Much slower valid convolution using complementary size of kernels.
조회 수: 1(최근 30일)
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
takes 5*5*8*(384-5+1)*(384-5+1)*(8-8+1) = 28880000 multiplications
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?
Bjorn Gustavsson 2020년 12월 21일
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.
Roshan Hingnekar 2020년 12월 22일
편집: Walter Roberson 2020년 12월 22일
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.