Functions : Recursion

31.1.2001 Sudeshna Sarkar, IIT Kharagpur1

Functions : Recursion

Lecture 124.2.2002


Recursion

a + (a+1) + (a+2) + ... + b

= a + ((a+1) + (a+2) + ... + b)

sum(a,b) = a + sum(a+1,b)

Base case ?

factorial (n) = n*factorial(n-1)

Base case ?


Recursion

int sum (int a, int b)

{

int temp;

if (a>b) return 0;

temp=sum(a+1,b);

return a+temp;

}

int factorial (int n)

{

int temp;

if (n<=1) return 1;

temp=factorial(n-1);

return n*temp;

}


Recursion

main () { . . .

x = sum(1,4); . . .}int sum (int a, int b){

int temp;if (a>b) return 0;temp=sum(a+1,b);return a+temp;

}

1 4sum(a,b) { temp=sum(2,4); return 1+temp;}

sum(2,4){ temp=sum(3,4); return 2+temp;}



sum(5,4){ return 0;}


Recursion

main () { . . .



}




sum(4,4){ temp=0; return 4;}


Recursion

main () { . . .



}



sum(3,4){ temp=4; return 3+4;}


Recursion

main () { . . .


int temp;if (a<=b) return 0;temp=sum(a+1,b);return a+temp;

}




Recursion

main () { . . .


int temp;if (a<=b) return 0;temp=sum(a+1,b);return a+temp;

}



void foo ()

{

char ch;

scanf (“%c”, &ch);

if (ch==‘\n’) return;

foo ();

printf (“%c”, &ch);

}

What will the programoutput on the followinginput ?PDS \n


Example

int sumSquares (int m, int n) {

if (m<n)

/* Recursion */

return sumSquares(m, n-1) + n*n;

else

/* Base Case */

return n*n ;

}

int sumSquares (int m, int n) {if (m<n)

/* Recursion */return m*m +

sumSquares(m+1, n);

else /* Base Case */return m*m ;

}


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 . . . 1087 . . .

mmidmid+1

n


Call Tree

sumSquares(5,10)sumSquares(5,10)

sumSquares(5,7) sumSquares(5,10)sumSquares(8,10)

sumSquares(5,6) sumSquares(7,7) sumSquares(8,9) sumSquares(10,10)



Annotated Call Tree

sumSquares(5,10)sumSquares(5,10)

sumSquares(5,7) sumSquares(5,10)sumSquares(8,10)



355

110

61 49

245

145 100

25 36 64 81

25 36 49 64 81 100


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


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.


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


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-aa| 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.


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);}


Tower of Hanoi

A B C


Tower of Hanoi

A B C


Tower of Hanoi

A B C


Tower of Hanoi

A B C


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) ;}


Disk 1 : A -> CDisk 2 : A -> BDisk 1 : C -> BDisk 3 : A -> CDisk 1 : B -> ADisk 2 : B -> CDisk 1 : A -> CDisk 4 : A -> BDisk 1 : C -> BDisk 2 : C -> ADisk 1 : B -> ADisk 3 : C -> BDisk 1 : A -> CDisk 2 : A -> BDisk 1 : C -> B

towers (4, ‘A’, ‘B’, ‘C’) ;


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) ;}


Call Tree

fib (5)

fib (3) fib (4)

fib (1)

fib (2)fib (1) fib (2)

fib (0)

fib (3)

fib (1)

fib (1) fib (2)

fib (0)

fib (0) fib (1)

0 1 0 1 1

1

10

1 1 1

111010

2

2

3

5

0 1


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 7x = 0 1 1 2 3 5lofib= 0 1 1 2 3 5 8hifib= 1 1 2 3 5 8 13


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 ?


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 ?

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Functions : Recursion