/*
*   SECANT ALGORITHM 2.4
*
*   To find a solution to the equation f(x) = 0
*   given initial approximations p0 and p1:
*
*   INPUT:   initial approximation p0, p1; tolerance TOL;
*            maximum number of iterations N0.
*
*   OUTPUT:  approximate solution p or
*            a message that the algorithm fails.
*
*/

#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   void OUTPUT(int *);

using namespace std;

int main()
{
      double P0,F0,P1,F1,P,FP,TOL;
      int I,NO,OK,FLAG;

      OK = true; 
      P0 = 1.0; 
      P1 = 2.0;
      TOL = 1.0e-10;
      NO = 200; 

      OUTPUT(&FLAG);
      /* STEP 1 */ 
      I = 2;      
      F0 = F(P0);
      F1 = F(P1);
      OK = true;                                
      /* STEP 2 */ 
      while ((I<=NO) && true == OK) {   
	 /* STEP 3 */ 
	 /* compute P(I) */ 
	 P = P1 - F1 * (P1 - P0) / (F1 - F0);

	 /* STEP 4 */  
	 FP = F(P);  
	 if (FLAG == 2)  
             cout<<setw(3)<<I <<"   " <<
             setiosflags(ios::fixed) << setprecision(8) << P <<"   "<<
             setiosflags(ios::fixed) << setprecision(8) << FP <<endl;
	 if (fabs(P - P1) < TOL) {
	    /* procedure completed successfully */
	    cout<<"\nSecant method succeeded\n";
            cout<<"\nApproximate solution P = " << setiosflags(ios::fixed) << setprecision(8) << P0<<endl;
            cout<< "with F(P) = "<< setiosflags(ios::fixed) << setprecision(8) << FP <<endl;
            cout<<"Number of iterations = "<<I <<endl;
            cout<<" Tolerance = "<< setiosflags(ios::fixed) << setprecision(10) << TOL <<endl;
	    OK = false;
	 }
	 else {
	    /* STEP 5 */ 
	    I++; 
	    
	    /* STEP 6 */ 
	    /* update P0, F0, P1, F1 */ 
	    P0 = P1;
	    F0 = F1;
	    P1 = P;
	    F1 = FP;
	 }
      }
      if (OK) 
      {
	 /* STEP 7 */ 
	 /* procedure completed unsuccessfully */       
	 cout<<"\nSecant method failed\n";
         cout<<"\nIteration number "<<NO<<endl;
	 cout<<"gave approximation "<< P <<endl;
	 cout<<"F(P) = " <<FP << " not within tolerance: " << TOL <<endl;
      }
      return 0;
}

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

   f = exp(X) + pow(2.0,-X)+2*cos(X)-6.0;

   return f;
}

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

   cout<<"Secant 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";
}