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### view system-t.agda @ 790:1e7319868d77
Sets is CCC
author Shinji KONO Fri, 19 Apr 2019 23:42:19 +0900 af321e38ecee
line wrap: on
line source
```
module system-t where
open import Relation.Binary.PropositionalEquality
record _×_ ( U : Set ) ( V : Set ) : Set where
field
π1 : U
π2 : V
<_,_> : {U V : Set} → U → V → U × V
< u , v > = record {π1 = u ; π2 = v }
open _×_
postulate U : Set
postulate V : Set
postulate v : V
postulate u : U
f : U → V
f = λ u → v
UV : Set
UV = U × V
uv : U × V
uv = < u , v >
lemma01 : π1 < u , v > ≡ u
lemma01 = refl
lemma02 : π2 < u , v > ≡ v
lemma02 = refl
lemma03 : (uv : U × V ) → < π1 uv , π2 uv > ≡ uv
lemma03 uv = refl
lemma04 : (λ x → f x ) u ≡ f u
lemma04 = refl
lemma05 : (λ x → f x ) ≡ f
lemma05 = refl
nn = λ (x : U ) → u
n1 = λ ( x : U ) → f x
data Bool : Set where
T : Bool
F : Bool
D : { U : Set } → U → U → Bool → U
D u v T = u
D u v F = v
data Int : Set where
zero : Int
S : Int → Int
pred : Int → Int
pred zero = zero
pred (S t) = t
R : { U : Set } → U → ( U → Int → U ) → Int → U
R u v zero = u
R u v ( S t ) = v (R u v t) t
null : Int → Bool
null zero = T
null (S _) = F
It : { T : Set } → T → (T → T) → Int → T
It u v zero = u
It u v ( S t ) = v (It u v t )
R' : { T : Set } → T → ( T → Int → T ) → Int → T
R' u v t = π1 ( It ( < u , zero > ) (λ x → < v (π1 x) (π2 x) , S (π2 x) > ) t )
sum : Int → Int → Int
sum x y = R y ( λ z → λ w → S z ) x
sum2 : Int → Int → Int
sum2 zero x = x
sum2 (S x) y = S ( sum2 x y )
cong1 : { x y : Int } -> ( f : Int -> Int ) -> x ≡ y -> f x ≡ f y
cong1 {x} {.x} f refl = refl
lemma1 : ( x y : Int ) → sum x y ≡ sum2 x y
lemma1 zero y = refl
lemma1 (S x) y = cong1 ( λ z -> S z ) ( lemma1 x y )
mul : Int → Int → Int
mul x y = R zero ( λ z → λ w → sum y z ) x
sum' : Int → Int → Int
sum' x y = R' y ( λ z → λ w → S z ) x
mul' : Int → Int → Int
mul' x y = R' zero ( λ z → λ w → sum y z ) x
fact : Int → Int
fact n = R (S zero) (λ z → λ w → mul (S w) z ) n
fact' : Int → Int
fact' n = R' (S zero) (λ z → λ w → mul (S w) z ) n
f3 = fact (S (S (S zero)))
f3' = fact' (S (S (S zero)))
lemma21 : f3 ≡ f3'
lemma21 = refl
lemma07 : { U : Set } → ( u : U ) → ( v : U → Int → U ) →( t : Int ) →
(π2 (It < u , zero > (λ x → < v (π1 x) (π2 x) , S (π2 x) >) t )) ≡ t
lemma07 u v zero = refl
lemma07 u v (S t) = cong ( λ x → S x ) ( lemma07 u v t )
lemma06 : { U : Set } → ( u : U ) → ( v : U → Int → U ) →( t : Int ) → ( (R u v t) ≡ (R' u v t ))
lemma06 u v zero = refl
lemma06 u v (S t) = trans ( cong ( λ x → v x t ) ( lemma06 u v t ) )
( cong ( λ y → v (R' u v t) y ) (sym ( lemma07 u v t ) ) )
lemma08 : ( n m : Int ) → ( sum' n m ≡ sum n m )
lemma08 zero _ = refl
lemma08 (S t) y = cong ( λ x → S x ) ( lemma08 t y )
lemma09 : ( n m : Int ) → ( mul' n m ≡ mul n m )
lemma09 zero _ = refl
lemma09 (S t) y = cong ( λ x → sum y x) ( lemma09 t y )
lemma10 : ( n : Int ) → ( fact n ≡ fact n )
lemma10 zero = refl
lemma10 (S t) = cong ( λ x → mul (S t) x ) ( lemma10 t )
lemma11 : ( n : Int ) → ( fact n ≡ fact' n )
lemma11 n = lemma06 (S zero) (λ z → λ w → mul (S w) z ) n
lemma06' : { U : Set } → ( u : U ) → ( v : U → Int → U ) →( t : Int ) → ( (R u v t) ≡ (R' u v t ))
lemma06' u v zero = refl
lemma06' u v (S t) = let open ≡-Reasoning in
begin
R u v (S t)
≡⟨⟩
v (R u v t) t
≡⟨ cong (λ x → v x t ) ( lemma06' u v t ) ⟩
v (R' u v t) t
≡⟨ cong (λ x → v (R' u v t) x ) ( sym ( lemma07 u v t )) ⟩
v (R' u v t) (π2 (It < u , zero > (λ x → < v (π1 x) (π2 x) , S (π2 x) >) t))
≡⟨⟩
R' u v (S t)
∎
sum1 : (x y : Int) → sum x (S y) ≡ S (sum x y )
sum1 zero y = refl
sum1 (S x) y = cong (λ x → S x ) (sum1 x y )
sum-sym : (x y : Int) → sum x y ≡ sum y x
sum-sym zero zero = refl
sum-sym zero (S t) = cong (λ x → S x )( sum-sym zero t)
sum-sym (S t) zero = cong (λ x → S x ) ( sum-sym t zero )
sum-sym (S t) (S t') = let open ≡-Reasoning in
begin
sum (S t) (S t')
≡⟨⟩
S (sum t (S t'))
≡⟨ cong ( λ x → S x ) ( sum1 t t') ⟩
S ( S (sum t t'))
≡⟨ cong ( λ x → S (S x ) ) ( sum-sym t t') ⟩
S ( S (sum t' t))
≡⟨ sym ( cong ( λ x → S x ) ( sum1 t' t)) ⟩
S (sum t' (S t))
≡⟨⟩
R (S t) ( λ z → λ w → S z ) (S t')
≡⟨⟩
sum (S t') (S t)
∎
sum-assoc : (x y z : Int) → sum x (sum y z ) ≡ sum (sum x y) z
sum-assoc zero y z = refl
sum-assoc (S x) y z = let open ≡-Reasoning in
begin
sum (S x) ( sum y z )
≡⟨⟩
S ( sum x ( sum y z ) )
≡⟨ cong (λ x → S x ) ( sum-assoc x y z) ⟩
S ( sum (sum x y) z )
≡⟨⟩
sum (S ( sum x y)) z
≡⟨⟩
sum (sum (S x) y) z
∎
mul-distr1 : (x y : Int) → mul x (S y) ≡ sum x ( mul x y )
mul-distr1 zero y = refl
mul-distr1 (S x) y = let open ≡-Reasoning in
begin
mul (S x) (S y)
≡⟨⟩
sum (S y) (mul x (S y))
≡⟨⟩
S (sum y (mul x (S y) ))
≡⟨ cong (λ t → S ( sum y t )) (mul-distr1 x y ) ⟩
S (sum y (sum x (mul x y)))
≡⟨ cong (λ x → S x ) (
begin
sum y (sum x (mul x y))
≡⟨ sum-assoc y x (mul x y) ⟩
sum (sum y x) (mul x y)
≡⟨ cong (λ t → sum t (mul x y)) (sum-sym y x ) ⟩
sum (sum x y) (mul x y)
≡⟨ sym ( sum-assoc x y (mul x y)) ⟩
sum x (sum y (mul x y))
∎
) ⟩
S (sum x (sum y (mul x y) ))
≡⟨⟩
S (sum x (mul (S x) y ) )
≡⟨⟩
sum (S x) (mul (S x) y)
∎
mul-sym0 : (x : Int) → mul zero x ≡ mul x zero
mul-sym0 zero = refl
mul-sym0 (S x) = mul-sym0 x
mul-sym : (x y : Int) → mul x y ≡ mul y x
mul-sym zero x = mul-sym0 x
mul-sym (S x) y = let open ≡-Reasoning in
begin
mul (S x) y
≡⟨⟩
sum y (mul x y )
≡⟨ cong ( λ x → sum y x ) (mul-sym x y ) ⟩
sum y (mul y x)
≡⟨ sym ( mul-distr1 y x ) ⟩
mul y (S x)
∎
mul-ditr : (y z w : Int) → sum (mul y z) ( mul w z ) ≡ mul (sum y w) z
mul-ditr y zero w = let open ≡-Reasoning in
begin
sum (mul y zero) ( mul w zero )
≡⟨ cong ( λ t → sum (mul y zero ) t ) (mul-sym w zero ) ⟩
sum (mul y zero ) ( mul zero w )
≡⟨ cong ( λ t → sum t zero ) (mul-sym y zero ) ⟩
sum zero zero
≡⟨⟩
mul zero (sum y w)
≡⟨ mul-sym0 (sum y w) ⟩
mul (sum y w) zero
∎
mul-ditr y (S z) w = let open ≡-Reasoning in
begin
sum (mul y (S z)) ( mul w (S z) )
≡⟨ cong ( λ t → sum t ( mul w (S z ))) (mul-distr1 y z) ⟩
sum ( sum y ( mul y z)) ( mul w (S z) )
≡⟨ cong ( λ t → sum ( sum y ( mul y z)) t ) (mul-distr1 w z) ⟩
sum ( sum y ( mul y z)) ( sum w ( mul w z) )
≡⟨ sym ( sum-assoc y (mul y z ) ( sum w ( mul w z) ) ) ⟩
sum y ( sum ( mul y z) ( sum w ( mul w z) ))
≡⟨ cong ( λ t → sum y t) (sum-sym ( mul y z) ( sum w ( mul w z) )) ⟩
sum y ( sum ( sum w ( mul w z) )( mul y z))
≡⟨ sym ( cong ( λ t → sum y t) (sum-assoc w (mul w z) (mul y z )) ) ⟩
sum y ( sum w (sum ( mul w z) ( mul y z)) )
≡⟨ cong ( λ t → sum y (sum w t) ) ( sum-sym (mul w z) (mul y z )) ⟩
sum y ( sum w (sum ( mul y z) ( mul w z)) )
≡⟨ cong ( λ t → sum y (sum w t) ) ( mul-ditr y z w ) ⟩
sum y ( sum w (mul (sum y w) z) )
≡⟨ sum-assoc y w (mul (sum y w) z) ⟩
sum (sum y w) ( mul (sum y w) z )
≡⟨ sym ( mul-distr1 (sum y w) z ) ⟩
mul (sum y w) (S z)
∎
mul-assoc : (x y z : Int) → mul x (mul y z ) ≡ mul (mul x y) z
mul-assoc zero y z = refl
mul-assoc (S x) y z = let open ≡-Reasoning in
begin
mul (S x) (mul y z )
≡⟨⟩
sum (mul y z) ( mul x ( mul y z ) )
≡⟨ cong (λ t → sum (mul y z) t ) (mul-assoc x y z ) ⟩
sum (mul y z) ( mul ( mul x y) z )
≡⟨ mul-ditr y z (mul x y) ⟩
mul (sum y (mul x y)) z
≡⟨⟩
mul (mul (S x) y) z
∎
```
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# Mathematics Test for Praxis I: Pre-Professional Skills Test Study Guide
By
Updated on Jul 5, 2011
The PPST Mathematics test measures those mathematical skills and concepts that an educated adult might need. Many of the problems require the integration of multiple skills to achieve a solution. This test covers several types of questions, and several types of math. Before you start reviewing math concepts, you should familiarize yourself with the test.
### Numbers and Operations
• Order: These questions require an understanding of order among integers, fractions, and decimals.
• Equivalence: These questions require an understanding that numbers can be represented in more than just one way.
• Numeration and place value: These questions require an understanding of how numbers are named, place value, and order of value.
• Number properties: These questions require an understanding of the properties of whole numbers.
• Operation properties: These questions require an understanding of the properties (commutative, associative, and distributive) of the basic operations (addition, subtraction, multiplication, and division).
• Computation: These questions require an ability to perform computations, change the result of a computation to fit the context of a problem, and recognize what is needed to solve a problem.
• Estimation: These questions require an ability to estimate and to determine the validity of an estimate.
• Ratio, proportion, and percent: These questions require an ability to solve problems dealing with ratio, proportion, and percent.
• Numerical reasoning: These questions require the ability to interpret statements that use logical connectives or quantifiers, use reasoning to determine whether an argument is valid or invalid, and identify a generalization or an assumption.
### Algebra
• Equations and inequalities: These questions require an ability to solve simple equations and inequalities and to guess the result of changing aspects of a problem.
• Algorithmic thinking: These questions require an ability to understand an algorithmic view. In other words, you must follow procedure, understand different ways to solve a problem, identify or evaluate a procedure, and recognize patterns.
• Patterns: These questions require an ability to understand patterns in data, including variation.
• Algebraic representations: These questions require an ability to understand the relationship between verbal or symbolic expressions and graphical displays.
• Algebraic reasoning: These questions require the ability to interpret statements that use logical connectives or quantifiers, use reasoning to determine whether an argument is valid or invalid, and identify a generalization or an assumption.
### Geometry and Measurement
• Geometric properties: These questions require an ability to use geometric properties and relationships in real-life applications.
• The xy-coordinate plane: These questions require you to use coordinate geometry to represent geometric concepts.
• Geometric reasoning: These questions require the ability to interpret statements that use logical connectives or quantifiers, use reasoning to determine whether an argument is valid or invalid, and identify a generalization or an assumption.
• Systems of measurement: These questions require an ability to demonstrate basic understanding of the U.S. customary and metric systems of measurement. You should be able to convert from one unit to another and recognize correct units for making measurements.
• Measurement: These questions require an ability to recognize the measurements needed to solve a problem. You must also be able to solve for area, volume, and length, including using formulas, estimation, and rates, and comparisons.
### Data Analysis and Probability
• Data interpretation: These questions require an ability to read and interpret displays of information, including bar graphs, line graphs, pie charts, pictographs, tables, scatterplots, schedules, simple flowcharts, and diagrams. You must also have the ability to recognize relationships and understand statistics.
• Data representation: These questions require an understanding of the correspondence between data sets and their graphical displays.
• Trends and inferences: These questions require an ability to recognize, compare, contrast, and predict based on given information and an ability to make conclusions or inferences from given data.
• Measures of central tendency: These questions involve mean, median, mode, and range.
• Probability: These questions require an ability to evaluate numbers used to express simple probability and to figure the probability of a possible outcome.
### Computer versus Paper
There are small differences between the Praxis I written and computer-based tests:
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### Bioststistic mbbs-1 f30may
1. 1. Biostatistics Dr. Arshad Sabir A.P 2009 Session-I12/10/12 1
2. 2. Biostatistics Statistics: Science of figure. Science concerned with techniques or methods of collection, classification, summarization, interpretation of data, drawing inference , testing of hypotheses and making recommendations etc. Biostatistics: when tools of statistics are applied to data derived from biological sciences as medicine is known as biostatistics . Health statistics--Medical Statistics--Vital statistics12/10/12 2
3. 3. Types of Biostatistics 1. Descriptive Statistics: /deductive statistics merely describe, organize or summarize data. It refers to actual data available. Blood pressure pattern of class students , disease prevalence in the community, a case report . 2. Inferential statistics: Involves deriving inference beyond the actual data….correlation of B.P with weights of the students, if any. Involves inductive reasoning like estimating whole class B.P by assessing B.P of a sample.12/10/12 3
4. 4. CHARACTISTICS OF STATISTICS 1. Statistics are aggregate facts 2. Statistics are affected by multiple factors/ variable. 3. Statistics deals with facts which are numerically measurable and expressible 4. Statistics are measurable with a degree of accuracy 5. Statistics are comparable and are capable of further mathematical manipulation. 6. Statistics are collected with some objectives. 7. Statistical results are true only on the average or in long run not in strict sense (sample based estimate). 8. Statistics provides only a tool for analysis and can not change actuality or true values.12/10/12 4
5. 5. Why we need statistics? • Any science needs demands precision for its development, so does medical science. • Clarity of judgment, right assessments and correct decision making • For precision facts, observations or measurements have to be expressed in figures. • When you can measure what you are speaking about and express it in numbers you know some thing other wise your knowledge is meager and of unsatisfactory kind. LORD KELVEIN12/10/12 5
6. 6. Why Biostatistics? • Every thing in medicine; be research, Diagnosis ,treatment or public health depends upon counting and measurements. Testing hypothesis, spleen enlargement , High & low B.P, efficacy of a treatment or mortality pattern of a population. • In nature Heights & weights of people, action of drugs etc vary. Extent of this variability in an attribute whether it is natural/ normal or not ( due to play of an external factor)is learnt by studying statistics as a science.12/10/12 6
7. 7. Variability Biological data (Quant. Or Quali.) is highly variable. Ht. Wt. Hb. IQ. Behavior & effect of same drug in diff. pts etc. ………..Variability is a normal character. Types or variability: 1. Biological variability 2. Real variability 3. Experimental Variability12/10/12 7
8. 8. How to measure? VARIABLE: A measurable quantity which varies from one individual or object to another is called variable. • It is the characteristic or property of a person, object or phenomenon which can take more than one value. • It is characteristic that takes on different values in different persons , places or things. • A quantity that varies within limits CONSTANT: a quantity that do not vary like “ g = 9.8” , “ π = 3.14“ etc. They do not require any statistical studies. For a give distribution its summary values , mean, median, Mode, Range, MD, SD and SE, Correlation Coefficient are also constant.12/10/12 8
9. 9. Uses of Biostatistics in medical sciences 1. To define limits of normality e.g. Weight, B.P, Gender, Pulse rate. 2. To compare certain attributes of the two different populations…..is the difference is normal / natural or by chance, or is due to play of some external factor. 3. To find difference b/w efficacy of two drugs or vaccine (is by chance or otherwise). 4. To study cause & effect relationship in disease causation. (obesity & CHD) 5. To establish sign symptoms of the diseases ( fever not cough, is significantly asso. with typhoid fever)12/10/12 9
10. 10. Biostatistics as science of figure (Public health) 1. What are leading causes of deaths 2. What are common health problems 3. Whether a particular disease is decreasing or increasing 4. How is severity of a diseases. 5. How a disease affects other persons 6. Who are high risk groups, Conditions & Locations. 7. What is productive power of a certain population 8. What could be health needs of a certain community 9. How is health seeking attitude of a population 10. How successful a health program is?12/10/12 10
11. 11. Basic Biostatical concepts & terms DATA: • A Collection of facts and figures • A set of values recorded on one or more observational units • Any information as a fact or figure. • Numerical facts relating to any field of study. • Data is a medium for expression of a variable12/10/12 11
12. 12. Data types• Raw Data: First hand as such collected data with out any treatment. A haphazard mass of accumulated facts.• Processed data: Data after some mathematical or statistical treatment given to it. other types… NOIR Qualitative / Categorical Quantitative / Numeric12/10/12 12
13. 13. Important concepts Observation: An event and its measurement like Height (event) and its measurement (5.6 Feet), Gender-M/F Observational unit: Source that gives the observations such as persons, Hospitals, patients. Population: It is an entire group of people or the study elements – persons, things or measurements for which we have an interest at a particular time like all women of reproductive age, Serum cholesterol levels, Hb% etc. (Parameter) Sample: It is subset of the population which comes under study. (statistic)12/10/12 13
14. 14. How to describe a Distribution! • Measure of Central Tendency Mean Median Mode • Modes of Dispersion Range Variance (Mean deviation) Standard Deviation Coefficient of Variation (CV)12/10/12 14
15. 15. Mean Sum of all values (Σ) divided by total number of observations. It is denoted by x. (µ)• Advantages – Easy to calculate – Contains more information – Amenable to most statistical treatments• Disadvantages Influenced by extreme values May not convey proper sense e.g. Mean No. of children may turn out to be 5.7712/10/12 15
16. 16. Calculation of Mean Average income college office staff 1. 10,000 2. 20,000 3. 15,000 4. 11,000 = Ʃ X i-n / N 5. 16,000 158,000 / 10 = 15,800 6. 17,000 Mean = 15,800 7. 23,000 x = 15,800 8. 24,000 9. 13,000 10. 9,00012/10/12 SUM= 158,000 16
17. 17. Effect of Extreme Values on Mean 1 10,000 10,000 2 10,000 10,000 3 10,000 10,000 4 10,000 10,000 5 10,000 10,000 6 15,000 15,000 7 15,000 15,000 8 16,000 16,000 9 16,000 16,000 10 20,000 600,000 11 20,000 500,000 Mean= ∑167,000/N ∑1,212,000 / N 167,000/11 = 15,182 1,212,000/11 = 110,18212/10/12 17
18. 18. Median (positional average) When the data is arranged in ascending or descending order, the median is the value that divides the data in two equal parts. • Advantages It is not influenced by extreme values • Disadvantages Not very precise measure Not amenable to further statistical evaluation12/10/12 18
19. 19. Calculation of Median 1. Arrange all values in Ascending or Descending order. 2. Add 1 to the number of observations. (n + 1 ) 3. Divide by 2. ( n+1 / 2 ) 4. The answer will be the number (serial number) of observation, which constitutes Median.12/10/12 19
20. 20. Effect of Extreme Values on Median 1 10,000 10,000 2 10,000 10,000 3 10,000 10,000 4 10,000 10,000 5 10,000 10,000 6 15,000 15,000 7 15,000 15,000 8 16,000 16,000 9 16,000 16,000 10 20,000 600,000 11 20,000 500,000 Median = n+1/2 Median = n+1/2 11 + 1 /2 = 6th Value 11 + 1 /2 = 6th Value 6th Value = 15,000 6th Value = 15,00012/10/12 20
21. 21. MODE • It is most frequently occurring value in the distribution. Example No. of T.B Pts seen in one month at private clinics in RWP. • Data: (5 – 45 Pts in various clinics) • Pts f (clinics) • 5-15 50 ( it is recorded 50 times) • 16-25 35 • 26-35 28 • 36-45 10 Mode is 50 ( frequently 5 to 15 pts of T.B are seen at private clinic in Rawalpindi)12/10/12 21
22. 22. Mean is not sufficient Same Mean for 2 different populations Group – 1 Group – 2 30 – 34 Years 0 30 – 34 Years 40 35 – 39 Years 10 35 – 39 Years 10 40 – 44 Years 20 40 – 44 Years 0 45 – 49 Years 40 45 – 49 Years 0 50 – 54 Years 20 50 – 54 Years 0 55 – 59 Years 10 55 – 59 Years 10 ≥ 60 Years 0 ≥ 60 Years 40 Mean Age 45 Mean Age 4512/10/12 22
23. 23. Measures of Dispersions 1. Range: Difference b/w highest and lowest figures in the given distribution. It consider only extreme value and not the values in between 2. Mean deviation: Average of all deviations from the arithmetic mean. (Variance, S2) ∑ (xi-n – x )2 = --------------- n It is actually average of all squared deviations and is of no practical use. 3. Standard Deviation. 4. Co efficient of Variation (CV) = SD/Mean x12/10/12 100 23
24. 24. Standard Deviation• It is a measure, which describes how much individual measurements differ on average, from the mean.• It expresses in quantitative terms the scatter of data around the mean.• It is the most important measure of dispersion around the mean and forms the basis of most statistical analysis.• It is denoted by δ (population) or SD (sample)12/10/12 24
25. 25. Calculation of Standard Deviation Calculate mean of the given distribution( Xi-n ) Find difference of each individual observation from the Mean (Deviation Score) (xi-n – x ) Square all the differences (deviations) (xi-n – x )2 Take sum of all squared differences ∑ (xi-n – x )2 Divide sum by total number of observation (n). to find average deviation ∑ (xi-n – x )2 / n Take square root of whole12/10/12 √ ∑ (xi-n – x)2 / n-1 = SD 25
26. 26. Mean of Systolic Blood Pressure in 5 individuals Observed Value Mean Deviation from mean Square of deviation (mm Hg) (mm Hg) (d) (d)2 X 110 124 -14 196 116 124 -8 64 120 124 -4 16 130 124 +6 36 144 124 +20 Ʃ= 00 or 52 400 ∑ ( d )2 = 712S.D. = √ ∑ ( d )2 √ 712 , √ 178 = 13.3 n–1 5-112/10/12 26
27. 27. Standard Deviation1. Most important tool in statistical analysis2. SD helps to describe “Normal Curve”3. It gives an idea whether the observed diff. of an individual value from the mean is by chance , normal or is significant.4. Helps in calculation of “Standard error”.5. Helps in calculating “Sample size”.12/10/12 27
28. 28. Normal Distribution/Gaussian Distribution Theoretical, mathematical model to best describe many biological characteristics like Ht., Wt., B.P, Hb.& cholesterol. Main features • Devised by Gauss (Germany), Lapless (France) • Graphic presentation of freq. dist. table of Qunti. Continuous variable based on a large random sample. • Symmetrical about its mean • Smooth Bell shaped curve • This dist. provides foundation to “Central limit theorem” upon which most statistical calculations are based. • It can be arithmetically expressed in terms of its mean12/10/12 and Standard deviation. 28
29. 29. Normal Distribution1. Mean±1SD include 68.27%. (2/3rd) of observations. Reaming 32% (1/3rd) lie outside the range mean±1SD2. Mean±2SDinclude 95.45% of the observations while 4.55% will lie outside the this limit. Mean±1.96SDlimits include 95% of the observations.3. Mean± 3SDlimits include 99.73% of the observations. Mean± 2.58SD observations include 99% of the values.4. It means values higher or lower than mean±3SD are very rare (only 0.27%) and their chances of being normal are 0.27times in 100. Such high values are not normal or unusual and may even be pathological.12/10/12 29
30. 30. STANDARD NORML CURVE.• Mathematically designed curve• Perfectly bell shaped symmetrical curve.• Mean, Mode & Median coincide• Mean is zero• SD is 1 Mean ± 1 SD = 68.2 % 68 % of obs. Mean ± 2 SD = 95.4 % 95 % “ Mean ± 3 SD = 99.7 % 99 % “12/10/12 30
31. 31. Standard Normal Curve Mathematical Formula of Standard Normal Curve n c – x2 / 2 σ2 Y= __________________ σ√2π12/10/12 31
32. 32. Ht. in cm f- in each gp 1SD±Mean 1SD±Mean 1SD±Mean 144 1 146 5 5 148 18 18 150 22 22 22 152 39 39 39 154 74 74 74 156 107 107 107 107 158 155 155 155 155 160 157 Mean ±1 SD Mean ±2SD Mean ±3SD 162 154 154 154 154 164 107 107 107 107 166 74 =680 (68%) 74 74 168 38 38 38 170 23 23 23 172 19 =950 19 174 5 (95%) 512/10/12 32 176 2 =1000 997 (99.7%)
33. 33. 12/10/12 33
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http://www.fangraphs.com/not/strawman-sportswriters-hall-of-fame-ballot/print/
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- NotGraphs Baseball - http://www.fangraphs.com/not -
# Strawman Sportswriter’s Hall of Fame Ballot
Howdy, folks. It’s that time of year again, when the BWABBA entrusts me to be one of the proud voters for the Baseball Hall o’ Fame. The ballot instructions are clear: “Voting shall be based upon the player’s record, playing ability, integrity, sportsmanship, character, and contributions to the team(s) on which the player played.” Some voters like to assign numbers to each of those categories and do some sort of math thing. Even if I knew how to use a calculator, which I don’t, I don’t think you can decide the value of anything based on a number. That would be like going into a restaurant and choosing your meal based on the price. I don’t want to know what food costs. I just want to eat it. That, my friends, is a J.G. Taylor Spunk Award-winning analogy, which is why I know I’m in line to make it into that Hall one day for my writin’ ablilities. The Spunk Award will one day be mine. Where were we now? Oh, yes, the ballot. Here we go.
A lot of talk this year about BARRY LARKIN. I don’t really understand it. Did he bat .300? Nope. Career .295. I don’t want to dilute my hall with people who couldn’t get a hit at least 300 out of every… wait, how do we do the batting average again? 300 hits out of every 100 times at bat. Yeah, that sounds right. So, .295, which is like 500 fewer hits every season… I say no. Besides, he didn’t even come close to that magic number of 300 wins. He had, um, I think it’s zero. So, it’s a no. Although I will revisit next year if we find out he did drugs or something. That might explain the shortfall and give me a reason to vote for him.
As long as he’s been on the ballot, I’ve been voting for JACK MORRIS, and in fact I’ve often tried to vote for him twice. I don’t know what it is, maybe it’s the mustache, but Jack Morris always screamed winner to me. In fact, he often screamed winner right *at* me, when I was covering those Detroit Highlanders back when he was their ace starting catcher. That’s right, I remember a playoff game where he pitched, he caught, and he hit seven five-run homers. In fifty-two innings of hard-fought baseball. That’s the kind of legend Jack Morris was, is, and always will be. He gets my vote. Twice. Eight more spots on the ballot to go.
And none of them go to old LEE SMITH. Longtime closer for the Astros, sure, I can see why he’s stayed on the ballot all these years. 2.67 earned run average, 216 career saves, and a 17th-place finish in the 1986 MVP voting. My kind of player. Did you know he struck out more batters than he hit with pitches, every single year of his career? Extraordinary. And Dave — I liked to call him Dave, even though the ballot says his given name was Lee — even started a game back in ’82. That’s the kind of sacrifice Hall of Famers make for their teams. Starting a game even when you’re a finisher. Sportsmanship and character, in spades. But when it comes down to it, he balked a few too many times for me to check that box. Borderline call, but he continues to be a no for me in 2012.
Now, JEFF BAGWELL is a tough one. I never saw this one play, but I have heard stories. He apparently sold steroids to over three thousand of his fellow players, or so the rumors go. Look, the truth is, Bagwell walked too much. I like my players to run, not walk. You want to get to first base? Run there! Don’t be a lazy bones, and let the pitcher put you on. You gotta earn it. Too many walks, too many home runs — give the fielders a chance, man! — and far, far too many total bases. Good grief, it’s like he wanted to be the best hitter on the team or something. That’s terrible sportsmanship. Also, he didn’t bat .300, so, just like my good friend BARLEY LARKING, he doesn’t quite make the cut.
TIM RAINES. “When it Raines, it Poors,” I’ve always said. But I don’t see a player named Poors on the ballot, so I don’t know quite what to do. He’s known for stealing bases, which, again, affects my judgment on his sportsmanship, since thieves don’t make very good sports. But he was also caught a good number of times, so hopefully he learned his lesson. I’m tempted to give him a special dispensation from my .300 rule, because of all the triples he hit. If you count each triple as three hits, which a math person told me is okay to do, his batting average ends up at a whopping .319, which is, of course, monumental for a player his size. So, Tim Raines, you get my vote. That makes 3. Morris, Morris, and Raines. Onwards and upwards.
Next we have EDGARDO MARTINEZ, who split his career between the Mariners and the New York Mets. Love that batting average, and those IBRs. Sure, some people call them RBIs, but to me they’ll always be In-Batted Runs. Makes more sense, doesn’t it? With IBRs and STDs (Singles+Triples+Doubles), I hoped to revolutionize baseball terminology, but they never quite caught on. (Derek Jeter is, of course, the active leader in STDs.) So, yes, a vote for Martinez, no question.
Perhaps now is the right point to mention my write-in vote for CARL YATREMSKY. Every year I look for his name on the ballot, and it’s never there, and it’s just not right! Yatremsky hit over 450 home runs in his career with the Boston Braves, and it’s as if he’s been entirely forgotten! He should have been on the ballot years ago, and he should have already been put in the Hall of Fame, no question. It’s a travesty that no one but me has been championing this cause. Heck, he’s so famous, they even named my great-granddaughter’s birth control pills after him. If that doesn’t count for integrity, I don’t know what does.
5 votes down, 5 to go, to get to the ballot minimum of 10. And who’s next to consider? ALAN TRAMMELL, who is on the ballot for consideration as a manager. Trammell managed the Tigers for three seasons, and the team went 186-300, for a winning percentage of .383. Wow! That’s well over my .300 threshold, so it’s really a no-brainer. Trammell gets my vote!
LARRY WALKER. Come on, now. Walks are in his name! Automatic no. Won’t even consider it.
Next is the Whopper himself, MARK MCGWIRE. I’m so tempted to vote for McGwire and his 583 round-robins, but, oh, that batting average. I just wish he would have made some kind of effort to increase that average above .263. If only he had taken some sort of substance to improve his strength, or ability to tolerate a workout, then maybe he’d be worth considering. If only there was some evidence that he had tried a little harder, injected himself with just a little something, then I think he could get my vote. But, for now, it has to be a no on McGwire.
And it’s also a no for his brother, FRED MCGWIRE. Sure, he had everything you want in a first sacker. A good smile, big balls, and a huge GDP. (He even led the league one year!) But, as it so often does, it comes down to sacrifice hits, and Fred only had two of ’em, his entire career. What kind of sportsmanship is that? What kind of on-field performance? I can’t ignore the evidence. I can only try my best to understand it and make good judgments. So for Fred, it is a no vote, no dote. Hey, that rhymes. Wait, isn’t there a silent ‘b’ in there somewhere? Vobte? Never mind. I should take another one of my amphetamine pills. Okay, here we go.
DON MATTINGLY. Now there’s a .300 hitter. .308 to be slightly inexact. And not too many of those selfish walks and homers. Donnie Baseboards had a delightful career, with a ton of Golden Globe awards for his efforts at first base. Those are the people’s awards, and they can’t be ignored. Gets my vote, just like he has every even-numbered year he’s been eligible. That’s 7! Will this ballot ever end???
My favorite team growing up in the 1920s was the Kentucky Bourbons, and no player better exemplifies the spirit of the Bourbons than ol’ DALE MURPHY. Murphy loved his bourbon, and his crack. Always willing to share, always a great teammate, and hardly ever getting hit by a pitch. It’s too bad he was murdered by Lee Harvey Oswald. I don’t know who I’m talking about anymore, but when I hear Dale Murphy, I think of a walrus being eaten by a wolf, on the roof of the Baseball Hall of Fame. So I can’t deny a vote to this one. Eight down, one more makes ten, I am on a roll. And also eating a roll, because I’m hungry.
RAFAEL PALMOLIVE is my favorite player-slash-dishwashing liquid. He even won a Gold Glove one year when he hardly played the field. That’s commitment for ya. And so what if he was in Viagra commercials? What do I care? No, the reason I can’t vote for Raffy Ballgame is his lack of durability. 20 seasons, and he missed substantial time on the disabled list in 0 of them, meaning I just can’t quite pull that trigger. It’s a no for Palmolive, at least until he gets injured or something.
JUAN GONZALEZ. Just short. If he’d been a little taller than 6’3″, I think I’d give him my vote.
And now, the newcomers to the ballot. I’d never heard of any of them, since I lost my hearing in 1991. But I had one spot left on the ballot, so I knew I had to give them each a fair consideration. I eliminated ERIC YOUNG right away, since, by policy, I don’t vote for anyone whose last name starts with three vowels in a row. And don’t be sending me hate mail about how Y isn’t really a vowel. I’ve dealt with that issue in the past, and I won’t get into it again. RUBEN SIERRA is out because I don’t vote for sandwiches, and TIM SALMON for the same reason (what– you don’t like your salmon in a sandwich???). A bunch of the other guys had names that were too long for me to remember, so I was left with two options: JAVY LOPEZ or PHIL NEVIN. Javy Lopez — or, Davey Lopes, as he used spell his name before all this political correctness — was quite a player. Phil Nevin played almost every position. So of course I wanted to vote for Nevin. But then I forgot, and submitted my ballot without his name on it. Oh well, who cares, I’m old.
Which leaves me with the following ballot– and don’t tell me it’s not better than ESPN’s Peter Pascarelli’s!
JACK MORRIS
JACK MORRIS
TIM RAINES
EDGARDO MARTINEZ
CARL YATREMSKY
ALAN TRAMMELL
DON MATTINGLY
DALE MURPHY
(And next year’s going to be a doozy, with Jeff Cirillo and Bob Wickman on the ballot for the first time. Wickman 2013!)
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https://security.stackexchange.com/questions/110143/asymmetric-encryption-for-multiple-recipients
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Asymmetric Encryption for Multiple Recipients?
Let's assume an arbitrary group of people who hold a key pair and who make their public key online accessible.
Let X, A, B and C be arbitrary people from that group.
Now X wants to encrypt a message once using A's, B's and C's public key so that only A, B and C will be able to decrypt the same ciphertext.
(Think of an encrypted FB/Diaspora where you would like to introduce asymmetric encryption but don't want to encrypt individually for each of your friends. And the content might be stored centralized - such that more people from the total user group have access to the data but shouldn't be able to decrypt it.)
Is that technically possible? If yes, does this asymmetric encryption method already have a name or known implementation?
• Does it need to be infeasible for X to cause A,B,C to not all get the same plaintext?
– user49075
Commented Jan 9, 2016 at 13:31
• Not entirely sure if I get the question. The scenario I have in mind could be something like a post on your (FB) wall / timeline which you only want to share with your friends or a sub-group of those. It would be the same plaintext for all of them. Commented Jan 9, 2016 at 13:33
• For the easiest way to get something like what you're asking about (namely, the way described in Steffen's answer), X could trivially cause A and B and C to each get different plain texts from the same ciphertext. Is that a problem?
– user49075
Commented Jan 9, 2016 at 13:35
• Not sure if that could cause a problem - but I don't see any until now. Commented Jan 9, 2016 at 13:37
In that case you generate a random symmetric key for each message and use that symmetric key to encrypt the message. Then you encrypt one copy of that symmetric key per recipient with the key of that recipient and attach these to the message. So the message will contain of:
``````symmetric key S encrypted with public key A
symmetric key S encrypted with public key B
symmetric key S encrypted with public key C
plaintext encrypted with symmetric key S
``````
This method is usually referred to as "hybrid encryption" and it is very common because most symmetric encryption and decryption systems are far faster than most asymmetric ones. So encrypting just a symmetric key with the expensive asymmetric algorithm and then using that symmetric key for the bulk of the message has far better performance, even when there is only one recipient.
Another useful side-effect is that it also prevents replay attacks and known-plaintext attacks. When you send the same message to the same recipient multiple times, the cyphertext will always be completely different because the symmetric key will be different. So an eavesdropper can't tell if a message is identical to a previous one. A man-in-the-middle can also not replay a previous message because the recipient could become suspicious if the sender uses the same symmetric key twice.
The usual way this is done in PGP etc is to create a symmetric key and encrypt this key with the public keys of each recipient. The original message itself will be encrypted with the symmetric key. Thus the result includes the encrypted message once (single key for all recipients) and then the encrypted symmetric key (short compared to the message) for each recipient.
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https://dmurdoch.github.io/rgl/reference/polygon3d.html
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This function takes a description of a flat polygon in x, y and z coordinates, and draws it in three dimensions.
polygon3d(x, y = NULL, z = NULL, fill = TRUE, plot = TRUE,
coords = 1:2, random = TRUE, ...)
## Arguments
x, y, z
Vertices of the polygon in a form accepted by xyz.coords.
fill
logical; should the polygon be filled?
plot
logical; should the polygon be displayed?
coords
Which two coordinates (x = 1, y = 2, z = 3) describe the polygon.
random
Should a random triangulation be used?
...
Other parameters to pass to lines3d or shade3d if plot = TRUE.
## Details
The function triangulates the two dimensional polygon described by coords, then applies the triangulation to all three coordinates. No check is made that the polygon is actually all in one plane, but the results may be somewhat unpredictable (especially if random = TRUE) if it is not.
Polygons need not be simple; use NA to indicate separate closed pieces. For fill = FALSE there are no other restrictions on the pieces, but for fill = TRUE the resulting two-dimensional polygon needs to be one that triangulate can handle.
## Value
If plot = TRUE, the id number of the lines (for fill = FALSE) or triangles (for fill = TRUE) that have been plotted. If plot = FALSE, then for fill = FALSE, a vector of indices into the XYZ matrix that could be used to draw the polygon. For fill = TRUE, a triangular mesh object representing the triangulation.
## Author
Duncan Murdoch
extrude3d for a solid extrusion of a polygon, triangulate for the triangulation.
## Examples
theta <- seq(0, 4*pi, len = 50)
r <- theta + 1
r <- c(r[-50], rev(theta*0.8) + 1)
theta <- c(theta[-50], rev(theta))
x <- r*cos(theta)
y <- r*sin(theta)
open3d()
plot(x, y, type = "n")
polygon(x, y)
polygon3d(x, y, x + y, col = "blue")
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2 1\n// File 1 is the vertex shader\n#ifdef GL_ES\n#ifdef GL_FRAGMENT_PRECISION_HIGH\nprecision highp float;\n#else\nprecision mediump float;\n#endif\n#endif\n\nattribute vec3 aPos;\nattribute vec4 aCol;\nuniform mat4 mvMatrix;\nuniform mat4 prMatrix;\nvarying vec4 vCol;\nvarying vec4 vPosition;\n\n#ifdef NEEDS_VNORMAL\nattribute vec3 aNorm;\nuniform mat4 normMatrix;\nvarying vec4 vNormal;\n#endif\n\n#if defined(HAS_TEXTURE) || defined (IS_TEXT)\nattribute vec2 aTexcoord;\nvarying vec2 vTexcoord;\n#endif\n\n#ifdef FIXED_SIZE\nuniform vec3 textScale;\n#endif\n\n#ifdef FIXED_QUADS\nattribute vec3 aOfs;\n#endif\n\n#ifdef IS_TWOSIDED\n#ifdef HAS_NORMALS\nvarying float normz;\nuniform mat4 invPrMatrix;\n#else\nattribute vec3 aPos1;\nattribute vec3 aPos2;\nvarying float normz;\n#endif\n#endif // IS_TWOSIDED\n\n#ifdef FAT_LINES\nattribute vec3 aNext;\nattribute vec2 aPoint;\nvarying vec2 vPoint;\nvarying float vLength;\nuniform float uAspect;\nuniform float uLwd;\n#endif\n\n\nvoid main(void) {\n \n#ifndef IS_BRUSH\n#if defined(NCLIPPLANES) || !defined(FIXED_QUADS) || defined(HAS_FOG)\n vPosition = mvMatrix * vec4(aPos, 1.);\n#endif\n \n#ifndef FIXED_QUADS\n gl_Position = prMatrix * vPosition;\n#endif\n#endif // !IS_BRUSH\n \n#ifdef IS_POINTS\n gl_PointSize = POINTSIZE;\n#endif\n \n vCol = aCol;\n \n#ifdef NEEDS_VNORMAL\n vNormal = normMatrix * vec4(-aNorm, dot(aNorm, aPos));\n#endif\n \n#ifdef IS_TWOSIDED\n#ifdef HAS_NORMALS\n /* normz should be calculated *after* projection */\n normz = (invPrMatrix*vNormal).z;\n#else\n vec4 pos1 = prMatrix*(mvMatrix*vec4(aPos1, 1.));\n pos1 = pos1/pos1.w - gl_Position/gl_Position.w;\n vec4 pos2 = prMatrix*(mvMatrix*vec4(aPos2, 1.));\n pos2 = pos2/pos2.w - gl_Position/gl_Position.w;\n normz = pos1.x*pos2.y - pos1.y*pos2.x;\n#endif\n#endif // IS_TWOSIDED\n \n#ifdef NEEDS_VNORMAL\n vNormal = vec4(normalize(vNormal.xyz/vNormal.w), 1);\n#endif\n \n#if defined(HAS_TEXTURE) || defined(IS_TEXT)\n vTexcoord = aTexcoord;\n#endif\n \n#if defined(FIXED_SIZE) && !defined(ROTATING)\n vec4 pos = prMatrix * mvMatrix * vec4(aPos, 1.);\n pos = pos/pos.w;\n gl_Position = pos + vec4(aOfs*textScale, 0.);\n#endif\n \n#if defined(IS_SPRITES) && !defined(FIXED_SIZE)\n vec4 pos = mvMatrix * vec4(aPos, 1.);\n pos = pos/pos.w + vec4(aOfs, 0.);\n gl_Position = prMatrix*pos;\n#endif\n \n#ifdef FAT_LINES\n /* This code was inspired by Matt Deslauriers' code in \n https://mattdesl.svbtle.com/drawing-lines-is-hard */\n vec2 aspectVec = vec2(uAspect, 1.0);\n mat4 projViewModel = prMatrix * mvMatrix;\n vec4 currentProjected = projViewModel * vec4(aPos, 1.0);\n currentProjected = currentProjected/currentProjected.w;\n vec4 nextProjected = projViewModel * vec4(aNext, 1.0);\n vec2 currentScreen = currentProjected.xy * aspectVec;\n vec2 nextScreen = (nextProjected.xy / nextProjected.w) * aspectVec;\n float len = uLwd;\n vec2 dir = vec2(1.0, 0.0);\n vPoint = aPoint;\n vLength = length(nextScreen - currentScreen)/2.0;\n vLength = vLength/(vLength + len);\n if (vLength > 0.0) {\n dir = normalize(nextScreen - currentScreen);\n }\n vec2 normal = vec2(-dir.y, dir.x);\n dir.x /= uAspect;\n normal.x /= uAspect;\n vec4 offset = vec4(len*(normal*aPoint.x*aPoint.y - dir), 0.0, 0.0);\n gl_Position = currentProjected + offset;\n#endif\n \n#ifdef IS_BRUSH\n gl_Position = vec4(aPos, 1.);\n#endif\n}","fragmentShader":"#line 2 2\n// File 2 is the fragment shader\n#ifdef GL_ES\n#ifdef GL_FRAGMENT_PRECISION_HIGH\nprecision highp float;\n#else\nprecision mediump float;\n#endif\n#endif\nvarying vec4 vCol; // carries alpha\nvarying vec4 vPosition;\n#if defined(HAS_TEXTURE) || defined (IS_TEXT)\nvarying vec2 vTexcoord;\nuniform sampler2D uSampler;\n#endif\n\n#ifdef HAS_FOG\nuniform int uFogMode;\nuniform vec3 uFogColor;\nuniform vec4 uFogParms;\n#endif\n\n#if defined(IS_LIT) && !defined(FIXED_QUADS)\nvarying vec4 vNormal;\n#endif\n\n#if NCLIPPLANES > 0\nuniform vec4 vClipplane[NCLIPPLANES];\n#endif\n\n#if NLIGHTS > 0\nuniform mat4 mvMatrix;\n#endif\n\n#ifdef IS_LIT\nuniform vec3 emission;\nuniform float shininess;\n#if NLIGHTS > 0\nuniform vec3 ambient[NLIGHTS];\nuniform vec3 specular[NLIGHTS]; // light*material\nuniform vec3 diffuse[NLIGHTS];\nuniform vec3 lightDir[NLIGHTS];\nuniform bool viewpoint[NLIGHTS];\nuniform bool finite[NLIGHTS];\n#endif\n#endif // IS_LIT\n\n#ifdef IS_TWOSIDED\nuniform bool front;\nvarying float normz;\n#endif\n\n#ifdef FAT_LINES\nvarying vec2 vPoint;\nvarying float vLength;\n#endif\n\nvoid main(void) {\n vec4 fragColor;\n#ifdef FAT_LINES\n vec2 point = vPoint;\n bool neg = point.y < 0.0;\n point.y = neg ? (point.y + vLength)/(1.0 - vLength) :\n -(point.y - vLength)/(1.0 - vLength);\n#if defined(IS_TRANSPARENT) && defined(IS_LINESTRIP)\n if (neg && length(point) <= 1.0) discard;\n#endif\n point.y = min(point.y, 0.0);\n if (length(point) > 1.0) discard;\n#endif // FAT_LINES\n \n#ifdef ROUND_POINTS\n vec2 coord = gl_PointCoord - vec2(0.5);\n if (length(coord) > 0.5) discard;\n#endif\n \n#if NCLIPPLANES > 0\n for (int i = 0; i < NCLIPPLANES; i++)\n if (dot(vPosition, vClipplane[i]) < 0.0) discard;\n#endif\n \n#ifdef FIXED_QUADS\n vec3 n = vec3(0., 0., 1.);\n#elif defined(IS_LIT)\n vec3 n = normalize(vNormal.xyz);\n#endif\n \n#ifdef IS_TWOSIDED\n if ((normz <= 0.) != front) discard;\n#endif\n \n#ifdef IS_LIT\n vec3 eye = normalize(-vPosition.xyz/vPosition.w);\n vec3 lightdir;\n vec4 colDiff;\n vec3 halfVec;\n vec4 lighteffect = vec4(emission, 0.);\n vec3 col;\n float nDotL;\n#ifdef FIXED_QUADS\n n = -faceforward(n, n, eye);\n#endif\n \n#if NLIGHTS > 0\n for (int i=0;i<NLIGHTS;i++) {\n colDiff = vec4(vCol.rgb * diffuse[i], vCol.a);\n lightdir = lightDir[i];\n if (!viewpoint[i])\n lightdir = (mvMatrix * vec4(lightdir, 1.)).xyz;\n if (!finite[i]) {\n halfVec = normalize(lightdir + eye);\n } else {\n lightdir = normalize(lightdir - vPosition.xyz/vPosition.w);\n halfVec = normalize(lightdir + eye);\n }\n col = ambient[i];\n nDotL = dot(n, lightdir);\n col = col + max(nDotL, 0.) * colDiff.rgb;\n col = col + pow(max(dot(halfVec, n), 0.), shininess) * specular[i];\n lighteffect = lighteffect + vec4(col, colDiff.a);\n }\n#endif\n \n#else // not IS_LIT\n vec4 colDiff = vCol;\n vec4 lighteffect = colDiff;\n#endif\n \n#ifdef IS_TEXT\n vec4 textureColor = lighteffect*texture2D(uSampler, vTexcoord);\n#endif\n \n#ifdef HAS_TEXTURE\n#ifdef TEXTURE_rgb\n vec4 textureColor = lighteffect*vec4(texture2D(uSampler, vTexcoord).rgb, 1.);\n#endif\n \n#ifdef TEXTURE_rgba\n vec4 textureColor = lighteffect*texture2D(uSampler, vTexcoord);\n#endif\n \n#ifdef TEXTURE_alpha\n vec4 textureColor = texture2D(uSampler, vTexcoord);\n float luminance = dot(vec3(1.,1.,1.), textureColor.rgb)/3.;\n textureColor = vec4(lighteffect.rgb, lighteffect.a*luminance);\n#endif\n \n#ifdef TEXTURE_luminance\n vec4 textureColor = vec4(lighteffect.rgb*dot(texture2D(uSampler, vTexcoord).rgb, vec3(1.,1.,1.))/3., lighteffect.a);\n#endif\n \n#ifdef TEXTURE_luminance_alpha\n vec4 textureColor = texture2D(uSampler, vTexcoord);\n float luminance = dot(vec3(1.,1.,1.),textureColor.rgb)/3.;\n textureColor = vec4(lighteffect.rgb*luminance, lighteffect.a*textureColor.a);\n#endif\n \n fragColor = textureColor;\n\n#elif defined(IS_TEXT)\n if (textureColor.a < 0.1)\n discard;\n else\n fragColor = textureColor;\n#else\n fragColor = lighteffect;\n#endif // HAS_TEXTURE\n \n#ifdef HAS_FOG\n // uFogParms elements: x = near, y = far, z = fogscale, w = (1-sin(FOV/2))/(1+sin(FOV/2))\n // In Exp and Exp2: use density = density/far\n // fogF will be the proportion of fog\n // Initialize it to the linear value\n float fogF;\n if (uFogMode > 0) {\n fogF = (uFogParms.y - vPosition.z/vPosition.w)/(uFogParms.y - uFogParms.x);\n if (uFogMode > 1)\n fogF = mix(uFogParms.w, 1.0, fogF);\n fogF = fogF*uFogParms.z;\n if (uFogMode == 2)\n fogF = 1.0 - exp(-fogF);\n // Docs are wrong: use (density*c)^2, not density*c^2\n // https://gitlab.freedesktop.org/mesa/mesa/-/blob/master/src/mesa/swrast/s_fog.c#L58\n else if (uFogMode == 3)\n fogF = 1.0 - exp(-fogF*fogF);\n fogF = clamp(fogF, 0.0, 1.0);\n gl_FragColor = vec4(mix(fragColor.rgb, uFogColor, fogF), fragColor.a);\n } else gl_FragColor = fragColor;\n#else\n gl_FragColor = fragColor;\n#endif // HAS_FOG\n \n}","players":[],"webGLoptions":{"preserveDrawingBuffer":true}},"evals":[],"jsHooks":[]}
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# Array Dimensions in Visual Basic
Visual Studio 2015
A dimension is a direction in which you can vary the specification of an array's elements. An array that holds the sales total for each day of the month has one dimension (the day of the month). An array that holds the sales total by department for each day of the month has two dimensions (the department number and the day of the month). The number of dimensions an array has is called its rank.
Note
You can use the Rank property to determine the how many dimensions an array has.
## Working with Dimensions
You specify an element of an array by supplying an index or subscript for each of its dimensions. The elements are contiguous along each dimension from index 0 through the highest index for that dimension.
The following illustrations show the conceptual structure of arrays with different ranks. Each element in the illustrations shows the index values that access it. For example, you can access the first element of the second row of the two-dimensional array by specifying indexes (1, 0).
One-dimensional array
Two-dimensional array
Three-dimensional array
### One Dimension
Many arrays have only one dimension, such as the number of people of each age. The only requirement to specify an element is the age for which that element holds the count. Therefore, such an array uses only one index. The following example declares a variable to hold a one-dimensional array of age counts for ages 0 through 120.
```Dim ageCounts(120) As UInteger
```
### Two Dimensions
Some arrays have two dimensions, such as the number of offices on each floor of each building on a campus. The specification of an element requires both the building number and the floor, and each element holds the count for that combination of building and floor. Therefore, such an array uses two indexes. The following example declares a variable to hold a two-dimensional array of office counts, for buildings 0 through 40 and floors 0 through 5.
```Dim officeCounts(40, 5) As Byte
```
A two-dimensional array is also called a rectangular array.
### Three Dimensions
A few arrays have three dimensions, such as values in three-dimensional space. Such an array uses three indexes, which in this case represent the x, y, and z coordinates of physical space. The following example declares a variable to hold a three-dimensional array of air temperatures at various points in a three-dimensional volume.
```Dim airTemperatures(99, 99, 24) As Single
```
### More than Three Dimensions
Although an array can have as many as 32 dimensions, it is rare to have more than three.
Note
When you add dimensions to an array, the total storage needed by the array increases considerably, so use multidimensional arrays with care.
## Using Different Dimensions
Suppose you want to track sales amounts for every day of the present month. You might declare a one-dimensional array with 31 elements, one for each day of the month, as the following example shows.
```Dim salesAmounts(30) As Double
```
Now suppose you want to track the same information not only for every day of a month but also for every month of the year. You might declare a two-dimensional array with 12 rows (for the months) and 31 columns (for the days), as the following example shows.
```Dim salesAmounts(11, 30) As Double
```
Now suppose you decide to have your array hold information for more than one year. If you want to track sales amounts for 5 years, you could declare a three-dimensional array with 5 layers, 12 rows, and 31 columns, as the following example shows.
```Dim salesAmounts(4, 11, 30) As Double
```
Note that, because each index varies from 0 to its maximum, each dimension of salesAmounts is declared as one less than the required length for that dimension. Note also that the size of the array increases with each new dimension. The three sizes in the preceding examples are 31, 372, and 1,860 elements respectively.
Note
You can create an array without using the Dim statement or the New clause. For example, you can call the CreateInstance method, or another component can pass your code an array created in this manner. Such an array can have a lower bound other than 0. You can always test for the lower bound of a dimension by using the GetLowerBound method or the LBound function.
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http://math.stackexchange.com/questions/235903/covers-and-compact-sets
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# Covers, and compact sets
Please go to this lecture(1). I have numbered my questions for organization and added the picture which I will constantly refer to. I also apologize in advance for this lengthy question
1. Right about here, he says that the point (the top one "enclosed" by the semi-circle) isn't open in $X$. I don't understand this. Why exactly isn't that open in $X$
2. Also he later talks about the semi-circle neighborhood being in $Y$ and that makes the point on the boundary an interior point. This confuses me. He left out the other part of the circle and just ignored that the part of the real "ball" lies outside $Y$ and therefore that ball isn't contained in $Y$.
3. Also are the sets $X$ and $Y$ even open sets in the lecture? Because later in the follow up video he writes $E = Y \cap G$. How can a metric space (is it even a set?) intersect with a set? Aren't they two different things? Isn't like saying $5 \cap [0,1]$? Sorry if the question is ignorant here.
