onkus

onkus t1_iwg7e0e wrote

I dont follow what you are showing with those matrices. You should actually ask a question.

Are you stating what H x I is? If so, you are using the same definition in your original question which i just said was wrong. I could have been more specific about H: each row and xolumn will have exactly one occurance of a 1 and N-1 zeros.

H =

0 0 1 0 0 0 0

0 0 0 1 0 0 0

0 0 0 0 1 0 0

0 0 0 0 0 1 0

0 0 0 0 0 0 1

1 0 0 0 0 0 0

0 1 0 0 0 0 0

for l =2,N=7

Do you know how to subtract matrices from one another? The second and third matrix you show suggests you don't.

Im not sure what to elaborate on regarding the rotation that the authors dont mention in their paper. I could talk for hours about rotations, pixel differences, numerical gradients etc. Im not just going to ramble on without context. What do you want to know?

3

onkus t1_iwfzulx wrote

Yes, I is the identity matrix.

The shift matrix, H, will not have a row or column with only zeros in it. If l is 2 and N is 7 then H(1, 3) (1 based) will be a 1 and the start of a diagonal.

You have similarly misunderstood equation 4. There will not be a row or column with only 0s in it.

The authors do not mention rotation at all in this paper. They do mention that gradients are computed along those directions by the pixel differences.

2