Download 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
Induction as a rule of Inference
P(0)
 k (P(k)  P(k+1))
  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
 k+1 – 1
[2 =  + 1]