Today. Climb an infinite ladder? Gauss and Induction i = 0 i = n ( - PowerPoint PPT Presentation

Today. Climb an infinite ladder? Gauss and Induction i = 0 i = n ( n + 1 ) Child Gauss: ( ∀ n ∈ N )( ∑ n ) Proof? 2 i = 0 i = k ( k + 1 ) Idea: assume predicate P ( n ) for n = k . P ( k ) is ∑ k . 2 Is predicate, P ( n ) true for n = k + 1? Principle of Induction.(continued.) i = 1 i )+( k + 1 ) = k ( k + 1 ) + k + 1 = k ( k + 1 )+ 2 ( k + 1 ) = ( k + 1 )( k + 2 ) ∑ k + 1 i = 0 i = ( ∑ k . 2 2 2 P ( n + 3 ) P ( 0 ) ∧ ( ∀ n ∈ N ) P ( n ) = ⇒ P ( n + 1 ) How about k + 2. Same argument starting at k + 1 works! P ( n + 2 ) Induction Step. P ( k ) = ⇒ P ( k + 1 ) . And we get... P ( 0 ) P ( n + 1 ) ∀ k , P ( k ) = ⇒ P ( k + 1 ) Is this a proof? It shows that we can always move to the next step. ( ∀ n ∈ N ) P ( n ) . P ( n ) P ( 0 ) = ⇒ P ( 1 ) = ⇒ P ( 2 ) = ⇒ P ( 3 ) ... Need to start somewhere. P ( 0 ) is ∑ 0 i = 0 i = ( 0 )( 0 + 1 ) ...Yes for 0, and we can conclude Yes for 1... Base Case. ( ∀ n ∈ N ) P ( n ) 2 and we can conclude Yes for 2....... Statement is true for n = 0 P ( 0 ) is true plus inductive step = ⇒ true for n = 1 ( P ( 0 ) ∧ ( P ( 0 ) = ⇒ P ( 1 ))) = ⇒ P ( 1 ) P(3) plus inductive step = ⇒ true for n = 2 ( P ( 1 ) ∧ ( P ( 1 ) = ⇒ P ( 2 ))) = ⇒ P ( 2 ) P(2) ... P(1) true for n = k = ⇒ true for n = k + 1 ( P ( k ) ∧ ( P ( k ) = ⇒ P ( k + 1 ))) = ⇒ P ( k + 1 ) P(0) ... Predicate, P ( n ) , True for all natural numbers! Proof by Induction. Your favorite example of forever..or the natural numbers... Another Induction Proof. Four Color Theorem. Two color theorem: example. Theorem: For every n ∈ N , n 3 − n is divisible by 3. (3 | ( n 3 − n ) ). Theorem: Any map can be colored so that those regions that share Any map formed by dividing the plane into regions by drawing straight an edge have different colors. lines can be properly colored with two colors. Proof: By induction. Base Case: P ( 0 ) is “ ( 0 3 ) − 0” is divisible by 3. Yes! B R Induction Step: ( ∀ k ∈ N ) , P ( k ) = ⇒ P ( k + 1 ) Induction Hypothesis: k 3 − k is divisible by 3. B R or k 3 − k = 3 q for some integer q . B R B R R B ( k + 1 ) 3 − ( k + 1 ) = k 3 + 3 k 2 + 3 k + 1 − ( k + 1 ) = k 3 + 3 k 2 + 2 k B R = ( k 3 − k )+ 3 k 2 + 3 k Subtract/add k (Poll!) = 3 q + 3 ( k 2 + k ) R B Induction Hyp. Factor. = 3 ( q + k 2 + k ) (Un)Distributive + over × R B R B Or ( k + 1 ) 3 − ( k + 1 ) = 3 ( q + k 2 + k ) . R B R B . ( q + k 2 + k ) is integer (closed under addition and multiplication). Check Out: “Four corners”. ⇒ ( k + 1 ) 3 − ( k + 1 ) is divisible by 3. Proper coloring: for each line segment the regions on the two sides = States connected at a point, can have same color. have different colors.1 Thus, ( ∀ k ∈ N ) P ( k ) = ⇒ P ( k + 1 ) Fact: Swapping red and blue gives another valid colors. Thus, theorem holds by induction. Quick Test: Which states? Utah. Colorado. New Mexico. Arizona.

