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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采纳的回答
Matt J
2021-11-30
编辑:Matt J
2021-11-30
It is because of edge effects. If you add sufficient zero padding you will see symmetric behavior.
3 个评论
Matt J
2021-11-30
编辑:Matt J
2021-11-30
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
更多回答(2 个)
Image Analyst
2021-11-30
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))
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