strong inductionirani/w17-6d/boardnotes/06_stronginductionpost.pdf• k-2 falls in the range 12…k,...
TRANSCRIPT
Strong Induction
ICS 6D Sandy Irani
Proof: Base case n = 0: h0 = (1/2)·50 - 2·0 – (5/2) = (1/2)-(5/2) = -2
Inductive Step: For k ≥ 0, if hk = (1/2)·5k – 2k – (5/2) then hk+1 = (1/2)·5k+1 – 2(k+1) – (5/2) hk+1= 5hk + 8(k+1) Recurrence relation defining the sequence
= 5((1/2)·5k – 2k – (5/2)) + 8(k+1) By the inductive hypothesis
= (1/2) 5 ·5k – 10k – (25/2) + 8k + 8 = (1/2) 5 ·5k – 10k – (25/2) + 8k + 8 = (1/2)·5k+1 – 2k – (9/2) = (1/2)·5k+1 – 2k – (4/2) – (5/2) = (1/2)·5k+1 – 2(k + 1) – (5/2)
Bound on Fibonacci Sequence
• f0 = 1 • f1 = 1 • fn = fn-1 + fn-2, for n ≥ 2
– Equivalently: fk+1 = fk + fk-1, for k ≥ 1
Theorem: for n ≥ 0, fn ≤ 2n
Inductive Step: For k ≥ 1, if fk ≤ 2k, then fk+1 ≤ 2k+1
Strong Induction for Bound on Fibonacci Sequence
• Base case: – f0 ≤ 20
– f1 ≤ 21
– Inductive Step: For k ≥ 1, • If: f0 ≤ 20 and f1 ≤ 21 and ….and fk ≤ 2k • then fk+1 ≤ 2k+1
Theorem: for n ≥ 0, fn ≤ 2n
• Base case: – f0 ≤ 20
– f1 ≤ 21
– Inductive Step: For k ≥ 1, • If fj ≤ 2j for any j in the range from 0 through k. • then fk+1 ≤ 2k+1
Principle of Strong Induction.
• Let P(n) be a statement that is parameterized by natural numbers n.
• Let a and b be constants, with a ≤ b. • If the following two conditions are true:
– P(a) and P(a +1)…and P(b) – For k ≥ b, [P(a) ˄ P(a +1) ˄… ˄ P(k)] → P(k +1)
• Equivalently: For k ≥ b, if P(j) is true for any j in the range from a through k, then P(k +1) is true.
• Then P(n) is true for all n ≥ a.
Principle of Strong Induction.
• Example a = 3, b = 6 – Base case: prove P(3), P(4), P(5) and P(6) – For k ≥ 6, if P(j) is true for any j in the range from
3 through k, then P(k+1) is true.
Strong vs. Weak Induction Weak Induction Strong Induction
Strong Induction Example: Buying Stamps
• Suppose that you can buy any number of 3-cent or 7-cent stamps. Show that you can buy exactly n-cents worth of stamps for any n ≥ 12.
• P(n): can buy n-cents worth of stamps by purchasing a combination of 3-cent and 7-cent stamps. – P(17) – P(11)
Strong Induction Example: Buying Stamps
• Base case: – P(12) (Can buy 12 cents worth of stamps by
buying a combination of 3-cent and 7-cent stamps).
– P(13) – P(14)
• Induction step: – For k ≥ 14, assume that you can buy j cents worth of
stamps for any j in the range from 12 through k. We will show that it is possible to buy k+1 cents worth of stamps.
– Since k ≥ 14, k-2 ≥ 12.
• k-2 falls in the range 12…k, so by the induction hypothesis, P(k-2) is true.
– Buy k-2 cents worth of stamps then buy one additional 3-
cent stamp. The value of the stamps is
k-2 + 3 = k+1
– Therefore, we can buy k+1 worth of stamps by purchasing only 3-cent or 7-cent stamps.
• What if we only had two values in the base case?
Chocolate Bar Theorem
• Rectangular chocolate bar consists of n squares. Can break along any horizontal or vertical line. How many breaks to break the chocolate bar into n separate squares? ( Call this number B(n) )
• Fact: Exactly n-1 breaks to break the chocolate bar into n squares. ( B(n) = n-1 )
• Proof: • Base case: n = 1. No breaks necessary.
• B(1) = 0 = 1-1 • Inductive step:
• Suppose that for k ≥ 1, a chocolate bar with j squares requires j-1 breaks for any j in the range from 1 through k.
• Then a chocolate bar with k+1 squares requires k breaks.
• Take a chocolate bar with k+1 squares and break it into two pieces with k1 and k2 squares
• Break the bar with k+1 squares into two pieces with k1 and k2 squares: k1 + k2 = k+1 • 1 ≤ k1 ≤ k • 1 ≤ k2 ≤ k
Sum of Powers of 2
• Theorem: Every positive integer can be written as a sum of distinct powers of 2.
n = 2j1 + 2j2 + … + 2jm
(the j1,…,jm are distinct)
17 = 31 = 22 =
Odd and Even integers
• If n is even, n = 2m for some integer m.
• If n is odd, n = 2m+1 for some integer m.
• Base case: – n = 1: 1 = 20
– n = 2: 2 = 21
• Inductive step: – For k ≥ 2, if every j in the range from 1 through k
can be written as a sum of distinct powers of 2. • then k+1 can be written as a sum of distinct powers of 2.
• Case 1: k+1 is even – k+1 = 2m for some integer m. – Need to show that 1 ≤ m ≤ k
• Case 1: k+1 is odd – k+1 = 2m+1 for some integer m. – Need to show that 1 ≤ m ≤ k