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 个评论
Bruno Luong
2020-12-24
编辑:Bruno Luong
2020-12-24
No not FLOPS. As you said the FLOPS are more or less indentical.
回答(3 个)
Bjorn Gustavsson
2020-12-21
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 个评论
Roshan Hingnekar
2020-12-22
编辑:Walter Roberson
2020-12-22
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
2020-12-22
I would suggest to do specific conv with MEX programing.
Not sure the chance to beat MATLAB though.
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