
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
Er.. I mean force . map head On Apr 3, 2016 1:14 AM, "David Feuer"
wrote: I choose the `force (map head)` attack. On Apr 3, 2016 1:04 AM, "Arseniy Alekseyev"
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
wrote: Of course, but something like take k . (!! m) will cut it down nicely.
Um, the result is exponential in size. A problem will emerge in any solution. :)
On 3 April 2016 at 05:38, David Feuer
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
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
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 On Sun, Apr 3, 2016 at 12:39 AM, Arseniy Alekseyev
wrote: 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@haskell.org >> http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe > >