# [Haskell-cafe] Re: another laziness blowup (this one unsolved)

Thomas Hartman tphyahoo at gmail.com
Thu Jul 16 20:25:35 EDT 2009

```solved. see the haskell wiki for spoiler:

2009/7/16 Thomas Hartman <tphyahoo at gmail.com>:
> 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
>
```