Question about repeatability in parfor-loop

Hi everyone,

I thought that simuation results from a for-loop would be identical to simuation results from a parfor-loop. I am wrong. Granted, the difference is very small (e.g. 1.6098e-15).

Does anyone know the reason for the difference? I am running both for-loop and parfor-loop on the same PC.

Here is my little example to show the difference.

Matrices X1 and X2 are identical since they are both generated by for-loops.

Matrices X1 and X3 are different since X1 comes from for-loop and X3 comes from parfor-loop.

Matrices X3 and X4 are identical since they are both generated by parfor-loops. If this is not true, then I am very confused.

kevin1336()

Elapsed time is 6.684938 seconds.
Elapsed time is 7.309150 seconds.

ans = 0

Elapsed time is 8.928830 seconds.

ans = 0

Elapsed time is 6.893095 seconds.

ans = 0

function kevin1336


%% Run for-loop the first time
N  = 100;
X1 = zeros(N, 5);
tic
for iter = 1:N
    rng(iter, 'twister')
    x           = core;
    X1(iter, :) = x.';
end
toc


%% Run the same for-loop the second time
X2 = zeros(N, 5);
tic
for iter = 1:N
    rng(iter, 'twister')
    x           = core;
    X2(iter, :) = x.';
end
toc
max(abs(X1(:) - X2(:)))


%% Run parfor-loop the first time
X3 = zeros(N, 5);
tic
parfor iter = 1:N
    rng(iter, 'twister')
    x           = core;
    X3(iter, :) = x.';
end
toc
max(abs(X1(:) - X3(:)))


%% Run parfor-loop the second time
X4 = zeros(N, 5);
tic
parfor iter = 1:N
    rng(iter, 'twister')
    x           = core;
    X4(iter, :) = x.';
end
toc
max(abs(X4(:) - X3(:)))


end



function x = core

M = 1e6;
A = rand(M,5);
b = rand(M,1);
x = A \ b;

end

Answer 1

[Walter Roberson]

Inside the parfor, by default each worker only gets a single core, so the \ operation is calculated using a single core.

Inside normal for loops, each worker gets the full complement of cores to automatically parallelize the \ operation over.

The difference in the number of cores results in slight differences in rounding.