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Writing in pointfree style f x = g x x

I am learning Haskell. I'm sorry for asking a very basic question but I cant seem to find the answer. I have a function f defined by :

f x = g x x

where g is an already defined function of 2 arguments. How do I write this pointfree style? Edit : without using a lambda expression.

Thanks

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user1188374 Avatar asked Feb 06 '12 23:02

user1188374


2 Answers

f can be written with Control.Monad.join:

f = join g

join on the function monad is one of the primitives used when constructing point-free expressions, as it cannot be defined in a point-free style itself (its SKI calculus equivalent, SIIap id id in Haskell — doesn't type).

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ehird Avatar answered Oct 03 '22 00:10

ehird


This is known as "W" combinator:

import Control.Monad
import Control.Monad.Instances
import Control.Applicative

f = join g       -- = Wg        (also, join = (id =<<))
  = (g `ap` id)  -- \x -> g x (id x) = SgI
  = (<*> id) g   --                  = CSIg
  = g =<< id     -- \x -> g (id x) x
  = id =<< g     -- \x -> id (g x) x

S,K,I are one basic set of combinators; B,C,K,W are another - you've got to stop somewhere (re: your "no lambda expression" comment):

_B = (.)     -- _B f g x = f (g x)     = S(KS)K
_C = flip    -- _C f x y = f y x       = S(S(K(S(KS)K))S)(KK)
_K = const   -- _K x y   = x
_W = join    -- _W f x   = f x x       = CSI = SS(KI) = SS(SK)
_S = ap      -- _S f g x = f x (g x)   = B(B(BW)C)(BB) = B(BW)(BBC)
   = (<*>)                                -- from Control.Applicative
_I = id      -- _I x     = x           = WK = SKK = SKS = SK(...)

{-
Wgx = gxx 
    = SgIx = CSIgx 
           = Sg(KIg)x = SS(KI)gx
    = gx(Kx(gx)) = gx(SKgx) = Sg(SKg)x = SS(SK)gx

-- _W (,) 5 = (5,5)
-- _S _I _I x = x x = _omega x         -- self-application, untypeable
-}
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Will Ness Avatar answered Oct 03 '22 01:10

Will Ness