On Mon, Nov 09, 2009 at 10:46:19PM +0100, legajid wrote:
{- Second solution -} futiles2 xx = [(x, y, z) | x <- xx, y <- xx, z <- xx, y < x, z < y] f2 = filter (\(x,y,z) -> (x+y+z)==19) (futiles2 nombres )
{- Third solution -} f3 = filter (\(x,y,z) -> (x+y+z)==19) ((\ xx -> [(x, y, z) | x <- xx, y <- xx, z <- xx, y < x, z < y]) nombres )
I think the second solution is best (the third solution seems hard to read). Shorter code is usually better, but avoid long lines that are hard to scan. Here's another possibility: f4 = filter (\(x,y,z) -> x+y+z == 19) [(x,y,z) | x <- [9,8..1], y <- reverse [1..x-1], z <- reverse [1..y-1]] This way you only generate (x,y,z) where x > y > z, and avoid all the wasted work of generating triples and then throwing them away. -Brent