Algorithms: Background

Transcription

1 Algorithms: Background Amotz Bar-Noy CUNY Amotz Bar-Noy (CUNY) Algorithms: Background 1 / 66

2 What is a Proof? Definition I: The cogency of evidence that compels acceptance by the mind of a truth or a fact. Definition II: The process or an instance of establishing the validity of a statement especially by derivation from other statements in accordance with principles of reasoning. Definition III: A sequence of statements, each of which is either validly derived from those preceding it or is an axiom or assumption, and the final member of which, the conclusion, is the statement of which the truth is thereby established. Definition IV: a chain of reasoning using rules of inference, ultimately based on a set of axioms, that lead to a conclusion. Amotz Bar-Noy (CUNY) Algorithms: Background 2 / 66

3 The Pythagorean Theorem c b a Theorem: c 2 = a 2 + b 2 for a right triangle with short sides of length a and b and long side of length c. Amotz Bar-Noy (CUNY) Algorithms: Background 3 / 66

4 Proof I b a a c c b b c c a a b (a + b) 2 = c ab 2 a2 + b 2 = c 2 Amotz Bar-Noy (CUNY) Algorithms: Background 4 / 66

5 Proof II c c b a b a a b a b c c c 2 = 4 ab 2 + (b a)2 c 2 = b 2 + a 2 Amotz Bar-Noy (CUNY) Algorithms: Background 5 / 66

6 Proof III a c c b a b a + b c2 (a + b) = ab 2 a2 + b 2 = c 2 Amotz Bar-Noy (CUNY) Algorithms: Background 6 / 66

7 More Proofs 118 proofs: 6 proofs without words: An experimental proof: More about the Theorem and its proofs: Amotz Bar-Noy (CUNY) Algorithms: Background 7 / 66

8 Prime Numbers Theorem: There are infinitely many prime numbers. Proof: Let p 1 < p 2 < < p n be a set of n primes. Let Q = p 1 p 2 p n + 1. If Q is a prime, then a new prime is found. Otherwise, Q is a product of two or more primes. The Fundamental Theorem of Arithmetic. None of these primes can be p 1,..., p n. Therefore, a new prime is found. This process can continue infinitely often to find infinitely many primes. Amotz Bar-Noy (CUNY) Algorithms: Background 8 / 66

9 A Proof by Contradiction Theorem: The number 2 is an irrational number. Proof: Assume towards contradiction that p q = 2 for two positive integers p and q that have no common divisor. Squaring the equation implies p 2 = 2q 2. Therefore, p must be even. Let p = 2r for a positive integer r. It follows that 2q 2 = (2r) 2 = 4r 2 q 2 = 2r 2. Therefore, q must be even. A contradiction since both p and q are even. Amotz Bar-Noy (CUNY) Algorithms: Background 9 / 66

10 The Numbers An integer number: One of the (counting) numbers..., 3, 2, 1, 0, 1, 2, 3,.... Z (Z +, Z ) The set of all (positive, negative) integers. A rational number: A number that can be expressed as a fraction p/q where p (the numerator) and q (the denominator) are integers and q 0. Q (Q +, Q ) The set of all (positive, negative) rationals. A real number: A number of the form n.b 1 b 2 b 3... where n is an integer and b i {0, 1} for all i = 1, 2,.... R (R +, R ) The set of all (positive, negative) reals. Amotz Bar-Noy (CUNY) Algorithms: Background 10 / 66

11 Equality Between Sets Definition: Two finite or infinite sets S and T have the same cardinality (S T ) if and only if there is a one-to-one mapping f : S T from S to T. Mapping: For all s S there exists t T such that f (s) = t. From S: s 1 s 2 f (s 1 ) f (s 2 ). To T : For all t T there exists s S such that f (s) = t. Remark: For finite sets S = T is equivalent to S T. Amotz Bar-Noy (CUNY) Algorithms: Background 11 / 66

