More Mathematical Induction. October 27, 2016

Transcription

1 More Mathematical Induction October 7, 016

2 In these slides... Review of ordinary induction. Remark about exponential and polynomial growth. Example a second proof that P(A) = A. Strong induction. Least integer principle.

3 Principle of Mathematical Induction To prove a property P holds for all integers n k : (Base step) Prove P holds for k. Induction step) Assume P holds for some n k (induction hypothesis) and deduce P holds for n + 1. More formally: From and you can conclude P(k) n((n k P(n)) P(n + 1)) n(n k P(n)).

4 Examples from the last class. For all n Z +, For all n 50, n j = j=1 n(n + 1). 1.3 n > n 3.

5 Aside exponential versus polynomial growth More generally, if c > 1, k > 0, then for all sufficiently large values of n, c n > n k. In fact, And similarly, n k lim n c n = 0. (log n) k lim n n = 0.

6 Consequences for efficiency of algorithms Algorithm 1 requires time T 1 (n) = c 1 n log n on inputs of length n. (e.g., mergesort). Algorithm requires time T (n) = c n on inputs of length n. (e.g., insertion sort). Since T 1 (n) lim n T (n) = 0, Algorithm 1 is much faster, as long as the input size is large enough.

7 Another example Earlier we showed that for any finite set A, P(A) = A. We did this by showing a bijection F : {0, 1} {0, 1} P({1,..., n}). }{{} n times Here we give a new proof of the same fact by mathematical induction.

8 Main idea of the argument Every time you add an element {1,, 3}... You double the number of subsets: {3} {1} {1, 3} {} {, 3} {1, } {1,, 3}

9 Official proof Statement: For all n 0, if A = n, then P(A) = n. Base step, if A = 0, then A =, so P(A) = { }, thus P(A) = 1 = 0. Inductive step: Let n 0 and suppose that every set with n elements has n subsets. (Inductive hypothesis.) Now let A = n + 1. We write where all the a i are distinct. Let A = {a 1,..., a n }. A = {a 1,..., a n, a n+1 },

10 Official proof continued P(A) is the disjoint union P(A ) {X A : a n+1 X}. P(A ) = n by inductive hypothesis. Y Y {a n+1 } gives a bijection between P(A ) and {X A : a n+1 X}, which thus also has n elements. Thus P(A) = P(A ) + {X A : a n+1 X} = n + n = n+1, so the claim holds for sets of cardinality n + 1.

11 Strong Induction

12 The unstacking game You have a stack of n bricks. On each move, you break a stack into two smaller stacks, and earn k points, where k is the product of the sizes of the two smaller stacks.

13 Example: =10 points

14 Example: =10 points

15 Conjecture: The unstacking strategy doesn t matter! No matter how you do it, the total number of points for completely breaking apart a stack of n disks is (n 1) + (n ) = How would you prove such a thing? n(n 1).

16 The basic reasoning principle Suppose your first move is to break a stack into two stacks with m 1 and m bricks, respectively. If the conjecture is correct for the smaller stacks, then we will get m 1 m + m 1(m 1 1) + m (m 1) points for completely breaking the original stack. Need to check that m 1 m + m 1(m 1 1) + m (m 1) = (m 1 + m )(m 1 + m 1)

17 Principle of strong mathematical induction To prove a property P holds for all n k, it suffices to prove: (Base step): P(k). (Inductive step): Assume for some n > k that P(m) holds for all m with k m < n, (not just n 1) and deduce P(n). We make a stronger inductive hypothesis.

18 Official strong induction proof of the the unstacking game theorem Statement: For all n 1, the number of points obtained by unstacking a stack of n bricks is n(n 1), independent of the strategy. Base step: If n = 1, there is no unstacking to do. You get points. 0 = 1(1 1) Inductive step: Let n > 1. Suppose that for all positive integers less m < n, you get m(m 1)/ points for unstacking a stack of m bricks. On the first move, you break the stack into two smaller stacks of sizes m and (n m), where 1 m < n. You get m(n m) points for this move.

19 Official proof, continued By the inductive hypothesis, since both m < n and n m < m, the total number of points for the complete unstacking is thus m(n m) + m(m 1) + (n m)(n m 1) 1 ((mn m ) + (m m) + (n mn + m n + m)) = n n n(n 1) So the formula holds for n bricks as well. This completes the proof. = =

20 revisited The same reasoning is behind the proof that there are no integers m, n > 0 with n = m (proof that is irrational). Clearly there is no positive integer m such that 1 = m. Let n > 1, and suppose that for every 1 k < n, there is no positive integer l such that k = l. If there were m Z + with n = m, we could argue as before that there is an integer l with m = l. Since m < n, this contradicts the inductive hypothesis, which says that no such l exists.

21 Least Integer Principle Used in this manner, the principle of strong induction is sometimes called the least integer principle: Every nonempty set of positive integers has a least element. If you want to use this to prove that some property P holds for no positive integer, suppose that P holds for some positive integer, and then set n to be the least positive integer for which this is true. Then obtain a contradiction by showing that n 1, and that if such an n exists, there must be a smaller value for which P holds.