# The Fibonacci sequence

(Difference between revisions)

#### 2.2.5 With state

import Control.Monad.State
{-
-- haskell translational of python algo
def fib(n):
a, b = 0, 1
for _ in xrange(n):
a, b = b, a + b
return a
print fib(1000)
-}
fib :: Integer -> Integer
fib n = flip evalState (0,1) $do forM (xrange n)$ \_ -> do
(a,b) <- get
put (b,a+b)
(a,b) <- get
return a

xrange n = [0..(n-1)]

## 3 Logarithmic operation 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 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

### 3.3 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

## 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 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.

### 5.1 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: $F^{(n)}_k := \begin{cases} 0 & \mbox{if } 0 \leq k < n-1 \\ 1 & \mbox{if } k = n-1 \\ \sum\limits_{i=k-n}^{(k-1)} F^{(n)}_i & \mbox{otherwise} \\ \end{cases}$

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:

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

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

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

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

$F^{(n)}_k := \begin{cases} 0 & \mbox{if } 0 \leq k < n-1 \\ 1 & \mbox{if } k = n-1 \\ 1 & \mbox{if } k = n \\ F^{(n)}_k := 2F^{(n)}_{k-1}-F^{(n)}_{k-(n+1)} & \mbox{otherwise} \\ \end{cases}$

(Notice the extra case that is needed)

Transforming this directly into Haskell gives us:

nfibs n = replicate (n-1) 0 ++ 1 : 1 :
zipWith (\b a -> 2*b-a) (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 let-binding and making the lambda pointfree results in:
nfibs n = let r = replicate (n-1) 0 ++ 1 : 1 : zipWith ((-).(2*)) (drop n r) r
in r