After reading Stack Overflow question Using vectors for performance improvement in Haskell describing a fast in-place quicksort in Haskell, I set myself two goals:
Implementing the same algorithm with a median of three to avoid bad performances on pre-sorted vectors;
Making a parallel version.
Here is the result (some minor pieces have been left for simplicity):
import qualified Data.Vector.Unboxed.Mutable as MV
import qualified Data.Vector.Generic.Mutable as GM
type Vector = MV.IOVector Int
type Sort = Vector -> IO ()
medianofthreepartition :: Vector -> Int -> IO Int
medianofthreepartition uv li = do
p1 <- MV.unsafeRead uv li
p2 <- MV.unsafeRead uv $ li `div` 2
p3 <- MV.unsafeRead uv 0
let p = median p1 p2 p3
GM.unstablePartition (< p) uv
vquicksort :: Sort
vquicksort uv = do
let li = MV.length uv - 1
j <- medianofthreepartition uv li
when (j > 1) (vquicksort (MV.unsafeSlice 0 j uv))
when (j + 1 < li) (vquicksort (MV.unsafeSlice (j+1) (li-j) uv))
vparquicksort :: Sort
vparquicksort uv = do
let li = MV.length uv - 1
j <- medianofthreepartition uv li
t1 <- tryfork (j > 1) (vparquicksort (MV.unsafeSlice 0 j uv))
t2 <- tryfork (j + 1 < li) (vparquicksort (MV.unsafeSlice (j+1) (li-j) uv))
wait t1
wait t2
tryfork :: Bool -> IO () -> IO (Maybe (MVar ()))
tryfork False _ = return Nothing
tryfork True action = do
done <- newEmptyMVar :: IO (MVar ())
_ <- forkFinally action (\_ -> putMVar done ())
return $ Just done
wait :: Maybe (MVar ()) -> IO ()
wait Nothing = return ()
wait (Just done) = swapMVar done ()
median :: Int -> Int -> Int -> Int
median a b c
| a > b =
if b > c then b
else if a > c then c
else a
| otherwise =
if a > c then a
else if b > c then c
else b
For vectors with 1,000,000 elements, I get the following results:
"Number of threads: 4"
"**** Parallel ****"
"Testing sort with length: 1000000"
"Creating vector"
"Printing vector"
"Sorting random vector"
CPU time: 12.30 s
"Sorting ordered vector"
CPU time: 9.44 s
"**** Single thread ****"
"Testing sort with length: 1000000"
"Creating vector"
"Printing vector"
"Sorting random vector"
CPU time: 0.27 s
"Sorting ordered vector"
CPU time: 0.39 s
My questions are:
It does not use quicksort; rather, it uses an efficient implementation of an algorithm called mergesort.
Whenever we want to sort the elements of a given list, then we make use of the sort function in Haskell programming language, and the name of the list consisting of elements that are to be sorted is passed as a parameter to the sort function and the sort function takes a list as the input and returns the sorted list as ...
Is Bubble Sort Stable? Yes, Bubble Sort is a stable sorting algorithm. We swap elements only when A is less than B. If A is equal to B, we do not swap them, hence relative order between equal elements will be maintained.
A better idea is to use Control.Parallel.Strategies
to parallelize quicksort. With this approach you will not create expensive threads for every code that can be executed in parallel. You can also create a pure computation instead an IO.
Then you have to compile according to the number of cores you have: http://www.haskell.org/ghc/docs/latest/html/users_guide/using-concurrent.html
For an example, look at this simple quicksort on lists, written by Jim Apple:
import Data.HashTable as H
import Data.Array.IO
import Control.Parallel.Strategies
import Control.Monad
import System
exch a i r =
do tmpi <- readArray a i
tmpr <- readArray a r
writeArray a i tmpr
writeArray a i tmpi
bool a b c = if c then a else b
quicksort arr l r =
if r <= l then return () else do
i <- loop (l-1) r =<< readArray arr r
exch arr i r
withStrategy rpar $ quicksort arr l (i-1)
quicksort arr (i+1) r
where
loop i j v = do
(i', j') <- liftM2 (,) (find (>=v) (+1) (i+1)) (find (<=v) (subtract 1) (j-1))
if (i' < j') then exch arr i' j' >> loop i' j' v
else return i'
find p f i = if i == l then return i
else bool (return i) (find p f (f i)) . p =<< readArray arr i
main =
do [testSize] <- fmap (fmap read) getArgs
arr <- testPar testSize
ans <- readArray arr (testSize `div` 2)
print ans
testPar testSize =
do x <- testArray testSize
quicksort x 0 (testSize - 1)
return x
testArray :: Int -> IO (IOArray Int Double)
testArray testSize =
do ans <- newListArray (0,testSize-1) [fromIntegral $ H.hashString $ show i | i <- [1..testSize]]
return ans
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