[Haskell-cafe] Parallel term reduction
John D. Ramsdell
ramsdell0 at gmail.com
Mon Feb 2 23:33:51 EST 2009
Luke,
I read your solution but didn't understand how it applies to my
problem. I must not have explained the problem well enough. Let me
try again.
I have a reduction system in which a rule takes a term and returns a
set of terms. The terms can be compared for equality, but they are
not ordered. The reduction system creates a tree that originates at a
starting value called the root using breadth first search. For most
problems, the reduction system terminates, but a step count limit
protects from non-termination. Rule application is expensive, so it
is essential that a rule is never applied to the same problem twice.
This check makes my program sequential, and I could find no place to
add parallel annotations that might help.
The key function in the enclosed code is the step function. It's
where all the time is spent. To allow effective use of more than one
core, doesn't one have to figure out how to allow parts of the seen
list to be searched simultaneously? Isn't that search the performance
bottleneck?
John
> module Main (main) where
> import System.Time (ClockTime(..), getClockTime)
> import Data.Tree (Tree(..), flatten)
> import Data.Maybe (isNothing)
The reduction system takes a rule, a step count, and an initial value,
and computes a tree of reductions.
> reduce :: (Eq a, Monad m) => (a -> [a]) -> Int -> a -> m (Tree a)
> reduce rule limit root =
> step rule limit [top] [top]
> where
> top = Item { item = root, parent = Nothing }
The Item data structure stores the information about a reduction step
in a form that can be used to construct the final answer as a tree.
> data Eq a => Item a
> = Item { item :: a,
> parent :: Maybe (Item a) }
> instance Eq a => Eq (Item a) where
> x == y = item x == item y
The step function is where nearly all of the time is used. It is
called as:
step rule limit seen todo
where seen is the items already seen, and todo is the items on the
queue. The order of the items in the seen list is irrelevant, because
the tree is assembled as a post processing step.
> step :: (Eq a, Monad m) => (a -> [a]) -> Int ->
> [Item a] -> [Item a] -> m (Tree a)
> step _ limit _ _
> | limit <= 0 = fail "Step limit exceeded"
> step _ _ seen [] = tree seen
> step rule limit seen (next : rest) =
> loop seen rest children
> where
> children = map child (rule (item next))
> child i = Item { item = i, parent = Just next }
> loop seen rest [] =
> step rule (limit - 1) seen rest
> loop seen rest (kid : kids) =
> if elem kid seen then
> loop seen rest kids
> else
> loop (kid : seen) (rest ++ [kid]) kids
The next two functions assemble the answer into a tree.
Sequential search is just fine for tree building.
> tree :: (Eq a, Monad m) => [Item a] -> m (Tree a)
> tree items =
> case filter (isNothing . parent) items of
> [root] -> return (build items (item root))
> _ -> fail "bad tree"
> build :: Eq a => [Item a] -> a -> Tree a
> build items root =
> Node { rootLabel = root,
> subForest = map (build items . item) children }
> where
> children = filter child items
> child i = maybe False ((== root) . item) (parent i)
A silly rule
> rule :: Int -> [Int]
> rule n = filter (>= 0) [n - 1, n - 2, n - 3]
> secDiff :: ClockTime -> ClockTime -> Float
> secDiff (TOD secs1 psecs1) (TOD secs2 psecs2)
> = fromInteger (psecs2 - psecs1) / 1e12 + fromInteger (secs2 - secs1)
> main :: IO ()
> main =
> do
> t0 <- getClockTime
> t <- reduce rule 20000 5000
> let ns = length (flatten t)
> t1 <- getClockTime
> putStrLn $ "length: " ++ show ns
> putStrLn $ "time: " ++ show (secDiff t0 t1) ++ " seconds"
The makefile
------------
PROG = reduce
GHCFLAGS = -Wall -package containers -fno-warn-name-shadowing -O
%: %.lhs
ghc $(GHCFLAGS) -o $@ $<
all: $(PROG)
clean:
-rm *.o *.hi $(PROG)
More information about the Haskell-Cafe
mailing list