GATE CS 2026 Set 1 Q.61: Functions, Recursion, Scope and Storage Classes

Consider the recursive functions represented by the following code segment:

int bar(int n){
     if (n == 1) return 0;
     else return 1 + bar(n/2);
}

int foo(int n){
     if (n == 1) return 1;
     else return 1 + foo(bar(n));
}

The smallest positive integer n for which foo(n) returns 5 is ______. (answer in integer)

Note: Ignore syntax errors (if any) in the function.

Your answer

Official answer and solution

Official answer: 65536
From: Official GATE 2026 answer key

Solution

Observe that bar(n) computes $\lfloor \log_2 n \rfloor$ , and work backwards from foo(1) = 1 to foo(n) = 5 by choosing the smallest $n$ at each step.

The function bar(n) counts how many times n is divided by 2 until it reaches 1, so $\text{bar}(n) = \lfloor \log_2 n \rfloor$ for all $n \ge 1$ . Notice that the smallest positive integer $n$ for which $\text{bar}(n) = k$ is $n = 2^k$ .
Working backwards for foo(n):
- $\text{foo}(n) = 1 \implies n = 1$
- $\text{foo}(n) = 2 \implies \text{foo}(\text{bar}(n)) = 1 \implies \text{bar}(n) = 1 \implies n_{\min} = 2^1 = 2$
- $\text{foo}(n) = 3 \implies \text{bar}(n) = 2 \implies n_{\min} = 2^2 = 4$
- $\text{foo}(n) = 4 \implies \text{bar}(n) = 4 \implies n_{\min} = 2^4 = 16$
- $\text{foo}(n) = 5 \implies \text{bar}(n) = 16 \implies n_{\min} = 2^{16} = 65536$ .

Final answer

65536

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 1 (CS1)
Question: 61 of 65
Section: Computer Science and Information Technology

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