The Golden Ratio

July 10, 2009

I was helping my daughter with her math homework recently, and came across this problem in continued fractions:

1 + \frac 1 { 1 + \frac 1 { 1 + \frac 1 { 1 + \frac 1 { 1 + \frac 1 \ldots } } } }

This fraction can be considered as a sequence of terms

G_0 = 1

G_1 = 1 + { 1 \over 1 } = 2

G_2 = 1 + { 1 \over 1 + { 1 \over 1 } } = { 3 \over 2 }

G_3 = 1 + { 1 \over 1 + { 1 \over 1 + { 1 \over 1 } } } = { 5 \over 3 }

Or, in general

G_{n+1} = 1 + \frac 1 { G_n }

The first ten elements of the sequence are 1, 2, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, and 89/55.

Your task is to write a program that evaluates the nth element of the sequence. What is the value of G_{200}? When you are finished, you may read or run a suggested solution, or post your own solution or discuss the exercise in the comments below.

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Posted by programmingpraxis
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22 Comments »

22 Responses to “The Golden Ratio”

  1. Programming Praxis – The Golden Ratio « Bonsai Code said
    July 10, 2009 at 11:09 AM

    […] Praxis – The Golden Ratio By Remco Niemeijer Today’s Programming Praxis problem is an easy one: all we have to do is make a function that calculates the […]

  2. Remco Niemeijer said
    July 10, 2009 at 11:09 AM

    My Haskell solution (see http://bonsaicode.wordpress.com/2009/07/10/programming-praxis-the-golden-ratio/ for a version with comments):

    import Data.Ratio

golden :: Int -> Rational
golden n = iterate (succ . recip) 1 !! n

main :: IO ()
main = print $ golden 200
  3. Matías Giovannini said
    July 10, 2009 at 12:02 PM

    Regular continued fractions have the nice property that they can be efficiently computed in the forward relation by a well-known recurrence:

    let cf f n =
	let module M = struct
		open Num
		let zero, one = Int 0, Int 1
		let rec go h k h' k' i =
			if i = n then string_of_num (h // k) else
			let x = num_of_int (f i) in
			go h' k' (h +/ h' */ x) (k +/ k' */ x) (succ i)
	end in
	M.go M.zero M.one M.one M.zero (-2)

    In this case, phi = cf (fun _ -> 1)

  4. brian said
    July 10, 2009 at 12:21 PM 
    import Data.Ratio ((%))

main :: IO ()
main = print (fromRational (g !! 200) :: Double)

g :: [Rational]
g = zipWith (%) (drop 2 fibs) (tail fibs)

fibs :: [Integer]
fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
  5. Connochaetes said
    July 10, 2009 at 2:49 PM

    Ruby:

    
def golden_mean(n)
  g = 1.0
  n.times { g = 1 + 1/g}
  return g
end

puts golden_mean(200)
  6. IChrisI said
    July 10, 2009 at 7:21 PM

    Python’s native bigint support makes this easy.

    >>> def golden(num):
... 	g = [1,1]
... 	for i in xrange(num):
... 		g.reverse() # calculate 1/g
... 		g[0] += g[1] # add 1
... 	return tuple(g)

>>> print '%i / %i' % golden(200)
  7. Darren said
    July 10, 2009 at 9:35 PM

    Couple of quick Clojure solutions, first a simple recursive solution:

    (defn g [n]
    (if (zero? n) 1
    (inc (/ 1 (g (dec n))))))

    And a more space efficient iterative solution:

    (defn g [n]
    (loop [x 1 n n]
    (if (zero? n) x
    (recur (inc (/ 1 x)) (dec n)))))

  8. Darren said
    July 10, 2009 at 9:46 PM

    Ooops, sorry I hadn’t read the “HOWTO” on posting source:

    [ EDIT: Clojure is not one of the supported languages. I changed the sourcecode tag to css. ]

    Recursive:

    (defn g [n]
        (if (zero? n) 1 
            (inc (g (dec n)))))

    Iterative:

    (defn g [n]
        (loop [x 1 n n]
          (if (zero? n) x
              (recur (inc (/ 1 x)) (dec n)))))
  9. Anne said
    July 10, 2009 at 11:41 PM

    Since this was posted for my math class (I’m the daughter mentioned in the problem), I was just wondering if anyone here can solve it without using computer programming. The reason I ask is that some interesting mathematical constants start to appear as you solve using algebra. If you don’t remember high school algebra, then solving the first ten numbers by had and leaving the fractions in improper form will reveal some sequences as well. That’s what the homework was actually designed to teach us. Just a thought.

  11. Chris said
    July 11, 2009 at 1:33 PM

    A little ruby solution…

    fib = Hash.new {|h,x| x == 0 ? h[x] = 0 : x == 1 ? h[x] = 1 : h[x] = h[x-1] + h[x-2]}
    g = lambda {|x| [fib[x+2],fib[x+1]]}
    puts g[200].join(‘ / ‘)

  12. jcs said
    July 11, 2009 at 5:12 PM

    Anne,

    As you noticed, the numerators and denominators of the terms are Fibonacci numbers. That’s because there is an intimate relationship between the Fibonacci numbers and the Golden Ratio. In particular, the ratio of n+1st and nth Fibonacci numbers tend to the Golden Ratio as n gets large. The same thing happens for the continued fraction as your dad illustrated. See http://www.friesian.com/golden.htm for an explanation of this that should be accessible to anyone with a little algebra background.

