CS70: Discrete Math and Probability And we get... Are we done? It - PowerPoint PPT Presentation

Induction Gauss and Induction Child Gauss: ( ∀ n ∈ N )( ∑ n i = 1 i = n ( n + 1 ) ) Proof? 2 Idea: assume predicate P ( n ) for n = k . P ( k ) is ∑ k i = 1 i = k ( k + 1 ) . 2 Is predicate, P ( n ) true for n = k + 1? i = 1 i )+( k + 1 ) = k ( k + 1 ) + k + 1 = ( k + 1 )( k + 2 ) ∑ k + 1 i = 1 i = ( ∑ k . Principle of Induction. 2 2 How about k + 2. Same argument starting at k + 1 works! P ( 0 ) ∧ ( ∀ n ∈ N ) P ( n ) = ⇒ P ( n + 1 ) Induction Step. P ( k ) = ⇒ P ( k + 1 ) . CS70: Discrete Math and Probability And we get... Are we done? It shows that we can always move to the next step. June 22, 2016 i = 0 i = 1 = ( 0 )( 0 + 1 ) ( ∀ n ∈ N ) P ( n ) . Need to start somewhere. P ( 0 ) is ∑ 0 Base Case. 2 ...Yes for 0, and we can conclude Yes for 1... Statement is true for n = 0 P ( 0 ) is true plus inductive step = ⇒ true for n = 1 ( P ( 0 ) ∧ ( P ( 0 ) = and we can conclude Yes for 2....... ⇒ P ( 1 ))) = ⇒ P ( 1 ) plus inductive step = ⇒ true for n = 2 ( P ( 1 ) ∧ ( P ( 1 ) = ⇒ P ( 2 ))) = ⇒ P ( 2 ) ... true for n = k = ⇒ true for n = k + 1 ( P ( k ) ∧ ( P ( k ) = ⇒ P ( k + 1 ))) = ⇒ P ( k + 1 ) ... Predicate, P ( n ) , True for all natural numbers! Is this a proof? Not really. Just an idea, not formal enough to be a proof yet 1 2 Induction Notes visualization Climb an infinite ladder? The canonical way of proving statements of the form ( ∀ k ∈ N )( P ( k )) Note’s visualization: an infinite sequence of dominos. • For all natural numbers n , 1 + 2 ··· n = n ( n + 1 ) . 2 • For all n ∈ N , n 3 − n is divisible by 3. P ( n + 3 ) • The sum of the first n odd integers is a perfect square. P ( n + 2 ) P ( 0 ) P ( n + 1 ) The basic form ∀ k , P ( k ) = ⇒ P ( k + 1 ) P ( 0 ) = ⇒ P ( 1 ) = ⇒ P ( 2 ) = ⇒ P ( 3 ) ... P ( n ) • Prove P ( 0 ) . “Base Case”. ( ∀ n ∈ N ) P ( n ) Prove they all fall down; • P ( k ) = ⇒ P ( k + 1 ) • P ( 0 ) = “First domino falls” • Assume P ( k ) , “Induction Hypothesis” P(3) • Prove P ( k + 1 ) . “Induction Step.” • ( ∀ k ) P ( k ) = ⇒ P ( k + 1 ) : “ k th domino falls implies that k + 1st domino falls” P(2) P ( n ) true for all natural numbers n !!! P(1) Get to use P ( k ) to prove P ( k + 1 ) ! ! ! ! P(0) 3 4 5

