Difference between revisions of "The Fibonacci sequence"
(Another fast fib) 
(→Fastest Fib in the West: primed names cause bad syntax highlighting) 

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−  Implementing the [http://en.wikipedia.org/wiki/Fibonacci_number 
+  Implementing the [http://en.wikipedia.org/wiki/Fibonacci_number Fibonacci sequence] is considered the "Hello, world!" of Haskell programming. This page collects Haskell implementations of the sequence. 
−  == Naive 
+  == Naive definition == 
+  The standard definition can be expressed directly: 

<haskell> 
<haskell> 

fib 0 = 0 
fib 0 = 0 

Line 8:  Line 9:  
fib n = fib (n1) + fib (n2) 
fib n = fib (n1) + fib (n2) 

</haskell> 
</haskell> 

+  This implementation requires ''O(fib n)'' additions. 

−  == Canonical zipWith implementation == 

+  == Linear operation implementations == 

+  
+  === With state === 

+  Haskell translation of python algo 

<haskell> 
<haskell> 

−  fib = 1 : 1 : zipWith (+) fib (tail fib) 

+  { def fib(n): 

+  a, b = 0, 1 

+  for _ in xrange(n): 

+  a, b = b, a + b 

+  return a } 

</haskell> 
</haskell> 

−  == 
+  ==== Tail recursive ==== 
+  Using accumulator argument for state passing 

<haskell> 
<haskell> 

−  fib = fix ((1:) . scanl (+) 1) 

+  {# LANGUAGE BangPatterns #} 

+  fib n = go n (0,1) 

+  where 

+  go !n (!a, !b)  n==0 = a 

+   otherwise = go (n1) (b, a+b) 

</haskell> 
</haskell> 

−  == 
+  ==== Monadic ==== 
+  <haskell> 

+  import Control.Monad.State 

+  fib n = flip evalState (0,1) $ do 

+  forM [0..(n1)] $ \_ > do 

+  (a,b) < get 

+  put (b,a+b) 

+  (a,b) < get 

+  return a 

+  </haskell> 

+  
+  === Using the infinite list of Fibonacci numbers === 

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

+  If <hask>fibs</hask> is the infinite list of Fibonacci numbers, one can define 

+  <haskell> 

+  fib n = fibs!!n 

+  </haskell> 

+  
+  ==== Canonical zipWith implementation ==== 

+  
+  <haskell> 

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

+  </haskell> 

+  
+  ==== With direct selfreference ==== 

+  
+  <haskell> 

+  fibs = 0 : 1 : next fibs 

+  where 

+  next (a : t@(b:_)) = (a+b) : next t 

+  </haskell> 

+  
+  ==== With scanl ==== 

+  
+  <haskell> 

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

+  fibs = 0 : scanl (+) 1 fibs 

+  </haskell> 

+  The recursion can be replaced with <hask>fix</hask>: 

+  <haskell> 

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

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

+  </haskell> 

+  
+  The <code>fix</code> used here has to be implemented through sharing, <code>fix f = xs where xs = f xs</code>, not code replication, <code>fix f = f (fix f)</code>, to avoid quadratic behaviour. 

+  
+  ==== With unfoldr ==== 

+  
+  <haskell> 

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

+  </haskell> 

+  
+  ==== With iterate ==== 

<haskell> 
<haskell> 

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

</haskell> 
</haskell> 

−  == A 
+  === A version using some identities === 
<haskell> 
<haskell> 

Line 40:  Line 54:  
</haskell> 
</haskell> 

−  == Another fast fib == 

+  This seems to use <math>O(log^2 n)</math> calls to <code>fib</code>. 

+  == Logarithmic operation implementations == 

+  
+  === Using 2x2 matrices === 

+  
+  The argument of <hask>iterate</hask> above is a [http://en.wikipedia.org/wiki/Linear_map linear transformation], so 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 [[Prelude_extensions#Matricessimple matrix implementation]] in [[Prelude extensions]], 

+  <haskell> 

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

+  </haskell> 

+  This technique works for any linear recurrence. 

+  
+  === Another fast fib === 

+  
+  (Assumes that the sequence starts with 1.) 

<haskell> 
<haskell> 

fib = fst . fib2 
fib = fst . fib2 

Line 55:  Line 83:  
</haskell> 
</haskell> 

−  == Fastest Fib in the West == 
+  === Fastest Fib in the West === 
This was contributed by [http://www.haskell.org/pipermail/haskellcafe/2005January/008839.html wli] 
This was contributed by [http://www.haskell.org/pipermail/haskellcafe/2005January/008839.html wli] 

+  (It assumes that the sequence starts with 1.) 

