the graph. We need to find a minimum cost f easible f low in the network flow graph (also known as the circulation graph). The problem is solvable in polynomial time for directed as well as un-directed graphs. and a cost function associated with flow. The lower bound on flow through every edge is exactly 1. (Please see Chapter 6 and 7 for more details)
The Chinese Postman Problem
We are given a graph D which is strongly connected if directed and connected if un-directed. The problem is solvable in polynomial time for directed graphs. There is a uniform cost of flow through every edge.
It may happen that out of these multiple known problems. thus showing that a new problem is indeed solvable. possible to make reductions play a positive role: In such cases we transform a problem into one of the solvable problems. the problems are taken from digital logic. The real challenge is thus to find an intelligent transformation into a simpler problem. As demonstrated in this chapter. Usually reductions are used in such a negative context especially in the field of complexity theory.1. There may be good chances that you are able to transform (or reduce) your unknown problem into one of the solvable problems in graph theory. A more desirable option is to better understand the problem in terms of reducing it into one of the known problems in computer science. molecular biology. one of the problems may be relatively simple to solve while the other known problem may be a hard one. thus showing that the new problem is as hard as some of the known hard problems. however.1
Introduction
When we face a real life problem then one possibility is to solve it right from scratch. We shall discuss a number of diverse problems in this chapter.
3.2
Reducing One Problem into Another
A reduction is a transformation of one problem into another problem. it is very much possible (and in fact desirable) to reduce an unknown problem into multiple known problems instead of just one known graph problem. Models & Graphs: Why Study Graph Theory
3. 3. yet each of them is transformed into a graph problem. 3.32
Problems. civil works. Transforming one problem into another requires that each instance of the new problem should be transformed into instances of the old problem. It is important to note the total time complexity of solving a new problem
. Once your problem is transformed into a known problem. Sometimes we reduce a known hard problem into the new problem. It is. and the field of operations research. distributed computing. We show such a transformation in Fig. Even if you end up reducing your problem into one of the hard (or unsolvable) problems – you certainly get a better insight. we then solve the old problem using a known algorithm and then again transform its results to obtain the final solution of the new problem. the complexity of solving your unknown problem will depend upon the complexity of your transformation as well as that of solving the known problem (see Fig.1). Almost all these problems do not seem to have a any relationship with graphs.
In complexity theory such problems are known as P problem. In complexity theory such problems are known as NP-complete problems. Reduction of an NP-complete problem into a P problem: Does it mean that an NP-complete problem has become a P problem? Why?
.Reducing One Problem into Another
Inputs
33
Outputs
Algorithm No. Reduction of a P problem into an NP-complete problem. The inputs for the new problem should be transformed into the inputs of the old problem. 2 Algorithm No. 2. Does it mean that a P problem has become an NP-complete problem? 4. Similarly the outputs should also be transformed. after reducing it into an old problem will be time complexity of the known algorithm (needed to solve the old problem) plus the time complexities of Algorithm No. We shall categorize a reduction into the following four categories: 1. 2. 1 and that of Algorithm No. 1
Inputs Outputs
Figure 3. Reduction of a hard problem (a problem for which) a polynomial time algorithm is not yet designed) into another hard problem. Reduction of a problem (with an existing) polynomial time algorithm into another such problem.1: Reducing a new problem into an old problem. 3.
3. 3.3
The Satisfiability Problem in Logic Circuits
Satisfiability in logic circuits is the problem of finding if we can assign 0 or 1 to the input variables so as to make the output of the logic circuit equal to 1. In the 3CNF Satisfiability problem (or the 3-SAT Problem). we are given a Boolean expression in disjunctive normal form (DNF). Models & Graphs: Why Study Graph Theory
We shall provide (or discuss) at least one example from each category in this chapter.
a
b c
a b c
a b c
a b c
Figure 3. the 3-Satisfiability problem is to find if it is possible to assign 0 or 1 to its inputs that will make the output equal to 1.3). If no assignment of input variables can make the output 1 then we claim that the logic formula (or the circuit) is not satisfiable.2). in terms of logic circuits it is the OR output of clauses of AND gates with exactly 3 inputs (see Fig. in simple words it is the AND output of clauses of OR gates with exactly three inputs (see Fig. we are given a boolean expression in 3-conjunctive normal form. have explored assignments of inputs for which the output of the logic circuit is 1 – but many
. Again we need to assign input variable such that the output of the circuit is 1. Given such an expression.
3. In the 3DNF Satisfiability Problem. A lot many of you might have played with logic circuits. The output is 1 for the selected inputs shown in orange color.34
Problems.2: A logic circuit consisting of OR gates (each with 3 inputs) and one AND gate.
Surprisingly applying DeMorgans laws it is possible to convert a 3-CNF Boolean expression in terms of a 3-DNF – giving a false impression that we can reduce an NP-complete problem into a P problem. A 3-CNF circuit is shown in Fig.1.The Satisfiability Problem in Logic Circuits
35
of you may not have realized (or truly appreciated) that the 3-CNF problem is one of the NP-complete problems while the 3-DNF is a solvable problem (in polynomial time). 3.
3.3: A logic circuit consisting of AND gates (each with 3 inputs) and one OR gate. We shall reduce the 3-SAT problem into the Independent Set Problem in graphs.1
Reducing a 3-SAT Problem into an Independent Set Problem
Let us start with an example from Category 1. Show that it is indeed possible to solve this problem in polynomial time (in fact in linear time).2. 3.1.2.1. A 3-DNF circuit is shown in Fig. Try to use a similar technique to solve the 3-CNF Satisfiability problem. In case of 3-DNF problem the output of the circuit will be high provided the output of any AND gate is high. Design an efficient algorithm to solve the 3-DNF Satisfiability problem.
OutPut
a
b
c
a
b c
a b c
a b c
Figure 3.3. the
.1.3.4. Remember in the 3-CNF problem one has to select inputs such that the output of each OR gate is high so as to satisfy the output. We shall discuss this issue (in detail) in the coming paragraphs. The output is 1 for the selected inputs shown in orange color. The purpose of this problem is to appreciate the inherent hardness of this problem. 3. Thus in Fig. It will not work. Problem Set 3. Problem 3. Problem 3.
the 3-Satisfiability problem is to find if it is possible to assign 0 and 1 to its inputs that will make the output equal to 1. Please note that at least one input from each OR gate should be 1 in order to pass the test for Satisfiability. The problem is to find if it is possible to select one vertex from each triangle such that no two selected vertices have an edge in common (they should not be adjacent).
. 3. in simple words it is the AND output of clauses of OR gates with exactly three inputs (see Fig.4 is the Independent Set problem in a special graph consisting of k triangles with edges connecting certain vertices within different triangles as shown in Fig.2). The figure also shows a combination of inputs (shown in orange color) for which the output of this logic circuit is 1. 3.36
Problems. The old problem in Fig.5.4: Reducing the 3-Satisfiability problem into Independent Set problem in a graph. It is obvious that for the size of the Independent Set to be equal to k we have to select exactly one vertex from each triangle such that no two selected vertices are adjacent. 3.
Inputs Outputs
Algorithm No. 2 Algorithm No. 1
Inputs Outputs
Figure 3. In other words we need to find if the size of the Independent Set (a set of vertices with no common edges) in this graph is equal to the number of triangles in the graph. Models & Graphs: Why Study Graph Theory
New Problem is the 3-CNF Satisfiability problem: we are given a Boolean expression in 3-conjunctive normal form. Given such an expression.
the output is high for the given logic circuit? Input : A Boolean formula (or circuit) in 3-CNF. Output: Yes/No. If Yes then a combination of inputs for which the output is high. Output: Yes or No. We consider this as a known problem in graph theory
. Input : A graph consisting of k triangles. We consider this as a new problem: You need to transform it into a known problem in graph theory?
Algorithm 2: The Independent Set Problem: Find if the size of the independent set in a given graph is k. 1.5: A graph consisting of k triangles where some vertices from different triangles are adjacent.
Algorithm 1: The 3-SAT Problem: Find for what inputs.The Satisfiability Problem in Logic Circuits
37
Figure 3. If Yes then select a vertex from each triangle such that no two selected vertices is adjacent. 1. We need to find if the size of the independent Set is equal to k.
.6: The 3-Satisfiability problem (top) is reduced into the Independent Set problem in the graph shown in the bottom.38
Problems. Models & Graphs: Why Study Graph Theory
OutPut AND
a ¬b c
a b ¬c
¬a ¬b c
¬a ¬b ¬c
c a
¬c
c
¬c
¬b
a
b
¬a
¬b
¬a
¬b
Figure 3. Orange vertices in the bottom graph provides a solution to the independent set problem in the bottom graph while their orange counterparts (input variables to the logic circuit) provides a satisfiable solution to the 3-SAT problem in the top circuit.
2
Reducing the 3-CNF Satisfiability Problem into the 3-DNF Satisfiability Problem
We know that it is possible to convert a 3-CNF Boolean formula into a 3DNF formula.2.The Satisfiability Problem in Logic Circuits
39
It is obvious that a possible solution of the Independent Set problem in the graph provides a combination of inputs for which the output will be high in the logic circuit. 3. 3.3.2.1) and you will be able to resolve this contradiction your self. Problem 3. We also know that the 3-CNF Satisfiability problem is NP-Complete while it is possible to solve the 3-DNF Satisfiability problem in polynomial time. Such a conversion is done after drawing the truth table or the K-Map of the 3-CNF expression as shown in Fig. Try to use a similar reduction to reduce the 2-SAT problem into the independent set problem in a graph. Problem 3. 2 as shown in this diagram. Is there a contradiction somewhere? Design the corresponding conversion algorithms (Algorithm No. Problem Set 3. Discuss why this may or may not be possible.2.
3. the solution of any one problem implies a possible solution for the other.6.1. In the next section we shall talk about a Category 2 reduction. Please design Algorithm No. 1 & 2 of Fig. Thus if the size of the Independent Set in the graph is equal to the number of OR gates in the logic circuit then it is possible to find a combinations of inputs which will make the output high.1 and Algorithm No. 3. Discuss why this may or may not be possible. Unfortunately both these problems belong to the class known as NP-Complete problems. and depicted in Fig.7. In fact the reduction goes in both directions. Try to use a similar reduction to reduce the 3-DNF problem into the independent set problem in a graph. Problem 3.3. 3.2.2.
.4. We have already talked about the 3-Sat problem and its reduction to independent set problem as modeled in Fig.
. For every OR gate in the Boolean expression with inputs x and y.The Satisfiability Problem in Logic Circuits
41
3. This problem and its possible solutions are discussed in detail in Chapter 9. in other words we need to check if there is a directed path from vertex x to vertex y and from vertex y back to vertex x. The 2-SAT problem is similar to the 3-SAT problem except that now each of the OR gates have two inputs rather than three. 2. x and y. In fact this problem can be reduced to another graph problem which is solvable in polynomial time.3
Reducing the 3-CNF Satisfiability Problem into another graph Problem
Here we intend to discuss a reduction of the 3-SAT problem into the Clique problem in graphs. It is left as an exciting exercise for the reader?
3. Path finding algorithms are described in Chapter 5. belong to a single strongly connected component. we claim that the given Boolean expression (or the logic circuit) is not satisfiable (that means for any combination of input variables) if and only if any vertex x and its complement vertex ¬x in graph D belongs to the same strongly connected component. We still need to find a combination of inputs for which the output of the circuit is 1.4
Reducing the 2-CNF Satisfiability Problem into a Graph Problem
We have already talked about reducing a 2-CNF Satisfiability problem into an Independent Set problem in graphs. For every variable x in the Boolean expression we create two vertices with labels x and ¬x in the directed graph D. we need to find if any two given vertices. We shall partially justify this reduction and leave the rest of the details as an interesting problem for the reader. Once we have a constructed directed graph D. we add two directed edges: One directed edge from vertex ¬x to vertex y and another directed edge from vertex ¬y to vertex x. we construct a directed graph D according to the following rules: 1. We shall describe this Category 2 reduction briefly in this section.3. Given the Boolean expression or the logic circuit.3. The graph problem deals with directed graphs.
It is transformed into a directed graph shown in
. Check if there is a path from vertex x to y and from y to x in the directed graph D.8: The 2-CNF Satisfiability problem is reduced into a graph problem. 1. and if the answer comes out to be NO in each case then the given Boolean expression is satisfiable otherwise not. We shall provide some hints in this regard and leave the rest as a problem for imagination of the reader. If any vertex x and its complement in the graph (shown in the right diagram) belongs to the same strongly connected component then the Boolean expression corresponding to the circuit (shown in the left diagram) is not satisfiable. This requires a deeper understanding and appreciation of different concepts involved in this reduction. Output: Yes/No. Models & Graphs: Why Study Graph Theory
¬a b ¬a ¬b c
OR
¬c
OR AND OR
Convert
a
b
a
¬c
¬b
¬a
a
OR
c
Figure 3. and two vertices x and y. Algorithm 3: Find if two given vertices belong to the same strongly connected component in a directed graph D. If yes then vertices x and y belong to the same strongly connected component otherwise not.42
Problems.9. We apply Algorithm 3 for every vertex x (and its complement ¬x) in graph D.
A 2-CNF expression consisting of a single OR/AND combination is shown in the left diagram of Fig. 3. Input : A directed graph D. If the Boolean expression is satisfiable then we have to find a combination of input variables for which the output of the logic circuit is 1.
If a = 0 then Output =1
¬a b
OR AND
For what values of input the output is 1
a
b
¬a ¬b
OR
If a = 1 then b=1 then ¬a =1
¬b
¬a
Figure 3. Note that if a = 1 then we face a contradiction in the implication graph.9: A 2-CNF expression with only one OR/AND gate is reduced into a directed graph.10: The 2-CNF Satisfiability problem is reduced into a graph problem. On the other hand if a = 0 then there is no such contradiction and the output of the logic circuit will be 1 for any value of b.
. The directed graph is in fact an implication graph with a truth table shown at the top. Note that the implication graph tells us that if a = 1 then b should be 1 also otherwise the output of the logic circuit will be zero.The Satisfiability Problem in Logic Circuits
43
a
0
Tr Lo uth gi Ta c C ble irc o ui f t
a→ b ¬a OR b ¬b → b ¬a
0 1 1 0 1 1 1 0 1 1 1 0
Im Trut pli h T ca a tio ble n Gr of ap h
0 1 1
¬a b
AND OR
For what values of input the output is 1
a
¬b
b
¬a
If a = 1 then b=1 for Output to be 1
Figure 3.
The truth table of this implication is exactly the same as that of the Boolean expression as shown in this figure. This transformation works both ways: if the circuit is satisfiable then we can select an independent set
. If a = 1 then b should be 1 so that the output of the top OR gate becomes 1. It will be interesting to find if the circuit is satisfiable (as we add more OR gates). But if b = 1 then according to the implication ¬a should be 1 so as to make the output of the second OR gate 1 – but that is a contradiction. We show the same 2-CNF logic circuit in Fig. Again the implication graph is telling us how it is possible to make the Boolean expression satisfiable. But if a = 0 then the implication does not dictate any thing – it means that the Boolean expression is satisfiable for any value of b. Models & Graphs: Why Study Graph Theory
the right diagram of this figure. it is possible to reach from vertex a to vertex ¬a.10. The implication tells us that in order to make the output 1 we shall make b = 1 if a = 1. it is also possible to reach from vertex ¬a to a as shown by the closed path shown in red color. We show various 2-CNF logic circuits for varying number of OR gates and the corresponding implication graphs in Fig.11. This makes sense because if both inputs of the OR gate are 0 then the expression will not be satisfiable. and in case it is then what inputs should be applied so that the output is 1 for each logic circuit shown in this figure. A directed edge in this graph from vertex x to vertex y implies that if the vertex x (or meaningfully) variable x is 1 then y is 1. The implication graph for a Boolean expression consisting of two OR gates are shown in Fig. Please note that in the bottom diagram of Fig. The Satisfiability problem in this logic circuit is transformed into a graph problem in the Implication graph shown in the bottom diagram. This implies that the logic circuit is not satisfiable. If it is possible to select one vertex from each adjacent pair such that no two selected vertices are adjacent then we claim that the logic circuit is satisfiable. As you should appreciate this directed graph is not just a directed graph – this is in fact an implication graph. 3.12.11. It should now be possible to design an efficient graph algorithm which operates on the implication graph but which finds the input combination for which the logic circuit output is 1. The Satisfiability problem is also transformed into an independent set problem where each OR gate in the logic diagram now corresponds to two adjacent vertices (known as an adjacent pair) instead of a triangular graph.44
Problems. 3. 3. 3.
In the bottom diagram this is not possible. In the top three graphs (shown on the right side).11: The 2-CNF Satisfiability problem is reduced into a graph problem.The Satisfiability Problem in Logic Circuits
45
a
¬a b
AND OR
Convert
b
¬b
¬a
¬a b
OR AND
Convert
a
b
¬a ¬b ¬a b
OR
¬b ¬c
¬a
OR
a
AND
Convert
b
¬a ¬b c
OR
¬b
¬a
a
¬a b ¬a ¬b c
OR
c
OR
¬c
OR AND OR
a
b
a
¬c
¬b
¬a
a
OR
Convert
c
Figure 3.
. vertex a and ¬a do not belong to the same strongly connected component – it means the output of the corresponding logic circuits is satisfiable.
This implies that the independent set problem (in some special graphs) can be reduced into an implication graph problem.46
Problems. Models & Graphs: Why Study Graph Theory
with size equal to the number of OR gates in the logic circuit. The same problem is reduced into a graph problem in an implication graph shown in the bottom diagram.12: The 2-CNF Satisfiability problem is reduced into an independent set problem shown in the top diagram.
an sf o
Tr an sf or m
rm
. It is now obvious that the independent set problem in the graph (which was derived from a 2-CNF logic circuit) can be transformed into a path finding problem in an Implication graph. Given a general graph how can you determine that this graph in fact represents a 2-CNF logic circuit? If it does represent such a logic circuit then we can solve this problem after reducing it into an implication graph?
¬a b ¬a ¬b c
OR
Transform
¬a
b
OR AND
¬a
¬b
a
¬c
OR
a
c
a
OR
a
¬c
Tr
¬c
a
b
¬b
¬a
c
Figure 3. Describe an efficient algorithm which finds an independent set in such a graph using an Implication graph.
the lower limit will be the length of the longest path in the graph. So that is an upper limit on time to complete the job. We also need to identify those activities which are critical: that means increasing the time duration of these activities will certainly increase the total completion time.e. In the table below. If we are unable to schedule any activity in parallel then the total time needed to construct the house will be the sum total of (the duration of) all activities.1. we have described a number of tasks. we assign the incoming edges the same weight as the vertex. Now for each vertex. 3. How will you model this problem in graph theoretic terms and then solve it? We start with an example of constructing a house as defined in Table 3.4
An Activity Scheduling Problem
The problem of activity scheduling is described in simple words. 3. their duration (in days) and what tasks must be completed before they can begin (prerequisites) ( the tasks they represent have no pre-requisites). Then we add a special vertex s and connect it to all such vertices that have no incoming edges (i. We need to find out the minimum possible time to complete a house according to the activities described in the table above. with the duration of the activity as a weight on the vertex. A longest path in this graph (see bottom diagram of Fig. We represent each activity in the Table usig a vertex.htm). The problem is then transformed into a graph problem as shown in Fig.An Activity Scheduling Problem
47
3.com/projex/PERT/aoa. we add a directed edge to all the vertices such that the task they represent are dependent on the task represented by v. Then there are activities which can run in parallel while some are strictly sequential. their IDs.15. On the other hand there are certain activities which if delayed do not necessarily increase the total completion time.
.waa-inc.15) corresponds to minimum amount of time needed to build the house? Why? The underlying assumption is that we should be able to schedule as many activities in parallel as possible by the pre-requisite relationships. For each Task vertex v.
8
Table 3. This kind of computing has a typical serial or chain like structure as shown in Fig. Algorithm 4: Find Minimum Time to Complete the Job.5
A Dual Machine Serial Processing Environment
Chain like program graphs is common in many digital signal processing applications.48 Task ID 1 2 3 4 5 6 7 8 9 10
Problems. A new problem: You need to transform it into a known problem in graph theory?
The challenge is to transform one problem (Algorithm 2.4) into a text book problem (Algorithm 2. In such applications each packet or frame of data may be processed through various transforms in a fixed sequence. Output: Minimum Time to complete the job. one module may run faster on one processor while the same module may take much longer time on a different machine in a typical distributed
.16. 9 2 6 5.1). and Pre-requisites (Table 2. 3. Duration. Models & Graphs: Why Study Graph Theory Task Description Clear land Lay foundations Build walls Electrical wiring Plastering Landscaping Gardening Interior work Roof Handover Duration 14 28 42 21 21 20 10 35 50 00 Prerequisites none 1 2 3 4. their IDs. Input : Tasks. Each program module in the chain structure may have a very different computation requirement. their duration (in days) and what tasks must be completed before they can begin (prerequisites). 9 3 7.1: We have tabulated a number of tasks.5)
3. 1.
3. Let us concentrate on a sub-problem in order to appreciate the intricacies of the problem. For example the execution cost of module 2 on processor A is 20 while it is 90 on processor B. it is zero if the two modules are assigned to the same machine.16.
. The execution cost of each module on either processor is indicated below each module. Models & Graphs: Why Study Graph Theory
10 6 20 21 21 35 00 7 00
14
1
28
2
42
3
4
5
8
10
50 9
21
35
Figure 3.50
Problems. In such a distributed heterogeneous computing environment we should somehow take advantage of the diverse and special characteristics of each machine as assigning all modules of the program to one machine may not be an optimal solution. A module can be processed on either processor A or on processor B but only one processor is active at any time. the total time of computation is the sum of total execution times plus the total communication times. A straight forward greedy solution would be to assign a module on a processor where it is least costly but then if two modules with a lot of communication traffic in between are assigned to different machines then it will again degrade the over all performance. The top diagram of the same figure shows a chain structured modular program consisting of four modules.15: Finding the longest path corresponds to the optimal scheduling of activities heterogeneous computing environment consisting of two processors as shown in the bottom diagram of Fig. The communication cost between module 2 and 3 is 50 provided the two modules are placed on different machines.
it is zero if the two modules are assigned to the same machine. The two machines have different capabilities. (Top) A chain structured modular program consisting of four modules (or nodes). Module 1 as well as 2 is processed on processor B: Then Cost of Execution will be 50+90 = 140. For example the execution cost of module 2 on processor A is 20 while it is 90 on processor B. The execution cost of each module on either processor is indicated below each module. 2. Module 1 is executed on machine A while module 2 is processed on
. 3. There are basically four possibilities for the first pair of adjacent modules to be assigned onto the dual processor system: 1.16: (Bottom) Processor A and processor B connected with a high speed communication link. Module 1 as well as 2 is processed on processor A: Then Cost of Execution will be 70+20 = 90. while there will be no Cost of Communication as the two modules are executed on the same machine.A Dual Machine Serial Processing Environment
51
Communication Costs
10 Execution Costs Processor B 50 Processor A 70 90 20
50 30 40
20
0 20 80
Processor
Processor
Figure 3. The communication cost between module 2 and 3 is 50 provided the two modules are placed on different machines. while there will be no cost of communication as the two modules are executed on the same machine.
3. while there will be a Communication Cost = 10. Problem 3.1. Total Cost = 80. Find the total cost if all modules are assigned to machine B. module 2 on processor A. Models & Graphs: Why Study Graph Theory B. For example assign module 1 on processor B. How to model this problem in graph theoretic terms? Problem 3. Cost of Execution will be 50 + 20 = 70. 4.6 and Algorithm 2. Cost of Execution will be 70 + 90 = 160. it is not a distributed assignment. Problem 3.3. The challenge is to find the optimal solution for the entire problem efficiently (without enumerating all possibilities as we did for the sub-problem). Find the Optimal Assignment by hit and trial (the size of this problem is small enough)? It should be less than (or equal to) the ones found in 1.3. Find the total cost if all modules are assigned to machine A. Please note that the cost consists of two parts. while there will be a Communication Cost = 10. How about if we (initially) ignore communication costs and assign a module on a machine where it is least costly. Now calculate the Total Cost of this assignment after taking into account the communication costs. Total Cost = 170. Problem 3.3. and the cost of communication between two adjacent modules provided the two modules are assigned to different machines.3. Module 1 is executed on machine B while module 2 is processed on A. Problem Set 3. 2.2.3.
. The total time of computation is the sum of total execution times plus the total communication times.3. This is also sequential processing but this time on machine B.5. we design Algorithm 2. cost of execution of a module on a processor. How about if instead of a chain structure.8. Hints are provided in the following figures. Problem 3. and 3. Problem 3. This is standard sequential processing on machine A. and so on.
The last option provides the optimal solution in terms of minimum completion time for the sub-problem.6. There is no communication cost if the two (adjacent) modules are assigned to the same machine.52
Problems.3.4.
This is equivalent to traversing a path which passes though the top vertices or the bottom vertices in a graph with two dummy vertices. (Bottom) The possibility of some modules assigned to processor A while other modules to processor B is shown with a zigzag path between the two dummy vertices.17: (top) We show the possibility of all modules assigned to either processor A or to processor B. A shortest path corresponds to an optimal assignment.A Dual Machine Serial Processing Environment
53
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6 and then use the text book Algorithm 2. ACAA.9 in order to solve the Minimum Cost Assugnment Problem. ACAG. T ACA. With a given unknown DNA sequence (of four letters A. CACA. it does not provide us about the order of the strings or their position in the DNA fragment. Fig. 3. The challenge is to tranform this problem into a know problem in graph algorithms.
3. 3)) will be equal to {CAT. AGT }. C.22. For example if the (unknown) sequence s = CAT GAGT then a set of all substrings of length 3 that s contains (known as Spectrum(s.6
Sequencing by Hybridization in Computational Biology
The Sequencing by Hybridization technique depends on the hybridization of target DNA fragment against a very large array of relatively short probes. an array (also known as the Gene Chip) tells us about all sub-strings of a fixed length that the DNA sequence contains. AT G. AACA}.7 in order to solve the Minimum Cost Assugnment Problem. G.20.Sequencing by Hybridization in Computational Biology
57
The challenge is to design Algorithm 2.21. 4) = {AT AC. 3. The following figures will provide hints to make multiple transformations. The sequencing by hybridization problem in molecular biology (or simply the spectrum problem) can be transformed either into a Hamiltonian path problem or an Euler path problem in directed graphs depending upon whether we map every element of the spectrum into a node or an edge of a directed graph respectively. T GA.
.8 and then use the text book Algorithm 2. 3. CAT A. Then we can find a sequence s as shown in Fig. The elements in the spectrum may not appear in the same order and the challenge is to find the string s. CAAC. ACAC. ACAT. In fact there are multiple strings possible (s1 = ACACAACAT ACAG & s2 = ACAT ACAACACAG) with the same spectrum as illustrated in these diagrams. and Fig. given its spectrum. GAG. Assume that Spectrum(s. & T ). Another challenge is to design Algorithm 2.
A new problem: You need to transform it into a known problem in graph theory?
Algorithm 11: Find a Hamiltonian Path in G Input : A graph G Output: A Hamiltonian Path in G 1. Output: Sequence s. A text book problem. see Chapter 6
Transform the sequencing problem into Hamiltonian path and then into an Eulerian path problems (see the figures below)
. l) Input : Spectrum(s. A text book problem. 1.58
Problems. l) of an unknown s. Models & Graphs: Why Study Graph Theory
Algorithm 10: Find a (Correct) Sequence s given its Spectrum(s. see Chapter 7
Algorithm 12: Find an Euler Trail in G Input : A graph G Output: An Euler path in G 1.
the edges are going to vertices labeled with ACAA.Sequencing by Hybridization in Computational Biology
59
Spectrum(s. T ACA. ACAT. CATA. ACAC. A directed edge exists from one vertex to another provided the last three letters of the first vertex match with the first three letters of the second vertex. CAAC.4) = {ATAC. ACAT. AACA} and show a directed graph in which every element of the spectrum was mapped onto a vertex. ACAC. ACAG. ACAG and ACAT . AACA}
ACAC ATAC ATAC CATA ACAT ACAG TACA
CACA
ATAC
TACA
ATAC TACA
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ACAG
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CAAC AACA
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Figure 3. 4) = {AT AC.20: We start with a Spectrum(s. ACAG. ACAC. ACAA. CACA. ACAA. CACA.
. CAT A. TACA. That is why there are four directed edges emanating from the vertex T ACA. CAAC.
We shall provide you below a certain sequence of questions to guide your search towards a suitable transformation.23: (Top) A Euler path in the graph of Fig. (Bottom) Another Euler path is shown in the same graph. For example in the first example
. Experience can help you a lot in moving forward.20. Does your problem have an inherent graphical structure? If yes then the transformation may be a lot simple. you should also try to find why or why not. this Euler path corresponds to the sequence s1 = ACACAACAT ACAG. we can only rely on heuristics. this Euler path corresponds to the sequence s2 = ACAT ACAACACAG. Note that most of these questions may not have a black and white answer. 3.62
Problems.7
Discussion & Problems
How can we exploit graph theory in order to solve an unknown problem? In other words how can we reduce the unknown problem into a graph problem? Finding the right reduction or transformation is not easy. Models & Graphs: Why Study Graph Theory
AACA
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Figure 3. Please note that in both the Euler paths shown in this diagram there is a vertex which is repeated several times while every edge is traversed exactly once. It is certainly not possible to design an exact algorithm to do this transformation. 1.
3.
3. it is rather impossible to capture its essence without directions. Do you map each element of your unknown problem into a vertex or an edge of a graph? Example 3 is a good illustration of this decision. Thus this example is also reduced to a directed graph problem. 4. Strangely this is equivalent to finding a longest path in a directed acyclic graph as shown in Fig.15. it is much easier to make a special case reduction. only a certain sequence (or order) of letters can correspond to a correct sequence s. In the first example we need to minimize the total completion time of building a house.Discussion & Problems
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we have activities and a pre-requisite relationship among themselves. For example in the second example it is much easier to visualize the special case (in terms of a
. 3. The second example is perhaps more interesting. this naturally leads to a directed graph. This problem is reduced to a decision problem (not an optimization problem) in graph theory whereby we decide if it is possible to find a Hamiltonian Path (or an Euler trail) in a directed graph. Is your unknown problem an optimization problem or otherwise? If it is an optimization problem where you need to maximize (or minimize) some parameter then you need to search for a graph optimization problem. Do components of your problem have a sequence (or an order) which can be transformed into a directed graph? The first example certainly implies a directed graph. l) is given. The activities have weights. The last example is not an optimization problem. in its first transformation it is transformed into a directed graph. Sometimes it is difficult to make a general transformation. these weights can be shifted to vertices if each vertex corresponds to an activity in your graph. 2. we need to find an unknown sequence s for which the Spectrum(s. In the second example we need to minimize the sum total of execution and communication costs of a modular program. If you decide to map each element of the Spectrum onto a vertex then an edge in this graph would represent what (and on what basis)? Will this be a directed edge or undirected? 5. The last example has an implicit directed structure. This problem is reduced to finding a shortest path in a directed graph. it is also reduced into a graph problem applicable to an un-directed graph.
The Marriage Problem is a well-known problem in mathematics as well as in any middle class conservative society.17. In one case you have to design an informal proof and in the other case you have to design a counter example.4. 3. Problem 3. It is a special graph in the sense that an edge between two vertices belonging to the same part (men or women) is not allowed. We show two reductions of this problem. The problem is to find out if it is possible to marry each woman to a man she likes or knows. This reduction depends upon the hypothesis that each vertex-disjoint path between vertex s and t in graph G corresponds to a matching edge in the bipartite graph. A liking between a man and a woman is represented by an edge between the two corresponding vertices.64
Problems. If it is not possible to find such a Perfect Matching then we should perhaps maximize the number of women each of whom is married to man whom she knows.1. (Please note that here we have not accounted for the amount of liking between a man and a woman – we just claim that a certain woman knows or likes or does not like a certain man. Models & Graphs: Why Study Graph Theory graph problem) when all modules are assigned either to processor A or processor B as shown in the top diagram of Fig. and the other is shown in Fig. and a B part consisting of women.25. In the first reduction. The quantitative aspects of this problem will be discussed in the coming problems. This reduction depends upon the hypothesis that each edge-disjoint path between vertex s and t in graph G corresponds to a matching edge in the bipartite graph.4. This special case more or less resembles the activity scheduling problem discussed in the first example.) This problem can be modeled by a graph consisting of each man and woman as a vertex. Please note that the reduction (which ever is correct) not only tells you if a
. The problem is to find out which reduction is correct and which one is false. we assume that we have a collection of men and an equal number of women. 3. One is shown in Fig.24. 3.
Problem Set 3. In the second reduction the problem is reduced to finding maximum vertex-disjoint paths between the same two vertices in the graph G. The graph is shown below. Please note that the graph consists of two parts – an A part consisting of men. each women knows (or likes) some of the men. we reduce the Marriage Problem into a problem of finding maximum edge-disjoint paths between vertex s and vertex t in a graph G. Such a graph is known as a bipartite graph.
27. An alternate way of finding a perfect matching in a bipartite graph is demonstrated in Fig. the reduction maximizes the number of marriages taking place. z) is part of a perfect matching in the bipartite graph. For example the P er(C) of the matrix shown in the figure below will be computed as follows:
P er(C) = a11 a22 a33 + a11 a23 a32 + a12 a21 a33 + a12 a23 a31 + a13 a22 a31 + a13 a21 a32 P er(C) = 0 + 1 + 0 + 1 + 1 + 0 = 3 As you can see each individual term in the expression for Permanent is a permutation – there is a one to one correspondence between each non zero term in this expression and a perfect matching in the bipartite graph. 1. Thus if P er(C) is zero then no perfect matching exists. On the other hand if it is non zero then there will be as many perfect matchings as the value of P er(C). In case a perfect matching is not possible. We reduce the problem into finding the permanent of the matrix C. 3. We need to find if a Perfect Matching exists in a given bipartite graph C. 2. it also tells us which woman to marry whom.Discussion & Problems
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perfect matching is possible. The said edge will be part of a perfect matching if and only if P er(Cyz) is non zero. If a perfect matching does not exist then we need to maximize the number of marriages as before. the problem is how to find one? Assume that the only operation that we can perform is to find a Permanent of the matrix after or before removing an edge of the bipartite graph. Problem 3. The permanent P er(C) of a matrix C is computed like the determinant of C except that the signs of all permutations are positive. We also need to identify which woman is marrying whom. Once we know that a perfect matching exists in the given bipartite graph C.4. How can you find a Perfect Matching in this graph? Discuss briefly. The problem of finding a perfect matching is thus reduced to finding the permanent of a matrix. Here we start with the adjacency matrix C of the bipartite graph C.2. We consider the same Marriage Problem as described before. On the
. We need to find if an edge (y.
We take the permanent of the adjacency matrix and claim that a perfect matching in the bipartite graph exists if and only if the value of the permanent is non zero.68
Problems.
. Models & Graphs: Why Study Graph Theory
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Figure 3.26: We show a bipartite graph and its adjacency matrix shown in the top diagrams.
why? How about if we put random weight on each edge of the bipartite graph? The probability of the value of the determinant becoming zero will depend upon the randomness of the weights assigned to the edges of the bipartite graph. Consider the (minimum weight) assignment problem. For the time being assume that the minimum weight perfect matching is unique. and then we take the permanent of matrix C. So hopefully the determinant of a bipartite graph will not be zero unless there is not a single perfect matching in the bipartite graph. How to find the edges of a perfect matching in this graph? How about using the previous technique?
Problem 3. We shall try to relax this condition later. We are given a complete balanced weighted bipartite graph.70
Problems. The highest power of 2 which divides the value of the permanent is the weight of the minimum weight perfect matching.
4.3. and we need to find a minimum weight perfect matching in this graph – that means a perfect matching in which the sum of weights of all (matched) edges is minimum.
.4. Models & Graphs: Why Study Graph Theory basis of this observation. We know that the determinant of a completely connected bipartite graph is zero.We use the same reduction in this problem as was used in the last problem with one slight modification. All edge weights in the adjacency matrix C are raised to the power of 2. It is much more complex to find the permanent of a matrix as compared to finding the determinant of a matrix – thus we are tempted to check if the determinant is as helpful as the permanent of a bipartite graph? How about if we find that the determinant of an adjacency matrix of a bipartite graph is non zero? Under such conditions. can we make a claim that a perfect matching exists in the graph? How about if the determinant is zero – can we claim that a perfect matching does not exist in the bipartite graph? Remember that the determinant of a completely connected bipartite graph is zero in spite of the fact that every permutation of the vertices of the bipartite graph is a perfect matching. design an efficient algorithm to find all edges belonging to the perfect matching.
3.
2. The value of the permanent will be divisible (with remainder equal to zero) by the weight of minimum weight perfect matching raised to the power of 2 provided we have a unique minimum weight perfect matching in the bipartite graph.Discussion & Problems
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P er(C) = 28+1+8 + 28+8+8 + 28+8+8 + 28+8+2 + 23+1+2 + 23+8+8 = 217 + 224 + 224 + 218 + 26 + 219 = 26 (1 + 217 + 224 + 224 + 218 + 219 ) Raising all weights to a power of two and then finding the permanent of the weighted matrix gives us a powerful reduction in which each individual term in the permanent is in fact the weight of a perfect matching raised to power of 2. What complications can arise in finding the value of minimum weight perfect matching provided such a matching is not unique? 3. Of course you can raise a number to the power of 2 or any other number of your choice. You need not find the value of the minimum weight perfect matching – just make an efficient check if it is unique or not?
. 1. Design an efficient algorithm to find edges belonging to the minimum weight perfect matching. Assume that now there is a possibility that the minimum weight perfect matching is not unique. The only reduction that you can use is to find permanent of a matrix.
Models & Graphs: Why Study Graph Theory
All other edge weights are 8
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Figure 3.28: A minimum weight assignment problem is reduced to evaluating the permanent of a matrix. In Chapter 5 we shall reduce this problem into a well known shortest path finding problem in any directed or un-directed graph.4. We show a weighted graph G in Fig. Using this pp matrix we
. Problem 3. Shortest path finding algorithms are relatively simple and are discussed in Chapter 4.4. Here in this problem we shall discuss a (bizarre) reduction in which we reduce the shortest path problem into the assignment problem.29 where we need to find a shortest path from vertex a to vertex d. The output also includes edges (or vertices) belonging to minimum weight perfect matching. 3.72
Problems. Consider the assignment problem in which we have a complete weighted bipartite graph and we need to find a minimum weight perfect matching. The weighted adjacency matrix of this graph is also shown in the top right diagram of this figure. We start with a black box which accepts a complete weighted bipartite graph as input and outputs the minimum weight perfect matching.
you will be designing Algorithm No. Design an efficient algorithm to transform the minimum weight perfect matching into a shortest path between the two given vertices. design Algorithm No.
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1.30. Design an efficient algorithm to transform this problem (of finding a shortest path between two vertices in a weighted graph) into a complete bipartite graph as shown in Fig.1. 1. 2. in terms of Fig.1.Discussion & Problems assignment problem. In terms of Fig.
. 2. 3. 3. 3.
2
A Mutual Friendship Graph
Assume that we have 5 persons in a hall. We shall discuss some necessary conditions for a degree sequence to be graphical. With each person in this diagram we show the number of people with whom he (or she) is friendly with. Now instead of a friendship sequence we have a degree sequence. C. If we note down this popularity number. The top left diagram in Fig. and C. This number in fact represents the amount of popularity a person enjoys. we shall define a walk. as discussed before we are talking about a symmetric relationship. The concept of graph connectedness will also be provided. We ask each person with whom he or she is friendly with and we get an answer like A is friendly with B only while B is friendly with A. this sequence is
. This so called mutual friendship is represented by the top left diagram in Figure 4. we need tools and techniques of graph theory.
4. D. a trail and a path in a connected graph. that means if person x knows y then it means that y knows x (for every pair (x. the relationship of friendship in the top left diagram is now transformed into an adjacency relationship in the graph shown in the top right diagram. Some of these people are friendly with each other while others are not.1
Introduction
We shall discuss a number of basic definitions in this chapter. D. & E.78
Basics of Graph Theory
4. Persons B and D are the most popular persons while persons A and E are the least popular. a double arrow edge is represented by an un-directed edge. We shall also talk about some special graphs at the end of this chapter. Please note that the double sided arrows emphasize the mutual friendship between two persons. We shall then talk about graph isomorphism and then come back to a discussion of necessary and sufficient conditions for a degree sequence to be graphical. We will be solving a couple of puzzles. B. we assume that their friendship is symmetric.1 is transformed into a graph shown in the top right corner of the same figure. y)). Let us name these people as A. and then sort this sequence we get the so called friendship sequence. Here each person is represented by a node or a vertex.1. thus A is friendly with just one person while B is friendly with three persons. 4. this sequence is also shown in the top left diagram. hopefully the students will realize that in such problems a stage comes when common sense alone is not sufficient to solve the puzzle.
1: We show five persons with a symmetric friendship relationship indicated by lines with double sided arrows in the top left diagram.3
Representation of a Graph
The graph shown in the top right corner in Fig. 4.Representation of a Graph
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exactly the same as the friendship sequence and is shown in the top right corner as well. The adjacency matrix representing the graph is shown in the bottom diagram. The friendship sequence as well as the degree sequence is also indicated along with the respective diagrams. The relationship is transformed into an un-directed graph as shown in the top right diagram.
Figure 4.1 is modeled by an adjacency matrix data structure as shown in the bottom diagram of Fig. In this adjacency matrix a 1 represents an adjacency relationship while a
.
4.1. 4.
1. The sum of all one's in a row corresponds to the degree of a vertex. Why? 3. Salma's husband Aslam (known as A) received the guest when they have arrived while Salma was in the kitchen preparing food. When we sort the numbers in the last column then we get the degree sequence or the friend-ship sequence. Please note that if these conditions are not met then the sequence will not be graphical. he answers zero (he does not shake hands with females). This is because in our understanding of friendship there is no room for self friendship. Is the husband telling the truth assuming the guests told the truth? How about if he answers 3 (he showed no discrimination). At least one number will be repeated in the sequence. the smallest number can not be less than zero. Problem 4.1) the largest number can not be larger than four. In addition to these we have other interesting limitations as discussed below. Why? 2. Find out the number of times the husband actually shook hands and more importantly with whom he shook hands and whom he has ignored.1.
.1. Now Salma asks the same question to her husband. she gets the following answer: B says 3. there is neither any room for negative friendship with some one which means animosity. for example the degree of vertex A is 1 while the degree of vertex B is 3.80
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zero (or an absence of 1) represents no adjacency relationship between the corresponding nodes in the graph. what does that mean? Under such conditions the sequence will neither be a friendship sequence. In a group of five persons (as shown in Fig. Salma invites three of her office colleagues for dinner. The sum of all numbers in the sequence will always be an even number. It will be interesting to explore the (special) structure of the friendship sequence (or the degree sequence). The odd numbers in the sequence will appear even number of times. Aslam shook hands with some of the guests. Why? Problem Set 4. from now onwards we shall use the term "the sequence". Obviously Salma was curious about whom Aslam shook hands and with whom he did not. She can not ask this question directly so she simple asks every colleague about the number of hands she had shaken. C says 2 and D says 1. 4. The degree of each vertex is also shown in the last column of the bottom diagram. what does that mean? 1.
No one shook hands with one self or with his or her spouse. Several handshakes took place when the guests have arrived.3.2.1. (a) Determine if you can seat these people around a round table so that every two neighbors are acquainted. The diagram on the right shows the next stage. The acquaintance graph of a group of eight people is given in the following diagram.Representation of a Graph
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Problem 4. Now Aslam is curious about the following: (a) Did a female shake hands with a male? (b) Did Salma shake hands with a male friend? (c) How many times Salma shake hands? (d) How many times Aslam shake hands? (e) Assume that Salma has not shaken hands with any male friend then did Aslam shake hands with a female? Under such conditions was there a husband who refused to shake hands with a female?
Figure 4. the motivation behind this seating strategy is to create a lively atmosphere.1. (b) If answer to (a) is no then is it possible to seat the people around a table so that every two neighbors are acquainted as far as possible? (c) Determine if you can seat these people around a table so that every two neighbors are not acquainted
. Problem 4. Surprisingly each person gave a different answer but Aslam does not know which answer belongs to whom.1. It is not yet obvious where Salma is and where Aslam is? Each bigger circle in these diagrams contains a husband and wife pair. The diagram on the left shows the graph in the making.2: We show some intermediate stages (and hints) in the solution of Problem 4.2. This time Aslam asked each person including his wife to write on a slip of paper how many hands he or she has shaken. Salma & Aslam has invited three married couples.
4 on the left most graph and the middle graph respectively Try to visualize a graph C having same number of vertices as in G or H and having an edge uv provided the edge uv exists either in G or in H.3: The acquaintance graph of a group of eight people is given.
4. We show a graph G and its complement H in Fig. 4. it will be useful and informative if in each problem you try to actually draw the underlying graph.82
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as far as possible. we should be able to draw an actual graph with the same degree sequence. Determine if you can seat these people around a round table so that every two neighbors are acquainted. We shall explore necessary and sufficient conditions for a sequence to be graphical but before we do that let us first discuss some more basic stuff (in general) and graph isomorphism (in particular). Now try to visualize a graph E having
. We have already discussed a number of necessary conditions for a sequence to be graphical. we shall need some of its (graph isomorphism) results for formulating necessary and sufficient conditions for a graphical sequence.
Figure 4. The first problem set (some parts of this at least) also exploits some of the necessary conditions.4
Complement of a Graph
Complement of a graph G is another graph H with the same number of vertices such that there is an edge uv in H if and only if there is no edge uv in G. the motivation behind this seating strategy is to encourage strangers to become acquainted with each other.
A completely connected graph is shown in the right diagram. Thus these two graphs are not equal. The graph on the extreme left and the graph on the extreme right are in fact equal.4: A Graph is shown in the left diagram and its complement is shown in the middle. This is because in one graph all the four degree vertices are connected in the form of a triangle while in the other graph this is not so. if you draw an adjacency matrix of each graph (after identical labeling) then you will find that the two adjacency matrices are exactly the same.5 after identical labeling then the two matrices comes out to be different as shown in Fig.5
Equal Graphs & Isomorphic Graphs
We show 4 graphs in Fig.5. If you closely look at the two middle graphs you realize that they are different graphs meaning that they are neither equal nor isomorphic. So there is a good possibility that they are all equal to each other. Note that vertices with the same degree in two different graphs are colored similarly.
4. The degree sequences of both these graphs are also indicated. in fact they all have the same degree sequence as shown in the same figure. and same number of edges. If now you draw the adjacency matrices of the graphs shown on the extreme left and middle left in Fig. the degree of each vertex is also indicated along with each vertex. they may still be isomorphic because we fail to find a visible difference
. 4. 4.
Figure 4.Equal Graphs & Isomorphic Graphs
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same number of vertices as in G or H and having an edge uv provided the edge uv exists in G and in H.6. 4. All these graphs have the same number of vertices.
6: Graph G shown in the top left corner is not equal to the graph H shown in the top right corner.84
Basics of Graph Theory
Figure 4.
. Out of these graphs at least two are equal graphs. Please note that vertices with the same degree are drawn in the same color. and at least two are isomorphic but not equal. There is still a possibility that the two graphs are isomorphic to each other as the two graphs have a number of similar qualities and no obvious differences. at least two are non isomorphic graphs (why?). This is evident from the adjacency matrices of the two graphs shown in the bottom diagram.5: We show four graphs each having the same degree sequence.
Figure 4. at least two are unequal graphs (why?).
5?
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These two graphs may still be isomorphic? What does that mean? Two graphs G and H are isomorphic provided they can be drawn with identical graph drawings. 4. this is shown in Fig. 4. It is quite obvious now that the drawing in the top left diagram is the same as the drawing shown in the top right corner. we rotate graph H of Fig. 4.6 is redrawn here in the top right corner after rotating it by an angle of 180 degrees. Then the two graphs are isomorphic provided there is an edge between f (u1 ) and f (u2 ) in the right graph if and only if there was an edge between u1 and u2 in the left graph for every two vertices in the two graphs as shown in the bottom diagram of this figure.
Figure 4.
. Let f be an isomorphism from the left graph (G) to the right graph (H). If. for example.2.7. 4.6 by an angle of 180 degrees then we get a drawing which is exactly the same as the graph G. Problem Set 4.7: Graph H shown in the top right diagram of Fig.Equal Graphs & Isomorphic Graphs like the one that we have found for the two middle graphs of Fig.
There is a possibility that the degree sequence may not be graphical? Under such conditions we claim that it is not possible to draw a graph for that sequence (why it is not possible?). 4. Find which two graphs are equal (and isomorphic). and which two are not isomorphic. Problem 4. Draw as many graphs as possible such that no two of them should be isomorphic to each other and each graph should have a degree sequence 332222.2. which are isomorphic but not equal and which are not isomorphic (and also not equal).8. Find an isomorphic function from a graph G to graph H in case graphs G and H are isomorphic to each other.2. and the same degree sequence. Draw as many graphs as possible such that no two of them should be isomorphic to each other and each graph should have a degree sequence 4443322. Problem 4.1. which two are isomorphic.6
The Degree Sequence
Assume that we are given a degree sequence of a graph and we need to find the corresponding graph provided the sequence is graphical.
4. Find a visible difference incase the two graphs are not isomorphic.
Figure 4. Find which two of them are equal.3.2. We show eight graphs with the same degree sequence in Fig.8: Eight graphs with same number of vertices. So we need to study necessary and sufficient conditions for a
.86
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Problem 4.2.
9.9. the new graph H will have number of vertices one less than G. Input : Original degree sequence SG (Example 4443322) Output: New degree sequence SH (Example 332222) 1. We apply the following procedure on this degree sequence SG and convert it into a new degree sequence SH which is equal to 332222. Subtract 1 from the first 4 (because maximum degree was 4) integers of the remaining sequence as shown in the top right diagram of Fig. 4.9 2. This is equivalent to reducing the degree by one of all adjacent vertices of v in
. we have already studied some necessary condition at the start of this chapter but those conditions were not sufficient (for a sequence to be graphical). Also assume that (the highest degree) vertex v is connected to the first u vertices (after v) in the degree sequence SG of G as shown in the bottom left diagram of Fig. Please note that removing the vertex v from G means that all edges emanating from G will also be removed and that amounts to operation number 2. Algorithm 13: Convert degree sequence SG into SH .
Algorithm 13 transforms a degree sequence into another degree sequence. 4.The Degree Sequence
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sequence to be graphical. Assume that we have a graph G in which the highest degree vertex is known as v and its degree is u. How about the degree sequence 543211? Let us assume we are given a degree sequence SG equal to 4443322. 4. this transformation makes no sense unless we visualize these operations as if they are performed on a graph. that amounts to removing the vertex v from the graph G. Now when we perform the first operation of removing the maximum degree from the degree sequence. Before moving forward find a sequence which satisfies all necessary conditions that we have discussed (earlier in this chapter) but it is impossible to draw a graph corresponding to this sequence. The new sequence SH will become 332222 shown in the same diagram. The maximum degree here is 4 and the total number of integers in the degree sequence is 7. The resulting sequence will be 443322 as shown in the top middle diagram of Fig. Remove the maximum degree (which is 4) from the degree sequence SG (thus reducing the length of the sequence from seven to six).
6. We claim (Havel-Hakimi) that the original sequence SG (corresponding to a graph G) is graphical if and only if the new sequence SH (corresponding to graph H) is graphical.88
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G. The maximum degree here is 4 and the total number of integers in the degree sequence is 7. Claim 4.
Figure 4.2. It is interesting to note that the above claims not only provides us a necessary and sufficient condition for a sequence to be graphical.1. You may have realized that the new graph will have a degree sequence equal to SH .9: We are given a degree sequence equal to 4443322 shown in the top left diagram. If the new sequence SH is graphical then the original sequence SG is graphical. The new sequence will be shorter by 1 as compared to the original sequence as shown in the top right diagram. and then find the actual graph corresponding to this sequence. Let us start with the same degree sequence. In order to prove this necessary and sufficient condition we have to make and prove two claims as follows: Claim 4. So before proving the above claims let first do the more interesting exercise of finding a graph. This meaningful interpretation is possible provided we have the crucial assumption: the highest degree vertex v in G is connected to the first u vertices (after v) in the degree sequence SG of G. We remove the maximum degree from the sequence and subtract 1 from the first 4 integers of the remaining sequence as shown in the top diagrams. verify that it is graphical. If the original sequence SG is graphical then the new sequence SH is graphical. We start with
.6. they also provide us means to draw a graph corresponding to a graphical sequence.
this has been illustrated by the bottom diagram of the same figure. It is very much possible to have two different graphs corresponding to the same degree sequence. Each iteration in the above procedure makes the degree sequence smaller in size. It is then converted into a 5 vertex graph. 1. and eventually into a 7 vertex graph as shown in the middle left diagram of Fig. you can draw the corresponding graph (output "Yes". If you get a number less than zero in the new sequence then the original sequence was not graphical (output "No".10. The four digit degree sequence can be recognized to be a graphical sequence. Either by carefully looking at the latest new sequence.
.The Degree Sequence
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a p length degree sequence. and terminate) otherwise there are two possibilities. and terminate) or if it is not possible then repeat step (1) on the latest new degree sequence (but sort it if it is not already sorted). The two graphs are neither equal nor isomorphic (why?). 2. and apply the following steps which outputs whether the sequence SG is graphical or not. transform the sequence into a new sequence using Algorithm 13 (by removing the first vertex v from the degree sequence with a degree equal to u. 4. a stage comes when it is very much possible to draw the corresponding graph as shown in Fig. In this figure the seven digit degree sequence is converted into a six digit degree sequence and then ultimately into a four digit degree sequence as shown in the top diagrams. 4.10. and subtracting 1 from the first remaining u degrees of the degree sequence). The resulting graph shown in the bottom left diagram is different from the middle left graph shown in the same figure. The corresponding 4 vertex graph is shown in the middle right diagram.
Let us now take up Claim 4.
. This seven vertex sequence is converted into a six vertex sequence. it says that if a new sequence SH is graphical then the original sequence SG is also graphical. We get a four vertex graph which is then converted into a five vertex graph and ultimately into a seven vertex graph as shown in the middle and bottom left diagrams.10: A seven vertex degree sequence is shown in the top left corner. This four vertex sequence is graphical as shown by the middle and bottom right diagrams.6. It is very much possible to get two different graphs from the same degree sequence. Let us design an algorithm to construct a graph G corresponding to an original sequence SG provided the new sequence SH and its corresponding graph H is given.1. Can you design a formal proof for this claim? Is your proof based on induction or is it proof by contradiction? Discuss briefly. For example if 332222 is graphical then 4443322 is graphical. and then ultimately into a four vertex sequence as shown in the top right corner.90
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Figure 4.
3. We are also given the new sequence SH and we need to show that it is graphical.
Let us now take up Claim 4. this means that we need to draw the corresponding graph H. Does Algorithm 14 perform the intended function correctly? Problem 4.1 then you should be able to prove this claim also.11. this is certainly a serious deficiency of this algorithm? Before correcting this deficiency let us look at it a bit more closely in the next part. It essentially means that now we are given a graph G corresponding to a degree sequence SG . Note down the highest degree in the degree sequence SG . it says that if the original sequence SG is graphical then the new sequence SH is graphical. Input : (1) New degree sequence SH (Example 332222). Problem 4.1.2. 2. Problem Set 4.The Degree Sequence Algorithm 14: Find a graph G corresponding to an original degree sequence SG provided the new sequence SH and its corresponding graph H is given. and its graph H is given. If the assumption is not true then this algorithm will not provide correct results. Does Algorithm 15 perform the intended function correctly? This is an important question because the answer may be no.6. (2) Original degree sequence SG is also given (Example: 4443322).
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1. If you can prove Claim 4.3. 4. Assume that we are given a degree sequence 4443322 and a graph G as shown in the left diagram of Fig. 4. just move backwards in Fig.10? Can you design an algorithm similar to (or almost a mirror image of) Algorithm 14. Output: Original Graph G corresponding to the degree sequence SG . Add a new vertex v in the given graph H. (the degree of v should be u (why?)) 3.3.3. Problem 4. let it be u.2.3. The said algorithm provides a correct solution provided we have the crucial assumption: the highest degree vertex v in G is connected to the first u vertices (after v) in the degree sequence SG of G. Add an edge between the vertex v and the appropriate u vertices in the degree sequence SH (why?). Is it possible to apply
.6.
and its graph G is given.3. 1. Remove the vertex v from the resulting graph and we obtain H (why?). Please note that there are three vertices in this graph with a highest degree equal to four. We claim that such a transformation is always possible. the maximum degree vertex v1 has a degree u. Output: Graph H corresponding to the degree sequence SH . Suppose that we are given a graph F and its degree sequence SF . 4.11.4. The following algorithm performs this transformation.92
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Algorithm 15: Find a graph H corresponding to a new degree sequence SH provided the original sequence SG and its corresponding graph G are given. Let this vertex be v and let its degree be equal to u. two of these vertices are not connected to the first four vertices in the degree sequence of this graph. Locate the vertex of highest degree in G. 3. What will be the result of our algorithm? Why our algorithm fails this time? Discuss briefly. 2. (2) New degree sequence SH is also given (Example: 332222). Now repeat the above part with the same degree sequences but this time G is as shown in the right diagram of Fig. and this vertex is not connected with the first u vertices in the degree sequence (after v1 ) of this graph. Remove all edges emanating from the vertex v. We need to convert the graph F into another graph G with the same degree sequence but in G the vertex v1 (having highest degree equal to u) is connected to the first u vertices in the degree sequence of this graph.
1
(Note that graph H corresponds to degree sequence SH ?)
Algorithm 15 in order to draw a graph with a degree sequence equal to 332222? Please note that there are three vertices with a highest degree equal to four. Problem 4. Why? Once it has been established by Algorithm 16 that a given graph F can
. none of these vertices are connected to the first four vertices in the degree sequence of this graph. Input : (1) Original degree sequence SG (Example 4443322).
vn ).The Degree Sequence
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Figure 4. the maximum degree vertex v1 has degree u. Output: A graph G with the same degree sequence. vk ) & (vj . locate a vertex vn such that vj is connected to vn while vk is not connected to vn . and this vertex is not connected with the first u vertices in the degree sequence (after v1 ) of this graph. vj ) & (vk . there is no need to do any thing else) 2. vn ). graph H is now transformed into graph G?
. Algorithm 16: Convert graph F with a degree sequence SF into a graph G with the same degree sequence (but with an important difference?) Input : A graph F and its degree sequence SF . and this vertex is connected to the first u vertices in the degree sequence of this graph. (If you can not find such vertices then graph F is already transformed into G. Please note that the two graphs are not isomorphic. remove edges (v1 . 1. In the graph H.11: Two graphs with the same degree sequence 4443322. Locate vertex vk and vertex vj in the graph F such that v1 is connected to vertex vk and not connected to vertex vj while the degree of vj is larger than that of vk . and insert edges (v1 . From graph H. the maximum degree vertex v1 has a degree u. (Why are you guaranteed to find such a vertex vn ? 3.
Let us summarize our recent findings: 1.13. 4. Algorithm 16 has also made it possible to design a constructive proof for Claim Number 2. We are given a graph G with a degree sequence 432221 as shown in the top left diagram of Fig. Problem 4. 2. Given a graph G in which the maximum degree vertex is not connected to the first u vertices in the degree sequence of this graph (where u is the degree of the highest degree vertex) it is not possible to apply Algorithm 15 in order to transform G into another graph H. It is possible to convert this graph G into another graph with the same degree sequence by using Algorithm 16 as shown in the top right diagram of Fig.4.4.2. We show the position of our claims and the respective proofs. Please note that the three algorithms perform a dual purpose. Under such conditions we first transform G into another graph G using Algorithm 16 and then Algorithm 15 will correctly transform G into H. Problem Set 4. it gives rises to a number of important conclusions. 4.14. What is that constructive proof? We show our strategy in handling the necessary & sufficient conditions for a degree sequence to be graphical in Fig.14. Given a graph G in which the maximum degree vertex is connected to the first u vertices in the degree sequence of this graph (where u is the degree of the highest degree vertex) it is possible to apply Algorithm 15 in order to transform G into another graph H. As shown in this figure the highest degree vertex is connected to a vertex of lowest degree while it is not connected to a vertex of a relatively higher degree. Problem 4. Concentrate on the third line of Algorithm 14: "Add an edge between the vertex v and the appropriate u vertices in the degree sequence SH (why?)". 3. they help us in transforming one graph into another but more importantly they provide crucial insight in designing constructive proofs for the two claims. The
. 4. Remember we have discussed this problem in the class but have not resolved it completely.4.1.94
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always be transformed into another graph G. What does appropriate mean? Carefully read your text book (page 17) and then make your decision appropriately.
The graph in the left diagram is converted into a graph shown in the right diagram by deleting two edges and by inserting two edges (shown in bold). All these three vertices (with a degree of 4) are connected to a vertex of minimum degree. it is not connected to a vertex of degree 2. The graph in the right diagram has the same degree sequence but there is an important difference. Here there is at least one vertex (v1 ) with a (highest degree equal to four) which is connected to vertices with higher degrees only.The Degree Sequence
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Figure 4.12: A seven vertex graph is shown in the left diagram with a degree sequence shown in the top of the graph. There are three vertices in this graph with a highest degree equal to four. v1 is connected to two vertices with degree 4 and two vertices with degree three. the minimum degree in this graph is 2.
.
the minimum degree in this graph is 2.96
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Figure 4.
. There are three vertices in this graph with a highest degree equal to four. All these three vertices (with a degree of 4) are connected to a vertex of minimum degree.13: A seven vertex graph is shown in the left diagram with a degree sequence shown in the top of the graph.
g. The new graph with the same degree sequence and the problem completely resolved is shown in the bottom diagram. g. We again apply the same algorithm to resolve the rest of the problem as shown in the middle right diagram. Do we know how to draw the graph for SH provided we have a graph for SG ? Discuss briefly. c. k and assume that a = 5. Problem 4. d.
. b.The Degree Sequence
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problem (that the highest degree vertex is connected to (relatively) lower degree vertices) is still not completely resolved as is evident from the middle left diagram of the same figure. i. The figure down below may be helpful for your imagination.3. As you may have noticed in this specific problem we have to apply the said algorithm twice to obtain the desired results. Let another sequence be SH = b − 1. j. f. f − 1.14: A graph with a given degree sequence shown in the top left diagram is converted into another graph with the same degree sequence shown in the bottom diagram. h. Let a degree sequence consisting of 11 numbers be SG = a. j. Is it possible to do some thing in this specific problem so that the problem is resolved just by applying the said algorithm only once? Can you generalize your findings? How much can you save in time in the worst case analysis?
Figure 4. c − 1. i.4. d − 1. h. e. k. e − 1.
Both these graphs have the same degree sequence which is 54444411111. e. b. Please verify if this is always right or wrong in general. j. k − 1. i. if it is graphical then we find and draw the actual graph G corresponding to this sequence.7.
Figure 4. Let a degree sequence consisting of 11 numbers be SG = a. the highest degree vertex v (with a degree equal to u) will always be connected to the first u vertices in the degree sequence SG ? Problem 4.15: Two graphs are shown in this figure. f. Discuss briefly. k and assume that a = 5. We check if the degree sequence is graphical.4.98
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Problem 4. h − 1. d.6. If the degree sequence is graphical then you are supposed to draw a graph G corresponding to this sequence such that the
. d. The figure down below may be helpful for your imagination. g − 1. f. We are given a degree sequence SG . Problem 4. e. Do we know how to draw the graph for SH provided we have a graph for SG ? Discuss briefly.4.4.4.4. Problem 4. The graph shown in the left diagram is a connected graph while the graph shown in the right diagram is a disconnected graph. In this graph G the highest degree vertex v (with a degree equal to u) will always be connected to the first u vertices in the degree sequence SG . c. h. i − 1. Let another sequence be SH = b. g.5. c. We check if the degree sequence is graphical. If the answer to the above problem is no then how can we do some thing to make sure that in the final graph. We are given a degree sequence SG . j − 1.
Is there a possibility that a degree sequence SG is graphical but it is impossible to draw a graph corresponding to this sequence such that the highest degree vertex v (with a degree equal to u) in the graph should be connected to the last u vertices in the degree sequence SG .4. Problem 4.4. Problem 4.9.4.The Degree Sequence
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highest degree vertex v (with a degree equal to u) should be connected to the last u vertices in the degree sequence SG . We are given a graph G (and its degree sequence SG ) in which the highest degree vertex v (with a degree equal to u) is connected to the first u vertices in the degree sequence SG .
. Discuss briefly if this is always possible and how will you do it? Problem 4.8. What are necessary & sufficient conditions for a degree sequence SG to be graphical such that in the resulting graph G the highest degree vertex v (with a degree equal to u) in the graph should be connected to the last u vertices in the degree sequence SG . Either prove or find a counter example. Is it always possible to convert it into another graph with the same degree sequence but now the highest degree vertex v (with a degree equal to u) is connected to the last u vertices in the degree sequence SG .10.
The graph shown in the left diagram is a connected graph while the graph shown in the right diagram is a disconnected graph.
4. Both these graphs have the same degree sequence which is 54444411111. & Paths
We show a walk from vertex a to vertex d in a graph shown in Fig.
.17. 4.100
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Figure 4.16: Two graphs are shown in this figure. A shortest path between the same two vertices is also shown. We also show a path between the same two vertices in this graph. Remember in a trail it is not allowed to traverse an edge more than once – but it is allowed to traverse a vertex several times. Trails.7
Walks. Please note that in a walk it is possible to traverse an edge (and therefore a vertex) several times. A walk or a trail can always be converted into a path as shown in this figure. We also show a trail between the same two vertices in the same graph. The path is shortest in terms of number of edges in between the two terminal vertices. You may have realized that in a path it is not allowed to traverse an edge or a vertex more than once.
the path consists of six edges. no edge is repeated. The shortest path from vertex a to d is shown in the bottom right diagram. & Paths
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Figure 4. A path is shown from vertex a to vertex d in the top right diagram.
.17: A walk from vertex a to vertex d consisting of eight edges as shown in the top left diagram. the trail consists of six edges. In this trail only a vertex is repeated. The trail in the top middle diagram is converted into a path from vertex a to d as shown in the bottom middle diagram. Trails. A trail from vertex a to d is shown in the top middle diagram.Walks. The walk in the top left corner (from vertex a to d) is converted into a four edge path as shown in the bottom left diagram. vertices as well as edges are repeated in the walk.
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4.e.. (c) Cycle graphs. Under special graphs we consider: (a) Completely connected graphs. We categorize graphs into three major categories: (1) Acyclic graphs (or trees). it consists of no self loops or parallel edges. (2) Bipartite graphs.18.
Figure 4.
.9
Broad Categories of Graphs & some Special Graphs
We shall talk about very broad categories of graphs and then some special graphs. 4. & (3) Cyclic graphs. It can be represented by an adjacency matrix.8
Multi-graphs and Pseudo-graphs
We show a simple graph. (d) Line graphs. & (e) Star graphs. i. This graph can be represented by an adjacency matrix also shown in the same Fig. We also show graphs with self loops and parallel edges. How about graphs which are not simple?
4.18: A simple graph containing no self loops and no parallel edges. (b) Regular graphs. graphs containing cycles.
Broad Categories of Graphs & some Special Graphs
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Figure 4.
.19: A cyclic graph which is neither bipartite nor acyclic is shown in the right diagrams. an acyclic graph (or a tree). shown in the left diagrams. is also a bipartite graph. A bipartite graph which is not acyclic is shown in the middle diagrams.
19 is not bipartite as it contains an odd cycle.
4. 4.1
Tree Graphs
A connected graph G is a tree provided it does not contain any cycles. A tree graph is shown in Fig. A graph is a line graph (or a chain graph) if the degree of every vertex is 2 except for two vertices where the degree is 1.19.3
Special Graphs
A graph G is k-regular if the degree of every vertex is exactly equal to k. Another bipartite graph is shown in the middle diagram of Fig. We show a number of non isomorphic trees with p larger than 1 and smaller than 6 in Fig. 4. If G does not contain a cycle then G is not only bipartite it is also a tree. A graph G having p vertices is completely connected if the degree of every vertex is p − 1 (please note that the degree of a vertex in a simple graph can not be more than p − 1).
4.2
Bipartite Graphs
A graph G is bipartite provided it does not contain odd cycles. The minimum vertex cover of this graph is also shown in this diagram. In other words every edge in a bipartite graph connects a vertex from set A to a vertex in Set B.9.19.9. it is bipartite because it does not contain any cycle at all. The graph shown in the right diagram of Fig.20.19 and 4.9. Please note that partite A is an independent set while partite B is also an independent set but neither of the two is a maximum independent set. The maximum independent set in this graph is shown in the right diagram of this figure. It may contain even cycles or no cycles at all. 4.21.104
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4. A bipartite graph is shown in the left diagram of Fig. A graph is a cycle graph if the degree of every vertex is exactly two. 4. it is bipartite because it does not contain any odd cycles. A line graph consisting of two vertices is a special case where the degree of both the vertices is 1.21.
. The A partite as well as the B partite are shown in the middle diagram where the bipartite graph is drawn with a different orientation to highlight the two parts. 4. We show another bipartite graph in Fig. The vertex set V (G) of a bipartite graph G can be partitioned into two disjoint sets A and B whereas both A as well as B are independent sets. 4.
The minimum (sized) vertex cover and the maximum (sized) independent set in graph G are shown in the right diagram. This graph is in fact a bipartite graph as shown in the middle diagram consisting of an A partite and a B partite.21: A graph G is shown in the left diagram.
.20: We show a number of non isomorphic trees with p larger than 1 and smaller than 6.
Figure 4.Broad Categories of Graphs & some Special Graphs
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Figure 4.
is always bipartite. A line graph is not regular unless it consists of a special case of a connected graph of two vertices. being acyclic. It is obvious from Fig. 4. The degree of every vertex is 1 except for one vertex where the degree is p − 1 in a star graph. The degree of every vertex is 2 except for two vertices where the degree is 1 in a line graph.
Figure 4. A start graph consisting of two vertices is a special case. The degree of every vertex is exactly 2 in a cycle graph. A line graph.22 that a completely connected graph is (p − 1)regular while a k-regular graph may not be completely connected.106
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A graph G is a star graph if the degree of one vertex is p − 1 while the degree of every other vertex is 1. A graph G is referred to as a forest if it contains a collection or set of trees. A cycle graph is also a 2-regular graph as shown in this figure. A star graph is not regular unless it has a size equal to 2 when it becomes a line graph which is 1-regular.22: The degree of every vertex in a completely connected graph is p − 1. The degree of every node is the same for every vertex in a regular graph.
.
being acyclic. 4.25. A number of cycle graphs are shown in Fig. A 3-regular graph is shown in the middle diagrams. The four partites are indicated in different colors in the bottom diagrams of this figure.23. We show a number of k-regular graphs in Fig. any edge in these graphs connects a vertex in one partite to any vertex in one of the other partites. 4. A 4-regular graph is shown in the right diagrams. We show a number of k-regular graphs in Fig.23. They are rather 4-partite graphs meaning that the vertex sets of each of these graphs can be partitioned into four disjoint sets of vertices (or partites).22. It is interesting to compare these graphs with the ones shown in Fig. All these regular graphs are bipartite as shown in the bottom diagrams. is always bipartite where one partite consists of size 1 while the other of size p − 1. 4. 4. Please note that all these graphs are bipartite as shown in the bottom diagrams of this figure.
Figure 4.Broad Categories of Graphs & some Special Graphs
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Also a star graph. These new graphs are regular and not bipartite.23: A 2-regular graph shown in the right diagrams. A curious reader might
.
while it is not bipartite if the cycle graph consists of odd number of vertices.
Figure 4. A 4-regular graph is shown in the middle diagrams.24: A 6-regular graph shown in the right diagrams.108
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appreciate the fact that a cycle graph is bipartite if it consists of even number of vertices. What may not be obvious is that an odd cycle graph will always be a 3-partite graph as shown in this figure. All these graphs are 3-partite as shown in the bottom diagrams.
. Another 4-regular graph is shown in the right diagrams.
27 above shows the same graph A as shown in Fig 4. It is a 7-regular graph consisting of eight vertices.5. graph B is a 6-regular graph. By deleting a different set of edges from graph A. Similarly graph C is derived from graph A by deleting another set of edges from B. The figure below shows a completely connected graph A in the left diagram.26. Similarly draw all possible non isomorphic graphs of 8 vertices which are 5regular. The middle graph shows graph B which is derived from graph A after deleting or subtracting a number of edges from graph A.2.Broad Categories of Graphs & some Special Graphs
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Figure 4. Please note that graph C is again a 5-regular graph consisting of 8 vertices.5. Problem 4. Similarly
. The Figure 4. we may obtain a graph which is not isomorphic to graph B. Problem Set 4.1.25: Cycle graphs of different sizes are shown. Problem 4. Graph E is obtained by deleting a vertex from graph A.5. Draw all possible graphs consisting of 8 vertices which are not isomorphic to B but which are 6-regular.
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Figure 4.26: Graphs B and C are derived from graph A by deleting certain edges.
.
Figure 4.27: Graphs E and F are derived from graph A by deleting certain vertices.
By carefully selecting and deleting certain edges of graph C shown in Fig. (b) Now concentrate on graph E.5. Try to match the graphs that you have obtained with the ones shown in the figure below. (c) By deleting another set of edges transform it into a 4-regular graph.3. (d) Draw all non isomorphic graphs which are 4-regular consisting of 7 vertices.28: Shows all non-isomorphic graphs consisting of 8 vertices which are 4-regular.26. Problem 4.5. 4. we can obtain a 4-regular graph of eight vertices. How many such non-isomorphic graphs we shall be able to obtain? The figure below shows all non-isomorphic graphs consisting of 8 vertices which are 4regular. Please note that both E and F are completely connected and regular graphs.
Figure 4. by deleting certain edges transform it into a 5-regular graph.Broad Categories of Graphs & some Special Graphs
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graph F is obtained by deleting another vertex of graph E. Look at the graphs consisting of 8 vertices and are 4regular as shown in the figure above. it is a 6-regular graph. (a) Is it possible to get a graph which is isomorphic to graph F by deleting any two vertices of graph A? Discuss briefly. (a) Find which of these graphs is a
.4. Problem 4.
(c) Draw all nonisomorphic graphs consisting of 12 vertices and are 6-regular.10
Integration of Concepts. and are regular consisting of eight vertices. and Action Items
We have talked about various concepts in graph theory in this chapter.29: Graphs G is derived from graph A by deleting certain edges.5. Draw all non-isomorphic graphs consisting of 8 vertices which are 5-regular. The graph G shown in the figure below is derived from graph A by subtracting certain edges from graph A. We have mainly confined ourselves to simple graphs in which there are no parallel edges and no self loops. we shall study directed graphs in detail in Chapter 8. a tree
. (d) Draw all non-isomorphic graphs consisting of eight vertices which are 3-regular. We have further limited our study to un-directed graphs in this chapter. Properties.
4.5. Note that graph G is 5-regular.
Figure 4. Problem 4. Connected graphs can also belong to certain categories like a line graph. Out of these graphs indicate which ones are bipartite graphs. This category of graphs can easily be represented by an adjacency matrix or an adjacency list data structure. a cycle graph.112
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bipartite graph? (b) Draw all non-isomorphic bipartite graphs which are connected. Un-directed graphs can be further classified into connected graphs and disconnected graphs.
a cyclic graph and a completely connected graph. like a Hamiltonian path or an Eulerian path in a graph. We have also talked about certain properties of graphs.Integration of Concepts.
. or a path between two vertices in a graph. Properties. and Action Items
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graph. a trail. We have also talked about a walk. It will be interesting if we integrate a couple of concepts with a number of properties.
The problem of how to construct a SC graph of fixed vertices will also be discussed. That means that both these graphs are self complementing. It can be seen that the left graph is the complement of the right graph as well as isomorphic to the right graph. The following concepts are used in a meaningful manner to advance our discussion on self complementing graphs. Regular & non regular graphs 4. Complement c(G) of a graph G 2. Graph isomorphism between two graphs G & H 3. We shall show how one SC graph can be transformed into another SC graph with less or more vertices? Some of the interesting properties of these graphs will also be elaborated. Some of the necessary conditions of SC graphs will be discussed next. Both of them are also Hamiltonian? Necessary Conditions
.11
Self Complementing Graphs
A self complementing (SC) graph is a graph G whose complement c(G) is isomorphic to itself. Eccentricity of a vertex 7. 1. Radius & Diameter of a graph This section is organized as follows. We shall discuss these graphs in some detail as they provide us a platform to connect a number of key concepts in graph theory & algorithms.114
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4.11. Degree sequence of a graph 5. We shall first show some SC graphs in order to give you a feel of such graphs. Deleting and inserting a vertex in a given graph 6.
4. We shall conclude with a number of interesting problems.1
Regular Self Complementing graphs
We show a number of non trivial (a graph of one vertex is a trivial example of such graphs) self complementing graphs in the following figure. These graphs are also regular as the degree of each vertex is 2.
When k = 2. Let us try to explore graphs with 9 vertices with the degree of each vertex equal to 4.Self Complementing Graphs
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Figure 4. p will become 9. If c(G) is isomorphic to G then it should have the same number of edges as G . Will all such graphs be isomorphic to each other? Will such graphs (or at
. We have already drawn such a graph with p = 5 when k = 1. Does that mean that p or p − 1 in an SC graph should be divisible by 4 4? Are these necessary or sufficient conditions for such a regular graph to be self complementing? It will be rewarding if you draw a couple of regular graphs with the property that the degree of each vertex is p−1 .30: A graph and its complement which is isomorphic to the original graph. Then the number of edges in the graph will be exactly 2 p(p−1) .that means the degree of every vertex in a regular self complementing graph should be p−1 . The complement c(G) of graph G is obtained by deleting all edges of graph G from the corresponding completely connected graph. Can you draw such a graph if p is even? 4 Can you draw such a graph if p is any odd number? It will be essential to answer the above questions before moving forward? If you have tried to draw such graphs then you will soon realize that a nregular graph (where n = p−1 ) is possible if and only if p = 4k + 1 where 4 k is an integer equal to or larger than one.
Does this mean that all regular graphs are self complementing? But a cycle graph of three or six vertices is not self complementing? There must be a class of regular graphs which will be self complementing? The following figure shows the same graph along with the corresponding completely connected graph of five vertices.
least some of them) self complementing? If you take the complement of any such graph it will certainly have the same degree sequence and same number of edges? Again it will be rewarding if you draw some of these graphs before arriving at a conclusion? We show three such graphs in the figure below. It will be exciting if you draw the complement of the left graph? You will soon realize that the top right graph is the complement of the top left graph while the bottom graph is isomorphic to the top right graph.31: A self complementing graph of 5 vertices (left diagram) and a completely connected graph of 5 vertices. What does this example tell us?
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Figure 4. Are any two of them isomorphic to each other? Is one complement of the other? Are they self complementing? Please try to answer these questions before moving forward.
Top right diagram shows the complement c(G) of graph G.
. The bottom graph is equal to c(G).Self Complementing Graphs
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Figure 4.32: Top left diagram shows a graph G.
How about imagining non isomorphic regular graphs having p = 13 and degree of each vertex equal to 6. There are (367860) six-regular non isomorphic graphs with 13 vertices.34: A regular self complementing graph of 13 vertices. A very few of them indeed transform into themselves when you take the complement. If we take the complement of one such graph it will transform either into one of such graphs or into itself. One such self complementing graph is shown below.
Figure 4.
. Most of them transform into one another if you take the complement of one such graph as shown above. One such graph is shown in the figure below.118
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If you actually draw all non isomorphic graphs with p = 9 and degree of each vertex equal to 4 you will realize that there are 16 such graphs possible.
Figure 4.33: A regular self complementing graph of 9 vertices.
& 32221. Now we have to verify if these degree sequences are really graphical.2
Non Regular Self Complementing graphs
One important question that should agitate you is every self complementing graph a regular graph? Is it possible to have a non regular graph to be self complementing? How about a line graph of four vertices? Its degree sequence is 2211. Necessary Conditions Is it possible to have a non regular graph of five vertices which is self complementary? Let us try out various degree sequences with the above necessary condition? There are only three choices possible under the conditions laid out before: 22222.11. The degree sequence of the complement of this line graph can be obtained by subtracting 2211 from 3333. it comes out to be the same as the four vertex line graph is self complementary.
Figure 4. and if they are then do they really belong to graphs which are self complementary? Let us start with the sequence 32221. 33211.35: A self complementary line graph of 4 vertices. Can you draw its graph and check out if the graph is self complementary? The sequence is graphical but is itself complementing?
. The other two belongs to non regular graphs and are of interest to us. The degree sequence of a completely connected graph of four vertices is 3333. This provides us a necessary condition for a non regular graph to be self complementary? The. The first degree sequence corresponds to a cycle graph which was regular and we have already seen it.Self Complementing Graphs
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4.
In fact we have earlier claimed that a regular SC graph can have only 4k + 1 vertices.120
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Now let us check the other sequence which is 33221. we can safely say that a SC graph of 4k vertices will always be non regular? Why?
. we have seen a non regular SC graph of 4 vertices in the last diagram.36: A 5 vertex non-regular self complementing graph. this indeed comes out to be graphical as well as self complementary as shown below:
Figure 4. it is not sufficient. We have already seen a regular SC graph of 4 + 1 = 5 vertices. Please note that the condition (the) is a necessary condition. we have also seen a non regular SC graph of 5 vertices. Thus this condition along with others (for example the number of odd degrees in a degree sequence should be even for a degree sequence to graphical) may narrow down our search for SC graphs but we always have to verify if a given graph G is a SC graph? These simple conditions tells us that an SC graph (which may or may not be regular) can only have p equal to 4k or 4k + 1 vertices where k can be equal to or larger than 1.
3
Constructing Self Complementary Graphs
Let us consider graphs where p = 8 and assume that its degree sequence satisfies the stated necessary conditions for a SC graph. If we connect graph G and its complement in the following configuration then we claim that the new graph will be a SC graph. 66443311. Some of the possible degree sequences are listed here: 66661111.we shall now discuss ways of constructing large SC graphs bypassing this tiring process? The new method will help us further in making meaningful connection between relevant concepts. If G contains a single vertex then this graph is certainly SC graph.37: Here vertex G is a graph while c(G) is complement of graph G. Let us actually construct such graph and see how it looks? Assume that G
. The resulting super graph is a self complementing graph.11. 55552222. But if G contains more than 1 vertex then why this composite graph is self complementary?
Figure 4. It will be interesting if you try to check if a graph corresponding to these degree sequences is indeed self complementary? This will certainly be a tiring process . An edge between G and c(G) means that every vertex G is connected to every vertex of c(G).Self Complementing Graphs
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4. Given any graph G we can always find its complement c(G). and G may contain more than one vertex. 44443333. Please note that a line between G and c(G) means that each vertex of G is connected to every vertex of c(G). 55443322.
Using these building blocks you can also construct another SC graph? If you have the patience of constructing such a graph you will realize that its degree sequence will be 55552222.here also G is any graph and is connected with its neighboring graph in the same fashion? Here x is just one vertex and we claim as before that the resulting graph will be self complementary. Another copy of G will consist of a line graph of two vertices .
Figure 4.then c(G) will simply consist of two isolated vertices. Another way to construct a SC graph is shown below . The degree sequence of this graph is 44443333 it is one of the sequences that we have predicted earlier in our discussion? We can also start with G equal to two isolated vertices .consisting of vertex 7 and vertex 8. How the above configuration would look like when it is actually drawn? If you look at the diagram below you will realize that the graph is indeed self complementary.and that will be two isolated vertices 3 & 4.
. Why? The number of vertices in the resulting SC graph will be 4p + 1 where p is the number of vertices in G.then c(G) will be a line graph of two vertices. The number of vertices in the SC graph will still be 4p where G has p vertices. Then c(G) will be its complement .38: Here graph G is a line graph of two vertices 1 & 2.122
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is a line graph of two vertices .
How about if we insert a self complementary graph H in the place of x as well as G in the above configuration? 4. Will the above configuration result into a regular or a non regular SC graph? The following diagrams will help you answer these interesting questions. Please find the degree sequence of each of the following graph? Can you figure out how these SC graphs are constructed?
. 1. An edge between G and c(G) means that every vertex G is connected to every vertex of c(G). Vertex x may be a single vertex graph. What would happen if the self complementary graph H in the above step is regular? How about if it is non regular? 5. The resulting super graph is a self complementing graph. There a number of interesting possibilities in the above configuration.Self Complementing Graphs
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Figure 4. Is it possible that we insert graph G in the place of x in the above configuration? Please note that G is any graph? 3.39: Here vertex G is a graph while vertex c(G) is the complement of graph G. How about if instead of single x vertex we have a graph H connected to its neighbors? In order to make the whole configuration SC should it be any graph H or a special graph H? What special property it should possess? As you can discover yourself H should be a self complementary graph for the above configuration to be self complementary? 2.
. It will contain 4k + 1 vertices.11.40: A line graph of 4 vertices is used as a building block in the left diagram.4
Transforming a SC graph with 4k + 1 vertices into another SC graph with 4k vertices
We know that a SC graph should have either 4k or 4k + 1 vertices? This immediately leads us to conjecture that if we have a SC graph of 4k + 1 vertices then by deleting one vertex we can convert it into another SC graph having 4k vertices? Is it straight forward or we need to devise an intelligent algorithm to do so? Similarly if we are given a SC graph having 4k vertices then is it possible to construct a SC graph having 4k + 1 vertices by inserting a new vertex and connecting it to some of the vertices of the original graph? Again do we need some thinking to do so or is it a trivial problem? In order to answer these questions let us start with a regular graph which is SC. A cycle graph of 5 vertices is used as a building block in the right diagram.124
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Figure 4. Can we delete any vertex and the resulting graph would stay a SC graph of 8 vertices? Please note that initially the degree sequence is 444444444. It becomes 44443333 after deleting that vertex.
4. By deleting any vertex from this graph is it possible to convert it into another SC graph having 4k vertices? Let us start with a simpler problem of a regular SC graph having 9 vertices as shown below.
.Self Complementing Graphs
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Figure 4. We remove vertex 2 from the same SC graph in the bottom diagram.41: We remove vertex 1 in the top diagram from a SC graph in the top diagram.
Is the resulting graph a SC graph? How about if we have a non regular SC graph of 4k + 1 vertices? Can we delete any vertex to make it another SC graph with 4k vertices? How about this graph? It has a degree sequence 332211.126
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What about if we delete any vertex from the 25 vertex graph shown earlier. It is no longer a trivial problem and we should devise an algorithm to solve the problem?
. Will the new graph be also self complementary? Why?
Figure 4.42: We remove a vertex from the 25 vertex SC graph.
Figure 4. This graph may also provide insights needed to design an intelligent algorithm to solve the above problem.after deleting one vertex the new degree sequence should possess certain properties if it represents a self complementing graph? For example it can be 66443311 or 77443300. Here we show another interesting graph which is not regular but is self complementing. Can we remove any vertex and still it remains a SC graph? Given a degree sequence 774444411 of a self complementing graph .44: A SC graph having 9 vertices.Self Complementing Graphs
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Figure 4.43: A SC graph having 5 vertices which is not regular. etc. The problem is to decide which vertex should be removed?
. Its degree sequence is 774444411.
The graph is redrawn in the bottom diagram to emphasize that it is indeed a beautiful regular graph?
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Should we always remove the vertex with degree equal to 4? Why? Should that vertex be connected with lower degrees or higher degrees?
4. Now start with the same original graph (with a degree sequence equal to 44443333) while the new graph should have a degree sequence equal to 444444444.
Figure 4.11.45: We insert a vertex x in a self complementing graph. The resulting graph is SC and is also regular . Again we shall show you a number of graphs and then provoke you to design an efficient algorithm to solve the problem. We show the same graph with a degree sequence 44443333 being converted into a new SC graph with degree sequence 555543333. Please note that we can also put an extra constraint that the resulting SC graph should be regular.although it has not been drawn in a suitable manner in the top diagram.5
Transforming a SC graph with 4k vertices into another SC graph with 4k + 1 vertices
Now let us look into the problem of inserting a vertex in a SC graph consisting of 4k vertices such that the new graph having 4k + 1 vertices is also selfcomplementing.
11.then can we solve the graph isomorphism problem? How?
.47: Inserting a vertex x in a SC graph?
4.this is c(G) and now check if G and c(G) are isomorphic.6
The Self Complementary problem and Graph Isomorphism
If we know how to find if G and H are isomorphic then we can always check if G is a self complementary graph? How? Take the complement of G .130
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So the problem is where to connect the inserted vertex and what should be its degree? Perhaps another example may provide you a solution? Please check the degree sequence of the graph before after inserting a new vertex?
Figure 4. But suppose we know how to check if G is a self complementary graph .
48: We need to check if graph G is isomorphic to graph H? Look at the figure above. We claim that if the top (or the bottom) graph is self complementary then graph G is isomorphic to graph H? Why? We can also use the following configuration to check if G and H are isomorphic also by substituting H in an appropriate place in this diagram?
Figure 4.
.Self Complementing Graphs
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Figure 4.7
A SC graph has diameter 2 or 3 .11.not less than 2 and not more than 3?
Instead of proving the above statement let us first do a simple one.49: We need to check if graph G is isomorphic to graph H?
4. Can we prove that for any graph G the following configuration will give us a SC graph with a diameter not more than 3. We need to check if a given graph G is isomorphic to another given graph H.
Let us see how the eccentricities of various vertices look like in a line graph of eight and five vertices as shown below. and the radius and diameter of a graph. The diameter in the top graph will be 7 while the radius will be 4 in the top diagram.132
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Figure 4. Similarly we show another five vertex graph where the
. The diameter and radius for the bottom line graph will be respectively 4 and 2.51: What is the diameter of this super graph? We shall now attack the more general problem? But before that let us refresh our knowledge regarding some old concepts: the eccentricity of a vertex.50: What is the diameter of this super graph? Again can we prove that for any graph G the following configuration will give a SC graph with diameter not more than 2?
Figure 4.
Thus its diameter is 2 while its radius is 1. Here the eccentricity of each vertex in G is 2 except for vertex 1. The five vertex line graph and its complement are shown in the bottom diagram.53: A completely connected graph G and its complement. Another five vertex graph G and its complement is shown below.52: The eccentricities of different vertices in a line graph. The eccentricity of each vertex is indicated in the line graph as well as in its complement.the diameter as well as the radius is 1 in this graph. You can well imagine that if the diameter in a graph G is more than 3 then
. eccentricity of each vertex is 1. Please note that the diameter in the line graph was 4 while it has reduced to 2 in the complement of the line graph. The diameter as well as the radius in the complement graph is 2.
Figure 4. It is a completely connected graph . Its diameter as well radius will be infinite.Self Complementing Graphs
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Figure 4. The complement of this graph is also shown in the bottom diagram. The diameter in the complement of graph G is infinite while its radius is also infinite.
54: A star graph and its complement.134
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Figure 4.55: A line graph and its complement.
.
Figure 4.
Please note that bipartite complement of a bipartite graph is different from this complement.thus all the A (and B) partite vertices in BP will become completely connected in the complement of BP . Here we first form a completely connected bipartite graph consisting of as many vertices in the A as well as B partites of bipartite graph BP . We assume that while taking the complement of graph BP we simply consider it a general graph . Let us call these category A SC bipartite graphs. Similarly if the radius or the diameter in a graph G is 1 then it cannot be isomorphic to its complement? Why? Does that mean that a SC graph can have diameter 2 or 3 . Let us call these category B SC bipartite graphs. Please note that this not a Category A SC bipartite graph? Why?
.we then remove those edges which are already present in BP . Alternatively in order to find complement of graph BP .11. That means a graph G with a diameter more than 3 cannot be self complementary.Self Complementing Graphs
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the diameter in its complement is reduced to 2. The following graph is a Category A SC graph. we first form a completely connected general graph consisting of all BP vertices . We then remove existing edges in BP from this completely connected bipartite graph .not more not less?
4.8
Bipartite self complementary graphs
Definition: Assume that the complement of a bipartite graph BP is isomorphic to bipartite graph BP .not more than 3 and not less than 2? How about the radius of a SC graph? Should it be always 2 .this will give us the complement of graph BP .the resulting graph will be a bipartite complement of bipartite graph BP . Now assume that the bipartite complement of bipartite graph B is isomorphic to bipartite graph B. Please note that it is not a Category B SC graph? Why? Can you draw another bipartite graph which is a Category A SC graph? Can you prove that such graphs are not possible if any partite contains more than two vertices? We show a category B SC graph BP in the figure below.
.56: A bipartite graph BP shown in the top left diagram.136
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Figure 4. The corresponding completely connected graph is shown in the top right diagram.57: A bipartite graph and its bipartite complement. The complement of graph BP is shown in the bottom diagram.
Figure 4.
58: Top left diagram shows a bipartite graph BP. The bottom diagrams once again shows the two bipartite graphs shown in the standard form.Self Complementing Graphs
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The same graph BP is drawn below in a different shape. Top middle diagram shows the corresponding completely connected bipartite graph. The degree sequence of a corresponding completely connected bipartite graph is also 44444444. Here we show another Category B SC graph with red vertices in one partite
.
Figure 4. As you can see this is a regular SC bipartite graph. The corresponding completely connected bipartite graph and the complement of bipartite graph BP is also shown here. Top right diagram shows the bipartite complement of the bipartite graph. The following bipartite graph is a non regular self complementing graph with a degree sequence 33222211. The degree sequence of the above graph is 22222222. While that of the completely connected bipartite graph is 44444444.
59: A SC bipartite graph which is not regular. From now on wards we shall consider only category B self complementary bipartite graphs . That is an important necessary condition for a bipartite graph to be self complementary. How many edges are there in a complete SC bipartite graph with m vertices in one partite and n vertices in another partite? That is equal to mn. The number of edges in a SC bipartite graph will be mn/2.we shall simply refer them as SC bipartite graphs.
.
Figure 4. and green vertices in another partite.60: A regular SC graph having 12 vertices.138
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Figure 4. Thus either m or n should be even.
Similarly there is a graph G(B) consisting of vertices in the set B.although vertex x provides a useful function of pinpointing certain vertices needed for the decomposition of the graph? Vertices in G − x which are adjacent to x belong to one partite while the rest of the vertices in G − x belong to the other partite. Thus even if we remove vertex x the resulting graph G − x will still be a SC graph as shown below . There will be a bipartite graph having edges going between the set A and the set B.9
Decomposition of a SC graph G
Here we shall talk about the decomposition of a SC graph G into a number of edge-disjoint graphs .
4.graph G(A).one of them is a SC bipartite graph? Initially we shall talk informally and provide some insight and then we shall discuss it more formally. graph G(B) and the vertex x connected to all
. We put vertices in G − x which are adjacent to x in set A while the rest of the vertices in G − x goes in the set B. bipartite graph BP .11. In addition to that there will be a graph inside G consisting of vertices belonging to the set A .61: A SC graph where the size of two partites is different. The original graph can thus be decomposed into edge-disjoint graphs .Self Complementing Graphs
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Figure 4.we call this the graph G(A). Please note that the following graph is a SC graph with or without the vertex x.
The graph (x. The two partites are balanced that is the size of A and B is the same.there is some relationship between these two graphs?
.
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Figure 4. 1. 2.62: A self complementing graph is decomposed into a number of edge-disjoint graphs. A) 3. The bipartite graph BP (A. B) will also be a SC bipartite graph.140 vertices in graph A. It can be observed that the following properties should hold for the original graph G to be SC. The graph G(A) and graph G(B) .
and 4 back to 1. The vertex 9 is mapped onto itself. This is a so called circular permutation in which we map 1 onto 2.63: A SC graph decomposed into a number of edge-disjoint graphs. 3 onto 4. This permutation permutes the vertices of G and produces a new graph which is not equal but isomorphic to the original graph G.
Figure 4. 4. Isomorphism.
4. In order to show that this new graph is isomorphic to the original graph G we have to find an isomorphic function (or a permutation) which maps the vertices of the new graph back onto the
.64. automorphism & Self Complementing Graphs
Any permutation of vertices of any graph G may create a different graph H.11.Self Complementing Graphs
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Let us consider another example to confirm/enhance our observations before formally discussing the decomposition of a SC graph.10
Permutation. This graph H will always be isomorphic to graph G but may not be equal to graph G (as the adjacency matrix may be different). 2 onto 3. For example consider the permutation p1 equal to (1234)(5678)(9) as shown in Fig.
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vertices of graph G such that adjacency as well as non adjacency is preserved in the two graphs. Any complementing
. The identity permutation is always an automorphism .
Figure 4. We may be more interested in the non trivial permutations of graph G? It may be possible for a certain category of graphs that the only automorphism is the trivial identity permutation? Please note that if p is an automorphism of graph G then the permutation p2 is always an automorphism of graph G If a permutation p of vertices of graph G creates a graph H which is the complement of graph G then the permutation p is known as the complementing permutation of graph G and graph G is a SC graph. The two graphs are not equal but they are isomorphic thus the permutation p1 is not an automorphism of graph G.64: Graph G (top left diagram) and another graph (top right diagram) where the vertices of G are permuted according to the given permutation p1 .it is known as a trivial permutation. In this case this permutation p2 will be (4321)(8765)(9). Please note that graph G is equal to the bottom graph If a permutation p of a graph G creates a graph H which is equal to graph G (that means the adjacency matrix will exactly be the same) then that permutation is known as an automorphism of graph G.
The two graphs are not only isomorphic but also equal thus the permutation p is an automorphism of graph G but it is not a complementing permutation of graph G. Thus every self complementing graph G has a complementing permutation p associated with that graph G.65: Graph G (left diagram) and another graph (right diagram) where the vertices of G are permuted according to the given permutation p. permutation of a graph G is certainly not an automorphism of graph G but the permutation p2 is always an automorphism of graph G if p is a complementing permutation. Whenever we claim that G is self complementing then we have to find out the self complementing permutation?
.Self Complementing Graphs
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Figure 4.
thus p2 (G) is an automorphism of graph G. The bottom diagram shows another graph where the vertices of p(G) are permuted once again according to the same permutation p.144
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Figure 4. This new graph is equal to graph G .
.that means the permutation p is a complementing permutation of graph G but it is not an automorphism of graph G. This new graph is also the complement of graph G .66: Graph G (top left diagram) and another graph (top right diagram) where the vertices of G are permuted according to the permutation p.
4.two of them are shown in the figures below.Self Complementing Graphs
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Let us now consider certain self complementary bipartite graphs and the related complementary permutations . But in the top permutation one of the partites is mapped onto the other partite in the lower permutation it is mapped onto the same partite?
.67 is a self complementing permutation. According to this permutation vertices of one partite are permuted onto vertices of the other partite in the complement of BP .
Figure 4.67: Bipartite graph BP and a complementing permutation. 4. The permutation in Fig.68 is also a self complementing permutation. The permutation in Fig.
The complementing permutation p = (1A3A)(2A)(4A)(1B3B)(2B)(4B).68: Bipartite graph BP and its complement. 38. In the decomposition or the synthesis effort to be explained in the coming examples we are more interested in those bipartite graphs which are similar to bipartite graphs similar to the one shown in Fig.146
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Figure 4. According to this complementing permutation vertices of one partite are permuted onto vertices of the same partite. Why? In the coming figures you find certain clues of synthesizing self complementing graphs using simple building blocks?
.
how can we check if a graph is cyclic. The original research papers were about encouraging students to discover and learn (graph algorithms) by themselves with minimal help provided by an instructor in the form of provocative questions. We have also studied that a graph is cyclic if it contains a cycle. we talked about a bridge edge and a cycle in a graph. 2. We have also studied a number of properties linking different concepts. and ofcourse computer science. Summer-Fall 2003.1). In this chapter we shall study a number of graph algorithms. and its title was Should We Teach Algorithms. mathematics.152
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Introduction
We have studied a number of concepts related to graph theory in the last chapter. engineering. For example. We also provide a number of powerful learning tools to understand and design various algorithms. biology. A four line procedure known as the Bucket Algorithm which can be molded into a number of useful and powerful graph algorithms based on greedy strategies. 2. and Yasser Hashmi and its title was Shortest Path Algorithms .
. Another four line procedure known as 2-edge Shortest Path Algorithm which finds shortest paths of length 2 from a given vertex to every other vertex in a weighted directed graph. For example. Vol.2. The later part of this chapter is based on one of our CS department research reports which was coauthored by Komal Syed. The paper was co-authored by Sara Tahir.Making and Breaking Connections. how can we check if an edge is a bridge edge. The recursiuon tree and the colored puzzle are some of these visual aids which facilitate a learner or a designer in his or her path towards discovery. Almost all of these algorithms are based on the following easy to understand and friendly to use buliding blocks: 1. we provide a detailed study of a number of graph algorithms that have applications in diverse fields like chemistry. In this chapter. The initial part of this chapter is based on one of our papers published in IJECE. how can we find an actual cycle in a cyclic graph (see Concept Map 5. No. For example an edge is a bridge edge provided its removal disconnects a graph. Again this building block can be used to design a number of sophisticated shortest path algorithms based on dynamic programming. social sciences.
there is no precise algorithm available that can be used to design new algorithms. and stimulating questions posed by the instructor. Our experience of teaching algorithms indicates that creativity in algorithm design depends. concerned with the design of algorithms. challenge. can sow the seeds that could blossom into the genius that produces efficient yet astonishingly simple algorithms. The second one. It is now
. The first objective.1
Design of Algorithms
Teaching the standard course "Analysis & Design of Algorithms" at an undergraduate level in a typical Computer Science program essentially has two objectives. rather incite. on how we deal with the analysis phase. The study of the methods and rules of discovery and invention is a field in its own right. dealing with analysis. Though there are rules of thumb that can be followed to help an individual design an algorithm. Polya [10] remembers the time when he was a student himself: he was always perturbed by the question: "Yes the solution seems to work. one can always encourage students to redesign an algorithm right from scratch. and suspense. students to create algorithms themselves using some very fundamental concepts. to a large extent. we should not formally teach anything. There is no guarantee that one who critiques literature can learn to write beautiful poetry. which is perhaps far more important.154
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5. We stress that while we are familiarizing students with existing algorithms. we believe that the instructor. Despite the fact that one cannot guarantee that a student could become an efficient algorithm designer. This second objective. excitement. following our approach and providing proper guidance. timely hints. and above all the confidence required in solving a non-textbook problem. is essentially a creative effort containing all the ingredients of a thriller: adventure. Similarly the ability to understand and analyze algorithms does not guarantee that one could become an efficient algorithm designer. The objective is that students should experience the thrill and excitement of discovery even during the initial phases of understanding existing algorithms. is to equip the students with the necessary tools and techniques. it appears to be correct. Instead we should encourage. is to familiarize students with existing algorithms. but how is it possible to invent such a solution? How could I invent or discover such things by myself ?" We feel that with the availability of some pre-requisite knowledge.
2
The Moore Method
R. The teacher. allowing his students to find the answers in their own ways". made and deployed. in our model. He also does not allow the use of any source material. directed set of decisions that are introduced. We.1. Many professors still use his teaching style not only in his subject of specialization (topology). [6]. The last definition suits our discovery based learning approach in which a teacher formulates a directed set of questions and hints in order to help his/her students design algorithms.2
The Bucket Algorithm
We start with a simple algorithm which we call the Bucket Algorithm (the bucket symbolizes a friendly container where a child puts every new toy or every new discovery) consisting of just four lines of pseudo code: We shall show how this primitive procedure can be used to reinvent a number
. Rine [13] defines design as "A systematic.
5. on the other hand. In the words of Hale [5]. and have advanced or modified the Moore Method in a number of ways [3]. solution. It is interesting to note that our approach is similar in some aspects with the so-called Moore Method of teaching and learning. mostly quiet.
5.1
What is Design?
According to the Webster's dictionary [1]. He sat in the back of the room. Moore was a professor of mathematics at the University of Texas. and other courses. or technology". "What was so special about his mode of teaching was that he did not lecture. "Design is the thought process comprising the creation of an entity". L. "Design is to conceive and plan out in the mind". and then actively guides the students in their path of discovery. does not allow collective effort on the part of the students inside or outside of class. but in analysis.1. algebra. occasionally asking a question. game theory. leading to an effective or efficient outcome. In the words of Miller [9]. encourage lively discussions inside as well as outside the classroom.
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5. starts with something (very simple). Taylor [18].The Bucket Algorithm important to find a good working definition of design (of algorithms). while characterizing (his version of) the Moore method of teaching. he did not profess.
.1: Two pictures of what the Bucket B will look like in the initial stages of the Bucket Algorithm.156
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Algorithm 17: The Bucket-Algorithm input : A Graph G output: A Bucket B 1 Put any vertex x of Graph G in the Bucket B.
j g h k f d e i g
j i h k
f d e
a b
c
a b
c
Figure 5. 2 while there are edges coming out of the Bucket B do 3 Select an edge connecting vertex u in B to v not in B. 4 Put v in B .
Find a spanning tree of a graph.1
Understanding the Bucket Algorithm
The Bucket Algorithm is simple and straightforward. Conduct a breadth first search in a graph. Find a path between two vertices in a graph provided a path exists.2. the students should develop a keen desire and ability to understand the motives behind. 2. [4]). With some encouragement from the instructor. 5. Find the number of connected components of a graph.
.The Bucket Algorithm
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of existing powerful algorithms in graph theory ([16].
5. Find a bridge in a graph. They would learn the ways and means of devising their own algorithms. 7. [17]. [2]. 6. Specifically the Bucket Algorithm would be used to solve the following problems: 1. It is just a 4-line algorithm with a simple while loop (with no conditional statements or recursive calls). because it is easy to understand and at the same time flexible enough to handle a variety of different problems. Conduct a depth first search in a graph. [14]. Solve the single-source shortest-paths problem: rediscover Dijkstra's Algorithm. and the procedures followed in order to arrive at innovative solutions. Find if a given graph is connected. 4. 9. We start with something simple but potentially very powerful. Find a minimum-spanning tree of a graph. 10. Find if a graph is a tree. 8. Rediscover Kruskal's Algorithm. 12. [15]. Simple. Rediscover Prim's Algorithm. [11]. 11. 3.
2). {b. the set {f. The second is the set of edges connecting vertices outside the bucket with each other: {{f.1). i. k}. d. h}. k}. g}. f }}. b}.2. {i.
5. and those outside the bucket. {i. The first is the set of edges connecting vertices inside the bucket with each other: {{a. {a. Next we choose any edge joining vertex a to any other node. The third is the set of edges (the "branches coming out of the Bucket B" in Step 2) connecting vertices inside the bucket to vertices outside the bucket (see the middle diagram of Figure 5.2(middle diagram) shows the bucket B after different iterations through the Bucket Algorithm.2.2. f }. in the graph (since all other nodes are currently outside of the bucket) and put b in the bucket. j. say node a. k}.
. {g. {d. {b. g.158
j g h k f d e a b e c f d i g h k a b c j i
Basics of Graph Algorithms
j g h k f d e a b c i
Figure 5. d}.2. k}. i}. This set of edges is equal to {{a. {a. These two different kinds of vertices give rise to three different kinds of edges. c. b. c}. Step 1 instructs us to put any node. {c.1 for a picture of what the Bucket B will look like at this stage. {d. Notice that there are two types of vertices: those inside the bucket represented by the set {a.2: Two pictures of the Bucket B after different iterations through the Bucket-Algorithm. say node b. c}. Figure 5. k}}. An edge belonging to this last set of edges is called a cross edge and is of most interest to us. we can implement numerous algorithms. j}. {h. See Figure 5. Now we have a set of nodes {a.2. {b. f }. Depending on the constraint we place on the selection of cross edges in Step 3. h. e}}. d}} to choose from in Step 3 as we iterate through the while loop. of the Graph G into the bucket.2
How does it Work?
We identify a Graph G and a Bucket B (See Figure 5. {b. e}. b} in the bucket giving rise to a set of edges {{a.
5
The Right Provocation
It is well known that a real understanding of the problem is a necessary condition to solve any problem. For
.2. It may be advisable to include it at a later stage.The Bucket Algorithm
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5.
5. and when would it be false? Does it make any difference if we have a different starting vertex? Note that there are situations when it really makes a difference. Out of a sequence of six questions posed by Skiena [16] in order to guide one to discover the right algorithm.3
Playing with the Algorithm
We strongly encourage our readers to play around with the Bucket Algorithm to get comfortable with it. What is the worst-case complexity of this algorithm? It is recommended that the instructor not involve the underlying data structure at this stage in order to tackle the issue of complexity.
5. v} we select in Step 3 that we discover the new vertex v. the first question is "Do I really understand the problem?" Then comes the role of the teacher in terms of how he/she states a problem and provokes (or guides) his/her students to solve it in a specified manner. After the students are confident that they understand the idea behind the Bucket Algorithm. "Half way home to solving a problem is a clear understanding of the problem". the instructor can start asking them to modify it to solve more complex problems.2. Students must come to realize the importance of cross edges: it is because of this cross edge {u. Such questions could be: Under what conditions would there be no edges coming out of the Bucket? Note that this condition should be met otherwise the algorithm would never terminate. Would all the vertices of the graph move into the bucket after the completion of the algorithm? When would this scenario be true.4
Solving Other Problems
The above questions would induce a deeper understanding amongst students about how the Bucket Algorithm works under different conditions and give some hints while solving more complex problems. It will be useful at this stage if the students are asked to derive the time complexity of the Bucket Algorithm.2. During this activity the instructor should ask thought provoking questions such that the students focus on multiple facets of the algorithm that would later help in designing new algorithms. According to David [12].
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example if a teacher is talking about Quick Sort. then graph G is connected. The teacher should first make the students appreciate the need for partitioning the array into halves such that all numbers in the first half are smaller than each number in the second half. The instructor in this case will have to make an extra effort to guide such students.2. Notice while students were becoming familiar with the Bucket Algorithm. there are still any nodes left outside the bucket then the graph is not connected..1).3
Finding if a Graph is Connected
Assuming that the students know what a connected graph is.e. he/she cannot expect his/her students to discover the said algorithm just after understanding the sorting problem. Once all students have understood the solution (having arrived at it on their own with well-timed prodding from the instructor) the instructor should start the discussion regarding cost calculation. Please see Concept Map 5. the complexity of the modified algorithm.3. after the Bucket Algorithm has been applied to a graph G.2. Not all students may be able to identify this property of the Bucket Algorithm.
. however. the instructor asked when there would be nodes left outside the bucket. The answer is simple: if. If. i. The understanding of the previous state of an abstract system and the (usefulness of the) final system state after the application of a so called fundamental operation [7] (for example the partitioning procedure in Quick Sort) is crucial in problem solving in computer science as in elsewhere. Brighter students would have been able to identify at that stage that some nodes will be left outside the Bucket B when a graph is not connected since cross edges do not exist connecting them to nodes inside the bucket (Figure 5. discovering (and even understanding) the said sorting algorithm. the instructor should ask the students: "Can you modify the Bucket Algorithm such that you may be able to determine whether a given graph G is connected?" The emphasis should be on using the existing techniques with minimum modification. Why we should do this and how should we do this are both equally important for designing. Please see Concept Map 5. all nodes come inside the bucket.
5.
The number of times we have to apply the Bucket-Algorithm depends upon the number of connected components. Applying the algorithm again with a new bucket would give us a new connected component. What would be the resulting time complexity of this algorithm? The
. The first problem is to check if a given edge is a bridge.3: A graph G that is not connected.2
Finding a Bridge in a Graph
A cut edge or bridge is one whose removal produces a graph with more connected components than the original. once the Bucket Algorithm terminates. This could be solved if we remove the given edge and then check the number of connected components in the resulting graph. Applying the Bucket Algorithm once on a graph with more than one connected component would tell us that the graph is not connected as all the vertices of the graph do not end up in the bucket.
5.
5. it is the job of the instructor to at least identify them for those students who cannot visualize the solution immediately.3. nodes i and j will be left outside the Bucket B. and this would determine the worst-case time complexity.Finding if a Graph is Connected
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g
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a b
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Figure 5. and so on and so forth.3.1
The Number of Connected Components
Once we understand how to find if an un-directed graph is connected the above problem becomes simple and very little imagination is needed to answer the above question. There are essentially two different problems here. The vertices that do end up in the bucket belong to a single connected component.
4.4
Finding if a Graph is a Tree
The algorithms that solve this problem depend on how we define a tree. which makes it a connected graph.4: A cut edge or bridge is one whose removal produces a graph with more connected components than the original graph. The problem is thus reduced to repeatedly applying the algorithm designed to
. Thus every edge in a tree is a bridge.1
Every Edge in a Tree is a Bridge
We know that a tree having n vertices consists of bare minimum number of edges. Solving a problem from different angles and then making a comparison is the single most important exercise for a student studying algorithms (Rawlins [11]). This implies that removing any edge would disconnect a tree. How many times the Bucket-Algorithm is applied and what is the resulting worst-case complexity of the algorithm?
g h j g i k h j i k c b e
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a
f d
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Figure 5. We already know how to check if a given edge is a bridge in a graph.
