NAG Library Manual, Mark 28.3
Interfaces:  FL   CL   CPP   AD 

NAG AD Library Introduction
Example description
/* C05RD_P0W_F C++ Header Example Program.
 *
 * Copyright 2022 Numerical Algorithms Group.
 * Mark 28.3, 2022.
 */

#include <iostream>
#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;

  cout << "C05RD_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;
  }

  // Call passive routine
  double diag[n], fjac[n * n], factor, fvec[n], qtf[n], r[n * (n + 1) / 2],
      rwsav[4 * n + 10], xtol;
  Integer irevcm, iwsav[17];

  xtol   = sqrt(X02AJC);
  factor = 100.0;
  for (int i = 0; i < n; ++i)
  {
    diag[i] = 1.0;
  }

  Integer           ifail = 0, mode = 2;
  nag::ad::handle_t ad_handle;
  irevcm = 0;
  do
  {
    nag::ad::c05rd(ad_handle, irevcm, n, x, fvec, fjac, xtol, mode, diag,
                   factor, r, qtf, iwsav, rwsav, ifail);

    switch (irevcm)
    {
    case 1:
      // Monitoring exit
      continue;
    case 2:
      for (int i = 0; i < n; ++i)
      {
        fvec[i] = (ruser[1] + ruser[2] * x[i]) * x[i] - ruser[4];
      }
      for (int i = 1; i < n; ++i)
      {
        fvec[i] = fvec[i] + ruser[0] * x[i - 1];
      }
      for (int i = 0; i < n - 1; ++i)
      {
        fvec[i] = fvec[i] + ruser[3] * x[i + 1];
      }
      break;
    case 3:
      for (int i = 0; i < n * n; ++i)
      {
        fjac[i] = 0.0;
      }
      fjac[0] = ruser[1] + 2.0 * ruser[2] * x[0];
      fjac[n] = ruser[3];
      for (int i = 1; i < n - 1; ++i)
      {
        int k       = i * n + i;
        fjac[k - n] = ruser[0];
        fjac[k]     = ruser[1] + 2.0 * ruser[2] * x[i];
        fjac[k + n] = ruser[3];
      }
      fjac[n * n - n - 1] = ruser[0];
      fjac[n * n - 1]     = ruser[1] + 2.0 * ruser[2] * x[n - 1];
      break;
    }
  } while (irevcm != 0);

  cout.setf(ios::scientific, ios::floatfield);
  cout.precision(3);
  cout << "           Solution:\n";
  for (int i = 0; i < n; ++i)
  {
    cout.width(8);
    cout << i + 1;
    cout.width(18);
    cout << x[i] << endl;
  }

  return exit_status;
}