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

Answer 1

The key observation for me is that infix operators can be written prefix:

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

Once you've gotten here, you have half a chance of recognizing that this is a composition, with (.) (!!) as the first argument and iterate . funcA as the second argument:

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

Now it's clear how to simplify this; after that, there's a lot of aesthetic choices about how to write it. For example, we might observe that (.) is associative, and so we can drop some parentheses; likewise, we can use operator sections to merge the unsightly ((.) (!!)) if you think it's more readable that way.

funcB = (  ((.) (!!)) . (iterate . funcA)  )
funcB = (.) (!!) . iterate . funcA -- uncontroversial parenthesis removal
funcB = ((!!) .) . iterate . funcA -- possibly controversial section rewrite

By the way, I don't think the beginning of your derivation is correct. You reached the right conclusion, but via incorrect middle steps. Corrected, it should look like this:

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