Download Introduction to the Theory of Computation

Introduction to the Theory
of Computation
AZADEH FARZAN
WINTER 2010
Monday, January 11, 2010
PROOFS
Proof by Contradiction
Proof by Construction
Jack sees Jill, who has just come in from outdoors
Proof by Induction
dry. Jack knows that it is not raining.
For any n, there exist n consecutive composite
integers.
His proof: if it were raining (assuming the statement
Proof:
here Jill
are the
n be
consecutive
composite integers
is false),
woud
wet (contradiction,
or false
for
any n:
consequence).
Therefore, it must not be raining.
(n+1)!+2, (n+1)!+3, ..., (n+1)!+n+1
Monday, January 11, 2010
INDUCTION
A method used to show that all elements of an infinite set
have a specific property.
Example: an arithmetic expression computes a desired quantity
for every assignment to its variables.
Example: program works correctly at all steps or for all input.
Every proof induction consists of two parts:
The basis.
The induction step.
Monday, January 11, 2010
IDEA
Let’s take the infinite set to be N = {1, 2, 3, . . . }.
Let’s call the property P.
The goal is to prove that P(k) is true for each natural
number k.
The basis: P(1) is true.
The induction step: if P(i) is true, then P(i + 1) is
true as well.
P(1) ⇒ P(2) ⇒ P(3) ⇒ . . . P(i) ⇒ . . .
Monday, January 11, 2010
Mathematical Induction is a way to show that for eac
we could eventually prove that P (n) is true.
FORMALLY: MATHEMATICAL INDUCTION PRINCIPLE
Definition 44 Principle of Mathematical Induction:
Let P () be a predicate defined on N. Then if
(1) P (0)
1 is true, and
(2) ∀n ∈ N [P (n) → P (n + 1)],
ntroduction to the
Theory
∀n ∈ N P (n) is true.
of Computation
then
7.1
Why it works
Suppose (1)AZADEH
and (2) hold.
Then
FARZAN
Monday, January 11, 2010
WINTER 2010
73
EXAMPLE
Prove
�n
i=1
i = n(n + 1)/2.
Proof by induction:
The basis (n=1): holds since 1 = 1 ×2/2.
�m
The induction step: if i=1 i = m(m + 1)/2, then
�m+1
i=1 i = (m + 1)(m + 2)/2.
m+1
�
i
=
i=1
i + (m + 1)
i=1
=
=
Monday, January 11, 2010
m
�
m(m + 1)/2 + (m + 1)
[m(m + 1) + 2(m + 1)] /2 = (m + 1)(m + 2)/2
LET’S FORMALIZE !
Monday, January 11, 2010
WELL-ORDERING
Principle of Well-ordering: Any non-empty subset of N has
minimum element.
∀A ⊆ N, s.t. A �= ∅, [∃a ∈ A s.t. (∀b ∈ A, a ≤ b)].
Is the principle true for Z?
Is the principle true for rational numbers in (0, 1)?
Monday, January 11, 2010
WHAT IS THE RELATION
BETWEEN WELL-ORDERING
AND INDUCTION?
Monday, January 11, 2010
Induction works as a proof method since:
Principle of well-ordering implies principle of mathematical induction.
proof on the board!
Monday, January 11, 2010
SIMPLE INDUCTION
Principle of Simple Induction: Let A be a set that satisfies
the following properties:
(i) 0 is in A;
(ii) for any i ∈ N, if i ∈ A, then i + 1 ∈ A.
Then N ⊆ A.
0 ∈ A ⇒ 1 ∈ A ⇒ 2 ∈ A ⇒ 3 ∈ A...
Monday, January 11, 2010
Problem: prove that every natural number greater than
1 can be written as a product of prime numbers.
LET’S LOOK AT A SLIGHTLY
DIFFERENT KIND OF
INDUCTION.
