Induction and Recursion Explicit definition Recursive definition Factorial

Induction and Recursion. Explicit definition Recursive definition • Factorial • Geometric progression • Power of relation on a set 1

• The recursive definition of a function makes reference to earlier versions of itself. • The main connection between recursion and induction is that objects are defined by means of a natural sequence. • Induction is usually the best (possibly the only) way to prove results about recursively defined objects. 2

How to find a closed form for a recursively defined function? In general there is no ready to use recipe. Some simple cases are • Linear function of integer n: g 1(n) = an+b g 1(n+1) = a(n+1)+b= g 1(n) + a • Quadratic function of integer n: g 2(n) = an 2+bn+c g 2(n+1) = a(n+1)2+b(n+1)+c = g 2(n) +2 an+(a+b) 3

Example. Prove that for all positive integers m and n, Proof. Let m be arbitrary positive integer and then prove by induction on n 1. 1) Basis. n=1: (by recursive definition of Rm+1) 2) IH: Assume that for some k 1 we have IS: We need to prove that ……. . definition of composition …. . by IH …. association of composition ……. . definition of composition 4

Transitive closures again. Theorem. The transitive closure of R is Proof. Denote. We need to prove that S satisfies the definition of transitive closure, i. e. i) S includes R, R S ii) S is transitive iii) S is the smallest relation satisfying i) and ii) assume (x, y) S and (y, z ) S to prove that (x, z) S. (x, y) S implies that (x, y) Rn , n is some integer, n Z+. (y, z ) S implies that (y, z ) Rm, m is some integer , n Z+. (x, z) Rno Rm= Rn+m, by the previous Theorem. (x, z) S , since Rn+m S 5

iii) Suppose T is any relation, satisfying i) R T and ii) T is transitive. and prove that S T. By the definition of S as , it suffices to prove that for any n Z+ (Rn T) We are going to prove it by induction on n 1. 1) Basis: when n=1, R 1=R T by assumption. 2) IH: Assume that for n=k, where k is some integer, k 1, Rk T. IS: Need to prove Rk+1 T Suppose (x, y) Rk+1=Rko R (by definition of composition). It implies that there exists some z A, that (x, z) Rk and (z, y) R. . (x, z) Rk implies (x, z) T , because Rk T by IH. (z, y) R implies (z, y) T , because R T by assumption. (x, z) T and (z, y) T imply (x, y) T by transitive property of T. We proved by induction that for any n 1, Rn T. 6

Fibonacci Numbers The sequence named for Fibonacci (1202) is introduced in terms of rabbits. Suppose that every month a breeding pair of rabbits produce a pair of offspring. The offspring in turn start breeding two months later, and so on. Denote Fn the number of pairs in month n. Suppose you buy a pair of rabbits in month 1 (F 0=0, F 1=1) then you still have one pair in month 2 (F 2=1), but in month 3 they start breeding, F 3=2. The sequence begins 0, 1, 1, 2, 3, 5, 8, 13, 21, … You can observe, that in a month n you have all pairs from the previous month (n 1), plus offspring pairs for all pairs from month (n 2), i. e. : , n 2 7

So, Fibonacci numbers can be defined recursively: These numbers have numerous applications in CS, mathematics, theory of games. Let’s prove the closed formula for the sum for all integers n 0 Proof by induction on n 0. 1) Basis. Take n=0 and check that ? By the recurrent formula for n =2 we have so F 2 1=0 and F 0=0. 8

2) IH: Assume that the summation formula is correct for n=k, where k is some integer k 0, i. e. IS: We need to prove that …………by IH ……. . by recurrence formula for (k+3) >1 9

• Sometimes we need stronger assumption to prove IS. • It happens when to prove IS for (n +1) you need to refer not only to the previous n , but to smaller numbers. Strong Induction Principle: Let A N denote a subset that satisfies the following two properties: 1) 0 A; 2) if 0, 1, …k A, then k+1 A. Then A=N. So, Strong Induction differs from Induction Principle in step 2): IP 2) n 0 [P (n) P (n+1)] Strong Induction 2) n 0 [( m n P (m)) P (n+1)] 10

Example. Prove that every amount of postage of 12 cents or more can be formed using only 4 cent and 5 cent stamps. We are going to prove by induction on n 12, that any postage n = 5 i + 4 j, where integers i, j 0. As we proceed n n +1, we can not derive case n +1 from the case n (n+1) 5 (n+1) 4 n n +1 We can infer n+1 = 5 i + 4 j either from (n+1) 5 = 5(i 1) + 4 j or from (n+1) 4 = 5 i + 4(j 1) We need to prove an extended basis here, that is to prove the basis on more then one point. 11

Proof. 1) Basis (extended): check that we can form postage of 12, 13, 14, and 15 cents using only 4 c and 5 c stamps. 12=4 3; 13=5+ 4 2; 14=5 2+4; 15=5 3. 2) IH: Assume that any postage 12, … k, where k 15 can be made using 4 c and 5 c stamps (strong induction) IS Prove that conjecture is true for the postage k +1. By IH any postage greater or equal then 12 c and less or equal k can be formed using 4 c and 5 c stamps. In particular it is true for k 3. Together with one more 4 c stamp it gives k +1 postage. 12

Theorem. The following three principles are equivalent: (a) the Induction Principle; (b) the Strong Induction Principle; and (c) the Well-Ordering principle. Proof. It suffices to prove (a) (b), (b) (c), and (c) (a). We already did this 13

Proof of (a) (b): Let A N and A satisfies the two properties: (1) 0 A; and (2) if 0, 1, …k A then k+1 A. We need to prove A=N. To make use of Induction Principle we define the auxiliary set C={m| m N and 0, 1, …m A}. Thus, C A N (3). Since 0 A by (1), 0 C by the definition of C. If k C, that is, 0, 1, …k A, then by the property (2) of the set A, k+1 A, which implies that k+1 C, because 0, 1, …k, k+1 A. Thus, set C satisfies the properties of the Induction Principle, so C=N. Therefore A=N because of (3). 14

Proof of (b) (c): Assume B N and B (4). Suppose the Well-Ordering Principle is false, i. e. suppose B does not have the smallest element (5). Prove that this leads to contradiction to the Strong Induction Principle. Let A=N B (6). By (5) 0 B, thus 0 A by (6). If the numbers 0, 1, …k A, then these numbers do not belong to B. Then k+1 does not belong to B, because otherwise it would be the smallest element of B contradicting to (5). Thus, k+1 A and by the Strong Induction Principle A=N, which implies that B = by (6), contradicting to (4). So, (c) is proved. 15

Example. Prove that n Fn+1 0 1 2 3 4 5 6 7 1 1 2 3 5 8 13 21 for all n 5. 1 1. 5 2. 25 3. 375 5. 0625 7. 59375 11. 390625 17. 0859375 Basis 16

Proof by induction on n 5 that Fn+1>(3/2)n 1) Basis (extended) n=5: F 6=8>(3/2)5 = 7. 59375 n=6: F 7= 13>(3/2)6 = 11. 390625 2) IH : Assume that for all n=5, …k, where k is some integer k 6 we have Fn+1>(3/2)n (strong induction). IS : We need to prove inequality Fn+1>(3/2)n for n = k +1, i. e. Fk+2>(3/2)k+1 Fk+2= Fk+1+ Fk ……recursive definition > (3/2)k + (3/2)k 1 ……. IH for n=k and n=k 1 = (3/2)k 1 ((3/2)+1)……algebra = (3/2)k 1 (5/2) = (3/2)k 1 (10/4) > (3/2)k 1 (9/4) = (3/2)k 1 (3/2)2 = (3/2)k+1 17