1. Lecture 12 Oct 13, 2008 Some simple recursive programs recursive problem solving, connection to induction Some examples involving recursion Announcements module final exam Oct 15 from 10 to 12 noon. project # 2 due Oct 20 (next Monday).

2. Problem 1 – iterative solution Write a program that takes as input as odd number k and print the product 1 x 3 x 5 x … k. If the input is NOT odd, the program should print –1. Iteration: int chainProduct(int k) { if (k % 2 == 0) return –1; int Result:= 1; for (j = 3; j <= k; j = j+2) { Result:= Result * j; Return Result; }

3. Induction and recursion These two are related concepts. Induction is a proof technique, recursion is a related programming concept. Induction proof: assume it is true for n – 1, and show it for n using the assumption. recursion: assume you know how to solve the problem when the input size n – 1, and design a solution for the case n using the solution for n - 1.

4. Induction Proof Example: Prove the formula 1 + 2 +... + n = n(n+1)/2 induction hypothesis: 1 + 2 +... + n – 1 = (n – 1)n/2 induction step: 1 + 2 +... + n = (1 + 2 +... + n – 1) + n = (n – 1)n /2 + n = n ( n +1)/2 and this completes the proof.

5. Key facts to remember about recursion If the problem involves only one input parameter n, then you should determine how to compute f(n) given the value of f(n – 1). When writing the code, you can call the function recursively to get the value of f(n – 1). Sometimes, f(n) may depend on f(n – 1) as well as f(n – 2) etc. This is fine, so long as f(n) does not depend on f(n+1) … Always provide exit from recursion. (The simplest base cases such as k = 1 does not need recursion and should be handled without recursive calls.)

6. Recursive Solution to the same problem int chainProduct(int k) { if (k % 2 == 0) return -1; else if (k == 1) return 1; else return k * chainProduct(k-2); } Here is the main program: int main() { // to compute f(15)‏ cout << “f(15) is “ << f(15) << endl; }

7. Another simple example The following problem is discussed in Chapter 7 of Weiss. Compute the sum 1 + 2 + … + n. Code is as shown: long s( int n ) { if( n == 1 ) return 1; else return s(n-1) + n; }

8. Recursive evaluation of x n Write a recursive procedure to compute g(x, n) = x n. A simple solution: g(x, n) = 1 if n = 0 x if n = 1 g(x, n – 1) * x else. This recursive solution is not as efficient as it can be. We will next see a faster recursive solution. This is the first example of the power of recursion. Iteratively implementing this idea harder and trickier.

9. g(x, n) = 1 if n = 0 x if n = 1 g(x, n – 1) * x if n > 1 and n is odd g(x, n/2) 2 if n > 1 and n is even int power(int x, int n) { if (n == 0) return 1; else if (n == 1) return x; else if (n % 2 == 1) return x * power(x, n-1); else { int temp = power(x, n/2); return temp*temp; } }

10. Prime number testing Write a program to print all the prime numbers below a given integer N. Recall that m is a prime if and only if there is no integer x, 2 <= x < m such that (m % x == 0). Write a recursive procedure isDivisible(m, k) that returns true if there is an integer x, k <= x < m such that (m % x == 0).

11. Prime number testing – recursive solution isDivisible(m, k) returns true if there is an integer x, k <= x < m such that (m % x == 0).

12. Question: Given the procedure isDivisible(x, k), how to write a procedure isPrime(m) that returns true if and only if m is a prime? static boolean isPrime(int n) { return !isDivisible(n, 2); }

13. Recursive version of selection sort Selection Sorting Algorithm: During the j-th pass (j = 0, 1, …, n – 2), examine the elements of the array a[j], a[j+1], …, a[n-1] and determine the smallest of them. Suppose the smallest of a[j], a[j+1], …, a[n-1] is in position min. Swap a[min] and a[j]. Recursive structure of the algorithm: selection_sort(int[] a, int j) // j-th pass { if (j = a.length) return; // only one element in array min = smallest of a[j], a[j+1], …, a[n-1]; swap a[min] and a[j]; selection_sort(a, j+1); }