numba cuda does not produce correct result with += (gpu reduction needed?)

Question

I am using numba cuda to calculate a function.

The code is simply to add up all the values into one result, but numba cuda gives me a different result from numpy.

numba code

 import math
 def numba_example(number_of_maximum_loop,gs,ts,bs):
        from numba import cuda
        result = cuda.device_array([3,])

        @cuda.jit(device=True)

    def BesselJ0(x):
        return math.sqrt(2/math.pi/x)

    @cuda.jit
    def cuda_kernel(number_of_maximum_loop,result,gs,ts,bs):
        i = cuda.grid(1)
        if i < number_of_maximum_loop:
            result[0] += BesselJ0(i/100+gs)
            result[1] += BesselJ0(i/100+ts)
            result[2] += BesselJ0(i/100+bs)

    # Configure the blocks
    threadsperblock = 128
    blockspergrid = (number_of_maximum_loop + (threadsperblock - 1)) // threadsperblock

    # Start the kernel 
    cuda_kernel[blockspergrid, threadsperblock](number_of_maximum_loop,result,gs,ts,bs) 

    return result.copy_to_host()

numba_example(1000,20,20,20) 

output:

array([ 0.17770302,  0.34166728,  0.35132036])

numpy code

import math
def numpy_example(number_of_maximum_loop,gs,ts,bs):
    import numpy as np
    result = np.zeros([3,])

    def BesselJ0(x):
        return math.sqrt(2/math.pi/x)

    for i in range(number_of_maximum_loop):
        result[0] += BesselJ0(i/100+gs)
        result[1] += BesselJ0(i/100+ts)
        result[2] += BesselJ0(i/100+bs)

    return result

numpy_example(1000,20,20,20) 

output:

array([ 160.40546935,  160.40546935,  160.40546935])

I don't know where I am being wrong. I guess I might use reduction. But it seems impossible to finish it with one cuda kernel.

Answer 1

Yes, a proper parallel reduction is needed to sum data from multiple GPU threads to a single variable.

Here's one trivial example of how it could be done from a single kernel:

$ cat t23.py
import math
def numba_example(number_of_maximum_loop,gs,ts,bs):
    from numba import cuda
    result = cuda.device_array([3,])

    @cuda.jit(device=True)
    def BesselJ0(x):
        return math.sqrt(2/math.pi/x)

    @cuda.jit
    def cuda_kernel(number_of_maximum_loop,result,gs,ts,bs):
        i = cuda.grid(1)
        if i < number_of_maximum_loop:
            cuda.atomic.add(result, 0, BesselJ0(i/100+gs))
            cuda.atomic.add(result, 1, BesselJ0(i/100+ts))
            cuda.atomic.add(result, 2, BesselJ0(i/100+bs))

# Configure the blocks
    threadsperblock = 128
    blockspergrid = (number_of_maximum_loop + (threadsperblock - 1)) // threadsperblock

 # Start the kernel
    init = [0.0,0.0,0.0]
    result = cuda.to_device(init)
    cuda_kernel[blockspergrid, threadsperblock](number_of_maximum_loop,result,gs,ts,bs)

    return result.copy_to_host()

print(numba_example(1000,20,20,20))
$ python t23.py
[ 162.04299487  162.04299487  162.04299487]
$

You can also do a proper reduction in numba directly with the reduce decorator as described here although I'm not sure 3 reductions can be done in a single kernel that way.

Finally, you could write an ordinary cuda parallel reduction using numba cuda as indicated here. It should not be difficult I think to extend that to performing 3 reductions in a single kernel.

These 3 different methods will likely have performance differences, of course.

As an aside, if you're wondering about the results discrepancy between my code above and your python code in the question, I can't explain it. When I run your python code I get results matching the numba cuda code in my answer:

$ cat t24.py
import math
def numpy_example(number_of_maximum_loop,gs,ts,bs):
    import numpy as np
    result = np.zeros([3,])

    def BesselJ0(x):
        return math.sqrt(2/math.pi/x)

    for i in range(number_of_maximum_loop):
        result[0] += BesselJ0(i/100+gs)
        result[1] += BesselJ0(i/100+ts)
        result[2] += BesselJ0(i/100+bs)

    return result

print(numpy_example(1000,20,20,20))
$ python t24.py
[ 162.04299487  162.04299487  162.04299487]
$