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lu.c
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#include <math.h>
#include <grass/gis.h>
#include <grass/gmath.h>
#define TINY 1.0e-20;
/*!
* \brief LU decomposition
*
* \param a double **
* \param n int
* \param indx int *
* \param d double *
*
* \return 0 on singular matrix, 1 on success
*/
int G_ludcmp(double **a, int n, int *indx, double *d)
{
int i, imax = 0, j, k;
double big, dum, sum, temp;
double *vv;
int is_singular = FALSE;
vv = G_alloc_vector(n);
*d = 1.0;
/* this pragma works, but doesn't really help speed things up */
/* #pragma omp parallel for private(i, j, big, temp) shared(n, a, vv,
* is_singular) */
for (i = 0; i < n; i++) {
big = 0.0;
for (j = 0; j < n; j++)
if ((temp = fabs(a[i][j])) > big)
big = temp;
if (big == 0.0) {
is_singular = TRUE;
break;
}
vv[i] = 1.0 / big;
}
if (is_singular) {
*d = 0.0;
return 0; /* Singular matrix */
}
for (j = 0; j < n; j++) {
for (i = 0; i < j; i++) {
sum = a[i][j];
for (k = 0; k < i; k++)
sum -= a[i][k] * a[k][j];
a[i][j] = sum;
}
big = 0.0;
/* not very efficient, but this pragma helps speed things up a bit */
#pragma omp parallel for private(i, k, sum, dum) shared(j, n, a, vv, big, imax)
for (i = j; i < n; i++) {
sum = a[i][j];
for (k = 0; k < j; k++)
sum -= a[i][k] * a[k][j];
a[i][j] = sum;
if ((dum = vv[i] * fabs(sum)) >= big) {
big = dum;
imax = i;
}
}
if (j != imax) {
for (k = 0; k < n; k++) {
dum = a[imax][k];
a[imax][k] = a[j][k];
a[j][k] = dum;
}
*d = -(*d);
vv[imax] = vv[j];
}
indx[j] = imax;
if (a[j][j] == 0.0)
a[j][j] = TINY;
if (j != n) {
dum = 1.0 / (a[j][j]);
for (i = j + 1; i < n; i++)
a[i][j] *= dum;
}
}
G_free_vector(vv);
return 1;
}
#undef TINY
/*!
* \brief LU backward substitution
*
* \param a double **
* \param n int
* \param indx int *
* \param b double []
*
* \return void
*/
void G_lubksb(double **a, int n, int *indx, double b[])
{
int i, ii, ip, j;
double sum;
ii = -1;
for (i = 0; i < n; i++) {
ip = indx[i];
sum = b[ip];
b[ip] = b[i];
if (ii >= 0)
for (j = ii; j < i; j++)
sum -= a[i][j] * b[j];
else if (sum)
ii = i;
b[i] = sum;
}
for (i = n - 1; i >= 0; i--) {
sum = b[i];
for (j = i + 1; j < n; j++)
sum -= a[i][j] * b[j];
b[i] = sum / a[i][i];
}
}