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second problem is to find or locate a bridge in a given graph. It highlights the fact that looking at various definitions or properties (which come from a study of graph theory) is sometimes extremely useful and it provides the seed for designing a number of very powerful algorithms.
5. Once the first problem is solved it should be a simple matter to handle it. This in not only true for this problem but is true for a majority of problems.
We know how to find if a given graph is connected using the Bucket Algorithm. Only when the students have gained confidence that they understand the basic problem and can find an efficient solution should we move to more complex problems such as finding whether a given graph is a forest. How complex is this problem? Is it possible to count the number of edges while we are checking if the given graph is connected? Would that perhaps reduce the complexity?
5. The number of times we would have to do this and finding the resulting complexity is an interesting exercise by itself. So the problem is reduced to counting the number of edges.4
A Comparison
A comparison of all these algorithms would be extremely beneficial to the students if they are encouraged to work it out independently. Thus the spanning tree of a tree would be exactly the same tree.2
The Number of Edges in a Graph
We can define a tree in a number of ways. In fact.Finding if a Graph is a Tree
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check if an edge is a bridge.
. This definition or property can be used to design an algorithm to check if a given graph is a tree.e.3
The Spanning Tree of a Tree
We know that a tree has the minimum number of edges required to connect a given number of vertices. For example. Once they have the answers it would again be stimulating for them to compare their findings with their colleagues within the classroom.
5.
5.4. a connected graph is a tree provided the number of edges in the graph is exactly equal to one minus the number of vertices in the graph. p − 1. i. A spanning tree of a given graph also satisfies this property.4. all of these definitions are equivalent implies. as it is a tree.4.. The catch is that the graph should be connected otherwise the definition would not apply (why?). Encouraging and initiating interesting discussions and even heated debates is one of the most important responsibilities of a teacher: (s)he must simply coordinate and make sure that the interaction is moving in the right direction.
In the second approach we should be careful not to create cycles in the graph.5.
5. What would be the worst-case time complexity of this algorithm?
5. It is also possible to identify some of the so-called cross edges (edges which are coming out of the Bucket).5
Finding a Spanning Tree of a Graph
The algorithm that we design to solve this problem depends on how we visualize the development of a spanning tree. which would constitute the spanning tree. If we just keep a record of all such edges we might get the spanning tree of the given
. In each case the Bucket Algorithm helps us. We can start with the original graph and start with pruning or removing edges until the graph becomes a tree. This idea would give birth to an algorithm: Remove all edges that do not disconnect the given graph. This approach is opposite to the one discussed above: instead of pruning we are growing edges. we might have noticed that every time we discover a new vertex it is because of a cross edge (step 3). The resulting complexity would change dramatically depending upon the approach used.5. Each approach has its merits and demerits and the comparison itself is very enlightening especially because each approach has more advanced applications. How many times we use the Bucket-Algorithm eventually decide the overall worst-case complexity.1
Cutting Edges
If we remove all redundant edges from a given graph and just keep edges essential to keep it connected the remaining graph would be a spanning tree of the given graph. Or we can start with no edges and start growing edges until we get a tree. and that the number of such cross edges would be exactly equal to p − 1. We add edges out of the edge pool of the graph such that the resulting graph remains a tree.
5. In the earlier approach we should be careful and should not disconnect the graph.5.3
Selecting Edges
While running the Bucket Algorithm.2
Growing Edges
We start with no edges at all but with p isolated vertices.164
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5.
While cutting edges we select the edge of maximum weight (provided it does not disconnect the graph).1
Cutting or Growing Edges: A Krushkal's like greedy algorithm
Each algorithm used to find a spanning tree in the previous section could be used with proper modification to find a minimum-spanning tree of a connected and weighted graph. We show in Concept Map 5.6. not feasible (although it is correct)?
5. which looks at all possible solutions and then selects the one of our choice. properties and the Bucket Algorithm.2. we can grow edges starting from the edge of minimum weight (making sure no cycle is created).6. It is very much possible to discover most of the algorithms (that we have presented in this chapter) in class once we have become comfortable using and manipulating the Bucket Algorithm. having first sorted the edges in descending order of weight. a number of systematic questions that if asked. thus giving rise to different spanning trees. 5. will provoke the learner to discover a number of interesting techniques.Finding a Minimum Spanning Tree of a Weighted Graph
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graph. It is very much possible to have multiple non isomorphic minimum spanning trees of a weighted graph but the weight
.
5. This would give rise to an algorithm very similar to Krushkal's.6
Finding a Minimum Spanning Tree of a Weighted Graph
There could be many non-isomorphic spanning trees possible for a given graph: each approach that we have described for finding a spanning tree of a graph was flexible and there was a lot of maneuvering possible within it.2) which integrates different concepts. which if refined will lead to a number of important algorithms.3.4
Integrating Concepts and Discovering Algorithms
We show a concept map (Concept Map 5. Similarly. What if we find all distinct spanning trees of a given graph using any approach and then select the one with minimum weight? Why is this approach. See Figure. How efficient would this be if compared with the algorithms described earlier?
5.5.
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Concept Map 5. The Concept Map and an iterative sequence (ascending order) of asking questions help student discover or understand a number of useful graph algorithms.2.
.
6. Without reading proofs given in the textbook they should come up with something of their own making.Finding a Minimum Spanning Tree of a Weighted Graph
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of each tree would be the same (why?).2
Selecting Edges: A Prim's like greedy Algorithm
While forming a spanning tree we can select any cross edge.5). select one with minimum weight connecting u in B to v not in B. they are not always optimal. They should also be asked to derive the time complexity
. Using this simple technique the Bucket-Algorithm can easily be modified to find a minimum spanning tree of a weighted directed or undirected graph (Figure 5. In order to form a minimum spanning tree. Please note that here we are not using any fancy data structure since the objective is not to have a complicated design. A lively discussion can be initiated to find the merits and demerits of individual work. although greedy approaches are relatively efficient (being based on local conditions only).6. We are lucky this time: a so-called greedy approach is working optimally and is in fact optimizing the global sum also. It is important that the minimum spanning tree problem is an optimization problem in which we intend to minimize the sum of weights of all edges in the spanning tree. unlike some textbooks.
It would be useful if the students were asked to prove that this greedy approach would actually find a minimum spanning tree. we should try to include edges of less weight thus excluding those of higher weight. put this edge in M ST . while there are edges coming out of the Bucket B do Out of all the edges coming out of B. It follows that among all cross edges that we may select we should pick the one of minimum weight. we are trying to minimize a local quantity. In order to minimize the global sum. However. Algorithm 18: Find Minimum Spanning Tree of a graph G input : A weighted Graph G output: Minimum Spanning Tree (MST) of Graph G
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Put any vertex x of Graph G in a Bucket B. We show here how the simple Bucket Algorithm can be reduced into a minimum spanning tree finding algorithm which resembles Prim's Algorithm.
5. Put v in B.
What is that magic? How and why it is working? Can this magic be used elsewhere and under what conditions? We shall discuss it later in this chapter. something magical which cuts down the time complexity for not so obvious reasons. However the time complexity of Prim's algorithm (as stated in most textbooks) is better.
5. 5. In the top diagrams we select an edge of minimum weight out of all edges coming out of the single bucket in which we have our minimum spanning tree growing.Finding a Minimum Spanning Tree of a Weighted Graph
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of this approach. we select an edge out of all edges (coming out of all buckets) in the graph and thus this technique grows various spanning trees in different buckets.6. This technique (based
.3
A Panoramic Picture of various MST Finding Techniques
It will be useful from a learning perspective if we make a comparison between various minimum spanning tree finding techniques. The top diagrams in this figure show how we grow a minimum spanning tree in a single bucket (this is a Prim's like algorithm). We should understand both the differences as well as the similarities. How about if we start Prim's algorithm from every vertex? Thus we may be able to avoid an expensive check (that the resulting graph should not be cyclic) each time we insert a new edge (as in Prim's algorithm) and at the same time we can exploit the inherent parallelism which was lacking in Prim's algorithm? We initially put every vertex in a separate bucket. the process of selecting the desired edge out of all edges coming out of the single bucket makes sure that no cycle is generated.6. The middle diagrams show a minimum spanning forest growing up in various buckets using a Kruskal's like algorithm. We show three different techniques in action on the same weighted graph in Fig. It would be useful if they compare this approach with Prim's algorithm. the two approaches look identical. Why? The reason is in fact more exciting because Prim's algorithm is not just greedy. In fact. there is something else. In the middle diagrams we can visualize that each vertex is initially in a separate bucket. In this algorithm (middle diagram) it is essential to check that the new edge that is selected should not form a cycle with the previously found minimum spanning tree. In the top algorithm there was no explicit need to check that this condition is true. for each bucket we select the edge of minimum weight coming out and thus grow various minimum spanning trees in different buckets.6.
The big question is do we need to make an expensive cycle check each time we insert an edge? It will be interesting to find under what conditions we need a cycle test and when we do not need such a test? This algorithm terminates when the number of edges selected becomes exactly equal to one less than the number of vertices in the graph.6. In the first case we have to make sure that no cycle is created and in the second case we should be worried about disconnecting the graph. This is because of the fact that we consider edges coming out of the bucket only. It will also
. Why? Do we need a cycle test in Kruskal's algorithm in case all edge weights are unique? We do not need a cycle test in Prim's algorithm in spite of the fact that edge weights are not unique? Why? 5. we grow a minimum spanning tree in each bucket just like Prim's algorithm. 5. We try to exploit the inherent parallelism in this scheme where each processor is running a Prim's like algorithm on each vertex. It will be useful to formally prove that in case of Prim's algorithm we do not need a cycle test and still no cycles will be formed. however we consider all edges incident on a vertex inside a bucket then we do need to be worried about the formation of a cycle. if. 2. In case of Kruskal's algorithm either we are growing a minimum spanning tree (by inserting edges starting from low cost ones) or we are cutting edges starting from high weight edges. 4.6. If all edge weights in a graph are unique then we do not need a cycle test in Boruvka's algorithm. In case of Prim's algorithm we need not be worried about the formation of a cycle.Finding a Minimum Spanning Tree of a Weighted Graph
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on the Boruvka's algorithm) is shown in action (on the same weighted graph) in the bottom diagrams of Fig. Prim's algorithm terminates when all vertices come inside the bucket. The algorithm terminates when all edges have been considered. the way we move forward makes sure that no cycle is formed. It is interesting to note the following about the three minimum spanning tree finding algorithms: 1. 3. But we do need a cycle test in case the edge weights are not unique. In case of Boruvka's algorithm.
We can make a small change in one of the Minimum Spanning Tree Algorithms in order to convert it into a Maximum Spanning Tree Algorithm as follows. 5. We shall address these and other related issues in a problem set.1. Problem Set 5. Algorithm 19: Find Maximum Spanning Tree of a graph G input : A weighted Graph G output: A Maximum Spanning Tree (MaxST) of Graph G
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Put any vertex x of Graph G in a Bucket B.
Problem 5. Apply the above algorithm to the following graph G consisting of positive as well as negative edge weights.
5. select one with maximum weight connecting u in B to v not in B. A curious reader may have realized by now that positive or negative edge weights in a weighted graph does not create any problem while evaluating the minimum or maximum spanning tree of a graph. and find the maximum spanning tree while showing the contents of the Bucket B after each step. We shall repeat this problem while finding the shortest path (or the longest path) in a graph having negative edge weights. Problem 5. Put v in B.1.6. We shall discuss that the shortest path problem becomes complicated if all edge weights are not positive.6. Do we face a similar problem while finding a minimum spanning tree of a graph having negative edge weights?
. put this edge in M axST .2.1.4
The Maximum Spanning Tree Problem
All the minimum spanning tree algorithms described here can easily be modified in order to find the maximum spanning tree of a weighted graph G. Do you think there will be any complication in finding a maximum spanning tree if the edge weights are negative (or if they are positive).1.7. A small change in the algorithm can do the job.176
Basics of Graph Algorithms be interesting to prove that in case of Boruvka's algorithm no cycles will be formed even if we do not make a cycle test provided all edge weights are unique. Apply this algorithm to the graph shown in Fig. while there are edges coming out of the Bucket B do Out of all the edges coming out of B.
Students should experience this confusion and the resultant backtracking. the remaining graph would be a "straight forward" path between the two vertices in the given graph (Fig. What would be the worst-case time complexity of this algorithm? Note that we have used a similar technique to find a spanning tree of a graph. we would eventually reach the root without any confusion (Fig. Suppose we apply the Bucket Algorithm starting with the given vertex: the spanning tree thus formed would originate from the given vertex since the given vertex would be the root. We also keep a record of the parent of every vertex in the spanning tree.1
Cutting Edges
If we remove all redundant edges from a given graph and just keep the edges essential to keep the two vertices connected.7. to the destination vertex? The answer is still "no" because a parent may have multiple children. and thus there still exist many diversions. 5.2).7
Finding a Path in a Graph
It is possible to find a path between two vertices provided the graph is connected. With this additional information would it be easier to find a path from the given vertex.178
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5.1). If we keep moving along the edges connecting one vertex to another within the graph.7.7. 5.
.2
Selecting Edges
Does the problem become simpler if we first find the spanning tree of the given graph? Now if we start moving from the given vertex to the destination vertex. Now instead of checking whether the graph is connected or not. since we might have to do a lot of backtracking.7. What is wrong with this approach? If there are cycles in the graph it is possible that we never reach our destination.
5. a time would come when we would reach our destination. However if we start from the destination vertex and keep selecting the parent vertex. but again we may start our journey in the wrong direction and would have to backtrack. would it be less confusing? Perhaps.
5. now the root. It would be useful to pinpoint the similarities as well as the differences. What if there are no cycles in the graph – what if we first make a spanning tree of the graph? Even now it would be difficult to find a path. we better check if the two given vertices belong to a single connected component.
5. would the path. would the problem become simpler? What if we first find a minimum spanning tree of a graph and then move backwards from the destination to the source vertex in order to find shortest paths as described earlier? It is obvious from Figure 5.2). There is indeed a delicate difference between the two – this difference should make us understand why minimum spanning tree algorithm fails to find a shortest path spanning tree and why a shortest path spanning tree algorithm fails to find a minimum spanning tree of a weighted graph G. The step by step working of the two algorithms is shown in the figure below (Fig.The Shortest Path Problem
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5.8. Now assume that the edge weights are different. Do we need a different algorithm from the one used to find a shortest path in a graph with uniform edge weights? Why? If somehow we remove edges of higher weights from the graph without disconnecting the two given vertices. 5.8. We shall come back to this problem after discussing graph traversal techniques in a later section. It should also be kept in mind that the minimum spanning tree can easily be modified to find a maximum spanning tree of a graph while it is not possible to do so in case of finding a longest path in a graph G (with positive edge weights). it will become obvious that the two algorithms are derived from a common ancestor – the Bucket Algorithm.8. be a shortest path? If not then what should be done to achieve our objective? Note that it is easier to find a shortest path in a graph with uniform edge weights.1. We assume that we are finding shortest paths from a single vertex to all other vertices. found using the algorithms of the previous section.8
The Shortest Path Problem
If all edges in the graph were to have the same weight. It is obvious that initially the two algorithms produce similar results but then they depart ultimately producing different results.
. that a minimum spanning tree of a weighted graph does not always provide shortest distances from a given vertex. so first we should solve this problem (which is simpler) before attacking a more complex one. It will still be interesting to investigate how a shortest path algorithm resembles and at the same time differs from a minimum spanning tree algorithm.1
Dijkstra's (like) Algorithm
We produce both the algorithms side by side.
Initialize Dist(x) = 0. v) is minimum where vertex u is in B and vertex v is outside the Bucket B. while there are edges coming out of the Bucket B do Out of all the edges coming out of B.
. and Dist(i) of every other vertex i equal to ∞ . Put edge (u. v) in SST . select one with minimum weight connecting u in B to v not in B. select the edge for which Dist(u) + w(u.
Algorithm 21: Find shortest distance of every vertex from a given vertex x in G & also the shortest path spanning tree of G from x input : A weighted Graph G. Put vertex v in B and edge (u. while there are edges coming out of the Bucket B do Out of all the edges coming out of B.The Shortest Path Problem
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Algorithm 20: Find Minimum Spanning Tree of a graph G input : A weighted Graph G output: A Minimum Spanning Tree M ST
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Put any vertex x of Graph G in a Bucket B. Put vertex v in B. Dist(v) = Dist(u) + w(u. a vertex x output: Shortest Distance Dist(i) of every vertex i from x
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Put vertex x in Bucket B. v) in M ST . v).
In the first iteration of the while loop. Out of all these vertices (some of which are at a distance of one edge and some at a distance of two edges).8. be violated if some edge weights in G are negative. In order to assert that Dijkstra's like algorithm is able to find correct results let us first present a scenario where the above mentioned algorithm fails to find shortest paths. i) goes in the shortest path spanning tree. In the second iteration of the while loop.184
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5. Out of all such vertices (at a distance of one edge from x).8. The cost of this action is equivalent to making p − 2 comparisons.
. however. we consider vertices which are directly connected to x. Again this action means that we have found a shortest path of vertex j from vertex x.3 where some edge weights are negative. This action is tantamount to a claim that we have found a shortest path from vertex x to vertex i. It is obvious from this figure that Dijsktra's algorithm fails to produce correct results. But before we do that let us discuss some of the salient features of this algorithm. Remember vertex i is the last vertex which went into the Bucket B. Remember that this algorithm works on a greedy strategy – it makes decisions on the basis of local conditions – but produces optimal results on a global basis provided all edge weights in a graph are positive. we select a vertex j which is at a shortest distance from vertex x. we consider vertices at a distance of one edge.2
Discussion on Dijkstra's (like) Algorithm
It was interesting to compare the described shortest path algorithm with the corresponding minimum spanning tree algorithm – but at the same time it will be a learning experience if we look into the working of the shortest path algorithm under conditions when it fails to provide optimal results. The cost of this action is equivalent to making p − 2 + p − 3 = 2p − 5 steps. the edge (x. The vertex j goes in the Bucket B and the corresponding edge goes in shortest path spanning tree. 2. 1. and which are indirectly connected to x through vertex i. We show a directed graph in the Figure 5. we select a vertex i which is at a shortest distance from x. and put it in Bucket B. If all edge weights are positive then this claim will be right – it will.
We show another weighted graph in the right diagram of this figure. 2.4). A k-edge shortest path (between two given vertices) requires that the path should be shortest but it should not consist of more than k edges.5. In the graph shown in the right diagram (Fig. The above observation is true for the left or the middle graph in the figure below (Fig. This graph is different from the one shown in the left (and middle) diagram in the following ways: 1. Such one edge paths are shown in the top left diagram in Figure 5.8. The weighted adjacency matrix of the given directed graph provides all one edge paths from vertex x.4) the above observations are not true.3
The Shortest Path Problem Redefined: The kedge Shortest Path Problem
In order to overcome the above mentioned complication we redefine the shortest path problem as follows.8. 5. Instead of finding the shortest path between two vertices we intend to find k-edge shortest paths from a given vertex to every other vertex in a directed graph containing negative edges – the graph may even contain negative weight cycles.8. In other words the shortest distance between any two vertices does not reduce if we move in a cyclic path. If there are no parallel edges in the graph then these paths will also be the shortest paths from vertex x to every vertex adjacent to x.
5. Once we have the 1-edge shortest paths – we can convert them into 2-edge shortest paths and then into 3-edge shortest paths according to the algorithm described below. 5.8. 5. The magnitude of this shortest path is finite. It is evident from the
. The shortest distances of the red vertices keep on decreasing if we move in a cyclic path as shown in red color with the result that the shortest distance of such vertices ultimately approaches minus infinity. The shortest distance path from vertex a to any other vertex is a simple path – no edge or vertex is repeated in this path.4.8.The Shortest Path Problem
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The same weighted graph and its (correct) shortest path spanning tree is shown in Fig. Please note that there are negative weight edges in this graph but no negative weight cycles: a negative weight cycle is a cycle in a graph where the net sum of the weight of the edges in the cycle is less than zero. This happens in any graph where there are negative weight cycles.
k-edge shortest distance of every vertex i from vertex x denoted by Distk (i) output: (k + 1)-edge Shortest distance of vertex i from vertex x. input : A weighted directed Graph G. Thus k keeps increasing and the distances keep going down. we claim that we have found the shortest paths.
.8. w(i.6 where we find different edge path from vertex x. Please note that the resulting shortest paths will be simple paths. When k increases the shortest path distances go down. if no improvement takes place in any path then we stop. It means that k would be less than or equal to p − 1 in case there are no negative weight cycles in the graph. j) + Distk (j)}.8. This graph is reproduced in Fig. 5.4. For the graph shown in the Fig. a vertex x. it is denoted by Distk+1 (i)
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for i = 1 to p do for j = 1 to p do Distk+1 (i) = min{Distk (i). now if we increase k to 6 there is no improvement in any shortest distance. no edge or vertex is repeated in these paths. Algorithm 22: Find (k+1)-edge shortest distance of every vertex from a given vertex x in a weighted directed graph G.8. This situation never becomes stable because of the presence of a negative weight cycle comprising three vertices shown in red color. 5. As soon as all the k-edge shortest paths become stable with increase in k. now we claim that we have found the shortest distance of every vertex from x. this happens when k = 5.
An Important Conclusion
If there are no negative weight cycles in a directed graph consisting of some negative weight edges then we can use our k-edge shortest path technique to find shortest paths.5. The shortest distances of the rest of the vertices become stable as soon as k approaches 5.The Shortest Path Problem
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figure below – as we move from k-edge to (k + 1)-edge shortest paths there is some improvement in the length of a shortest path. 5.
Let us explore the consequences of the presence of negative weight cycles in a graph like the one shown in the right diagram of Fig.
Find k-edge shortest distances of all vertices with respect to vertex a. Remember if we increase k beyond 5 while finding k-edge shortest paths then no (shortest) distance (with respect to vertex a) changes with k.1.6. Find the value of k at which the shortest distances become stable. We show another weighted graph as shown in the right diagram of Figure 5.8. i.2.2. As we increase k the longest path of certain vertices increases.
.8. 5. Find how and at what value of k.4.4
The k-edge Longest Path Problem
The k-edge longest path problem is quite similar to the k-edge shortest path problem. vary k from 1 to 11.8. the shortest distances become stable at a finite value of k? Why? What do you think are necessary and sufficient conditions for the shortest distances to become stable for a finite value of k in a weighted graph? Problem 5. there are positive weight cycles in the graph then we shall be caught in an infinite loop and the longest distances (of at least some vertices) will keep increasing with increase in k. Please note that there is a big negative weight equal to -40 associated with the edge (d. 5. it may become stable after some time as shown in Fig. however. We need to explore if shortest distances of all vertices with respect to vertex a become stable when k increases while finding k-edge shortest paths. 5. the shortest paths becomes stable.7 where we operate on the same graph of Fig.2. Problem Set 5..8.8.3.8.e. Problem 5. If. We now change the magnitude of the weight associated with the edge (f. This is quite expected in view of our prior experience: if there are negative weight cycles in a graph then the shortest path (of certain vertices) keep increasing with k. it is just like maximum spanning tree versus minimum spanning tree problem. 5.8. Find k-edge shortest paths with respect to vertex a while k changes from 1 to 10.8. 5. g) from 4 to 40 as shown in the middle diagram of this figure. Please recall the weighted graph of Fig. In spite of a large negative weight in the graph shown in the right diagram of Fig.2.2.The Shortest Path Problem
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5. Problem 5. Problem 5.8.5. b). Describe an efficient algorithm (based on finding k-edge shortest distances) to find if a directed graph contains negative weight cycles.8. if we further increase k then there is no change in the shortest distance of any vertex with respect to vertex a.2. the same weighted graph is reproduced below in the left diagram of Fig.
The undirected graph is converted into a directed graph as shown in the right diagram of this figure.8.8.8.8.2.9. 5. we have created a two-edge negative weight cycle in the directed graph.2.8.5. 5. Problem 5. Find if the longest paths become stable with increasing value of k. When we apply our k-edge shortest path algorithm to the directed graph (shown in the right diagram of Fig. We show an undirected graph in left diagram of Fig. The working of the k-edge shortest path technique on this graph is shown in Fig. What do you think are necessary and sufficient conditions for the longest distances to become stable for a finite value of k in a weighted graph? Problem 5.8.6. and we get incorrect answers for shortest distances. Problem 5.2.The Shortest Path Problem
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Problem 5. 5.8.8. With negative edge weights (but no negative weight cycles) we expect that our kedge shortest path technique will work with the hope that the k-edge shortest paths will stabilize for a finite value of k and we shall get the optimal answer.2.9) we run into a complication. Comment on the claim that the longest path problem is a hard problem while the shortest path problem is a solvable problem.2. 5. The
. Find k-edge longest paths for each vertex with respect to vertex a in the graph shown in the left diagram of Fig.10. Repeat the above problem for the graph shown in the right diagram of Fig.5
The Shortest Path Problem in Undirected Graphs with Negative Weights
We consider the shortest path problem in undirected graphs with negative edge weights but no negative weight cycles. Remember that the so called technique was designed for directed graphs and in order to apply it to undirected graphs we have to first convert the undirected graph into a directed one.9.
5. We already know that if there are no negative weights then a simple greedy strategy (like that of Dijskrta's Algorithm) will solve the shortest path problem in undirected graphs. It is quite evident that k-edge shortest distances do not stabilize in this graph with increase in k. Is it possible to use the above algorithm to find if an undirected graph contains negative weight cycles? Problem 5. 5.8. The correct shortest paths & distances with respect to vertex a are indicated in the middle diagram.7.
Out of the remaining paths we claim that Distk+1 (i) = min{Distk (i). challenge is how to handle this complication? Please understand that we are limiting our focus on a directed graph which is derived from an undirected graph as shown in Fig.9. input : A weighted directed Graph G.
. 5. w(i. Algorithm 23: Find (k + 1)-edge shortest path & distance of every vertex from a given vertex x in a weighted directed graph G. it is denoted by SP athk+1 (i).196
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Figure 5. j) + Distk (j)} . output: (k + 1)-edge Shortest path of vertex i from vertex x. k-edge shortest path of every vertex j from vertex x denoted by SP athk (j).22: We show an undirected graph with negative edge weights but no negative weight cycles in the left diagram. Now we are in a position to find (k + 1)-edge shortest path of a vertex i from vertex x in a graph using the following algorithm. Its weight is represented by Distk (j). Each such shortest path starts from vertex x and terminates at vertex j. Let us represent a k-edge shortest path of any vertex j from vertex x by SP athk (j). The shortest distances with respect to vertex a are indicated in the middle diagram. a vertex x. The last vertex in a shortest path SP athk (j) is vertex yj before terminating at j. its weight is equal to Distk+1 (i)
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We need to find SP athk+1 (i): We consider k-edge shortest path for every vertex j in the graph except for j where yj = i.8. assume that the last vertex in this shortest path is yj before terminating at j. The undirected graph is converted into a directed graph as shown in the right diagram.
While introducing this algorithm we purposely did not disclose the implementation details ignoring the underlying data structure required to program the algorithm. Please note that this graph contains negative weight cycles.11 for values of k in the range from 1 to 8. Will it be possible to provide a warning that the given graph contains negative weight cycles? What modifications are needed in our existing algorithms to solve this problem?
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5.198 Problem Set 5.1.9
Graph Traversal Techniques
It is possible to traverse a graph in a haphazard manner. Problem 5. 5.
Basics of Graph Algorithms
Problem 5. The objective was to highlight the basic idea and initially suppress the programming details. Apply the above Algorithm to the graph shown in Fig.1).9 and verify that it provides correct results for values of k in the range of 1 to 5.8. You might have noticed that the Bucket-Algorithm is essentially a graph traversal algorithm. 5. The Bucket Algorithm is simple because it is more abstract and flexible.1
Traditional Techniques & the Bucket-Algorithm
It is interesting to note that the Breadth as well as Depth First Searches are two different implementations of the Bucket-Algorithm (Fig. Consider an undirected graph containing negative weight cycles. 5.3. Baase [2] uses JAVA to describe algorithms and this may be one reason why the book is relatively difficult to read even if students have prior knowledge of the language.
. Problem 5. Compare your results with the correct shortest paths given in the middle diagram of Fig.3. Cormen [17] and Skiena [16] use a pseudo programming language and operate at a slightly higher level. We must make every move in a systematic manner to ensure that we do not miss out any vertex belonging to the same connected component [3].8. 5. However.9. Apply the above Algorithm to the graph shown in Fig.9. you would only find some very specific techniques like the Breadth and Depth First Search traversal algorithms – more specialized and less flexible than our Bucket Algorithm. Check if k-edge shortest paths stabilize for a finite value of k. in most of the current textbooks. Efficiency demands that we do not visit the same vertex again and again.3.3.3.8.9.2.
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.24: We show an un-directed graph (left diagram) with a negative weight cycle. The undirected graph is converted into a directed graph as shown in the right diagram.
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The Underlying Data Structure
We know that we use a cross edge to discover a new vertex in the BucketAlgorithm (step 3). all vertices at a distance of one edge from the starting vertex are selected (or goes in the Bucket). In the first iteration of the BFS traversal.10. in the k th iteration of the BFS traversal.9. Some of these cross edges come from vertices that entered the bucket earlier. others from vertices that are new comers in the bucket.1.
. The non spanning tree edges are shown in pink color by thin lines.26: A Graph G shown in the left diagram. Using a Last in First Out (LIFO or a stack) or a First In First Out (FIFO or a queue) data structure to store the already discovered vertices would make all the difference: a stack implementation would convert the Bucket-Algorithm into a Depth First Search while a queue would transform it into a Breadth First Search. its BFS spanning tree is shown in the middle while a DFS spanning tree of this graph is shown in the right diagram.10
Some Graph Theoretic Claims
We make the following claims about the BFS traversal in an un-directed and connected graph G: Claim 5.1 below. Please see the Figure 5.10. all vertices at a distance of exactly k edges from the starting vertex will be discovered. The way we decide which vertex to choose would convert the BucketAlgorithm into a Breadth First Search. Depth First Search.
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5. or a combination of the two.
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.27: Pictures of the Bucket B while performing a BFS traversal in a graph starting from a vertex a shown in the top diagrams.
5. If every edge of G connects a vertex x to vertex y in the BFS spanning tree of G such that Dist(x)−Dist(y) is exactly equal to 1 then graph G is a bipartite graph. Claim 5. Please see Fig.10. (Remember Dist(x) is the distance of vertex x from the starting vertex in the BFS spanning tree). 5.2.2.10.30 below.10.
. If there is an edge in G other than the ones in the BFS spanning tree of G connecting vertex x and vertex y in the BFS spanning tree such that Dist(x) − Dist(y) is one then there will be an even cycle in graph G.10.6.29 below. Claim 5. If there is no edge in G other than the ones in the BFS spanning tree of G then G is a tree provided G is a connected graph.10.10.4.10. Claim 5. Claim 5. BFS spanning tree of a graph G is a minimum distance spanning tree in terms of number of edges between the starting vertex and any other vertex in graph G.10.7. Similarly if there is an edge of G connecting vertex x to vertex y in the BFS spanning tree of G such that Dist(x) − Dist(y) is exactly equal to 0 then graph G is not a bipartite graph. If a graph G is a tree then there will be a unique path between every pair of vertices of G. 5.9. Similarly if there is a unique path between every pair if vertices in a connected graph G then G is a tree.Some Graph Theoretic Claims
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Claim 5.5. Please see Fig. If there is an edge in G other than the ones in the BFS spanning tree of G connecting vertex x and vertex y in the BFS spanning tree such that Dist(x) − Dist(y) is zero then there will be an odd cycle in graph G.10. Claim 5. (Remember Dist(x) is the distance of vertex x from the starting vertex in the BFS spanning tree). A graph G is bipartite if and only if it does not contain any odd cycles. If there is an edge in G other than the ones in the BFS spanning tree of G then this edge will connect two vertices x and y in the BF S spanning tree such that Dist(x) − Dist(y) is either zero or one whereas Dist(x) is the distance of vertex x from the starting vertex in the BFS spanning tree. Please see the Fig. Claim 5.2 Claim 5.10.8.3. 5. Please see Fig.
As we know the distance of vertex h from vertex a. these two paths along with edge j. we also know a path between vertex h and vertex a as shown in green color in the top right diagram. h makes a cycle. It will be an odd cycle as vertices h and j are at the same distance from vertex a. We also know a path between vertex j and vertex a.
.28: BFS spanning tree of a graph G is shown. The edge between vertex h and vertex j creates an odd cycle in the graph as both these vertices are at the same distance from vertex a.204
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.29: BFS spanning tree of a graph G. The edge between vertex h and vertex i creates an even cycle in the graph G as vertex h is at a distance one larger than the distance of i from a.
5.Some Graph Theoretic Claims
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Claim 5. the given vertex goes in the bucket first.31: We show the contents of the Bucket at different times while we make a BFS traversal of a graph. one more vertex and an edge go in Bucket. In the first iteration of this algorithm. When we grow a BFS spanning tree in a Bucket starting from a given vertex. we discover a new vertex because of an edge going out of the Bucket. Initially there will be only one vertex and no spanning edge as shown in the top left corner. Finally there will be 10 spanning edges and 11 vertices in the Bucket. At any point in time there will be k vertices and k − 1 edges in the Bucket.
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Figure 5. Please see the Fig.10.10.31.
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. This is about single source shortest path algorithms assuming that all edge weights are positive in the given weighted graph. Finding k-edge shortest or longest paths in a weighted graph. 2. 3.10. The interesting thing about this style of design is that we shall be using a single building block (sometimes the Bucket Algorithm and some times a 2-edge Shortest path Algorithm) to design or describe an algorithm. If G is a connected graph and if the number of edges is one less than the number of vertices in G then G is a tree. We shall further modify the Bucket algorithm to achieve our objectives. Here we assume that there may be negative edge weights in the given weighted directed graph. Design of a shortest path algorithm for directed acyclic graphs with negative as well positive edge weights.12.11. 2. 1.11
Shortest Path Algorithms
We have already done the following in earlier sections of this chapter. Modifying the Bucket algorithm to find a shortest path spanning tree (SST) or shortest distances in a weighted graph from a given vertex. If G is acyclic & number of edges in G is one less than the number of vertices then G is a tree. 3. Claim 5.
5. Modifying the Bucket algorithm to find a minimum spanning tree (MST). Analyze the existing minimum spanning tree & shortest path finding algorithms and improve their efficiency as far as possible.10. We shall again modify the Bucket algorithm to achieve our objective. On the basis of that precious prior knowledge we can design interesting shortest path algorithms.208
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Claim 5. We also assume that the directed graph may have cycles. Design of single source shortest path algorithms based on our prior knowledge of finding k-edge shortest paths from a given vertex. We intend to do the following in this section: 1.
5. Algorithm 24: (Crude-Dijkstra): Find shortest distance of every vertex from a given vertex a in G & also the shortest path spanning tree (SST ) of G from vertex a input : A weighted Graph G. We apply the shortest path algorithm and find the shortest distances from vertex a in this graph where a = 1 also shown in Fig.
5.Shortest Path Algorithms
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4. Design of all pair shortest path algorithms including the slow all pair. 5. Put vertex k in B. That is why the shortest paths found by the crude Dijkstra's algorithm are not correct. k).
Food for thought: The graph shown in Fig. and Dist(k) of every other vertex k from vertex a equal to ∞.32.it is fixed and finalized.32 has negative edge weights. a vertex a output: Shortest distance Dist(k) of every vertex k from vertex a
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Put vertex a in Bucket B.11. Dist(k) = Dist(j) + w(j. faster all pair.32. Floyd-Warshall and Johnsons shortest path algorithms. We have already witenessed that this algorithm does not always provide correct
. Initialize Dist(a) = 0. The visual tool demonstrates at what stage and when a vertex enters the Bucket while the shortest path algorithm moves forward. 5. We shall study this algorithm using different tools before improving its time complexity. Put edge (j. The shortest distances from vertex 1 as found by this algorithm are also indicated in the bottom diagram. k) in SST . while there are edges coming out of the Bucket B do Select the edge for which Dist(j) + w(j. As soon as a vertex enters the bucket its color changes from blue to orange and then its distance from the start vertex cannot change . We introduce a visual tool known as "opening up the graph" as shown in Fig. k) is minimum where vertex j is in B and vertex k is outside the Bucket.1
Single Source Shortest Path Algorithms with positive edge weights
We copy here the modified Bucket algorithm designed in the last section to find shortest paths from a given start vertex.
32: We use a new tool "Opening up the graph" as shown here. As soon as a vertex enters the bucket its color changes from blue to orange and then its distance from the start vertex cannot change .
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Figure 5.it is fixed and finalized. It also shows how and when the distance of a vertex changes from the start vertex. It shows at what stage (edge distance) a vertex enters the Bucket while executing crude shortest path algorithm.
Although it provides correct results for positive edge weights. But we also know that this very algorithm provides correct results when all edge weights are positive. In fact there are two complications with this algorithm: 1.3. The issue of negative edge weights will be handled in the next sub-section.Shortest Path Algorithms
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results for graphs with negative weight edges. What we essentially do is to change the graph itself at each step as shown in Fig.this will automatically cut down the time complexity of this single source shortest path algorithm to p2 . The number of edges coming out of the Bucket B at any time is proportional to p2 as illustrated in Fig.11. Its time complexity is too high.
Figure 5. The outcome
.33. 5. Surprisingly this can be done with a slight modification in the crude algorithm as shown below. Somehow we should reduce the number of relevant edges coming out of the Bucket to as small as p . We shall reduce the time complexity in the following refined version of shortest path algorithm. 2. 5. It does not provide correct results for negative edge weights. As this loop runs as many times as p so the overall time complexity will be p3 of this algorithm. This requires as many comparison steps to move forward in the while loop. Now we consider the edges coming out of the Bucket from (only) vertex a instead of edges coming out of the Bucket from all vertices in the Bucket.33: The number of edges coming out of the Bucket B at any time will be proportional to p2 under worst case conditions.
You can yourself make suitable modifications in the algorithm according to these modifications.
. k)
of this step is that the number of edges (which really matters) coming out of the bucket B from vertex a is limited by p and not p2 . a vertex a output: Modified adjacency matrix of graph G in which the row corresponding to vertex a gives shortest distances of every vertex from vertex a. j) + w(j. It is interesting to note how the shortest distances are provided in the output in this new algorithm.
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Put vertex a in Bucket B while there are edges coming out of the Bucket B from vertex a do Select the edge for which w(a. Again it will be interesting to understand that this algorithm will provide the weight of the minimum spanning tree of a weighted graph . k). 5. We have provided a hint in the same figure. We can recover shortest paths by adding a parent table as shown in Fig. Put vertex j in B for every edge coming out of B from vertex j do assign min{w(a. w(a. If we can recover shortest paths then we should also be able to recover a minimum spanning tree using the refined minimum spanning tree algorithm. k)} to w(a.38 determining the actual shortest paths in a directed graph is an interesting problem.not the minimum spanning tree itself as shown in the diagrams below. It is important to note that we may be able to make similar modifications in the crude minimum spanning tree algorithm to make it more efficient.212
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Algorithm 25: (Refined-Dijkstra):Find shortest distance of every vertex from a given vertex a in weighted graph G input : A weighted Graph G. The refined version of that algorithm is given below . however. Food for thought: Given a parent table and shortest distances as shown in Fig. could not be found without an extra effort using this refined algorithm.its time complexity also reduces from O(p3 ) to O(p2 ). See how by increasing memory or space requirements we can reduce the time complexity of an algorithm. j) is minimum where vertex j is outside the Bucket B. The shortest paths.38.37 & 5. 5.
We can find shortest paths in such graphs in the presence of negative edge weights and we can also find longest paths in the presence of positive edge weights.35: We modify the graph as we move forward in the minimum spanning tree algorithm.214
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Figure 5.2
Single Source Shortest Path Algorithms for Directed Acyclic Graphs
We know that by definition a directed acyclic graph contains no cycles. The Bucket Algorithm (described earlier) can easily be modified to create an algorithm which can find shortest paths from any given vertex in a very efficient manner. We can even solve the Hamiltonian Path problem in this very restricted class of graphs. The crucial question is which will be the next vertex to go in the bucket and on what basis.11. The intuition of this algorithm comes from the observation that we can always arrange the vertices of a DAG such that all edges in the DAG move from left to right as shown in the figure below. If we need to find shortest paths from a source vertex a then we should put that vertex in the bucket first. An answer to this question will not
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5. This observation is explained in detail in Chapter 8 while discussing directed acyclic graphs.
Shortest distances with respect to vertex a are indicated in the bottom left graph. Weight of the minimum spanning tree is indicated in the bottom right graph. Please note that the shortest paths and minimum spanning tree are not easily available here.36: The original graph is shown in the top diagram.
.Shortest Path Algorithms
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Figure 5.
38: How to recover shortest paths with the help of a parent table?
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Figure 5.216
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Figure 5.37: We need to add a parent table array in order to find shortest paths in addition to shortest distances.
In the bottom diagram we arrange its vertices such that all edges move from left to right.11.it will also decide its time complexity. So coming back to the important question: which vertex (and on what basis) should enter the bucket after putting the start vertex in the bucket? The interesting observation is that after arranging vertices of a DAG (such that all edges move from left to right) the next vertex to enter the bucket has already been decided. Please see the concept map in Fig. We need no comparisons or extra steps to make this decision.39: A directed acyclic graph D shown in the top diagram. The numbers inside each vertex is the start time and finish times obtained during a depth first search of the directed graph. Recall how we select the next candidate vertex which enters into the bucket in case of Dijkstra like algorithm or in case of Prim's like algorithm.
. 5. The corresponding algorithm is described below. Its working is shown in Fig. It is the next left vertex in the new arrangement of the vertices of the graph. 5. In both these algorithms we made certain comparisons to select the next entrant into the bucket.40.Shortest Path Algorithms
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only determine the character of this algorithm .10.
Figure 5. It will be an interesting challenge to derive the time complexity of this elegant shortest path algorithm.
w(a.40: A concept map depicting which vertex should next enter the bucket in different algorithms. k). a vertex a output: Modified graph D in which the weighted edges coming out of vertex a provides shortest distances from this vertex.
Algorithm 27: Find shortest distance of every vertex from a given vertex a in a directed acyclic graph D input : A directed acyclic and weighted Graph D.
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Arrange vertices of the DAG such that all edges move from left to right Put given vertex a in Bucket B while there is an edge going out of the Bucket B do Select the next right vertex j. j) + w(j. k)} to w(a.218
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Figure 5. k)
. Put vertex j in B for every edge coming out of B from vertex j do assign min{w(a.
11.10. How about if there are negative weight edges in the graph? Would this algorithm still provide correct results? 4. Can we use a similar algorithm to find a Hamiltonian Path in a directed acyclic graph provided it exists?
. Can we still apply this algorithm without any modification to find shortest paths? How about if we need to find shortest paths from a vertex other than the source vertex in a DAG? 3.Shortest Path Algorithms Food for Thought
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1. Assume that we have a DAG with multiple source vertices. We need to find shortest paths from vertex a in this graph. The graph is already drawn such that all edges are going from left to right. If we had applied Dijkstra's algorithm to this graph then it would have selected vertex c (instead of vertex b) as the edge joining vertex a with vertex c having the minimum weight. How about if we need to find longest paths instead of shortest paths from a given vertex in this or any other directed acyclic graph? Which vertex will next end up in the bucket? What changes are needed in this algorithm to find longest paths? Please see figure 5. Do you think Dijkstra's like algorithm will also find correct shortest paths in a DAG? And at what cost? 2. Consider the directed acyclic graph shown in Fig. 5. The next vertex which goes in the bucket is vertex b in spite of the fact that the edge joining vertex a with vertex b is the heaviest edge in this graph.42.
. Weight of every edge not indicated in the diagram is equal to 1. The vertices of the DAG have been arranged such that all edges move from left to right.220
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Figure 5.41: We need to find shortest distances from vertex a in this graph. Now the next vertex to move in the bucket will always be the next left vertex.
Shortest Path Algorithms
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Figure 5. The vertices of the DAG have been arranged such that all edges move from left to right. Now the next vertex to move in the bucket will always be the next left vertex.
.42: We need to find longest distances from vertex a. Weight of every edge not indicated in the diagram is equal to 1.
k) + Disti (j)}
A 2-edge shortest path algorithm
Let us make this building block consistent with our earlier policy of modifying the graph as we move forward in the algorithm. i-edge shortest distance of every vertex k from vertex a denoted by Disti (k) output: (i + 1)-edge Shortest distance of vertex k from vertex a.11. Let us recall the (i + 1)-edge shortest path algorithm described earlier in this chapter.3
Single Source Shortest Path Algorithms for directed graphs with negative edge weights
We have discussed directed acyclic graphs and the ease with which we can find shortest paths in such graphs in the last section. it is denoted by Disti+1 (k)
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for each (terminating) vertex k do for each (intermediate) vertex j do Disti+1 (k) = min{Disti (k). k) represents the initial one edge path of vertex k from
. We shall follow this terminology throughout this section. a vertex a. Here w(a. Remember we have already reduced the time complexity of our modified shortest path algorithm from p3 to p2 . We assume that all distances are to be measured and minimized with respect to a start vertex a. In this building block we just convert the one edge distances into 2-edge shortest distances as given in the following algorithm. Now is the time to get rid of the bucket as it is hindering our way to handle negative edge weights. input : A weighted directed Graph G. That was Dijkstra's like algorithm which can handle positive edge weights in a graph with cycles but does not provide us with correct results if there are negative edge weights in the graph. Then we need to revert back to algorithms discussed earlier. If a graph is cyclic then we cannot use the simplicity and elegance of this algorithm. w(j. It has been copied below. The terminating vertex is k and the intermediate vertex is indicated by vertex j.222
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5. This will make it friendlier to use as a building block. Algorithm 28: (Bellman-Ford Building Block-A): Find (i + 1)-edge shortest distance of every vertex k from a given vertex a in a weighted directed graph G.
1-edge shortest distance of every vertex k from vertex a provided by the row corresponding to vertex a in the adjacency matrix G output: Modified graph G in which row corresponding to vertex a provides 2-edge shortest distance of every vertex k from vertex a
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for each (terminating) vertex k of graph G do for each (intermediate) vertex j of graph G do Find a 2-edge path from vertex a to k passing through j.1)
w(a. w(a. k)}
Algorithm 29: (Bellman-Ford Building Block-B): Find 2-edge shortest distance of every vertex k from a given vertex a in a weighted directed graph G. k). j) + w(j. a vertex a. input : A weighted directed Graph G. j) + w(j. k). and assign min{w(a. k)} to w(a. The 2-edge shortest distance w(a. k) = min{w(a. w(a.Shortest Path Algorithms
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vertex a. k) from vertex a to vertex k will then be given by the following recursive equation:
(5. k)
.
13). Please note that it is immaterial whether you execute the j-loop first and the k-loop later or vice versa.44: We execute the k-loop first and then the j-loop in the 2-edge shortest path algorithm. Here intermediate vertex j is fixed at 4 and the destination vertex k is varied first.11.Shortest Path Algorithms
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Figure 5. 5. The end result remains the same.it also becomes a powerful tool for the algorithm designer (Fig.
. This switching of two loops is an exciting idea in the development of shortest path algorithms .
As you can understand this 2-edge shortest path algorithm becomes a building block for so many shortest path algorithms. k). While vertex (13) in the top row of the recursion tree corresponds to the two edge shortest distance from vertex 1 to vertex 3 in the original graph. k)} is also shown in the bottom diagram. The number of edges in the recursion tree corresponds exactly to the number of steps performed by the two edge shortest distance algorithm. A vertex (13) in the bottom row of the recursion tree corresponds to one edge distance of vertex 3 from vertex 1 in original graph. 5. A recursion tree corresponding to the equation w(a. w(a. j)+ w(j.45.226
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Figure 5.45: Execution of the 2-edge shortest path algorithm is illustrated. Please note that the number of edges in the recursion tree is exactly equal to the number of steps performed by the algorithm.
The Recursion Tree
We illustrate the 2-edge shortest path algorithm by one additional tool known as the recursion tree (although strictly speaking it is a DAG) as shown in Fig.
. k) = min{w(a.
The outer most loop in the Bellman-Ford Algorithm (known as Algorithm 30) is represented by blue color in Fig. 5. The other is the intermediate vertex j loop represented by orange color in the puzzle. k)} to w(a. Here we introduce another tool to study an algorithm. and assign min{w(a. k) := min{w(a. k)
This algorithm resembles the so called Bellman-Ford algorithm with a time complexity of p3 as we have three nested for loops. See without this blue box the remaining two boxes (brown and orange) represents the 2-edge shortest path algorithm known as Algorithm 29. One is the terminating vertex k loop represented by brown color in the colored puzzle. Algorithm 30: (Bellman-Ford1): Find shortest distance of every vertex k from a given vertex a in G input : Adjacency matrix of weighted Graph G and a start vertex a output: Modified adjacency matrix G in which row corresponding to vertex a provides shortest distance of every vertex k from vertex a in G
1 2 3 4
for w(a. k)}. k).Shortest Path Algorithms Bellman-Ford like Shortest Path Algorithm
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In order to find a shortest distance in a weighted graph with negative edge weights we just have to run the 2-edge shortest path building block sufficient number of times as shown in the following algorithm. w(a.
The Colored Puzzle
The colored puzzle highlights the fact that Algorithm 29 can be used to become a building block for Bellmam-Ford Algorithm. Please note that the nesting of f or loops in these algorithm can be nicely
. This outer loop executes Algorithm 29 p − 1 number of times and minimizes the value of w(a. k) according the following equation: w(a. Algorithm 29 has two f or loops. j) + w(j. k). j) + w(j. Fig.46.
5.47 along with the corresponding colored puzzles. It is possible to run the intermediate vertex j-loop (orange color) first and then the destination vertex k-loop (brown color) or vice versa. Other shortest path algorithms can also be represented by this puzzle with the addition of the start vertex a loop shown by green color in coming figures. It is interesting to note that each such algorithm appears in pairs .
. cannot be changed without adversely affecting the performance of the algorithm. The outer most loop in the algorithm is represented by blue color.
for i =1 to p-1 for each vertex k for each vertex j for i =1 to p-1 for each vertex j for each vertex k
Figure 5.46: A colored puzzle depicting the positioning of the different for loops in the shortest distance finding algorithm.46. The positioning of the j and k loops in the Bellman-Ford Algorithm can be interchanged without affecting the outcome of this algorithm as shown in Fig. however. 5. The position of the blue loop.thanks to the colored puzzle which provokes a learner to note this interesting property. The recursion tree corresponding to the Bellman-Ford Algorithm is shown in Fig.228
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captured by nested colored boxes in the colored puzzle.
which is equal to p3 . Time Complexity of Bellman-Ford shortest path algorithm We have witnessed that the time complexity of Bellman-Ford1 shortest path algorithm is O(p3 ) with an adjacency matrix data structure. It is also possible to rephrase this algorithm so that we do not need any
.47: The recursion tree corresponding to Bellman-Ford like algorithm. How about a sparse graph and if we represent it using an adjacency list representation? It may be a good idea to see if the time complexity can be reduced to at least O(pq) which will be less than O(p3 ) for a sufficiently sparse graph. The colored puzzle corresponding to this algorithm is again shown in the bottom diagram.Shortest Path Algorithms
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Figure 5. The complexity of algorithm is equal to the number of edges in this recursion tree .
Its working is explained in the figure below. k)}
Dist(k) = min{ Dist(k). Algorithm 31: (Bellman-Ford2): Find shortest distance of every vertex k from a given vertex a in G input : Directed and weighted graph D. In this version of Bellman-Ford the time complexity will always be pq with an adjacency list data structure and without an extra intelligence in its implementation. k). k) in graph D do Dist(k) = min{Dist(k). k)} a
Dist(k) Dist(j)
j
a
Dist(k) = 6 Dist(j) = 4
w(j.48: How and when the (shortest) distance Dist(k). Dist(j) + w(j.k) = 3
k
j
w(j. Line 3 is the basic building block of this algorithm. with respect to vertex a changes when we consider the directed edge (j.k) = 1
k
New Dist(k) = 6
New Dist(k) = 5
Figure 5. The rephrased algorithm is shown below.
1 2 3
for i=1 to p-1 do for every directed edge (j.230
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extra intelligence (in its implementation) to make it a O(pq) algorithm.
.k)
k
a
Dist(k) = 6 Dist(j) = 4
j
w(j. Distance Dist(a) of vertex a from itself is zero and from any other vertex it is infinite. output: Distance array Dist(k) storing minimum distances of every vertex k from vertex a. Dist(j) + w(j.
11. They will all use a single building block which is the 2-edge shortest path algorithm described earlier. k)
The Faster All Pair shortest Path Algorithm It is interesting to note that if we switch the blue p-loop with the green a-loop in the colored puzzle as shown in the diagram below then we end up with another all pair shortest path algorithm. k)} to w(a. The slow all pair shortest path algorithm has a O(p4 ) complexity. and assign min{w(a. The Slow All Pair shortest Path Algorithm The Bellman-Ford like algorithm finds shortest distances from a fixed start vertex in a graph. w(a. k). Algorithm 32: Slow All Pair: Find shortest distance of every vertex from every vertex in G input : Adjacency matrix of weighted Graph G output: Shortest distance of every vertex from every vertex in G
1 2 3 4 5
for each (start) vertex a of graph G do for j) + w(j. faster all pair shortest path algorithm has a O(p3 log2 p) complexity while Floyd-Warshall all pair shortest path algorithm has O(p3 ) time complexity under worst case conditions. The working of this algorithm is shown in the diagram below.Shortest Path Algorithms
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5. It will be interesting to derive its time complexity and compare it with that of slow all pair shortest path algorithm. This algorithm is depicted pictorially by the colored puzzle shown in the left diagram of the figure below. known as faster all pair shortest path algorithm.4
All Pair Shortest Path Algorithms
We shall describe now three all pair shortest path algorithms.
. If we run this algorithm for every vertex in the graph then we end up with the so called slow all pair shortest path algorithm with a time complexity of O(p4 ).
k)
. w(a. w(a. The left diagram depicts the slow all pair shortest path algorithm while the right diagram represents the faster all pair shortest path algorithm.232
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for each vertex a for i =1 to p-1 for each vertex j for each vertex k
for i =1 to p-1 for each vertex a for each vertex j for each vertex k
w(a.
Algorithm 33: Faster All Pair: Find shortest distance of every vertex from every vertex in G input : Adjacency matrix of weighted Graph G output: Shortest distance of every vertex from every vertex in G
1 2 3 4 5
for i=1 to p-1 do for each (start) vertex a of graph G do for each (intermediate) vertex j of graph G do for each (terminating) vertex k of graph G do Find a two edge path from vertex a to vertex k in G passing through an intermediate vertex j.k) = min{w(a. There is another p-loop . and assign min{w(a.49: The start vertex loop is represented by green color. k). j)+w(j.k).it is represented by blue color. k)} to w(a. k)}
Figure 5. j) + w(j. The diagram shows the effect of switching between the blue and the green loops. The intermediate vertex j loop is represented by orange color and the terminating vertex k loop is represented by brown color.
50: Working of the faster all pair shortest path algorithm is shown.
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Figure 5. Please note that from two edge all pair shortest distances we jump to 4-edge and then to 8-edge shortest distances.
51: The recursion tree and the colored puzzle corresponding to faster all pair shortest path algorithm. Here it is also obvious that from 2-edge shortest distances we directly jump to 4-edge shortest distances.234
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Figure 5.
.
52. Is the time complexity of this algorithm any better than that of slow all pair shortest path algorithm? If yes then why? 2.51. Please note that now we have only three loops and there is in fact no need to have the fourth loop .
. 3. Its time complexity is as good (or as bad) as that of Bellman-Ford algorithm (which is a single source shortest path algorithm) for non sparse graphs. 5.Shortest Path Algorithms Food for thought
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1. We know it can be represented by the colored puzzle as shown in Fig.53 & 5. In the slow all pair shortest path algorithm the blue loop should run from 1 to p and not from 1 to log2 p? Why? All Pair (Floyd-Warshall) Shortest Path Algorithm Consider the faster all pair shortest path algorithm. 5.54. You may have noticed that in this algorithm (FAster All Pair) the number of edges in shortest paths jumps in the powers of 2 as shown in Figure 5.53. If the intermediate vertex j loop (also known as the orange loop in the colored puzzle) becomes the outermost loop while the source vertex a loop (green) and destination vertex k loop (brown) are inner loops (in any order) then (surprisingly) we end up with one of the most efficient all pair shortest path algorithms as shown in Fig. There are essentially four for loops in this algorithm represented by four nested rectangles in the multi colored puzzle. See Fig. 5. The basic building block of this algorithm is the same recursive equation that we used in other algorithms.the so called blue loop in the colored puzzle. The time complexity will now be O(p3 ) as there are only three loops. The recursive equation used as a building block is also indicated in this figure. What does that mean? The outer most loop (the blue colored loop in the colored puzzle) should run from 1 to p or from 1 to log2 p? How will it affect the time complexity of this so called faster all pair shortest path algorithm.
w(a.k). j)+w(j.52: The colored puzzles corresponding to faster all pair shortest path algorithm. k)} to w(a. k). j) + w(j.
