GATE CS 2014 Set 3 Q.20 (CS Q.10): Pointers, Arrays and Strings

Let $A$ be a square matrix of size $n \times n$ . Consider the following pseudocode. What is the expected output?

C = 100;
for i = 1 to n do
    for j = 1 to n do
    {
        Temp = A[i][j] + C;
        A[i][j] = A[j][i];
        A[j][i] = Temp - C;
    }
for i = 1 to n do
    for j = 1 to n do
        output (A[i][j]);

The matrix A itself

Transpose of the matrix A

Adding 100 to the upper diagonal elements and subtracting 100 from lower diagonal elements of A

None of the above

Official answer and solution

Official answer: A
From: Official GATE 2014 answer key

Solution

Trace the nested loop over the entire matrix to analyze the swapping behavior.

Examine the inner loop body for each pair $(i, j)$ :
```
Temp = A[i][j] + C;
A[i][j] = A[j][i];
A[j][i] = Temp - C;
```
Since $\text{Temp} - C = (A[i][j] + C) - C = A[i][j]$ , this block performs a swap of the elements $A[i][j]$ and $A[j][i]$ .
Trace diagonal elements ( $i = j$ ):
When $i = j$ , $A[i][i]$ is swapped with itself, leaving its value unchanged.
Trace off-diagonal elements ( $i \ne j$ ):
- When the outer loop visits $(i, j)$ where $i < j$ , the elements $A[i][j]$ and $A[j][i]$ are swapped.
- Later, when the outer loop visits the symmetric position $(j, i)$ where $j > i$ , the exact same code executes, swapping $A[j][i]$ and $A[i][j]$ a second time.
Swapping any two values twice restores each element to its original value and position.

Hence, the matrix remains completely unchanged, producing The matrix A itself (option A).

Final answer

Common trap

Mistaking this code for an in-place matrix transposition, overlooking that the loops iterate over all indices rather than j > i.

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

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 3
Question: 20 of 65, printed as CS Q.10
Section: Computer Science and Information Technology

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