GATE CS 2020 Q.58 (CS Q.48): Pointers, Arrays and Strings

Consider the following C functions.

int tob(int b, int* arr) {
    int i;
    for (i = 0; b > 0; i++) {
        if (b % 2) arr[i] = 1;
        else arr[i] = 0;
        b = b / 2;
    }
    return (i);
}

int pp(int a, int b) {
    int arr[20];
    int i, tot = 1, ex, len;
    ex = a;
    len = tob(b, arr);
    for (i = 0; i < len; i++) {
        if (arr[i] == 1)
            tot = tot * ex;
        ex = ex * ex;
    }
    return (tot);
}

The value returned by pp(3, 4) is _______.

Your answer

Official answer and solution

Official answer: 81
From: Official GATE 2020 answer key

Solution

Trace the binary conversion function tob and the modular exponentiation loop in pp.

Trace of tob(4, arr):

$b = 4$ : $b \% 2 = 0 \implies \text{arr}[0] = 0$ , $b = 2$ , $i = 1$ .
$b = 2$ : $b \% 2 = 0 \implies \text{arr}[1] = 0$ , $b = 1$ , $i = 2$ .
$b = 1$ : $b \% 2 = 1 \implies \text{arr}[2] = 1$ , $b = 0$ , $i = 3$ .
Loop terminates with $i = 3$ . Returns $\text{len} = 3$ .
Array arr holds binary digits of 4 in little-endian order: $\text{arr}[0] = 0, \text{arr}[1] = 0, \text{arr}[2] = 1$ (representing $4 = 0 \cdot 2^0 + 0 \cdot 2^1 + 1 \cdot 2^2$ ).

Trace of pp(3, 4):

Initialization: tot = 1, ex = a = 3, len = 3.
Iteration $i = 0$ $i = 0$ :
- $\text{arr}[0] == 0$ : tot remains 1.
- ex = ex * ex = 3 * 3 = 9 ( $3^2$ ).
Iteration $i = 1$ $i = 1$ :
- $\text{arr}[1] == 0$ : tot remains 1.
- ex = ex * ex = 9 * 9 = 81 ( $3^4$ ).
Iteration $i = 2$ $i = 2$ :
- $\text{arr}[2] == 1$ : tot = tot * ex = 1 * 81 = 81.
- ex = ex * ex = 81 * 81 = 6561 ( $3^8$ ).
Loop finishes ( $i = 3$ ). Returns tot = 81.

This function computes $a^b = 3^4 = 81$ using repeated squaring.

Final answer

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

Source

Paper: GATE 2020 Computer Science and Information Technology (CS)
Question: 58 of 65, printed as CS Q.48
Section: Computer Science and Information Technology

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