使用 cublasSgetriBatched 对 GPU 上的两个矩阵求逆

如何解决使用 cublasSgetriBatched 对 GPU 上的两个矩阵求逆

我是 cublas 的新手。我想在 GPU 上并行计算两个矩阵的逆。矩阵是 [4 8;3 9] 和 [5 2;1 7]。是否可以使用 cublasSgetriBatched 这样做？这是我的代码，我得到了不正确的结果。这里我取了 2x2 矩阵，但我想找到一种方法来解决多个 mxm matices 的问题。

#include <stdio.h>

#include "cuda_runtime.h" 
#include "device_launch_parameters.h"

#include "cublas_v2.h"



int main() {

const unsigned int N = 2; 

const unsigned int Nmatrices = 2;

cublasHandle_t handle;
cublasCreate(&handle);

// --- Matrices to be inverted 
float *h_A = new float[N*N*Nmatrices];
float *r_A = new float[N*N*Nmatrices];//result 

h_A[0] = 4.f;  
h_A[1] = 3.f;
h_A[2] = 8.f;
h_A[3] = 9.f;
h_A[4] = 5.f; 
h_A[5] = 1.f; 
h_A[6] = 2.f; 
h_A[7] = 7.f;


// --- Allocate device matrices 
float *d_A; cudaMalloc((void**)&d_A,N*N*Nmatrices*sizeof(float));
float *c_A; cudaMalloc((void**)&c_A,N*N*Nmatrices*sizeof(float));

// --- Move the matrix to be inverted from host to device
cudaMemcpy(d_A,h_A,N*N*Nmatrices*sizeof(float),cudaMemcpyHostToDevice);

// --- Creating the array of pointers needed as input to the batched getrf
float **h_inout_pointers = (float **)malloc(Nmatrices*sizeof(float *));
//for (int i=0; i<Nmatrices; i++) h_inout_pointers[i]=(float *)((char*)d_A+i*((size_t)N*N)*sizeof(float));
*h_inout_pointers=d_A;

float **d_inout_pointers;
cudaMalloc((void**)&d_inout_pointers,Nmatrices*sizeof(float *));
cudaMemcpy(d_inout_pointers,h_inout_pointers,Nmatrices*sizeof(float *),cudaMemcpyHostToDevice);
//free(h_inout_pointers);

float **r_inout_pointers = (float **)malloc(Nmatrices*sizeof(float *));
//for (int i=0; i<Nmatrices; i++) h_inout_pointers[i]=(float *)((char*)d_A+i*((size_t)N*N)*sizeof(float));
*r_inout_pointers=c_A;

float **rd_inout_pointers;
cudaMalloc((void**)&rd_inout_pointers,Nmatrices*sizeof(float *));
cudaMemcpy(rd_inout_pointers,r_inout_pointers,cudaMemcpyHostToDevice);

int *d_PivotArray; cudaMalloc((void**)&d_PivotArray,N*Nmatrices*sizeof(int));
int *d_InfoArray;  cudaMalloc((void**)&d_InfoArray,Nmatrices*sizeof(int));

int *h_PivotArray = (int *)malloc(N*Nmatrices*sizeof(int));
int *h_InfoArray  = (int *)malloc(  Nmatrices*sizeof(int));

cublasSgetrfBatched(handle,N,d_inout_pointers,d_PivotArray,d_InfoArray,Nmatrices);
//cublasSafeCall(cublasSgetrfBatched(handle,NULL,Nmatrices));

//gpuErrchk(cudaMemcpy(h_InfoArray,Nmatrices*sizeof(int),cudaMemcpyDeviceToHost));
cublasSgetriBatched(handle,rd_inout_pointers,Nmatrices);


cudaMemcpy(h_A,d_A,N*N*sizeof(float),cudaMemcpyDeviceToHost);
cudaMemcpy(r_A,c_A,cudaMemcpyDeviceToHost);
//gpuErrchk(cudaMemcpy(h_PivotArray,N*Nmatrices*sizeof(int),cudaMemcpyDeviceToHost));

for (int i=0; i<N*N*Nmatrices; i++) printf("A[%i]=%f\n",i,r_A[i]);


return 0;
}

解决方法

您的代码至少有 3 个使用问题：

1. 您没有正确设置 d_A 的指针数组。
2. 您没有正确设置 c_A 的指针数组。
3. 您对 cudaMemcpy 的最后 r_A 语句仅传输一个矩阵的数据 (N*N)。