Again: Simple induction proof. Try it yourself! Homework, Exam For all natural numbers n , 0 2 + 1 2 + 2 2 ··· n 2 = 1 6 n ( n + 1 )( 2 n + 1 ) Define predicate p ( n ) as 0 2 + 1 2 + 2 2 ··· n 2 = 1 6 n ( n + 1 )( 2 n + 1 ) for n ∈ N Theorem: For all natural numbers n , 0 + 1 + 2 ··· n = n ( n + 1 ) 2 Base case: For n = 0, 0 2 = 1 6 ∗ 0 ∗ 1 ∗ 1 = 0, p ( 0 ) is true. Base Case: Does 0 = 0 ( 0 + 1 ) ? Yes. 2 Induction hypothesis: assume p ( k ) is true for some natural number k . Induction Hypothesis: P ( k ) is true: 1 + ··· + k = k ( k + 1 ) 2 Inductive steps: need to prove p ( k ) = ⇒ p ( k + 1 ) Induction Step: Show ∀ k ≥ 0 , P ( k ) = ⇒ P ( k + 1 ) We will use some problems from homework in our exams, 0 2 + 1 2 + 2 2 ··· + k 2 +( k + 1 ) 2 ( 0 2 + 1 2 + 2 2 ··· + k 2 )+( k + 1 ) 2 k ( k + 1 ) = 1 + ··· + k +( k + 1 ) = +( k + 1 ) 2 with some modifications like the question we just saw. 1 6 k ( k + 1 )( 2 k + 1 )+( k + 1 ) 2 = k 2 + k + 2 ( k + 1 ) = Take homework seriously, and study the solutions carefully after we release them. 2 ( k + 1 )( 1 = 6 k ( 2 k + 1 )+( k + 1 )) k 2 + 3 k + 2 = 2 1 6 ( k + 1 )( 2 k 2 + k + 6 k + 6 ) = ( k + 1 )( k + 2 ) = 2 1 = 6 ( k + 1 )( k + 2 )( 2 k + 3 ) 1 P ( k + 1 ) !. By principle of induction... = 6 ( k + 1 )( k + 2 )( 2 ( k + 1 )+ 1 ) p ( k + 1 ) is true.By principle of induction... 6 7 8 Another Induction Proof. Four Color Theorem. Two color theorem: example. Any map formed by dividing the plane M into regions by drawing straight lines can be colored with two colors so that those regions share an edge have different colors. Theorem: For every n ∈ N , n 3 − n is divisible by 3. (3 | ( n 3 − n ) ). Theorem: Any map can be 4-colored so that those regions that share an edge have different colors. B R Proof: By induction. Base Case: P ( 0 ) is “ ( 0 3 ) − 0” is divisible by 3. Yes! Induction Hypothesis: k 3 − k is divisible by 3. R B or k 3 − k = 3 q for some integer q . Induction Step: ( ∀ k ∈ N ) , P ( k ) = ⇒ P ( k + 1 ) 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 = ( k 3 − k )+ 3 k 2 + 3 k Subtract/add k R B = 3 q + 3 ( k 2 + k ) Induction Hyp. Factor. = 3 ( q + k 2 + k ) (Un)Distributive + over × R B Or ( k + 1 ) 3 − ( k + 1 ) = 3 ( q + k 2 + k ) . ( q + k 2 + k ) is integer (closed under addition and multiplication). R B R B ⇒ ( k + 1 ) 3 − ( k + 1 ) is divisible by 3. = R B R B Thus, ( ∀ k ∈ N ) P ( k ) = ⇒ P ( k + 1 ) Not gonna prove it. . Thus, theorem holds by induction. Fact: Swapping red and blue gives another valid colors. 9 10 11

Two color theorem: proof illustration. Strenthening Induction Hypothesis. Tiling Cory Hall Courtyard. s Use these L -tiles. C A r o l o B R B R c h c Theorem: The sum of the first n odd numbers is a perfect square. t i w Theorem: The sum of the first n odd numbers is n 2 . s To Tile this 4 × 4 courtyard. B R s w i t c k th odd number is 2 ( k − 1 )+ 1. B B R R B R R h R B B R R B B B E Base Case 1 (1th odd number) is 1 2 . switch R R R R Alright! C E Induction Hypothesis Sum of first k odds is perfect square a 2 = k 2 . 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 B R R R B with a center hole. 2. Sum of the first k + 1 odds is B B B B B B B R R R R R B a 2 + 2 k + 1 = k 2 + 2 k + 1 B Base Case. 3. k 2 + 2 k + 1 = ( k + 1 ) 2 1. Add line. D ... P(k+1)! 2. Get inherited color for split regions A 3. Switch on one side of new line. D (Fixes conflicts along line, and makes no new ones.) Algorithm gives P ( k ) = ⇒ P ( k + 1 ) . Can we tile any 2 n × 2 n with L -tiles (with a hole) for every n ! 12 13 14 Hole have to be there? Maybe just one? Hole in center? Hole can be anywhere! Theorem: Can tile the 2 n × 2 n to leave a hole adjacent anywhere. Theorem: Can tile the 2 n × 2 n square to leave a hole adjacent to the center. Theorem: Any tiling of 2 n × 2 n square has to have one hole. Proof: Better theorem ...better induction hypothesis! Proof: The remainder of 2 2 n divided by 3 is 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. Base case: true for k = 0. 2 0 = 1 Induction Hypothesis: Flipping the orientation can leave hole anywhere. Any 2 n × 2 n square can be tiled with a hole at the center. Ind Hyp: 2 2 k = 3 a + 1 for integer a . Induction Hypothesis: “Any 2 n × 2 n square can be tiled with a hole anywhere. ” 2 n + 1 Consider 2 n + 1 × 2 n + 1 square. 2 2 k ∗ 2 2 2 2 ( k + 1 ) = 4 ∗ 2 2 k = = 4 ∗ ( 3 a + 1 ) = 12 a + 3 + 1 2 n + 1 What to do now??? Use induction hypothesis in each. = 3 ( 4 a + 1 )+ 1 2 n a integer = ⇒ ( 4 a + 1 ) is an integer. 2 n Use L-tile and ... we are done. 15 16 17