tlmdif_.c

该程序实现了非线性最小二乘问题和非线性方程组的解法

/*      driver for lmdif example. */

#include <stdio.h>
#include <math.h>
#include <minpack.h>

void fcn(const int *m, const int *n, const double *x, double *fvec, int *iflag);

int main()
{
  int i, j, m, n, maxfev, mode, nprint, info, nfev, ldfjac;
  int ipvt[3];
  double ftol, xtol, gtol, epsfcn, factor, fnorm;
  double x[3], fvec[15], diag[3], fjac[15*3], qtf[3], 
    wa1[3], wa2[3], wa3[3], wa4[15];
  int one=1;
  double covfac;

  m = 15;
  n = 3;

/*      the following starting values provide a rough fit. */

  x[1-1] = 1.;
  x[2-1] = 1.;
  x[3-1] = 1.;

  ldfjac = 15;

  /*      set ftol and xtol to the square root of the machine */
  /*      and gtol to zero. unless high solutions are */
  /*      required, these are the recommended settings. */

  ftol = sqrt(dpmpar_(&one));
  xtol = sqrt(dpmpar_(&one));
  gtol = 0.;

  maxfev = 800;
  epsfcn = 0.;
  mode = 1;
  factor = 1.e2;
  nprint = 0;

  lmdif_(&fcn, &m, &n, x, fvec, &ftol, &xtol, &gtol, &maxfev, &epsfcn, 
	 diag, &mode, &factor, &nprint, &info, &nfev, fjac, &ldfjac, 
	 ipvt, qtf, wa1, wa2, wa3, wa4);

  fnorm = enorm_(&m, 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");
  ftol = dpmpar_(&one);
  covfac = fnorm*fnorm/(m-n);
  covar_(&n, fjac, &ldfjac, ipvt, &ftol, wa1);
  printf("      covariance\n");
  for (i=1; i<=n; i++) {
    for (j=1; j<=n; j++)
      printf("%s%15.7g", j%3==1?"\n     ":"", fjac[(i-1)*ldfjac+j-1]*covfac);
  }
  printf("\n");
  return 0;
}

void fcn(const int *m, const int *n, const double *x, double *fvec, int *iflag)
{

/*      subroutine fcn for lmdif example. */

  int i;
  double tmp1, tmp2, tmp3;
  double y[15]={1.4e-1, 1.8e-1, 2.2e-1, 2.5e-1, 2.9e-1, 3.2e-1, 3.5e-1,
		3.9e-1, 3.7e-1, 5.8e-1, 7.3e-1, 9.6e-1, 1.34, 2.1, 4.39};

  if (*iflag == 0)
    {
      /*      insert print statements here when nprint is positive. */
      return;
    }
  for (i = 1; i <= 15; i++)
    {
      tmp1 = i;
      tmp2 = 16 - i;
      tmp3 = tmp1;
      if (i > 8) tmp3 = tmp2;
      fvec[i-1] = y[i-1] - (x[1-1] + tmp1/(x[2-1]*tmp2 + x[3-1]*tmp3));
    }
  return;
}