# Much slower valid convolution using complementary size of kernels.

조회 수: 1(최근 30일)
Shen Zhao 2020년 12월 13일
편집: Shen Zhao 2020년 12월 24일
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?
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Bruno Luong 2020년 12월 24일
No not FLOPS. As you said the FLOPS are more or less indentical.

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### 답변(3개)

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.
HTH
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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.
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Shen Zhao 2020년 12월 24일
I read the corredponding webpages, could you explain more on why the same dimensional convolution will be faster?

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Bruno Luong 2020년 12월 22일
I would suggest to do specific conv with MEX programing.
Not sure the chance to beat MATLAB though.
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Shen Zhao 2020년 12월 24일
We once tried to write some C++ code to compete with matlab convn especially for 'valid' convolution, we found it really hard to beat Matalb convn. The matlab convn is really well-optimized.

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R2020b

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