Golden ratio/Convergence

The golden ratio can be defined as the continued fraction

\phi =1+{1 \over {1+{1 \over {1+{1 \over {1+\cdots }}}}}}

Thus $\phi =1+{1/\phi }$ . Multiplying both sides by $\phi$ and solving the resulting quadratic equation for its positive solution, one gets $\phi =(1+{\sqrt {5}})/2\approx 1.61803398875$ .

The golden ratio has the slowest convergence of any continued fraction, as one might guess by noting that the denominators are made of the smallest positive integer. This task treats the problem of convergence in a somewhat backwards fashion: we are going to iterate the recursion $\phi _{n+1}=1+{1/\phi _{n}}$ , $\phi _{0}=1$ , and see how long it takes to get an answer.

Task

Iterate $\phi _{n+1}=1+{1/\phi _{n}}$ , $\phi _{0}=1$ , until $|\phi _{n+1}-\phi _{n}|\leq 1\times 10^{-5}$ . Report the final value of $\phi _{n+1}$ , the number of iterations required, and the error with respect to $(1+{\sqrt {5}})/2$ .

ATS

#include "share/atspre_staload.hats"

%{^
#include <math.h>
%}

extern fn sqrt : double -<> double = "mac#sqrt"

fun
iterate {n   : nat}
        (phi : double,
         n   : int n) : @(double, intGte 1) =
  let
    val phi1 = 1.0 + (1.0 / phi)
    and n1 = succ n
  in
    if abs (phi1 - phi) <= 1.0e-5 then
      @(phi1, n1)
    else
      iterate {n + 1} (phi1, n1)
  end

implement
main0 () =
  let
    val @(phi, n) = iterate {0} (1.0, 0)
    val _ = $extfcall (int, "printf",
                       "Result: %.10f after %d iterations\n",
                       phi, n)
    val _ = $extfcall (int, "printf",
                       "The error is approximately %.10f\n",
                       phi - (0.5 * (1.0 + sqrt (5.0))))
  in
  end

Output:

Result: 1.6180327869 after 14 iterations
The error is approximately -0.0000012019

C

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

static void
using_float ()                  /* C2x does not require "void". */
{
  int count = 0;
  float phi0 = 1.0f;
  float phi1;
  float difference;
  do
    {
      phi1 = 1.0f + (1.0f / phi0);
      difference = fabsf (phi1 - phi0);
      phi0 = phi1;
      count += 1;
    }
  while (1.0e-5f < difference);

  printf ("Using type float --\n");
  printf ("Result: %f after %d iterations\n", phi1, count);
  printf ("The error is approximately %f\n",
          phi1 - (0.5f * (1.0f + sqrtf (5.0f))));
}

static void
using_double ()                 /* C2x does not require "void". */
{
  int count = 0;
  double phi0 = 1.0;
  double phi1;
  double difference;
  do
    {
      phi1 = 1.0 + (1.0 / phi0);
      difference = fabs (phi1 - phi0);
      phi0 = phi1;
      count += 1;
    }
  while (1.0e-5 < difference);

  printf ("Using type double --\n");
  printf ("Result: %f after %d iterations\n", phi1, count);
  printf ("The error is approximately %f\n",
          phi1 - (0.5 * (1.0 + sqrt (5.0))));
}

int
main ()                         /* C2x does not require "void". */
{
  using_float ();
  printf ("\n");
  using_double ();
}

Output:

Using type float --
Result: 1.618033 after 14 iterations
The error is approximately -0.000001

Using type double --
Result: 1.618033 after 14 iterations
The error is approximately -0.000001

Common Lisp

Translation of: Scheme

Note that, although standard Scheme guarantees a tail recursion will act like a GOTO rather than an ordinary subroutine call, Common Lisp does not. Therefore this implementation, translated from the Scheme, may require stack space. The amount will be very little.

(You could use this recursive method in C and many other languages where tail recursions are not guaranteed to be "proper". The compiler's optimizer may or may not turn the tail call into a GOTO.)

