/*
*   STEFFENSEN'S ALGORITHM 2.6
*
*   To find a solution to g(x) = x
*   given an initial approximation p0:
*
* This sample is to solve x^3 + 4x^2 - 10 = 0 by fixed point iteration with Aitken's acceleration.
* g(x) = (10/(x+4))^{1/2}
*
*   INPUT:   initial approximation p0; tolerance TOL;
*            maximum number of iterations N0.
*
*   OUTPUT:  approximate solution p or
*            a message that the method fails.
*/

#include<stdio.h>
#include<math.h> 
static   void OUTPUT(int *);
static   double G(double); 
#define ZERO 1.0E-20
#define true 1
#define false 0

main()
{
      double TOL,P0,P1,P2,P,D;
      int I,NO,FLAG,OK;
      char AA;

      P0 = 1.5; 
      TOL = 1.0e-11;
      NO = 200;
      OK = true; 
   
      OUTPUT(&FLAG);

      /* STEP 1 */ 
      I = 1; OK = true;         
      /* STEP 2 */ 
      while((I<=NO) && OK) {

	 /* STEP 3 */ 
	 /* compute P(1) with superscript (I-1) */      
	 P1 = G(P0);
	 /* compute P(2) with superscript (I-1) */      
	 P2 = G(P1);
	 if (fabs(P2-2.0*P1+P0) < ZERO) {
	    FLAG = 1;
	    D = 10.0;
	    printf("Denominator = 0, method fails\n");
	    printf("best possible is P2(%2d) = %15.10f\n",I,P2);
	    OK = false;
	 }
	 else {
	    D = (P1 - P0) * (P1 - P0) / (P2 - 2.0 * P1 + P0);
	 }
	 /* compute P(0) with superscript (I - 1) */
	 P = P0 - D;
	 if (FLAG == 2) printf("%3d   %15.10e\n", I, P);
	 /* STEP 4 */ 
	 if (fabs(D) < TOL) { 
	    /* procedure completed successfully */
	    printf("\nApproximate solution = %12.10f\n", P);
	    printf("Number of iterations = %3d", I);
	    printf("    Tolerance = %15.10e\n",TOL);
	    OK = false;
	 }
	 else {
	    /* STEP 5 */ 
	    I++;        
	    /* STEP 6 */ 
	    /* update P0 */ 
	    P0 = P; 
	 }
      }
      if (OK) {
	 /* STEP 7 */ 
	 /* procedure completed unsuccessfully */ 
	 printf("\nIteration number %3d", NO);
	 printf(" gave approximation %12.10f\n", P);
	 printf("not within tolerance %14e\n",TOL);
      }
}

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

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

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\n");
   scanf("%d", FLAG);
   printf("STEFFENSENS METHOD\n"); 
   if (*FLAG == 2) printf("  I                 P\n");
}