Question

我目前正在尝试了解如何在Haskell中并行编程。我正在关注Simon Peyton Jones和Satnam Singh的论文“Haskell中的并行和并发编程教程”。源代码如下：

module Main where
import Control.Parallel
import System.Time

main :: IO ()
main = do
      putStrLn "Starting computation....."
      t0 <- getClockTime
      pseq r1 (return())
      t1 <- getClockTime
      putStrLn ("sum: " ++ show r1)
      putStrLn ("time: " ++ show (secDiff t0 t1) ++ " seconds")
      putStrLn "Finish."

r1 :: Int
r1 = parSumFibEuler 38 5300

-- This is the Fibonacci number generator
fib :: Int -> Int
fib 0 = 0
fib 1 = 1
fib n = fib (n-1) + fib (n-2)

-- Gets the euler sum
mkList :: Int -> [Int]
mkList n = [1..n-1]

relprime :: Int -> Int -> Bool
relprime x y = gcd x y == 1

euler :: Int -> Int
euler n = length $ filter (relprime n) (mkList n)

sumEuler :: Int -> Int
sumEuler = sum.(map euler).mkList

-- Gets the sum of Euler and Fibonacci (NORMAL)
sumFibEuler :: Int -> Int -> Int
sumFibEuler a b = fib a + sumEuler b

-- Gets the sum of Euler and Fibonacci (PARALLEL)
parSumFibEuler :: Int -> Int -> Int
parSumFibEuler a b =
  f `par` (e `pseq`(f+e))
    where
    f = fib a
    e = sumEuler b

-- Measure time
secDiff :: ClockTime -> ClockTime -> Float
secDiff (TOD secs1 psecs1) (TOD secs2 psecs2)
  = fromInteger (psecs2 -psecs1) / 1e12 + fromInteger (secs2- secs1)

我使用以下命令编译它：

ghc --make -threaded Main.hs

a）使用1核心：

./Main +RTS -N1

b）使用2核心：

./Main +RTS -N2

然而，一个核心运行53.556秒。然而，两个核心运行73.401秒。我不明白2核如何实际运行速度慢于1核。也许传递开销的消息对于这个小程序来说太大了？与矿山相比，该论文的结果完全不同。以下是输出细节。

1核心：

Starting computation.....
sum: 47625790
time: 53.556335 seconds
Finish.
  17,961,210,216 bytes allocated in the heap
      12,595,880 bytes copied during GC
         176,536 bytes maximum residency (3 sample(s))
          23,904 bytes maximum slop
               2 MB total memory in use (0 MB lost due to fragmentation)

                                    Tot time (elapsed)  Avg pause  Max pause
  Gen  0     34389 colls,     0 par    2.54s    2.57s     0.0001s    0.0123s
  Gen  1         3 colls,     0 par    0.00s    0.00s     0.0007s    0.0010s

  Parallel GC work balance: -nan (0 / 0, ideal 1)

                        MUT time (elapsed)       GC time  (elapsed)
  Task  0 (worker) :    0.00s    (  0.00s)       0.00s    (  0.00s)
  Task  1 (worker) :    0.00s    ( 53.56s)       0.00s    (  0.00s)
  Task  2 (bound)  :   50.49s    ( 50.99s)       2.52s    (  2.57s)

  SPARKS: 0 (0 converted, 0 overflowed, 0 dud, 0 GC'd, 0 fizzled)

  INIT    time    0.00s  (  0.00s elapsed)
  MUT     time   50.47s  ( 50.99s elapsed)
  GC      time    2.54s  (  2.57s elapsed)
  EXIT    time    0.00s  (  0.00s elapsed)
  Total   time   53.02s  ( 53.56s elapsed)

  Alloc rate    355,810,305 bytes per MUT second

  Productivity  95.2% of total user, 94.2% of total elapsed

gc_alloc_block_sync: 0
whitehole_spin: 0
gen[0].sync: 0
gen[1].sync: 0

对于2核心：

Starting computation.....
sum: 47625790
time: 73.401146 seconds
Finish.
  17,961,210,256 bytes allocated in the heap
      12,558,088 bytes copied during GC
         176,536 bytes maximum residency (3 sample(s))
         195,936 bytes maximum slop
               3 MB total memory in use (0 MB lost due to fragmentation)

                                    Tot time (elapsed)  Avg pause  Max pause
  Gen  0     34389 colls, 34388 par    7.42s    4.73s     0.0001s    0.0205s
  Gen  1         3 colls,     3 par    0.01s    0.00s     0.0011s    0.0017s

  Parallel GC work balance: 1.00 (1432193 / 1429197, ideal 2)

                        MUT time (elapsed)       GC time  (elapsed)
  Task  0 (worker) :    1.19s    ( 40.26s)      16.95s    ( 33.15s)
  Task  1 (worker) :    0.00s    ( 73.40s)       0.00s    (  0.00s)
  Task  2 (bound)  :   54.50s    ( 68.67s)       3.66s    (  4.73s)
  Task  3 (worker) :    0.00s    ( 73.41s)       0.00s    (  0.00s)

  SPARKS: 0 (0 converted, 0 overflowed, 0 dud, 0 GC'd, 0 fizzled)

  INIT    time    0.00s  (  0.00s elapsed)
  MUT     time   68.87s  ( 68.67s elapsed)
  GC      time    7.43s  (  4.73s elapsed)
  EXIT    time    0.00s  (  0.00s elapsed)
  Total   time   76.31s  ( 73.41s elapsed)

  Alloc rate    260,751,318 bytes per MUT second

  Productivity  90.3% of total user, 93.8% of total elapsed

gc_alloc_block_sync: 12254
whitehole_spin: 0
gen[0].sync: 0
gen[1].sync: 0

Answer 1

r1 = sumFibEuler 38 5300

我相信你的意思

r1 = parSumFibEuler 38 5300

在我的配置上（使用parSumFibEuler 45 8000并且只有一次运行）：

当N1 = 126.83s
当N2 = 115.46s

我怀疑fib功能比sumEuler更耗费CPU。这可以解释-N2的低改善。在你的情况下不会有一些偷工作的东西。

通过记忆，你的斐波纳契功能会好得多，但我认为这不是你想要尝试的。

编辑：正如评论中所提到的，我认为对于-N2，你有很多中断，因为你有两个核心可用。
我的配置示例（4核）sum $ parMap rdeepseq (fib) [1..40]

与-N1需要~26s
与-N2需要~16s
与-N3需要~13s
使用-N4需要大约30秒（好吧，Haskell程序并不是一个人在这里）

来自here：

使用机器中的所有处理器时要小心：如果有的话您的处理器正在被其他程序使用，这实际上可能会造成伤害表现而不是改善它。

为什么我的程序更快，一个核心而不是两个核心？

1 个答案: