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The Fibonacci sequence

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* [http://comments.gmane.org/gmane.comp.lang.haskell.cafe/19623 Discussion at haskell cafe]
 
* [http://comments.gmane.org/gmane.comp.lang.haskell.cafe/19623 Discussion at haskell cafe]
 
* [http://www.cubbi.org/serious/fibonacci/haskell.html Some other nice solutions]
 
* [http://www.cubbi.org/serious/fibonacci/haskell.html Some other nice solutions]
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* In [http://projecteuler.net/ Project Euler], some of the problems involve Fibonacci numbers. There are some solutions in Haskell ('''Spoiler Warning:''' Do not look at solutions to Project Euler problems until you have solved the problems on your own.):
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** [[Euler_problems/1_to_10#Problem_2|Problem 2]]
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** [[Euler_problems/21_to_30#Problem_25|Problem 25]]
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** [[Euler_problems/101_to_110#Problem_104|Problem 104]]
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** [[Euler_problems/131_to_140#Problem_137|Problem 137]]
   
 
[[Category:Code]]
 
[[Category:Code]]

Revision as of 10:46, 6 November 2007

Implementing the Fibonacci sequence is considered the "Hello, world!" of Haskell programming. This page collects Haskell implementations of the sequence.

Contents

1 Naive definition

fib 0 = 0
fib 1 = 1
fib n = fib (n-1) + fib (n-2)

2 Linear-time implementations

One can compute the first n Fibonacci numbers with O(n) additions.

If
fibs
is the infinite list of Fibonacci numbers, one can define
fib n = fibs!!n

2.1 Canonical zipWith implementation

fibs = 0 : 1 : zipWith (+) fibs (tail fibs)

2.2 With scanl

fibs = fix ((0:) . scanl (+) 1)

2.3 With unfoldr

fibs = unfoldr (\(f1,f2) -> Just (f1,(f2,f1+f2))) (0,1)

2.4 With iterate

fibs = map fst $ iterate (\(f1,f2) -> (f2,f1+f2)) (0,1)

3 Log-time implementations

3.1 Using 2x2 matrices

The argument of
iterate
above is a linear transformation,

so we can represent it as matrix and compute the nth power of this matrix with O(log n) multiplications and additions. For example, using the simple matrix implementation in Prelude extensions,

fib n = head (apply (Matrix [[0,1], [1,1]] ^ n) [0,1])

This technique works for any linear recurrence.

3.2 A fairly fast version, using some identities

fib 0 = 0
fib 1 = 1
fib n | even n         = f1 * (f1 + 2 * f2)
      | n `mod` 4 == 1 = (2 * f1 + f2) * (2 * f1 - f2) + 2
      | otherwise      = (2 * f1 + f2) * (2 * f1 - f2) - 2
   where k = n `div` 2
         f1 = fib k
         f2 = fib (k-1)

3.3 Another fast fib

fib = fst . fib2
 
-- | Return (fib n, fib (n + 1))
fib2 0 = (1, 1)
fib2 1 = (1, 2)
fib2 n
 | even n    = (a*a + b*b, c*c - a*a)
 | otherwise = (c*c - a*a, b*b + c*c)
 where (a,b) = fib2 (n `div` 2 - 1)
       c     = a + b

3.4 Fastest Fib in the West

This was contributed by wli (It assumes that the sequence starts with 1.)

import Data.List
 
fib1 n = snd . foldl fib' (1, 0) . map (toEnum . fromIntegral) $ unfoldl divs n
    where
        unfoldl f x = case f x of
                Nothing     -> []
                Just (u, v) -> unfoldl f v ++ [u]
 
        divs 0 = Nothing
        divs k = Just (uncurry (flip (,)) (k `divMod` 2))
 
        fib' (f, g) p
            | p         = (f*(f+2*g), f^2 + g^2)
            | otherwise = (f^2+g^2,   g*(2*f-g))

4 Constant-time implementations

The Fibonacci numbers can be computed in constant time using Binet's formula. However, that only works well within the range of floating-point numbers available on your platform.

Beyond that, you can use unlimited-precision floating-point numbers, but the result will probably not be any better than the log-time implementations above.

4.1 Using Binet's formula

fib n = round $ phi ** fromIntegral n / sq5
  where
    sq5 = sqrt 5 :: Double
    phi = (1 + sq5) / 2

5 See also