如何格式化 CUBLAS 例程 cublasdtbsv 的 A 矩阵？

问题描述

我是 Cuda 库的新手，想求解对称带状矩阵方程。我找到了使用 LU Factorization 解决这个问题的示例代码。我现在正在尝试使用 cudaBlas 例程 cublasdtbsv 来求解方程。我无法找到此函数的示例代码，并已将我自己的解决方案放在一起。我认为我遇到的问题是我不理解为该例程输入 A 矩阵的正确方法。这是我的一个非常简单的 3x3 矩阵的示例代码，其中一个右手边。它包括使用 LU Factorization 的正确解决方案以及我尝试使用 cublasdtbsv 例程：

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <cuda_runtime.h>
#include <cusolverDn.h>

void test();
void printMatrix2(int m,int n,const double* A,int lda,const char* name);
int LUFactorizationSolver2();
int TriBandedSymSolver2();

int main(int argc,char* argv[])
{
    test();

    return 0;

}

void test()
{
    LUFactorizationSolver2();
    TriBandedSymSolver2();

    return;
}

int TriBandedSymSolver2()
{
    printf("\n****      example of cublasDtbsv \n\n");

    const int n = 3;
    const int ldm = n;
    const int k = 1;// n - 1;
    const int lda = n;
    const int nrhs = 1;
    const int incx = 1;

    double M[ldm * n] = { 1.0,0.0,2.0,3.0,4.0
                        };

    double A[lda * n] = { 1.0,4.0,0.0
                        };

    double x[n * nrhs] = { 00.0,40.0,00.0
                         };

    cublasHandle_t cublasHandle = NULL;
    cudaStream_t stream = NULL;

    cublasstatus_t cublasstatus = CUBLAS_STATUS_SUCCESS;
    cudaError_t cudaStat1 = cudaSuccess;
    cudaError_t cudaStat2 = cudaSuccess;
    cudaError_t cudaStat3 = cudaSuccess;
    cudaError_t cudaStat4 = cudaSuccess;

    double* d_A = NULL; /* device copy of A */
    double* d_x = NULL; /* device copy of x */

    printf("example of tbsv \n");

    printf("A = (matlab base-1)\n");
    printMatrix2(n,n,A,lda,"A");
    printf("=====\n");

    printf("x (b) = (matlab base-1)\n");
    printMatrix2(n,nrhs,x,"x");
    printf("=====\n");

    /* step 1: create cusolver handle,bind a stream */
    cublasstatus = cublasCreate(&cublasHandle);
    assert(CUBLAS_STATUS_SUCCESS == cublasstatus);

    /* step 2: copy A to device */
    cudaStat1 = cudamalloc((void**)&d_A,sizeof(double) * lda * n);
    cudaStat2 = cudamalloc((void**)&d_x,sizeof(double) * n * nrhs);
    assert(cudaSuccess == cudaStat1);
    assert(cudaSuccess == cudaStat2);

    cudaStat1 = cudamemcpy(d_A,sizeof(double) * lda * n,cudamemcpyHostToDevice);
    cudaStat2 = cudamemcpy(d_x,sizeof(double) * n * nrhs,cudamemcpyHostToDevice);
    assert(cudaSuccess == cudaStat1);
    assert(cudaSuccess == cudaStat2);

    /*
     * step 5: solve A*x = b
     *
     */
    cublasstatus = cublasDtbsv(cublasHandle,CUBLAS_FILL_MODE_LOWER,CUBLAS_OP_N,CUBLAS_DIAG_NON_UNIT,k,d_A,d_x,incx
    );
    cudaStat1 = cudaDeviceSynchronize();
    assert(CUBLAS_STATUS_SUCCESS == cublasstatus);

    cudaStat1 = cudamemcpy(x,cudamemcpyDevicetoHost);
    assert(cudaSuccess == cudaStat1);

    printf("X = (matlab base-1)\n");
    printMatrix2(n,"x");
    printf("=====\n");

    /* free resources */
    if (d_A) cudaFree(d_A);
    if (d_x) cudaFree(d_x);

    if (cublasHandle) cublasDestroy(cublasHandle);
    if (stream) cudaStreamDestroy(stream);

    cudaDeviceReset();

    return 0;

