[Haskell-cafe] Re: Looking for pointfree version

[George Pollard]

On Thu, 2009-02-12 at 11:08 +0100, Heinrich Apfelmus wrote:
> Benja Fallenstein wrote:
> > Kim-Ee Yeoh wrote:
> >>
> >> On the same note, does anyone have ideas for the following snippet? Tried the
> >> pointfree package but the output was useless.
> >>
> >> pointwise op (x0,y0) (x1,y1) = (x0 `op` x1, y0 `op` y1)
> > 
> > import Control.Monad.Reader  -- for the (Monad (a ->)) instance
> > import Control.Bifunctor  -- package category-extras
> > 
> > dup = join (,)
> > mapPair = uncurry bimap
> > pointfree = (mapPair .) . mapPair . dup
> > 
> > Or if you're not afraid of *some* points, and want to avoid the imports:
> > 
> > dup x = (x,x)
> > mapPair (f,g) (x,y) = (f x, g y)
> > pointfree op = mapPair . mapPair (dup op)
> > 
> > That what you're looking for? :-)
> 
> The pairs are of course an applicative functor
> 
>   (<*>)  = uncurry (***)  -- from Control.Arrow
>   pure x = (x,x)
> 
>   pointwise op x y = pure op <*> x <*> y
> 
> 
> Regards,
> apfelmus

Concretely (this might do with a few laziness notations):

> import Control.Applicative
> 
> data Pair a = a :*: a
> 
> instance Functor Pair where
> 	f `fmap` (x :*: y) = f x :*: f y
> 
> instance Applicative Pair where
> 	(f :*: g) <*> (x :*: y) = f x :*: f y
> 	pure x = x :*: x
> 
> pointfree :: (a -> b -> c) -> Pair a -> Pair b -> Pair c
> --pointfree o x y = pure o <*> x <*> y
> pointfree = ((<*>) .) . (<*>) . pure
> -- in the applicative paper notation:
> --pointfree o x y = [| o x y |]

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