Bottom-up natural mergesort in Haskell

msort.hs

module Msort (msortBy, msort) where

msortBy :: (a -> a -> Ordering) -> [a] -> [a]
msortBy orderOp =
  foldr merge [] . foldr mergeStack [] . runs
  where
    -- mergeStack :: [a] -> [[a]] -> [[a]]
    --   mergeStack "k" [ "" "ij" "" "abcdefgh" ] = [ "k" "ij" "" "abcdefgh" ]
    --   mergeStack "l" [ "k" "ij" "" "abcdefgh" ] = [ "" "" "ijkl" "abcdefgh" ]
    mergeStack x ([]:s) = x:s
    mergeStack x (y:s) = []:mergeStack (merge x y) s
    mergeStack x [] = [x]
    -- merge :: [a] -> [a] -> [a]
    merge xx@(x:xs) yy@(y:ys)
      | orderOp x y /= GT = x:merge xs yy
      | otherwise = y:merge xx ys
    merge x [] = x
    merge [] y = y
    -- runs :: Ord a => [a] -> [[a]]
    runs (x:xs) = collectRun x x (x:) xs
    runs [] = []
    -- collectRun :: Ord a => a -> a -> ([a] -> [a]) -> [a] -> [[a]]
    collectRun mn mx f (x:xs)
      | orderOp x mn == LT = collectRun x mx (\y -> x:(f y)) xs -- prepend
      | orderOp x mx /= LT = collectRun mn x (\y -> f (x:y)) xs -- append
    collectRun mn mx f x = f [] : runs x

msort :: Ord a => [a] -> [a]
msort = msortBy compare

Modified to clarify the algorithm and emphasize the relation to Timsort.

Amusingly, the function still sorts if you replace

foldr merge [] . foldr mergeStack [] . runs

with

foldr merge [] . runs

But in that case you're just merging all the runs (effectively an insertion sort) which is O(n*m) where, again, m is the number of runs. Best case O(n), worst case O(n^2). "foldr mergeStack []" ensures that merges are scheduled as they are in conventional mergesort.

Collect both ascending and descending runs (reversed).

But here's a funny thing. Timing the function on a big string

let big = take 10000000 $ cycle "the quick brown fox jumps over the lazy dog."

msort is actually faster when timed in GHCi without foldr mergeStack [] (basically just merging all the runs) and nearly as fast as the built-in sort. I have no explanation for that at all.

UPDATE: Explained below.

Moreover, while msort and sort are designed to sort already sorted data quickly, in both cases sorting and then sorting again is unexpectedly much slower!

*Main> let big = take 10000000 $ cycle "the quick brown fox jumps over the lazy dog."
*Main> take 1 $ sort big
" "
(10.03 secs, 4,525,157,720 bytes)
*Main> take 1 $ sort $ sort big
" "
(128.88 secs, 61,507,910,736 bytes)

My sort fares worse

*Main> take 1 $ msort big
" "
(20.07 secs, 7,257,693,600 bytes)
*Main> take 1 $ msort $ msort big
" "
(239.14 secs, 111,587,668,512 bytes)

For lack of a better theory, I chalk this up to some weird effect of laziness.

UPDATE: Explained below.

Yes, it was an effect of laziness. "take 1" did not force the sort function to sort the whole data set. It sufficed to compare the first element of each run and in that case, for all sorts, the cost was linear. It's amusing that non-strict evaluation is able to "discover" a cheaper way to find the least element while (partially) evaluating a sort function!

Taking the "maximum" of the sorted list forces complete evaluation and expected run times. Sorting the result of a sort adds only a trivial O(n) cost, as expected.

You can uncomment your local type signatures if you enable {-# LANGUAGE ScopedTypeVariables #-} and use msortBy :: forall a . (a -> a -> Ordering) -> [a] -> [a].