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📄 tlmstr_.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
💻 C
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/*      driver for lmstr example. */#include <stdio.h>#include <math.h>#include <minpack.h>void fcn(const int *m, const int *n, const double *x, double *fvec, double *fjrow, int *iflag);int main(){  int j, m, n, ldfjac, maxfev, mode, nprint, info, nfev, njev;  int ipvt[3];  double ftol, xtol, gtol, factor, fnorm;  double x[3], fvec[15], fjac[3*3], diag[3], qtf[3],     wa1[3], wa2[3], wa3[3], wa4[15];  int one=1;  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 = 3;  /*      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 = 400;  mode = 1;  factor = 1.e2;  nprint = 0;  lmstr_(&fcn, &m, &n, x, fvec, fjac, &ldfjac, &ftol, &xtol, &gtol, 	&maxfev, diag, &mode, &factor, &nprint, &info, &nfev, &njev, 	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("      number of Jacobian evaluations%10i\n\n", njev);  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(const int *m, const int *n, const double *x, double *fvec, double *fjrow, int *iflag){  /*      subroutine fcn for lmstr example. */  int i;  double tmp1, tmp2, tmp3, tmp4;  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;    }  if (*iflag < 2)    {      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));}			 }else  {    i = *iflag - 1;    tmp1 = i;    tmp2 = 16 - i;    tmp3 = tmp1;    if (i > 8) tmp3 = tmp2;    tmp4 = (x[2-1]*tmp2 + x[3-1]*tmp3); tmp4 = tmp4*tmp4;    fjrow[1-1] = -1.;    fjrow[2-1] = tmp1*tmp2/tmp4;    fjrow[3-1] = tmp1*tmp3/tmp4;  }  return;}

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