Hi, I'm trying to understand pointfree style better, but it's not coming along as well as I'd like it to. The thing I can't get to work is to reduce an argument that is used more than once in a function. My function looks like this now (which works like it should): f x = g ((h . i) x) x But I'd like to reduce the last argument x. I've looked at the wiki[1] but I couldn't find a systematic way to obtain pointfree functions when they get more complicated. Any pointers to pages or papers with more examples of obtaining pointfree functions are appreciated. Thanks, Bas van Gijzel [1]http://www.haskell.org/haskellwiki/Pointfree
Am Dienstag, 11. November 2008 12:40 schrieb Bas van Gijzel:
Hi,
I'm trying to understand pointfree style better, but it's not coming along as well as I'd like it to. The thing I can't get to work is to reduce an argument that is used more than once in a function.
My function looks like this now (which works like it should): f x = g ((h . i) x) x
But I'd like to reduce the last argument x. I've looked at the wiki[1] but I couldn't find a systematic way to obtain pointfree functions when they get more complicated. Any pointers to pages or papers with more examples of obtaining pointfree functions are appreciated.
lambdabot can pointfree your functions, the command is @pl: lambdabot> @pl \x -> \y -> (y,x) flip (,) lambdabot> @pl \x -> g ((h . i) x) x g =<< h . i (the monad here is ((->) a), where x :: a, I think, you need Control.Monad.Instances for that to be in scope). I'm not sure reading lambdabot's sources is a good way to learn how to systematically pointfree functions, but looking at lambdabot's results should help.
Thanks,
Bas van Gijzel
Cheers, Daniel
On Tue, Nov 11, 2008 at 12:40:19PM +0100, Bas van Gijzel wrote:
Hi,
I'm trying to understand pointfree style better, but it's not coming along as well as I'd like it to. The thing I can't get to work is to reduce an argument that is used more than once in a function.
My function looks like this now (which works like it should): f x = g ((h . i) x) x
But I'd like to reduce the last argument x. I've looked at the wiki[1] but I couldn't find a systematic way to obtain pointfree functions when they get more complicated. Any pointers to pages or papers with more examples of obtaining pointfree functions are appreciated.
If you're doing this just to learn, great. If you're doing this because you think a pointfree style is somehow 'better', you should know that there are limits. =) In the case of your function f above (and, in general, with any function that uses its argument more than once) I would leave it as it is. The really important things to know are composition, i.e. f x = g (h (x)) becomes f x = (g . h) x and eta-reduction, i.e. f x = foo x becomes f = foo. There are other things that can be nice, such as flip, and various Arrow combinators (such as (&&&), (***)) for the (->) instance of Arrow [1] for use with functions involving tuples. Going much beyond that is often just obfuscation, IMO. But to answer your question, a systematic way to transform functions which use their argument more than once into pointfree versions is to use the ((->) e) (aka reader) monad [2]: f x = g (h x) x becomes f x = (h >>= g) x Essentially, in the ((->) e) monad, (>>=) is a combinator to do exactly what you are asking about -- compose two functions with a duplicated input. Of course, if you dig into the implementation of (>>=) for the ((->) e) monad, you will eventually find a function which is not point-free -- this is unavoidable at some level; you can't actually duplicate an arbitrary thing without giving it a name. Applying this to your example: f x = g ((h . i) x) x f x = ((h . i) >>= g) x f = (h . i) >>= g -Brent [1] http://haskell.org/ghc/docs/latest/html/libraries/base/Control-Arrow.html [2] http://haskell.org/ghc/docs/latest/html/libraries/base/Control-Monad-Instanc...
Wow, thanks for the thorough answer :). I already understood quite a bit
about composition and eta reduction, but your response made it clear for me
that this example couldn't be solved that easily.
Thanks for the information, it really helps.
Bas
On Tue, Nov 11, 2008 at 7:51 PM, Brent Yorgey
On Tue, Nov 11, 2008 at 12:40:19PM +0100, Bas van Gijzel wrote:
Hi,
I'm trying to understand pointfree style better, but it's not coming along as well as I'd like it to. The thing I can't get to work is to reduce an argument that is used more than once in a function.
My function looks like this now (which works like it should): f x = g ((h . i) x) x
But I'd like to reduce the last argument x. I've looked at the wiki[1] but I couldn't find a systematic way to obtain pointfree functions when they get more complicated. Any pointers to pages or papers with more examples of obtaining pointfree functions are appreciated.
If you're doing this just to learn, great. If you're doing this because you think a pointfree style is somehow 'better', you should know that there are limits. =) In the case of your function f above (and, in general, with any function that uses its argument more than once) I would leave it as it is. The really important things to know are composition, i.e.
f x = g (h (x)) becomes f x = (g . h) x
and eta-reduction, i.e.
f x = foo x becomes f = foo.
There are other things that can be nice, such as flip, and various Arrow combinators (such as (&&&), (***)) for the (->) instance of Arrow [1] for use with functions involving tuples. Going much beyond that is often just obfuscation, IMO.
But to answer your question, a systematic way to transform functions which use their argument more than once into pointfree versions is to use the ((->) e) (aka reader) monad [2]:
f x = g (h x) x becomes f x = (h >>= g) x
Essentially, in the ((->) e) monad, (>>=) is a combinator to do exactly what you are asking about -- compose two functions with a duplicated input. Of course, if you dig into the implementation of (>>=) for the ((->) e) monad, you will eventually find a function which is not point-free -- this is unavoidable at some level; you can't actually duplicate an arbitrary thing without giving it a name.
Applying this to your example:
f x = g ((h . i) x) x f x = ((h . i) >>= g) x f = (h . i) >>= g
-Brent
[1] http://haskell.org/ghc/docs/latest/html/libraries/base/Control-Arrow.html [2] http://haskell.org/ghc/docs/latest/html/libraries/base/Control-Monad-Instanc... _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners
participants (3)
-
Bas van Gijzel -
Brent Yorgey -
Daniel Fischer