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

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].