/* nag_check_deriv(c05zbc) 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 <nagc05.h>

#ifdef __cplusplus
extern "C" {
#endif
static void f(Integer n, double xc[], double fvecc[],
              double fjacc[], Integer tdj, Integer *userflag);
#ifdef __cplusplus
}
#endif

int main(void)
{
#define NMAX 5

  double fjac[NMAX][NMAX], fvec[NMAX], x[NMAX];
  Integer i, j, n, tdfjac;
  static NagError fail;

  fail.print = TRUE;
  Vprintf("c05zbc Example Program Results\n");
  n = 3;
  tdfjac = NMAX;

  /* Set up an arbitrary point at which to check the 1st derivatives */
  x[0] = 9.2e-01;
  x[1] = 1.3e-01;
  x[2] = 5.4e-01;
  Vprintf("The test point is ");
  for (j=0; j<n; ++j)
    Vprintf("%13.3e", x[j]);
  Vprintf("\n\n");
  c05zbc(n, x, fvec, &fjac[0][0], tdfjac, f, &fail);
  if (fail.code != NE_NOERROR) return EXIT_FAILURE;
  Vprintf("1st derivatives are consistent with residual values.\n\n");
  Vprintf("At the test point, f() gives\n\n");
  Vprintf("    Residuals            1st derivatives\n\n");
  for (i=0; i<n; ++i)
    {
      Vprintf("%13.3e", fvec[i]);
      for (j=0; j<n; ++j)
        Vprintf("%13.3e", fjac[i][j]);
      Vprintf("\n");
    }
  return EXIT_SUCCESS;
}

static void f(Integer n, double x[], double fvec[], double fjac[],
              Integer tdfjac, Integer *userflag)
{
#define FJAC(I,J) fjac[((I))*tdfjac+(J)]
  Integer j, k;

  if (*userflag != 2)
    {
      /* Calculate the function values */
      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;
        }
    }
  else
    {
      /* Calculate the corresponding first derivatives */
      for (k=0; k<n; k++)
        {
          for (j=0; j<n; j++)
            FJAC(k,j)=0.0;
          FJAC(k,k) = 3.0 - x[k] * 4.0;
          if (k>0)
            FJAC(k,k-1) = -1.0;
          if (k<n-1)
            FJAC(k,k+1)= -2.0;
        }
    }
}