问题
Since CUDA 5.5, the CUBLAS library contains routines for batched matrix factorization and inversion (cublas<t>getrfBatched and cublas<t>getriBatched respectively).
Getting guide from the documentation, I wrote a test code for inversion of an N x N matrix using these routines. The code gives correct output only if the matrix has all non zero pivots. Setting any pivot to zero results in incorrect results. I have verified the results using MATLAB.
I realize that I am providing row major matrices as input while CUBLAS expects column major matrices, but it shouldn't matter as it would only transpose the result. To be sure, I also tested on column major input, but getting same behavior.
I am confused as, cublas<t>getriBatched expects pivot exchange information array P as input, which is the output from cublas<t>getrfBatched. So, if any zero pivots are eliminated by row exchange, then the inversion routine should handle it automatically.
How to perform inversion of matrices which contain a zero pivot using CUBLAS?
Following is a self contained compile-able example with different test cases:
#include <cstdio>
#include <cstdlib>
#include <cuda_runtime.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_device(float* src_d, float* dst_d, int n)
{
cublasHandle_t handle;
cublascall(cublasCreate_v2(&handle));
int batchSize = 1;
int *P, *INFO;
cudacall(cudaMalloc<int>(&P,n * batchSize * sizeof(int)));
cudacall(cudaMalloc<int>(&INFO,batchSize * sizeof(int)));
int lda = n;
float *A[] = { src_d };
float** A_d;
cudacall(cudaMalloc<float*>(&A_d,sizeof(A)));
cudacall(cudaMemcpy(A_d,A,sizeof(A),cudaMemcpyHostToDevice));
cublascall(cublasSgetrfBatched(handle,n,A_d,lda,P,INFO,batchSize));
int INFOh = 0;
cudacall(cudaMemcpy(&INFOh,INFO,sizeof(int),cudaMemcpyDeviceToHost));
if(INFOh == n)
{
fprintf(stderr, "Factorization Failed: Matrix is singular\n");
cudaDeviceReset();
exit(EXIT_FAILURE);
}
float* C[] = { dst_d };
float** C_d;
cudacall(cudaMalloc<float*>(&C_d,sizeof(C)));
cudacall(cudaMemcpy(C_d,C,sizeof(C),cudaMemcpyHostToDevice));
cublascall(cublasSgetriBatched(handle,n,A_d,lda,P,C_d,lda,INFO,batchSize));
cudacall(cudaMemcpy(&INFOh,INFO,sizeof(int),cudaMemcpyDeviceToHost));
if(INFOh != 0)
{
fprintf(stderr, "Inversion Failed: Matrix is singular\n");
cudaDeviceReset();
exit(EXIT_FAILURE);
}
cudaFree(P), cudaFree(INFO), cublasDestroy_v2(handle);
}
void invert(float* src, float* dst, int n)
{
float* src_d, *dst_d;
cudacall(cudaMalloc<float>(&src_d,n * n * sizeof(float)));
cudacall(cudaMemcpy(src_d,src,n * n * sizeof(float),cudaMemcpyHostToDevice));
cudacall(cudaMalloc<float>(&dst_d,n * n * sizeof(float)));
invert_device(src_d,dst_d,n);
cudacall(cudaMemcpy(dst,dst_d,n * n * sizeof(float),cudaMemcpyDeviceToHost));
cudaFree(src_d), cudaFree(dst_d);
}
void test_invert()
{
const int n = 3;
//Random matrix with full pivots
float full_pivots[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 zero_pivot_col_major[n*n] = { 0, 1, 4,
3, 3, 9,
4 , 10, 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 singular[n*n] = {1,2,3,
4,5,6,
7,8,9};
//Select matrix by setting "a"
float* a = zero_pivot;
fprintf(stdout, "Input:\n\n");
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
fprintf(stdout,"%f\t",a[i*n+j]);
fprintf(stdout,"\n");
}
fprintf(stdout,"\n\n");
invert(a,a,n);
fprintf(stdout, "Inverse:\n\n");
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
fprintf(stdout,"%f\t",a[i*n+j]);
fprintf(stdout,"\n");
}
}
int main()
{
test_invert();
int n; scanf("%d",&n);
return 0;
}
回答1:
There seems to be a bug in the current CUBLAS library implementation of cublas<t>getrfBatched for matrices of dimension (n) such that 3<=n<=16, when there is a "zero pivot" as you say.
A possible workaround is to "identity-extend" your A matrix to be inverted, when n<17, to a size of 17x17 (using matlab nomenclature):
LU = getrf( [A 0 ; 0 I]);
continuing, you can then use cublas<t>getriBatched in an "ordinary" fashion:
invA = getri( LU(1:3,1:3) )
(You can also leave everything at n=17, call getri that way, and then extract the result as the first 3x3 rows and columns of invA.)
