Taking the transpose of a matrix in C with 1D arrays - c

Question

I have a matrix that is represented by a one dimensional array,
example:
the matrix
0 1 2 3
4 5 6 7
8 9 10 11
the array
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
now give the dimensions of this matrix and the array I want to find the transpose, i.e.
0, 4, 8, 1, 5, 9, 2, 6, 10, 3, 7, 11
I'm working in C and here is my code
#include <stdlib.h>
#include <stdio.h>
void transpose(int *array, int m, int n){
int new_array[12];
for (int i=0; i<m*n; i++) {
new_array[i] = ??;
}
for (int i=0; i<m*n; i++) {
array[i] = new_array[i];
}
}
void print_array(int array[], int size){
for (int i=0; i<size; i++) {
printf("%d\n",array[i]);
}
}
int main(){
int array[12];
for (int i=0; i<12; i++) {
array[i]=i;
}
print_array(array,12);
transpose(array,3,4);
print_array(array,12);
return 0;
}
I've tried a dozen times and failed. Is there a simple way to do this that I have missed?

Answer 1

Use couple of for loops to make the code easier to follow.
void transpose(int *array, int m, int n){
int new_array[12];
for (int i = 0; i < m; ++i )
{
for (int j = 0; j < n; ++j )
{
// Index in the original matrix.
int index1 = i*n+j;
// Index in the transpose matrix.
int index2 = j*m+i;
new_array[index2] = array[index1];
}
}
for (int i=0; i<m*n; i++) {
array[i] = new_array[i];
}
}

Answer 2

void transpose(int *array, int m, int n){
int new_array[12];
int k = 0;
for(int i = 0; i < n; i++){
for (int j = 0; j < m; j++){
new_array[k++] = array[j*n + i];
}
}
for (int i=0; i<m*n; i++) {
array[i] = new_array[i];
}
}

Answer 3

Write the transpose in terms of a double loop over range [0..n) and [0..m) and calculate the indexes corresponding to the old position and the new position:
#include <stdio.h>
static void transpose(int *array, int m, int n)
{
int new_array[m * n];
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
int old_idx = i * n + j;
int new_idx = j * m + i;
new_array[new_idx] = array[old_idx];
}
}
for (int i = 0; i < m * n; i++)
{
array[i] = new_array[i];
}
}
static void print_array(int array[], int size)
{
for (int i = 0; i < size; i++)
{
printf(" %d", array[i]);
}
putchar('\n');
}
int main(void)
{
int array[12];
for (int i = 0; i < 12; i++)
{
array[i] = i;
}
print_array(array, 12);
transpose(array, 3, 4);
print_array(array, 12);
return 0;
}
The functions have to be pre-declared or made static to compile with my default compilation options. The transpose() function shown will work with any shape (size) of matrix (whereas the original won't work if the product of the dimensions is more than 12). I flattened the output from the array printer, too (though I'd probably make it print matrix shaped output if it were for 'production' use). I do assume that you have C99, or C11 with VLA support.
$ ./trans
0 1 2 3 4 5 6 7 8 9 10 11
0 4 8 1 5 9 2 6 10 3 7 11
$

Answer 4

I have changed:
for (int i = 0 ; i < m*n ; i++) {
new_array[i] = ??;
}
to:
int ctr = 0;
for (int i = 0 ; i <= m*n ; i++) {
if (ctr > m*n)
ctr -= m*n - 1;
new_array[i] = array[ctr];
ctr += n;
}
The logic is quite simple. Each row is n integers long and so the number below a num in the 2-D matrix form would be n+num

Answer 5

You can rewrite your transpose function as like this:
static void transpose(int *array, int m, int n)
{
int *temp=malloc(m*n*sizeof(int)); //need to create a temporary array.
memcpy(temp,array,m*n*sizeof(int));
int i, j;
for (i = 0; i < m; ++i)
{
for (j = 0; j < n; ++j)
{
array[j*m+i]=temp[i*n+j];
}
}
free(temp);
}

Answer 6

You almost had it, I think this would do it...
void transpose(int *array, int m, int n){
int new_array[12];
int count;
count = 0;
for (int i=0; i < n; i++) {
for (int j=0; j < m; j += n) {
new_array[count++] = i + j;
}
}
for (int i=0; i < m * n; i++) {
array[i] = new_array[i];
}
}

Answer 7

This algorithm works for any number of rows an columns:
for (std::size_t i = 0; i < col; i++) {
for (std::size_t j = 0; j < row; j++) {
transpose[c++] = array[j * col + i];
}
}