# The Fibonacci sequence

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− | == Linear-time implementations == |
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One can compute the first ''n'' Fibonacci numbers with ''O(n)'' additions. |
One can compute the first ''n'' Fibonacci numbers with ''O(n)'' additions. |
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− | == Log-time implementations == |
+ | == Logarithmic operation implementations == |

=== Using 2x2 matrices === |
=== Using 2x2 matrices === |
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The Fibonacci numbers can be computed in constant time using Binet's formula. |
The Fibonacci numbers can be computed in constant time using Binet's formula. |
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However, that only works well within the range of floating-point numbers |
However, that only works well within the range of floating-point numbers |
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− | available on your platform. |
+ | 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 [http://haskell.org/haskellwiki/Applications_and_libraries/Mathematics#Real_and_rational_numbers unlimited-precision floating-point numbers], |
Beyond that, you can use [http://haskell.org/haskellwiki/Applications_and_libraries/Mathematics#Real_and_rational_numbers unlimited-precision floating-point numbers], |

## Revision as of 14:32, 29 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 operation implementations

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

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 Logarithmic operation implementations

### 3.1 Using 2x2 matrices

The argument ofso we can represent it as matrix and compute the *n*th 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. 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.

### 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

- 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.):