Two color theorem: proof illustration. Strenthening Induction Hypothesis. Tiling Cory Hall Courtyard. s r o Theorem: The sum of the first n odd numbers is a perfect square. l o C c R R B B Use these L -tiles. A Theorem: The sum of the first n odd numbers is n 2 . h c t i w B R s k th odd number is 2 ( k − 1 )+ 1. s To Tile this 4 × 4 courtyard. w i t c B R R R B R B R B R B R B B B h Base Case 1 (first odd number) is 1 2 . switch E R R R R Induction Hypothesis Sum of first k odds is perfect square a 2 = k 2 . C E Alright! B B Tiled 4 × 4 square with 2 × 2 L -tiles. B Induction Step 1. The ( k + 1)st odd number is 2 k + 1. R R R B R B 2. Sum of the first k + 1 odds is with a center hole. B B B B B B B R R R R R B a 2 + 2 k + 1 = k 2 + 2 k + 1 Base Case. B 1. Add line. ???? D 3. k 2 + 2 k + 1 = ( k + 1 ) 2 2. Get inherited color for split regions A 3. Switch on one side of new line. ... P(k+1)! D (Fixes conflicts along new line, and makes no new ones along previous line.) Can we tile any 2 n × 2 n with L -tiles (with a hole) for every n ! Algorithm gives P ( k ) = ⇒ P ( k + 1 ) . Hole have to be there? Maybe just one? Hole in center? Hole can be anywhere! Theorem: Can tile the 2 n × 2 n square to leave a hole adjacent to the Theorem: Can tile the 2 n × 2 n to leave a hole adjacent anywhere. Theorem: Any tiling of 2 n × 2 n square has to have one hole. center. Proof: The remainder of 2 2 n divided by 3 is 1. Proof: Better theorem ...better induction hypothesis! Base case: true for k = 0. 2 0 = 1 Base case: A single tile works fine. Base case: Sure. A tile is fine. The hole is adjacent to the center of the 2 × 2 square. Ind Hyp: 2 2 k = 3 a + 1 for integer a . Flipping the orientation can leave hole anywhere. Induction Hypothesis: Any 2 n × 2 n square can be tiled with a hole at the center. Induction Hypothesis: “Any 2 n × 2 n square can be tiled with a hole anywhere. ” 2 2 k ∗ 2 2 2 2 ( k + 1 ) Consider 2 n + 1 × 2 n + 1 square. = 2 n + 1 4 ∗ 2 2 k = = 4 ∗ ( 3 a + 1 ) = 12 a + 3 + 1 Use induction hypothesis in each. 2 n + 1 What to do now??? = 3 ( 4 a + 1 )+ 1 2 n a integer = ⇒ ( 4 a + 1 ) is an integer. Use L-tile and ... we are done. 2 n

Strong Induction. Well Ordering Principle and Induction. Well ordering principle. Theorem: Every natural number n > 1 can be written as a (possibly If ( ∀ n ) P ( n ) is not true, then ( ∃ n ) ¬ P ( n ) . trivial) product of primes. Definition: A prime n has exactly 2 factors 1 and n . Consider smallest m , with ¬ P ( m ) , m ≥ 0 Base Case: n = 2. P ( m − 1 ) = ⇒ P ( m ) must be false (assuming P ( 0 ) holds.) Induction Step: Thm: All natural numbers are interesting. P ( n ) = “ n can be written as a product of primes. “ This is a proof of the induction principle! 0 is interesting... Either n + 1 is a prime or n + 1 = a · b where 1 < a , b < n + 1 . I.e., Let n be the first uninteresting number. P ( n ) says nothing about a , b ! ( ¬∀ n ) P ( n ) = ⇒ (( ∃ n ) ¬ ( P ( n − 1 ) = ⇒ P ( n )) . But n − 1 is interesting and n is uninteresting, so this is the first uninteresting number. Strong Induction Principle: If P ( 0 ) and (Contrapositive of Induction principle (assuming P ( 0 ) ) But this is interesting. ( ∀ k ∈ N )(( P ( 0 ) ∧ ... ∧ P ( k )) = ⇒ P ( k + 1 )) , It assumes that there is a smallest m where P ( m ) does not hold. Thus, there is no smallest uninteresting natural number. then ( ∀ k ∈ N )( P ( k )) . The Well ordering principle states that for any subset of the natural Thus: All natural numbers are interesting. numbers there is a smallest element. P ( 0 ) = ⇒ P ( 1 ) = ⇒ P ( 2 ) = ⇒ P ( 3 ) = ⇒ ··· Examples: even numbers, odd numbers, primes, non-primes, etc.. Strong induction hypothesis: “ a and b are products of primes” True for rational numbers? Poll. = ⇒ “ n + 1 = a · b = ( factorization of a )( factorization of b ) ” Note: can do with different definition of smallest n + 1 can be written as the product of the prime factors! Tournaments have short cycles Tournament has a cycle of length 3 if at all. Tournaments have long paths. Def: A round robin tournament on n players : all pairs p and q play, Assume the the smallest cycle is of length k . and either p → q ( p beats q ) or q → p ( q beats q .) Case 1: Of length 3. Done. Def: A Hamiltonian path : a sequence Def: A round robin tournament on n players : every player p plays p 1 ,..., p n , ( ∀ i , 0 ≤ i < n ) p i → p i + 1 . Case 2: Of length larger than 3. every other player q , and either p → q ( p beats q ) or q → p ( q beats ··· 2 1 7 p .) p 2 Base: True for two vertices. 1 2 Def: A cycle : a sequence of p 1 ,..., p k , p i → p i + 1 and p k → p 1 . (Also for one, but two is more fun as base case!) Case 1: “ p 3 → p 1 ” = ⇒ 3 cycle B Tournament on n + 1 people, D Remove arbitrary person → yield tournament on n − 1 people. p 1 p 3 Contradiction. By induction hypothesis: There is a sequence p 1 ,..., p n A C contains all the people where p i → p i + 1 p k p 4 = ⇒ k − 1 length cycle! Case 2: “ p 1 → p 3 ” Theorem: Any tournament that has a cycle has a cycle of length 3. ··· ··· a a m c c W b b ··· ··· Contradicts assumption of smallest k ! ··· ··· If p is big winner, put at beginning. Big loser at end. ··· If neither, find first place i , where p beats p i . p 1 ,..., p i − 1 , p , p i ,... p n is Hamiltonion path.