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. -Brent _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