Monday, January 11, 2010
7.3 Strong Mathematical Induction
FORMALLY: STRONG INDUCTION PRINCIPLE
Definition 46 Strong Mathematical Induction Principle
If P (0) and
∀n ≥ 0 [(P (0) ∧ P (1) ∧ . . . ∧ P (n)) → P (n + 1)]
then ∀n P (n).
Using the above (stronger) principle, to prove P (n + 1) we can rely on inductive
hypothesis that all of P (0), P (1), . . . , P (n) are true.
The reason this principle works, and the proof that it does (based on wellordering principle) is identical to the weak induction.
EXAMPLE ! Given n black points and n white points in the plane, no three
WAIT,
WHY
of them
on a line,
proveDOES
that thereTHIS
exists aONE
way toWORK?
match them one-to-one, such
that the line segments between the matched pairs do not intersect.
Proof. Base Case. n = 1.
Just join them together; there is only one segment so clearly it doesn’t intersect
Monday, Januaryany
11, 2010others.
COMPLETE INDUCTION
Principle of Complete Induction: Let A be a set that satisfies
the following properties:
(*) for any i ∈ N, if ∀j < i, j ∈ A, then i ∈ A.
Then N ⊆ A.
, 1∈A ⇒
, 2∈A⇒
, 3 ∈ A ⇒ ...
0∈A ⇒
Monday, January 11, 2010
THREE EQUIVALENT NOTIONS
Theorem. The principles of (a) well-ordering, (b) simple
induction, and (c) complete induction are equivalent.
Monday, January 11, 2010
ALL DEFINITIONS
Principle of Well-ordering: Any non-empty subset of N has
minimum element.
Principle of Simple Induction: Let A be a set that satisfies
the following properties:
(i) 0 is in A;
(ii) for any i ∈ N, if i ∈ A, then i + 1 ∈ A.
Then N ⊆ A.
Principle of Complete Induction: Let A be a set that satisfies
the following properties:
(*) for any i ∈ N, if ∀j < i, j ∈ A, then i ∈ A.
Then N ⊆ A.
Monday, January 11, 2010
EXAMPLE
For every natural number n, the power set of a set with n
elements has 2n elements.
Basis: The power set of ∅ (with 0 elements) has 20 = 1 element
(∅ itself).
Induction step: if n element sets have 2n subsets, then n + 1
element sets have 2n+1 subsets.
Let |X| = n + 1 and x ∈ X and let Y = X − {x}. Then
|Y | = n. |X| has two types of subsets: (i) those which contain
x, and (ii) those which do not.
• Subsets of type (i) are subsets of Y as well, and there are
2n of them by IH.
• Subsets of type (ii) are subsets of Y if we remove x from
them, and there are 2n of them by IH.
Total number of subsets: 2n + 2n = 2n+1 .
Monday, January 11, 2010
NON-ZERO BASES
For all n ≥ 6, we have 2n > 10n.
Basis: n = 6, clearly 26 = 64 > 10.6 = 60.
Induction Step: For an arbitrary i ≥ 6, 2i > 10i ⇒ 2i+1 > 10(i + 1).
2
i+1
= 2.2
i
> 2.10i
= 10i + 10i
> 10i + 10
= 10(i + 1)
Monday, January 11, 2010
A VARIATION
Sometimes it is easier to have an explicit basis for complete
induction.
Basis: Prove that P(0) is true.
Induction Step: Prove that , for each natural number
i > 0, if P(j) holds for all natural numbers j < i, then
P(i) holds as well.
Monday, January 11, 2010
EXAMPLE
P(n): If a full binary tree has n nodes, then n is an odd
number.
Prove P(n) is true for all n ≥ 1.
Basis: n = 1. A full binary tree with one node, and one
is odd.
Induction Hypothesis: Assume we have a binary tree
with n > 1 nodes.
If T1 is the left subtree and T2 is the right subtree of
our full binary tree T , then we have:
nodes(T )=nodes(T1 )+nodes(T2 )+1
Monday, January 11, 2010
READING
Chapter 0: to remind yourself of the basic notions and
definitions.
Chapters 1 for this week’s discussions on induction.
Monday, January 11, 2010