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Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Test  >  Question Bank for GATE Computer Science Engineering  >  Test: Recursion- 3 - Computer Science Engineering (CSE) MCQ

Recursion- 3 - Free MCQ Practice Test with solutions, GATE CSE (CSE)


MCQ Practice Test & Solutions: Test: Recursion- 3 (20 Questions)

You can prepare effectively for Computer Science Engineering (CSE) Question Bank for GATE Computer Science Engineering with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Recursion- 3". These 20 questions have been designed by the experts with the latest curriculum of Computer Science Engineering (CSE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 55 minutes
  • - Number of Questions: 20

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Test: Recursion- 3 - Question 1
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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);

Detailed Solution: Question 1

  • The function fun(x, y) is a recursive function.
  • Base case: If x == 0, it returns y.
  • Recursive case: It calls fun(x - 1, x + y).
  • For fun(4, 3):
    • fun(4, 3) = fun(3, 4 + 3) = fun(3, 7)
    • fun(3, 7) = fun(2, 3 + 7) = fun(2, 10)
    • fun(2, 10) = fun(1, 2 + 10) = fun(1, 12)
    • fun(1, 12) = fun(0, 1 + 12) = fun(0, 13)
    • fun(0, 13) = 13 (base case) 
  • Thus, the value of fun(4, 3) is 13.

Test: Recursion- 3 - Question 2
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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;

Detailed Solution: Question 2

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

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Test: Recursion- 3 - Question 3
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What does the following function do?
int fun(int x, int y)
{
if (y == 0)   return 0;
return (x + fun(x, y-1));

Detailed Solution: Question 3

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

Test: Recursion- 3 - Question 4
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What does the following function print for n = 25?

void fun(int n)

{

if (n == 0)

return;

printf("%d", n%2);

fun(n/2);

Detailed Solution: Question 4

The function mainly prints binary representation in reverse order.

Test: Recursion- 3 - Question 5
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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;
}

Detailed Solution: Question 5

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 4000 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.

Test: Recursion- 3 - Question 6
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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));
}

Detailed Solution: Question 6

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

Test: Recursion- 3 - Question 7
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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;
}

Detailed Solution: Question 7

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.

Test: Recursion- 3 - Question 8
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Find the number of elements in the following array.

float f [6] [4] [2]; 

Detailed Solution: Question 8

A number of elements present in the array can be found by calculating the length of the array.

Algorithm:

Step 1: Start

Step 2: Initialize arr[] = {1, 2, 3, 4, 5, 6}

Step 3: length = sizeof(arr) / sizeof(arr[0])

Step 4: Print "number of elements present in given array" by assigning length.

Step 5: Return 0

Step 6: End

Analysis:- float f [6] [4] [2]

Number of element = 6 × 4 × 2 = 48.

So, the number of elements in the array, float f [6] [4] [2] is 48.

Test: Recursion- 3 - Question 9
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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);
}

Detailed Solution: Question 9

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.

Test: Recursion- 3 - Question 10
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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) ?

Detailed Solution: Question 10

Test: Recursion- 3 - Question 11
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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)?

Detailed Solution: Question 11

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.

Test: Recursion- 3 - Question 12
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Output of following program?
#include <stdio.h>
int fun(int n, int *f_p)
{
    int t, f;
    if (n <= 1)
    {
        *f_p = 1;
        return 1;
    }
    t = fun(n- 1,f_p);
    f = t+ * f_p;
    *f_p = t;
    return f;
}
 
int main()
{
    int x = 15;
    printf (" %d n", fun(5, &x));
    return 0;
}

Detailed Solution: Question 12

Let x is stored at location 2468 i.e. &x = 2468

(dots are use just to ensure alignment)
x = 15
fun(5, 2468)
...{t = fun(4, 2468)
.......{t = fun(3, 2468)
...........{t = fun(2,2468)
...............{t = fun(1, 2468)
...................{// x=1
........................return 1}
................t = 1
................f = 2 //1+1 //since *f_p is x
................x = t = 1
................return 2}
...........t = 2
...........f = 2+1
...........x = t = 2
...........return 3}
........t = 3
........f = 3+2
........x = t = 3
........return 5}
....t = 5
....f = 5+3
....x = t = 5
....return 8}

which implies fun (5,2468) is 8.

Test: Recursion- 3 - Question 13
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#include<stdio.h>
int f(int *a, int n)
{
  if(n <= 0) return 0;
  else if(*a % 2 == 0) return *a + f(a+1, n-1);
  else return *a - f(a+1, n-1);
}
  
int main()
{
  int a[] = {12, 7, 13, 4, 11, 6};
  printf("%d", f(a, 6));
  getchar();
  return 0;
}

Detailed Solution: Question 13

f() is a recursive function which adds f(a+1, n-1) to *a if *a is even. If *a is odd then f() subtracts f(a+1, n-1) from *a. See below recursion tree for execution of f(a, 6). .
 f(add(12), 6) /*Since 12 is first element. a contains address of 12 */
    |
    |
 12 + f(add(7), 5) /* Since 7 is the next element, a+1 contains address of 7 */
        |
        |
     7 - f(add(13), 4)
              |
              |
           13 - f(add(4), 3)
                     |
                     |
                  4 + f(add(11), 2)
                             |
                             |
                           11 - f(add(6), 1)
                                    |
                                    |
                                 6 + 0
So, the final returned value is 12 + (7 – (13 – (4 + (11 – (6 + 0))))) = 15

Test: Recursion- 3 - Question 14
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Consider the C function given below.
int f(int j)
{
  static int i = 50;
  int k;
  if (i == j)
  {
    printf(“something”);
    k = f(i);
    return 0;
  }
  else return 0;
}

Which one of the following is TRUE?

