# [Haskell-cafe] Fastest way to calculate all the ways to interleave two lists

David Feuer david.feuer at gmail.com
Sun Apr 3 04:41:45 UTC 2016

```Of course, but something like take k . (!! m)   will cut it down nicely.

On Sun, Apr 3, 2016 at 12:39 AM, Arseniy Alekseyev
<arseniy.alekseyev at gmail.com> wrote:
> Um, the result is exponential in size. A problem will emerge in any
> solution. :)
>
> On 3 April 2016 at 05:38, David Feuer <david.feuer at gmail.com> wrote:
>>
>> Your lists are very short. Pump them up to thousands of elements each
>> and you will see a problem emerge in the naive solution.
>>
>> On Sun, Apr 3, 2016 at 12:16 AM, Arseniy Alekseyev
>> <arseniy.alekseyev at gmail.com> wrote:
>> > I measure the following naive solution of interleave2 beating yours in
>> > performance:
>> >
>> > i2 [] ys = [ys]
>> > i2 xs [] = [xs]
>> > i2 (x : xs) (y : ys) =
>> >   fmap (x :) (i2 xs (y : ys)) ++ fmap (y :) (i2 (x : xs) ys)
>> >
>> > The program I'm benchmarking is:
>> >
>> > main = print \$ sum \$ map sum \$ interleavings
>> > [[1,2,3,4],[5,6,7,8],[9,10,11,12],[1,1,1]]
>> >
>> >
>> >
>> > On 3 April 2016 at 04:05, David Feuer <david.feuer at gmail.com> wrote:
>> >>
>> >> I ran into a fun question today:
>> >> http://stackoverflow.com/q/36342967/1477667
>> >>
>> >> Specifically, it asks how to find all ways to interleave lists so that
>> >> the order of elements within each list is preserved. The most
>> >> efficient way I found is copied below. It's nicely lazy, and avoids
>> >> left-nested appends. Unfortunately, it's pretty seriously ugly. Does
>> >> anyone have any idea of a way to do this that's both efficient and
>> >> elegant?
>> >>
>> >> {-# LANGUAGE BangPatterns #-}
>> >> import Data.Monoid
>> >> import Data.Foldable (toList)
>> >> import Data.Sequence (Seq, (|>))
>> >>
>> >> -- Find all ways to interleave two lists
>> >> interleave2 :: [a] -> [a] -> [[a]]
>> >> interleave2 xs ys = interleave2' mempty xs ys []
>> >>
>> >> -- Find all ways to interleave two lists, adding the
>> >> -- given prefix to each result and continuing with
>> >> -- a given list to append
>> >> interleave2' :: Seq a -> [a] -> [a] -> [[a]] -> [[a]]
>> >> interleave2' !prefix xs ys rest =
>> >>   (toList prefix ++ xs ++ ys)
>> >>      : interleave2'' prefix xs ys rest
>> >>
>> >> -- Find all ways to interleave two lists except for
>> >> -- the trivial case of just appending them. Glom
>> >> -- the results onto the given list.
>> >> interleave2'' :: Seq a -> [a] -> [a] -> [[a]] -> [[a]]
>> >> interleave2'' !prefix [] _ = id
>> >> interleave2'' !prefix _ [] = id
>> >> interleave2'' !prefix xs@(x : xs') ys@(y : ys') =
>> >>   interleave2' (prefix |> y) xs ys' .
>> >>       interleave2'' (prefix |> x) xs' ys
>> >>
>> >> -- What the question poser wanted; I don't *think* there's
>> >> -- anything terribly interesting to do here.
>> >> interleavings :: [[a]] -> [[a]]
>> >> interleavings = foldr (concatMap . interleave2) [[]]
>> >>
>> >>
>> >> Thanks,
>> >> David
>> >> _______________________________________________