4. Later in here, he tries to prove that $E = Y \cap G \iff E$ is open. He says that if a neighbourhood is contained in G, then the intersection of that neighbourhood and $Y$ is a neighbourhood in both $Y$ and $E$. But what happens if he takes a point that's not even in $Y$? And draw a neighbourhood around it such that it doesn't even intersect $Y$ or $E$? And suppose $Y$ really is a set (assuming (3) had been answered and clarifed), How does that imply this mysterious set intersection with an open set gives me and open set? That can't be true in general. What if $(-2,2)\cap [-1,1] = [-1,1]$? That doesn't give me an open set
5. Finally, I don't understand the idea and strategy in the forward direction of the proof. Like at all...
6. I found this proof (part of (5)) , in one line they say that $d(p,q) < r_p, q \in Y \implies q \in E$. How does that make sense? What if $q$ is outside of $E$?
-
1. He's saying that the semi-circular set is open in $Y$ but not in $X$. He's not talking about the point itself. The reason this set isn't open in $X$ is because points along the top edge of the rectangle $Y$ have the property that every open ball contains a point in the set $Y$ and a point not in the set $Y$ (draw a little circle around a point lying on that edge, you'll see what I mean).
2. I'm not entirely sure what you mean. Could you like to a time in the video?
3. $X$ is a metric space to start with, which is a set "endowed" with a metric. Then $Y$ starts out life as an open set in $X$, which we then 'turn into' a metric space by giving it the restricted metric. As sets, $Y\subset X$. For $E=Y\cap G$, you think of this intersection as plain old ordinary set intersection - $Y$ is still a set even though we've given it a metric.
4. (and 5.) It's definitely important to note that he's showing $E$ is open in $Y$. For the forward direction, we're actually constructing the set $G$ so that (a) it is open in $X$ and (b) $E=Y\cap G$. You simply construct this by drawing small enough neighborhoods around all the points in $E$. These neighborhoods might 'spill over' into $X$, but that's okay. Then, since neighborhoods are open, we can take an arbitrary union of them (in this case, take the union over all points in $E$) and the result will be an open set. You can then check that (a) and (b) hold. For the reverse direction, it doesn't matter if he picks a point that isn't in $Y$. All we need to show is that each point that is in $E$ has a neighborhood that is contained in $E$. $x\in E$ means that $x\in G$, and $G$ is open so we have our neighborhood. Like he says, this neighborhood intersected with $Y$ is a neighborhood in $Y$, which is what we need.
5. That such an $r_p$ exists is guaranteed by the assumption that $E$ is open relative to $Y$. I.e. $q$ can't be outside of $E$ because I've said that $d(p,q)$ is small enough so that it isn't. Try drawing a picture.
-
For some reason it's numbering things weird. The last item should be numbered 6. – icurays1 Nov 12 '12 at 20:41
1. That's what I mean, I drew the circle. But the "big" box is $X$, anything interior point in $Y$ is an interior in $X$ no? 2.See edit 3. answered, thank you. I think (1) needs to be clarified before I can understand the others. Thank you – Hawk Nov 12 '12 at 23:34
Since $Y\subset X$ it is true that interior points of $Y$ are also interior points of $X$. However the crux here is thinking about $Y$ in two different ways: as a subset of $X$ and as its very own space. If you look at $Y$ as a subset of $X$, those points along the edge are boundary points of $Y$, not interior points. If you look at $Y$ 'by iteself', they become interior points. Every time you use the word 'interior' you should really add 'with respect to $Y$' or 'with respect to $X$' – icurays1 Nov 12 '12 at 23:46
@icuraysi, let's just say then we do look at $Y$ by itself, (so in our heads, the whole metric X disappears from our picture). Then the set containing a point on the boundary of $Y$ contains points in $Y$ and other things outside our universe, $Y$. Yet this weird ball isn't solely contained in $Y$, it's containing some empty space outside Y. Is the problem I have here that i am not treating $Y$ as my "universe"? – Hawk Nov 12 '12 at 23:49
Correct, any time you look at things 'relative to $Y$', $Y$ becomes your universe. There is nothing outside of $Y$, so thats why these neighborhoods 'on the boundary' look a little funny. – icurays1 Nov 12 '12 at 23:57
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# Mathematics - the Cosmic Eye of Humanity
by Eberhard Zeidler
## Part IV
The theory of dynamic systems ranges from the movements of the planets and comets all the way to processes in the human brain and how epidemics like aids spread. One of the most important applications of mathematics in all conceivable areas is being able to precisely determine the optimal design of processes. Take the example of the space shuttle. When it landed on the moon, it had to be piloted using the minimum amount of fuel while the Apollo capsule had to return to earth heating up the heat shield as little as possible. It was not possible to test that experimentally, so that it had to be calculated on computers with the methods of optimal control developed by the blind Russian mathematician Lew Pontrjagin (1908-1988) in 1955. All of the fundamental physical processes function so that the action, i.e. the product of energy and time, must be a minimum (or at least critical). In the framework of his search for a universal law of radiation, Max Planck (1858-1947) made the epoch-making discovery in 1900 that there exists a smallest amount of action in nature -- Planck's quantum of action -- and he received the Nobel Prize in Physics in 1918 for this. Later physicists and mathematicians developed general methods to construct theories with discrete action from theories with continuous action. This is a process we call quantization and we encounter mathematical objects here (take vectors in infinite-dimensional Hilbert spaces) that have both the nature of a wave and a particle. The prototype for this are Einstein's light quanta (photons) that he received the Nobel Prize in Physics in 1921 for.
It is only possible to have such an abundance of shapes and forms in nature because there are stable states. There are significant jumps in the time evolution of a system when the system loses its stability when acted upon by an outside force. The bifurcation theory enables mathematics to calculate the limits of stability and the resulting new structures. That is very closely related to phase transitions, an area of intensive research in physics and mathematics. When water cools off, it makes a transition from water to ice, creating bizarre flowery patterns made of ice on your window. But when the universe cooled off after the Big Bang, there were a series of phase transitions and physicists believe that the originally uniform force was broken down into a strong, weak and electromagnetic interaction. The mathematics of symmetry, also called group theory by mathematicians, is very important in developing an understanding of our world. In 1918, the mathematician Emmy Noether (1882-1935) published a celebrated paper, in which she stated that conservation laws follow from symmetries. For example, the theorem of the conservation of energy applies to any physical system subject to laws that remain the same when time is shifted. This holds for the motions of the planets. The standard model of elementary particles maintains that our world consists of 12 basic particles: 6 quarks and 6 leptons (such as the electron and the neutrino). There are also their antiparticles (such as the positron as the antiparticle of the electron). The interactions between these 12 basic particles are mediated with 12 exchange particles: the massless photon (light), three heavy-mass vector bosons (for such things as radioactive decay) and eight massless gluons (for such things as nuclear forces). The Higg's hypothetical particle is responsible for the mass of the vector bosons.
In 1964, the American physicist Murray Gell-Mann (born in 1929) predicted that the proton consists of three quark particles based upon mathematical symmetry deliberations. Gell-Mann received the Nobel Prize in Physics in 1969 for his theory of unitary symmetry. This type of symmetry is difficult to imagine since it is not something we experience in everyday life, but we encounter it in the world of elementary particles. Unitary symmetry uses the imaginary number i in its mathematical deliberations. In contrast to real numbers, it possesses the remarkable property that
.
The Italian mathematician Raffael Bombielli introduced it formally in 1550 to be able to solve certain equations. We often notice in the history of mathematics that constructions initially introduced for purely innermathematical considerations later develop surprising applications. Take the example of Fermat's theorem of number theory over three hundred and fifty years old that we still use today to very simply and reliably encode information in banks. Gauss said:
Science should be the friend of applications, not its slave.
A number of phenomena in nature and technology involve a violation of symmetries, which is also known as breaking of symmetries. The ice flowers on your window have a significantly lower degree of symmetry than the homogeneous water that is not frozen. We can explain the phenomenon of forbidden spectral lines in the spectra of molecules mathematically with symmetry violations. Experiments show that processes of the weak interaction violate reflection symmetry and the reflected process is prohibited for certain processes such as the beta decay of cobalt. Lee and Yang received the Nobel Prize in Physics in 1957 for the corresponding theory of parity violation in the weak interaction. A special example of this is the fact that amino acids may be either right-handed or left-handed. Amazingly enough, there are only left-handed amino acids in living matter. That might have to do with the fact that right-handed amino acids are destroyed by ultraviolet radiation. We believe that the early universe had a fundamental supersymmetry between elementary particles with half-integral and integral spin. This supersymmetry is no longer observed today, however there are hopes of observing relicts of it at the high-energy acceleration experiments on the CERN beginning in 2006 and providing undoubted proof of the still missing Higgs particle for the standard model. One of the important differences between living and non-living matter is the fact that living processes cannot be reversed in time, meaning they are always irreversible and symmetry under time reflection is violated.
The key method for describing nature with mathematics is setting up mathematical models. An efficient model leaves non-essential details out while concentrating on the essentials. Also, it is important for every model to be aware of the limits of its validity. A case in point is the typical energy scale of the model. A feature of complicated processes like phase transitions is the fact that you have to link several scales with one another. The American physicist Kenneth Wilson received the Nobel Prize in Physics in 1982 for his theory of the renormalization group. Changing scales is also important in the mathematical theory of microstructures of materials and when setting up effective computer algorithms (multigrid methods). Quantum chemistry is familiar with Schrödinger's equation, the mathematical equation for all molecules. However, this equation is not very helpful with large molecules since it is not possible to do so many calculations. Quantum chemists created an extraordinary useful rough model in their density functional method and Walter Kohn received the Nobel Prize in Chemistry in 1999 for this together with John Pople. In the next few years, we will be able to receive gravitational waves from space with the methods of quantum optics (laser technology). We are expecting to gather information on the collision of black holes from this new window into space. However, in order to decode this information, it is necessary to have extraordinarily complicated computer simulations for 4-dimensional shock waves. But, our computers and scientific calculations are at their limits of the performance capability with the 3-dimensional shock waves from supersonic aeroplanes. If we combine computer simulations for molecular dynamics with the knowledge of an experienced chemist, we can eliminate costly experiments in chemistry when manufacturing new medicines. There is a great deal of mathematical knowledge on geometric models behind every image on a computer screen.
Developing a useful model is an art that has to be learned. A mathematician is dependant upon experience gained from experiments and the intuition of the natural scientist and engineer. It is not always easy to maintain this dialogue since different ways of thinking come into collision with one another. And it is therefore an important task for the future to educate young people who are capable of not only thinking mathematically, but also scientifically.
In 1984, the American Mathematical Society published a report on the future of mathematics and Arthur Jaffe from Harvard University wrote at that time:
Mathematical research should be as broad and as original as possible, with very long-range goals. We expect history to repeat itself: we expect that the most profound and useful future applications of mathematics cannot be predicted today, since they will arise from mathematics yet to be discovered.
As with every science, mathematics also has its limits and only religion is in a position to give answers to the profound existential questions of mankind. The quotation of the German poet, Johann Wolfgang von Goethe on the Harnack House of the Max Planck Society in the Dahlem suburb of Berlin sums it up as follows: "The greatest joy of a thinking man is to have explored the explorable and just to admire the unexplorable."
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# source:trunk/Cbc/src/CbcSymmetry.cpp@2049
Last change on this file since 2049 was 2049, checked in by forrest, 6 years ago
File size: 36.3 KB
Line
1/* \$Id: CbcSymmetry.cpp 925 2012-11-27 19:11:04Z stefan \$
2 *
3 * hacked from CouenneSymmetry.cpp
4 * Name: CbcSymmetry.cpp
5 * Author: Jim Ostrowski (the good bits - rest JJHF)
6 * Purpose: methods for exploiting symmetry
7 * Date: October 13, 2010
8 *
10 */
11
12#include <stdio.h>
13
14#ifdef COIN_HAS_NTY
15
16#include <cassert>
17#include <vector>
18#include <algorithm>
19#include <ostream>
20#include <iterator>
21
22#include "CbcSymmetry.hpp"
23#include "CbcBranchingObject.hpp"
24#include "CoinTime.hpp"
25/* Deliberately not threadsafe to save effort
26 Just for statistics
27 and not worth gathering across threads
28 can redo later
29 */
30static int nautyBranchCalls_ = 0;
31static int nautyBranchSucceeded_ = 0;
32static int nautyFixCalls_ = 0;
33static int nautyFixSucceeded_ = 0;
34static double nautyTime_ = 0.0;
35static double nautyFixes_= 0.0;
36static double nautyOtherBranches_ = 0.0;
37
38void Node::node(int i, double c , double l, double u, int cod, int s){
39 index = i;
40 coeff = c;
41 lb = l;
42 ub = u;
43 color = -1;
44 code = cod;
45 sign = s;
46}
47
48inline bool CbcSymmetry::compare (register Node &a, register Node &b) const {
49 if(a.get_code() == b.get_code() )
50 if(a.get_coeff() == b.get_coeff() )
51 if(a.get_sign() == b.get_sign() )
52 if( fabs ( a.get_lb() - b.get_lb() ) <= COUENNE_HACKED_EPS )
53 if( fabs ( a.get_ub() - b.get_ub() ) <= COUENNE_HACKED_EPS )
54 return 1;
55 return 0;
56}
57
58void CbcSymmetry::Compute_Symmetry() const{
59
60
61 std::sort(node_info_. begin (), node_info_. end (), node_sort);
62
63 for (std::vector <Node>:: iterator i = node_info_. begin (); i != node_info_. end (); ++i)
64 (*i).color_vertex(-1);
65
66 int color = 1;
67 for (std::vector <Node>:: iterator i = node_info_. begin (); i != node_info_. end (); ++i) {
68 if( (*i).get_color() == -1){
69 (*i).color_vertex(color);
70#ifdef PRINT_MORE
71 printf ("Graph vertex %d is given color %d\n", (*i).get_index(), color);
72#endif
73 nauty_info_ -> color_node((*i).get_index(), color);
74 for (std::vector <Node>:: iterator j = i+1; j != node_info_. end (); ++j)
75 if( compare( (*i) , (*j) ) ==1){
76 (*j).color_vertex(color);
77 nauty_info_ -> color_node((*j).get_index(),color);
78#ifdef PRINT_MORE
79 printf ("Graph vertex %d is given color %d, the same as vertex %d\n", (*j).get_index(), color, (*i).get_index());
80#endif
81 }
82 // else
83 // j = node_info_. end();
84 color++;
85 }
86 }
87
88 //Print_Orbits ();
89 nauty_info_ -> computeAuto();
90 //Print_Orbits ();
91}
92
93int
94CbcSymmetry::statsOrbits(CbcModel * model, int type) const
95{
96 char general[200];
97 int returnCode=0;
98 if (type) {
99 double branchSuccess=0.0;
100 if (nautyBranchSucceeded_)
101 branchSuccess = nautyOtherBranches_/nautyBranchSucceeded_;
102 double fixSuccess=0.0;
103 if (nautyFixSucceeded_)
104 fixSuccess = nautyFixes_/nautyFixSucceeded_;
105 sprintf(general,"Orbital branching tried %d times, succeeded %d times - average extra %7.3f, fixing %d times (%d, %7.3f) - %.2f seconds",
106 nautyBranchCalls_,nautyBranchSucceeded_,branchSuccess,
107 nautyFixCalls_,nautyFixSucceeded_,fixSuccess,nautyTime_);
108 } else {
109 returnCode = nauty_info_->getNumGenerators();
110 if (returnCode)
111 sprintf (general,"Nauty: %d orbits, %d generators, group size: %g - dense size %d, sparse %d - going %s",
112 nauty_info_->getNumOrbits(),
113 nauty_info_ -> getNumGenerators () ,
114 nauty_info_ -> getGroupSize (),
115 whichOrbit_[0],whichOrbit_[1],nauty_info_->isSparse() ? "sparse" : "dense");
116 else
117 sprintf(general,"Nauty did not find any useful orbits");
118 }
119 model->messageHandler()->message(CBC_GENERAL,
120 model->messages())
121 << general << CoinMessageEol ;
122 return returnCode;
123}
124
125void CbcSymmetry::Print_Orbits () const {
126
127 //printf ("num gens = %d, num orbits = %d \n", nauty_info_ -> getNumGenerators(), nauty_info_ -> getNumOrbits() );
128
129 std::vector<std::vector<int> > *new_orbits = nauty_info_->getOrbits();
130
131 printf ("Nauty: %d generators, group size: %.0g",
132 // nauty_info_->getNumOrbits(),
133 nauty_info_ -> getNumGenerators () ,
134 nauty_info_ -> getGroupSize ());
135
136 int nNonTrivialOrbits = 0;
137
138 for (unsigned int i = 0; i < new_orbits -> size(); i++) {
139 if ((*new_orbits)[i].size() > 1)
140 nNonTrivialOrbits++;
141 else continue;
142
143 // int orbsize = (*new_orbits)[i].size();
144 // printf( "Orbit %d [size: %d] [", i, orbsize);
145 // copy ((*new_orbits)[i].begin(), (*new_orbits)[i].end(),
146 // std::ostream_iterator<int>(std::cout, " "));
147 // printf("] \n");
148 }
149
150 printf (" (%d non-trivial orbits).\n", nNonTrivialOrbits);
151
152#if 1
153 if (nNonTrivialOrbits) {
154
155 int orbCnt = 0;
156
157 std::vector<std::vector<int> > *orbits = nauty_info_ -> getOrbits ();
158
159 for (std::vector <std::vector<int> >::iterator i = orbits -> begin (); i != orbits -> end (); ++i) {
160
161 printf ("Orbit %d: ", orbCnt++);
162
163 for (std::vector<int>::iterator j = i -> begin (); j != i -> end (); ++j)
164 printf (" %d", *j);
165
166 printf ("\n");
167 }
168 }
169#endif
170
171
172#if 0
173 if (nNonTrivialOrbits)
174 for (int i=0; i< nVars (); i++) {
175
176 std::vector< int > *branch_orbit = Find_Orbit (i);
177
178 if (branch_orbit -> size () > 1) {
179 printf ("x%04d: ", i);
180
181 for (std::vector<int>::iterator it = branch_orbit -> begin (); it != branch_orbit -> end (); ++it)
182 printf ("%d ", *it);
183 printf ("\n");
184 }
185 }
186#endif
187 delete new_orbits;
188}
189void
190CbcSymmetry::fillOrbits()
191{
192 for (int i=0;i<numberColumns_;i++)
193 whichOrbit_[i]=-1;
194 numberUsefulOrbits_=0;
195
196 std::vector<std::vector<int> > *orbits = nauty_info_ -> getOrbits ();
197
198 for (std::vector <std::vector<int> >::iterator i = orbits -> begin (); i != orbits -> end (); ++i) {
199 int nUseful=0;
200 int jColumn=-2;
201 for (std::vector<int>::iterator j = i -> begin (); j != i -> end (); ++j) {
202 int iColumn=*j;
203 if (iColumn<numberColumns_) {
204 whichOrbit_[iColumn]=numberUsefulOrbits_;
205 nUseful++;
206 jColumn=iColumn;
207 }
208 }
209 if (nUseful>1) {
210 numberUsefulOrbits_++;
211 } else if (jColumn>=0) {
212 assert (nUseful);
213 whichOrbit_[jColumn]=-2;
214 }
215 }
216 delete orbits;
217}
218int
219CbcSymmetry::largestOrbit(const double * lower, const double * upper) const
220{
221 int * counts = new int[numberUsefulOrbits_];
222 memset(counts,0,numberUsefulOrbits_*sizeof(int));
223 for (int i=0;i<numberColumns_;i++) {
224 int iOrbit=whichOrbit_[i];
225 if (iOrbit>=0) {
226 if (lower[i]==0.0&&upper[i]==1.0)
227 counts[iOrbit]++;
228 }
229 }
230 int iOrbit=-1;
231 int maxOrbit=0;
232 for (int i=0;i<numberUsefulOrbits_;i++) {
233 if (counts[i]>maxOrbit) {
234 maxOrbit=counts[i];
235 iOrbit=i;
236 }
237 }
238 delete [] counts;
239 return iOrbit;
240}
241
242std::vector<int> *CbcSymmetry::Find_Orbit(int index) const{
243
244 std::vector<int> *orbit = new std::vector <int>;
245 int which_orbit = -1;
246 std::vector<std::vector<int> > *new_orbits = nauty_info_->getOrbits();
247
248 for (unsigned int i = 0; i < new_orbits -> size(); i++) {
249 for (unsigned int j = 0; j < (*new_orbits)[i].size(); j++) {
250 // for (std::vector <int>:: iterator j = new_orbits[i].begin(); new_orbits[i].end(); ++j){
251 if( (*new_orbits)[i][j] == index)
252 which_orbit = i;
253 }
254 }
255
256 // for (std::vector <int>:: iterator j = new_orbits[which_orbit].begin(); new_orbits[which_orbit].end(), ++j)
257 for (unsigned int j = 0; j < (*new_orbits)[which_orbit].size(); j++)
258 orbit -> push_back ((*new_orbits)[which_orbit][j]);
259
260 delete new_orbits;
261
262 return orbit;
263}
264
265
266void CbcSymmetry::ChangeBounds (const double * new_lb, const double * new_ub,
267 int num_cols, bool justFixedAtOne) const {
268 if (justFixedAtOne)
269 nautyFixCalls_++;
270 else
271 nautyBranchCalls_++;
272 std::sort(node_info_. begin (), node_info_. end (), index_sort);
273
274 for (int i = 0; i < num_cols; i++) {
275 // printf("Var %d lower bound: %f upper bound %f \n", i, new_lb[i], new_ub[i]);
276
277 assert (node_info_[i].get_index () == i);
278 double newLower = new_lb[i];
279 double newUpper = new_ub[i];
280 if (justFixedAtOne) {
281 // free up all fixed at zero
282 if (!newLower)
283 newUpper = 1.0;
284 }
285 node_info_[i ].bounds ( newLower , newUpper );
286 //printf("Var %d INPUT lower bound: %f upper bound %f \n", i, node_info_[i].get_lb(), node_info_[i].get_ub());
287 }
288}
289void CbcSymmetry::setupSymmetry (const OsiSolverInterface & solver) {
290 const double *objective = solver.getObjCoefficients() ;
291 const double *columnLower = solver.getColLower() ;
292 const double *columnUpper = solver.getColUpper() ;
293 int numberColumns = solver.getNumCols() ;
294 int numberRows = solver.getNumRows();
295 int iRow, iColumn;
296
297 // Row copy
298 CoinPackedMatrix matrixByRow(*solver.getMatrixByRow());
299 const double * elementByRow = matrixByRow.getElements();
300 const int * column = matrixByRow.getIndices();
301 const CoinBigIndex * rowStart = matrixByRow.getVectorStarts();
302 const int * rowLength = matrixByRow.getVectorLengths();
303
304 const double * rowLower = solver.getRowLower();
305 const double * rowUpper = solver.getRowUpper();
306 // // Find Coefficients
307
308 /// initialize nauty
309
310 int num_affine = 0;
311
312 for (iColumn = 0; iColumn < numberColumns; iColumn++) {
313 if (objective[iColumn] && objective[iColumn]!=1.0)
314 num_affine++;
315 }
316 for (iRow = 0; iRow < numberRows; iRow++) {
317 for (CoinBigIndex j = rowStart[iRow];
318 j < rowStart[iRow] + rowLength[iRow]; j++) {
319 int jColumn = column[j];
320 double value = elementByRow[j];
321 if (value!=1.0)
322 num_affine++;
323 }
324 }
325
326 // Create Nauty object
327
328 int coef_count= numberRows + numberColumns +1;
329 int nc = num_affine + coef_count;
330 // create graph (part 1)
331
332 for (iColumn = 0; iColumn < numberColumns; iColumn++) {
333 Node var_vertex;
334 var_vertex.node(iColumn,0.0,columnLower[iColumn],columnUpper[iColumn],-1,-1 );
335 node_info_.push_back(var_vertex);
336 }
337 // objective
338 int index = numberColumns;
339 {
340 Node vertex;
341 vertex.node( index , 0.0 , -COIN_DBL_MAX, COIN_DBL_MAX,
342 COUENNE_HACKED_EXPRGROUP, 0);
343 node_info_.push_back( vertex);
344 }
345
346 // compute space for sparse
347 size_t * v = NULL;
348 int * d = NULL;
349 int * e = NULL;
350 bool sparse=false;
351 double spaceDense = ((nc+WORDSIZE-1)*(nc+WORDSIZE-1))/WORDSIZE;
352 int spaceSparse = 0;
353 {
354 size_t numberElements = 0;
355 for (iColumn = 0; iColumn < numberColumns; iColumn++) {
356 double value = objective[iColumn];
357 if (value) {
358 if (value==1.0) {
359 numberElements+=2;
360 } else {
361 numberElements+=4;
362 coef_count ++;
363 }
364 }
365 }
366 for (iRow = 0; iRow < numberRows; iRow++) {
367 for (CoinBigIndex j = rowStart[iRow];
368 j < rowStart[iRow] + rowLength[iRow]; j++) {
369 int jColumn = column[j];
370 double value = elementByRow[j];
371 if (value==1.0) {
372 numberElements+=2;
373 } else {
374 numberElements+=4;
375 coef_count ++;
376 }
377 }
378 }
379 spaceSparse = 2*nc+numberElements;
380 //printf("Space for sparse is %d for dense %g\n",
381 // spaceSparse,spaceDense);
382#ifdef NTY_TRACES
383 bool goSparse = true;
384#else
385 bool goSparse = (spaceSparse<spaceDense);
386#endif
387 // for now always sparse
388 goSparse=true;
389 if (goSparse) {
390 sparse=true;
391 v = new size_t [nc+1];
392 d = new int [nc];
393 e = new int [numberElements];
394 size_t * counts = new size_t [coef_count+1];
395 memset(counts,0,coef_count*sizeof(size_t));
396 coef_count= numberRows + numberColumns +1;
397 // create graph (part 2)
398 for (iColumn = 0; iColumn < numberColumns; iColumn++) {
399 double value = objective[iColumn];
400 if (value) {
401 if (value==1.0) {
402 counts[index]++;
403 counts[iColumn]++;
404 } else {
405 counts[index]++;
406 counts[coef_count]+=2;
407 counts[iColumn]++;
408 coef_count ++;
409 }
410 }
411 }
412 index++;
413 for (iRow = 0; iRow < numberRows; iRow++) {
414 for (CoinBigIndex j = rowStart[iRow];
415 j < rowStart[iRow] + rowLength[iRow]; j++) {
416 int jColumn = column[j];
417 double value = elementByRow[j];
418 if (value==1.0) {
419 counts[index]++;
420 counts[jColumn]++;
421 } else {
422 counts[index]++;
423 counts[coef_count]+=2;
424 counts[jColumn]++;
425 coef_count ++;
426 }
427 }
428 index++;
429 }
430 // create graph (part 3)
431 assert (nc==coef_count);
432 numberElements=0;
433 v[0]=0;
434 for (int i=0;i<nc;i++) {
435 int count=counts[i];
436 d[i]=count;
437 counts[i]=v[i];
438 numberElements+=count;;
439 v[i+1]=numberElements;
440 }
441 index = numberColumns;
442 coef_count= numberRows + numberColumns +1;
443 for (iColumn = 0; iColumn < numberColumns; iColumn++) {
444 double value = objective[iColumn];
445 if (value) {
446 int where;
447 if (value==1.0) {
448 where = counts[index];
449 counts[index]++;
450 e[where]=iColumn;
451 where = counts[iColumn];
452 counts[iColumn]++;
453 e[where]=index;
454 } else {
455 Node coef_vertex;
456 coef_vertex.node( coef_count, value, value, value, -2, 0 );
457 node_info_.push_back(coef_vertex);
458 where = counts[index];
459 counts[index]++;
460 e[where]=coef_count;
461 where = counts[coef_count];
462 counts[coef_count]++;
463 e[where]=index;
464 where = counts[coef_count];
465 counts[coef_count]++;
466 e[where]=iColumn;
467 where = counts[iColumn];
468 counts[iColumn]++;
469 e[where]=coef_count;
470 coef_count ++;
471 }
472 }
473 }
474 index++;
475 for (iRow = 0; iRow < numberRows; iRow++) {
476 Node vertex;
477 vertex.node( index , 0.0 , rowLower[iRow], rowUpper[iRow],
478 COUENNE_HACKED_EXPRGROUP, 0);
479 node_info_.push_back( vertex);
480 for (CoinBigIndex j = rowStart[iRow];
481 j < rowStart[iRow] + rowLength[iRow]; j++) {
482 int jColumn = column[j];
483 double value = elementByRow[j];
484 int where;
485 if (value==1.0) {
486 where = counts[index];
487 counts[index]++;
488 e[where]=jColumn;
489 where = counts[jColumn];
490 counts[jColumn]++;
491 e[where]=index;
492 } else {
493 Node coef_vertex;
494 coef_vertex.node( coef_count, value, value, value, -2, 0 );
495 node_info_.push_back(coef_vertex);
496 where = counts[index];
497 counts[index]++;
498 e[where]=coef_count;
499 where = counts[coef_count];
500 counts[coef_count]++;
501 e[where]=index;
502 where = counts[coef_count];
503 counts[coef_count]++;
504 e[where]=jColumn;
505 where = counts[jColumn];
506 counts[jColumn]++;
507 e[where]=coef_count;
508 coef_count ++;
509 }
510 }
511 index++;
512 }
513 delete [] counts;
514 }
515 }
516
517 nauty_info_ = new CbcNauty(nc,v,d,e);
518 delete [] v;
519 delete [] d;
520 delete [] e;
521 if (!sparse) {
522 // create graph (part 2)
523 coef_count= numberRows + numberColumns +1;
524 for (iColumn = 0; iColumn < numberColumns; iColumn++) {
525 double value = objective[iColumn];
526 if (value) {
527 if (value==1.0) {
530 } else {
531 Node coef_vertex;
532 coef_vertex.node( coef_count, value, value, value, -2, 0 );
533 node_info_.push_back(coef_vertex);
538 coef_count ++;
539 }
540 }
541 }
542 index++;
543 for (iRow = 0; iRow < numberRows; iRow++) {
544 Node vertex;
545 vertex.node( index , 0.0 , rowLower[iRow], rowUpper[iRow],
546 COUENNE_HACKED_EXPRGROUP, 0);
547 node_info_.push_back( vertex);
548 for (CoinBigIndex j = rowStart[iRow];
549 j < rowStart[iRow] + rowLength[iRow]; j++) {
550 int jColumn = column[j];
551 double value = elementByRow[j];
552 if (value==1.0) {
555 } else {
556 Node coef_vertex;
557 coef_vertex.node( coef_count, value, value, value, -2, 0 );
558 node_info_.push_back(coef_vertex);
563 coef_count ++;
564 }
565 }
566 index++;
567 }
568 }
569 numberColumns_ = numberColumns;
570 whichOrbit_ = new int [2*numberColumns_];
571 nautyBranchCalls_ = 0;
572 nautyBranchSucceeded_ = 0;
573 nautyFixCalls_ = 0;
574 nautyFixSucceeded_ = 0;
575 nautyTime_ = 0.0;
576 nautyFixes_= 0.0;
577 nautyOtherBranches_ = 0.0;
578 Compute_Symmetry ();
579 //Print_Orbits ();
580 // stats in array
581 whichOrbit_[0]=spaceDense;
582 whichOrbit_[1]=spaceSparse;
583}
584// Fixes variables using orbits (returns number fixed)
585int
586CbcSymmetry::orbitalFixing(OsiSolverInterface * solver)
587{
588 int numberColumns = solver->getNumCols();
589 char * status = new char [numberColumns];
590 ChangeBounds(solver->getColLower(),
591 solver->getColUpper(),
592 solver->getNumCols(),true);
593 Compute_Symmetry();
594 fillOrbits();
595 int n=0;
596 //#define PRINT_MORE 1
597 const int * alternativeOrbits = whichOrbit();
598 if (alternativeOrbits) {
599 for (int i=0;i<numberColumns;i++) {
600 char type='0';
601 if (solver->getColUpper()[i]) {
602 if (solver->getColLower()[i]) {
603 type='1';
604 } else {
605 double value=solver->getColSolution()[i];
606 if (value<0.0001)
607 type='L';
608 else if (value>0.9999)
609 type='U';
610 else
611 type='X';
612 }
613 }
614 status[i]=type;
615 }
616 n=0;
617 for (int i=0;i<numberColumns;i++) {
618 if (status[i]!='0'&&status[i]!='1') {
619 int iOrbit=alternativeOrbits[i];
620 for (int j=i+1;j<numberColumns;j++) {
621 if (status[j]=='0'&&alternativeOrbits[j]==iOrbit) {
622#if PRINT_MORE>1
623 printf("In alternative orbit %d - %d free (%c), %d fixed to 0\n",
624 iOrbit,i,status[i],j);
625#endif
626 status[i]='0'; // can fix on both branches
627 solver->setColUpper(i,0.0);
628 n++;
629 break;
630 }
631 }
632 }
633 }
634 }
635 delete [] status;
636 if (n) {
637 nautyFixSucceeded_++;
638 nautyFixes_ += n;
639#if PRINT_MORE
640 printf("%d orbital fixes\n",n);
641#endif
642 }
643 return n;
644}
645// Default Constructor
646CbcSymmetry::CbcSymmetry ()
647 : nauty_info_(NULL),
648 numberColumns_(0),
649 numberUsefulOrbits_(0),
650 whichOrbit_(NULL)
651{
652}
653// Copy constructor
654CbcSymmetry::CbcSymmetry ( const CbcSymmetry & rhs)
655{
656 node_info_ = rhs.node_info_;
657 nauty_info_ = new CbcNauty(*rhs.nauty_info_);
658 numberUsefulOrbits_ = rhs.numberUsefulOrbits_;
659 numberColumns_ = rhs.numberColumns_;
660 if (rhs.whichOrbit_)
661 whichOrbit_=CoinCopyOfArray(rhs.whichOrbit_,numberColumns_);
662 else
663 whichOrbit_ = NULL;
664}
665
666// Assignment operator
667CbcSymmetry &
668CbcSymmetry::operator=( const CbcSymmetry & rhs)
669{
670 if (this != &rhs) {
671 delete nauty_info_;
672 node_info_ = rhs.node_info_;
673 nauty_info_ = new CbcNauty(*rhs.nauty_info_);
674 delete [] whichOrbit_;
675 numberColumns_ = rhs.numberColumns_;
676 numberUsefulOrbits_ = rhs.numberUsefulOrbits_;
677 if (rhs.whichOrbit_)
678 whichOrbit_=CoinCopyOfArray(rhs.whichOrbit_,numberColumns_);
679 else
680 whichOrbit_ = NULL;
681 }
682 return *this;
683}
684
685// Destructor
686CbcSymmetry::~CbcSymmetry ()
687{
688 delete nauty_info_;
689 delete [] whichOrbit_;
690}
691
692CbcNauty::CbcNauty(int vertices, const size_t * v, const int * d, const int * e)
693{
694 //printf("Need sparse nauty - wordsize %d\n",WORDSIZE);
695 n_ = vertices;
696 m_ = (n_ + WORDSIZE - 1)/WORDSIZE;
697 if (v)
698 nel_ = v[n_];
699 else
700 nel_ = 0;
701
702 //printf ("size of long = %d (%d)\nwordsize = %d\nn,m = %d,%d\n",
703 // SIZEOF_LONG, sizeof (long), WORDSIZE, n_, m_);
704
705 nauty_check (WORDSIZE, m_, n_, NAUTYVERSIONID);
706
707 /// Apparently sizes are skewed on 64bit machines
708
709#define MULTIPLIER 1
710
711 if (!nel_) {
712 G_ = (graph *) malloc(MULTIPLIER * m_ * n_ * sizeof(int));
713 GSparse_ = NULL;
714 } else {
715 G_ = NULL;
716 GSparse_ = (sparsegraph *) malloc(sizeof(sparsegraph));
717 SG_INIT(*GSparse_);
718 SG_ALLOC(*GSparse_,n_,nel_,"malloc");
719 GSparse_->nv = n_; /* Number of vertices */
720 GSparse_->nde = nel_;
721 }
722 lab_ = (int *) malloc(MULTIPLIER * n_ * sizeof(int));
723 ptn_ = (int *) malloc(MULTIPLIER * n_ * sizeof(int));
724 active_ = NULL;
725 orbits_ = (int *) malloc(MULTIPLIER * n_ * sizeof(int));
726#ifndef NTY_TRACES
727 options_ = (optionblk *) malloc(MULTIPLIER * sizeof(optionblk));
728 stats_ = (statsblk *) malloc(MULTIPLIER * sizeof(statsblk));
729#else
730 options_ = (TracesOptions *) malloc(MULTIPLIER * sizeof(TracesOptions));
731 stats_ = (TracesStats *) malloc(MULTIPLIER * sizeof(TracesStats));
732#endif
733 worksize_ = 100*m_;
734 workspace_ = (setword *) malloc(MULTIPLIER * worksize_*sizeof(setword));
735 canonG_ = NULL;
736 if ((G_ == 0&&GSparse_ == 0) || lab_ == 0 || ptn_ == 0 ||
737 orbits_ == 0 || options_ == 0 || stats_ == 0 ||
738 workspace_ == 0) assert(0);
739
740 // Zero allocated memory
741 if (G_) {
742 memset(G_, 0, m_*n_*sizeof(int));
743 } else {
744 //for (int i=0;i<n_;i++) {
745 //GSparse_->v[i]=v[i];
746 //}
747 memcpy(GSparse_->v,v,n_*sizeof(size_t));
748 memcpy(GSparse_->d,d,n_*sizeof(int));
749 memcpy(GSparse_->e,e,nel_*sizeof(int));
750 }
751 memset(lab_, 0, n_*sizeof(int));
752 memset(ptn_, 0, n_*sizeof(int));
753 memset(orbits_, 0, n_*sizeof(int));
754 memset(workspace_, 0, worksize_*sizeof(setword));
755#ifndef NTY_TRACES
756 memset(options_, 0,MULTIPLIER * sizeof(optionblk));
757#else
758 memset(options_, 0,MULTIPLIER * sizeof(TracesOptions));
759#endif
760
761 // Set the options you want
762#ifndef NTY_TRACES
763 options_->getcanon = FALSE;
764 options_->digraph = FALSE;
765 options_->writeautoms = FALSE;
766 options_->writemarkers = FALSE;
767 options_->defaultptn = TRUE;
768 options_->cartesian = FALSE;
769 options_->linelength = 78;
770 options_->outfile = NULL;
771 options_->userrefproc = NULL;
772 options_->userautomproc = NULL;
773 options_->userlevelproc = NULL;
774 options_->usernodeproc = NULL;
775 // options_->usertcellproc = NULL;
776 options_->invarproc = NULL;
777 options_->tc_level = 100;
778 options_->mininvarlevel = 0;
779 options_->maxinvarlevel = 1;
780 options_->invararg = 0;
781 options_->dispatch = &dispatch_graph;
782#else
783 options_->getcanon = FALSE;
784 options_->writeautoms = FALSE;
785 options_->cartesian = FALSE;
786 options_->digraph = FALSE;
787 options_->defaultptn = TRUE;
788 options_->linelength = 78;
789#endif
790 if (G_) {
791 // Make an empty graph
792 for (int j = 0; j < n_; j++) {
793 set *gv = GRAPHROW(G_, j, m_);
794 EMPTYSET(gv, m_);
795 }
796 }
797
798 vstat_ = new int[n_];
799 clearPartitions();
800 afp_ = NULL;
801}
802
803CbcNauty::~CbcNauty()
804{
805 if (G_) free(G_);
806 if (GSparse_) {
807 SG_FREE(*GSparse_);
808 free(GSparse_);
809 }
810 if (lab_) free(lab_);
811 if (ptn_) free(ptn_);
812 if (active_) free(active_);
813 if (orbits_) free(orbits_);
814 if (options_) free(options_);
815 if (stats_) free(stats_);
816 if (workspace_) free(workspace_);
817 if (canonG_) free(canonG_);
818 if (vstat_) delete [] vstat_;
819}
820// Copy constructor
821CbcNauty::CbcNauty ( const CbcNauty & rhs)
822{
823 n_ = rhs.n_;
824 m_ = rhs.m_;
825 nel_ = rhs.nel_;
826 G_ = NULL;
827 GSparse_ = NULL;
828 if (!nel_) {
829 G_ = (graph *) malloc(MULTIPLIER * m_ * n_ * sizeof(int));
830 } else {
831 GSparse_ = (sparsegraph *) malloc(sizeof(sparsegraph));
832 SG_INIT(*GSparse_);
833 SG_ALLOC(*GSparse_,n_,nel_,"malloc");
834 GSparse_->nv = n_; /* Number of vertices */
835 GSparse_->nde = nel_;
836 }
837 lab_ = (int *) malloc(MULTIPLIER * n_ * sizeof(int));
838 ptn_ = (int *) malloc(MULTIPLIER * n_ * sizeof(int));
839 orbits_ = (int *) malloc(MULTIPLIER * n_ * sizeof(int));
840#ifndef NTY_TRACES
841 options_ = (optionblk *) malloc(MULTIPLIER * sizeof(optionblk));
842 stats_ = (statsblk *) malloc(MULTIPLIER * sizeof(statsblk));
843#else
844 options_ = (TracesOptions *) malloc(MULTIPLIER * sizeof(TracesOptions));
845 stats_ = (TracesStats *) malloc(MULTIPLIER * sizeof(TracesStats));
846#endif
847 worksize_ = 100*m_;
848 workspace_ = (setword *) malloc(MULTIPLIER * worksize_*sizeof(setword));
849 vstat_ = new int[n_];
850 canonG_ = NULL;
851 if ((G_ == 0 && GSparse_ == 0) || lab_ == 0 || ptn_ == 0 ||
852 orbits_ == 0 || options_ == 0 || stats_ == 0 ||
853 workspace_ == 0) assert(0);
854
855 // Copy allocated memory
856 if (G_) {
857 memcpy(G_, rhs.G_, m_*n_*sizeof(int));
858 } else {
859 memcpy(GSparse_->v,rhs.GSparse_->v,n_*sizeof(size_t));
860 memcpy(GSparse_->d,rhs.GSparse_->d,n_*sizeof(int));
861 memcpy(GSparse_->e,rhs.GSparse_->e,nel_*sizeof(int));
862 }
863 memcpy(lab_, rhs.lab_, n_*sizeof(int));
864 memcpy(ptn_, rhs.ptn_, n_*sizeof(int));
865 memcpy(orbits_, rhs.orbits_, n_*sizeof(int));
866 memcpy(workspace_, rhs.workspace_, worksize_*sizeof(setword));
867#ifndef NTY_TRACES
868 memcpy(options_, rhs.options_, MULTIPLIER * sizeof(optionblk));
869 memcpy(stats_, rhs.stats_, MULTIPLIER * sizeof(statsblk));
870#else
871 memcpy(options_, rhs.options_,MULTIPLIER * sizeof(TracesOptions));
872 memcpy(stats_, rhs.stats_, MULTIPLIER * sizeof(TracesStats));
873#endif
874 memcpy(vstat_,rhs.vstat_,n_*sizeof(int));
875
876 // ? clearPartitions();
877 active_ = NULL;
878 afp_ = rhs.afp_; // ? no copy ?
879}
880
881// Assignment operator
882CbcNauty &
883CbcNauty::operator=( const CbcNauty & rhs)
884{
885 if (this != &rhs) {
886 if (G_) free(G_);
887 if (GSparse_) {
888 SG_FREE(*GSparse_);
889 free(GSparse_);
890 }
891 if (lab_) free(lab_);
892 if (ptn_) free(ptn_);
893 if (active_) free(active_);
894 if (orbits_) free(orbits_);
895 if (options_) free(options_);
896 if (stats_) free(stats_);
897 if (workspace_) free(workspace_);
898 if (canonG_) free(canonG_);
899 if (vstat_) delete [] vstat_;
900 {
901 n_ = rhs.n_;
902 m_ = rhs.m_;
903 nel_ = rhs.nel_;
904 G_ = NULL;
905 GSparse_ = NULL;
906 if (!nel_) {
907 G_ = (graph *) malloc(MULTIPLIER * m_ * n_ * sizeof(int));
908 } else {
909 GSparse_ = (sparsegraph *) malloc(sizeof(sparsegraph));
910 SG_INIT(*GSparse_);
911 SG_ALLOC(*GSparse_,n_,nel_,"malloc");
912 GSparse_->nv = n_; /* Number of vertices */
913 GSparse_->nde = nel_;
914 }
915 lab_ = (int *) malloc(MULTIPLIER * n_ * sizeof(int));
916 ptn_ = (int *) malloc(MULTIPLIER * n_ * sizeof(int));
917 orbits_ = (int *) malloc(MULTIPLIER * n_ * sizeof(int));
918#ifndef NTY_TRACES
919 options_ = (optionblk *) malloc(MULTIPLIER * sizeof(optionblk));
920 stats_ = (statsblk *) malloc(MULTIPLIER * sizeof(statsblk));
921#else
922 options_ = (TracesOptions *) malloc(MULTIPLIER * sizeof(TracesOptions));
923 stats_ = (TracesStats *) malloc(MULTIPLIER * sizeof(TracesStats));
924#endif
925 worksize_ = 100*m_;
926 workspace_ = (setword *) malloc(MULTIPLIER * worksize_*sizeof(setword));
927 vstat_ = new int[n_];
928 canonG_ = NULL;
929 if ((G_ == 0 && GSparse_ == 0) || lab_ == 0 || ptn_ == 0 ||
930 orbits_ == 0 || options_ == 0 || stats_ == 0 ||
931 workspace_ == 0) assert(0);
932
933 // Copy allocated memory
934 if (!nel_) {
935 memcpy(G_, rhs.G_, m_*n_*sizeof(int));
936 } else {
937 memcpy(GSparse_->v,rhs.GSparse_->v,n_*sizeof(size_t));
938 memcpy(GSparse_->d,rhs.GSparse_->d,n_*sizeof(int));
939 memcpy(GSparse_->e,rhs.GSparse_->e,nel_*sizeof(int));
940 }
941 memcpy(lab_, rhs.lab_, n_*sizeof(int));
942 memcpy(ptn_, rhs.ptn_, n_*sizeof(int));
943 memcpy(orbits_, rhs.orbits_, n_*sizeof(int));
944 memcpy(workspace_, rhs.workspace_, worksize_*sizeof(setword));
945#ifndef NTY_TRACES
946 memcpy(options_, rhs.options_, MULTIPLIER * sizeof(optionblk));
947 memcpy(stats_, rhs.stats_, MULTIPLIER * sizeof(statsblk));
948#else
949 memcpy(options_, rhs.options_,MULTIPLIER * sizeof(TracesOptions));
950 memcpy(stats_, rhs.stats_, MULTIPLIER * sizeof(TracesStats));
951#endif
952 memcpy(vstat_,rhs.vstat_,n_*sizeof(int));
953
954 // ? clearPartitions();
955 active_ = NULL;
956 afp_ = rhs.afp_; // ? no copy ?
957 }
958 }
959 return *this;
960}
961
962void
964{
965 // Right now die if bad index. Can throw exception later
966 //printf("addelement %d %d \n", ix, jx);
967 assert(ix < n_ && jx < n_);
968 if(ix != jx){ //No Loops
969 set *gv = GRAPHROW(G_, ix, m_);
971 set *gv2 = GRAPHROW(G_, jx, m_);
973 autoComputed_ = false;
974 }
975}
976
977void
978CbcNauty::clearPartitions()
979{
980 for (int j = 0; j < n_; j++) {
981 vstat_[j] = 1;
982 //printf("vstat %d = %d", j, vstat_[j]);
983 }
984
985 autoComputed_ = false;
986}
987
988void
989CbcNauty::computeAuto()
990{
991
992 // if (autoComputed_) return;
993
994 double startCPU = CoinCpuTime ();
995
996 options_->defaultptn = FALSE;
997
998 // Here we only implement the partitions
999 // [ fix1 | fix0 (union) free | constraints ]
1000 int ix = 0;
1001
1002 for( int color = 1; color <= n_; color++){
1003 for (int j = 0; j < n_; j++) {
1004 if (vstat_[j] == color) {
1005 lab_[ix] = j;
1006 ptn_[ix] = color;
1007 ix++;
1008 }
1009 }
1010 if (ix > 0) ptn_[ix-1] = 0;
1011 }
1012
1013 /*
1014 for (int j = 0; j < n_; j++)
1015 printf("ptn %d = %d lab = %d \n", j, ptn_[j], lab_[j]);
1016 */
1017
1018 // Should be number of columns
1019 assert(ix == n_);
1020 // Now the constraints if needed
1021
1022 // Compute Partition
1023
1024 if (G_) {
1025#ifndef NTY_TRACES
1026 nauty(G_, lab_, ptn_, active_, orbits_, options_,
1027 stats_, workspace_, worksize_, m_, n_, canonG_);
1028#else
1029 abort();
1030#endif
1031 } else {
1032#ifndef NTY_TRACES
1033 options_->dispatch = &dispatch_sparse;
1034 sparsenauty(GSparse_, lab_, ptn_, orbits_, options_,
1035 stats_, NULL);
1036#else
1037 //options_->dispatch = &dispatch_sparse;
1038 Traces(GSparse_, lab_, ptn_, orbits_, options_,
1039 stats_, NULL);
1040#endif
1041 }
1042 autoComputed_ = true;
1043
1044 double endCPU = CoinCpuTime ();
1045
1046 nautyTime_ += endCPU - startCPU;
1047 // Need to make sure all generators are written
1048 if (afp_) fflush(afp_);
1049}
1050
1051void
1052CbcNauty::deleteElement(int ix, int jx)
1053{
1054 // Right now die if bad index. Can throw exception later
1055 assert(ix < n_ && jx < n_);
1056 set *gv = GRAPHROW(G_, ix, m_);
1057 if (ISELEMENT(gv, jx)) {
1058 DELELEMENT(gv, jx);
1059 }
1060 autoComputed_ = false;
1061}
1062
1063double
1064CbcNauty::getGroupSize() const
1065{
1066 if (!autoComputed_) return -1.0;
1067 return( stats_->grpsize1 * pow(10.0, (double) stats_->grpsize2) );
1068}
1069
1070int
1071CbcNauty::getNumGenerators() const
1072{
1073 if (!autoComputed_) return -1;
1074 return(stats_->numgenerators);
1075}
1076
1077int
1078CbcNauty::getNumOrbits() const
1079{
1080 if (!autoComputed_) return -1;
1081 return(stats_->numorbits);
1082}
1083
1084std::vector<std::vector<int> >
1085*CbcNauty::getOrbits() const
1086{
1087 std::vector<std::vector<int> > *orb = new std::vector<std::vector<int> >;
1088 if (!autoComputed_) return orb;
1089 orb -> resize(getNumOrbits());
1090 std::multimap<int, int> orbmap;
1091 std::set<int> orbkeys;
1092 for (int j = 0; j < n_; j++) {
1093 orbkeys.insert(orbits_[j]);
1094 orbmap.insert(std::make_pair(orbits_[j], j));
1095 }
1096
1097 int orbix = 0;
1098 for (std::set<int>::iterator it = orbkeys.begin();
1099 it != orbkeys.end(); ++it) {
1100 std::multimap<int, int>::iterator pos;
1101 for (pos = orbmap.lower_bound(*it);
1102 pos != orbmap.upper_bound(*it); ++pos) {
1103 (*orb)[orbix].push_back(pos->second);
1104 }
1105 orbix++;
1106 }
1107
1108 assert(orbix == getNumOrbits());
1109 return orb;
1110}
1111
1112void
1113CbcNauty::getVstat(double *v, int nv)
1114{
1115 assert(nv == n_);
1116 memcpy(v, vstat_, nv * sizeof(VarStatus));
1117}
1118
1119/*
1120bool
1121CbcNauty::isAllFixOneOrbit(const std::vector<int> &orbit) const
1122{
1123
1124 for(std::vector<int>::const_iterator it = orbit.begin();
1125 it != orbit.end(); ++it) {
1126 if (*it >= n_) return false;
1127 if (vstat_[*it] != FIX_AT_ONE) return false;
1128 }
1129 return true;
1130}
1131
1132bool
1133CbcNauty::isAllFreeOrbit(const std::vector<int> &orbit) const
1134{
1135 for(std::vector<int>::const_iterator it = orbit.begin();
1136 it != orbit.end(); ++it) {
1137 if (*it >= n_) return false;
1138 if (vstat_[*it] != FREE) return false;
1139 }
1140 return true;
1141}
1142
1143bool
1144CbcNauty::isConstraintOrbit(const std::vector<int> &orbit) const
1145{
1146 for(std::vector<int>::const_iterator it = orbit.begin();
1147 it != orbit.end(); ++it) {
1148 if (*it >= n_) return true;
1149 }
1150 return false;
1151
1152}
1153
1154bool
1155CbcNauty::isMixedFreeZeroOrbit(const std::vector<int> &orbit) const
1156{
1157 bool containsFree = false;
1158 bool containsZero = false;
1159
1160 for(std::vector<int>::const_iterator it = orbit.begin();
1161 it != orbit.end(); ++it) {
1162 if (*it >= n_) return false;
1163 if (vstat_[*it] == FREE) containsFree = true;
1164 if (vstat_[*it] == FIX_AT_ZERO) containsZero = true;
1165 if (containsFree && containsZero) break;
1166 }
1167 return (containsFree && containsZero);
1168}
1169*/
1170
1171void
1172CbcNauty::setWriteAutoms(const std::string &fname)
1173{
1174 afp_ = fopen(fname.c_str(), "w");
1175 options_->writeautoms = TRUE;
1176#ifndef NTY_TRACES
1177 options_->writemarkers = FALSE;
1178#endif
1179 options_->outfile = afp_;
1180
1181}
1182
1183void
1184CbcNauty::unsetWriteAutoms()
1185{
1186 fclose(afp_);
1187 options_->writeautoms = FALSE;
1188}
1189
1190// Default Constructor
1191CbcOrbitalBranchingObject::CbcOrbitalBranchingObject()
1192 : CbcBranchingObject(),
1193 column_(-1),
1194 numberOther_(0),
1195 numberExtra_(0),
1196 fixToZero_(NULL)
1197{
1198}
1199
1200// Useful constructor
1201CbcOrbitalBranchingObject::CbcOrbitalBranchingObject (CbcModel * model, int column,
1202 int way ,
1203 int numberExtra,
1204 const int * extraToZero)
1205 : CbcBranchingObject(model, -1, way, 0.5),
1206 column_(column),
1207 numberOther_(0),
1208 numberExtra_(0),
1209 fixToZero_(NULL)
1210{
1211 CbcSymmetry * symmetryInfo = model->symmetryInfo();
1212 assert (symmetryInfo);
1213 // Filled in (hopefully)
1214 const int * orbit = symmetryInfo->whichOrbit();
1215 int iOrbit=orbit[column];
1216 assert (iOrbit>=0);
1217 int numberColumns = model->getNumCols();
1218 numberOther_=-1;
1219 for (int i=0;i<numberColumns;i++) {
1220 if (orbit[i]==iOrbit)
1221 numberOther_++;
1222 }
1223 assert (numberOther_>0);
1224 nautyBranchSucceeded_++;
1225 nautyOtherBranches_ += numberOther_;
1226 numberExtra_ = numberExtra;
1227 fixToZero_ = new int [numberOther_+numberExtra_];
1228 int n=0;
1229 for (int i=0;i<numberColumns;i++) {
1230 if (orbit[i]==iOrbit && i != column)
1231 fixToZero_[n++]=i;
1232 }
1233 for (int i=0;i<numberExtra;i++) {
1234 fixToZero_[n++]=extraToZero[i];
1235 }
1236}
1237
1238// Copy constructor
1239CbcOrbitalBranchingObject::CbcOrbitalBranchingObject (const CbcOrbitalBranchingObject & rhs)
1240 : CbcBranchingObject(rhs),
1241 column_(rhs.column_),
1242 numberOther_(rhs.numberOther_),
1243 numberExtra_(rhs.numberExtra_)
1244{
1245 fixToZero_ = CoinCopyOfArray(rhs.fixToZero_,numberOther_+numberExtra_);
1246}
1247
1248// Assignment operator
1249CbcOrbitalBranchingObject &
1250CbcOrbitalBranchingObject::operator=( const CbcOrbitalBranchingObject & rhs)
1251{
1252 if (this != &rhs) {
1253 CbcBranchingObject::operator=(rhs);
1254 delete [] fixToZero_;
1255 column_=rhs.column_;
1256 numberOther_=rhs.numberOther_;
1257 numberExtra_=rhs.numberExtra_;
1258 fixToZero_ = CoinCopyOfArray(rhs.fixToZero_,numberOther_+numberExtra_);
1259 }
1260 return *this;
1261}
1262CbcBranchingObject *
1263CbcOrbitalBranchingObject::clone() const
1264{
1265 return (new CbcOrbitalBranchingObject(*this));
1266}
1267
1268
1269// Destructor
1270CbcOrbitalBranchingObject::~CbcOrbitalBranchingObject ()
1271{
1272 delete [] fixToZero_;
1273}
1274
1275double
1276CbcOrbitalBranchingObject::branch()
1277{
1278 decrementNumberBranchesLeft();
1279 if (model_->logLevel()>1)
1280 print();
1281 OsiSolverInterface * solver = model_->solver();
1282 if (way_ < 0) {
1283 solver->setColUpper(column_,0.0);
1284 for ( int i = 0; i < numberOther_+numberExtra_; i++) {
1285 solver->setColUpper(fixToZero_[i],0.0);
1286 }
1287 way_ = 1; // Swap direction
1288 } else {
1289 solver->setColLower(column_,1.0);
1290 for (int i = numberOther_; i < numberOther_+numberExtra_; i++) {
1291 solver->setColUpper(fixToZero_[i],0.0);
1292 }
1293 way_ = -1; // Swap direction
1294 }
1295 return 0.0;
1296}
1297/* Update bounds in solver as in 'branch' and update given bounds.
