c++ - CUDA 中的稀疏矩阵 vector 乘法

Question

我正在尝试在 GPU 上实现矩阵 vector 乘法(使用 CUDA)。

在我的 C++ 代码 (CPU) 中，我将矩阵加载为密集矩阵，然后使用 CUDA 执行矩阵 vector 乘法。我还使用共享内存来提高性能。

1. 如何在知道我的矩阵是稀疏矩阵的情况下以高效方式加载矩阵？

下面是我加载矩阵的 C++ 函数:

int readMatrix( char* filename, float* &matrix, unsigned int *dim = NULL, int majority = ROW_MAJOR )
{
    unsigned int w, h, x, y, num_entries;

    float val;

    std::ifstream file( filename );

    if ( file )
    {
        file >> h >> w >> num_entries;
        cout << w << " " << h << " " << num_entries << "\n";

        assert( w == h || w == 1 || h == 1 );

        if( dim != NULL ) *dim = std::max( w, h );

        matrix = new float[ w * h ];

        unsigned int i;
        for( i = 0; i < num_entries; i++ ){

            if( file.eof() ) break;

            file >> y >> x >> val;

            if( majority == ROW_MAJOR ){

                matrix[ w * y + x ] = val;

            } else if( majority == COLUMN_MAJOR ){

                matrix[ h * x + y ] = val;
            }
        }
        file.close();

        if( i == num_entries )
            std::cout << "\nFile read successfully\n"; 
        else
            std::cout << "\nFile read successfully but seems defective:\n num entries read = " << i << ", entries epected = " << num_entries << "\n"; 

        // print first few elements
        if( w == h ){
            for( unsigned int i = 0; i < w; i++ ){

                printf("\n");
                for( unsigned int j = 0; j < h; j++ ){

                    printf("%.2f ", matrix[ j + w * i ] );
                }
            }   
        }
        else{   

            printf("\n");
            for( unsigned int j = 0; j < h; j++ ){

                printf("%.2f ", matrix[ j ] );
            }
        }

    } else {

        std::cout << "Unable to open file\n";
        return false;
    }

    return true;
}

下面是我处理矩阵 vector 乘法的 CUDA 内核函数:

__global__ void
_cl_matrix_vector_( float *A, float *b, float *x, int dim )
{
    extern __shared__ float vec[];
    unsigned int idx = blockIdx.x * blockDim.x + threadIdx.x;
    float temp = 0.0;
    int vOffs = 0;

    //load vector into shared memory
    for (int i = 0; i < (dim/blockDim.x) + 1 ; ++i, vOffs+= blockDim.x) {
        vec[vOffs + threadIdx.x] = b[vOffs + threadIdx.x];
    }

    //make sure all threads are synchronized
     __syncthreads();

    if (idx < dim) {
        temp = 0.0;
        //dot product (multiplication)
        for (int i = 0; i < dim; i++){
            temp += A[idx * dim + i] * vec[i];
        }
         x[idx] = temp;
    } 

}

1. 考虑到我的矩阵是稀疏矩阵，我必须对我的 CUDA 代码进行哪些必要的更改？
2. 我从一个论坛上发现我们也可以使用填充来优化性能，但这需要我改变读取矩阵/对矩阵排序的方式。任何想法如何以我读取矩阵和执行计算的方式实现此填充？

Answer 1

这是一篇非常古老的帖子，我想强调一下 cuSPARSE (从现在开始)使稀疏矩阵之间或稀疏矩阵与密集 vector 之间的乘法例程可用。

对于 csr格式，稀疏矩阵和密集 vector 相乘的相关例程是cusparse<t>csrmv .下面是一个完整的示例，展示了它的用法。

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <assert.h>

#include "Utilities.cuh"

#include <cuda_runtime.h>
#include <cusparse_v2.h>

/********/
/* MAIN */
/********/
int main()
{
    // --- Initialize cuSPARSE
    cusparseHandle_t handle;    cusparseSafeCall(cusparseCreate(&handle));

    /**************************/
    /* SETTING UP THE PROBLEM */
    /**************************/
    const int N     = 4;                // --- Number of rows and columns

    // --- Host side dense matrices
    double *h_A_dense = (double*)malloc(N * N * sizeof(double));
    double *h_x_dense = (double*)malloc(N *     sizeof(double));
    double *h_y_dense = (double*)malloc(N *     sizeof(double));

    // --- Column-major ordering
    h_A_dense[0] = 0.4612;  h_A_dense[4] = -0.0006;     h_A_dense[8]  = 0.3566;     h_A_dense[12] = 0.0; 
    h_A_dense[1] = -0.0006; h_A_dense[5] = 0.4640;      h_A_dense[9]  = 0.0723;     h_A_dense[13] = 0.0; 
    h_A_dense[2] = 0.3566;  h_A_dense[6] = 0.0723;      h_A_dense[10] = 0.7543;     h_A_dense[14] = 0.0; 
    h_A_dense[3] = 0.;      h_A_dense[7] = 0.0;         h_A_dense[11] = 0.0;        h_A_dense[15] = 0.1; 

    // --- Initializing the data and result vectors
    for (int k = 0; k < N; k++) {
        h_x_dense[k] = 1.;
        h_y_dense[k] = 0.;
    }