}

int LUFactorizationSolver2()
{
    printf("\n****      example of cusolverDnDgetrs \n\n");

    cusolverDnHandle_t cusolverH = NULL;
    cudaStream_t stream = NULL;

    cusolverStatus_t status = CUSOLVER_STATUS_SUCCESS;
    cudaError_t cudaStat1 = cudaSuccess;
    cudaError_t cudaStat2 = cudaSuccess;
    cudaError_t cudaStat3 = cudaSuccess;
    cudaError_t cudaStat4 = cudaSuccess;

    const int m = 3;
    const int lda = m;
    const int nrhs = 1; // number of right-hand sides
    const int ldb = m;


    double A[lda * m] =  { 1.0,4.0
                         };

    double B[m * nrhs] = { 00.0,00.0
                         };
    double X[m * nrhs]; /* X = A\B */
    double LU[lda * m]; /* L and U */
    int Ipiv[m];      /* host copy of pivoting sequence */
    int info = 0;     /* host copy of error info */

    double* d_A = NULL; /* device copy of A */
    double* d_B = NULL; /* device copy of B */
    int* d_Ipiv = NULL; /* pivoting sequence */
    int* d_info = NULL; /* error info */
    int  lwork = 0;     /* size of workspace */
    double* d_work = NULL; /* device workspace for getrf */

    const int pivot_on = 0; // 1;

    if (pivot_on) {
        printf("pivot is on : compute P*A = L*U \n");
    }
    else {
        printf("pivot is off: compute A = L*U (not numerically stable)\n");
    }

    printf("A = (matlab base-1)\n");
    printMatrix2(m,m,"A");
    printf("=====\n");

    printf("B = (matlab base-1)\n");
    printMatrix2(m,B,ldb,"B");
    printf("=====\n");

    /* step 1: create cusolver handle,bind a stream */
    status = cusolverDnCreate(&cusolverH);
    assert(CUSOLVER_STATUS_SUCCESS == status);

    cudaStat1 = cudaStreamCreateWithFlags(&stream,cudaStreamNonBlocking);
    assert(cudaSuccess == cudaStat1);

    status = cusolverDnSetStream(cusolverH,stream);
    assert(CUSOLVER_STATUS_SUCCESS == status);

    /* step 2: copy A to device */
    cudaStat1 = cudamalloc((void**)&d_A,sizeof(double) * lda * m);
    cudaStat2 = cudamalloc((void**)&d_B,sizeof(double) * m * nrhs);
    cudaStat2 = cudamalloc((void**)&d_Ipiv,sizeof(int) * m);
    cudaStat4 = cudamalloc((void**)&d_info,sizeof(int));
    assert(cudaSuccess == cudaStat1);
    assert(cudaSuccess == cudaStat2);
    assert(cudaSuccess == cudaStat3);
    assert(cudaSuccess == cudaStat4);

    cudaStat1 = cudamemcpy(d_A,sizeof(double) * lda * m,cudamemcpyHostToDevice);
    cudaStat2 = cudamemcpy(d_B,sizeof(double) * m * nrhs,cudamemcpyHostToDevice);
    assert(cudaSuccess == cudaStat1);
    assert(cudaSuccess == cudaStat2);

    /* step 3: query working space of getrf */
    status = cusolverDnDgetrf_bufferSize(
        cusolverH,&lwork);
    assert(CUSOLVER_STATUS_SUCCESS == status);

    cudaStat1 = cudamalloc((void**)&d_work,sizeof(double) * lwork);
    assert(cudaSuccess == cudaStat1);

