/* nag_zero_nonlin_eqns_expert (c05qcc) Example Program.
 *
 * Copyright 2011 Numerical Algorithms Group.
 *
 * Mark 23, 2011.
 */

#include <nag.h>
#include <nagx04.h>
#include <stdio.h>
#include <nag_stdlib.h>
#include <math.h>
#include <nagc05.h>
#include <nagx02.h>

#ifdef __cplusplus
extern "C" {
#endif
static void NAG_CALL fcn(Integer n, const double x[], double fvec[],
                         Nag_Comm *comm, Integer *iflag);
#ifdef __cplusplus
}
#endif

int main(void)
{

  Integer  exit_status = 0, i, j, n = 9, maxfev, ml, mu, nfev, nprint;
  NagError fail;
  Nag_Comm comm;
  Nag_ScaleType scale_mode;
  double   *diag = 0, *fjac = 0, *fvec = 0, *qtf = 0, *r = 0, *x = 0;
  double   epsfcn, factor, xtol;

  INIT_FAIL(fail);

  printf("nag_zero_nonlin_eqns_expert (c05qcc) Example Program Results\n");
  if (n > 0)
    {
      if (!(diag = NAG_ALLOC(n, double)) ||
          !(fjac = NAG_ALLOC(n*n, double)) ||
          !(fvec = NAG_ALLOC(n, double)) ||
          !(qtf = NAG_ALLOC(n, double)) ||
          !(r = NAG_ALLOC(n*(n+1)/2, double)) ||
          !(x = NAG_ALLOC(n, double)))
        {
          printf("Allocation failure\n");
          exit_status = -1;
          goto END;
        }
    }
  else
    {
      printf("Invalid n.\n");
      exit_status = 1;
      goto END;
    }

  /* The following starting values provide a rough solution. */
  for (j = 0; j < n; j++)
    x[j] = -1.0;

  /* nag_machine_precision (x02ajc).
   * The machine precision
   */
  xtol = sqrt(nag_machine_precision);

  for (j = 0; j < n; j++)
    diag[j] = 1.0;

  maxfev = 2000;
  ml = 1;
  mu = 1;
  epsfcn = 0.0;
  scale_mode = Nag_ScaleProvided;
  factor = 100.0;
  nprint = 0;

  /* nag_zero_nonlin_eqns_expert (c05qcc).
   * Solution of a system of nonlinear equations (function
   * values only)
   */
  nag_zero_nonlin_eqns_expert(fcn, n, x, fvec, xtol, maxfev, ml, mu,
                              epsfcn, scale_mode, diag, factor, nprint, &nfev,
                              fjac, r, qtf, &comm, &fail);

  if (fail.code == NE_NOERROR)
    {
      printf("Final approximate solution\n\n");
      for (j = 0; j < n; j++)
        printf("%12.4f%s", x[j], (j%3 == 2 || j == n-1)?"\n":" ");
    }
  else
    {
      printf("Error from nag_zero_nonlin_eqns_expert (c05qcc).\n%s\n",
             fail.message);
      if (fail.code == NE_TOO_MANY_FEVALS ||
          fail.code == NE_TOO_SMALL ||
          fail.code == NE_NO_IMPROVEMENT)
        {
          printf("Approximate solution\n\n");
          for (i = 0; i < n;  i++)
            printf("%12.4f%s", x[i], (i%3 == 2 || i == n-1)?"\n":" ");
          exit_status = 2;
        }
    }
 END:
  if (diag) NAG_FREE(diag);
  if (fjac) NAG_FREE(fjac);
  if (fvec) NAG_FREE(fvec);
  if (qtf) NAG_FREE(qtf);
  if (r) NAG_FREE(r);
  if (x) NAG_FREE(x);
  return exit_status;
}
static void NAG_CALL fcn(Integer n, const double x[], double fvec[],
                         Nag_Comm *comm, Integer *iflag)
{
  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;
    }
}