(defun iterate (phi n)
  ;; This is a tail recursive definition, copied from the
  ;; Scheme. Common Lisp does not guarantee proper tail calls, but the
  ;; depth of recursion will not be too great.
  (let ((phi1 (1+ (/ phi)))
        (n1 (1+ n)))
    (if (<= (abs (- phi1 phi)) 1/100000)
        (values phi1 n1)
        (iterate phi1 n1))))

(multiple-value-bind (phi n) (iterate 1 0)
  (princ "Result: ")
  (princ phi)
  (princ " (")
  (princ (* 1.0 phi))
  (princ ") after ")
  (princ n)
  (princ " iterations")
  (terpri)
  (princ "The error is approximately ")
  (princ (- phi (* 0.5 (+ 1.0 (sqrt 5.0)))))
  (terpri))

Output:

Result: 987/610 (1.6180328) after 14 iterations
The error is approximately -1.1920929e-6

EasyLang

phi0 = 1
repeat
   phi = 1 + 1 / phi0
   until abs (phi - phi0) < 1e-5
   phi0 = phi
   iter += 1
.
numfmt 10 0
print "Iterations: " & iter
print "Result: " & phi
print "Error: " & phi - (1 + sqrt 5) / 2

Mercury

Translation of: Scheme

%%% -*- mode: mercury; prolog-indent-width: 2; -*-

:- module golden_ratio_convergence_mercury.

:- interface.
:- import_module io.
:- pred main(io::di, io::uo) is det.

:- implementation.
:- import_module float.
:- import_module int.
:- import_module math.
:- import_module string.

:- pred iterate(float::in, float::out, int::in, int::out) is det.
iterate(!Phi, !N) :-
  Phi1 = (1.0 + (1.0 / !.Phi)),
  N1 = !.N + 1,
  (if (abs(Phi1 - !.Phi) =< 1.0e-5)
   then (!:Phi = Phi1, !:N = N1)
   else (iterate(Phi1, !:Phi, N1, !:N))).

main(!IO) :-
  iterate(1.0, Phi, 0, N),
  print("Result: ", !IO),
  print(from_float(Phi), !IO),
  print(" after ", !IO),
  print(from_int(N), !IO),
  print(" iterations", !IO),
  nl(!IO),
  print("The error is approximately ", !IO),
  print(from_float(Phi - (0.5 * (1.0 + (sqrt(5.0))))), !IO),
  nl(!IO).

:- end_module golden_ratio_convergence_mercury.

Output:

Result: 1.6180327868852458 after 14 iterations
The error is approximately -1.2018646491362972e-06

ObjectIcon

import io, util

procedure main ()
  local phi0, phi1, count

  count := 1
  phi0 := 1.0
  while abs ((phi1 := 1.0 + (1.0 / phi0)) - phi0) > 1.0e-5 do
  {
    phi0 := phi1
    count +:= 1
  }
  io.write ("Result: ", phi1, " after ", count, " iterations")
  io.write ("The error is approximately ",
            phi1 - (0.5 * (1.0 + Math.sqrt (5.0))))
end

Output:

Result: 1.618032787 after 14 iterations
The error is approximately -1.201864649e-06

RATFOR

program grconv
  integer count
  real phi0, phi1, diff

  count = 0
  phi0 = 1.0
  diff = 1e+20
  while (1e-5 < diff)
    {
      phi1 = 1.0 + (1.0 / phi0)
      diff = abs (phi1 - phi0)
      phi0 = phi1
      count = count + 1
    }

  write (*,'("Result:", F9.6, " after", I3, " iterations")') _
    phi1, count
  write (*,'("The error is approximately ", F9.6)') _
          phi1 - (0.5 * (1.0 + sqrt (5.0)))
end

Output:

Result: 1.618033 after 14 iterations
The error is approximately -0.000001

Scheme

Translation of: ATS

This will run without modification in R5RS Scheme implementations, among others.

(define (iterate phi n)
  (let ((phi1 (+ 1 (/ phi)))
        (n1 (+ n 1)))
    (if (<= (abs (- phi1 phi)) 1/100000)
        (values phi1 n1)
        (iterate phi1 n1))))

(call-with-values (lambda () (iterate 1 0))
  (lambda (phi n)
    (display "Result: ")
    (display phi)
    (display " (")
    (display (* 1.0 phi))
    (display ") after ")
    (display n)
    (display " iterations")
    (newline)
    (display "The error is approximately ")
    (display (- phi (* 0.5 (+ 1.0 (sqrt 5.0)))))
    (newline)))

Output:

Result: 987/610 (1.618032786885246) after 14 iterations
The error is approximately -1.2018646489142526e-6

Task

See also