Solving a linear system with Lapack's dgeqrf_

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?

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 QT to get QT*b as in R*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 colsA entries of b contain the solution vector x. 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

标签

c++

math

visual-studio-2012

linear-algebra

lapack