12 Examples Z Z + although it seems like Z is twice as large as Z +. Q Z although there are infinite number of rational numbers between any two consecutive integers. R Q although there are infinite number of rational numbers between any two real numbers. Since Q R (every rational number is also a real number), it follows that Q R from a cardinality point of view. Amotz Bar-Noy (CUNY) Algorithms: Background 12 / 66

13 Z Z + The mapping f : Z Z + : f (n) = 2n for a positive integer n. f ( n) = 2n + 1 for a positive integer n. f (0) = Amotz Bar-Noy (CUNY) Algorithms: Background 13 / 66

14 Z + Q + The mapping: From Z + to Q + by following the arrows and skipping the already assigned fractions. 1/1 2/1 3/1 4/1 5/1... 1/2 2/2 3/2 4/2 5/2... 1/3 2/3 3/3 4/3 5/3... 1/4 2/4 3/4 4/4 5/4... 1/5.. 2/5... 3/5.. 4/5... 5/ Amotz Bar-Noy (CUNY) Algorithms: Background 14 / 66

15 Z + R + Assume that a mapping f : Z + R + exists. f (1) = n 1.b1b 1 1b bi 1... f (2) = n 2.b2b 1 2b bi 2... f (3) = n 3.b3b 1 3b bi f (i) = n i.bi 1 bi 2 bi 3... bi i... Let x = n 1. b 1 1 b 2 2 b b i i... where b j j = 1 b j j. x f (i) for any i = 1, 2,.... Therefore f is not a valid mapping: a contradiction. Amotz Bar-Noy (CUNY) Algorithms: Background 15 / 66

16 A False Proof a = b + c (a b)a = (a b)(b + c) a 2 ab = ab + ac b 2 bc a 2 ab ac = ab b 2 bc a(a b c) = b(a b c) a = b Amotz Bar-Noy (CUNY) Algorithms: Background 16 / 66

17 Another False Proof i = i 1 = = 1 = = = 1 Amotz Bar-Noy (CUNY) Algorithms: Background 17 / 66

18 Induction Let P n be a statement for all the positive integer (n = 1, 2, 3,...). If the following 2 properties hold: Property I: P 1 is true. Property II: P k+1 is true if P k is true for each positive integer k. Then P n is true for all n. Amotz Bar-Noy (CUNY) Algorithms: Background 18 / 66

19 Why Induction Works? Let S be the set of all numbers n for which P n is false. Let k be the minimum number in S. k > 1 since by Property I of the induction definition, P 1 is true. By the minimality of k, P k 1 is true and P k is false. A contradiction to Property II of the induction definition. Amotz Bar-Noy (CUNY) Algorithms: Background 19 / 66

20 A Summation Problem Prove that for an integer n 1: Define: n = n(n+1) 2. L(n) = n. R(n) = n(n+1) 2. Prove that L(n) = R(n) for n 1. Amotz Bar-Noy (CUNY) Algorithms: Background 20 / 66

21 Verifying the Claim Check correctness for small values of n: n L(n) R(n) = 1 2 = = 3 2 = = 6 2 = = 55 2 = 55 Amotz Bar-Noy (CUNY) Algorithms: Background 21 / 66

22 A Direct Proof Idea: Compute the value of 2L(n). Example: 2( ) = ( ) + ( ) = (1 + 3) + (2 + 2) + (3 + 1) = 3 4 = 12 = Amotz Bar-Noy (CUNY) Algorithms: Background 22 / 66

23 A Direct Proof 2 L(n) = (1 + + n) + (n + + 1) = (1 + n) + (2 + (n 1)) + + (n + 1) = (n + 1) + (n + 1) + + (n + 1) = n(n + 1) This implies that: L(n) = n(n+1) 2 = R(n). Amotz Bar-Noy (CUNY) Algorithms: Background 23 / 66

24 A Proof by Induction Prove correctness for n + 1 assuming correctness for n: L(n + 1) = n + (n + 1) = L(n) + (n + 1) = R(n) + (n + 1) n(n + 1) = + (n + 1) 2 (n + 1)(n + 2) = 2 = R(n + 1) Amotz Bar-Noy (CUNY) Algorithms: Background 24 / 66

