void qr_solve(double x[], int m, int n, double a[], double b[])
/******************************************************************************/
/*
Purpose:
QR_SOLVE solves a linear system in the least squares sense.
Discussion:
If the matrix A has full column rank, then the solution X should be the
unique vector that minimizes the Euclidean norm of the residual.
If the matrix A does not have full column rank, then the solution is
not unique; the vector X will minimize the residual norm, but so will
various other vectors.
Licensing:
This code is distributed under the GNU LGPL license.
Modified:
11 September 2012
Author:
John Burkardt
Reference:
David Kahaner, Cleve Moler, Steven Nash,
Numerical Methods and Software,
Prentice Hall, 1989,
ISBN: 0-13-627258-4,
LC: TA345.K34.
Parameters:
Input, int M, the number of rows of A.
Input, int N, the number of columns of A.
Input, double A[M*N], the matrix.
Input, double B[M], the right hand side.
Output, double QR_SOLVE[N], the least squares solution.
*/
{
double a_qr[n * m], qraux[n], r[m], tol;
int ind, itask, jpvt[n], kr, lda;
r8mat_copy(a_qr, m, n, a);
lda = m;
tol = r8_epsilon() / r8mat_amax(m, n, a_qr);
itask = 1;
ind = dqrls(a_qr, lda, m, n, tol, &kr, b, x, r, jpvt, qraux, itask); UNUSED(ind);
}
void test04 ( )
/******************************************************************************/
/*
Purpose:
TEST04 tests DQRLS.
Licensing:
This code is distributed under the GNU LGPL license.
Modified:
11 September 2012
Author:
John Burkardt
*/
{
double *a;
double b[5] = { 1.0, 2.3, 4.6, 3.1, 1.2 };
int i;
int ind;
int itask;
int j;
int *jpvt;
int kr;
int m = 5;
int n = 3;
double *qraux;
double tol;
double *x;
a = ( double * ) malloc ( m * n * sizeof ( double ) );
jpvt = ( int * ) malloc ( n * sizeof ( int ) );
qraux = ( double * ) malloc ( n * sizeof ( double ) );
x = ( double * ) malloc ( n * sizeof ( double ) );
/*
Set up least-squares problem
quadratic model, equally-spaced points
*/
printf ( "\n" );
printf ( "TEST04\n" );
printf ( " DQRLS solves a linear system A*x = b in the least squares sense.\n" );
for ( i = 0; i < m; i++ )
{
a[i+0*m] = 1.0;
for ( j = 1; j < n; j++ )
{
a[i+j*m] = a[i+(j-1)*m] * ( double ) ( i + 1 );
}
}
tol = 1.0E-06;
r8mat_print ( m, n, a, " Coefficient matrix A:" );
r8vec_print ( m, b, " Right hand side b:" );
/*
Solve least-squares problem
*/
itask = 1;
ind = dqrls ( a, m, m, n, tol, &kr, b, x, b, jpvt, qraux, itask );
/*
Print results
*/
printf ( "\n" );
printf ( " Error code = %d\n", ind );
printf ( " Estimated matrix rank = %d\n", kr );
r8vec_print ( n, x, " Least squares solution x:" );
r8vec_print ( m, b, " Residuals A*x-b" );
free ( a );
free ( jpvt );
free ( qraux );
free ( x );
return;
}
