numpy ReduceL2算子的朴素cuda核

ivqmmu1c  于 2023-08-05  发布在  其他
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我试图实现一个cuda内核的操作ReduceL2的教育目的。例如,我有一个形状为400*500*256的三维Tensor,它应该沿着第一和第二轴被'l2'约化。根据定义,结果是1*1*256Tensor。下面对numpy库的调用演示了操作符正在做什么:

reduced = np.sqrt(np.sum(a=np.square(input), axis=tuple((0,1)), keepdims=1))

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为了在cuda内核中实现它,我首先在python中实现了它,如下所示:

reduce_first = np.zeros([500, 256], dtype=np.float32)
reduce_second = np.zeros([256], dtype=np.float32)

res *= res

for i in range(res.shape[1]): #500
    for j in range(res.shape[2]): #256
        sum = 0
        for k in range(res.shape[0]): #400
            sum += res[k][i][j]
        reduce_first[i][j] = sum
    
#print(reduce_first.shape) #(500, 256)

for i in range(reduce_first.shape[1]): #256
    sum = 0
    for j in range(reduce_first.shape[0]): #500
        sum += reduce_first[j][i]
    reduce_second[i] = np.sqrt(sum)


reduce_secondTensor被验证为与上述reducedTensor相同。然后,我将python代码简单地翻译为下面的cuda内核

__global__ void ReduceL2Kernel(float* out, float* in, unsigned int first_dim, unsigned int second_dim, unsigned int third_dim){

    //400 * 500 * 256 -> 1 * 1 * 256
    //idz * idy * idx
    const int idz = blockIdx.z;//0~399
    const int idy = blockIdx.y * blockDim.y + threadIdx.y;//0~499
    const int idx = blockIdx.x * blockDim.x + threadIdx.x;//0~255

    if(idz >= first_dim || idy >= second_dim || idx >= third_dim) return;

    const int index = (idz * second_dim + idy) * third_dim + idx; //400*500*256
    in[index] *= in[index];

    //reduce along '400' axis
    float sum_first = 0.f;
    for(unsigned int i = 0; i < first_dim; i++)
        sum_first += in[(i * second_dim + idy) * third_dim + idx];
    out[idy * third_dim + idx] = sum_first;

    __threadfence();

    //reduce along '500' axis
    float sum_second = 0.f;
    for(unsigned int i = 0; i < second_dim; i++)
        sum_second += out[i * third_dim + idx];
    out[idx] = sqrtf(sum_second);
}


下面是调用CUDA内核的CPP端代码

inline int CeilDiv(int a, int b) { return (a + b - 1) / b;}

dim3 block_dims{64, 16, 1};
dim3 grid_dims{CeilDiv(third_dim, 64), CeilDiv(second_dim, block_dims.y), first_dim};
ReduceL2Kernel<<<grid_dims, block_dims, 0, stream>>>((float*)out, (float*)in, first_dim, second_dim, third_dim);


outTensor与python代码中的reducedreduce_secondTensor不同。我不知道为什么,任何更正都是值得的。

gzszwxb4

gzszwxb41#

简单地实现reducel2算子的正确方法如下:

__global__ void ReduceL2Kernel(float* out, float* in, unsigned int first_dim, unsigned int second_dim, unsigned int third_dim){

    //400 * 500 * 256 -> 1 * 1 * 256
    //idz * idy * idx
    const int idz = blockIdx.z;//0~399
    const int idy = blockIdx.y * blockDim.y + threadIdx.y;//0~499
    const int idx = blockIdx.x * blockDim.x + threadIdx.x;//0~255

    if(idz >= first_dim || idy >= second_dim || idx >= third_dim) return;

    const int index = (idz * second_dim + idy) * third_dim + idx; //400*500*256
    //in[index] *= in[index];

    //reduce along '400' axis
    float sum_first = 0.f;
    for(unsigned int i = 0; i < first_dim; i++)
        sum_first += in[(i * second_dim + idy) * third_dim + idx] * in[(i * second_dim + idy) * third_dim + idx];
    out[idy * third_dim + idx] = sqrtf(sum_first);

    __threadfence();

    //reduce along '500' axis
    float sum_second = 0.f;
    for(unsigned int i = 0; i < second_dim; i++)
        sum_second += out[i * third_dim + idx] * out[i * third_dim + idx];
    out[idx] = sqrtf(sum_second);
}

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