/*
*   NEWTON-RAPHSON ALGORITHM 2.3
*
*   To find a solution to f(x) = 0 given an
*   initial approximation p0:
*
*   INPUT:   initial approximation p0; tolerance TOL;
*            maximum number of iterations NO.
*
*   OUTPUT:  approximate solution p or a message of failure
*/

#include <stdio.h>
#include <math.h>
#include <iostream>
#include <iomanip>

#define true    1
#define false   0
#define PI      3.1415926535897932384626433832795

static   double F(double);
static   double FP(double);
static   void OUTPUT(int *);

using namespace std;

int main()
{
      double TOL,P0,D,F0,FP0;
      int OK,I,NO,FLAG;

      P0 = PI/4.0;
      TOL = 1.0e-12;
      NO = 200;

      OUTPUT(&FLAG);
      F0 = F(P0);

      /* STEP 1 */ 
      I = 1;      
      OK = true;        
      /* STEP 2 */ 
      while ((I<=NO) && true == OK) 
      {  
	 /* STEP 3 */ 
	 /* compute P(I) */ 
	 FP0 = FP(P0);
	 D = F0/FP0;
	 /* STEP 6 */ 
	 P0 = P0 - D;  
	 F0 = F(P0);
	 if (FLAG == 2) 
             cout<<setw(3)<<I <<"   " <<
           setiosflags(ios::fixed) << setprecision(8) << P0 <<"   "<<
           setiosflags(ios::fixed) << setprecision(8) << F0 <<endl;
	 /* STEP 4 */ 
	 if (fabs(D) < TOL) { 
	    /* procedure completed successfully */
	    cout<<"\nNewton's method successed\n";
            cout<<"\nApproximate solution P = " << setiosflags(ios::fixed) << setprecision(8) << P0<<endl;
            cout<< "with F(P) = "<< setiosflags(ios::fixed) << setprecision(8) << F0 <<endl;
            cout<<"Number of iterations = "<<I <<endl;
            cout<<" Tolerance = "<< setiosflags(ios::fixed) << setprecision(10) << TOL <<endl;
	    OK = false;
	 }
	 /* STEP 5 */ 
	 else I++;      
      }
      if (true == OK) {
	 /* STEP 7 */ 
	 /* procedure completed unsuccessfully */
	 cout<<"\nNewton's method failed"<<endl;
	 cout<<"\nIteration number "<<NO<<endl;
	 cout<<" gave approximation "<<P0<<endl;
	 cout<<"with F(P) = "<< F0 << " not within tolerance "<<TOL<<endl;
      }
      return 0;
}

/* Change functions F and FP for a new problem */
static double F(double X)
{
   double f; 

   f = cos(X) - X;
   return f;
}
/*  FP is the derivative of F */
static double FP(double X)
{
   double fp; 

   fp = -sin(X) - 1;
   return fp;
}

static void OUTPUT(int *FLAG)
{
   char NAME[30];

   cout<<"Newton's Method"<<endl;
   cout<<"Select amount of output\n";
   cout<<"1. Answer only\n";
   cout<<"2. All intermeditate approximations\n";
   cout<<"Enter 1 or 2: ";

   scanf("%d", FLAG);
   if (*FLAG == 2) cout<<"  I           P           F(P)\n";
}