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@kkraus14

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@kkraus14 kkraus14 commented Apr 10, 2019

Note: the GPU solution cheats a bit by operating in place. I wasn't sure how best to do this with the stencil operation on the CPU.

That's not "a bit". A memory allocation is likely more expensive than the compute kernel entirely, especially in the case of cudaMalloc. If we wanted to be fair, you should allocate a new GPU output array and have the kernel populate that array based on the input array. You could also try running the CPU kernel inplace by passing the input array into the out parameter of the stencil decorator: https://numba.pydata.org/numba-doc/dev/user/stencil.html#out.

threadsperblock = (16, 16)
blockspergrid_x = math.ceil(x_gpu.shape[0] / threadsperblock[0])
blockspergrid_y = math.ceil(x_gpu.shape[1] / threadsperblock[1])
blockspergrid = (blockspergrid_x, blockspergrid_y)

%timeit smooth_gpu[blockspergrid, threadsperblock](x_gpu)

For 1d arrays you can use .forall(input.size) to have it handle the threadperblock and blockpergrid sizing under the hood but this doesn't exist for 2d+ arrays unfortunately. The current 16 threads per block seems really low where typically you see 128 or 256 so I'm not sure if this is best practice sans for a minimal documentation example. Maybe someone else can comment on a better threads per block and blocks per grid setting based on the 10k x 10k input array.

702 ms ± 66.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

This may be capturing JIT compilation on each run since it's only 1 loop, would recommend running an equal number of loops between CPU and GPU to be fair with regards to caching the JIT kernel compilation.

@pentschev

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@pentschev pentschev commented Apr 10, 2019

out_gpu = cupy.ones((10000, 10000), dtype='int8')

This should be cupy.zeros(), since you're not computing image borders.

@mrocklin

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@mrocklin mrocklin commented Apr 11, 2019

A memory allocation is likely more expensive than the compute kernel entirely, especially in the case of cudaMalloc

This is why we have pool memory allocators, yes?

You could also try running the CPU kernel inplace by passing the input array into the out parameter of the stencil decorator:

I tried this briefly and wasn't able to get it to work. I also ended up timing the allocation on the CPU side and it was only 40-50ms, which is about 10% of the total compute time. I agree though that this would be useful to investigate further if someone does a real benchmark here (that is not my intention for this particular notebook).

This may be capturing JIT compilation on each run since it's only 1 loop,

I've rerun it several times within the same process (to avoid the JIT compilation) and didn't notice any difference.

@Peacekeep3r

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@Peacekeep3r Peacekeep3r commented Jul 23, 2020

this example is great and seems to be everywhere on the internet, but I think there is a bug in using cupy-arrays. For one thing, you should get identical (?) performance feeding Numpy-Arrays, since the calculations are both done on gpu anyway. More importantly, I think that using cupy-arrays causes timeit to show only the kernel invocation time - nothing has actually been calculated. Can you please check this again? This is a top Google search result for numpy gpu stencils. Try to print the output, and the calculation will actually run. I get around 160 ms!

sadly the cpu version using parallel computing is still faster even for big arrays! (60 ms). The original stencil function is just slow in numba. Better do it manually:

@njit(nopython=True,parallel=True)
def smooth_cpu(x, out_cpu):

    for i in prange(1,np.shape(x)[0]-1):
        for j in range(1,np.shape(x)[1]-1):
            out_cpu[i, j] =  (x[i - 1, j - 1] + x[i - 1, j] + x[i - 1, j + 1] + x[i    , j - 1] + x[i    , j] + x[i    , j + 1] +x[i + 1, j - 1] + x[i + 1, j] + x[i + 1, j + 1]) / 9

edit: it seems I was wrong and it's mostly because of data transfer times as the cupy arrays are already on the GPU. I still think it needs a "cuda.synchronize()" for a fair comparison which increase running time quite alot.

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