问题
I am trying to factorize a matrix with the QR factorization in C++, using Lapack's functions in order to solve a system of linear equations (Ax=b)
As far as I understood, dgeqrf computes the QR factorization and overwrites the input matrix. The output clearly contains values for L (upper triangular), but how do I obtain Q?
I tried dormqr, which is said to calculate Q from dgeqrf's output, but the result is the same matrix as in the previous call.
Here's my complete code:
boost::numeric::ublas::matrix<double> in_A(4, 3);
in_A(0, 0) = 1.0;
in_A(0, 1) = 2.0;
in_A(0, 2) = 3.0;
in_A(1, 1) = -3.0;
in_A(1, 2) = 2.0;
in_A(1, 3) = 1.0;
in_A(2, 1) = 2.0;
in_A(2, 2) = 0.0;
in_A(2, 3) = -1.0;
in_A(3, 1) = 3.0;
in_A(3, 2) = -1.0;
in_A(3, 3) = 2.0;
boost::numeric::ublas::vector<double> in_b(4);
in_b(0) = 2;
in_b(1) = 4;
in_b(2) = 6;
in_b(3) = 8;
int rows = in_A.size1();
int cols = in_A.size2();
double *A = (double *)malloc(rows*cols*sizeof(double));
double *b = (double *)malloc(in_b.size()*sizeof(double));
//Lapack has column-major order
for(size_t col=0; col<in_A.size2(); ++col)
{
for(size_t row = 0; row<in_A.size1(); ++row)
{
int D1_idx = col*in_A.size1() + row;
A[D1_idx] = in_A(row, col);
}
b[col] = in_b(col);
}
integer m = rows;
integer n = cols;
integer info = 0;
integer k = n; /* k = min(m,n); */
integer lda = m; /* lda = max(m,1); */
integer lwork = n; /* lwork = max(n,1); */
int max = lwork; /* max = max(lwork,1); */
double *work;
double *tau;
char *side = "L";
char *TR = "T";
integer one = 1;
int i;
double *vec;
work = (double *) malloc( max * sizeof( double ) );
tau = (double *) malloc( k * sizeof( double ) );
vec = (double *) malloc( m * sizeof( double ) );
memset(work, 0, max * sizeof(double));
memset(tau, 0, k * sizeof(double));
std::cout << std::endl;
for(size_t row = 0; row < rows; ++row)
{
for(size_t col = 0; col < cols; ++col)
{
size_t idx = col*rows + row;
std::cout << A[idx] << " ";
}
std::cout << std::endl;
}
dgeqrf_(&m, &n, A, &lda, tau, work, &lwork, &info);
//printf("tau[0] = %f tau[1] = %f\n",tau[0],tau[1]);
std::cout << std::endl;
for(size_t row = 0; row < rows; ++row)
{
for(size_t col = 0; col < cols; ++col)
{
size_t idx = col*rows + row;
std::cout << A[idx] << " ";
}
std::cout << std::endl;
}
memset(vec, 0, m * sizeof(double));
vec[2] = 1.0;
dormqr_(side, TR, &m, &one, &k, A, &lda, tau, vec, &lda, work, &lwork, &info);
free(vec);
free(tau);
free(work);
What's wrong with my code?
How can I factorize a matrix and solve a corresponding system of linear equations?
回答1:
According to the documentation in
(http://www.netlib.org/lapack/explore-html/da/d82/dormqr_8f.html)
you are computing in vec the product Q^T*e3, where e3 is the third canonical basis vector (0,0,1,0,0,...,0). If you want to compute Q, then vec should contain a matrix sized array filled with the unit matrix, and TRANS should be "N".
dormqr (SIDE, TRANS, M, N, K, A, LDA, TAU, C, LDC, WORK, LWORK, INFO)
SIDE = "L" for the normal QR decomposition with Q left,
TRANS = "N" to return QC in the place of C
A has layout LDA x K in memory, of which the upper M x K block is used and encodes K reflectors
tau contains the factors for the K reflectors
C has layout LDC x M in memory, of which the upper M x N block will be used to hold the result QC
For C to hold Q on return, C must be a square M x M matrix initialized as identity, i.e., with diagonal entries all 1.
You might consider to use the lapack numeric bindings provided for ublas, as in
(http://boost.2283326.n4.nabble.com/How-to-use-the-qr-decomposition-correctly-td2710159.html)
However, this project may be defunct or resting by now.
Lets start again from first principles: The aim is to solve Ax=b, or at least to minimize |Ax-b|+|x|. For that to be consistent one needs colsA=rowsx and rowsA=rowsb.
Now for the discussed code to work A has to be square or a tall rectangular matrix, colsA<=rowsA, so that the system is overdetermined.
Computation steps
Solve
Q*R=A: (http://www.netlib.no/netlib/lapack/double/dgeqrf.f)DGEQRF( rowsA, colsA, A, rowsA, TAU, WORK, LWORK, INFO )
Multiply by
QTto getQT*bas inR*x=QT*b(http://www.netlib.no/netlib/lapack/double/dormqr.f)DORMQR( 'L', 'T', rowsA, 1, colsA, A, rowsA, TAU, b, rowsA, WORK, LWORK, INFO )
Use back-substitution using the upper right part of
A(http://www.netlib.no/netlib/lapack/double/dtrtrs.f)DTRTRS( 'U', 'N', 'N', colsA, 1, A, rowsA, b, rowsA, INFO )
Now the first
colsAentries ofbcontain the solution vectorx. The euclidean norm of the remaining entries at index colsA+1 and thereafter is the error |A*x-b| of the solution.
