imgaussfilt asymmetry as a linear operator
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The Gaussian image blurring operation imgaussfilt is not a symmetric linear operator, as the test below shows. This is surprising to me based on my understanding of Gaussian blurring. I am wondering (a) why it is unsymmetric, and (b) given that it is unsymmetric, how I can obtain its adjoint operator. In other words, I would like a function imgaussfilt_adjoint such that sum(sum(X.*imgaussfilt(Y,k)))==sum(sum(imgaussfilt_adjoint(X,k).*Y)). Note that I am not referring to the 'symmetry' boundary option.
>> X = randn(200,200);
>> Y = randn(200,200);
>> Xblur = imgaussfilt(X,100);
>> Yblur = imgaussfilt(Y,100);
>> sum(sum(X.*Yblur))
ans =
3.6114
>> sum(sum(Xblur.*Y))
ans =
4.5137
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Matt J
le 30 Nov 2021
Modifié(e) : Matt J
le 30 Nov 2021
It is because of edge effects. If you add sufficient zero padding you will see symmetric behavior.
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Matt J
le 30 Nov 2021
Modifié(e) : Matt J
le 30 Nov 2021
It also appears that symmetry is present with any of the padding modes except for 'replicate'.
X = randn(200,200);
Y = randn(200,200);
for p=["symmetric","circular","replicate"]
Xblur = imgaussfilt(X,200,'Padding',p);
Yblur = imgaussfilt(Y,200,'Padding',p);
asymmetry=sum(X.*Yblur, 'all')-sum(Xblur.*Y , 'all')
disp ' '
end
Plus de réponses (2)
Image Analyst
le 30 Nov 2021
It's because they're random numbers. X and Y are not equal, and neither are the blurred versions. So why would you expect two different random matrices multiplied by each other element by element to have the identical integrated gray value? Look, I don't even get the same numbers as you
X = randn(200,200);
Y = randn(200,200);
Xblur = imgaussfilt(X,100);
Yblur = imgaussfilt(Y,100);
sum(sum(X.*Yblur))
sum(sum(Xblur.*Y))
Matt J
le 30 Nov 2021
Modifié(e) : Matt J
le 30 Nov 2021
(b) given that it is unsymmetric, how I can obtain its adjoint operator.
In the asymmetric case, a brute force solution would be to use func2mat to get the matrix form of the operator,
A=func2mat( @(X) imgaussfilt(X,10), ones(200)).';
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