If I understand the problem correctly... Prelude> let diag = concat . diags where diags ((x:xs):xss) = [x] : zipWith (:) xs (diags xss) Prelude> take 10 $ diag [[ (m,n) | n <- [1..]] | m <- [1..]] [(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1)] Sebastian Fischer wrote on 15.04.2009 14:32:
Fancy some Codegolf?
I wrote the following function for list diagonalization:
diag l = foldr (.) id ((sel l . flip sel) ((:[]).(:))) [] where sel = foldr (\a b c -> id : mrg (a c) (b c)) (const []) . map (flip id)
mrg [] ys = ys mrg xs [] = xs mrg (x:xs) (y:ys) = (x.y) : mrg xs ys
Self explanatory, isn't it? Here is a test case:
*Main> take 10 $ diag [[ (m,n) | n <- [1..]] | m <- [1..]] [(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1)]
I was trying to golf it down [^1] but my brain explodes. If you succeed in reducing keystrokes, I'd be happy to know!
Cheers, Sebastian
[^1]: http://codegolf.com/ _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe