# Mutex for Increment in Matlab Parfor?

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Mikhail Kandel on 25 Mar 2012
I am trying to use Matlab's parfor for the following code:
parfor j=1:10
if (vtw(dta{j},A2,Mu2,Sigma2,A5,Mu5,Sigma5)==1)
results(1,1)=results(1,1)+1;
end
end
Matlab doesn't like this code saying parfor cannot be used because of the way results is being used.
To me this is a bit confusing as in C++ I would consider this a very separable operation and simply insert a mutex around the results++ (which doesn't really need one as increments are atomic). How do I fix this?

Edric Ellis on 26 Mar 2012
It's only the indexing into results that PARFOR doesn't understand. The following should work I think:
results = 0;
parfor j=1:10
if (vtw(dta{j},A2,Mu2,Sigma2,A5,Mu5,Sigma5)==1)
results = results + 1;
end
end
Edric Ellis on 26 Jun 2013
There is no race condition - PARFOR understands the reduction using '+' and how to perform that in parallel.

Walter Roberson on 25 Mar 2012
parfor j = 1 : 10
results_j(j) = vtw(dta{j},A2,Mu2,Sigma2,A5,Mu5,Sigma5)==1;
end
results = sum(results_j);
Walter Roberson on 26 Mar 2012
Misha, sorry, I do not have the appropriate toolbox to test this code, so I cannot say why MATLAB might not like it. If you post the error message someone might recognize it. Might be something as simple as initializing results_j ahead of time.

owr on 26 Mar 2012
Be careful using if statements without matching elses inside parfor loops. Ive experienced some strange behaviour in both 2011a and 2011b when this is done with linear indexing.
This example is a bit contrived, but it illustrates the issue:
matlabpool open;
[x,y] = meshgrid(1:10,1:10);
results = nan(10);
parfor i =1:100
if( x(i) <= y(i) )
results(i) = 2;
end
end
>> results
results =
2 0 0 0 0 0 NaN NaN NaN NaN
2 2 0 0 0 0 NaN NaN NaN NaN
2 2 2 0 0 0 NaN NaN NaN NaN
2 2 2 2 0 0 NaN NaN NaN NaN
2 2 2 2 2 0 0 NaN NaN NaN
2 2 2 2 2 2 0 0 NaN NaN
2 2 2 2 2 2 2 0 NaN NaN
2 2 2 2 2 2 2 2 0 NaN
2 2 2 2 2 2 2 2 2 0
2 2 2 2 2 2 2 2 2 2
Where did the zeros come from?
Edric Ellis on 27 Mar 2012
Hm, that looks like a bug. Thanks for reporting this.

Jan on 25 Mar 2012
When different PARFOR loops process this line:
results(1,1) = results(1,1) + 1;
The following can happen:
1. Thread 1: results(1,1) is evaluated and stored temporarily
3. Thread 2: results(1,1) is evaluated and stored temporarily
4. Thread 1: results(1,1) is updated
5. Thread 2: results(1,1) is updated
Now the results(1,1) is increased by 1, and not by 2. This happens because there is no mutex to block simultaneous access to result.
If the increment is atomic, there would be no need to implement InterlockedIncrement(). While the 32 bit increment is usually atomic on modern compilers and processors, this is not guaranteed for 64 bit types as int64 and doubles. Using a critical section is recommended, or InterlockedIncrement().
Edric Ellis on 26 Mar 2012
Actually, PARFOR understands various reduction operations by taking advantage of the mathematical properties of the expression. So, this addition can be done, but the syntax needs tweaking.

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