<haskell> 
<haskell> 

−  import System.Environment 

import Data.List 
import Data.List 

−  +  fib1 n = snd . foldl fib_ (1, 0) . map (toEnum . fromIntegral) $ unfoldl divs n 

where 
where 

unfoldl f x = case f x of 
unfoldl f x = case f x of 

Line 72:  Line 100:  
divs k = Just (uncurry (flip (,)) (k `divMod` 2)) 
divs k = Just (uncurry (flip (,)) (k `divMod` 2)) 

−  +  fib_ (f, g) p 

 p = (f*(f+2*g), f^2 + g^2) 
 p = (f*(f+2*g), f^2 + g^2) 

 otherwise = (f^2+g^2, g*(2*fg)) 
 otherwise = (f^2+g^2, g*(2*fg)) 

+  </haskell> 

−  main = getArgs >>= mapM_ (print . fib . read) 

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

+  <haskell> 

+  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 [s1,s2..0]] 

+  where 

+  fib_ (f, g) p 

+   p = (f*(f+2*g), ss) 

+   otherwise = (ss, g*(2*fg)) 

+  where ss = f*f+g*g 

+  foldl_ = foldl'  ' 

</haskell> 
</haskell> 

−  == See also == 

+  == Constanttime implementations == 

−  Discussion at haskell cafe: 

+  The Fibonacci numbers can be computed in constant time using Binet's formula. 

+  However, that only works well within the range of floatingpoint 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 [http://haskell.org/haskellwiki/Applications_and_libraries/Mathematics#Real_and_rational_numbers unlimitedprecision floatingpoint numbers], 

+  but the result will probably not be any better than the [[#Logtime_implementationslogtime implementations]] above. 

+  
+  === Using Binet's formula === 

+  
+  <haskell> 

+  fib n = round $ phi ** fromIntegral n / sq5 

+  where 

+  sq5 = sqrt 5 :: Double 

+  phi = (1 + sq5) / 2 

+  </haskell> 

+  
+  == Generalization of Fibonacci numbers == 

+  The numbers of the traditional Fibonacci sequence are formed by summing its two preceding numbers, with starting values 0 and 1. Variations of the sequence can be obtained by using different starting values and summing a different number of predecessors. 

+  
+  === Fibonacci ''n''Step Numbers === 

+  The sequence of Fibonacci ''n''step numbers are formed by summing ''n'' predecessors, using (''n''1) zeros and a single 1 as starting values: 

+  <math> 

+  F^{(n)}_k := 

+  \begin{cases} 

+  0 & \mbox{if } 0 \leq k < n1 \\ 

+  1 & \mbox{if } k = n1 \\ 

+  \sum\limits_{i=kn}^{(k1)} F^{(n)}_i & \mbox{otherwise} \\ 

+  \end{cases} 

+  </math> 

+  
+  Note that the summation in the current definition has a time complexity of ''O(n)'', assuming we memoize previously computed numbers of the sequence. We can do better than. Observe that in the following Tribonacci sequence, we compute the number 81 by summing up 13, 24 and 44: 

+  
+  <math> 

+  F^{(3)} = 0,0,1,1,2,4,7,\underbrace{13,24,44},81,149, \ldots 

+  </math> 

+  
+  The number 149 is computed in a similar way, but can also be computed as follows: 

+  
+  <math> 

+  149 = 24 + 44 + 81 = (13 + 24 + 44) + 81  13 = 81 + 81  13 = 2\cdot 81  13 

+  </math> 

+  
+  And hence, an equivalent definition of the Fibonacci ''n''step numbers sequence is: 

+  
+  <math> 

+  F^{(n)}_k := 

+  \begin{cases} 

+  0 & \mbox{if } 0 \leq k < n1 \\ 

+  1 & \mbox{if } k = n1 \\ 

+  1 & \mbox{if } k = n \\ 

+  F^{(n)}_k := 2F^{(n)}_{k1}F^{(n)}_{k(n+1)} & \mbox{otherwise} \\ 

+  \end{cases} 

+  </math> 

+  
+  ''(Notice the extra case that is needed)'' 

+  
+  Transforming this directly into Haskell gives us: 

+  <haskell> 

+  nfibs n = replicate (n1) 0 ++ 1 : 1 : 

+  zipWith (\b a > 2*ba) (drop n (nfibs n)) (nfibs n) 

+  </haskell> 

+  This version, however, is slow since the computation of <hask>nfibs n</hask> is not shared. Naming the result using a letbinding and making the lambda pointfree results in: 

