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# Diagonal elements

Hi,

I have to set all the diagonal elements of a 2D matrix to a given value (i.e. zero). This is easy if the matrix is square, but I'm having problems with non-square matrices.

This is what I've tried so far:

int N , M; //Number of rows and columns

float* fData = new float [N * M];

float fValue = 0.0f;

float fSlope  = (float)M / (float)N;

for (register int i = 0; i < M; i++)
{

int j = (int) floor( i / fSlope + 0.5);

*fData = fValue;

int nOffset = (int)floor (nCols * slope + 0.5) +1;

fData += nOffset;

}

This works fine if the matrix is square but fails in the non-square case

elito
0
elito
1 Solution

Commented:
Please make clear How the selection criteria

Actually, Only square matrix have only diagonal elements.

If you want this is a particular application, so you have a criteria right...

Roshmon
0

Commented:
As you are allocating the matrix as a linear array, you could just

int N , M; //Number of rows and columns
int size = N * M;
float* fData = new float [size];

float fValue = 0.0f;

for (register int i = 0; i < size; i++) {

*(fData + i) = fValue;
}

0

Commented:
Explain for example what do you want to have in case of 3x4 matrix.

Something like this?
#000
0##0
000#
0

Commented:
The problem is that a non-square matrix, by definition, does not have a diagonal. The best you can do is an approximation of the diagonal. So first decide how you would estimate this diagonal, let us know, and then we can help you with the code.

Just a question (out of curiosity): Why do you need to do that? I am not aware of any matrix operations/calculations that need that for
0

Commented:
The projection of two coordinate (x,y) onto a one dimensional array f[N*M] is easy: f[M*y+x]. Your
'diagonal' traversal is equivalent to a raster graphics
line drawing problem, i.e. which pixels to choose when
there is no exact match. Have a look at the Bresenham

kind regards
0

Commented:
It seems to me that there are many diagonals:
#00000
0#0000
00#000
and
0#0000
00#000
000#00
and
00#000
000#00
0000#0
and
000#00
0000#0
00000#
But I am one of few people I know who enjoy playing tick-tack-toe in a rectagular grid, so please shoot me.

-- Dan

0

Author Commented:
sorry for the late response. I tried Bresenham algorithm and yes, that was exactly what I needed.
Thanks for the help,
elito
0

Author Commented:
Dan,
I meant the main diagonal of the matrix, the one that connects point (0,0) with point (M, N).
0
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