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

David Feuer david.feuer at gmail.com
Sun Apr 3 05:14:54 UTC 2016

```I choose the `force (map head)` attack.
On Apr 3, 2016 1:04 AM, "Arseniy Alekseyev" <arseniy.alekseyev at gmail.com>
wrote:

> I see! At this point I'd say that you probably have the wrong type: there
> are ways to produce n'th interleaving much faster, but let's continue
> optimizing for the hell of it!
>
> i2 :: ([a] -> [b]) -> [a] -> [a] -> [[b]] -> [[b]]
> i2 f [] ys = (f ys :)
> i2 f xs [] = (f xs :)
> i2 f (x : xs) (y : ys) =
>   i2 (f . (x :)) xs (y : ys) . i2 (f . (y :)) (x : xs) ys
>
> interleave2 xs ys = i2 id xs ys []
>
> Seems faster than your original solution on examples I tried it on and it
> has fewer characters. :)
>
> On 3 April 2016 at 05:41, David Feuer <david.feuer at gmail.com> wrote:
>
>> 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
>> >> >> _______________________________________________
>> >> >> Haskell-Cafe mailing list
>> >> >
>> >> >
>> >
>> >
>>
>
>
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