1. Simple Recursion

2. COMP104 Lecture 35 / Slide 2 Recursion: Example 0 * What does the following program do? #include using namespace std; int fac(int n){ // Assume n >= 0 int product; if(n <= 1) return 1; product = n * fac(n-1); return product; } void main(){// driver function int number; while(true){ cout << "Enter integer (negative to stop): "; cin >> number; if(number < 0) break; cout << fac(number) << endl; }

3. COMP104 Lecture 35 / Slide 3 Recursion: Example 0 * Assume the number typed is 3. fac(3) : 3 <= 1 ? No. product 3 = 3 * fac(2) fac(2) : 2 <= 1 ? No. product 2 = 2 * fac(1) fac(1) : 1 <= 1 ? Yes. return 1 product 2 = 2 * 1 = 2 return product 2 product 3 = 3 * 2 = 6 return product 3 fac(3) has the value 6

4. COMP104 Lecture 35 / Slide 4 Recursion * Recursion is one way to decompose a task into smaller subtasks. * At least one of the subtasks is a smaller example of the same task. * The smallest example of the same task has a non-recursive solution. Example: The factorial function n! = n * (n-1) * (n-2) *... * 1 or n! = n * (n-1)! and 1! = 1

5. COMP104 Lecture 35 / Slide 5 Recursion * A recursive solution may be simpler to write (once you get used to the idea) than a non-recursive solution.  But a recursive solution may not be as efficient as a non-recursive solution of the same problem.

6. COMP104 Lecture 35 / Slide 6 Writing Recursive Function * A recursive function always consists of two parts: n basis case (stopping criterion)  e.g. F(n) = 1 when n = 1 n inductive case  e.g. F(n) = n * F(n-1) * It has the same mathematical meaning as induction

7. COMP104 Lecture 35 / Slide 7 Iterative Factorial // Non-recursive factorial function // Compute the factorial using a loop int fac(int n){ // Assume n >= 0 int k, product; if(n <=1) return 1; product = 1; for(k=1; k<=n; k++) product*= k; return product; }

8. COMP104 Lecture 35 / Slide 8 Other Recursive Applications * Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34,... where each number is the sum of the preceding two. * Recursive definition: n F(0) = 0 n F(1) = 1 n F(n) = F(n-1) + F(n-2)

9. COMP104 Lecture 35 / Slide 9 Other Recursive Applications * Binary search: n Compare search element with middle element of the array: If not equal, then apply binary search to half of the array (if not empty) where the search element would be.

10. COMP104 Lecture 35 / Slide 10 Recursion General Form * How to write recursively? int rec(1-2 parameters){ if(stopping condition) return stopping value; // second stopping condition if needed return value/rec(revised parameters) +-*/ rec(revised parameters); }

11. COMP104 Lecture 35 / Slide 11 Recursion: Example 1  How to write exp(int x, int y) recursively? int exp(int x, int y){ if(y==0) return 1; return x * exp(x, y-1); }

12. COMP104 Lecture 35 / Slide 12 Recursion: Example 2 * Write a recursive function that takes a double array and its size as input and returns the sum of the array: double asum(int a[], int size){ if(size==0) return 0; return asum(a, size-1)+a[size-1]; }

13. COMP104 Lecture 35 / Slide 13 Recursion: Example 3 * Write a recursive function that takes a double array and its size as input and returns the product of the array: double aprod(int a[], int size){ if(size==0) return 1; return aprod(a, size-1)*a[size-1]; }

14. COMP104 Lecture 35 / Slide 14 Recursion: Example 4 * Write a recursive function that counts the number of zero digits in a non-negative integer  zeros(10200) returns 3 int zeros(int n){ if(n==0) return 1; if(n < 10) return 0; if(n%10 == 0) return 1 + zeros(n/10); else return zeros(n/10); }

15. COMP104 Lecture 35 / Slide 15 Recursion: Example 5 * Write a recursive function to determine how many factors m are part of n. For example, if n=48 and m=4, then the result is 2 (48=4*4*3). int factors(int n, int m){ if(n%m != 0) return 0; return 1 + factors(n/m, m); }