[Haskell-cafe] Eta Reduction
Alexander Solla
alex.solla at gmail.com
Tue Apr 1 17:32:54 UTC 2014
On Mon, Mar 31, 2014 at 11:54 PM, Dan Doel <dan.doel at gmail.com> wrote:
> In the past year or two, there have been multiple performance problems in
> various areas related to the fact that lambda abstraction is not free,
> though we
> tend to think of it as so. A major example of this was deriving of
> Functor. If we
> were to derive Functor for lists, we would end up with something like:
>
> instance Functor [] where
> fmap _ [] = []
> fmap f (x:xs) = f x : fmap (\y -> f y) xs
>
> This definition is O(n^2) when fully evaluated,, because it causes O(n) eta
> expansions of f, so we spend time following indirections proportional to
> the
> depth of the element in the list. This has been fixed in 7.8, but there are
> other examples. I believe lens, [1] for instance, has some stuff in it that
> works very hard to avoid this sort of cost; and it's not always as easy to
> avoid
> as the above example. Composing with a newtype wrapper, for instance,
> causes an
> eta expansion that can only be seen as such at the core level.
>
> The obvious solution is: do eta reduction. However, this is not
> operationally
> sound currently. The problem is that seq is capable of telling the
> difference
> between the following two expressions:
>
> undefined
> \x -> undefined x
>
> The former causes seq to throw an exception, while the latter is considered
> defined enough to not do so. So, if we eta reduce, we can cause terminating
> programs to diverge if they make use of this feature.
>
> Luckily, there is a solution.
>
> Semantically one would usually identify the above two expressions. While I
> do
> believe one could construct a semantics that does distinguish them, it is
> not
> the usual practice. This suggests that there is a way to not distinguish
> them,
> perhaps even including seq. After all, the specification of seq is
> monotone and
> continuous regardless of whether we unify ⊥ with \x -> ⊥ x or insert an
> extra
> element for the latter.
>
> The currently problematic case is function spaces, so I'll focus on it. How
> should:
>
> seq : (a -> b) -> c -> c
>
> act? Well, other than an obvious bottom, we need to emit bottom whenever
> our
> given function is itself bottom at every input. This may first seem like a
> problem, but it is actually quite simple. Without loss of generality, let
> us
> assume that we can enumerate the type a. Then, we can feed these values to
> the
> function, and check their results for bottom. Conal Elliot has prior art
> for
> this sort of thing with his unamb [2] package. For each value x :: a,
> simply
> compute 'f x `seq` ()' in parallel, and look for any successes. If we ever
> find
> one, we know the function is non-bottom, and we can return our value of c.
> If we
> never finish searching, then the function must be bottom, and seq should
> not
> terminate, so we have satisfied the specification.
>
Love it. I have always been a fan of "fast and loose" reasoning in
Haskell. Abstracting on seq, and treating it as a bottom if it evaluates
to one, fits in perfectly.
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