pointfree - Currying 3 Arguments in Haskell

Question

I'm having trouble with currying a function to remove three arguments in Haskell.

Disclaimer: Not Coursework, I was asked this question by someone struggling with this today and it's been bugging me.

The custom types/functions we were given were (can only remember types)

type MyThing
  = (Char, String)
type MyThings
  = [MyThing]

funcA :: MyThings -> String -> String
funcB :: MyThings -> String -> Int -> String

We started with:

funcB as str n = iterate (funcA as) str !! n

And reduced it down as follows:

funcB as str n = iterate (funcA as) str !! n
funcB as str = (!!) . (iterate (funcA as)) str
funcB as = (!!) . (iterate (funcA as))
funcB as = (!!) . (iterate . funcA) as

Then, stuck. We just can't figure out how to avoid using the last argument. I know I've seen a similar situation somewhere before and there was a solution.

Hoping some Haskell genius can point out why I'm being an idiot...

Answer 1

All you need here is the following three "laws" of operator sections:

(a `op` b) = (a `op`) b = (`op` b) a = op a b
          (1)          (2)          (3)

so that the operand goes into the free slot near the operator.

For (.) this means that: (a . b) = (a .) b = (. b) a = (.) a b. So,

f (g x) y !! n      
= (!!) (f (g x) y) n              by (3) 
= ((!!) . f (g x)) y n
= ((!!) . (f . g) x) y n
= ((!!) .) ((f . g) x) y n        by (1)
= (((!!) .) . (f . g)) x y n
= (((!!) .) . f . g) x y n

You should only do as much pointfree transformation as you're comfortable with, so that the resulting expression is still readable for you - and in fact, clearer than the original. The "pointfree" tool can at times produce unreadable results.

It is perfectly OK to stop in the middle. If it's too hard for you to complete it manually, probably it will be hard for you to read it, too.

((a .) . b) x y = (a .) (b x) y = (a . b x) y = a (b x y) is a common pattern that you will quickly learn to recognize immediately. So the above expression can be read back fairly easily as

(!!) ((f . g) x y) n = f (g x) y !! n

considering that (.) is associative:

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