[Haskell-cafe] Mysterious factorial
Daniel Fischer
daniel.is.fischer at web.de
Wed Dec 30 11:34:01 EST 2009
Am Mittwoch 30 Dezember 2009 11:57:28 schrieb Artyom Kazak:
> 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.
If you follow the evaluation of fact2, it is basically the same as
fact3 n = binaryMult [1 .. n]
where
binaryMult [p] = p
binaryMult xs = binaryMult (pairMult xs)
pairMult (x:y:zs) = x*y : pairMult zs
pairMult xs = xs
, just without the list construction, but with a few more ones [aside: You should subtract
one from the exponent of y. As it is, in the first call to f, the second factor is always
1 because x < 2*y. Doesn't make much of a difference regarding performance, but it seems
cleaner.]. Perhaps fact3 is a little easier to follow.
Looking at fact3 (2^k), we see that in the first iteration of binaryMult, 2^(k-1) products
of small integers (<= k+1 bits) are carried out. These multiplications are fast.
In the second iteration, we have 2^(k-2) products of still small integers (<= 2*k bits).
These multiplications are a tad slower, but still fast.
In the third iteration, we have 2^(k-3) products of integers of (<= k*2^2 bits) and so on.
We see that the overwhelming majority of the 2^k-1 multiplications carried out don't
involve huge numbers and thus are relatively fast (for k = 32, no multiplication in the
first five iterations involves a number with more than 1000 bits, so no more than 3% of
the multiplications involve large numbers; for k = 20, the first product with more than
100 bits is produced in the sixth iteration, so less than 20000 multiplications involve a
number with more than 1000 bits, less than 1000 multiplications have a factor with more
than 10000 bits, less than 128 have a factor with more than 100000 bits).
If the factorial is computed sequentially, like
fact0 n = foldl' (*) 1 [2 .. n]
-- or product [2 .. n], just don't build huge thunks like in fact above
, you have many multiplications involving one huge number (and one small), since fact k
has of the order of k*log k bits. 1000! has about 8500 bits, 10000! has about 120000 bits
and 100000! has about 1.5 million bits.
So that way, computing the factorial of 2^20 needs more than 990000 multiplications where
one factor has more than 100000 bits and over 900000 multiplications where one factor has
more than one million bits.
Since multiplications where one factor is huge take long, even if the other factor is
small, we see why a sequential computation of a factorial is so much slower than a tree-
like computation.
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