[Haskell-cafe] does the compiler optimize repeated calls?
Brian Hulley
brianh at metamilk.com
Wed Sep 6 13:03:16 EDT 2006
John Hughes wrote:
> The trouble is that this isn't always an optimisation. Try these two
> programs:
>
> powerset [] = [[]]
> powerset (x:xs) = powerset xs++map (x:) (powerset xs)
>
> and
>
> powerset [] = [[]]
> powerset (x:xs) = pxs++map (x:) pxs
> where pxs = powerset xs
>
> Try computing length (powerset [1..n]) with each definition. For small
> n, the second is faster. As n gets larger, the second gets slower and
> slower, but the first keeps chugging along. The problem is that the
> second has exponentially higher peak memory requirements than the
> first. Round about n=25, on my machine, all other programs stop responding
> while the second one runs. You don't really want a compiler to make
> that kind of "pessimisation" to your program, which is why it's a
> deliberate decision to leave most CSE to the programmer. You can
> still write the second version, and suffer the consequences, but at least
> you know
> it's your own fault!
Thanks for the above example. I found it quite difficult to understand why
the second is worse than the first for large n, but I think the reason is
that you're using the second def in conjunction with (length). Therefore it
is the *combination* of the cse'd (powerset) with (length) that is less
efficient, because (length) just reads its input as a stream so there is no
need for the whole of (powerset xs) to exist in memory thus the non cse'd
version gives a faster (length . powerset).
Ideally it would be great if the compiler could make use of the context in
which a function is being applied to produce optimized code across function
boundaries. In the above example of (length . powerset), (length) has no
interest in the contents of the powerset itself so could the compiler not
fuse (length . powerset) into the following function:
lengthPowerset [] = 1
lengthPowerset (x:xs) = 2 * lengthPowerset xs
The compiler would need to analyse the definition of (++) and (map) to
discover that
length (x ++ y) === length x + length y
length (map f y) === length y
and with that knowledge I imagine the steps could be something like:
lengthPowerset [] = length (powerset []) = length ([[]]) = 1
lengthPowerset (x:xs)
= length (powerset xs ++ map (:x) (powerset xs))
= length (powerset xs) + length (map (:x) (powerset xs))
= length (powerset xs) + length (powerset xs)
= lengthPowerset xs + lengthPowerset xs
= 2 * lengthPowerset xs
After getting the function (lengthPowerset) as above, I'd also expect the
compiler to apply a transformation into a tail recursive function:
lengthPowerset y = lengthPowerset' y 1
where
lengthPowerset' [] i = i
lengthPowerset' (_:xs) i = lengthPowerset' xs $! 2*i
resulting in a tightly coded machine code loop to rival, or greatly
exceed(!), the best efforts of C.
In the meantime I tend to code in Haskell just expecting these kind of
optimizations to be done (unless I'm writing a really time-critical piece of
code that can't wait), knowing of course that they might not be done just at
the moment but at least some time in the (hopefully not too distant)
future... ;-)
Regards, Brian.
--
Logic empowers us and Love gives us purpose.
Yet still phantoms restless for eras long past,
congealed in the present in unthought forms,
strive mightily unseen to destroy us.
http://www.metamilk.com
More information about the Haskell-Cafe
mailing list