25 The Pigeonhole Principle Simple form: If k + 1 objects are placed into k boxes then there exists at least one box containing 2 or more objects. Dual form: If k objects are placed into k + 1 boxes then there exists at least one empty box. Generalized form: If K objects are placed into k boxes then there exists at least one box containing K /k or more objects. Amotz Bar-Noy (CUNY) Algorithms: Background 25 / 66

26 Hand Shaking Problem: In a group of n people handshaking takes place. No-one shakes their own hand and no pair of people shake hands more than once. Show that there must be two people who have shaken the same number of hands. Amotz Bar-Noy (CUNY) Algorithms: Background 26 / 66

27 Hand Shaking Problem: In a group of n people handshaking takes place. No-one shakes their own hand and no pair of people shake hands more than once. Show that there must be two people who have shaken the same number of hands. Proof: For 1 i n, person P i shakes 0 h i n 1 hands. There cannot be 2 people P i and P j such that h i = 0 and h j = n 1. There are only n 1 possible values for h 1, h 2,..., h n. By the pigeonhole principle at least 2 people have shaken the same number of hands. Amotz Bar-Noy (CUNY) Algorithms: Background 26 / 66

28 Friends and Enemies Problem: In a group of 6 people, each pair of individuals consists of 2 friends or 2 enemies. Show that there are either 3 mutual friends or 3 mutual enemies in the group. Amotz Bar-Noy (CUNY) Algorithms: Background 27 / 66

29 Friends and Enemies Problem: In a group of 6 people, each pair of individuals consists of 2 friends or 2 enemies. Show that there are either 3 mutual friends or 3 mutual enemies in the group. Proof: Let A be one of the 6 people. By the generalized pigeonhole principle either A has 3 friends or A has 3 enemies. Assume A has 3 friends B, C, D. If any 2 of B, C, D are friends then these 2 and A are 3 mutual friends. Otherwise B, C, D are 3 mutual enemies. Similar arguments hold if B, C, D are the enemies of A. Amotz Bar-Noy (CUNY) Algorithms: Background 27 / 66

30 Discussing Topics Problem: 17 people discuss 3 topics where each pair discuss only 1 topic. Prove that there are at least 3 people who discuss among themselves the same topic. Amotz Bar-Noy (CUNY) Algorithms: Background 28 / 66

31 Discussing Topics Problem: 17 people discuss 3 topics where each pair discuss only 1 topic. Prove that there are at least 3 people who discuss among themselves the same topic. Proof: Let A be one of the 17 people. By the generalized pigeonhole principle there exists a topic T discussed by A with at least 6 people. If among these 6 people there are B and C who discuss T, then A, B, C discuss T among themselves. Otherwise, these 6 people discuss only the other 2 topics. By the previous problem, there are 3 people who discuss among themselves the same topic. Amotz Bar-Noy (CUNY) Algorithms: Background 28 / 66

32 Ramsey s Numbers Friends and Enemies: R(3, 3) Discussing Topics: R(3, 3, 3) Amotz Bar-Noy (CUNY) Algorithms: Background 29 / 66

33 Induction vs. Pigeonhole Principle Induction:... then for all... Pigeonhole Principle:... then there exists... Amotz Bar-Noy (CUNY) Algorithms: Background 30 / 66

34 The Subsequence Theorem Theorem: Let x 1,..., x n 2 +1 be any sequence of distinct numbers. Then there exists either an increasing or a decreasing (n + 1)-subsequence Amotz Bar-Noy (CUNY) Algorithms: Background 31 / 66

35 The Subsequence Theorem Theorem: Let x 1,..., x n 2 +1 be any sequence of distinct numbers. Then there exists either an increasing or a decreasing (n + 1)-subsequence Without loss of generality the numbers are 1, 2,..., n Trivial for n = 1: the only possible sequences of length n = = 2 are the increasing sequence (1, 2) and the decreasing sequence (2, 1) of length n + 1 = 2. Amotz Bar-Noy (CUNY) Algorithms: Background 31 / 66