+  <haskell> 

+  nfibs n = let r = replicate (n1) 0 ++ 1 : 1 : zipWith (().(2*)) (drop n r) r 

+  in r 

+  </haskell> 

+  
+  
+  == See also == 

−  http://comments.gmane.org/gmane.comp.lang.haskell.cafe/19623 

+  * [http://cgi.cse.unsw.edu.au/~dons/blog/2007/11/29 Naive parallel, multicore version] 

+  * [[Fibonacci primes in parallel]] 

+  * [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] 

+  * 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.): 

+  ** [[Euler_problems/1_to_10#Problem_2Problem 2]] 

+  ** [[Euler_problems/21_to_30#Problem_25Problem 25]] 

+  ** [[Euler_problems/101_to_110#Problem_104Problem 104]] 

+  ** [[Euler_problems/131_to_140#Problem_137Problem 137]] 

[[Category:Code]] 
[[Category:Code]] 
Latest revision as of 17:37, 5 November 2016
Implementing the Fibonacci sequence is considered the "Hello, world!" of Haskell programming. This page collects Haskell implementations of the sequence.
Contents
Naive definition
The standard definition can be expressed directly:
fib 0 = 0
fib 1 = 1
fib n = fib (n1) + fib (n2)
This implementation requires O(fib n) additions.
Linear operation implementations
With state
Haskell translation of python algo
{ def fib(n):
a, b = 0, 1
for _ in xrange(n):
a, b = b, a + b
return a }
Tail recursive
Using accumulator argument for state passing
{# LANGUAGE BangPatterns #}
fib n = go n (0,1)
where
go !n (!a, !b)  n==0 = a
 otherwise = go (n1) (b, a+b)
Monadic
import Control.Monad.State
fib n = flip evalState (0,1) $ do
forM [0..(n1)] $ \_ > do
(a,b) < get
put (b,a+b)
(a,b) < get
return a
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 direct selfreference
fibs = 0 : 1 : next fibs
where
next (a : t@(b:_)) = (a+b) : next t
With scanl
fibs = scanl (+) 0 (1:fibs)
fibs = 0 : scanl (+) 1 fibs
The recursion can be replaced with fix
:
fibs = fix (scanl (+) 0 . (1:))
fibs = fix ((0:) . scanl (+) 1)
The fix
used here has to be implemented through sharing, fix f = xs where xs = f xs
, not code replication, fix f = f (fix f)
, to avoid quadratic behaviour.
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)
A 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 (k1)
This seems to use calls to fib
.
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*fg))
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 [s1,s2..0]]
where
fib_ (f, g) p
 p = (f*(f+2*g), ss)
 otherwise = (ss, g*(2*fg))
where ss = f*f+g*g
foldl_ = foldl'  '
Constanttime implementations
The Fibonacci numbers can be computed in constant time using Binet's formula. However, that only works well within the range of floatingpoint 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 unlimitedprecision floatingpoint numbers, but the result will probably not be any better than the logtime implementations above.
Using Binet's formula
fib n = round $ phi ** fromIntegral n / sq5
where
sq5 = sqrt 5 :: Double
phi = (1 + sq5) / 2
Generalization of Fibonacci numbers
The numbers of the traditional Fibonacci sequence are formed by summing its two preceding numbers, with starting values 0 and 1. Variations of the sequence can be obtained by using different starting values and summing a different number of predecessors.
Fibonacci nStep Numbers
The sequence of Fibonacci nstep numbers are formed by summing n predecessors, using (n1) zeros and a single 1 as starting values:
Note that the summation in the current definition has a time complexity of O(n), assuming we memoize previously computed numbers of the sequence. We can do better than. Observe that in the following Tribonacci sequence, we compute the number 81 by summing up 13, 24 and 44:
The number 149 is computed in a similar way, but can also be computed as follows:
And hence, an equivalent definition of the Fibonacci nstep numbers sequence is:
(Notice the extra case that is needed)
Transforming this directly into Haskell gives us:
nfibs n = replicate (n1) 0 ++ 1 : 1 :
zipWith (\b a > 2*ba) (drop n (nfibs n)) (nfibs n)
This version, however, is slow since the computation of nfibs n
is not shared. Naming the result using a letbinding and making the lambda pointfree results in:
nfibs n = let r = replicate (n1) 0 ++ 1 : 1 : zipWith (().(2*)) (drop n r) r
in r
See also
 Naive parallel, multicore version
 Fibonacci primes in parallel
 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.):