# Benchmarks Game/Parallel/BinaryTrees

### From HaskellWiki

DonStewart (Talk | contribs) |
m (Shootout/Parallel/BinaryTrees moved to Benchmarks Game/Parallel/BinaryTrees: The name of the benchmarks site has changed) |

(4 intermediate revisions by one user not shown) |

## Latest revision as of 22:26, 22 January 2012

## [edit] 1 Binary Trees

### [edit] 1.1 2009-03-01: Current Entry

Submitted: http://alioth.debian.org/tracker/index.php?func=detail&aid=311523&group_id=30402&atid=411646

Also filed a bug ticket with GHC to find out if the GC growth strategy can be improved (so that -H240M isn't required): http://hackage.haskell.org/trac/ghc/ticket/3061

{-# OPTIONS -funbox-strict-fields #-} {-# LANGUAGE BangPatterns #-} -- -- The Computer Language Benchmarks Game -- http://shootout.alioth.debian.org/ -- -- Contributed by Don Stewart -- Modified by Stephen Blackheath to parallelize (a very tiny tweak) -- -- Compile with: -- -- > ghc -O2 -fasm -threaded --make -- -- Run with: -- -- > ./A +RTS -N4 -H300M -RTS 20 -- -- Where '4' is the number of cores. and "set your -H value high (3 or -- more times the maximum residency)", as per GHC User's Guide: -- -- <http://haskell.org/ghc/docs/6.10.1/html/users_guide/runtime-control.html#rts-options-gc> -- -- -H "provides a “suggested heap size” for the garbage collector. The -- garbage collector will use about this much memory until the program -- residency grows and the heap size needs to be expanded to retain -- reasonable performance." -- import System import Data.Bits import Text.Printf import Control.Parallel.Strategies -- -- an artificially strict tree. -- -- normally you would ensure the branches are lazy, but this benchmark -- requires strict allocation. -- data Tree = Nil | Node !Int !Tree !Tree minN = 4 io s n t = printf "%s of depth %d\t check: %d\n" s n t main = do n <- getArgs >>= readIO . head let maxN = max (minN + 2) n stretchN = maxN + 1 -- stretch memory tree let c = check (make 0 stretchN) io "stretch tree" stretchN c -- allocate a long lived tree let !long = make 0 maxN -- allocate, walk, and deallocate many bottom-up binary trees let vs = parMap rnf id $ depth minN maxN mapM_ (\((m,d,i)) -> io (show m ++ "\t trees") d i) vs -- confirm the the long-lived binary tree still exists io "long lived tree" maxN (check long) -- generate many trees depth :: Int -> Int -> [(Int,Int,Int)] depth d m | d <= m = (2*n,d,sumT d n 0) : depth (d+2) m | otherwise = [] where n = 1 `shiftL` (m - d + minN) -- allocate and check lots of trees sumT :: Int -> Int -> Int -> Int sumT d 0 t = t sumT d i t = sumT d (i-1) (t + a + b) where a = check (make i d) b = check (make (-i) d) -- traverse the tree, counting up the nodes check :: Tree -> Int check Nil = 0 check (Node i l r) = i + check l - check r -- build a tree make :: Int -> Int -> Tree make i 0 = Node i Nil Nil make i d = Node i (make (i2-1) d2) (make i2 d2) where i2 = 2*i; d2 = d-1

### [edit] 1.2 Parallel Strategies: parMap

- Status: submitted.

Flags:

$ ghc -O2 --make -fasm -threaded Parallel.hs $ ./Parallel 20 +RTS -N5 -A350M

This is a version of the Haskell GHC binary-trees benchmark, annotated for parallelism, using parallel strategy combinators. When compiled with the -threaded flag, and run with +RTS -N5 -RTS, it will exploit all cores on the quad-core machine, dramatically reducing running times.

On my quad core, running time goes from,

* single core, 26.997s * quad core, 5.692s

The following flags should be used:

Compile time:

ghc -O2 -fasm --make Parallel2.hs -threaded

Runtime:

./Parallel2 20 +RTS -N5 -A350M -RTS

The -N5 flag asks the Haskell runtime to use 5 capabilites, which map onto the underlying cores.

Here is the result on my quad core,

$ time ./Parallel2 20 +RTS -N5 -A350M -RTS stretch tree of depth 21 check: -1 2097152 trees of depth 4 check: -2097152 524288 trees of depth 6 check: -524288 131072 trees of depth 8 check: -131072 32768 trees of depth 10 check: -32768 8192 trees of depth 12 check: -8192 2048 trees of depth 14 check: -2048 512 trees of depth 16 check: -512 128 trees of depth 18 check: -128 32 trees of depth 20 check: -32 long lived tree of depth 20 check: -1 ./Parallel2 20 +RTS -N5 -A350M -RTS 15.80s user 1.52s system 304% cpu 5.692 total

Which is a satisfying result, as the parallelisation strategy is super simple.

Code:

{-# OPTIONS -fbang-patterns -funbox-strict-fields #-} -- -- The Computer Language Shootout -- http://shootout.alioth.debian.org/ -- -- Contributed by Don Stewart and Thomas Davie -- -- This implementation uses a parallel strategy to exploit the quad core machine. -- For more information about Haskell parallel strategies, see, -- -- http://www.macs.hw.ac.uk/~dsg/gph/papers/html/Strategies/strategies.html -- import System import Data.Bits import Text.Printf import Control.Parallel.Strategies import Control.Parallel -- -- an artificially strict tree. -- -- normally you would ensure the branches are lazy, but this benchmark -- requires strict allocation. -- data Tree = Nil | Node !Int !Tree !Tree minN = 4 io s n t = printf "%s of depth %d\t check: %d\n" s n t main = do n <- getArgs >>= readIO . head let maxN = max (minN + 2) n stretchN = maxN + 1 -- stretch memory tree let c = check (make 0 stretchN) io "stretch tree" stretchN c -- allocate a long lived tree let !long = make 0 maxN -- allocate, walk, and deallocate many bottom-up binary trees let vs = (parMap rnf) (depth' maxN) [minN,minN+2..maxN] mapM_ (\((m,d,i)) -> io (show m ++ "\t trees") d i) vs -- confirm the the long-lived binary tree still exists io "long lived tree" maxN (check long) -- generate many trees depth' :: Int -> Int -> (Int,Int,Int) depth' m d = (2*n,d,sumT d n 0) where n = 1 `shiftL` (m - d + minN) -- allocate and check lots of trees sumT :: Int -> Int -> Int -> Int sumT d 0 t = t sumT d i t = sumT d (i-1) (t + a + b) where a = check (make i d) b = check (make (-i) d) -- traverse the tree, counting up the nodes check :: Tree -> Int check Nil = 0 check (Node i l r) = i + check l - check r -- build a tree make :: Int -> Int -> Tree make i 0 = Node i Nil Nil make i d = Node i (make (i2-1) d2) (make i2 d2) where i2 = 2*i; d2 = d-1