Algorithm 34: (Floyd-Warshall): Find shortest distance of every vertex from every vertex in G input : Adjacency matrix of weighted Graph G output: Shortest distance of every vertex from every vertex in G
1 2 3 4
for each (intermediate) vertex j of graph G do for each (start) vertex a of graph G do for each (terminating) vertex k of graph G do Find a two edge path from vertex a to vertex k in G passing through an intermediate vertex j.k) = min{w(a. k)
. w(a. and assign min{w(a. k)}
for i =1 to log2p for each vertex a for each vertex j for each vertex k for i =1 to log2p for each vertex a for each vertex k for each vertex j
Figure 5.236
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w(a.
Please appreciate the fact that for the entire galaxy of such algorithms we use a single building block .54. In this subsection we shall study another all pair shortest path algorithm which works faster than O(p3 ) for graphs which are sufficiently sparse. Note that the Bucket algorithm is the ancestor of most of these algorithms.Shortest Path Algorithms
for i =1 to for each vertex j p-1 for each vertex k for each vertex a for each vertex j for each vertex a for each vertex k
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Figure 5.55. Food for Thought: What is the trick or intuition behind this algorithm? Is this not surprising that without a fourth loop we can design an all pair shortest path algorithm? A spectrum of single source and all pair shortest path algorithms We show a spectrum of shortest path algorithms in Fig.the 2-edge shortest path algorithm. 5. Required Prior Knowledge: First we shall talk about the prior knowledge required to understand this
. Once the j-loop becomes the outer most loop then it does not matter if the order of the green loop and that of the brown loop is interchanged. This is the best performance seen so far for an all pair shortest path algorithm.5
Johnson's all Pair Shortest Path Algorithm
We have studied in the last section that all pair shortest path algorithm (Floyd-Warshall) has a worst case time complexity of O(p3 ). We also show connections or links between different algorithms.53: We show the possibility of having the intermediate vertex j loop become the outer most loop.11.
5. There is no fourth loop in this diagram and surprisingly there is no need for it. 5. This panorama of shortest path algorithms is also depicted by the colored puzzle shown in Fig.
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Figure 5.54: The recursion tree corresponding to Floyd-Warshall all pair shortest path algorithm. The number of edges in the recursion tree is equal to the number of steps performed by the said algorithm - and this is equal to O(p3 ).
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Figure 5.55: We show a concept map of various single source and all pair shortest path algorithms.
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w(a,k) = min{w(a,k), w(a, j)+w(j, k)}
2-edge shortest path algorithm
Bellman-Ford (single source)
Slow All Pair
Faster All Pair
Floyd-Warshall (all pair)
Figure 5.56: We show the galaxy of single pair and all pair shortest path algorithms. Each algorithm can be represented by a different color arrangement in the rectangular puzzle. It is interesting to note that each shortest path algorithm has at least one dual with the same performance and output.
Figure 5.57: It will be interesting to see if any of these colored arrangements represents one of the already discussed shortest path algorithms.
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algorithm. If that becomes clear then it is almost trivial to appreciate the innovation behind this algorithm. It is interesting to note that this is not entirely a new algorithm - it innovatively combines two shortest path algorithms (Dijkstra + Bellman-Ford) and creates an all pair shortest path algorithm such that the overall time complexity becomes better than that of best known shortest path algorithm - Floyd-Warshall (p3 ) under certain conditions. 1. We know that Dijkstra's shortest path algorithm finds correct shortest paths from a single vertex provided all edge weights are positive. We also know that its time complexity is O(p2 ) if we use an adjacency matrix as a data structure. Its time complexity can be improved with an adjacency list data structure provided we use a minimum heap to locate the next vertex which goes in the bucket. The improved time complexity is O(qlogp). This can further be improved to O(plogp + q) if we use a Fibonacci heap to implement the minimum priority queue. If all edge weights are positive then we can apply Dijkstra's algorithm p times to find all pair shortest paths in O(p2 logp + pq). This time complexity is better than O(p3 ) (Floyd-Warshall) in sufficiently spare graphs. But if there are negative edge weights then we shall get incorrect results. 2. We know that Bellman-Ford algorithm can find shortest paths from a single source vertex in time O(p3 ) even if there are negative edge weights in a directed graph. The time complexity of this algorithm is O(pq) provided we use an adjacency list as a data structure to represent the input graph. Again this is an improvement over O(p3 ) provided we have a sufficiently sparse graph. If we convert this O(pq) Bellman-Ford algorithm into an all pair shortest path algorithm then its time complexity would become O(p2 q) which is worse than O(p3 ) for Floyd-Warshall. So we need to do something more? Something very innovative? Johnsons'Algorithm = Bellman-Ford + Innovation + Dijkstra Johnson's algorithm first uses Bellman-Ford to check if there are any negative weight cycles. If there are no negative cycles then this algorithm somehow uses results of Bellman-Ford algorithm to convert negative edge weights into positive without disturbing the relative path lengths. This requires time proportional to pq. Once all edge weights are made positive we can use
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the improved Dijkstra's algorithm to find all pair shortest paths in time O(p2 logp + pq). Consider the Bellman-Ford algorithm described earlier and reproduced here. It was already discussed that its worst case time complexity will be O(pq). All distances are measured with respect to vertex a. We initialize Dist(a) = 0 and set Dist(x) of every other vertex x from vertex a equal to infinity. Algorithm 35: Find shortest distances of every vertex k from vertex a in D input : Directed and weighted graph D. Distance Dist(a) of vertex a from itself is zero. output: Distance array Dist(k) storing minimum distances of every vertex k from vertex a.
1 2 3
for i = 1 to p − 1 do for every directed edge (j, k) in the graph D do Dist(k) = min{Dist(k), Dist(j) + w(j, k)};
Checking Negative Weight Cycles in a directed graph: The above algorithm finds shortest paths correctly in case there are no negative weight cycles in the directed graph. If there are negative weight cycles then complications arise as already discussed. Under such conditions this algorithm should at least inform us that in the given graph there are negative weight cycles. What modification is needed in this algorithm for this extra intelligence? The required modification is simple and elegant? We run the outer most loop p − 1 times and store distance of each vertex x from the start vertex a. We then run this loop one more time to check if any distance changes. If it does then it means we have negative weight cycles in the directed graph reachable from vertex a as shown in the figure below. If it does not then there are no negative cycles reachable from vertex a in the directed graph. Please see the figure below. Food for Thought 1. How about if the distance of no vertex changes after an initial iteration i when i < p − 1. Does that mean there are no negative weight cycles and should we stop without further iterating? See Fig. 5.59.
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Figure 5.58: If we find i-edge distances in this graph from vertex a then we observe that some distances will change when i goes from 4 to 5. This does not confirm that there is a negative weight cycle. But when a distance of a vertex from vertex a changes when i goes from 5 to 6 then that is a confirmation that there is indeed a negative weight cycle in this graph. 2. How about if we apply Bellman-Ford at a vertex which is not reachable to a negative cycle. See Fig. 5.60. Does that mean we have to apply Bellman-Ford algorithm at each vertex to find if there are any negative weight cycles in the graph? But that will be very costly?
Figure 5.59: After finding 3-edge shortest paths in this graph from vertex a there will be no change in distance calculations. Does that mean that we should stop here and declare that there are no negative weight cycles in this graph? Applying Bellman-Ford Algorithm once to determine Negative Weight Cycles The problem is how can we apply Bellman-Ford algorithm just once from a (special) vertex and check if there are any negative weight cycles in the
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Figure 5.60: If we apply Bellman-Ford algorithm in this graph to find shortest distances from vertex e then we should never be able to confirm that there will be negative weight cycles in this graph. If, however, we apply this algorithm from vertex a then it is possible to verify that indeed there are negative weight cycles in this graph. directed graph? The answer to this problem is given in elegant transformation shown in Fig. 5.61. The application of Bellman-Ford algorithm in the transformed graph just once (from the newly added vertex x) provides us the following information: 1. Whether there are any negative weight cycles in the graph. If there are any negative weight cycle then the algorithm should not move forward and should terminate. 2. If there are no negative weight cycles then what are the shortest distances of each vertex from the newly added vertex x. This information will further be used to convert negative edge weights into positive edge weights. If the graph does not contain any negative weight cycles but contains negative weight edges then we should somehow try to make the negative weight edges positive and then apply Dijsksta's algorithm from each vertex as already planned. How about adding a big positive number in each edge weight such that every (edge) weight becomes positive? See what complication would arise if we do so as shown in Fig. 5.62. Thus we cannot add an arbitrary positive number in each edge weight as it will disturb the relative path lengths in the new graph as shown in the following figure. Converting negative weight edges into positive weight edges Please note that we need to simultaneously fulfill the following two objectives:
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Figure 5.61: The transformation in this diagram allows us to apply BellmanFord algorithm just once and verify if there are any negative weight cycles in the graph. All edges coming from vertex x have a zero weight. We apply Bellman-Ford algorithm to this transformed graph and find shortest distances from vertex x.
Figure 5.62: We can always convert negative edge weights by adding a positive number in each weight so that we can apply a more efficient algorithm to find shortest paths in a graph with positive edge weights. But adding a constant in each edge weight disturbs the relative weights of different paths and leads to a wrong answer.
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1. All negative edge weights should be converted into positive weights. 2. All relative distances between two vertices should remain the same in the modified graph with positive edge weights. First we shall devise a simple scheme to alter edge weights such that relative distances between any two vertices do not change in the given graph. Then we shall modify this scheme so as to convert all negative edge weights into positive edge weights thus fulfilling both the above objectives. Consider a directed edge (j, k) with a weight equal to w. How about associating any arbitrary number Label(j) with a vertex j and another arbitrary number Label(k) with vertex k. Now the edge weight w(j, k) is changed according to a formula where wnew = wold +Label(j)−Label(k). If we associate an arbitrary number with each vertex of the graph and change edge weights according to the formula described then we claim that relative path lengths between any two vertices will be the same in the new graph as compared to the old graph. See figure 5.63. In fact the new path length from a vertex a to a vertex b will be the old path length plus Label(a) − Label(b). Putting arbitrary labels solves only one problem, namely the relative distances between two vertices remain the same in the new graph as compared to the old graph. The other problem (converting all negative edge weights into positive) remains to be solved. It requires that vertices should not be labeled arbitrariliy but with some intelligence as described below. So now the problem is reduced to finding an appropriate number Label(j) to be associated with a vertex j and another number Label(k) to be associated with vertex k (assuming that there is an edge from j to k with a weight w(j, k)) such that if w(j, k)old is negative then w(j, k)new = w(j, k)old + Label(j) − Label(k) becomes positive.This would require that wjk (old) + Label(j)−Label(k) is greater than or equal to zero. It means that wjk (old)+ Label(j) is not less than Label)k). In other words Label(k) should be at least equal or more negative than wjk (old)+Label(j) assuming that wjk (old), Label(j), and Label(k) are all negative. Before finding a systematic scheme of providing labels to each vertex let us try to work out a simple example. What numbers are desirable and should be associated with two adjacent vertices j and k are indicated in Fig. 5.64. Here vertex x is a vertex added to the given directed graph just like the one shown in Fig. 5.61. In the top diagram we try to initially label k such that
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b
11 11-6=5 -6
a
-2 -2+3-4+7=4
c
7 -4
-3
d
3
e
f
0 11
b
-6 11-6=5 -4
a
-3 -2 -2+3-4+7=4
c
b
0 11+0+3 11-6+0+4=9 -6-3+4
7 -4
d
-4 -1
3
e
-2
f
-5
a
-2+0+1
c
7-5+4
-2+3-4+7+0+4=8
d
-1
3-1+2
e
-2
-4-2+5
f
-5
Figure 5.63: How relative distances remain the same if we associate an arbitrary number with a vertex and then add a number in the edge weight of each edge according to a fixed formula.
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weight Now coming back to the general question: What systematic scheme should be applied which guarantee the allocation of desirable numbers to vertices of a graph such that all edge weights become positive while relative distances between any two vertices remain the same? After looking at the last figure you must have some idea of what is going on or what should be done. If you are still undecided then read the following paragraphs and look at the coming figure.
Figure 5.64: We assume that vertices j1 , j2 , and j3 are already labelled while we need to label k so as to make edge weights positive. In the top diagram we try to initially label k such that weight
k)} x
Dist(k) Dist(j)
j x
Dist(k) = -10 Dist(j) = -20
w(j.
Dist(k) = min{ Dist(k).65 for a demonstration of this updating.
.65: While executing Bellman-Ford algorithm we use a basic building block as shown above. Please see once again Fig. 5.Shortest Path Algorithms
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Let us recall the basic building block of Bellman-Ford algorithm and how the shortest distance of a vertex k from a source vertex x is updated.65 where the shortest distance of vertex j is updated.k)
k x
Dist(k) = -70
j
w = -40
k
New Dist(k) = -60
j
Dist(j) = -20
w = -40
k
New Dist(k) = -70
Figure 5. Dist(j) + w(j. The bottom diagrams show how and when the shortest distance of vertex k from a source vertex x is updated.66.
Summary
So now we are in a position to describe Johnson's algorithm in the following meaningful manner as applied on a directed graph D. k) will eventually become positive if it was initially negative. 5. 5. The interesting thing is that we need not spend extra time in finding these labels as they have already been found while checking if the given graph has negative weight cycles. See Fig. It should become quite evident now that if Label(j) is the shortest distance of vertex j from vertex x and Label(k) is the shortest distance of vertex k from vertex x then the required inequality would be satisfied and the edge weight for edge (j. 5.64 where the label of a vertex is updated and now loook at Fig. See Fig.
66: How to determine which labels to associate with each vertex.250
x
Transform
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0
b
8 -2 -4 -4 1 -2 2
b 0 -2 a
-2+2 8-2 1+0 2+4 -4+4
a
c
-4 c
d
0
3
e
0
d 0
3+0
e 0
Figure 5. If there are no negative weight cycles in graph D then move to step 2 otherwise terminate. 5.
5. Check if there are any negative weight cycles in the graph D using Bellman-Ford algorithm applied to the graph after inserting a source vertex x as shown in Fig. According to Polya [10]: If the student is left alone with his problem without any help or with insufficient help.66. If the teacher helps too much. That shortest distance Dist(k) become Label(k) of vertex k. All negative edges become positive as shown in Fig. 3. nothing is left to the students.66. Now draw another copy of graph D with all edge weights modified. he may make no progress at all. 2. The teacher should help. but not too much and not
.12
Discussion
The most important task of a teacher should enable the students to discover and acquire experience of independent work. 5. Now apply Dijksta's shortest path algorithm at every vertex in the modified graph (with positive edge weights) to find all pair shortest distances and paths. 1. In step 1 we have already found shortest distance Dist(k) of each vertex k from newly inserted vertex x.
H. S. Mian. The theory of NP-Completeness connects all problems that are NP-Complete: it is also possible to find a useful relationship among solvable problems and this is what we have attempted to do in this chapter. We have also shown that making comparisons between various techniques and solutions provides a deep insight which itself is very useful in solving otherwise difficult problems [8]. At times it is almost impossible to solve a given problem while it is easy to solve a related problem (the shortest path problem is solvable while the longest path problem is unsolvable). To learn means to cause your mind to function in a different way: new memories are created and/or new connections are forged." These relationships provide the algorithm designer a perspective that proves invaluable when solving new problems and analyzing old one's. Maud. Skiena. T. According to Hale [5]: "There are different kinds of learning. J. We wish to specially thank R. Baase. A. Khan for providing motivation as well as inspiration for this project. starting with something seemingly simplistic yet capable of being transformed into a number of powerful algorithms with minor modifications. Similarly it is difficult to solve a problem in its original form while it is easier to solve it while placing certain restrictions (the graph isomorphism problem is solvable for trees but is difficult to solve in general). Alvi. Lahore University of Management Sciences for providing support for this research.
.Discussion
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too little. Mahkari. Ikram. Jadoon for their help and encouragement. It is extremely useful to find why a certain technique works under certain conditions and why it fails in others (greedy methods provide optimal solutions in finding the shortest path but fail to find the longest path). We also wish to thank S. S. K. In this paper we have demonstrated how a teacher can help students discover a number of graph algorithms with some initial help. Fahd. M.
Acknowledgement
We are thankful to the Department of Computer Science. so that the student shall have a reasonable share of the work. We have shown that by asking thought provoking questions it becomes possible for the teacher to guide the students while solving complex problems. but I refer here to the intellectual kind.
Our intention (and desire) in this chapter.2
Definitions & Prior Knowledge
Set Cover: Given a set of subsets S of a Universal Set U . where the maximum degree of each vertex is one. no edge has a common end point. i. What is the smallest subset of vertices of the graph that covers all edges? Independent (Vertex) Set: What is the largest subset S of vertices of a graph such that no pair of vertices in S has an edge in between? Is there a connection between the vertex cover and the independent set problem? Matching (Independent Edge Set): A sub-graph of G.
6.e. the maximum flow at minimum cost problem. Specifically we shall be describing Menger's Theorem which relates maximum number of vertex-disjoint (or edge-disjoint) paths with minimum number of vertices (or edges). That is why we first provide a unified picture and then go deeper in order to analyze each area in detail. and last but not the least the Circulation problem. we shall be designing algorithms to solve a number of related problems. Vertex Cover: The Universal set U is the set of all edges in a graph. We shall also discuss Konig's Theorem which relates the size of the vertex cover to the size of maximum matching in a bipartite graph. In addition to making formal proofs for a number of theorems. We shall also discuss the matching problem in bipartite graphs. We shall also be discussing the network flow problem. Connectivity and Matching Problems
6. We use a single building block in this entire chapter for designing almost every algorithm. We have already witnessed the subset sum problem (in previous courses) in which we have to select integers (out of a set of integers) such that the sum of the selected integers is equal to a given constant.
. The Marriage (Hall's) theorem provides necessary and sufficient conditions for a bipartite graph to have a perfect matching. what is the smallest subset T of S such that the union of all these sub sets in T covers all elements of U . which if removed will disconnect a special node from another special node in a graph.1
Introduction
We shall first address the problem of vertex or edge connectivity in general graphs. will be to integrate concepts so that the enabled learner is able to appreciate the bigger picture where one relationship implies another and one theorem can be used to prove the other.254
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Note that every non leaf vertex is a cut vertex and every edge is a cut edge or a bridge in a tree graph. while any vertex cover will include non leaf vertex v.Definitions & Prior Knowledge
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.1: A perect binary tree graph is shown. Any independent set will include leaf vertex u.
no two edges in the set share a common vertex. A Perfect Binary Tree: We know that a tree is a connected graph where each edge is a bridge edge. A perfect binary tree (shown in Fig.1. every vertex of the graph is incident to exactly one edge of the matching. Every perfect matching is both maximum and hence maximal.1 has 8 leaf vertices. In some literature. The root vertex is shown as the top most vertex in Fig. It has a single vertex with degree equal to 2. Maximum Matching: It is a matching in a graph with maximum possible size? How bad can a maximal matching become as compared to maximum matching? Edge Cover: The Universal set is the set of all vertices in a graph. 6. It has exactly 2h leaf vertices. that is.
. 2. It has exactly 2h+1 − 1 vertices where h is the path length (in terms of number of edges) between the root vertex and any leaf veretex in the perfect binary tree. It has exactly 2h − 1 non leaf vertices including the root vertex. Let us call it the root vertex. 6. As described before each leaf vertex in a perfect binary tree has a path length equal to h from the root vertex. the term complete matching is used for it. There are 7 non leaf vertices in tree shown in Fig.1) has the following features: 1. 6. Find a simple algorithm to find a maximal matching in a line graph. The perfect binary tree shown in Fig. 4. which covers all vertices? How small (or big) can the size of the edge cover become. 3.1. Here we define a perfect binary tree. All non leaf vertices other than the root vertex has a degree exactly equal to 3.256
Network Flows. What is the smallest subset of edges. Connectivity and Matching Problems
These edges are also known as independent edges.1. as compared to the number of vertices in a graph? A Perfect Matching is a matching which covers all vertices of the graph. Maximal Matching: This is a matching in which more edges cannot be added to increase the size of this matching. 6. The perfect binary tree shown in Fig. 6. A binary tree is a tree where the degree of each vertex is less than or equal to 3. Perhaps a more meaningful definition will be a set of non-adjacent edges. That is.
Definitions & Prior Knowledge
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Figure 6. The maximum matching in this graph is not a perfect matching. The rest of the vertices belong to the independent set in this diagram as shown in the top right corner. in brown bold. The bottom left diagram highlights. The bottom right diagram shows the edges belonging to the edge cover. as all vertices here are not matched as shown by black circled vertices. You may have noticed that the size of the maximum matching is equal to the size of the vertex cover in this graph. these edges are shown in bold in green. the maximum matching edges in the given tree. You may have also noticed some relationship between the size of the vertex cover and the size of the independent set?
.2: The top left diagram shows the vertices belonging to the vertex cover highlighted by double circled vertices.
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Network Flows.1. Edges belonging to the edge cover are shown in bold in the bottom right diagram of Fig.3. It may thus be possible to find a maximum matching given a vertex cover in a tree graph.1. meaning a maximum matching in a perfect binary tree? Problem 6. How can we efficiently find an edge cover in a perfect binary tree? Problem 6.1. How can we find an independent (vertex) set in a perfect binary tree? Problem 6.1. Problem 6.1. Connectivity and Matching Problems
Problem Set 6. 6. 6. We are given a perfect binary tree G (see Fig. In a perfect binary tree graph G. We know that in a tree every edge is a cut edge or bridge while every (non leaf) vertex is a cut vertex.2 shows a maximum matching in the perfect binary tree graph of Fig. How can we find efficiently if a perfect matching exists in a perfect binary tree?
.1) and we intend to solve a number of problems related to connectivity and matching. Using this logic all leaf vertices will be part of the independent set and will not be part of the vertex cover. How can we find an independent (edge) set. Independent set and vertex cover vertices are shown in Fig. a leaf vertex u will always part of an independent set (why?). In a perfect binary tree graph an edge connecting a leaf vertex u with a non leaf vertex v (see Fig.1.2. Please note that a perfect binary tree graph is a fairly restricted structure and our existing prior knowledge of graph theory and algorithms is sufficient to solve these problems. 6. 6. How can we efficiently find a vertex cover in a perfect binary tree? Problem 6.2 for the perfect binary tree graph of Fig.1. While a non leaf vertex v which is adjacent to a leaf vertex will always be part of a vertex cover (why?). 6.1.2.4. We shall observe later in this chapter that there is indeed a relationship between graph connectivity and number of paths in a graph (Menger's Theorem). We shall observe later that in a bipartite graph (such as a tree) the size of a maximum matching is equal to the size of vertex cover (Konig's Theorem). 6. We also know that there is always a unique path between every two vertices in a tree.5. 6. It also indicates matched and unmatched vertices and edges.1.1) will always be part of the edge cover. Based upon the above observations it is possible to design efficient algorithms to solve the following problems in a tree graph. The bottom left diagram of Fig.
.1.Definitions & Prior Knowledge
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Concept Map 6. A concept map showing a number of relevant concepts and a number of theorems which relate different concepts.
6. If the two halves are not of the same size.8. The edge connectivity for a completely connected graph is of highest value.1. What are the necessary and sufficient conditions for a complete matching to exist in a tree graph? Problem 6. while a tree graph T has λ(T ) = 1. will disconnect it so that all paths between vertices s and t are destroyed. Show that a tree cannot have two different perfect or maximum matching? Edge Connectivity λ(G) of a Graph G is the minimum number of edges which if removed will disconnect the graph G (see Disconnected Graph). that is it may be broken up into more connected
. For a perfect matching to exist in a tree graph. Thus λ(s. A complete matching requires that all vertices belonging to the smaller half are matched to vertices in the larger half. Connectivity and Matching Problems
Problem 6.1. 6. Similarly we define κ(s. t) as the minimum number vertices which if removed from G. Edge-disjoint paths do not share any edge. vertices s and t of G would now belong to different connected components. Vertex Connectivity κ(G) of a Graph G is the minimum number of vertices which if removed will disconnect the graph G (see Disconnected Graph). Problem 6. while vertex-disjoint paths do not share any vertex except the terminal vertices. where both s and t belonged to G.2 to see if a complete matching exists in a given tree.1. then a perfect matching cannot exist. Sometimes we not only want the graph G to be disconnected but also want to make sure that a special vertex s is separated from another special vertex t. Check Fig. what conditions are necessary and which are sufficient? Problem 6. however a complete matching may exist.260
Network Flows. A disconnected graph G has λ(G) = 0.1. We know that a tree is a (restricted) bipartite graph consisting of two halves.9. t) is the minimum number of edges which if removed will destroy all paths between vertices s and t in G. So our new requirement is that when G is disconnected (by removing certain edges) then vertices s and t should belong to different connected components of G. Vertex-disjoint paths are also edge-disjoint but it may not be true the other way round. A perfect matching may exist for the tree graph if the two halves have the same size.7. Disconnected Graph: A graph G may simply be disconnected into two or more connected components.
3: There are several paths between vertices s and t in this graph. By removing certain edges (vertices) it is possible to remove all these paths between the two vertices.Definitions & Prior Knowledge
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Figure 6.
. the resulting graph will be a disconnected graph.
4.4. then the graph G is disconnected but vertex s is still accessible to vertex t. We intend to explore different paths between these two vertices and see how we can destroy these paths in G. There is no cut vertex in this graph.
Four Edge and Six Edge Paths a g a g
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. a}. There is only one bridge edge in this graph. 6.2. Draw all 4 edge paths between vertex s and t.4: Four edge and six edge paths from vertex s to vertex t in graph G. Draw all 5 and 6 edge paths between the two vertices. Connectivity and Matching Problems
components or we may require that G would be disconnected such that two special vertices s and t of G should lie in separate connected components. We are given a graph G with two special vertices s and t as shown in Fig. 6. We have to remove a lot more than one vertex in order to disconnect G so as to destroy all paths between vertex s and t.262
Network Flows.3. See Fig. if we remove this edge {s. 6.
that other possible paths have disappeared? Why? What are its ramifications in designing an algorithm to find all possible edge-disjoint paths in a graph?
.2.2.Definitions & Prior Knowledge
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Problem 6.2. Note that if we remove all edges in this 6 edge path. The left diagram of Fig. Draw a 4 edge path between the two vertices such that not more than one additional edge-disjoint path is possible between the two vertices.2. What is the maximum number of such paths? Problem 6.1 and Problem 6. Problem 6.2. What is the maximum number of such paths?
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Figure 6.7. Draw more than one edge-disjoint paths between the two vertices such that at least one path should be of 6 edge length.2 short list the ones that are edge-disjoint.6. Problem 6.2.1 and Problem 6.8.5: Some edge-disjoint paths.2 short list the ones that are vertex-disjoint. these paths are not vertex-disjoint? Problem 6. then all the paths between vertex s and t are destroyed.3.2. We know that there existed more than one path between the two vertices.2. but only because we have selected one wrong path initially.2. Out of all paths that you have listed in Problem 6.5. 6.7 shows a 6 edge path (shown in bold) between vertices s and t in the same graph.4. Problem 6. Draw a 6 edge path between the two vertices such that no additional edge-disjoint (or vertex-disjoint) path is possible between the two vertices.2. Out of all paths that you have listed in Problem 6.
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Network Flows.6: A six edge path and a couple of four edge-disjoint paths. Connectivity and Matching Problems
Figure 6.
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What is a minimal subset of these edges which will do the same job of destroying the connectivity between s and t? One such subset is shown in the right diagram of Fig. This subset consists of 4 edges unlike three in the last part? Problem 6. 6. Once these paths are selected. The left diagram of Fig.7 in order to destroy connectivity between s and t. The blue (green) cut cuts those edges which. Problem 6. A minimal subset of these edges shown by a cut is sufficient to do the job. In fact it is not essential to remove all edges in the single 6 edge path as shown in Fig. The right diagram shows a minimum sized subset of vertices which if removed will disconnect s and t from each other. t) for this graph?
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.11. Can you find another subset of three edges which when removed will disconnect s and t? What is λ(s. List down these edges.7 shows two 4 edge-disjoint paths (shown in bold) between vertex s and t. 6. if removed will disconnect s from t.Definitions & Prior Knowledge
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Problem 6. which if removed will destroy the connectivity between vertices s and t. The right diagram of Fig.2.8 shows a minimum sized subset of edges. it will no longer possible to find an extra edge-disjoint path between the two vertices. The right diagram shows two 4 edge-disjoint paths (shown in bold) between vertex s and t. Do these three edges corresponds to a minimum number of edges which if removed will destroy all paths between vertices s and t.2. 6.9. 6. You can remove all edges belonging to the paths and see for yourself that the connectivity between s and t is destroyed.
The right diagram shows a minimum sized subset of vertices (shown by double circles) which if removed will disconnect s and t from each other. if removed will destroy the connectivity between vertices s and t.
.8: The left diagram shows a minimum sized subset of edges (shown in bold) which.266
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6. It states that the maximum number of vertex-disjoint paths between vertex s and t are equal to minimum number of vertices which.9 shows the same bipartite graph with two dummy nodes designated as s and t. A relationship between (minimum) vertex cover and maximum matching in a bipartite graph (Konig's Theorem) is transformed into a relationship be-
. The vertex cover of this bipartite graph is indicated by double circled vertices. we shall first provide a panoramic picture of how different concepts are interrelated. if removed will destroy all paths between vertex s and t in the right diagram (why?). Does this mean that there will be three edge-disjoint paths in this graph? Find these three paths. in bold. 6. The left diagram of Fig. Similarly each matched edge (in the left diagram) will correspond to a vertex-disjoint path between vertex s and t in the right diagram (why?). it also shows a maximum matching with the maximum matching edges shown in bold. Now we need to find three edge-disjoint paths between the same vertices. How one theorem implies another and can be derived from each other.Konig's Theorem.3
Konig's Theorem. 6. All vertices in the first partite of the bipartite graph are connected to s while all vertices in the second partite are connected to the dummy vertex t as shown in the right diagram of Fig. Be careful as this is the same cut shown in the left diagram of Fig. The vertex cover in the bipartite graph (shown in the left diagram) becomes the minimum sized subset of vertices which. The right diagram of Fig. if removed from the graph will disconnect s and t (see Concept Map 6.9 shows a bipartite graph with partite A and partite B. Menger's Theorem & Hall's (Marriage) Theorem
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two vertices. and the above theorems in detail in subsequent sections. Konig's theorem tells us that the cardinality of a maximum matching in a bipartite graph is equal to the (minimum) size of the vertex cover (see Concept Map 6.9. 6. two vertex-disjoint paths between s and t. 6.7 where there existed a single path of six edges (with no more room for additional paths).8 is an illustration of Menger's Theorem. Menger's Theorem & Hall's (Marriage) Theorem
Before discussing connectivity.1). matching. The right diagram of Fig. Is this a coincidence that the (maximum) number of vertex-disjoint paths is exactly equal to (minimum) number of vertices which if removed will disconnect s and t? The left diagram shows an edge cut of size three. e and f . the diagram also shows. 6.1).
b2 } N (a2 . a2 ) = {b1 . a4 ) = {b1 .4
Menger's Theorem
Connectivity of an un-directed graph is expressed in terms of minimum number of edges or minimum number of vertices which if removed will disconnect the graph G. a3 . Had a perfect matching existed then the size of the vertex cover would have been equal to the size of A (or B) partite. only three satisfy the condition that |N (S)| ≥ |S|. Menger's Theorem relates the
. Under such conditions the size of the maximum matching and that of the vertex cover will be equal to the size of remaining set A which is equal to three. What should be the size of the neighborhood N (S) for every subset S of A which will guarantee a perfect matching (or a vertex cover equal to the size of A) is the subject of Marriage (Hall's) Theorem. (see Concept map 6. b4 } N (a1 .
6. You may have realized yourself that this necessary and sufficient condition (applicable to the bipartite graph shown on the left of Fig.268
Network Flows. 6. a3 ) = {b1 .9 (Menger's Theorem). Connectivity and Matching Problems
tween κ(s.1). b2 . In other words a perfect matching does not exist in this bipartite graph.9) implies that the (maximum) number of vertex-disjoint paths (and κ(s. b3 . a3 ) = {b2 } N (a2 .9. For a better understanding of the Marriage Theorem a few neighborhood subsets are indicated below for the bipartite graph of Fig. Note that all vertices in the A partite are not matched to all vertices in the B partite although the size of A partite is equal to size of B partite in the given bipartite graph. while in the remaining two the said condition is violated. 6. It is interesting to note that if we remove vertex a3 from A and vertex b3 from B then the condition |N (S)| ≥ |S| is true for every S ⊆ A. t) and (maximum) number of vertex-disjoint paths between vertex s and t in the right diagram of Fig. b3 . a2 . a4 ) = {b2 . N (a1 . a3 . 6.9 would be equal to the size of A (and B) partite in the graph. t)) in the graph shown on the right side of Fig. b2 } N (a1 . a2 . It states that a bipartite graph (of equal halves) have a perfect matching if and only if |N (S)| ≥ |S| for every S ⊆ A. 6. b4 } Out of the five neighborhood subsets N (S).
Given a directed graph D how can we find a minimum number of edges (vertices) which if removed will destroy all paths from vertex s to t in D? 3. the maximum number of edge-disjoint paths from vertex s to t will be equal to λ(s. How about devising the following common sense algorithm to find the maximum number of edge-disjoint paths from vertex s to t in a directed graph D? Let us apply this simple algorithm to solve the problem in the graph of Fig.
6.
. how can we efficiently find the maximum number of edge-disjoint paths from vertex s to vertex t? Let U be the minimum sized subset of the edge set of D. We shall also address the following algorithmic problems: 1. the minimum number of edges (belonging to Edge Cutset) or minimum number of vertices (belonging to Vertex Cutset) which if removed will disconnect the graph into two or more connected components? The graph theoretic aspects of the above problems and a proof for Menger's Theorem will be presented after a better appreciation of the issues involved. equivalent to finding a minimum sized edge set which if removed from D will disconnect t from s.4. t). Thus the problem of finding maximum number of paths from s to t is. we can find a directed path P1 .1
Maximum Edge-Disjoint Paths in Directed Graphs
Given a directed graph D and two specific vertices s and t. Then according to Menger's Theorem.10. Using any path finding algorithm. The size of U will obviously be designated by λ(s. Given a directed graph D containing two special vertices s and t. such that D − U does not contain any directed path from vertex s to t. We shall start this section with directed graphs and then generalize our results for un-directed graphs. in fact. Connectivity of a directed graph D also poses similar problems. t).270
Network Flows. Given an un-directed graph how can we find its connectivity. Connectivity and Matching Problems
maximum number of edge-disjoint (vertex-disjoint) paths between vertex s and t to minimum number of edges (vertices) which if removed will disconnect s from t. how can we find maximum edge-disjoint (vertex-disjoint) paths from s to t in a directed graph D? 2. 6.
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Algorithm 36: Find Maximum edge-disjoint paths from s to t in D.Menger's Theorem
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Concept Map 6.
. A concept map showing a number of concepts related to vertex connectivity and edge connectivity and some important relationships that we shall explore in this section.2. Remove all edges in the path P and. and vertices s & t. otherwise exit the algorithm. input : Directed graph D. If you are successful in finding a path then keep a record of this path. go to step 1 . output: (Maximum) edge-disjoint paths from s to t in D.
(s → b → d → t). If path P1 = s → a → d → t (as shown by bold lines in Fig. d) we can move from 'a' to 'd' but not from 'd' to 'a'. or (s → d → t).and do something else with these edges. We are in dire need of some innovation? There are potentially two ways to fix this problem: 1. It can not be (s → d → a → c → t) or (s → b → d → a → c → t) because on directed edge (a. 6. and the other (s → b → d → t or s → d → t). Algorithm 36 fails to find maximum number of edge-disjoint paths in a graph but still it finds maximal number of edge-disjoint paths in a graph? 2. (where an initial wrong choice will make things hard) a general phenomenon in almost all graphs or is this a problem in a certain class of graphs? 3. Is it possible to convert one class of graphs into another? If this is possible then after the conversion we can use our (stupid) Algorithm 36 to find out maximum edge-disjoint paths in a graph?
. If we are lucky. we can move forward in Algorithm 36.but this may not be an efficient solution? Why? 2. But if we are unlucky. Let us look at what are different possibilities for P1 ? It may be (s → a → c → t). But before finding another edge-disjoint path we should first remove the edges of P1 (why?) and then find another path P2 . we shall land in a difficult situation as depicted in Fig. (s → a → d → t). Is this complication. Thus an initial wrong choice will make things hard for us. 6. Once we have selected an initial path P1 . Connectivity and Matching Problems
from vertex s to t in D. Do not delete the edges of the chosen path . we shall be able to find the two edge-disjoint paths. Is there a class of directed or un-directed graphs where an initial wrong selection would not create any complication? How the class of graphs (where an initial wrong selection really matters) structurally different from the other class (where an initial wrong selection does not matter)? 4. What is that some thing else? Before we actually do an innovation consider the following: 1. one being (s → a → c → t).10. it has now become impossible to find another path in this graph. Pick the initial path more intelligently .272
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It is rule number 2. A directed edge (x.11 before arriving at a conclusion. given a set of pseudo edge-disjoint paths in a graph it is possible to find an equal sized set of edge-disjoint paths in a graph. A directed edge (x. The status of such an edge when traversed by a single path from x to y will be "used". Finding edge-disjoint paths in a general directed graph may be hard to find. The in-degree is equal to the out-degree for every vertex i of D other than vertex s and t. in fact. We shall later show in this section that the number of pseudo edgedisjoint paths between two vertices in a graph is exactly equal to the number of edge-disjoint paths. The status of such an edge.274
Network Flows. Show that if you apply Algorithm 36 to this graph then you will be able to find maximum number of edge-disjoint paths without a complication? How about if the directed graph D is cyclic but the in-degree as well as the out-degree of every node are equal? You should consider both options: when vertices s and t are not part of any cycle and when they are also part of some cycles? See Fig. which is traversed by two paths in opposite directions will be "not used by any path" or "unused". Please note that rule number 1 is common in pseudo edge-disjoint as well as edge-disjoint paths. y). already traversed by a path from x to y. We are given a directed acyclic graph D with a source vertex s and a sink vertex t. Connectivity and Matching Problems
5. y). Finding maximum number of pseudo edge-disjoint paths is a problem which can be solved in class using the exciting process of discovery based learning. in case of pseudo edge-disjoint paths we follow the following rules of the game: 1. 2. let us try ourselves to solve a simpler problem (finding pseudo edge-disjoint paths) and then use our newly found experience and confidence to solve the harder problem (finding edge-disjoint paths). which makes the two categories different. y) can not be shared by more than one path. In edge-disjoint paths an edge (x. So the current problem is how to maximize the number of pseudo edgedisjoint paths in D? Instead of deleting all edges in a (recently found) path
. can be traversed by a new path from y to x and not from x to y. 6. not already in use by a path. can be traversed (or used) by a path from x to y and not from y to x. expecting an innovation to achieve this objective may be unrealistic.
Menger's Theorem
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Figure 6. This diagram is provided by Khawaja Fahd.
.11: A directed graph D in which the in-degree of every vertex is equal to the out-degree of every vertex including vertices s and t.
Let us execute this simple four line algorithm on the graph of Fig. Connectivity and Matching Problems
(as is done in step number 2 of Algorithm 36) we reverse the direction of each edge in the path. and then go back to step number 2. In this process (Algorithm 37) we convert graph D into F .
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Example 6. In step number 2 we find a path from vertex s to t in D using any path finding algorithm.12. Graph F helps us identify the set P using Algorithm 38.4. If you are successful in finding a path P then keep a record of the number of paths found so far otherwise exit with graph F as output. input : Directed graph D. It also outputs the Status(e) of every edge e of D. Reverse the direction of edge e and then go back to step 2.1. for every edge e in path P do set Status(e) = used if initially it was unused otherwise set Status(e) = unused. Algorithm 37 is based on this innovative idea. In step number 4 (instead of removing every edge in the path as was done in Algorithm 36) we reverse the direction of every edge in the path found in step number 1. We apply the above algorithms on the graph D shown below. Using this set P we identify edges of D which belong to the minimum cut. Find a directed path P from vertex s to vertex t in F . Let us carefully look at this new algorithm which claims to find maximum number of pseudo edge-disjoint paths from s to t in a directed graph D. is not occupied by any path and thus the Status of every edge is unused. Note that Algorithm 37 (unlike Algorithm 36) allows an edge to be used by two different edge-disjoint paths moving in opposite directions? Algorithm 37: Find Maximum pseudo edge-disjoint paths from s to t in directed graph D. The Paths P1 (green) and P2 (blue) that we have found during the execution
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Network Flows. As soon as an unused edge is occupied by a path its Status is changed from unused to used in step number 4. Initially every edge of D. and vertices s & t. We first find maximum pseudo edge-disjoint paths from vertex s to vertex t in this graph. output: Maximum pseudo edge-disjoint paths from s to t in D and a Status(e) for every directed edge e of D. Status(e) = unused for every directed edge e in D.
2. This Algorithm outputs maximum pseudo edge-disjoint paths and the Graph F . The diagram on the bottom right of Fig. this new graph has some very desirable properties? The Minimum Cut is found in Fig.
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of this algorithm are not edge-disjoint paths (see Fig.278
Network Flows. We execute Algorithm 37 once again on a directed graph
. 6. Before jumping to any conclusions and in order to gain more confidence let us solve another example. 6.14. yet this algorithm tells us the maximum number of possible edge-disjoint paths in D (Why and how?).13 shows the original graph D with used edges only. The most useful result of this algorithm is the Status of every edge e in D after its termination.
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Figure 6.14: The top left diagram shows the Original Graph D. the top right diagram shows the final graph F with a couple of paths found and reversed.
. The bottom left diagram shows the final graph F where the Min Cut is found and then it is applied to the original graph D on the bottom right diagram.
4. TC = {a. Connectivity and Matching Problems
D shown in Fig. The bottom left diagram shows the edges of the path P3 shown in bold. This is because in the Cut B the vertex 'a' belongs to SB while vertex d belonged to TB as shown in the bottom left diagram of Fig. The top right diagram shows the edges of another path P2 shown in bold.16.
6. TA ) = {(a. b. a. 6. 6. Let us also consider the graph D (of Example 6.2
The Concept of a Minimum Cut in Directed Graphs
Let us define the concept of a cut more formally.280
Network Flows.15. For Cut A : SA = {s. Let S be the set of vertices of D containing s but not t and let the subset T = V (D) − S. t)} It is interesting to note that the edge (a.1) minus the unused edges as shown in the bottom of Fig. d) in the Cut C does not belong to the set (SC . 6. d)} For Cut C : SC = {s. d). (SA . Thus the edge (a.4. (SC . (b. (d. Out of many possible cuts we show only three cuts in the directed graph shown in the top diagram of Fig. TB = {c. (a. t}. Please note that it is the same directed graph for which we have found maximum number of edge-disjoint paths as shown in Fig. d. TA = {c. (a.15. (SB . Looking at this problem from a different angle:
. c).4. d}. T ) will be a subset of E(D) and is known as a cut in the graph D. d). T ) is a set of directed edges from a vertex in S to a vertex in T . Let us again consider graph D (of Example 6. TC ) = {(s. d}. Then (S. they both belong to a (desirable) class? You can apply (the very stupid) Algorithm 36 to this class of graphs and you will find maximum sized subset of edge-disjoint paths without any complication.16. Again Algorithm 37 provides the maximum number of edge-disjoint paths present in this graph although the paths P1 . The original graph D is shown in the bottom right diagram with used edges shown in bold. 6. a). 6. a. While in the Cut C.16.15. d) is physically cut by Cut B as well as Cut C but it is only included in the Cut B and not in Cut C. t)} For Cut B : SB = {s.13. t}. vertex 'a' belongs to TC and vertex d belongs to SC . The set (S. c). t}. 6.2) minus the unused edges shown in Fig. P2 & P3 found by this algorithm are not edge-disjoint as shown in Fig. 6. Both these graphs share something in common. b.10. TC ) as shown in the bottom right diagram of Fig. b}. The top left diagram shows the directed graph D with a selected path P1 from vertex s to vertex t shown in bold. (d. c. TB ) = {(s. 6.
The original graph D is shown in the bottom right diagram with used edges shown in bold. The top right diagram shows the edges of another path P2 shown in bold.
.15: The top left diagram shows a directed graph D with a selected path P1 from vertex s to vertex t shown in bold.Menger's Theorem
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Figure 6. The bottom left diagram shows the edges of the path P3 shown in bold.
d) is not contributing to a path from s to t in D and is therefore not required to be removed.
. TA ) is equal to U (which is the minimum sized subset of edges which. the edge (a. Thus λ(s. (SA . According to Menger's Theorem the maximum number of edge-disjoint paths in this graph will equal λ(s. How to find the minimum cut systematically? We shall discuss it in a proof of Menger's Theorem. By a little inspection in this graph it can easily be verified that the Cut A is indeed one of the minimum cuts. which is 2.16: We are given a directed graph D and we intend to find maximum number of edge-disjoint (directed) paths from vertex s to vertex t. The bottom left diagram shows the subset SB along with the Cut B edges shown in bold. While in the Cut C you have to remove only two edges present in this cut and the vertex s is disconnected from t. The bottom right diagram shows the subset SC along with the Cut C edges shown in bold. The size of Cut B is larger then the size of Cut A and that of Cut C. TA )| = 2. Connectivity and Matching Problems
In Cut B you have to remove all four edges present in this cut in order to destroy all paths from vertex s to t in D. t) = |(SA . The top diagram also shows three cuts. t). if removed will destroy all paths from s to t in D).
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6. T )|min where the minimum is taken over all cuts (S. 6.3. there can not be more than k pseudo edge-disjoint paths in D.
.4. As we have seen before there may be different cuts possible with different sizes in the same graph as shown in Fig. i.3
A Proof of Menger's Theorem and Finding the Minimum Cut in Directed Graphs
We are given a directed graph D and two special nodes s and t.2. the minimum cut. Claim 6. The number of real (not pseudo) edge-disjoint paths from s to t in D can not exceed the number of (pseudo) edge-disjoint paths from s to t in D. Claim 6. The maximum number of the two paths will be exactly equal (given one. Algorithm 37 finds k pseudo edge-disjoint paths from s to t in D before it terminates.4.1.4. We intend to prove Menger's Theorem in terms of edge connectivity of a directed graph.4. The maximum number of edge-disjoint paths from vertex s to t is equal to the minimum number of edges of D which if removed will destroy all paths from s to t in D. the minimum number of vertex-disjoint paths from s to t is equal to the cardinality of minimum sized vertex cut-set which will disconnect t from s in D. If we try to maximize the number of paths from s to t they will only be limited by the bottleneck in the graph which will be the minimum cut. TX ) where vertex s belongs to the subset SX while vertex t belongs to TX as defined earlier in this section. We know that number of edge-disjoint paths from vertex s to t in a graph can not exceed the size of a cut (SX .e. We shall provide a constructive proof. T ) in D. Menger's Theorem tells us that the two quantities (maximum number of paths and size of the minimum cut) are exactly equal. using this proof one can also find the maximum number of edge-disjoint paths as well as the minimum cut. We shall make the following claims here: Claim 6. we should be able to find the other). This proof technique can easily be adapted for un-directed graphs and can also be used to prove the vertex form of Menger's Theorem which states that.. That means the maximum number of edge-disjoint paths from s to t in D will not exceed |(S. The maximum number of pseudo edge-disjoint paths from vertex s to t is equal to the minimum number of edges of D which if removed will destroy all paths from s to t in D.
may consist of unused edges (not used by any path so far) or used edges (occupied by a previous path in the opposite direction) as shown in the top right corner of Fig.4.1 & 6. Although no (additional) path from vertex s to t exists anymore. It is obvious that if there is a (pseudo) path left (in addition to the one's that are already found) then Algorithm 37 will find it before its exits. what we need to show now is that there will actually be a cut whose size will be equal to the maximum number of paths from s to t in D.15. Please note that each (pseudo) path found in step number 2. Connectivity and Matching Problems
Proof for Claim 6. it shows the number of vertices (enclosed in the shaded area) which are reachable from s in D after the termination of Algorithm 37. The bottom diagram shows the modified graph D with used edges shown in bold.14 is duplicated in the top left corner. and go back to step number 2 to find a new path (see Fig. 6.4. We have seen that the maximum number of pseudo edge-disjoint paths can not exceed the size of a minimum cut.284
Network Flows. 6.2: In step number 2 of Algorithm 37 we find a path from vertex s to t in D using any path finding algorithm. 6.13). The maximum number of pseudo edge-disjoint paths is shown in bold in the top right diagram.17: The bottom left diagram of Fig. We keep on doing this until we are no longer able to find a path from s to t in D and then the algorithm terminates. Let us assume that while executing this algorithm we have reached a stage where we are no longer able to find a new path and the Algorithm 37 terminates. but still it may be possible to reach a number of vertices in
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6. Q) is in fact the Minimum Cut equal to the maximum number of edge-disjoint paths from vertex s to t in D. We define subset Q to be equal to V (D) − P . Q) consists of edges all of which are already occupied by existing paths as shown in the bottom right corner of Fig. 3. It is important to note that the vertex t will belong to Q otherwise it will be in P and then we can find an extra path from vertex s to vertex t in D. Under such conditions vertex y will also be reachable from vertex s which contradicts our initial assumption. When Algorithm 37 is unable to find an extra path it terminates as shown in the bottom left corner of Fig. The Cut (P. It will be a directed acyclic graph. Q) will consist of edges such that each edge which is part of this cut will be occupied by a unique path from vertex s to t in D. The Cut (P. Can the number of paths be smaller than the size of this cut? If the number of paths is smaller than the size of the cut then you can identify at least one edge (x.14. If we remove all edges of D which have a Status = unused from D then the resulting graph D would have the following properties: 1. In fact it is possible to find the (real) edge-disjoint paths from the modified graph D (as shown in the bottom of Fig.14. The in-degree of every node other than vertex s and t will be equal to its out-degree. y) which is part of the cut but not part of any path from s to t in D.Menger's Theorem
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the modified graph D from the vertex s.3: We have seen that the Algorithm 37 not only provides maximum number of pseudo edge-disjoint paths (see Fig. The above three properties guarantee that the modified graph D have as many (real) edge-disjoint paths from s to t as the number of pseudo edgedisjoint paths in the original graph D. The Cut (P.14.4. The number of paths can not be larger than the size of this cut as discussed before.15. Thus each edge of the Cut (P. The out-degree of s will be equal to the in-degree of t in D. Let P be a nonempty set (a subset of V (D)) containing vertex s and all other vertices which are reachable from s.
.) but also the status of every edge is provided by it. 6. This situation is depicted in Fig.13) using the (very simple) Algorithm 36. The set of vertices which are reachable from s (known as P ) are shown shaded in this diagram. Proof for Claim 6. 6. 6. 2. 6. 6. Q) will be part of exactly one pseudo path from vertex s to t in D.13 & Fig.
Connectivity and Matching Problems
6. 6. An efficient algorithm to find maximum number of vertex-disjoint paths
.18 shown below shows a directed graph H in the top left diagram. Please note that removal of certain vertices does not destroy all paths from vertex s to vertex t in this graph (see the top right and bottom left diagrams) while the removal of some other vertices does indeed destroy the s − t connectivity in this graph. It also shows how this graph would look like if we remove a number of vertices.4
Finding Maximum Vertex-Disjoint Paths & Minimum Vertex Cut in Directed Graphs
We have shown earlier that the maximum number of edge-disjoint paths (from vertex s to vertex t) in a directed graph is equal to the minimum number of edges (which if removed will destroy all paths from vertex s to t). The vertex form of Menger's Theorem equates maximum number of vertexdisjoint paths (from vertex s to vertex t) to minimum number of vertices which if removed will destroy all paths from vertex s to t in a directed graph.4. The Fig.
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Figure 6. There are basically two issues that we would like to tackle: 1.286
Network Flows.18: Various vertices are removed to show that the graph does not necessarily become disconnected by their removal and only specific ones make the graph disconnected.
19 using Algorithm 39. 6. the corresponding
. We split each vertex x (excluding vertex s and t) of directed graph H into x1 and x2 in D. A proof that maximum number of vertex-disjoint paths (from vertex s to vertex t) is equal to the minimum number of vertices which if removed will destroy all paths from vertex s to t in a directed graph. 2. There is a path from vertex s to t in D corresponding to any path from s to t in H. Algorithm 39: Transform directed graph H into directed graph D. Thus for the new graph D we have V (D) = 2V (H) − 2 & E(D) = E(H) + V (H) − 2. Every path passing from vertex s to t in H has to pass through a number of k intermediate nodes (nodes other than s and t). For every edge (s. input : Directed graph H with special vertices s & t. Let us call these edges external edges. we call these edges internal edges. y1 ) in D.
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2. Interestingly both these problems can be resolved using our prior experience provided we make a couple of transformations on a given directed graph H (left diagram) and convert it into another directed graph D (right diagram) as shown in Fig. y) in H is transformed into a directed edge (x2 . there will be a corresponding edge in D as shown in brown color in the right diagram of Fig. output: Directed graph D with internal & external edges marked
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Convert H into D by splitting all the vertices (except s and t) as described. x2 ). Similarly for every edge (x.
Once the directed graph H is transformed into D it has now become possible to appreciate the following: 1. The corresponding path in D passes through external as well as internal edges as shown in Fig. Thus for every edge in H. We insert an extra edge between x1 and x2 in D for every vertex x in H except for s and t as shown here. 6.Menger's Theorem from vertex s to t and a minimum cut in terms of vertex cut-set. 6.19. Identify internal and external edges of the graph D. t) in H insert a directed edge from x2 to t in D.20 (top left corner). x) in H insert a directed edge from s to x1 in D. In addition to these edges we have edges of the form (x1 . each directed edge (x.
the working of this approach on the directed graph D of Fig. If it passes through (some of the) external edges then the minimum cut will not correspond to a minimum vertex cut. 6. A Minimum (edge) Cut in graph D will correspond to a Minimum (vertex) Cut in graph H provided the Minimum (edge) Cut in D passes through internal edges only. It is important to appreciate that the minimum cut should (be forced to) pass through internal edges only. 3. the Minimum Cut as found by the original Algorithm 37.19: Each vertex of the directed graph of shown in the left diagram is split up into two vertices. and thus any number of edge-disjoint paths in D will correspond to the same number of vertexdisjoint paths in the directed graph H. In order to do so we make a one line modification in our earlier approach in terms of Algorithm 40 as described
. 6. We can construct (an almost) similar constructive proof that the maximum number of vertex-disjoint paths from s to t in H is exactly equal to the minimum number of vertices which if removed from H will destroy all paths from s to t in H.20. External edges are shown in brown color while internal edges are shown in bold orange color. We can use Algorithm 39 to convert graph H into graph D and then use Algorithm 37 without any change to find the maximum edge-disjoint paths and the Minimum Cut. as shown in the right diagram.19 is shown in Fig.288
Network Flows. There is just one problem to be resolved. may pass through some of the external edges as shown in the middle right diagram in Fig. Connectivity and Matching Problems
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Figure 6. The above observations provide us enough insight to find maximum edgedisjoint paths in the directed graph D which will correspond to maximum vertex-disjoint paths in H. 6.20. path in D will pass through k internal edges.
y) of D Where x belongs to P and y belongs to V (D) − P . 6.4. The path between s and t in G (Fig. input : Directed graph D with special vertices s & t and internal & external edges marked output: Internal Edges of D that belong to the Minimum Cut.5
Menger's Theorem for Un-directed Graphs
We shall consider Menger's Theorem in terms of edge connectivity (Menger's Theorem in terms of vertex connectivity can also be derived from in a similar fashion). 6. We can in fact use Algorithm 37 without any changes to find maximum number of edge-disjoint paths from s to t as shown in Fig.