Detailed Solution: Question 14

When j is 50, the function would call itself again and again as neither i nor j is changed inside the recursion.

Test: Recursion- 3 - Question 15
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Which of the following statements are CORRECT?

1) Static allocation of all data areas by a compiler makes it impossible to implement recursion.

2) Automatic garbage collection is essential to implement recursion

3) Dynamic allocation of activation records is essential to implement recursion

4) Both heap and stack are essential to implement recursion

Detailed Solution: Question 15

Statement 1 – True

Dynamic memory allocation is necessary for implementing recursion because the number of function calls is not known in advance, requiring the use of a function call stack.

Statement 2 – False

Automatic garbage collection is not essential for recursion implementation, as languages provide specific functions for manual memory management.

Statement 3 – True

Dynamic memory allocation is required for recursion since the exact number of function calls cannot be predetermined.

Statement 4 – False

Recursion requires a stack structure. If a heap is used, it must be simulated as a stack for recursion purposes. The heap is generally used for user-driven dynamic memory allocation and is not needed for function calls. Thus, both stack and heap are not required simultaneously; either one is sufficient to implement recursion.

Therefore, statements 1 and 3 are true.

Test: Recursion- 3 - Question 16
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Consider the following function
double f(double x){
  if (abs(x*x - 3) < 0.01) return x;
  else return f(x/2 + 1.5/x);
}
Give a value q (to 2 decimals) such that f(q) will return q:_____.

Detailed Solution: Question 16

anil_ds_50

anil_ds_50_1

Test: Recursion- 3 - Question 17
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Consider the following C function.
int fun (int n)
{
  int x=1, k;
  if (n==1) return x;
  for (k=1; k<n; ++k)
     x = x + fun(k) * fun(n – k);
  return x;
}
The return value of fun(5) is __________.

Detailed Solution: Question 17

fun(5) = 1 + fun(1) * fun(4) + fun(2) * fun(3) + 
             fun(3) * fun(2) + fun(4) * fun(1)
       = 1 + 2*[fun(1)*fun(4) + fun(2)*fun(3)]

Substituting fun(1) = 1
       = 1 + 2*[fun(4) + fun(2)*fun(3)]

Calculating fun(2), fun(3) and fun(4)
fun(2) = 1 + fun(1)*fun(1) = 1 + 1*1 = 2
fun(3) = 1 + 2*fun(1)*fun(2) = 1 + 2*1*2 = 5
fun(4) = 1 + 2*fun(1)*fun(3) + fun(2)*fun(2)
       = 1 + 2*1*5 + 2*2 = 15

Substituting values of fun(2), fun(3) and fun(4)
fun(5) = 1 + 2*[15 + 2*5] = 51 

Test: Recursion- 3 - Question 18
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Consider the following recursive C function. If get(6) function is being called in main() then how many times will the get() function be invoked before returning to the main()?
void get (int n)
{
   if (n < 1) return;
   get(n-1);
   get(n-3);
   printf("%d", n);
}

Detailed Solution: Question 18

get(6) [25 Calls]
                              /      \
               [17 Calls] get(5)       get(3) [7 Calls]
                        /     \
                    get(4)    get(2)[5 Calls]
                   /    \ 
     [7 Calls] get(3)  get(1)[3 Calls]
                /     \
             get(2)   get(0)
            /    \
[3 Calls]get(1)  get(-1) 
   /  \
get(0) get(-2)
We can verify the same by running below program. [sourcecode language="CPP"] # include int count = 0; void get (int n) { count++; if (n < 1) return; get(n-1); get(n-3); } int main() { get(6); printf("%d ", count); } [/sourcecode] Output: 25

Test: Recursion- 3 - Question 19
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The function f is defined as follows:
int f (int n) {
    if (n <= 1) return 1;
    else if (n % 2  ==  0) return f(n/2);
    else return f(3n - 1);
}
Assuming that arbitrarily large integers can be passed as a parameter to the function, consider the following statements.
1. The function f terminates for finitely many different values of n ≥ 1.
ii. The function f terminates for infinitely many different values of n ≥ 1.
iii. The function f does not terminate for finitely many different values of n ≥ 1.
iv. The function f does not terminate for infinitely many different values of n ≥ 1.
Which one of the following options is true of the above?

Detailed Solution: Question 19

The function terminates for all values having a factor of 2 {(2.x)2==0}
So, (i) is false and (ii) is TRUE.
Let n = 3, it will terminate in 2nd iteration.
Let n=5, it will go like 5 - 14 - 7 - 20 - 10 - 5 – and now it will repeat.
And any number with a factor of 5 and 2, there are infinite recursions possible.
So, (iv) is TRUE and (iii) is false.

Test: Recursion- 3 - Question 20
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Match the pairs in the following questions:

2222

Detailed Solution: Question 20

Correct matching is A – 3, B – 1, C – 4, D – 2 Option (C) is correct.

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This test contains 20 MCQs on Recursion- 3 with detailed solutions for each question. It is part of the Question Bank for GATE Computer Science Engineering course on EduRev, designed to align with the current Computer Science Engineering (CSE) syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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