Have you encountered different results using "parfor" vs. "for" loop using Matlab 2019b

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Charles Nguyen
Charles Nguyen 2023-9-12
评论： Charles Nguyen 2023-9-12
Hello,
I would like to ask if you have different output results using "parfor" vs. "for" loop especically using the random generator (rng command) with fix seed (both "parfor" and "for" have the same seed).
If so, why is that as I expect the same results for both methods.
Thank you,
Charles

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回答（1 个）

Rik
Rik 2023-9-12
Ran in:
Put simply, each worker in the parallel pool inherits the workspace from before the loop.
So if you put your rng before the loop, you should expect different results. If you put the rng call inside the loop, you should expect the same results.
The reason is that each call to a random function influence the current seed. So if your loop contents use random generated data, that means that the iterations are not independent on their order. You should think of a parfor loop as doing this:
N=5;
SomeCode=@(n)disp(n);
v=1:N;
for n=v(randperm(end))
SomeCode(n)
end
4 5 3 2 1
If the results of this loop are stable, that means the iterations are independ, which is a requirement for a parfor loop.
With this example it easy why the results are not stable if you put rng outside the loop:
v=v(randperm(end));
old_seed = rng(0);
results=zeros(1,N);
for n=v
results(n)=rand+n;
end
results
results = 1×5
1.9058 2.9134 3.8147 4.1270 5.6324
rng(old_seed)
v=v(randperm(end));
old_seed = rng(0);
results=zeros(1,N);
for n=v
results(n)=rand+n;
end
results
results = 1×5
1.9058 2.1270 3.6324 4.8147 5.9134
Now with rng in the loop:
rng(old_seed)
results=zeros(1,N);
for n=v(randperm(end))
rng(n)
results(n)=rand+n;
end
results
results = 1×5
1.4170 2.4360 3.5508 4.9670 5.2220
results=zeros(1,N);
for n=v(randperm(end))
rng(n)
results(n)=rand+n;
end
results
results = 1×5
1.4170 2.4360 3.5508 4.9670 5.2220
  4 个评论
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Rik
Rik 2023-9-12
@Charles Nguyen
My point was that your code should assume such a randperm call. If it doesn't work if you do that, then your code will not have stable results. Did you try my examples with parfor n=1:N?
@Edric Ellis
Thank you for the correction/clarification. While my answer was technically incorrect, can you confirm my intuition that the random state not advancing continuously is likely the source of the issue of OP?
Charles Nguyen
Charles Nguyen 2023-9-12
@Rik,
Please see below
parfor n=1:N
rng(n)
results(n)=rand+n;
end
results
The results are below which is different than yours.
Starting parallel pool (parpool) using the 'local' profile ...
Connected to the parallel pool (number of workers: 4).
results =
1.1404 2.6097 3.8122 4.0277 5.3973

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