Functions : Recursion

Functions : Recursion Lecture 11 4.2.2002 Sudeshna Sarkar, IIT Kharagpur

Announcements • Class Test – I on 7th February, 6pm • Section 1 & 2 at S-301 • Section 5 & 6 at S-302 Sudeshna Sarkar, IIT Kharagpur

Call by value Note : The parameter of a function can be a constant, variable, expression – anything that has a value. Only the value is passed to the function. void printDouble (int x) { printf (“Double of %d “, x); x *= 2; printf (“is %d\n”, x) ; } void main () { int num=15; printDouble (num); printf (“ num = %d\n”, num); } Sudeshna Sarkar, IIT Kharagpur

Recursion • Def : A function is recursive if it calls itself. int rfun (int x) { . . . rfun (2*a) ; . . . } • Questions : • How does recursion work ? • Why do I write a recursive function ? Sudeshna Sarkar, IIT Kharagpur

Thinking Recursively Learning Objectives • To learn to think recursively • To learn how strategies for recursion involve both base cases and recursion cases • To learn how to search for different ways of decomposing a problem into subproblems. • To understand how to use call trees and traces to reason about how recursive programs work Sudeshna Sarkar, IIT Kharagpur

Program vs Process • Program = a set of instructions • akin to the blueprint of an information factory • Process = activity performed by computer when obeying the instructions • akin to operation of a working factory • can create multiple factories as needed from the same blueprint • need to allocate different local variables Sudeshna Sarkar, IIT Kharagpur

Factorial via Recursion /* 0! = 1! = 1, for n>1, n! = n* (n-1)! */ int factorial (int n) { int t; if (n <=1) t=1; else t = n * factorial (n-1) ; return t; } Sudeshna Sarkar, IIT Kharagpur

Review : Function Basics • Tracing recursive functions is apparent if you remember the basics about functions : • Formal parameters and variables declared in a function is local to it. • Allocated (created) on function entry • De-allocated (destroyed) on function return • Formal parameters initialized by copying value of actual parameter. Sudeshna Sarkar, IIT Kharagpur

Factorial factorial (4) = 4 * factorial (3) = 4 * 3 * factorial (2) = 4 * 3 * 2 * factorial (1) = 4 * 3 * 2 * 1 = 24 Sudeshna Sarkar, IIT Kharagpur

Factorial Trace n t n n t t n t n t Sudeshna Sarkar, IIT Kharagpur

int findmax (int n) { int i, num, max; scanf ("%d", &num) ; max = num; for (i=1; i<n; i++) { scanf ("%d", &num) ; if (num > max) max = num; } return max; } int findmax (int n, int max) { int num; if (n==0) return max; scanf ("%d", &num) ; if (num > max) max = num; max = findmax (n-1, max) ; return max; } Sudeshna Sarkar, IIT Kharagpur

Iteration vs. Recursion • Any iterative algorithm can be re-worked to use recursion, and vice-versa. • Some algorithms are more naturally written with recursion. • But naive applications of recursion can be inefficient. Sudeshna Sarkar, IIT Kharagpur

When to use Recursion ? • Problem has 1 or more simple cases. • These have a straightforward non-recursive solution. • Other cases can be re-defined in terms of problems that are closer to simple cases • By applying this redefn process repeatedly one gets to one of the simple cases. Sudeshna Sarkar, IIT Kharagpur

Example int sumSquares (int m, int n) { int i, sum; sum = 0; for (i=m; i<=n; i++) sum += i*i; return sum; } Sudeshna Sarkar, IIT Kharagpur

int sumSquares (int m, int n) { if (m<n) /* Recursion */ return sumSquares(m, n-1) + n*n; else /* Base Case */ return n*n ; } Example int sumSquares (int m, int n) { if (m<n) /* Recursion */ return m*m + sumSquares(m+1, n); else /* Base Case */ return m*m ; } Sudeshna Sarkar, IIT Kharagpur

Example int sumSquares (int m, int n) { int middle ; if (m==n) return m*m; else { middle = (m+n)/2; return sumSquares(m,middle) + sumSquares(middle+1,n) ; } 5 . . . 7 8 . . . 10 n m mid mid+1 Sudeshna Sarkar, IIT Kharagpur

Call Tree sumSquares(5,10) sumSquares(5,10) sumSquares(5,10) sumSquares(8,10) sumSquares(5,7) sumSquares(10,10) sumSquares(8,9) sumSquares(7,7) sumSquares(5,6) sumSquares(9,9) sumSquares(6,6) sumSquares(8,8) sumSquares(5,5) Sudeshna Sarkar, IIT Kharagpur

Annotated Call Tree 355 sumSquares(5,10) sumSquares(5,10) 245 110 sumSquares(5,10) sumSquares(8,10) sumSquares(5,7) 100 49 145 61 sumSquares(10,10) sumSquares(8,9) sumSquares(7,7) sumSquares(5,6) 36 81 25 64 sumSquares(9,9) sumSquares(6,6) sumSquares(8,8) sumSquares(5,5) 49 100 36 81 25 64 Sudeshna Sarkar, IIT Kharagpur

