> For a more efficient Haskell implementation of fibonacci numbers, try
>
> fibs :: [Integer]
> fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
>
> fib n = fibs !! n
This is uglier, but just to keep using just plain recursion:
fib = fib' 0 1
where
fib' a b 0 = a
fib' a b n = fib' b (a+b) (n-1)
You may want "fib' a b n | a `seq` b `seq` n `seq` False = undefined"
for strictness if the compiler isn't smart enough to figure out
(sorry, didn't test it).
And, *please* correct me if I said something stupid =).
See ya,
--
Felipe.
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