Efficient bitstreams in Haskell

问题:

In an ongoing endeavour to efficiently fiddle with bits (e.g. see this SO question) the newest challenge is the efficient streaming and consumption of bits.

As a first simple task I choose to find the longest sequence of identical bits in a bitstream generated by /dev/urandom. A typical incantation would be head -c 1000000 </dev/urandom | my-exe. The actual goal is to stream bits and decode an Elias gamma code, for example, i.e. codes that are not chunks of bytes or multiples thereof.

For such codes of variable length it is nice to have the take, takeWhile, group, etc. language for list manipulation. Since a BitStream.take would actually consume part of the bistream some monad would probably come into play.

The obvious starting point is the lazy bytestring from Data.ByteString.Lazy.

A. Counting bytes

This very simple Haskell program performs on par with a C program, as is to be expected.

import qualified Data.ByteString.Lazy as BSL

main :: IO ()
main = do
    bs <- BSL.getContents
    print $ BSL.length bs

B. Adding bytes

Once I start using unpack things should get worse.

main = do
    bs <- BSL.getContents
    print $ sum $ BSL.unpack bs

Suprisingly, Haskell and C show the almost same performance.

C. Longest sequence of identical bits

As a first nontrivial task the longest sequence of identical bits can be found like this:

module Main where

import           Data.Bits            (shiftR, (.&.))
import qualified Data.ByteString.Lazy as BSL
import           Data.List            (group)
import           Data.Word8           (Word8)

splitByte :: Word8 -> [Bool]
splitByte w = Prelude.map (\i-> (w `shiftR` i) .&. 1 == 1) [0..7]

bitStream :: BSL.ByteString -> [Bool]
bitStream bs = concat $ map splitByte (BSL.unpack bs)

main :: IO ()
main = do
    bs <- BSL.getContents
    print $ maximum $ length <$> (group $ bitStream bs)

The lazy bytestring is converted to a list [Word8] and then, using shifts, each Word is split into the bits, resulting in a list [Bool]. This list of lists is then flattened with concat. Having obtained a (lazy) list of Bool, use group to split the list into sequences of identical bits and then map length over it. Finally maximum gives the desired result. Quite simple, but not very fast:

# C
real    0m0.606s

# Haskell
real    0m6.062s

This naive implementation is exactly one order of magnitude slower.

Profiling shows that quite a lot of memory gets allocated (about 3GB for parsing 1MB of input). There is no massive space leak to be observed, though.

From here I start poking around:

• There is a bitstream package that promises "Fast, packed, strict bit streams (i.e. list of Bools) with semi-automatic stream fusion.". Unfortunately it is not up-to-date with the current vector package, see here for details.
• Next, I investigate streaming. I don't quite see why I should need 'effectful' streaming that brings some monad into play - at least until I start with the reverse of the posed task, i.e. encoding and writing bitstreams to file.
• How about just fold-ing over the ByteString? I'd have to introduce state to keep track of consumed bits. That's not quite the nice take, takeWhile, group, etc. language that is desirable.

And now I'm not quite sure where to go.

Update:

I figured out how to do this with streaming and streaming-bytestring. I'm probably not doing this right because the result is catastrophically bad.

import           Data.Bits                 (shiftR, (.&.))
import qualified Data.ByteString.Streaming as BSS
import           Data.Word8                (Word8)
import qualified Streaming                 as S
import           Streaming.Prelude         (Of, Stream)
import qualified Streaming.Prelude         as S

splitByte :: Word8 -> [Bool]
splitByte w = (\i-> (w `shiftR` i) .&. 1 == 1) <$> [0..7]

bitStream :: Monad m => Stream (Of Word8) m () -> Stream (Of Bool) m ()
bitStream s = S.concat $ S.map splitByte s

main :: IO ()
main = do
    let bs = BSS.unpack BSS.getContents :: Stream (Of Word8) IO ()
        gs = S.group $ bitStream bs ::  Stream (Stream (Of Bool) IO) IO ()
    maxLen <- S.maximum $ S.mapped S.length gs
    print $ S.fst' maxLen

This will test your patience with anything beyond a few thousand bytes of input from stdin. The profiler says it spends an insane amount of time (quadratic in the input size) in Streaming.Internal.>>=.loop and Data.Functor.Of.fmap. I'm not quite sure what the first one is, but the fmap indicates (?) that the juggling of these Of a b isn't doing us any good and because we're in the IO monad it can't be optimised away.