1298 branchState is -1 for 'down' +1 for 'up' */
1299void
1300CbcOrbitalBranchingObject::fix(OsiSolverInterface * solver,
1301 double * lower, double * upper,
1302 int branchState) const
1303{
1304 if (branchState < 0) {
1305 upper[column_]=0.0;
1306 for ( int i = 0; i < numberOther_+numberExtra_; i++) {
1307 upper[fixToZero_[i]]=0.0;;
1308 }
1309 } else {
1310 lower[column_]=1.0;
1311 for (int i = numberOther_; i < numberOther_+numberExtra_; i++) {
1312 upper[fixToZero_[i]]=0.0;;
1313 }
1314 }
1315}
1316// Print what would happen
1317void
1318CbcOrbitalBranchingObject::print()
1319{
1320 if (way_ < 0) {
1321 printf("Orbital Down - to zero %d",column_);
1322 for ( int i = 0; i < numberOther_+numberExtra_; i++) {
1323 printf(" %d",fixToZero_[i]);
1324 }
1325 } else {
1326 printf("Orbital Up - to one %d, to zero",column_);
1327 for (int i = numberOther_; i < numberOther_+numberExtra_; i++) {
1328 printf(" %d",fixToZero_[i]);
1329 }
1330 }
1331 printf("\n");
1332}
1333
1334/** Compare the original object of \c this with the original object of \c
1335 brObj. Assumes that there is an ordering of the original objects.
1336 This method should be invoked only if \c this and brObj are of the same
1337 type.
1338 Return negative/0/positive depending on whether \c this is
1339 smaller/same/larger than the argument.
1340*/
1341int
1342CbcOrbitalBranchingObject::compareOriginalObject
1343(const CbcBranchingObject* brObj) const
1344{
1345 const CbcOrbitalBranchingObject* br =
1346 dynamic_cast<const CbcOrbitalBranchingObject*>(brObj);
1347 assert(!br);
1348 abort();
1349 return 0;
1350}
1351
1352/** Compare the \c this with \c brObj. \c this and \c brObj must be os the
1353 same type and must have the same original object, but they may have
1354 different feasible regions.
1355 Return the appropriate CbcRangeCompare value (first argument being the
1356 sub/superset if that's the case). In case of overlap (and if \c
1357 replaceIfOverlap is true) replace the current branching object with one
1358 whose feasible region is the overlap.
1359*/
1360CbcRangeCompare
1361CbcOrbitalBranchingObject::compareBranchingObject
1362(const CbcBranchingObject* brObj, const bool replaceIfOverlap)
1363{
1364 const CbcOrbitalBranchingObject* br =
1365 dynamic_cast<const CbcOrbitalBranchingObject*>(brObj);
1366 assert(!br);
1367 abort();
1368 return CbcRangeDisjoint;
1369}
1370#endif
1371
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# Astrobiobound! The Search for Life in the Solar System
This project engages students in the science and engineering processes used by NASA Astrobiologists as they explore our Solar System and try to answer the compelling question, "Are we Alone?" Students will identify science mission goals and select... (View More)
# Carbon Now
Unit two of the "Carbon Connections: The Carbon Cycle and the Science of Climate" curriculum examines the role of carbon and the carbon cycle in current climate. Students discover how carbon in Earth's system is monitored and also investigate the... (View More)
# Life in Icy Places
This is a lesson about the field of astrobiology, the study of life in the universe, and ice as a preservative for evidence of life. Learners will consider the relationship between ice and life as they investigate the conditions required for life to... (View More)
# Using Mathematical Models to Investigate Planetary Habitability: Activity B Making a Simple Mathematical Mode
In this activity, students build a simple computer model to determine the black body surface temperature of planets in our solar system: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto. Experiments altering the luminosity and... (View More)
Audience: High school
Materials Cost: Free per group of students
# Using Mathematical Models to Investigate Planetary Habitability: Activity C The Role of Actual Data in Mathematical Models
Students explore how mathematical descriptions of the physical environment can be fine-tuned through testing using data. In this activity, student teams obtain satellite data measuring the Earth's albedo, and then input this data into a... (View More)
Audience: High school
Materials Cost: Free per student
# Can Venus and Mars Be Made Habitable?
This activity is about planetary climate. Once familiar with the factors that determine a planet's surface temperature, learners will use an interactive spreadsheet model of a planet's atmosphere to determine if greenhouse gases, luminosity of the... (View More)
Audience: High school
Materials Cost: Free
# Temperature Variations and Habitability: Activity A Observing, Describing, and Adapting to Environmental Variations
In this activity, student teams design and conduct a scientific investigation in which they explore the conditions necessary for life. They conduct observations of environmental conditions both indoor and outdoor, and determine the range of... (View More)
Audience: High school
Materials Cost: Over \$20 per group of students
# How do Atmospheres Affect Planetary Temperatures? Activity B How do Atmospheres Produce their Effect Upon Surface Temperatures?
In this kinesthetic activity, the concept of energy budget is strengthened as students conduct three simulations using play money as units of energy, and students serve as parts of a planetary radiation balance model. Students will determine the... (View More)
Audience: High school
Materials Cost: Free per student
# Bird Beak Accuracy Assessment
In this activity, students quantitatively evaluate the accuracy of a classification and understand a simple difference/error matrix. Students sort birds into three possible classes based on each bird’s beak: carnivores (meat eaters), herbivores... (View More)
# What is Industrial Agriculture?
Highly productive commercial agriculture, known as industrial agriculture, is the focus of this investigation. Students will analyze and compare the inputs and outputs of industrial agriculture after reading background information, examining photos... (View More)
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# Recent content by zz_
1. ### Complex 2-Move Generator Reduction Method (c2gr)
First I took into account solved cube with base line on its place. I devided remaining corners into two cycles: two corners at DR and four corners at U and started generating and drawing all possible situations. Then I devided four corners' cycle into two cycles with two corners but the...
2. ### Complex 2-Move Generator Reduction Method (c2gr)
Micki Let me use my colour scheme as an example: red at front and yellow at top. First let us look at the solved cube. According to the pictures of the solved cube, described in "New concepts" part, there are two pictures which represent two possibilities of three vectors of two corners each...
3. ### Complex 2-Move Generator Reduction Method (c2gr)
Pyjam Your situation is D8 after making "a" move. Pay special attention that every two corners make vectors, not pairs - they are directed and you must take under consideration the vectors' beginnings and ends (white-black or black-white). Finally your case (only CP) is: (-) (a U R U' a') (U R...
4. ### Complex 2-Move Generator Reduction Method (c2gr)
Hello Micki I had doubts on my description if it will be understood, but here I explain it to you. The picture D1, similarly to all other pictures, does not mean "<corner> should go to <position>"; it is a picture of vectors described in "New concepts" part, where you can find two versions of...
5. ### Complex 2-Move Generator Reduction Method (c2gr)
Hello Pyjam, First primo: I am not Mr, but just zz. Second primo: I think this method should be easier and moves for the whole cube should be more ergonomic than in old zz. I am only afraid the second step will show up the weakness. zz
6. ### Complex 2-Move Generator Reduction Method (c2gr)
Thank you for your answer. So you are able to predict all the moves for EOPair and CPLine in this short time of pre-inspection for every scramble? I am really impressed. I thought you predict only EOPair. Regards zz
7. ### Complex 2-Move Generator Reduction Method (c2gr)
Hello I want to share with you with my new speedcubing method, which I have described at my OneDrive: https://1drv.ms/w/s!Am1bZghSQmingyzL3Msq9ZW2LMc5 Generally the idea is to predict in pre-inspection time the moves leading the cube to <U,u,R,r> group situation. I have called this Complex...
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All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Time for some Prime (Posted on 2013-05-05)
Prove that if m=5^n+3^n+1 is a prime, then 12 divides n.
No Solution Yet Submitted by Danish Ahmed Khan No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
possible solution Comment 1 of 1
The expression is divisible by 3 if n is odd, and divisible by 5 if n is of the form 4n+2. The net effect is that n must be a multiple of 4. But even then, the expression is divisible by 7, unless the said multiple is also divisible by 3. Hence, if the expression is prime, then 12 divides n.
n=0, n=12, n=36, n=48, are prime solutions.
n=24 is 103×457×61331×20646607.
n=60 is 3144853×3847314239×680874654589×105287080263229
Posted by broll on 2013-05-06 01:29:17
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https://www.physicsforums.com/threads/determine-initial-velocity-max-ht-for-diver-proj-motion-hw.289244/
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Determine Initial Velocity, Max Ht for Diver Proj. Motion HW
• TheShehanigan
In summary, The diver leaves the end of a five meter high diving board and strikes the water three meters beyond the end. His initial velocity, height, and velocity when entering the water can all be determined using the homework equations.
TheShehanigan
Homework Statement
A high diver leaves the end pf a 5.0-m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Determine his inicial velocity, his maximum height and the velocity with which he enters water. (Projectile Motion).
*See details at the end*
Homework Equations
y = yo + vot + 1/2gt^2
The Attempt at a Solution
I have worked on this exercise, and can fully answer it using the water as the origin with the y-axis pointing upward. However, I wanted to try it with the diver leaving the board as the origin with the y-axis pointing downward (thus making g positive) and it's not giving me the correct result. Once I clear y = yo + vot + 1/2gt^2 for vo, I obtain vo = (y - yo + 1/2gt^2)/t. Since I am using the y-axis downward, I am making y = 5.00m and g = 9.8 m/s^2. I left the +1/2 gt^2 0 upon clearing the formula for vo since the acceleration is always downward (thus positive always).
However, it's not giving me the right result. Someone see anything wrong with my logic?
*I don't need help with the problem per-se, but with why the y component of velocity isn't giving me the right number. *
The correct answer should be 2.5 i for the y component of vo.
When you choose down as positive y then the sense of velocity and acceleration are both changed. Picking a proper origin is also complicating.
It should all work out the same, but you will need to take care in what is y=0 and with +/- for Vyo as well as g.
So a = - g for upward becomes a = g for downward, (with g = 9.80 m/s^2) and what would be vo = + for upward becomes vo = - for downward, or what? Also, y = 0 is the exact last point of the diving board in the way I'm trying to make the exercise.
I uploaded a diagram for reference.
Attachments
• Diagram.JPG
13.1 KB · Views: 843
Last edited:
If Vo is against the direction you chose as positive, then of course it is negative.
What is the formula for determining initial velocity in projectile motion?
The formula for determining initial velocity in projectile motion is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
How do you find the maximum height for a projectile in motion?
The maximum height for a projectile in motion can be found using the formula h = (u^2sin^2θ)/2g, where h is the maximum height, u is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
What is the role of gravity in determining the initial velocity and maximum height of a projectile?
Gravity plays a crucial role in determining the initial velocity and maximum height of a projectile. It is the force that causes the projectile to accelerate towards the ground, and it affects the trajectory of the projectile. The acceleration due to gravity is also used in the formulas for determining initial velocity and maximum height.
How does air resistance affect the initial velocity and maximum height of a projectile?
Air resistance can decrease the initial velocity and maximum height of a projectile as it acts against the motion of the projectile. The force of air resistance increases as the velocity of the projectile increases, so it has a greater impact on high-velocity projectiles.
Can the initial velocity and maximum height of a projectile be determined without considering air resistance?
Yes, the initial velocity and maximum height of a projectile can be determined without considering air resistance. This is because air resistance only has a significant impact on projectiles with high velocities. For low-velocity projectiles, the effect of air resistance is negligible, and the calculations can be done without considering it.
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## FANDOM
2,497 Pages
The Old Safe is a puzzle in Professor Layton and the Diabolical Box.
## Puzzle
US Version
Find the four-number code that opens the safe. You can use the numbers zero through five in your answer, but each number can be used only once. The small lights next to each row of numbers are the key to finding the code, as they tell you how much in common that row has with the final code. Each white light indicates a number that matches one in the code but is in different place within the sequence. An orange light indicates a number that is in the code and in its correct spot. Tap the numbers at the bottom to enter the code.
UK Version
Find the four-digit code that opens the safe. You can use the digits 0 to 5 in your answer, but each digit can only be used once. The small lights next to each row of digits are the key to finding the code, as they tell you how much in common that row has with the code. A white light indicates a digit that is in the code but in a different place in the sequence. An orange light indicates a digit that is in the code and in the correct spot. Touch the digits at the bottom of the screen to change them, then touch Submit when you have your answer.
## Hints
Click a Tab to reveal the Hint.
US Version
Start by figuring out which numbers belong in the code, regardless of order.
UK Version
Start by working out which digits belong in the code, regardless of order.
US Version
You may have already noticed, but every number in the code is in the very top row of numbers. Two of the numbers in 4150 are even in the correct spot relative to where they should be in the code.
UK Version
You may have already noticed, but every digit in the code is in the very top row of digits. Two of the digits in 4150 are even in the right place.
US Version
Take a look at the third row of numbers.
You can see that two numbers math ones in the code and are in the correct position. Since you know the code uses 4, 1, 5, and 0, you can assume the 0 and 1 from this row are part of the code and in their appropriate spots.
Now just figure out where 4 and 5 go.
UK Version
Take a look at the third row of digits.
You can see that two digits match ones in the and are in the correct position. Since you know the code uses 4, 1, 5 and 0, you can assume the 0 and 1 from this row are part of the code and in the right place.
Now work out where 4 and 5 go.
## Solution
### Incorrect
Give the puzzle another try.
### Correct
That's right!
US Version
The code is 0154. Some people solve this by entering numbers haphazardly until they stumble upon the solution, but it's much more fun to think your way to the answer.
UK Version
The code is 0154. You could solve this by entering numbers haphazardly until you stumble upon the solution, but it's much more fun to think your way to the answer.
A big thanks to http://professorlayton2walkthrough.blogspot.com
Community content is available under CC-BY-SA unless otherwise noted.
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### Home > CCAA8 > Chapter 10 Unit 11 > Lesson CCA: 10.3.3 > Problem10-137
10-137.
Consider the quadratic inequality $x^2+2x+1<4$.
1. Solve for the boundary point(s). How many boundary points are there?
Rearrange the inequality so that you can factor it.
How many solutions do you get when you do this?
There are $2$ boundary points. $x=1$ and $x=−3$.
2. Place the boundary point(s) on a number line. How many regions do you need to test?
If there are $2$ boundary points, how many regions are there surrounding those two points on a number line?
3. Test each region and determine which one(s) make the inequality true. Identify the solution algebraically and on the number line.
Substitute a point from each one of these regions into the inequalities.
Does that point make the inequality true? If not, then it is not included.
$−3.
Make sure to draw this on a number line.
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# physics
posted by on .
A 0.10 kilogram ball dropped vertically from a height of 1.0 meter above the floor bounces back to a height of 0.80 meter. The mechanical energy lost by the ball as it bounces is aprox.
a. 0.080 J
b. 0.20 J
c. 0.30 J
d. 0.78 J
• physics - ,
E=mgh1-mgh2=mg(h1-h2)=
=0.1•9.8(1-0.8)=0.78 J
• physics - ,
B.) (.10kg)(10 m/s^2)(0.20m) = .20J
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# The Electric Field The electric field is present
• Slides: 12
The Electric Field The electric field is present in any region of space if there exists electric forces on charges. These electric forces can be detected using a test charge. Test charges are theoretical positive charges that do not alter the electric field to be detected. Electric field at a point in space is defined as the electric force per unit test charge placed at that point. The SI unit for the electric field is the newton per coulomb (N/C). Electric field is a vector.
Electric Force
Electric Field Lines • • • Positive point charge Negative point charge Dipole Two like charges Parallel plate Capacitor
Effects of Electric Field
Charge moving in an Electric Field [CJ 10 P 52, Chap 18] The right drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 5. 71 × 106 m/s. The capacitor is 2. 00 cm long, and its plates are separated y 0. 150 cm. a. When the capacitor plates are not charged, as in the left drawing, draw the subsequent motion of the electron, and determine how long the electron takes to cross the plates. b. Assume that the electric field between the plates is uniform everywhere in the right drawing, and find its magnitude.
The Electric Field of a Point Charge
P 45 Chap 18 Two charges are located on the x axis: q 1 = +6. 0 µC at x 1 = +4. 0 cm, and q 2 = +6. 0 µC at x 2 = -4. 0 cm. Two other charges are located on the y axis: q 3 = +3. 0 µC at y 3 = +5. 0 cm, and q 4 = -8 µC at y 4 = +7. 0 cm. Find (a) the magnitude and (b) the direction of the net electric field at the origin.
18. 8. The Electric Field Inside a Conductor: Shielding At electrostatic equilibrium: 1. Any excess charge resides on the surface. 2. The electric field is zero inside the conductor.
Conductor in Electric Field Under electrostatic equilibrium: 1. The conductor shields the electric field. 2. The electric field just outside the surface a conductor is perpendicular to the surface. Sensitive electronic circuits are often enclosed within metal boxes that provide shielding from external fields.
18. 10 Copying Machine
Electrostatic Precipitator (a) Schematic of an electrostatic precipitator. Air is passed through grids of opposite charge. The first grid charges airborne particles, while the second attracts and collects them. (b) The dramatic effect of electrostatic precipitators is seen by the absence of smoke from this power plant. (credit: Cmdalgleish, Wikimedia Commons)
Gauss' Law: The electric flux, ΦE through a Gaussian surface is equal to the net charge Q enclosed by the surface divided by, ϵ 0 the permittivity of free space:
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# 150 students at seward high school. 66 play baseball, 45
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150 students at seward high school. 66 play baseball, 45 [#permalink]
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24 Jul 2007, 21:36
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150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?
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Re: PS: high school sports [#permalink]
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25 Jul 2007, 01:15
ArvGMAT wrote:
150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?
I'm getting 57 by using venn diagram method.
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### Show Tags
25 Jul 2007, 03:23
66 + 45 + 42 - 2*27 - 3*3 = 153 - 60 = 93 play atleast one.
Ans : 57.
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25 Jul 2007, 11:35
I get 21 ...
65+45+42-27+3=129
150-129=21 dont play anything
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### Show Tags
25 Jul 2007, 11:44
fresinha12 wrote:
I get 21 ...
65+45+42-27+3=129
150-129=21 dont play anything
I agree with the method and the answer. I made a calculation error. it has to be 21
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25 Jul 2007, 11:47
Hi fresinha12
Why have you added the triple? I think you should minus two times tripple. What do you think?
In the condition we have:
"27 play exactly 2 sports and 3 play all 3 sports"
I think that in 27 we don't have triple.
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### Show Tags
25 Jul 2007, 12:08
66 + 45 + 42 - 2*27 - 3*3 = 153 - 60 = 93 play atleast one.
Ans : 57.
I agree with this rationale, but I get 60.
66+45+42=153
2*27+3*3=63
153-63=90 that play one sport.
Therefore 150-90 = 60 that play no sport.
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25 Jul 2007, 15:40
emoryhopeful wrote:
66 + 45 + 42 - 2*27 - 3*3 = 153 - 60 = 93 play atleast one.
Ans : 57.
I agree with this rationale, but I get 60.
66+45+42=153
2*27+3*3=63
153-63=90 that play one sport.
Therefore 150-90 = 60 that play no sport.
That was a stupid one. I guess I got influenced by the previous post. Yeah... 60 looks correct.
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25 Jul 2007, 16:19
23 play no sports at all.
each of the given sports have a total of 10 players that play that and at least one other sport. these add up to 127 sports participants.
150 - 127 = 23 play nothing.
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Re: PS: high school sports [#permalink]
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25 Jul 2007, 18:34
ArvGMAT wrote:
150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?
Using a Venn diagram, there are 3 areas that overlap 2 sports, and 1 area that overlaps all three sports. Put 9 into each of the "double" areas (for the 27 that play 2 sports), and 3 into the center of the diagram. (Assuming that the 27 are equally divided among the sports, but that really doesn't matter). Then you can see that for each sport, you need to subtract 21. That means that, of the students that only play one sport, you have 45 in baseball, 24 in basketball, and 21 in soccer. That, plus the 27 (2 sports) and 3(3 sports)= 120 students that play sports. So 30 don't.
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Re: PS: high school sports [#permalink]
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25 Jul 2007, 21:16
Robin in NC wrote:
ArvGMAT wrote:
150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?
Using a Venn diagram, there are 3 areas that overlap 2 sports, and 1 area that overlaps all three sports. Put 9 into each of the "double" areas (for the 27 that play 2 sports), and 3 into the center of the diagram. (Assuming that the 27 are equally divided among the sports, but that really doesn't matter). Then you can see that for each sport, you need to subtract 21. That means that, of the students that only play one sport, you have 45 in baseball, 24 in basketball, and 21 in soccer. That, plus the 27 (2 sports) and 3(3 sports)= 120 students that play sports. So 30 don't.
OMG. This looks correct too. I think the number I found in the first post is actually the no. of people who play just one sport. I should have added the rest too.
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### Show Tags
25 Jul 2007, 21:55
Set up a venn diagram with the following:
#(all three) = 3
#(soccer and baseball) = z
So x+y+z = 27
#(none) = n
63-x-z + 42-x-y + 39-y-z + x + y + z + 3 + n= 150
147-x-y-z+n= 150
147 - (x+y+z) + n = 150
147 - (27) + n = 150
n = 30
Sorry, too tired and lazy to draw it out =(
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26 Jul 2007, 01:11
ywilfred wrote:
Set up a venn diagram with the following:
#(all three) = 3
#(soccer and baseball) = z
So x+y+z = 27
#(none) = n
63-x-z + 42-x-y + 39-y-z + x + y + z + 3 + n= 150
147-x-y-z+n= 150
147 - (x+y+z) + n = 150
147 - (27) + n = 150
n = 30
Sorry, too tired and lazy to draw it out =(
Great explanation. Thanks a lot. Helped me correct my mistake with the venn diagram.
Using the formula for 3 overlapping sets we get 30 as well.
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Re: PS: high school sports [#permalink]
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26 Jul 2007, 16:09
Robin in NC wrote:
ArvGMAT wrote:
150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?
Using a Venn diagram, there are 3 areas that overlap 2 sports, and 1 area that overlaps all three sports. Put 9 into each of the "double" areas (for the 27 that play 2 sports), and 3 into the center of the diagram. (Assuming that the 27 are equally divided among the sports, but that really doesn't matter). Then you can see that for each sport, you need to subtract 21. That means that, of the students that only play one sport, you have 45 in baseball, 24 in basketball, and 21 in soccer. That, plus the 27 (2 sports) and 3(3 sports)= 120 students that play sports. So 30 don't.
duh. what a dumb mistake I made. great write-up.
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27 Jul 2007, 13:19
you minus the doubles cause you have counted them twice, once when we are given the number of basket ball player, football etc..you add the triple cause triple +single -double will give you the total number of players...
try it with a simple example..
Andrey2010 wrote:
Hi fresinha12
Why have you added the triple? I think you should minus two times tripple. What do you think?
In the condition we have:
"27 play exactly 2 sports and 3 play all 3 sports"
I think that in 27 we don't have triple.
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27 Jul 2007, 13:27
ywilfred wrote:
Set up a venn diagram with the following:
#(all three) = 3
#(soccer and baseball) = z
So x+y+z = 27
#(none) = n
63-x-z + 42-x-y + 39-y-z + x + y + z + 3 + n= 150
147-x-y-z+n= 150
147 - (x+y+z) + n = 150
147 - (27) + n = 150
n = 30
Sorry, too tired and lazy to draw it out =(
nothing to add. but i do differently
total = x + y + z - (xy + yz + xz) - 2(xyz) + n
150 = 66 + 45 + 42 - (27) - 2(3) + n
n = 30
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31 Jul 2007, 03:12
Himalayan wrote:
ywilfred wrote:
Set up a venn diagram with the following:
#(all three) = 3
#(soccer and baseball) = z
So x+y+z = 27
#(none) = n
63-x-z + 42-x-y + 39-y-z + x + y + z + 3 + n= 150
147-x-y-z+n= 150
147 - (x+y+z) + n = 150
147 - (27) + n = 150
n = 30
Sorry, too tired and lazy to draw it out =(
nothing to add. but i do differently
total = x + y + z - (xy + yz + xz) - 2(xyz) + n
150 = 66 + 45 + 42 - (27) - 2(3) + n
n = 30
Hi Himalayan,
However, could you plz tell me if it is the standard formula which you have used to solve the problem.
Regards
Nikhil
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02 Aug 2007, 10:14
I am still sticking with 21.....
whats the OA??
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16 Jun 2008, 05:35
[quote="ywilfred"]Set up a venn diagram with the following:
#(all three) = 3
#(soccer and baseball) = z
So x+y+z = 27
#(none) = n
63-x-z + 42-x-y + 39-y-z + x + y + z + 3 + n= 150
147-x-y-z+n= 150
147 - (x+y+z) + n = 150
147 - (27) + n = 150
n = 30
Sorry, too tired and lazy to draw it out =([/quote]
Great explanation, was stuck in this ques for a while. Thanks.
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16 Jun 2008, 14:02
ioiio wrote:
66 + 45 + 42 - 2*27 - 3*3 = 153 - 60 = 93 play atleast one.
Ans : 57.
ioiio, you are correct except you don't multiply the 27 by 2. The 27 figure encompasses all 2-sport individuals and thus no manipulation is needed. The formula is:
A + B + C -AB - AC - BC - 2ABC + Neither = Total
In this problem, it states that AB + AC + BC = 27 --> don't need to multiply by 2.
Plus, the 3 figure is multiplied by 2, not 3.
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Re: [#permalink] 16 Jun 2008, 14:02
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Hey guys...
I created a uber-basic brute force program(not exactly 'brute force'), where in the program opens a 'log.txt' file and reads the contents. I created a list of alphabets A-Z and numbers 0-9....what I wanted the program to do was, try out different combinations with the given set of characters until it matches the given set of characters in the log.txt file, and then display an appropriate message...here's the code I used:
killzone2(in the log.txt file)
``````import random
combos=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','v','w','x','y','z']
combos.append(range(10))
choice=random.choice(combos)
log=open('log.txt','r')
choice=random.choice(combos)
print choice``````
Thanks for the help in advance all you guys! ;P
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Last Post by vegaseat
When comparing two strings you want to make sure they are the same length. They will never be equal if they are different lengths, so add a statement which prints if they are not equal,
``````log_fp=open('log.txt','r')
print "choice length", choice, len(choice), \
choice=random.choice(combos)
print choice``````
When comparing two strings you want to make sure they are the same length. They will never be equal if they are different lengths, so add a statement which prints if they are not equal,
``````log_fp=open('log.txt','r')
print "choice length", choice, len(choice), \
choice=random.choice(combos)
print choice``````
Hey, I got to know one more thing(I think I'm correct) that the 'choice' command only picks a *single* entity from the list....so since my actual text is 'killzone2', the program will go on for ever and never end, because every time, only a single character will be matched with the given text! :( Any idea on how to actually pick items randomly from a list? Thnks
Even if you use the modified code below, it will take almost forever ...
``````import random
import string
combos = list(string.ascii_lowercase + string.digits)
#log=open('log.txt','r')
# for test only
search = 'killzone2'
combo = ''
while combo != search:
# build up string combo the same length as search
for n in search:
combo += random.choice(combos)
# test
print combo,
# reset
combo = ''
print combo``````
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# Finding locations with a distance between 1 and 5000
I'm wondering how I could refactor this code. It may look a little reiterative and probably there's a shorter method to accomplish the same.
def self.location(distance=100,location)
if distance.is_a? Integer
if distance.between?(1,5000)
distance = distance
elsif distance < 1
distance = 1
elsif distance > 5000
distance = 5000
end
else
distance = 100
end
if location
within(distance, :origin => location)
else
find(:all)
end
end
## migrated from stackoverflow.comApr 26 '12 at 3:30
This question came from our site for professional and enthusiast programmers.
• begin by splitting in two methods: one with distance condition, other with conditionnal scope – apneadiving Apr 25 '12 at 22:11
• I think checking if it's Integer is a smell. What if I pass 1.0 to the function? Why would it be replaced with 100? – Alexey Apr 29 '12 at 18:57
Shorter, and while min/max was tricky, IMO, this is easier to understand:
def self.location(distance=100,location)
distance = 100 unless distance.is_a?(Integer)
distance = 1 if distance < 1
distance = 5000 if distance > 5000
if location
within(distance, :origin => location)
else
find(:all)
end
end
I've replaced the part where you take care that distance is between 1 and 5000 with distance = [1, [distance, 5000].min].max .
def self.location(distance=100,location)
if distance.is_a? Integer
distance = [1, [distance, 5000].min].max
else
distance = 100
end
if location
within(distance, :origin => location)
else
find(:all)
end
end
• whats exactly happening here [1, [distance, 5000].min].max? Is it picking the lower value from distance and 5000 then the highest value from the product of the earlier comparison and 1? – Benjamin Udink ten Cate Apr 25 '12 at 23:05
• we are comparing the minimum between distance and 5000 with 1, and returning the maximum there. If distance is greater than 5000, we will compare 5000 with 1, and return 5000, else if distance is less than 5000, we compare it with 1. Now if distance is less than 1, we return 1, else we return the distance wich is between 1 and 5000 – gabitzish Apr 25 '12 at 23:08
First, I would not set the distance to 100 if it's not an integer. What if someone uses a Fixnum such as 150.0? I would check to see if it responds to to_i.
Second, I would break the distance out into a separate method because it's easier to follow outside the context of another method.
def self.location(distance, location)
return find(:all) unless location
distance = normalize_distance(distance)
within(distance, :origin => location)
end
def self.normalize_distance(distance=100)
return 100 unless distance.respond_to? :to_i
return 1 if distance < 1
return 5000 if distance > 5000
distance
end
• Just because an object responds to to_i doesn't mean it represents an integer. – Mark Thomas Apr 26 '12 at 1:24
• @MarkThomas let me clarify the point I was trying to make. Checking an object's class rather than its behavior is going to make the code less flexible. If a Fixnum or Float were passed into the location method, why should it behave any different? 100.0 is equivalent to 100 but they are different classes and only one is an integer. This sort of flexibility is called duck typing: en.wikipedia.org/wiki/Duck_typing – Peter Brown Apr 26 '12 at 2:17
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## Related posts of "Pooh Coloring Book"
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#### Composition Of Functions Worksheet Answers
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CC-MAIN-2021-25
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latest
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https://www.teacherspayteachers.com/Product/Eric-Carle-Papa-Please-Get-the-Moon-for-Me-Common-Core-Math-Board-Game-791578
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text/html
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crawl-data/CC-MAIN-2017-09/segments/1487501171632.91/warc/CC-MAIN-20170219104611-00021-ip-10-171-10-108.ec2.internal.warc.gz
| 896,939,165
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Total:
\$0.00
# Eric Carle Papa, Please Get the Moon for Me Common Core Math Board Game
Subjects
Resource Types
Product Rating
3.8
File Type
PDF (Acrobat) Document File
92.64 MB | 8 pages
### PRODUCT DESCRIPTION
Inspired by Eric Carle’s Papa, Please Get the Moon For Me!, this engaging board game challenges students to move forward and backward from 1-99 on a number board. There are 3 differentiated spinners so students of all levels can play the game together. Students can practice number sequence, addition, and subtraction. They also apply reading skills to follow a series of fun directions on the game board. Great way to integrate math and literacy!
**This product is included in our Eric Carle Author Study--a Complete Integrated Unit! Check out this bundle and save a bundle:
Link-Eric Carle Author Study Unit: Integrated, Differentiated, CCSS Aligned
Learning experiences are directly correlated with the Common Core State Standards:
Know number names and the count sequence.
CCSS.Math.Content.K.CC.A.1 Count to 100 by ones and by tens.
CCSS.Math.Content.K.CC.A.2 Count forward beginning from a given number within the known sequence (instead of having to begin at 1).
CCSS.Math.Content.K.OA.A.1 Represent addition and subtraction with objects, fingers, mental images, drawings1, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
CCSS.Math.Content.K.OA.A.2 Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
Extend the counting sequence.
CCSS.Math.Content.1.NBT.A.1 Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
CCSS.Math.Content.1.OA.C.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
CCSS.Math.Content.1.OA.C.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Use place value understanding and properties of operations to add and subtract.
CCSS.Math.Content.1.NBT.C.4 Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
CCSS.Math.Content.1.NBT.C.5 Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
CCSS Informational Reading 7. Use the illustrations and details in a text to describe its key ideas.
CCSS Foundational Reading 4. Read with sufficient accuracy and fluency to support comprehension.
Total Pages
8
N/A
Teaching Duration
N/A
### Average Ratings
4.0
Overall Quality:
3.8
Accuracy:
3.8
Practicality:
3.8
Thoroughness:
3.8
Creativity:
3.8
Clarity:
3.8
Total:
8 ratings
\$3.00
User Rating: 3.9/4.0
(256 Followers)
\$3.00
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| 3.671875
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CC-MAIN-2017-09
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http://oeis.org/A190318
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text/html
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crawl-data/CC-MAIN-2019-43/segments/1570986705411.60/warc/CC-MAIN-20191020081806-20191020105306-00107.warc.gz
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A190318 Number of obtuse isosceles triangles on a (n X n)-grid. 4
0, 0, 0, 4, 36, 100, 256, 496, 968, 1672, 2736, 4092, 6188, 8764, 12144, 16464, 22224, 28928, 37400, 47076, 59244, 73580, 90344, 109000, 132048, 158000, 187528, 220716, 259348, 301388, 350088, 402792, 463176, 529720, 602888, 683092, 774476, 872100, 978232 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,4 COMMENTS Place all bounding boxes of A280639 that will fit into the n X n grid in all possible positions, and the proper rectangles in two orientations: a(n) = Sum(i=1..n, Sum(j=1..i, k * (n-i+1) * (n-j+1) * A280639(i,j))) where k=1 when i=j and k=2 otherwise. - Lars Blomberg, Mar 02 2017 LINKS Lars Blomberg, Table of n, a(n) for n = 1..10000 (the first 100 terms from Chai Wah Wu) Nathaniel Johnston, C program for computing terms FORMULA a(n) = A186434(n) - A190317(n) - A187452(n). CROSSREFS Cf. A186434, A187452, A190317, A280639. Sequence in context: A193183 A152760 A016826 * A193874 A254939 A038688 Adjacent sequences: A190315 A190316 A190317 * A190319 A190320 A190321 KEYWORD nonn AUTHOR Martin Renner, May 08 2011 EXTENSIONS a(10)-a(39) from Nathaniel Johnston, May 09 2011 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified October 20 05:49 EDT 2019. Contains 328247 sequences. (Running on oeis4.)
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| 3.421875
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CC-MAIN-2019-43
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http://www.aqua-calc.com/one-to-all/torque/preset/kilogram-force-meter/15
| 1,409,718,852,000,000,000
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| 632,368,483
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# One to all
## Torque units conversions
Unit
name value symbol reference
dyne centimeter 1 470 997 504.5 dyn⋅cm
dyne decimeter 147 099 750.45 dyn⋅dm
dyne meter 14 709 975.05 dyn⋅m
dyne millimeter 14 709 975 045 dyn⋅mm
gram force centimeter 1 500 000 gf-cm
gram force decimeter 150 000 gf-dm
gram force meter 15 000 gf-m
gram force millimeter 15 000 000 gf-mm
kilogram force centimeter 1 500 kgf-cm
kilogram force decimeter 150 kgf-dm
kilogram force meter 15 kgf-m
kilogram force millimeter 15 000 kgf-mm
newton centimeter 14 709.98 N⋅cm
newton decimeter 1 471 N⋅dm
newton meter 147.1 N⋅m
newton millimeter 147 099.75 N⋅mm
pound force foot 108.5 lbf-ft
pound force inch 1 301.94 lbf-in
#### Foods, Nutrients and Calories
Cereals ready-to-eat, KELLOGG, KELLOGG'S CORN POPS weigh(s) 31.7 gram per (metric cup) or 1.06 ounce per (US cup), and contain(s) 387 calories per 100 grams or ≈3.527 ounces [ calories | weight to volume | volume to weight | density ]
#### Gravels and Substrates
CaribSea, Freshwater, African Cichlid Mix, Sahara Sand density is equal to 1473.7 kg/m³ or 92 lb/ft³ with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylinderquarter cylinder or in a rectangular shaped aquarium or pond
#### Materials and Substances
Chloroform weigh(s) 1.489 gram per (cubic centimeter) or 0.861 ounce per (cubic inch) [ weight to volume | volume to weight | density ]
#### What is gram per square millimeter?
Gram per square millimeter (g/mm²) is a metric measurement unit of surface or areal density. The surface density is used to measure the thickness of paper, fabric and other thin materials.
#### What is energy or heat measurement?
for energy units details see below
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| 2.796875
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CC-MAIN-2014-35
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https://atcoder.jp/contests/tenka1-2013-quala/submissions/92043
| 1,618,246,991,000,000,000
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text/html
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crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00096.warc.gz
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Contest Duration: - (local time) (120 minutes) Back to Home
Submission #92043
Source Code Expand
Copy
```#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include<cassert>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include<bitset>
#define REP(i,b,n) for(int i=b;i<(int)n;i++)
#define rep(i,n) REP(i,0,n)
#define ALL(C) (C).begin(),(C).end()
#define FOR(it,o) for(__typeof((o).begin()) it=(o).begin(); it!=(o).end(); ++it)
#define dbg(x) cout << __LINE__ << ' ' << #x << " = " << (x) << endl
typedef long long ll;
using namespace std;
int B, L, N;
int cnt, pos;
string in ;
char res[1050];
int num(){
int ret = 0;
while(isdigit(in[pos])){
ret *= 10;
ret += in[pos] - '0';
pos++;
}
return ret-1;
}
pair<int, int> parse(){
if(in[pos]=='@')return make_pair(cnt, cnt);
pair<int, int> ret;
ret.first = cnt;
while(1){
if(isalpha(in[pos])){
res[cnt++] = in[pos];
pos++;
}else if(in[pos] == '('){
pos++;
pair<int, int> hoge = parse();
pos++;
int n = num();
rep(i, n){
REP(j, hoge.first, hoge.second){
res[cnt++] = res[j];
}
}
}
else{
break;
}
}
ret.second = cnt;
return ret;
}
int main(){
while(cin >> B >> L >> N){
assert(L <=100 && N <=100);
cnt = pos = 0;
cin >> in;
pair<int, int> pii = parse();\
if(B < 0){
B += pii.second;
}
REP(i, B, B+L){
cout << res[i];
}
cout << endl;
}
return 0;
}
```
Submission Info
Submission Time 2013-08-03 20:42:22+0900 D - 天下一展開 shioshiota C++ (G++ 4.6.4) 50 1691 Byte WA 379 ms 816 KB
Judge Result
Set Name small medium large
Score / Max Score 50 / 50 0 / 50 0 / 30
Status
AC × 34
AC × 36 WA × 20 RE × 10
AC × 36 WA × 40 RE × 13
Set Name Test Cases
small small, small/01_manual1, small/01_manual2, small/01_sample2, small/10_manual3, small/11_small00, small/11_small01, small/11_small02, small/11_small03, small/11_small04, small/11_small05, small/11_small06, small/11_small07, small/11_small08, small/11_small09, small/11_small10, small/11_small11, small/11_small12, small/11_small13, small/11_small14, small/11_small15, small/11_small16, small/11_small17, small/11_small18, small/11_small19, small/11_small20, small/11_small21, small/11_small22, small/11_small23, small/11_small24, small/11_small25, small/11_small26, small/11_small27, small/11_small28, small/11_small29
medium small, small/01_manual1, small/01_manual2, small/01_sample2, small/10_manual3, small/11_small00, small/11_small01, small/11_small02, small/11_small03, small/11_small04, small/11_small05, small/11_small06, small/11_small07, small/11_small08, small/11_small09, small/11_small10, small/11_small11, small/11_small12, small/11_small13, small/11_small14, small/11_small15, small/11_small16, small/11_small17, small/11_small18, small/11_small19, small/11_small20, small/11_small21, small/11_small22, small/11_small23, small/11_small24, small/11_small25, small/11_small26, small/11_small27, small/11_small28, small/11_small29, medium, medium/20_manual5, medium/20_manual6, medium/21_medium30, medium/21_medium31, medium/21_medium32, medium/21_medium33, medium/21_medium34, medium/21_medium35, medium/21_medium36, medium/21_medium37, medium/21_medium38, medium/21_medium39, medium/21_medium40, medium/21_medium41, medium/21_medium42, medium/21_medium43, medium/21_medium44, medium/21_medium45, medium/21_medium46, medium/21_medium47, medium/21_medium48, medium/21_medium49, medium/21_medium50, medium/21_medium51, medium/21_medium52, medium/21_medium53, medium/21_medium54, medium/21_medium55, medium/21_medium56, medium/21_medium57, medium/21_medium58, medium/21_medium59
large large, medium, small, large/30_manual4, large/30_manual7, large/31_large60, large/31_large61, large/31_large62, large/31_large63, large/31_large64, large/31_large65, large/31_large66, large/31_large67, large/31_large68, large/31_large69, large/32_large70, large/32_large71, large/32_large72, large/32_large73, large/32_large74, large/32_large75, large/32_large76, large/32_large77, large/32_large78, large/32_large79, large/33_large80, medium/20_manual5, medium/20_manual6, medium/21_medium30, medium/21_medium31, medium/21_medium32, medium/21_medium33, medium/21_medium34, medium/21_medium35, medium/21_medium36, medium/21_medium37, medium/21_medium38, medium/21_medium39, medium/21_medium40, medium/21_medium41, medium/21_medium42, medium/21_medium43, medium/21_medium44, medium/21_medium45, medium/21_medium46, medium/21_medium47, medium/21_medium48, medium/21_medium49, medium/21_medium50, medium/21_medium51, medium/21_medium52, medium/21_medium53, medium/21_medium54, medium/21_medium55, medium/21_medium56, medium/21_medium57, medium/21_medium58, medium/21_medium59, small/01_manual1, small/01_manual2, small/01_sample2, small/10_manual3, small/11_small00, small/11_small01, small/11_small02, small/11_small03, small/11_small04, small/11_small05, small/11_small06, small/11_small07, small/11_small08, small/11_small09, small/11_small10, small/11_small11, small/11_small12, small/11_small13, small/11_small14, small/11_small15, small/11_small16, small/11_small17, small/11_small18, small/11_small19, small/11_small20, small/11_small21, small/11_small22, small/11_small23, small/11_small24, small/11_small25, small/11_small26, small/11_small27, small/11_small28, small/11_small29
Case Name Status Exec Time Memory
large/30_manual4 RE 379 ms 788 KB
large/30_manual7 RE 247 ms 788 KB
large/31_large60 WA 20 ms 680 KB
large/31_large61 WA 20 ms 788 KB
large/31_large62 WA 19 ms 656 KB
large/31_large63 WA 18 ms 792 KB
large/31_large64 WA 18 ms 816 KB
large/31_large65 WA 18 ms 780 KB
large/31_large66 WA 19 ms 784 KB
large/31_large67 WA 20 ms 668 KB
large/31_large68 WA 19 ms 788 KB
large/31_large69 WA 20 ms 688 KB
large/32_large70 WA 19 ms 664 KB
large/32_large71 WA 19 ms 788 KB
large/32_large72 WA 18 ms 788 KB
large/32_large73 WA 19 ms 788 KB
large/32_large74 WA 19 ms 788 KB
large/32_large75 WA 18 ms 784 KB
large/32_large76 WA 19 ms 792 KB
large/32_large77 WA 18 ms 792 KB
large/32_large78 WA 19 ms 816 KB
large/32_large79 WA 21 ms 656 KB
large/33_large80 RE 247 ms 768 KB
medium/20_manual5 RE 246 ms 636 KB
medium/20_manual6 RE 250 ms 788 KB
medium/21_medium30 RE 252 ms 784 KB
medium/21_medium31 WA 20 ms 780 KB
medium/21_medium32 RE 256 ms 788 KB
medium/21_medium33 WA 19 ms 788 KB
medium/21_medium34 AC 22 ms 636 KB
medium/21_medium35 RE 244 ms 792 KB
medium/21_medium36 RE 246 ms 696 KB
medium/21_medium37 RE 246 ms 788 KB
medium/21_medium38 AC 20 ms 788 KB
medium/21_medium39 RE 253 ms 792 KB
medium/21_medium40 WA 21 ms 640 KB
medium/21_medium41 WA 20 ms 792 KB
medium/21_medium42 WA 20 ms 692 KB
medium/21_medium43 WA 20 ms 788 KB
medium/21_medium44 WA 19 ms 788 KB
medium/21_medium45 WA 19 ms 792 KB
medium/21_medium46 WA 20 ms 788 KB
medium/21_medium47 RE 256 ms 788 KB
medium/21_medium48 WA 18 ms 792 KB
medium/21_medium49 RE 244 ms 796 KB
medium/21_medium50 WA 18 ms 788 KB
medium/21_medium51 WA 20 ms 688 KB
medium/21_medium52 WA 20 ms 712 KB
medium/21_medium53 WA 20 ms 692 KB
medium/21_medium54 WA 19 ms 788 KB
medium/21_medium55 WA 19 ms 784 KB
medium/21_medium56 WA 20 ms 788 KB
medium/21_medium57 WA 20 ms 792 KB
medium/21_medium58 WA 20 ms 788 KB
medium/21_medium59 WA 20 ms 788 KB
small/01_manual1 AC 20 ms 784 KB
small/01_manual2 AC 19 ms 792 KB
small/01_sample2 AC 20 ms 792 KB
small/10_manual3 AC 21 ms 788 KB
small/11_small00 AC 21 ms 636 KB
small/11_small01 AC 21 ms 784 KB
small/11_small02 AC 21 ms 784 KB
small/11_small03 AC 21 ms 784 KB
small/11_small04 AC 21 ms 788 KB
small/11_small05 AC 20 ms 768 KB
small/11_small06 AC 24 ms 772 KB
small/11_small07 AC 21 ms 788 KB
small/11_small08 AC 20 ms 784 KB
small/11_small09 AC 18 ms 784 KB
small/11_small10 AC 20 ms 784 KB
small/11_small11 AC 19 ms 788 KB
small/11_small12 AC 19 ms 660 KB
small/11_small13 AC 20 ms 792 KB
small/11_small14 AC 20 ms 780 KB
small/11_small15 AC 19 ms 788 KB
small/11_small16 AC 21 ms 800 KB
small/11_small17 AC 19 ms 788 KB
small/11_small18 AC 20 ms 788 KB
small/11_small19 AC 19 ms 784 KB
small/11_small20 AC 21 ms 788 KB
small/11_small21 AC 19 ms 784 KB
small/11_small22 AC 20 ms 780 KB
small/11_small23 AC 18 ms 784 KB
small/11_small24 AC 19 ms 792 KB
small/11_small25 AC 22 ms 780 KB
small/11_small26 AC 21 ms 788 KB
small/11_small27 AC 20 ms 784 KB
small/11_small28 AC 19 ms 792 KB
small/11_small29 AC 21 ms 632 KB
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| 2.515625
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http://exxamm.com/QuestionSolution15/3D+Geometry/The+equation+ax+a+2+y+z+0+bx+b+2+y+z+0+and+cx+c+2+y+z+0+have+only+solution+0+0+0+If+t/1277412386
| 1,553,222,736,000,000,000
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text/html
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crawl-data/CC-MAIN-2019-13/segments/1552912202589.68/warc/CC-MAIN-20190322014319-20190322040319-00155.warc.gz
| 81,955,331
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|
The equation ax + a^2 y + z = 0 , bx + b^2 y + z = 0 and cx + c^2 y + z = 0 have only solution (0, 0, 0) . If t
### Question Asked by a Student from EXXAMM.com Team
Q 1277412386. The equation ax + a^2 y + z = 0 , bx + b^2 y + z = 0 and cx + c^2 y + z = 0 have only solution (0, 0, 0) . If the coefficients a, b, c are in G. P. , then the common ratio of G.P. cannot be equal to
A
-2
B
2
C
-1
D
-3
#### HINT
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https://se.mathworks.com/matlabcentral/answers/1451249-how-to-write-to-solve-this-type-of-system-of-equations
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# how to write to solve this type of system of equations ?
1 view (last 30 days)
NIRUPAM SAHOO on 11 Sep 2021
Commented: NIRUPAM SAHOO on 12 Sep 2021
##### 1 CommentShowHide None
NIRUPAM SAHOO on 12 Sep 2021
please anyone solve this . here u and v are functions of r.
Wan Ji on 12 Sep 2021
Hey friend
Just expand the left items of the two equations, extract u'' and v'', then an ode45 solver is there for you.
Walter Roberson on 12 Sep 2021
Edited: Walter Roberson on 12 Sep 2021
I was not able to figure out what is being raised to 10/9 . I used squiggle instead.
syms u(r) v(r)
syms N squiggle real
assume(r, 'real')
du = diff(u);
dv = diff(v);
left1 = diff(r^(N-1)*du^3)
left1(r) =
right1 = r^(N-1) * sqrt(u) * sqrt(v) / (3*r^(2/3) * sqrt(1 + 9*squiggle^(10/9)/(10*(3*N-2)^(1/3))))
right1(r) =
left2 = diff(r^(N-1)*dv^3)
left2(r) =
right2 = r^(N-1) * u * v / (3*r^(2/3))
right2(r) =
eqn1 = left1 == right1
eqn1(r) =
eqn2 = left2 == right2
eqn2(r) =
ic = [u(0) == 1, v(0) == 1, du(0) == 0, dv(0) == 0]
ic =
sol = dsolve([eqn1, eqn2, ic])
Warning: Unable to find symbolic solution.
sol = [ empty sym ]
string(eqn1)
ans = "3*r^(N - 1)*diff(u(r), r)^2*diff(u(r), r, r) + r^(N - 2)*(N - 1)*diff(u(r), r)^3 == (r^(N - 1)*u(r)^(1/2)*v(r)^(1/2))/(3*r^(2/3)*((9*squiggle^(10/9))/(10*(3*N - 2)^(1/3)) + 1)^(1/2))"
string(eqn2)
ans = "3*r^(N - 1)*diff(v(r), r)^2*diff(v(r), r, r) + r^(N - 2)*(N - 1)*diff(v(r), r)^3 == (r^(N - 1)*u(r)*v(r))/(3*r^(2/3))"
string(ic)
ans = 1×4 string array
"u(0) == 1" "v(0) == 1" "subs(diff(u(r), r), r, 0) == 0" "subs(diff(v(r), r), r, 0) == 0"
Lack of a symbolic solution means that you would have to do numeric solutions -- but you cannot do a numeric solution to infinity, and you certainly would not get a formula out of it.
NIRUPAM SAHOO on 12 Sep 2021
syms p(t) m(t) t Y
Eqns = [diff((t^(100-1))*(diff(p(t),t))) == (t^(100-1))*(t-1)*exp(t)*p(t)*m(t);
diff((t^(100-1))*(diff(m(t),t))) == (t^(100-1))*(t-1)*exp(t)*p(t)^(1/2)*m(t)^(1/2)]
[DEsys,Subs] = odeToVectorField(Eqns);
DEFcn = matlabFunction(DEsys, 'Vars',{t,Y});
tspan = [0,100];
y0 = [0 0 0 0];
[t,Y] = ode45(DEFcn, tspan, y0);
plot(t,Y)
## i write this but the graph does not shows . please help me
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https://tallahasseescene.com/2019/11/09/michael-c-hall-gets-an-interesting-astrology-analysis-11-09-2019/
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# Michael C. Hall Gets An Interesting Astrology Analysis (11/09/2019)
How will Michael C. Hall perform on 11/09/2019 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is not scientifically verified – don’t get too worked up about the result. I will first calculate the destiny number for Michael C. Hall, and then something similar to the life path number, which we will calculate for today (11/09/2019). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology people.
PATH NUMBER FOR 11/09/2019: We will take the month (11), the day (09) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 09 we do 0 + 9 = 9. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 9 + 12 = 23. This still isn’t a single-digit number, so we will add its digits together again: 2 + 3 = 5. Now we have a single-digit number: 5 is the path number for 11/09/2019.
DESTINY NUMBER FOR Michael C. Hall: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Michael C. Hall we have the letters M (4), i (9), c (3), h (8), a (1), e (5), l (3), C (3), H (8), a (1), l (3) and l (3). Adding all of that up (yes, this can get tedious) gives 51. This still isn’t a single-digit number, so we will add its digits together again: 5 + 1 = 6. Now we have a single-digit number: 6 is the destiny number for Michael C. Hall.
CONCLUSION: The difference between the path number for today (5) and destiny number for Michael C. Hall (6) is 1. That is smaller than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But this is just a shallow analysis! As mentioned earlier, this is not scientifically verified. If you want really means something, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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https://gateoverflow.in/43578/gate2003-79
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6.5k views
A processor uses $\text{2-level}$ page tables for virtual to physical address translation. Page tables for both levels are stored in the main memory. Virtual and physical addresses are both $32$ bits wide. The memory is byte addressable. For virtual to physical address translation, the $10$ most significant bits of the virtual address are used as index into the first level page table while the next $10$ bits are used as index into the second level page table. The $12$ least significant bits of the virtual address are used as offset within the page. Assume that the page table entries in both levels of page tables are $4$ bytes wide. Further, the processor has a translation look-aside buffer (TLB), with a hit rate of $\text{96%}$. The TLB caches recently used virtual page numbers and the corresponding physical page numbers. The processor also has a physically addressed cache with a hit rate of $\text{90%}$. Main memory access time is $10$ ns, cache access time is $1$ ns, and TLB access time is also $1$ ns.
Suppose a process has only the following pages in its virtual address space: two contiguous code pages starting at virtual address $0x00000000$, two contiguous data pages starting at virtual address $0x00400000$, and a stack page starting at virtual address $0xFFFFF000$. The amount of memory required for storing the page tables of this process is
1. $\text{8 KB}$
2. $\text{12 KB}$
3. $\text{16 KB}$
4. $\text{20 KB}$
edited | 6.5k views
0
Just one variation in the given question -->
Although it is given in the question that page table entry size is 4B. But assume that it is not given, then what will be page table entry size ?
PS: For hint check - https://gateoverflow.in/490/gate2008-67
0
In that case I can think of 1 page in page table 1 is pointing to 3 pages in page table two as entries are consecutive so I can assume that to utilize spatial locality of reference consecutive locations are loaded into same page making it 1+ 3= 4 pages in all. And offset bits are 12 that gives me 4 kb for each page. Hence 4 X 4kb makes it 16 kb.
Can I think like this while evaluating such questions? Or this is not correct approch if page table entry size is not given.
0
What would be the answer if had it been a 3-level paging ?
Then will 2nd level page tables be all possible count of it and only 3 page tables for 3rd level ?
@MiNiPanda
@Ahwan
0
Can anyone explain me the concept?
I stuck in the inner page and address locations!
First level page table is addressed using $10$ bits and hence contains $2^{10}$ entries. Each entry is $4$ bytes and hence this table requires $4$ KB. Now, the process uses only $3$ unique entries from this $1024$ possible entries (two code pages starting from $0x00000000$ and two data pages starting from $0x00400000$ have same first $10$ bits). Hence, there are only $3$ second level page tables. Each of these second level page tables are also addressed using $10$ bits and hence of size $4$ KB. So,
total page table size of the process
= 4 KB + 3 * 4 KB
= 16 KB
Correct Answer: $C$
answered by Veteran (414k points)
edited
0
What would be the change if the memory were nibble addressable ?
+38
@pc see this video.
0
Sir,I am confused that next or 2nd code page address will be 0x00000001 or 0x00001000 as first 20 bits will be taken page no. in virtual memory or physical memory. page size or 12 bits of lsb wont be changed.
What do you suggest??
