GATE IT 2004 | Question: 58

Question

Consider the following C program which is supposed to compute the transpose of a given $4 \times 4$ matrix $M$. Note that, there is an $X$ in the program which indicates some missing statements. Choose the correct option to replace $X$ in the program.

#include<stdio.h>
#define ROW 4
#define COL 4
int M[ROW][COL] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16};
main()
{
    int i, j, t;
    for (i = 0; i < 4; ++i)
    {
        X
    }
    for (i = 0; i < 4; ++i)
        for (j = 0; j < 4; ++j)
            printf ("%d", M[i][j]);
}

• A.

  for(j = 0; j < 4; ++j){
       t = M[i][j];
       M[i][j] = M[j][i];
       M[j][i] = t;
  }
  

• B.

  for(j = 0; j < 4; ++j){
       M[i][j] = t;
       t = M[j][i];
       M[j][i] = M[i][j];
  }
  

• C.

  for(j = i; j < 4; ++j){
       t = M[i][j];
       M[i][j] = M[j][i];
       M[j][i] = t;
  }
  

• D.

  for(j = i; j < 4; ++j){
       M[i][j] = t;
       t = M[j][i];
       M[j][i] = M[i][j];
  }

Comments

@Arjun I think the option c is wrong [just my thought correct if I'm wrong] so when we initializing j=i it still start with as j=0 and loop runs again and remain the same but if we initializing j=i+1 then it swaps upper triangular matrix with lower triangular matrix 

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It is correct, j's initial value is updated in each iteration. if i = 0, j= 0 (0th row first 0 elements are skipped) i = 1, j = 1(1st row 1 element is skipped) and so on. this ensures that swaps for transpose is done only once (upper triangular). try to dry run for some iterations and you'll understand it better

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nope!

Answer 1

Option C:

look at the initial value of $j$, if $j$ starts with $0$, then double for loop will swap  $ M[i][j]$ with $M[j][i]$ and also $M[j][i]$ and $M[i][j]$ so the matrix $M$ will remain unchanged, so to avoid this double swapping we need to initialize $j = i$ and swap only upper triangular matrix with lower triangular matrix.

for(j = i; j < 4; ++j){
 // code for swapping M[i][j] with M[j][i]
     t = M[i][j];
     M[i][j] = M[j][i];
     M[j][i] = t;
}

Comments

Its correct, and output is transposed matrix.
http://ideone.com/DGjfQU

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In option C it was t = M[j][i]; so option a is answer.

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@Ravi, 3 lines inside inner for loop are to swap M[i][j] and M[j][i]. It doesn't matter if you assign t value of M[i][j] or M[j][i].

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i think option A is just doing some redundant work but still it gives the right output.

Answer 2

When swapping, this is the correct assignment for the very first location

t = M[i][j];

and not this

M[i][j] = t;

We want to save our first value in a temporary variable; and not overwrite our first value by the temporary variable.

So, Options B and D are eliminated straightaway.

 

Now, if column 0 and row 0 are already swapped, we don't wanna swap it again.

So, j shouldn't start from 0 every time. It should start with value == i.

So, Option C

Comments

nicely explained

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Really Nicely explained

Answer 3

option C

Comments

there was a typo in the choices, can you check your answer again?

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option D is non-sensical, right?

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Sir what is typo

Answer 4

Let me help you solve this quesion in a faster way and take around 1min 
so option B & D are directly eliminated as they are not storing intial value which will make difficulty to restore old value  of array and assign garbage values.

• B for(j = 0; j < 4; ++j){
       M[i][j] = t;//this will assign garbage value so B &d Not possible
       t = M[j][i];
       M[j][i] = M[i][j];
  }

Now look  at option A & B 
take a 2d array
 

1	2
3	4

Now apply the code  given in A option you will see you get the original arrray 
and we wanted transpose  so A worng 
SO C is  the answer

 

 

Answer 5

If (j) starts at 0, each pair is swapped twice (mean same element swap again and again) , returning them to their original positions; therefore, j must start at i to swap each pair only once.

So -  for(j=i;  j<4; j++){

       }

Answer 6

Answer: OPTION C