CUDA--cublas--矩阵的逆(0)

　　用CUDA求解矩阵的逆，有多种方法，也可以自己编写内核函数去实现，我查阅CSDN上用

cublas求解矩阵逆的方法，但是作者写的比较繁琐，其他观看学习的人会觉得比难懂。所以我

决定自己写一个。我采用的是LU分解法，cublas提供了相应的函数。代码如下：

#include <stdio.h>
#include <stdlib.h>
#include<malloc.h>
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <cublas_v2.h>

#define cudacall(call)                                                                                                          \
    do                                                                                                                          \
    {                                                                                                                           \
        cudaError_t err = (call);                                                                                               \
        if(cudaSuccess != err)                                                                                                  \
        {                                                                                                                       \
            fprintf(stderr,"CUDA Error:\nFile = %s\nLine = %d\nReason = %s\n", __FILE__, __LINE__, cudaGetErrorString(err));    \
            cudaDeviceReset();                                                                                                  \
            exit(EXIT_FAILURE);                                                                                                 \
        }                                                                                                                       \
    }                                                                                                                           \
    while (0)


#define cublascall(call)                                                                                        \
    do                                                                                                          \
    {                                                                                                           \
        cublasStatus_t status = (call);                                                                         \
        if(CUBLAS_STATUS_SUCCESS != status)                                                                     \
        {                                                                                                       \
            fprintf(stderr,"CUBLAS Error:\nFile = %s\nLine = %d\nCode = %d\n", __FILE__, __LINE__, status);     \
            cudaDeviceReset();                                                                                  \
            exit(EXIT_FAILURE);                                                                                 \
        }                                                                                                       \
                                                                                                                \
    }                                                                                                           \
    while(0)

void invert(float** src, float** dst, int n, const int batchSize)
{
    cublasHandle_t handle;
    cublascall(cublasCreate_v2(&handle));

    int *P, *INFO;

    cudacall(cudaMalloc(&P, n * batchSize * sizeof(int)));
    cudacall(cudaMalloc(&INFO, batchSize * sizeof(int)));

    int lda = n;

    float **A = (float **)malloc(batchSize * sizeof(float *));
    float **A_d, *A_dflat;
    cudacall(cudaMalloc(&A_d, batchSize * sizeof(float *)));
    cudacall(cudaMalloc(&A_dflat, n*n*batchSize * sizeof(float)));
    A[0] = A_dflat;

    for (int i = 1; i < batchSize; i++)
        A[i] = A[i - 1] + (n*n);
    cudacall(cudaMemcpy(A_d, A, batchSize * sizeof(float *), cudaMemcpyHostToDevice));
    for (int i = 0; i < batchSize; i++)
        cudacall(cudaMemcpy(A_dflat + (i*n*n), src[i], n*n * sizeof(float), cudaMemcpyHostToDevice));

    cublascall(cublasSgetrfBatched(handle, n, A_d, lda, P, INFO, batchSize));

    int *INFOh=new int[batchSize];
    //int INFOh[batchSize];
    cudacall(cudaMemcpy(INFOh, INFO, batchSize * sizeof(int), cudaMemcpyDeviceToHost));

    for (int i = 0; i < batchSize; i++)
        if (INFOh[i] != 0)
        {
            fprintf(stderr, "Factorization of matrix %d Failed: Matrix may be singular\n", i);
            cudaDeviceReset();
            exit(EXIT_FAILURE);
        }

    float **C = (float **)malloc(batchSize * sizeof(float *));
    float **C_d, *C_dflat;
    cudacall(cudaMalloc(&C_d, batchSize * sizeof(float *)));
    cudacall(cudaMalloc(&C_dflat, n*n*batchSize * sizeof(float)));
    C[0] = C_dflat;
    for (int i = 1; i < batchSize; i++)
        C[i] = C[i - 1] + (n*n);
    cudacall(cudaMemcpy(C_d, C, batchSize * sizeof(float *), cudaMemcpyHostToDevice));
    cublascall(cublasSgetriBatched(handle, n, (const float **)A_d, lda, P, C_d, lda, INFO, batchSize));

    cudacall(cudaMemcpy(INFOh, INFO, batchSize * sizeof(int), cudaMemcpyDeviceToHost));


    for (int i = 0; i < batchSize; i++)
        if (INFOh[i] != 0)
        {
            fprintf(stderr, "Inversion of matrix %d Failed: Matrix may be singular\n", i);
            cudaDeviceReset();
            exit(EXIT_FAILURE);
        }
    for (int i = 0; i < batchSize; i++)
        cudacall(cudaMemcpy(dst[i], C_dflat + (i*n*n), n*n * sizeof(float), cudaMemcpyDeviceToHost));
    cudaFree(A_d); cudaFree(A_dflat); free(A);
    cudaFree(C_d); cudaFree(C_dflat); free(C);
    cudaFree(P); cudaFree(INFO); cublasDestroy_v2(handle); delete[]INFOh;
    
}

void test_invert()
{
    const int n = 3;
    const int mybatch = 4;

    //Random matrix with full pivots
    float full_pivot[n*n] = { 0.5, 3, 4,
                                1, 3, 10,
                                4 , 9, 16 };
    //Almost same as above matrix with first pivot zero
    float zero_pivot[n*n] = { 0, 3, 4,
                              1, 3, 10,
                              4 , 9, 16 };

    float another_zero_pivot[n*n] = { 0, 3, 4,
                                      1, 5, 6,
                                      9, 8, 2 };

    float another_full_pivot[n * n] = { 22, 3, 4,
                                        1, 5, 6,
                                        9, 8, 2 };
    float *result_flat = (float *)malloc(mybatch*n*n * sizeof(float));
    float **results = (float **)malloc(mybatch * sizeof(float *));
    for (int i = 0; i < mybatch; i++)
        results[i] = result_flat + (i*n*n);
    float **inputs = (float **)malloc(mybatch * sizeof(float *));
    inputs[0] = zero_pivot;
    inputs[1] = full_pivot;
    inputs[2] = another_zero_pivot;
    inputs[3] = another_full_pivot;

    for (int qq = 0; qq < mybatch; qq++) {
        fprintf(stdout, "Input %d:\n\n", qq);
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
                fprintf(stdout, "%f\t", inputs[qq][i*n + j]);
            fprintf(stdout, "\n");
        }
    }
    fprintf(stdout, "\n\n");
    invert(inputs, results, n, mybatch);

    for (int qq = 0; qq < mybatch; qq++) {
        fprintf(stdout, "Inverse %d:\n\n", qq);
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
                fprintf(stdout, "%f\t", results[qq][i*n + j]);
            fprintf(stdout, "\n");
        }
    }
}


int main(void) {

    test_invert();
    return 0;
}

inverse matrix

运行结果：

 

标签：--,batchSize,float,++,int,cublas,CUDA,cudacall,sizeof
来源： https://www.cnblogs.com/xuelanga000/p/13358449.html