Mutex for Increment in Matlab Parfor?

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?

Answer 1

[Edric Ellis]

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

[Mikhail Kandel]

Out of curiousity, is this incompatibility due to parfor's checking or is matlab unsure of results(1,1) location at run/parsetime?

[Mikhail Kandel]

That is to say, is results(1,1) contantly revaluated.Is something like &results+i+wj constantly evaluated? Or does matlab immediatly optomize it?

[Edric Ellis]

PARFOR analyses your code before execution, and it doesn't consider "results(1,1)" to be a valid target for a reduction.

[Abel]

so parfor will actively prevent the race-condition??

[Edric Ellis]

There is no race condition - PARFOR understands the reduction using '+' and how to perform that in parallel.

Answer 2

[Walter Roberson]

parfor j = 1 : 10
  results_j(j) = vtw(dta{j},A2,Mu2,Sigma2,A5,Mu5,Sigma5)==1;
end
results = sum(results_j);

[Mikhail Kandel]

Thank you for the solution, but why does matlab not like this code?

[Geoff]

Even in C/C++, I would have a preference for Walter's solution over one that uses calls to the kernel.

[Walter Roberson]

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.

Answer 3

[owr]

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]

Hm, that looks like a bug. Thanks for reporting this.

Answer 4

[Jan]

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
2. Thread 2: 1 is added to the temporary variable
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]

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.