/*
* FIXED-POINT ALGORITHM 2.2
*
* To find a solution to p = g(p) given an
* initial approximation p0
* This code is to solve Illustration problem on Page 61 of the textbook. 
* Formulation (d) is used.
*
* INPUT:  initial approximation; tolerance TOL;
*         maximum number of iterations NO.
*
* OUTPUT: approximate solution p or 
*         a message that the method fails.
*/

#include <stdio.h>
#include <math.h> 
#include <iostream>
#define true 1
#define false 0

static   void      OUTPUT(int *); /* Set up output files for saving the answer */
static   double    G(double );             /* function g */

using namespace std;

int main()
{
      double TOL, P0, P;
      int I,   NO,  FLAG, OK;
      char AA;

      P0 = 1.5;              // init. guess 
      TOL = 1.0e-14;         // tolerance for |p-p0|
      NO =  200;             // max. number of iterations

      OUTPUT(&FLAG);

      /* STEP 1 */ 
      I = 1;  
      /* STEP 2 */ 
      OK = true;
      while(I<=NO && OK) 
      {
         /* STEP 3 */ 
         /* compute P(I) */  
         P = G(P0);
         if (FLAG == 2) printf("%3d   %15.8e\n", I, P);
         /* STEP 4 */ 
         if (fabs(P-P0) < TOL) { 
            /* procedure completed successfully */
            printf("\nApproximate solution P = %12.8f\n", P);
            printf("Number of iterations = %3d", I);
            printf("    Tolerance = %14.8e\n",TOL);
            printf("|p-p0| = %14.8e\n", fabs(P-P0));
            OK = false;
         }
         else {
            /* STEP 5 */ 
            I++;
            /* STEP 6 */ 
            /* update P0 */ 
            P0 = P; 
         }
      }
      if (true == OK) {
         /* STEP 7 */  
         /* procedure completed unsuccessfully */ 
         printf("\nIteration number %3d", NO);
         printf(" gave approximation %12.8f\n", P);
         printf("not within tolerance %14.8e\n",TOL);
      }

   return 1;
}
/* Change function G for a new problem */
static double G(double X)
{
   double g; 

   g = sqrt(10.0 / (4.0 + X)); 
   return g;
}

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

   printf("Select amount of output\n");
   printf("1. Answer only\n");
   printf("2. All intermeditate approximations\n");
   printf("Enter 1 or 2: ");
   scanf("%d", FLAG);
   printf("FIXED-POINT METHOD\n"); 
   if (*FLAG == 2) printf("  I                 P\n");
}