GPU 上的自定义缩减与 CPU 产生不同的结果

Question

为什么我在 GPU 上看到的结果与顺序 CPU 不同？

import numpy
from numba import cuda
from functools import reduce

A = (numpy.arange(100, dtype=numpy.float64)) + 1
cuda.reduce(lambda a, b: a + b * 20)(A) 
# result 12952749821.0
reduce(lambda a, b: a + b * 20, A) 
# result 100981.0

import numba
numba.__version__
# '0.34.0+5.g1762237'

使用 Java 流 API 并行化缩减 CPU 时会发生类似的行为：

int n = 10;
float inputArray[] = new float[n];
ArrayList<Float> inputList = new ArrayList<Float>();
for (int i=0; i<n; i++)
{
    inputArray[i] = i+1;
    inputList.add(inputArray[i]);
}
Optional<Float> resultStream = inputList.stream().parallel().reduce((x, y) -> x+y*20);
float sequentialResult = array[0];
for (int i = 1; i < array.length; i++)
{
    sequentialResult = sequentialResult + array[i] * 20;            
}
System.out.println("Sequential Result "+sequentialResult); 
// Sequential Result 10541.0
System.out.println("Stream Result "+resultStream.get()); 
// Stream Result 1.2466232E8

Answer 1

看来，正如 Numba's team, lambda a, b: a + b * 20 isn't associative and commutative 缩减函数所指出的那样， 产生了这个意想不到的结果。

GPU 上的自定义缩减与 CPU 产生不同的结果

Custom reduction on GPU vs CPU yield different result

python

lambda

gpu

reduction

numba