+2
@gateasp17
The last 12 bits are offset so addresses for pages are
For Code pages:
0x00000000 and 0x00001000 (Note: only 10 bits are for addressing 2nd level page table). So these 10 bits will be:
For code page 0x00000000 2nd level PT offset is (00 0000 0000)2
For code page 0x00001000 2nd level PT offset is (00 0000 0001)2
For Data pages
0x00400000 and 0x00401000
For data page 0x00400000 2nd level PT offset is (00 0000 0000)2
For data page 0x00401000 2nd level PT offset is (00 0000 0001)2
:)
0
Yaahh.... I got it THAT here for page size or frame size 12 bits (LSB). So remaining 32-12=20 bits will be assigned for no. Of pages that's why changes will be made to these 20 bits only, for contiguous pages of code data and stack..
Thanks.. :-)
+22
..........................
Links from second level page tables are not pointers rather they represents coverage in the logical address space.
0
@Debashish Deka why there a red arrow from 3 block of first to the logical address space?Second level 1st page table should have only two enteries as it is used to point to two pages.What is thirst arrow representing?
0
Second level 1st page table should have only two enteries as it is used to point to two pages.What is thirst arrow representing?
Yes, it has two contiguous words within the same page,i.e the first page in logical address space but the 3rd arrow indicates last address that is covered by the 1st inner page table. It is shown just to indicate that the 1st inner page table cannot access page at address: 0x00400000
0
I've a doubt.
Is the sequence of Page no-frame no entry in Page table same as the sequence of pages stored in Main memory?
0
@Tuhin.I did not get the point.Even for second level second table has only one arrow.It should have two.
0
@Rahul, I told about the blue arrow from address 0x00400000. Regarding your query about 2 links or arrows from the 2nd inner page table, yes there should have been two arrows but I think it is not shown because of keeping clarity in the diagram. The arrow from the last PTE of the 1st inner page table shows that it points to the starting location of the last page that the 1st inner page table can address, just for our understanding. Now you should have observed the green colored consecutive pages shown in logical address space. For each of these pages, inner page tables must have pointers to them.
0
If I want to figure out what will be page size based on given data?? How will I do that? Or can I simply say page size is nos of offset bits?
12 bits are used for offset hence page size is 2^12 , 4 kb?
0
There could be variable partition used in main memory
0x00400000 can be pointed by inner page table.
0
@Arjun Sir
Sir, you have considered full outer page table & partial inner page table(pages for which the second set of 10 most significant digit is changing) while calculating the memory required to stored the page table for the process.Can we load partial inner page table?
if we can load partial page tables then is it only the inner most tables that we can load partial or at any level considering an 'n' level paging.
Thanks
+1
@Arjun Sir,
my doubt got cleared after referring the solved problem enclosed in ur expiation.
:)
0
Nice explanation
0
Now, the process uses only 33 unique entries from this 10241024 possible entries (two code pages starting from 0x00000000 and two data pages starting from 0x00400000 have same first 1010 bits).
IN ABOVE IMAGE T1 IS TABLE1 AND T2 IS TABLE2
answered by Active (1.6k points)
0
Is this correct solution or the cancelled ?
0
How u figured out what is page size??
0
"The 12 least significant bits of the virtual address are used as offset within the page."
given in ques.
Since page offset is of size 12 bit and memory is byte addressable. So we can say that page size=2^12 Byte=4KB.
Now process needs 5 pages of page no.
0x 00000 and 0x 00001 for code,
0x 00400 and 0x 00401 for data, and
0x FFFFF for stack.
Now for mapping above pages in to frames we need one outer page table and two inner page tables(One inner page table needed for mapping all these four pages
0x 00000, 0x 00001, 0x 00400 and 0x 00401, and another inner page needed for mapping 0x FFFFF).
So, total no of page table needed for above process is 3.
The amount of memory required for storing the page tables of this process is 3*4KB = 12KB
answered by (33 points)
edited
+6
0x 00000, 0x 00001 are in one inner page table and 0x 00400 and 0x 00401 in another inner page table,because their first 10 bits are different (4 is 0100 and so 01 is the part of ist 10 bits).So a total of 3 inner tables and 1 outer page table are required. so 4*4KB=16KB .
0
Hi navnit, can you plz explain me how inner and outer page tables can be decided??
1
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# HDU 4786Fibonacci Tree
2018-02-22 10:48:40来源:cnblogs.com作者:自为风月马前卒人点击
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5934 Accepted Submission(s): 1845
Problem Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to do some research on Spanning Tree. So Coach Pang decides to solve the following problem:
Consider a bidirectional graph G with N vertices and M edges. All edges are painted into either white or black. Can we find a Spanning Tree with some positive Fibonacci number of white edges?
(Fibonacci number is defined as 1, 2, 3, 5, 8, ... ) Input The first line of the input contains an integer T, the number of test cases.
For each test case, the first line contains two integers N(1 <= N <= 105) and M(0 <= M <= 105).
Then M lines follow, each contains three integers u, v (1 <= u,v <= N, u<> v) and c (0 <= c <= 1), indicating an edge between u and v with a color c (1 for white and 0 for black). Output For each test case, output a line “Case #x: s”. x is the case number and s is either “Yes” or “No” (without quotes) representing the answer to the problem. Sample Input24 41 2 12 3 13 4 11 4 05 61 2 11 3 11 4 11 5 13 5 14 2 1 Sample OutputCase #1: YesCase #2: No Source2013 Asia Chengdu Regional Contest RecommendWe have carefully selected several similar problems for you: 6263 6262 6261 6260 6259 和昨天ysy讲的那道题差不多而且这道题在题目中直接给提示了——》黑边为0，白边为1这样的话我们做一个最小生成树和一个最大生成树如果在这两个值的范围内有斐波那契数，就说明满足条件 简单证明：
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https://www.teacherspayteachers.com/Product/January-Calendar-Morning-Meeting-Routines-for-Smartboard-1054281
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Total:
\$0.00
# January Calendar & Morning Meeting Routines for Smartboard
Resource Type
Product Rating
4.0
6 Ratings
File Type
NOTEBOOK (SMARTboard) File
19 MB|37 pages
Share
Product Description
My morning meeting continues to be an integral part of the day. We spend time on ELA and math skills during this time. The children really enjoy the interactive slides. I have some great new winter themes this month.
For January there are 37 slides, which include:
-Lunch chart that you can personalize for your students
-Classroom jobs to personalize
Daily morning meeting-
-January calendar (new)
-Morning Message to write the news of the day. Have 1 child write the letter while other children are doing morning work(new)
Weather chart that can be customized for your town. Click on the weather people to link to your local accu- weather forecast
-Weather bar graph to go along with the weather of the day- you can keep track for the month and compare most, least, equal to
-Odd or Even number chart (new background and picture)
-How Many days in January.- ten frame & tally mark chart to practice these very important math skills each day (new snowman theme)
-How many days in school hundred board counting with a number bond and two number sentences to review addition and subtraction
-Days in school place value chart using one and tens base ten blocks
-Class bank to continue money recognition
Math Review skill slides including:
Ways to make a number- more and less slide- choose a new number each day and fill in 1 more, 1 less, 10 more and 10 less , find a number on the number line, highlight the number in the 100 board. Find the number in the hundred board. (new)
-Count by 2’s how many days we are in school and write a number sentence(new)
-Pick any number and have the students show in tens and ones and write a number sentence(new)
-Guess my number game (new)
-Make a 10-give a number sentence and have children show how to make a 10 to solve (new)
-Fact families- write a new family each day and have children write all related number sentences(new)
-Subtraction practice(new)
-Time practice(new)
-Zero the Hero- every 10th day the students can watch this fun video about Zero!
-Count by 10, 5, 2 songs to help children with these skills
Word work slides-
- first grade detectives use a magnifying glass to read all your high frequency words you can put any words you are currently working on.
-Handwriting page to practice letter formation
-Build a word with vowel teams- oo, ou, oi, aw
5 slides with a penguin theme (new)
-4 new Poems & Songs for the month- Children read the poems each day for the week and then on Friday they get a copy to print out and illustrate in their poetry books.
I hope your students enjoy their morning meeting as much as mine do.
I keep this program open the entire day. I find that I use different parts at different times in the day including small group time, math and center work.
My best,
Eileen Cording
Total Pages
37 pages
N/A
Teaching Duration
N/A
Report this Resource
\$8.00
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# Chemistry and Biology
Topics: Asymptote, Limit of a function, Asymptotic curve Pages: 19 (1190 words) Published: March 16, 2013
CHAPTER 7 LIMITS AND
CONTINUITY
Focus on Exam 7
1 (a) |x + 3| =
{
-x - 3,
x < -3,
x + 3,
x ≥ -3.
(x + 1)(-x - 3)
x+3
= -x - 1
(x + 1)(x + 3)
For x ≥ -3, f (x) =
x+3
=x+1
Hence, in the non-modulus form,
-x - 1, x < -3,
f (x) =
x + 1,
x ≥ -3.
For x < -3, f (x) =
{
(b) The graph of f(x) is as shown below.
y = −x − 3
y
2
y=x+1
1
−3
−2
−1
x
O
−1
−2
(c) lim f (x) = 2
x → -3-
lim f (x) = -2
x → -3+
(d) lim f (x) does not exist because lim f (x) ≠ lim f (x). x → -3
2 (a) lim h(x) = 2
x → -1
-1 + p = 2
p=3
x → -3
x → -3+
x = -1 is in the range -3 ≤ x < 0, so the
part of the function x + p is used.
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Chap-07-FWS.indd 1
10/18/2012 9:48:51 AM
2
ACE AHEAD Mathematics (T) Second Term
(b) Since lim h(x) exists,
x → -3
lim h(x) = lim h(x)
x → -3-
x → -3+
x2 - k
(-3)2 - k = -3 + 3
k=9
Since lim h(x) exists,
x+3
x→0
lim h(x) = lim h(x)
x→0-
x→0+
x+3
0+3=e
0-q
ex - q
ln 3 = -q
q = -ln 3
= ln 3-1
= ln 1
3
(c) The graph of y = h(x) is as shown below.
{
ex
eq
ex
ln
y = 3e x
1
e3
ex
=
1
3
= 3e x
3 (a) f o g = f [g(x)]
1
=f
x-3
1
1
+
3
3
y = x2 − 9
x
=
(1, 8.2)
2
=
e x-q =
y
y
h(x) =
x < -3,
-3 ≤ x < 0,
x ≥ 0.
x 2 - 9,
x + 3,
3e x,
1
−4 −3 −2 −1 O
x
1
2
2
2
1
1
x-3
= 2(3 + x - 3)
= 2x
The domain of f o g is the same as the domain of g, i.e. {x : x ∈ R, x ≠ 3}. Because the domain cannot take the value 3, the range of f o g cannot take the value 2x = 2(3)
= 6.
Hence, the range of f o g is {y : y ∈ R, y ≠ 6}.
(b) The graph of
y = f g(x)
= 2x, x ≠ 3
is as shown below.
=2 3+
1
2
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Chap-07-FWS.indd 2
10/18/2012 9:48:52 AM
Fully Worked Solution
3
y
6
x
O
3
y = 2x
(c) lim f g(x) = 2(3)
x→3
=6
and
lim f g(x) = 2(3)
-
x→3+
=6
Since lim f g(x) = lim f g(x)
x→3+
x→3+
=6
then lim f g(x) = 6
x→3
4 In the non-modulus form,
f (x) =
{
x 2 - 1,
-x2 + 1,
(x - 2)(x - 3),
x < -1,
-1 ≤ x < 1,
x ≥ 1.
The graph of y = f (x) is as shown below.
y
2
2
y=x −1
y = (x − 2)(x − 3)
1
−1 O
1
2
3
4
x
y = −x 2 + 1
(b) (i) lim f (x) = 12 - 1
x→-1
=0
lim f (x) = -12 + 1
x→-1
=0
f (-1) = -12 + 1
=0
-
+
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Chap-07-FWS.indd 3
10/18/2012 9:48:52 AM
4
ACE AHEAD Mathematics (T) Second Term
Since lim f (x) = lim f (x)
x→-1-
x→-1+
= f (-1)
= 0, then f (x) is continuous at x = -1.
(ii) lim f (x) = -12 + 1
x→1-
=0
lim f (x) = (1 - 2)(1 - 3)
x→1+
=2
Since lim f (x) ≠ lim f (x), then lim f (x) does not exist. x→1-
x→1+
x→1
Hence, f (x) is not continuous at x = 1.
5 (a) In the non-modulus form,
f (x) =
{
x2 , x < 0,
-x
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What is a positive correlation. Positive Correlation Examples
What is Correlation?
When you draw a scattergram it doesn't matter which variable goes on the x-axis and which goes on the y-axis. Strong Correlation: A weak correlation means that as one variable increases or decreases, there is a lower likelihood of there being a relationship with the second variable. If the stocks of eBay, Amazon and Best Buy pick up due to increased online revenue, it is likely that PayPal will experience a similar boost as its fee-driven income picks up and positive earnings reports encourage investors. In case, markets outperform when stocks are not performing well. Inverse correlations describe two factors that seesaw relative to each other. Example: Ice Cream Sales The local ice cream shop keeps track of how much ice cream they sell versus the temperature on that day, here are their figures for the last 12 days: Ice Cream Sales vs Temperature Temperature °C Ice Cream Sales 14. A correlation of +1 indicates a perfect positive correlation, meaning that both variables move in the same direction together.
Nächster
How Are Correlations Are Used in Psychology Research
Learn how to visualize correlation with a! Here, the x-values are depicted on the horizontal axis and y-values are on the vertical axis. Correlation Estimation There are different ways of calculating or estimating the correlation between variables. For example, a person spending more will see a decline in his or her bank balance, or the more a person drives a car, the less will be its gas mileage. These are nothing but positive correlation and negative correlation. There do appear to be some interesting outliers around a wind speed of 10-15 meters per hour as well, where further investigation would be required.
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Positive Correlation Examples
It is not necessary for causation to take place if correlation exists. Wind speed is strongly correlated with wave heights in from Pacific Ocean buoys for one month, for example. How To Calculate How did I calculate the value 0. Negative Correlation Correlation in the opposite direction is called a negative correlation. Correlations also leave room for additions that can skew these tendencies. If two variables sometimes but not always change in tandem, the correlation is expressed as greater than zero but less than +1. Stereotypes are a good example of illusory correlations.
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What is Correlation?
Special Considerations Positive Correlation in Finance A simple involves the use of an interest-bearing savings account with a set interest rate. A correlation of —1 indicates a perfect negative correlation, meaning that as one variable goes up, the other goes down. It gets so hot that people aren't going near the shop, and sales start dropping. Remember, in correlations we are always dealing with paired scores, so the values of the 2 variables taken together will be used to make the diagram. Correlation allows the researcher to clearly and easily see if there is a relationship between variables. The relationship is good but not perfect. For example suppose we found a positive correlation between watching violence on T.
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What is Correlation?
Figure a shows a correlation of nearly +1, Figure b shows a correlation of —0. Values between -1 and 1 denote the strength of the correlation, as shown in the example below. If an employee workers for 25 hours instead of 20 hours, his or her salary will increase accordingly. If correlation does not mean causation, then what does it mean? A positive correlation is not an indicator of advantage or growth. A strong negative correlation, on the other hand, would indicate a strong connection between the two variables, but that one goes up whenever the other one goes down. In statistical terms, a perfectly positive correlation signifies the correlation coefficient value of +1. For example, complementary demand of product.
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Correlation vs. Causation
Correlation By , updated 2018 Correlation means association - more precisely it is a measure of the extent to which two variables are related. This is because businesses that have very different operations and will produce different products and services using different inputs. This post will define positive and negative correlations, illustrated with examples and explanations of how to measure correlation. This is so because two different companies will manufacture and deal in different products and services utilizing distinct resources. Another problem with correlation is that it summarizes a linear relationship. Misinterpreting correlations Just about all the common problems that can render statistical analysis meaningless can occur with correlations. A Word From Verywell Correlations play an important role in.
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What Does it Mean if the Correlation Coefficient is Positive, Negative, or Zero?
In other words, the values cannot exceed 1. Here if one variable increases the other decreases and vice versa. Research has shown that people tend to assume that certain groups and traits occur together and frequently overestimate the strength of the association between the two variables. The Correlation Coefficient Remember, correlation strength is measured from -1. Finally, some pitfalls regarding the use of correlation will be discussed.
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what is the meaning of area in mathematical terms
# what is the meaning of area in mathematical terms
Get an answer for What is the meaning of MATHEMATICS? and find homework help for other Math questions at eNotes.Contact Us. Terms of Use. Pricing. The level of mathematical precision and generality is appropriate to a serious upper-level undergraduate course.area one. The vertical subdivisions represent the proportions of males and females.(In fact, this is the meaning of normal in geometry.) You can see from Figure 2.4 wherein the revisionist sense that we must so construe the meaning of mathematical propositions as to eliminate the ap-parent reference to mathematicalFor example, in L is interpreted in terms of the mathematical there exists.70 The Provenance of Pure Reason. our knowledge in dierent areas. Mathematics is a Greek word, and, by origin or etymologically, it means "something that must be learnt or understood", perhaps "acquiredWhat is mathematics in the modern sense of the term, its implications and connotations?Mathematical theorems must be deductively established and proved. This tells us, in practical terms, that, for every one unit that the x-variable increases (that is, moves over to the right), the y-variable increases (that is, goes up) by three-fifths of a unit.What is the meaning of this y-value? The intercept value tells me that, in 1960 (when they started counting), the average Harmonic analysis is a wellspring of ideas and applicability that has our-ished, developed, and deepened over time within many disciplines and by means of creative cross-fertilization with diverse areas.In mathematical terms, the Fourier transform of a continuous-time signal. Mathematical language also includes many technical terms such as homeomorphism and integrable that have no meaning outside of mathematics.
In these traditional areas of mathematical statistics, a statistical-decision problem is formulated by minimizing an objective function, like expected loss or What is a mathematical structure?The Greek root of the word "mathematics" is mathma, which means "what is learned."Besides, Piaget was not a mathematician and the "elegance" of putting his theory into these mathematical terms was likely not as obvious to him as it would have been to, say Mathematicians work to expand their new pictorial mathematical language into other areas. A newly discovered prime number makes its debut. Oct 13, 2008 2. Werg22. Re: What does "finite" mean in mathematical terms? First, finite/infinite only applies to sets. Space is an ambiguous term in mathematics and, in any case, more commonly might be expected to be used when referring to the VOLUME of an object, (i.e. a 3 dimensional object). The Area relates specifically to the size of something in two dimensions. In math, area represents the size or total amount of space taken up by a two-dimensional surface, or the amount of space inside the boundaries of a flat object.What is the definition of "volume" in math terms? Why is geometry so important? Historically, the theo-rems of some areas of mathematics, such as non-Euclidean geometry, were arrived at through purely deductive means [11].Students are traditionally in-troduced to the term mathematical proof in high school geometry classes, where they are taught that proofs are highly But often mathematics inspired by one area proves useful in many areas, and joins the general stock of mathematical concepts.Technical terms such as homeomorphism and integrable have precise meanings in mathematics. Mathematical Conventions Figure 1. 5.
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# Variable High-voltage Capacitor
2,823
17
## Introduction: Variable High-voltage Capacitor
after my microwave capacitor died iv'e been looking for a good replacement, id heard of the overhead slide and tin foil ones but i figured i may as well laminate tin foil if i'm going to make that, after messing around a bit with different dielectrics, i made one that worked so well, i decided to make my own instructable on it.
this works great for Tesla coils, Marx generators (if you have the patience to make enough of them), and other high voltage aplications
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Stuff
items consumed:
1/2" PVC (and a little 4" but that's optional)
aluminum foil
black plastic
duct tape (unless you say its a tool) <ans store:A>
tools:
<insert "duct tape", if A=1>
scissors
tape measure
work space: id recommend sweeping first, you will get better results.
<don't freak out if you don't know any programming languages, its just a joke.>
## Step 2: Lets Get Started!
this is the second one i have built and yes it is quite a bit larger, the first pictures are the original and the last one is the first step in the construction process
lay out the plastic and put the tin foil on it
## Step 3: "its Got a Light Side and a Dark Side and It Holds the Universe Together"
no you will not be taping the whole tin foil surface, only the ends on this one, i made an improvement, anyway tape the ends MAKE SURE ITS FLAT
measure out twice the with of the tinfoil + 2" overhang and cut off the rest of the plastic then fold it in half
## Step 4: Wire It!
you can duct tape them, or use foil tape like i did
then tape the ends of the plastic and fold it in half the other way
## Step 5: Roll It Up
get some 1/2" PVC and roll it as tight as possible then you can string it shut, stick it in a tube (that's where the 4" PVC came from) or whatever, but make sure it can be re-opened
## Step 6: "hey I Thought You Said This Was Adjustable!"
it is, remember i wanted you to make it so it could still be opened? well just stuff some more black plastic (or newspaper, plastic, fiberglass, flux, etc) in there and you can adjust to the desired voltage capabilities/ capacitance.
## Recommendations
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# Schaum strength of materials pdf
This fifth edition of Schaum's Strength of Materials book has been substantially modified by the So, Statics always precedes the study of Strength of Materials. Nash, William A. Schaumk outllt'le ofthcory and problems of strength at" materials J. William A. Nash. - 4th ed. p. cm. - (Schaum's outline series} includes index. Abstract: Study faster, learn better, and get top gradesModified to conform to the current curriculum, Schaum's Outline of Strength of Materials complements.
from this work, please submit a written request to Pearson Education, Inc.,.. when something that was difficult to und Schaum's Outline of Discrete. Schaum's Outline of Strength of Materials, 6th Edition (Schaum's Outlines) [ William Nash] on nbafinals.info *FREE* shipping on qualifying offers. Tough Test . Editorial Reviews. About the Author. William Nash, was a professor of civil engineering at the University of Massachusetts, Amherst. Merle Potter is Professor.
Ductile and Brittle Materials Metallic engineering materials are commonly classified as either ductile or brittle materials. A ductile material is one having a relatively large tensile strain up to the point of rupture for example, structural steel or aluminum whereas a brittle material has a relatively small strain up to this same point. An arbitrary strain of 0. Cast iron and concrete are examples of brittle materials. The quantity E, i.
P are at the same level before the uniform rigid block weighing 40 kips is attached. Each steel bar has a length of 3 ft, and area of 1. For the bronze bar, the area is 1. Determine a the length of the bronze bar so that the load on each steel bar is twice the load on the bronze bar, and b the length of the bronze that will make the steel stress twice the bronze stress. Solution a Condition: P has negligible mass and rests on two steel bars, each The center bar is aluminum and The center-line spacing between the holes is 30 ft in the two outer bars, but 0.
Find the shearing stress developed in the drip pins. Neglect local deformation at the holes.
## Schaum's Outline Strength of material
Solution Middle bar is 0. P, three steel wires, each 0. Their unstressed lengths are For steel: P consists of a light rigid bar AB, pinned at O, that is attached to the steel and aluminum rods. Compute the stress in the aluminum rod when the lower end of the steel rod is attached to its support.
It carries an axial load P applied as shown in Fig. Solution to Problem Statically Indeterminate A homogeneous bar with a cross sectional area of mm2 is attached to rigid supports. Determine the stress in segment BC. Use the results of Prob. Then use the principle of superposition to compute the reactions when both loads are applied.
Solution From the results of Solution to Problem P is firmly attached to unyielding supports. P is stress-free before the axial loads P 1 and P 2 are applied. Solution From the FBD of each material shown: The inside diameter of the aluminum tube is mm, and the wall thickness of each tube is 2. Compute the contact pressure and tangential stress in each tube when the aluminum tube is subjected to an internal pressure of 5. Solution Internal pressure of aluminum tube to cause contact with the steel: Find the stresses if the nut is given one additional turn.
How many turns of the nut will reduce these stresses to zero? P are identical except for length. Before the load W was attached, the bar was horizontal and the rods were stress-free. P is pinned at B and connected to two vertical rods. P, a rigid beam with negligible weight is pinned at one end and attached to two vertical rods.
Find the vertical movement of W. P, a rigid bar with negligible mass is pinned at O and attached to two vertical rods. Assuming that the rods were initially stress-free, what maximum load P can be applied without exceeding stresses of MPa in the steel rod and 70 MPa in the bronze rod. P is a section through a balcony. The total uniform load of kN is supported by three rods of the same area and material.
Compute the load in each rod. Assume the floor to be rigid, but note that it does not necessarily remain horizontal. Assuming that there was no slack or stress in the rods before the load was applied, find the stress in each rod. Initially, the assembly is stress free. Horizontal movement of the joint at A is prevented by a short horizontal strut AE. If temperature deformation is permitted to occur freely, no load or stress will be induced in the structure. In some cases where temperature deformation is not permitted, an internal stress is created.
The internal stress created is termed as thermal stress. For a homogeneous rod mounted between unyielding supports as shown, the thermal stress is computed as: If the wall yields a distance of x as shown, the following calculations will be made: Take note that as the temperature rises above the normal, the rod will be in compression, and if the temperature drops below the normal, the rod is in tension.
Solution to Problem Thermal Stress A steel rod with a cross-sectional area of 0. At what temperature will the stress be zero? At what temperature will the rails just touch? What stress would be induced in the rails at that temperature if there were no initial clearance? Solution a Without temperature change: See figure above. Find the temperature at which the compressive stress in the bar will be 35 MPa. Assume that the supports are unyielding and that the bar is suitably braced against buckling.
Neglect the deformation of the wheel caused by the pressure of the tire. P is pinned at B and attached to the two vertical rods. Initially, the bar is horizontal and the vertical rods are stress-free. Neglect the weight of bar ABC. Solution Contraction of steel rod, assuming complete freedom: In terms of aluminum, this movement is by ratio and proportion: P, there is a gap between the aluminum bar and the rigid slab that is supported by two copper bars.
Solution Assuming complete freedom: Solution Before temperature change: Assume the coefficients of linear expansion are If the system is initially stress-free.
Calculate the temperature change that will cause a tensile stress of 90 MPa in the brass rod. Assume that both rods are subjected to the change in temperature. Each bar has a cross-sectional area of mm2. Torsion 2. Flanged Bolt Couplings 3. Torsion of Thin-Walled Tubes 4. Such a bar is said to be in torsion.
For solid cylindrical shaft: Determine the maximum shearing stress and the angle of twist. What maximum shearing stress is developed?
What power can be transmitted by the shaft at 20 Hz? If the shearing stress is limited to 12 ksi, determine the maximum horsepower that can be transmitted.
Solution Hollow circular shaft: Solution to Problem Torsion An aluminum shaft with a constant diameter of 50 mm is loaded by torques applied to gears attached to it as shown in Fig. Determine the maximum length of the shaft if the shearing stress is not to exceed 20 ksi. What will be the angular deformation of one end relative to the other end? Solution Based on maximum allowable shearing stress: At the right end, 30 kW are removed and another 20 kW leaves the shaft at 1. Determine the maximum permissible value of T subject to the following conditions: The two shafts are then fastened rigidly together at their ends.
What torque can be applied to the composite shaft without exceeding a shearing stress of psi in the bronze or 12 ksi in the steel? Determine the maximum shearing stress in each segment and the angle of rotation of the free end. P is attached to rigid supports. What torque T is required? Solution From Solution P, to a solid shaft with built-in ends. How would these values be changed if the shaft were hollow?
Equations 1 and 2a: Therefore, the values of T1 and T2 are the same no change if the shaft were hollow. Solution to Problem Torsion A solid steel shaft is loaded as shown in Fig. Solution Based on maximum allowable shear: Determine the maximum shearing stress developed in each segment. Solution Stress developed in each segment with respect to TA: The rotation of B relative to A is zero. For the bronze segment AB, the maximum shearing stress is limited to psi and for the steel segment BC, it is limited to 12 ksi.
P, each with one end built into a rigid support have flanges rigidly attached to their free ends. The shafts are to be bolted together at their flanges.
Determine the maximum shearing stress in each shaft after the shafts are bolted together. See figure. For rigid flanges, the shear deformations in the bolts are proportional to their radial distances from the shaft axis. Determine the torque capacity of the coupling if the allowable shearing stress in the bolts is 40 MPa. Determine the torque capacity of the coupling if the allowable shearing stress in the bolts is psi. What torque can be applied without exceeding a shearing stress of 60 MPa in the bolts?
Solution For one bolt in the outer circle: Determine the shearing stress in the bolts. What torque can be applied without exceeding psi in the steel or psi in the aluminum? Solution to Problem Flanged bolt couplings A plate is fastened to a fixed member by four mm-diameter rivets arranged as shown in Fig.
Compute the maximum and minimum shearing stress developed. P to the fixed member. Using the results of Prob.
What additional loads P can be applied before the shearing stress in any rivet exceeds psi? Solution Without the loads P: P is fastened to the fixed member by five mm-diameter rivets. Compute the value of the loads P so that the average shearing stress in any rivet does not exceed 70 MPa. Solution Solving for location of centroid of rivets: Determine the wall thickness t so as not to exceed a shear stress of 80 MPa.
What is the shear stress in the short sides? Neglect stress concentration at the corners. Solution to Problem Torsion of thin-walled tube A tube 0.
What torque will cause a shearing stress of psi? Determine the smallest permissible dimension a if the shearing stress is limited to psi. Assume that the shearing stress at any point is proportional to its radial distance.
J J Helical Springs When close-coiled helical spring, composed of a wire of round rod of diameter d wound into a helix of mean radius R with n number of turns, is subjected to an axial load P produces the following stresses and elongation: For heavy springs and considering the curvature of the spring, a more precise formula is given by: Use Eq.
Compute the number of turns required to permit an elongation of 4 in. P supports a load P. The upper spring has 12 turns of mm-diameter wire on a mean radius of mm. The lower spring consists of 10 turns of mm diameter wire on a mean radius of 75 mm.
If the maximum shearing stress in either spring must not exceed MPa, compute the maximum value of P and the total elongation of the assembly. Compute the equivalent spring constant by dividing the load by the total elongation. Determine the maximum load W that may be supported if the shearing stress in the springs is limited to 20 ksi. Each spring consists of 20 turns of mm wire having a mean diameter of mm.
Compute the maximum shearing stress in the springs, using Eq. Neglect the mass of the rigid bar.
P, a homogeneous kg rigid block is suspended by the three springs whose lower ends were originally at the same level. Compute the maximum shearing stress in each spring using Eq.
According to determinacy, a beam may be determinate or indeterminate. Statically Determinate Beams Statically determinate beams are those beams in which the reactions of the supports may be determined by the use of the equations of static equilibrium. The beams shown below are examples of statically determinate beams.
Statically Indeterminate Beams If the number of reactions exerted upon a beam exceeds the number of equations in static equilibrium, the beam is said to be statically indeterminate. In order to solve the reactions of the beam, the static equations must be supplemented by equations based upon the elastic deformations of the beam.
The degree of indeterminacy is taken as the difference between the umber of reactions to the number of equations in static equilibrium that can be applied.
Types of Loading Loads applied to the beam may consist of a concentrated load load applied at a point , uniform load, uniformly varying load, or an applied couple or moment. These loads are shown in the following figures. Assume that the beam is cut at point C a distance of x from he left support and the portion of the beam to the right of C be removed. The portion removed must then be replaced by vertical shearing force V together with a couple M to hold the left portion of the bar in equilibrium under the action of R1 and wx.
The couple M is called the resisting moment or moment and the force V is called the resisting shear or shear. The sign of V and M are taken to be positive if they have the senses indicated above.
Write shear and moment equations for the beams in the following problems. In each problem, let x be the distance measured from left end of the beam. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam in each problem. See the instruction. Solution From the load diagram: In segment AB, the shear is uniformly distributed over the segment at a magnitude of —30 kN. In segment BC, the shear is uniformly distributed at a magnitude of 26 kN.
In segment CD, the shear is uniformly distributed at a magnitude of —24 kN. To draw the Moment Diagram: At segment AB, the shear is uniformly distributed at lb.
A shear of — lb is uniformly distributed over segments BC and CD. Note that the maximum moment occurs at point of zero shear. For segment AB, the shear is uniformly distributed at 20 kN.
The shear for segment CD is uniformly distributed at —40 kN. Solution Segment AB: Solution to Problem Shear and Moment Diagrams Cantilever beam carrying a distributed load with intensity varying from wo at the free end to zero at the wall, as shown in Fig. To draw the Moment diagram: The shear is uniformly distributed at — lb along segments CD and DE.
## coiskepagti.tk
Shear is uniform along segment CD at —20 kN. To draw the Moment Diagram 1. Solution By symmetry: The other half of the diagram can be drawn by the concept of symmetry. Solution to Problem Shear and Moment Diagrams A total distributed load of 30 kips supported by a uniformly distributed reaction as shown in Fig.
For the next half of the beam, the shear diagram can be accomplished by the concept of symmetry. P if a the load P is vertical as shown, and b the load is applied horizontally to the left at the top of the arch. Properties of Shear and Moment Diagrams The following are some important properties of shear and moment diagrams: The area of the shear diagram to the left or to the right of the section is equal to the moment at that section. The slope of the moment diagram at a given point is the shear at that point.
The slope of the shear diagram at a given point equals the load at that point. The maximum moment occurs at the point of zero shears. This is in reference to property number 2, that when the shear also the slope of the moment diagram is zero, the tangent drawn to the moment diagram is horizontal.
When the shear diagram is increasing, the moment diagram is concave upward. When the shear diagram is decreasing, the moment diagram is concave downward.
## Schaum’s Outline of Strength of Materials, Fifth Edition
Sign Convention The customary sign conventions for shearing force and bending moment are represented by the figures below. A force that tends to bend the beam downward is said to produce a positive bending moment. An easier way of determining the sign of the bending moment at any section is that upward forces always cause positive bending moments regardless of whether they act to the left or to the right of the exploratory section.
Without writing shear and moment equations, draw the shear and moment diagrams for the beams specified in the following problems. Give numerical values at all change of loading positions and at all points of zero shear. Note to instructor: Problems to may also be assigned for solution by semi-graphical method describes in this article.
Solution To draw the Shear Diagram 1. Solving for x: For segment BC, the location of zero moment can be accomplished by symmetry and that is 5 ft from B. Solving for point of zero moment: This point is the appropriate location for construction joint of concrete structures. Location of zero moment at segment BC: By squared property of parabola: Location of zero shear: P consists of two segments joined by a frictionless hinge at which the bending moment is zero.
The location of zero moment in segment BH can easily be found by symmetry. P consists of two segments joined by frictionless hinge at which the bending moment is zero. Draw shear and moment diagrams for each of the three parts of the frame.
It is subjected to the loads shown in Fig. P, which act at the ends of the vertical members BE and CF. These vertical members are rigidly attached to the beam at B and C. Draw shear and moment diagrams for the beam ABCD only. Shear in segments AB and BC is zero. Moment in segment AB is zero 2. Location of zero shear C: The shear in AB is a parabola with vertex at A, the starting point of uniformly varying load. The load in AB is 0 at A to downward wo or —wo at B, thus the slope of shear diagram is decreasing.
For decreasing slope, the parabola is open downward. The shear diagram is second degree curve, thus the moment diagram is a third degree curve. The maximum moment highest point occurred at C, the location of zero shear. VBC is also parabolic since the load in BC is linear. MAC is third degree because the shear diagram in AC is second degree. The shear from A to C is decreasing, thus the slope of moment diagram from A to C is decreasing. The shear diagram in AB is second degree curve.
The shear in AB is from —wo downward wo to zero or increasing, thus, the slope of shear at AB is increasing upward parabola. The shear diagram in BC is second degree curve. The shear in BC is from zero to —wo downward wo or decreasing, thus, the slope of shear at BC is decreasing downward parabola To draw the Moment Diagram 1.
The shear diagram from A to C is decreasing, thus, the moment diagram is a concave downward third degree curve. To draw the Shear Diagram 1. From the load diagram: The location of zero shear is obviously at the midspan or 2 m from B.
Load and moment diagrams for a given shear diagram Instruction: I hope we'll have a deeper understanding of the relationship between gravity and quantum mechanics. It seems like black holes are sort of the frontier in trying to do that.
## Schaum’s Outline of Strength of Materials PDF
Unlike codes, which are fixed and stable, moral economies permit us to grasp the changes in time and the appropriations by agents: Jump to Navigation. There are no events today. Please visit our full calendar. View Full Calendar. April 09, March 28, April 10, What if black holes behave like ordinary quantum mechanical objects—and information about them is not lost, as previously thought, but retained on their horizons?
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## ››Convert arpent [Canada] to digit
arpent digit
Did you mean to convert arpent [Canada] arpent [France] to digit
How many arpent in 1 digit? The answer is 0.00032495296733368.
We assume you are converting between arpent [Canada] and digit.
You can view more details on each measurement unit:
arpent or digit
The SI base unit for length is the metre.
1 metre is equal to 0.017102787754404 arpent, or 52.631578947368 digit.
Note that rounding errors may occur, so always check the results.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of arpent to digit
1 arpent to digit = 3077.36842 digit
2 arpent to digit = 6154.73684 digit
3 arpent to digit = 9232.10526 digit
4 arpent to digit = 12309.47368 digit
5 arpent to digit = 15386.84211 digit
6 arpent to digit = 18464.21053 digit
7 arpent to digit = 21541.57895 digit
8 arpent to digit = 24618.94737 digit
9 arpent to digit = 27696.31579 digit
10 arpent to digit = 30773.68421 digit
## ››Want other units?
You can do the reverse unit conversion from digit to arpent, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Digit
unit of length equal to about 3/4 of an inch
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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# Placement | Cubes
A type of cuboid in which all the sides i.e length, breadth & height are equal. All faces of cubes are of the same area.
• In a cube, there are 8-corners/vertices
• There are 6-faces(all equal in area)
• There are 12 edges(all equal in length)
`No. of edges = (No. of vertices) + (No. of facees) - 2`
Primarily question in cubes are based on the painting of faces of cubes & then cutting the painted cube into identical cubelets. Because question are based on the way cube is painted & how cuts are being made, hence we don’t recommend any thumb rule. Still however because some basic rules are always applicable which we will be using as building blocks for answering the question.
• If n equidistant cuts are made(all parallel to the same surface), the cube will be divided into (n + 1) identical cuboidal pieces with each such cut there will 2a2 new surface area will be generated which will be unpainted.
• If we want to cut our bigger cube into identical n3 cubelets, using minimum number cuts, we need total 3(n – 1) cuts, such that (n – 1) cuts parallel to each of these faces which are joining to corner.
• If the number of cuts is not multiple of three then cube can never be cut into identical cubes but still it can be cut into the maximum number of identical cuboidal pieces. To maximize such number of pieces we need to split the number of cuts into three parts which are closest.
Directions for Example (1-4):
If a cube is cut into n3 identical cubelets using minimum no. of cuts, after painting all faces of the cube with white color, then answer the following questions.
Example-1: What is the maximum number of possible painted faces in once of such of cubelets?
Solution – Maximum faces painted will be three in any such cubelets, which will be in case of cubelets coming out of corners of big cube after cutting.
Example-2: What is the minimum number of cuts required?
Solution – Total number cuts required in 3(n – 1).
(n – 1) equidistant cuts parallel to each of 3 faces which are joining to corner.
Example-3: How many cubelets will have at most 2 faces painted?
Solution – To find out the number of cubelets with at most 2 faces painted, we need to remove all those cubes which have exactly 3 faces painted
= Total no. of cubes – No. of cubes with 3 faces painted
= (n3 – 8)
Example-4: How many cubelets will have at least 1 face painted?
Solution – To find out the number of cubelets with at least one face painted, we need to remove all the cubelets which have no face painted
= n3 – (n – 2)3
Directions for Example (5-6):
A cube is divided into 343 identical cubelets.Each cut is made parallel to some surface of the cube. But before doing that the cube is colored with green color on one set of adjacent faces, red on the second and blue on the third set.
Example-5: How many minimum cuts you have made?
(a)15
(b)18
(c)21
(d)9
Solution –
```n3 = 343 = 73
==> n = 7 ```
Minimum number of cuts = 3(n -1)
```= 3(7 - 1)
= 3 X 6
= 18```
Example-6: How many cubelets are coloured with exactly two colours?
(a)59
(b)63
(c)51
(d)54
Solution –
```n3 = 343 = 73
==> n = 7 ```
Number of cubelet with no face painted will
```= (n - 2)3
= (7 - 2)3
= 125 ```
No. of cubelet with 2 color
= Total no. of cubelet – [cubelets with one color + cubelets with no color + cubelets with three color]
```= 343 - [125 + 2 + 165]
= 51```
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# 1.2: The Analytical Perspective
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Having noted that each area of chemistry brings a unique perspective to the study of chemistry, let’s ask a second deceptively simple question: What is the analytical perspective? Many analytical chemists describe this perspective as an analytical approach to solving problems.
For different viewpoints on the analytical approach see (a) Beilby, A. L. J. Chem. Educ. 1970, 47, 237-238; (b) Lucchesi, C. A. Am. Lab. 1980, October, 112-119; (c) Atkinson, G. F. J. Chem. Educ. 1982, 59, 201-202; (d) Pardue, H. L.; Woo, J. J. Chem. Educ. 1984, 61, 409-412; (e) Guarnieri, M. J. Chem. Educ. 1988, 65, 201-203, (f) Strobel, H. A. Am. Lab. 1990, October, 17-24.
Although there likely are as many descriptions of the analytical approach as there are analytical chemists, it is convenient to define it as the five-step process shown in Figure 1.2.1 .
Three general features of this approach deserve our attention. First, in steps 1 and 5 analytical chemists have the opportunity to collaborate with individuals outside the realm of analytical chemistry. In fact, many problems on which analytical chemists work originate in other fields. Second, the heart of the analytical approach is a feedback loop (steps 2, 3, and 4) in which the result of one step requires that we reevaluate the other steps. Finally, the solution to one problem often suggests a new problem.
Analytical chemistry begins with a problem, examples of which include evaluating the amount of dust and soil ingested by children as an indicator of environmental exposure to particulate based pollutants, resolving contradictory evidence regarding the toxicity of perfluoro polymers during combustion, and developing rapid and sensitive detectors for chemical and biological weapons. At this point the analytical approach involves a collaboration between the analytical chemist and the individual or agency working on the problem. Together they determine what information is needed and clarify how the problem relates to broader research goals or policy issues, both essential to the design of an appropriate experimental procedure.
These examples are taken from a series of articles, entitled the “Analytical Approach,” which for many years was a regular feature of the journal Analytical Chemistry.
To design the experimental procedure the analytical chemist considers criteria, such as the required accuracy, precision, sensitivity, and detection limit, the urgency with which results are needed, the cost of a single analysis, the number of samples to analyze, and the amount of sample available for analysis. Finding an appropriate balance between these criteria frequently is complicated by their interdependence. For example, improving precision may require a larger amount of sample than is available. Consideration also is given to how to collect, store, and prepare samples, and to whether chemical or physical interferences will affect the analysis. Finally a good experimental procedure may yield useless information if there is no method for validating the results.
The most visible part of the analytical approach occurs in the laboratory. As part of the validation process, appropriate chemical and physical standards are used to calibrate equipment and to standardize reagents.
The data collected during the experiment are then analyzed. Frequently the data first is reduced or transformed to a more readily analyzable form and then a statistical treatment of the data is used to evaluate accuracy and precision, and to validate the procedure. Results are compared to the original design criteria and the experimental design is reconsidered, additional trials are run, or a solution to the problem is proposed. When a solution is proposed, the results are subject to an external evaluation that may result in a new problem and the beginning of a new cycle.
Chapter 3 introduces you to the language of analytical chemistry. You will find terms such accuracy, precision, and sensitivity defined there. Chapter 4 introduces the statistical analysis of data. Calibration and standardization methods, including a discussion of linear regression, are covered in Chapter 5. See Chapter 7 for a discussion of how to collect, store, and prepare samples for analysis. See Chapter 14 for a discussion about how to validate an analytical method.
As noted earlier some scientists question whether the analytical approach is unique to analytical chemistry. Here, again, it helps to distinguish between a chemical analysis and analytical chemistry. For an analytically-oriented scientist, such as a physical organic chemist or a public health officer, the primary emphasis is how the analysis supports larger research goals that involve fundamental studies of chemical or physical processes, or that improve access to medical care. The essence of analytical chemistry, however, is in developing new tools for solving problems, and in defining the type and quality of information available to other scientists.
##### Exercise 1.2.1
As an exercise, let’s adapt our model of the analytical approach to the development of a simple, inexpensive, portable device for completing bioassays in the field. Before continuing, locate and read the article
“Simple Telemedicine for Developing Regions: Camera Phones and Paper-Based Microfluidic Devices for Real-Time, Off-Site Diagnosis”
by Andres W. Martinez, Scott T. Phillips, Emanuel Carriho, Samuel W. Thomas III, Hayat Sindi, and George M. Whitesides. You will find it on pages 3699-3707 in Volume 80 of the journal Analytical Chemistry, which was published in 2008. As you read the article, pay particular attention to how it emulates the analytical approach and consider the following questions:
1. What is the analytical problem and why is it important?
2. What criteria did the authors consider in designing their experiments? What is the basic experimental procedure?
3. What interferences were considered and how did they overcome them? How did the authors calibrate the assay?
4. How did the authors validate their experimental method?
5. Is there evidence that steps 2, 3, and 4 in Figure 1.2.1 are repeated?
6. Was there a successful conclusion to the analytical problem?
Don’t let the technical details in the paper overwhelm you; if you skim over these you will find the paper both well-written and accessible.
What is the analytical problem and why is it important?
A medical diagnoses often relies on the results of a clinical analysis. When you visit a doctor, they may draw a sample of your blood and send it to the lab for analysis. In some cases the result of the analysis is available in 10-15 minutes. What is possible in a developed country, such as the United States, may not be feasible in a country with less access to expensive lab equipment and with fewer trained personnel available to run the tests and to interpret the results. The problem addressed in this paper, therefore, is the development of a reliable device for rapidly performing a clinical assay under less than ideal circumstances.
What criteria did the authors consider in designing their experiments?
In considering a solution to this problem, the authors identify seven important criteria for the analytical method: (1) it must be inexpensive; (2) it must operate without the need for much electricity, so that it can be used in remote locations; (3) it must be adaptable to many types of assays; (4) its must not require a highly skilled technician; (5) it must be quantitative; (6) it must be accurate; and (7) it must produce results rapidly.
What is the basic experimental procedure?
The authors describe how they developed a paper-based microfluidic device that allows anyone to run an analysis simply by dipping the device into a sample (synthetic urine, in this case). The sample moves by capillary action into test zones containing reagents that react with specific species (glucose and protein, for this prototype device). The reagents react to produce a color whose intensity is proportional to the species’ concentration. A digital photograph of the microfluidic device is taken using a cell phone camera and sent to an off-site physician who uses image editing software to analyze the photograph and to interpret the assay’s result.
What interferences were considered and how did they overcome them?
In developing this analytical method the authors considered several chemical or physical interferences. One concern was the possibility of non-specific interactions between the paper and the glucose or protein, which might lead to non-uniform image in the test zones. A careful analysis of the distribution of glucose and protein in the text zones showed that this was not a problem. A second concern was the possibility that particulate materials in the sample might interfere with the analyses. Paper is a natural filter for particulate materials and the authors found that samples containing dust, sawdust, and pollen do not interfere with the analysis for glucose. Pollen, however, is an interferent for the protein analysis, presumably because it, too, contains protein.
How did the author’s calibrate the assay?
To calibrate the device the authors analyzed a series of standard solutions that contained known concentrations of glucose and protein. Because an image’s intensity depends upon the available light, a standard sample is run with the test samples, which allows a single calibration curve to be used for samples collected under different lighting conditions.
How did the author’s validate their experimental method?
The test device contains two test zones for each analyte, which allows for duplicate analyses and provides one level of experimental validation. To further validate the device, the authors completed 12 analyses at each of three known concentrations of glucose and protein, obtaining acceptable accuracy and precision in all cases.
Is there any evidence of repeating steps 2, 3, and 4 in Figure 1.2.1?
Developing this analytical method required several cycles through steps 2, 3, and 4 of the analytical approach. Examples of this feedback loop include optimizing the shape of the test zones and evaluating the importance of sample size.
Was there a successful conclusion to the analytical problem?
Yes. The authors were successful in meeting their goals by developing and testing an inexpensive, portable, and easy-to-use device for running clinical samples in developing countries.
This exercise provides you with an opportunity to think about the analytical approach in the context of a real analytical problem. Practice exercises such as this provide you with a variety of challenges ranging from simple review problems to more open-ended exercises. You will find answers to practice exercises at the end of each chapter.
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## Wednesday, November 30, 2016
### Check your lotto chances: what do the numbers say?
This post is a part 2 of using probability to determine our chances to win the lottery. In the previous post we checked the probability of drawing the numbers.
Oh the lottery win, could it be? We have determined that our chances are very slim, but we also know that the lotto people have to use a machine or a bunch of balls in a rolling drum to get the numbers. Maybe there could be an issue with the balls or the machine? or the environment its in? Maybe its not really random? Maybe there will be some patterns to how they get drawn.
Lets get all the numbers from previous draws and see what comes out of that. In Canada the lottery published the list of winning numbers, so we are going to get the files, and find out if there is (just maybe) a way to predict the winning numbers. Maybe one number is more likely to be drawn then the others.
To find out, lets load up the numbers from a CSV file and find out whats there:
import os, csv
from collections import Counter
#'Atlantic649.csv'#'Lotto649.csv'#
class LottoStats(object):
def __init__(self, file_name='Lotto649.csv'):
'''
We want to get the numbers out of a spreadsheet, its colums are like so:
Draw Date,No1,No2,No3,No4,No5,No6,Bonus,
'''
lotto_file = open(os.path.abspath(file_name), 'rb')
self.lotto_numbers = []
self.draws = 0
self.draws += 1
for i, cell in enumerate(row):
if (i != 0 and i != 7 and cell):
self.lotto_numbers.append(cell)
'''
Python has the handy Counter object that
will return the elements and their counts
'''
self.counts = Counter(self.lotto_numbers)
def get_numbers(self):
return self.lotto_numbers
def get_counts(self):
return self.counts.items()
def count_most_common(self, n=9):
return self.counts.most_common(n)
def count_least_common(self, n=9):
return self.counts.most_common(len(self.lotto_numbers))[:-n-1:-1]
def get_draws(self):
return self.draws
stats = LottoStats('Lotto649.csv')
# lets validate the expected counts with a test
assert len(stats.get_counts()) == 49
print stats.get_draws() # 2950
print stats.count_most_common(6) #[('34', 406), ('31', 405), ('45', 397), ('43', 394), ('40', 393), ('47', 386)]
print stats.count_least_common(6) #[('28', 331), ('14', 333), ('15', 339), ('16', 340), ('22', 341), ('13', 342)]
counts = stats.get_counts()
print(counts)
occurances = [occ[1] for occ in counts]
print(str(occurances))
From our observation by just checking the common and least common, we see that some numbers seem to be drawn more than others. Is that a big deal?
From the occurrences, we have the number 34 appearing 406 times, and we have 28 in there 331 times. That seems like it could be a big spread, but remember that we had 2950 draws.
We need to determine the chances that the number can be drawn at all. Out of 2950 draws, it's a 406/2950 or .1376 or 13% chance of happening and 28 has .1122.
Whats a good tool to add some insight here? What does this data tell us? There are two really valuable statistical tools to give us some understanding.
One is sample standard deviation which just indicates how far apart the values are. This isn't really useful on its own, but because our data has the properties it does it can be combined with the average to find.the coefficient of variance, which tells us how close all the values are to the average.
A set like [10,10,10] has a cv of 0; there is no difference. What does the set of our probabilities tell us?
from stats import standard_deviation
num_chances = [(num / float(2950)) for num in occurances]
print (num_chances)
sd = standard_deviation(num_chances, False) #The standard deviation is 0.00614129698134.
mean = sum(num_chances) / len(num_chances)
print("Mean "+str(mean))
variance = sd / float(mean)
print("coefficient of variance: "+str(variance)) #0.0501539253476
The standard deviation is 0.00614129698134, and the Mean (or average) is 0.122448979592, so the coefficient of variance: 0.0501539253476
The 0.05 is extremely low, getting close to 0 and that tells us the values of the numbers are all very, very close together. This means the actual drawing of the numbers seems to have no meaningful effect on the odds. It pretty much a random draw, so we can conclude that the spinning drum with the balls flying around do a pretty good job.
### Predictions?
So we have described the situation, so lets try some Predictive or Prescriptive analytics on this issue.
Can we predict the lottery? With the long odds, and the measured fairness of the process, it's just not accurately predictable enough to pursue specific numbers. The prescriptive then follows that you just shouldn't buy a lottery ticket expecting to make money doing it. But what are the odds of winning over 10 or 20 years, even just once? They are the same each week, they don't change.
Statistics tell a story about data. There has to be some understanding about the data itself, and
the types of questions you have understood to apply the methods and tools of stats properly.
### Aside
There is a way to make money from the lottery. Let's calculate the cost if played over 10 years, every week playing 100 draws a year. I think its $2. Lets say I bought an investment that gave 5% every year. year_cost = 2 * 52 * 2 #$208 per year
total_gain = 0
for y in range(10):
total_gain += year_cost * 1.05
print(total_gain)
If you kept that under your mattress, you would have $2080 or$2184 if you bought the investment and the best part is that you have 100% chance of getting that money, instead of the 0.00000000715112% chance of the lotto.
Maybe think of it this way: what are the chances that you will need a spare \$2000 in the future?
## Saturday, November 26, 2016
### Checking my lottery chances with Probability
Check your lottery chances with Statistics and Probability, maybe we can find a way to get some lucky numbers!
In this post we show how methods of statistics and probability can be used to
understand the chances of actually winning the lottery. The various methods will be worked out in python.
### Analytics
This is really 'Descriptive' analytics, we are just telling the story of what's there.
After, we can get into predictive and prescriptive analytics but before that the foundation needs some building.
First lets get an understanding of the odds of winning the lottery, and how we find out what those odds are.
There are actually 2 interpretations of probability; 2 different approaches to solve a how likely an event will happen. One is called "classic" and has evolved a bit to be termed "frequentist". The other is usually called "subjective" or "bayesian". Lets see how these approaches compare with our lotto chances problem.
### Counting
First, the classic example. We have a set of 49 numbers and we want to find out what the chances are of finding a set of 6 unique numbers. This we can check as a combination, it's 49 'choose' 6:
Combinations are expressed as:
$$\dfrac{n!}{(n-r!)*(r!)}$$
We use combinations instead of permutations because the actual order of the result doesn't matter. If we draw a 1, 20, 33, 41, 5, 6, or 1, 33, 41, 20, 6, 5. It doesn't matter, we just need 6 unique numbers.
A solution in python could be something like:
n = 49
r = 6
import math
denominator = math.factorial(n-r) * math.factorial(r)
num_combos = math.factorial(n) / denominator
print(num_combos) #13983816
So that's about 14 million, so you have a 1 in 14 million chances to get those numbers.
### Bayesian
Now let's use a Bayesian probability approach to find out how good our chances are. In Bayesian probability only the current likelihood is known and that prior knowledge is taken into account to determine the probability of the next event. Remember that the entire set of data doesn't need to be known, you are just venturing forward with what you have for each new case.
Do the numbers drawn previously matter? They do: we are looking for unique numbers, so we can't draw the same number twice. These are dependent events. Two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the second so that the probability is changed.
When two events, A and B, are dependent, the probability of both occurring is expressed by:
P(A and B) = P(A)*P(B|A)
Its the probability of the first event, combined with second, and that result it combined with the third, and so on. The order of the result still doesn't matter.
In python it could be expressed like so:
select = 49
choose = 6
total_prob = choose / float(select)
for x in range(5):
select = select - 1
choose = choose - 1
prob = choose / float(select)
total_prob = total_prob * prob
odds = 1 / float(total_prob)
print str(odds) # 13983816.0
Just under 14 million again, but when you look closer you see that the result is exactly the same (13983816).
It's the same result, just a different way of going about it.
What is interesting about the Bayesian probability interpretation solution is that we loop until we get to the end; but don't need to know how big the initial set is. That way interpreting probability lends itself nicely to computation models and is why Bayesian probability methods are a big part of the emerging field of Data Science. This isn't so obvious in the descriptive part of analytics, but the Bayesian methods have really shined when data isn't complete, or when the data set is extremely large.
Using frequentest methods with large data sets you need to take samples and understand what constitutes a sample set that is significant enough.
Is one better than the other? I don't think it's a better or worse, but just different. They both have advantages considering the problem and conditions around it, but it's interesting to see how different methods can have the same results.
To get really into it, read this very good book:
• http://xcelab.net/rm/statistical-rethinking/
Its also a good idea just to read posts from those that do this all the time:
• http://www.fharrell.com/2017/02/my-journey-from-frequentist-to-bayesian.html
Save the theory for another day, we have to get back to our Lottery problem!