36 Sequences of Length n 2 The sequence (7, 8, 9, 4, 5, 6, 1, 2, 3) is an example why the theorem is not correct for n 2 numbers for n = 3. The sequence (4, 3, 2, 1, 8, 7, 6, 5, 12, 11, 10, 9, 16, 15, 14, 13) is an example why the theorem is not correct for n 2 numbers for n = 4. Amotz Bar-Noy (CUNY) Algorithms: Background 32 / 66

37 Sequences of Length n 2 The sequence (7, 8, 9, 4, 5, 6, 1, 2, 3) is an example why the theorem is not correct for n 2 numbers for n = 3. The sequence (4, 3, 2, 1, 8, 7, 6, 5, 12, 11, 10, 9, 16, 15, 14, 13) is an example why the theorem is not correct for n 2 numbers for n = 4. The following two sequences are examples why the theorem is not correct for n 2 numbers for any n 1: ([n 2 n + 1, n 2 n + 2,..., n 2 ], [n 2 2n + 1, n 2 2n + 2,..., n 2 n],..., [1, 2,..., n]) ([n, n 1,..., 1], [2n, 2n 1,..., n + 1],..., [n 2, n 2 1,..., n 2 n + 1]) Amotz Bar-Noy (CUNY) Algorithms: Background 32 / 66

38 n = 2: Sequences of Length 5 Check all the 120 permutations of 1, 2, 3, 4, 5 and show either an increasing or a decreasing 3-subsequence. Amotz Bar-Noy (CUNY) Algorithms: Background 33 / 66

39 n = 2: Sequences of Length 5 Check all the 120 permutations of 1, 2, 3, 4, 5 and show either an increasing or a decreasing 3-subsequence. Assume 3 is in position 3 in the sequence (,, 3,, ): If 1 is in positions 4 or 5, then either 4 or 5 must be in positions 1 or 2, creating a decreasing 3-subsequence. E.g., (5,, 3,, 1). If 1 is in positions 1 or 2, then either 4 or 5 must be in positions 4 or 5, creating an increasing 3-subsequence. E.g., (, 1, 3,, 4). Similar arguments find a decreasing or an increasing 3-subsequence if 2, 3, or 4 are in positions 2, 3, or 4. The proof follows since by the pigeonhole principle one of 2, 3, and 4 must be in one of the positions 2, 3, or 4. Amotz Bar-Noy (CUNY) Algorithms: Background 33 / 66

40 Proofs Five Proofs of the Subsequence Theorem By William Gasarch gasarch/blogpapers/subseq4proofs.pdf The three proofs by induction are more involved than the two proofs that use the pigeonhole principle. However, all the proofs use similar arguments. All the proofs fail on the example of a sequence of length n 2 that does not have an n + 1-subsequence. Amotz Bar-Noy (CUNY) Algorithms: Background 34 / 66

41 First Proof from the Book Let x 1,..., x n 2 +1 be any sequence of distinct numbers. Let f (i) be the length of the longest increasing subsequence that ends at x i. If there exists i such that f (i) n + 1, then an increasing (n + 1)-subsequence exists. Otherwise, f (i) n for all i. There are n indices, therefore by the generalized pigeonhole principle, there exists 1 v n and at least n + 1 indices i 1, i 2,..., i n+1 for which v = f (i 1 ) = f (i 2 ) = = f (i n+1 ). x ij > x ik for any 1 j < k n + 1 since otherwise f (k) f (j) + 1. Therefore, x i1 > x i2 > > x in+1 is a decreasing (n + 1)-subsequence. Amotz Bar-Noy (CUNY) Algorithms: Background 35 / 66

42 Second Proof from the Book Let x 1,..., x n 2 +1 be any sequence of distinct numbers. Let f (i) be the length of the longest increasing subsequence that ends at x i and let g(i) be the length of the longest decreasing subsequence that ends at x i. Assume towards contradiction that both f (i), g(i) n for all i. Then there are at most n 2 different pairs (f (i), g(i)). Since there are n indices, by the pigeonhole principle, there exist indices 1 j < k n such that f (j) = f (k) and g(j) = g(k). If x j < x k then f (k) f (j) + 1; a contradiction. If x j > x k then g(k) g(j) + 1; a contradiction. Amotz Bar-Noy (CUNY) Algorithms: Background 36 / 66

43 Properties of Logarithms For real numbers a > 1, b > 1, x > 0, and y > 0: log a x = z a z = x log a 1 = 0 log a a = 1 log a (x y ) = y log a x log a (xy) = log a x + log a y log a (x/y) = log a x log a y a log b x = x log b a log a b = 1 log b a log a x = log b x log b a = log a b log b x Amotz Bar-Noy (CUNY) Algorithms: Background 37 / 66

44 Summation Formulas n i=1 i0 = = n 1 1 n1 n i=1 i1 = n = n(n+1) n2 n i=1 i2 = n 2 = n(n+1)(2n+1) n3 n i=1 ik = 1 k + 2 k + + n k 1 k+1 nk+1 Amotz Bar-Noy (CUNY) Algorithms: Background 38 / 66

45 Summation Formulas Finite sums for a > 1: n i=0 ai = 1 + a + + a n = an+1 1 a 1 n i=0 2i = n = 2 n+1 1 Finite sums for a < 1: n i=0 ai = 1 + a + + a n = 1 an+1 n i=0 1 a ( 1 ) i 2 = = 2 1 n 2 n Infinite sums for a < 1: i=0 ai = 1 + a + a 2 + = 1 ( 1 i=0 2 1 a ) i = = 2 Amotz Bar-Noy (CUNY) Algorithms: Background 39 / 66

46 Summation Formulas n (i 2 i ) = n 2 n = (n 1)2 n i= n 2 n 2 n 2 n 2 n 2(2 n 1) 4(2 n 1 1) 8(2 n 2 1) 16(2 n 3 1) 2 n (2 1). n n n (i 2 i ) = 2 i (2 n i+1 1) = (2 n+1 2 i ) i=1 i=1 i=1 = n2 n+1 2(2 n 1) = (n 1)2 n Amotz Bar-Noy (CUNY) Algorithms: Background 40 / 66

47 Summation Formulas i=1 i 2 i = i 2 i + = 2 1/2 1/4 1/4 1/8 1/8 1/8... 1/2 i 1/2 i 1/2 i 1/2 i /2 1/4 1/2 i 1 i=1 i 2 i = i=0 1 2 i = 2. Amotz Bar-Noy (CUNY) Algorithms: Background 41 / 66

48 Summation Formulas n log 2 i = log log log 2 n i=1 = log 2 (1 2 n) = log 2 (n!) n log 2 n Upper bound: n log 2 i < i=1 n log 2 n = n log 2 n i=1 Lower bound: n n log 2 i > log 2 (n/2) = n/2 log 2 (n/2) n/2 i=1 i=n/2+1 Amotz Bar-Noy (CUNY) Algorithms: Background 42 / 66

49 Summation Formulas For n = 2 k : log 2 n i=0 log 2 ( n 2 i ) = log 2 ( n 1 ) ( n ) ( n ) + log log 2 2 n = log 2 n = log 2 n(log 2 n + 1) 2 log2 2 n 2 Amotz Bar-Noy (CUNY) Algorithms: Background 43 / 66

50 Sum Manipulation Rules u i=l ca i = c u i=l a i u i=l (a i ± b i ) = u i=l a i ± u i=l b i u i=l a i = m i=l a i + u i=m+1 a i for l m u u i=l+1 (a i a i 1 ) = a u a l Amotz Bar-Noy (CUNY) Algorithms: Background 44 / 66

51 Bounding Summations by Integrals For a monotonic non-decreasing function f (x): u l 1 f (x)dx u f (i) i=l u+1 l f (x)dx For a monotonic non-increasing function f (x): u+1 l f (x)dx u f (i) i=l u l 1 f (x)dx Amotz Bar-Noy (CUNY) Algorithms: Background 45 / 66

