1. CSE 321 Discrete Structures Winter 2008 Lecture 12 Induction

2. Announcements Readings –Monday, Wednesday Induction and recursion –4.1-4.3 (5 th Edition: 3.3-3.4) –Midterm: Friday, February 8 In class, closed book Estimated grading weight: –MT 12.5%, HW 50%, Final 37.5%

3. Induction Example Prove 3 | 2 2n -1 for n  0

4. Induction as a rule of Inference P(0)  k (P(k)  P(k+1))   n P(n)

5. 1 + 2 + 4 + … + 2 n = 2 n+1 - 1

6. Harmonic Numbers

7. Cute Application: Checkerboard Tiling with Trinominos Prove that a 2 k  2 k checkerboard with one square removed can be tiled with:

8. Strong Induction P(0)  k ((P(0)  P(1)  P(2)  …  P(k))  P(k+1))   n P(n)

9. Player 1 wins n  2 Chomp! Winning strategy: chose the lower corner square Theorem: Player 2 loses when faced with an n  2 board missing the lower corner square

10. Induction Example A set of S integers is non-divisible if there is no pair of integers a, b in S where a divides b. If there is a pair of integers a, b in S, where a divides b, then S is divisible. Given a set S of n+1 positive integers, none exceeding 2n, show that S is divisible. What is the largest subset non-divisible subset of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }.

11. If S is a set of n+1 positive integers, none exceeding 2n, then S is divisible Base case: n =1 Suppose the result holds for n –If S is a set of n+1 positive integers, none exceeding 2n, then S is divisible –Let T be a set of n+2 positive integers, none exceeding 2n+2. Suppose T is non-divisible.

12. Proof by contradiction Claim: 2n+1  T and 2n + 2  T Claim: n+1  T Let T* = T – {2n+1, 2n+2}  {n+1} If T is non-divisible, T* is also non-divisible /

13. Recursive Definitions F(0) = 0; F(n + 1) = F(n) + 1; F(0) = 1; F(n + 1) = 2  F(n); F(0) = 1; F(n + 1) = 2 F(n)

14. Fibonacci Numbers f 0 = 0; f 1 = 1; f n = f n-1 + f n-2

15. Bounding the Fibonacci Numbers Theorem: 2 n/2  f n  2 n for n  6

16. Recursive Definitions of Sets Recursive definition –Basis step: 0  S –Recursive step: if x  S, then x + 2  S –Exclusion rule: Every element in S follows from basis steps and a finite number of recursive steps

17. Recursive definitions of sets Basis: 6  S; 15  S; Recursive: if x, y  S, then x + y  S; Basis: [1, 1, 0]  S, [0, 1, 1]  S; Recursive: if [x, y, z]  S,  in R, then [  x,  y,  z]  S if [x 1, y 1, z 1 ], [x 2, y 2, z 2 ]  S then [x 1 + x 2, y 1 + y 2, z 1 + z 2 ] Powers of 3

18. Strings The set  * of strings over the alphabet  is defined –Basis:   * ( is the empty string) –Recursive: if w   *, x  , then wx   *

19. Families of strings over  = {a, b} L 1 –  L 1 –w  L 1 then awb  L 1 L 2 –  L 2 –w  L 2 then aw  L 2 –w  L 2 then wb  L 2

20. Function definitions Len( ) = 0; Len(wx) = 1 + Len(w); for w   *, x   Concat(w, ) = w for w   * Concat(w 1,w 2 x) = Concat(w 1,w 2 )x for w 1, w 2 in  *, x  

21. Well Formed Fomulae Basis Step –T, F, and s, where is a propositional variable are in WFF Recursive Step –If E and F are in WFF then (  E), (E  F), (E  F), (E  F) and (E  F) are in WFF

22. Tree definitions A single vertex r is a tree with root r. Let t 1, t 2, …, t n be trees with roots r 1, r 2, …, r n respectively, and let r be a vertex. A new tree with root r is formed by adding edges from r to r 1,…, r n.

23. Extended Binary Trees The empty tree is a binary tree. Let r be a node, and T 1 and T 2 binary trees. A binary tree can be formed with T 1 as the left subtree and T 2 as the right subtree. If T 1 is non-empty, there is an edge from the root of T 1 to r. Similarly, if T 2 is non-empty, there is an edge from the root of T 2 to r.

24. Full binary trees The vertex r is a FBT. If r is a vertex, T 1 a FBT with root r 1 and T 2 a FBT with root r 2 then a FBT can be formed with root r and left subtree T 1 and right subtree T 2 with edges r  r 1 and r  r 2.

25. Simplifying notation ( , T 1, T 2 ), tree with left subtree T 1 and right subtree T 2  is the empty tree Extended Binary Trees (EBT) –   EBT –if T 1, T 2  EBT, then ( , T 1, T 2 )  EBT Full Binary Trees (FBT) –   FBT –if T 1, T 2  FBT, then ( , T 1, T 2 )  FBT

26. Recursive Functions on Trees N(T) - number of vertices of T N(  ) = 0; N(  ) = 1 N( , T 1, T 2 ) = 1 + N(T 1 ) + N(T 2 ) Ht(T) – height of T Ht(  ) = 0; Ht(  ) = 1 Ht( , T 1, T 2 ) = 1 + max(Ht(T 1 ), Ht(T 2 )) NOTE: Height definition differs from the text Base case H(  ) = 0 used in text

27. More tree definitions: Fully balanced binary trees  is a FBBT. if T 1 and T 2 are FBBTs, with Ht(T 1 ) = Ht(T 2 ), then ( , T 1, T 2 ) is a FBBT.

28. And more trees: Almost balanced trees  is a ABT. if T 1 and T 2 are ABTs with Ht(T 1 ) -1  Ht(T 2 )  Ht(T 1 )+1 then ( , T 1, T 2 ) is a ABT.

29. Is this Tree Almost Balanced?

30. Structural Induction Show P holds for all basis elements of S. Show that if P holds for elements used to construct a new element of S, then P holds for the new element.

31. Prove all elements of S are divisible by 3 Basis: 21  S; 24  S; Recursive: if x, y  S, then x + y  S;

32. Prove that WFFs have the same number of left parentheses as right parentheses

33. Well Formed Fomulae Basis Step –T, F, and s, where is a propositional variable are in WFF Recursive Step –If E and F are in WFF then (  E), (E  F), (E  F), (E  F) and (E  F) are in WFF

34. Fully Balanced Binary Tree If T is a FBBT, then N(T) = 2 Ht(T) - 1

35. Binary Trees If T is a binary tree, then N(T)  2 Ht(T) - 1 If T =  : If T = ( , T 1, T 2 ) Ht(T 1 ) = x, Ht(T 2 ) = y N(T 1 )  2 x, N(T 2 )  2 y N(T) = N(T 1 ) + N(T 2 ) + 1  2 x – 1 + 2 y – 1 + 1  2 Ht(T) -1 + 2 Ht(T) – 1 – 1  2 Ht(T) - 1

36. Almost Balanced Binary Trees Let  = (1 + sqrt(5))/2 Prove N(T)   Ht(T) – 1 Base case: Recursive Case: T = ( , T 1, T 2 ) Let Ht(T) = k + 1 Suppose Ht(T 1 )  Ht(T 2 ) Ht(T 1 ) = k, Ht(T 2 ) = k or k-1

37. Almost Balanced Binary Trees N(T) = N(T 1 ) + N(T 2 ) + 1   k – 1 +  k-1 – 1 + 1   k +  k-1 – 1 [   =  + 1]   k+1 – 1