More multiplication operations require less time

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Matt J
Matt J 2023-8-29
评论： Matt J 2023-8-31
Ran in:
I would expect the execution times for the 3 operations below to get longer and longer. Where have I misled myself? Is it an issue with tic/toc as the timing method, or something else?
A=rand(500,500,500);
tic;
A.*A;
toc;
Elapsed time is 0.256989 seconds.
tic;
A.*A.*A;
toc;
Elapsed time is 0.124557 seconds.
tic;
A.*A.*A.*A.*A.*A;
toc;
Elapsed time is 0.099304 seconds.

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Walter Roberson
Walter Roberson 2023-8-29
Ran in:
It is because you are not recording the output.
I introduced T0 here because I was noticing that in my tests, T1 (the first operation) was consistently slower than T2 (the second operation), and I suspected that time to parse or something similar was being allocated against the first operation. With the T0 introduced, the measured time for A.*A reduces.
A=rand(500,500,500);
tic;
T0 = A;
toc;
Elapsed time is 0.001954 seconds.
tic;
T1 = A.*A;
toc;
Elapsed time is 0.199039 seconds.
tic;
T2 = A.*A.*A;
toc;
Elapsed time is 0.201588 seconds.
tic;
T3 = A.*A.*A.*A.*A.*A;
toc;
Elapsed time is 0.208999 seconds.
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Walter Roberson
Walter Roberson 2023-8-30
Oh, right, it makes sense for James to have done that work! (But it would also have made sense for John to have done it as part of his high precision packages.)
Matt J
Matt J 2023-8-31
Part of the reason that the cost doesn't grow linearly here is that several .* calls in one operation are optimized so that the results are computed in one go, without intermediate arrays being constructed.
I see. Well, that seems like a very well-intentioned optimization, but hazardous for users trying to compare algorithms. It makes it impossible for the user to know if the operations they code are implemented literally.

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