6. An un-directed graph G is shown in Fig. We first add the external edges of graph D in graph F and then find the appropriate cut as shown in the bottom diagrams of Fig. 6. 6.21) is also shown in bold in the top left diagram of Fig. The maximum number of directed edge-disjoint paths from s to t in D will be equal to the maximum number of un-directed edge-disjoint paths between s and t in G. Minimum Cut will consist of edges (x. It states that the maximum number of edge-disjoint paths between two specific nodes in an un-directed graph G is equal to minimum number of edges needed to destroy all paths between the two specific vertices. x) but not both. x) in D.20 (thereby forcing the entire cut to pass through internal edges alone). Each un-directed edge {x. For every un-directed path between vertex s and t in G there is a corresponding directed path from s to t (or from t to s) in D. The un-directed graph G of Fig.22. y} of G is thus transformed into two directed edges (x. 6.21 is transformed into a directed graph D as shown in Fig. Add all the external edges of Graph D to F . Algorithm 40: Find a Minimum Cut in a directed graph D passing through internal edges only. The minimum cut corresponding to minimum number of edges required for destroying all
. y) or (y.22. y) and (y.Menger's Theorem
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below. 6.
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Convert D into F using Algorithm 37.22.21 along with two vertices s and t and a path from s to t shown by bold lines. Let P be the set of vertices (of F ) reachable from vertex s in F . 6. It is reasonable to assume that any directed path from s to t in D will consume either (x.
Now when we reverse selected edges of this path then it is no longer possible to find a path from vertex s to t as shown in the middle right diagram. We again find a path (blue) as shown in the middle left diagram.
. The top right diagram Graph H shows selected edges of the path P1 reversed. This is the final Graph F . to which we add the external edges of Graph D and find the Minimum Cut.20: The top left diagram shows a directed graph with a selected c a path P1 from vertex s to vertex t shown in pink. Connectivity and Matching ProblemsGraph D and initial Graph F
Graph FFinal Graph F: Cut passing through an external edge a1 a2 c1 c2 a1 a2 c1 c2
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Figure 6. The minimum (vertex) cut is indicated by a bold line. The reason for the step in the bottom right diagram is to ensure the Minimum Cut passes through internal edges only.290
Network Flows.
4. for every vertex x there is a corresponding two vertices {x1 . Vertex Connectivity κ(G) of a Graph G is the minimum number of vertices which if removed will disconnect graph G.
.22.25 show two un-directed graphs. 6. x) in D. 6. Thus the total number of vertices and edges are almost doubled.23.23 and reverse the edges.23. Thus each un-directed edge {x.6
Edge Connectivity and Vertex Connectivity for Un-directed Graphs
Edge Connectivity λ(G) of an un-directed graph G is the minimum number of edges which if removed will disconnect graph G. This way in the bottom-left diagram of Fig. 6. y) and (y. An un-directed graph G is shown at the top of Fig. x2 }. y} of G is transformed into two directed edges (x. 6.Menger's Theorem
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paths between s and t can also be found using similar techniques as shown in the bottom left diagram of Fig. 6.
6. The only difference from the previous procedure is that we apply Algorithm 39 to find the Minimum Cut for vertexdisjoint paths.23 is transformed into a directed graph D as shown in the middle diagram of Fig. 6. The top graph shows its edge connectivity. The diagrams in Fig.23 along with two vertices s and t. The un-directed graph G of Fig. 6. We then apply the standard algorithm for finding edgedisjoint paths from vertex s to vertex t as shown in the top-right diagram of Fig.23 are added where they do not exist in graph F (not shown in the Figure) and the minimum cut is found and applied to Graph G. 6. Then all the vertices except for s and t are split into two as shown in the bottom diagram of Fig. 6. V (D) = 2V (G) − 2 and E(D) = 2E(G). The bottom diagram shows the vertex connectivity of the bottom graph. Menger's Theorem states that the maximum number of vertex-disjoint paths between two specific nodes in an un-directed graph G is equal to minimum number of vertices needed to destroy all paths between the two specific vertices.
Maximum Vertex-disjoint Paths & Minimum Vertex Cut in un-directed graphs
Now considering Menger's Theorem in terms of vertex connectivity of a graph where s and t are already given. The problem is to efficiently find both edge connectivity and the vertex connectivity of an un-directed graph G.24 the external edges of Graph D of Fig.
.292
Network Flows. Connectivity and Matching Problems
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Figure 6. if we initially select a path as shown by bold red lines and then remove its edges from G then it becomes impossible to find another path in the remaining un-directed graph (right diagram).21: This is Graph G (left diagram).
Now when we reverse the edges of this path then it is no longer possible to find a path from vertex s to t as shown in the middle right diagram.Menger's Theorem
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Figure 6. The top right diagram shows the edges of the path P1 reversed.22: The top left diagram shows an un-directed graph with a selected path P1 from vertex s to vertex t shown in bold.
. We again find a path as shown in the middle left diagram. The minimum cut is indicated by a shaded region in the bottom diagrams.
t) is equal to the edge connectivity of graph G. 6. t) and claim that λ(s.Menger's Theorem
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Our prior knowledge & expertise tells us that given an un-directed graph G and two already selected special vertices s and t we can efficiently find λ(s. t) (because of O(p2 ) distinct pairs of s − t vertices) we select the one with minimum value and that will be the edge connectivity λ(G) of the graph.27. t). The next logical step should be to find ways to reduce the complexity. Is it possible to arbitrarily select s and t in the graphs below. 6. t) pairs in the graph G? As the graph G consists of p nodes. What is then the way out? Perhaps you should consider all possible (s. only O(p) pairs will be sufficient for calculating edge connectivity as shown in Fig. We know that the maximum number of edge-disjoint paths is equal to λ(s. For each such pair (we call it an s − t pair) we find the maximum number of edge-disjoint paths from vertex s to vertex t in the un-directed graph G. Out of all the O(p2 ) values for λ(s. The problem of efficiently finding vertex connectivity is slightly more complex as we shall explain in the coming section. You may have realized that there is no need to consider all O(p2 ) distinct pairs of vertices. There is a possibility that a wrong choice for vertex s and vertex t may give you an incorrect result as shown in these diagrams.
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Figure 6. t) which is equal to minimum number of edges which if removed will disconnect s from vertex t in G. that it is λ(G)? The problem may not be that simple as it is evident from the diagram below (Fig. find λ(s. It will be a useful experience to derive the overall time complexity of this simple algorithm.
.26).26: Various Min-cuts in the same graph. there could be O(p2 ) distinct pairs of vertices. (We actually do it by finding maximum number of edgedisjoint paths from vertex s to t in a graph G).
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Network Flows. t) for every different value of t in the un-directed graph as shown in the figure below. We select a vertex s and keep it fixed throughout the working of the algorithm. Out of all results we select the minimum. we may fix s arbitrarily but then compute λ(s.27: When vertex s moves from one place to another? Edge Connectivity of an un-directed Graph We have already hinted before that in order to efficiently compute edge connectivity of an un-directed graph.3.7. The total time complexity will thus be O(p2 q).7.28.3. Connectivity and Matching Problems
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If s is fixed at 2 then t=5. This intelligent observation will certainly cut down the time complexity of our earlier technique as described in the algorithm given below. or 8 will provide the optimal answer If s is fixed at 1 then t=5.6. Problem Set 6. 6. We select a different vertex t in each of the p − 1 graphs and in each such graph find maximum number of edge-disjoint paths from vertex s to vertex t in time equal to O(pq). We repeat this process as many number of times as the number of graphs.2. or 8 will provide the optimal answer If s is fixed at 8 then t=1.
. We thus have to apply the maximum edge-disjoint paths finding algorithm for all possible p − 1 pairs where s is fixed while t has every possible value as shown in Fig.6. or 4 will provide the optimal answer
Figure 6.
How about finding actual edges belonging to minimum sized set of edges which if removed will disconnect graph G? Problem 6. In the j th copy of graph G we select vertex j as t where 2 ≤ j ≤ p. In every copy of graph G we select vertex 1 as s. Assume that instead of finding edge connectivity (which is an optimization problem) we intend to solve the corresponding decision problem: Is the edge connectivity of graph G is less than or equal to k where k is an arbitrary number and is always less than p (why?)?
Of course you can use the edge-connectivity finding algorithm to solve this problem but then it will be overkill.3. Please try to derive the time complexity of this algorithm.300
Network Flows. The modified Algorithm 43 is described below.3. We now know how to find the magnitude of λ(G). Algorithm 42: Find if the edge connectivity λ(G) of a graph G less than or equal to k? input : An un-directed graph G output: Yes/No
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Let us now try to make the previous algorithm more efficient. We make p − 1 copies of graph D numbered 2 to p and then fix vertex s and t in each copy. In fact the problem becomes a search problem in a finite search space.1.2. Now instead of finding Max-Paths or the Min-Cut in each graph we find k edge-disjoint paths in each copy of graph G starting from k equal to 1. If we can find k edge-disjoint paths in all copies then we move forward otherwise we exit out with the edge connectivity of the graph G.
. So let us design an efficient algorithm to solve this problem. Once we have solved the decision problem it becomes almost trivial to solve the corresponding optimization problem of finding the (minimum) edge connectivity of a graph. You have the option of making a linear search or a more efficient binary search in order to find the edge connectivity of a graph. Please try to derive the time complexity of this algorithm. Connectivity and Matching Problems
Problem 6.
How about if we use the same complexity cutting strategy used in finding the edge connectivity. If we find a κ(s.
Assume that the above algorithm is applied to the graph shown below.
Vertex Connectivity of an un-directed Graph We know how to find κ(s. 6. In every copy of graph G select vertex 1 as s. Find κ(s. The outcome of the above algorithm when applied to this graph (Fig.302
Network Flows. we shall get the vertex connectivity κ(G) for graph G. output: Edge Connectivity λ(G) of graph G. t) for every possible pair of vertices in the graph G and then select the minimum (out of all O(p2 ) possible values). that is minimum number of vertices which if removed will destroy all paths between vertex s and t in a given un-directed graph G. The minimum value of all κ(s. If you can not find k edge-disjoint paths in any one of p − 1 graphs then exit with edge connectivity λ(G) = k − 1. input : Un-directed Graph G with vertices numbered from 1 to p. In the j th copy of graph G select vertex j as t where 2 ≤ j ≤ p. Connectivity and Matching Problems
Algorithm 43: Find edge connectivity λ(G) of graph G. t). In the j th copy of graph G select vertex j as t where 2 ≤ j ≤ p.31.
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Construct p − 1 copies of graph G numbered from 2 to p. In every copy of graph G select vertex 1 as s. 6. Find k edge-disjoint path from vertex s to vertex t in each of the p − 1 graphs. input : Un-directed Graph G with vertices numbered from 1 to p.30) is shown in Fig. select s arbitrarily but allow t to have all possible values? Please concentrate on the algorithm described below: Algorithm 44: Find vertex connectivity κ(G) of graph G. t) in every graph. t)'s will be the answer. Let k = 1. output: Vertex Connectivity κ(G) of graph G?
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Construct p − 1 copies of graph G numbered from 2 to p. Increment k and go to step 3.
. The vertex connectivity κ(G) of this graph is also indicated here.
That means we should change s and again apply this algorithm but we need not do this repetition more than the latest output of our algorithm.
. if we select s among vertices which belong to the minimum sized vertex set (which if removed will disconnect graph G) then our algorithm will provide an incorrect estimate for vertex connectivity. Connectivity and Matching Problems
The figure shows that if s is initially selected as g (or h) in this graph then the output of this algorithm will be 3 which is a wrong answer for vertex connectivity. Any other choice for vertex s will always provide us the correct answer in this graph. You can easily generalize these observations. How this intelligent strategy cuts down the time complexity will be interesting to explore.304
Network Flows. The output of the above algorithm will not be equal to the vertex connectivity of the given graph but it will still give us an upper bound on vertex connectivity.
We shall. refer to these paths as paths only and not vertex-disjoint or edgedisjoint paths. The minimum cut in terms of number of edges will be equal to the minimum cut in terms of minimum number of vertices. Menger's Theorem & Hall's (Marriage) Theorem Revisited
We have earlier proved Menger's Theorem which states that the maximum number of edge-disjoint (vertex-disjoint) paths from vertex s to t in a directed graph D is equal to the minimum number of edges (vertices) of G which need to be removed in order to disconnect vertex t from s. The directed graph D with maximum number of edge-disjoint paths is transformed into a directed graph F after the direction of each path in D is reversed as shown in the top left diagram of Fig. 6. Every edge-disjoint path from vertex s to t in the graph D is in fact a vertex-disjoint path between the same two vertices.
. Connectivity and Matching Problems
6. Thus maximizing number of paths in D is equivalent to finding maximum matching in B. 6. Thus the maximum number of paths between vertex s and t in G will be equal to the size of the maximum matching in the bipartite graph. namely Konig's and Hall's Theorem. therefore. Each path from vertex s to vertex t in this diagram corresponds to a matching edge in the bipartite graph B as shown in the top right diagram of this figure. This theorem is applicable to any graph while the other two theorems (Hall's Theorem and Konig's Theorem) are applicable to bipartite graphs only.5
Konig's Theorem. Let us now start with a (un-directed) bipartite graph B we transform this graph into a directed graph D after adding vertices s and t as shown in the left diagrams in Fig. 3. We find maximum edge-disjoint paths from s to t in the graph D as shown in the bottom right diagram of this figure.32. If you can not find an additional path from vertex s to vertex t in graph D then it means that you can not improve the size of the existing matching in the bipartite graph. 2.306
Network Flows. 1.33. The following observations will help us in proving the remaining two theorems. There will be a path between vertex s and t in D corresponding to every matching edge in the bipartite graph B.
. The set X contains those vertices which are common in the set P and the partite A. The set Y is in fact the vertex cover in the bipartite graph shown in the bottom right diagram. However it is still possible to reach some vertices from s. Connectivity and Matching Problems
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Network Flows. The middle right diagram shows the minimum cut. The bottom left diagram shows the set Y containing (minimum number of) vertices which if removed will destroy all paths from vertex s to t in the directed graph.
The Fig. the resulting directed graph F is shown in the in the top right diagram of this figure. Obviously P will not contain t (why?). The minimum cut in terms of vertex-disconnecting set will be A − X + N (X).4. The neighborhood N (X) of X is shown in the middle right diagram of this figure. 6. The minimum cut in terms of edge-disconnecting set will be edges from s to A − X and from N (X) to t as shown in the middle right diagram of Fig. If the set P does not contain any vertex of A then it implies that the number of paths from s to t is equal to the degree of node s in the graph. The size of the minimum cut will thus be equal to the size of A − X and N (X). 6.34.4. The minimum sized vertex-disconnecting set in the graph D will correspond to the vertex cover in the bipartite graph B. The same set Y is in fact the vertex cover in the bipartite graph in the bottom right diagram. There are two major cases to consider as shown in Fig. 7. It is first transformed into a directed graph D as described before.33. Under such conditions the size of the maximum matching will be (at least one) less than the size of partite A. 6. Let P represent the vertices which are reachable from vertex s once we have found maximum number of paths and it is no longer possible to find an additional path from s to t in D as shown in the top left diagram of Fig. 8. If the set P contains a vertex x belonging to the partite A then obviously vertex x is not matched to a vertex in the B partite in the maximum matching of the bipartite graph. Menger's Theorem & Hall's (Marriage) Theorem Revisited
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4. 9. This set is denoted by Y in the bottom left diagram of Fig.33.35 below shows a bipartite graph B in the top left diagram. 5.1.33. Problem Set 6. 6.Konig's Theorem. Problem 6. 6. Each matching edge in the bipartite graph corresponds to
. Let X represents vertices which are common between P and the partite A as shown in the middle ledt diagram of the same figure. we find maximum edge-disjoint paths in D. It also implies that every vertex belonging to partite A is matched to a vertex in the B partite in the corresponding bipartite graph. then reverse the direction of each path. 6.
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Problem 6. 6. Find the sets X. N (X). The figure shown below (Fig. X. It is interesting to note that all vertices belonging to P . The corresponding graph F showing reversed paths is shown on the right side of each bipartite graph. 6. It is quite evident from these diagrams that for different maximum matching's in the same bipartite graph we get the same set P . Under such conditions vertex s will be part of a directed cycle. A − X. For each bipartite graph (with a maximum matching indicated) draw the corresponding graph F .2. Also find the minimum cut in terms of vertices as well as edges. Problem 6. there is a directed path from vertex t to vertex s corresponding to each matching edge in B. and A − X in the graph F . You know that in graph F . Prove or give a counter example.5.4. It further implies that we have found the maximum matching in the bipartite graph B. It is obvious that the maximum matching in this bipartite graph is not a perfect matching – all vertices of partite A are not matched.36 shows a bipartite graph with maximum matching edges indicated in different colors.4. The right diagram shows the same bipartite graph with two extra vertices s and t added
. We show a bipartite graph with two different maximum matching's in Fig. It is quite obvious that now it is no longer possible to find an additional path in the graph F . The set of vertices reachable from vertex s. such that the number of vertices of partite A in the cycle will always be one larger than the number of vertices of partite B belonging to the same cycle.Konig's Theorem.4. known as the set P is also indicated in each graph F .38) shows a bipartite graph B with maximum matching edges shown in different colors. Problem 6. are part of a directed cycle containing vertex s.35) shows two different maximum matching's of the same bipartite graph B in the bottom diagrams. 6. and N (X). Can we prove this observation in general or is this localized to this graph only? What are the implications of this observation? Is this something to do with the proof of Hall's Theorem? Problem 6.4. Menger's Theorem & Hall's (Marriage) Theorem Revisited
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a directed path from vertex t to s in F . The left diagram of Fig. The figure shown above (Fig.37. All vertices which are still reachable from vertex s belong to the set P which is also indicated in the right diagram. Find the set P.3.4. Is this a coincidence in this bipartite graph or will it always be true. Find the minimum Cut in terms of edges as well as vertices of F . 6.
It is obvious that corresponding to every matching edge in this bipartite graph.1
Network Flows
Finding Maximum Edge-Disjoint Paths in MultiGraphs
We consider here the problem of finding maximum number of edge-disjoint paths (and a minimum cut in terms of edges) from a vertex s to a vertex t in a directed multi-graph. We can use our earlier algorithms to solve this problem. We find maximum edge-disjoint (or vertexdisjoint) paths between vertex s and vertex t in graph G.Network Flows
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to it. what will be the resulting time complexity would be interesting to find? We can certainly make adjustments in order to increase the efficiency of our earlier approach. For example if P = (s → a → d → t)
. Problem 6. We reproduce an earlier algorithm below for a ready reference. Assume that we convert a bipartite graph (any bipartite graph – not necessarily the one shown in Fig. 6. there is a path from between vertex s and vertex t in the graph G (Is this one to one correspondence between a matching edge in the bipartite graph and an edge-disjoint path in graph G a general phenomena or is it restricted to this bipartite graph – we shall address this issue in the next problem).4. Note that the bipartite graph B in this problem is transformed into an un-directed graph G instead of a directed graph D. Whenever we find a path P from vertex s to a vertex t according to step number 2 of this algorithm we now have an idea of the vertices through which P passes. all paths parallel to P (that means passing through the same set of vertices) can be discovered right away. We designed this algorithm to find maximum edge-disjoint paths in a directed graph.
6. there will be a corresponding matching edge in the bipartite graph B? Either prove or give a counter example.6.6.38) into an un-directed graph G as shown in the figure above. Do you think that corresponding to every edge-disjoint path between vertex s and t in this undirected graph. We assume that there are no self loops but parallel edges are allowed in the directed multi-graph.6
6. This may be possible without damaging the original character of our algorithm. Find the minimum edge cut & minimum vertex cut in graph G and show that it is equal to the vertex cover in this bipartite graph B.
It will be useful if we spend some time on the selection of a suitable data structure to represent a multi-graph. The left diagram of Fig.39 then we have not one but three edge-disjoint paths passing through the same vertices.39.40 reproduces the multi-directed graph of Fig. How can we exploit this data structure in order to use Algorithm 45 (efficiently) to solve the edge connectivity problem? When you derive the time complexity of Algorithm 45 (or its modified version) you may realize that the complexity expression may depend upon the graph edge weights in addition to the size
. and vertices s & t.
as shown in the top left diagram of Fig. The weight of each edge (x. Please note that it is an un-weighted graph. input : Directed graph D. Reverse the direction of edge e and then go back to step 2. 6. How this will cut down the time complexity of our modified approach? We have asked you to derive the time complexity of this algorithm (with or without modification) when applied to Multi-graphs. y) in this weighted graph corresponds to the number of edges from vertex x to vertex y in the un-weighted multi-directed graph shown in the left diagram. for every edge e in path P do set Status(e) = used if initially it was unused otherwise set Status(e) = unused. Instead of finding these paths sequentially (strictly according to this algorithm) we should be able to do it in one go as shown in the same figure.Network Flows
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Algorithm 45: Find Maximum pseudo edge-disjoint paths from s to t in directed graph D. Find a directed path P from vertex s to vertex t in F . Status(e) = unused for every directed edge e in D. and can conveniently be represented by a weighted adjacency list or adjacency matrix data structure. 6. The right diagram shows how it can be represented by a weighted graph.
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Copy Graph D into F . output: Maximum pseudo edge-disjoint paths from s to t in D and a Status(e) for every directed edge e of D. If you are successful in finding a path P then keep a record of the number of paths found so far otherwise exit with graph F as output. 6. Note that this graph (shown in the right diagram) is a simple graph with no parallel edges.
The minimum cut is also indicated in the bottom diagrams. We then find another path and reverse the direction of its edges.
. The top right diagram shows the edges of these paths reversed.318
Network Flows.39: The top left diagram shows a multi-directed graph with three selected paths from vertex s to vertex t shown in bold. The bottom right diagram shows a stage when it is no longer possible to find an additional path from s to t in this graph. Connectivity and Matching Problems
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If (u. v). There are a number of following related problems and issues with network flows: 1. c(u. v) is a directed edge then the capacity of this edge is denoted by c(u. All vertices other than the source or the sink (known as intermediate vertices) can neither generate any flow nor consume any flow.
6. The network flow problem is to find the maximum flow which can take place from the source vertex to the sink vertex such that the flow f (u. v). We assume that flow can originate from the source vertex and can be consumed by the sink vertex. v) taking place in any edge (u. the weight of each edge (x.6.2
The Maximum Flow & the Minimum Cut
We are given a network which is essentially a weighted directed graph with two special vertices (a source vertex) s and (a sink vertex) t as shown in Fig. it implies that your algorithm is no longer an algorithm but a technique? Why has this happened? How can we over come this short coming?
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of the problem (that means number of vertices and edges). the weight of the edge is known as the capacity of the edge.40: The left diagram shows the un-weighted multi-directed graph of Fig.39. The network flow f (N ) taking place in a network N is equal to the net flow coming out of the source vertex or the net flow consumed in the
. 6. v) should not exceed the capacity of this edge.
6. 5. Connectivity and Matching Problems
2. The flow f (N ) in the network is bounded by the expression: f (N ) ≤ min{c(S. the curious reader may have realized that the algorithm for finding maximum edge-disjoint paths for multi-graphs can be used to find maximum flow as well the minimum cut in the network.
Network Flows. T ) in N . T ) and some flow in the direction from t to s which is represented by f (T. There is however. 3. This relationship is described by the famous MinCut-MaxFlow Theorem. which means that whatever flow goes into any vertex is equal to the flow coming out of that vertex and no new flow is generated by the vertex itself. Here we apply our earlier techniques of finding maximum edge-disjoint paths in a directed graph with just one important difference – whenever we find a path from vertex s to vertex t. we use breadth first search – thus ensuring that
. If (S. S)). T )). and minimizing the flow in the opposite direction (that is f (T. one serious problem regarding (the complexity of) this algorithm. 6. the network flow f (N ) is the difference between the two. Hence every cut will have some flow in the direction from s to t represented by f (S. T )−f (T. Whereas in the case of s and t the flow going out of s is absorbed by the flow going into t. The maximum flow in a network is achieved through maximizing the flow in one direction (that is f (S. T )} where the minimum is taken over all cuts (S. T ) is a cut in the network then the network flow f (N ) is given by the equation: f (N ) = f (S. The Fig.3
Algorithmic Issues & Complexity Calculations
Although we have not formally described an algorithm to find a maximum flow in a network in the last section. S). In every network the value of the maximum flow is equal to the capacity of a minimum cut. The net flow coming in or going out of a vertex other than s to t is zero. S). 4. 6.320 sink.42 shows a network graph with upper bounds on edge flow. we have briefly talked about this issue before and shall try to settle it now. Given a network it is possible to efficiently find the maximum flow and the minimum cut in the network?
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We find another shortest path from s to 2 to t in the second step – again a flow of 3 units is possible in this path.41: The left diagram shows a network flow graph with the capacity of each directed edge shown. therefore rely on our earlier techniques. In the shortest path we move from vertex s to 1 and then to t – thus ensuring a network flow of 3 units from the source to the sink as shown in the top right diagram of this figure. We reverse its edges and then take another path – this
. Please note that if the upper bound on flow in every edge is 3000 instead of 3 – even then we have to apply the same number of steps to find the maximum flow. 6. Fig. This time we take a longer path from vertex s to vertex t. In fact it can be proved that application of BFS (a minor change in our algorithm) ensures (something really big) that the time complexity of the resulting algorithm will not depend upon the magnitudes of the upper bounds on flow – it will only depend upon the size of the problem.
we find a shortest path in terms of number of edges. Let us first find out what extra price we have to pay if we do not use BFS. In other words the time complexity does not depend upon the magnitude of the upper bound on flow at least in this example. We apply our earlier technique of finding any path (not necessarily shortest) from vertex s to vertex t in this network. The Fig.43 below shows the same network graph.Network Flows
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Figure 6. It also shows the maximum flow and the minimum cut. The maximum flow and the minimum cut are found after applying BFS twice in this graph. The path goes from s to 1 to 2 and then to t.43 shows various stages of the working of our algorithm while it is trying to find a maximum flow in the network. 6. Unfortunately the said proof and the resulting time complexity calculations are beyond the scope of this book – we. The right diagram shows the actual flow taking place in an edge divided by the capacity of that edge.
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time it goes from vertex s to 2 to 1 to t. Remember the (size of the) Min-Cut specifies the maximum number of (edge-disjoint) paths where each edge is traversed at most once. every cut cuts a number of edges which if removed will destroy all paths from vertex s to vertex t. Please note that we have already described an efficient algorithm to find a Min-Cut or maximum edge-disjoint paths in a graph. The (size of the) Max-Cut specifies the minimum number of paths from vertex s to t where each path passes through every edge of the graph at least once (see the right diagram of Fig. The intellectual exercise of reducing one problem into another is always an exciting venture – especially when the similarity between the two problems is not so obvious. The Max-Cut is indicated in the diagram above and is replicated again in Fig. 6. The algorithm finally converges and we get the correct answer but after passing through a number of iterations proportional to k where k is an upper bound on flow in the network. Instead of devising an entirely new algorithm let us explore if we can solve the problem using existing or modified algorithms. We also show a number of cuts in this graph. The size of the Min-Cut in this graph is only two and thus there are only two edge-disjoint paths in this graph which are also indicated in the right diagram shown below.45. Here the time complexity has become dependent not only on the size of the problem but the magnitude of the numbers involved. 6. where each edge is traversed at least once. The Min-Cut passes through minimum number of edges while the Max-Cut passes through maximum number of edges.6. We now address the problem of how to find Max-Cut and minimum number of paths between vertex s and vertex t in a given directed acyclic graph. As it is clear from its name Max-Cut passes through maximum number of edges which if removed will destroy all paths between vertex s and t in the graph.45).44 below. We have earlier discussed the network flow problem where each edge has an upper bound on the amount of the flow that can take place.4
Lower Bounds on Edge Flows and the Max-Cut
We show a directed graph with two special vertices known as s and t in the Fig. The size of the MinCut specifies the maximum number of edge-disjoint paths between s and t in a graph (according to Menger's Theorem). 6. and we need to maximize the total flow taking place in the
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6. Interestingly the Max-Cut in a graph has also an important significance.
44: The left diagram shows various cuts which disconnect graph and the right diagram shows the Min-Cut for this graph.Network Flows
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Min-Cut
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.45: The left diagram shows the Max-Cut for the graph and the right diagram shows the minimum number of paths where each path passes through every edge of the graph at least once. Connectivity and Matching Problems
Max-Cut
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Figure 6.326
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entire network. The minimum flow and the corresponding maximum cut are indicated in the right diagram of Fig. After we have found an acceptable flow from vertex s to vertex t in the network graph D. 6. A zero flow through every edge is a possible answer. Instead of minimizing this
. 6. Algorithm 46 finds an acceptable flow through the network D.48 is not a minimum flow taking place from vertex s to t in the network. 6.47. Please note that an acceptable (or legal flow) shown in the right diagram of Fig. In other words we need to find the maximum cut in the network graph D that will disconnect vertex t from vertex s in D. Now when we have lower bounds on flow that can take place through any edge then we should start with a large acceptable flow – large enough that the lower bound (or limit) on flow through any edge is not violated. Remember when we have (only) an upper bound on flow that can take place through any edge in a network then we start with a small acceptable flow – so small that it can take place through every edge without violating any bounds. It requires that the total flow coming towards a vertex should be exactly equal to the total flow going out of that vertex (except for vertices s and t in the network).46 shows a network graph D with the maximum capacity of each edge indicated. Then we try to increase and maximize the flow from vertex s to vertex t in the network. The Fig. 6.47 except that each weight associated with an edge in this network graph signifies not the upper bound but a lower bound on the flow that can take place through that edge. We should be careful about one thing – the conservation of the flow taking place in the network. we assume that the lower bound on flow through each edge is equal to 1. 6. The input as well as the output networks of this algorithm is shown in Fig.48. It may have become obvious now that if we can solve the problem of finding a Max-Cut and Min-Flow in a network graph with lower bounds on edge capacities then we can also solve the problem of finding minimum number of paths between vertex s and vertex t in a given directed graph where each edge is traversed at least once. The problem is to find a minimum flow in this graph such that the flow taking place through any edge does not go below the lower bound of that edge. We show the same network graph with the same edge capacities in Fig. now we need to minimize it. The maximum flow and the corresponding minimum cut are shown in the right diagram of the same figure.
The weight of each edge signifies an actual and acceptable flow taking place through that edge. Connectivity and Matching Problems
Algorithm 46: Find an acceptable (not necessarily minimum) flow from vertex s to vertex t in a given network graph D. input : A weighted directed network graph D with vertices s and t. output: A weighted network graph D.330
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for every directed edge (x. The weight of any edge signifies the lower bound of flow that can take place through that edge. b) in path P do Push an additional flow in the entire path P equal to the lower bound on edge (x. y) in graph D do Find a path P from vertex s to vertex t in graph D passing through edge (x.48: The left diagram shows the input graph to Algorithm 46 and the right diagram shows one of the possible output for the algorithm.
49. the size of the two cuts are however different as is evident from Fig. The maximum flow taking place through any edge (y. y) − m(y. For every edge (x. The weight of each edge in this diagram shows how much actual flow is taking place in this network without violating any lower bound on flow taking place in any edge. x) in this graph is represented by m(y. y) is the lower bound on flow that can take place in the edge (x. there is an additional directed edge (y. 6.49.49 below. The Max-Cut in the graph D from vertex s to vertex t is shown in the bottom right diagram. The network graph D is converted into a directed graph F as follows: 1.
. 6. 6. y) is equal to w(x. We copy graph D into graph F without any edge weights. The Min-Cut in graph F from vertex t to vertex s is shown in the middle right diagram. The corresponding minimum flow in the graph D is shown in the bottom left diagram. Please note that the Max-Cut in D from s to t cuts the same edges as the Min-Cut in graph F from vertex t to vertex s.Network Flows
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flow directly we do so indirectly – by pushing an opposite flow taking place from vertex t to vertex s in D as shown in the Fig. We find the maximum flow in graph F taking place from vertex t to vertex s as shown in the middle left diagram. where c(x. 2. 6. x). It means that all vertices as well as directed edges of D are copied in directed graph F .49. Here the weight of an edge (x. y). We start with an acceptable flow in the network graph D as shown in the top left diagram of Fig. x). x) in graph F with a weight equal to w −c(x. y) in the graph D. y) in D with a weight equal to w. The resulting graph F is shown in the top right diagram of Fig.
We need to find minimum number of paths from vertex s to vertex t such that each vertex of D is traversed at least once by any of the s − t paths.5. prove that this algorithm always finds the correct result or find a counter example.5. 1. Either prove that Musharraf is right or find a counter example.4. We are given a directed network graph D with special vertices s and t. Problem 6.5.6. yes in case an acceptable flow is possible and no in case it is not possible? A special case of this problem is when the fixed flow through every edge is exactly one. We are given a network graph D with special vertices s and t as in Fig.5. Such graphs (where an acceptable flow) is possible have a special structure as we shall study in coming chapters.1.Network Flows Problem Set 6.2.How about if the lower bound is a positive number (it may be different for different edges) while the upper bound on flow is the same for all edges? Repeat (1) for this problem. Problem 6.5.
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Problem 6. Once we find an acceptable flow – it can always be maximized or minimized. We need to make sure that a same fixed amount of flow should take place through every directed edge in that graph.50. 2. (b) An acceptable flow exists if and only if the flow through any edge does not exceed its upper bound.5. How can you design an efficient algorithm which will output either yes or no.5. We are given a network graph D with special vertices s and t.
. (c) If an acceptable flow is possible then find a minimum flow in this network. Each edge in this network has a lower as well as an upper bound on edge flow. Musharraf designed the following intelligent algorithm to find an acceptable flow provided it exists in the network graph of the above problem: (a) Initially ignore the upper bound on each edge and find a minimum flow in the network (from vertex s to vertex t) keeping into account the lower bounds on flow through each edge. Now Musharraf insists that his algorithm will find an optimal answer for Problem No. we assume that the lower bound as well as upper bound is the same for each edge although the two bounds may be different from each other. Problem 6. (b) If an acceptable flow is possible and is provided to you then find a maximum flow in this network. Problem 6. Problem 6. 6.5. Kashif finds the following counter example for the above problem as shown in the figure below.3. (a) Design an efficient algorithm to find if an acceptable flow is possible in this network.
shown in red color. We apply Algorithm 45 on this graph. Please recall the directed graph D shown in top left diagram of Fig. We are given a network graph D with special vertices s and t. 6. Connectivity and Matching Problems
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.8.15 (also shown below Fig.50: Upper/Lower limit for flow in each edge is shown. this condition is false in a network graph (that means the in-degree is at most two times the out-degree of any vertex other than vertex s and t) then a feasible flow will exist in the network graph? Can you counter this argument? Can you now design an efficient algorithm to find a feasible flow in the graph?
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D without the unused edges.10. Here we shall address the matching problem in un-weighted and then weighted
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The Matching Problem in Bipartite Graphs
We have already talked about the Matching problem in different contexts. Can you think of a directed graph where property (a) is applicable but (b) is not? Can you think of a graph where (b) is applicable but (a) is not? Problem 6. We claim that in any network graph D without the unused edges. it is possible to push a flow of exactly one unit in every directed edge. either prove this or give a counter example.5.5. A directed network graph D without unused edges has two important properties: (a) every edge is part of a directed path from vertex s to vertex t in D.52:
6. (a) Is it possible to push a flow of exactly one unit through every directed edge in this graph? (b) If it is possible then does it mean that every edge in this graph will be part of a directed path from vertex s to vertex t in this graph? (c) If (a) is possible then can we find maximum edge-disjoint paths from s to t in this graph using the stupid Algorithm 36?
Vertex 1 is s and 7 is t
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Figure 6. Problem 6. The two graphs are reproduced in the figure below for ready reference. & (b) We can use (the so called stupid) Algorithm 36 to find maximum edge-disjoint paths in this graph. Consider a directed graph D where the in-degree of every vertex (other than s and t) is equal to the corresponding out-degree.
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Maximum Matching in Un-weighted bipartite graphs
Given a maximal matching. We shall devise efficient algorithms to find maximum matching in un-weighted bipartite graphs and weighted maximum matching in a weighted bipartite graph. in fact any two matched edges in the bipartite graph will correspond to two edge-disjoint directed paths from vertex s to vertex t in the corresponding directed graph. 6. As explained before each matched edge in the bipartite graph (top left corner) corresponds to a path in the directed graph (top right corner).53) we can find the corresponding matched edges in the bipartite graph as shown in the bottom right diagram of the same figure. Unfortunately it is difficult to adapt this algorithm to find a maximum weighted matching in a weighted bipartite graph. Once we find the maximum number of paths in the directed graph (see the bottom left diagram of Fig.53. The algorithm described above works well for finding a maximum matching in an un-weighted bipartite graph. We are given a bipartite graph having a B partite (consisting of a number of boys) and a G partite (consisting of a number of girls). 6. Connectivity and Matching Problems
bipartite graphs. present (in the coming sub-section) a slightly different version of this algorithm to find maximum matching in an un-weighted bipartite graph? Problem Set 6. The matching problem is converted into a connectivity problem by adding two dummy vertices s and t as shown in the top right corner of Fig. An edge between a boy and a girl shows a degree of compatibility. please note that the direction of each such edge is from the partite A to partite B. it can
. We add directed edges from vertex s to every vertex in partite A.7.53. this completes the transformation from a bipartite graph of the top left corner into the directed graph shown in the top right diagram.336
Network Flows. We need to find maximum pairs of boys and girls such that the boy and girl in every pair are compatible. We.6. therefore.
6. This problem is also known as the Marriage Problem. We also add directed edges from every vertex of the partite B to vertex t. 6. The problem of finding maximum matching in a bipartite graph is thus transformed into the problem of finding maximum number of edge-disjoint (or vertex-disjoint) paths in the directed graph. We convert un-directed edges of the bipartite graph (top left diagram) into directed edges as shown in the top right corner. how can we find a maximum matching in a bipartite graph? A maximal matching in a bipartite graph is shown by bold lines in the top left corner of Fig.
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Figure 6.4. each girl can marry a single boy.338
Network Flows. Assume that a boy can marry four girls.1. Connectivity and Matching Problems
be modeled by an un-weighted bipartite graph as shown below (Fig.54: A bipartite graph showing a B partite consisting of a set of boys.6. each boy can marry a single girl. Compatibility between a boy and a girl is indicated by an edge between the corresponding vertices.6. a set of girls. we intend to maximize the number of marriages taking place. Problem 6.2. Assume that we intend to solve the decision problem in which we intend to find a yes/no answer corresponding to the question: Is it possible to marry all boys? (Or is it possible to marry all girls?) Problem 6. How will you model this problem in terms of a known graph problem? Under what conditions it will be possible? Problem 6. 6. How will you model this
.3.6.6. How will you model this problem in terms of a known graph problem? Under what conditions it will be possible? Problem 6. Assume that we are required to solve the marriage problem: namely we intend to marry maximum number of girls. a G Partite showing.54). The problem can be solved by techniques similar to the ones that we have just studied. Assume that we are required to solve the marriage problem: namely we intend to marry maximum number of boy.
Here we shall describe another algorithm to find the maximum mathing -
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problem in terms of a known graph problem? Under what conditions will it be possible? Problem 6. How will you model this problem in terms of a known graph problem? Under what conditions will it be possible? Problem 6. We are given a bipartite graph with edge weights equal to either zero or one. How about if we remove all directions from the directed graph D. Problem 6. Discuss how you will solve this problem efficiently using similar techniques.5. Problem 6.7. Can we still claim that the maximum matching in the bipartite graph is equal to the maximum number of edge-disjoint paths between vertex s and vertex t in the un-directed graph D.7. Discuss with the help of an example. We need to find maximum weighted matching in this weighted bipartite graph. 6. We intend to maximize the number of married boys and girls subject to the condition that a boy can marry four girls.6.2.7. In other words the problem of finding maximum matching in a bipartite graph is transformed into the problem of finding maximum number of edge-disjoint (or vertex-disjoint) paths in a directed graph D. Assume that a boy can marry four girls. It can be found using our expertise gained in the last section.7. thus converting it into an undirected graph.6. Problem Set 6.6.
6.6.2
Maximum Matching in Complete (Binary) Weighted Bipartite Graphs
We are given an un-weighted balanced bipartite graph as shown in left diagram of Fig. A maximum matching in this graph is also shown in this diagram. How will you model this problem in terms of a known graph problem? Under what conditions will it be possible? Problem 6.55.1. Assume that a boy can marry four girls. we intend to maximize the number of girls who are married. The maximum matching in a bipartite graph is equal to the maximum number of edge-disjoint paths in a directed graph D. we intend to maximize the number of boys who are married.7.
We start with a complete weighted bipartite graph G of size k (it means there will be k vertices in the A partite as well as k vertices in the B partite).340
Network Flows. A maximum weighted matching in the right graph of this figure will be a maximum matching in the original bipartite graph. We now describe a (not very efficient but useful algorithm) which works on the directed graph D and outputs the maximum weighted matching in the bipartite graph. An edge in the original bipartite graph has a weight of 1 in the complete bipartite graph while every other edge has a weight equal to zero. The bipartite graph is converted into a completely connected binary weighted bipartite graph as shown in the right diagram. The results of this algorithms can be used to find maximum matching in a bipartite graph. We describe in the following paragraphs a useful algorithm to solve the maximum weighted matching problem in a complete bipartite graph with binary weights. As shown in Fig.
. 6.55 we convert this bipartite graph into a complete weighted bipartite graph .see the right diagram of this figure. Thus the complete bipartite graph has binary weights.
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Figure 6. Connectivity and Matching Problems
this new algorithm may not be as efficient as the one described before but it has the added advantage of being flexible. It can handle (with or without minor modification) the more general problem of finding maximum weighted matching in a bipartite graph. The algorithm can also be used to find maximum weighted in a complete bipartite graph with non binary weights as described in the next section. We convert the bipartite graph into a weighted directed graph D after adding vertices s and t according to the rules described previously.55: A balanced bipartite graph G with maximum matching of size 3 is shown in the left diagram. Edge weights not shown in the right diagram are equal to zero.
If you can not find such a path then terminate. the modifications should be such that the maximum matching in the new bipartite is equal to the maximum matching in the original graph D.1. Problem 6. y) of path P in graph D do Reverse the direction of edge (x. input : A complete balanced (binary) weighted bipartite graph.8. for each edge (x. Problem 6.
. Whereas in the case of Algorithm 47 for a complete weighted bipartite graph. Connectivity and Matching Problems
Algorithm 47: Find a maximum weighted matching in a complete balanced bipartite graph G. y) with a negative sign and go back to Step 2
Problem Set 6.
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Transform the bipartite graph G into a directed graph D after adding vertices s and t to G according to the rules already defined. also it is applied p (no of vertices) times on the graph until all the paths have been found. What is the time complexity of Algorithm 47 and the previous algorithm described for un-weighted graphs in the last section? Note that we are applying a simple path finding algorithm for un-weighted edges in the previous algorithm of the last section and its time complexity was O(p + q).2. Find a longest path P from vertex s to t with a weight equal to or larger than zero.8. y) Multiply the weight of edge (x. Would we be able to get the maximum matching in a bipartite graph if the graph is neither complete nor balanced by applying Algorithm 47? Problem 6. How is Algorithm 47 applicable to find maximum matching in a general weighted graph D with weights greater than 1? Discuss the possible modification that need to made to the general graph D before we can use it as input for the Algorithm.8. output: A maximum weighted matching in G. the Bellman-Ford algorithm is applied and its time complexity is O(p3 ) and again it is applied p number of times on the given network graph.342
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48 requires us to find the maximum weighted matching in the first x layers. Similarly in order to implement Algorithm No. Before discussing the details of the two procedures we first need to transform the weighted bipartite un-directed graph G into a directed graph D with additional vertices s and t according to the following rules as shown in the diagram below. It is obvious from these diagrams that the intermediate results may be different but the end results are same for the two algorithms. 1.
. Do not move forward before understanding Fig. b) in bipartite graph G add a direction going from vertex b to a in the directed graph D. Note that vertex a belongs to A partite while b belongs to the partite B of the bipartite graph G.The Matching Problem in Bipartite Graphs
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The working of the two algorithms is shown in Fig. Now add another vertex t such that there is a directed edge from every unmatched vertex in the B partite to vertex t with a weight equal to 0. 1 & 2. 2. 4. All other edge weights retain their original signs in D as there are in G. 1: Given a maximum weighted matching in the first x layers of G it finds the maximum weighted matching in the first x + 1 layers of G. Add vertex s to G such that in the resulting graph D there is a directed edge from vertex s to every unmatched vertex in the A partite with a weight equal to 0. 3. Algorithm No. For every unmatched edge (a. In order to implement this algorithm efficiently we need to design an efficient procedure which performs the following function: Procedure No. The sign of every weight w for every matched edge in G is changed to a minus sign in the directed graph D. 6. 49 efficiently we need to design an efficient procedure which performs the following function: Procedure No. 6.58. We shall now discuss the details of Procedure No. For every matched edge (a.58. b) in G add a direction going from vertex a to b in the directed graph D. 2: Given a maximum weighted matching of size x in G it finds the maximum weighted matching of size x + 1 in G.
6.1. Please comment. The weight of the longest path from vertex s to t in D is equal to the gain in the weight of maximum matching when the size of matching is increased from x to x + 1 in the bipartite graph G.1.61. 1) then we need to reverse edges belonging to Path 1. Claim 6.62. we need to find a longest path from vertex s to vertex t.2.9.7. 6. If on the other hand we reverse edges in Path 2 then a cycle with a net positive value at most equal to y − z will be formed as shown in Fig. Assume that we are given a complete balanced weighted bipartite graph with positive edge weights. however.9. Describe an efficient algorithm to find a maximum matching in the very first layer of the bipartite graph. Claim 6. The left diagram of Fig.7.
. Please note that there are 2k + 2 vertices in this directed graph and your answer should be independent of edge weights in the graph. Some one thinks that the longest path problem is NP-Complete. Problem 6. 1 (& Procedure No. We shall never encounter a situation where there will be a positive weight cycle in D (provided we follow steps given in Procedure No. 6.1 or Procedure No. We need to find what will be the maximum length (in terms of the number of edges) of the longest path passing through the first two layers of the graph. 2). Negative weight cycles will. Then the reversed edges of Path 1 plus edges belonging to Path 2 will form a cycle in the graph with a net negative value at most equal to y − z. Carefully derive the time complexity of this algorithm.The Matching Problem in Bipartite Graphs
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Claim 6. Problem 6.7. we now need to extend this maximum matching in the first x+1 layers of G as shown in Fig.2. 2). Assume that we have already found a maximum matching in the first x layers of graph G. We transform the bipartite graph into a directed graph also shown in Fig.3. Problem Set 6.65 shows a directed graph in which there is only one edge going from a vertex b to a vertex a.9.3. If we select Path 1 (as dictated by Procedure No. Suppose we have found two paths (one longest and one relatively shorter) from vertex s to vertex t. 6. Path 1 is the longest path and has a weight equal to y and Path 2 is a relatively shorter path and has a weight equal to z. Problem 6. In Procedure No.61. be present in the directed graph D.9.
9. Problem 6. Carefully derive the time complexity of Procedure No.9.
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Problem 6.5. 6.9. Design an efficient algorithm to solve this problem.6. Is it possible to use an existing text book algorithm (without any modification) in order to solve the previous problem? Discuss briefly. Problem 6. We are working under the assumption that there are no positive weight cycles in the directed graph D.
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Figure 6. The right diagram shows another directed graph D where there are x edges going from a vertex b to a vertex a.65: The left diagram shows a directed graph D in which there is only one edge going from a vertex b to a vertex a. 1 and then the time complexity of Algorithm No.9.63).65 shows a directed graph in which there are x edges going from a b vertex to an a vertex. The right diagram of Fig. 48 (see Fig. Problem 6. Now we need to find a 5-edge longest path from s to t in the same graph consisting of first x + 1 layers of graph D. Carefully derive the time complexity of this algorithm in terms of the size of the problem. Discuss briefly if your algorithm is a greedy algorithm or does it use dynamic programming. Assume that we have found a 3-edge longest path from vertex s to vertex t in a directed graph consisting of the first x + 1 layers of graph D (see right diagram of Fig.4. We need to find what will be the maximum length (in terms of number of edges) of the longest path passing through the first x + 1 layers of the graph. 6. 6.65). Please note that there are 2k + 2 vertices in this directed graph and your answer should be independent of edge weights in the graph.7.
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Problem 6. Problem 6. Suppose we need to find a best possible matching of size x in a given complete balanced weighted bipartite graph G of size k and x is much smaller than k. Suppose we need to find a best possible matching for the first x members of partite B in a given complete balanced weighted bipartite graph G. Problem 6. Discuss how you will solve this problem and carefully derive the time complexity of your algorithm in terms of x and k. 48 or 49 to solve this problem? How can you design a better algorithm to solve this problem? Discuss briefly.14.9.16. Discuss if we can use Algorithm No.15. 48 or 49 to solve this problem? How can you design a better algorithm to solve this problem? Discuss briefly.9. Discuss if we can use Algorithm No. Suppose we need to find a best possible matching for the first x members of partite A in a given complete balanced weighted bipartite graph G.
and also remind ourselves of some of the powerful techniques which will be useful in solving this problem. We also need to make sure that the maximum flow is taking place at minimum cost.
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The Max-Flow Min-Cost Problem
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Maximum Flow (3 units) at Minimum Cost (8+5+4+2)
Figure 6. We need to find maximum flow coming out of vertex s and being absorbed by vertex t.8. The maximum flow at minimum cost is indicated by colored lines in the right diagram. We need to push a maximum flow in this network graph (starting from vertex s and ending at vertex t) at minimum cost. A possible solution of this network graph is shown in the right diagram of the same figure. 6. It is interesting to note
.8. we are pushing a maximum flow of 3 units at a total cost of 19. We assume that the capacity as well cost per unit flow through every edge is a positive integer.67: We show a directed graph D (left diagram) having two special vertices s and t.
6. it is indicated with each edge in the figure below.8
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The Max-Flow Min-Cost Problem
Introduction
We consider the Maximum Flow at Minimum Cost problem in a network graph as shown in Fig. The capacity/cost of an edge is shown along with each edge.358
Network Flows.2
Finding a Maximum Flow or Finding a Shortest Path?
Before answering the above question it is important to revise the relevant prior knowledge. The problem is to efficiently find maximum flow at minimum cost in a given network graph.67. Connectivity and Matching Problems
6.
If we start minimizing the cost by finding shortest paths. We need to minimize the cost of (unit) flow starting from vertex s and ending at vertex t. 6. So let us list down the general problem (once again) and its special cases: Category 1: Given a network flow graph as shown in Fig. finding shortest paths without considering capacities will create complications. We need to maximize the flow and we already know how to do it (but we should also remember our short comings and limitations).67 how can we find maximum flow at minimum cost from a source vertex s to a sink vertex t? We assume that edge capacities are integers while per unit cost of flow through any edge may be a real number. then how will we be able to tackle the problem of maximizing flow? Different edges have different capacities. 2. Please note that this problem is equivalent to finding maximum edge-disjoint paths from vertex s to vertex t at minimum cost. Category 2: Given a network flow graph as shown in Fig. Category 3: The network flow graph is derived from a complete balanced and weighted bipartite graph. Here we add a source vertex s to the A partite
.e. Before we solve this general problem we shall try to reflect what similar problems we have already solved and what are some of the special cases of this general problem which can be resolved using our prior knowledge? We shall then extend or modify specialized solutions to solve this general problem. 6. and this somehow looks like finding a shortest path from vertex s to t (we are aware of a number of simple and efficient algorithms to solve shortest path problems). We know how to solve the above two problems in isolation but how to fulfill the two requirements simultaneously? If we start finding flows without looking at costs then we may end up with a maximum flow but at higher cost. i. It is certainly a very exciting mixture of two important problems.The Max-Flow Min-Cost Problem
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that this challenging problem (which we call a Category 1 problem) has two requirements: 1. the sum of edge costs in all edge-disjoint (shortest) paths should be as small as possible.68 how can we find maximum flow at minimum cost from a source vertex s to a sink vertex t? We assume that capacity of each edge is exactly 1 while per unit cost of flow through any edge may be a real number.
69 we were essentially solving the maximum flow at maximum cost problem as shown in a network flow graph in Fig.8. Both problems have their applications in graph theory and elsewhere. All edge capacities are 1. 6. We assume that the capacity of each edge is one in the network flow graph. Here the problem is to find maximum flow at minimum cost in the network flow graph shown in the right diagram. 6. It is possible to recognize that finding maximum flow (or maximum edge-disjoint paths) at minimum cost in the right diagram is equivalent to finding a minimum cost perfect matching in the bipartite graph shown in the left diagram.70.69: In a Category 3 problem a network flow graph is derived from a weighted bipartite graph by inserting a source vertex and a sink vertex.69. 6. A related problem in a bipartite graph would be to find a maximum cost perfect matching .that would require us to find maximum flow (or maximum edge-disjoint paths) at maximum cost.
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and a sink vertex t to the B partite as shown in Fig.3
Category 3 network flow Problems
We shall now try attacking these problems starting from Category 3. costs not shown are equal to 1
Category 3: The network flow graph is derived from a bipartite graph. Similarly while we were solving the minimum weighted perfect matching
. A curious reader might have noticed that while solving the maximum cost perfect matching problem in the bipartite graph shown in Fig. All edge weights associated with source vertex s and sink vertex t are zero
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We have in fact solved the Category 3 network flow problem without explicitly saying so as our primary objective was to find a perfect matching of maximum (or minimum) cost. 6. 6.362
Network Flows. The algorithm terminates when it is no longer possible to find a path from the source to the sink as shown in the bottom right diagram of Fig. Connectivity and Matching Problems
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Category 3: Maximum Flow at Maximum Cost
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problem in a bipartite graph we are essentially solving the minimum cost maximum flow problem. As we shall show later this algorithm is powerful enough to handle Category 2 and 1 network flow problems.71.69. 6. Maximum Flow at Maximum Cost in Category 3 problems Please recall Algorithm 47 which was designed to find a maximum weighted matching in a complete bipartite graph. We reverse the direction of each edge in path P and also multiply weight of each edge in the path by minus 1. The step by step working of this algorithm is shown in Fig. The problem is how to recover maximum flow at minimum cost from this (final) graph F ?
. It is slightly modified as shown below (Algorithm 52). What we essentially do here is to find a longest path P from vertex s to vertex t in the given graph. We apply this algorithm to the network flow graph shown in the right diagram of Fig.70: A maximum weighted perfect matching in bipartite graph (left diagram) corresponds to a maximum flow at maximum cost in the right diagram.71. We then again find a longest path from the source vertex to the sink vertex in the modified graph.
6. y) with a negative sign and go back to Step 2
The solution to this problem is again consistent to our earlier approach of deleting unused edges from the network flow graph. We intend to solve the maximum flow at minimum cost problem using multiple algorithms in order to provide a better insight to the problem and its possible solutions. Find a flow of one unit through a longest path P (in terms of edge costs) from vertex s to t in F . y) Multiply the weight of edge (x. If you are successful in finding a flow then keep a record of the path found otherwise exit with graph F as output. But in order to do that we need
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Algorithm 52: Find a Maximum Flow at maximum cost from vertex s to vertex t in a directed network graph D belonging to Category 3 input : Directed & Weighted graph D. Maximum Flow at Minimum Cost in Category 3 Problems The above algorithm can easily be adopted to find maximum flow at minimum cost or in other words maximum edge-disjoint paths at minimum cost in a Category 3 network flow graph. output: Maximum Flow at minimum cost from s to t in D
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Category 2 (and 1) network flow Problems
Here we shall consider Category 2 (and Category 1) network flow problems. for each edge (x. We shall first show that Algorithm 53 can be used solve these problems without any modification.72. 6.8.70. The only change that we need to do is to replace longest path in line 2 of this algorithm by shortest path as shown in the following algorithm (Algorithm 53).
6. The unused edges in final graph F are those which have positive weights as shown in Fig. By deleting these edges it is possible to find maximum flow at maximum cost in a Category 3 problem as shown in Fig. One would like to compare this answer with the one obtained while finding a maximum weighted perfect matching in a bipartite graph shown in Fig. y) of path P in graph F do Reverse the direction of edge (x. 6.
