> fact2 is O(log n) while fact is O(n).
> fact2 is partitioning the problem.
But fact2 sparks off two recursive calls. If we assume that
multiplication is a constant-time operations, both algorithms
have the same asymptotic running time (try a version that
operates on Ints). For Integers, this assumption is, of course,
> On Wed, Dec 30, 2009 at 08:57, Artyom Kazak <[email protected]> wrote:
> > Why fact2 is quicker than fact?!
> > fact2 :: Integer -> Integer
> > fact2 x = f x y
> > where
> > f n e | n < 2 = 1
> > | e == 0 = n * (n - 1)
> > | e > 0 = (f n (e `div` 2)) * (f (n - (e * 2)) (e `div` 2))
> > y = 2 ^ (truncate (log (fromInteger x) / log 2))
> > fact :: Integer -> Integer
> > fact 1 = 1
> > fact n = n * fact (n - 1)
> > I tried to write tail-recursive fact, fact as "product [1..n]" - fact2 is
> > quicker!
> > fact2 1000000 == fact 1000000 - I tested.
> > _______________________________________________
> > Haskell-Cafe mailing list
> > [email protected]
> > http://www.haskell.org/mailman/listinfo/haskell-cafe
Haskell-Cafe mailing list