We now know that the chances are long, no matter how they are measured. How about the actual drawing of the numbers? In the next post, we are going to get some historical data on the numbers that have been drawn, and see if we can find some patterns to better our odds.
## Tuesday, November 15, 2016
### Algorithms intro: Profiling python code
When developing software and implementing algorithms its always important to be as efficient as possible, to have the most elegant solution that is concise, understandable to others, but most importantly uses the resources it has in the most efficient way.
This post extends the previous http://jseller.blogspot.ca/2016/11/algorithms-intro-implementing-math-proof.html and profiles the algorithms we created to implement Euclids GCD method
### Profiling
With python and most every language has some profiling tools to check various computer resources; most frequently this is the memory used and CPU cycles consumed. For our algorithm post we are just concerned with the time any of our algorithms take to execute.
import cProfile, pstats, io
#you can run any code as a string to be executed:
cProfile.run('fill_rectangle_with_squares(Rectangle(24000,19453))')
#but the profile object can also be enabled and disabled to run many lines.
pr = cProfile.Profile()
pr.enable()
fill_rectangle_with_squares(Rectangle(2400000,194530))
pr.disable()
pr.print_stats()
### With
If there are many lines, this can get a little difficult keeping the first two and last two in the same spot, or accidentally deleted the cleanup part. Python has a 'with' statement to make this nice and clean. http://effbot.org/zone/python-with-statement.htm
import cProfile, pstats, io
class profile_block:
def __enter__(self):
self.profile = cProfile.Profile()
self.profile.enable()
#set things up
#return thing
def __exit__(self, type, value, traceback):
#tear things down
self.profile.disable()
self.profile.print_stats()
Try running a few versions and look at the output. It will probably take some larger numbers until you can see a change. The times depends on the machine, but by using the 'with' block we know the disable and printing will always happen, and the code is nicer.
with profile_block():
fill_rectangle_with_squares(Rectangle(24000000,19453003))
with profile_block():
fill_rectangle_with_squares_iterative(24000000,19453003)
Running the different approaches with the large example show a 3x difference with the optimized version.
with profile_block():
fill_rectangle_with_squares_iterative(2400000223334219423423424234240,1945303434230234234242322)
with profile_block():
greatest_common_divisor(2400000223334219423423424234240,1945303434230234234242322)
The output. This is a big improvement between our two versions.
fill 2400000223334219423423424234240, 1945303434230234234242322 with squares
smallest square found 2, 2
6 function calls in 0.989 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 0.000 0.000 euclid.py:104(__exit__)
1 0.000 0.000 0.000 0.000 euclid.py:22(__init__)
2 0.000 0.000 0.000 0.000 euclid.py:25(__str__)
1 0.989 0.989 0.989 0.989 euclid.py:55(fill_rectangle_with_squares_iterative)
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
greatest common divisor 2
3 function calls in 0.386 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 0.000 0.000 euclid.py:104(__exit__)
1 0.386 0.386 0.386 0.386 euclid.py:77(greatest_common_divisor)
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
### Wrap it up
From the first part of this post we solved the problem with computation and in this case it was a classic, Euclid Greatest Common Denominator method. His proof in Elements uses only geometry, but we also used geometry as a handy tool to visualize and understand the proof.
What we wanted to find out though, was how to validate this method with computation, and understand which approach works best.
As recursive implementations can more closely model the real world scenario, we have to keep in mind that this is being solved with computation, so our iterative approach is able to take advantage of the computing power at hand.
Finding out how well that actually performed is the job of profiling, and a well measured system, or just a small part of it, can go a long way to understanding how to write efficient code.
### Algorithms intro: implementing a math proof
What's a nice way of showing a visual example to better understand how a math proof can be implemented as an algorithm in code? Here we are showing an example of implementing an algorithm, comparing iterative and recursive algorithms for solving a method; and validating the results with profiling.
In problem solving, a great methodology to use is: Understand, Plan, Execute and Review.
We will use a bit of geometry to understand the problem, and then plan how to solve it; then we will implement code to execute that plan and review how things worked out.
### Understand
The Euclidean algorithm calculates the greatest common divisor (GCD) of two natural numbers a and b. (or x and y).
We can use some geometry to visualize this, check out this video of it in action. From that, it looks like a matter of:
1. Fill with squares along the shortest side of the rectangle
2. For the remaining rectangle, fill with squares again, until smallest square is found.
### Plan
Lets write down the steps in more detail:
• fill rectangle with squares
• find shortest sid
• divide long side by square side, get remainder
• if remainder is 0, smallest square is found
• for remainder, fill rectangle with squares
### Execute
To model the visualization, lets make a Rectangle object, and use that with a function. Lets start with what we know, the Rectangle:
class Rectangle(object):
def __init__(self, x, y):
self.x = x
self.y = y
def __str__(self):
return "%s, %s" % (self.x, self.y)
Now lets plan how to fill the rectangle just like the visualization happens, the next rectangle is filled, and so on until the end. This style of implementation is recursive.
• fill_rectangle_with_squares(2,4)
• fill_rectangle_with_squares(2,2)
• 2 is smallest square
We keep calling the fill_rectangle_with_squares function, until the x and y sides are equal. This is the stopping condition.
def fill_rectangle_with_squares(rectangle):
print("Fill " + str(rectangle))
# check the sides, if they aren't equal make a new rectangle with the remaining
if rectangle.x == rectangle.y:
print('smallest square found %s ' % str(rectangle))
return rectangle.x
elif rectangle.x > rectangle.y:
return fill_rectangle_with_squares(Rectangle(rectangle.x - rectangle.y, rectangle.y))
elif rectangle.y > rectangle.x:
return fill_rectangle_with_squares(Rectangle(rectangle.x, rectangle.y - rectangle.x))
Now we call the function with an assert to test the result
To test to see if this function returns what we want, we use assert to check. We want to assert that a 2 by 4 rectangle is anything other than 2. Lets test some others as well
assert fill_rectangle_with_squares(Rectangle(2,2)) == 2
assert fill_rectangle_with_squares(Rectangle(140,68)) == 4
assert fill_rectangle_with_squares(Rectangle(24,12)) == 12
### Review
Lets keep reviewing our solution with other sizes, like: Rectangle(143,62) and Rectangle(270,192). These work fine, but what happens when larger values are involved?
Rectangle(2400000223334219423423424234240,1945303434230234234242322)
When this would fail would vary from machine to machine, but on my home laptop I get a:
RuntimeError: maximum recursion depth exceeded
This means the call stack limit is reached, and this limit is bound by the design of the python language itself. Some languages like LISP don't have this limitation, it's just ends up using what resources it has. The limit is imposed because the amount of callable memory is a finite space on the machine. If you are using a large amount of recursion depth, and not getting the results you need then there is way to only bound by time, and the running of the CPU.
How can we make this better? Lets implement the algorithm using iteration.
def fill_rectangle_with_squares_iterative(x,y):
rectangle = Rectangle(x,y)
print('fill %s with squares ' % str(rectangle))
found = False
found = rectangle.x == rectangle.y
if rectangle.x > rectangle.y:
rectangle.x = rectangle.x - rectangle.y
elif rectangle.y > rectangle.x:
rectangle.y = rectangle.y - rectangle.x
print('smallest square found %s ' % str(rectangle))
So far, so good, and the tests work with fill_rectangle_with_squares_iterative(24000000,19453003)
Can the iterative version take the massive difficult rectangle that blew the call stack with the recursive version?
fill_rectangle_with_squares_iterative(2400000223334219423423424234240,1945303434230234234242322)
It does, but takes a couple seconds.
### Optimize
I know it took a couple seconds because the console lets me know how long any code execution takes. What's really happening here? To find out more I need to profile the code which will will in the second part of this post.
Before we do though, lets take a crack at optimizing the code even farther.
When you look at the iterative function, see that we are just modifying the Rectangle objects x and y values. We can just keep track of the x and y and not bother making the Rectangle object at all. We could maybe make one object and pass it instead of making a new object for each call, but lets go the step further. It's only 2 variables that makes it concise.
We used the Rectangle as a way to visualize the problem, but nice terse notation is possible and ends up being very clear, so lets boil down the description to the original:
Given two (natural) numbers not prime to one another, to find their greatest common measure.
(The Elements: Book VII: Proposition 2)
def greatest_common_divisor(x,y):
while not x == y:
if x > y:
x = x - y
elif y > x:
y = y - x
return x
print('greatest common divisor found %s' % (str(greatest_common_divisor(24,10))))
That's much more straightforward, and can probably be more concise, but I'd like to stop here to keep it readable and easier to understand.
### Wrap it up
Lets review what we did in the post. we understood the problems and used some geometry to show a solution. Since it was Euclids proof, it seems like the right thing to do. We then planned a solution using recursion as it matched our real world model.
This worked to a point, but we reviewed our options and implemented an iterative method that could handle much larger rectangles. Once we had that solved we optimized for the computational abilities of the language and the machine we are using.
Our last implementation works for all the tests above, and certainly seems to run faster. How much faster?
We will profile all these functions and find out in the next post: http://jseller.blogspot.ca/2016/11/algorithms-intro-profiling-python-code.html
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# Variance of Random Variable and Normal Variable
Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean -1 and variance unknown. If $$P(X\leq-1)=P(Y\geq2)$$ then standard deviation of Y is
My Solution:
$E(X)=\mu=1, V(X)=\sigma^2=4$ as
$\sigma^2=E[X^2]-(E[X])^2=E[X^2]-\mu^2$
$4=E[X^2]-1^2$
$\implies E[X^2]=5$
Now How can we calculate $\sigma^2$ for Y
• Your "solution" does not use the given equality. So - as to be expected - is not fruitful. – drhab Jan 23 '15 at 12:22
We will use the fact that $P(Z\le-1)=P(Z\ge1)$, where $Z$ is a standard normal random variable. We have that $$P(X\le-1)=P\biggl(\frac{X-1}2\le-1\biggr),$$ $$P(Y\ge2)=P\biggl(\frac{Y+1}\sigma\ge\frac3\sigma\biggr)$$ and $$P\biggl(\frac{X-1}2\le-1\biggr)=P\biggl(\frac{Y+1}\sigma\ge\frac3\sigma\biggr).$$ $(X-1)/2$ and $(Y+1)/\sigma$ are both standard normal random variables. Hence, $3/\sigma$ must be equal to $1$ and $\sigma$ to $3$.
The random variables $X$ and $Y$ can be written as $X=2U+1$ and $Y=\sigma V-1$ where $U$ and $V$ both have standard normal distribution.
$P\{U\geq 1\}=P\{U\leq -1\}=P\{X\leq -1\}=P\{Y\geq2\}=P\{V\geq\frac{3}\sigma\}=P\{U\geq\frac{3}\sigma\}$
The first equality is a consequence of the fact that the standard normal distribution is symmetric. The last equality is a consequence of the fact that $U$ and $V$ have the same distribution.
So $1=\frac{3}\sigma$, so that $\sigma^2=9$.
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##### A polling company reported that 57% of 1018 surveyed adults said that rising gas prices are...
A polling company reported that 57% of 1018 surveyed adults said that rising gas prices are quite annoying a. What is the exact value that is 57% of 1018? The exact value is Complete parts a through d below Type an integer or a decimal.) b. Could the result from part (a) be the actual number of adul...
##### (5). This problem involves the mapping w(z)-,(z + z") between the z-plane and the w-plane. The tw...
(5). This problem involves the mapping w(z)-,(z + z") between the z-plane and the w-plane. The two parts can be solved independently. 2 (a). Identify all of the values of z for which the mapping w(z) fails to be conformal. In each case, explain why the mapping is not conformal at that value of z...
##### 8:33 Search I LIE Module 2 - Chapter 3 - Excercise 3.... 1. Listed below are...
8:33 Search I LIE Module 2 - Chapter 3 - Excercise 3.... 1. Listed below are accounts to use for transactions (a) through each identified by a number. Following this list are the transactions. You are to indicate for each transaction the accounts that should be debited and credited by placing the ac...
##### Cool down question number 1. Please make sure to include not only the magnitude, but also...
Cool down question number 1. Please make sure to include not only the magnitude, but also the DIRECTION. Please show all work, detailed, step by step. Thank you. Cool Down(s) 1. Think about your work during the recent electric field lab and then consider the plot shown in figure Q21.6 on page 694 o...
##### If a hole was dug 5 feet deep and 5 feet in diameter, what would be the volume of dirt removed from the hole
If a hole was dug 5 feet deep and 5 feet in diameter, what would be the volume of dirt removed from the hole?...
##### 13. Would a debit or a credit increase the balance in the rent expense account when...
13. Would a debit or a credit increase the balance in the rent expense account when making a journal entry? 14. Would a debit or a credit decrease the balance in the Account Payable account when making a journal entry?...
##### What will the following laboratory tests show in a patient with chronic renal failure? (Write increased...
What will the following laboratory tests show in a patient with chronic renal failure? (Write increased or decreased, metabolic acidosis or alkalosis) 1. Urea nitrogen 2. Creatinine 3. Arterial blood gas 4. GFR 5. Hemoglobin & hematocrit_ 6. Potassium...
##### For these problems, words are assumed to be 4 bytes, and the references are word-addresses. Thus,...
For these problems, words are assumed to be 4 bytes, and the references are word-addresses. Thus, the words in memory are located in word-addresses 0, 1, 2, ... As a comparison, note that byte-addresses for words are multiples of 4. Thus, the byte-addresses for words are 0, 4, 8, 12,.... Note that t...
##### The total translational kinetic energy of the molecules of a sample of gas at 489 K...
The total translational kinetic energy of the molecules of a sample of gas at 489 K is 2.47 × 104 J. How many moles does the sample comprise? Also, find the average translational kinetic energy of a single molecule....
Complete the following Flexible Budget Performance Report. Interpret your results. Remember on the variances to put Favorable (F) or Unfavorable (U) Flexible i e Flexible Volume Master Actual Budget B. Budget Variance Budget Variance Sales 57,500 57,500 53,000 Volume Sales Revenue $206,500 ($3.50 pe...
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# 1972 AHSME Problems/Problem 20
We start by letting $\tan x = \frac{\sin x}{\cos x}$ so that our equation is now: $$\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}$$ Multiplying through and rearranging gives us the equation: $$\cos x = \frac{a^2-b^2}{2ab} * \sin x$$ We now apply the Pythagorean identity $\sin ^2 x + \cos ^2 x =1$, using our substitution: $$\left(\frac{a^2-b^2}{2ab} * \sin x \right)^2 + \sin ^2 x =1$$ We can isolate $\sin x$ without worrying about division by $0$ since $a \neq b \neq 0$ our final answer is $(E) \frac{2ab}{a^2+b^2}$
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http://turkishlaborlaw.com/news/business-in-turkey/2022-minimum-wage-computation/
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Breaking News
# 2022 Minimum Wage Computation
From new year’s onward, the net minimum wage has increased from 2,825.90 TL to 4,253.40 TL, with a 50% hike. A different exemption is also introduced this year: the exemption of minimum wage earners from income tax and stamp tax. There will be no income tax and stamp duty in the calculation of the new minimum wage from gross to net. Only the employee’s insurance premiums will be deducted over the minimum wage.
#### Gross to Net Minimum Wage Calculation
The minimum wage amount for the year 2022, which was determined by the suggestions of TÜİK, the demands of the labor unions, and the expectations by the employer, was 5,004.00 TL in gross.
Traditionally we have always talked about minimum wage in gross and net terms as insurance premiums and taxes were deducted from the wages. In other words, the minimum wage amounts determined in gross became clear after all necessary deductions were made. The amount that goes into the pocket of the worker is the amount after the minimum living allowance is added to this net amount. The calculation of the minimum wage that will go from gross to net without income tax and stamp duty in 2022 will be as follows:
Gross to Net Minimum Wage Calculation:
#### The Cost of the Minimum Wage to the Employer
While the net amount of the minimum wage is the earnings of the employees, the gross amount is the amount out of the employer’s pocket. Since income tax and stamp duty will not be calculated on the minimum wage announced on the net amount this year, the increase rate of approximately 50% in the net amount will not reflect the same rate on the gross amount.
The cost of employers includes the insurance premiums for which the employer is responsible for that worker in addition to the worker’s gross wage. In other words, the cost of a minimum wage to the employer is not only the gross minimum wage but also the employer’s insurance shares.
Cost of Minimum Wage to the Employer (without incentive):
Cost of Minimum Wage to the Employer (with incentive):
#### Employer’s SSI Ceiling Cost
One of the parameters that change with the minimum wage is the amount of premiums calculated for people whose total earnings, subject to premium, exceed the SSI Ceiling Basis..
With the minimum wage for 2022, the SSI Base and Base Amounts are determined as follows:
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https://qsstudy.com/third-equation-dimensional-motion/
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Physics
# Third Equation of Dimensional Motion
Third Equation of Dimensional Motion
In one dimensional motion a body moves along a straight line. So quantities associated with motion, for example displacement, velocity, acceleration etc., have only one component (moving along X-axis will have X-component and Y and Z components will be zero). In deriving equations of linear motion we will consider that the body is moving along X-axis. In that case subscripts associated with different quantifies of motion may be omitted. Normally, vx will be represented by v and ax by a.
Third Equation
Equation of motion relating position or displacement, acceleration and time: S =v0t + ½ st2.
or, x = x0 + vx0t + ½ axt2
If a body is moving with uniform acceleration its average velocity is equal to half of the sum of the initial velocity v0 and the final velocity v.
Thus, average velocity v = (v0 + v) / 2
But we know, v = v0 + at
So, average velocity, v = (v0 + v0 + at) / 2
= v0 + ½ at … … … (1)
Again we know average velocity,
v = s/t = displacement/time
or, s = vt … … … (2)
Now putting v from equation (1) to equation (2); we get:
S = (v0 + ½ at) t = v0t + ½ at2 … … … (3)
For one dimensional motion, say along X-axis, let the initial position at t = 0 is x0 and at time t = t. the position is x. So, the change of position, displacement, s = x – x0. Now, if initial velocity is vx0 and the final velocity v and the acceleration a, then from equation (3) we get;
S = x – x0 = vx0t + ½ axt2
or, x = x0 + vx0t + ½ axt2
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# Cantilever beam with load at free end
Calculate the Cantilever beam with load at free end with our free online tool using the input parameters.
Determine the slope and deflection of a cantilever beam subjected to a uniformly distributed load using this convenient online calculator.
lbs
Elastic Modulus
lbs
Area moment of inertia
inches
Length of the beam
inches
Send the result to an email
2 Number of calculations
• Slope at free end = PL3 / 6EI
• Deflection at any section = Px2( x3 + 6L2 – 4Lx ) / 24EI
The variables used in the formula are:
P is the externally applied load,
E is the Elastic Modulus,
I is the Area moment of Inertia,
L is the Length of the beam and
x is the position of the load
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# Walkthrough of an Interactive Zero-Knowledge Proof for Sudoku Puzzle
## 🔏 Learn how to prove possession of a Sudoku solution with Zero-Knowledge and build a PoC with pure Python.
Feb 17 · 11 min read
# 📖 Introduction
Without a doubt, we currently live in a data-driven society where in fact, data has become a more valuable resource than gold or oil. Seriously, consider the amount of personal data that we share online every minute, on a daily basis. Location, feelings, preferences, passwords, messages… and the list keeps growing…
Fortunately for us, symmetric and asymmetric modern crypto made it possible to protect our data against malicious adversaries attempting to eavesdrop on our communication channels. But what about the data controllers — the guys who we legitimately send the data to? How can consumers make sure that their data are not mishandled or abused? One way for sure is refusing to send data in the first place. But in reality, is not that simple. It’s an exchange. We exchange a bit of privacy for some kind of service that they provide right? Still, on many occasions, this exchange is not a fair one from the consumers perspective, since data handlers ask for way more data than what’s actually needed.
Yet again, cryptography may have a solution for this. What if I told you that it’s possible to avoid sharing your data after all. For instance, instead of sending a complete salary overview and job details to a letting agency for a credit check, send only a proof that you earn more than 40k per annum. This is exactly what Zero-Knowledge proofs (ZKP) provide! ( It becomes slightly more complicated than that in practice for whole lot other reasons. Like how do we trust the origin of the data etc… but you get the idea 😅 )
Although there are many ways in which you can construct ZKP, in this post, I will be giving a walkthrough (with full code snippets) on how to create a ZKP for a Sudoku puzzle solution using only hash-functions as commitments. ZKP can be quite daunting to understand initially, thus I truly believe that Sudoku puzzle is a good example to understand how ZKPs work at a very high level. Plus, Sudoku is something that most people are familiar with. But let’s just cut to the chase.
# 💡 Background Knowledge
Zero-Knowledge protocols originate back to the 80s when they were first proposed by Shafi Goldwasser, Silvio Micali and Charles Rackoff in MIT [2]. More precisely, they described an interactive proof system that involves one party (‘Prover’) to prove to a second party (‘Verifier’) that a statement holds.
Contextualising this into the usual Alice/Bob example, we can consider the following scenario: Alice stumbled upon an online competition with a puzzle to be solved with a prize stake of £100. She asks for Bob’s help and they agree to split the price if any of them solves it. Not much time later, Bob claims that he has solved the problem and Alice asks him to share the solution. However, Bob is reluctant to sharing because he thinks that Alice may submit the solution by herself without sharing the price. Consequently, he is looking for a secure way to prove to Alice that he has the solution but without sharing it directly with her. Yeah, you guessed right! A ZKP is exactly what he is after!
But let’s define more thoroughly what is meant by the term “secure way”. In general, a ZKP must provide the following properties (at least with very high confidence!):
Completeness: Every invalid proof must be rejected by the verifier
Soundness: Every valid proof must be accepted by the verifier
Zero-Knowledge: Verifier does not learn any information about the statement other than its assertion.
Note that this is a very basic definition of interactive ZKPs. There a wide variety of more sophisticated interactive/non-interactive ZKPs but for the present walkthrough, this definition suffices.
Commitment schemes are a crucial ingredient of ZKPs and frankly, of cryptography in general. In simple terms, they are cryptographic primitives that allow a party to commit to a specific value without revealing the actual value while providing a way to reveal the value and verify the commitment in a later phase.
More formally, let C be a commitment scheme, it must provides these two properties:
Hiding: Hard to distinguish between commitments of different values. i.e:
Binding: There should be no way for a person who commits to one bit, to claim that he has committed to another value later:
One way to create a commitment scheme is by using one-way hash functions and cryptographically secure random bytes. It should be noted that in such case the security of the scheme is governed by the cryptographic assumptions of the hash function itself (i.e. it’s truly one-way).
To add more clarity, let’s walk through an example with the usual suspects. Alice and Bob decide to play a game of rock, paper, scissors digitally. They both make their choices and exchange them such that a winner is decided. Naturally, in a digital world, one of them has to share his/her pick first, which brings her/him in a disadvantageous position as he/she can just share a different choice after reviewing what the other player picked. This is exactly the kind of problems that commitment schemes solve!
Alternatively, they can create a commitment based on their choices and share the commitment instead of their actual choice! For instance:
Let a set:
S = {“Rock”, ”Paper”,”Scissors”}
Bob and Alice both randomly pick P and Pᴮ from S respectively. Now they calculate: ( || -> represents concatenation)
Cᴬ = sha256(Rᴬ || Pᴬ) and Cᴮ = sha256(Rᴮ || Pᴮ)
Bob shares Cᴬ with Alice and Alice Cᴮ with Bob. Note that by now, they both committed to these values.
Finally, they share their original choices and random bytes Pᴬ, Rᴬ and Pᴮ, Rᴮ. With this information, each party can verify the commitment by hashing P || R and assert their equivalence. Based on their picks the winner can be decided and none of them could have altered their initial choice since the hashes wouldn’t match.
# 🧩 Sudoku ZKP
It’s time for the main part of this article. Sudoku is a very well known puzzle that is also known to be an NP problem (in fact NP-Complete [4]) and is proven that there is a ZKP for any problem in NP [1]. Sudoku ZKP is by no means something new but I have yet to find an intuitive and clear explanation of the protocol with code examples so at the very least this is what this article aims to provide. Actually, the protocol described here is the implementation of this very interesting work from Gradwohl et al. [3] therefore for more formal details you can refer there. Interestingly, in their paper, they also described a physical protocol to perform the proof using a deck of cards which is fun if you want to demonstrate ZPK physically but let’s stick to the digital proof for now.
Before jumping into the code let’s see a high-level plain-English description of the protocol itself.
Alice wants to prove to Bob that she has a solution to a Sudoku Puzzle but Bob doesn’t believe her. Assume she on hold of the following puzzle and solution.
To avoid confusion let’s follow the proof in steps:
1. Alice creates a permutation of the sudoku digitis which effective is a one-to-one mapping for each digit. i.e. 1 -> 3, 2 ->8 ….
2. Additionally, she generates a random byte sequence (nonce) for each sudoku cell. This leads to 81 random nonces.
3. She applies the mapping on the numbers of her sudoku solution to obtain a masked solution. Note that by permuting the values the numbers still appear just once since it is a one-to-one mapping.
Alice splits the masked solution to sets of numbers from each Row, Column, Subgrid, and a set of the known numbers that were part of the puzzle definition in the first place. In effect, she ends up with 27+1 sets of numbers where the 27 derived from the rows, columns, subgrids must satisfy the sudoku requirement. i.e. All digits from 1–9 appear just once.
Then, she creates a commitment to the masked solution by hashing each cell with the corresponding nonce, sends that commitment to Bob and asks Bob to choose a row, columns, subgrid or to the set of known numbers.
Bob picks one and Alice sends him the nonces and permuted numbers that correspond to Bob’s choice.
In the case that Bob has picked the list of known numbers Alice also sends him the one-to-one mapping, she generated initially.
Bob then verifies that the permuted values indeed appear just once, and recreates the commitment using the nonces to verify Alice’s commitment.
In the case that Bob has picked the list of known numbers he also checks that the mappings are also correct. i.e. Let M be the mapping and n an integer 1–9: M(n) == Received List of known numbers.
These 9 steps summarize one iteration of the proof!
💡 Steps 1, 3 are executed such that the Verifier does not learn anything about the solution of the puzzle.
💡Step 2, the nonces are used to stop the Verifier from performing a known plain text attack. Without nonces, Bob could just hash all numbers (1–9) and match the corresponding digest with the commitments to reveal the values.
💡 Step 5, by creating a commitment and sending it to Bob, Alice commits to her solution and therefore cannot change it (or the mapping).
💡 Steps 6,7,8,9 are the hardest to understand. By letting Bob choose any of those Alice provides a chance to the Verifier to confirm that the solution is correct. Recall that she can’t change the mapping since she committed to it. Additionally, you may be thinking: “This proves that she may have just a solution, not that particular one”. You are 100% correct! That is the reason that the choice of requesting the initial numbers given with the puzzle exists. That way, Alice cannot cheat because at any time Bob may choose to request those numbers and obviously they won’t match if it’s a different solution.
💡More importantly, it is necessary to stress the soundness error of this approach. Recall that soundness is the property of ZKP Verifier to accept every valid proof. In this case, since there are 28 choices (3n+1) that the Verifier can pick from, this means the protocol is only 1/28 % sound, which implies a 1–1/28 soundness error! In other words, after one iteration the Verifier is only 1/28 sure that this is a valid proof. That’s not very high, is it? For that reason, the steps above can be performed for many iterations to decrease the soundness error to a negligible value. (Bayes Rule can be used to infer the soundness rates per iteration)
I hope this makes more sense now. Let’s proceed to develop a small PoC proof using Python. You can find the full code here:
# 💻 Python ZKP PoC
Since generating and solving the sudoku is not in the scope of this post I will just use a static one for the time being but feel free to replace this with code to generate a new sudoku puzzle each time:
The sudoku puzzle is stored as a flat python list for easier processing.
Next, let’s add the code for permuting the sudoku solution using the one-to-one mapping and for creating the random nonces:
The zero element, is added after the shuffle because I don’t want it to be part of the mapping. It’s just there such the index of the numbers starts from 1.
In addition, below is the function to create a commitment of a sudoku solution:
As shown above, for commitments the SHA256 hashing algorithm is used. Furthermore, below is the code for split the puzzle into rows, columns and subgrids:
Finally here is a check to verify that all numbers from 1–9 exists and appear just once:
Now we can stitch all these functions together to create a Proof of Concept of the protocol and verify it’s correctness:
# Improvements
The code is just for demonstration purposes and is may not be cryptographically secure so please don’t use as-is in production. There are several optimizations and improvements that could be added so if you ever feel the urge, don’t hesitate to open a Pull Request in the GitHub repo.
# Conclusion
Hopefully, this blog post provided some general basic insights about ZKPs and more particularly how to prove possession of a Sudoku solution with a Zero-Knowledge proof. In future posts, we will explore more sophisticated proofs that do not require interaction between the Prover and the Verifier and also employ more complex cryptographic primitives. Stay tuned!
# References
[1] Oded Goldreich, Silvio Micali, and Avi Wigderson. Proofs that yield nothing but their validity or all languages in np have zero-knowledge proof systems. Journal of the ACM(JACM), 38(3):690–728, 1991.
[2] Shafi Goldwasser, Silvio Micali, and Charles Rackoff. The knowledge complexity of interactive proof systems. SIAM Journal on Computing, 18(1):186–208, 1985.
[3] Ronen Gradwohl, Moni Naor, Benny Pinkas, and Guy N Rothblum. Cryptographic and physical zero-knowledge proof systems for solutions of sudoku puzzles. In International Conference on Fun with Algorithms pages 166–182. Springer, 2007.
[4] Takayuki Yato and Takahiro Seta. Complexity and completeness of finding another solution and its application to puzzles. IEICE transactions on fundamentals of electronics, communications, and computer sciences, 86(5):1052–1060, 2003.
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# Your question: How long does it take to burn 1000 calories walking?
Contents
How long you have to walk to burn 1000 calories depends on your weight and speed of walking. For example, a 180-pound person would have to walk for a little over 13.1 miles to burn 1000 calories, which means that they would have to walk for 238 minutes at the very least.
## How long do you have to walk to burn 1000 calories?
How much do I have to walk to burn 1000 calories a day? You need to walk for about 120 minutes at 6 mph to burn 1000 calories in a day.
## Can you burn 1000 calories in an hour?
Most often, an hour of cardio was in the 500-calorie-an-hour range. Keeping in mind that most gym classes are less than an hour—forty to fifty minutes is a more realistic range—it becomes clear that while “up to 1,000” an hour may be true, it’s not going to be true for you.
IT IS INTERESTING: Best answer: Can I lose weight if I workout at home?
## How many calories does a 30 minute walk burn?
Calories burned in 30-minute activities
Walking: 3.5 mph (17 min/mi) 120 178
Badminton: general 135 200
Walking: 4 mph (15 min/mi) 135 200
Kayaking 150 222
## How many calories do you burn in a five mile walk?
5 miles. Walking 5 miles a day, the average person will burn 350 to 600 calories. Walking 5 miles a day for 5 days a week will burn an extra 1,750 to 3,000 calories. Eating the same as before you are likely to lose ½ to close to 1 pound of fat per week.
## What is the fastest way to burn calories?
Running is the winner for most calories burned per hour. Stationary bicycling, jogging, and swimming are excellent options as well. HIIT exercises are also great for burning calories. After a HIIT workout, your body will continue to burn calories for up to 24 hours.
## How many calories does 4 hours walking burn?
For example, a 155-pound person burns approximately 232 calories walking at a moderate (3.5 mph) pace on a flat surface for one hour. That same person could burn up to 439 calories an hour hiking a more mountainous trail while wearing a weighted backpack.
## How many calories will 500 skips burn?
You can burn 107 calories in 500 jump ropes if you do it at a slow pace. It can improve your agility, coordination, balance, stamina, reflexes, and heart health. But you have to also take a healthy and controlled diet. Because you should burn the number of calories you take in, to maintain your fitness.
## How many calories does 1 hour of exercise burn?
Well, based on the fact that one pound of fat equates to around 3,500 calories, a daily workout of one hour where you burn 500 calories, should help you lose a pound a week, as long as your diet is sensible.
IT IS INTERESTING: Do you burn more calories when you are fit?
## How much can you lose on 800 calories a day?
Some people go on a very low-calorie diet for rapid weight loss, often consuming only 800 calories a day. This type of diet usually includes special foods such as shakes, bars, or soups to replace meals and for added vitamins. Very low-calorie diets can help a person achieve weight loss of up to 3 to 5 pounds per week.
## Does walking 1 hour a day help lose weight?
Walking can help you lose weight
Walking 1 hour each day can help you burn calories and, in turn, lose weight. In one study, 11 moderate-weight women lost an average of 17 pounds (7.7 kg), or 10% of their initial body weight, after 6 months of brisk daily walking ( 3 ).
## Can you lose belly fat by walking?
Walking might not be the most strenuous form of exercise, but it is an effective way to get in shape and burn fat. While you can’t spot-reduce fat, walking can help reduce overall fat (including belly fat), which, despite being one of the most dangerous types of fat, is also one of the easiest to lose.
## How long does it take to burn 500 calories walking?
Individuals weighing 155 pounds burn 500 calories walking 4 miles per hour for 90 minutes, or walking at a pace of 3.5 miles per hour for about 100 minutes, according to Harvard Health Publications. Boosting your walking speed up to 4.5 miles per hour means you’ll expend 500 calories in just 81 minutes.
## How can I lose 1 pound a day?
You need to burn 3500 calories a day to lose one pound a day, and you need anywhere between 2000 and 2500 calories in a day if you are doing your routine activities. That means you need to starve yourself the whole day and exercise as much as to lose the remaining calories.
IT IS INTERESTING: How many calories do you burn walking 20 minutes?
## Does walking 10000 steps help lose weight?
Completing an extra 10,000 steps each day typically burns about 2000 to 3500 extra calories each week. One pound of body fat equals 3500 calories, so depending on your weight and workout intensity, you could lose about one pound per week simply by completing an extra 10,000 steps each day.
## How can I burn 1000 calories a day?
Be well hydrated and have a small breakfast. Walk on a treadmill at an incline for an hour. I am 6′ and 200 lbs, and when I walk at 4 mph and a 6% incline, I burn about 1,000 calories an hour. So one way to reach your goal is to do this for 5 hours (adjusting for your calorie burn based on your own research).
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## 6.15. Number summary
The sumti which refer to numbers consist of the cmavo li (of selma'o LI) followed by an arbitrary Lojban mekso, or mathematical expression. This can be anything from a simple number up to the most complicated combination of numbers, variables, operators, and so on. Much more information on numbers is given in Chapter 18. Here are a few examples of increasing complexity:
Example 6.95.
li vo the-number four 4
Example 6.96.
li re su'i re the-number two plus two 2 + 2
Example 6.97.
li .abu bi'epi'i xy. bi'ete'a re su'i by. bi'epi'i xy. su'i cy. the-number a times x to-power 2 plus b times x plus c ax2 + bx + c
An alternative to li is me'o, also of selma'o LI. Number expressions beginning with me'o refer to the actual expression, rather than its value. Thus Example 6.95 and Example 6.96 above have the same meaning, the number four, whereas
Example 6.98.
me'o vo the-expression four
“4”
and
Example 6.99.
me'o re su'i re the-expression two plus two
“2+2”
refer to different pieces of text.
The implicit quantifier for numbers and mathematical expressions is su'o, because these sumti are analogous to lo descriptions: they refer to things which actually are numbers or pieces of text. In the case of numbers (with li), this is a distinction without a difference, as there is only one number which is 4; but there are many texts 4, as many as there are documents in which that numeral appears.
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4 months ago
# CLASS_11_MATHS_SOLUTIONS_NCERT
Class XI Chapter 1 – Sets Maths ______________________________________________________________________________ Exercise 1.1 Book Name: NCERT Solutions Question 1: Which of the following are sets? Justify our answer. (i) The collection of all months of a year beginning with the letter J. (ii) The collection of ten most talented writers of India. (iii) A team of eleven best-cricket batsmen of the world. (iv) The collection of all boys in your class. (v) The collection of all natural numbers less than 100. (vi) A collection of novels written by the writer Munshi Prem Chand. (vii) The collection of all even integers. (viii) The collection of questions in this chapter. (ix) A collection of most dangerous animals of the world. Solution 1: (i) The collection of all months of a year beginning with the letter J is a well-defined collection of objects because one can definitely identity a month that belongs to this collection. Hence, this collection is a set. (ii) The collection of ten most talented writer of India is not a well-defined collection because the criteria for determining a writer’s talent vary from person to person. Hence, this collection is not a set. (iii) A team of eleven best cricket batsmen of the world is not a well-defined collection because the criteria for determining a batsman’s talent may vary from person to person. Hence, this collection is not a set. (iv) The collection of all boys in your class is a well-defined collection because you can definitely identify a boy who belongs to this collection. Hence, this collection is a set. (v) The collection of all natural numbers less than 100 is a well-defined collection because one can definitely identify a number that belongs to this collection. Hence, this collection is a set. (vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined collection because one can definitely identify a book that belongs to this collection. Hence, this collection is a set. (vii) The collection of all even integers is a well-defined collection because one can definitely identify an even integer that belongs to this collection. Hence, this collection is a set. (viii) The collection of questions in this chapter is a well-defined collection because one can definitely identify a question that belongs to this chapter. Hence, this collection is a set. Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
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Original - University of Toronto Libraries
LaPromozionale.it - Mascotte
Bukhovtsev-et-al-Problems-in-Elementary-Physics
basic_engineering_mathematics0
Teaching & Learning Plans - Project Maths
Untitled
Maths on track! - STEPS | Engineers Ireland
103 Trigonometry Problems
t - Eötvös Loránd University
The Philosophical magazine; a journal of theoretical ... - Index of
ICfJff' - Al Kossow's Bitsavers
MARYLAND Orbital Maneuvering
Bell J.L. A primer of infinitesimal analysis (2ed., CUP, 2008)(ISBN 0521887186)(O)(138s)_MCat_
Stability Of Solitary Waves Of A Generalized Two-Component ...
THE MATHS TEACHER'S HANDBOOK - Arvind Gupta
View report - eResearch SA
March 2011 - Career Point
IIT-JEE 2010 - Career Point
PDF file
The Cubic Spline - STEM2
SOME ELEMENTARY INEQUALITIES IN GAS DYNAMICS EQUATION
XT â MATHS Grade 11 â Equations
Convergence to Stochastic Integrals involving Non-linear Functions
Elliptic variational problems with critical exponent
A Nonlinear Heat Equation with Temperature-Dependent ... - UFRJ
Optimal Bounds on the Kuramoto-Sivashinsky Equation Felix Otto ...
Rudin's Principles of Mathematical Analysis: Solutions to ... - MIT
Einstein's 1905 paper
thief of ice crime
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• # StanBij.mws
• Referenced in 10 articles [sw06645]
• combinatorial proof of a partition identity of Andrews and Stanley In his paper ... certain partition identity analytically and asks for a combinatorial proof. This paper provides the requested...
• # Tame Graphs
• Referenced in 16 articles [sw28578]
• Proofs. [CUP 2012]. The values of the constants in the definition of tameness are identical ... refers to the original version of Hales’ proof, the ITP 2011 paper by Nipkow refers...
• # BijTools
• Referenced in 3 articles [sw12643]
• finding combinatorial bijections. Consider a combinatorial identity that can be proved by induction. In this ... general method for translating the inductive proof into a recursive bijection. Furthermore, we will demonstrate ... often results in a bijective proof of the identity that helps illuminate the underlying combinatorial...
• # Examples
• Referenced in 3 articles [sw12644]
• finding combinatorial bijections. Consider a combinatorial identity that can be proved by induction. In this ... general method for translating the inductive proof into a recursive bijection. Furthermore, we will demonstrate ... often results in a bijective proof of the identity that helps illuminate the underlying combinatorial...
• # Fibonacci
• Referenced in 3 articles [sw12645]
• finding combinatorial bijections. Consider a combinatorial identity that can be proved by induction. In this ... general method for translating the inductive proof into a recursive bijection. Furthermore, we will demonstrate ... often results in a bijective proof of the identity that helps illuminate the underlying combinatorial...
• # TransMethodZeck
• Referenced in 3 articles [sw12646]
• finding combinatorial bijections. Consider a combinatorial identity that can be proved by induction. In this ... general method for translating the inductive proof into a recursive bijection. Furthermore, we will demonstrate ... often results in a bijective proof of the identity that helps illuminate the underlying combinatorial...
• # ZeckFibBijections
• Referenced in 3 articles [sw12647]
• finding combinatorial bijections. Consider a combinatorial identity that can be proved by induction. In this ... general method for translating the inductive proof into a recursive bijection. Furthermore, we will demonstrate ... often results in a bijective proof of the identity that helps illuminate the underlying combinatorial...
• # OrientedSwaps
• Referenced in 2 articles [sw35044]
• Knuth and Burge correspondences. A second probabilistic identity, relating those two vectors to a vector ... give a computer-assisted proof of this identity for n≤6 after first reformulating...
• # MJ
• Referenced in 12 articles [sw24342]
• which are sufficiently small to facilitate formal proofs of key properties. However many ... compact, MJ models features such as object identity, field assignment, constructor methods and block structure ... operational semantics of MJ, and give a proof of type safety. In order to demonstrate...
• # Regular_Algebras
• Referenced in 3 articles [sw32237]
• regular expressions as induced by regular language identity. We use Isabelle/HOL for a detailed systematic ... relationships between these classes, formalise a soundness proof for the smallest class (Salomaa ... provide a large collection of regular identities in the general setting of Boffa’s axiom ... algebra hierarchy in the Archive of Formal Proofs; we have not aimed at an integration...
• # SicoTHEO
• Referenced in 8 articles [sw09982]
• exploited by competition: on each processor, an identical copy of SETHEO tries to prove ... soon as one processor finds a proof, the entire system is stopped. Three different versions...
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# Using the expression language
An expression is a sequence of operands, operators, and parentheses that the runtime system can evaluate as a single value.
The RTL Expression language follows the Java™ syntax for expressions and evaluation semantics. Expressions can include these components:
## Operators
Table 1. RTL Operators
Operator Description Example Precedence
``%``
Arithmetic: Modulus
``x % 2``
8
``*``
Multiplication
``x * y``
7
``/``
Division
``x / y``
7
``+``
``x + y``
7
``-``
Subtraction
``x - y``
6
``+``
Concatenation
``string + string ``
5
``<``
Relational/Boolean: Less than
``numeric < 100``
4
``<=``
Less then or equal to
``numeric <= 100``
4
``>``
Greater than
``numeric > 100``
4
``>=``
Greater than or equal to
``numeric >= 100``
4
``==``
Equal to
``numeric == 100``
4
``!=``
Not equal to
``numeric <> 100``
4
``!``
Logical inverse (NOT)
``!(x = y )``
3
``&&``
Logical intersection (AND)
``expr1 && expr2``
2
``||``
Logical union (OR)
``expr1 || expr2``
1
The first column in the table describes the precedence order of the operators, listed highest to lowest. For example, the `%` modulus operator has a higher precedence than the `*` operator. Parentheses can be used to overwrite the precedence of operators.
## Conditional Expressions
Conditional expressions allow you to express IF/ELSE statements.
Syntax:
`` Boolean-expression?expression-1:expression-2``
The `?` operator indicates that this expression is conditional; the return value is dependent on the result of the Boolean expression. If the Boolean expression is `TRUE`, the first expression is the return value; otherwise, the second expression is the return value.
You can use the `null` keyword in the ternary conditional operator. The “if then” and “if else” operands can be either expressions or the keyword `null`. A property whose RTL expression yields “null” is not set. This is useful in cases where a property should be set only when a certain condition is met. Consider the case where the background color of a WORDBOX should be set to red when a variable value x drops below a value of 10. The expression for this would be:
``x<10?Color.RED:null``
## Operands
Operands include:
• Literal values
• Other expressions
• FGL Variables
• RTL Class Members
• Objects
• Methods (returning a single value)
A literal value for a string in an expression should be delimited by double quotes: "Test".
## FGL Variables
The data types of FGL variables are taken into account when constructing expressions. For every FGL variable an object is created that is either an instance of a FGLNumericVariable or an FGLStringVariable. These objects hold the value of the FGL variable, and at the same time they contain a member variable value which also contains the value. For this reason, it is legal to write "order_line.itemprice" in your expression as a shortcut for "order_line.itemprice.value". Both types of objects have these specific member variables defined as in Table 2.
Table 2. Member Variables for FGLNumericVariable and FGLStringVariable
Name Description
`value` The value of the FGL variable.
`fglValue` (FGLNumericVariable only) The value of the field as formatted by the DVM.
`name` A String specifying the name of the field.
`caption` A String specifying the title of the field.
`type` A String specifying the RTL type of the field.
`isoValue` The locale and formatting-independent representation of the value of the variable.
The conversion table lists FGL data types and the type into which they are converted within an RTL expression, as in Table 3.
Table 3. FGL data types and the type into which they are converted within an RTL expression
FGL type Corresponding RTL type
CHAR, VARCHAR, STRING, TEXT, DATE, DATETIME, and INTERVAL FGLStringVariable
INTEGER, SMALLINT, FLOAT, SMALLFLOAT, DECIMAL and MONEY FGLNumericVariable, limited to 15 significant digits. The value of a number larger than 15 digits will be truncated, and the resulting number is rounded. For example, `12345678901234567` will be rounded to `123456789012346`.
Important:
To avoid problems with inaccurate totals on a report due to rounding, do not perform RTL arithmetic on either the individual values or the total value; calculate the value of the item in the BDL program instead, before passing the value to the report.
## Examples
For the purpose of these examples, `order_line` has been replaced with `order`.
1. To add 10% to the itemprice: `order.itemprice*1.10`
The data item order_line.itemprice is converted to a Numeric type, so we can use the Numeric operators. In order to display the result of a Numeric expression in a Word Box, we must convert the result to a String. See Example 1 in the Using RTL Classes section.
2. Let's add 10% to the item price conditionally, depending on the value: `order.itemprice<100?order.itemprice*1.10:order.itemprice`
The condition in this Boolean expression tests whether the itemprice is greater than 100; if so, the value returned is 110% of the itemprice; otherwise, the value returned is simply the itemprice.
3. To set the font of a report item to italic when the BDL variable order_line.lineitemprice exceeds \$20, we must create an expression for the fontItalic property: `order.lineitemprice>20`
The property fontItalic is of type boolean, so any RTL expression that we use for that property must return a boolean value (`TRUE`/`FALSE`). Any of the relational operators yields a boolean, so the type of the returned value of this expression is a boolean (The expression will return `TRUE` if the lineitemprice exceeds 20).
Note:
A numeric value by itself is not a boolean value as it is in some programming languages.
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1. ## Differential equation
Hey guys i'm banging my head against the wall with this question.
dy/dx = (e^(x+y) ) / y-1
i've tried multiplying top & bottom by e^-y and seperating, but I cant seem to match the answer in the back of the book which is e^x + ye^-y = C
The fact that the equation is equal to C gave me the idea that it might be exact, but when I tried the partial derivative test I couldn't get a single variable by itself (I might have screwed up there somewhere also)
2. Hello, ubique!
You started off great . . .
$\displaystyle \frac{dy}{dx} \:=\:\frac{e^{x+y}}{y-1}$
i've tried multiplying top & bottom by $\displaystyle e^{-y}$ and separating . Good!
but I can't seem to match the answer in the book: .$\displaystyle e^x + ye^{-y} \:=\: C$
We have: .$\displaystyle \frac{dy}{dx} \:=\:\frac{e^x\cdot e^y}{y-1}$
Separate variables: .$\displaystyle \frac{y-1}{e^y}\,dy \;=\;e^x\,dx$ .[1]
On the left, we can integrate by parts:
. . $\displaystyle \begin{array}{ccccccc}u &=& y-1 & & dv &=& e^{-y}\,dy \\ du &=& dy & & v &=& \text{-}e^{-y} \end{array}$
And we have: .$\displaystyle -e^{-y}(y-1) + \int e^{-y}\,dy \;=\;-e^{-y}(y-1) - e^{-y}+C_1\;=\;-ye^{-y} +C_1$
Then [1] becomes: .$\displaystyle -ye^{-y} +C_1\;=\;e^x+ C_2 \quad\Rightarrow\quad\boxed{ e^x + ye^{-y} \:=\:C}$
3. Thanks a lot Soroban
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# Definition:Entropy (Probability Theory)
It has been suggested that this page or section be merged into Definition:Uncertainty. (Discuss)
## Definition
Let $X$ be a discrete random variable that takes on the values of $\set {x_1, x_2, \ldots, x_n}$ and has a probability mass function of $\map p {x_i}$.
Then the entropy of $X$ is:
$\ds \map H X := -\sum_{i \mathop = 1}^n \map p {x_i} \log_2 \map p {x_i}$
and is measured in units of bits.
By convention $0 \log_2 0 = 0$, which is justified since $\ds \lim_{x \mathop \to 0^+} x \log_2 x = 0$.
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### Explaining a Fastball’s Unexpected Twist
#### Investigators at Utah State University determine how pitches vary depending on spin speed, spin axis and ball orientation.
Article ID: 703864
Released: 13-Nov-2018 12:20 AM EST
Source Newsroom: American Physical Society's Division of Fluid Dynamics
• Credit: Sakib and Smith
The baseball in this illustration of a knuckleball pitch is moving to the left, leaving a wake trailing behind (i.e., to the right). The blue-colored air is rotating clockwise; the red air is rotating counterclockwise. The drag on the ball depends on the wake size, which depends in turn on the distance between the uppermost blue and lowermost red points. The upward wake means the ball is being pushed downward.
Newswise — WASHINGTON, D.C., November 18, 2018 -- An unexpected twist from a four-seam or a two-seam fastball can make the difference in a baseball team winning or losing the World Series. However, “some explanations regarding the different pitches are flat-out wrong,” said Barton Smith, a professor of mechanical and aerospace engineering at Utah State University who considers himself a big fan of the game.
He and his doctoral student, Nazmus Sakib, are conducting experiments to explain how baseballs move. Sakib and Smith will present their findings at the American Physical Society’s Division of Fluid Dynamics 71st Annual Meeting, which will take place Nov. 18-20 at the Georgia World Congress Center in Atlanta, Georgia.
A baseball is asymmetric owing to the figure-eight stitching pattern, and the way a baseball moves through the air depends on the degree and direction of its spin and its orientation when the hand releases it. The Magnus effect, or the force on a spinning object moving through a fluid like air, pushes in the direction that the front of the ball is spinning. So it causes a ball with topspin to drop and a ball with backspin to gain some lift -- enough to slow its fall, but not enough to overcome gravity.
This well-studied phenomenon affects most pitches except for the virtually spin-free knuckle ball, which is gripped with the thumb and fingertips. The two-seam fastball, which is gripped by the middle and index fingers along the seams, seemed to also behave in a way not explained by the Magnus effect.
Sakib and Smith focus on these two pitches, which are influenced by forces other than the Magnus effect. In their study, the researchers set up a pitching machine that hurls fastballs and knuckleballs through a smoky path. Automatic photographs, triggered by laser sensors, captured two images of the ball and smoke after release. Then, using a technique called particle image velocimetry, Sakib and Smith tracked the movements of the smoke particles to compute the velocity field around the ball and the direction of the rotating air at a given spot.
Then, they computed the “boundary layer separation” by identifying the portions of the ball’s surface where the layer of air surrounding the ball had separated to form the wake. While the boundary layer separation varies differently for the two fastball pitches as the ball rotates, the net effect is the same.
Sakib and Smith found that the two-seam pitch has a tilted spin axis due to the fact that one finger leaves the seam before the other, which can cause the ball to move sideways, unlike a four-seam fastball. In the case of the knuckleball, the point of separation can change midflight, causing the ball to randomly shift directions.
Smith is now “hoping to meet a major league pitcher who wants to use what we’ve learned through fluid dynamics to throw a better pitch.”
###
Presentation E17.1, “Velocity fields of pitched baseballs using Particle Image Velocimetry” by Nazmus Sakib and Barton Smith, will be Sunday, Nov. 18, 5:10 p.m. in Room B304 of the Georgia World Congress Center in Atlanta. Abstract: http://meetings.aps.org/Meeting/DFD18/Session/E17.1
###
----------------------- MORE MEETING INFORMATION -----------------------
Main meeting website: https://www.apsdfd2018.org/
Meeting technical program: http://meetings.aps.org/Meeting/DFD18/SessionIndex2
Invited talks: http://meetings.aps.org/Meeting/DFD18/APS_Invited
Hotel information: https://www.apsdfd2018.org/hotels/
GALLERY OF FLUID DYNAMICS
At the Annual Meeting, The Gallery of Fluid Motion will consist of posters and videos submitted by attendees illustrating the science and beauty of fluid motion. More information can be found here: https://gfm.aps.org/
PRESS REGISTRATION
We will grant free registration to credentialed journalists and professional freelance journalists. If you are a reporter and would like to attend, contact Rhys Leahy or the AIP Media Line (media@aip.org, 301-209-3090). We can also help with setting up interviews and obtaining images, sound clips or background information.
LIVE MEDIA WEBCAST
A press briefing featuring a selection of newsworthy research will be webcast live from the conference Monday, Nov. 19. Times and topics to be announced. Members of the media should register in advance at http://apswebcasting.com/webcast/registration/aps1118.php
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# Calculate Area with Shapes
Whenever I took out manipulatives to use while teaching math, students undoubtedly used them as a creative outlet and building tool. Why not ask learners to build first, then ask some powerful wonder and notice questions afterwards?
Really thinking about how learning through play never grows old.
THE LESSON RECAP:
Ask students to build a robot with the shapes allowed to them- it can be as simple or complex as they would like.
Scaffold building a landscape, a personal logo, and a 'free choice' creation.
Once the build and creations are complete, use the Robot build to ask questions about how many centimetres it would take to walk around your Robot. How do you know? Can you support your answer? Discuss with the students the variance in perimeter and why they have differences. Ask for some predictions about the calculation of Area.
A QUICK TIP
• Group/Break apart- you can group and break apart shapes you’re using to move creations around easily and duplicate shapes
There are so many ways to tweak this lesson to make it cross-curricular. Your build can also support any themes or subjects you are studying in other areas. As a challenge, limit the target area or perimeter desired. As a class, build a community with specific parameters included. Have your learners decide what they would like to collaborate on creating as a community of creators.
Learn how, and explore more Everyone Can Create Projects >
## Attachments
3 replies
August 13, 2023
Really love the focus on the inquiry process here -- what a fun and playful way to get students thinking critically about math concepts!
August 13, 2023
TamiB, Thank you for sharing. What fun use of grouping/ungrouping shapes. I wonder, would using screenshots of buildings from Maps and then overlaying shapes on the screenshot add any additional value?
December 02, 2023
Great idea. And it could also be done in Numbers to be able to add a table to calculate in the same file and leave traces of their learning.
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# OCML代写 | CSE 595 – Homework I
CSE 595 – Homework I
1 λ Calculus
1. In the basic untyped lambda calculus, the boolean true is encoded as λx.λy.x (i.e., it takes in two (3)
arguments and returns the first), and false is encoded as λx.λy.y (i.e., takes in two arguments and
returns the second). These definitions of the boolean constants may seem strange, but they are designed
to work with the if-then-else expression. The if-then-else expression is defined as λx.λy.λz.((xy)z).
Verify that these definitions do, indeed, make sense, by evaluating the following:
(a) (((λx.λy.λz.((xy)z)(λu.λv.u))A)B)
(b) (((λx.λy.λz.((xy)z)(λu.λv.v))A)B)
2. Reduce the following λ-expressions to their fullest possible extent: (3)
(a) (λx.xx)(λy.yx)z
(b) (((λx.λy.(xy))(λy.y))w)
3. Given the definitions of true and false as in the very question, and not defined as λx.((x false) true), (4)
and or defined as λx.λy.((x true) y), prove the following:
(a) not (not true) = true
(b) or false true = true
2 Recursion and Higher-order Functions
In this section, you may not use any functions available in the OCaml library that already solves all or most
of the question. For example, OCaml provides a List.rev function, but you may not use that in this section.
1. Write a recursive function pow, which takes two integer parameters x and n, and returns x (5) n
. Also write a
function float pow, which does the same thing, but for x being a float. n is still a non-negative integer.
2. Write a function compress to remove consecutive duplicates from a list. (5)
# compress [“a”;”a”;”b”;”c”;”c”;”a”;”a”;”d”;”e”;”e”;”e”];;
– : string list = [“a”; “b”; “c”; “a”; “d”; “e”]
3. Write a function remove if of the type ‘a list -> (‘a -> bool) -> ‘a list, which takes a list and (5)
a predicate, and removes all the elements that satisfy the condition expressed in the predicate.
# remove_if [1;2;3;4;5] (fun x -> x mod 2 = 1);;
– : int list = [2; 4]
4. Some programming languages (like Python) allow us to quickly slice a list based on two integers i and (5)
j, to return the sublist from index i (inclusive) and j (not inclusive). We want such a slicing function
in OCaml as well.
