solved. see the haskell wiki for spoiler: http://haskell.org/haskellwiki/Blow_your_mind 2009/7/16 Thomas Hartman :

Is it possible to fix alternate' (the second version), or otherwise define a fast stepwise alternate that doesn't blow up on long lists?

alternate just breaks up [1,2,3,4,5] into ([1,3,5],[2,4])

Thanks!

{-# LANGUAGE BangPatterns #-} import Data.List import Control.Arrow import Control.Parallel.Strategies

t = (last *** last) $! alternate $ [1..(10^6)] t' = (last *** last) $! alternate' $ [1..(10^6)] t'' = (last *** last) $! alternate'' $ [1..(10^6)]

-- finishes reasonably fast, but does a separate computation for the list and its tail rather than just rip through it alternate x = (skip1 x,(skip1 $ tail x)) skip1 = skip 1 skip n xs =  let (a,b) = splitAt (n+1) xs  in case a of     [] -> []     x:_ -> x : skip n b

-- this one overflows on million element list, even after fiddling with strictness on input args. can this be fixed? alternate' xs =  let f3 :: Int -> (([Int],[Int]),Int) -> (([Int],[Int]),Int)      f3 x ((a,b),n) = -- rnf (x,((a,b),n)) `seq`        let nxtn = n+1        in if n `mod` 2 == 0            then ((x:a,b),nxtn)            else ((a,x:b),nxtn)  in fst . foldr f3 (([],[]),0) $ xs

-- no overflow, goes through the list stepwise, but it's actually slightly slower than the first alternate because of the reverses alternate'' xs =  let f3 ((a,b),n) x =        (let nxtn = n+1        in if n `mod` 2 == 0            then ((x:a,b),nxtn)            else ((a,x:b),nxtn) )  in (reverse *** reverse) . fst . foldl' f3 (([],[]),0) $ xs