52 Bounding Summations by Integrals Amotz Bar-Noy (CUNY) Algorithms: Background 46 / 66

53 Harmonic Numbers H(n) = n i=1 1 i. f (x) = 1/x is a monotonic non-increasing function. Upper Bound: H(n) 1 + n dx 1 x Lower bound: H(n) n+1 dx 1 x = 1 + ln(n) ln(1) = 1 + ln(n). = ln(n + 1) ln(1) = ln(n + 1). Exact bound: H n = ln(n) + γ + 1/(2n) 1/(12n 2 ) + ɛ n /(120n 4 ) for 0 < ɛ n < 1 and γ (Euler s constant). Amotz Bar-Noy (CUNY) Algorithms: Background 47 / 66

54 Harmonic Numbers Amotz Bar-Noy (CUNY) Algorithms: Background 48 / 66

55 Floor and Ceiling Formulas For real x and integer n: x 1 < x x x < x + 1 x + n = x + n x + n = x + n n/2 + n/2 = n log 2 (n + 1) = 1 + log 2 n Amotz Bar-Noy (CUNY) Algorithms: Background 49 / 66

56 Modular Arithmetic For integers n and m and a positive integer p: (n + m) mod p = (n mod p + m mod p) mod p. (n m) mod p = (n mod p m mod p) mod p. Amotz Bar-Noy (CUNY) Algorithms: Background 50 / 66

57 Stirling s Formula n! = ( n ) n ( ( )) 1 2πn 1 + Θ e n n n! = ( n ) n 2πn e log 2 (n!) = Θ(n log 2 n) Amotz Bar-Noy (CUNY) Algorithms: Background 51 / 66

58 Fibonacci Numbers The sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, Recursive Definition: 0 for k = 0, F k = 1 for k = 1, F k 1 + F k 2 for k 2. Wikipedia: Amotz Bar-Noy (CUNY) Algorithms: Background 52 / 66

59 Fibonacci Numbers - The Original Problem Story: A young pair of rabbits (one of each sex) is placed on an island. A pair of rabbits does not breed until they are 2 months old. After they are 2 months old, each pair of rabbits produces another pair each month. Problem: Find a recurrence relation for the number of pairs of rabbits on the island after k months, assuming that no rabbits ever die. Solution: F k pairs of rabbits on the island after k months. Amotz Bar-Noy (CUNY) Algorithms: Background 53 / 66

60 The Golden Ratio Find a rectangle of size x 1 such that after removing a 1 1 square the remaining rectangle has size 1 1 x. x 1 = 1 x. Amotz Bar-Noy (CUNY) Algorithms: Background 54 / 66

61 Solving the Equation x = x Start with x 1 = 1. Define x i = x i 1 for i > 1. x 2 = = 2. x 3 = = 3 2 = 1.5. x 4 = /2 = x 5 = /3 = 8 5 = 1.6. x 6 = /5 = 13 8 = x 7 = /8 = x 8 = /13 = x 9 = /21 = Amotz Bar-Noy (CUNY) Algorithms: Background 55 / 66

62 Solving the Equation x = x 1 x = Amotz Bar-Noy (CUNY) Algorithms: Background 56 / 66

63 Solving the Equation x = x Proposition: x i = F i+1 F i. Proof: Initially x 1 = 1 = 1 1 = F 2 F 1. Assume the statement is correct for x i and prove it by induction for x i+1. x i+1 = = 1 + x i F i+1 /F i = 1 + F i F i+1 = F i+1 + F i F i+1 = F i+2 F i+1. Amotz Bar-Noy (CUNY) Algorithms: Background 57 / 66

64 The Golden Ratio Definition: The positive root of the equation x 2 x 1 = 0. Remark: This equation is equivalent to x = x. Solution to the quadratic equation: x = 1± 5 2. The positive root: φ = (1 + 5)/ The negative root: ˆφ = (1 5)/2 = 1 φ Amotz Bar-Noy (CUNY) Algorithms: Background 58 / 66

