Shared Memory Reduction example assuming 1 block

This example (can’t add link …/examples.html#shared-memory-reduction) in the documentation seems to assume 1 block.

import numpy as np
from numba import cuda
from numba.types import int32

# generate data
a = cuda.to_device(np.arange(1024))
nelem = len(a)

def array_sum(data):
    tid = cuda.threadIdx.x
  size = len(data)

  if tid < size:
      i = cuda.grid(1)

        # Declare an array in shared memory
        shr = cuda.shared.array(nelem, int32)
        shr[tid] = data[i]

        # Ensure writes to shared memory are visible
        # to all threads before reducing

        s = 1
        while s < cuda.blockDim.x:
            if tid % (2 * s) == 0:
                # Stride by `s` and add
                shr[tid] += shr[tid + s]
            s *= 2

        # After the loop, the zeroth  element contains the sum
        if tid == 0:
            data[tid] = shr[tid]

array_sum[2, 512](a)
print(a[0])                  # 523776
print(sum(np.arange(1024)))  # 523776

With array_sum[2, 1024](a), the result is 0.

I am confused by this part:

   tid = cuda.threadIdx.x
   size = len(data)

    if tid < size:
        i = cuda.grid(1)

If there is one block, then i simply equals tid.

Here is the equivalent implementation that also works with non-multiples of 2.

import numpy as np
from numba import cuda, int32

a = np.arange(1_000)
a_gpu = cuda.to_device(a)
n = len(a)

def sum_reduce_gpu(a): # assume 1 block
  # thread id within a 1D block (position relative to a block)
  thread_id = cuda.threadIdx.x
  # Create an array in shared memory
  shared_array = cuda.shared.array(n, int32) # shape needs to be a "simple constant", len(a) won't work
  # each thread is responsible for copying one value
  shared_array[thread_id] = a[thread_id]
  # wait for all threads to finish
  step_size = 1
  while step_size < n:
    if thread_id % (2 * step_size) == 0 and thread_id + step_size < n:
      shared_array[thread_id] += shared_array[thread_id + step_size]
    step_size *= 2
    # wait for all threads to finish to make sure shared_array[thread_id + step_size] has been updated
  # one thread is responsible for writing the result
  if thread_id == 0:
    # store the result in a
    a[thread_id] = shared_array[0]

sum_reduce_gpu[1, n](a_gpu)
print(a.sum(), a_gpu[0])

If I wanted to use multiple blocks, would I need a 2 step process? Each block will be responsible for one part of the array, then at the end, one of the threads will combine the results of each block.

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Yes, that’s generally the way to do it. The example is intended to demonstrate shared memory, rather than a really good example of a reduction. For the second step, you could use an atomic addition: Supported Atomic Operations — Numba 0.56.4+0.g288a38bbd.dirty-py3.7-linux-x86_64.egg documentation - if

You can also use the @reduce decorator to generate a reduction kernel: GPU Reduction — Numba 0.56.4+0.g288a38bbd.dirty-py3.7-linux-x86_64.egg documentation

1 Like