Pourquoi est-il plus rapide de résumer un Data.Sequence par diviser pour mieux régner, même sans parallélisme?

Question

Je jouais avec une réduction parallèle sur un Data.Sequence.Seq , et j'ai remarqué que la division et la conquête donnent un avantage de vitesse même sans parallélisme. Quelqu'un sait-il pourquoi?

Voici mon code:

 import qualified Data.Sequence as S
import qualified Data.Foldable as F

import System.Random
import Control.DeepSeq
import Criterion.Main
import Test.QuickCheck
import Control.Exception ( evaluate )

instance (Arbitrary a) => Arbitrary (S.Seq a) where
    arbitrary = fmap S.fromList arbitrary

instance NFData a => NFData (S.Seq a) where
    rnf = F.foldr seq ()

funs :: [(String, S.Seq Int -> Int)]
funs =
    [ ("seqDirect"   , seqDirect)
    , ("seqFoldr"    , seqFoldr)
    , ("seqFoldl'"   , seqFoldl')
    , ("seqSplit  1" , (seqSplit  1))
    , ("seqSplit  2" , (seqSplit  2))
    , ("seqSplit  4" , (seqSplit  4))
    , ("seqSplit  8" , (seqSplit  8))
    , ("seqSplit 16" , (seqSplit 16))
    , ("seqSplit 32" , (seqSplit 32)) ]

main :: IO ()
main = do
    mapM_ (\(_,f) -> quickCheck (\xs -> seqDirect xs == f xs)) funs
    gen <- newStdGen
    let inpt = S.fromList . take 100000 $ randoms gen
    evaluate (rnf inpt)
    defaultMain [ bench n (nf f inpt) | (n,f) <- funs ]

seqDirect :: S.Seq Int -> Int
seqDirect v = case S.viewl v of
    S.EmptyL -> 0
    x S.:< xs -> x + seqDirect xs

seqFoldr :: S.Seq Int -> Int
seqFoldr = F.foldr (+) 0

seqFoldl' :: S.Seq Int -> Int
seqFoldl' = F.foldl' (+) 0

seqSplit :: Int -> S.Seq Int -> Int
seqSplit 1 xs = seqFoldr xs
seqSplit _ xs | S.null xs = 0
seqSplit n xs =
    let (a, b) = S.splitAt (S.length xs `div` n) xs
        sa = seqFoldr a
        sb = seqSplit (n-1) b
    in  sa + sb
 

Et les résultats:

 $ ghc -V
The Glorious Glasgow Haskell Compilation System, version 7.0.4

$ ghc --make -O2 -fforce-recomp -rtsopts seq.hs
[1 of 1] Compiling Main             ( seq.hs, seq.o )
Linking seq ...

$ ./seq +RTS -s
./seq +RTS -s
+++ OK, passed 100 tests.
+++ OK, passed 100 tests.
+++ OK, passed 100 tests.
+++ OK, passed 100 tests.
+++ OK, passed 100 tests.
+++ OK, passed 100 tests.
+++ OK, passed 100 tests.
+++ OK, passed 100 tests.
+++ OK, passed 100 tests.
warming up
estimating clock resolution...
mean is 5.882556 us (160001 iterations)
found 2368 outliers among 159999 samples (1.5%)
  2185 (1.4%) high severe
estimating cost of a clock call...
mean is 85.26448 ns (44 iterations)
found 4 outliers among 44 samples (9.1%)
  3 (6.8%) high mild
  1 (2.3%) high severe

benchmarking seqDirect
mean: 23.37511 ms, lb 23.01101 ms, ub 23.77594 ms, ci 0.950
std dev: 1.953348 ms, lb 1.781578 ms, ub 2.100916 ms, ci 0.950

benchmarking seqFoldr
mean: 25.60206 ms, lb 25.39648 ms, ub 25.80034 ms, ci 0.950
std dev: 1.030794 ms, lb 926.7246 us, ub 1.156656 ms, ci 0.950

benchmarking seqFoldl'
mean: 10.65757 ms, lb 10.29087 ms, ub 10.99869 ms, ci 0.950
std dev: 1.819595 ms, lb 1.703732 ms, ub 1.922018 ms, ci 0.950

benchmarking seqSplit  1
mean: 25.50376 ms, lb 25.29045 ms, ub 25.71225 ms, ci 0.950
std dev: 1.075497 ms, lb 961.5707 us, ub 1.229739 ms, ci 0.950

benchmarking seqSplit  2
mean: 18.15032 ms, lb 17.62943 ms, ub 18.66413 ms, ci 0.950
std dev: 2.652232 ms, lb 2.288088 ms, ub 3.044585 ms, ci 0.950

benchmarking seqSplit  4
mean: 10.48334 ms, lb 10.14152 ms, ub 10.87061 ms, ci 0.950
std dev: 1.869274 ms, lb 1.690063 ms, ub 1.997915 ms, ci 0.950

benchmarking seqSplit  8
mean: 5.737956 ms, lb 5.616747 ms, ub 5.965689 ms, ci 0.950
std dev: 825.2361 us, lb 442.1652 us, ub 1.232003 ms, ci 0.950

benchmarking seqSplit 16
mean: 3.677038 ms, lb 3.669035 ms, ub 3.685547 ms, ci 0.950
std dev: 42.18741 us, lb 36.57112 us, ub 49.93574 us, ci 0.950

benchmarking seqSplit 32
mean: 2.855626 ms, lb 2.849962 ms, ub 2.862226 ms, ci 0.950
std dev: 31.25475 us, lb 26.49104 us, ub 37.18611 us, ci 0.950
  25,154,069,064 bytes allocated in the heap
   4,120,506,464 bytes copied during GC
      32,344,120 bytes maximum residency (446 sample(s))
       4,042,704 bytes maximum slop
              78 MB total memory in use (0 MB lost due to fragmentation)

