[Haskell-cafe] Re: Looking for pointfree version (Kim-Ee Yeoh)
(02/12)
Luke Palmer
lrpalmer at gmail.com
Fri Feb 20 05:19:27 EST 2009
On Fri, Feb 20, 2009 at 3:10 AM, Robert Vollmert <rvollmert-lists at gmx.net>wrote:
> On Feb 20, 2009, at 06:07, Gene Arthur wrote:
>
> Kim-Ee Yeoh said:
>>> 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)
>>>
>>
>> First sorry for the delay in getting to this.. been behind on
>> projects so had some days of mail piled up. Here is what
>> I came up with, using one arrow operator,
>> so you would have to import: Control.Arrow ((***)) at minimum
>> to use this solution:
>>
>> pointfree ::
>> forall t t1 c. (t -> t1 -> c) -> (t, t1) -> (t, t1) -> (c, c)
>>
>> pointfree op =
>> curry $ (\(a,b) -> a `op` b) *** (\(a,b) -> a `op` b)
>>
>> examples of use, that were executed using:
>>
>> ghci -fglasgow-exts -farrows Control.Arrow
>>
>> *>pointfree (*) (3,5) (12,12)
>>
>> (15,144)
>>
>
> That's not quite what pointwise above does, though. The following works
> using (***):
>
> tr :: ((a,b),(c,d)) -> ((a,c),(b,d))
> tr ((x,y),(z,w)) = ((x,z),(y,w))
>
tr = (fst *** fst) &&& (snd *** snd)
>
> diag :: (a -> a -> b) -> (a -> b)
> diag f x = f x x
diag = join -- for Monad (->) a, in Control.Monad.Instances
>
> pointfree :: (a -> b -> c) -> (a,a) -> (b,b) -> (c,c)
> pointfree = curry . (. tr) . diag (***) . uncurry
>
> Do tr or diag exist in the libraries?
>
> Cheers
> Robert
>
>
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