Discrepancy between convolution and filtering
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AlexRD
el 5 de Abr. de 2021
Comentada: AlexRD
el 5 de Abr. de 2021
I've been trying to optimize the convolution process for my neural network to make it work on a big matrix. I figured out the logic, but the implementation is giving me different results from the conv2 function.
I've written a code to exemplify this:
% The image is a 28x28 picture, and the filter 5x5. The code i've written
% here is not the one i've optimized for matrix multiplication, but it
% shows the same logic where you sweep the image, multiply it by the
% filter and sum it.
image = data(:,:,1);
filter = rand(5, 5);
a = conv2(image, filter, 'valid');
b = zeros(24, 24);
for j=1:24
for i=1:24
b(i, j) = sum(filter .* image(i:i+4, j:j+4), 'all');
end
end
Although a and b are very similar:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/573317/image.png)
They're usually off by a non-trivial amount:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/573322/image.png)
Is there something i'm doing wrong here? Or maybe the conv2 algorithm does something to speed up the process that changes the result in a significant way?
Also, when the filter is all ones (like ones(5,5) instead of rand), a and b are identical.
2 comentarios
Rik
el 5 de Abr. de 2021
I believe filter and conv2 are equivalent except for flipping the second input. That would explain why the result is the same for a symmetrical kernel.
Did you check the documentation?
Respuesta aceptada
Image Analyst
el 5 de Abr. de 2021
Your manual way is not flipping the kernel like conv2() does. So when the kernel is symmetric, the results will be the same and when the kernel is not symmetric, the results will be different. Try flipping the kernel in your "manual" way and you'll find they are the same.
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