平行“折叠"在哈斯克尔 [英] parallel "Folding" in Haskell

问题描述

我有一个类型如下的函数:

I have a function with type below:

union :: a -> a -> a

并且 a 具有 可加性 属性.所以我们可以把union看成是(+)

And a has additivity property. So we can regard union as a version of (+)

比如说，我们有[a]，想要进行一个并行的"folding"，对于非并行的折叠我们只能做:

Say, we have [a], and want to perform a parallel "folding", for non-parallel foldling we can do only:

foldl1' union [a]

但是如何并行执行呢?我可以演示 Num 值和 (+) 函数的问题.

But how to perform it in parallel? I can demonstrate problem on Num values and (+) function.

例如，我们有一个列表 [1,2,3,4,5,6] 和 (+)同时我们应该拆分

For example, we have a list [1,2,3,4,5,6] and (+) In parallel we should split

[1,2,3] (+) [4,5,6]
[1,2] (+) [3] (+) [4,5] (+) [6]
([1] (+) [2]) (+) ([3] (+) [4]) (+) ([5] (+) [6])

然后每个(+)我们要并行执行的操作，并结合起来回答

then each (+) operation we want to perform in parallel, and combine to answer

[3] (+) [7] (+) [11] = 21

请注意，由于a 可加性，我们拆分列表或以任何顺序执行操作.

Note, that we split list, or perform operations in any order, because of a additivity.

有没有办法使用任何标准库来做到这一点?

Is there any ways to do that using any standard library?

推荐答案

您需要将 union 推广到任何关联二元运算符 ⊕使得 (a ⊕ b) ⊕c == a ⊕(b ⊕ c).如果同时你甚至有一个相对于 ⊕ 中性的单位元素，你就有了幺半群.

You need to generalize your union to any associative binary operator ⊕ such that (a ⊕ b) ⊕ c == a ⊕ (b ⊕ c). If at the same time you even have a unit element that is neutral with respect to ⊕, you have a monoid.

关联性的一个重要方面是您可以将列表中的连续元素块任意分组，并且 ⊕它们以任何顺序排列，因为 ⊕(b ⊕ (c ⊕ d)) == (a ⊕ b) ⊕(c ⊕ d) - 每个括号可以并行计算；然后你需要减少"所有括号的总和"，并且你已经对你的 map-reduce 进行了排序.

The important aspect of associativity is that you can arbitrarily group chunks of consecutive elements in a list, and ⊕ them in any order, since a ⊕ (b ⊕ (c ⊕ d)) == (a ⊕ b) ⊕ (c ⊕ d) - each bracket can be computed in parallel; then you'd need to "reduce" the "sums" of all brackets, and you've got your map-reduce sorted.

为了使这种并行化有意义，您需要比 ⊕ 更快的分块操作.- 否则，做 ⊕顺序优于分块.一种这样的情况是当您有一个随机访问列表" - 例如一个数组.Data.Array.Repa 具有大量并行折叠功能.

In order for this parallellisation to make sense, you need the chunking operation to be faster than ⊕ - otherwise, doing ⊕ sequentially is better than chunking. One such case is when you have a random access "list" - say, an array. Data.Array.Repa has plenty of parallellized folding functions.

如果你想练习自己实现一个，你需要选择一个好的复杂函数⊕如此好处就会显现出来.

If you are thinking of practicising to implement one yourself, you need to pick a good complex function ⊕ such that the benefit will show.

例如:

import Control.Parallel
import Data.List

pfold :: (Num a, Enum a) => (a -> a -> a) -> [a] -> a
pfold _ [x] = x
pfold mappend xs  = (ys `par` zs) `pseq` (ys `mappend` zs) where
  len = length xs
  (ys', zs') = splitAt (len `div` 2) xs
  ys = pfold mappend ys'
  zs = pfold mappend zs'

main = print $ pfold (+) [ foldl' (*) 1 [1..x] | x <- [1..5000] ]
  -- need a more complicated computation than (+) of numbers
  -- so we produce a list of products of many numbers

这里我特意使用了一个仅在本地称为 mappend 的关联操作，以表明它可以用于比幺半群更弱的概念——只有关联性对并行性很重要；因为并行只对非空列表有意义，所以不需要 mempty.

Here I deliberately use a associative operation, which is called mappend only locally, to show it can work for a weaker notion than a monoid - only associativity matters for parallelism; since parallelism makes sense only for non-empty lists anyway, no need for mempty.

ghc -O2 -threaded a.hs
a +RTS -N1 -s

总运行时间为 8.78 秒，而

Gives 8.78 seconds total run time, whereas

a +RTS -N2 -s

在我的双核笔记本电脑上的总运行时间为 5.89 秒.显然，在这台机器上尝试超过 -N2 是没有意义的.

Gives 5.89 seconds total run time on my dual core laptop. Obviously, no point trying more than -N2 on this machine.

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