Write a function slice as follows: given a list and two indices, i and j, extract the slice of the list
containing the elements from the i
th (inclusive) to the j
th (not inclusive) positions in the original list.
# slice [“a”;”b”;”c”;”d”;”e”;”f”;”g”;”h”] 2 6;;
– : string list = [“c”; “d”; “e”; “f”]
Invalid index arguments should be handled gracefully. For example,
# slice [“a”;”b”;”c”;”d”;”e”;”f”;”g”;”h”] 3 2;;
– : string list = []
# slice [“a”;”b”;”c”;”d”;”e”;”f”;”g”;”h”] 3 20;
– : string list = [“d”;”e”;”f”;”g”;”h”];
You do not, however, need to worry about handling negative indices.
5. Write a function equivs of the type (‘a -> ‘a -> bool) -> ‘a list -> ‘a list list, which par- (7)
titions a list into equivalence classes according to the equivalence function.
# equivs (=) [1;2;3;4];;
– : int list list = [[1];[2];[3];[4]]
# equivs (fun x y -> (=) (x mod 2) (y mod 2)) [1; 2; 3; 4; 5; 6; 7; 8];;
– : int list list = [[1; 3; 5; 7]; [2; 4; 6; 8]]
6. Goldbach’s conjecture states that every positive even number greater than 2 is the sum of two prime (7)
numbers. E.g., 18 = 5 + 13, or 42 = 19 + 23. It is one of the most famous conjectures in number theory.
It is unproven, but verified for all integers up to 4 × 1018. Write a function goldbachpair : int ->
int * int to find two prime numbers that sum up to a given even integer. The returned pair must have
a non-decreasing order.
# goldbachpair 10;; (* must return (3, 7) and not (7, 3) *)
– : int * int = (3, 7)
Note that the decomposition is not always unique. E.g., 10 can be written as 3+7 or as 5+5, so both
(3, 7) and (5, 5) are correct answers.
7. Write a function called equiv on, which takes three inputs: two functions f and g, and a list lst. It (7)
returns true if and only if the functions f and g have identical behavior on every element of lst.
# let f i = i * i;;
val f : int -> int = <fun>
# let g i = 3 * i;;
val g : int -> int = <fun>
# equiv_on f g [3];;
– : bool = true
# equiv_on f g [1;2;3];;
– : bool = false
8. Write a functions called pairwisefilter with two parameters: (i) a function cmp that compares two (7)
elements of a specific T and returns one of them, and (ii) a list lst of elements of that same type T. It
returns a list that applies cmp while taking two items at a time from lst. If lst has odd size, the last
element is returned “as is”.
# pairwisefilter min [14; 11; 20; 25; 10; 11];;
– : int list = [11; 20; 10]
# (* assuming that shorter : string * string -> string = <fun> already exists *)
# pairwisefilter shorter [“and”; “this”; “makes”; “shorter”; “strings”; “always”; “win”];;
– : string list = [“and”; “makes”; “always”; “win”]
E-mail: vipdue@outlook.com 微信号:vipnxx
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# Log-log plot with error band that has negative numbers
6 Ansichten (letzte 30 Tage)
L'O.G. am 16 Jan. 2023
Kommentiert: dpb am 17 Jan. 2023
How do I make an error band? I am attaching some sample data. The fill function generates negative numbers, so that when I try either this:
figure;
loglog(x,y)
fill([x; flipud(x)],[y-std_y;flipud(y+std_y)],'k','FaceAlpha',0.5);
or this:
figure;
plot(x,y)
fill([x; flipud(x)],[y-std_y;flipud(y+std_y)],'k','FaceAlpha',0.5);
set(gca, 'XScale','log', 'YScale','log')
The error band doesn't show up.
##### 0 Kommentare-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden
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### Akzeptierte Antwort
dpb am 16 Jan. 2023
Verschoben: dpb am 16 Jan. 2023
You simply can't plot negative numbers on a log axis...they don't exist (well, they do, but they're complex).
It's not the fill function's fault; it's that abs(std_y)>abs(y) for at least some of your y values. You'll get the same result if you just plot() y-std_y.
That's why I constrained the error magnitude in the previous example as was noted in the comments of the sample code.
##### 1 KommentarKeine anzeigenKeine ausblenden
dpb am 17 Jan. 2023
You could illustrate the general effect if you were to choose a lower bound decade at which to cut off the lower limit and then clip the y_lo values at that point. Remember, however, that log(0) --> -Inf so there are an infinite number of decades to desplay to cover the full range even to get to the crossing point so the best you can hope for is a very crude representation.
In that regards, drawing the boundary line could be a better alternative; while it will also not show past where the data go to zero, it will display the points that are positive which is something. fill() is a one object for the defined region; the only way to draw it would be to only define regions for which the result is positive for the lower bound area and draw multiple areas -- the positive error above the data could be one, then depending upon the shape of the data, one or more to cover the areas on the lower side that are all greater than zero. This will leave one with a truncated area on the bottom as compared to the top, of course, but not really anything can do about that other than the aforementioned truncation of values to keep all positive -- but that misrepresents the values so don't really recommend it. The best solution probably is to just forego the log axis for linear if the data really do go negative.
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Journal of Function Spaces and Applications
Volume 2013 (2013), Article ID 143686, 9 pages
http://dx.doi.org/10.1155/2013/143686
Research Article
## Suzuki-Type Fixed Point Results in Metric-Like Spaces
1Department of Mathematics, Babol Branch, Islamic Azad University, Babol, Iran
2Department of Mathematics, Qaemshahr Branch, Islamic Azad University, Qaemshahr, Iran
3Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
Received 16 May 2013; Accepted 17 July 2013
Copyright © 2013 Nabiollah Shobkolaei et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### Abstract
We demonstrate a fundamental lemma for the convergence of sequences in metric-like spaces, and by using it we prove some Suzuki-type fixed point results in the setup of metric-like spaces. As an immediate consequence of our results we obtain certain recent results in partial metric spaces as corollaries. Finally, three examples are presented to verify the effectiveness and applicability of our main results.
#### 1. Introduction
There are a lot of generalizations of Banach fixed-point principle in the literature. So far several authors have studied the problem of existence and uniqueness of a fixed point for mappings satisfying different contractive conditions (e.g., [120]). In 2008, Suzuki introduced an interesting generalization of Banach fixed-point principle. This interesting fixed-point result is as follows.
Theorem 1 (see [19]). Let be a complete metric space, and let be a mapping on . Define a nonincreasing function from into by
Assume that there exists , such that for all , then there exists a unique fixed-point of . Moreover, for all .
Suzuki proved also the following version of Edelstein's fixed point theorem.
Theorem 2. Let be a compact metric space. Let be a self-map, satisfying for all , the condition
Then has a unique fixed point in .
This theorem was generalized in [3].
In addition to the above results, Kikkawa and Suzuki [8] provided a Kannan type version of the theorems mentioned before. In [14], Chatterjea type version is provided. Popescu [15] presented a Cirić type version. Recently, Kikkawa and Suzuki also provided multivalued versions which can be found in [9, 10].
Very recently Hussain et al. [4] have extended Suzuki's Theorems 1 and 2, as well as Popescu's results from [15] to the case of metric type spaces and cone metric type spaces (see also [57, 11]).
The aim of this paper is to generalize the above-mentioned results. Indeed we prove a fixed point theorem in the set up of metric-like spaces and derive certain new results as corollaries. Finally, three examples are presented to verify the effectiveness and applicability of our main results.
In the rest of this section, we recall some definitions and facts which will be used throughout the paper. First, we present some known definitions and propositions in partial metric and metric-like spaces.
A partial metric on a nonempty set is a mapping such that for all , if and only if ,, , .
A partial metric space is a pair such that is a nonempty set and is a partial metric on . It is clear that if , then from () and () . But if , may not be . A basic example of a partial metric space is the pair , where for all .
Lemma 3 (see [17]). Let and be a metric space and partial metric space, respectively. Then(i)the function defined by is a partial metric;(ii)let be defined by ; then is a partial metric on , where is an arbitrary function;(iii)Let be defined by ; then is a partial metric on ;(iv)Let be defined by ; then is a partial metric on , where .
Other examples of the partial metric spaces which are interesting from a computational point of view may be found in [7, 11, 12, 18].
Each partial metric on generates a topology on which has as a base the family of open -balls , where for all and .
Let be a partial metric.
A sequence in a partial metric space converges to a point if and only if .
A sequence in a partial metric space is called a Cauchy sequence if there exists (and is finite) .
A partial metric space is said to be complete if every Cauchy sequence in converges, with respect to , to a point such that .
Suppose that is a sequence in partial metric space ; then we define .
The following example shows that every convergent sequence in a partial metric space may not be a Cauchy sequence. In particular, it shows that the limit is not unique.
Example 4 (see [17]). Let and . Let Then clearly it is a convergent sequence, and for every we have , hence . But does not exist; that is, it is not a Cauchy sequence.
Definition 5 (see [2]). A metric-like on a nonempty set is a mapping such that for all ,, , .
The pair is called a metric-like space. Then a metric-like on satisfies all of the conditions of a metric except that may be positive for . Each metric-like on generates a topology on whose base is the family of open -ball, , where for all and .
A sequence in a metric-like space converges to a point if and only if .
A sequence in a metric-like space is called a -Cauchy sequence if there exists (and is finite) .
A metric-like space is said to be complete if every -Cauchy sequence in converges, with respect to , to a point such that
Every partial metric space is a metric-like space. Below we give some examples of a metric-like space.
Example 6. Let ; then mapping defined by is a metric-like on .
Example 7. Let ; then mappings defined by are metric-like space on , where and .
#### 2. Main Results
We start our work by proving the following crucial lemma.
Lemma 8. Let be a metric-like space, and suppose that is -convergent to . Then for every , one has In particular, if , then one has .
Proof. Using the triangle inequality in a metric-like space, it is easy to see that
Taking the upper limit as in the first inequality and the lower limit as in the second inequality, we obtain the desired result.
Theorem 9. Let be a complete metric-like space. Let be a self-map, and let be defined by
If there exists such that for each Then has a unique fixed point , and for each , the sequence converges to .
Proof. Putting in (10), hence from it follows for every . Let be arbitrary and form the sequence by and for . By (12), we have Also, by the condition of the definition of metric-like space, for all , we have
Hence, is a -Cauchy sequence.
Since is -complete, there exists such that That is, . We prove that . Putting in (12), we get that holds for each (where ). It follows by induction that
Let us prove now that holds for each . Since and by Lemma 8, it follows that there exists such that holds for every . Assumption (10) implies that for such , thus as , we get that We will prove that for each . For , this relation is obvious. Suppose that it holds for some . If , then and . If , then applying (18) and the induction hypothesis; we get that and (21) is proved by induction.
In order to prove that , we consider two possible cases.
Case I. (and hence ). We will prove first that for . For , it follows from (16). Suppose that (23) holds for some . Then which implies . Using (17) we obtain Assumption (10) and relation (21) imply that So relation (23) is proved by induction.
Now and (23) implies that for each . Hence, (18) imply that Hence , thus and; using Lemma 8 in (23), we have as which implies that , a contradiction.
Case II. (and so ). We will prove that there exists a subsequence of such that holds for each . From (12) we know that holds for each . Suppose that hold for some . Then which is impossible. Hence one of the following holds for each : or In particular, or In other words, there is a subsequence of such that (28) holds for each . But then assumption (10) implies that or Passing to the limit when , we get that , which is possible only if .
Thus, we have proved that is a fixed point of . The uniqueness of the fixed point follows easily from (10). Indeed, if and are two fixed points of such that , then from (18) we have which is a contradiction.
According to Theorem 9, we get the following result.
Corollary 10 (see [19]). Let be a complete metric space, and let be a mapping on . Define a nonincreasing function from into by
Assume that there exists , such that for all ; then there exists a unique fixed-point of . Moreover, for all .
Proof. Using a similar argument given in Theorem 9 for , the desired result is obtained.
Now, in order to support the useability of our results, let us introduce the following example.
Example 11. Let . Define by for all . Then is a complete metric-like space. Define a map by
for . Then for each , we have
On the other hand, we have Thus satisfies all the hypotheses of Theorem 9, and hence has a unique fixed point. Indeed, , , and is the unique fixed point of .
Theorem 12. Let be a complete metric-like space. Let , be two self-mappings. Suppose that there exists such that for every and that for every with that is not a common fixed point of and . Then there exists such that . Moreover, if , then .
Proof. Let be arbitrary, and define a sequence by Then if is odd, we have
If is even, then by (44), we have Thus for any positive integer , it must be the case that
Repeating (49), we obtain So, if , then
Thus .
That is, is a -Cauchy sequence in the metric-like space . Since is -complete, there exist such that
Assume that is not a common fixed point of and . Then by hypothesis which is a contradiction. Therefore, .
If for some , then which gives that .
Example 13. Let be a metric-like space where and . Define by , , and and , , and . Then for , we have It is easy to see that the above inequality is true for and for . Also, for every with y is not a common fixed point of and . This shows that all conditions of Theorem 12 are satisfied and 0 is a common fixed point for and .
Corollary 14. Let be a complete metric-like space, and let be a mapping. Suppose that there exists such that for every and that for every with . Then there exists such that . Moreover, if , then .
Proof. Taking in Theorem 12, the conclusion of the corollary follows.
Theorem 15. Let be a complete metric-like space. Let , be mappings from onto itself. Suppose that there exists such that for every and that for every with that is not a common fixed point of and . Then there exists such that . Moreover, if , then .
Proof. Let be arbitrary. Since is onto, there is an element satisfying . Since is also onto, there is an element satisfying . Proceeding in the same way, we can find that and for . Therefore, and for . If , then using (59) If , then using (59) Thus for any positive integer , it must be the case that which implies that Let ; then since .
Now, (64) becomes So, if , then
Thus . That is, is a -Cauchy sequence in the metric-like space . Since is -complete, there exists such that Assume that is not a common fixed point of and . Then by hypothesis which is a contradiction. Therefore, .
If for some , then which gives that .
Corollary 16. Let be a complete metric-like space, and let be an onto mapping. Suppose that there exists such that for every and that for every with . Then there exists such that . Moreover, if , then .
Proof. Taking in Theorem 15, we have the desired result.
Definition 17. Let and be metric-like spaces. Then is said to be a continuous mapping, if implies that .
Corollary 18. Let be a complete metric-like space, and let be a mapping of into itself. If there is a real number with satisfying for every and is onto and continuous, then has a fixed point.
Proof. Assume that there exists with and Then there exists a sequence such that So, we have and as . Since, , hence as . Now, Since is continuous, we have This is a contradiction. Hence if , then which is condition (71) of Corollary 16. By Corollary 16, there exists such that .
Now we give an example to support our result.
Example 19. Let and . Define by .
Obviously is onto and continuous. Also for each , we have where . Thus satisfies the conditions given in Corollary 18, and is the fixed point of .
Corollary 20. Let be a complete metric-like space, and be a mapping of into itself. If there is a real number with satisfying for every and is onto and continuous, then has a fixed point.
Proof. Replacing by in (79), we obtain for all .
Without loss of generality, we may assume that . Otherwise has a fixed point. Since , it follows from (80) that for every . By the argument similar to that used in Corollary 18, we can prove that if , then which is condition (71) of Corollary 16. So, Corollary 16 applies to obtain a fixed point of .
According to Theorem 12, we get the following result.
Corollary 21 (see [17, Theorem 1]). Let be a complete partial metric space. Let , be two self-mappings. Suppose that there exists such that for every and that for every with that is not a common fixed point of and . Then there exists such that . Moreover, if , then .
Proof. Using a similar argument given in the Theorem 12 for , the desired result is obtained, where is a partial metric on .
Also, according to Theorem 15, we get Theorem 2 from [17].
#### References
1. A. Aghajani, S. Radenović, and J. R. Roshan, “Common fixed point results for four mappings satisfying almost generalized $\left(S,T\right)$-contractive condition in partially ordered metric spaces,” Applied Mathematics and Computation, vol. 218, no. 9, pp. 5665–5670, 2012.
2. A. Amini-Harandi, “Metric-like spaces, partial metric spaces and fixed points,” Fixed Point Theory and Applications, vol. 2012, article 204, 10 pages, 2012.
3. D. Đorić, Z. Kadelburg, and S. Radenović, “Edelstein-Suzuki-type fixed point results in metric and abstract metric spaces,” Nonlinear Analysis: Theory, Methods & Applications, vol. 75, no. 4, pp. 1927–1932, 2012.
4. N. Hussain, D. Đorić, Z. Kadelburg, and S. Radenović, “Suzuki-type fixed point results in metric type spaces,” Fixed Point Theory and Applications, vol. 2012, article 126, 12 pages, 2012.
5. N. Hussain and M. H. Shah, “KKM mappings in cone $b$-metric spaces,” Computers & Mathematics with Applications, vol. 62, no. 4, pp. 1677–1684, 2011.
6. N. Hussain, M. H. Shah, A. Amini-Harandi, and Z. Akhtar, “Common fixed point theorems for generalized contractive mappings with applications,” Fixed Point Theory and Applications, vol. 2013, article 169, 17 pages, 2013.
7. N. Hussain, Z. Kadelburg, S. Radenović, and F. Al-Solamy, “Comparison functions and fixed point results in partial metric spaces,” Abstract and Applied Analysis, vol. 2012, Article ID 605781, 15 pages, 2012.
8. M. Kikkawa and T. Suzuki, “Some similarity between contractions and Kannan mappings,” Fixed Point Theory and Applications, vol. 2008, Article ID 649749, 8 pages, 2008.
9. M. Kikkawa and T. Suzuki, “Some notes on fixed point theorems with constants,” Bulletin of the Kyushu Institute of Technology, no. 56, pp. 11–18, 2009.
10. M. Kikkawa and T. Suzuki, “Three fixed point theorems for generalized contractions with constants in complete metric spaces,” Nonlinear Analysis: Theory, Methods & Applications, vol. 69, no. 9, pp. 2942–2949, 2008.
11. M. A. Kutbi, J. Ahmad, N. Hussain, and M. Arshad, “Common fixed point results for mappings with rational expressions,” Abstract and Applied Analysis, vol. 2013, Article ID 549518, 11 pages, 2013.
12. S. G. Matthews, “Partial metric topology,” in Papers on General Topology and Applications, vol. 728 of Annals of the New York Academy of Sciences, pp. 183–197, New York Academy of Sciences, New York, NY, USA, 1994.
13. V. Parvaneh, J. R. Roshan, and S. Radenović, “Existence of tripled coincidence points in ordered b-metric spaces and an application to a system of integral equations,” Fixed Point Theory and Applications, vol. 2013, article 130, 19 pages, 2013.
14. O. Popescu, “Fixed point theorem in metric spaces,” Bulletin of the Transilvania University of Braşov, vol. 150, pp. 479–482, 2008.
15. O. Popescu, “Two fixed point theorems for generalized contractions with constants in complete metric space,” Central European Journal of Mathematics, vol. 7, no. 3, pp. 529–538, 2009.
16. J. R. Roshan, V. Parvaneh, S. Sedghi, N. Shobkolaei, and W. Shatanawi, “Common fixed points of almost generalized ${\left(\psi ,\phi \right)}_{s}$-contractive mappings in ordered b-metric spaces,” Fixed Point Theory and Applications, vol. 2013, article 159, 23 pages, 2013.
17. S. Sedghi and N. Shobkolaei, “Common fixed point of maps in complete partial metric spaces,” East Asian Mathematical Journal, vol. 29, no. 1, pp. 1–12, 2013.
18. N. Shobkolaei, S. Sedghi, J. R. Roshan, and I. Altun, “Common fixed point of mappings satisfying almost generalized $\left(S,T\right)$-contractive condition in partially ordered partial metric spaces,” Applied Mathematics and Computation, vol. 219, no. 2, pp. 443–452, 2012.
19. T. Suzuki, “A generalized Banach contraction principle that characterizes metric completeness,” Proceedings of the American Mathematical Society, vol. 136, no. 5, pp. 1861–1869, 2008.
20. T. Suzuki, “A new type of fixed point theorem in metric spaces,” Nonlinear Analysis: Theory, Methods & Applications, vol. 71, no. 11, pp. 5313–5317, 2009.
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# What are the Asterisks or Star Symbols '*' and '**' in Python?
Gain a solid understanding of the various ways you can use the asterisk or star * symbol
## Table of Content
When you've been exposed to Python codes, you would have come across one or more of the asterisk symbols '*' or '**'.
As confusing and almost irrelevant as they might look, they are actually a really important part of Python programming.
Good knowledge of them might help you write efficient and more generic solutions when solving problems using Python language. Think of the DRY principle.
The asterisk (star) operator is used in Python with more than one meaning attached to it.
After going through this article, you will gain a solid understanding of the various ways you can use the asterisk or star * symbol and the counterpart double-asterisk or double-star * symbol in Python programming.
## What are the Asterisks or Star Symbols * and ** in Python?
The single asterisk or star (*) and the double-asterisk or double-star (**) symbols are both used primarily for numeric multiplication and exponential calculation respectively in Python.
They also hold the same conventional use case in other programming languages.
## 5 Different Ways to Use the Asterisks or Star Symbols '*' and '**' in Python
In Python programming language, the single asterisk or star (*) and the double-asterisk or double-star (**) symbols can be used for other functionality that is different from the conventional use case.
Here are five different ways you can use them in Python programming:
## 1. Perform calculations on numeric data types
This is more conventional use in programming.
In Python programming language, as with most programming languages, the single asterisk or star symbol (*) is used as a multiplication operator to get the product of two numeric values or variables.
The double asterisk or star symbol (**) is used as an exponential operator and is used to raise the value or variable on the left to the power of the value or variable on the right.
## 2. Unpack an arbitrary number of arguments in a function parameter
There is a chance you should have come across the *args and **kwargs in a function argument in Python codes.
These symbols specify that the function could receive an unspecified and a possibly unlimited number of arguments or keyword arguments.
If you are not sure how many arguments and keyword arguments will be passed, you can use a single asterisk or star (*) and the double-asterisk or double-star (**) symbols, respectively.
For arguments, the values are received as a tuple and all properties of a tuple can be performed on it:
You can also pass the arguments as a list, tuple, or set. For a dictionary, to pass the keys only, use `*`, to pass the values only, use `**`.
For keyword arguments, the values are received as a dictionary, and all properties of a dictionary can be performed on it:
## 3. Force all arguments passed to a function to be keyword-only arguments
Another way we can use an asterisk in a function is to ensure that a function can only receive keyword arguments.
To do that, simply pass a single asterisk or star `*` in the parameter list and the following arguments must be passed as keyword arguments.
Also, we can restrict a few arguments to be keyword-only by just putting the positional arguments before the asterisk.
## 4. Unpack Elements of an iterable into a new iterable
One way we can use asterisks to make our programs clear and elegant is when we need to combine different iterable such as lists, tuples, and sets into a new list.
An obvious solution for this is to use for-loops to iterate all items and add them to a new list one after the other.
Even though the code block above accomplishes our mission, the code looks so long and is not very “Pythonic”.
We could use list comprehensions to make the code a lot better.
Using list comprehension takes us a step closer to achieving a great solution, but we can do even better than that by using asterisks.
This method works really well for iterable such as lists, tuples, and sets.
However, it's different for dictionaries.
If we use a single asterisk `*` as a prefix to unpack `dict`, its keys will be unpacked. And if we use double asterisks `**` as a prefix, its values will be unpacked.
But, we need the keys of a dictionary to receive the unpacked values, which makes unpacking a dictionary to be inconvenient and uncommon.
### Extended unpacking of Iterable
In Python 3, it is possible to use `*l` on the left side of an assignment as an Extended Iterable Unpacking proposed in PEP 3132.
However, the variable with the '*' becomes a list instead of a tuple in this context:
## 5. Repeat a string multiple times
The single asterisk (*) can be used to repeat values of variables in sequences such as string, list, and tuple.
## Wrap Off
So far, we've covered the Asterisk(*) of Python. It was interesting to be able to do various operations with one operator, and most of those above are the basics for writing Pythonic code.
We. hope you've gained a solid understanding of the various ways you can use the asterisk or star * symbol and the counterpart double-asterisk or double-star * symbol in Python programming.
If you learned from this tutorial, or it helped you in any way, please consider sharing and subscribing to our newsletter.
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+0
# re-post
+5
174
6
+302
This question was already answered but i did'nt understand. please check my post and give me a better explaination:)
From a point P on the ground the angle of elevation of the top of a tower is 30°and tht of the top of a flagstaff fixed on top of the tower is 60°. If the length of the flagstaff is 5m, find the height of tower.
SARAHann Mar 12, 2017
#3
+10657
+10
From the diagram, x= tower height
tan 30 = x / d
tan 60 = (x+5)/d re arrange d= (x+5)/tan60 = .57735x + 2.8868 Substtitute this into the first equation
tan30 = .57735 = x/(.57735x+2.8868) solve for x
.33333x +1.6666= x
x= 2.5 m 'tower' height (short !)
ElectricPavlov Mar 12, 2017
Sort:
#1
+91229
+10
Hi Sarah,
Do you have the address of your original question so that we can go to it?
(do you know what I mean? - just copy the address and paste it into a new post here)
It is good that you say when you do not understand but it is important that you say it on the original thread because the answerer want your feedback.
People often give trivial anwers because they do not think anyone is really trying to learn from it. We get so little feedback here. Many askers only want to copy the answer and do not really want to learn.
Anyway, if you have that original question can you please copy the address and post it in here and then I will help you some more. :)
Melody Mar 12, 2017
#4
+302
+5
- http://web2.0calc.com/questions/applications-of-trigonometry
I did say in my original 1 tht i did'nt understand. But no one answered. Could u check my feedback in the org one? I hv said wut i did'nt understand based on CPhil's answers. Thanks a lot!
SARAHann Mar 12, 2017
#2
+91229
0
I found all these ones, that CPhill answered, but the one you are asking now does not seem to be there.
http://web2.0calc.com/questions/applications-to-trignometry_1
Melody Mar 12, 2017
#3
+10657
+10
From the diagram, x= tower height
tan 30 = x / d
tan 60 = (x+5)/d re arrange d= (x+5)/tan60 = .57735x + 2.8868 Substtitute this into the first equation
tan30 = .57735 = x/(.57735x+2.8868) solve for x
.33333x +1.6666= x
x= 2.5 m 'tower' height (short !)
ElectricPavlov Mar 12, 2017
#5
+91229
+10
Hi again Sara, thank you for giving me what I asked for. I can see that you did ask for more help before too, just like you said :)
It is very easy for things like this to be overlooked so it was good that you reposted. :)
You have lots of answers now but if it is still not clear to you then say so. :))
If you would like me, or someone else to look at it specifically for you, you can also message us privately, don't forget to include the address as well though so we can find it :)
If you have specific questions about someone's answer then aks them on the forum and maybe back it up with a private message :)
Melody Mar 13, 2017
#6
+302
+1
Hey, melody! thank you for helping me. I understood how to solve the question when my doubt was cleared. I had made a mistake while I was working out the problem. thank you so much for looking out for me!
cheers!
SARAHann Mar 15, 2017
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# predication
predication, in logic, the attributing of characteristics to a subject to produce a meaningful statement combining verbal and nominal elements. Thus, a characteristic such as “warm” (conventionally symbolized by a capital letter W) may be predicated of some singular subject, for example, a dish—symbolized by a small letter d, often called the “argument.” The resulting statement is “This dish is warm”; i.e., Wd. Using ∼ to symbolize “not,” the denial ∼Wd can also be predicated. If that of which “warm” is predicated is indefinite, a blank may be left for the predicate, W—, or the variable x may be employed, Wx, thus producing the propositional function “x is warm” instead of a definite proposition. By quantifying the function by (∀x), meaning “For every x . . . ,” or by (∃x), meaning “There is an x such that . . . ,” it is transformed into a proposition again, either general or particular instead of singular, which predicates warmness (or its negation) of several or many subjects of a kind. The predication is identical if it characterizes every referent (x); it is disparate if it fails to characterize some or all of the referents. The predication is formal if the subject necessarily entails (or excludes) the predicate; it is material if the entailment is contingent.
Philosophers have long debated what predicates really are. In the early Middle Ages, they were usually treated as having a being beyond all linguistic and mental entities and thus were viewed as metaphysical. Garland the Computist, the author of an early system of logic, however, viewed predication as mere utterance (vox). Peter Abelard, the foremost dialectician of the 12th century, amended this view to include significatio as well as vox.
Logicians have long distinguished the existential statement “x is” from the predicational statement “x is Y.” Franz Brentano, a precursor of Phenomenology prior to World War I, argued that they are both existential, that “x is Y ” means “xY is”; e.g., “Some fish have four eyes” means “Four-eyed fish exist.” An exactly opposite approach was taken by Alexander Bain, a Scottish philosopher and psychologist, who held that all existential statements have complex subjects from which a predicate can be extracted.
The limitations of predication as a logical form are increasingly evident. The predicate logic is now seen to be but one species of the logic of terms—the others being the logic of classes, the logic of relations, and the logic of identity; and the entire logic of terms, in turn, is distinct from the propositional logic, which deals with whole or unanalyzed statements. In the logic of relations, it is even questionable whether there is any predicate at all, since all of the terms can be regarded as subjects on the same footing (as in “Jane is the sister of Edith is the sister of Rachel”). Moreover, logics that distribute the predicate (with the quantifiers “all,” “some,” etc.) have also been explored.
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*NEWLY UPDATED FEBRUARY 2018!* *NOW with 40 task cards! TEKS Aligned: 7.9A (Readiness Standard): The student is expected to solve problems involving the volume of rectangular prisms, triangular prisms,rectangular pyramids, and triangular pyramids. __________________________________________________
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This foldable organizes a total of 10 examples for finding the volume of the following geometric solids: • Rectangular Prism • Triangular Prism • Trapezoidal Prism (2nd option does NOT include this tab) • Rectangular Pyramid • Triangular Pyramid Now with TWO options- with or without trapezoidal pr
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This foldable provides students with a total of 10 examples for finding the volume of the 3-D figures listed below: Prisms (Rectangular & Triangular) Pyramids (Rectangular & Triangular) Cylinders Cones Spheres This is a great way to keep students organized and provides them with a great s
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Included in this set are 4 volume of rectangular prism cards, 4 volume of triangular prims cards, 4 volume of square pyramids cards and 4 volume word problem cards. A recording sheet is included. Students are to find the volume of the figures and find the answers on another card. Students are finish
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Your students will love this March Madness style volume activity. Includes 16 different prisms and pyramids grouped into 4 different sections of the bracket- rectangular prism, triangular prism, square pyramid, and triangular pyramid. The different 3D shapes will compete against each other on a br
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In this activity, students cut out 12 task cards. They choose a task card and solve the problem on the lower half. Then they find the card with the matching answer. This leads them to their next problem. Students can glue the cards onto a recording sheet (included). Problems include the followin
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This bundle of activities provide students with the opportunity to practice using surface area and volume formulas of 12 polyhedra. Each activity includes a diagram and formulas of surface area and volume for each specific polyhedron. An interesting extension is to ask the students to see if they
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This file contains all you need for a fun filled class period reviewing volume of geometric figures including prisms, cubes, and pyramids It is a multiple step hands on activity in which students will determine the appropriate formulas provided a given shape either from memory or a formula chart.
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#### Videos matching "volume of prisms and pyramids" (BETA)
Math Teacher's Got Problems presents a fun, engaging way for students to practice finding the surface area and volume of prisms and pyramids. NO students like to sit and solve problems, but as math teachers we know practice is the key to mastery. This is a favorite activity of my students. (Yes, mat
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This quiz covers volume of rectangular and triangular prisms and square pyramids. Diagrams are provided for most problems. Some problems require students to find a dimension when given the volume and other dimensions. Part of the TEKS quiz series, available for all 7th and 8th grade math TEKS. Each
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This set of Volume Task Cards 7th-Grade focus on Prisms and Pyramids. Volume Task Cards can be used to support small group guided math instruction, individual seat work, learning stations, or for whole class instruction. There are 16 Open-Ended Task Card questions that align to STAAR. Perfect for 7t
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Math Teacher's Got Problems's presents you with a print ready foldable and activity for teaching students how to calculate the volume of rectangular and triangular prisms and pyramids. The foldable breaks down the process by requiring students to calculate big B(area of the base) as the first step,
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Students practice identifying the correct formula to use when given a word problem and then solve. Answer key and student work pages included! Directions: This is a loop game for practice using formulas with real world problems. Students can start on any card. They will then find the answer at
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4 ratings
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PDF (2.35 MB)
I use this as a self-check for my 7th graders after learning about prisms and pyramids. Students will determine the volume of rectangular and triangular prisms and pyramids, then follow the direction of the correct answers until arriving at the finish. I have students show their work on notebook pap
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2 ratings
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Students will discover the definition of prisms and pyramids while learning how to find volume for rectangular & triangular prisms and rectangular & triangular pyramids.
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This foldable, which includes pictures, can be used to go over volume of prisms, pyramids, and cubes. It includes an answer key.
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Introduce your class to Volume in a simple way with these guided notes! The notes aim to help tackle misconceptions like: - the location of the base(s). - the height of the 3D figure. - the difference in "b" and "B" in the formula. Key included.
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This set of 20 task cards engages students as they study volume with prisms and pyramids. There are 11 task cards for pyramids and 9 task cards for prisms. Student answer sheets and an answer key are provided. Task cards are a wonderful tool to use in a classroom. Give your students the chance to wo
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I use this handout to review Prisms and Pyramids with my 8th graders. Students will discuss the differences between the two, the formulas, and work out several examples with and without a diagram. This should be printed 2-sided flipped on the short edge. Students will cut on the dashed lines and fo
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As our EOC format changes from multiple choice to computer-based and performance-based testing, our assessments must change as well. This rigorous quiz bundle contains 4 quiz versions and assesses student ability to solve problems involving volumes of prisms and pyramids. +++++++++++++++++++++++++
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Here is a foldable that can be used for Lesson 4 and Lesson 5 for Chapter 8 of the Glencoe Math course 2 Volume 2 Book. You would print them out as they are and then cut them down the middle and then staple them at the top to create a flip book effect.
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Notes and practice problems on finding volume of prisms, pyramids and cylinders.
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Volume problems include prisms, pyramids and cylinders.
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Volume and Surface Area Coloring Activity (Prisms, Pyramids, Cylinders, Cones) This is a fun way for students to practice finding both volume and surface area of 3D figures. There are 12 problems total, 6 volume and 6 surface area. This includes rectangular prisms, cylinders, cones, and pyramids.
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Surface Area of Prisms and Pyramids Interactive Notebook Pages, Nets and Guided Notes, CCS:6.G.4 Included in this product: Three Dimensional Figures Guided Notes (1 and 2 per page) Venn Diagram Foldable (Compare and Contrast Volume and Surface Area) Nets Exit Slip (2 per page) 8 Shutter Prisms and
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This set of worksheets can be used to introduce or review finding the surface area and volume of rectangular prisms, triangular prisms, and trapezoidal prisms and finding the surface area of triangular and square pyramids. Two problems are included for each concept (only whole number lengths are us
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Skills practice focused on solving for volume, total surface area, or lateral surface areas of pyramids and prisms using 3-D shapes or their nets. This activity is great for getting your kids moving and working together. You can find the bundle here-->Scavenger hunts for Middle School Math - Gr
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Surface Area and Volume Lab--No Materials Required! What's Included: Surface Area and Volume Notes Formulas for Surface Area of Prisms (Rectangular, Triangular, Pentagonal, Hexagonal, Octagonal), Surface Area of Pyramids (Square, Rectangular, Triangular, Pentagonal, Hexagonal, and Octagonal) and Vol
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This volume and surface area discovery worksheet promotes interdisciplinary learning (marketing). Students engage in a series of higher order thinking tasks and experience prisms in a real context with true relevance. This product is part of the Discovery-Based Worksheet Series. Discovery-Based W
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As our EOC format changes from multiple choice to computer-based and performance-based testing, our assessments must change as well. Those of you who already have computer-based EOCs will find this bundle worth its weight in gold. This rigorous quiz bundle contains 44 quizzes, each of which has 4 v
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Volume Interactive Notebook: Create this reference guide with your students! Volume of Spheres, Cones, Cylinders, Prisms, Pyramids...my 8th graders have tested and approved this resource! Please note: this product is included in a money saving bundle! Check it out: Voluminous Volume Growing Bundl
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My 7th Grade Math Task Cards Mega Bundle is Complete! There are a total of TWENTY-FOUR sets of task cards! There are 520 task cards total in this bundle! What is included? Unit Rates Task Cards (7.RP.1) ♥ ratios, rates, unit rates (vocabulary) ♥ unit rates with whole numbers, decimals, and compl
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\$46.00
\$35.00
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These stations make great practice or review for volume and surface area of prisms and pyramids. Students will be able to solve mathematical and real world problems. Students practice by working collaboratively in groups and moving around the classroom where they find a new set of practice problems
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\$4.00
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3.8
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In this Volume of Rectangular Prisms and Pyramids Loop Activity, students will solve 10 word problems that involve challenging applications of the Volume formulas for prisms and pyramids. They can choose any of the 10 to start with, and the answer to that problem will be atop another problem. That w
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\$2.25
5 ratings
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PDF (1.91 MB)
This file contains all you need for a fun filled class period reviewing volume of geometric figures including prisms, cubes, and pyramids. It is a multiple step hands on activity in which students will determine the appropriate steps to determine surface area provided a given shape either from mem
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\$1.50
10 ratings
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Highly engaging! 2 Olympic Math Mosaics included! Each student's worksheet is different, ensuring individual accountability. Each worksheet represents a small section of the big picture, providing collaborative motivation! Engage your class in an Olympic mosaic involving real-life volume applica
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\$9.95
\$5.00
2 ratings
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ZIP (2.02 MB)
Students will enjoy finding the volume of pyramids and cones with these READY TO PRINT task cards! There are so many uses in your classroom! In this set of task cards, students will find the volume of various cones and pyramids. Problems with diagrams and "word" problems are included. There are
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\$3.00
7 ratings
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PDF (3.44 MB)
Students will find the surface areas and volumes of prisms, pyramids, cylinders, cones, and spheres. Answers are represented in terms of pi rather than using 3.14 for pi. A page of formulas is included. Cut the 16 puzzle pieces apart, then paste them into place so that each diagram matches its an
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\$2.00
50 ratings
4.0
PDF (713.09 KB)
Students will love creating a paper chain to decorate your classroom while practicing volume calculations. Simply print both pages (on color paper if possible) and have students solve each problem. There is plenty of room to show work in the shapes. They will then cut out the shapes and attach the
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Volume of Prisms and Cylinders Lesson In this lesson, students will learn the formula for volume of prisms, including cylinders. Included: • Warm-Up - The warm-up is a problem where students must find the lateral area and surface area of a square pyramid. • Guided Notes - Two versions are includ
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\$3.00
1 rating
4.0
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Math Bowl is a new series that I have created to help keep your students "on a roll" in math! (Cheesy, I know!) In this activity, students compete against each other either individually or with a partner as they work to solve word problems. The objective of this game is for students to achieve a “st
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\$2.50
1 rating
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PDF (1.9 MB)
A structured approach to teaching volume can make it so much easier for students to understand! These materials lend themselves to use in interactive notebooks. These foldable notes, examples, and practice problems feature a structured step by step format to guide student thinking as they calculat
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\$3.00
51 ratings
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PDF (14 MB)
Volume of Prisms Clip Art This file contains 15 different shaped prisms in 6 different colors both in solid colors and outline trace, plus blacklines. Clip art specifically created for calculating volume. Clip art includes height measurement line as an added convenience to help quickly create pro
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\$7.00
3 ratings
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Geometry - Volume of Prisms and Cylinders Common Core Aligned Lesson with Homework This lesson includes: -Lecture Notes (PDF, SMART Notebook, and PowerPoint) -Blank Lecture Notes (PDF and SMART Notebook) -Homework (PDF) -Answer Key (PDF) You do not need to have SMART Notebook or PowerPoint to rece
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The Surface Area of Prisms, Pyramids, and Cylinder Scavenger Hunt was created for a 7th grade math/pre-algebra classroom for the measuring figures unit. It contains one scavenger hunt which has 20 problems to practice finding the surface area of triangular/rectangular prisms, triangular/rectangular
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This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Printing should be: LANDSCAPE and DOUBLE-SIDED, with the Flip being along the 'SHORT' edge or side Want MORE Power for your Dollar? Give Brocci Bundles a Try before you buy! If you wanna see some B
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In this activity, students will find the volume of prisms and pyramids. This ladder activity is a great alternative to a worksheet. Students will cut the problems apart along the lines and work the card labeled "Start" and find their answer on the top of another card. They tape their cards togeth
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This download includes a foldable that guides students through the process of determining the Volume of Triangular Pyramids. Three word problems are provided for teachers who like to teach by giving students scenarios to work though versus skill-based problems. The first problem shows students how t
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This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Printing should be: LANDSCAPE and DOUBLE-SIDED, with the Flip being along the 'SHORT' edge or side Want MORE Power for your Dollar? Give Brocci Bundles a Try before you buy! If you wanna see some B
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*****This product is now available in a Bundle Pack! Follow the link to check out the Complete 7th Grade Math Bundle Pack available in my TpT Store! ***** https://www.teacherspayteachers.com/Product/Complete-7th-Grade-Math-INB-Bundle-Pack-7th-Grade-TEKS-2405745 Interactive Math Journal Pages that a
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By purchasing this resource you get my Complete Measurement Unit: Perimeter, Surface Area and Volume including a Unit Test, 2 Quizzes, a Task Card Review Activity, an assignment on Optimization and 5 Lessons. The Unit Includes: - 2 --> 3 weeks of teaching resources - Everything is in Word and
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Need a fun activity to review surface area and volume? These task cards will sharpen those skills! Get this in my MEGA bundle!! 7th Grade Task Cards Mega Bundle Task Cards will cover the following: ♥ definitions of surface area and volume ♥ calculating surface area of prisms and pyramids ♥ calcu
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Related searches for volume of prisms and pyramids
showing 1-52 of 703 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.
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# Asymmetric Normal Probability Distribution
I'm looking for a continuous probability distribution a little bit like the normal distribution but asymmetric. In my opinion this distribution applies to phenomenons related to response time in environments marked by resource contention.
Examples I have in mind are:
• In real life: the time it takes for my bus to go from my home to my office in the morning. In average it's around 15 minutes. However the way this duration varies each side of the mean value is asymmetric: it can hardly be 10 minutes less than the average but can easily take 10 minutes more.
• In computer capacity planning (which is the real domain where I want to use it ;-). One transaction needs to take place in 5 seconds (average) but my QoS constraint is that 90% of the time it takes less than 15 seconds.
Here is a diagram to illustrate my ideal distribution.
In this last example I could approximate the distribution to a Gaussian distribution and decide that 90% is roughly equivalent to a 1.5 standard deviation.
However I'm curious to know whether there is probability distribution more adapted to my problem.
The end goal is to deduce what percentage of my resources should be free (e.g. each CPU core should in average be at least 50% free, disk controllers bandwidth should be 50% free, etc) in order to satisfy the 90% threshold constraints.
## Edit
Going back to the example of my Bus journey, there is a minimum travel time which depends on propagation law limits (dictated by highway code or physics).
Similarly in a computer system, when my request runs unhampered by concurrent usage of the available physical resource, one can probably observe consistently close response times. I term this minimum latency and I ascribe the variations above this minimum latency time to other concurrent requests in real life.
The important thing here is that when contention increases, mean, median and mode values all increase when $\sigma$ increases.
Here is another diagram to illustrate what I mean.
So it looks like the Rayleigh distribution seems closer to what I need. However it also looks like it lacks some kind of "$\mu$" parameter since I have three sizing conditions to satisfy:
• average response time: 5s.
• In the CDF when the cumulated probability = 0.9 then the response time is 15s.
-
Try a Gamma distribution, maybe? – Dilip Sarwate Aug 31 '12 at 11:50
I think when you fit your distribution to the Rayleigh distribution buy some kind of fitting criterion like minimum mean squared error, there should be somehow one to one correspondance between your parameters and the parameters provided by the fit, namely by that specific Rayleigh distribution. – Seyhmus Güngören Aug 31 '12 at 16:58
I suggest Rayleigh distribution as it is quite similar to your figure, however it starts from zero. But one can shift it how he/she wants to. It is the distribution of the amplitude of the complex gaussian random variable.
http://en.wikipedia.org/wiki/Rayleigh_distribution
-
It looks like this is the closest thing to what I need. I've edited my post accordingly. Thx. – Alain Pannetier Aug 31 '12 at 14:45
@AlainPannetier it is my pleasure. – Seyhmus Güngören Aug 31 '12 at 14:48
Response accepted. The reservations I had were due to the fact that I wrongly assumed that both conditions ("1. average response time: 5s" and "2/ In the CDF when the cumulated probability = 0.9 then the response time is 15s") were to be satisfied exactly. Instead these are two different ways of computing the standard deviation. To size the system I need to select the most severe value. Experimentations on the real platform will show whether this distribution needs to be calibrated. – Alain Pannetier Aug 31 '12 at 16:37
If $X$ is normal, consider $Y=e^X$. It will be distributed as $$\frac{1}{x}e^{-\ln(x)^2} = \frac{1}{x^{1+\ln(x)}}$$
This is quite similar to your picture.
As Sasha said, it's the log-normal distribution.
-
Of course you may add that $Y$ is known as a log-normal random variable. – Sasha Aug 31 '12 at 12:05
@Xoff, From what I understand this log-normal distribution does not work completely. What I need is a distribution in which the mode shifts rightwards when $\sigma$ increases. In Log-normal distributions mode, median and mean values shift leftwards when $\sigma$ increases. Thanks anyway. Already upvoted. – Alain Pannetier Aug 31 '12 at 14:50
If your asymmetric random variable is defined on $\mathbb{R}$, as opposed to $\mathbb{R}^+$, you may want to look into Azzallini's skew-normal distribution. It is implemented in R (package sn), and in Mathematica (ref-page).
-
Try the log-normal distribution. It is probably the simpliest distribution that mimic the behavior you are searching for. It is easily implementable and has been successfully applied to many applied mathematics fields.
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# Remove list element without mutation
Assume you have a list
```>>> m = ['a','b','c']
```
I'd like to make a new list n that has everything except for a given item in m (for example the item 'a'). However, when I use
```>>> m.remove('a')
>>> m
m = ['b', 'c']
```
the original list is mutated (the value 'a' is removed from the original list). Is there a way to get a new list sans-'a' without mutating the original? So I mean that m should still be [ 'a', 'b', 'c' ], and I will get a new list, which has to be [ 'b', 'c' ].
I assume you mean that you want to create a new list without a given element, instead of changing the original list. One way is to use a list comprehension:
```m = ['a', 'b', 'c']
n = [x for x in m if x != 'a']
```
n is now a copy of m, but without the 'a' element.
Another way would of course be to copy the list first
```m = ['a', 'b', 'c']
n = m[:]
n.remove('a')
```
If removing a value by index, it is even simpler
```n = m[:index] + m[index+1:]
```
You can create a new list without the offending element with a list-comprehension. This will preserve the value of the original list.
```l = ['a','b','c']
[s for s in l if s!='a']
```
Another approach to list comprehension is numpy:
```>>> import numpy
>>> a = [1, 2, 3, 4]
>>> list(numpy.remove(a,a.index(3)))
[1, 2, 4]
```
There is a simple way to do that using built-in function :filter .
Here is ax example:
```a = [1, 2, 3, 4]
b = filter(lambda x: x != 3,a)
```
### Differences between TCP sockets and web sockets, one more time
Trying to understand as best as I can the differences between TCP socket and websocket, I've already found a lot of useful information within these questions:
### Does Scala have static imports like Java?
Does Scala support static imports, like Java does?
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Total:
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Math Review Files - End of the Year Math Test Review Bundle
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9 MB|60 pages
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Product Description
YEAR END REVIEW!
This is a collection of ALL of my Math File Review task Card (180 task cards!), representing all five math strands. This collection was designed as a year end review for 5th grade, but could be used with upper end 4th graders (or as a preview for 5th!) or to review some 6th grade skills.
There are 5 sets of cards here - 180 task cards in all. Save \$5.00 by purchasing this bundle (\$14.00 for all five sets!). That's like getting two of the smaller sets for free!
The following review cards are included in this pack:
Number Sense – 56 cards covering: identifying fractions on a number line; simplifying fractions; factors & multiples; mixed numbers and improper fractions; adding/subtracting fractions & mixed numbers; representing decimals on a grid; naming fractions, decimals, and percent's; addition/subtraction/multiplication/division of whole and decimal numbers; comparing & ordering fractions & decimals; rounding decimal and whole numbers; using expressions; place value up to millions; ratios; elapsed time and multiple step word problems.
Data & Graphs – 28 cards covering: reading and creating a variety of charts and graphs – including line, circle, pictograph, and bar graphs; median, mode, range, and mean; probability
Measurement – 28 cards covering: Using a ruler to the nearest ½ inch and nearest cm; area & perimeter; converting measurements in both customary and metric systems – weight, volume, length; estimating appropriate choice of measurement.
Geometry – 40 cards covering: polygons, triangles, quadrilaterals, angles, finding measurements of interior angles, congruent & similar shapes, symmetry, constructing and deconstructing shapes, translations, rotations, reflections; volume, nets.
Algebra – 28 cards covering: evaluating expressions, variables, finding the rule, solving the missing part of a pattern (using addition, subtraction, multiplication, and addition), 2 step patterns, order of operations, solving simple equations, inequalities
Ideas for using these task cards:
1) Use in a center.
2) Use for small group instruction.
3) Use as assessment for checking student's ability to move on to the next skill.
4) Set out for use by those who finish their work early.
5) Use for partner work. Students choose a card and their partner works the problem out on a white board or scrap paper. Check each other's answers.
6) Use these as a part of whole class lessons, interactive bulletin boards, etc.
7) Play games such as Jeopardy and Scoot!
The individual cards can be found here:
Math Review Files - Numbers and Operations Review Task Cards
Math Review Files - Data and Graphs Review Task Cards
Math Review Files - Measurement Review Task Cards
Math Files Review - Geometry Review Task Cards
Math Files - Algebra Review Task Cards
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http://www.gurufocus.com/term/e10/CAMP/E10/CalAmp%2BCorp
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GuruFocus has detected 3 Warning Signs with CalAmp Corp \$CAMP.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
CalAmp Corp (NAS:CAMP)
E10
\$-0.40 (As of Nov. 2016)
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. E10 is the average of the inflation adjusted earnings of a company over the past 10 years.
CalAmp Corp's adjusted earnings per share data for the three months ended in Nov. 2016 was \$-0.040. Add all the adjusted EPS for the past 10 years together and divide 10 will get our e10, which is \$-0.40 for the trailing ten years ended in Nov. 2016.
As of today, CalAmp Corp's current stock price is \$16.69. CalAmp Corp's E10 for the quarter that ended in Nov. 2016 was \$-0.40. CalAmp Corp's Shiller P/E Ratio of today is .
During the past 13 years, the highest Shiller P/E Ratio of CalAmp Corp was 300.00. The lowest was 0.00. And the median was 293.00.
Definition
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. When we calculate the today's Shiller P/E ratio of a stock, we use todays price divided by E10.
What is E10? How do we calculate E10?
E10 is the average of the inflation adjusted earnings of a company over the past 10 years. Lets use an example to explain.
If we want to calculate the E10 of Wal-Mart (WMT) for Dec. 31, 2010, we need to have the inflation data and the earnings from 2001 through 2010.
We adjusted the earnings of 2001 earnings data with the total inflation from 2001 through 2010 to the equivalent earnings in 2010. If the total inflation from 2001 to 2010 is 40%, and Wal-Mart earned \$1 a share in 2001, then the 2001s equivalent earnings in 2010 is \$1.4 a share. If Wal-Mart earns \$1 again in 2002, and the total inflation from 2002 through 2010 is 35%, then the equivalent 2002 earnings in 2010 is \$1.35. So on and so forth, you get the equivalent earnings of past 10 years. Then you add them together and divided the sum by 10 to get E10.
For example, CalAmp Corp's adjusted earnings per share data for the three months ended in Nov. 2016 was:
Adj_EPS = Earnigns per Share / CPI of Nov. 2016 (Change) * Current CPI (Nov. 2016) = -0.04 / 241.353 * 241.353 = -0.040
Current CPI (Nov. 2016) = 241.353.
CalAmp Corp Quarterly Data
201408 201411 201502 201505 201508 201511 201602 201605 201608 201611 per share eps 0.09 0.11 0.18 0.11 0.1 0.11 0.14 -0.07 0.01 -0.04 CPI 237.852 236.151 234.722 237.805 238.316 237.336 237.111 240.229 240.849 241.353 Adj_EPS 0.091 0.112 0.185 0.112 0.101 0.112 0.143 -0.07 0.01 -0.04 201202 201205 201208 201211 201302 201305 201308 201311 201402 201405 per share eps 0.05 0.14 0.12 0.14 1.09 0.05 0.08 0.12 0.09 0.07 CPI 227.663 229.815 230.379 230.221 232.166 232.945 233.877 233.069 234.781 237.9 Adj_EPS 0.053 0.147 0.126 0.147 1.133 0.052 0.083 0.124 0.093 0.071 200908 200911 201002 201005 201008 201011 201102 201105 201108 201111 per share eps -0.17 -0.05 -0.05 -0.09 -0.03 -0.01 0.01 0.02 0.05 0.06 CPI 215.834 216.33 216.741 218.178 218.312 218.803 221.309 225.964 226.545 226.23 Adj_EPS -0.19 -0.056 -0.056 -0.1 -0.033 -0.011 0.011 0.021 0.053 0.064 200702 200705 200708 200711 200802 200805 200808 200811 200902 200905 per share eps 0.03 -0.48 -0.19 -2.49 -0.37 -0.02 -0.06 -0.07 -1.86 -0.16 CPI 203.499 207.949 207.917 210.177 211.693 216.632 219.086 212.425 212.193 213.856 Adj_EPS 0.036 -0.557 -0.221 -2.859 -0.422 -0.022 -0.066 -0.08 -2.116 -0.181
Add all the adjusted EPS together and divide 10 will get our e10.
Explanation
If a company grows much fast than inflation, E10 may underestimate the company's earnings power. Shiller P/E Ratio can seem to be too high even the actual P/E is low.
For the Shiller P/E, the earnings of the past 10 years are inflation-adjusted and averaged. The result is used for P/E calculation. Since it looks at the average over the last 10 years, the Shiller P/E is also called PE10.
The Shiller P/E was first used by professor Robert Shiller to measure the valuation of the overall market. The same calculation is applied here to individual companies.
CalAmp Corp's Shiller P/E Ratio of today is calculated as
Shiller P/E Ratio = Share Price / E10 = 16.69 / -0.40 =
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
During the past 13 years, the highest Shiller P/E Ratio of CalAmp Corp was 300.00. The lowest was 0.00. And the median was 293.00.
Be Aware
Shiller P/E Ratio works better for cyclical companies. It gives you a better idea on the company's real earnings power.
Related Terms
Shiller P/E Ratio
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
CalAmp Corp Annual Data
Feb07 Feb08 Feb09 Feb10 Feb11 Feb12 Feb13 Feb14 Feb15 Feb16 e10 0.03 -0.29 -0.48 -0.48 -0.55 -0.59 -0.49 -0.51 -0.51 -0.54
CalAmp Corp Quarterly Data
Aug14 Nov14 Feb15 May15 Aug15 Nov15 Feb16 May16 Aug16 Nov16 e10 -0.52 -0.52 -0.51 -0.52 -0.53 -0.54 -0.54 -0.38 -0.39 -0.40
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# Dr. Erdal Ozkan PhD Petroleum Engineering
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This course covers the fundamental aspects of horizontal well reservoir and production engineering with special emphasis on the new developments. Each topic covered highlights the concepts that are generic to horizontal wells and draws attention to the pitfalls of applying conventional concepts to horizontal wells without critical evaluation. There is no set prerequisite for the course, but basic knowledge on general reservoir engineering concepts is useful.
#### PEGN 601A: APPLIED MATHEMATICS OF FLUID FLOW IN POROUS MEDIA
This course is intended to expose petroleum-engineering students to the special mathematical techniques used to solve transient flow problems in porous media. Bessel’s equation and functions, Laplace and Fourier transformations, the method of sources and sinks, Green’s functions, and boundary integral techniques are covered. Numerical evaluation of various reservoir engineering solutions, numerical Laplace transformation and inverse transformation are also discussed.
