Transforming a function in Haskell to point free notation

user_10513295 0
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I have a function in haskell on paper as an example:

function2 a b c = (a * b) + c

and I'm required to write the example in point free notation. I'm really bad at working with point free style as I find it really confusing with no proper guide through it so i gave it a try:

function2 a b c = (a * b) + c
function2 a b c = ((*) a b) + c #operator sectioning
function2 a b c = (+) ((*) a b)c #operator sectioning once more
#I'm stuck here now

I wasnt sure what should come up next as that was the limit i could think of for this example. Would appreciate some help on this.

--Second example:

function3 a b = a `div` (g b)
function3 a b = `div` a (g b) --operator sectioning
function3 a b = (`div` a) (g b) --parentheses
function3 a b = ((`div` a g).)b --B combinator
function3 a   = ((`div` a g).) --eta conversion
function3 a   = ((.)(`div` a g)) --operator sectioning
function3 a   = ((.)flip(`div` g a))
function3 a   = ((.)flip(`div` g).a) --B combinator
function3     = ((.)flip(`div` g)) --eta conversion (complete)
pointfree haskell
2 Answers

You can apply the B combinator (i.e. (f . g) x = f (g x)) there:

function2 a b c = (a * b) + c
function2 a b c = ((*) a b) + c    -- operator sectioning
function2 a b c = (+) ((*) a b) c  -- operator sectioning once more
      = (+) (((*) a) b) c          -- explicit parentheses
      = ((+) . ((*) a)) b c        -- B combinator
      = ((.) (+) ((*) a)) b c      -- operator sectioning 
      = ((.) (+) . (*)) a b c      -- B combinator

Indeed the types are the same:

> :t let function2 a b c = (a * b) + c in function2
let function2 a b c = (a * b) + c in function2
  :: Num a => a -> a -> a -> a

> :t ((.) (+) . (*))
((.) (+) . (*)) :: Num b => b -> b -> b -> b

We work by extricating the arguments one by one in the correct order, to end up with 

function2 a b c = (......) a b c

so that the eta-contraction can be applied to get rid of the explicit arguments,

function2       = (......)

Our tools in this, which we get to apply in both directions, are

S a b c  =  (a c) (b c)  =  (a <*> b) c
K a b    =  a            =  const a b
I a      =  a            =  id a
B a b c  =  a (b c)      =  (a . b) c
C a b c  =  a c b        =  flip a b c
W a b    =  a b b        =  join a b
U a      =  a a          -- not in Haskell: `join id` has no type

There's also (f =<< g) x = f (g x) x = join (f . g) x.

Some more, useful patterns that emerge when we work with pointfree for a while are:

((f .) .) g x y = f (g x y)
(((f .) .) .) g x y z = f (g x y z)
.....
((. g) . f) x y = f x (g y)
((. g) . f . h) x y = f (h x) (g y)

(update.) There's an error near the start in your second example, invalidating all the following steps after it:

function3 a b = a `div` (g b)
function3 a b = -- `div` a (g b)     -- wrong syntax, you meant
                div a (g b)
function3 a b = -- (`div` a) (g b)   -- wrong; it is
                (a `div`) (g b) --operator sectioning
function3 a b = ((a `div`) . g) b --B combinator
function3 a   = (div a . g) --eta conversion; back with plain syntax
function3 a   = (.) (div a) g --operator sectioning
function3 a   = flip (.) g (div a) --definition of flip
function3 a   = (flip (.) g . div) a --B combinator
function3     = (flip (.) g . div) --eta conversion
              = (.) (flip (.) g) div  --operator section

So yeah, some of the steps were in the right direction.

user_8498910

We can continue with:

function2 a b = (+) ((*) a b)     [eta-reduction]
function2 a = (+) . (*) a         [using a dot, to eliminate the b]
function2 = ((.) . (.)) (+) (*)   ["blackbird" operator to pass two parameters]

Here the "blackbird operator" is a combination of three (.) :: (b -> c) -> (a -> b) -> a -> c operators. It is functionally equivalent to:

((.) . (.)) :: (c -> d) -> (a -> b -> c) -> a -> b -> d
((.) . (.)) f g x y = f (g x y)

see here for a derivation.

user_675790
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