Is it possible to find transpose of a matrix without using extra array.I mean in the below program if I have not to use Transpose[ ][ ] array then how can I find the solution.

Question

5.00/5 (1 vote)

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C++

#include< stdio.h >

int main()
{
    int m, n, c, d, matrix[10][10], transpose[10][10];

    printf ("Enter the number of rows and columns of matrix: \n");
    scanf ("%d%d",&m,&n);
    printf ("Enter the elements of matrix: \n");

    for(c = 0; c < m; c++)
    {
        for(d = 0; d < n; d++)
        {
            scanf("%d",&matrix[c][d]);
        }
    }

    for(c = 0; c < m; c++)
    {
        for(d = 0; d < n; d++)
        {
            transpose[d][c] = matrix[c][d];
        }
    }

    printf("Transpose of entered matrix: \n");

    for(c = 0; c < n; c++)
    {
        for(d = 0; d < m; d++)
        {
            printf("%d\t",transpose[c][d]);
        }
        printf("\n");
    }

    return 0;
}

Posted 23-Nov-13 7:22am

kumar_11

Updated 11-Oct-17 20:22pm

enhzflep

v2

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enhzflep · Answer 1 · 2013-11-23T08:20:00

Solution 1

Sure it is. You're so very close!

Just remember what the transposition of a matrix actually involves - it involves reflecting the elements across the diagonal axis that runs from top-left to bottom-right.

That is to say:

[0 1]    [0 2]
[2 3]    [1 3]

[0 1 2]  [0 3 6]
[3 4 5]  [1 4 7]
[6 7 8]  [2 5 8]

In each case, the row & column indicess of an element are swapped unless the row index == column index.

So, you just need a single temporary variable of whatever type your matrix holds. Its just a series of (N^2 - N) swap operations, where N is the number of rows or columns - assuming a square matrix.

Put the contents of [row][column] into tmpVal
Set the contents of [row][column] to be [column][row]
Set the contents of [column][row] to be tmpVal

Naturally, this doesn't take into account the fact that we don't have to copy elements if column==row. Its actually probably quicker just to blast through, never comparing the column and the row. Theres relatively few element this will be true for - worst case is 50%, when you've got a 2x2 matirx - or for a 10x10 matrix, it will be 10 elements. This can be calculated (for a square matrix) by N/(N^2) So, as the matrix grows larger, the potential savings get ever smaller.

Yet, if you try to do this 'optimization' you'll find yourself comparing the row to the column for 100% of the elements.

Posted 23-Nov-13 8:20am

enhzflep

Comments

kumar_11 23-Nov-13 14:27pm

I really appreciate your way of making things clear.
Just one doubt of mine,Is it doable where the condition arises like raw!=column.

enhzflep 23-Nov-13 14:41pm

That's okay, you're really welcome.

A tuna sandwich is less valuable than a fishing lesson. A tuna sandwich and a fishing lesson can often be even better again!

Yep, it is doable when row != column, in fact - the only time that you _need_ to swap is when row != column. It will be quicker to swap regardless of whether row==column or not (no conditional branching or pipeline stalls).

Since [1][1] is always going to be [1][1], and [2][2] always goes to [2][2], you can safely swap [row=1][column=1] with [column=1][row=1] etc, etc.

Probably best you forget all about this mention of an optimization for now. It will only make the code less clear to read, for very, very little gain. Often optimized code is harder to read than incorrect, yet simply implemented code.

kumar_11 24-Nov-13 3:47am

Thank you so much for the answer.
I request you to write the code here because I tried so hard but still not getting result.

enhzflep 24-Nov-13 3:55am

Code or it didn't happen! :p

If you add what you've tried, I'll try to guide you further.

kumar_11 24-Nov-13 4:16am

Here is what I have written, Kindly correct me where I am wrong.

enhzflep 24-Nov-13 4:49am

Ah hah! Something just occured to me you know.

I'm guessing that your 'transposed' matrix looks just the same as your original matrix, right?
I totally overlooked something before, I was that tired. I feel like a bit of a dummy, to be honest.

Each element is being swapped twice..
E.g
Row 1, Column 3 gets swapped with Row 3, Column 1
then later on,
Row3, Column 1 gets swapped with Row 1, Column 3

You simply need to change the upper limit of your loop through each column.

instead of covering the whole matrix, we only need to cover the bottom-left corner or the top-right half.

So, in your swapping loop -

run (j=0; j< j++) as you already do.

but instead of running through i=0 to i=n, run from i=0 to i=j. Fixed!

kumar_11 1-Dec-13 16:26pm

Let me try it then I will be able to tell you anything.

Member 9863088 30-Sep-15 8:05am

didnt worked :/

Member 13460419 · Answer 2 · 2017-10-11T20:22:00

//transpose of matrix 12/10/2017 ashu jain
#include<stdio.h>
int main()
{
int arr[10][10],i,j,row,col,temp;
printf("how many rows and columns you want to take");
scanf("%d %d",&row,&col);
printf("enter the elements of matrix");
for(i=0;i<row;i++)
for(j=0;j<col;j++)
scanf("%d",&arr[i][j]);
printf("transpose of matrix is\n");
if(row<col)
{
for(i=0;i<row;i++)
for(j=0;j<col;j++)
{

if(i<j)
{
temp=arr[i][j];
arr[i][j]=arr[j][i];
arr[j][i]=temp;
}
}}
else

for(i=0;i<row;i++)
for(j=0;j<col;j++)
{

if(i>j)
{
temp=arr[i][j];
arr[i][j]=arr[j][i];
arr[j][i]=temp;
}}
for(i=0;i<col;i++)
{
for(j=0;j<row;j++)
printf("%d ",arr[i][j]);
printf("\n");
}
}