How to transpose a matrix without using another matrix? - arrays

Question

Write a program which will accept 2-dimensional square matrix and find out the transpose of it.
Program should not make use of another matrix
Hi I am trying to transpose a 2*2 matrix without using another matrix.
Is there anything wrong with my transpose logic?
I am a newbie
#include <stdio.h>
int main()
{
int mat[2][2];
int i, j, temp;
for (i = 0; i < 2; i++) {
printf("\nEnter elements of %d row of first matrix: ", i + 1); //i+1 so that it can print 1 row, 2 row, 3 row etc
for (j = 0; j < 2; j++) { //loop inside to loop to get value for a[0][0],a[0][1],a[0][2]
scanf("%d", &mat[i][j]);
}
}
printf("The matrix\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
//transpose logic using same matrix
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
temp = mat[i][j];
mat[i][j] = mat[j][i];
mat[j][i] = temp;
}
}
printf("The transpose of the matrix is\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
}
EDIT: I found an easier way to do it however I still don't understand why my transpose logic by using this
temp = mat[i][j];
mat[i][j] = mat[j][i];
mat[j][i] = temp;
cannot get it to transpose.
Below is my corrected answer
#include <stdio.h>
int main()
{
int mat[2][2];
int i, j, temp;
for (i = 0; i < 2; i++) {
printf("\nEnter elements of %d row of first matrix: ", i + 1);//i+1 so that it can print 1 row, 2 row, 3 row etc
for (j = 0; j < 2; j++) {//loop inside to loop to get value for a[0][0],a[0][1],a[0][2]
scanf("%d", &mat[i][j]);
}
}
printf("The matrix\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
printf("The transpose of the matrix is\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[j][i]);
}
printf("\n");
}
}

Answer 1

Your program is a good attempt, but transposing the matrix is like reversing an array: you must stop half way to avoid swapping the transposed values twice, leading to the original matrix as you observe.
You should stop the inner loop when j == i, hence change the inner loop to:
for (j = 0; j < i; j++) { // j < i instead of j < 2
Here is a modified version:
#include <stdio.h>
int main() {
int mat[2][2];
int i, j, temp;
for (i = 0; i < 2; i++) {
printf("\nEnter elements of %d row of first matrix: ", i + 1);
for (j = 0; j < 2; j++) {
if (scanf("%d", &mat[i][j]) != 1)
return 1;
}
}
printf("The matrix:\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
//transpose logic using same matrix
for (i = 0; i < 2; i++) {
for (j = 0; j < i; j++) {
temp = mat[i][j];
mat[i][j] = mat[j][i];
mat[j][i] = temp;
}
}
printf("The transpose of the matrix is:\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
return 0;
}
Your corrected answer does not transpose the matrix at all, it merely outputs the transposed matrix. The matrix mat in memory is unchanged.