why square takes longer than multiplication

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Jerry
Jerry 2011 年 12 月 6 日
My question is why in matlab square takes more time than multiplication. I read an article, say, square actually can be done a little bit quicker than multiplication. But the test code below doesn't support this. So how does Matlab implement square(and other integer power, which I'm also interested in)? The test code is following:
a=1; tic; for k=1:10000 b=a*a; end toc;
Then, without any changes except "b=a*a" is changed into "b=a^2":
a=1; tic; for k=1:10000 b=a^2; end toc;
Consequently, multiplication takes about 77 microseconds while square takes 500+.
Thanks in advance!! Jerry

採用された回答

Walter Roberson
Walter Roberson 2011 年 12 月 6 日
The .^ operator is implemented in terms of pow(), which can involve logs. .^ does not appear to optimize .^2
If I recall correctly, there is a File Exchange contribution that decomposes integer powers to create the optimal multiplication series.
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Jerry
Jerry 2011 年 12 月 6 日
Thank you very much. Do you know if there is anyway to find how Matlab writes its built-in function?
Walter Roberson
Walter Roberson 2011 年 12 月 6 日
The way to find the code for a built-in MATLAB function is to get a code development job with MathWorks.
I did at one time write code (for my work) that decomposed integer powers into multiplications. Eventually, though, I removed that code again, as I was able to show that the result was lower precision than .^ was able to get.

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その他の回答 (1 件)

Jan
Jan 2011 年 12 月 6 日
Squaring is cheaper than the multiplication if it is performed in C and if it is implemented using the multiplication (!):
double a = 3.14159265, b;
b = a * a; // Standard
a *= a; // Slightly faster inplace squaring
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Jerry
Jerry 2011 年 12 月 6 日
Thank you! I hope to accept your answer as well, but unexpectedly I can accept just one.

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