Is there faster way to apply `det` function along the third dimension?

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wei zhang 2020-7-17

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https://ww2.mathworks.cn/matlabcentral/answers/566382-is-there-faster-way-to-apply-det-function-along-the-third-dimension

评论： wei zhang 2020-7-19

I am trying to calculate the det of many 4*4 matrix. I store the data in a matrix with shape 4*4*n. (n =4000000). I am using the for loop to get the result as below. Is there any way to accelerate the progress? Like bsxfun or arrayfun, or some simillar ideas?

v = zeros(length(m),1);
for i=1:size(m,3)
    v(i)=det(m(:,:,i));
end

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Walter Roberson 2020-7-17

在 MATLAB Online 中打开

function dets = det4(x)
   in1 = reshape(x,[],size(x,3));
   dets = in1(1,:).*in1(6,:).*in1(11,:).*in1(16,:)-in1(1,:).*in1(6,:).*in1(15,:).*in1(12,:)-in1(1,:).*in1(10,:).*in1(7,:).*in1(16,:)+in1(1,:).*in1(10,:).*in1(15,:).*in1(8,:)+in1(1,:).*in1(14,:).*in1(7,:).*in1(12,:)-in1(1,:).*in1(14,:).*in1(11,:).*in1(8,:)-in1(5,:).*in1(2,:).*in1(11,:).*in1(16,:)+in1(5,:).*in1(2,:).*in1(15,:).*in1(12,:)+in1(5,:).*in1(10,:).*in1(3,:).*in1(16,:)-in1(5,:).*in1(10,:).*in1(15,:).*in1(4,:)-in1(5,:).*in1(14,:).*in1(3,:).*in1(12,:)+in1(5,:).*in1(14,:).*in1(11,:).*in1(4,:)+in1(9,:).*in1(2,:).*in1(7,:).*in1(16,:)-in1(9,:).*in1(2,:).*in1(15,:).*in1(8,:)-in1(9,:).*in1(6,:).*in1(3,:).*in1(16,:)+in1(9,:).*in1(6,:).*in1(15,:).*in1(4,:)+in1(9,:).*in1(14,:).*in1(3,:).*in1(8,:)-in1(9,:).*in1(14,:).*in1(7,:).*in1(4,:)-in1(13,:).*in1(2,:).*in1(7,:).*in1(12,:)+in1(13,:).*in1(2,:).*in1(11,:).*in1(8,:)+in1(13,:).*in1(6,:).*in1(3,:).*in1(12,:)-in1(13,:).*in1(6,:).*in1(11,:).*in1(4,:)-in1(13,:).*in1(10,:).*in1(3,:).*in1(8,:)+in1(13,:).*in1(10,:).*in1(7,:).*in1(4,:)
end

This will vectorize the calculation.

Walter Roberson 2020-7-18

Bruno's criticism of the precision problems and the high length of the formula for increasing n, are valid criticisms. Generally speaking, it is often the case that making code faster comes at the price of making it less accurate towards the margins. Theoretical definitions that suppose infinite precision get substituted for more nuanced checks that deal with floating-point realities, and time gets saved by not making the checks to figure out what compensation is needed for each case.

If you have some time you should look at how hypot() (finding the length of a hypotenuse) has to be implemented in practice in order to maintain accuracy. Consider sqrt(A^2 + B^2) under the circumstance that A or B is smaller than sqrt(realmin) and so squaring it might underflow to 0...

wei zhang 2020-7-19

@Bruno Luong Thank you for introducing the formula in wiki.

@Walter Roberson.

1. I am glad to know the matlabfunction command. It is much faster than combination of subs and syms.

2. Thank you for introducing the cause of error. It seems the error is increasing fast with the size of matrix increasing. May cause terrible results.Wow.

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