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CSE 321 Discrete StructuresPowerPoint Presentation

CSE 321 Discrete Structures

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Announcements

- Readings
- Monday, Wednesday
- Induction and recursion
- 4.1-4.3 (5th Edition: 3.3-3.4)

- Induction and recursion
- Midterm:
- Friday, February 8
- In class, closed book
- Estimated grading weight:
- MT 12.5%, HW 50%, Final 37.5%

- Monday, Wednesday

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

Cute Application: Checkerboard Tiling with Trinominos

Prove that a 2k 2k checkerboard with one

square removed can be tiled with:

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 then S is divisible

- 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 then S is divisible

- 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 then S is divisible

- f0 = 0; f1 = 1; fn = fn-1 + fn-2

Bounding the Fibonacci Numbers then S is divisible

- Theorem: 2n/2 fn 2n for n 6

Recursive Definitions of Sets then S is divisible

- 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 then S is divisible

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 then S is divisible

- The set * of strings over the alphabet is defined
- Basis: * ( is the empty string)
- Recursive: if w *, x , then wx *

Families of strings over then S is divisible = {a, b}

- L1
- L1
- w L1 then awb L1

- L2
- L2
- w L2 then aw L2
- w L2 then wb L2

Function definitions then S is divisible

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 then S is divisible

- 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 then S is divisible

- 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 then S is divisible

- 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 then S is divisible

- 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 then S is divisible

- (, 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 then S is divisible

- 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 then S is divisible

- 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 then S is divisible

- 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? then S is divisible

Structural Induction then S is divisible

- 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 then S is divisible

- 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 right parentheses

- 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 right parentheses

- If T is a FBBT, then N(T) = 2Ht(T) - 1

Binary Trees right parentheses

- 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 right parentheses

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 right parentheses

N(T) = N(T1) + N(T2) + 1

k – 1 + k-1 – 1 + 1

k + k-1 – 1 [2 = + 1]

k+1 – 1

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