以下代码修复了项目，并且运行时不会出现运行时错误，产生非零结果。如果您认为这些结果是错误的，您应该说明原因以及您认为正确的结果应该是什么。

#include <stdio.h>

#include <cublas_v2.h>



int main() {

const unsigned int N = 2;

const unsigned int Nmatrices = 2;

cublasHandle_t handle;
cublasCreate(&handle);

// --- Matrices to be inverted
float *h_A = new float[N*N*Nmatrices];
float *r_A = new float[N*N*Nmatrices];//result

h_A[0] = 4.f;
h_A[1] = 3.f;
h_A[2] = 8.f;
h_A[3] = 9.f;
h_A[4] = 5.f;
h_A[5] = 1.f;
h_A[6] = 2.f;
h_A[7] = 7.f;


// --- Allocate device matrices
float *d_A; cudaMalloc((void**)&d_A,N*N*Nmatrices*sizeof(float));
float *c_A; cudaMalloc((void**)&c_A,N*N*Nmatrices*sizeof(float));

// --- Move the matrix to be inverted from host to device
cudaMemcpy(d_A,h_A,N*N*Nmatrices*sizeof(float),cudaMemcpyHostToDevice);

// --- Creating the array of pointers needed as input to the batched getrf
float **h_inout_pointers = (float **)malloc(Nmatrices*sizeof(float *));
//for (int i=0; i<Nmatrices; i++) h_inout_pointers[i]=(float *)((char*)d_A+i*((size_t)N*N)*sizeof(float));
h_inout_pointers[0]=d_A;
h_inout_pointers[1]=d_A+N*N;

float **d_inout_pointers;
cudaMalloc((void**)&d_inout_pointers,Nmatrices*sizeof(float *));
cudaMemcpy(d_inout_pointers,h_inout_pointers,Nmatrices*sizeof(float *),cudaMemcpyHostToDevice);
//free(h_inout_pointers);

float **r_inout_pointers = (float **)malloc(Nmatrices*sizeof(float *));
//for (int i=0; i<Nmatrices; i++) h_inout_pointers[i]=(float *)((char*)d_A+i*((size_t)N*N)*sizeof(float));
r_inout_pointers[0]=c_A;
r_inout_pointers[1]=c_A+N*N;

float **rd_inout_pointers;
cudaMalloc((void**)&rd_inout_pointers,Nmatrices*sizeof(float *));
cudaMemcpy(rd_inout_pointers,r_inout_pointers,cudaMemcpyHostToDevice);

int *d_PivotArray; cudaMalloc((void**)&d_PivotArray,N*Nmatrices*sizeof(int));
int *d_InfoArray;  cudaMalloc((void**)&d_InfoArray,Nmatrices*sizeof(int));

int *h_PivotArray = (int *)malloc(N*Nmatrices*sizeof(int));
int *h_InfoArray  = (int *)malloc(  Nmatrices*sizeof(int));

cublasSgetrfBatched(handle,N,d_inout_pointers,d_PivotArray,d_InfoArray,Nmatrices);
//cublasSafeCall(cublasSgetrfBatched(handle,NULL,Nmatrices));

//gpuErrchk(cudaMemcpy(h_InfoArray,Nmatrices*sizeof(int),cudaMemcpyDeviceToHost));
cublasSgetriBatched(handle,rd_inout_pointers,Nmatrices);


cudaMemcpy(h_A,d_A,N*N*sizeof(float),cudaMemcpyDeviceToHost);
cudaMemcpy(r_A,c_A,Nmatrices*N*N*sizeof(float),cudaMemcpyDeviceToHost);
//gpuErrchk(cudaMemcpy(h_PivotArray,N*Nmatrices*sizeof(int),cudaMemcpyDeviceToHost));

for (int i=0; i<N*N*Nmatrices; i++) printf("A[%i]=%f\n",i,r_A[i]);


return 0;
}
$ nvcc -o t133 t133.cu -lcublas
$ cuda-memcheck ./t133
========= CUDA-MEMCHECK
A[0]=0.750000
A[1]=-0.250000
A[2]=-0.666667
A[3]=0.333333
A[4]=0.212121
A[5]=-0.030303
A[6]=-0.060606
A[7]=0.151515
========= ERROR SUMMARY: 0 errors
$

（FWIW 我相信结果是正确的，用计算器测试 here）