并行或多线程运行haskell操作

Question

I have just started learning functional programming and Haskell and am I not even close to understanding the syntax and how it works because it is very different from what I have been used to before (ruby, php, C# or node for instance, which are OOP languages - more or less).

I am am trying to put together a small program that checks for solutions for Brocard's Problem or so called Brown Numbers and I first created a draft in ruby, but I discovered that Haskell is much more appropriate for doing mathematical expressions. I've asked a question previously and I got an answer pretty quick on how to translate my ruby code to Haskell:

results :: [(Integer, Integer)] --Use instead of `Int` to fix overflow issue
results =  [(x,y) | x <- [1..1000], y <- [1..1000] , 1 + fac x == y*y]
    where fac n = product [1..n]

I changed that slightly so I can run the same operation from and up to whatever number I want, because the above will do it from 1 up to 1000 or any hardcoded number but I would like to be able to decide the interval it should go through, ergo:

pairs :: (Integer, Integer) -> [(Integer, Integer)]
pairs (lower, upper) =  [(m, n) | m <- [lower..upper], n <- [lower..upper], 1 + factorial n == m*m] where factorial n = product [1..n]

The reason I ask this questions is that I was thinking if I could run this operations in parallel (and/or on multiple threads and/or cores) and append the results to the same variable.

I have not yet understood the parallel processing in Haskell or the threading model, but in ruby a good example of how the threading model works is something like this :

def func1
  i=0
  while i<=2
    puts "func1 at: #{Time.now}"
    sleep(2)
    i=i+1
  end
end

def func2
  j=0
  while j<=2
    puts "func2 at: #{Time.now}"
    sleep(1)
    j=j+1
  end
end

puts "Started At #{Time.now}"
t1=Thread.new{func1()}
t2=Thread.new{func2()}
t1.join
t2.join
puts "End at #{Time.now}"

And in my case, the func1 and func2 would be the same function results computed on different intervals ( [1..1000] -> [i..j] ).

I would appreciate some help with this as I am not capable of doing it myself at this point :)

Answer 1

Parallel and Concurrent Programming in Haskell has a lot of good information, and async is a good library for this stuff.

At the bottom level though, you'll find forkIO to start a new lightweight thread.

Of course that's concurrency , not deterministic parallelism, parallel is the library for that, and also covered in the book.

Your example translates to:

import Data.Time.Clock (getCurrentTime)

main = do
    start <- getCurrentTime
    putStr "Started At " >> print start
    _ <- forkIO func1
    _ <- forkIO func2
    end <- getCurrentTime
    putStr "End at " >> print end

func1 = helper "func1" 2

func2 = helper "func2" 1

helper name sleepTime = go 0
 where
    go 3 = return ()
    go n = do
        now <- getCurrentTime
        putStr name >> putStr " at: " >> print now
        threadDelay sleepTime
        go $ succ n

I recommend learning the parallel and/or async libraries mentioned above, though, instead of writing your own threading stuff, at least initially.

Here's a not-so-great example of running tests on 8-ish processors using parallel:

import Control.Parallel.Strategies

factorial = product . enumFromTo 1
pairs (lower, upper) =  map fst . filter snd . withStrategy sparkTest
                     $ [ ((m, n), b)
                       | m <- [lower..upper]
                       , n <- [lower..upper]
                       , let b = 1 + factorial n == m*m
                       ]
sparkTest = evalBuffer 8 $ evalTuple2 rseq rpar