/* nag_zero_nonlin_eqns(c05nbc) Example Program
 *
 * Copyright 1991 Numerical Algorithms Group.
 *
 * Mark 2, 1991.
 * Mark 7 revised, 2001.
 */

#include <nag.h>
#include <stdio.h>
#include <nag_stdlib.h>
#include <math.h>
#include <nagc05.h>
#include <nagx02.h>

#ifdef __cplusplus
extern "C" {
#endif
static void f(Integer n, double x[], double fvec[], Integer *userflag);
#ifdef __cplusplus
}
#endif

#define NMAX 9
int main(void)
{

  double fvec[NMAX];
  Integer i, j;
  double x[NMAX];
  double xtol;
  static NagError fail;
  Integer n = NMAX;

  Vprintf("c05nbc Example Program Results\n");
  /* The following starting values provide a rough solution. */
  for (j=0; j<n; j++)
    x[j] = -1.0;
  xtol = sqrt(X02AJC);
  c05nbc(n, x, fvec, f, xtol, &fail);
  if (fail.code == NE_NOERROR)
    {
      Vprintf("Final approximate solution\n\n");
      for (j=0; j<n; j++)
        Vprintf("%12.4f%s",x[j], (j%3==2 || j==n-1) ? "\n" : " " );
      return EXIT_SUCCESS;
    }
  else
    {
      Vprintf("%s\n", fail.message);
      if (fail.code == NE_TOO_MANY_FUNC_EVAL ||
          fail.code == NE_XTOL_TOO_SMALL ||
          fail.code ==  NE_NO_IMPROVEMENT)
        {
          Vprintf("Approximate solution\n\n");
          for (i=0; i<n;  i++)
            Vprintf("%12.4f%s",x[i], (i%3==2 || i==n-1) ? "\n" : " " );
        }
      return EXIT_FAILURE;
    }
}
static void f(Integer n, double x[], double fvec[], Integer *userflag)

{
  Integer k;

  for (k=0; k<n; ++k)
    {
      fvec[k] = (3.0-x[k]*2.0)*x[k]+1.0;
      if (k>0)
        fvec[k] -= x[k-1];
      if (k<n-1)
        fvec[k] -= x[k+1]*2.0;
    }
}