Point Free problems in Haskell

Question

I am trying to convert the following Haskell code to point free style, to no avail.

bar f g xs = filter f (map g xs )

I'm new to Haskell and any help would be great.

Answer 1

Converting to pointfree style can be done entirely mechanically, though it's hard without being comfortable with the fundamentals of Haskell syntax like left-associative function application and x + y being the same as (+) x y. I will assume you are comfortable with Haskell syntax; if not, I suggest going through the first few chapters of LYAH first.

You need the following combinators, which are in the standard library. I have also given their standard names from combinator calculus.

id :: a -> a                                   -- I
const :: a -> b -> a                           -- K
(.) :: (b -> c) -> (a -> b) -> (a -> c)        -- B
flip :: (a -> b -> c) -> (b -> a -> c)         -- C
(<*>) :: (a -> b -> c) -> (a -> b) -> (a -> c) -- S

Work with one parameter at a time. Move parameters on the left to lambdas on the right, e.g.

f x y = Z

becomes

f = \x -> \y -> Z

I like to do this one argument at a time rather than all at once, it just looks cleaner.

Then eliminate the lambda you just created according to the following rules. I will use lowercase letters for literal variables, uppercase letters to denote more complex expressions.

1. If you have \x -> x, replace with id
2. If you have \x -> A, where A is any expression in which x does not occur, replace with const A
3. If you have \x -> A x, where x does not occur in A, replace with A. This is known as "eta contraction".
4. If you have \x -> A B, then
   1. If x occurs in both A and B, replace with (\x -> A) <*> (\x -> B).
   2. If x occurs in just A, replace with flip (\x -> A) B
   3. If x occurs in just B, replace with A . (\x -> B),
   4. If x does not occur in either A or B, well, there's another rule we should have used already.

And then work inward, eliminating the lambdas that you created. Lets work with this example:

f x y z = foo z (bar x y)
-- Move parameter to lambda:
f x y = \z -> foo z (bar x y)
-- Remember that application is left-associative, so this is the same as
f x y = \z -> (foo z) (bar x y)
-- z appears on the left and not on the right, use flip
f x y = flip (\z -> foo z) (bar x y)
-- Use rule (3) 
f x y = flip foo (bar x y)

-- Next parameter
f x = \y -> flip foo (bar x y)
-- Application is left-associative
f x = \y -> (flip foo) (bar x y)
-- y occurs on the right but not the left, use (.)
f x = flip foo . (\y -> bar x y)
-- Use rule 3
f x = flip foo . bar x

-- Next parameter
f = \x -> flip foo . bar x
-- We need to rewrite this operator into normal application style
f = \x -> (.) (flip foo) (bar x)
-- Application is left-associative
f = \x -> ((.) (flip foo)) (bar x)
-- x appears on the right but not the left, use (.)
f = ((.) (flip foo)) . (\x -> bar x)
-- use rule (3)
f = ((.) (flip foo)) . bar
-- Redundant parentheses
f = (.) (flip foo) . bar

There you go, now try it on yours! There is not really any cleverness involved in deciding which rule to use: use any rule that applies and you will make progress.