65 k as a Function of the Fibonacci Number F k F k = (φ k ˆφ k )/ 5. F k+1 = φf k + ˆφ k. ˆφ < 1 F k = φk 5 rounded to the nearest integer. Corollary: log φ (F k ) + 1 k log φ (F k ) + 2 Corollary: k = Θ(log(F k )). Amotz Bar-Noy (CUNY) Algorithms: Background 59 / 66

66 Interesting Facts F k 1 F k+1 F 2 k = ( 1)k for k 1. F 5 F 7 F 2 6 = = 1. F 6 F 8 F 2 7 = = 1. F mk is a multiple of F m and F k for m, k 1. F 15 = 610 is a multiple of F 5 = 5 and F 3 = 2. gcd(f m, F k ) = F gcd(m,k). gcd(f 12, F 18 ) = gcd(144, 2584) = 8 = F 6 = F gcd(18,12). Amotz Bar-Noy (CUNY) Algorithms: Background 60 / 66

67 A Paradox? The area of the left rectangle is 8 8 = 64. The area of the right rectangle is 5 13 = 65. But both rectangles are composed of the same 4 shapes! Animation: Amotz Bar-Noy (CUNY) Algorithms: Background 61 / 66

68 A Paradox? The area of the left rectangle is 8 8 = 64. The area of the right rectangle is 5 13 = 65. But both rectangles are composed of the same 4 shapes! Animation: Hint: F k 1 F k+1 F 2 k = ( 1)k. Amotz Bar-Noy (CUNY) Algorithms: Background 61 / 66

69 A Paradox? The area of the left rectangle is 8 8 = 64. The area of the right rectangle is 5 13 = 65. But both rectangles are composed of the same 4 shapes! Animation: Hint: F k 1 F k+1 F 2 k = ( 1)k. Solution: Amotz Bar-Noy (CUNY) Algorithms: Background 61 / 66

70 A Paradox? The area of the left rectangle is 8 8 = 64. The area of the right rectangle is 5 13 = 65. But both rectangles are composed of the same 4 shapes! Animation: Hint: F k 1 F k+1 F 2 k = ( 1)k. Solution: The missing square puzzle: Amotz Bar-Noy (CUNY) Algorithms: Background 61 / 66

71 Kilometers and Miles One mile is approximately φ kilometers. F k1 + F k2 + + F kh kilometers are approximately F k1 1 + F k F kh 1 miles. 32 = = F 8 + F 6 + F 4 kilometers are approximately F 7 + F 5 + F 3 = = 20 miles. 32 = = F 8 + F 6 + F 4 miles are approximately F 9 + F 7 + F 5 = = 52 kilometers. Amotz Bar-Noy (CUNY) Algorithms: Background 62 / 66

72 Combinatorics Number of permutations of an n-element set: n! Number of k-combinations of an n-element set: ( ) n k = n! k!(n k)! Number of subsets of an n-element set: 2 n Number of binary strings of length n: 2 n Number of k-ary strings of length n: k n Amotz Bar-Noy (CUNY) Algorithms: Background 63 / 66

73 Binomial Coefficients C(n, k) = ( n k) : The number of size k subsets from the set {1, 2,..., n}. The coefficient of x k in the expansion of (1 + x) n : (1 + x) n = ( ( n 0) + n ( 1) x + n ) 2 x ( ) n n x n Recursive Definition: C(n, k) = ( ) n k = 1 for k = 0 1 for k = n ( n 1 ) ( k 1 + n 1 ) k for 0 < k < n 0 otherwise Amotz Bar-Noy (CUNY) Algorithms: Background 64 / 66

74 The Pascal Triangle Amotz Bar-Noy (CUNY) Algorithms: Background 65 / 66

75 Binomial Coefficients Some Facts ( n ) k = n! k!(n k)! = n(n 1) (n k+1) k(k 1) 1 ( n ( k) = n ) n k ( n k ) k n k n 1 k 1 n k+1 1 (n k + 1) k ( n ) k = Θ(n k ) for a fixed k Amotz Bar-Noy (CUNY) Algorithms: Background 66 / 66