    /* step 4: LU factorization */
    if (pivot_on) {
        status = cusolverDnDgetrf(
            cusolverH,d_work,d_Ipiv,d_info);
    }
    else {
        status = cusolverDnDgetrf(
            cusolverH,NULL,d_info);
    }
    cudaStat1 = cudaDeviceSynchronize();
    assert(CUSOLVER_STATUS_SUCCESS == status);
    assert(cudaSuccess == cudaStat1);

    if (pivot_on) {
        cudaStat1 = cudamemcpy(Ipiv,sizeof(int) * m,cudamemcpyDevicetoHost);
    }
    cudaStat2 = cudamemcpy(LU,cudamemcpyDevicetoHost);
    cudaStat3 = cudamemcpy(&info,d_info,sizeof(int),cudamemcpyDevicetoHost);
    assert(cudaSuccess == cudaStat1);
    assert(cudaSuccess == cudaStat2);
    assert(cudaSuccess == cudaStat3);

    if (0 > info) {
        printf("%d-th parameter is wrong \n",-info);
        exit(1);
    }
    if (pivot_on) {
        printf("pivoting sequence,matlab base-1\n");
        for (int j = 0; j < m; j++) {
            printf("Ipiv(%d) = %d\n",j + 1,Ipiv[j]);
        }
    }
    printf("L and U = (matlab base-1)\n");
    printMatrix2(m,LU,"LU");
    printf("=====\n");

    /*
     * step 5: solve A*X = B
     *
     */
    if (pivot_on) {
        status = cusolverDnDgetrs(
            cusolverH,/* nrhs */
            d_A,d_B,d_info);
    }
    else {
        status = cusolverDnDgetrs(
            cusolverH,d_info);
    }
    cudaStat1 = cudaDeviceSynchronize();
    assert(CUSOLVER_STATUS_SUCCESS == status);
    assert(cudaSuccess == cudaStat1);

    cudaStat1 = cudamemcpy(X,cudamemcpyDevicetoHost);
    assert(cudaSuccess == cudaStat1);

    printf("X = (matlab base-1)\n");
    printMatrix2(m,X,"X");
    printf("=====\n");

    /* free resources */
    if (d_A) cudaFree(d_A);
    if (d_B) cudaFree(d_B);
    if (d_Ipiv) cudaFree(d_Ipiv);
    if (d_info) cudaFree(d_info);
    if (d_work) cudaFree(d_work);

    if (cusolverH) cusolverDnDestroy(cusolverH);
    if (stream) cudaStreamDestroy(stream);

    cudaDeviceReset();

    return 0;

}

void printMatrix2(int m,const char* name)
{
    printf("%18s","");
    for (int col = 0; col < n; coL++) { printf("%7s(*,%2d)       ",name,col + 1); }
    printf("\n");

    for (int row = 0; row < m; row++) {
        printf("%4s(%2d,*) = ",row + 1);
        for (int col = 0; col < n; coL++) {
            double Areg = A[row + col * lda];
            printf("%20.9f",Areg);
        }
        printf("\n");
    }
    return;
}

正确答案应该是：

   0.00
-160.00
 120.00

但我明白了：

 0.000000000
         inf
        -inf

我正在使用 Visual Studio 2019 在 Windows 10 上进行开发。

我遗漏了什么，或者有人可以指出我 cublasDtbsv 例程的工作示例吗？

解决方法

CUBLAS tbsv 是带状三角形求解器。它期望您的 M 矩阵是带状和三角形的。如果您想看看它是什么样子，this 是一个很好的参考。

您的 M 矩阵不是三角形。三角矩阵的上三角部分（不包括主对角线）或下三角部分（不包括主对角线）全为零。您的 M 矩阵不符合该定义。

带状三角矩阵 M 可能如下所示：

    | 2.0 0.0 0.0 |
M = | 1.0 1.0 0.0 |
    | 0.0 1.0 1.0 |

让我们以它为例，RHS 为 | 2.0 2.0 2.0 |，并使用为 A 矩阵格式 here 给出的建议。在这种情况下，我们的 A 矩阵将如下所示：

A = | 2.0 1.0 1.0 |    (the main diagonal of M)
    | 1.0 1.0 0.0 |    (the first sub-diagonal of M)

