Lecture 2 Aug 28 goals: Introduction to recursion

1. Lecture 2 Aug 28 • goals: • Introduction to recursion • examples of recursive programs

2. Recursion • Recursive program • A function that calls itself. • e.g. f(x)= 2 * f(x – 1) • Recursive definition • defining a concept in terms of itself. • e.g. an arithmetic expression is either a number, or is of the form X op Y where X and Y are arithmetic expressions and op is an operator. • Recursive data structure • a structure whose substructure is of the same type as the structure

3. Recursion • Recursive data structure • a structure whose substructure is of the same type as the structure • e.g. a binary tree is either a single node or a node with two children both of which are also binary trees.

4. Recursion - rules Rule 1: Provide exit from recursion. (focus on base cases – some times, more than one base case is needed.) Rule 2: Make sure that the successive recursive calls progress towards the base case. Rule 3: Assume that recursive calls work. Rule 4: Compound interest rule: Avoid redundant calls.

5. Recursion – simple examples • Compute n! • n! = 1 x 2 … x n = (n – 1)! x n • 2) Compute 1 + 2 + … + n • f(n) = n + f(n – 1) • 3) Compute the n-th Fibonacci number • f(n) = f(n – 1) + f(n – 2)

6. Example: (from Chapter 2 of text) Given x and n, we want to compute xn. Obvious iterative solution: int exp(int x, int n) { int temp = x; for (int j= 1; j < n; ++j) temp*= x; return temp; } The number of multiplications performed is n – 1. We will now see that a recursive algorithm will provide a much faster solution. Faster algorithm is crucial for RSA encryption algorithm.

7. Idea behind the algorithm • Rule of exponents: • xn = (x2)n/2 if n is even • x * (x2)(n-1)/2 if n is odd • base case n = 1  return x • even n  call exp(x*x , n/2) and return the result. • odd n  call exp(x*x, (n – 1)/2) and multiply the result of call by x and return the output.

8. Code for recursive exponent algorithm int exp (int x, int n) { if (n == 1) return x; else if (n%2 == 0) return exp(x*x, n/2); else return x* exp(x*x, (n – 1)/2); } Fact: When called to compute exp(x,100000), the iterative algorithm performs 99999 multiplications while the recursive algorithm will performs 21 multiplications.

9. Recursive algorithm for modular exponentiation • In cryptography (RSA algorithm), the following problem arises. • Given x, n and m, compute xn (mod m). • In addition to the recursive algorithm we described above, we need one more idea to make this computation feasible. • In practice, x, n and m are extremely large numbers, with hundreds of digits in it.

10. Modular exponentiation algorithm int exp (int x, int n, int m) { if (n == 1) return x % m; else if (n%2 == 0) return exp((x*x) % m, n/2); else return (x* exp((x*x) % m, (n – 1)/2)) % m; }

11. An example involving 1-d array Given two segments of sorted arrays A and B, output the result of merging them as a single sorted array. B A merge(A, 2, 3, B, 9, 13) should return 0 1 2 3 4 5 6 1 2 2 3 4 4 5 merge(A, 2, 2, B, 9,10) should return the array 1 2 4 merge(A, 3, 4, B, 9, 8) should return 5 7 since the second array is empty. (high index < low index means the array is empty.)

12. What are the base cases?

13. What are the base cases? • one array segment is empty. • In this case, what is the output?

14. What are the base cases? • one array segment is empty. • In this case, what is the output? • The other array segment, so we just have to copy the segment to the output.

15. What are the base cases? • one array segment is empty. • In this case, what is the output? • The other array segment, so we just have to copy the segment to the output. • what if both segments are not empty? We need to make recursive call.

16. Before we proceed, we need to make a change to the prototype of the function merge. Why? • We need to add two more parameters - the name of the array and the starting index in the output array at which we want to write.

17. A B merge(A, 2, 3, B, 9, 13, C, 3) should return 3 4 5 6 7 8 9 C 1 2 2 3 4 4 5

18. merge(A, low1,high1, B, low2, high2, C, low) B A Case 1: A[low1] < B[low2] Example: merge(A, 9, 13, B, 2, 3, C, 0) 0 1 2 3 4 5 6 C 1 2 2 3 4 4 5

19. merge(A, low1,high1, B, low2, high2, C,low) B A Case 1: A[low1] < B[low2] Example: merge(A, 9, 13, B, 2, 3, C, 0) Step1: move A[low1] to C[low]. Now a recursive call to merge will get the rest of the job done. What call? 0 1 2 3 4 5 6 1 2 2 3 4 4 5

20. merge(A, low1,high1, B, low2, high2) B A Case 1: A[low1] < B[low2] Example: merge(A, 9, 13, B, 2, 3, C, 0) Step1: move A[low1] to Out[0]. Now a recursive call to merge will get the rest of the job done. What call? merge(A, low1+1, high1, B, low2,high2, C, low + 1) 0 1 2 3 4 5 6 1 2 2 3 4 4 5

21. Case 2: A[low1] > B[low2]. Move B[low2] to C[low] and make a recursive call to merge(A,low1, high1, B, low2+1, high2, C, low+1) The complete code is shown below:

22. void merge(int A[], int low1, int high1, int B[], int low2,int high2, int C[], int low) { if (high1 < low1) copy(B, low2, high2, C,low); else if (high2 < low2) copy(A, low1, high1, C,low); else if (A[low1] < B[low2]) {C[low] = A[low1]; merge(A,low1+1, high1, B, low2, high2, C,low + 1);} else {C[low] = B[low2]; merge(A,low1, high1, B, low2+1, high2, C,low + 1); } } void copy(int A[], int s, int t, int B[], int s1) { for (int j= s; j<= t; ++j) B[s1 +j - s] = A[j]; }

23. Last example: recursion and a linked list. Suppose we want to insert a key x as a last item of a list. a c g k x = ‘b’ a c g k b

24. void insertLast(int k) { // insert k as the last item of the list Node* tmp = new Node(k); if (head == 0) head = tmp; else if (head->next == 0) { head->next = tmp;} else {List rest; rest.head = head->next; rest.insertLast(k); } }