在 CUDA 的平铺矩阵乘法中访问矩阵作为其转置

Posted 2023-03-23

【中文标题】在 CUDA 的平铺矩阵乘法中访问矩阵作为其转置

【英文标题】：Access an matrix as its tranpose in tiled matrix mutliplication in CUDA

【发布时间】：2020-05-11 19:29:45

【问题描述】：

我目前正在试验 CUDA，我从矩阵乘法的答案中发现了这个内核：https://***.com/a/18856054/7867026

我不想做 A*B 来做 A_Transpose*A 但不保存 A_Transpose（只有矩阵 A 作为内核的输入）。我必须正确设置索引，但我对这个矩阵表示感到困惑。任何帮助将不胜感激。

【参考方案1】：

您需要的大部分内容是here 和here。

在第一个链接中确定 AxAT 涉及获取矩阵 A 的行的内积，类似地，ATxA 将涉及获取矩阵 A 的列的内积。还要注意对称性声明。在第二个链接中（在编程指南中从该点向下滚动一点），您将找到一个完整的平铺矩阵乘法。您只需按列对两个图块进行索引。

这是一个工作示例，使用来自 the SO answer you linked 的代码：

$ cat t1654.cu
#include <iostream>
#include <cstdio>
#include <cstdlib>

const int TILE_DIM = 32;
template <typename T>
__global__ void ATA(const T * __restrict__  A, T * __restrict__  C, int ARows, int ACols)

    T CValue = 0;

    int Row = blockIdx.y*TILE_DIM + threadIdx.y;
    int Col = blockIdx.x*TILE_DIM + threadIdx.x;

    __shared__ T As[TILE_DIM][TILE_DIM];
    __shared__ T Bs[TILE_DIM][TILE_DIM];

    for (int k = 0; k < (TILE_DIM + ARows - 1)/TILE_DIM; k++) 

         if (k*TILE_DIM + threadIdx.y < ARows && blockIdx.y*blockDim.y+threadIdx.x < ACols)
             As[threadIdx.y][threadIdx.x] = A[(k*TILE_DIM + threadIdx.y)*ACols + blockIdx.y*blockDim.y+threadIdx.x];
         else
             As[threadIdx.y][threadIdx.x] = 0.0;

         if (k*TILE_DIM + threadIdx.y < ARows && Col < ACols)
             Bs[threadIdx.y][threadIdx.x] = A[(k*TILE_DIM + threadIdx.y)*ACols + Col];
         else
             Bs[threadIdx.y][threadIdx.x] = 0.0;

         __syncthreads();

         for (int n = 0; n < TILE_DIM; ++n)
             CValue += As[n][threadIdx.y] * Bs[n][threadIdx.x];

         __syncthreads();
    

    if (Row < ACols && Col < ACols)
        C[((blockIdx.y * blockDim.y + threadIdx.y)*ACols) +
           (blockIdx.x * blockDim.x)+ threadIdx.x] = CValue;


template <typename T>
__global__ void transpose_naive(const T * __restrict__ in, T * __restrict__ out, const int dim)
  int col = threadIdx.x+blockDim.x*blockIdx.x;
  int row = threadIdx.y+blockDim.y*blockIdx.y;
  if ((col < dim) && (row < dim)) out[col*dim+row] = in[row*dim+col];


template <typename T>
__global__ void mm_naive(const T * __restrict__ A, const T * __restrict__ B, T * __restrict__ C, const int rowA, const int colA, const int colB)
  int col = threadIdx.x+blockDim.x*blockIdx.x;
  int row = threadIdx.y+blockDim.y*blockIdx.y;
  if ((row < rowA) && (col < colB))
    T Cval = 0;
    for (int i = 0; i < colA; i++) Cval += A[row*colA+i]*B[i*colB+col];
    C[row*colB+col] = Cval;



typedef float mt;
int main()

  mt *d_A, *d_B, *d_C, *h_A, *h_C, *h_C1;
  int m = 64;
  int n = 64;
  h_A  = new mt[m*n];
  h_C  = new mt[n*n];
  h_C1 = new mt[n*n];
  cudaMalloc(&d_A, m*n*sizeof(d_A[0]));
  cudaMalloc(&d_B, m*n*sizeof(d_A[0]));
  cudaMalloc(&d_C, n*n*sizeof(d_C[0]));
  // test 1
  for (int i = 0; i < m; i++)
    for (int j = 0; j < n; j++)
      h_A[i*n+j] = (i==j)?1.0f:0.0f;
  cudaMemcpy(d_A, h_A, m*n*sizeof(d_A[0]), cudaMemcpyHostToDevice);
  dim3 block(TILE_DIM, TILE_DIM);
  dim3 grid((n+block.x-1)/block.x, (n+block.y-1)/block.y);
  ATA<<<grid,block>>>(d_A, d_C, m, n);
  cudaMemcpy(h_C, d_C, n*n*sizeof(d_C[0]), cudaMemcpyDeviceToHost);
#ifdef DEBUG
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      std::cout << h_C[i*n+j] << " ";
    std::cout << std::endl;
  std::cout << std::endl;
#endif
  // test 2
  for (int i = 0; i < m; i++)
    for (int j = 0; j < n; j++)
      h_A[i*n+j] = rand()%10;
  cudaMemcpy(d_A, h_A, m*n*sizeof(d_A[0]), cudaMemcpyHostToDevice);
  ATA<<<grid,block>>>(d_A, d_C, m, n);
  cudaMemcpy(h_C, d_C, n*n*sizeof(d_C[0]), cudaMemcpyDeviceToHost);
#ifdef DEBUG
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      std::cout << h_C[i*n+j] << " ";
    std::cout << std::endl;
  std::cout << std::endl;
#endif
  transpose_naive<<<grid,block>>>(d_A, d_B, n);
  mm_naive<<<grid,block>>>(d_B, d_A, d_C, n, n, n);
  cudaMemcpy(h_C1, d_C, n*n*sizeof(d_C[0]), cudaMemcpyDeviceToHost);
#ifdef DEBUG
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      std::cout << h_C1[i*n+j] << " ";
    std::cout << std::endl;
  std::cout << std::endl;
#endif
  for (int i = 0; i < n*n; i++) if (h_C[i] != h_C1[i]) std::cout << "mismatch at: " << i << " was: " << h_C[i] << " should be: " << h_C1[i] << std::endl; return 0;

$ nvcc -o t1654 t1654.cu
$ cuda-memcheck ./t1654
========= CUDA-MEMCHECK
========= ERROR SUMMARY: 0 errors
$

请注意，在两种情况下加载 Bs 磁贴是相同的。主要变化在于加载As 磁贴，同时注意计算Cvalue 时的索引变化。这些更改对于在这两种情况下按列编制索引都是必需的。

可能仍然存在错误。我没有测试过非正方形的情况，也没有测试过矩阵大小不是块大小倍数的情况。此外，我没有利用输出中的对称性。但是，这应该有助于索引。