# Image Processing - Absolute sum of the differences employing a weighted kernel

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Dear all,

I hope you can help. I am trying to replicate an alghoritm (for target detection) found in a research paper:

Unfortunately, I am stuck with an equation (see Eq. 20 and 21 in the link above), and I would grately appreciate your help.

I have an image, let's call it I (I=5x5).

I performed a Top-hat transform and now I need to apply a 'Local Difference Criterion'.

I created four direction vectors centered at I(i,j), where i and j are the coordinates of the central pixel. The four vectors are defined as:

L1=[I(i-2,j-2),I(i-1,j-1),I(i+1,j+1),I(i+2,j+2)];

L2=[I(i,j-2),I(i,j-1),I(i,j+1),I(i,j+2)];

L3=[I(i+2,j-2),I(i+1,j-1),I(i-1,j+1),I(i-2,j+2)];

L4=[I(i-2,j),I(i-1,j),I(i+1,j),I(i+2,j)];

Now, I need to calculate the sum of the differences in gray values between I(i+x,j+y) and I(i,j) as follows:

where Wx,y is the weighted kernel to describe the absolute difference between I(i+x,j+y) and I(i,j), and is equal to:

Would anyone be able to help me with writing a code to accomplish the above?

Thanks a lot in advance

##### 3 Comments

Image Analyst
on 1 May 2023

Edited: Image Analyst
on 1 May 2023

### Accepted Answer

Matt J
on 1 May 2023

Edited: Matt J
on 1 May 2023

There are problems with the mathematical formulation, as noted in the comments above. However, for the general task of computing shift-invariant weighted differences, I would set it up like the following:

L(1).deltaX=[-2 -1 1 2]

L(1).deltaY=[-2 -1 1 2]

L(1).weight=rand(1,4);

L(2).deltaX=[0 0 0 0]

L(2).deltaY=[-2 -1 1 2]

L(2).weight=rand(1,4);

...

clear d

for i=numel(L):-1:1

tmp=0;

for j=1:4

x=L(i).deltaX(j); y=L(i).deltaY(j); wxy=L(i).weight(j);

tmp=tmp+wxy*abs(circshift(I,[x,y])-I);

end

d(:,:,i)=tmp;

end

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