Trace sumSq(5,10) = (sumSq(5,7) + sumSq(8,10)) = (sumSq(5,6) + (sumSq(7,7)) + (sumSq(8,9) + sumSq(10,10)) = ((sumSq(5,5) + sumSq(6,6)) + sumSq(7,7)) + ((sumSq(8,8) + sumSq(9,9)) + sumSq(10,10)) = ((25 + 36) + 49) + ((64 + 81) + 100) = (61 + 49) + (145 + 100) = (110 + 245) = 355 Sudeshna Sarkar, IIT Kharagpur

Recursion : The general idea • Recursive programs are programs that call themselves • to compute the solution to a subproblem having these properties : 1. the subproblem is smaller than the overall problem or, simpler in the sense that it is closer to the final solution 2. the subproblem can be solved directly (as a base case) or recursively by making a recursive call. 3. the subproblem’s solution can be combined with solutions to other subproblems to obtain the solution to the overall problem. Sudeshna Sarkar, IIT Kharagpur

Think recursively • Break a big problem into smaller subproblems of the same kind, that can be combined back into the overall solution : Divide and Conquer Sudeshna Sarkar, IIT Kharagpur

Exercise 1. Write a recursive function that computes xn, called power (x, n), where x is a floating point number and n is a non-negative integer. 2. Write an improved recursive version of power(x,n) that works by breaking n down into halves, squaring power(x, n/2), and multiplying by x again if n is odd. 3. To calculate the square root of x (a +ve real) by Newton’s method, we start with an initial approximation a=x/2. If |x-aa| epsilon, we stop with the result a.Otherwise a is replaced with the next approximation, (a+x/a)/2. Write a recursive method, sqrt(x) to compute square root of x by Newton’s method. Sudeshna Sarkar, IIT Kharagpur

Common Pitfalls • Infinite Regress : a base case is never encountered • a base case that never gets called : fact (0) • running out of resources :each time a function is called, some space is allocated to store the activation record. With too many nested calls, there may be a problem of space. int fact (int n) { if (n==1) return 1; return n*fact(n-1); } Sudeshna Sarkar, IIT Kharagpur

Tower of Hanoi A B C Sudeshna Sarkar, IIT Kharagpur

void towers (int n, char from, char to, char aux) { if (n==1) { printf (“Disk 1 : %c -> &c \n”, from, to) ; return ; } towers (n-1, from, aux, to) ; printf (“Disk %d : %c  %c\n”, n, from, to) ; towers (n-1, aux, to, from) ; } Sudeshna Sarkar, IIT Kharagpur

Disk 1 : A -> C Disk 2 : A -> B Disk 1 : C -> B Disk 3 : A -> C Disk 1 : B -> A Disk 2 : B -> C Disk 1 : A -> C Disk 4 : A -> B Disk 1 : C -> B Disk 2 : C -> A Disk 1 : B -> A Disk 3 : C -> B Disk 1 : A -> C Disk 2 : A -> B Disk 1 : C -> B towers (4, ‘A’, ‘B’, ‘C’) ; Sudeshna Sarkar, IIT Kharagpur

Recursion may be expensive ! • Fibonacci Sequence: • fib (n) = n if n is 0 or 1 • fib (n) = fib (n-2) + fib(n-1) if n>= 2. int fib (int n) { if (n==0 or n==1) return 1; return fib(n-2) + fib(n-1) ; } Sudeshna Sarkar, IIT Kharagpur

Call Tree 5 fib (5) 2 3 fib (3) fib (4) 1 2 1 1 fib (1) fib (2) fib (2) fib (3) 1 1 1 0 1 1 0 fib (0) fib (1) fib (1) fib (2) fib (0) fib (1) 1 1 1 1 0 0 0 fib (1) fib (0) 1 0 Sudeshna Sarkar, IIT Kharagpur

Iterative fibonacci computation int fib (int n) { if (n <= 1) return n; lofib = 0 ; hifib = 1 ; for (i=2; i<=n; i++) { x = lofib ; lofib = hifib ; hifib = x + lofib; } return hifib ; } i = 2 3 4 5 6 7 x = 0 1 1 2 3 5 lofib= 0 1 1 2 3 5 8 hifib= 1 1 2 3 5 8 13 Sudeshna Sarkar, IIT Kharagpur

Merge Sort • To sort an array of N elements, • 1. Divide the array into two halves. Sort each half. • 2. Combine the two sorted subarrays into a single sorted array • Base Case ? Sudeshna Sarkar, IIT Kharagpur

Binary Search revisited • To search if key occurs in an array A (size N) sorted in ascending order: • mid = N/2; • if (key == N/2) item has been found • if (key < N/2) search for the key in A[0..mid-1] • if (key > N/2) search for key in A[mid+1..N] • Base Case ? Sudeshna Sarkar, IIT Kharagpur

Functions : Recursion