/* nag_opt_lsq_check_deriv (e04yac) 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 <nage04.h>

#ifdef __cplusplus
extern "C" {
#endif
static void lsqfun(Integer m, Integer n, double x[], double fvec[],
                   double fjac[], Integer tdj, Nag_Comm *comm);
#ifdef __cplusplus
}
#endif

int main(void)
{
#define MMAX 15
#define NMAX 3
#define Y(I) comm.user[I]
#define T(I,J) comm.user[(I)*NMAX + (J) + MMAX]

  double fjac[MMAX][NMAX], fvec[MMAX], x[NMAX];
  double work[MMAX + MMAX*NMAX];
  Integer i, j, m, n, tdj;
  Nag_Comm comm;
  static NagError fail;

  Vprintf("e04yac Example Program Results\n");
  Vscanf(" %*[^\n]"); /* Skip heading in data file */

  n = 3;
  m = 15;
  tdj = NMAX;

  fail.print = TRUE;

  /* Allocate memory to communication array */
  comm.user = work;

  /* Observations t (j = 0, 1, 2) are held in T(i, j)
   * (i = 0, 1, 2, . . . , 14) */
  for (i = 0; i < m; ++i)
    {
      Vscanf("%lf", &Y(i));
      for (j = 0; j < n; ++j) Vscanf("%lf", &T(i,j));
    }

  /* Set up an arbitrary point at which to check the 1st derivatives */
  x[0] = 0.19;
  x[1] = -1.34;
  x[2] = 0.88;
  Vprintf("\nThe test point is ");
  for (j = 0; j < n; ++j)
    Vprintf(" %9.3e", x[j]);
  Vprintf("\n");

  fail.print = TRUE;
  e04yac(m, n, lsqfun, x, fvec, &fjac[0][0], tdj, &comm, &fail);

  if (fail.code != NE_NOERROR) return EXIT_FAILURE;

  Vprintf("\nDerivatives are consistent with residual values.\n");
  Vprintf("\nAt the test point, lsqfun() gives\n\n");
  Vprintf("      Residuals                   1st derivatives\n");
  for (i = 0; i < m; ++i)
    {
      Vprintf("     %9.3e  ", fvec[i]);
      for (j = 0; j < n; ++j)
        Vprintf("     %9.3e", fjac[i][j]);
      Vprintf("\n");
    }
  return EXIT_SUCCESS;
}

static void lsqfun(Integer m, Integer n, double x[], double fvec[],
                   double fjac[], Integer tdj, Nag_Comm *comm)
{
  /* Function to evaluate the residuals and their 1st derivatives. */

#define YC(I) comm->user[(I)]
#define TC(I,J) comm->user[(I)*NMAX + (J) + MMAX]
#define FJAC(I,J) fjac[(I)*tdj + (J)]

  Integer i;
  double denom, dummy;

  for (i = 0; i < m; ++i)
    {
      denom = x[1]*TC(i,1) + x[2]*TC(i,2);
      if (comm->flag != 1)
        fvec[i] = x[0] + TC(i,0)/denom - YC(i);
      if (comm->flag != 0)
        {
          FJAC(i,0) = 1.0;
          dummy = -1.0 / (denom * denom);
          FJAC(i,1) = TC(i,0)*TC(i,1)*dummy;
          FJAC(i,2) = TC(i,0)*TC(i,2)*dummy;
        }
    }
}                               /* lsqfun */