Here's some code I wrote a while back for computing the nth Fibonacci number. It has O(log n) time complexity rather than O(n). It isn't the most elegant example, but it should be one of the fastest approaches.
import Data.Bits (shiftR, xor, (.|.), (.&.)) import Data.Word (Word32)
fibo :: Word32 -> Integer fibo n = loop (highestBitMask n) 1 0 where loop :: Word32 -> Integer -> Integer -> Integer loop i a b | i == 0 = b | n .&. i /= 0 = (loop (shiftR i 1) $! a*(2*b + a)) $! a*a + b*b | otherwise = (loop (shiftR i 1) $! a*a + b*b) $! b*(2*a - b)
highestBitMask :: Word32 -> Word32 highestBitMask x = case (x .|. shiftR x 1) of x -> case (x .|. shiftR x 2) of x -> case (x .|. shiftR x 4) of x -> case (x .|. shiftR x 8) of x -> case (x .|. shiftR x 16) of x -> (x `xor` (shiftR x 1))
Cheers,
Spencer Janssen
On 6/15/06, Vladimir Portnykh
Fibonacci numbers implementations in Haskell one of the classical examples. An example I found is the following:
fibs :: [Int] fibs = 0 : 1 : [ a + b | (a, b) <- zip fibs (tail fibs)]
To get the k-th number you do the following: Result = fibs !! k
It is elegant but creates a list of all Fibonacci numbers less than k-th, and the code is not very readable :).
I wrote my own Fibonacci numbers generator:
fib :: Int -> [Int] fib 0 = [0,0] fib 1 = [1,0] fib n = [sum prevFib, head prevFib] where a = fib (n - 1)
To get the k-th number you do the following:
result = head (fib k)
It does not generate full list of Fibonacci numbers, but keeps only 2 previous numbers, and has only one recursive call. Because the list always has only 2 elements using the functions head and sum is a bit overkill.
Can we do better?
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