onkus t1_iwfw2jo wrote
Reply to comment by eternal-abyss-77 in Can someone explain me the math behind this paper and tell me whether the way I have understood this paper is right or not? by eternal-abyss-77
How can anyone answer this?
If you are wanting people to tell you if you've understood it correctly, you need to tell us what your understanding is.
eternal-abyss-77 OP t1_iwfxk5b wrote
Ok I'll ask you exactly what i don't understand.
In equations 2 and 4 there is I which represents Identity matrix right?
So if let's say l = 2 and N = 7
Now will the shift matrix be like
[ 0 0 0 0 0 1 0 ]
[ 0 0 0 0 0 0 1 ]
[ 0 0 0 0 0 0 0 ]
[ 0 0 0 0 0 0 0 ]
[ 0 0 0 0 0 0 0 ]
[ 1 0 0 0 0 0 0 ]
[ 0 1 0 0 0 0 0 ]
If yes, then
[ I -I ]
[-I I ]
Should be of the form,
[ 1 0 0 0 0 -1 0 ]
[ 0 1 0 0 0 0 -1 ]
[ 0 0 0 0 0 0 0 ]
[ 0 0 0 0 0 0 0 ]
[ 0 0 0 0 0 0 0 ]
[ -1 0 0 0 0 1 0 ]
[ 0 -1 0 0 0 0 1 ]
Or
[ 1 0 0 0 0 -1 0 ]
[ 0 1 0 0 0 0 -1 ]
[ 0 0 1 0 0 0 0 ]
[ 0 0 0 1 0 0 0 ]
[ 0 0 0 0 1 0 0 ]
[ -1 0 0 0 0 1 0 ]
[ 0 -1 0 0 0 0 1 ]
?
Be it horizontal shift or vertical shift.
And what do they mean by rotations here : 0°, 45°, 90°, 135° ?
Because I'm extending this idea, so I am asking community help, perspectives, opinions and understandings, so I may not be wrongly understanding math.
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.
eternal-abyss-77 OP t1_iwg0e31 wrote
Please see dm
onkus t1_iwg3rav wrote
Much better to keep going here. Others can see and learn and/or contribute. Everyone is better off.
Do you have more questions?
eternal-abyss-77 OP t1_iwg59kg wrote
Yes
eternal-abyss-77 OP t1_iwg5fm0 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.
Hl is this
[ 0 0 0 0 0 1 0 ]
[ 0 0 0 0 0 0 1 ]
[ 0 0 0 0 0 0 0 ]
[ 0 0 0 0 0 0 0 ]
[ 0 0 0 0 0 0 0 ]
[ 1 0 0 0 0 0 0 ]
[ 0 1 0 0 0 0 0 ]
I - Hl is ?
[ 1 0 0 0 0 -1 0 ]
[ 0 1 0 0 0 0 -1 ]
[ 0 0 0 0 0 0 0 ]
[ 0 0 0 0 0 0 0 ]
[ 0 0 0 0 0 0 0 ]
[ -1 0 0 0 0 1 0 ]
[ 0 -1 0 0 0 0 1 ]
Or
[ 1 0 0 0 0 -1 0 ]
[ 0 1 0 0 0 0 -1 ]
[ 0 0 1 0 0 0 0 ]
[ 0 0 0 1 0 0 0 ]
[ 0 0 0 0 1 0 0 ]
[ -1 0 0 0 0 1 0 ]
[ 0 -1 0 0 0 0 1 ]
?
Show me how the matrix is written
And elaborate this:
> 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.
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?
eternal-abyss-77 OP t1_iwg7mjm wrote
I am asking what is
I - H is?
I get the same H as you say, but what is the matrix we get after I - H? Is it a mirror of H? As in paper, they said
I -I
-I I
So, the I in I-H is, normal identity matrix where major diagonal elements are 1 or is it mirror of H
onkus t1_iwja7ea wrote
What do you mean by “mirror of H"? I is always the identity matrix here. Why do you think it could be the "mirror" of H?
eternal-abyss-77 OP t1_iwjvpds wrote
So I - HL will be
[ 1 0 0 0 0 -1 0 ]
[ 0 1 0 0 0 0 -1 ]
[ 0 0 1 0 0 0 0 ]
[ 0 0 0 1 0 0 0 ]
[ 0 0 0 0 1 0 0 ]
[ -1 0 0 0 0 1 0 ]
[ 0 -1 0 0 0 0 1 ]
This. And same goes for I - VL
Fine.
What do they mean by those rotations?
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