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/* driver for hybrj1 example. */
#include <stdio.h>
#include <math.h>
#include <minpack.h>
void fcn(int *n, double *x, double *fvec, double *fjac, int *ldfjac,
int *iflag);
int main()
{
int j, n, ldfjac, info, lwa;
double tol, fnorm;
double x[9], fvec[9], fjac[9*9], wa[99];
int one=1;
n = 9;
/* the following starting values provide a rough solution. */
for (j=1; j<=9; j++)
{
x[j-1] = -1.;
}
ldfjac = 9;
lwa = 99;
/* set tol to the square root of the machine precision. */
/* unless high solutions are required, */
/* this is the recommended setting. */
tol = sqrt(dpmpar_(&one));
hybrj1_(&fcn, &n, x, fvec, fjac, &ldfjac, &tol, &info, wa, &lwa);
fnorm = enorm_(&n, fvec);
printf(" final l2 norm of the residuals%15.7g\n\n", fnorm);
printf(" exit parameter %10i\n\n", info);
printf(" final approximate solution\n");
for (j=1; j<=n; j++) printf("%s%15.7g", j%3==1?"\n ":"", x[j-1]);
printf("\n");
return 0;
}
void fcn(int *n, double *x, double *fvec, double *fjac, int *ldfjac,
int *iflag)
{
/* subroutine fcn for hybrj1 example. */
int j, k;
double one=1, temp, temp1, temp2, three=3, two=2, zero=0, four=4;
if (*iflag != 2)
{
for (k = 1; k <= *n; k++)
{
temp = (three - two*x[k-1])*x[k-1];
temp1 = zero;
if (k != 1) temp1 = x[k-1-1];
temp2 = zero;
if (k != *n) temp2 = x[k+1-1];
fvec[k-1] = temp - temp1 - two*temp2 + one;
}
}
else
{
for (k = 1; k <= *n; k++)
{
for (j = 1; j <= *n; j++)
{
fjac[k-1 + *ldfjac*(j-1)] = zero;
}
fjac[k-1 + *ldfjac*(k-1)] = three - four*x[k-1];
if (k != 1) fjac[k-1 + *ldfjac*(k-1-1)] = -one;
if (k != *n) fjac[k-1 + *ldfjac*(k+1-1)] = -two;
}
}
return;
}
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