CSE 321 Discrete Structures

CSE 321 Discrete Structures Winter 2008 Lecture 12 Induction

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

Induction Example • Prove 3 | 22n -1 for n  0 Examples to show its true P(0) P(k)  P(k+1)

Induction as a rule of Inference P(0)  k (P(k)  P(k+1))  n P(n) Formal steps Show P(0) Assume P(k), Prove P(k+1), Conclude P(k) P(k+1) Conclude  k (P(k) P(k+1)) Conclude  n P(n)

1 + 2 + 4 + … + 2n = 2n+1 - 1

Harmonic Numbers

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

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

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

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 }.

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.

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 /

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) = 2F(n)

Fibonacci Numbers • f0 = 0; f1 = 1; fn = fn-1 + fn-2

Bounding the Fibonacci Numbers • Theorem: 2n/2 fn 2n for n  6

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

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 [x1, y1, z1], [x2, y2, z2]  S then [x1 + x2, y1 + y2, z1 + z2] Powers of 3

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

Families of strings over  = {a, b} • L1 •  L1 • w  L1 then awb  L1 • L2 •  L2 • w  L2 then aw  L2 • w  L2 then wb  L2

Function definitions Len() = 0; Len(wx) = 1 + Len(w); for w *, x  Concat(w, ) = w for w * Concat(w1,w2x) = Concat(w1,w2)x for w1, w2 in *, x 

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

Tree definitions • A single vertex r is a tree with root r. • Let t1, t2, …, tn be trees with roots r1, r2, …, rn respectively, and let r be a vertex. A new tree with root r is formed by adding edges from r to r1,…, rn.

Extended Binary Trees • The empty tree is a binary tree. • Let r be a node, and T1 and T2 binary trees. A binary tree can be formed with T1 as the left subtree and T2 as the right subtree. If T1 is non-empty, there is an edge from the root of T1 to r. Similarly, if T2 is non-empty, there is an edge from the root of T2 to r.

Full binary trees • The vertex r is a FBT. • If r is a vertex, T1 a FBT with root r1 and T2 a FBT with root r2 then a FBT can be formed with root r and left subtree T1 and right subtree T2 with edges r r1 and r r2.

Simplifying notation • (, T1, T2), tree with left subtree T1 and right subtree T2 •  is the empty tree • Extended Binary Trees (EBT) •  EBT • if T1, T2 EBT, then (, T1, T2)  EBT • Full Binary Trees (FBT) •  FBT • if T1, T2 FBT, then (, T1, T2)  FBT

Recursive Functions on Trees • N(T) - number of vertices of T • N() = 0; N() = 1 • N(, T1, T2) = 1 + N(T1) + N(T2) • Ht(T) – height of T • Ht() = 0; Ht() = 1 • Ht(, T1, T2) = 1 + max(Ht(T1), Ht(T2)) NOTE: Height definition differs from the text Base case H() = 0 used in text

More tree definitions: Fully balanced binary trees •  is a FBBT. • if T1 and T2 are FBBTs, with Ht(T1) = Ht(T2), then (, T1, T2) is a FBBT.

And more trees: Almost balanced trees •  is a ABT. • if T1 and T2 are ABTs with Ht(T1) -1  Ht(T2)  Ht(T1)+1 then (, T1, T2) is a ABT.

Is this Tree Almost Balanced?

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.

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

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

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

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

Binary Trees • If T is a binary tree, then N(T)  2Ht(T) - 1 If T = : If T = (, T1, T2) Ht(T1) = x, Ht(T2) = y N(T1)  2x, N(T2)  2y N(T) = N(T1) + N(T2) + 1  2x – 1 + 2y – 1 + 1  2Ht(T) -1 + 2Ht(T) – 1 – 1  2Ht(T) - 1

Almost Balanced Binary Trees Let  = (1 + sqrt(5))/2 Prove N(T) Ht(T) – 1 Base case: Recursive Case: T = (, T1, T2) Let Ht(T) = k + 1 Suppose Ht(T1)  Ht(T2) Ht(T1) = k, Ht(T2) = k or k-1

Almost Balanced Binary Trees N(T) = N(T1) + N(T2) + 1 k – 1 + k-1 – 1 + 1 k + k-1 – 1 [2 =  + 1] k+1 – 1

CSE 321 Discrete Structures