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 Kommentare
Bruno Luong
am 24 Dez. 2020
Bearbeitet: Bruno Luong
am 24 Dez. 2020
No not FLOPS. As you said the FLOPS are more or less indentical.
Antworten (3)
Bjorn Gustavsson
am 21 Dez. 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 Kommentare
Roshan Hingnekar
am 22 Dez. 2020
Bearbeitet: Walter Roberson
am 22 Dez. 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
am 22 Dez. 2020
I would suggest to do specific conv with MEX programing.
Not sure the chance to beat MATLAB though.
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