Haskell：顺序斐波那契比并行更快

Question

So, I'm experimenting with parallelism in Haskell. I took a classic example of implementing a Fibonacci sequence method both in sequential and in parallel. here is my Main.hs file:

module Main where
import Control.Parallel
main = print (fib 47)
fib :: Int -> Int
fib n
| n <=1 = n
| otherwise = fib (n-1) + fib (n-2)

I compile with ghc -O2 --make Main.hs -threaded -rtsopts and execute with time ./Main +RTS -N4 which gives me:

2971215073
63.23user 13.03system 0:20.30elapsed 375%CPU (0avgtext+0avgdata 3824maxresident)k
0inputs+0outputs (0major+276minor)pagefaults 0swaps

So with normal Fibonacci it takes about 20 seconds.

Now if I change my fib method to

pfib :: Int -> Int
pfib n
    | n <= 1 = n
    | otherwise = n1 `par` (n2 `par` n1 + n2)
        where
            n1 = pfib (n - 1)
            n2 = pfib (n - 2)

Compiling and running as above, time takes way longer and finishes with the output:

2971215073
179.50user 9.04system 0:53.08elapsed 355%CPU (0avgtext+0avgdata 6980maxresident)k
0inputs+0outputs (0major+1066minor)pagefaults 0swaps

Further modifying my pfib to use pseq instead of the second par , time gives:

2971215073
113.34user 3.42system 0:30.91elapsed 377%CPU (0avgtext+0avgdata 7312maxresident)k
0inputs+0outputs (0major+1119minor)pagefaults 0swaps

Is there an issue with my code? why do I have that illogical time difference between the various implementations?

Answer 1

From the documentation for par :

Also it is a good idea to ensure that a is not a trivial computation, otherwise the cost of spawning it in parallel overshadows the benefits obtained by running it in parallel.

An addition and a couple of subtractions are a trivial computation. If you only run a few levels of depth in parallel, you'll see benefits:

module Main where
import Control.Parallel

main = print (pfib 16 47)

fib :: Int -> Int
fib n
    | n <= 1 = n
    | otherwise = fib (n-1) + fib (n-2)

pfib :: Int -> Int -> Int
pfib 1 n = fib n
pfib p n
    | n <= 1 = n
    | otherwise = n1 `par` (n2 `par` n1 + n2)
        where
            n1 = pfib (p - 1) (n - 1)
            n2 = pfib (p - 1) (n - 2)