关于二次对角线的转置（翻转）矩阵

Question

I'm trying to write a program to transpose a square matrix about it's secondary diagonal. I know how to transpose it normally (Along it's normal diagonal), but I am not able to figure out how to do it about the secondary axis.

What is wrong in the loop? I know that I have to run it till 'N/2' and change the initialization of 'i' and 'j', it does not work even if I do that though.

void transpose(int a[][N]) // Transposes matrix along the secondary diagonal 
{
    for (int i = 0; i < N; i++)
        for (int j = 0; j < i; j++)
        {
            int tmp = a[i][j];
                a[i][j] = a[N - 1 - j][N - 1 - i];
                a[N - 1 - j][N - 1 - i] = tmp;
        }

}

Answer 1

Try a simple 3x3 matrix first. You only have to access (N^2 - N)/2 elements.

Here is a visual.

   0 1 2
0  * * /
1  * / x
2  / x x

You only need to access (9-3)/2 = 3 elements. Specifically (0,0),(0,1), and (1,0), so a nested loop for N = 3 should look something like

for(int i = 0; i < 2; i++)
    for(int j = 0; j < 2-i; j++)
        //etc..

Hope this helps.

Answer 2

So, I figured out the loop after a lot of time. The correct code should access the upper triangular elements in the square matrix, and swap them with the corresponding elements in the lower triangular matrix. The secondary diagonal is left unchanged.

Here is the tested, working code:

void swapEl(int mat[][n])
{
    for (int i = 0; i < (n - 1); i++)
        for (int j = 0; j < (n - 1) - i; j++)
        {
            int tmp = mat[i][j];
                mat[i][j] = mat[(n - 1) - j][(n - 1) - i];
                mat[(n - 1) - j][(n - 1) - i] = tmp;
        }       
}

Answer 3

The code accesses the upper triangular elements in the square matrix and swaps them with the corresponding element in the lower triangular matrix. The secondary diagonal is left unchanged.

for(int i=0;i<(n-1);i++)
{
    for(int j=0;j<(n-i-1);j++)
    {
        int t = a[i][j];
        a[i][j] = a[n-j-1][n-i-1];
        a[n-j-1][n-i-1] = t;
    }
}