/* CS 1711B Class; 1998/10/2

To find the square root of a positive number A by iteration of
x_n = x_(n-1)/2 + A/(2*x_(n-1).

file: iter-rt.cpp
*/

#include <iostream.h>
#include <math.h>
#include <stdlib.h>


main()
{ double A,
		 xzero,     // initial estimate
		 tolerance,
		 xnow, xprev;  // intermediate values

  int    max_iter, // maximum and counts of iterations
		 iter;

  cout << endl << endl;
  cout << "Approximates the square root of A from xzero within tolerance\n";
  cout << "using at most max iterations\n\n";
  cout << "Input A, xzero, tolerance, max iterations>" << endl;
  cin >> A>> xzero>> tolerance>> max_iter;

  // bad input trap
  while ( (A < 0.0) || (xzero < 0.0) || (tolerance < 0.0) || (max_iter < 0))
  {
	cout << "All values must be positive, try again!\n";
	cout << "Input A, xzero, tolerance, max iters>" << endl;
	cin >> A>> xzero>> tolerance>> max_iter;
  }


  iter = 0;
  xprev = xzero;
  xnow = xzero;


  while ( (fabs(A - xnow*xnow) >= tolerance) && (iter < max_iter))
   { iter++;
	 xnow = xprev/2.0 + A/(2.0*xprev);
	 xprev = xnow;
   }

  if ( iter >= max_iter )
   cout << "\nFailure: too many iterations used, but ..." << endl;

  cout << "\nThe calculated approximation after " << iter << " iterations " <<endl;
  cout <<  "is " << xnow << endl << endl;
  cout << "Its square is " << xnow *xnow << " which differs from " << A ;
  cout << " by " << fabs(A - xnow*xnow) << endl ;


  return(0);
}