  Generation 0: 42092 collections,     0 parallel,  6.57s,  6.57s elapsed
  Generation 1:   446 collections,     0 parallel,  2.62s,  2.62s elapsed

  INIT  time    0.00s  (  0.00s elapsed)

  MUT   time   18.57s  ( 18.58s elapsed)
  GC    time    9.19s  (  9.19s elapsed)
  EXIT  time    0.00s  (  0.00s elapsed)
  Total time   27.76s  ( 27.77s elapsed)

  %GC time      33.1%  (33.1% elapsed)

  Alloc rate    1,354,367,579 bytes per MUT second

  Productivity  66.9% of total user, 66.9% of total elapsed
 

Answer 1

Note: Cette réponse n'est pas réellement répondre à la question. Il ne reformule la question d'une manière différente. La raison précise pour laquelle Data.Sequence.foldr ralentit la séquence est de plus en plus gros à est encore inconnue.

---

Votre code

seqFoldr :: S.Seq Int -> Int
seqFoldr = F.foldr (+) 0

a non-linéaire des performances en fonction de la longueur de la séquence. Jetez un oeil à cette référence:

./seq-customized +RTS -s -A128M

[Length] [Performance of function seqFoldr]
25000:   mean: 1.096352 ms, lb 1.083301 ms, ub 1.121152 ms, ci 0.950
50000:   mean: 2.542133 ms, lb 2.514076 ms, ub 2.583209 ms, ci 0.950
100000:  mean: 6.068437 ms, lb 5.951889 ms, ub 6.237442 ms, ci 0.950
200000:  mean: 14.41332 ms, lb 13.95552 ms, ub 15.21217 ms, ci 0.950

À l'aide de la ligne de 25000 comme base nous donne le tableau suivant:

[Length] [Performance of function seqFoldr]
1x:      mean: 1.00 = 1*1.00
2x:      mean: 2.32 = 2*1.16
4x:      mean: 5.54 = 4*1.39
8x:      mean: 13.15 = 8*1.64

Dans le tableau ci-dessus, la non-linéarité est démontré par la série 1.00, 1.16, 1.39, de 1,64.

Voir aussi http://haskell.org/haskellwiki/Performance#Data.Sequence_vs._lists

---

En supposant que la longueur initiale de la séquence Seq xs est 100000 et n 32, votre code

seqSplit n xs =
let (a, b) = S.splitAt (S.length xs `div` n) xs
    sa = seqFoldr a
    sb = seqSplit (n-1) b
in  sa + sb

seront de passage à un peu plus courte Seqs à la fonction seqFoldr. Les longueurs successives de la Seqs passé depuis le code ci-dessus pour la fonction seqFoldr ressemble:

(length xs)/n = (length a)
--------------------------
100000/32 = 3125
(100000-3125)/31 = 3125
(100000-2*3125)/30 = 3125
...
(100000-30*3125)/2 = 3125

Basé sur la première partie de ma réponse (où nous avons vu que la performance est non-linéaire), [32 appels d' seqFoldr avec Seq de longueur 3125] exécutera plus rapidement que [1 appel à l' seqFoldr avec un seul Seq de longueur 32*3125=100000].

Ainsi, la réponse à votre question est: Parce qu' foldr sur Data.Sequence est plus lente que la séquence est d'obtenir plus grand.

Answer 2

Essayez d'utiliser foldr' au lieu de foldr . Je parie que c'est par un comportement paresseux de foldr qui conduit à allouer du thunk pour chaque donnée en séquence et à évaluer à la fin.

Éditer:

Donc, utiliser foldr' moitié a pris du temps dans le cas du mien mais encore plus lentement même alors foldl' . Ce qui signifie qu'il y a un problème de complexité dans l'implémentation de Data.Sequence.fold* .

 benchmarking seqFoldr
collecting 100 samples, 1 iterations each, in estimated 2.516484 s
bootstrapping with 100000 resamples
mean: 24.93222 ms, lb 24.72772 ms, ub 25.15255 ms, ci 0.950
std dev: 1.081204 ms, lb 938.4503 us, ub 1.332666 ms, ci 0.950
found 1 outliers among 100 samples (1.0%)
variance introduced by outliers: 0.999%
variance is unaffected by outliers

benchmarking seqFoldr'
collecting 100 samples, 1 iterations each, in estimated 902.7004 ms
bootstrapping with 100000 resamples
mean: 11.05375 ms, lb 10.68481 ms, ub 11.42519 ms, ci 0.950
std dev: 1.895777 ms, lb 1.685334 ms, ub 2.410870 ms, ci 0.950
found 1 outliers among 100 samples (1.0%)
variance introduced by outliers: 1.000%
variance is unaffected by outliers

benchmarking seqFoldl'
collecting 100 samples, 1 iterations each, in estimated 862.4077 ms
bootstrapping with 100000 resamples
mean: 10.35651 ms, lb 9.947395 ms, ub 10.73637 ms, ci 0.950
std dev: 2.011693 ms, lb 1.875869 ms, ub 2.131425 ms, ci 0.950
variance introduced by outliers: 1.000%
variance is unaffected by outliers