The Fibonacci sequence
Implementing the Fibonacci sequence is considered the "Hello, world!" of Haskell programming. This page collects Haskell implementations of the sequence.
fib 0 = 0 fib 1 = 1 fib n = fib (n-1) + fib (n-2)
One can compute the first n Fibonacci numbers with O(n) additions.
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)
fibs = fix ((0:) . scanl (+) 1)
fibs = unfoldr (\(f1,f2) -> Just (f1,(f2,f1+f2))) (0,1)
fibs = map fst $ iterate (\(f1,f2) -> (f2,f1+f2)) (0,1)
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.
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)
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
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))
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.
Using Binet's formula
fib n = round $ phi ** fromIntegral n / sq5 where sq5 = sqrt 5 :: Double phi = (1 + sq5) / 2
- Discussion at haskell cafe
- Some other nice solutions
- In 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.):