在这种情况下，我们的 A 矩阵有 2 行，因此 A 的前导维度由 lda = 2 给出

如果我们将所有这些都放入您的测试框架中，它似乎会给出正确的结果：

$ cat t158.cu
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <cuda_runtime.h>
#include <cublas_v2.h>

void printMatrix2(int m,int n,const double* A,int lda,const char* name)
{
    printf("%18s","");
    for (int col = 0; col < n; col++) { printf("%7s(*,%2d)       ",name,col + 1); }
    printf("\n");

    for (int row = 0; row < m; row++) {
        printf("%4s(%2d,*) = ",row + 1);
        for (int col = 0; col < n; col++) {
            double Areg = A[row + col * lda];
            printf("%20.9f",Areg);
        }
        printf("\n");
    }
    return;
}



int main(int argc,char* argv[])
{
    printf("\n****      example of cublasDtbsv \n\n");

    const int n = 3;
//    const int ldm = n;
    const int k = 1;
    const int lda = k+1;
    const int nrhs = 1;
    const int incx = 1;
/*
    double M[ldm * n] = { 2.0,0.0,1.0,1.0
                        };
*/
    double A[lda * n] = { 2.0,0.0
                        };

    double x[n * nrhs] = { 2.0,2.0,2.0
                         };

    cublasHandle_t cublasHandle = NULL;

    cublasStatus_t cublasStatus = CUBLAS_STATUS_SUCCESS;
    cudaError_t cudaStat1 = cudaSuccess;
    cudaError_t cudaStat2 = cudaSuccess;

    double* d_A = NULL; /* device copy of A */
    double* d_x = NULL; /* device copy of x */

    printf("example of tbsv \n");

    printf("A = (matlab base-1)\n");
    printMatrix2(n,n,A,lda,"A");
    printf("=====\n");

    printf("x (b) = (matlab base-1)\n");
    printMatrix2(n,nrhs,x,"x");
    printf("=====\n");

    /* step 1: create cublas handle */
    cublasStatus = cublasCreate(&cublasHandle);
    assert(CUBLAS_STATUS_SUCCESS == cublasStatus);

    /* step 2: copy A to device */
    cudaStat1 = cudaMalloc((void**)&d_A,sizeof(double) * lda * n);
    cudaStat2 = cudaMalloc((void**)&d_x,sizeof(double) * n * nrhs);
    assert(cudaSuccess == cudaStat1);
    assert(cudaSuccess == cudaStat2);

    cudaStat1 = cudaMemcpy(d_A,sizeof(double) * lda * n,cudaMemcpyHostToDevice);
    cudaStat2 = cudaMemcpy(d_x,sizeof(double) * n * nrhs,cudaMemcpyHostToDevice);
    assert(cudaSuccess == cudaStat1);
    assert(cudaSuccess == cudaStat2);

    /*
     * step 5: solve A*x = b
     *
     */
    cublasStatus = cublasDtbsv(cublasHandle,CUBLAS_FILL_MODE_LOWER,CUBLAS_OP_N,CUBLAS_DIAG_NON_UNIT,k,d_A,d_x,incx
    );
    cudaStat1 = cudaDeviceSynchronize();
    assert(CUBLAS_STATUS_SUCCESS == cublasStatus);

    cudaStat1 = cudaMemcpy(x,cudaMemcpyDeviceToHost);
    assert(cudaSuccess == cudaStat1);

    printf("X = (matlab base-1)\n");
    printMatrix2(n,"x");
    printf("=====\n");
    /* free resources */
    if (d_A) cudaFree(d_A);
    if (d_x) cudaFree(d_x);

    if (cublasHandle) cublasDestroy(cublasHandle);

    return 0;
}
$ nvcc -o t158 t158.cu -lcublas
$ ./t158

****      example of cublasDtbsv

example of tbsv
A = (matlab base-1)
                        A(*,1)             A(*,2)             A(*,3)
   A( 1,*) =          2.000000000         1.000000000         1.000000000
   A( 2,*) =          1.000000000         1.000000000         0.000000000
   A( 3,*) =          1.000000000         1.000000000         0.000000000
=====
x (b) = (matlab base-1)
                        x(*,1)
   x( 1,*) =          2.000000000
   x( 2,*) =          2.000000000
   x( 3,*) =          2.000000000
=====
X = (matlab base-1)
                        x(*,*) =          1.000000000
   x( 2,*) =          1.000000000
   x( 3,*) =          1.000000000
=====
$