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Fundamentals of shale-reservoir engineering and special topics of production from shale reservoirs are covered. The question of what makes shale a producing reservoir is explored. An unconventional understanding of shale-reservoir characterization is Geological, geomechanical, and engineering aspects of shale reservoirs are explained. Well completions with emphasis on hydraulic fracturing and fractured horizontal wells are discussed from the view-point of reservoir engineering. Darcy flow, diffusive flow, and desorption in shale matrix are covered. Contributions of hydraulic and natural fractures are discussed and the stimulated reservoir volume concepts is introduced. Interactions of flow between fractures and matrix are explained within the context of dual-porosity modeling. Applications of pressure-transient, rate-transient, decline-curve and transient-productivity analyses are covered. Field examples are studies.
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My Math Forum Goldbach's conjecture proof
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September 13th, 2016, 01:32 AM #1 Newbie Joined: Sep 2016 From: algeria Posts: 2 Thanks: 0 Goldbach's conjecture proof Here I will proof Goldbach's conjecture by a simple way: to prove the conjecture, we will prove that any odd natural number greater than (3) is a sum of a prime number and an even number, we take that an even number is a sum of two odd numbers, from the first proof we decompose the result to a sum of two prime numbers and two even numbers, then we substitute the sum of the two even numbers from the original even number we get an even number equal a sum of two prime numbers and that is the Goldbach's conjecture. For the interval [10, +∞ [ we have: Let ({prime, even, odd}, +, ×) be a field with three elements. All odd numbers greater than 3 is a sum of a prime number and an even number: (Odd= prime + even)……………………………………… ………………………….(1) We have that every even number is a sum of two odd numbers: Even = odd + odd From (1) we get: that every even number is a sum of two prime numbers and two even numbers: Even = prime + even + prime + even Sash as the sum of two even numbers is an even number we get: => Even = prime + prime + even => Even – even = prime + prime Sash as the substitution of two even numbers is an even number we get: Even = prime + prime……………………………………… …………………...……………….(2) Sash as the set of even numbers and prime numbers are both an infinite sets the equation is correct for every even number belong to the interval [10, +∞] . For the interval ]2, 10[: We have only three even numbers : 4,6,8 : And we know that: 4=2+2……………………………………… …………………………...…………(3) 6=3+3……………………………………… …………………………...…………(4) 8=5+3……………………………………… …………………………...…………(5) Conclusion: From (2) , (3) , (4) , (5) we have that every even number belong to the interval ]2, +∞ [ is a sum of two prime numbers and the Goldbach's conjecture is correct. . . Last edited by skipjack; September 13th, 2016 at 11:38 AM.
September 13th, 2016, 03:32 AM #2 Member Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 4 This is nice work. Good luck with publishing it. Thanks from redos Last edited by 1ucid; September 13th, 2016 at 03:57 AM.
September 13th, 2016, 04:46 AM #3
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Quote:
Originally Posted by redos => Even = prime + prime + even
In this line, you need to prove that the left-hand side is any even number. At the moment your derivation shows only that any even number less some particular even number (that depends on the choice of prime) is equal to the right-hand side.
Last edited by skipjack; September 13th, 2016 at 11:37 AM.
September 13th, 2016, 05:52 AM #4
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Originally Posted by v8archie you need to prove that the left hand side is any even number.
I think is already proven by the proofs of Olivier Ramaré and Harald Helfgott which is saying that any even number is the sum of 6 primes, that means:
even = prime + prime + (prime + prime + prime + prime)
we know that (prime + prime + prime + prime) is an even number.
But I still think that is proven by the derivation I wrote above.
Last edited by skipjack; September 13th, 2016 at 11:36 AM.
September 13th, 2016, 10:06 AM #5 Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra That idea suffers from the same problem as your proof, which explains why you think it is correct. Perhaps you should ask Olivier Ramaré and Harald Helfgott whether they think they have proved Goldbach. I can guarantee that if they thought they had, their paper giving the six prime proof would have had Goldbach as its main focus. Last edited by skipjack; September 13th, 2016 at 11:35 AM.
September 18th, 2016, 08:15 AM #6 Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 Can you please give a numerical example, illustrating all steps of obtaining (even=prime+prime)?
September 18th, 2016, 09:21 AM #7 Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra The Goldbach Conjecture is a statement of the existence of such primes, not a claim that there exists any special method for finding them. Thanks from ProofOfALifetime
September 11th, 2017, 06:14 AM #8 Newbie Joined: Jan 2010 Posts: 9 Thanks: 0 goldbach proof The formula that generates whole prime numbers : y=(2^(x-1)-1)/x ( formula-1) If y is an integer , then x must be absolutely a prime number . the set of x for any value of integer y ; x = { 3,5,7,11,13,....} and it generates all the prime numbers . The question is that for the set of prime numbers ( x1 , x2) does the formula generates all the even numbers or not ? y1 = (2^(x1-1) -1 ) /x1 + (2^(x2-1) -1 ) /x2 for ( x1 , x2) = ( 3,3) then y1 = 2 ; for ( x1 , x2) = ( 3,5) then y2 = 4 ; for ( x1 , x2) = ( 5,5) then y3 = 6 ; for ( x1 , x2) = (5,7) then y4 = 8 ; .................... The result for whole prime sets of ( x1 , x2) then you can generate all the even number's set . P.S.: For the proof of formula-1 and to learn more about it please contact me . For example formula-1 must be always divided by 3 . METE UZUN TEL: +905315540733 e-mail: meteuzun@hotmail.com
September 11th, 2017, 07:44 AM #9
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Quote:
Originally Posted by meteuzun If y is an integer, then x must be absolutely a prime number.
That's incorrect. For example, y is an integer if x = 341 = 11 × 31.
September 11th, 2017, 09:12 AM #10 Newbie Joined: Jan 2010 Posts: 9 Thanks: 0 Thank you skipjack. Can you test this formula, please? (2^(x-1) + (x-1)!)/x Last edited by skipjack; June 25th, 2019 at 08:37 PM.
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# Any serious policy discussion about acceptable levels of risk in conne
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Any serious policy discussion about acceptable levels of risk in connection with explosions is not well served if the participants fail to use the word “explosion” and use the phrase “energetic disassembly” instead. In fact, the word “explosion” elicits desirable reactions, such as a heightened level of attention, whereas the substitute phrase does not. Therefore, of the two terms, “explosion” is the one that should be used throughout discussions of this sort.
Which of the following is an assumption on which the argument above depends?
(A) In the kind of discussion at issue, the advantages of desirable reactions to the term “explosion” outweigh the drawbacks, if any, arising from undesirable reactions to that term.
(B) The phrase “energetic disassembly” has not so far been used as a substitute for the word “explosion” in the kind of discussion at issue.
(C) In any serious policy discussion, what is said by the participants is more important than how it is put into words.
(D) The only reason that people would have for using “energetic disassembly” in place of “explosion” is to render impossible any serious policy discussion concerning explosions.
(E) The phrase “energetic disassembly” is not necessarily out of place in describing a controlled rather than an accidental explosion.
Originally posted by amirdubai1982 on 09 Sep 2010, 03:23.
Last edited by Bunuel on 08 Nov 2018, 02:43, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Any serious policy discussion about acceptable levels of risk in conne [#permalink]
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09 Sep 2010, 05:24
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I used POE to get to A.
C, D & E are straight out, sort of irrelevant. Re-checking A & B, even B appeared irrelevant.
Plus, the statement emphasizes that the desirable impact is dependent on the word that is used; explosion or energetic disassembly". Hence, it is closer to A, which is emphasizing that advantages (desirable reactions).
The argument simply means - use of the word explosion has "desirable reactions" and "energetic ...." doesn't have the same impact. .............. one is more advantageous than the other ........
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20 Apr 2011, 04:04
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amirdubai1982 wrote:
Any serious policy discussion about acceptable levels of risk in connection with explosions is not well served if the participants fail to use the word “explosion” and use the phrase “energetic disassembly” instead. In fact, the word “explosion” elicits desirable reactions, such as a heightened level of attention, whereas the substitute phrase does not. Therefore, of the two terms, “explosion” is the one that should be used throughout discussions of this sort.
Which of the following is an assumption on which the argument above depends?
(A) In the kind of discussion at issue, the advantages of desirable reactions to the term “explosion” outweigh the drawbacks, if any, arising from undesirable reactions to that term.
(B) The phrase “energetic disassembly” has not so far been used as a substitute for the word “explosion” in the kind of discussion at issue.
(C) In any serious policy discussion, what is said by the participants is more important than how it is put into words.
(D) The only reason that people would have for using “energetic disassembly” in place of “explosion” is to render impossible any serious policy discussion concerning explosions.
(E) The phrase “energetic disassembly” is not necessarily out of place in describing a controlled rather than an accidental explosion
I am not able to paraphrase the argument pls help????
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OA: A
Paraphrase is: If you have to really discuss risks related to explosions, call explosions 'explosions' and not 'energetic dissembly', because only word explosion creates in audience reactions like serious attention which are required for serious discussion.
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06 Jun 2015, 05:52
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The Argument doesnt talk anything about the types of explosion..hence its quite out of scope to include them
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07 Jun 2015, 22:31
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Use the word “Explosion” = desirable reactions = discussed well served
Use the word “Energetic disassembly” = NO desirable reactions = discussion Not well served
Conclusion: Use the word "Explosion" throughout in the discussion.
Possible assumptions:
1) Only those terms concerning "Desirable reactions" find place throughout the discussion.
2) Only those discussions that are well served are used throughout the discussion.
Choice A is correct.
P.S.: I had initially chosen choice C , but then I realised that it does not bridge the gap between the premise and conclusion, which is why to use the word "Explosion" throughout in the discussion.
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27 Aug 2018, 09:22
The key here is to look at what would happen if the opposite of (A) was true - would the argument still stand? The argument itself is that the word explosion gets more of a reaction than "energetic disassembly," and that the increased reaction is a good thing.
If (A) were to be negated, it would read something along the lines of "the advantages of the desirable reactions do NOT outweigh the disadvantages of the undesirable reactions." If this was true, the entire argument for using the word explosion would fall apart. You wouldn't use a word if the undesirable reactions were more intense than the desirable reactions. Because negating that statement seriously harms the argument's conclusion, it must be an assumption that the argument is built on.
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27 Aug 2018, 09:33
LauraOrion wrote:
The key here is to look at what would happen if the opposite of (A) was true - would the argument still stand? The argument itself is that the word explosion gets more of a reaction than "energetic disassembly," and that the increased reaction is a good thing.
If (A) were to be negated, it would read something along the lines of "the advantages of the desirable reactions do NOT outweigh the disadvantages of the undesirable reactions." If this was true, the entire argument for using the word explosion would fall apart. You wouldn't use a word if the undesirable reactions were more intense than the desirable reactions. Because negating that statement seriously harms the argument's conclusion, it must be an assumption that the argument is built on.
Thank you very much...!!
I got confused/rather got sink in words.
A. In the kind of discussion at issue, the advantages of desirable reactions to the term “explosion” outweigh the drawbacks, if any, arising from undesirable reactions to that term.
Although, I read this twice or thrice, my mind was interpreting advantages of desirable reactions outweigh disadvantages of the same thing.
And then I went on thinking that option A is out of scope as argument never talk of disadvantages of desirable results.
Thanks for the quick response.
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27 Aug 2018, 09:38
amirdubai1982 wrote:
Any serious policy discussion about acceptable levels of risk in connection with explosions is not well served if the participants fail to use the word “explosion” and use the phrase “energetic disassembly” instead. In fact, the word “explosion” elicits desirable reactions, such as a heightened level of attention, whereas the substitute phrase does not. Therefore, of the two terms, “explosion” is the one that should be used throughout discussions of this sort.
Which of the following is an assumption on which the argument above depends?
(A) In the kind of discussion at issue, the advantages of desirable reactions to the term “explosion” outweigh the drawbacks, if any, arising from undesirable reactions to that term.
(B) The phrase “energetic disassembly” has not so far been used as a substitute for the word “explosion” in the kind of discussion at issue.
(C) In any serious policy discussion, what is said by the participants is more important than how it is put into words.
(D) The only reason that people would have for using “energetic disassembly” in place of “explosion” is to render impossible any serious policy discussion concerning explosions.
(E) The phrase “energetic disassembly” is not necessarily out of place in describing a controlled rather than an accidental explosion.
CONCLUSION: EXPLOSION should be used in discussions of this sort (Serious)
PREMISE 1: Serious policy discussion is served better with Explosion than Energetic Disassembly
PREMISE 2: EXPLOSION also invokes attention
Assumption: Lets say conclusion does not hold
which means "explosion should not be used in discussions"
But why,
What if explosion ,though it brings attention, is still not required in serious policy discussion. In discussions we need positive mindset and seek solutions where energetic disassembly can score. Energetic disassembly brings positive connotation and people become more solution oriented ie. to say it has more advantages than "explosion"
A possible assumption is : Such scenario will not exist. i.e Explosion does not have any drawbacks or positive outweighs its drawbacks...
Hence answer is A
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Re: Any serious policy discussion about acceptable levels of risk in conne &nbs [#permalink] 27 Aug 2018, 09:38
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# Any serious policy discussion about acceptable levels of risk in conne
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# How do you graph y=2^-x?
Feb 9, 2018
here is the graph
graph{2^-x [-10, 10, -5, 5]}
#### Explanation:
Start by changing $y = {2}^{-} x$ to $y = {\left(\frac{1}{2}\right)}^{x}$ to have an untransformed graph.
Here are some characteristics of $y = {\left(\frac{1}{2}\right)}^{x}$ to make sure you plot the points correctly:
-it will have y-intercept (0,1)
-the domain is all reals
-has $0 <$base$< 1$
-is a decreasing function
-the horizontal asymptote is $y = 0$, therefore the range is $y > 0$
To graph, start with the point (0,1) and multiply $y$ by $\frac{1}{2}$ to get the positive coordinates of $x$. To get the negative coordinates of $x$, start with (0,1) and multiply $y$ by 2. Keep in mind the restrictions and guidelines above.
then graph accordingly making sure to get close to zero, but never actually reach it.
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Search i-have-an-equation-for-a-value-of-y-i-need-to-rearrange-the-equation-to-find-x-and-then
# I have an equation for a value of y i need to rearrange the equation to find x and then
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1.AU MAT 120 Systems of Linear Equations and Inequalities Discussion
mathematicsalgebra Physics
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# Fast Formula Evaluation
Like in the “oldie, but goldie” Fractint, an auto-evaluator would be nice in XaoS as well. I.e., when a user enters a formula like
`z=z^2+c`
then XaoS automagically evaluates this formula for each c pixel and draws the image. The main point here is to do this reasonably fast.
For this, the formconv library could help. We have very nice benchmarks of this library soon, for details go to the Real Time Zooming Math Engine (rtzme) web page. However, rtzme is written in C++ (and wxWidgets), it is also possible to extend XaoS capabilities inside the C programming style (so you are not enforced to use C++ and/or rewrite the whole XaoS code into C++).
Here is a sample code which illustrates the technique (cut and pasted from the `test/complex.cpp` file from the formconv library):
```/*
* Demo program which shows the power of the libformconv library.
* Written by Z. Kovacs, modifications by G. Bakos.
* Copyright (C) 2004-2005 Z. Kovacs, G. Bakos.
*/
#include <h/fastcompute.h>
int
main (int argc, char **argv)
{
char myFormula[50];
formula < FastComplexCompute < std::complex < double > > >*f;
complex < double >c;
double i, j, re, im;
strcpy (myFormula, "sinz^2");
f =
formconv_prepare < FastComplexCompute < std::complex < double
> > >(myFormula);
for (i = 0; i < 1; i += 0.3)
{
for (j = 0; j < 1; j += 0.3)
{
c = complex < double >(i, j);
c =
formconv_evaluate < FastComplexCompute < std::complex <
double > > >(f, c);
re = real (c);
im = imag (c);
printf ("%5.3f+%5.3f*I ", re, im);
}
printf ("\n");
}
formconv_free < FastComplexCompute < std::complex < double > > >(f);
}```
The formconv library is also useful for intuitively entered formulas. So the user is allowed to type
`z=sin^2z+c`
`z=cpow(sin(z),2)+c`
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# Re: Aggregates
From: Andy Seaborne <andy.seaborne@epimorphics.com>
Date: Thu, 08 Dec 2011 09:02:56 +0000
Message-ID: <4EE07D40.2080208@epimorphics.com>
```
On 07/12/11 13:57, Steve Harris wrote:
> On 2011-12-07, at 13:00, Andy Seaborne wrote:
>
>>
>>
>> On 06/12/11 22:40, Steve Harris wrote:
>>> Hi all,
>>>
>>> I've now got the aggregates in a state where I think all the information is carried through from one end of the query to the other… but I've thought that before :)
>>>
>>> I also think ORDER BY is covered.
>>>
>>> Here's a sketch of what I think should be happening:
>>>
>>> Data
>>>
>>> <a> <p> 1 .
>>> <a> <p> 2 .
>>> <b> <p> 3 .
>>>
>>> Query
>>>
>>> SELECT (MAX(?o) AS ?max) (MIN(?o) AS ?min)
>>> WHERE { ?s ?p ?o }
>>> GROUP BY ?s
>>> ORDER BY AVG(?o)
>>>
>>>
>>> Ω = Sol ?s ?p ?o
>>> μ1<a> <p> 1
>>> μ2<a> <p> 2
>>> μ3<b> <p> 3
>>>
>>> G = Group((?s), Ω)
>>> = { ((<a>), { μ1, μ2 }), ((<b>), { μ3 }) }
>>>
>>> Q = SELECT agg1 agg2
>>> WHERE { ?s ?p ?o }
>>> GROUP BY ?s
>>> ORDER BY agg3
>>>
>>> E = { (?max, agg1), (?min, agg2) }
>>>
>>> A1 = Aggregation((?o), Max, {}, G)
>>> A2 = Aggregation((?o), Min, {}, G)
>>> A3 = Aggregation((?o), Avg, {}, G)
>>>
>>> J = AggregateJoin(A) =
>>> { { (agg1, 2), (agg2, 1), (agg3, 1.5) }
>>> { (agg1, 3), (agg2, 3), (agg3, 3) } }
>>
>>
>> This is the evaluation of AggregateJoin at execution time.
>>
>> I don't understand this step: how does it know the variables are agg1, agg2, and agg3? There could be other agg_i from other query levels. And why this order not agg3, agg2, agg1?
>
> From the A, A has members 1, 2, and 3 in this case. A1 pairs with agg1 for e.g.
>
> If it were a lower query level it might have members 4, 5, and 6 for e.g.
Not quite: i is reset on every SELECT processed
"""
# Note, i is global for the query, defaults to 1
Let i := 1
"""
The comment might have that intent, but, to me, "Let" introduces a
variable each time.
It is workable as a definition but rather unclear to me. The fact it
works relies on scoping features of variables so that the use of agg_1
twice does not fall apart.
Also, the variable names are regenerated. How does AggregateJoin know
the variable is called "agg1" not "__gen1" because the query really does
use ?agg1 in teh user written part?
I think it would have been easier to do it all in the translation and
not have AggregateJoin which is really just a form of "extend" assigning
Ai to agg_i.
Just rewriting to
extend( ?agg1 := Aggregation((?o), Max, {}, G),
?agg2 := Aggregation((?o), Min, {}, G),
?agg3 := Aggregation((?o), Avg, {}, G) )
would keep the aggregation and the variable together.
If the reviewers are comfortable with the form in the doc, then I can
live with it but I think it works by relying on human reading and
associating text.
Andy
```
Received on Thursday, 8 December 2011 09:03:31 UTC
This archive was generated by hypermail 2.3.1 : Wednesday, 7 January 2015 15:01:05 UTC
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# Convergence Distribution and Probability [closed]
Suppose that $$|X_n - Y_n|$$ converges in probability to 0, and that $$X_n$$ converges in distribution to X. Show that $$Y_n$$ converges in distribution to X.
Goal: show $$P(Y_n \le t) \to P(X \le t)$$ [for $$t$$ at which $$F_X(t):=P(X \le t)$$ is continuous].
• $$P(Y_n \le t) = P(Y_n \le t, |X_n - Y_n| > \epsilon) + P(Y_n \le t, |X_n - Y_n| \le \epsilon)$$.
• $$P(Y_n \le t, |X_n - Y_n| > \epsilon) \le P(|X_n - Y_n| > \epsilon)$$. What does the right-hand side converge to?
• $$P(Y_n \le t, |X_n - Y_n| \le \epsilon) \le P(X_n \le t + \epsilon)$$. What does the right-hand side converge to?
• Use the assumption that $$F_X(t)$$ is continuous at $$t$$.
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| 9,014
|
VJ193259 C Asa's Chess Problem
# VJ193259 C Asa's Chess Problem
Donny
October 29, 2017
2192
-
Tags in blue are handcrafted tags; Tags in green are generated using AutoTag.
## Problem
N X N 的棋盘和棋子。棋子有白有黑。同行或同列的某两个棋子是一对。要求通过交换配对的棋子，使得行和列的黑色棋子数在给定的范围。
2
0 0
1 1
2 2
0 1
0 2
0 2
1 1 2 1
1 2 2 2
4
0 0 1 1
0 0 1 1
1 1 0 0
1 1 0 0
2 2
1 3
1 3
2 2
0 1
3 4
0 1
3 4
1 1 1 3
1 2 1 4
2 1 2 3
2 2 2 4
3 1 3 4
3 2 3 3
4 1 4 4
4 2 4 3
2
4
## Code
``````#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <cmath>
#define memz(_a) memset(_a,0,sizeof _a)
const int INF=0x7FFFFFFF;
const int maxn=55*2;
const int maxm=2*maxn*maxn;
struct EDG
{
int u,v,cap,f,c;
} es[maxm];
int fst[maxn],nxt[maxm];
int coe;
int V;
int D[maxn];
int SS,TT;
inline void init() {
coe=1; memz(fst); V=0;
memz(D);
}
inline void mkedg(int u,int v,int cap,int c)
{
es[++coe]={u,v,cap,0,c};
nxt[coe]=fst[u];
fst[u]=coe;
}
inline void mkedgulb(int u,int v,int cl,int cu,int cost)
{
D[u]-=cl; D[v]+=cl;
if (cu-cl==0) return;
mkedg(u,v,cu-cl,cost),mkedg(v,u,0,-cost);
}
inline void refactor()
{
SS=V+1; TT=V+2;
for (int v=1;v<=V;++v)
{
if (D[v]<0) mkedg(v,TT,-D[v],0),mkedg(TT,v,0,0);
if (D[v]>0) mkedg(SS,v,D[v],0),mkedg(v,SS,0,0);
}
}
int d[maxn];
int p[maxn];
int a[maxn];
bool inq[maxn];
bool spfa()
{
std::fill(d,d+maxn,INF);
memz(inq);
std::queue<int> Q;
d[SS]=0; p[SS]=0; a[SS]=INF;
Q.push(SS); inq[SS]=1;
while (!Q.empty())
{
int u=Q.front(); Q.pop();
inq[u]=0;
for (int k=fst[u];k;k=nxt[k])
{
EDG& e=es[k];
if (e.cap>e.f && d[e.v]>d[u]+e.c)
{
d[e.v]=d[u]+e.c;
p[e.v]=k;
a[e.v]=std::min(a[u],e.cap-e.f);
if (!inq[e.v]) {
Q.push(e.v); inq[e.v]=1;
}
}
}
}
return d[TT]!=INF;
}
int mincost()
{
int flow=0,cost=0;
while (spfa())
{
flow+=a[TT];
cost+=d[TT]*a[TT];
for (int u=TT;u!=SS;u=es[p[u]].u) {
es[p[u]].f+=a[TT];
es[p[u]^1].f-=a[TT];
}
}
return cost;
}
bool checkulb()
{
for (int k=fst[SS];k;k=nxt[k])
{
if (es[k].f!=es[k].cap) return false;
}
return true;
}
int N;
int rowc[maxn],colc[maxn];
int G[maxn][maxn];
int r[maxn],c[maxn];
int S,T;
int main()
{
int x;
while (~scanf("%d",&N))
{
init();
memz(rowc); memz(colc);
S=++V; T=++V;
// S: 1; T: 2; rows: 3~N+2; cols:N+3~2N+2
for (int i=1;i<=N;++i)
for (int j=1;j<=N;++j)
{
scanf("%d",&G[i][j]);
if (G[i][j]) ++rowc[i],++colc[j];
}
int lc,uc;
for (int i=1;i<=N;++i)
{
scanf("%d%d",&lc,&uc);
r[i]=++V;
mkedgulb(S,r[i],rowc[i],rowc[i],0);
mkedgulb(r[i],T,lc,uc,0);
}
for (int j=1;j<=N;++j)
{
scanf("%d%d",&lc,&uc);
c[j]=++V;
mkedgulb(S,c[j],colc[j],colc[j],0);
mkedgulb(c[j],T,lc,uc,0);
}
int x1,x2,y1,y2;
for (int i=1;i<=N*N/2;++i)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
if (G[x1][y1]==G[x2][y2]) continue;
if (!G[x1][y1]) std::swap(x1,x2),std::swap(y1,y2);
if (y1==y2) mkedgulb(r[x1],r[x2],0,1,1);
if (x1==x2) mkedgulb(c[y1],c[y2],0,1,1);
}
mkedg(T,S,INF,0); mkedg(S,T,0,0);
refactor();
int ans=mincost();
if (!checkulb()) ans=-1;
printf("%d\n",ans);
}
return 0;
}
``````
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## college101 2 years ago A grandmother deposited $5,000 in an account that pays 5% per year compounded annually when her granddaughter was born. What will the value of the account be when the granddaughter reaches her 14th birthday? a)$9,879.66 b) $9,869.66 c)$9,899.66 d) $9,929.66 e)$9,889.66 f) None of the above.
1. tkhunny
$$5,000\cdot 1.05^{14}$$?
2. tkhunny
What did you get? Was there anything about that little formula that you could not have done on your own? Let's see another one and show your work.
3. college101
c) \$9,899.66
4. tkhunny
Sweet. What else have you?
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| 3.09375
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http://gmatclub.com/forum/1b-2b-3c-4b-5c-6b-7e-8b-9c-10b-11c-12e-13d-14d-15d-16a-5577.html
| 1,485,105,258,000,000,000
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| 114,369,768
| 51,149
|
1B 2B 3C 4B 5C 6B 7E 8B 9C 10B 11C 12E 13D 14D 15D 16A : GMAT Sentence Correction (SC)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
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| 646
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https://physics.stackexchange.com/questions/661884/how-does-sum-k-psi-k-vecr-psi-k-vecr-delta-vecr-vecr-exp
| 1,723,003,665,000,000,000
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| 369,224,374
| 40,862
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# How does $\sum_k \psi_k^*(\vec{r})\psi_k(\vec{r}')=\delta(\vec{r}-\vec{r}')$ express the completeness of a basis?
The annihilation field operator is defined as $$\hat{\psi}(\vec{r})=\sum_k \hat{b}_k \psi_k(\vec{r})$$ Two of these operators satisfy the commutation relations $$[\hat{\psi}(\vec{r}),\hat{\psi}^{\dagger}(\vec{r}')]=\delta(\vec{r}-\vec{r}')$$ $$[\hat{\psi}(\vec{r}),\hat{\psi}(\vec{r}')]=[\hat{\psi}^{\dagger}(\vec{r}),\hat{\psi}^{\dagger}(\vec{r}')]=0$$
My textbook (Nazarov and Danon) states that these can be derived from the definition of the field operator if one notes the completeness of the basis which is expressed by $$\sum_k \psi_k^*(\vec{r})\psi_k(\vec{r}')=\delta(\vec{r}-\vec{r}')\tag{1}$$ How is the above equation an expression of completeness though? Similarly, given a complete basis $$\{\psi_k\}$$, how would one go about proving that condition (1) is satisfied for the given complete basis? The way that I would usually express the completeness of a basis is with the relation $$\sum_k |\psi_k\rangle \langle \psi_k |=\hat{1}$$ But this expression is seemingly nothing alike the expression in eq 1. So how does the condition shown in eq 1 express the completeness of a basis?
If $$\sum_k |\psi_k\rangle \langle \psi_k|= {\mathbb I}$$ then, sandwiching this expression between $$\langle x|$$ and $$|x'\rangle$$, we have $$\sum_k \langle x|\psi_k\rangle \langle \psi_k|x'\rangle= \langle x|x'\rangle$$ or, since $$\langle x|\psi_k\rangle\equiv \psi_k(x)$$ and $$\langle \psi_k|x'\rangle = (\langle x|\psi_k\rangle)^*$$, and also $$\langle x|x'\rangle= \delta(x-x')$$, your eq 1 becomes
$$\sum_k \psi_k(x) \psi^*_k(x')= \delta(x-x')$$
• Thanks for the great answer! All cleared up now. In line with what Dvij D.C. said though, is there a way to intuitively understand why eq (1) implies completeness? If I imagine the basis set to be the basis of the infinite square well $\{\psi_n\}={\sqrt(2/a)\sin(n\pi x/a)}$, Its not immediately obvious to me that applying eq (1) to this set will yield zero for $x\neq x'$. I can see how the summation over all n will yield infinity when $x=x'$ though. Commented Aug 28, 2021 at 13:07
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# [ncl-talk] slice plot
David Brown dbrown at ucar.edu
Thu Oct 2 15:56:26 MDT 2014
```Hi Bedassa,
You can't plot something with a 2D coordinate against a 1D coordinate in a
2D plot. The lon2d gives the longitude values for a horizontal 2D field.
You have to make some choices to get a vertical plot -- you need to reduce
the latitude dimension to a single element somehow. The simplest thing to
do would be to just pick one index along the y axis of the lon coordinate
-- e.g. lon2d(100,:). However, because the coordinates are 2d, the chances
are that this would not be a constant latitude, so you could not say for
instance that your plot represents height vs longitude at 10 degrees North
latitude.
But with proper labeling you could describe what this plot does represent.
I suggest you look at the NCL Examples page for Pressure/Height vs
Longitude plots and specifically at the last example on the page called
narr_6.ncl. It shows height plots for data where the coordinates are 2D
(aka "curvilinear" coordinates). Notice that the X axis for these plots has
both lat/lon values for the tickmark labels.This should give you some ideas
of what the issues are and how to solve them.
Creating a plot where the latitude is constant would require a bit more
work. But if you wanted to try you would probably want to use the function
getind_latlon2d. This might actually be a good topic for a future example.
-dave
On Wed, Oct 1, 2014 at 4:24 PM, Bedassa Regassa <beregassa at gmail.com> wrote:
> I am new to ncl. I am to plot the vertical Height vs longitude but I got
> warning and I tried to a lot to find my error.
>
> here is the wraning
>
> warning:ScalarFieldSetValues: 2d coordinate array sfXArray has an
> incorrect dimension size: defaulting sfXArray
>
> here is my data information ncl_filedump and below alos my scrip
>
> dimensions:
> x = 186
> y = 166
> pressure = 9
> time = 1 // unlimited
> nb2 = 2
> variables:
> float lon ( y, x )
> standard_name : longitude
> long_name : longitude
> units : degrees_east
> _CoordinateAxisType : Lon
>
> float lat ( y, x )
> standard_name : latitude
> long_name : latitude
> units : degrees_north
> _CoordinateAxisType : Lat
>
> float pressure ( pressure )
> standard_name : air_pressure
> long_name : pressure
> units : Pa
> positive : down
> axis : Z
>
> double time ( time )
> standard_name : time
> long_name : time
> bounds : time_bnds
> units : seconds since 2009-05-01 00:00:00
> calendar : proleptic_gregorian
>
> double time_bnds ( time, nb2 )
> units : seconds since 2009-05-01 00:00:00
> calendar : proleptic_gregorian
>
> float U ( time, pressure, y, x )
> standard_name : grid_eastward_wind
> long_name : U-component of wind
> units : m s-1
> coordinates : lon lat
>
> ;************************************************
> ;************************************************
> ;************************************************
> begin
> ;************************************************
> ;open file and read in data
> ;************************************************
> ;************************************************
> ; get variables infromation and print them cclm
> ;************************************************
> vars = getfilevarnames(f)
> print(vars)
> ;************************************************
> var=f->U(0,:,15,:)
> lon2d=f->lon
> lev=f->pressure
>
> var at lon2d=lon2d
> var at lev=lev
> ;*******************************
> ;create plot
> ;************************************************
>
> ;-- define workstation
> wks = gsn_open_wks("png","plot_slices")
> gsn_define_colormap(wks,"ncl_default") ;-- set the colormap to be used
> ;-- set resources
> res = True
> res at tiMainString = "DKRZ NCL Tutorial Example: Slice plot at 40N"
> res at cnFillOn = True ;-- turn on color fill
> res at cnLineLabelsOn = False ;-- turns off contour line labels
> res at cnInfoLabelOn = False ;-- turns off contour info label
> res at lbOrientation = "vertical" ;-- vertical label bar
> res at tiYAxisString = var at long_name+" [hPa]"
> ;-- append units to y-axis label
> ;res at tfDoNDCOverlay = True
> res at sfXArray = lon2d ;-- uses lon_t as plot x-axis
> res at sfYArray = lev/100 ;-- uses lev_t in hPa as plot y-axis
> res at gsnYAxisIrregular2Linear = True ;-- converts irreg depth to linear
> res at trYReverse = True ;-- reverses y-axis
>
> ;-- generate the plot
> plot = gsn_csm_contour(wks,var,res)
> end
>
> _______________________________________________
> ncl-talk mailing list
> List instructions, subscriber options, unsubscribe:
> http://mailman.ucar.edu/mailman/listinfo/ncl-talk
>
>
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# rotating frame of reference
Register Blogs Members List Search Today's Posts Mark Forums Read
June 5, 2007, 15:44 rotating frame of reference #1 madhuri Guest Posts: n/a hi iam madhuri.i am doing a project relatedto swirl generator.so my geometry or domain has 230 small tubes where in air flows inside tubes.the tubes are to be rotated at 100-800 rpm.for this geometry when iam setting moving frame of reference option in fluent to rotate the tubes.the question is should i give rotation to fluid or walls of the tubes or both together.i thought if i give rotation to fluid it would be sufficient.but there is even option for the walls of tube to be stationary or rotating .what should i do to this problem. thanks in advance Madhuri seela
June 5, 2007, 17:09 Re: rotating frame of reference #2 Ketan Guest Posts: n/a Hi As I understood that there are 230 small tube in the space and they are rotating,also there is inlet and out let for each tube. and you have a rotating referance frame. If you give a rotating referance frame in the BC->Fluid-> Moving reference fame and provide the rotation speed this means that the whole fluid domain starts rotating with the rotation you supplied. And if the wall is also rotating with the same rotation as of fluid and i.e they have same rotation rate then give the motion of wall as moving and in motion type relative to cell zone , rotation and then give value as "0" in speed. Does it make sense? If I am not very clear please reply with ur question cleary.
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PDB_Stat_100_Lecture_13
# PDB_Stat_100_Lecture_13 - STA 100 Lecture 13 Paul Baines...
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STA 100 Lecture 13 Paul Baines Department of Statistics University of California, Davis February 2nd, 2011
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Admin for the Day I Please pick up Midterm+old homeworks I Homework 3 due today by 5pm!
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Admin for the Day I Please pick up Midterm+old homeworks I Homework 3 due today by 5pm! I !!!!!! PLEASE TURN IN R OUTPUT FOR HWK 3 !!!!!
Admin for the Day I Please pick up Midterm+old homeworks I Homework 3 due today by 5pm! I !!!!!! PLEASE TURN IN R OUTPUT FOR HWK 3 !!!!! I Homework 4 will be posted later today, due next Wednesday
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Admin for the Day I Please pick up Midterm+old homeworks I Homework 3 due today by 5pm! I !!!!!! PLEASE TURN IN R OUTPUT FOR HWK 3 !!!!! I Homework 4 will be posted later today, due next Wednesday I For Hwk 2 you can get back the points you lost for not turning in your R output back if. . . You print out your R output, staple it to your graded homework and bring it to me by Friday !
Admin for the Day I Please pick up Midterm+old homeworks I Homework 3 due today by 5pm! I !!!!!! PLEASE TURN IN R OUTPUT FOR HWK 3 !!!!! I Homework 4 will be posted later today, due next Wednesday I For Hwk 2 you can get back the points you lost for not turning in your R output back if. . . You print out your R output, staple it to your graded homework and bring it to me by Friday !
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Admin for the Day I Please pick up Midterm+old homeworks I Homework 3 due today by 5pm! I !!!!!! PLEASE TURN IN R OUTPUT FOR HWK 3 !!!!! I Homework 4 will be posted later today, due next Wednesday I For Hwk 2 you can get back the points you lost for not turning in your R output back if. . . You print out your R output, staple it to your graded homework and bring it to me by Friday ! References for Today: Rosner, Ch 5.1-5.6 (7th Ed.) References for Friday: Rosner, Ch 5.7-5.9, Ch 6.5 (7th Ed.)
Key Ideas from Last Time 1. How to conduct a statistical analysis
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Key Ideas from Last Time 1. How to conduct a statistical analysis I Plot, Assume, Estimate, Check, Answer, Predict.
Key Ideas from Last Time 1. How to conduct a statistical analysis I Plot, Assume, Estimate, Check, Answer, Predict. 2. Motivation
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Key Ideas from Last Time 1. How to conduct a statistical analysis I Plot, Assume, Estimate, Check, Answer, Predict. 2. Motivation I Why do we care about estimating the parameters (e.g., λ )? (b/c they tell use everything we need!)
Key Ideas from Last Time 1. How to conduct a statistical analysis I Plot, Assume, Estimate, Check, Answer, Predict. 2. Motivation I Why do we care about estimating the parameters (e.g., λ )? (b/c they tell use everything we need!) I Why do we need to estimate the parameters e.g., λ ? (b/c we don’t know them!)
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Key Ideas from Last Time 1. How to conduct a statistical analysis I Plot, Assume, Estimate, Check, Answer, Predict. 2. Motivation I Why do we care about estimating the parameters (e.g., λ )?
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College Algebra with Corequisite Support 2e
# 3.5Transformation of Functions
College Algebra with Corequisite Support 2e3.5 Transformation of Functions
## Learning Objectives
In this section, you will:
• Graph functions using vertical and horizontal shifts.
• Graph functions using reflections about the x-axis and the y-axis.
• Determine whether a function is even, odd, or neither from its graph.
• Graph functions using compressions and stretches.
• Combine transformations.
## Corequisite Skills
### Learning Objectives
• Identify graphs of basic functions, (IA 3.6.2)
• Graph quadratic functions using transformations, (IA 9.7.4)
### Objective 1: Identify graphs of basic functions, (IA 3.6.2)
Basic functions have unique shapes, characteristics, and algebraic equations. It will be helpful to recognize and identify these basic or “toolkit functions” in our work in algebra, precalculus and calculus. Remember functions can be represented in many ways including by name, equation, graph, and basic tables of values.
#### Practice Makes Perfect
Use a graphing program to help complete the following. Then, choose three values of x to evaluate for each. Add the x and y to the table for each exercise.
1.
Name Equation Graph
Constant y=c, where c is a constant
Choose 3 values of x to evaluate for each.
x y
2.
Name Equation Graph
Identity y=x
Choose 3 values of x to evaluate for each.
x y
3.
Name Equation Graph
Absolute Value y=|x|
Choose 3 values of x to evaluate for each.
x y
4.
Name Equation Graph
Choose 3 values of x to evaluate for each.
x y
5.
Name Equation Graph
Cubic y=x3
Choose 3 values of x to evaluate for each.
x y
6.
Name Equation Graph
Reciprocal $y=1xy=1x$
Choose 3 values of x to evaluate for each.
x y
7.
Name Equation Graph
Square Root $y=xy=x$
Choose 3 values of x to evaluate for each.
x y
8.
Name Equation Graph
Cube Root $y=x3y=x3$
Choose 3 values of x to evaluate for each.
x y
9.
Name Equation Graph
Exponential $y=exy=ex$
Choose 3 values of x to evaluate for each.
x y
### Objective 2: Graph quadratic functions using transformations (IA 9.7.4)
When we modify basic functions by adding, subtracting, or multiplying constants to the equation, very systematic changes take place. We call these transformations of basic functions. Here we will investigate the effects of vertical shifts, horizontal shifts, vertical stretches or compressions, and reflections on quadratic functions. We could use any basic function to illustrate transformations, but quadratics work nicely because we can easily keep track of a point called the vertex.
#### Practice Makes Perfect
The graphs of quadratic functions are called parabolas. Use a graphing program to graph each of the following quadratic functions. For each graph find the vertex (the minimum or maximum value) of the parabola and list its coordinates. Most importantly use the patterns observed to answer each of the given questions.
10.
In general, what effect does adding or subtracting a constant have on the graph of $f(x)=x2f(x)=x2$ ?
$f(x)=x2f(x)=x2$ vertex: ________ $f(x)=x2+2f(x)=x2+2$ vertex: ________ $f(x)=x2–4f(x)=x2–4$ vertex: ________
11.
In general, what effect does adding or subtracting a value to x before it is squared have on the graph of $f(x)=x2f(x)=x2$ ?
$f(x)=(x-2)2f(x)=(x-2)2$ vertex: ________ $f(x)=(x-4)2f(x)=(x-4)2$ vertex: ________ $f(x)=(x+3)2f(x)=(x+3)2$ vertex: ________
12.
In general, what effect does multiplying by a constant have on the graph of $f(x)=x2f(x)=x2$ ?
$f(x)=2x2f(x)=2x2$ vertex: ________ $f(x)=5x2f(x)=5x2$ vertex: ________ $f(x)=110x2f(x)=110x2$ vertex: ________
13.
In general, what effect does multiplying by a negative constant have on the graph of $f(x)=x2f(x)=x2$ ?
$f(x)=–x2f(x)=–x2$ vertex: ________ $f(x)=–3x2f(x)=–3x2$ vertex: ________ $y=(x–2)2y=(x–2)2$ vertex: ________
14.
$f(x)=(x+3)2+2f(x)=(x+3)2+2$ vertex: ________ $f(x)=(x+3)2–2f(x)=(x+3)2–2$ vertex: ________ $f(x)=(x–3)2+2f(x)=(x–3)2+2$ vertex: ________ $f(x)=(x–3)2–2f(x)=(x–3)2–2$ vertex: ________
15.
Answer each of the following based on the changes you saw in the graphs above.
Based on your observations from the previous graphs, what are the coordinates of the vertex of the parabola $f(x)=(x+200)2-67f(x)=(x+200)2-67$ ? Do not attempt to graph!
Based on your observations from the previous graphs, what are the coordinates of the vertex of the parabola $f(x)=12(x+6)2+111f(x)=12(x+6)2+111$ ? Do not attempt to graph!
16.
Fill in the blanks:
If c > 0, the graph of $y=f(x)+cy=f(x)+c$ is obtained by shifting the graph of $y=f(x)y=f(x)$ to the ________ a distance of c units. The graph of $y=f(x)–cy=f(x)–c$ is obtained by shifting the graph of $y=f(x)y=f(x)$ to the ________ a distance of c units.
If c > 0, the graph of $y=f(x)–cy=f(x)–c$ is obtained by shifting the graph of $y=f(x)y=f(x)$ to the ________ a distance of c units. The graph of $y=f(x)+cy=f(x)+c$ is obtained by shifting the graph of $y=f(x)y=f(x)$ to the ________ a distance of c units.
17.
Apply what you have learned in this skill sheet regarding transformations. Write the equation of a quadratic function that has been transformed in the each of the ways described in parts ⓐ and ⓑ below. Write equations in f(x) = form. After writing an equation check your answer using a graphing program and graph below. Be sure to label the vertex as an ordered pair. Does your graph match the description?
Flipped upside down and shifted to the right 3 units and down 2 units.
f(x) =
Stretched vertically by a factor of 4 and shifted left 6 units and up 5 units.
f(x) =
18.
Remember the basic transformations investigated in this activity apply to all basic functions. Apply what you have learned in this lab about transformations. Write the equation of a function that has been transformed in the following ways. Write equations in f(x) = form. After writing an equation check your answer using a graphing program and graph below. Be sure to label a point on the graph. Does your graph match the description?
Begin with a basic square root function. Reflect the graph over the x-axis and shift it to the right 2 units and down 1 unit.
f(x) =
Begin with an absolute value function. Stretch the graph vertically by a factor of 3 and shift it left 4 units and up 5 units.
f(x) =
Figure 1 (credit: "Misko"/Flickr)
We all know that a flat mirror enables us to see an accurate image of ourselves and whatever is behind us. When we tilt the mirror, the images we see may shift horizontally or vertically. But what happens when we bend a flexible mirror? Like a carnival funhouse mirror, it presents us with a distorted image of ourselves, stretched or compressed horizontally or vertically. In a similar way, we can distort or transform mathematical functions to better adapt them to describing objects or processes in the real world. In this section, we will take a look at several kinds of transformations.
## Graphing Functions Using Vertical and Horizontal Shifts
Often when given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs, and equations. One method we can employ is to adapt the basic graphs of the toolkit functions to build new models for a given scenario. There are systematic ways to alter functions to construct appropriate models for the problems we are trying to solve.
### Identifying Vertical Shifts
One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function $g(x)=f(x)+k, g(x)=f(x)+k,$ the function $f( x ) f( x )$ is shifted vertically $k k$ units. See Figure 2 for an example.
Figure 2 Vertical shift by $k=1 k=1$ of the cube root function $f(x)= x 3 . f(x)= x 3 .$
To help you visualize the concept of a vertical shift, consider that $y=f( x ). y=f( x ).$ Therefore, $f( x )+k f( x )+k$ is equivalent to $y+k. y+k.$ Every unit of $y y$ is replaced by $y+k, y+k,$ so the y-value increases or decreases depending on the value of $k. k.$ The result is a shift upward or downward.
## Vertical Shift
Given a function $f( x ), f( x ),$ a new function $g(x)=f(x)+k, g(x)=f(x)+k,$ where $k k$ is a constant, is a vertical shift of the function $f( x ). f( x ).$ All the output values change by $k k$ units. If $k k$ is positive, the graph will shift up. If $k k$ is negative, the graph will shift down.
## Example 1
### Adding a Constant to a Function
To regulate temperature in a green building, airflow vents near the roof open and close throughout the day. Figure 3 shows the area of open vents $V V$ (in square feet) throughout the day in hours after midnight, $t. t.$ During the summer, the facilities manager decides to try to better regulate temperature by increasing the amount of open vents by 20 square feet throughout the day and night. Sketch a graph of this new function.
Figure 3
## How To
Given a tabular function, create a new row to represent a vertical shift.
1. Identify the output row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down.
## Example 2
### Shifting a Tabular Function Vertically
A function $f( x ) f( x )$ is given in Table 2. Create a table for the function $g(x)=f(x)−3. g(x)=f(x)−3.$
$x x$ 2 4 6 8 $f(x) f(x)$ 1 3 7 11
Table 2
### Analysis
As with the earlier vertical shift, notice the input values stay the same and only the output values change.
## Try It #1
The function $h(t)=−4.9 t 2 +30t h(t)=−4.9 t 2 +30t$ gives the height $h h$ of a ball (in meters) thrown upward from the ground after $t t$ seconds. Suppose the ball was instead thrown from the top of a 10-m building. Relate this new height function $b(t) b(t)$ to $h(t), h(t),$ and then find a formula for $b(t). b(t).$
### Identifying Horizontal Shifts
We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift, shown in Figure 5.
Figure 5 Horizontal shift of the function $f(x)= x 3 . f(x)= x 3 .$ Note that $(x+1)(x+1)$ means $h=–1 h=–1$, which shifts the graph to the left, that is, towards negative values of $x. x.$
For example, if $f(x)= x 2 , f(x)= x 2 ,$ then $g(x)= (x−2) 2 g(x)= (x−2) 2$ is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in $f. f.$
## Horizontal Shift
Given a function $f, f,$ a new function $g( x )=f( x−h ), g( x )=f( x−h ),$ where $h h$ is a constant, is a horizontal shift of the function $f. f.$ If $h h$ is positive, the graph will shift right. If $h h$ is negative, the graph will shift left.
## Example 3
### Adding a Constant to an Input
Returning to our building airflow example from Figure 3, suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function.
### Analysis
Note that $V(t+2) V(t+2)$ has the effect of shifting the graph to the left.
Horizontal changes or “inside changes” affect the domain of a function (the input) instead of the range and often seem counterintuitive. The new function $F( t ) F( t )$ uses the same outputs as $V( t ), V( t ),$ but matches those outputs to inputs 2 hours earlier than those of $V( t ). V( t ).$ Said another way, we must add 2 hours to the input of $V V$ to find the corresponding output for $F:F(t)=V(t+2). F:F(t)=V(t+2).$
## How To
Given a tabular function, create a new row to represent a horizontal shift.
1. Identify the input row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each input cell.
## Example 4
### Shifting a Tabular Function Horizontally
A function $f(x) f(x)$ is given in Table 4. Create a table for the function $g(x)=f(x−3). g(x)=f(x−3).$
$x x$ 2 4 6 8 $f(x) f(x)$ 1 3 7 11
Table 4
### Analysis
Figure 7 represents both of the functions. We can see the horizontal shift in each point.
Figure 7
## Example 5
### Identifying a Horizontal Shift of a Toolkit Function
Figure 8 represents a transformation of the toolkit function $f(x)= x 2 . f(x)= x 2 .$ Relate this new function $g(x) g(x)$ to $f(x), f(x),$ and then find a formula for $g(x). g(x).$
Figure 8
### Analysis
To determine whether the shift is $+2 +2$ or $−2 −2$ , consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, $f(0)=0. f(0)=0.$ In our shifted function, $g(2)=0. g(2)=0.$ To obtain the output value of 0 from the function $f, f,$ we need to decide whether a plus or a minus sign will work to satisfy $g(2)=f(x−2)=f(0)=0. g(2)=f(x−2)=f(0)=0.$ For this to work, we will need to subtract 2 units from our input values.
## Example 6
### Interpreting Horizontal versus Vertical Shifts
The function $G(m) G(m)$ gives the number of gallons of gas required to drive $m m$ miles. Interpret $G(m)+10 G(m)+10$ and $G(m+10). G(m+10).$
## Try It #2
Given the function $f(x)= x , f(x)= x ,$ graph the original function $f(x) f(x)$ and the transformation $g(x)=f(x+2) g(x)=f(x+2)$ on the same axes. Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units?
### Combining Vertical and Horizontal Shifts
Now that we have two transformations, we can combine them. Vertical shifts are outside changes that affect the output (y-) values and shift the function up or down. Horizontal shifts are inside changes that affect the input (x-) values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and left or right.
## How To
Given a function and both a vertical and a horizontal shift, sketch the graph.
1. Identify the vertical and horizontal shifts from the formula.
2. The vertical shift results from a constant added to the output. Move the graph up for a positive constant and down for a negative constant.
3. The horizontal shift results from a constant added to the input. Move the graph left for a positive constant and right for a negative constant.
4. Apply the shifts to the graph in either order.
## Example 7
### Graphing Combined Vertical and Horizontal Shifts
Given $f(x)=| x |, f(x)=| x |,$ sketch a graph of $h(x)=f(x+1)−3. h(x)=f(x+1)−3.$
## Try It #3
Given $f(x)=| x |, f(x)=| x |,$ sketch a graph of $h(x)=f(x−2)+4. h(x)=f(x−2)+4.$
## Example 8
### Identifying Combined Vertical and Horizontal Shifts
Write a formula for the graph shown in Figure 11, which is a transformation of the toolkit square root function.
Figure 11
### Analysis
Note that this transformation has changed the domain and range of the function. This new graph has domain $[1,∞) [1,∞)$ and range $[2,∞). [2,∞).$
## Try It #4
Write a formula for a transformation of the toolkit reciprocal function $f( x )= 1 x f( x )= 1 x$ that shifts the function’s graph one unit to the right and one unit up.
## Graphing Functions Using Reflections about the Axes
Another transformation that can be applied to a function is a reflection over the x- or y-axis. A vertical reflection reflects a graph vertically across the x-axis, while a horizontal reflection reflects a graph horizontally across the y-axis. The reflections are shown in Figure 12.
Figure 12 Vertical and horizontal reflections of a function.
Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis. The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis.
## Reflections
Given a function $f(x), f(x),$ a new function $g(x)=−f(x) g(x)=−f(x)$ is a vertical reflection of the function $f(x), f(x),$ sometimes called a reflection about (or over, or through) the x-axis.
Given a function $f(x), f(x),$ a new function $g(x)=f(−x) g(x)=f(−x)$ is a horizontal reflection of the function $f(x), f(x),$ sometimes called a reflection about the y-axis.
## How To
Given a function, reflect the graph both vertically and horizontally.
1. Multiply all outputs by –1 for a vertical reflection. The new graph is a reflection of the original graph about the x-axis.
2. Multiply all inputs by –1 for a horizontal reflection. The new graph is a reflection of the original graph about the y-axis.
## Example 9
### Reflecting a Graph Horizontally and Vertically
Reflect the graph of $s(t)= t s(t)= t$ (a) vertically and (b) horizontally.
## Try It #5
Reflect the graph of $f(x)=|x−1| f(x)=|x−1|$ (a) vertically and (b) horizontally.
## Example 10
### Reflecting a Tabular Function Horizontally and Vertically
A function $f(x) f(x)$ is given as Table 6. Create a table for the functions below.
1. $g(x)=−f(x) g(x)=−f(x)$
2. $h(x)=f(−x) h(x)=f(−x)$
$x x$ 2 4 6 8 $f(x) f(x)$ 1 3 7 11
Table 6
## Try It #6
A function $f(x) f(x)$ is given as Table 9. Create a table for the functions below.
1. $g(x)=−f(x) g(x)=−f(x)$
2. $h(x)=f(−x) h(x)=f(−x)$
$x x$ −2 0 2 4 $f(x) f(x)$ 5 10 15 20
Table 9
## Example 11
### Applying a Learning Model Equation
A common model for learning has an equation similar to $k(t)=− 2 −t +1, k(t)=− 2 −t +1,$ where $k k$ is the percentage of mastery that can be achieved after $t t$ practice sessions. This is a transformation of the function $f(t)= 2 t f(t)= 2 t$ shown in Figure 15. Sketch a graph of $k(t). k(t).$
Figure 15
### Analysis
As a model for learning, this function would be limited to a domain of $t≥0, t≥0,$ with corresponding range $[0,1). [0,1).$
## Try It #7
Given the toolkit function $f(x)= x 2 , f(x)= x 2 ,$ graph $g(x)=−f(x) g(x)=−f(x)$ and $h(x)=f(−x). h(x)=f(−x).$ Take note of any surprising behavior for these functions.
## Determining Even and Odd Functions
Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the toolkit functions $f(x)= x 2 f(x)= x 2$ or $f(x)=| x | f(x)=| x |$ will result in the original graph. We say that these types of graphs are symmetric about the y-axis. A function whose graph is symmetric about the y-axis is called an even function.
If the graphs of $f(x)= x 3 f(x)= x 3$ or $f(x)= 1 x f(x)= 1 x$ were reflected over both axes, the result would be the original graph, as shown in Figure 17.
Figure 17 (a) The cubic toolkit function (b) Horizontal reflection of the cubic toolkit function (c) Horizontal and vertical reflections reproduce the original cubic function.
We say that these graphs are symmetric about the origin. A function with a graph that is symmetric about the origin is called an odd function.
Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, $f(x)= 2 x f(x)= 2 x$ is neither even nor odd. Also, the only function that is both even and odd is the constant function $f(x)=0. f(x)=0.$
## Even and Odd Functions
A function is called an even function if for every input $x x$
$f(x)=f(−x) f(x)=f(−x)$
The graph of an even function is symmetric about the $y- y-$ axis.
A function is called an odd function if for every input $x x$
$f(x)=−f(−x) f(x)=−f(−x)$
The graph of an odd function is symmetric about the origin.
## How To
Given the formula for a function, determine if the function is even, odd, or neither.
1. Determine whether the function satisfies $f(x)=f(−x). f(x)=f(−x).$ If it does, it is even.
2. Determine whether the function satisfies $f(x)=−f(−x). f(x)=−f(−x).$ If it does, it is odd.
3. If the function does not satisfy either rule, it is neither even nor odd.
## Example 12
### Determining whether a Function Is Even, Odd, or Neither
Is the function $f(x)= x 3 +2x f(x)= x 3 +2x$ even, odd, or neither?
### Analysis
Consider the graph of $f f$ in Figure 18. Notice that the graph is symmetric about the origin. For every point $( x,y ) ( x,y )$ on the graph, the corresponding point $( −x,−y ) ( −x,−y )$ is also on the graph. For example, (1, 3) is on the graph of $f, f,$ and the corresponding point $(−1,−3) (−1,−3)$ is also on the graph.
Figure 18
## Try It #8
Is the function $f(s)= s 4 +3 s 2 +7 f(s)= s 4 +3 s 2 +7$ even, odd, or neither?
## Graphing Functions Using Stretches and Compressions
Adding a constant to the inputs or outputs of a function changed the position of a graph with respect to the axes, but it did not affect the shape of a graph. We now explore the effects of multiplying the inputs or outputs by some quantity.