We are given a directed & weighted graph D. Negative Weight Cycles & Improvement in Cost of Flow We show a flow of one unit in the network D shown in the top left diagram of Fig. If on the other hand there are no negative weight cycles then it is not possible to reduce the cost of existing flow. Connectivity and Matching Problems
to do some serious graph theoretic work in terms of claims and some hints for their proofs. We now reverse the direction of each edge in path P in graph F . Please note that this flow of one unit is not taking place on a shortest path from vertex s to vertex t. The value of net weight in this cycle is -2. Claim 6. Now we claim that as directed graph D does not contain any negative weight cycle then graph F will also not contain any negative weight cycle. Now we claim that as directed graph D does not contain any negative edges then graph F will not contain any negative weight cycle.8. It contains a source vertex s and a sink vertex t. Claim 6. We find a flow of one unit from vertex s to vertex t. By redirecting the flow in the negative weight cycle it is possible to reduce the cost of flow by an amount exactly equal to 2 as shown in the bottom diagram of this figure. If there is a negative weight cycle in graph F with a net weight equal to -k (graph F is derived from a network graph D after reversing the edges in the direction of the flow and multiplying the costs in
. We then find a negative weight cycle in this new graph F highlighted by orange color.8. We copy this graph in graph F . We also multiply the weight of each edge in path P by minus one in graph F . We also multiply the weight of each edge in path P by minus one in graph F . 6. Given a network flow graph D.2. We are given a directed & weighted graph D with no negative weight cycles but it may contain negative weight edges.73. It contains a source vertex s and a sink vertex t. We now reverse the direction of each edge in path P in graph F .74. We find a shortest path P from source vertex s to the sink vertex t in directed graph F . The graph D may contain cycles but it does not contain any negative weight edge.8. Such a scenario is shown in Fig. We copy this graph in graph F .1.3. We find a shortest path P from source vertex s to the sink vertex t in directed graph F . We reverse the direction of the edges used by the flow and multiply the cost of these edges by minus 1 as shown in graph F (see top right diagram). Claim 6. 6.366
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Network Flows. we keep a record of the length of each such path and call these paths existing paths in D. If there is no negative weight cycle in graph F (derived from a network graph D after reversing the edges in the direction of the flow and multiplying the costs in these edges by minus 1) then it is not possible to reduce the cost of the existing flow by re-adjusting the flow in any direction.
. Connectivity and Matching Problems
these edges by minus 1) then it is possible to reduce the cost of the existing flow by k by re-adjusting the flow in the direction of the negative cycle. Claim 6.8. We know the path taken by each unit of flow from vertex s to vertex t in D.8.75. Claim 6. If (after reversing the edges in the direction of the flow) there are no negative weight cycles then the cost of flow can not be further reduced. 6.4. Given a graph D and a finite flow taking place from vertex s to vertex t. Please see Fig.c
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Figure 6. If there is no improvement possible in the cost of existing flow then there will be no negative weight cycle in graph F (derived from a network graph D after reversing the edges in the direction of the flow and multiplying the costs in these edges by minus 1).74: We show a flow of two units in the network shown in the top left diagram.5.
75: We show two existing paths from s to t in graph D (top left diagram).
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Graph F: we have reversed the edges in the direction of the flow in graph D and have multiplied the edge costs by -1
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Figure 6. The paths are reversed in graph F as shown in the top right diagram.The Max-Flow Min-Cost Problem
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A flow of 2 units in graph D The red path has length equal to 4.
and vertices s & t.70.370
Network Flows. Reverse the edges in the path of every flow. Problem 6.76. Our last three claims support the argument that the following algorithm would be able to correctly solve maximum flow at minimum cost problem in Category 1 network flow graphs. but it may give rise to another negative weight cycle. If you find one then go to step 4 else output the maximum flow at minimum cost and exit. First we find maximum flow.
.1. output: Maximum Flow from s to t in D at Minimum Cost
1 2
3
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Copy graph D into a graph F .
We now describe an alternate algorithm (Algorithm 54) to find a maximum flow at minimum cost for Category 1 problems. multiply the costs corresponding to these edges with minus 1. If there are any negative weight cycles in the graph then we have to remove every negative weight cycle. now go to step 3. The process of finding maximum flow at minimum cost may be accelerated if we some how find a negative weight cycle of higher value and remove it as shown in Fig. Find if there is a negative weight cycle in graph F.10. multiply the cost of these edges with minus one. Find Maximum Flow from vertex s to t in F (ignoring costs). 6. we then reverse the edges in the direction of the flow. Connectivity and Matching Problems
Finding Maximum Flow at Minimum Cost In view of the above claims it is obvious that Algorithm 53 can be used to find maximum flow at minimum cost in Category 2 as well as Category 1 network flow problems. Algorithm 54: Find Maximum Flow at Minimum Cost from s to t in a directed graph D in a Category 1 problem input : Directed & Weighted graph D. 6. We have demonstrated the working of an algorithm (to find maximum flow at minimum cost in Category 3 problems) in Fig. Remove the negative weight cycle by adjusting the flow accordingly. It will be interesting to derive the time complexity of that algorithm for both these categories.10.77. The process is repeated until there are no more negative weight cycles left in the graph. the removal of a negative weight cycle will certainly reduce the cost of flow by at least one unit. The step by step detailed working of the above algorithm on a Category 1 problem is shown in Fig. 6. Problem Set 6.
. Carefully derive the time complexity of this algorithm. The capacity of every edge in the D − s − t graph is infinite while the cost is a positive integer indicated in the diagram. Algorithm 54 can be used to find maximum flow at minimum cost in Category 1 problems.80: We show a special network graph D. In this special category of graphs the sum of capacities of all edges coming out of s is equal to the sum of capacities of all edges going into vertex t.The Max-Flow Min-Cost Problem
375
Describe this algorithm (known as Algorithm 53) in your own words and carefully derive its time complexity. Is it possible to solve problems belonging to this category using Algorithm 53 even if the cost of a unit flow in an edge is a real number? Discuss briefly. You may design a better algorithm if you want and if you can?
a
4/0
∞/8 ∞/3 ∞/7
b
5/0
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t
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Figure 6. Discuss briefly. Find if you can use any earlier techniques (Algorithm 53) to solve this interesting problem. The cost of every edge coming out of s and going into t is zero. The sum of capacities of all edges coming out of source s is exactly equal to the sum of capacities of edges going in sink t while the costs of these edges are zero.10. The step by step working of this algorithm is demonstrated in Fig.10. Is this possible to use this technique (without any appreciable change) to find the minimum cost for a fixed flowthe amount of fixed flow may not be the maximum flow in the network. Problem 6.2. 6.80. Problem 6. Here edge capacities in graph D − s − t are all infinite.3.76. 6. We show an interesting special case of Category 4 in Fig. We need to find maximum flow at minimum cost using an efficient algorithm.
In this category all edge capacities are equal to 1. 6.
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Equivalent
Category 3
The Max flow Min-Cost Problem
Category 3
The Max flow Min-Cost Problem
Figure 6.72.376
Network Flows.8.78.
.
6. this special case is shown in the bottom diagram of Fig.81: We show that a Category 3 problem (where edges adjacent to s and t have non-zero cost) is as difficult (or as easy) to solve as a Category 3 problem where the s and t edges have zero costs (top diagram). On the other extreme Category 2 & 3 problems can be solved using very efficient algorithms. Design a (very) efficient algorithm to find a maximum flow at minimum cost in this special category. 6.72. Some of these problems were addressed in the last problem set. The bottom diagram shows an interesting variation of Category 3 problem where we need to minimize cost not for a maximum flow but for a fixed flow in the network. We have not yet devised an exact algorithm to find maximum flow (even if we ignore costs) in such problems.5
A Panoramic Picture of Similar Problems & Solutions (once again)
Please note that Category 1 problems are the most general. Describe an efficient algorithm to handle Category 2 problems. In between these two extremes there is an exciting range of problems to be explored. 6. Because of its restricted nature. Connectivity and Matching Problems
Problem 6. we expect to solve it using a simple and a very efficient algorithm. thus maximum flow here corresponds to maximum edge-disjoint paths as shown in Fig.4. Now consider a special case of this category where costs are expressed by binary numbers. We have applied Algorithm 54 to solve a Category 2 problem as shown in Fig.10. applying Algorithm 54 seems to be overkill.
Understanding of these transformations is a must for appreciating the new knowledge described here. how can we find a maximum flow from a source vertex to a sink vertex? The maximum flow as well as the minimum cut is indicated in the graph shown below. maximum flow at minimum cost. a finite upper
.82: Left diagram shows a network flow graph with upper bound on flow in each edge.
6.
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Figure 6. We are given a network flow graph with zero lower bound. It is interesting to note that this section relies on old concepts like network flows. We start this section with the required prior knowledge which is essential to understand the theory and the practice described here.9.9
Network flows with lower & upper bounds on flow and the Circulation Problem
We shall discuss the Circulation Problem and the related problems of finding feasible flow in network flow graphs. We shall also list down the specific problems that we shall address here. 2. These theorems as well as algorithms depend upon a number of powerful transformations. Given a network flow graph with zero lower bound and a finite upper bound on flow in each edge. etc. maximum flow and minimum cut. The lower bound on flow through every edge is zero. Please note that the flow is conserved at every vertex except the source and the sink vertex.1
Prior knowledge:
1. The right diagram shows the maximum flow and the minimum cut in the network shown in the left diagram. We shall introduce one new concept and that is of a Circulation Graph? We provide a systematic and step by step treatment of a number of theorems and algorithms.Network flows with lower & upper bounds on flow and the Circulation Problem
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6.
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Figure 6. Please note that the flow is conserved at every vertex except the source and the sink vertex. Connectivity and Matching Problems bound on flow in each edge. The lower bound on flow through every edge is zero. 3.83: A network flow graph with upper bound on flow in each edge as well as per unit cost of flow through that edge. With the above two problems solved we should also be able to solve a slightly different problem in which we can find in terms of yes or no if a fixed flow of k units can be pushed in a network flow graph from a source vertex and taken back into a sink vertex from the network. The problem is to find maximum flow at minimum cost.
6. We are also given a cost per unit flow in each edge of the network graph.that means the actual flow entering a vertex is equal to the actual flow coming out of it.9. In a network flow graph. the source vertex has the capability to produce an infinite flow while the sink vertex has the capability to sink an infinite flow.2
New concepts
In a network flow graph we assume that there is a single source vertex and a single sink vertex.378
Network Flows. Every edge in a circulation graph may have a
. Thus in a circulation graph there is no source and no sink vertex. and now we need to find a maximum (or a fixed) flow at minimum cost from a source vertex to a sink vertex in the network. In a Circulation Graph the law of conservation of flow should hold for every vertex. For example in the following network flow graph the minimum cost of a flow of one unit from the source to the sink vertex is 4. in contrast. For the rest of the vertices the law of conservation of flow holds .
and perunit cost on flow through that edge.
6. We need to find a feasible flow in a network graph with a source and a sink vertex.9.84: A Circulation graph with lower as well as upper bound on flow in each edge. Every edge has a lower bound as well as an upper bound on flow (see diagram below).Network flows with lower & upper bounds on flow and the Circulation Problem
379
nonzero lower bound and an upper bound on flow through this edge. Please see the solution of the Circulation problem in the following diagram. Please note that if the lower bound on flow in each edge is zero then a zero flow will always be a feasible flow from the source to the sink.
2.7
a 5
3.9
s
4. Every edge has an associated lower bound. We need to find a minimum cost feasible flow in a Circulation graph. the lower bound on flow in
. A feasible flow in the graph is also indicated.7
b
Figure 6. We need to know how to find a feasible flow in a Circulation Graph where every edge has a lower bound as well as an upper bound on flow. 3. For example a feasible flow of 5 is possible in the circulation graph shown below. Please note that the flow is conserved at every vertex without any exception. If. an upper bound. There may also be per-unit cost associated with each edge.3
New Problems
1. Usually this problem is known as the minimum cost Circulation Problem or the Circulation Problem. however.
2.5
t
5.
Please see the solution of this problem in the diagram below. We need to find a maximum or a fixed flow at minimum cost in a network graph with a source and a sink vertex. A minimum cost feasible flow is indicated in the right diagram. any edge is nonzero then finding a feasible flow is not a trivial problem as the zero flow is not a correct answer.
. A feasible flow in the circulation graph is indicated in the middle diagram. 4.380
Network Flows. and per unit cost on flow through this edge. The cost of a unit flow through every edge is equal to one in this graph. an upper bound. Please note that the flow is conserved at every vertex without any exception.85: A Circulation graph (left diagram) with lower bound equal to 1 and upper bound equal to five on flow in each edge. Connectivity and Matching Problems
2
a e d
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3 3
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Figure 6. Every edge has an associated lower bound.
an upper bound. Also note that the feasible flow taking place may not be a maximum flow (or a minimum flow) from the source to the sink vertex. per unit cost on flow.4.6/3
t
Figure 6.2. Please note that the flow is conserved at every vertex except the source and the sink vertex.9/2
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.87: Left graph is a network flow with a flow of 4 units taking place from the source to the sink vertex.5/2
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1. and actual flow in each edge is indicated in the respective order. and actual flow taking place in that edge. Each edge has an associated lower bound.8/2
z
Figure 6.86: A network flow graph with lower bound. upper bound.7/3
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3. The right diagram shows the same amount of flow .Network flows with lower & upper bounds on flow and the Circulation Problem
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w
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1.4/2
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2.that is 4 units .3.15/2
1.6/2
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s
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s
0. A flow of 4 units is a feasible flow but it is taking place at a higher cost in the left diagram.taking place at minimum cost.3.5/3
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1.
Here all edges except one have a lower bound equal to zero. Connectivity and Matching Problems
Before finding feasible flow in a Network graph and in a Circulation graph we should again note that in a Circulation graph the flow is conserved at every vertex while in a network flow graph it is not conserved at the source as well as the sink vertex. Right graph is a circulation graph. We transform the left circulation graph into the right network flow graph and then claim that a feasible flow exists in the circulation graph if and only if we can push a specified flow (equal to 2 in this case) from the source vertex x and pull the same amount of flow from vertex y in the network flow graph.2
a
2
0. Obviously a feasible flow does not exist in this example because of obvious reasons. It will be interesting to formally prove this claim.
.1
s
0.2
a
No Feasible Flow
0.
2.88: Left graph is a network flow graph with a source and sink vertex.5
t
0. The edge (s.382
Network Flows. a) has a lower bound equal to the upper bound on flow and they are both equal to 2.9.4
Finding a feasible flow in a Circulation graph with one special edge
Consider the graph shown in the left diagram. here a feasible flow does not exist because of obvious reasons.1
b
Figure 6. Thus it may be possible that a feasible flow does not exist in a Circulation graph (see the right diagram) while a feasible flow exists in the same graph having a source and sink vertex as shown below in the left diagram. here a feasible flow exists.9
s
0.
6.5
t b
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2.
6.9.89: Left graph is a Circulation graph with associated lower bound and upper bound on flow in each edge.9
s
0.90).2
a
0. 6. The edge (s. Here the source vertex x is trying to push a flow of 2 units in the network graph while the sink vertex y is trying to pull 2 units from the network graph.5
Finding a feasible flow in a network flow graph with one special edge
Consider the network flow graph shown in the left diagram with a designated source and a sink vertex (Fig.1
t b
0. Please note that except for one edge all edges have lower bounds equal to zero. The Circulation graph is converted into a network flow graph shown in the right diagram.9
y
2
a s
0. We claim that a feasible flow exists in the network flow graph (left diagram) if and only if a feasible flow exists in the circulation graph shown in the middle. It will be interesting to formally prove this claim.1
Figure 6.5
0. a) has a lower bound equal to upper bound equal to 2.
.91. 6. Once this relationship is established then we can use earlier results to find a feasible flow provided it exists as shown in Fig.Network flows with lower & upper bounds on flow and the Circulation Problem
383
x
2
2.5
t b
0. We transform this network flow graph into a circulation graph as shown in the middle diagram.
1
0.5
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0.5
t b
0. We find a feasible flow in the Circulation graph and then convert it into a feasible flow in the original network flow graph shown in the bottom diagram.∞
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2.∞
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0.90: Left diagram is a network flow graph with associated lower bound and upper bound on flow in each edge.384
Network Flows.1
t
Figure 6.1
Figure 6.9
s
0. First it is converted into a Circulation graph (top right diagram).2
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0. First it is converted into a Circulation graph (middle diagram) and then we find a feasible flow in the circulation graph.9
s
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a 2
0. Connectivity and Matching Problems
x
2
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x
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2.∞
0.91: Top left diagram is a network flow graph with associated lower bound and upper bound on flow in each edge.
9
a
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0. a) has a lower bound equal to 2 and an upper bound equal to 7.Network flows with lower & upper bounds on flow and the Circulation Problem
385
6.7
t
b
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x
2.but now the upper bound is higher than a non zero lower bound on flow through this edge.5
t b
0. We transform this circulation graph (left diagram) into another circulation graph shown in the middle diagram and claim that a feasible flow in the left circulation graph exists if and only if a feasible flow exists in the middle circulation graph.5
a
0.9
s
0.7 s
0. Here the source vertex x is trying to push a flow of 2 units in the network graph while the sink vertex y is trying to pull 2 units from the network graph. It will be interesting to formally prove this claim.92: Left graph is a Circulation graph with associated lower bound and upper bound on flow in each edge. The edge (s.
2 2.7
s
0. Please note that except for one edge all edges have lower bounds equal to zero.2 0.5
2 b
0.
.6
When upper bound is higher than a non zero lower bound
We again consider a circulation graph (shown in the left diagram) where every edge has a lower bound equal to zero except for one special edge .5
y a
0. 6.9.9
2.7
t
Figure 6. This edge is split into two edges as shown in the middle graph.92. Once this relationship is established then we can use earlier results to find a feasible flow provided it exists as shown in Fig. The Circulation graph is converted into a network flow graph shown in the right diagram.5
2
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a
0.5 t s
0.
386
Network Flows.5
t b
5.2
4
5 x
2
3
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1 x
Figure 6. The Circulation graph is converted into a network flow graph shown in the middle diagram. Connectivity and Matching Problems
Finding feasible flow in a general Circulation graph? Here we consider a circulation graph where each edge may have a nonzero lower bound and a different upper bound on flow through this edge.2
s
0. Here the source vertex x is trying to push a flow of number of units in the network graph while the sink vertex y is trying to pull the same amount of units from the network graph.7
s
0.5
s
4.6
2
a
3.93: Left graph is a Circulation graph with associated lower bound and upper bound on flow in each edge.1
t b
0.7
y 5
0.
y 2
2.1
t b
0.5
3 a
4
0. We claim that a feasible flow in the circulation graph exists if and only if we can push a specified amount of flow from vertex x in the middle or the right diagram.6
1 a
0.9
0. The middle network flow diagram is in turn transformed into another simplified network flow diagram shown in the right diagram. Using earlier transformations we convert the circulation graph shown in the left diagram into a network flow graph shown in the middle diagram.
.
1/1 0.7
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2
2.in this example it is equal to 1.5
t b
5+0 5. How to solve the minimum cost Circulation Problem? Consider the following circulation problem where the lower bound for each edge is 1 while the upper bound on flow through each edge is 3. The middle diagram shows an intermediate stage.7
s
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.9 3+2
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4. This is per unit flow cost for each edge .Network flows with lower & upper bounds on flow and the Circulation Problem
y 1
2.2/0
t b
s
4+1 4. In addition to lower and upper bounds we have a cost associated with each edge.7
2
1 x
Figure 6.5
t b
5.7
a
3.5/3
0.9
a
0. We need to find a feasible flow of minimum cost in this circulation. The right diagram shows a feasible flow in the circulation graph.94: Left graph is a Circulation graph with associated lower bound and upper bound on flow in each edge.
Here the source vertex x is trying to push a flow of 2 units in the network graph while the sink vertex y is trying to pull the same amount of units from the network graph.96. But we know that this feasible flow may not be the minimum cost feasible flow.388
Network Flows. We try to push 2 units of flow from vertex x.95: The left diagram is a Circulation graph with associated lower bound and upper bound on flow in each edge. 6. If we just want to find a feasible flow (not the minimum cost feasible flow) then we know what to do? We convert the circulation (left diagram) into a network flow graph as shown in the right diagram.
. The Circulation graph is converted into a network flow graph shown in the right diagram. Such a feasible flow is shown in the left diagram of Fig. Connectivity and Matching Problems
x
Find a Feasible Flow
1 1
Push 2 units of Flow
-1
+1
-1
+1
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Upper Bound is 3 while Lower bound is 1
Upper Bound is 2 while Lower bound is 0 for black edges
1
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Figure 6. The cost of per-unit flow through each edge is equal to 1. If we are successful then it means that a feasible flow exists in the circulation.
. 6. we should push 2 units of flow from vertex x at a minimum cost as shown in the right diagram of Fig.96: The left diagram shows that it is possible to push 2 units of flow from the source vertex x at some cost.96. This provides us a solution to the (minimum cost) Circulation Problem. The right diagram shows that it is possible to push the same amount of flow from vertex x at a lower cost in fact at a minimum cost.95.Network flows with lower & upper bounds on flow and the Circulation Problem
389
x
1 1
Push 2 units of Flow at any Cost
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1 1
Push 2 units of Flow at Minimum Cost
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1 Upper Bound is 2 while Lower bound is 0 for black edges 1 Upper Bound is 2 while Lower bound is 0 for black edges 1
+1
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Figure 6. 6. In order to find a feasible flow at minimum cost in the circulation graph of Fig. 6.97. This flow of 2 units at minimum cost is then translated into a feasible flow at minimum cost as shown in the right diagram of Fig.
97: The left diagram shows the source vertex x pushing two units of flow at minimum cost. Connectivity and Matching Problems
x
1 1
Push 2 units of Flow at Minimum Cost
Minimum Cost Feasible Flow
-1
+1
-1
+1
1 Upper Bound is 2 while Lower bound is 0 for black edges 1
y
Upper Bound is 3 while Lower bound is 1
Figure 6. We call this Problem 1. Here the flow in each edge is 1 except for the red bold edges where the flow is 2 units.390
Network Flows.7
How to solve the Circulation Problem for undirected graphs?
We need to find a (or size of) maximum cut in an undirected network flow graph having vertices s and t. We know that minimum flow from vertex s to vertex t is equal to max cut in a network
.
6. Our Strategy of Solving the Circulation & other related problems Our strategy of solving the Circulation and other related problems are summarized in the following four figures which are self explainatory.9. The cut should separate vertex s from vertex t. The right diagram shows the resulting solution of the minimum cost Circulation Problem. But we know that this problem is a hard problem? Assume that we need to find minimum flow from vertex s to vertex t in a network flow graph which is un-directed. We call this Problem 2.
3
s
0.6
0.3
w t
2.9
2.5 1.
. The minimum flow is 5 and thus the max cut will also have same size.but instead a negative result.4
s
0.9
2.392
Network Flows. The top figure shows that if we need to find max cut in this directed graph (Problem 1) then we should find the min flow from vertex s to vertex t (Problem 2).8
z
Find Maximum Flow at Minimum Cost
The Circulation Problem
Figure 6.4
t
2.7
x
3.8
z
y
2.101: The network flow problem is first converted into a circulation problem.∞
y
2.3 1.that will give us a minimum flow from s to t in the network flow graph as shown below. Thus we moved in the following fashion to find the max cut in a network flow graph? In case of un-directed graphs we have the same sort of strategy .5 1.7
x
3.3
5. In order to find minimum flow we need to make it a circulation as shown above and then find a minimum cost circulation (Problem 3) in this graph . Connectivity and Matching Problems
flow graph? Suppose we can solve the circulation problem (Problem 3) at minimum cost in an un-directed graph? Then we can use the solution of Problem 3 in order to solve Problem 2? We can then use this solution to find a solution to Problem 1?
w
1.6
5.
If you are successful in finding a path then keep a record of this path otherwise exit the algorithm. and vertices s & t. Thus we start with a panoramic picture. the in-degree is equal
. Eulerian graphs belong to one of these categories which we discuss in detail. Remove all edges in the path P and go to step 1. for the rest of the vertices of D. We end this chapter with a detailed study of the Chinese Postman problem for both directed as well as un-directed graphs. Exploiting our prior knowledge about this category we define a number of new categories of graphs which are to some extent similar and at the same time different from it.396
Eulerian Graphs & the Chinese Postman Problem
Introduction
We shall start with a special category of graphs which was earlier discovered in the last chapter. Algorithm 55: Find Maximum edge-disjoint paths from s to t in D input : Directed graph D.
Consider the directed graph D shown in Fig. for a directed graph we shall assume that the underlying un-directed graph is connected. We come back to our categories of graphs and look at these in the light of our newly acquired experience about Eulerian graphs. and then come back to the panoramic picture with new tools and techniques. We have claimed earlier that there is a class of graphs where even an unintelligent algorithm (like Algorithm 36) can efficiently find the maximum edge-disjoint paths.
7.1
A Special Class of Graphs
We have studied the problem of finding maximum number of edge-disjoint paths from a source vertex to a sink vertex in a graph in the last chapter. We make a number of inter-related claims about such graphs and then show how the proof of one claim can lead to the proof of another. We shall study this and similar classes of graphs in detail in this chapter. We shall consider connected un-directed graphs. output: (Maximum) edge-disjoint paths from s to t in D
1
2
Find a directed path P from vertex s to vertex t in D. 7. Please recall Algorithm 36 which is again reproduced below. move in depth with one category.1) In this graph vertex a is a source vertex having only out-degree while the vertex d is a sink vertex having only in-degree.
3.2. There is. the in-degree of every vertex is equal to the corresponding out-degree. Every edge of the graph is covered by one of the edge-disjoint trails (or paths) from the source vertex to the sink vertex. We also assume that vertex a has an out-degree larger than the in-degree. All directed graphs fulfilling the above properties are known as Class A graphs. Please note that each edge of this graph is covered either by the trail or by the path as shown in this diagram. however. etc) in order to execute the first step of this algorithm. How is the class B different from class A and in what respect they are similar? Try to answer this question before moving forward. The maximum (number of) edge-disjoint paths in the directed graph D are shown in the left diagram of Fig. The maximum number of edge-disjoint paths from the source vertex to the sink vertex can be found by Algorithm 55. For the rest of the vertices in D the in-degree is equal to the respective out-degree. Depth First Search. while the vertex d has an in-degree larger than the out-degree. We can use Algorithm 55 to find maximum number of edge-disjoint paths from vertex a to d in this graph (there is no need to use the more sophisticated Algorithm 37).A Special Class of Graphs
397
to the corresponding out-degree. Let us now define a Class B category of graphs: In this category of directed graphs we again have two special nodes a and d. Again we shall discuss the proofs later. The right diagram shows the same graph with one trail and one path from vertex a to vertex d in the given graph. Class
. an important difference this time: Vertex a has an out-degree but also an indegree while the vertex d has an in-degree but also an out-degree. In fact all class A directed graphs possesses the following properties: 1. In this special class of graphs the in-degree of vertex d is always equal to the out-degree of vertex a (why?). 7. Class C deals with directed graphs in which there are no special vertices. Please note that we can use any traversal algorithm (Breadth First Search. We shall prove these properties later in this chapter. 2. Thus the vertex a resembles a source vertex while vertex d resembles a sink vertex. The maximum number of edge-disjoint paths from the source vertex to the sink vertex is equal to the out-degree of the source vertex or the in-degree of the sink vertex.
We can find maximum edge-disjoint paths from vertex a to d in this class of directed graphs using Algorithm 55.1: A directed graph D with two special nodes a and d.398
Eulerian Graphs & the Chinese Postman Problem
z
e
f
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a
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Figure 7.
. The indegree is equal to the out-degree for every node in this graph except a and d.
There are two edge-disjoint trails in this graph such that each edge of the graph is covered exactly once by either of the two trails as shown in the right diagram.
Concept Map 7. A concept map showing various classes of some special graphs.1. For every other node in this graph the in-degree is equal to the corresponding out-degree.
.A Special Class of Graphs
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Figure 7. There are two edgedisjoint paths from vertex a to vertex d in the graph D as shown in the left diagram.2: We show a directed graph of Class A: Vertex a has only outdegree while vertex d has only in-degree.
One such property is shown in Fig. We shall now make a number of simple claims and it is quite possible for you to prove them yourself:
. We can also find trails in this graph such that each edge of this graph is covered exactly once by one of the trails? We show a Class C directed graph in the left diagram of Fig.400
Eulerian Graphs & the Chinese Postman Problem
C may also include un-directed graphs where the degree of every vertex is even. How other properties of a graph change (or do not change) after such a transformation is interesting to explore. Class E deals with un-directed graphs where the degree of every vertex is even except for two special vertices a and d where the degree is an odd number. the in-degree is equal to the corresponding out-degree.3. 7. If however we remove (all edges in) a path from any vertex x to a vertex y in a Class C graph then the new graph will not be a Class C graph any more. it will become a Class A or Class B graph. 7.4. Both the graphs shown in this diagram can be partitioned into edge disjoint cycles (or circuits) shown in different colors. the rest of the p − 2k vertices have an even degree.4. Remember the number of vertices having odd degree in a graph can not be odd (why?). There is no special vertex in this graph. the two paths are shown in Fig. the out-degree of vertex x is 1 while its in-degree is equal to 3. the in-degree of every vertex is equal to the corresponding out-degree. Class F deals with a more general category of un-directed graphs in which every graph G (having p vertices) have 2k nodes with an odd degree. The out-degree of vertex f is 3 while the in-degree is equal to 1. 7. if we now remove edges of this cycle from the original graph then the new graph will also belong to the same class (why?). We can use any traversal algorithm to find a cycle in such graphs. We can use Algorithm 55 to find the two edge-disjoint paths from vertex f to vertex x in D. As you should discover yourself Class C directed and un-directed graphs have some special and exciting properties.3. It is interesting to note that a directed graph (where the in-degree of each node is equal to the corresponding outdegree) belongs to the same class as an un-directed graph where the degree of each vertex is even (why?). for the rest of the nodes. An un-directed graph belonging to the same category is shown in the left diagram of this figure. We show a Class B directed graph in Fig. The degree of each node in this un-directed graph is even. Similarly if we add a path between vertex a and d (these are the only two vertices having an odd degree) in a Class E graph then it will be transformed into a Class C category. 7.
For every other node in this graph the in-degree is equal to the corresponding out-degree.
.3: A directed graph of Class B: The out-degree of node f is larger than the in-degree while the in-degree of x is larger than its out-degree.A Special Class of Graphs
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Eulerian Circuits and Graphs
A graph G is Eulerian provided it contains an Eulerian circuit. What are necessary and sufficient conditions for a (connected) graph to be cyclic? 2. What about if each vertex in a graph have a degree at least equal to two? Remember we are considering connected graphs only as mentioned earlier in this section. The edge set of these graphs can be partitioned into edge-disjoint cycles (shown by different colors) which if combined together will create a circuit consisting of all edges of the graph. Can we make the above claim if the degree of every vertex is even? 5. Can we claim that in such a graph every vertex will lie on some cycle? 4. a circuit which contains every edge of G.402
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7. please remember that in a circuit it is possible to visit a vertex several times but we are allowed to traverse an edge only once. What can you say about other properties of this special graph? 3. Let us consider a special case of Class C graphs where the degree of each vertex is not only even it is exactly two. 1. An un-directed graph where the degree of each vertex is even is shown in the right diagram.
. Both these graphs belong to our Class C category.4: Every node in the directed graph has an in-degree equal to the corresponding out-degree as shown in the left diagram.
If this circuit is a cycle then the proof is complete otherwise it will consist of several cycles. then we should be able to prove that the degree of each vertex of G is even (you should be able to do it easily). A graph G is Eulerian if and only if the edge set of G can be decomposed into edge-disjoint cycles. If the graph is Eulerian then there will be an Eulerian Circuit inside that graph as shown in left diagram of Fig. so the graph will be Eulerian. Thus the cycles forming the Eulerian circuit will be edge-disjoint as shown in the left diagram of Fig. Let us do it now. The degree of each vertex in a graph G is even if and only if the edge set of G can be decomposed into edge-disjoint cycles. Assume that a graph G can be decomposed into edge-disjoint cycles. 2. We shall start with proving Claim Number 4 and then work backwards in order to prove earlier claims. A graph G is Eulerian if and only if the degree of each vertex is even. Remember in a circuit a vertex may be repeated but an edge can not be repeated. 4.5. This part of the proof is done and let us attempt the other part: if a graph is Eulerian then (we shall be able to prove that) it can be partitioned into edge-disjoint cycles. If the edge set of a graph can be partitioned into edge-disjoint cycles then we can always combine these cycles to create a circuit which will cover every edge of the graph exactly once. A graph G is Eulerian if and only if every edge of G lies on an odd number of cycles. As the degree of each vertex in G is even thus we can find a cycle C in G using any traversal algorithm. Let us now tackle Claim Number 3. If we remove all edges belonging to C from G then in the resulting graph the degree of each vertex will again be even (why?) but this new graph will have fewer edges as compared to the original graph G. 3. So we assume that the degree of each vertex is even and now we should be able to prove that the graph G can be decomposed into edge-disjoint cycles. 7. 7.Eulerian Circuits and Graphs
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Now we make a number of claims (it will become evident from these claims that an Eulerian graph belongs to our Class C category): 1.5. It is possible to build logic on this observation in order to design a formal proof.
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5: An Eulerian graph containing an Eulerian circuit is shown in the left diagram. 7. Now assume that the graph G is Eulerian. Please remember that in a trail we may repeat vertices but we can not repeat edges while in a path neither vertices nor edges can be repeated. We start from a (in Fig.6. The circuit can be decomposed into edge-disjoint cycles as shown in the right diagram.404
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Figure 7. This may not be very obvious so we shall prove this after first finding the number of distinct trails between the two vertices. It means that the degree of node a as well as that of b will be even. thus G will be an Eulerian graph.
. in fact there will always an odd number of ways out (why?).6) and arrive at the adjacent vertex e. as the degree of every vertex in G is even so if you can enter a vertex then you can leave it also. Proving that edge ab will be part of an odd number of cycles is equivalent to proving that there are an odd number of paths between vertex a and vertex b (why?). First assume that every edge ab in graph G is part of an odd number of cycles. In Claim Number 3 we have proved that a graph G is Eulerian if and only if the edge set of G can be decomposed into edge-disjoint cycles. we have to prove that every edge ab in G will be part of an odd number of cycles as shown in Fig. 7. We have earlier proved in Claim Number 4 that a graph can be partitioned into edge-disjoint cycles if and only if the degree of each vertex in the graph is even. Combining the two we can prove that a Graph is Eulerian if and only if the degree of each node in the graph is even. Let us now concentrate on Claim Number 1. As the graph G is Eulerian thus the degree of each node will be even. thus the degree of node a as well as that of node b will be even. Proving that there is an odd number of paths between vertex a and vertex b is in fact equivalent to proving that there is an odd number of trails from vertex a to vertex b.
Consider vertex a which is adjacent to vertex b in this graph. Please note that if there is a counter clockwise cycle from vertex f to vertex y to vertex z and back to f then there is clockwise cycle from vertex f to z to y and back to vertex f .6: An Eulerian graph is shown.Eulerian Circuits and Graphs
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. We intend to prove that the edge ab is part of an odd number of cycles in this graph.
There are a number of algorithmic issues apart from the above (theoretical) claims and their respective proofs: 1.Eulerian Circuits and Graphs
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This means that if you draw a tree of all possible trails then the out-degree of every node in this tree will be an odd number as shown in Fig.6 will be equal to the number of leaf vertices in the tree of Fig. 7. How can you generalize the four claims that we have made for multigraphs. How can you efficiently find edge-disjoint cycles in an Eulerian graph? There are a number exciting theoretical problems which you should attempt before moving forward: 1. in multi-graphs we allow parallel edges and self loops? (Hint: Can you convert a multi-graph into a simple graph?) 2.7 which will be an odd number (why?).7. 7. 7. Every cycle in the graph contributes to two trails as shown in Fig. The total number of trails in the graph of Fig.7. How can you generalize (or modify) the four claims in case of directed graphs?
. How can you efficiently find an Eulerian circuit in an Eulerian graph? 3. 7. Thus the total number of paths between vertex a and vertex b will be an odd number (why?). 7. How can you efficiently check if a graph G is Eulerian using different (necessary & sufficient) conditions for a graph to be Eulerian? 2.6 and Fig.
6. Problem 7. Problem 7. Please read the following algorithm which is primarily designed to find edge-disjoint cycles in a graph G where the degree of every vertex is even. Problem 7. Algorithm 56: Find Cycles in an un-directed graph G where degree of every vertex is even input : Un-directed graph G where degree of every vertex is even output: Edge-disjoint cycles in graph G
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Find a Cycle C in graph G. Some one claims that the algorithm outputs maximal number of edge-disjoint cycles in a graph G. How the solution of earlier problems will change or not change?
. Some one claims that the algorithm outputs maximum number of edge-disjoint cycles in a graph G.1.4. Problem 7. How the algorithm would behave in this type of graph? What will be the output of the algorithm? How the output will be different in this case? Discuss briefly. Apply the above algorithm on the graph G as shown in Fig. Problem 7.1. 7. and specify each cycle outputted by the algorithm. Assume that we apply the above algorithm to graph H which is different from graph G (in what respect?).2. Some one claims that the algorithm outputs edge disjoint cycles? Discuss why or why not.1.8. It will be useful to design the following algorithms for an un-directed graph G.3. Prove or give counter example. Show or give counter example.1. Problem 7.1.7. How about if we have a directed graph in which the indegree of every vertex is equal to its out-degree. and keep a record of it.1.1.408
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Problem Set 7.5. Remove all edges in the Cycle C from graph G. If some edges are still left in G then go to step 1 otherwise exit.1.1.
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We can use Algorithm 56 to find out efficiently one such set of edge-disjoint cycles. Several sets of edge-disjoint cycles are shown here.9: A graph (with every vertex having even degree) can be split up into edge-disjoint cycles.
. Note that every edge is part of an edge-disjoint cycle.410
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10: We have seen that if the degree of every vertex is even then the graph can be split into edge-disjoint cycles again shown in the top diagram. There will be an Eulerian circuit in the reconstructed graph.
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Figure 7. On the other hand if a graph can be split into edge-disjoint cycles or if the edge-disjoint cycles of a graph are given then we can reconstruct the original graph as shown in the middle diagram. The original graph will have all nodes with even degree? The reconstruction algorithm will help you reconcile with this claim.
These graphs are reproduced in Fig. Such a graph is shown in the left diagram of Fig.11.11. 7.3. How can we prove this and how can we find an Eulerian trail? Perhaps we can find a constructive proof which will solve both the problems.2 and 7. 7. 7. Its counter part in directed graphs is shown in the right diagram of Fig.412
Eulerian Graphs & the Chinese Postman Problem
Algorithm 57: Find graph G given a set of edge-disjoint cycles of G input : Edge-disjoint cycles in an unknown graph G output: Un-directed graph G (degree of every vertex in G will be even)
Algorithm 58: Find an Eulerian circuit in graph G given a set of edge-disjoint cycles of G input : Edge-disjoint cycles in an unknown graph G output: An Eulerian circuit in G
7. a trail in which every edge of the graph is covered (exactly once). for the rest of the vertices the in-degree is equal to the corresponding out-degree. 7.12 for comparison Algorithm 59: Find a set of edge-disjoint cycles of G given an Eulerian circuit in a graph G input : An Eulerian circuit in G output: Edge-disjoint cycles in an unknown graph G
. here the outdegree of one special vertex is larger than its in-degree by one.3
Eulerian Trails and Related Problems
Let us now consider Class E un-directed graphs where the degree of every vertex is even except for two special vertices f and x where the degree is an odd number (will the degree of the two odd vertices be the same? Why?). Again we should appreciate that we can use Algorithm 55 to find maximum number of edge-disjoint paths from a source to a sink vertex? We have earlier presented a Class B and a Class A category directed graphs in Fig. while it is the other way round for the other special vertex. In both these graphs it is possible to find an Eulerian trail from vertex f to vertex x.
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Figure 7.13. the in-degree is equal to the out-degree for every node in this graph except f and x. the rest of the p − 2k vertices have an even degree. Such a graph is shown in the left diagram of Fig.13. This observation should lead you to design a formal proof for the above claim. this is a Class E un-directed graph. 7. We have made certain claims about Class A graphs earlier in this chapter. 7. A directed graph D with two special nodes f and x is shown in the right diagram. the out-degree of vertex f is one larger than its in-degree and it is the other way round for vertex x in this directed graph. This class comprises of un-directed graphs having 2k nodes with an odd degree.Eulerian Trails and Related Problems
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with graphs in Fig.11.11: An un-directed graph shown in the left diagram. It will then become possible to find an Eulerian circuit in the resulting graph which is also shown in the right diagram of Fig. The right diagram shows the same graph where we add an extra edge between two odd vertices converting this graph into a Class C category where the degree of each vertex is even. 7. with odd vertices shown in bold. The claim for such a graph G is that the edge set of G can be partitioned into k trails where each trail is connecting two odd vertices.
. These claims can be generalized with some interesting modifications for Class B directed graphs and then proved using our newly acquired experience of Eulerian Graphs? We now present one last claim which is applicable to Class F un-directed graphs. the degree of vertex f and x is odd while the degree of every other vertex is even.
This problem is formally defined as the Chinese Postman Problem: We need to find a shortest closed walk in a graph G which passes through every edge of G at least once. it is the other way round for vertex x. the out degree of vertex f is three while its in-degree is 1. vertex a has only out-degree while vertex d has only in-degree. he would certainly like to traverse each lane at least once while making sure that the traversals of the same lane should be minimized. in a weighted graph.4
Eulerian Walk and the Chinese Postman Problem
We know that if the degree of every vertex is even in an un-directed graph G then we can find an Eulerian Circuit in G by traversing each edge exactly once. the degree of every vertex is not even then the problem is to traverse each edge at least once (not exactly once) and making sure that the number of edges. In the figure below. we need to minimize the total sum of edge weights in a closed walk which covers every edge of the graph at least once.
7. Such a walk (walk because some edges will be traversed more than once) is also known as an Eulerian walk in the graph G. A directed graph of Class A category is shown in the right diagram. This problem is faced by any post man delivering letters in houses along lanes or a sweeper who is sweeping roads. There are
.12: A directed graph of Class B category is shown in the left diagram. however. In the left diagram of this figure we have an un-weighted graph while the right diagram shows a weighted graph. If. the in-degree is equal to the out-degree for every node in this graph except a and d. As you can well imagine. traversed more than once.414
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Figure 7. b.13: An un-directed graph of Class F shown in the left diagram. y and x is odd while the degree of every other vertex is even. the degree of vertex f .
Each resulting Eulerian graph H will have different number of edges. 1.16. We show a simple technique of converting a graph with odd vertices into a graph with all vertices even in Fig.
. We show different Eulerian graphs corresponding to a weighted graph G in Fig. 7. consists of more edges than the graph shown in the middle diagram. Please note that the degree of the two terminal odd vertices in the path will become even while the degree of an even vertex in the middle of the path will stay even. It is interesting to note that optimal graph. All these possible paths gives rise to different number of duplicated edges required to convert a graph into an Eulerian graph. 7. In other words the Eulerian graph H should be of minimum size in terms of number of edges or in terms of sum of edge weights of H. Indeed there could be paths of different edge lengths between the two odd vertices in a graph as shown in Fig. We find a path from one odd vertex to another odd vertex and duplicate every edge encountered in that path. 2. 7. shown in the right diagram. We need to convert a given graph G into an Eulerian graph H by duplicating some existing edges of G. The resulting graph will be Eulerian as shown in the right diagram of this figure. We need to make sure that the Eulerian trail that we have created in the first part is shortest in terms of number of edges involved or in terms of edge weights.15.17.416
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basically two problems that we intend to solve simultaneously but first we need an understanding of these problems in isolation. we need to select the one with the minimum number. The Chinese Postman Problem for an un-weighted graph can thus be rephrased: We need to convert a graph G (having some odd vertices) into an Eulerian graph H (having all vertices with even degree) by duplicating minimum number of existing edges of graph G. in graph G we shall be traversing an edge twice when in graph H we shall be traversing a duplicated edge. Now an Eulerian circuit in H will correspond to a closed Eulerian Walk in G. For a weighted graph we need to convert G into an Eulerian graph by duplicating certain edges such that the sum total of edge weighted corresponding to duplicated edges is minimized.
The right diagram shows a weighted graph. Both graphs have two vertices with an odd degree.14: An un-directed graph and un-weighted graph (left diagram).Eulerian Walk and the Chinese Postman Problem
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Figure 7.15: Shows how can we convert a graph G having two odd vertices into a graph H where the degree of each vertex is even. The duplicated edges are shown in red color in the graph H. An Eulerian Circuit in H is equal to an Eulerian Walk in G.
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Optimal Solution Only one edge (red) is duplicated to create an Eulerian Circuit
Figure 7.16: It is possible to convert a graph having two odd vertices into different Eulerian graphs with varying number of edges. The Eulerian circuit in each graph is also indicated.
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Optimal Solution Two edges are duplicated Cost is 2+4 = 6
Figure 7.17: It is possible to convert a graph having two odd vertices into different Eulerian graphs with varying number of edges. The Eulerian circuit in each graph is also indicated. It is quite obvious now that a shortest path between the two odd vertices provides us an optimal solution to the Chinese Postman Problem provided a graph has only two odd vertices. What about if a graph G consists of more than 2 odd vertices as shown in Fig. 7.18? How about if we do the same trick of finding a shortest path between vertices belonging to different pairs of odd vertices? (Remember the total number of odd vertices in any graph will always be even). Let us start with an arbitrary selection of vertices in the three pairs as shown in Fig. 7.19. The resulting Eulerian graph is shown in the right diagram of the same figure. The total number of edges duplicated is also indicated in this diagram. Finding
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Figure 7.18: A graph G consisting of 6 odd vertices.
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shortest paths for a fixed set of pairs of odd vertices is certainly helps to reduce the cost of making the degree of each vertex even - but a different set of pair of odd vertices may help us in further reducing this cost as shown in Fig. 7.20. Thus the problem is reduced to finding the set of pairs of odd vertices which minimizes the cost of duplicating the edges. Do we have to enumerate all possible sets of pairs of odd vertices in the graph G? How to do that systematically and estimate what is the total number of possibilities? Can we use a brute force approach or have to become cleverer in order to efficiently solve this problem.
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Figure 7.20: We find shortest paths between a different set of odd vertices in graph G. The resulting Eulerian graph H is shown in the right diagram. Given 2k items (corresponding to 2k odd vertices) numbered from 1 to 2k; let us draw a completely connected graph consisting of 2k = 6 vertices (of G) as shown in Fig. 7.21. A perfect matching in this graph provides the desired set of all pairs of odd vertices. It is possible to put weights on edges of the completely connected graph as shown in Fig. 7.22. A minimum weight perfect matching in the completely connected weighted graph will provide an efficient solution to our problem as shown in the following figures.
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Figure 7.21: A perfect matching in a completely connected graph K of odd vertices of graph G. An efficient solution of the Chinese Postman Problem in an un-directed graph G is thus reduced to the following steps: 1. Identify odd vertices in graph G. 2. Create a completely connected graph K; every odd vertex of G is a vertex of this completely connected graph K. The weight of an edge between two vertices in graph K is equal to the weight of the shortest path between the corresponding odd vertices in graph G. 3. A minimum cost perfect matching in the completely connected graph K provides us the desired pairs of vertices (x, y) in graph K. For every such pair (x, y), we duplicate edges along the shortest path between vertex x and vertex y in G.
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Figure 7.23: We enumerate all distinct perfect matching in a completely connected graph consisting of 6 odd vertices of G. Please note that looking at all enumerations will be very costly; we have to find a better solution?
Figure 7.24: The minimum weight perfect matching in a completely connected graph K among all odd vertices in G corresponds to minimum number of edges of G which if duplicated will convert graph G into an Eulerian graph H. Without enumerating all perfect matchings in a graph we can still find efficiently the minimum cost perfect matching in a graph.
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The Chinese Postman Problem for Directed Graphs
We discuss here the Chinese Postman Problem for a directed graph. If a directed graph G have all vertices where the in-degree is equal to the corresponding out-degree then the graph G is Eulerian. If, however, some vertices have out-degree larger or smaller than the in-degree then we need to convert graph G into an Eulerian graph H by duplicating minimum number of existing edges of G. We show an un-directed graph G in the top left corner of Fig. 7.25. By putting directions on edges of G, this graph is converted into a directed graph as shown in the top right diagram of the same figure. The un-directed graph G has even as well as odd vertices - that is why graph G is not Eulerian. In contrast, the directed graph D has three types of vertices as described below. 1. The out-degree is equal to the corresponding in-degree. The difference ∆ (equal to out-degree minus in-degree) for any vertex is zero for such vertices. Two such vertices exist in the directed graph D and are shown in green color. 2. The out-degree is larger than the corresponding in-degree; hence ∆ is positive. There are three such vertices in graph D, all shown in red. 3. The out-degree is less than the corresponding in-degree; hence ∆ is negative. There are four such vertices in graph D, all shown in orange. If all vertices in a directed graph D have ∆ equal to zero then graph D is Eulerian. If, however there are vertices with +ive and/or -ive ∆ then we face the challenge of solving the Chinese Postman Problem. A solution to this problem for an un-directed graph G is not helpful even if the directed graph is derived from the same un-directed graph as shown in the bottom left and right diagrams of Fig. 7.25. We show a directed graph G where there is only one vertex with positive ∆ as shown in Fig. 7.26. Here the problem is simple: Finding a path from each of the three orange vertices to the only red vertex and duplicating edges in these paths will convert the graph G into an Eulerian graph. Finding shortest paths from each orange vertex (with ∆ negative) to the only red vertex (with ∆ positive) will provide us the optimal solution to
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We duplicate certain edges as shown in red color. The resulting multi-graph H has now become Eulerian.
Figure 7.25: We again show an un-directed graph G in the top left diagram. It is converted into a directed graph D as shown in the top right diagram. It is obvious that graph G is not Eulerian. After duplicating certain edges in graph G it is transformed into an Eulerian graph as shown in the bottom left diagram. An Eulerian walk in G is shown in the bottom right diagram. The Eulerian walk in G is not an Eulerian walk in the directed graph D.
. All other vertices have either out-degree equal to in-degree (shown in green) or less than in-degree (shown in orange).26: We show a directed graph G where there is only one vertex (shown in red) where the out-degree is larger than the in-degree.The Chinese Postman Problem for Directed Graphs
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this will make the new directed graph Eulerian. 7. 7. The maximum flow in this graph (as shown in the bottom diagram) will provide the required information. Please note that this will be a complete bipartite graph. An efficient solution of the Chinese Postman Problem in a directed graph G is thus reduced to the following steps:
. We create a bipartite graph B consisting of an A partite (consisting of all orange vertices) and a B partite (consisting of red vertices) as shown in the middle left diagram of Fig. duplicating edges on shortest paths ensures that the number of extra edges (added) is minimized. vertex 9 has ∆ equal to +2.28. Finding a maximum flow efficiently (in polynomial time) in this (special) network graph N is by itself an interesting problem.27). Further adding vertices s and t and edge capacities (middle right diagram) the problem is converted into a flow problem. This has been illustrated in Fig. 7. Similarly there are three vertices with ∆ positive. In order to make delta of every vertex zero (thus converting the graph G into an Eulerian graph) we should find out which path originating from an orange vertex should terminate at which red vertex. the problem is to minimize the cost also? The solution is simple: we create a weighted bipartite graph B where the weight of an edge from an orange vertex x to a red vertex y in B signifies the weight of the shortest path between the orange vertex x to red vertex y in graph G. all shown in red color. Vertex 4 has ∆ equal to −3 while vertex 10 has ∆ equal to −1. Similarly two paths should be terminating at vertex 9 to convert its delta from +2 to zero (please see Fig. Finding a maximum flow with minimum cost in a network graph N will ensure that the graph becomes Eulerian and the number of edges (which are) duplicated are minimized at the same time.430
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the Chinese Postman problem as shown in the same figure. Please note that all the three shortest paths originating from vertices with ∆ equal to minus 1 are terminating at a single vertex. there will be an edge from every orange vertex to every red vertex in B signifying that there will be a path from every orange vertex to every red vertex in graph G. Please note that there should be three paths coming out of vertex 4 in order to increase its delta from −3 to zero. There are two vertices (4 & 10) with ∆ negative shown in orange color. The graph G will become Eulerian but the cost in terms of number of edges of G that are duplicated may be high. We show a directed graph in the top diagram of Fig.27.27. that single vertex has ∆ exactly equal to 3. 7.
The weight of an edge from an orange vertex x to a red vertex y in B signifies the weight of the shortest path between the orange vertex x to red vertex y in graph G. Problem Set 7. Problem 7.29.432
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1. it means that there should be a path between every pair of vertices in graph D.2. while y belongs to partite B of red vertices). Problem 7.2. y) pair (x belongs to partite A consisting of orange vertices.2. The cost of any other edge in N will be the corresponding weight in the bipartite graph. A minimum cost maximum flow in the network graph N will provide us the desired pairs of orange/red vertices in graph G. We show a directed graph in the top left diagram of Fig. Show that a necessary & sufficient condition for a directed graph D to have a solution to the Chinese Postman Problem is that graph D should be strongly connected.1. 2.2. The costs of these edges will be zero. Identify orange and red vertices in G. Problem 7. The examples of directed graphs that we have considered in this section have one thing in common: If we sum delta of all vertices in a graph it comes out to be zero.3. Corresponding to every such (x. duplicate edges in the shortest path from vertex x to vertex y in graph G. 3. Convert bipartite graph B into a network flow graph after adding vertices s & t. 7. Is this a coincidence or every directed graph will possess this property? Discuss briefly. the capacity of these edges will be infinite. The capacity of an edge from a red vertex y to t will be equal to ∆(y). 4. It will be interesting to derive the time complexity of the above algorithm and compare it with that of the algorithm used to solve the Chinese Postman Problem for un-directed graphs.2. Create a weighted bipartite graph B consisting of orange vertices as an A partite and red vertices as the B partite. The capacity of an edge from s to an orange vertex x is equal to ∆(x). Here there are four vertices with ∆ negative (shown in orange color) while
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Green vertices have out-degree equal to the in-degree.29: We show a directed graph G where there are three vertices. We describe how we can convert a directed graph G into an Eulerian graph H by duplicating minimum number of edges of G.434
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We have duplicated edges in each path corresponding to the minimum cost perfect matching: The ∆ for each vertex becomes 0: The new graph H has now become Eulerian
Figure 7.
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We demonstrate an alternate scheme to solve the Chinese Postman problem.2. Describe an algorithm behind this demonstration and calculate the time complexity of this algorithm in comparison with that of the algorithm discussed in the text (based on minimum cost maximum flow algorithm). Derive the time complexity of the algorithm (that we have described in the text) which can be used to solve the Chinese Postman prob-
.4.2. Problem 7.30. We need to convert these graphs into Eulerian graphs by duplicating minimum number of existing edges of these graphs.30: We show two directed graphs which are not Eulerian.5. two vertices where in-degree is larger than out-degree
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there are three red vertices with ∆ positive. Solve the Chinese Postman problem for both these graphs using an efficient algorithm.
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There are two vertices where out-degree is larger than in-degree. We show two directed graphs in Fig. Problem 7.
it is neither strongly connected yet it may be possible for the Chinese Postman to distribute mail in (only red) streets while traversing minimum number of extra graph edges. if that is not possible then we should allow him (the Chinese Postman) to traverse the red edges at least once and come back after traversing a minimum number of edges in this graph. 7. Please describe an efficient algorithm to solve this problem.7.436
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lem for directed graphs.2. Compare its value with the time complexity of the algorithm used for un-directed graphs. This graph is not Eulerian. The Chinese Postman is supposed to deliver mail in red streets only.31. Problem 7. The Chinese Postman is supposed to distribute mail on streets shown in red color only.31: We show two directed graphs.8.32. We need to find an Eulerian circuit consisting of edges corresponding to red streets alone. 7. We show another directed graph in the right diagram of Fig. show the detailed working of this algorithm on this graph. Problem 7. thus the graph is Eulerian. The Chinese Postman is supposed to start delivering his mail from a specific
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Figure 7. Problem 7.31. We show a directed graph in the left diagram of Fig. the graph shown in the left diagram is Eulerian while the one shown in the right diagram is neither Eulerian nor it is strongly connected. We show a directed graph G in the left diagram of Fig. please note that the in-degree of every vertex in this diagram is exactly equal to its corresponding out-degree.
32 provide an optimal solution to our problem for the graph shown in the left diagram? Discuss briefly?
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point and end his delivery job at another point as shown in the figure.32: We show a directed graph G in the left diagram. The right diagram of this figure shows a transformation of graph G into another graph where certain edges of G are duplicated. 7.
. the red vertices have in-degree smaller than the corresponding out-degree while the orange vertices have in-degree larger than the out-degree. If this is not possible then he should traverse each edge at least once and also minimize the number of edges which are traversed more than once. The right diagram shows a transformation of graph G into another graph where it is possible to find an Eulerian Path from vertex 8 to vertex 3. Show that an Euler trail is possible from vertex 8 to vertex 3 in this graph. Does the right diagram of Fig.
