Parfor performance are too good to be true

Question

Noam 2015-11-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/252782-parfor-performance-are-too-good-to-be-true

编辑： Thorsten 2015-11-4

Code:

function X
z=ones(1000);
tic
parfor j = 1:200
  for k = 1:200
      z = z+1;
  end
end;
toc
zz=ones(1000);
tic
for j = 1:200
  for k = 1:200
      zz = zz+1;
  end
end;
toc
isequal(z,zz)

Output:

Elapsed time is 0.091983 seconds.
Elapsed time is 12.810199 seconds.
ans =
       1

I have two questions:

1. I have only 4 cores, how does parfor run more than 100 times faster?

2. Is the code legal? z = z+1 from 4 cores simultaneously can cause unwanted result (two +1 operations in the same time can result only one increment).

Thanks

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Answer 1

Thorsten 2015-11-4

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https://ww2.mathworks.cn/matlabcentral/answers/252782-parfor-performance-are-too-good-to-be-true#answer_198413

编辑：Thorsten 2015-11-4

1. I guess the code is optimized, such that, e.g., the for loop is replaced by z = z + 200;

2. That's valid, because the computation of z can be done in any order. It's a simplified version of the last example in http://de.mathworks.com/help/distcomp/parfor.html

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Noam 2015-11-4

1. I guess you are right, but it's strange that in the other case it's not optimised.

2. So I understand that we do not need semaphors etc. as in C.

Thorsten 2015-11-4

编辑：Thorsten 2015-11-4

1. In a parfor Matlab has to ensure that the results of each iteration are independent. When this is analyzed, the optimization is probably done as a by-product.

2.Right.

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