haskell - subsequences of length n from list performance

Question

I implemented a version of this answer https://stackoverflow.com/a/9920425/1261166 (I don't know what was intended by the person answering)

sublistofsize 0 _        = [[]]
sublistofsize _ []       = []
sublistofsize n (x : xs) = sublistsThatStartWithX ++ sublistsThatDontStartWithX
  where sublistsThatStartWithX = map (x:) $ sublistofsize (n-1) xs
        sublistsThatDontStartWithX = sublistofsize n xs

what I'm unsure of is sublistsThatStartWithX = map (x:) $ sublistofsize (n-1) xs

I assume that map (x:) gives a problem performance wise, but not sure of how to solve it. I've done profiling on print $ length $ sublistofsize 5 $ primesToTakeFrom 50

COST CENTRE                                  MODULE                                        no.     entries  %time %alloc   %time %alloc
sublistofsize                             Main                                          112     4739871   46.9   39.9    96.9  100.0
 sublistofsize.sublistsThatDontStartWithX Main                                          124     2369935    2.2    0.0     2.2    0.0
 sublistofsize.sublistsThatStartWithX     Main                                          116     2369935   47.8   60.1    47.8   60.1

Did I implement it in a good way? Are there any faster ways of doing it?

Answer 1

I assume that map (x:) gives a problem performance wise

No. map is coded efficiently and runs in linear time, no problems here.

However, your recursion might be a problem. You're both calling sublistofsize (n-1) xs and sublistofsize n xs, which - given a start list sublistofsize m (_:_:ys) - does evaluate the term sublistofsize (m-1) ys twice, as there is no sharing between them in the different recursive steps.

So I'd apply dynamic programming to get

subsequencesOfSize :: Int -> [a] -> [[a]]
subsequencesOfSize n xs = let l = length xs
                          in if n>l then [] else subsequencesBySize xs !! (l-n)
 where
   subsequencesBySize [] = [[[]]]
   subsequencesBySize (x:xs) = let next = subsequencesBySize xs
                             in zipWith (++) ([]:next) (map (map (x:)) next ++ [[]])

Not that appending the empty lists is the most beautiful solution, but you can see how I have used zipWith with the displaced lists so that the results from next are used twice - once directly in the list of subsequences of length n and once in the list of subsequences of length n+1.

Testing it in GHCI with :set +s, you can see how this is drastically faster than the naive solutions:

*Main> length $ subsequencesOfSize 7 [1..25]
480700
(0.25 secs, 74132648 bytes)
(0.28 secs, 73524928 bytes)
(0.30 secs, 73529004 bytes)
*Main> length $ sublistofsize 7 [1..25] -- @Vixen (question)
480700
(3.03 secs, 470779436 bytes)
(3.35 secs, 470602932 bytes)
(3.14 secs, 470747656 bytes)
*Main> length $ sublistofsize' 7 [1..25] -- @Ganesh
480700
(2.00 secs, 193610388 bytes)
(2.00 secs, 193681472 bytes)
*Main> length $ subseq 7 [1..25] -- @user5402
480700
(3.07 secs, 485941092 bytes)
(3.07 secs, 486279608 bytes)