[Haskell-beginners] Factorials using foldx
ajb at spamcop.net
ajb at spamcop.net
Fri Aug 8 01:00:23 EDT 2008
G'day all.
Quoting Daniel Fischer <daniel.is.fischer at web.de>:
> facr will build a thunk of the form
> 1 * (2 * (3 * (4 * (... * (n * 1) ...)))),
[...]
> facl will build a thunk of the form
> (...(((1 * 1) * 2) * 3) * ...) * n, which will also blow the stack if n is
> large enough.
[...]
> You can get that behaviour without relying on the optimiser by using the
> strict left fold
>
> foldl'
>
> from Data.List, which forces evaluation at each step.
There's one more possibility you should be aware of, assume you're
trying to compute large factorials, and that's to use a binary
tree-style recursion pattern. This one is bottom-up, but you could
also do top-down:
<<
{-# SPECIALIZE binaryProduct :: [Integer] -> Integer #-}
binaryProduct :: (Num a) => [a] -> a
binaryProduct [] = 1
binaryProduct [x] = x
binaryProduct xs
= binaryProduct (bip' xs)
where
bip' (x1:x2:xs) = x1*x2 : bip' xs
bip' xs = xs
fac1 :: Integer -> Integer
fac1 n = foldl' (*) 1 [1..n]
fac2 :: Integer -> Integer
fac2 n = binaryProduct [1..n]
>>
The speed difference for large n is remarkable:
<<
Factorial> fac1 100000 == fac2 100000
True
Factorial> :set +s
Factorial> fac1 100000 `seq` ()
()
(6.27 secs, 9304407772 bytes)
Factorial> fac2 100000 `seq` ()
()
(0.31 secs, 20527124 bytes)
>>
(The `seq` () idiom, by the way, forces computation of the result
without incurring the expense of calling "show". The "show" function
on Integers is quite slow for very large numbers, and the factorial
of 100,000 is a very large number.)
As you can probably guess, the speedup is not due to lazy evaluation.
A good exercise for bright students who know something about computer
arithmetic: Why is the binary tree-style version so much faster?
Cheers,
Andrew Bromage
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