    // --- Create device arrays and copy host arrays to them
    double *d_A_dense;  gpuErrchk(cudaMalloc(&d_A_dense, N * N * sizeof(double)));
    double *d_x_dense;  gpuErrchk(cudaMalloc(&d_x_dense, N     * sizeof(double)));
    double *d_y_dense;  gpuErrchk(cudaMalloc(&d_y_dense, N     * sizeof(double)));
    gpuErrchk(cudaMemcpy(d_A_dense, h_A_dense, N * N * sizeof(double), cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_x_dense, h_x_dense, N     * sizeof(double), cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_y_dense, h_y_dense, N     * sizeof(double), cudaMemcpyHostToDevice));

    // --- Descriptor for sparse matrix A
    cusparseMatDescr_t descrA;      cusparseSafeCall(cusparseCreateMatDescr(&descrA));
    cusparseSafeCall(cusparseSetMatType     (descrA, CUSPARSE_MATRIX_TYPE_GENERAL));
    cusparseSafeCall(cusparseSetMatIndexBase(descrA, CUSPARSE_INDEX_BASE_ONE));  

    int nnzA = 0;                           // --- Number of nonzero elements in dense matrix A

    const int lda = N;                      // --- Leading dimension of dense matrix

    // --- Device side number of nonzero elements per row of matrix A
    int *d_nnzPerVectorA;   gpuErrchk(cudaMalloc(&d_nnzPerVectorA, N * sizeof(*d_nnzPerVectorA)));
    cusparseSafeCall(cusparseDnnz(handle, CUSPARSE_DIRECTION_ROW, N, N, descrA, d_A_dense, lda, d_nnzPerVectorA, &nnzA));

    // --- Host side number of nonzero elements per row of matrix A
    int *h_nnzPerVectorA = (int *)malloc(N * sizeof(*h_nnzPerVectorA));
    gpuErrchk(cudaMemcpy(h_nnzPerVectorA, d_nnzPerVectorA, N * sizeof(*h_nnzPerVectorA), cudaMemcpyDeviceToHost));

    printf("Number of nonzero elements in dense matrix A = %i\n\n", nnzA);
    for (int i = 0; i < N; ++i) printf("Number of nonzero elements in row %i for matrix = %i \n", i, h_nnzPerVectorA[i]);
    printf("\n");

    // --- Device side sparse matrix
    double *d_A;            gpuErrchk(cudaMalloc(&d_A, nnzA * sizeof(*d_A)));

    int *d_A_RowIndices;    gpuErrchk(cudaMalloc(&d_A_RowIndices, (N + 1) * sizeof(*d_A_RowIndices)));
    int *d_A_ColIndices;    gpuErrchk(cudaMalloc(&d_A_ColIndices, nnzA * sizeof(*d_A_ColIndices)));

    cusparseSafeCall(cusparseDdense2csr(handle, N, N, descrA, d_A_dense, lda, d_nnzPerVectorA, d_A, d_A_RowIndices, d_A_ColIndices));

    // --- Host side sparse matrices
    double *h_A = (double *)malloc(nnzA * sizeof(*h_A));        
    int *h_A_RowIndices = (int *)malloc((N + 1) * sizeof(*h_A_RowIndices));
    int *h_A_ColIndices = (int *)malloc(nnzA * sizeof(*h_A_ColIndices));
    gpuErrchk(cudaMemcpy(h_A, d_A, nnzA * sizeof(*h_A), cudaMemcpyDeviceToHost));
    gpuErrchk(cudaMemcpy(h_A_RowIndices, d_A_RowIndices, (N + 1) * sizeof(*h_A_RowIndices), cudaMemcpyDeviceToHost));
    gpuErrchk(cudaMemcpy(h_A_ColIndices, d_A_ColIndices, nnzA * sizeof(*h_A_ColIndices), cudaMemcpyDeviceToHost));

    printf("\nOriginal matrix A in CSR format\n\n");
    for (int i = 0; i < nnzA; ++i) printf("A[%i] = %f ", i, h_A[i]); printf("\n");

    printf("\n");
    for (int i = 0; i < (N + 1); ++i) printf("h_A_RowIndices[%i] = %i \n", i, h_A_RowIndices[i]); printf("\n");

    printf("\n");
    for (int i = 0; i < nnzA; ++i) printf("h_A_ColIndices[%i] = %i \n", i, h_A_ColIndices[i]);  

    printf("\n");
    for (int i = 0; i < N; ++i) printf("h_x[%i] = %f \n", i, h_x_dense[i]); printf("\n");

    const double alpha = 1.;
    const double beta  = 0.;
    cusparseSafeCall(cusparseDcsrmv(handle, CUSPARSE_OPERATION_NON_TRANSPOSE, N, N, nnzA, &alpha, descrA, d_A, d_A_RowIndices, d_A_ColIndices, d_x_dense, 
                                    &beta, d_y_dense));

    gpuErrchk(cudaMemcpy(h_y_dense,           d_y_dense,            N * sizeof(double), cudaMemcpyDeviceToHost));

    printf("\nResult vector\n\n");
    for (int i = 0; i < N; ++i) printf("h_y[%i] = %f ", i, h_y_dense[i]); printf("\n");

}