NAG Library Manual, Mark 30.3
Interfaces:  FL   CL   CPP   AD 

NAG AD Library Introduction
Example description
/* C05QS_P0W_F C++ Header Example Program.
 *
 * Copyright 2024 Numerical Algorithms Group.
 * Mark 30.3, 2024.
 */

#include <iostream>
#include <dco.hpp>
#include <math.h>
#include <nag.h>
#include <nagad.h>
#include <nagx02.h>
#include <nagx04.h>
#include <stdio.h>
#include <string>
using namespace std;

int main()
{
  // Scalars
  int           exit_status = 0;
  const Integer n = 7, lrcomm = 12 + 3 * n, licomm = 8 * n + 19 + 3 * n;

  cout << "C05QS_P0W_F C++ Header Example Program Results\n\n";

  // problem parameters and starting value
  double ruser[5], x[7];

  ruser[0] = -1.0;
  ruser[1] = 3.0;
  ruser[2] = -2.0;
  ruser[3] = -2.0;
  ruser[4] = -1.0;

  for (int i = 0; i < n; ++i)
  {
    x[i] = -1.0;
  }

  Integer           ifail = 0;
  nag::ad::handle_t ad_handle;
  double            fvec[n], rcomm[lrcomm], xtol;
  Integer           icomm[licomm];
  logical           init = Nag_TRUE;

  xtol = sqrt(X02AJC);
  auto fcn = [&](nag::ad::handle_t &     ad_handle,
                const Integer &         n,
                const Integer &         lindf,
                const Integer           indf[],
                const double *x,
                double *fvec,
                Integer &               iflag)
              {
                for (int ind = 0; ind < lindf; ++ind)
                {
                  int i   = indf[ind] - 1;
                  fvec[i] = (ruser[1] + ruser[2] * x[i]) * x[i] - ruser[4];
                  if (i > 0)
                    fvec[i] = fvec[i] + ruser[0] * x[i - 1];
                  if (i < n - 1)
                    fvec[i] = fvec[i] + ruser[3] * x[i + 1];
                }
                iflag = 0;
              };
              
  nag::ad::c05qs(ad_handle, fcn, n, x, fvec, xtol, init, rcomm, lrcomm, icomm, licomm, ifail);

  cout.setf(ios::scientific, ios::floatfield);
  cout.precision(4);
  cout << "           Solution:\n";
  for (int i = 0; i < n; ++i)
  {
    cout.width(10);
    cout << i + 1;
    cout.width(20);
    cout << x[i] << endl;
  }

  return exit_status;
}