Remark: For the pure solution process there is no reason to compute 'Q' explicitly or to invoke the generic matrix multiplication DGEMM. These should be reserved for experiments to check if A-QR is sufficiently close to zero.
Remark: Explore the optimal allocation of the WORK array by performing a dry run with LWORK=-1.
To conclude some code that works, however, the connection between ublas and lapack seems suboptimal
#include "boost/numeric/ublas/matrix.hpp"
#include "boost/numeric/ublas/vector.hpp"
typedef boost::numeric::ublas::matrix<double> bmatrix;
typedef boost::numeric::ublas::vector<double> bvector;
namespace lapack {
extern "C" {
void dgeqrf_(int* M, int* N,
double* A, int* LDA, double* TAU,
double* WORK, int* LWORK, int* INFO );
void dormqr_(char* SIDE, char* TRANS,
int* M, int* N, int* K,
double* A, int* LDA, double* TAU,
double* C, int* LDC,
double* WORK, int* LWORK, int* INFO );
void dtrtrs_(char* UPLO, char* TRANS, char* DIAG,
int* N, int* NRHS,
double* A, int* LDA,
double* B, int* LDB,
int* INFO );
}
int geqrf(int m, int n,
double* A, int lda, double *tau) {
int info=0;
int lwork=-1;
double iwork;
dgeqrf_(&m, &n, A, &lda, tau,
&iwork, &lwork, &info);
lwork = (int)iwork;
double* work = new double[lwork];
dgeqrf_(&m, &n, A, &lda, tau,
work, &lwork, &info);
delete[] work;
return info;
}
int ormqr(char side, char trans, int m, int n, int k,
double *A, int lda, double *tau, double* C, int ldc) {
int info=0;
int lwork=-1;
double iwork;
dormqr_(&side, &trans, &m, &n, &k,
A, &lda, tau, C, &ldc, &iwork, &lwork, &info);
lwork = (int)iwork;
double* work = new double[lwork];
dormqr_(&side, &trans, &m, &n, &k,
A, &lda, tau, C, &ldc, work, &lwork, &info);
delete[] work;
return info;
}
int trtrs(char uplo, char trans, char diag,
int n, int nrhs,
double* A, int lda, double* B, int ldb
) {
int info = 0;
dtrtrs_(&uplo, &trans, &diag, &n, &nrhs,
A, &lda, B, &ldb, &info);
return info;
}
}
static void PrintMatrix(double A[], size_t rows, size_t cols) {
std::cout << std::endl;
for(size_t row = 0; row < rows; ++row)
{
for(size_t col = 0; col < cols; ++col)
{
// Lapack uses column major format
size_t idx = col*rows + row;
std::cout << A[idx] << " ";
}
std::cout << std::endl;
}
}
static int SolveQR(
const bmatrix &in_A, // IN
const bvector &in_b, // IN
bvector &out_x // OUT
) {
size_t rows = in_A.size1();
size_t cols = in_A.size2();
double *A = new double[rows*cols];
double *b = new double[in_b.size()];
//Lapack has column-major order
for(size_t col=0, D1_idx=0; col<cols; ++col)
{
for(size_t row = 0; row<rows; ++row)
{
// Lapack uses column major format
A[D1_idx++] = in_A(row, col);
}
b[col] = in_b(col);
}
for(size_t row = 0; row<rows; ++row)
{
b[row] = in_b(row);
}
// DGEQRF for Q*R=A, i.e., A and tau hold R and Householder reflectors
double* tau = new double[cols];
PrintMatrix(A, rows, cols);
lapack::geqrf(rows, cols, A, rows, tau);
PrintMatrix(A, rows, cols);
// DORMQR: to compute b := Q^T*b
lapack::ormqr('L', 'T', rows, 1, cols, A, rows, tau, b, rows);
PrintMatrix(b, rows, 1);
// DTRTRS: solve Rx=b by back substitution
lapack::trtrs('U', 'N', 'N', cols, 1, A, rows, b, rows);
for(size_t col=0; col<cols; col++) {
out_x(col)=b[col];
}
PrintMatrix(b,cols,1);
delete[] A;
delete[] b;
delete[] tau;
return 0;
}
int main() {
bmatrix in_A(4, 3);
in_A(0, 0) = 1.0; in_A(0, 1) = 2.0; in_A(0, 2) = 3.0;
in_A(1, 0) = -3.0; in_A(1, 1) = 2.0; in_A(1, 2) = 1.0;
in_A(2, 0) = 2.0; in_A(2, 1) = 0.0; in_A(2, 2) = -1.0;
in_A(3, 0) = 3.0; in_A(3, 1) = -1.0; in_A(3, 2) = 2.0;
bvector in_b(4);
in_b(0) = 2;
in_b(1) = 4;
in_b(2) = 6;
in_b(3) = 8;
bvector out_x(3);
SolveQR( in_A, in_b, out_x);
return 0;
}
来源：https://stackoverflow.com/questions/21970510/solving-a-linear-system-with-lapacks-dgeqrf