Here is a fully worked example, borrowing from the code you supplied, showing the inversion of your supplied 3x3 zero_pivot matrix, using the zero_pivot_war matrix as an "identity-extended" workaround:
$ cat t340.cu
#include <cstdio>
#include <cstdlib>
#include <cuda_runtime.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_device(float* src_d, float* dst_d, int n)
{
cublasHandle_t handle;
cublascall(cublasCreate_v2(&handle));
int batchSize = 1;
int *P, *INFO;
cudacall(cudaMalloc<int>(&P,17 * batchSize * sizeof(int)));
cudacall(cudaMalloc<int>(&INFO,batchSize * sizeof(int)));
int lda = 17;
float *A[] = { src_d };
float** A_d;
cudacall(cudaMalloc<float*>(&A_d,sizeof(A)));
cudacall(cudaMemcpy(A_d,A,sizeof(A),cudaMemcpyHostToDevice));
cublascall(cublasSgetrfBatched(handle,17,A_d,lda,P,INFO,batchSize));
int INFOh = 0;
cudacall(cudaMemcpy(&INFOh,INFO,sizeof(int),cudaMemcpyDeviceToHost));
if(INFOh == 17)
{
fprintf(stderr, "Factorization Failed: Matrix is singular\n");
cudaDeviceReset();
exit(EXIT_FAILURE);
}
float* C[] = { dst_d };
float** C_d;
cudacall(cudaMalloc<float*>(&C_d,sizeof(C)));
cudacall(cudaMemcpy(C_d,C,sizeof(C),cudaMemcpyHostToDevice));
cublascall(cublasSgetriBatched(handle,n,A_d,lda,P,C_d,n,INFO,batchSize));
cudacall(cudaMemcpy(&INFOh,INFO,sizeof(int),cudaMemcpyDeviceToHost));
if(INFOh != 0)
{
fprintf(stderr, "Inversion Failed: Matrix is singular\n");
cudaDeviceReset();
exit(EXIT_FAILURE);
}
cudaFree(P), cudaFree(INFO), cublasDestroy_v2(handle);
}
void invert(float* src, float* dst, int n)
{
float* src_d, *dst_d;
cudacall(cudaMalloc<float>(&src_d,17 * 17 * sizeof(float)));
cudacall(cudaMemcpy(src_d,src,17 * 17 * sizeof(float),cudaMemcpyHostToDevice));
cudacall(cudaMalloc<float>(&dst_d,n * n * sizeof(float)));
invert_device(src_d,dst_d,n);
cudacall(cudaMemcpy(dst,dst_d,n * n * sizeof(float),cudaMemcpyDeviceToHost));
cudaFree(src_d), cudaFree(dst_d);
}
void test_invert()
{
const int n = 3;
//Random matrix with full pivots
/* float full_pivots[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 zero_pivot_col_major[n*n] = { 0, 1, 4,
3, 3, 9,
4 , 10, 16 };
*/
float zero_pivot_war[17*17] = {
0,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
1,3,10,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
4,9,16,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1 };
/*
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 singular[n*n] = {1,2,3,
4,5,6,
7,8,9};
*/
float result[n*n];
//Select matrix by setting "a"
float* a = zero_pivot_war;
fprintf(stdout, "Input:\n\n");
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
fprintf(stdout,"%f\t",a[i*17+j]);
fprintf(stdout,"\n");
}
fprintf(stdout,"\n\n");
invert(a,result,n);
fprintf(stdout, "Inverse:\n\n");
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
fprintf(stdout,"%f\t",result[i*n+j]);
fprintf(stdout,"\n");
}
}
int main()
{
test_invert();
// int n; scanf("%d",&n);
return 0;
}
$ nvcc -arch=sm_20 -o t340 t340.cu -lcublas
$ cuda-memcheck ./t340
========= CUDA-MEMCHECK
Input:
0.000000 3.000000 4.000000
1.000000 3.000000 10.000000
4.000000 9.000000 16.000000
Inverse:
-0.700000 -0.200000 0.300000
0.400000 -0.266667 0.066667
-0.050000 0.200000 -0.050000
========= ERROR SUMMARY: 0 errors
$
The above result appears to me to be correct based on a simple test elsewhere.
I don't have any further technical details about the nature of the possible bug in CUBLAS. From what I can tell, it is present in both CUDA 5.5 and CUDA 6.0 RC. Detailed bug discussions for NVIDIA-supplied assets (e.g. CUBLAS library) should be taken up on the NVIDIA developer forums or directly at the bug filing portal on developer.nvidia.com (you must be a registered developer to file a bug).
来源：https://stackoverflow.com/questions/22887167/cublas-incorrect-inversion-for-matrix-with-zero-pivot