我有一个看似微不足道的并行快速排序实现，代码如下：
import System.Random
import Control.Parallel
import Data.List
quicksort :: Ord a => [a] -> [a]
quicksort xs = pQuicksort 16 xs -- 16 is the number of sparks used to sort
-- pQuicksort, parallelQuicksort
-- As long as n > 0 evaluates the lower and upper part of the list in parallel,
-- when we have recursed deep enough, n==0, this turns into a serial quicksort.
pQuicksort :: Ord a => Int -> [a] -> [a]
pQuicksort _ [] = []
pQuicksort 0 (x:xs) =
let (lower, upper) = partition (< x) xs
in pQuicksort 0 lower ++ [x] ++ pQuicksort 0 upper
pQuicksort n (x:xs) =
let (lower, upper) = partition (< x) xs
l = pQuicksort (n `div` 2) lower
u = [x] ++ pQuicksort (n `div` 2) upper
in (par u l) ++ u
main :: IO ()
main = do
gen <- getStdGen
let randints = (take 5000000) $ randoms gen :: [Int]
putStrLn . show . sum $ (quicksort randints)
我使用以下编译器
ghc --make -threaded -O2 quicksort.hs
并且跑步
./quicksort +RTS -N16 -RTS
无论我做什么，都不能使这个程序在多个CPU上运行得比在单个CPU上运行的简单顺序实现快。
- 能否解释一下为什么这个程序在多个CPU上运行得比单个CPU上慢那么多？
- 是否有可能通过某些技巧使其至少与CPU数量呈次线性缩放？
编辑：@tempestadept暗示快速排序本身就是问题所在。为了验证这一点，我按照上面的示例实现了一个简单的归并排序。它具有相同的行为，添加的处理能力越多，性能表现越差。
import System.Random
import Control.Parallel
splitList :: [a] -> ([a], [a])
splitList = helper True [] []
where helper _ left right [] = (left, right)
helper True left right (x:xs) = helper False (x:left) right xs
helper False left right (x:xs) = helper True left (x:right) xs
merge :: (Ord a) => [a] -> [a] -> [a]
merge xs [] = xs
merge [] ys = ys
merge (x:xs) (y:ys) = case x<y of
True -> x : merge xs (y:ys)
False -> y : merge (x:xs) ys
mergeSort :: (Ord a) => [a] -> [a]
mergeSort xs = pMergeSort 16 xs -- we use 16 sparks
-- pMergeSort, parallel merge sort. Takes an extra argument
-- telling how many sparks to create. In our simple test it is
-- set to 16
pMergeSort :: (Ord a) => Int -> [a] -> [a]
pMergeSort _ [] = []
pMergeSort _ [a] = [a]
pMergeSort 0 xs =
let (left, right) = splitList xs
in merge (pMergeSort 0 left) (pMergeSort 0 right)
pMergeSort n xs =
let (left, right) = splitList xs
l = pMergeSort (n `div` 2) left
r = pMergeSort (n `div` 2) right
in (r `par` l) `pseq` (merge l r)
ris :: Int -> IO [Int]
ris n = do
gen <- getStdGen
return . (take n) $ randoms gen
main = do
r <- ris 100000
putStrLn . show . sum $ mergeSort r
pseq来使其表现更好，即使使用sum清除了任何可能的thunks。也许涉及到完全不同的问题。 ——由于我已经删除了我的答案，所以在这里再次作为评论：1.将该函数命名为quicksort可能会引起混淆，因为您不会期望这样的函数接受额外的并行性参数；2.对于顶级函数，始终使用类型签名，尤其是当它们的工作方式略有不同于名称所示时；3.如果可能，请使用库函数，例如partition。 ——顺便说一句，这是一个很好的问题。 - leftaroundaboutl `par` u `pseq`(u ++ l)。(2) 虽然你在并行运行子计算，但它们直到需要时才真正被评估。因此，你应该强制每个子列表到 NF（或至少其完整结构），类似于forceList l `par` forceList u `pseq` (u ++ l)，其中forceList是（你自己的）强制评估列表的函数。另外，为了进行适当的基准测试，我建议使用criterion。 - Petr