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/* driver for hybrd example. */
#include <stdio.h>
#include <math.h>
#include <minpack.h>
void fcn(int *n, double *x, double *fvec, int *iflag);
int main()
{
int j, n, maxfev, ml, mu, mode, nprint, info, nfev, ldfjac, lr;
double xtol, epsfcn, factor, fnorm;
double x[9], fvec[9], diag[9], fjac[9*9], r[45], qtf[9],
wa1[9], wa2[9], wa3[9], wa4[9];
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;
lr = 45;
/* set xtol to the square root of the machine precision. */
/* unless high solutions are required, */
/* this is the recommended setting. */
xtol = sqrt(dpmpar_(&one));
maxfev = 2000;
ml = 1;
mu = 1;
epsfcn = 0.;
mode = 2;
for (j=1; j<=9; j++)
{
diag[j-1] = 1.;
}
factor = 1.e2;
nprint = 0;
hybrd_(&fcn, &n, x, fvec, &xtol, &maxfev, &ml, &mu, &epsfcn,
diag, &mode, &factor, &nprint, &info, &nfev,
fjac, &ldfjac, r, &lr, qtf, wa1, wa2, wa3, wa4);
fnorm = enorm_(&n, fvec);
printf(" final l2 norm of the residuals %15.7g\n\n", fnorm);
printf(" number of function evaluations %10i\n\n", nfev);
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, int *iflag)
{
/* subroutine fcn for hybrd example. */
int k;
double one=1, temp, temp1, temp2, three=3, two=2, zero=0;
if (iflag == 0)
{
/* insert print statements here when nprint is positive. */
return;
}
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;
}
return;
}
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