That means this (q ≥ p − 1) will be a necessary condition for a connected graph. Such a graph is known a tree where each vertex is a cut vertex and each edge is a bridge edge. But then there are graphs where the number of edges is more than this critical number but still these graphs are not connected.2
Necessary Conditions for a Hamiltonian Graph
A graph can not be Hamiltonian unless it is connected. It is thus a necessary condition. there should be at least p − 1 edges otherwise the graph will be guaranteed to be disconnected. We shall also be considering certain conditions (necessary and then sufficient) for a graph to be Hamiltonian. This means that no vertex should be a cut vertex and no edge is a
. One of the problems (of learning) is confusion about necessary and sufficient conditions. It will be interesting to find sufficient conditions for a graph to be connected.1
Introduction
We shall address the problem of finding a Hamiltonian cycle and a Hamiltonian Path in a graph.2
8. Thus a Hamiltonian graph should be more than a tree graph. It should be connected and there should be a cycle spanning all vertices.
8. A tree graph is connected but it can not be Hamiltonian as it has no cycles (it can. For a graph (with p vertices) to be connected. there are necessary and sufficient conditions. Having edges less than this limit will be a sufficient condition for a graph to be disconnected. but we shall discuss this issue later in this topic.
8. contain a Hamiltonian Path).1
Prior Knowledge
Necessary Conditions for a Connected Graph
Let us start with a familiar and simple example of a connected graph. In case of Hamiltonian Graphs we are unaware of such a characterization.2. In case of Eulerian Graphs where we have a neat classification.2. although a loose necessary condition. We shall be using constructive proof techniques for certain sufficient conditions. however. we shall be talking about necessary and sufficient conditions separately. the same techniques can be used to find a Hamiltonian Cycle in a graph where a Hamiltonian Cycle is guaranteed (by the sufficient conditions) to exist.442
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1). We can certainly find sharper necessary conditions as are explained in most of the text books.2. A tree graph is just connected and belongs to one side of the extreme of connected graphs.
8. It will be useful if a learner solves this problem by himself or herself. Any permutation of the vertices of this graph will give you a Hamiltonian Cycle (see Fig. let us solve a much familiar and also simpler problem of finding a sufficient condition for a graph to be connected. The other side of the spectrum is a completely connected graph where every vertex is connected to every other vertex. This loose sufficient condition will be used in later pages to derive sharper sufficient conditions. 8.3
A Loose Sufficient Condition for a Hamiltonian Graph
Let us think of a familiar graph where a Hamiltonian Cycle is guaranteed to exist and we can find this cycle using a trivial algorithm. On the other hand if a graph contains a bridge edge (or a cut vertex) then it is certainly not Hamiltonian.Prior Knowledge
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bridge edge in a Hamiltonian graph.2.
8. It is interesting to note that complete connectedness is not a sufficient condition for a graph to be Eulerian.4
Sufficient Condition for a Connected Graph
Before finding a tighter sufficient condition for a Hamiltonian graph. and Hamiltonian (see Concept Map
.5
A Concept Map
We show a concept map indicating some necessary and sufficient conditions for a graph to be connected. Thus the completely connected property is a sufficient condition for a graph to be Hamiltonian. 2 It can also be shown that if the maximum degree of any vertex in a graph is equal to (p−2) then you can always draw a disconnected graph (How?) which 2 implies that the condition (minimum degree ≥ (p−1) ) is certainly a sharp 2 condition for connectedness. The minimum degree of any vertex in a graph should be at least (p−1) for a graph to be connected. Again these will be necessary conditions as there are graphs which satisfy these conditions but are not Hamiltonian.
8. although it is perhaps an overkill and thus not a sharp sufficient condition. Eulerian.2. Obviously a completely connected condition would be just too loose for a graph to be just connected.
Summary It will be interesting to summarize of what is possible (or not possible) at what cost. A simple Breadth First Search (or any traversal algorithm) can solve this problem efficiently. each cycle corresponds to a different permutation of vertices of the graph. Subsequently we can design efficient algorithms to actually find an Euler Cycle in a graph provided the graph satisfies the given conditions.1: A completely connected graph of 6 vertices. Two Hamiltonian Cycles are shown in the completely connected graph. If the number of edges is less than p − 1 then the graph will certainly be disconnected. We shall 2
. Interestingly a minimum degree equal to (p−1) is a sufficient condition for 2 graph to contain a Hamiltonian Path. And a minimum degree slightly larger than (p−1) is a sufficient condition for a graph to be Hamiltonian. In case of connected or Hamiltonian graphs it is not possible to find necessary and sufficient conditions.444
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1). In case of Eulerian graphs we can find a nice characterization that is necessary and sufficient for a graph to be Eulerian. We can still draw a disconnected graph if the minimum degree of a node in the graph is less than or equal to (p−2) . Thus for a general graph which does not satisfy the sufficient conditions for a Hamiltonian Cycle it is extremely hard to design an algorithm which can find a Hamiltonian Cycle (provided such a cycle exists) or out puts in negative if a Hamiltonian Cycle does not exist. in fact we can draw a graph with the above property which is clearly not Hamiltonian. In case of connected property again it is not possible to devise necessary and sufficient conditions. however it is possible to design efficient algorithms which can determine if a given graph is connected or not.
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now provide constructive proof techniques to prove a number of sufficient conditions for a graph to be Hamiltonian.
8.3
Hamiltonian Graphs
We shall make a drastically different start. Instead of teaching we shall make you solve a number of related but simple problems. You just need some prior knowledge in addition to some common sense. It is important for the learner to gain confidence: You can discover and create new knowledge. We shall also encourage you to solve a puzzle; the experience that you will gain will provide you powerful tools that we shall use in this chapter in solving various problems. Deciding when a specific tool can be used and where it can not be used is certainly a valuable learning experience. Problem Set 8.1. Let us start with simpler problems. For each of the problem you are supposed to design a formal proof; Problem 8.1.1. Suppose G is a star graph then G2 has a Hamiltonian Cycle. Please note that the diameter of a star graph is equal to 2 irrespective of the order of the graph. Problem 8.1.2. Suppose G is a line graph then G2 has a Hamiltonian Cycle. Please note that the diameter of a line graph is p − 1, and as it is obvious it is a function of the order of the graph. No graph (other than a line graph) can have a diameter as large as p − 1. For each of the above two problems it will be useful if the learner actually draws such graphs and then discovers the answer himself or herself. The k th power Gk of a graph G is a graph with same number of vertices as in G in which two vertices are adjacent if and only if they are at most d distance apart from each other in G. If the diameter of a graph G is k then Gk will be a completely connected graph and that is why G2 of a star graph is Hamiltonian. The diameter of a line graph is proportional to the size of the graph but still G2 is Hamiltonian. It will be interesting to explore if this is a more general result: G2 of any tree is Hamiltonian? Either show it or find a counter example. Let us now try to solve a problem which belongs to the so called critical activity section. You must solve this problem before moving forward. Please see Fig. 8.2 and Fig. 8.3.
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Concept Map 8.1. A concept map of certain necessary and sufficient conditions for a graph to be connected, Eulerian, and Hamiltonian.
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8.3.1
A Puzzle:
We are given a line graph with p nodes, the starting and end node of the line graph are designated as u1 and up respectively. The intermediate nodes are labeled as ui , where 2 ≥ i ≥ p − 1 as shown in Fig. 8.2. You are allowed to insert new edges between u1 and any of the intermediate nodes. Similarly you can insert edges between up and any of the intermediate nodes. While inserting edges you should keep the degree of u1 and up exactly the same. You can not insert an edge between the two terminal nodes otherwise a Hamiltonian Cycle will immediately be formed (we already have a Hamiltonian Path between the two terminal vertices). While inserting an edge from u1 to an intermediate vertex (and then an edge from up to again an intermediate vertex) try your best that a Hamiltonian Cycle is not formed. The problem is to find out (by an actual trial and error drawing) the minimum degree of the two terminal vertices when it will be impossible to resist a Hamiltonian Cycle. That will be a sufficient condition for this graph to be Hamiltonian. If, instead of resisting a Hamiltonian Cycle, you are adamant to form one as soon as the first opportunity arises then what will be the minimum degree of the two terminal vertices which will guarantee a Hamiltonian cycle? Now come back to the previous problem where we try to resist a Hamiltonian Cycle as far as possible. But let us relax the condition that the degree of the two terminal vertices should be the same. Now the degree of one terminal vertex will be larger and the other smaller. Again find what will be the minimum sum of the two degrees (of the terminal vertices) when it will be impossible to resist a Hamiltonian Cycle. Again this will be a sufficient condition for the modified graph to be Hamiltonian. You may like to solve the puzzle for the graph shown in Fig. 8.3
8.3.2
Actually Finding a Hamiltonian Cycle in the Puzzle:
Let us look at the same problem from an algorithmic point of view: You are given a graph G which contains a Hamiltonian path between two non adjacent vertices u and v. Assume that the sum of the degrees of u and v is equal to or larger than p. Also assume that the Hamiltonian Path between u and v is already provided as an input. You are supposed to design an efficient algorithm which will output the actual Hamiltonian Cycle in G.
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Figure 8.2: A line graph with six nodes. We should add extra edges in this graph but do our best to avoid a Hamiltonian Cycle.
Use the knowledge and expertise that you have gained while solving the puzzle;
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Figure 8.4: We are given three graphs where a Hamiltonian path exists between two special vertices. The Hamiltonian path between the two vertices is also indicated by shaded lines in each graph. In the top graph the degree sum of the two special vertices is equal to the number of nodes in the graph; in the bottom graphs the degree sum of the two special vertices is less than the number of nodes in the graph. Please note that a Hamiltonian Cycle exists in each of these graphs.
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Apply your algorithm on the graphs given in Fig. 8.4 in order to find a Hamiltonian Cycle in each graph; please note that each of the graphs does indeed contain a Hamiltonian Cycle (which can easily be found by hit and trial method). Let us see where and why your algorithm fails and where it does find a Hamiltonian Cycle. Is this not strange that a graph contains a Hamiltonian Cycle but your algorithm can not find it? Let us now describe the details of the algorithm which solves the problem outlined above. The working of the algorithm is illustrated in Fig. 8.5. Algorithm 61: Find a Hamiltonian Cycle in an un-directed graph G where deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G; a Hamiltonian path between u and v is also given. input : A Hamiltonian Path between vertices u and v where deg(u) + deg(v) ≥ p in an un-directed graph G. output: A Hamiltonian Cycle in G
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It is interesting to note that our algorithm also has a serious short coming: a graph may have a Hamiltonian Cycle but we can not find it (as shown in Fig. 8.6). Interestingly there may have graphs where the condition (that deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G) is not met yet our algorithm will be able to find a Hamiltonian cycle as shown in the same figure.
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Figure 8.5: Intermediate stages of how Algorithm 61 works to find a Hamiltonian Cycle in a graph G where the condition (that deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G) is met and where there is a Hamiltonian path between vertices u and v in G.
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Figure 8.6: A graph G shown (left diagram) where the condition (that deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G) is not met and the Algorithm 61 will NOT be able to find a Hamiltonian cycle; but a Hamiltonian Cycle exists in this graph as shown by blue lines. A graph G shown where the condition (that deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G) is not met yet the Algorithm 61 will be able to find a Hamiltonian cycle as shown in the right diagram.
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8.3.3
Basic Intuition
Let us carefully look at the ramifications of Algorithm 61. It provides a couple of very powerful theoretical results and useful algorithmic tools which can be used to find a Hamiltonian cycle in a certain category of graphs. Assume that a graph G has two vertices where the degree sum of these two vertices is more than p as shown in the figure below. We insert an edge between these two vertices and assume that in the resulting graph there is a Hamiltonian Cycle as shown in the top left diagram of Fig. 8.7. This essentially means that there will be a Hamiltonian Path starting from vertex 1 and terminating at vertex p in the original graph G. Then, we claim, that there will be a Hamiltonian Cycle in graph G (without that extra edge) as shown in the right diagrams of the Fig. 8.7. Now assume that after inserting that extra edge between the two vertices the resulting graph does not have a Hamiltonian Cycle. Then, we claim, that there will not be a Hamiltonian Cycle in the graph G (without that extra edge) as shown in the left diagrams of the same figure. Consider a graph G where the degree sum of any pair of vertices in G is equal to or larger than p as shown in Fig. 8.8. Now visualize a hypothetical Hamiltonian Cycle in this graph starting from 1, passing through 2, 3, 4, . . . p and then going back to 1. If this hypothetical Hamiltonian Cycle does not pass through original graph edges then insert Extra edges as shown in (red) in this figure to create a cycle consisting of some graph and some Extra edges. You know that we can always delete Extra edges one by one (using the techniques that we have recently acquired) forcing the Hamiltonian Cycle to divert through only graph edges. This intuition provides a powerful technique which can be used in most of Hamiltonian finding algorithms in Hamiltonian graphs.
8.3.4
Bondy & Chvatal's Theorem:
Now let us come back to a text book theorem which says that a graph G is Hamiltonian if and only if G + uv is Hamiltonian provided vertices u and v are non adjacent and deg(u) + deg(v) ≥ p. Before attempting to prove this theorem let us look at the problem more closely and try to relate it to the previous knowledge that we have acquired. G+uv is Hamiltonian means that there is a Hamiltonian Cycle in the graph G+uv. There are two possibilities:
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We insert a direct edge from u1 to up Assume that now a HAM Cycle does exist in the resulting graph
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Now we remove the direct edge from u1 to up A HAM Cycle will NOT exist in graph G?
Now we remove the direct edge from u1 to up A HAM Cycle will still exist in graph G?
Figure 8.7: We make a number of claims in this diagram: A graph G is Hamiltonian if and only if the graph G plus an edge between two non adjacent vertices of G is Hamiltonian provided certain conditions are met.
u1 up u1 up
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Consider a graph where the degree sum of every pair of vertices is equal to or larger than p
We force a Hamiltonian Cycle in this graph: If there no graph edges available then we insert Extra edges as shown in red
Figure 8.8: We make a number of claims in this diagram: A graph G is Hamiltonian if and only if the graph G plus an edge between two non adjacent vertices of G is Hamiltonian provided certain conditions are met.
Under such conditions a Hamiltonian Path between u and v may not exist. that the graph G is not Hamiltonian while the graph G + uv is Hamiltonian. does not exist. Again notice the figure of eight in this diagram. If there is a graph G in which there are two adjacent nodes u and v such that deg(u) + deg(v) ≥ p then G is Hamiltonian if and only if G + uv is Hamiltonian.9.
.5
Summary
Let us look at what we have really understood so far. It will be instructive to actually draw such a graph. Such a possibility is indicated in the left diagram of Fig.
8.3. The edge uv does not make any difference. Then there will certainly be a Hamiltonian Path between the vertex u and vertex v. As the edge uv is not part of the Hamiltonian Cycle so it is possible to remove this edge and still a Hamiltonian Cycle will exist in G.3. 8.456
Hamiltonian Graphs
1. This means the possibility. The Hamiltonian Cycle does not pass through the edge uv.6
Closure of a Graph:
A closure c(G) of a graph G of order p is a graph obtained from G by recursively joining pairs of non adjacent vertices u and v whose degree sum is more than or equal to p until no such pair remains in G. How about if we are supposed to design an algorithm to find a Hamiltonian Cycle in the graph G provided the above stated conditions are true. This also means that if we remove the edge uv even then there will be a Hamiltonian Cycle in G provided there was a Hamiltonian Cycle in G + uv. The Hamiltonian Cycle passes through the edge uv.
8.9. Again it will be interesting to relate this algorithmic problem with the last such problem that we have discussed. Such a possibility is shown in the right diagram of Fig. 8. 2. Under such conditions we can map this problem to the last problem that we have discussed except that the actual Hamiltonian Path between u and v is not provided here. It can now be proved that a graph G is Hamiltonian if and only if its closure c(G) is Hamiltonian. its inclusion does not convert a non Hamiltonian graph into Hamiltonian and its removal does not convert a Hamiltonian graph into a non Hamiltonian graph provided other conditions are also met.
8. 2. The right diagram shows a graph where the Hamiltonian Cycle does not pass through the edge between vertex u1 and u2 .Hamiltonian Graphs
u5 u4 u5 u4
457
u6
u3
u6
u3
u1
u2
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Figure 8. You can verify that a Hamiltonian path does not exist between these two vertices in the right diagram. We have also stated that for a completely connected graph. It can also be used constructively
. Edges which were originally present in G. But it may also pass through some of the extra edges belonging to the second type. then the closure c(G) will be obviously be a completely connected graph. But remember the Hamiltonian Cycle that we have found belonged to the closure of G. finding a Hamiltonian Cycle is a trivial problem. The following algorithm can be used to find a Hamiltonian Cycle in graph G.3. These edges were not part of G. that means E(G). that is. We call these edges extra edges. The Hamiltonian Cycle in c(G) will pass through some of the original edges of the graph. We have already stated that complete connectedness is a very loose sufficient condition for a graph to be Hamiltonian.9: A graph containing a Hamiltonian Cycle which passes through the edge between vertex u1 and u2 is shown in the left diagram. How to find a Hamiltonian Cycle which passes through E(G) alone is an interesting problem. degu + v ≥ p.7
Ore's Theorem:
The usefulness of the above result becomes obvious when in a graph G every pair of non adjacent vertices satisfies the above mentioned condition. the same Hamiltonian Cycle may not exist in the actual graph G? This is because of the fact that c(G) consists of two types of edges: 1. Extra edges inserted between very pair of non adjacent vertices u and v where deg(u) + (v) ≥ p.
We find a Hamiltonian Cycle in graph H.2. and last but not the least discuss where Algorithm 62 have to be modified in order to find a Hamiltonian Cycle.458
Hamiltonian Graphs
to prove that if deg(u) + deg(u)v ≥ p for every non adjacent pair of vertices in a graph G then G is Hamiltonian.this closure is our starting graph H. input : An un-directed Graph G where deg(u) + deg(v) ≥ p for very pair of non adjacent vertices u & v in G. Draw three different graphs. We know that deg(u) + deg(v) ≥ p. while there are extra edges in the Hamiltonian Cycle in graph H do Remove an extra edge uv from the cycle and from graph H thereby creating a Hamiltonian Path between two vertices u and v in the new graph H.
. Also pinpoint the graph where it may not be easy to find a Hamiltonian Cycle using the knowledge that we have acquired until now.1. in the first graph G the degree sum of each pair of non adjacent vertices should be at least p. Problem 8. This is certainly a tighter sufficient condition for a graph G to be Hamiltonian as compared to the complete connectedness of a graph. In the third graph J.
Problem Set 8. Algorithm 62: Find a Hamiltonian Cycle in a graph G where deg(u)+ deg(v) ≥ p for very pair of non adjacent vertices u & v in G. This cycle may pass through some of the graph G original edges and may also pass through some of the edges not present in G (but are present in H) known as extra edges. Discuss briefly where it is possible to use Algorithm 62 without any modification to find a Hamiltonian Cycle. thus if there is a Hamiltonian Path between u and v in H then we can find a new Hamiltonian cycle in H which does not pass through the edge uv (use Algorithm 61). the degree sum of each pair of non adjacent vertices is not equal to p. This sufficient condition was originally discovered by Ore. In the second graph H the degree sum of each pair of non adjacent vertices is not p but the closure of H is a completely connected graph. output: A Hamiltonian Cycle in G
1
2 3
4
We find a closure of graph G known as c(G) .2.
The closure of this graph thus contains a Hamiltonian
. The closure of this graph is a completely connected graph as shown in the right diagram of the same figure. Note that the closure is a completely connected graph containing a Hamiltonian Cycle also shown in the right diagram. a completely connected graph as shown in the top right diagram. 8. Note that the Hamiltonian Cycles passes entirely through extra edges.11.13. We remove extra edges one by one and each time find a new Hamiltonian Cycle. Example 2: We show a graph G in the top left diagram of Fig. The closure of this graph is. The Hamiltonian Cycle is also indicated in this diagram. Why? The order will. 8. however. A formal proof of this theorem can easily be derived from the above discussion. become important in the next example. the degree of some of the nodes is also indicated in this diagram.10. the extra edges removed are indicated by dotted lines while a Hamiltonian Cycle is shown by bold lines in Fig. The closure of this graph is shown in the right diagram. the degree sum of a pair of non adjacent vertices is less than 6 in this graph. 8. interestingly the Hamiltonian Cycle passes entirely through extra edges.Hamiltonian Graphs
459
Dirac's Theorem: A graph G where the degree of each node is more than or equal to p/2 is Hamiltonian. however.11. A constructive proof can also be designed on similar lines. Example 1: We show a graph G where the degree of each node is equal to p/2 in the left diagram of Fig. We apply Algorithm 62 to this graph and show a number of intermediate results in Fig.10: A graph G where the degree sum of each pair of non adjacent vertices is equal to p. 8.
u5 u4 u5 u4
u6
u3
u6
u3
u1
u2
u1
u2
Figure 8. the graph is shown in the left diagram. Please note that the order in which extra edges are removed is not very important in this example.
Initially the Hamiltonian Cycle passes through all the extra edges as shown in the top left diagram. passes through No Extra Edge
Figure 8.
.12: It is possible to remove more than one extra edge (in a single step) from the graph. Another Extra Edge can NOT be removed
Redraw the graph and remove another Extra Edge
Hamiltonian Cycle shown in blue.Hamiltonian Graphs
461
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Extra edges are shown in red. Another Extra Edge can be removed
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Redraw the graph and remove the last Extra Edge
Hamiltonian Cycle. shown in blue. Another Extra Edge can be removed
Redraw the graph and remove another Extra Edge
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Hamiltonian Cycle shown in blue. We remove one Extra Edge
Hamiltonian Cycle shown in blue. At the end it passes through the graph edges as shown in the bottom right diagram.
one should be very careful about the order in which they are removed otherwise one may fall into a trap as shown in the bottom diagrams. remove other Extra Edges
Figure 8.
.13: While removing Extra edges.462
Hamiltonian Graphs
4 u5
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We start with a graph where degree sum of a few pairs of vertices is not p
Add Extra Edges where the degree sum condition is satisfied. The closure is complete
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The Hamiltonian Cycle does not pass through any Extra Edge
Find a Hamiltonian Cycle in the remaining graph as shown in blue color
Delete an Extra Edge
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?
u1 u2
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Redraw the graph and try to remove the last Extra Edge through which the Hamiltonian Cycle passes NOT Possible?
Find a different Hamiltonian Cycle in the remaining graph as shown in blue color. It passes through one Extra Edge. This will increase the degree?
Add Extra Edges where the degree sum condition is recently satisfied.
14: There is no need to make the closure of a graph complete.Hamiltonian Graphs
463
A given graph where degree sum condition for every pair of vertices is satisfied
We add Extra Edges so that a Hamiltonian Cycle is formed. we can add just enough extra edges such that a Hamiltonian Cycle becomes possible in the graph.
. the closure is not yet complete
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It is possible to add just one Extra Edge so as to get a Hamiltonian Cycle
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Figure 8.
10? It is essential for you to answer these questions before moving forward.13.15: The top left diagram shows a graph in which the degree sum of every pair of non adjacent vertices is not equal to (or larger than) the number of nodes in the graph. Let us solve puzzle No.4
Bipartite Hamiltonian Graphs
We are now in a position to design or discover similar sufficient conditions for a bipartite graph to be Hamiltonian (It will also be exciting to discover some of the necessary conditions for a bipartite graph to be Hamiltonian).1 once again.3. an edge uv exists provided
.
8. 8.
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Figure 8.464
Hamiltonian Graphs
Cycle. Be careful this time. The bottom right diagram shows that a difficult situation arises if we remove extra edges in the wrong order. Why this order has suddenly become important and why it was not important in the graph of Fig. The closure of this graph is still a completely connected graph as shown in the bottom left diagram. while inserting edges you should keep in mind that in a bipartite graph. If we now remove the extra edges in the wrong order then we may end up with a difficult situation as depicted in the bottom right diagram of Fig. 8. 8.
If we insert an edge between the two end vertices then a Hamiltonian Cycle will immediately be formed. Fig.17 and Fig. You can not insert an edge between the two terminal nodes otherwise a Hamiltonian Cycle will immediately be formed (we already have a Hamiltonian Path between the two terminal vertices).
8. Please see Fig. The problem is to find out (by an actual trial and error drawing) the minimum degree of the two terminal vertices when it will be impossible to resist a Hamiltonian Cycle.5
Some Theoretical Claims
We shall describe a number of special Hamiltonian graphs and then present a number of graph theoretical claims. 8.16.
.16: We show a bipartite graph consisting of eight vertices containing a Hamiltonian Path between two end vertices. You should also keep the degree of u1 and up exactly the same while inserting new edges. It is always Moorish Connected.Some Theoretical Claims
465
vertex u and vertex v belong to different partites. 8. That will be a sufficient condition for this graph to be Hamiltonian. Hamiltonian Graphs: p-closure is complete in G: A Hamiltonian Cycle exists in G.18
u1 u2
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Figure 8. While inserting an edge from u1 to an intermediate vertex (and then an edge from up to again an intermediate vertex) try your best that a Hamiltonian Cycle is not formed. Hamiltonian Connected: (p+1)-Closure is complete: A Hamiltonian Path passes through every pair of vertices. 8.
Notice the bi-colored vertices. Edges not allowed in a bipartite graph are indicated in both the diagrams in the form of dotted lines.16. In a bipartite graph. 8. similar color vertices belong to one partite while all the other color vertices are in the second partite.
.466
Hamiltonian Graphs
u1
up
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u3 u4 u5
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up
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Figure 8. an edge uv exists provided u and v belong to different partite.17: Here we show a Hamiltonian Path between two vertices in the bipartite graph shown in Fig.
. If the closure of G + x is complete then there will be a HAM cycle in G + x and a HAM path in G between vertex u and vertex v of G. Claim No. If vertex u and v are adjacent in graph G then there will be a HAM cycle in G passing through u and then v. 1: Given a graph G (having p vertices). we add a vertex x to G and connect it to vertex u and vertex v of G. Now if p + 1 closure of H (having p + 1 vertices) is complete then there will be a HAM path in G with end points u and v in G.468
Hamiltonian Graphs
Moorish Connected: A Hamiltonian Path passes through every pair of adjacent vertices.
k k
G
Degree of every vertex larger than or equal to k k k k
u 2
k+1
k k k
x
k+1
v
If p+1 closure is complete in graph G+x then there will be a HAM cycle in the graph G+x.19: We show a graph G. the new graph is known as graph H as shown in the figure below.
u
k
k
G
k
k Then there will be a HAM cycle passing through vertex u & v provided u and v are adjacent in G
v
k
Figure 8. we add and connect a vertex x to vertices u and v of G as shown in the top diagrams. It may or may not be Hamiltonian Connected.
2: Assume that in a graph G. Such a graph G (where a HAM path exists between every pair of vertices in G) is known as Hamiltonian Connected. Degree sum of 4 and 5 becomes 6
Connect 4 & 5. 1: We show a graph G in Fig. we add and connect a vertex x to vertices 4 and 5 of G as shown in the top diagram. It is somewhat surprising to note that a p closure of graph G is not complete yet a p + 1 closure of graph G + x is complete. in fact it is more than that: every edge u. Degree sum of other pairs becomes 6
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All vertices in graph G+x are now completely connected
3
All vertices in G are completely connected
There is a HAM path between vertex 4 and 5
Figure 8. v of G. v will be part of some HAM cycle.20: We show a graph G. We are lucky in this example as we do find a HAM path between the two given vertices. The closure of G + x is complete and there will be a HAM cycle in G + x and a HAM path in G between vertex 4 and 5 of G.20 and we need to check if our newly acquired knowledge can really confirm if there is a HAM path between two vertices 4 & 5 in graph G. We know before hand that a HAM path exists between these vertices but we also know that there is a possibility that a HAM path exists in a graph but we are unable to confirm or find it. Claim No. the p + 1 closure is complete then there will be a HAM path between every pair u.
. 8.Some Theoretical Claims
469
Example No.
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Connect vertices where degree sum is equal to 6
Connect x to 4 and 5. It follows that such a graph G is Hamiltonian.
21.22 with end vertices 3 and 5. 3: Given a graph G (having p vertices).
. now the p + 1 closure of G + x becomes complete as shown in the rest of the diagrams. 8. thus we can not confirm that a HAM path exists between vertex 3 and 5 in G. If the closure of G + x is complete then there will be a HAM path in G. we add and connect a vertex x to every vertex of G as shown in the top diagrams.21: We show a graph G. 4: Assume that in a graph G. 2: We show a chain graph G in Fig. We add a vertex x and a vertex y to G such that vertex x is connected to vertices u and v of G. v. 8. Example No. the p − 1 closure is complete then there will be a HAM path in G. and w such that u is adjacent to v and v is adjacent to w. we add a new vertex x to G and connect it to every vertex of G. Now if p + 1 closure of H (having p + 1 vertices) is complete then there will be a HAM path in G. The p + 1 closure of graph G + x (having p + 1 vertices) is not complete. 5: Given a graph G (having p vertices) with three vertices u.
k+1
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Degree of every vertex larger than or equal to k
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AH AM pat h in
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If p+1 closure of G+x is complete then there will be a HAM cycle in G+x
Figure 8. thus confirming that a HAM path do exist in this graph. Claim No. Then we connect vertex x to every vertex in G. We add a vertex x to G and connect it to vertex 3 and 5 as shown in the top left diagram.470
Hamiltonian Graphs
Claim No. We know that a HAM path exists between vertex 3 and 5 in this graph. Claim No. the new graph is known as the graph H as shown in Fig.
the new graph G + x + y is known as graph H as shown in the figure below. Claim No. 6: Assume that in a graph G.
k
v
G
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k
There will be a HAM cycle passing through u then v and then w in G provided u is adjacent to v and v is adjacent to w. Now if p + 2 closure of H (having p+2 vertices) is complete then there will be a HAM cycle in G which will be passing from vertex u to v and then to vertex w (see Fig.23: If the p + 2 closure of graph G + x + y (having p + 2 vertices) is complete then there will be a HAM cycle in G passing from vertex u to vertex v and then to vertex w of G.
w
Figure 8. 7: In a regular graph G where the degree of every vertex is p/2. a HAM path exists between vertex u and vertex v of G if and only if p + 1
. w} in G. 8. the p + 2 closure is complete then some HAM cycle will pass through every pair of adjacent edges {u. v} and {v.472
Hamiltonian Graphs
and vertex y is connected to vertex v and w of G. Claim No. and so on.
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2
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If p+2 closure is complete in G+x+y then there will be a HAM cycle passing from u to x to v to y and then to w.23).
u
p/2 p/2
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Degree of every vertex is exactly p/2
p/2 p/2
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There will be a HAM path passing through vertex u & v in G
If and only if
p+1 closure is complete in graph G+x
Figure 8. Problem 8.Some Theoretical Claims closure of G + x is complete (please see Fig. 8: In a regular graph G where the degree of each vertex is p/2. 8. We add a vertex x to G and connect x to vertex u and v in G as shown in the right diagram. v} will be part of a HAM cycle.24 in this regard). every edge {u. If some of the graphs do not satisfy sufficient conditions for Hamiltonian graphs then there is a possibility that a Hamiltonian Cycle exists but our expertise is not able to find it. 9: A regular graph G. Problem Set 8. 8.3.25.24). How can you modify the algorithm (which was earlier used to find a Hamiltonian Cycle in a Hamiltonian graph) which makes sure that the resulting Hamiltonian Cycle passes through a given edge. Look at the graphs in Fig.
.24: We show a regular graph G.2. 8. Problem 8. the degree of every vertex is p/2. You can use our past experience of finding a Hamiltonian cycle in these graphs. Claim No.3. where the degree of each vertex is p/2.
473
Claim No. is Hamiltonian Connected (that means there is a HAM Path between every pair of vertices in G) if and only if p + 1 closure of G + x is complete for every pair u and v of G (please see Fig.3.1.
Draw a graph which is not Hamiltonian (that means it does not satisfy the sufficient conditions) but in which there is a Hamiltonian path between every pair of vertices.474
Hamiltonian Graphs
1.6
A Categorization of Hamiltonian Graphs
Category A: There is not a Hamiltonian path between every pair of non adjacent vertices in G and G is Hamiltonian (that means it satisfies the sufficient conditions). Problem 8. Draw a graph which is not Hamiltonian (that means it does not satisfy the sufficient conditions) but in which there is a Hamiltonian path between every pair of non adjacent vertices. Problem 8. Draw a graph which is Hamiltonian (that means it satisfies the sufficient conditions) but in which there is not a Hamiltonian path between every pair of non adjacent vertices. We call it category E graphs.7. Also consider the option when it is Hamiltonian Connected (B2). check if there is a Hamiltonian cycle and also check if our expertise can find it or can not find it. Draw a graph which is Hamiltonian (that means it satisfies the sufficient conditions) but in which there is a Hamiltonian path between every pair of vertices. Design an efficient algorithm to find a Hamiltonian path in these graphs between any two vertices. We call it category D graphs. Problem 8.3.
. 3. Category B: There is a Hamiltonian path between every pair of non adjacent vertices in G and G is Hamiltonian (that means it satisfies the sufficient conditions) and it is not Hamiltonian Connected (B1).3.3.3. Design an efficient algorithm to find a Hamiltonian path in these graphs between two given (not any two but two special) vertices. We call it category B graphs. We call it category C graphs.3. 2.
8.3. Problem 8. Problem 8.6. 8.5.4. We call it category A graphs.25. For every graph in Fig. Draw a graph which is Hamiltonian (that means it satisfies the sufficient conditions) but in which there is a Hamiltonian path between every pair of non adjacent vertices.
25: You may find some graphs here which you were supposed to draw in earlier problems.
.A Categorization of Hamiltonian Graphs
475
Figure 8.
12.15. v and w in G (see the bottom right diagram in Fig.27). You can relax this condition later and cater to the condition when the degree of every vertex may be larger than p/2. 8. Prove that G3 of a graph G is always Hamiltonian.13.10. Try to draw a graph G where G2 is not Hamiltonian while G3 is Hamiltonian Connected.27).3.14. Problem 8.11. Problem 8. 8. and the degree of every vertex is larger than or equal to p/2. We need to determine k such that a Hamiltonian Path (or a cycle) passes through every two vertices u and v in G (see the bottom left diagram in Fig.3.3. Problem 8. A graph G where the minimum degree is k is given as shown below.16. Problem 8.9. Problem 8.3. Initially you can assume that G is a regular graph and thus the degree of every vertex is the same.3.3. We need to determine conditions under which there will be a Hamiltonian Path in G (see the top diagrams in Fig.A Categorization of Hamiltonian Graphs
477
Category C: G is Hamiltonian (that means it satisfies the sufficient conditions) & Hamiltonian Connected (C1). Problem 8. Category D: There is a Hamiltonian path between every pair of non adjacent vertices in G and G is not Hamiltonian (that means it does not satisfy the sufficient conditions). Also consider the option when G is not Hamiltonian Connected (C2).3. Try to draw a connected graph G where G3 is not Hamiltonian Connected. We need to find (that is design an efficient algorithm to solve this
.27). Try to draw a graph G where G2 is not Hamiltonian while G3 is not Hamiltonian Connected. We need to determine k such that a Hamiltonian Path (or a cycle) passes through three vertices u. Category E: G is Hamiltonian Connected but is not Hamiltonian. Assume that we are given a graph G with number of vertices equal to p where p is even. Problem 8. Problem 8.3. Try to draw a graph G where G2 is not Hamiltonian while G3 is Hamiltonian.8.3. Problem 8. 8.
and (b) when they are not adjacent.A Categorization of Hamiltonian Graphs
479
problem) if there is a Hamiltonian path between two given vertices.
.28 may provide some food for thought. If there is really a Hamiltonian path then our algorithm should be able to output this path (we should provide a proof for this). 8. Consider two options separately: (a) when the two given vertices are adjacent. The following graphs in Fig.
.28: We show here a number of regular graphs where the degree of each vertex is exactly p/2 and p is even.480
Hamiltonian Graphs
Figure 8.
When we make a transformation on a graph we create a graph with new properties. It will be interesting to examine how different properties of a graph change (or do not change) if we transform a graph into a new directed graph using
. unilaterally connected directed graphs and tournaments. and if not then what are its strongly connected components. directed graphs can also be classified into different categories. As with un-directed graphs. We have also discussed a connected graph which was not a tree that means a graph which was cyclic. An undirected graph is known as strongly orient-able if it has the potential of becoming a strongly connected directed graph. and then we had a just barely connected graph known as a tree. and how they are connected. We shall discuss and prove (again using constructive proof strategies) necessary and sufficient conditions for an undirected graph to be strongly orient-able. Graphs are also judged by different properties that they possess. We have also talked about certain actions on graphs like the square of a graph or the complement of a graph.482
Strongly Connected Directed Graphs and Tournaments
Introduction
We shall talk about some specific features of directed graphs in this chapter.1
Concepts. This is how we shall study directed graphs in this chapter. directed acyclic graphs. A completely connected graph is always Hamiltonian. We shall be talking about strongly connected directed graphs. and a classification of un-directed and directed graphs are shown in a concept map (see Concept map 1). We shall also describe efficient algorithms to find if a directed graph is strongly connected. some action items (or so called transformations). We have seen in the last chapter that if we take the square of a line graph or a star graph then the resulting graph is Hamiltonian. We had an un-directed graph which was not connected. On the other extreme we have seen the completely connected graph. Each category of directed graph possesses some specific properties. Some of the properties of graphs. We shall start with different categories (shown as concepts in Concept map 1) of directed graphs. Properties & Actions
We have earlier classified an un-directed graph on the basis of connectedness. For example a line graph which is acyclic and is not Hamiltonian.
9. These actions on a graph G transform G into another graph H.
1.
. some action items or transformations. Please note that the action items are represented by elliptical objects while the properties are shown by square boxes. and a classification of undirected and directed graphs. A concept map showing some of the properties of graphs.Concepts. Properties & Actions
483
Concept Map 9.
Completely connected un-directed graphs are both unilaterally orient-able and strongly orient-able. directed graphs can also be classified on the basis of connectedness. If we remove directions from a directed graph D then the resulting graph will become undirected and is known as the underlying undirected graph. A directed graph having no cycles is called an directed acyclic graph or a DAG. It is obvious that a tournament directed graph may be acyclic
. Not all un-directed graphs can be converted into a strongly connected directed graph. A directed graph formed by putting arbitrary directions in a completely connected undirected graph has a special name that is a tournament.1 shows three different directed graphs. there is a path from u to v or a path from v to u. One of these graphs is acyclic while the rest are cyclic. When we put directions in an un-directed (or bi-directed) graph G then it becomes a directed graph D. That means a strongly connected graph is always a unilaterally connected graph but not vise versa. Both strongly connected as well as unilaterally connected directed graphs are always weakly connected but again it may not be true the other way round. A directed graph D is unilaterally connected provided for every pair of vertices u and v. One of them is strongly connected. 9.
Concepts
As you can see in Concept map 1. If the underlying undirected graph of D is connected then D is known as a weekly connected directed graph. and is therefore not strongly connected. The third is neither unilaterally connected nor strongly connected. This means that if we intelligently put directions in a completely connected un-directed graph then the resulting directed graph will be strongly connected (or unilaterally connected).484
Strongly Connected Directed Graphs and Tournaments
some of the action items shown in Concept map 1. The other is unilaterally connected but not strongly connected. Un-directed graphs which can be converted into a unilaterally connected directed graph are known as unilaterally orient-able undirected graphs. A directed graph where you can reach any vertex from any other vertex is known as a strongly connected graph. Fig. Please try to pinpoint which one is which. It is also possible to put directions in a completely connected graph in such a manner that the resulting directed graph is acyclic. In a strongly connected directed graph there is a path from u to v and a path from v to u. Those which can be converted into a strongly connected graph are known as strongly orient-able un-directed graphs.
Problem 9. Problem Set 9.1. or may contain a cycle.2. Only two edges in this graph are directed while the rest are left un-directed or bi-directed. A graph consisting of six vertices is shown in Fig.1.1. Properties & Actions
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Figure 9.4. Problem 9.Concepts.1. in fact it may contain a Hamiltonian Cycle.1. You are supposed to put directions in these undirected edges in order to fulfill some objective as defined below. Problem 9. The resulting graph should be acyclic and not unilaterally connected. One of them is strongly connected. as we shall discuss in detail in this chapter.1. Why?
. After putting appropriate directions draw the directed graph D.1: Three different directed graphs.3. The resulting graph should be acyclic and unilaterally connected. We have not asked you to draw a graph which should be acyclic and strongly connected. Problem 9. The resulting graph should be cyclic and not unilaterally connected. 9.2.
You are supposed to make it a directed graph by putting directions on un-directed edges in order to fulfill certain objectives Problem Set 9. Problem 9. Problem 9. Draw a directed graph D in order to fulfill the following objectives: Problem 9. The resulting graph should be unilaterally connected but not strongly connected.1. Problem 9. A directed graph in which every node lies on a directed cycle but the graph is not strongly connected. The directed graph should be weekly connected. and D is not strong. The resulting directed graph should be cyclic but not strongly connected.2.2.2. not strongly connected but cyclic.
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.2: A graph in which only two edges are directed while the rest are left un-directed.2.7.1.6.2.1. The resulting directed graph should be cyclic and strongly connected.1.3.5.486
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Problem 9. A directed graph in which if there is a path from u to v then there is a path from v to u for every pair of vertices u and v in D.1.
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Figure 9. Will the resulting graph always be unilaterally connected? Discuss briefly. Problem 9. Problem 9. A directed graph in which every node lies on a directed cycle but the graph is not strongly connected.8. The directed graph may not be weekly connected. The resulting graph should be unilaterally connected.
or acyclic? Discuss briefly.2.
Properties
A directed graph D is Hamiltonian provided there is a spanning cycle in the graph. Will the resulting graph be strongly connected. unilaterally connected. if there is a path from u to v then there is no path from v to u. The Reachable Relation graph of D can be represented by an adjacency matrix A in which A(u. An Eulerian Trail is a closed trail which spans all edges exactly once. A directed graph D is transitive if and only if there is an edge from u to v provided
. Similarly the Reachable Relation matrix of a strongly connected directed graph will also contain all 1's.Concepts. We know that a spanning cycle passes through every vertex exactly once. v) = 1 provided there is a directed path from u to v in D. A closed spanning walk in a directed graph D passes through every node of D and comes back from where it has started. This definition is almost the same as given in the last chapter for un-directed graphs. The Reachable Relation (or the transitive closure) of a directed graph D is another directed graph in which there is an edge from vertex u to vertex v provided v is reachable from u in D. A directed graph D is n − cyclic if it contains a directed cycle consisting of n nodes where n ≤ p.4.
Action Items
Let us now discuss some of the action items or so called transformations. The and it is zero otherwise. The transpose of a directed graph D is another directed graph E in which there is an edge from u to v if and only if there is an edge from v to u in D. A weekly connected directed graph in which for every pair of vertices u and v. The Reachable Relation matrix of an undirected connected graph will contain all 1's. As we know in a walk a node as well as an edge may be traversed several times. An Eulerian Trail may pass through a vertex several times but it should pass through every edge exactly once. If n = p then the graph D is Hamiltonian. We shall introduce more action items as the need arises. Properties & Actions
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Problem 9. A Hamiltonian Path or a spanning path spans every vertex only once but is not closed so you can not come back from where you have started.
3. Draw D2 .
. Problem Set 9. answer the following: 9.1. Find a closed spanning walk in this graph. Problem 9. Problem 9. Problem 9. For the directed graph D shown in Fig.3. Draw the transpose of the graph. Problem 9. Is this graph strongly connected? Why? Problem 9.3.3.3.3.3.4. try to
Problem 9.3.3.7.2.3. Find a Reachable Relation Matrix for this graph.6.488
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there is an edge from u to any vertex w and an edge from w to v for every pair of vertices u and v in D. Is the transpose also strongly connected? Why? Problem 9.3: A directed graph D for the problem set. Is D or D2 transitive? Discuss briefly.
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Figure 9. Problem 9.3. Is this graph Hamiltonian? Discuss briefly.5. Is D2 Hamiltonian? Discuss briefly.8.
Closed spanning walk 4. the Reachable Relationship of a directed graph. and thus such a graph will be strongly connected. When a graph is strongly connected then it is possible to reach every vertex from every vertex in that graph. Is it the other way round also. a trail passes through every edge exactly once but may pass through a vertex several times. Hamiltonian Cycle 2. this implies that every vertex is reachable on this cycle. Does it also imply that if the Reachable Relation matrix of a graph contains all 1's then the graph is strongly connected? I think it is quite obvious that the answer is yes. that is every strongly connected graph is Hamiltonian? I think you should be able to find a counter example quite easily.2
Strongly Connected Directed Graphs
Let us focus on a strongly connected graph (we consider it as the current concept). These properties are: 1. Eulerian Trail 3. Let us now focus on Eulerian directed graphs. Please do it before moving forward.2). In such graphs it is possible to find a Eulerian trail. Why it can not be an acyclic graph? Because otherwise it will not be possible to reach from v to u if it is possible to reach from u to v. n-cyclic 5. This directly implies that the Reachable Relation matrix of a strongly connected graph will contain all 1's (see Concept map 9.Strongly Connected Directed Graphs
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9. that is. Let me now pick one action item. Keep in mind a number of properties a directed graph can possess. Thus a strongly connected graph should never be acyclic. A directed graph which is Hamiltonian contains a Hamiltonian Cycle. As every vertex is visited at least once and we come back from where we have started thus every Eulerian graph
. Again it is not the other way round that is every cyclic graph may not be strongly connected (we have already witnessed this in a problem set). Transitive. A strongly connected graph may not be Hamiltonian but it perhaps should always contain a cycle of length n where n < p.
We also know that in a strongly connected graph it is possible to move from any vertex to any other vertex. Let us call this a closed walk U . be not a closed spanning walk. When we merge U and W we shall get a closed spanning walk in the graph D. Let us summarize of what we have gained so far. We repeat this process until the closed walk becomes a closed spanning walk (see Fig. it means that you can always come back to a vertex from where you have started. Once we have the Reachable Relation we have some information about the nature of the directed graph. As U does not span all vertices thus there will be a vertex u in U which will be connected to a vertex v of D which is not part of U (why?). Algorithm 63 answers if a given directed graph is strongly connected (or not) in a polynomial time algorithm. that will in fact be a closed walk. 9. It uses the Reachable Relation of a graph as an intermediate or a so called bridging concept. We know that it is possible to traverse a node or even an edge several times in a walk. go to vertex v.
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will be strongly connected but not vise versa. This will result in another closed walk W which when merged in U gives us a closed walk with more vertices that in U .4). it will still be possible to come back to vertex u.
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Figure 9. let us look at a relatively more general graph D which contains a closed spanning walk. We are now in a position to design algorithms to solve a number of related problems. Another closed walk W is also shown in the same directed graph. A graph D is strongly connected if and only if we can find a closed spanning walk in D or D is strongly connected if and only if the Reachable Relationship of D contains all 1's. it may. however. We can start from u.4: A closed walk U is indicated in a directed graph D. After looking at Hamiltonian or Eulerian graphs which are quite restrictive in nature.
Eulerian.2. It is interesting to note that the Reachable Relation graph of a Hamiltonian.Strongly Connected Directed Graphs
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Algorithm 63: Check if directed graph D is strongly connected input : Directed graph D output: Graph D is Strongly Connected or not
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Transform the Directed Graph D into a Reachable Relation graph by applying any traversal algorithm on every vertex of D. A map showing the concept of a strongly connected graph. If the Reachable Relation matrix contains all 1's then the graph D is strongly connected otherwise not.
. a couple of properties and an action item which is the Reachable Relation.
Concept Map 9. and strongly connected directed graph is exactly the same.
Algorithm 64: Find a cycle in a directed graph D input : A directed graph D which is not acyclic output: A cycle in D
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Find a closed walk u in D.
Khawaja Fahd says that we should just check the diagonal boxes in the Reachable Matrix and there is no need to check any other box. We shall address this problem once we acquire the required knowledge. Algorithm 65: Find a closed spanning walk in a strong graph D input : A strongly connected directed graph D output: A closed spanning walk in D
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Find a closed walk u in D. What do you think?
It is interesting to note that it is possible to design a much more efficient algorithm to solve the same problem but then we need some more concepts and a deeper insight into the problem. It will be interesting to derive the time complexity of this algorithm and to make it more efficient if possible. Increase its size until it becomes a closed spanning walk as previously explained in this section. The given directed graph D may not be strongly connected but it should contain a cycle. Convert it into a cycle.
Note that this algorithm is directly related to the (constructive) proof which proves that a directed graph D is strongly connected if and only if D contains
. If D is strongly connected then it will certainly contain a cycle.492
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If the relation matrix contains all 1's then the graph D is strongly connected otherwise it is not. A closed walk can always be converted into a cycle.
Algorithm 64 finds a cycle in a directed graph D. In this algorithm first we find a closed walk in graph D using any traversal algorithm.
We next consider a class of un-directed graphs (known as strongly orient-able) where if we put a direction intelligently on each edge then the resulting directed graph will be strongly connected.4). Is this also a necessary condition for a graph G to be strongly orient-able? The following algorithm converts an un-directed graph with no bridge edges into a strongly connected directed graph by putting appropriate direction on each edge (see Fig. The algorithm is quite similar to the algorithm
. we have taken the transpose of D. that is. 9. Now assume that we reverse the direction of each edge in graph D. This implies that a graph G should not contain any bridge edge for G to be strongly orient-able. It is obvious that the un-directed G should be connected. Will it be still possible to reach any vertex from any vertex in this graph? If yes then the transpose of D will always be strongly connected provided D was strongly connected. every directed edge should be part of a directed cycle otherwise it will be impossible to ensure reach ability from any vertex to every other vertex. On the other hand there are un-directed graphs where it is impossible to convert them into strongly connected directed graphs. One of the necessary conditions for a graph to be Hamiltonian is that there should be no cut vertex. The proof was earlier presented informally in this section. Let us discuss strongly connected graphs in terms of one more property (transitive) and one more transformation (transpose). We should remember that the desired directed graph should not only be cyclic. It is interesting (or shocking) to note that the underlying un-directed graph of a strongly connected graph may not be strongly orient-able (why?).Strongly Connected Directed Graphs
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a closed spanning walk. It should not be acyclic otherwise it will be impossible to create a directed cyclic graph out of it which is strongly connected. It will again be interesting to derive the time complexity of this algorithm and to make it more efficient if possible.
Strongly orient-able Un-directed Graphs
We need to discover necessary and sufficient conditions for an undirected graph G to be transformed into a strongly connected graph. Is a strongly connected graph D transitive? Should it never be transitive? If it is transitive then will it be a special strongly connected graph? If a strongly connected graph D is not transitive then will D2 be transitive? Under what conditions D2 be transitive? We know that in a strongly connected graph D it is possible to reach any vertex from any vertex in D.
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Directed Acyclic Graphs (DAG's)
In a strongly connected directed graph every vertex is reachable from every other vertex. that is. a DAG is quite the apposite. We have yet to prove that if there is a bridge edge in a graph G then it is impossible to convert G into a strongly connected graph D. Can a DAG be unilaterally connected? Can a DAG be not unilaterally connected? Should a DAG be always transitive or always not transitive? These are some of the questions
. We have seen in the last chapter that G2 of a line graph is Hamiltonian and is therefore strongly orient-able.494
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used to find a closed spanning walk in a strongly connected directed graph. Algorithm 66: Convert an un-directed G into a strongly connected D input : An un-directed graph G output: A strongly connected directed graph D
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Find a closed spanning walk u in G. Increase the size of the spanning walk until it becomes a closed spanning walk as previously explained. an n-Cycle.
The algorithm can also be used as a constructive proof to prove that if there are no bridge edges in an un-directed graph G then it is possible to convert G into a strongly connected directed graph D. a vertex u is reachable from v and the vertex v is reachable from u for every vertex u and v in D. a Hamiltonian Cycle. here if a vertex v is reachable from u then the vertex u is not reachable from v for every pair of vertices u and v in a DAG. That will in fact complete the proof of a theorem which states that a connected graph G is strongly orient-able if and only if G contains no bridge edges. Thus a DAG can never be strongly connected. an Eulerian Trail.
9. As opposed to a strongly connected graph. We have also seen that G2 of a star graph is also Hamiltonian. It will be worthwhile to conjecture that if G is connected but not strongly orient-able then G2 is certainly strongly orient-able. or a closed walk can not exist in a DAG. In other words there is no cycle in a DAG and that is why the name: directed acyclic graph. You should either prove this conjecture or find a counter example. Put appropriate directions. As it should be evident.
and a number of intermediate vertices. Try your luck and design an algorithm which can transform an un-directed completely connected graph into a DAG (see Fig.Directed Acyclic Graphs (DAG's)
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which should trouble your mind. Looking from the other way round you can start from any undirected graph G and put a direction on each edge such that the resulting directed graph becomes a DAG (in fact it can be converted into several different DAG's). There are certain nodes in a DAG which have some special features not noticed in other directed graphs. as this DAG is derived from a completely connected un-directed graph so it is known as a tournament DAG.10). Perhaps you may like to find a non tournament DAG which is also transitive? Let us look at a DAG along with an action item. the direction of each edge is reversed only. a sink vertex.3. How about if we take the square of a DAG? Will the resulting graph also a DAG? How the roles of different vertices change or remain the same in the square graph? Will the resulting graph be transitive? If we draw a Reachable Relation matrix A for a DAG. But then a DAG consists of a source vertex. What happens to these vertices when we take a transform of a DAG? A source node has no in-degree but when we reverse directions it will have no out-degree so it will be transformed into a sink as shown in Concept map 9. 9. For example every DAG will have at least one source node and at least one sink node in addition to other ordinary (sometimes called intermediate nodes). the transpose of a DAG. Will the transpose of a DAG always a DAG? Why? It will obviously contain the same number of vertices and exactly the same number of edges. We shall shortly provide an efficient algorithm to solve this problem. An intermediate node has both in degree as well as out degree. A tournament DAG is always transitive (why?). how the matrix will tell us that it represents the reachable relation for a directed acyclic graph?
. Similarly the sink of the original DAG will be transformed into a source in the transpose of the DAG. On the other hand if an un-directed graph is completely connected then we have to be careful in selecting the direction of an edge in order to transform G into a DAG. It is interesting to note that the underlying undirected graph of a DAG will be an undirected graph. A source node is defined as a node with no in degree while a sink node has no out degree. Answer these questions before moving forward. If the undirected graph is a tree then we may put an arbitrary direction on each edge and the resulting directed graph will guaranteed to be a DAG.
We know that a directed graph is acyclic if and only if a DFS of the directed graph does not produce a back edge. and cross edges if any. you must have become familiar with Depth First Search of a graph while studying algorithms. In this so called topological sort.4. Perform a DFS from any vertex u in D. Cross edges 4. Back edges 5. We also know that if a directed graph is a DAG then the vertex with maximum finishing time will always be (one of) the source vertices of the DAG. Find the vertex of maximum finishing time. Start a DFS from a vertex other than u and again locate a vertex with the maximum finishing time. The source vertex will always be on the extreme left while the sink will be on the extreme right on the horizontal line (see Fig. the vertices of the DAG can be so arranged on a horizontal line such that all directed edges go from left to right.496
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If we take the transpose of a DAG then obviously the Reachable Relation matrix for the transpose will be different from that of the original graph.3. DFS spanning tree consisting of tree edges 2. back edges.1. 9.3). Problem 9.4. Finishing times of all vertices (please see Concept map 9. do the following: Problem 9. It is interesting to note that the vertices of a cyclic graph can not be so arranged on a horizontal line such that all edges go in only one direction. it should still tell us somehow that it represents a directed acyclic graph? Let us now introduce a new action item. The output of the DFS are: 1. Problem 9.2.
. identify tree edges. This visualization of a DAG may encourage us in devising a scheme to convert an un-directed graph into an acyclic directed graph. Problem Set 9.6).4. 9. More importantly it is beneficial to sort the vertices of the DAG on the basis of decreasing finishing times.5.4. the directed edges should always go only from left to right. forward edges. For the directed acyclic graph D shown in Fig. Forward edges 3.
A concept map showing the concept of a directed acyclic graph.
.3.Directed Acyclic Graphs (DAG's)
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Concept Map 9. a couple of properties and two action items which are Reachable Relation and transpose of a DAG.
Find if a given directed graph is acyclic.5. Find if a given directed graph (not necessarily a DAG) has a node u such that you can reach every vertex from u. Problem 9. Discuss the outcome?
