1. Induction and Recursion

2. 4.1 Mathematical Induction Introduction • Mathematical Induction is used to show that P(n) is true for every positive integer n. • Used only to prove results obtained in some other way. • Example: Suppose we have an infinite ladder, and we want to know whether we can reach every step on the ladder. We know two things 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we can reach the next rung. Can we conclude that we can reach every rung? By 1, we can reach the first rung. By 2, we can reach the second rung. Apply 2 again, we can reach the third rung… After 100 uses of 2, we can reach the 101st rung….

3. 4.1 Mathematical Induction To complete the inductive step of a proof using the principle of mathematical induction, we assume that P(k) is true for an arbitrary positive integer k and show that under this assumption, P(k+1) must also be true. The assumption that P(k) is true is called the inductive hypothesis. The proof technique is stated as [P(1)Λk(P(k) →P(k+1) )] → nP(n) where the domain is the set of positive integers Mathematical Induction PRINCIPLE OF MATHEMATICAL INDUCTION To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, we complete two steps. BASIS STEP: We verify that P(1) is true. INDUCTIVE STEP: We show that the conditional statement P(k) →P(k+1) is true for all positive integers k. 3

4. 4.1 Mathematical Induction Show that if n is a positive integer, then Solution: BASIS STEP: P(1) is true, because 1 = INDUCTIVE STEP: For the inductive hypothesis we assume that P(k) holds for an arbitrary positive integer k. That is, we assume that Under this assumption, it must be shown that P(k+1) is true, i.e.: When we add k+1 to both sides of the equation in P(k), we obtain This shows that P(k+1) is true under the assumption that P(k) is true. This complete the inductive step. Examples of Proofs by Mathematical Induction 4

5. 4.1 Mathematical Induction Use mathematical induction to prove that n3 – n is divisible by 3 whenever n is a positive integer. Solution: BASIS STEP: The statement P(1) is true because 13 – 1 = 0 is divisible by 3. INDUCTIVE STEP: For the inductive hypothesis we assume that P(k) is true; that is we assume that k3 – k is divisible by 3. To complete the inductive step, we must show that when we assume the inductive hypothesis, the statement (k+1)3 – (k+1) is also divisible by 3. (k+1)3 – (k+1) = (k3 + 3k2 + 3k + 1) – (k + 1) = (k3 – k) + 3(k2 + k) By inductive hypothesis, we know that (k3 – k) is divisible by 3; and the second term is also divisible by 3. This completes the inductive step. 5

6. 4.3 Recursive Definitions and Structural Induction Introduction 6

7. 4.3 Recursive Definitions and Structural Induction Sometimes it’s easier to define an object in terms of itself. This process is called Recursion. Example: The sequence of powers of 2 is given by an = 2n for n = 0, 1, 2, …. This sequence can also be defined by giving the first term of the sequence, namely, a0 = 1, and a rule finding a term of the sequence from the previous one, namely, an+1 = 2an, for n = 0, 1, 2, …. 7

8. 4.3 Recursive Definitions and Structural Induction Use two steps to define a function with the set of nonnegative integers as its domain: BASIS STEP: Specify that value of the function at zero. RECURSIVE STEP: Give a rule for finding its value at an integer from its values at smaller integers. Such a definition is called a recursive or inductive definition. Examples: Suppose that f is defined recursively by f(0) = 3, f(n+1) = 2f(n) + 3 Find f(1), f(2), f(3), and f(4). Solution: f(1) = 2f(0) + 3 = 2*3 + 3 = 9 f(2) = 2f(1) + 3 = 2*9 + 3 = 21 f(3) = 2f(2) + 3 = 2*21 + 3 = 45 f(4) = 2f(3) + 3 = 2*45 + 3 = 93 Recursively Defined Functions 8

9. 4.3 Recursive Definitions and Structural Induction Examples: Give an inductive definition of the factorial function F(n) = n!. Solution: Basis Step: F(0) = 1 Inductive Step: F(n+1) = (n+1)F(n) E.g. Find F(5). F(5) = 5F(4) = 5*4F(3) = 5*4*3F(2) = 5 * 4 * 3 * 2F(1) = 5 * 4 * 3 * 2 * 1F(0) = 5 * 4 * 3 * 2 * 1 * 1 = 120 Give a recursive definition of . Solution: Basis Step: Inductive Step: 9

10. Example: Find the Fibonacci numbers f2, f3, f4, f5, and f6. Solution: f2 = f1 + f0 = 1 + 0 = 1, f3 = f2 + f1 = 1 + 1 = 2, f4 = f3 + f2 = 2 + 1 = 3, f5 = f4 + f3 = 3 + 2 = 5, f6 = f5+ f4 + 5 + 3 = 8. Example: Find the Fibonacci numbers f7. Solution ? 4.3 Recursive Definitions and Structural Induction DEFINITION 1 The Fibonacci numbers, f0, f1, f2, …, are defined by the equations f0= 0, f1 = 1, and fn = fn-1 + fn-2 for n = 2, 3, 4, …. 10

11. Example: Give a recursive algorithm for computer n!, when n is a nonnegative integer. Review: Basis Step: F(0) = 1 Inductive Step: F(n+1) = (n+1)F(n) E.g. Find F(5). F(5) = 5F(4) = 5*4F(3) = 5*4*3F(2) = 5 * 4 * 3 * 2F(1) = 5 * 4 * 3 * 2 * 1F(0) = 5 * 4 * 3 * 2 * 1 * 1 = 120 4.4 Recursive Algorithms Introduction DEFINITION 1 An algorithm is called recursive if it solves a problem by reducing it to an instance of the same problem with smaller input. ALGORITHM 1 A Recursive Algorithm for Computing n!. procedure factorial(n: nonnegative integer) if n = 0 then factorial(n):=1 else factorial(n):=n*factorial(n-1) 11

12. Example: Give a recursive algorithm for computer an, where a is a nonzero number and n is a nonnegative integer 4.4 Recursive Algorithms ALGORITHM 2 A Recursive Algorithm for Computing an. procedure power(a: nonzero real number, n: nonnegative integer) if n = 0 then power(a,n):=1 else power(a,n):=a*power(a, n-1) 12

13. 4.4 Recursive Algorithms f4 f3 f2 fn+1 -1 addition to find fn f2 f0 f1 f1 (f5-1) addition to find f4 = (5-1) = 4 additions f0 f1 13