[Haskell-cafe] Pretty little definitions of left and right folds
Brent Yorgey
byorgey at seas.upenn.edu
Sat Jun 21 22:48:59 EDT 2008
On Sat, Jun 21, 2008 at 09:36:06PM -0500, Derek Elkins wrote:
> On Sat, 2008-06-21 at 21:11 -0400, Brent Yorgey wrote:
> > On Fri, Jun 20, 2008 at 09:52:36PM -0500, Derek Elkins wrote:
> > > On Fri, 2008-06-20 at 22:31 -0400, Brent Yorgey wrote:
> > > > On Fri, Jun 20, 2008 at 06:15:20PM -0500, George Kangas wrote:
> > > > >
> > > > > foldright (+) [1, 2, 3] 0 == ( (1 +).(2 +).(3 +).id ) 0
> > > > > foldleft (+) [1, 2, 3] 0 == ( id.(3 +).(2 +).(1 +) ) 0
> > > > >
> > > >
> > > > Hi George,
> > > >
> > > > This is very cool! I have never thought of folds in quite this way
> > > > before. It makes a lot of things (such as the identities you point
> > > > out) obvious and elegant.
> > > >
> > > > > We can also see the following identities:
> > > > >
> > > > > foldright f as == foldright (.) (map f as) id
> > > > > foldleft f as == foldright (flip (.)) (map f as) id
> > > > >
> > > > > I like that second one, after trying to read another definition of
> > > > > left fold in terms of right fold (in the web book "Real World Haskell").
> > > > >
> > > > > The type signature, which could be written (a -> (b -> b)) -> ([a] ->
> > > > > (b -> b)), suggests generalization to another type constructor C: (a ->
> > > > > (b -> b)) -> (C a -> (b -> b)). Would a "foldable" typeclass make any
> > > > > sense?
> > > >
> > > > As Brandon points out, you have rediscovered Data.Foldable. =) There's
> > > > nothing wrong with that, congratulations on discovering it for
> > > > yourself! But again, I like this way of organizing the type
> > > > signature: I had never thought of a fold as a sort of 'lift' before.
> > > > If f :: a -> b -> b, then foldright 'lifts' f to foldright f :: [a] ->
> > > > b -> b (or C a -> b -> b, more generally).
> > > >
> > > > > Okay, it goes without saying that this is useless dabbling, but have
> > > > > I entertained anyone? Or have I just wasted your time? I eagerly await
> > > > > comments on this, my first posting.
> > > >
> > > > Not at all! Welcome, and thanks for posting.
> > >
> > > Look into the theory of monoids, monoid homomorphisms, M-sets and free
> > > monoids.
> >
> > Thanks for the pointers! Here's what I've come up with, after
> > re-reading some Barr-Wells lecture notes.
> >
> > First, given finite sets A (representing an 'alphabet') and S
> > (representing 'states'), we can describe a finite state machine by a
> > function phi : A x S -> S, which gives 'transition rules' giving a new
> > state for each combination of alphabet character and state. If we
> > squint and wave our hands and ignore the fact that types aren't
> > exactly sets, and most of the types we care about have infinitely many
> > values, this is very much like the Haskell type (a,s) -> s, or
> > (curried) a -> s -> s, i.e. a -> (s -> s). So we can think of a
> > Haskell function phi :: a -> (s -> s) as a sort of 'state machine'.
> >
> > Also, for a monoid M and set S, an action of M on S is given by a
> > function f : M x S -> S for which
> >
> > (1) f(1,s) = s, and
> > (2) f(mn,s) = f(m,f(n,s)).
> >
> > Of course, in Haskell we would write f :: m -> (s -> s),
>
> This change is not completely trivial.
Hmm... why is that? Is it because of the types-aren't-really-sets
thing? Or are there other reasons as well?
>
> > and we would
> > write criteria (1) and (2) as
> >
> > (1) f mempty = id
> > (2) f (m `mappend` n) = f m . f n
>
> So what does this make f? Hint: What is (s -> s)?
Aha! f is a monoid homomorphism to the monoid of endomorphisms on s!
Right?
-Brent
More information about the Haskell-Cafe
mailing list