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Figure 9.5.7.498
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Problem 9. Find the source and the sink node in a DAG.3.4.5.1. reverse the direction of the edge between vertex 2 and vertex 4.5: A directed graph D for Problem Set. Repeat the above three steps (1. & 3) on the same graph with one change. Convert a completely connected un-directed graph into a DAG. Convert a connected graph into a DAG. Problem 9. Find if a given DAG has a node u such that you can reach every other vertex of the DAG from u.6. Problem 9.4.5.2. Let us try to address the following problems: Problem Set 9. Problem 9.5.5. 2. Problem 9.5.5.4.
. We are now in a position to address a number of algorithmic issues. Problem 9. Design an efficient algorithm to solve the following: Problem 9.5. Find if a DAG has a node u such that you can reach u from every other vertex of the DAG.
Algorithm 67 used DFS as a bridge to solve a couple of problems.3 for further hints. input : A directed graph D output: Yes/No. if yes then the source and the sink vertices
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Find a Reachable Relation R for the given graph D. find the source and the sink. How about if instead of a DAG we have an ordinary directed graph which may contain cycles? There
. The source vertex can be found directly by identifying the vertex with maximum finishing time. find the source and the sink. Algorithm 67 can also be slightly modified to locate a vertex u in a special DAG from where it is possible to reach every other vertex.
You should try to solve these problems yourself before moving forward. Algorithm 68: Check if D is a DAG. Source: Vertex with maximum reachability in R?. if yes then the source and the sink vertices
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Apply DFS on graph D. If no back edge then D is a DAG. This is because the Reachable Relation Matrix R is expensive as compared to a single DFS. if all vertices of D are spanned by the BFS then the DAG possesses the desired characteristics (Does it mean that D is strongly connected?). Source vertex will be the one with maximum finishing time. Using a Reachable Relation Matrix we can easily find if there is a node u in D from where it is possible to reach every vertex of D. But then the Reachable Relation provides much more information about a graph. It is not possible to find the sink node directly. input : A directed graph D output: Yes/No. If all diagonal entries in R are zero then D is a DAG. Perhaps you should take the transpose of D and then again perform a DFS now on the transpose of D. Sink: Vertex with zero reachabiliy?. Look at Concept map 9.Directed Acyclic Graphs (DAG's) Algorithm 67: Check if D is a DAG. In fact it can answer this question even if the graph D is not acyclic. this will be a source vertex.
You might have noticed that Algorithm 68 is much more costly than Algorithm 67. First we can find the vertex with maximum finishing time. Now we can start a new BFS from this very vertex. Sink vertex: You find out.
Arrange the Hamiltonian Path as a Horizontal line. and if it does then the algorithm outputs the actual Hamiltonian Path. Perhaps in an ordinary graph.500
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may still be a node u in D from where it is possible to reach every where. We can easily design an efficient algorithm to check if such a path exists. now the vertices of the DAG can be so arranged on a horizontal line such that the source vertex will always be on the extreme left while the sink will be on the extreme right on the horizontal line (see Fig.
Algorithm 69 can be used to put appropriate directions on a completely connected un-directed graph in order to convert it into a DAG (see Fig. we should first put directions arbitrarily and then remove cycles by reversing back edges? Or put edge directions such that no cycle is created in the first place? Or why not convert the given un-directed graph into a completely connected un-directed graph and then use Algorithm 69. Put back the remaining edges of G with directions going from left to right only. It is left as an exercise for the learner to design an algorithm in order to convert an ordinary connected graph into a DAG. For a Hamiltonian path to exist in this graph there should be a directed edge
. This node u is not a conventional source vertex as it may have an in-degree. All edges of the graph will go from left to right in this arrangement otherwise it will not be a directed acyclic graph. Please note that if the undirected graph G is not completely connected then it will be hard to find a Hamiltonian Path in G (as is done in Algorithm 69.6). 9. will the maximum finishing time be able to locate this u in D? It is important for you to verify or contradict this conjecture at this very stage before moving forward. Put directions on this line going from left to right only. The first step should be to do a topological sort on the vertices of the given directed graph. But then we have to remove the extra edges that we have inserted into the original un-directed graph? So what? It is interesting to note that finding a Hamiltonian Path in a DAG is not a hard problem as it is in other graphs. How about if we perform a DFS in this directed graph. 9.10). Algorithm 69: Convert a completely connected graph into a DAG input : An undirected completely connected graph G output: A DAG D (with same number of nodes and edges)
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Find a Hamiltonian Path in G.
A directed edge between two consecutive vertices (8/9 & 1/6) is missing. The vertices of D are topologically sorted on the basis of decreasing finishing times as shown in the bottom diagram.6 shows a directed acyclic graph D. thus it is not possible to find a directed path passing through all the topologically sorted vertices. Algorithm 70: Find a Hamiltonian path in a DAG input : A directed acyclic graph D output: A Hamiltonian path in D provided it exists
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Perform a topological sort on vertices of D.Directed Acyclic Graphs (DAG's)
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(from left to right) between every two consecutive vertices. this observation confirms that D is a directed acyclic graph. The resulting start / finishing times are indicated along with each vertex in this diagram.. 9. As one can see all edges of this graph are going from left to right. It implies that a Hamiltonian path does not exist in this directed acyclic graph. We start a Depth First Search in this graph starting from a vertex designated as a start vertex in the top diagram. A path passing through all topological sorted vertices means a Hamiltonian path exists otherwise not. a Hamiltonian path in D will essentially be a path between topological sorted vertices in D (in the topological sorted order).
. Fig.
The directed graph D is acyclic thus all edges move from left to right in the bottom diagram.
.6: A depth first search is conducted on the directed acyclic graph D starting from a vertex labeled as start node as shown in the top diagram.502
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Figure 9. The start/finishing times of each vertex is also indicated here. Vertices of D are topologically sorted on the basis of decreasing finishing times as shown in the bottom diagram. A direct edge between two consecutive vertices is missing in the bottom diagram thus a Hamiltonian path is not possible in this graph.
v in D.4). the resulting graph is no longer independently connected.Strongly Connected Components
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9. Thus if we start a BFS from any vertex belonging to any strongly connected component in an independently connected graph then the search is confined to that strongly connected component. In the bottom diagram of Fig. & 2. it is in fact an efficient way of determining vertices belonging to a strongly connected
. An independently connected graph D which is not strongly connected is shown in the top of Fig. 9.4).7 and Concept map 9. It is important to appreciate that in an independently connected directed graph there are no edges among strongly connected components (see top diagram of Fig. we add an edge between component E and F . a vertex u is reachable to vertex v and the vertex v is reachable to vertex u for every pair of vertices u. Let us now consider a (related) class of directed graphs which we call independently connected directed graphs. In such a graph every vertex u in D may not be reachable to every vertex v in D but if u is reachable to v then v is reachable to u for every pair of vertices u.4
Strongly Connected Components
We know that in a strongly connected graph D.7. it can not enter into another strongly connected component. 9. 9. The concept of independently connected graphs is introduced as it is much simpler to solve the above problem in such graphs. Nodes belonging to a strongly connected component. v in D. The two sub-graphs in Fig. 9. both of these sub-graphs are strongly connected while graph D as a whole is not strongly connected. It is quite evident from this figure that graph D consists of two sub-graphs E and F .7 are in fact two strongly connected components. An independently connected graph may not be strongly connected but a strongly connected directed graph is always independently connected (see Concept map 9.7. in the top diagram there is no edge between component E and F . The number of strongly connected components in a directed graph. It is now natural to define a new concept before moving forward: a strongly connected component in a directed graph D is a maximal set of vertices of a directed graph in which vertices u and v are reachable from each other for every pair of vertices u and v in that maximal set. The problem that we intend to address in this section is to understand how we can efficiently find the following: 1.
9. In a directed graph which is neither strongly connected nor independently connected there are edges between different strongly connected components. The question
. 9. In the bottom diagram there is an edge from E to F and the graph D is neither strongly connected nor independently connected.7 If you start a BFS in this graph then the search may not be confined to only one strongly connected component. it may also enter another strongly connected component.504
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Figure 9. If some how we can remove edges connecting different strongly connected components in a directed graph (like the edge (3.7: Sub-graph E and sub-graph F are both strongly connected components. In the top diagram. 9. note the directed edge from component E to component F in the directed graph D shown in the bottom of Fig. component (it is as simple as determining nodes belonging to a connected component in an un-directed graph).7) thereby converting it into an independently connected graph then the problem of determining nodes belonging to a strongly connected component is much simplified. For example if you start a BFS from vertex 1 belonging to component E then the search will traverse vertices belonging to component E as well as F (see Fig.5) in D shown in the bottom of Fig.7) Thus the problem of determining nodes belonging to a strongly connected component becomes harder if a directed graph is neither strongly connected nor independently connected. there is no edge between the two components and graph D is independently connected.
9. The corresponding equivalence classes (that the relationship generates) are in fact the strongly connected components E and F of the independently connected graph D.7. symmetric as well as transitive.
.7 is shown by the bold 1's in any of the table shown in Fig. 9. If you carefully look at the resulting Reachable matrix you will notice that the relationship it depicts on this graph is reflexive. Once we have the Reachable matrix of an independently connected graph. 9.Strongly Connected Components is how to locate and then remove these edges?
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It is important at this learning stage to apply the Reachable transformation to a number of independently connected directed graphs and draw the corresponding Reachable Matrices. We shall certainly encourage you to draw the Reachable matrix for the independently connected graph shown at the top of Fig.8. it is not reachable to vertices belonging to other strongly connected components. it is an equivalence relationship on the vertices of the independently connected graph. it becomes trivial to pinpoint vertices belonging to a strongly connected component. In simple words it means that in an independently connected graph a vertex is reachable only to vertices belonging to the same strongly connected component. The Reachable Relation Matrix for the independently connected graph of Fig.
The Reachable relation is not symmetric.8 is again equivalent to removing the edge (3. The reachability across strongly connected components is certainly affected by taking the transpose of a directed graph. It is interesting to note that the graph D is different from its transpose but the Reachable relation of the two graphs will be exactly the same as demonstrated by bold 1's in the two tables of Table shown in Fig. It is thus no longer trivial to determine vertices in a strongly connected component in such directed graphs? If somehow we can convert it back into a symmetric relation without disturbing the connectivity inside any strongly connected component and without changing the number of strongly connected components then the problem will become much simpler? This means that some how we should be able to remove the small 1's in the Reachable Relation Matrix A of table in Fig. It is quite obvious in the tables that the reachability within a strongly connected component is not affected by the transpose transformation (see the bold 1's in the two tables in Fig. this is indicated by the small 1's in the two tables given in the figure. You might have noticed that removing those small ones in the table shown in Fig. 9. we take the transpose of a directed graph which is not independently connected then things will be different. 9.5) in the bottom diagram of Fig.7.8. 9.
. 9. For example if you draw the Reachable matrix of the graph shown in the bottom of Fig. however. 9. 9. The question is again how to do it? Let us again consider an independently connected directed D and assume that we take a transpose of D. A careful look at the two tables will provide the required answer.8). We show the same directed graph D as depicted in the bottom of Fig.7 as well as its transpose on the top of Table in Fig. although vertex 1 belongs to strongly connected component E (please see Reachable Relation Matrix A in the table shown in Fig. The Reachable Relation Matrix A of D and the Reachable Relation Matrix B of the transpose of D are also shown in this table. 9.7 thereby converting it again into an independently connected directed graph. 9. hence it is not an equivalence relationship.506
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The picture is not so rosy for a directed graph which is not independently connected. 9.8.8. you will realize the inherent complication. 9. The problem of determining strongly connected components in a directed graph is thus reduced to recovering the bold 1's (or throwing away the small 1's) in the Reachable Relation Matrices of Fig.8). If. you can reach all vertices belonging to component E as well as component F from vertex 1.8.
9. Take AND of Matrix A & B. 9.Strongly Connected Components
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For each of the graphs shown in Fig.
Algorithm 71: Find strongly connected components of graph D input : A directed graph D output: Strongly connected components of graph D
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Find Reachable Relation Matrix A of D. Find Transpose T (D) of D. Graph I is the same as the transpose of D in Fig.8 except that the edge (3.8 except for one common edge. It uses two transformations: Reachable Relation as well as the Transpose as intermediate building blocks of this algorithm. 9. 9. Graph J is almost the same as graph D in Fig. Again it is quite different from graph D in Fig.
Algorithm 71 is a straight forward algorithm to find all strongly connected components in a directed graph as already illustrated in the Table shown in Fig.5).3). 9.8 except that the edge (5. It may not be very efficient (still a polynomial time algorithm) but certainly a good start for a new learner in this field. Please note that: 1. We shall describe hints for designing a more efficient algorithm to solve a similar problem in the coming
. Find Reachable Relation Matrix B of T (D).6) in I is reversed in K thus creating a new cycle in the directed graph.8 except for one similar edge and that is (5. find the Reachable Relation Matrix and compare this matrix with the ones given in Table in Fig.8 except the connectivity inside strongly connected component E in D is altered. Graph K is almost the same as graph I except that the edge (2. Graph H is the same as graph D in Fig. 3.8.9.5) in D is reversed. 9.8 (this comparison will certainly be a useful learning experience). 9. 2.3) is reversed and another edge (6. (please see Fig.2) is added. 9. 4.8). It is very different from the transpose of D in Fig. that is (3. 9.
When you insert edges you will soon realize that in order to achieve the above mentioned objective you should be careful not to create a cycle otherwise all strongly connected components within that cycle will collapse into one strongly connected component and the above condition will be violated.
.. Call BFS on T starting from u. 9. let us describe another algorithm (Algorithm 73) which will also output a strongly connected component of F . This further implies that there will be at least one source and one sink strongly connected component. We know there are no edges between strongly connected components in an independently connected graph. You are supposed to insert extra edges between the k strongly connected components such that the new directed graph should still have as many as k strongly connected components. Again consider an independently connected directed graph D consisting of K strongly connected components. This looks very similar but there are interesting differences between the two algorithms. Graph I has two strongly connected components in spite of the extra edges (2.
Algorithm 72: Find a strongly connected components of graph D input : A directed graph D output: One Strongly Connected Component of D
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Call DFS on D. the component E was a source while the component F was a sink. 9.8. Let us for the time being restrict ourselves to find one strongly connected component of a directed graph. Please recall graph I & K in Fig. A comparison of the two will certainly be an exciting learning experience. Find Transpose T of D. This immediately suggests that strongly connected components in a directed graph are always connected in the form of a directed acyclic graph (see Concept map 9. component F was a source and E was a sink strongly connected component.
Before discussing this algorithm (Algorithm 72). While in the top right diagram of the same figure.2).6) while Graph K is reduced into one strongly connected component because of a cycle created by the edge (6. find vertex u of maximum finishing time.2). Output all vertices traversed by the BFS.9. In the top left diagram of Fig.508
Strongly Connected Directed Graphs and Tournaments
paragraphs.
. It is interesting to note that the Reachable Relation graph of a Hamiltonian. Call BFS on D starting from u. Call DFS on T : find vertex u of maximum finishing time. and strongly connected directed graph is exactly the same. A map showing the concept of a strongly connected graph.Strongly Connected Components
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Concept Map 9. a couple of properties and an action item which is the Reachable Relation. Eulerian. Output all vertices traversed by the BFS.4.
Algorithm 73: Find a strongly connected component of graph D input : A directed graph D output: One Strongly Connected component of D
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Find Transpose T of D.
If we start a traversal from this node in the graph F then the traversal will not only traverse vertices belonging to the source strongly connected component. But how to determine if a directed graph possesses this property? How about if the sink strongly connected component consists of all vertices of the directed graph? We can determine this in O(p + q) instead of O(p3 )
.9: Directed graphs for the problem set As you may have noticed that Algorithm 72 locates a vertex u of maximum finishing time after doing a DFS on D. What does that mean in terms of our newly acquired knowledge of strongly connected components? A strongly connected graph is just one strongly connected component. The source strongly connected component in D will become a sink in the transpose of D. Algorithm 73 first takes the transpose of D and then locates the vertex u of maximum finishing time. it will also enter and traverse vertices belonging to other strongly connected components. the output will thus be all vertices belonging to the source component in D as shown in Concept map 9.Strongly Connected Components
Graph H 4 E 1 2 3 5 6 F 1 2 7 E 3 5 6 F 4 Graph I
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Figure 9. We therefore take the transpose of D and then start a traversal from vertex u.5. Vertex u will belong to the source strongly connected component of directed graph D. This vertex will belong to the source component of the transpose of D. this time a traversal will be contained in the sink component of the transpose of D.5)? Please recall the algorithm in which we have tried to solve the problem of determining if a given directed graph is strongly connected. The output of this algorithm will thus be the source or the sink strongly connected component of directed graph D (see Concept map 9.
A directed graph F consists of a DAG of its strongly connected components.512
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Concept Map 9. The source strongly connected component belonging to F is the same as the sink strongly connected component of the transpose of F .
.5. Transpose of F also consists of a (different) DAG of strongly connected components.
10 determine the corresponding strongly connected components and the underlying DAG connecting the strongly connected components. The bottom left diagram of this figure is one strongly connected directed graph or component.Strongly Connected Components
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for Algorithm 63. For the two bottom graphs in Fig.
. Fig.10: A completely connected undirected graph of five vertices is shown in the top diagram.10 (top) shows a completely connected undirected graph. 9. The bottom of this figure shows two directed graph derived from the top graph by putting a directions on edges of the un-directed complete graph (such a directed graph is known as a tournament). Please note that all edges in the bottom left diagram are not directed. 9. You may put any directions on these edges (indicated by thin lines).1. Directions are added in this undirected graph in order to convert it into an acyclic directed graph shown in the bottom right. Problem Set 9. Different directions are added in the completely connected un-directed graph to convert it into a strongly connected graph shown in the bottom left. It is interesting to note that the directed graph (bottom right) is a directed acyclic graph as all edges are going from left to right (see Algorithm 69). Please solve the following:
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Figure 9. if BFS traverses all vertices of the directed graph D then D is strongly connected.6. It should read. What we need to do is to slightly modify the last line of Algorithm 72 or 73. Problem 9.6. the graph still remains a strongly connected graph.
Group all tournaments which correspond to the same DAG.7. How many non-isomorphic DAG's are possible corresponding to all possible tournaments of 5 vertices. please see Fig. 9. Is it possible to convert the top un-directed graph Fig. 9. Problem Set 9.4. Convert the completely connected un-directed graph in Fig.1. Is it possible to draw a tournament of 5 nodes which consists of 4 strongly connected components? Discuss briefly.6. Draw all these DAGs.10 into a directed graph consisting of two strongly connected components such that one strongly connected component should consist of two nodes and the other will consist of the remaining three nodes? Discuss briefly how this is possible or why it is not possible. Problem 9.7.
. it shows two tournaments (top and middle) with the same DAG and one with a different DAG (bottom).11.2. What are their respective sizes in terms of number of nodes? Please remember that each node in a DAG corresponds to a strongly connected component. Problem 9. 9. Problem 9. Find and draw all non isomorphic directed graphs derived from the completely connected un-directed graph shown in the figure.11 show three non-isomorphic directed graphs (tournaments) derived by putting directions on edges of the completely connected un-directed graph of Fig.6. Draw the corresponding DAG of strongly connected components for each of the tournament graph that you have drawn.7. 9. Problem 9. 9. The DAG of strongly connected components of the bottom graph is not isomorphic to any of the two. Problem 9.5.3. Note that the top two directed graphs are not isomorphic to each other yet the DAG's of strongly connected components are isomorphic to each other.514
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Problem 9. Find its strongly connected components and determine the DAG connecting these components.10.6. Carefully examine each of non-isomorphic tournaments that you have drawn in the last problem set. The corresponding DAG of strongly connected components of each tournament graph is drawn in front of it.10 into a directed graph such that the resulting directed graph (which will be another tournament graph) is neither strongly connected nor acyclic. Fig.6. and the respective DAG of strongly connected components of each tournament.2.
Note that all edges are not directed as shown by thin lines in the bottom graph.4
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1. the strongly connected components and the corresponding DAG will still remain the same. you may put any directions on these edges.
3.4. Why? Only because it is a tournament? Recall how a tournament is constructed from a completely connected un-directed graph and also recall the definition of a unilaterally connected directed graph?
9. Assume that we have a tournament of p nodes and we have already found a path P of length k in this tournament. We know
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Tournaments
A Panoramic Picture and a Concept Map
A directed graph obtained by putting directions on edges in a completely connected un-directed graph is known as a tournament (see Fig.2
A Hamiltonian Path in a Tournament
Every tournament has a Hamiltonian path. we can design an efficient algorithm to find that path.6). a tournament is always unilaterally connected.5.5. We shall study these and many other interesting properties of tournaments in this section. 9.5
9. Similarly a tournament is transitive if and only if it is acyclic (why?). Within these two extremes there is a lot of variety of possible tournaments consisting of various strongly connected components (see Concept map 9. 9. While playing with tournaments in previous problems.516
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Problem 9. 9. 9.7. The directed graphs that you have studied in Fig.12).12 are in fact all tournaments. Is it possible to draw a DAG of 4 strongly connected components corresponding to a tournament of size 4? Problem 9.
9.7. Derive an exact expression for the number of non isomorphic DAGs of strongly connected components of tournaments of size p.7. you might have noticed that a tournament can be strongly connected while another tournament may be a directed acyclic graph again shown in Fig. Is the DAG of strongly connected components of a tournament also a tournament? Why? Problem 9.12 . We can take any node u in T which is not part of P . Let us start with some very simple properties.5. There are several non-isomorphic tournaments possible which are not acyclic (Fig. The goal is to find a Hamiltonian path in the tournament which will be a simple path of length p − 1.13) but there is always a unique acyclic tournament for a fixed number of vertices (why?).
An un-directed and completely connected graph can be converted into a directed graph by putting directions on each edge in an arbitrary fashion. Such a directed graph is known as a Tournament.6. it possesses some very special properties as shown in this concept map.Tournaments
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Concept Map 9.
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The extended Hamiltonian path is shown in bold in the bottom diagrams.14: A directed path P of length k is shown in a tournament. Vertex u will be connected to every vertex in P. you are at liberty to put any directions in the top diagram you will always end up with an extended path P of length k+1.520
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. Any exercise to put directions so that the path length does not increase will fail as shown in the bottom diagrams. Consider any vertex u in T which is not included in P.
How about if we use a similar approach used earlier for extending the path length.
Using Algorithm 74 it is possible to find a Hamiltonian path in a tournament.5. and a cycle C of length k in T output: A cycle of length k + 1 in T
Before providing an answer to the above problem. 9.Tournaments
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that u is connected to all nodes in T (why?) as shown in Fig. We will now show that T + u
. Also assume that now we add another vertex u to T such that T +u is strongly connected and is also a tournament.
9.14 (bottom). Include u in P and output the new path P . thus it may not be strongly connected. Algorithm 75: Extend cycle length in T from k to k + 1 input : A strong Tournament T. it will not contain a Hamiltonian cycle. We now prove that if a tournament is strongly connected then it will always be Hamiltonian. Algorithm 74: Extend path length in P from k to k + 1 input : A Tournament T . you are bound to get an increase in the path length of P from k to k + 1. we first prove a simpler hypothesis: if there is a cycle C of length k in a tournament then it is possible to find a cycle of length k +1 in that tournament. See if you can find the Basic Idea and Time Complexity of Algorithm 75 as shown below. and a simple path P of length k in T output: A simple path P of length k + 1 in T
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Consider any node u in T not already in the path P .14 (top). you may put any directions on these edges. The algorithm also serves as a constructive proof that there is always a Hamiltonian path in any tournament. let us look at a related problem: Assume that we are given a Hamiltonian cycle in a tournament T of size k. 9.3
A Hamiltonian Cycle in a Strong Tournament
We know that a tournament can be acyclic. It will be an interesting (but futile) exercise to put directions such that the path length does not increase by one as is demonstrated in Fig. Before proving that general result. Try your luck and check if it is possible.
it is possible that a tournament T is not strongly connected but T + u is a strongly connected tournament.15. It is thus always possible to find a Hamiltonian cycle in T + u as shown in bold in the left diagram.15). How the above hypothesis will help us in finding a Hamiltonian cycle in a strongly connected tournament or in proving that a strongly connected tournament is Hamiltonian? How about finding a cycle of length 3 in a strong tournament. 9. It means that if T is Hamiltonian then T + u will also be Hamiltonian provided T + u is a strong tournament. This type of approach will be very similar to Algorithm 74 where we extend the path length incrementally and finally output the Hamiltonian path.
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Figure 9. 9. The vertex u will have at least one incoming edge and one outgoing edge (right). Thus it will always be possible to find a Hamiltonian cycle in T + u as shown in Fig. the vertex u will have a finite in-degree and a finite out-degree as T + u is strongly connected (see Fig. 9. In fact this special case is simple to solve as there is only one way to convert a directed acyclic tournament graph (we already
.16 before making up your mind.15: A Hamiltonian cycle in a tournament T of size k is shown. and so on until we find a Hamiltonian cycle. that is certainly not a hard problem? Then we should locate a vertex u such that the cycle length increases from 3 to 4. As you can see there is a serious complication here. A new vertex u is added to T such that T + u is a tournament and is also strongly connected. As T + u is a tournament thus u will be connected to every vertex of T. Is it feasible or not? Carefully look at Fig.522
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is Hamiltonian. how about if T is a directed acyclic tournament but T +u is a strongly connected. Then how can we show that T + u will be Hamiltonian? Let us look at the extreme.
A cycle of length 4 is shown in the bottom right. we have also considered the possibility when T is a directed acyclic graph. but there is a possibility when T is neither strongly connected nor acyclic. How to cater to that class of graph. Before proving this hypothesis let us look at its repercussions. Please note that vertex 4 has all incoming edges from vertices belonging to the cycle C while vertex 5 has all out going edges to vertices belonging to the cycle C. in fact it is more than that. both these vertices are outside the cycle C and are double circled. thus a strongly connected tournament is Hamiltonian. Let us come back to the problem of designing Algorithm 75 or proving that if a cycle C of length k exists in a strong tournament then a cycle of length k + 1 also exists in the strong tournament. it will be interesting for you to explore?
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Figure 9. We have already considered the possibility when T is strongly connected.
. this will certainly result in a Hamiltonian graph (why?). Consider vertices 4 and 5 in this figure.Tournaments
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know that it will have a Hamiltonian path inside it) into a strongly connected graph is to insert the extra node u is to connect it with the source node of the DAG and then connect the sink with node u. If it is right then the tournament will have a cycle of every possible length until we have a Hamiltonian cycle. A cycle of length 5 is shown in bottom left diagram.16: A directed cycle C of length 3 is shown (enclosed in a dotted circle) in a strongly connected tournament of size 5 (top).
Now we intend to find a cycle of length k + 1 in the same tournament. If you can not find a vertex u with the above property then you will certainly find a vertex u and a vertex v such that (a) all edges to u from every vertex of C are incoming towards u. You can find a vertex u such that it has at least one incoming and one out going edge connecting u to vertices already in the given cycle C. 9. Now it is possible to find a cycle of length k + 2 (see the bottom left of Fig.16. If (c) is not true then the tournament T will not be strongly connected. This scenario is depicted in Fig. it is possible to convert this cycle into a cycle of length k + 1 by a simple manipulation (see the bottom right of Fig. 9.
The Concept of a Rip Vertex
Let us define a new term before designing an alternate strategy to find a Hamiltonian cycle in a strong tournament. 9. You are already familiar with the concept of a cut vertex in an un-directed graph (a vertex v in an undirected connected graph G is a cut vertex provided G − v is not connected). 2. Note that if (a) or (b) is not true then we shall end up with the first possibility. 9. You can certainly find a cycle of length K +1 in this situation by extending the cycle length of C from k to k + 1.16). Let us now come to the proof of the above hypothesis. and (c) there is an edge from u to v.15. This also provides an efficient algorithm to find a cycle from length 3 to all the way to a Hamiltonian cycle of length p − 1. (b) all edges from v to every vertex of the cycle C are out going. 1. There are essentially two possibilities. This possibility is quite similar to the one illustrated in Fig.16and also Fig.524
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a cycle of length k is possible in a strongly connected tournament where k is equal to or more than 3. 9. Assume that we have found a cycle C of length k in a strongly connected tournament. This completes the constructive proof that you can extend the cycle length in a strong tournament of p nodes from 3 to p−1.17 ). Corresponding to a cut vertex in a connected un-directed graph there is a counter part concept in a (strongly) connected directed graph: a vertex v is a rip vertex (in a strongly connected directed graph D) provided D − v is not
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17: Extending a 3-cycle into a 5-cycle in a strongly connected tournament.
. It can be extended into a 5-cycle directly
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8. In the earlier algorithm we grow a cycle within the original tournament from a small size to p. We can do this step of removing a non rip vertex recursively until the original tournament is reduced to such a small sized tournament where it will be trivial to find a Hamiltonian cycle. Problem 9. Draw a strong tournament in which no vertex is a rip vertex.2. Problem 9. Now we can start inserting back the removed vertices one by one in the last removed first inserted order.8. How can you efficiently find a non rip vertex in a strong tournament?
. Now assume that we are given a strongly connected tournament T of size p.3. the Hamiltonian cycle will grow incrementally (with the graph) as shown in Fig.8. Problem Set 9.5. The resulting graph T − u will be a tournament (why?) and a strong tournament (because u was not a rip vertex) of size p − 1. In the later algorithm we remove non rip vertices from a tournament one by one until it is possible to find a Hamiltonian cycle in the reduced size tournament. Problem 9. 9.8. Problem 9. Prove that there always exists at least one non rip vertex in a strong tournament. Problem 9. It will be a learning experience to compare the working of the two algorithms that we have described to find a Hamiltonian cycle in a strong tournament. 9. It will be interesting to derive and compare the time complexities of the two algorithms. we then insert the removed vertices one by one and Hamiltonian cycle also increases incrementally with the size of the graph. It is important for you to answer the following in order to meaningfully understand the last algorithm that we have described to find a Hamiltonian cycle in a strong tournament.8.8.19.4.526
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(strongly) connected. until you find the Hamiltonian cycle in the original tournament T . Draw a strong tournament in which there are at least 3 rip vertices in a tournament of size 5 or 6 .16. Indicate which vertices are rip vertices and which vertices are not rip vertices in Fig. On the basis of these ideas it is possible to construct an alternate algorithm to find a Hamiltonian cycle in a strong tournament. We first locate and then remove a non rip vertex u.1.
A variety of non-isomorphic directed acyclic graphs are possible which will produce such a Reachable Relation matrix. Can you visualize a tournament graph where there is no rip vertex? Please see Fig. in fact it will be strongly connected.8. In a strongly connected directed graph there is a path from u to v and a path from v to u.6. If instead of one both A[i. Thus a directed graph D will be unilaterally connected if A[i. Each of the matrices. 9. We have also seen that a tournament graph is also unilaterally connected.21 corresponds to a unilaterally connected graph which is also directed acyclic (why?) assuming that all other entries in the matrix are equal to zero. That means a strongly connected graph is always a unilaterally connected graph but not vise versa. It is clear from the definition of such directed graphs that either A[i.7. This means that the Reachable Relation Matrix A for a unilaterally connected directed graph will have all 1's either on upper side of the diagonal (or on the lower side of the diagonal) as shown in Table.Unilaterally Connected Directed Graphs:
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9.8.8. j in D then D will be more than unilaterally connected. i] = 1 for every pair of vertices i. 9.8. j] = 1 or A[j.21. If instead of removing a non rip vertex in a strong tournament.8. What will be the overall time complexity of finding a Hamiltonian cycle in a strong tournament using the above algorithm.6
Unilaterally Connected Directed Graphs:
A directed graph D is unilaterally connected provided for every pair of vertices u and v. i] are equal to 1 for every pair of vertices i. shown in Table 9. It will be useful at this stage to draw a couple of such
.9. we remove a rip vertex then what will be the complication in reconstructing the Hamiltonian cycle in T ? Problem 9. j] = 1 for every pair of vertices i. Problem 9. In this section we shall discuss various properties of unilaterally connected directed graphs. j] and A[j. j in D if j > i. j in D. Let us first address the problem of how a Reachable Relation Matrix A of a unilaterally connected graph would look like. Why we insist that a non rip vertex should be removed instead of a rip vertex? Problem 9. there is a path from u to v or a path from v to u.
a couple of properties and an action item which is the Reachable Relation. A tree graph is unilaterally orient-able provided it is a path graph.
.7.Unilaterally Connected Directed Graphs:
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Concept Map 9. A map showing the concept of a unilaterally connected directed graph. An un-directed connected graph which can be converted into a unilaterally connected directed graph is known as a unilaterally orient-able graph.
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Figure 9.
. Both theses graphs are directed acyclic.22: A unilaterally connected directed graph D having minimum number of edges with all 1's in the upper side of the diagonal and 0's else where as shown in the left diagram.21: The Reachable Relation Matrix A for a unilaterally connected directed graph should have all 1's either on the upper side of the diagonal (left) or on the lower side as shown in the right table.
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9. 9. Let us summarize our observations before making a number of formal proofs. such a unilaterally connected graphs. Under such conditions one would expect that the resulting directed graph will be unilateral. v} in D either u is reachable to v or v is reachable to u. We are dealing with a unilaterally connected directed graph D. This means there will be a direct edge between every two consecutive vertices (again from left to right). 9. thus if we do a topological sort on vertices of D then the ordering of the vertices of the graph D along a horizontal line is such that all directed edges will be going from left to right as shown in Fig. as well as cyclic containing a Hamiltonian path (as before) as shown in the middle diagram of Fig.23. this would imply a Hamiltonian path in D (it will
. 9. In fact we shall prove that a directed graph is unilaterally connected if and only if it contains an open spanning walk. Assume that in addition to all 1's on the upper side of the diagonal there are some additional 1's as shown (in bold) in the top diagram of Fig. We have observed that a unilaterally connected graph contains a Hamiltonian path if it is acyclic (see Fig. We shall soon prove that a unilaterally connected directed acyclic graph D will always contain a Hamiltonian path inside D.Unilaterally Connected Directed Graphs:
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graphs with the above property and the assumption that all other entries in the matrix are zero. assume that u is on the left of v in linear ordering then there must be a direct edge from u to v otherwise the condition stated above will be violated or there will be an edge going from right to left. What is quite unexpected is that the Hamiltonian path may disappear because of the extra 1's that we have added in the Reachable Relation matrix as shown in the bottom diagram of Fig. If there are cycles in the unilaterally connected graph then there may or may not be a Hamiltonian path inside the graph (see Fig. What we have not observed until now is that whether a unilaterally connected directed graph contains a Hamiltonian path or not it will certainly contain an open spanning walk. 9.23. one having minimum number of edges (left) and the other having maximum number of edges as shown on the right of this diagram. But before that let us try to imagine what will happen if there are more 1's then are absolutely essential for a graph to be unilaterally connected. There will not be any edge going from right to left because otherwise graph D will be cyclic.23. Let us first find a constructive proof that a unilaterally connected directed acyclic graph will always contain a Hamiltonian path inside it.22. we show in Fig. 9.22.22). 9. thus for every consecutive vertex pair {u. We are dealing with a directed acyclic graph D.23).
We also know that there is a closed walk spanning all vertices inside each strongly connected component.Unilaterally Connected Directed Graphs: essentially be a path between topological sorted vertices in D). Let us first handle the hypothesis that if D is unilaterally connected then there will be an open spanning walk inside D. however. now we claim that a directed graph D is unilaterally connected if and only if it contains an open spanning walk. 9.23) and so it may not have a Hamiltonian path. thus there will be (not only a spanning walk) but a spanning path inside D. We have already observed that if D is a unilaterally connected directed acyclic graph then there will be a Hamiltonian path inside D. If.24: A unilaterally connected directed graph D having 5 strongly connected components.
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Figure 9. Let us now consider the possibility when a unilaterally connected directed graph is not directed acyclic (see Fig. D contains cycles then we can always find strongly connected components of D and we know that the graph interconnecting its strongly connected components will be a directed acyclic graph. Also note that the directed acyclic graph connecting all strongly connecting components should be a line graph otherwise D will not be a unilaterally oriented directed graph. Thus there will be a Hamiltonian path passing through all strongly connected components of D as shown in Fig.24. We can again make a topological sort on strongly connected components of D and argue that there will be an edge going from left to right between every two consecutive strongly connected components of D. Note that inside each strongly connected component there is a closed walk. 9. thus there will be an open spanning
.
We have already considered the first possibility when G is a tree and every edge of G is a bridge edge. Every edge of G is a bridge edge.
Strongly Connected Directed Graphs and Tournaments
9. ??. 3. Every edge of G is a non-bridge edge. Let us consider the second possibility when every edge of G is a non bridge edge.6. We also know that a directed acyclic unilaterally connected graph D always contain a Hamiltonian path. Now we shall consider the third possibility when some edges of G are bridge edges while some other edges are non bridges as shown in Fig.1
Unilaterally orient-able Un-directed Graphs
We know that if we put directions on edges in an un-directed acyclic connected graph G (that means a tree) then no matter what is our direction scheme the resulting graph D would be a directed acyclic graph. A tree of ordinary vertices can be unilaterally
. If you recall this is a sufficient condition for a graph to be strongly orient-able. If the graph G contains K bridge edges and if we remove all these K edges then G will be decomposed into k + 1 connected components. Consider each connected component as a single (super) vertex. Now let us find necessary and sufficient conditions for a general connected graph G (G is no longer restricted to be a tree and may contain cycles) to be unilaterally orient-able. Thus a tree graph is unilaterally orient-able if and only if it is a line graph (or a path graph). Some edges are non bridge edges while some are bridges. The only tree graph which contains a Hamiltonian path is in fact a line graph. If G is strongly orient-able then it will also be unilaterally orient-able. This would require that there should be an un-directed Hamiltonian path inside the tree graph G in the first place otherwise it would have been impossible to convert G into D with a directed Hamiltonian path inside it. 2. There are essentially three possibilities: 1. Thus when we put directions on edges in a tree graph in order to make it unilaterally connected then the resulting directed graph should also have a Hamiltonian path inside it.536 walk inside D. If we put back the bridge edges then these (super) vertices will be connected in the form of a tree (why?).
25: An un-directed connected graph G with 4 bridge edges shown in bold in the top diagram.Unilaterally Connected Directed Graphs:
537
3
1 2
4
5
3
1
2
4
5
Figure 9. If we remove all bridge edges then the un-directed graph is disconnected into 5 connected components shown in shaded circles. The connected components are always connected in the form of a tree as shown in the bottom diagram.
.
Thus each connected component is strongly orient-able (and will subsequently become a strongly connected component in the directed graph). This completes the proof that if a graph G is unilaterally orient-able then it will contain an open spanning walk.538
Strongly Connected Directed Graphs and Tournaments
orient-able if and only if the tree is a line graph. A tree of super vertices can be unilaterally orient-able provided: (1) The tree is a line graph and (2) We can find a closed spanning walk inside each connected component (why?). You may prove the converse as an exercise?
. So once again concentrate on a single connected component. there will be a closed spanning walk inside each component. no edge inside the component will be a bridge edge (why?). | 677.169 | 1 |
This introduction to the ideas and methods of linear functional analysis shows how familiar and useful concepts from finite-dimensional linear algebra can be extended or generalized to infinite-dimensional spaces. Aimed at advanced undergraduates in mathematics and physics, the book assumes a standard background of linear algebra, real analysis (including the theory of metric spaces), and Lebesgue integration, although an introductory chapter summarizes the requisite material.
The initial chapters develop the theory of infinite-dimensional normed spaces, in particular Hilbert spaces, after which the emphasis shifts to studying operators between such spaces. Functional analysis has applications to a vast range of areas of mathematics; the final chapters discuss the particularly important areas of integral and differential equations.
Further highlights of the second edition include:
a new chapter on the Hahn–Banach theorem and its applications to the theory of duality. This chapter also introduces the basic properties of projection operators on Banach spaces, and weak convergence of sequences in Banach spaces - topics that have applications to both linear and nonlinear functional analysis;
extended coverage of the uniform boundedness theorem;
plenty of exercises, with solutions provided at the back of the book.
Review:
"The authors write with a strong narrative thrust and a sensitive appreciation of the needs of the average student so that, by the final chapter, there is a real feeling of having "gotten somewhere worth getting" by a sensibly paced, clearly signposted route." Mathematical Gazette, 2000
"It is a fine book, with material well-organized and well-presented. A particularly useful feature is the material on compact operators and applications to differential equations." CHOICE magazine
"The presentation is quite elementary, and there are sufficiently many illuminating examples and exercises... this nice textbook perfectly fits the readership, i.e., undergraduate students in mathematics and physics... It may be recommended to all students who want to get in touch with the basic ideas of functional analysis and operator theory for the first time." Zentralblatt MATH
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Math For Life And Food Service
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Requiring a basic knowledge of arithmetic, this book familiarizes users with some of mathematical skills involved in the food service industry. It also focuses on the discipline and organization needed to achieve success using mathematics in everyday life. Chapter topics include a fractional, decimal, and algebra review; fractions and percents; interest: simple, compound, credit cards.; pie and bar graphs; checking accounts; price lists/requisitions/purchase orders/invoices; guest checks, tips, guestimation; pay checks, business income statement; converting; adding weights and measures; costing: menus, markups, food cost percent; recipes: yields, costing, converting; and bakers formulas. For individuals preparing for success in the food service industry—and life.
About The Author
Anneke Kleppe is a consultant and adviser at Klasse Objecten, which she founded in 1995 to train and coach companies on the use of object technology, modeling, and MDA. She was intensively involved in the development of the Unified Modeling Language (UML) and the new UML 2.0 standard. The author of several books, Anneke started a knowl...
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Title:Math For Life And Food ServiceFormat:PaperbackPublished:November 7, 2001Publisher:Pearson EducationLanguage:Englishtoday
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An Introduction to the Theory ... of Plane and Spherical
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General Investigation of the Convergence of Trigonometric Series Into Which Arbitrary Functions Are Expanded and Some New Applications of the Same
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Pre-Calculus Mathematics - Fundamentals of Functions
How to represent all fundamental math functions graphically quick overview of the fundamentals of mathematical functions. By taking and finishing this course you are going to understand what the basic functions mean and how you can represent them graphically. Therefore you are going to know how you can draw functions which are connected to the basic functions.
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038790199Elementary Algebraic Geometry (Graduate Texts in Mathematics)
This book was written to make learning introductory algebraic geometry as easy as possible. It is designed for the general first- and second-year graduate student, as well as for the nonspecialist; the only prerequisites are a one-year course in algebra and a little complex analysis. There are many examples and pictures in the book. One's sense of intuition is largely built up from exposure to concrete examples, and intuition in algebraic geometry is no exception. I have also tried to avoid too much generalization. If one under stands the core of an idea in a concrete setting, later generalizations become much more meaningful. There are exercises at the end of most sections so that the reader can test his understanding of the material. Some are routine, others are more challenging. Occasionally, easily established results used in the text have been made into exercises. And from time to time, proofs of topics not covered in the text are sketched and the reader is asked to fill in the details. Chapter I is of an introductory nature. Some of the geometry of a few specific algebraic curves is worked out, using a tactical approach that might naturally be tried by one not familiar with the general methods intro duced later in the book. Further examples in this chapter suggest other basic properties of curves. In Chapter II, we look at curves more rigorously and carefully | 677.169 | 1 |
This text provides an invaluable introduction to the mathematical tools that undergraduate economists need. The coverage is comprehensive, ranging from elementary algebra to more advanced material, whilst focusing on all the core topics that are usually taught in undergraduate courses on mathematics for economistsAll the mathematical tools that an economist needs are provided in this worldwide bestseller.
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Knut Sydsæter is an Emeritus Professor of Mathematics in the Economics Department at the University of Oslo, where he has been teaching mathematics for economists since 1965.
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Synopsis
In the traditional curriculum, students rarely study nonlinear differential equations and nonlinear systems due to the difficulty or impossibility of computing explicit solutions manually. Although the theory associated with nonlinear systems is advanced, generating a numerical solution with a computer and interpreting that solution are fairly elementary. Bringing the computer into the classroom, Ordinary Differential Equations: Applications, Models, and Computing emphasizes the use of computer software in teaching differential equations. Providing an even balance between theory, computer solution, and application, the text discusses the theorems and applications of the first-order initial value problem, including learning theory models, population growth models, epidemic models, and chemical reactions. It then examines the theory for n-th order linear differential equations and the Laplace transform and its properties, before addressing several linear differential equations with constant coefficients that arise in physical and electrical systems.
The author also presents systems of first-order differential equations as well as linear systems with constant coefficients that arise in physical systems, such as coupled spring-mass systems, pendulum systems, the path of an electron, and mixture problems. The final chapter introduces techniques for determining the behavior of solutions to systems of first-order differential equations without first finding the solutions. Designed to be independent of any particular software package, the book includes a CD-ROM with the software used to generate the solutions and graphs for the examples. The appendices contain complete instructions for running the software. A solutions manual is available for qualifying instructors | 677.169 | 1 |
Developmental Math Fall 2014
Developmental Math
Fall 2014
Math Essentials (MTE)
You spoke and we listened!
Click here for a 15 minute
narrated video of the PowerPoint.
What are Developmental Math courses?
• Developmental Math courses are prerequisites to entry-level college math
courses.
• The Virginia Placement Test (VPT) must be taken by new students to see if any
Developmental Math courses are needed. The VPT is free, and can be taken in the
Testing Center on either campus.
• Course material is divided into nine separate units, referred to as modules. The
VPT allows you to test out of modules you have already mastered.
• By dividing material into modules, students only need to complete material not
already mastered and that is required for their curriculum. This will reduce the
amount of time needed to prepare for the entry-level college math course in an
academic program.
•
A student who provides official evidence of a satisfactory mathematics score of
520 on the SAT or a mathematics score of 22 on the ACT taken within the last two
years is not required to take JTCC Developmental Math courses.
Structure of Developmental Math:
• Math Essentials (MTE) courses will be offered for Developmental Math.
Each MTE is a 4-week course that covers one module for one credit.
• Both live-lecture and video-lecture courses are offered, and both use
computer-based software to complete all assignments.
• Students are strongly encouraged to spend additional time each week in
the Math Lab, which is staffed by instructors who will answer any
questions. There is a Math Lab on the Midlo and Chester campuses.
• A maximum of four MTE courses can be taken in one semester: one each
in the first 4-week session, the second 4-week session, the third 4-week
session, and the fourth 4-week session.
• Click here for the Master Developmental Math Faster flier.
(Use Slide Show mode for link to work.)
MTE Courses
The course name corresponds to the module number. Go to the Schedule
of Classes and look under Math Essentials . Each MTE course is 1 credit.
MTE 1
MTE 2
MTE 3
MTE 4
MTE 5
MTE 6
MTE 7
MTE 8
MTE 9
Module 1 - Operations with Positive Fractions
Module 2 - Operations with Positive Decimals and Percents
Module 3 - Algebra Basics
Module 4 - First Degree Equations and Inequalities in One Variable
Module 5 - Linear Equations, Inequalities, and Systems of Linear
Equations in Two Variables
Module 6 - Exponents, Factoring, and Polynomial Equations
Module 7 - Rational Expressions and Equations
Module 8 - Rational Exponents and Radicals
Module 9 - Functions, Quadratic Equations, and Parabolas
Registering for the correct MTE course(s):
See a counselor or your advisor for help.
Step 1 Establish a "Target Class". Your Target Class is the entry-level college math
class needed for your program of study. Once you know the Target Class, find out
what modules are prerequisite.
Step 2 Decide which modules you have credit for already:
a) New and returning students: Credit from the Virginia Placement Test (VPT)
Returning Students:
a) Credit from prior MTH 2/3/4
b) Credit from a specific Pearson test
Step 3 Choose a delivery method and register for MTE courses.
a) Delivery method: choose on-campus (live or video lecture) or virtual (online).
b) Register for MTE course(s) still needed to meet the prerequisite.
Click here for MTE Advising Worksheet.
(Use Slide Show mode for link to work.)
Step 1
Establish a "Target Class" to find out which modules are prerequisite.
• A student will need Modules 1-5 before taking MTH 158.
• A student will need Modules 1-9 before taking MTH 163/166.
• If a student is not planning to transfer to a four-year college and a student's
degree program lists MTH 120 or MTH 103, then the student will only
need Modules 1-3.
Important question to ask: Does the degree/major a student will be pursuing
require a lot of math, such as engineering, science, pre-med, or business?
• If so, then the student will need Modules 1-9.
• If not, then the student will probably only need Modules 1-5. There are
some exceptions to this, so a student must check with their advisor about
the transfer college requirements.
Step 2
Decide which modules you have credit for already.
a. Credit from prior Developmental Math courses:
If you took classes prior to Spring 2012, look on MyTyler unofficial transcripts
to see if you have an 'S' in any classes below.
•
•
•
An 'S' in MTH 2 gives credit for Modules 1-3
An 'S' in MTH 3 gives credit for Modules 1-6
An 'S' in MTH 4 gives credit for Modules 1-9
Note: If you successfully passed MTH 2/3/4 more than 5 years ago, you
are strongly encouraged to take the VPT and use those scores for placement.
Step 2
b. Check the Virginia Placement Test scores
• The Virginia Placement Test (VPT) has been used for Math Placement since November
2011.
• You were given a handout after finishing the VPT that listed the Modules for which you
got credit. If you lost the handout, you must go to a counselor or your advisor to get
that information.
• If you took a placement test prior to November 2011 and have not enrolled in a
Developmental Math class, you must take the VPT and use those scores for placement.
This is free, and done at the Testing Center on either campus.
(Use Slide Show mode for link to work.)
c. Verify any modules you got credit for by taking a Pearson test
and scoring 75% or higher from Spring 2012-Summer 2014.
Step 3
Choose…
Live Lecture on campus
o
o
o
o
All students in the classroom are in the same module.
An instructor will prepare and deliver a lecture on all content.
Students will work in groups, as well as do assignments using computer-based software.
Choose section numbers 01-29. Examples: MTE 5 – 03A, MTE 5 – M19B
Video Lecture on campus
o
o
o
o
Students watch a video lecture and read the textbook to master material in a specific module.
Students work independently, doing assignments using computer-based software.
An instructor is in the classroom to help as needed.
Choose section numbers 31-99. Examples: MTE 5 – 37A, MTE 5 – M62B
Video Lecture online (virtual)
o Identical to on-campus except the instructor is available online.
o The final test must be proctored on campus.
o Choose section numbers with an N. Examples: MTE 5 – N01A, MTE 5 – N05B
Register for MTE course(s) still needed to
meet the prerequisite.
• Each MTE course is 1 credit.
• Students on Financial Aid that are taking courses during the 16-week
semester must register for all MTE courses at the beginning of the semester.
• The course name corresponds to the module number. Go to the
Schedule of Classes and look under Math Essentials .
Useful Links
Click here for a pdf of MTE courses available in Fall 2014.
(File/download/print)
Click to watch a video to see how to go to the Schedule of Classes and make
sure you register for the proper MTE in the correct four-week session.
Financial Aid Implications of MTE
Students on Financial Aid that are taking courses during the 16-week
semester must register for all MTE courses at the beginning of the semester.
Financial Aid will not pay for a BSK: Module 0 course.
Students who fail an MTE course will only "risk" 1 credit at a time.
Students who fail an MTE course will have to be closely monitored by their
math instructor and Admissions and Records to properly change their
schedule to repeat a failed course.
If students try to change their own schedule, they will end up paying out of
pocket. Do not do this!!!
The MTE swap must be done by Admissions and Records
personnel using a special process, so students should NOT
follow the swap guidelines in the Student Services brochure.
If you have any questions, please contact a
Math Department Co-Chair
Sue McBride on the Midlothian campus at [email protected]
Dr. Jodie Miller on the Chester campus at [email protected]
Links used in this PowerPoint (Use Slide Show mode for links to work)
Click here for the Master Developmental Math Faster flier. (File/download/print)
Click here for MTE Advising Worksheet. (File/download/print)
Click here for a pdf of MTE courses available in Fall 2014. (File/download/print)
Click here to watch a video on how to register on the JTCC Schedule of Classes.
Click here for a 15 minute narrated video of this PowerPoint.
For Students: Why Do Math First?
• Math abilities bring your performance in science and business courses up to a whole
new level. Why not learn something so helpful as soon as possible? Take the math
placement test and then complete your required math courses one right after the
other.
• Avoid unpleasant confrontations with reality during your "last" semester. Students
who put off math until the last possible moment may put themselves under a lot of
unnecessary stress. Failing means not graduating at this point!
• Why work so hard on the basics in your Developmental Math courses and then allow
your skills to grow rusty before taking MTH 158/163? Dive right in while your math
skills are at their peak! | 677.169 | 1 |
mathematical software
mathematical software
[¦math·ə¦mad·ə·kəl ′sȯft'wer]
(computer science)
The set of algorithms used in a computer system to solve general mathematical problems.
Mathematical software
The collection of computer programs that can solve equations or perform mathematical manipulations. The developing of mathematical equations that describe a process is called mathematical modeling. Once these equations are developed, they must be solved, and the solutions to the equations are then analyzed to determine what information they give about the process. Many discoveries have been made by studying how to solve the equations that model a process and by studying the solutions that are obtained.
Before computers, these mathematical equations were usually solved by mathematical manipulation. Frequently, new mathematical techniques had to be discovered in order to solve the equations. In other cases, only the properties of the solutions could be determined. In those cases where solutions could not be obtained, the solutions had to be approximated by using numerical calculations involving only addition, subtraction, multiplication, and division. These methods are called numerical algorithms. These algorithms are often straightforward, but they are usually tedious and require a large number of calculations, usually too many for a human to perform. There are also many cases where there are too many equations to write down. SeeAlgorithm, Numerical analysis
The advent of computers and high-level computer languages has allowed many of the tedious calculations to be performed by a machine. In the cases where there are too many equations, computer programs have been written to manipulate the equations. A numerical algorithm carried out by a computer program can then be applied to these equations to approximate their solutions. Mathematical software is usually divided into two categories: the numerical computation environment and the symbolic computation environment. However, many software packages exist that can perform both numerical and symbolic computation.
Mathematical software that does numerical computations must be accurate, fast, and robust. Accuracy depends on both the algorithm and the machine on which the software is run. Most mathematical software uses the most advanced numerical algorithms. Robustness means that the software checks to make sure that the user is inputting reasonable data, and provides information during the performance of the algorithm on the convergence of the calculated numbers to an answer. Mathematical software packages can approximate solutions to a large range of problems in mathematics, including matrix equations, nonlinear equations, ordinary and partial differential equations, integration, and optimization. Mathematical software libraries contain large collections of subroutines that can solve problems in a wide range of mathematics. These subroutines can easily be incorporated into larger programs.
Early computers were used mainly to perform numerical calculations, while the mathematical symbolic manipulations were still done by humans. Now software is available to perform these mathematical manipulations. Most of the mathematical software packages that perform symbolic manipulations can also perform numerical calculations. Software can be written in the package to perform the numerical calculations, or the calculations can be performed after the symbolic manipulations by putting numbers into the symbolic formulas. Mathematical software that is written to solve a specific problem using a numerical algorithm is usually computationally more efficient than these software environments. However, these software environments can perform almost all the commonly used numerical and symbolic mathematical manipulations. SeeSymbolic computing
Supply and training of personnel user of software for 3d modeling of building and experimentation expected to be achieved at the eli-np, analysis of stress and strain of the various components will be designed in the eli-np, analysis constructive elements kinematics of the various assemblies necessary experiments, analysis in terms of access and installation phases of various experiments and mathematical software, software for numerical simulations and software coupled multidisciplinary field of computed tomography.
Using mathematical software for engineering calculations MathCAD the graphics profiles of bending of the circular plate along the radius with rigid and pin joint support along the contour are plotted (Figs.
Among specific topics are supporting mathematical communication through technology, the design and implementation of computational modeling for learning mathematical concepts, supporting pattern exploration and algebraic reasoning through the use of spreadsheets, supporting teachers' instrumental genesis with dynamic mathematical software, and using digital resources to support elementary school teachers who are implementing the Standards.
Even though past research has indicated how the use of ICT could best be used to assist in mathematical concept development, the data shows that teachers are still not utilising dedicated mathematical software to assist students to develop understanding of mathematical concepts.
The effectiveness of solving problems in modern management systems is to a great extent determined by the quality of the specialised mathematical software of the decision-making process support systems (Vasin et al.
We plan to use two key applications as the test beds for the research uEtsunami modelling and plasma physics uEand build on ANU's expertise in advanced mathematical techniques including wavelets and high-dimensional approximations to deliver advanced mathematical software for petascale and future advanced supercomputers.
London, United Kingdom, Nov 3, 2010 - (ABN Newswire) - Intertek Group plc (LON:ITRK), a leading provider of quality and safety solutions serving a wide range of industries around the world, has signed an agreement to acquire Profitech (UK) Ltd, a technology provider of advanced mathematical software modeling services serving the oil and gas | 677.169 | 1 |
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