We can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each change has a specific effect that can be seen graphically.
### Vertical Stretches and Compressions
When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is between 0 and 1, we get a vertical compression. Figure 19 shows a function multiplied by constant factors 2 and 0.5 and the resulting vertical stretch and compression.
Figure 19 Vertical stretch and compression
## Vertical Stretches and Compressions
Given a function $f(x), f(x),$ a new function $g(x)=af(x), g(x)=af(x),$ where $a a$ is a constant, is a vertical stretch or vertical compression of the function $f(x). f(x).$
• If $a>1, a>1,$ then the graph will be stretched.
• If $0 then the graph will be compressed.
• If $a<0, a<0,$ then there will be combination of a vertical stretch or compression with a vertical reflection.
## How To
Given a function, graph its vertical stretch.
1. Identify the value of $a. a.$
2. Multiply all range values by $a. a.$
3. If $a>1, a>1,$ the graph is stretched by a factor of $a. a.$
If $0 the graph is compressed by a factor of $a. a.$
If $a<0, a<0,$ the graph is either stretched or compressed and also reflected about the x-axis.
## Example 13
### Graphing a Vertical Stretch
A function $P( t ) P( t )$ models the population of fruit flies. The graph is shown in Figure 20.
Figure 20
A scientist is comparing this population to another population, $Q, Q,$ whose growth follows the same pattern, but is twice as large. Sketch a graph of this population.
## How To
Given a tabular function and assuming that the transformation is a vertical stretch or compression, create a table for a vertical compression.
1. Determine the value of $a. a.$
2. Multiply all of the output values by $a. a.$
## Example 14
### Finding a Vertical Compression of a Tabular Function
A function $f f$ is given as Table 10. Create a table for the function $g(x)= 1 2 f(x). g(x)= 1 2 f(x).$
$x x$ 2 4 6 8 $f(x) f(x)$ 1 3 7 11
Table 10
### Analysis
The result is that the function $g(x) g(x)$ has been compressed vertically by $1 2 . 1 2 .$ Each output value is divided in half, so the graph is half the original height.
## Try It #9
A function $f f$ is given as Table 12. Create a table for the function $g(x)= 3 4 f(x). g(x)= 3 4 f(x).$
$x x$ 2 4 6 8 $f(x) f(x)$ 12 16 20 0
Table 12
## Example 15
### Recognizing a Vertical Stretch
The graph in Figure 22 is a transformation of the toolkit function $f(x)= x 3 . f(x)= x 3 .$ Relate this new function $g(x) g(x)$ to $f(x), f(x),$ and then find a formula for $g(x). g(x).$
Figure 22
## Try It #10
Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units.
### Horizontal Stretches and Compressions
Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function.
Figure 23
Given a function $y=f(x), y=f(x),$ the form $y=f(bx) y=f(bx)$ results in a horizontal stretch or compression. Consider the function $y= x 2 . y= x 2 .$ Observe Figure 23. The graph of $y= ( 0.5x ) 2 y= ( 0.5x ) 2$ is a horizontal stretch of the graph of the function $y= x 2 y= x 2$ by a factor of 2. The graph of $y= ( 2x ) 2 y= ( 2x ) 2$ is a horizontal compression of the graph of the function $y= x 2 y= x 2$ by a factor of $1212$.
## Horizontal Stretches and Compressions
Given a function $f(x), f(x),$ a new function $g(x)=f(bx), g(x)=f(bx),$ where $b b$ is a constant, is a horizontal stretch or horizontal compression of the function $f(x). f(x).$
• If $b>1, b>1,$ then the graph will be compressed by $1 b . 1 b .$
• If $0 then the graph will be stretched by $1 b . 1 b .$
• If $b<0, b<0,$ then there will be combination of a horizontal stretch or compression with a horizontal reflection.
## How To
Given a description of a function, sketch a horizontal compression or stretch.
1. Write a formula to represent the function.
2. Set $g(x)=f(bx) g(x)=f(bx)$ where $b>1 b>1$ for a compression or $0 for a stretch.
## Example 16
### Graphing a Horizontal Compression
Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, $R, R,$ will progress in 1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population.
### Analysis
Note that the effect on the graph is a horizontal compression where all input values are half of their original distance from the vertical axis.
## Example 17
### Finding a Horizontal Stretch for a Tabular Function
A function $f(x) f(x)$ is given as Table 13. Create a table for the function $g(x)=f( 1 2 x ). g(x)=f( 1 2 x ).$
$x x$ 2 4 6 8 $f(x) f(x)$ 1 3 7 11
Table 13
### Analysis
Because each input value has been doubled, the result is that the function $g(x) g(x)$ has been stretched horizontally by a factor of 2.
## Example 18
### Recognizing a Horizontal Compression on a Graph
Relate the function $g(x) g(x)$ to $f(x) f(x)$ in Figure 26.
Figure 26
### Analysis
Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of $1 4 1 4$ in our function: $f( 1 4 x ). f( 1 4 x ).$ This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching.
## Try It #11
Write a formula for the toolkit square root function horizontally stretched by a factor of 3.
## Performing a Sequence of Transformations
When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the original function gets stretched when we stretch first.
When we see an expression such as $2f(x)+3, 2f(x)+3,$ which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of $f(x), f(x),$ we first multiply by 2, causing the vertical stretch, and then add 3, causing the vertical shift. In other words, multiplication before addition.
Horizontal transformations are a little trickier to think about. When we write $g(x)=f(2x+3), g(x)=f(2x+3),$ for example, we have to think about how the inputs to the function $g g$ relate to the inputs to the function $f. f.$ Suppose we know $f(7)=12. f(7)=12.$ What input to $g g$ would produce that output? In other words, what value of $x x$ will allow $g(x)=f(2x+3)=12? g(x)=f(2x+3)=12?$ We would need $2x+3=7. 2x+3=7.$ To solve for $x, x,$ we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a horizontal compression.
This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function.
$f(bx+p)=f( b( x+ p b ) ) f(bx+p)=f( b( x+ p b ) )$
Let’s work through an example.
$f( x )= ( 2x+4 ) 2 f( x )= ( 2x+4 ) 2$
We can factor out a 2.
$f( x )= ( 2( x+2 ) ) 2 f( x )= ( 2( x+2 ) ) 2$
Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way allows us to horizontally stretch first and then shift horizontally.
## Combining Transformations
When combining vertical transformations written in the form $af(x)+k, af(x)+k,$ first vertically stretch by $a a$ and then vertically shift by $k. k.$
When combining horizontal transformations written in the form $f(bx-h), f(bx-h),$ first horizontally shift by $hb hb$ and then horizontally stretch by $1 b . 1 b .$
When combining horizontal transformations written in the form $f(b(x-h)), f(b(x-h)),$ first horizontally stretch by $1 b 1 b$ and then horizontally shift by $h. h.$
Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are performed first.
## Example 19
### Finding a Triple Transformation of a Tabular Function
Given Table 15 for the function $f(x), f(x),$ create a table of values for the function $g(x)=2f(3x)+1. g(x)=2f(3x)+1.$
$x x$ 6 12 18 24 $f(x) f(x)$ 10 14 15 17
Table 15
## Example 20
### Finding a Triple Transformation of a Graph
Use the graph of $f( x ) f( x )$ in Figure 27 to sketch a graph of $k(x)=f( 1 2 x+1 )−3. k(x)=f( 1 2 x+1 )−3.$
Figure 27
## Media
Access this online resource for additional instruction and practice with transformation of functions.
## 3.5 Section Exercises
### Verbal
1.
When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal shift from a vertical shift?
2.
When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal stretch from a vertical stretch?
3.
When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal compression from a vertical compression?
4.
When examining the formula of a function that is the result of multiple transformations, how can you tell a reflection with respect to the x-axis from a reflection with respect to the y-axis?
5.
How can you determine whether a function is odd or even from the formula of the function?
### Algebraic
For the following exercises, write a formula for the function obtained when the graph is shifted as described.
6.
$f(x)= x f(x)= x$ is shifted up 1 unit and to the left 2 units.
7.
$f(x)=| x | f(x)=| x |$ is shifted down 3 units and to the right 1 unit.
8.
$f(x)= 1 x f(x)= 1 x$ is shifted down 4 units and to the right 3 units.
9.
$f(x)= 1 x 2 f(x)= 1 x 2$ is shifted up 2 units and to the left 4 units.
For the following exercises, describe how the graph of the function is a transformation of the graph of the original function $f. f.$
10.
$y=f(x−49) y=f(x−49)$
11.
$y=f(x+43) y=f(x+43)$
12.
$y=f(x+3) y=f(x+3)$
13.
$y=f(x−4) y=f(x−4)$
14.
$y=f(x)+5 y=f(x)+5$
15.
$y=f(x)+8 y=f(x)+8$
16.
$y=f(x)−2 y=f(x)−2$
17.
$y=f(x)−7 y=f(x)−7$
18.
$y=f(x−2)+3 y=f(x−2)+3$
19.
$y=f(x+4)−1 y=f(x+4)−1$
For the following exercises, determine the interval(s) on which the function is increasing and decreasing.
20.
$f(x)=4 (x+1) 2 −5 f(x)=4 (x+1) 2 −5$
21.
$g(x)=5 (x+3) 2 −2 g(x)=5 (x+3) 2 −2$
22.
$a(x)= −x+4 a(x)= −x+4$
23.
$k(x)=−3 x −1 k(x)=−3 x −1$
### Graphical
For the following exercises, use the graph of $f(x)= 2 x f(x)= 2 x$ shown in Figure 31 to sketch a graph of each transformation of $f(x). f(x).$
Figure 31
24.
$g(x)= 2 x +1 g(x)= 2 x +1$
25.
$h(x)= 2 x −3 h(x)= 2 x −3$
26.
$w(x)= 2 x−1 w(x)= 2 x−1$
For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions.
27.
$f(t)= (t+1) 2 −3 f(t)= (t+1) 2 −3$
28.
$h(x)=|x−1|+4 h(x)=|x−1|+4$
29.
$k(x)= (x−2) 3 −1 k(x)= (x−2) 3 −1$
30.
$m(t)=3+ t+2 m(t)=3+ t+2$
### Numeric
31.
Tabular representations for the functions $f,g, f,g,$ and $h h$ are given below. Write $g(x) g(x)$ and $h(x) h(x)$ as transformations of $f(x). f(x).$
$x x$ −2 −1 0 1 2 $f(x) f(x)$ −2 −1 −3 1 2
$x x$ −1 0 1 2 3 $g(x) g(x)$ −2 −1 −3 1 2
$x x$ −2 −1 0 1 2 $h(x) h(x)$ −1 0 −2 2 3
32.
Tabular representations for the functions $f,g, f,g,$ and $h h$ are given below. Write $g(x) g(x)$ and $h(x) h(x)$ as transformations of $f(x). f(x).$
$x x$ −2 −1 0 1 2 $f(x) f(x)$ −1 −3 4 2 1
$x x$ −3 −2 −1 0 1 $g(x) g(x)$ −1 −3 4 2 1
$x x$ −2 −1 0 1 2 $h(x) h(x)$ −2 −4 3 1 0
For the following exercises, write an equation for each graphed function by using transformations of the graphs of one of the toolkit functions.
33.
34.
35.
36.
37.
38.
39.
40.
For the following exercises, use the graphs of transformations of the square root function to find a formula for each of the functions.
41.
42.
For the following exercises, use the graphs of the transformed toolkit functions to write a formula for each of the resulting functions.
43.
44.
45.
46.
For the following exercises, determine whether the function is odd, even, or neither.
47.
$f(x)=3 x 4 f(x)=3 x 4$
48.
$g(x)= x g(x)= x$
49.
$h(x)= 1 x +3x h(x)= 1 x +3x$
50.
$f(x)= (x−2) 2 f(x)= (x−2) 2$
51.
$g(x)=2 x 4 g(x)=2 x 4$
52.
$h(x)=2x− x 3 h(x)=2x− x 3$
For the following exercises, describe how the graph of each function is a transformation of the graph of the original function $f. f.$
53.
$g(x)=−f(x) g(x)=−f(x)$
54.
$g(x)=f(−x) g(x)=f(−x)$
55.
$g(x)=4f(x) g(x)=4f(x)$
56.
$g(x)=6f(x) g(x)=6f(x)$
57.
$g(x)=f(5x) g(x)=f(5x)$
58.
$g(x)=f(2x) g(x)=f(2x)$
59.
$g(x)=f( 1 3 x ) g(x)=f( 1 3 x )$
60.
$g(x)=f( 1 5 x ) g(x)=f( 1 5 x )$
61.
$g(x)=3f( −x ) g(x)=3f( −x )$
62.
$g(x)=−f(3x) g(x)=−f(3x)$
For the following exercises, write a formula for the function $g g$ that results when the graph of a given toolkit function is transformed as described.
63.
The graph of $f(x)=|x| f(x)=|x|$ is reflected over the $y y$ -axis and horizontally compressed by a factor of $1 4 1 4$ .
64.
The graph of $f(x)= x f(x)= x$ is reflected over the $x x$ -axis and horizontally stretched by a factor of 2.
65.
The graph of $f(x)= 1 x 2 f(x)= 1 x 2$ is vertically compressed by a factor of $1 3 , 1 3 ,$ then shifted to the left 2 units and down 3 units.
66.
The graph of $f(x)= 1 x f(x)= 1 x$ is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units.
67.
The graph of $f(x)= x 2 f(x)= x 2$ is vertically compressed by a factor of $1 2 , 1 2 ,$ then shifted to the right 5 units and up 1 unit.
68.
The graph of $f(x)= x 2 f(x)= x 2$ is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units.
For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation.
69.
$g(x)=4 (x+1) 2 −5 g(x)=4 (x+1) 2 −5$
70.
$g(x)=5 (x+3) 2 −2 g(x)=5 (x+3) 2 −2$
71.
$h(x)=−2|x−4|+3 h(x)=−2|x−4|+3$
72.
$k(x)=−3 x −1 k(x)=−3 x −1$
73.
$m(x)= 1 2 x 3 m(x)= 1 2 x 3$
74.
$n(x)= 1 3 |x−2| n(x)= 1 3 |x−2|$
75.
$p( x )= ( 1 3 x ) 3 −3 p( x )= ( 1 3 x ) 3 −3$
76.
$q( x )= ( 1 4 x ) 3 +1 q( x )= ( 1 4 x ) 3 +1$
77.
$a(x)= −x+4 a(x)= −x+4$
For the following exercises, use the graph in Figure 32 to sketch the given transformations.
Figure 32
78.
$g(x)=f(x)−2 g(x)=f(x)−2$
79.
$g(x)=−f(x) g(x)=−f(x)$
80.
$g(x)=f(x+1) g(x)=f(x+1)$
81.
$g(x)=f(x−2) g(x)=f(x−2)$
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Comparing probability and dimension forumulas, concidence?
So we know that if given two finite vector spaces $V,W$ then the $\dim(V\cup W)=\dim(V)+\dim(W)-\dim(V\cap W)$ This curiously corresponds with the Probability formula of, given two probabilities $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
Question, is this a mere coincidence or is there something deeper going on? Probabilities are always numbers between 0 and 1, the dimension of a vector space is, I assume, always a natural number. So anything special going on?
-
I wonder what it is that you call the dimension of $V\cup W$. – Did Oct 14 '12 at 22:29
Perhaps what's going on is that for any measure $\mu$ you get this equation, and dimension is a sort of counting measure on vector spaces. – Gerry Myerson Oct 14 '12 at 22:29
@did, remind me to look for hidden meanings in your comments in the future. – Gerry Myerson Oct 14 '12 at 22:51
3 Answers
WARNING: This does not work for sums of more than two subspaces.
$$\dim(U+V+W)$$ may differ from $$\dim U+\dim V+\dim W-\dim(U\cap V)-\dim(U\cap W)-\dim(V\cap W)+\dim(U\cap V\cap W).$$ For example, it doesn't work if $U,V,W$ are distinct one-dimensional subspaces of a two-dimensional space.
-
This is a consequence of the additivity property, both of the dimension of a subspace and of probability, as Gerry commented, they both are some kind of measures.
If $U,V$ are 'disjoint' subspaces (i.e. $U\cap V=\{0\}$), then $\dim(U+V)=\dim U+\dim V$.
If $A,B$ are disjoint 'events' (subsets of the probability space), then $P(A\cup B)=P(A)+P(B)$.
This is the basic rule for area/volume, and any measure. There are also Projection valued measures, which can be used for some kind of probability using (projections to) subspaces of a Hilbert space..
-
I think this point of view may mislead people into thinking that this works for more than two subspaces, i.e. that in general the dimension of the sum of three subspaces is $\dim U+\dim V+\dim W-\dim(U\cap V)-\dim(U\cap W)-\dim(V\cap W)+\dim(U\cap V\cap W)$. But that is false. – Michael Hardy Oct 15 '12 at 13:05
I guess you are looking for the Inclusion–exclusion principle.
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# What does the volume of water mean?
The amount of space occupied by a three-dimensional object or region of space, expressed in cubic units. b. The capacity of such a region or of a specified container, expressed in cubic units. a. Amount; quantity: a low volume of business; a considerable volume of lumber.
Similarly, it is asked, how do I measure volume of water?
Length times width gives the surface area of the pool. Multiplying that by the average depth gives the volume in cubic feet. Since there are 7.5 gallons in each cubic foot, multiply the cubic feet of the pool by 7.5 to arrive at the volume of the pool (expressed in gallons).
How do you find the volume of a liquid?
Procedure
• Find the mass of an empty graduated cylinder.
• Pour 100 mL of water into the graduated cylinder.
• Weigh the graduated cylinder with the water in it.
• Find the mass of only the water by subtracting the mass of the empty graduated cylinder.
• Use the mass and volume of the water to calculate density.
• How do I calculate how many Litres?
You can use the conversion 1 liter = 1,000 cubic centimeters. To convert from liters to cubic centimeters, you would multiply by 1,000. For example, if a cube has a volume of 34 liters, to find the volume in cubic centimeters, multiply by 1,000: 34 x 1,000 = 34,000 cubic centimeters.
## How do you calculate volume in liters?
You can use the conversion 1 liter = 1,000 cubic centimeters. To convert from liters to cubic centimeters, you would multiply by 1,000. For example, if a cube has a volume of 34 liters, to find the volume in cubic centimeters, multiply by 1,000: 34 x 1,000 = 34,000 cubic centimeters.
## What is volume equal to?
You know that density equals mass divided by volume. That relationship can then be expressed in many ways by carefully rearranging the equation. Mass equals volume times density. Volume equals mass divided by density.
## What is the volume of a book mean?
A volume is a physical book. It may be printed or handwritten. The term is commonly used to identify a single book that is part of a larger collection. A publisher may also separately publish a volume out of previously published issues; this is common with graphic novels.
## What is the volume of a stock mean?
In the context of a single stock trading on a stock exchange, the volume is commonly reported as the number of shares that changed hands during a given day. The average volume of a security over a longer period of time is the total amount traded in that period, divided by the length of the period.
## How do you figure out the volume of an object?
Measure the length, width, and height of the object. Multiply these dimensions together to get the volume. To find the volume of a cylinder, multiply the square of its radius by pi, or 3.14. Then multiply this value by the height of the cylinder.
## What does it mean when water is displaced?
In fluid mechanics, displacement occurs when an object is immersed in a fluid, pushing it out of the way and taking its place. Thus buoyancy is expressed through Archimedes’ principle, which states that the weight of the object is reduced by its volume multiplied by the density of the fluid.
## What is the displacement of water?
Water Displacement. Download as SWF (.swf) Volume is a measure of the amount of space an object takes up. When a cylinder is submerged in the water it pushes water out of the way. If you measure the amount the water level increases, you can find the volume of the water pushed out of the way.
## What is the definition of volume in science?
Volume is the quantity of three-dimensional space occupied by a liquid, solid, or gas. Common units used to express volume include liters, cubic meters, gallons, milliliters, teaspoons, and ounces, though many other units exist.
## How do I measure volume of water?
Length times width gives the surface area of the pool. Multiplying that by the average depth gives the volume in cubic feet. Since there are 7.5 gallons in each cubic foot, multiply the cubic feet of the pool by 7.5 to arrive at the volume of the pool (expressed in gallons).
## What is the measure of volume?
Volume. Volume is a measure of the amount of space that a substance or an object takes up. The basic SI unit for volume is the cubic meter (m3), but smaller volumes may be measured in cm3, and liquids may be measured in liters (L) or milliliters (mL).
## How do you find the mass of water?
Procedure
• Find the mass of an empty graduated cylinder.
• Pour 100 mL of water into the graduated cylinder.
• Weigh the graduated cylinder with the water in it.
• Find the mass of only the water by subtracting the mass of the empty graduated cylinder.
• Use the mass and volume of the water to calculate density.
• ## Is the water displacement method more accurate?
The change in water level equals the volume of the submerged object. This method is more accurate than measuring water that has spilled out an overflowing beaker. You may need to remind students of the need for accuracy, not only in the weighing of the objects, but also in measuring the volume of displaced water.
## What is the definition of volume in math?
Math Term Definition. Volume. Volume is the measure of the amount of space inside of a solid figure, like a cube, ball, cylinder or pyramid. It’s units are always “cubic”, that is, the number of little element cubes that fit inside the figure.
## How do you determine the density of an object from a volume and mass?
Density is the mass of an object divided by its volume. Density often has units of grams per cubic centimeter (g/cm3). Remember, grams is a mass and cubic centimeters is a volume (the same volume as 1 milliliter). Density is a fundamental concept in the sciences; you will see it throughout your studies.
## How do you find the volume?
To find the volume of any cube you need to know the length, width and height. The formula to find the volume multiplies the length by the width by the height. The good news for a cube is that the measure of each of these dimensions is exactly the same. Therefore, you can multiply the length of any side three times.
## What is the formula for water displacement?
A submerged object displaces a volume of liquid equal to the volume of the object. One milliliter (1 mL) of water has a volume of 1 cubic centimeter (1cm3). Density equals the mass of the object divided by its volume; D = m/v.
## What does volume mean in sound?
Definition of volume. 1 : the degree of loudness or the intensity of a sound; also : loudness. 2 : the amount of space occupied by a three-dimensional object as measured in cubic units (such as quarts or liters) : cubic capacity — see Metric System Table, Weights and Measures Table.
## What is the mass of a molecule of water?
The gram molecular mass of water is 18 grams per mole. This isbecause a water molecule contains two hydrogen atoms (one protoneach) and one oxygen atom (8 protons and 8 neutrons), for a totalof 18 nucleons. Avogadro’s number is the number of water moleculesneeded to obtain a mass of 18 grams.
## What is the definition of volume in music?
Volume. Volume. Definition and background: Musically speaking, volume can be defined as the loudness of a sound, as measured in decibels. A common, non-technical term meaning sound pressure level, and also meaning audio voltage level.
## What is mass?
Mass (symbolized m) is a dimensionless quantity representing the amount of matter in a particle or object. The standard unit of mass in the International System (SI) is the kilogram (kg). The mass of an object can be calculated if the force and the acceleration are known.
Originally posted 2022-03-31 05:06:30.
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# Rabbit Algebra
(Difference between revisions)
Revision as of 03:11, 8 July 2007 (edit) (→Fun Facts)← Older edit Revision as of 03:12, 8 July 2007 (edit) (undo) (→Fun Facts)Newer edit → Line 5: Line 5: *If the problem seen in Peasant's Quest Preview is solved, it would yield $x = \pm 9$. *If the problem seen in Peasant's Quest Preview is solved, it would yield $x = \pm 9$. *The score, $\log_x 14$, is dependent on ''x''. *The score, $\log_x 14$, is dependent on ''x''. - **Setting ''x'' to 14 or any integer root of 14 (e.g. square root, cube root, etc) yields an interger score. + **Setting ''x'' to 14 or any integer root of 14 (e.g. square root, cube root, etc) yields an integer score. **If $x = 9$ ''(see above)'', the score is 1.2011. **If $x = 9$ ''(see above)'', the score is 1.2011. **If $x = - 9$, the score is the [[Wikipedia:Complex number|complex number]] $0.3945 - 0.5641i$. **If $x = - 9$, the score is the [[Wikipedia:Complex number|complex number]] $0.3945 - 0.5641i$.
## Revision as of 03:12, 8 July 2007
"Solve for X!!"
Rabbit Algebra is a Videlectrix educational title, shown in the Peasant's Quest Preview. It is a "game" where algebra is taught by a rabbit. This is a parody of The Learning Company's Reader Rabbit and Math Rabbit, where various animals teach elementary subjects. It is not listed on Videlectrix's website and cannot be played.
## Fun Facts
• If the problem seen in Peasant's Quest Preview is solved, it would yield $x = \pm 9$.
• The score, logx14, is dependent on x.
• Setting x to 14 or any integer root of 14 (e.g. square root, cube root, etc) yields an integer score.
• If x = 9 (see above), the score is 1.2011.
• If x = − 9, the score is the complex number 0.3945 − 0.5641i.
• If x = 3 (see below), the score is 2.4022.
• The score cannot be zero.
• In the number of mans, 2x = 6, x is equal to 3.
• The version on HomestarRunner.com PAY PLUS! has the problem x2 + 28 = 44, which yields $x = \pm 4$.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dchrval Structured version Unicode version
Theorem dchrval 23352
Description: Value of the group of Dirichlet characters. (Contributed by Mario Carneiro, 18-Apr-2016.)
Hypotheses
Ref Expression
dchrval.g DChr
dchrval.z ℤ/n
dchrval.b
dchrval.u Unit
dchrval.n
dchrval.d mulGrp MndHom mulGrpfld
Assertion
Ref Expression
dchrval
Distinct variable groups: , , , , ,
Allowed substitution hints: () ()
Proof of Theorem dchrval
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 dchrval.g . 2 DChr
2 df-dchr 23351 . . . 4 DChr ℤ/n mulGrp MndHom mulGrpfld Unit
32a1i 11 . . 3 DChr ℤ/n mulGrp MndHom mulGrpfld Unit
4 fvex 5881 . . . . 5 ℤ/n
54a1i 11 . . . 4 ℤ/n
6 ovex 6319 . . . . . . 7 mulGrp MndHom mulGrpfld
76rabex 4603 . . . . . 6 mulGrp MndHom mulGrpfld Unit
87a1i 11 . . . . 5 ℤ/n mulGrp MndHom mulGrpfld Unit
9 dchrval.d . . . . . . . . . . 11 mulGrp MndHom mulGrpfld
109ad2antrr 725 . . . . . . . . . 10 ℤ/n mulGrp MndHom mulGrpfld
11 simpr 461 . . . . . . . . . . . . . . . . 17
1211fveq2d 5875 . . . . . . . . . . . . . . . 16 ℤ/n ℤ/n
13 dchrval.z . . . . . . . . . . . . . . . 16 ℤ/n
1412, 13syl6reqr 2527 . . . . . . . . . . . . . . 15 ℤ/n
1514eqeq2d 2481 . . . . . . . . . . . . . 14 ℤ/n
1615biimpar 485 . . . . . . . . . . . . 13 ℤ/n
1716fveq2d 5875 . . . . . . . . . . . 12 ℤ/n mulGrp mulGrp
1817oveq1d 6309 . . . . . . . . . . 11 ℤ/n mulGrp MndHom mulGrpfld mulGrp MndHom mulGrpfld
1916fveq2d 5875 . . . . . . . . . . . . . . 15 ℤ/n
20 dchrval.b . . . . . . . . . . . . . . 15
2119, 20syl6eqr 2526 . . . . . . . . . . . . . 14 ℤ/n
2216fveq2d 5875 . . . . . . . . . . . . . . 15 ℤ/n Unit Unit
23 dchrval.u . . . . . . . . . . . . . . 15 Unit
2422, 23syl6eqr 2526 . . . . . . . . . . . . . 14 ℤ/n Unit
2521, 24difeq12d 3628 . . . . . . . . . . . . 13 ℤ/n Unit
2625xpeq1d 5027 . . . . . . . . . . . 12 ℤ/n Unit
2726sseq1d 3536 . . . . . . . . . . 11 ℤ/n Unit
2818, 27rabeqbidv 3113 . . . . . . . . . 10 ℤ/n mulGrp MndHom mulGrpfld Unit mulGrp MndHom mulGrpfld
2910, 28eqtr4d 2511 . . . . . . . . 9 ℤ/n mulGrp MndHom mulGrpfld Unit
3029eqeq2d 2481 . . . . . . . 8 ℤ/n mulGrp MndHom mulGrpfld Unit
3130biimpar 485 . . . . . . 7 ℤ/n mulGrp MndHom mulGrpfld Unit
3231opeq2d 4225 . . . . . 6 ℤ/n mulGrp MndHom mulGrpfld Unit
3331, 31xpeq12d 5029 . . . . . . . 8 ℤ/n mulGrp MndHom mulGrpfld Unit
3433reseq2d 5278 . . . . . . 7 ℤ/n mulGrp MndHom mulGrpfld Unit
3534opeq2d 4225 . . . . . 6 ℤ/n mulGrp MndHom mulGrpfld Unit
3632, 35preq12d 4119 . . . . 5 ℤ/n mulGrp MndHom mulGrpfld Unit
378, 36csbied 3467 . . . 4 ℤ/n mulGrp MndHom mulGrpfld Unit
385, 37csbied 3467 . . 3 ℤ/n mulGrp MndHom mulGrpfld Unit
39 dchrval.n . . 3
40 prex 4694 . . . 4
4140a1i 11 . . 3
423, 38, 39, 41fvmptd 5961 . 2 DChr
431, 42syl5eq 2520 1
Colors of variables: wff setvar class Syntax hints: wi 4 wa 369 wceq 1379 wcel 1767 crab 2821 cvv 3118 csb 3440 cdif 3478 wss 3481 csn 4032 cpr 4034 cop 4038 cmpt 4510 cxp 5002 cres 5006 cfv 5593 (class class class)co 6294 cof 6532 cc0 9502 cmul 9507 cn 10546 cnx 14499 cbs 14502 cplusg 14567 MndHom cmhm 15817 mulGrpcmgp 16990 Unitcui 17137 ℂfldccnfld 18267 ℤ/nℤczn 18386 DChrcdchr 23350 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1601 ax-4 1612 ax-5 1680 ax-6 1719 ax-7 1739 ax-9 1771 ax-10 1786 ax-11 1791 ax-12 1803 ax-13 1968 ax-ext 2445 ax-sep 4573 ax-nul 4581 ax-pr 4691 This theorem depends on definitions: df-bi 185 df-or 370 df-an 371 df-3an 975 df-tru 1382 df-ex 1597 df-nf 1600 df-sb 1712 df-eu 2279 df-mo 2280 df-clab 2453 df-cleq 2459 df-clel 2462 df-nfc 2617 df-ne 2664 df-ral 2822 df-rex 2823 df-rab 2826 df-v 3120 df-sbc 3337 df-csb 3441 df-dif 3484 df-un 3486 df-in 3488 df-ss 3495 df-nul 3791 df-if 3945 df-sn 4033 df-pr 4035 df-op 4039 df-uni 4251 df-br 4453 df-opab 4511 df-mpt 4512 df-id 4800 df-xp 5010 df-rel 5011 df-cnv 5012 df-co 5013 df-dm 5014 df-res 5016 df-iota 5556 df-fun 5595 df-fv 5601 df-ov 6297 df-dchr 23351 This theorem is referenced by: dchrbas 23353 dchrplusg 23365
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# [Solved] Given an array, rotate the array to the right by k steps, where k is non-negative.
## Question
Given an array, rotate the array to the right by `k` steps, where `k` is non-negative.
Example 1:
```Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]
```
Example 2:
```Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]
Explanation:
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]
```
Constraints:
• `1 <= nums.length <= 105`
• `-231 <= nums[i] <= 231 - 1`
• `0 <= k <= 105`
• Could you do it in-place with `O(1)` extra space?
```class Solution:
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## Seating Arrangement For SBI PO : Set – 06
(1-5) Study the information carefully and answer the following questions:
Eight friends – Romil, Ramesh, Rakesh, Rohit, Rahul, Abhijit, Abhishek and Anil – are sitting around a circular table, not necessarily in the same order. Four of them are facing inside others are facing outside. They are belong to eight different cities – Bhopal, Patna, Kolkata, Delhi, Gwalior, Bengaluru, Chennai and Rajkot, but not necessarily in the same order.Abhijit faces the Centre and sits third to the right of Rakesh. Rohit belongs to Kolkata and faces the person who belongs to Bengaluru. Abhishesk sits third to the right of Ramesh, who stays in Bhopal. The persons who belong to Delhi and Gwalior are facing to the same direction (inside or outside). Rahul is sitting between the person who belongs to Kolkata and the one from Rajkot respectively. Romil belongs to Gwalior and Rakesh belongs to Patna. The person who belongs to Chennai is facing outward and immediate neighbor of Rajkot. Anil is immediate neighbor of the persons who belong to Gwalior and Chennai. Rahul is immediate left of Rohit.
1. Who belongs to Bengaluru?
1) Romil
2) Rohit
3) Anil
4) Abhishek
5) Rahul
2. Immediate neighbors of Romil?
1) Romesh and Rakesh
2) Rahul and Rohit
3) Anil and Ramesh
4) Abhishek and Rahul
5) Rohit and Abhijeet
3. Which of the following pair is true?
1) Romil – Bhopal
2) Anil – Bengaluru
3) Rohit – Delhi
4) Abhijeet – Delhi
5) Abhishek – Patna
4. Position of Ramesh with respect to Rahul?
1) Third to the right
2) Fourth to the left
3) Fifth to the right
4) Second to the right
5) Third to the left
5. If Rohit and Anil interchange their positions and similarly Abhijit and Rakesh interchange their positions then what is the position of Romil with respect to Rakesh?
1) Third to the left
2) Third to the right
3) Second to the left
4) Immediate left
5) None of these
1) 3
2) 3
3) 2
4) 5
5) 1
(7-12): Read the following information carefully and answer the questions that follow:
Seven friends A, B, C, D, E, F and G are sitting around a circular table facing either the centre or outside. Each one of them belongs to a different department viz. Finance, Marketing, Sales, HR, Corporate Finance, Investment Banking and Operations but not necessarily in the same order.C sits third to the right of G. G faces the centre. Only one person sits between C and the person working in HR department. Immediate neighbours of C face outside. Only one person sits between F and D. Both F and D face the centre. D does not work in HR or Finance department.A, works in Investment Banking department and faces the centre. Two people sit between the persons who work in Investment Banking and Marketing departments. The person who works in Corporate Finance sits to the immediate left of E. C faces same direction as E. The person who works in Corporate Finance sits to the immediate right of the person who works for Operations department.
7. B, works in which of the following departments?
(1) Finance
(2) Marketing
(3) HR
(4) Corporate Finance
(5) Operations
8.What is the position of B with respect to the person who works sales for Sales department?
(1) Immediate right
(2) Third to the left
(3) Second to the right
(4) Second to the left
(5) Fourth to the right
9.Who sits to the immediate right of E?
(1) The person who works in Marketing department
(2) C
(3) B
(4) The person who works for HR department
(5) A
10.Who amongst the following sits exactly between C and the person who works for HR department?
(1) B
(2) The person who works for Marketing department
(3) The person who works for Operations department
(4) D
(5) G
11.Who amongst the following sit between the persons who work for Marketing and Investment Banking departments when counted for the left hand side of the person working for Marketing department?
(1) F & G
(2) E & C
(3) C & B
(4) F & D
(5) B & D
12.How many people sit between the person who works for Operation department and A, when counted from the right hand side of A?
(1) One
(2) Two
(3) Three
(4) Four
(5) More than four
7) 5
8) 4
9) 2
10) 2
11) 4
12) 5
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Steering Behaviors working too quickly
I am trying to implement steering behaviours in my XNA game and mostly there but could use a hand with some little things. The behaviours work perfectly well, my issues lie in the timing and magnitudes of the updates.
(I am using SharpSteer (almost entirely at this point) it can be found here:
The Update() method is where I am struggling. Currently I am using this method (straight from SharpSteer)
`````` // apply a given steering force to our momentum,
// adjusting our orientation to maintain velocity-alignment.
public void ApplySteeringForce(Vector3 force, float elapsedTime)
{
// enforce limit on magnitude of steering force
// compute acceleration and velocity
Vector3 newAcceleration = (clippedForce / Mass);
Vector3 newVelocity = Velocity;
// damp out abrupt changes and oscillations in steering acceleration
// (rate is proportional to time step, then clipped into useful range)
if (elapsedTime > 0)
{
float smoothRate = Utilities.Clip(9 * elapsedTime, 0.15f, 0.4f);
Utilities.BlendIntoAccumulator(smoothRate, newAcceleration, ref acceleration);
}
// Euler integrate (per frame) acceleration into velocity
newVelocity += acceleration * elapsedTime;
// enforce speed limit
newVelocity = Vector3Helpers.TruncateLength(newVelocity, MaxSpeed);
// update Speed
Speed = (newVelocity.Length());
// Euler integrate (per frame) velocity into position
Position = (Position + (newVelocity * elapsedTime));
// regenerate local space (by default: align vehicle's forward axis with
// new velocity, but this behavior may be overridden by derived classes.)
RegenerateLocalSpace(newVelocity, elapsedTime);
// maintain path curvature information
MeasurePathCurvature(elapsedTime);
// running average of recent positions
Utilities.BlendIntoAccumulator(elapsedTime * 0.06f, // QQQ
Position,
ref smoothedPosition);
}
``````
This works, but the changes from frame to frame are huge due to the elapsed time multiplier (I think SharpSteer uses it's own game clock and not gameTime, I am using gameTime)
I am looking for a way to slow this down a lot, and to use speed values in a greater range than 1.0f.
Currently I am using a speed of 1.0f and a maxForce of 1.0f. I have noticed that Velocity is always (0,0,0) and elapsedTime is 16 (this is called 60 times per second) smoothRate is always 0.4f, also because elapsedTime is too large.
If anyone could help be balance the speed of the vehicle somewhat I'd appreciate it.
-
increasing the mass will dampen (slowdown) acceleration for the same amount of force coming into the method. Are you sure you have the appropriate mass value set?
Also, is your velocity representing units per second? if so, then an elapsed time of 16 milliseconds should be 0.016;
-
Look carefully through the code and mark/rename each time and speed related variables/paremeters with corresponding measurement units. Most likely you'll find that you either mixing seconds and milliseconds or some sort of distance/per second mismatch.
Sample names:
``````public void ApplySteeringForce(Vector3 force, float elapsedTimeInSeconds)
Vector3 newVelocity = Velocity; // Pixel per Millisecond
``````
-
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# Events & Promotions
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# UNC MAC Class of 2013
Author Message
Intern
Joined: 06 Feb 2012
Posts: 13
Location: United States
GPA: 3.5
WE: Corporate Finance (Investment Banking)
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
08 Feb 2012, 22:59
HI,
How did you guys pay the $500 deposit ? Please could someone give the me the exact payment details? I tried searching through the cashiers web page but to no avail. Any detail would do - whom to make a credit card payment to OR whose name should be put on a bank draft. Thanks !! _________________ Virtus junxit mors non separabit Intern Joined: 06 Feb 2012 Posts: 13 Location: United States GPA: 3.5 WE: Corporate Finance (Investment Banking) Followers: 0 Kudos [?]: 6 [0], given: 0 Re: UNC MAC Class of 2013 [#permalink] ### Show Tags 09 Feb 2012, 00:35 Any idea when the Student Personal Identification Number is created ? (student PID) The cashier's webpage says that a PID is needed for a wire transfer. _________________ Virtus junxit mors non separabit Intern Joined: 16 Oct 2011 Posts: 38 Followers: 1 Kudos [?]: 2 [0], given: 3 Re: UNC MAC Class of 2013 [#permalink] ### Show Tags 09 Feb 2012, 01:31 ronnycpa13 wrote: HI, How did you guys pay the$500 deposit ?
Please could someone give the me the exact payment details? I tried searching through the cashiers web page but to no avail.
Any detail would do - whom to make a credit card payment to OR whose name should be put on a bank draft.
Thanks !!
I just did a cashier's check from the post office. I checked with Stacey beforehand and she said it was fine.
Intern
Joined: 24 Aug 2011
Posts: 22
Followers: 2
Kudos [?]: 1 [0], given: 0
Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
09 Feb 2012, 17:11
The new Academic Calendar is up
http://www.kenan-flagler.unc.edu/progra ... c-calendar
2012-13*
Summer Sequence
May 18 Mandatory Orientation & Test Out
May 21 Summer Classes Begin
May 28 Memorial Day Holiday — no classes
July 26 Last Day of Summer Classes
July 27 - August 12 Summer Break
Mod I
August 13 - 16 Mandatory Orientation
August 17 Module 1 Classes Begin
September. 3 Labor Day Holiday — no classes
October 12 Module 1 Classes End
October 13 - 21 Fall Break
Mod II
October 22 Module 2 Classes Begin
November 21 - 25 Thanksgiving Holiday - no classes
December 19 Module 2 Classes End
December 20 - January 6 Winter Break
Mod III
January 7 Module 3 Classes Begin
January 21 Martin Luther King Holiday - no classes
March 28 Module 3 Classes End
March 1 - 10 Spring Break & Global Immersion Elective
Mod IV
March 11 Module 4 Classes Begin
March 29 Easter Holiday - no classes
May 6 Module 4 Classes End
May 8 MAC Commencement!
*Dates subject to change
Intern
Joined: 21 Jan 2012
Posts: 3
Location: United States
GPA: 3.51
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
09 Feb 2012, 19:57
Thanks for posting, Scott. On a different note, I got an email today from UNC confirming that my $500 enrollment deposit was posted to my account. I'm guessing that those who have sent in their checks will get a similar note soon--or at least see their checks clear. I'm curious as to what you all think of Carolina's decision to go to the UAE for the Global Immersion Elective. Has anyone here traveled there before? I'm glad to see it has been moved to the spring--which should mean that the GIE won't interfere with recruiting... Intern Joined: 16 Oct 2011 Posts: 38 Followers: 1 Kudos [?]: 2 [0], given: 3 Re: UNC MAC Class of 2013 [#permalink] ### Show Tags 09 Feb 2012, 20:15 Youinsee wrote: Thanks for posting, Scott. On a different note, I got an email today from UNC confirming that my$500 enrollment deposit was posted to my account. I'm guessing that those who have sent in their checks will get a similar note soon--or at least see their checks clear.
I'm curious as to what you all think of Carolina's decision to go to the UAE for the Global Immersion Elective. Has anyone here traveled there before? I'm glad to see it has been moved to the spring--which should mean that the GIE won't interfere with recruiting...
UAE? I didn't see that. I've NEVER been to that part of the world. Should be very exciting.
Intern
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Posts: 22
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
09 Feb 2012, 20:20
I have spent a good amount of time in the Middle East but never in the UAE. When did you hear about their decision? I knew that it was going to be in the Spring but I heard that they were thinking of going to South Africa again. While I would have enjoyed that opportunity as well, I had noticed that they had not traveled to anywhere in the Middle East for the GIE. I will be interested to see how many people will choose to attend with it being so late in the year. I'm definitely interested.
Youinsee wrote:
Thanks for posting, Scott. On a different note, I got an email today from UNC confirming that my \$500 enrollment deposit was posted to my account. I'm guessing that those who have sent in their checks will get a similar note soon--or at least see their checks clear.
I'm curious as to what you all think of Carolina's decision to go to the UAE for the Global Immersion Elective. Has anyone here traveled there before? I'm glad to see it has been moved to the spring--which should mean that the GIE won't interfere with recruiting...
Intern
Joined: 24 Aug 2011
Posts: 22
Followers: 2
Kudos [?]: 1 [0], given: 0
Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
09 Feb 2012, 20:23
I see it now. They snuck the information into the GIE description. I hope that they decide to go to multiple countries in the region and not just the UAE. I haven't seen anything about my check yet but I guess it will arrive soon.
Intern
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Posts: 3
Location: United States
GPA: 3.51
Followers: 0
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
09 Feb 2012, 20:30
You can find the GIE listed under the "Curriculum" tab on the admitted students page. The UAE is listed there. One of my college roommates was from Pakistan, and raved about the UAE--specifically Dubai. I'm interested too, barring any major deterioration in the area.
Intern
Joined: 06 Feb 2012
Posts: 13
Location: United States
GPA: 3.5
WE: Corporate Finance (Investment Banking)
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
13 Feb 2012, 20:13
Guys,
Any luck on the housing front ?
Also, I am planning to do some prep work from Becker's CPA books, i guess Becker's is one of the leading CPA coaching classes.
Anyone got any views on the same ? (i have 2.5 years of experience as a M&A analyst -- but almost zero accounting knowledge).
_________________
Virtus junxit mors non separabit
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Re: UNC MAC Class of 2013 [#permalink]
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20 Feb 2012, 20:35
I hope I'm not the only one who has enjoyed bandwagoning UNC basketball. I'm already learning how to properly hate Duke. Austin Rivers trained at the gym for my undergrad... to see him rip my heart out like that was pretty painful. My school couldn't care less about sports, it will be nice to go to a school that is passionate about them. Can't wait to get moved in and started.
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Re: UNC MAC Class of 2013 [#permalink]
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20 Feb 2012, 20:36
I grew up a Maryland fan so I don't think I'll ever be a true Tar Heel, but hey, at least hating Duke comes naturally to me
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
21 Feb 2012, 15:18
Hi everyone. Thanks for the information on Sunstone Apartments, Scott, they look really nice and the location sounds great. Is anyone else looking for a roommate? I'm a female, but I don't really care either way to live with a guy or a girl. Are you guys replying to the roommate spreadsheet or are you contacting people on your own?
Last edited by dailyomnivore on 21 Feb 2012, 15:32, edited 1 time in total.
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Re: UNC MAC Class of 2013 [#permalink]
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21 Feb 2012, 17:06
2012 class schedule is posted, looks like a busy summer!
Intern
Joined: 25 Jul 2011
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
22 Feb 2012, 21:03
ronnycpa13 wrote:
Any idea when the Student Personal Identification Number is created ? (student PID)
The cashier's webpage says that a PID is needed for a wire transfer.
http://pid.unc.edu/PIDLookup.aspx
dailyomnivore, I'm not actively looking for a roommate right now, but I wouldn't mind having one. I'll be bringing a car so that might be a plus. what is this spreadsheet you're referring to?
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
22 Feb 2012, 21:20
ugen64 wrote:
ronnycpa13 wrote:
Any idea when the Student Personal Identification Number is created ? (student PID)
The cashier's webpage says that a PID is needed for a wire transfer.
http://pid.unc.edu/PIDLookup.aspx
dailyomnivore, I'm not actively looking for a roommate right now, but I wouldn't mind having one. I'll be bringing a car so that might be a plus. what is this spreadsheet you're referring to?
Upon admission, one of the forms you could send in was the roommate-interest form or something like that. Just some basic info that they said they would have up by March 1st. Maybe it's up early?
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
23 Feb 2012, 07:34
UNC PID - I got mine from the admissions department.
Guys -- The Week starting May 14th 2012 -- IS NOT for our batch (i.e. class of 2013)
Class of 2013 has its mandatory summer orientation on the 18th of May 2012 (Saturday) and classes begin on the 20th of May (Monday)
The timetable given on the MAC website is for the class of 2012 - its a review course for the CPA examination.
@ dailyomnivore
I am looking for a flatmate and I have filled up the roommate form.
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
23 Feb 2012, 14:30
Like Visirale said, the roommate spreadsheet is on the MAC Admitted Students website under Roommates and Housing. They say the list should be posted March 1.
@ronnycpa13 and @ugen64: Right now I'm looking at sharing a two bedroom/two bathroom apartment at SunStone Apartments after Scott's tip. It's within walking/biking distance to campus and it's affordable. If either of you are interested in looking into it feel free to private message me!
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
26 Feb 2012, 20:11
Hey yall.
My name is Chris and I will be attending the UNC MAC Program starting in May. I was actually a UNC undergrad (graduated in December) so I know the ins and outs of Chapel Hill already. I already sent in the acceptance check and I have all the FAFSA forms filled out so I guess my next step is the Supplemental Application. Does anyone know how we find out what classes to take? So far I feel like this is very unorganized. It actually reminds me a lot of my undergraduate career here.
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Re: UNC MAC Class of 2013 [#permalink]
### Show Tags
26 Feb 2012, 20:18
EugeneTheTJ wrote:
Hey yall.
My name is Chris and I will be attending the UNC MAC Program starting in May. I was actually a UNC undergrad (graduated in December) so I know the ins and outs of Chapel Hill already. I already sent in the acceptance check and I have all the FAFSA forms filled out so I guess my next step is the Supplemental Application. Does anyone know how we find out what classes to take? So far I feel like this is very unorganized. It actually reminds me a lot of my undergraduate career here.
Hey Chris,
Guess you could be a big help to all of us regarding housing and much more.
Which class do you wish to take ? -- Are you inclined towards taxation or auditing ?
_________________
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Re: UNC MAC Class of 2013 [#permalink] 26 Feb 2012, 20:18
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SOLIDWORKS Simulation Pin Connector
Article by Irfan Zardadkhan, PhD, CSWE updated March 17, 2015
Article
A Pin Connector connects a solid or a shell body to another solid or shell body. The selection entities can be cylindrical faces or circular edges from the same body or two different bodies.
The Pin connector is represented with a very stiff beam. Each end of the Pin is located at the center of the cylindrical face (on the axis, and halfway across the height). Each end is then connected with perfectly rigid bars to each node of the cylindrical face it corresponds to.
• The pin remains straight (it does not bend)
• Each face maintains its original shape but can move as a rigid body
• All faces defining the pin connector remain coaxial
The With retaining ring and With key options are implemented by freeing some DOFs between the rigid bars (translation for With retaining ring and rotation With key)
The Material properties and the Tensile Stress Area are not used to realistically represent the stiffness of the pin.
Instead, the Pin forces are calculated based on assumption that the Pin is very stiff. The axial and shear forces and the bending and torsional moments in the Pins are calculated during the solution based on that assumption. Then, from the obtained values, the defined Strength data is used to calculate the factor of safety.
The pin connectors can be defined between two cylindrical faces from two separate parts. However, if you want to define between three (or more) components, you need to define two pin connectors (or more). Before you define the pin connectors, split the cylindrical surface of the middle component in two cylindrical faces (each of half the initial face’s height) and use them for each pin connector definitions (first cylindrical face with one part and the second cylindrical face with the other part).
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Stitz-Zeager_College_Algebra_e-book
# Graphically we have7 y 1 x 1 6 lastly we construct a
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Unformatted text preview: is no way any environment can support infinitely many bacteria, so shortly before t = 5 the environment would collapse. Now that we have thoroughly investigated vertical asymptotes, we now turn our attention to horizontal asymptotes. The next theorem tells us when to expect horizontal asymptotes. Theorem 4.2. Location of Horizontal Asymptotes: Suppose r is a rational function and ( x) r(x) = p(x) , where p and q are polynomial functions with leading coefficients a and b, respectively. q • If the degree of p(x) is the same as the degree of q (x), then y = of the graph of y = r(x). a b is thea horizontal asymptote • If the degree of p(x) is less than the degree of q (x), then y = 0 is the horizontal asymptote of the graph of y = r(x). • If the degree of p(x) is greater than the degree of q (x), then the graph of y = r(x) has no horizontal asymptotes. a The use of the definite article will be justified momentarily. 4.1 Introduction to Rational Functions 239 Like Theorem 4.1, Theorem 4.2 is proved using Calculus. Nevertheless, we can understand the idea x− behi...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.
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# How to fsum on only one variable?
5 views (last 30 days)
Fred on 8 Jul 2013
Hi, I want to find the following Z(i)s but inside the sum is only for j. How can I write following script with sum on only j? Sorry if my question is very basic stuff
for i=1:n
for j=1:m
Z(i)=Z(i)-sum(S(j).*Z(i-j))
end
end
Fred on 9 Jul 2013
H=filter([1,-S],0 , Z);
Image Analyst on 8 Jul 2013
I have no idea what you're trying to do. First of all, Z(1) will depend on Z(1) which hasn't been defined yet so that will bomb. Next, i-j will equal zero when both i and j are 1, so that will bomb also. Beyond that I have no idea whatsoever what you're trying to do. And what is S?
#### 1 Comment
Fred on 8 Jul 2013
Sorry for confusion; I have two vectors Z and S (Z is my data and S are coefficients from another formula) and I want to adjust Z values by some type of differencing. So old Z values are replaced by new ones (I used the same name as Z but I can use a new name too like Z2). about i-j I think that I should think of a condition It is some think like this:
Z(i)=Z(i)-0.3*Z(i-1)-0.105*Z(i-2)-0.059*Z(i-3)-...
or
Z2(i)=Z(i)-0.3*Z(i-1)-0.105*Z(i-2)-0.059*Z(i-3)-...
where S=[0.3 0.105 0.059 ...]
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https://www.gurufocus.com/term/Total+Liabilities/NG/Total+Liabilities/Novagold+Resources%252C+Inc
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Switch to:
Novagold Resources Inc (AMEX:NG) Total Liabilities: \$113.01 Mil (As of Aug. 2017)
Novagold Resources Inc's total liabilities for the quarter that ended in Aug. 2017 was \$113.01 Mil.
Novagold Resources Inc's quarterly total liabilities increased from Feb. 2017 (\$107.89 Mil) to May. 2017 (\$109.39 Mil) and increased from May. 2017 (\$109.39 Mil) to Aug. 2017 (\$113.01 Mil).
Novagold Resources Inc's annual total liabilities declined from Nov. 2014 (\$119.43 Mil) to Nov. 2015 (\$104.29 Mil) but then increased from Nov. 2015 (\$104.29 Mil) to Nov. 2016 (\$108.00 Mil).
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Novagold Resources Inc Annual Data
Nov07 Nov08 Nov09 Nov10 Nov11 Nov12 Nov13 Nov14 Nov15 Nov16 Total Liabilities 208.43 113.04 119.43 104.29 108.00
Novagold Resources Inc Quarterly Data
Nov12 Feb13 May13 Aug13 Nov13 Feb14 May14 Aug14 Nov14 Feb15 May15 Aug15 Nov15 Feb16 May16 Aug16 Nov16 Feb17 May17 Aug17 Total Liabilities 107.23 108.00 107.89 109.39 113.01
Calculation
Total liabilities are the liabilities that the company has to pay others. It is a part of the balance sheet of a company that shareholders do not own, and would be obligated to pay back if the company liquidated.
Novagold Resources Inc's Total Liabilities for the fiscal year that ended in Nov. 2016 is calculated as
Total Liabilities = Total Current Liabilities + Total Long Term Liabilities = Total Current Liabilities + (Long-Term Debt & Capital Lease Obligation + Other Long Term Assets = 3.051 + (84.812 + 3.5527136788E-15 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 20.135 + 0 + 0) = 108.00
Total Liabilities = Total Assets (A: Nov. 2016 ) - Total Equity (A: Nov. 2016 ) = 408.261 - 300.263 = 108.00
Novagold Resources Inc's Total Liabilities for the quarter that ended in Aug. 2017 is calculated as
Total Liabilities = Total Current Liabilities + (Total Long Term Liabilities) = Total Current Liabilities + (Long-Term Debt & Capital Lease Obligation + Long-Term Debt & Capital Lease Obligation = 2.745 + (88.693 + -3.5527136788E-15 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 21.57 + 0 + 0) = 113.01
Total Liabilities = Total Assets (Q: Aug. 2017 ) - Total Equity (Q: Aug. 2017 ) = 413.318 - 300.31 = 113.01
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
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Replies: 49 Last Post: Jan 13, 2012 2:37 PM
Messages: [ Previous | Next ]
Wayne Bishop Posts: 5,465 Registered: 12/6/04
Posted: Dec 18, 2011 6:46 PM
At 08:26 PM 12/17/2011, Dan Christensen wrote:
>Do your "garden variety mathematical theories" include things like
>number theory, abstract algebra and geometry? While designed as
>primarily a learning aid, I see no reason that DC Proof could not be
>used to generate most if not all of current mathematical theory --
>that is, theories based on some underlying set, e.g. the set of
>natural numbers in number theory, or the set of points in a plane for geometry.
>
but I must be missing something. I couldn't see how to make it prove
anything at all or even to verify a proof in, say, plane
geometry. Suppose, for example, I have a proof that, in neutral
(a.k.a. absolute) geometry (so we don't have to worry about
consequences of unique parallels), any triangle can be circumscribed.
(The usual "Euclidean" tools of compass and straightedge are allowed
in neutral geometry). What do I do to verify the accuracy of the proof?
Wayne
Date Subject Author
12/11/11 kirby urner
12/12/11 Dan Christensen
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12/12/11 Dan Christensen
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12/13/11 Dan Christensen
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12/15/11 Joe Niederberger
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12/16/11 Dan Christensen
12/16/11 Joe Niederberger
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12/18/11 Wayne Bishop
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