  13. Mark VandeWettering said
    July 12, 2009 at 11:14 PM

    Pretty straightforward in Python. Simultaneous assignment makes it look pretty.

    #!/usr/bin/env python
# compute approximations to the golden ratio using the Fibonacci sequence
# https://programmingpraxis.com/2009/07/10/the-golden-ratio/

def golden(n):
    a, b = 1, 1
    for x in range(n):
        a, b = a+b, a
    return (a, b)

print golden(200)
  14. Scott Pedersen said
    July 16, 2009 at 1:26 AM

    Before I really thought about the problem very much I tried coding it up in C#, which was easy enough. It surprised me when it overflowed a 64-bit unsigned integer. The reason why it caused an overflow became obvious once I took a look at the comments here. The code is just calculating a Fibonacci number. G(n) = F(n+2)/F(n+1). A true solution would require a bigint implementation. Without resorting to that, I rewrote the code using doubles. I thought it was interesting to discover that after G(40), the error in the approximation of the golden ratio became too small to store in a double.

    http://inscrutable.pastebin.com/f2dc90711

  15. Eric H said
    July 18, 2009 at 4:22 PM

    PLT Scheme:

    (define/contract (continued-fraction-term N)
    (-> (and/c (not/c negative?)
    integer?)
    number?)
    (if (zero? N)
    1.
    (add1 (/ (continued-fraction-term (sub1 N))))))

  16. veer said
    July 19, 2009 at 3:29 PM

    Trying to get hold of prolog :)

    g(0,1).
g(N,G) :-
	N > 0,M is N-1,
	g(M,H),
	G is 1+rdiv(1,H).


g1(0,1).
g1(N,G) :-
	N > 0,M is N-1,
	g1(M,H),
	G is 1+1/H.
  17. jcs said
    July 20, 2009 at 3:23 PM

    A solution in O(log n) time

    It’s nice to see this thread still active. Here’s a solution that runs
    in log n time. It’s based on Anne’s observation that Gn = Fibn+1
    / Fibn (easily shown by a simple induction) and the fast-fib routine
    from Exercise 1.19 of SICP.

    (define golden-ratio
  (lambda (n)
    (let iter ((n n) (p 0) (q 1) (a 1) (b 1))
        (cond ((= n 0) (/ a b))
              ((even? n) (iter (/ n 2) (+ (* p p) (* q q))
                               (+ (* 2 p q) (* q q)) a b))
              (else (iter (- n 1) p q (+ (* b q) (* a q) (* a p))
                          (+ (* b p) (* a q))))))))
  18. daelin said
    October 1, 2009 at 8:00 PM 
    -- Golden ratio as a recursive fraction
phiRatio :: Int -> Double
phiRatio n
    | n > 0 =  1 + ( 1 / phiRatio (n - 1) )
    | otherwise = 1

-- Golden ratio as a recursive square root
phiRoot n
    | n > 0 = sqrt ( 1 + phiRoot (n - 1) )
    | otherwise = 1

    I also did one that uses the Fibonacci sequence:

    fib :: (Num a, Num b) => a -> b
fib n = fibGen 0 1 n

fibGen :: (Num a, Num b) => b -> b -> a -> b
fibGen a b n = case n of
                 0 -> a
                 n -> fibGen b (a + b) (n - 1)

fib :: (Num a, Num b) => a -> b
fib n = fibGen 0 1 n

fibGen :: (Num a, Num b) => b -> b -> a -> b
fibGen a b n = case n of
                 0 -> a
                 n -> fibGen b (a + b) (n - 1)
  19. Uros Dimitrijevic said
    November 22, 2009 at 3:38 AM

    As for Clojure, there’s an even shorter solution than Darren’s:

    (def G (iterate (comp inc /) 1)) ;; G is a lazy sequence of Ratios

(def phi (double (nth G 200)))   ;; 1.618033988749895, (an approximation of) the golden ratio!

    We could have also BigDecimals, for more precision, like so:

    (with-precision 100
  (bigdec (nth G 1000)))
;; Result: 1.6180339887498948482045868343656381177203091798058M
  20. Graham said
    May 19, 2011 at 5:59 PM

    I’m working my way through old exercises in an effort to learn Common Lisp.

    (defun golden-iter (n)
  (loop for i from 0 below n with g = 1
        do (setf g (1+ (/ g)))
        finally (return g)))

(format t "~S~%" (golden-iter 200))

(defun golden-rec (n)
  (if (zerop n)
      1
      (1+ (/ (golden-rec (1- n))))))

(format t "~S~%" (golden-rec 200))
  21. Heath said
    October 28, 2012 at 6:25 PM

    (defun fibonacci (n &optional (a 0) (b 1))
    “Accepts a number and fetches the corresponding number in the Fibonacci sequence”
    (if (< n 2)
    (* n b)
    (fibonacci (- n 1) b (+ a b))))

    (defun goldenr (n)
    "Accepts a number and approximates the golden ratio to that step of certainty"
    (/ (fibonacci n) (fibonacci (- n 1))))

  22. Steven said
    November 5, 2014 at 4:44 PM

    Haskell, using the definitions supplied in the top portion (G0 = 1, G1 = 2 … )

    golden :: Int -> Rational
    golden 0 = 1
    golden 1 = 2
    golden n = 1 + (1 / (golden (n-1)))

    OUTPUT:
    ———–
    ghci> golden 200
    734544867157818093234908902110449296423351 % 453973694165307953197296969697410619233826

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