Is there a better way to express the absolute error function in point-free notation?

Question

In pointful notation:

absoluteError x y = abs (x-y)

An unclear example in pointfree notation:

absoluteError\' = curry (abs . uncur

Answer 1

Here's how you could derive it yourself, in small steps:

absoluteError x y = abs (x-y) = abs ((-) x y) = abs ( ((-) x) y) 
                  = (abs . (-) x) y = ( (abs .) ((-) x) ) y = 
                  = ( (abs .) . (-) ) x y

so, by eta-reduction, if f x y = g x y we conclude f = g.

Further, using _B = (.) for a moment,

(abs .) . (-) = _B (abs .) (-) = _B (_B abs) (-) = (_B . _B) abs (-)
              = ((.) . (.)) abs (-)

Answer 2

Here's a handful of ways.

1. the old-fashioned: absoluteError = (abs .) . (-)
2. use the so-called "boobs operator", or "owl operator" absoluteError = ((.) . (.)) abs (-)

3. name the boobs operator something more politically correct (and what the heck, generalize it at the same time)

   (.:) = fmap fmap fmap
   absoluteError = abs .: (-)
   

4. using semantic editor combinators:

   result :: (o1 -> o2) -> (i -> o1) -> (i -> o2)
   result = (.)
   
   absoluteError = (result . result) abs (-)
   

Of course, these are all the same trick, just with different names. Enjoy!