On Mon, 2011-04-11 at 11:22 +0200, Adam Krauze wrote:
f :: (Num a) => [a] -> [a] -> [(a,a)] // f takes two lists and zips them into one in some special way g :: (Num a) => a -> [(a,a)] -> [a] // g using some Num value calculates list of singletons from list of pairs
Prelude> let h x y = (g 0 (f x y))
How to do pointfree definition of h?
You can eliminate the second point, y, pretty easily by just using function composition: let h x = g 0 . f x To eliminate x, we can first rewrite this expression using a section of the (.) operator. let h x = (g 0 .) (f x) and then introduce another function composition: let h = (g 0 .) . f Whether that's clearer than the pointed definition is up for debate, but there it is. Just keep in mind that sections of (.) are very confusing to some people. -- Chris Smith