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

Arseniy Alekseyev arseniy.alekseyev at gmail.com
Sun Apr 3 06:05:35 UTC 2016

```That (specifically, the benchmark below) shows your thing is faster, but
I'm not sure why. Maybe it's because Seq is cheaper than a closure, but
maybe it's something more meaningful than that. Looks like you've guided
myself roughly to your original solution now so I'm giving up. :)

main = print \$ sum \$ map head \$ take 1000000 \$ interleavings
[[1..100],[5..100]]

On 3 April 2016 at 06:20, David Feuer <david.feuer at gmail.com> wrote:

> Er.. I mean force . map head
> On Apr 3, 2016 1:14 AM, "David Feuer" <david.feuer at gmail.com> wrote:
>
>> 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
>>>> >> >
>>>> >> >
>>>> >
>>>> >
>>>>
>>>
>>>
-------------- next part --------------
An HTML attachment was scrubbed...