Recursion

Recursion : The process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called as recursive function…. Read More

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Question 1

Predict output of following program

#include <stdio.h>

int fun(int n)
{
    if (n == 4)
       return n;
    else return 2*fun(n+1);
}


int main()
{
   printf("%d ", fun(2));
   return 0;
}

A	4
B	8
C	16
D	Runtime Error

Recursion
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Question 1 Explanation:

Fun(2) = 2 * Fun(3) and Fun(3) = 2 * Fun(4) ....(i)
Fun(4) = 4 ......(ii)
From equation (i) and (ii),
Fun(2) = 2 * 2 * Fun(4)
Fun(2) = 2 * 2 * 4
Fun(2) = 16. 

So, C is the correct answer

Question 2

Consider the following recursive function fun(x, y). What is the value of fun(4, 3)

int fun(int x, int y) 
{
  if (x == 0)
    return y;
  return fun(x - 1,  x + y);
}

A	13
B	12
C	9
D	10

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Question 2 Explanation:

The function fun() calculates and returns ((1 + 2 … + x-1 + x) +y) which is x(x+1)/2 + y.

Question 3

What does the following function print for n = 25?

void fun(int n)
{
  if (n == 0)
    return;

  printf("%d", n%2);
  fun(n/2);
}

A	11001
B	10011
C	11111
D	00000

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Question 3 Explanation:

The function mainly prints binary representation in reverse order.

Question 4

What does the following function do?

int fun(int x, int y)
{
    if (y == 0)   return 0;
    return (x + fun(x, y-1));
}

A	x + y
B	x + x*y
C	x*y
D	x^y

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Question 4 Explanation:

The function adds x to itself y times which is x*y.

Question 5

What does fun2() do in general?

int fun(int x, int y)
{
    if (y == 0)   return 0;
    return (x + fun(x, y-1));
}

int fun2(int a, int b)
{
    if (b == 0) return 1;
    return fun(a, fun2(a, b-1));
}

A	x*y
B	x+x*y
C	x^y
D	y^x

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Question 5 Explanation:

The function multiplies x to itself y times which is x^y.

Question 6

Output of following program?

#include<stdio.h>
void print(int n)
{
    if (n > 4000)
        return;
    printf("%d ", n);
    print(2*n);
    printf("%d ", n);
}

int main()
{
    print(1000);
    getchar();
    return 0;
}

A	1000 2000 4000
B	1000 2000 4000 4000 2000 1000
C	1000 2000 4000 2000 1000
D	1000 2000 2000 1000

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Question 6 Explanation:

First time n=1000 Then 1000 is printed by first printf function then call print(2*1000) then again print 2000 by printf function then call print(2*2000) and it prints 40000 next time print(4000*2) is called. Here 8000 is greater than 4000 condition becomes true and it return at function(2*4000). Here n=4000 then 4000 will again print through second printf. Similarly print(2*2000) after that n=2000 then 2000 will print and come back at print(2*1000) here n=1000, so print 1000 through second printf. Option (B) is correct.

Question 7

What does the following function do?

int fun(unsigned int n)
{
    if (n == 0 || n == 1)
        return n;

    if (n%3 != 0)
        return 0;

    return fun(n/3);
}

A	It returns 1 when n is a multiple of 3, otherwise returns 0
B	It returns 1 when n is a power of 3, otherwise returns 0
C	It returns 0 when n is a multiple of 3, otherwise returns 1
D	It returns 0 when n is a power of 3, otherwise returns 1

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Question 7 Explanation:

Lets solve with example, n = 27 which power of 3. First time if condition is false as n is neither equal to 0 nor equal to 1 then 27%3 = 0. Here, again if condition false because it is equal to 0. Then fun(27/3) will call. Second time if condition is false as n is neither equal to 0 nor equal to 1 then 9%3 = 0. Here again if condition false because it is equal to 0. Then fun (9/3) will call and third time if condition is false as n is neither equal to 0 nor equal to 1 then 3%3 = 0. Here again if condition false because it is equal to 0. Then fun(3/3) will call here n==1 if condition gets true and it return n i.e. 1. Option (B) is correct.

Question 8

Predict the output of following program

#include <stdio.h>
int f(int n)
{
    if(n <= 1)
        return 1;
    if(n%2 == 0)
        return f(n/2);
    return f(n/2) + f(n/2+1);
}


int main()
{
    printf("%d", f(11));
    return 0;
}

A	Stack Overflow
B	3
C	4
D	5

Recursion
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Question 8 Explanation:

On successive recursion F(11) will be decomposed into F(5) + F(6) -> F(2) + F(3) + F(3) -> F(1) + 2 * [F(1) + F(2)] -> 1 + 2 * [1 + F(1)] -> 1 + 2 * (1 + 1) -> 5. Hence , option D is the correct answer i.e, 5.

Question 9

Consider the following recursive C function that takes two arguments

 unsigned int foo(unsigned int n, unsigned int r) {
  if (n  > 0) return (n%r +  foo (n/r, r ));
  else return 0;
}

What is the return value of the function foo when it is called as foo(345, 10) ?

A	345
B	12
C	5
D	3

Recursion GATE CS 2011
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Question 9 Explanation:

The call foo(345, 10) returns sum of decimal digits (because r is 10) in the number n. Sum of digits for 345 is 3 + 4 + 5 = 12.

This solution is contributed by Pranjul Ahuja

Question 10

Consider the same recursive C function that takes two arguments

unsigned int foo(unsigned int n, unsigned int r) {
  if (n  > 0) return (n%r +  foo (n/r, r ));
  else return 0;
}

What is the return value of the function foo when it is called as foo(513, 2)?

A	9
B	8
C	5
D	2

Recursion GATE CS 2011
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Question 10 Explanation:

foo(513, 2) will return 1 + foo(256, 2). All subsequent recursive calls (including foo(256, 2)) will return 0 + foo(n/2, 2) except the last call foo(1, 2) . The last call foo(1, 2) returns 1. So, the value returned by foo(513, 2) is 1 + 0 + 0…. + 0 + 1. The function foo(n, 2) basically returns sum of bits (or count of set bits) in the number n.

There are 26 questions to complete.

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