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

David Feuer david.feuer at gmail.com
Sun Apr 3 05:17:29 UTC 2016


By the way, I have a bit of an unfair advantage because a lot of these
things came up in the redesign of Data.List.inits. I know I can improve my
constant factors by using a simpler, less flexible queue implementation
than Data.Sequence, but that's a whole different issue.
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
>>> >> >> Haskell-Cafe at haskell.org
>>> >> >> http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
>>> >> >
>>> >> >
>>> >
>>> >
>>>
>>
>>
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