Re: Looking for pointfree version (Kim-Ee Yeoh) (02/12)
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) *> pointfree (++) ("This ","That") ("Old"," Man") ("This That","Old Man") Hope that helps. -- gene ============================== website: http://haskblog.cloud.prohosting.com ============================== If Helen Keller is alone in a forest and falls, does she make a sound? ____________________________________________________________ Receive Notifications of Incoming Messages Easily monitor multiple email accounts & access them with a click. Visit http://www.inbox.com/notifier and check it out!
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)) diag :: (a -> a -> b) -> (a -> b) diag f x = f x x 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
On Fri, Feb 20, 2009 at 3:10 AM, Robert Vollmert
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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participants (3)
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Gene Arthur -
Luke Palmer -
Robert Vollmert