分治算法的并行性

Question

我面临着让我的代码并行运行的问题.它是一个3D Delaunay生成器,使用名为DeWall的分治算法.

主要功能是:

deWall::[SimplexPointer] -> SetSimplexFace -> Box -> StateT DeWallSets IO ([Simplex], [Edge])
deWall p afl box = do
   ...
   ...
   get >>= recursion box1 box2 p1 p2 sigma edges
   ...
   ...

它调用可能会调用dewall函数的"递归"函数.正是在这里出现了平行机会.以下代码显示了顺序解决方案.

recursion::Box -> Box -> [SimplexPointer] -> [SimplexPointer] -> [Simplex] -> [Edge] -> DeWallSets -> StateT DeWallSets IO ([Simplex], [Edge])    
recursion box1 box2 p1 p2 sigma edges deWallSet
        | null afl1 && null afl2 = return (sigma, edges)
        | (null) afl1 = do
            (s, e) <- deWall p2 afl2 box2
            return (s ++ sigma, e ++ edges)
        | (null) afl2 = do
            (s,e) <- deWall p1 afl1 box1
            return (s ++ sigma, e ++ edges)
        | otherwise   = do
            x <- get
            liftIO $ do
                (s1, e1) <- evalStateT (deWall p1 afl1 box1) x
                (s2, e2) <- evalStateT (deWall p2 afl2 box2) x
                return (s1 ++ s2 ++ sigma, e1 ++ e2 ++ edges)

        where   afl1 = aflBox1 deWallSet
                afl2 = aflBox2 deWallSet

状态和IO monad用于管道状态并为使用MVar找到的每个四面体生成UID.我的第一次尝试是添加一个forkIO,但它不起作用.由于在合并部分期间缺乏控制而不等待两个线程完成,因此输出错误.我不知道如何让它等待它们.

            liftIO $ do
                let 
                    s1 = evalStateT (deWall p1 afl1 box1) x
                    s2 = evalStateT (deWall p2 afl2 box2) x
                    concatThread var (a1, b1) = takeMVar var >>= \(a2, b2) -> putMVar var (a1 ++ a2, b1 ++ b2)
                mv <- newMVar ([],[])
                forkIO (s1 >>= concatThread mv)
                forkIO (s2 >>= concatThread mv)
                takeMVar mv >>= \(s, e) -> return (s ++ sigma, e ++ edges)

因此,我的下一次尝试是使用更好的并行策略"par"和"pseq",它提供了正确的结果,但根据threadScope没有并行执行.

        liftIO $ do
            let
                s1 = evalStateT (deWall p1 afl1 box1) x
                s2 = evalStateT (deWall p2 afl2 box2) x
                conc = liftM2 (\(a1, b1) (a2, b2) -> (a1 ++ a2, b1 ++ b2))
            (stotal, etotal) = s1 `par` (s2 `pseq` (s1 `conc` s2))
            return (stotal ++ sigma, etotal ++ edges)

我究竟做错了什么？

更新:不知何故,这个问题似乎与IO monads的存在有关.在没有IO monad的其他(旧)版本中,只有State monad,并行执行用'par'和运行'pseq'.GHC -sstderr给出了SPARKS: 1160 (69 converted, 1069 pruned).

recursion::Box -> Box -> [SimplexPointer] -> [SimplexPointer] -> [Simplex] -> [Edge] -> DeWallSets -> State DeWallSets ([Simplex], [Edge])  
recursion p1 p2 sigma deWallSet
     | null afl1 && null afl2 = return sigma
     | (null) afl1 = do
         s <- deWall p2 afl2 box2
         return (s ++ sigma)
     | (null) afl2 = do
         s <- deWall p1 afl1 box1
         return (s ++ sigma)
     | otherwise   = do
                     x <- get
                     let s1 = evalState (deWall p1 afl1 box1) x
                     let s2 = evalState (deWall p2 afl2 box2) x
                     return $ s1 `par` (s2 `pseq` (s1 ++ s2 ++ sigma))
     where   afl1 = aflBox1 deWallSet
             afl2 = aflBox2 deWallSet

云有人解释一下吗？

Answer 1

完成这项工作的最简单方法是使用如下内容：

liftIO $ do
            let 
                s1 = evalStateT (deWall p1 afl1 box1) x
                s2 = evalStateT (deWall p2 afl2 box2) x
            mv1 <- newMVar ([],[])
            mv2 <- newMVar ([],[])
            forkIO (s1 >>= putMVar mv1)
            forkIO (s2 >>= putMVar mv2)
            (a1,b1) <- takeMVar mv1
            (a2,b2) <- takeMVar mv2
            return (a1++a2++sigma, b1++b2++edges)

这可行，但有一些不必要的开销。更好的解决方案是：

liftIO $ do
            let 
                s1 = evalStateT (deWall p1 afl1 box1) x
                s2 = evalStateT (deWall p2 afl2 box2) x
            mv <- newMVar ([],[])
            forkIO (s2 >>= putMVar mv2)
            (a1,b1) <- s1
            (a2,b2) <- takeMVar mv2
             return (a1++a2++sigma, b1++b2++edges)

或者，如果结果没有按照您希望的方式进行评估，则可能会出现这种情况：

liftIO $ do
        let 
            s1 = evalStateT (deWall p1 afl1 box1) x
            s2 = evalStateT (deWall p2 afl2 box2) x
        mv <- newMVar ([],[])
        forkIO (s2 >>= evaluate >>= putMVar mv2)
        (a1,b1) <- s1
        (a2,b2) <- takeMVar mv2
         return (a1++a2++sigma, b1++b2++edges)

（这些是我在 #haskell 中给发帖者的答案，我认为在这里也有用）

编辑：删除了不必要的评估。