From … to Induction CS 270 Math Foundations of CS Jeremy Johnson

From … to Induction

CS 270 Math Foundations of CSJeremy Johnson

Objective

• To illustrate the shortcomings to the proof techniques seen so far and to introduce the principle of induction which allows the proof of infinitely many cases at once.

• With practice, students should be able to carry out simple inductive proofs following the lecture.

Outline

1. Motivating Example2. Induction Principle3. Inductive Proofs

1. Distributive Law2. DeMorgan’s Law3. Sums

4. Counting Clauses

Atmost One

• Recall the condition for at most one of the variables P1,…,Pt to be true

P1 (P2 Pt)

…Pt-2 (Pt-1 Pt)

Pt-1 Pt

• We see a pattern and fill in the dots

Atmost One

• When converting to CNF we used a generalized version of the distributive law

P1 (P2 Pt)

P1 (P2 Pt)

(P1 P2) (P1 Pt)

• Again we fill in the dots and assume the distributive law generalizes appropriately

Generalized Distributive Law

• A (B C) (A B) (A C)

• What about• A (B C D) (A B) (A C) (A D)

• What does this mean?• (B C D) and (A B) (A C) (A D)

Generalized Distributive Law

• A (B C D) º A ((B C) D) º (A (B C)) (A D) º ((A B) (A C)) (A D) º (A B) (A C) (A D)

Generalized Distributive Law

• A (B C D E) º A ((B C D) E) º (A (B C D)) (A E) º ((A B) (A C) (A D)) (A E) º (A B) (A C) (A D) (A E)

Generalized Distributive Law

• A (B C D E F) º A ((B C D E) F) º (A (B C D E)) (A F) º ((A B) (A C) (A D) (A D))

(A F) º (A B) (A C) (A D) (A E) (A F) • …

Generalized Distributive Law

• Define

Induction Principle

• Let S(n) be a statement paramterized by a non-negative integer n

• If S(0) is true and S(n) S(n+1) then S(n) holds for all non-negative integers.• S(0), S(0) S(1) S(1)• S(1), S(1) S(2) S(2)• S(2), S(2) S(3) S(3)• …

• This allows a proof of infinitely many cases

Inductive Proofs

• Let S(n) be a statement paramterized by n a nonnegative integer. To prove S(n) holds for all non-negative integers.

1. Prove S(0) [Base case]2. Assume S(n) [inductive hypothesis] and

prove S(n+1). This proves S(n) S(n+1)

• Can start with a positive integer k and show S(n) holds for all integers k.

Generalized Distributive Law

• Theorem. )

• Proof by induction on n.

• Base case ()• )

Generalized Distributive Law

• Theorem. )• Assume Inductive Hypothesis (IH)• )

º [by definition of ]º [by distrib. law]º ) [by IH]º ) [by definition of ]

Exercise Generalized DeMorgan’s Law

• Prove by induction on n that

• Base case.

• Inductive Hypothesis

Solution

• Prove by induction on n that

• Base case. ()

Solution

• Inductive Hypothesis• Assume and show that

• [by def ]º [by DeMorgan’s law]º ) [by IH]º [by def of ]

Counting Clauses

• We would like a formula for the number of clauses in the N-queens problem.

• Count the number of clauses for atmost_one, atleast_one and exactly_one

• Sum over all rows, columns and diagonals

Number of Clauses in atmost_one

• The number of clauses in atmost_one(P1,…,Pt) is equal to

•

P1 (P2 Pt)

…Pt-2 (Pt-1 Pt)

Pt-1 Pt

Summation Formula

• Theorem. • Proof by induction on n.• Base case (n=1). • Assume [IH] and show

Summation Formula

• Prove by induction on n that • Base Case• Inductive Hypothesis

Counting Clauses

• M(n) number of clauses for atmost_one• L(n) number of clauses for atleast_one• E(n) number of clauses for exactly_one

• M(n) = n(n-1)/2• L(n) = 1• E(n) = M(n)+E(n)

Clauses in N-Queens SAT Problem

Constraints Exactly one queen per row

N*E(N) = N(N(N-1)/2+1) = ½*N*(N2-N+2)

Exactly one queen per column N*E(N) = N(N(N-1)/2+1) = ½*N*(N2-N+2)

At most one queen on diagonal 2*(2 + M(N)) = 4*+2M(N) = 2+ N(N-1) = 2= 1/3N(N-1)(2N-1)

Clauses in N-Queens SAT Problem

C(N) = R(N) + C(N) + D(N) =

C(3) = 34, C(4) = 84, C(100) = 1,646,900