From … to Induction
CS 270 Math Foundations of CS
Jeremy 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 Example
2. Induction Principle
3. Inductive Proofs
1. Distributive Law
2. DeMorgan’s Law
3. 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)
A  ((B  C)  D)
(A  (B  C))  (A  D)
((A  B)  (A  C))  (A  D)
(A  B)  (A  C)  (A D)
Generalized Distributive Law
•




A  (B  C  D  E)
A  ((B  C  D)  E)
(A  (B  C  D))  (A  E)
((A  B)  (A  C)  (A D))  (A  E)
(A  B)  (A  C)  (A D)  (A  E)
Generalized Distributive Law
A  (B  C  D  E  F)
A  ((B  C  D  E)  F)
(A  (B  C  D  E))  (A  F)
((A  B)  (A  C)  (A D)  (A D))
 (A  F)
 (A  B)  (A  C)  (A D)  (A  E)  (A 
F)
• …
•



Generalized Distributive Law
• Define
•
𝑛
𝑖=1 𝐴𝑖
𝐴1 ∧ ⋯ ∧ 𝐴𝑛 =
=
𝑛−1
𝑖=1 𝐴𝑖
𝐴1
𝑛
𝑖=1 𝐴𝑖
∧ 𝐴𝑛
𝑛>1
𝑛=1
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(2)
S(2), S(2)  S(3)
…
 S(1)
 S(2)
 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. 𝐴 ∨
𝑛
𝑖=1 𝐵𝑖
≡
𝑛
𝑖=1(𝐴
∨ 𝐵𝑖 )
• Proof by induction on n.
• Base case (𝑛 = 1)
• 𝐴∨
1
𝑖=1 𝐵𝑖
≡ (𝐴 ∨ 𝐵1 ) ≡
1
𝑖=1(𝐴
∨ 𝐵𝑖 )
Generalized Distributive Law
• Theorem. 𝐴 ∨ 𝑛𝑖=1 𝐵𝑖 ≡ 𝑛𝑖=1(𝐴 ∨ 𝐵𝑖 )
• Assume Inductive Hypothesis (IH)
• 𝐴∨



≡
𝑛−1
𝑖=1 (𝐴
∨ 𝐵𝑖 )
𝑛
𝑖=1 𝐵𝑖
𝐴 ∨ (( 𝑛−1
]
𝑖=1 𝐵𝑖 ) ∧ 𝐵𝑛 ) [by definition of
(𝐴 ∨ ( 𝑛−1
𝑖=1 𝐵𝑖 ) ∧ (𝐴 ∨ 𝐵𝑛 ) [by distrib. law]
𝑛−1
𝑖=1 (𝐴 ∨ 𝐵𝑖 ) ∧ (𝐴 ∨ 𝐵𝑛 ) [by IH]
𝑛
]
𝑖=1(𝐴 ∨ 𝐵𝑖 ) [by definition of
• 𝐴∨

𝑛−1
𝑖=1 𝐵𝑖
Exercise Generalized
DeMorgan’s Law
• Prove by induction on n that
𝑛
𝑛
¬ 𝑖=1 𝐴𝑖 ≡ ⋁𝑖=1 ¬𝐴𝑖
• Base case.
• Inductive Hypothesis
Solution
• Prove by induction on n that
𝑛
𝑛
¬ 𝑖=1 𝐴𝑖 ≡ ⋁𝑖=1 ¬𝐴𝑖
• Base case. (𝑛 = 1)
• ¬
1
𝑖=1 𝐴𝑖
= ¬𝐴1
• ⋁1𝑖=1 ¬𝐴𝑖 = ¬𝐴1
Solution
• Inductive Hypothesis
𝑛−1
• Assume ¬ 𝑛−1
𝐴
≡
⋁
𝑖=1 𝑖
𝑖=1 ¬𝐴𝑖 and show
𝑛
𝑛
that ¬ 𝑖=1 𝐴𝑖 ≡ ⋁𝑖=1 ¬𝐴𝑖
• ¬



𝑛
𝑖=1 𝐴𝑖
≡¬
𝑛−1
𝑖=1 𝐴𝑖
∧ 𝐴𝑛 [by def
𝑛
𝑖=1 𝐴𝑖 ]
¬ 𝑛−1
𝑖=1 𝐴𝑖 ∨ ¬𝐴𝑛 [by DeMorgan’s
(⋁𝑛−1
𝑖=1 ¬𝐴𝑖 ) ∨ ¬𝐴𝑛 [by IH]
𝑛
𝑛
⋁𝑖=1 ¬𝐴𝑖 [by def of ⋁𝑖=1 ]
law]
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
•
𝑡−1
𝑖=1 𝑖
=
𝑡(𝑡−1)
2
P1  (P2    Pt)
…
Pt-2  (Pt-1  Pt)
Pt-1  Pt
Summation Formula
𝑛
𝑖=1 𝑖
• Theorem.
=
𝑛(𝑛+1)
2
• Proof by induction on n.
1
𝑖=1 𝑖
• Base case (n=1).
• Assume
•
•
𝑛
𝑖=1 𝑖
=
𝑛+1
𝑖=1 𝑖
𝑛
𝑖=1 𝑖
𝑛(𝑛+1)
2
2(𝑛+1)
2
=
𝑛+1 =
+
=1=
𝑛(𝑛+1)
2
[IH] and show
+ 𝑛+1 =
=
1(1+1)
2
𝑛+1 (𝑛+2)
2
𝑛(𝑛+1)
+
2
Summation Formula
• Prove by induction on n that
𝑛(𝑛+1)(2𝑛+1)
6
• Base Case
• Inductive Hypothesis
𝑛
2
𝑖
𝑖=1
=
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
𝑁−1
𝑁−1
𝑑=1 𝑀(𝑑) + M(N)) = 4* 𝑑=1 𝑀(𝑑)+2M(N)
= 2 𝑁−1
𝑑=1 𝑑 𝑑 − 1 + N(N-1) =
2 − 2 𝑁−1 𝑑 + 𝑁(𝑁 − 1)= 1/3N(N-1)(2N-1)
2 𝑁−1
𝑑
𝑑=1
𝑑=1
 2*(2


Clauses in N-Queens SAT
Problem
 C(N) = R(N) + C(N) + D(N) =
5𝑁3 −6𝑁2 +7
3
 C(3) = 34, C(4) = 84, C(100) = 1,646,900