Difference between revisions of "The Fibonacci sequence"

From HaskellWiki
Jump to: navigation, search
("fairly fast version" is actually linear in n)
m (Another fast fib assumes that the sequence starts with 1)
Line 78: Line 78:
 
=== Another fast fib ===
 
=== Another fast fib ===
   
  +
(Assumes that the sequence starts with 1.)
 
<haskell>
 
<haskell>
 
fib = fst . fib2
 
fib = fst . fib2

Revision as of 10:28, 19 August 2008

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

Naive definition

The standard definition can be expressed directly:

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

This implementation requires O(fib n) additions.

Linear operation implementations

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)

Using the infinite list of Fibonacci numbers

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

Canonical zipWith implementation

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

With scanl

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

The recursion can be replaced with fix:

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

With unfoldr

fibs = unfoldr (\(a,b) -> Just (a,(b,a+b))) (0,1)

With iterate

fibs = map fst $ iterate (\(a,b) -> (b,a+b)) (0,1)

Logarithmic operation implementations

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.

Another fast fib

(Assumes that the sequence starts with 1.)

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

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))

An even faster version, given later by wli on the IRC channel.

import Data.List
import Data.Bits

fib :: Int -> Integer
fib n = snd . foldl' fib' (1, 0) $ dropWhile not $
            [testBit n k | k <- let s = bitSize n in [s-1,s-2..0]]
    where
        fib' (f, g) p
            | p         = (f*(f+2*g), ss)
            | otherwise = (ss, g*(2*f-g))
            where ss = f*f+g*g

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. Implementing Binet's formula in such a way that it computes exact results for all integers generally doesn't result in a terribly efficient implementation when compared to the programs above which use a logarithmic number of operations (and work in linear time).

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.

Using Binet's formula

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

See also