[Haskell-cafe] Pretty little definitions of left and right folds
derek.a.elkins at gmail.com
Fri Jun 20 22:52:36 EDT 2008
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
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