Much slower valid convolution using complementary size of kernels.
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I am using the valid convolution using convn( T, a, 'valid').
I have run the code below:
T = randn(384,384,8);
a = randn(5,5,8);
b = randn(380,380,1);
tic; convn(T,a,'valid'); toc
tic; convn(T,b,'valid'); toc
The reuslt in my computer is
Elapsed time is 0.002837 seconds.
Elapsed time is 0.016301 seconds.
Thus the the latter is much slower compared to fomer one.
However, in terms of flops, or only in terms of multiplications
convn(T,a,'valid')
takes 5*5*8*(384-5+1)*(384-5+1)*(8-8+1) = 28880000 multiplications
convn(T,b,'valid')
also takes 380*380*1*(384-380+1)*(384-380+1)*(8-1+1) = 28880000 multiplications
So why are the two computing time so different?
And is there some ways to implement the convn(T,b,'valid') much faster?
3 comentarios
Bruno Luong
el 24 de Dic. de 2020
Editada: Bruno Luong
el 24 de Dic. de 2020
No not FLOPS. As you said the FLOPS are more or less indentical.
Respuestas (3)
Bjorn Gustavsson
el 21 de Dic. de 2020
No, n-dimensional fourier-transforms, multiplication of the Fourier-transforms of 5-5-8 a with T will be a fair bit faster than the multiplication of the 380-by-380-by-1 b with T.
HTH
0 comentarios
Roshan Hingnekar
el 22 de Dic. de 2020
Editada: Walter Roberson
el 22 de Dic. de 2020
T and 'a' are 3 dimensional where as 'b' is 2 dimensional, convolution of 3-dimensional with 2-dimensional will be slower than a 3-dimensional with a 3-dimensional.
refer to the below links for further insight on randn and convn functions.
Bruno Luong
el 22 de Dic. de 2020
I would suggest to do specific conv with MEX programing.
Not sure the chance to beat MATLAB though.
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