## Using reduction variables on the GPU: arrayfun or other options

### D. Plotnick (view profile)

on 24 Sep 2018
Latest activity Commented on by D. Plotnick

on 5 Dec 2018

### Joss Knight (view profile)

Hello all,
I am trying to figure out whether/how to use a reduction variable as the output of an arrayfun performed on the GPU. The basic problem is something of the form.
x = zeros(bigNumber,1);
for i = 1:I
x = x + f(i,argin);
end
where bigNumber makes x a large vector, f is some function, and argin are the arguments of that function. Now, using parfor, I can actually run this on the cpu in a manner where x is a reduction variable. However, I want to be able to run this on the GPU. Normally I would do something like:
function x = mainFun()
x = gpuArray.zeros(bigNumber,1);
v = arrayfun(@(i) loopedFun(gpuArray(i),args), 1:N,'UniformOutput',false)
v = sum(cat(2,v{:}),2);
x = x + v
end
function v = loopedFun(i,args)
v = someFunction(i,args);
end
This typically works, but since we are creating a [bigNumber,N] matrix in the intermediate stage this can rapidly fill up memory and become a huge computational burden in concatenation and summation stage. It should be far more efficient if I can initialize x as a reduction variable, and allow arrayfun to write to it in the same OOp independent manner that parfor would. However, I cannot find any examples of this type of operation. SO:
• Is using reduction variables possible on the gpu, using something like arrayfun,cellfun, etc.?
• What is the syntax, if it is? Reduction variables in parfor are pretty sensitive to syntax.
Thanks, -Dan

2018 年 9 月 25 日

### Joss Knight (view profile)

2018 年 12 月 4 日
採用された回答

I suppose it depends on what f is, is it a scalar operation for each element of x? If so you can move your loop over I inside your arrayfun function and, as long as bigNumber is suitably large, the GPU will do a great job.
function a = nMyFunc(a, j)
for i = 1:I
a = a + f(i, j, argin);
end
end
x = gpuArray.zeros(bigNumber,1);
x = arrayfun(@nMyFunc, x, (1:bigNumber)');
j identifies which element of x is being processed.
If f is a vector operation then you need to inspect it to see what you can do to vectorize it.
The only other thing that comes to mind is, if you're willing to write your own CUDA kernel, you could use atomic operations to accumulate into x as a large bigNumber-by-I kernel.

D. Plotnick

### D. Plotnick (view profile)

2018 年 12 月 3 日
With regards to your above answer, I thought Matlab did not have an equivalent to +=. A similar version of the above code returns the error:
"a" was previously used as a variable, conflicting with its use here as the name of a function or command.
Joss Knight

### Joss Knight (view profile)

2018 年 12 月 4 日
Yes, I was being lax with my code. I'll correct it.
D. Plotnick

### D. Plotnick (view profile)

2018 年 12 月 5 日
Thanks. I thought I might be missing something key for a minute there.