I also have the streaming equivalent of the byte adder here: SumBytesStream.hs, which is slightly slower than the simple lazy ByteString implementation, but still decent. Since streaming-bytestring is proclaimed to be "bytestring io done right" I expected better. I'm probably not doing it right, then.

In any case, all these bit-computations shouldn't be happening in the IO monad. But BSS.getContents forces me into the IO monad because getContents :: MonadIO m => ByteString m () and there's no way out.

Update 2

Following the advice of @dfeuer I used the streaming package at master@HEAD. Here's the result.

longest-seq-c       0m0.747s    (C)
longest-seq         0m8.190s    (Haskell ByteString)
longest-seq-stream  0m13.946s   (Haskell streaming-bytestring)

The O(n^2) problem of Streaming.concat is solved, but we're still not getting closer to the C benchmark.

Update 3

Cirdec's solution produces a performance on par with C. The construct that is used is called "Church encoded lists", see this SO answer or the Haskell Wiki on rank-N types.

Source files:

All the source files can be found on github. The Makefile has all the various targets to run the experiments and the profiling. The default make will just build everything (create a bin/ directory first!) and then make time will do the timing on the longest-seq executables. The C executables get a -c appended to distinguish them.

回答1:

Intermediate allocations and their corresponding overhead can be removed when operations on streams fuse together. The GHC prelude provides foldr/build fusion for lazy streams in the form of rewrite rules. The general idea is if one function produces a result that looks like a foldr (it has the type (a -> b -> b) -> b -> b applied to (:) and []) and another function consumes a list that looks like a foldr, constructing the intermediate list can be removed.

For your problem I'm going to build something similar, but using strict left folds (foldl') instead of foldr. Instead of using rewrite rules that try to detect when something looks like a foldl, I'll use a data type that forces lists to look like left folds.

-- A list encoded as a strict left fold.
newtype ListS a = ListS {build :: forall b. (b -> a -> b) -> b -> b}

Since I've started by abandoning lists we'll be re-implementing part of the prelude for lists.

Strict left folds can be created from the foldl' functions of both lists and bytestrings.

{-# INLINE fromList #-}
fromList :: [a] -> ListS a
fromList l = ListS (\c z -> foldl' c z l)

{-# INLINE fromBS #-}
fromBS :: BSL.ByteString -> ListS Word8
fromBS l = ListS (\c z -> BSL.foldl' c z l)

The simplest example of using one is to find the length of a list.

{-# INLINE length' #-}
length' :: ListS a -> Int
length' l = build l (\z a -> z+1) 0

We can also map and concatenate left folds.

{-# INLINE map' #-}
-- fmap renamed so it can be inlined
map' f l = ListS (\c z -> build l (\z a -> c z (f a)) z)

{-# INLINE concat' #-}
concat' :: ListS (ListS a) -> ListS a
concat' ll = ListS (\c z -> build ll (\z l -> build l c z) z)

For your problem we need to be able to split a word into bits.

{-# INLINE splitByte #-}
splitByte :: Word8 -> [Bool]
splitByte w = Prelude.map (\i-> (w `shiftR` i) .&. 1 == 1) [0..7]

{-# INLINE splitByte' #-}
splitByte' :: Word8 -> ListS Bool
splitByte' = fromList . splitByte

And a ByteString into bits

{-# INLINE bitStream' #-}
bitStream' :: BSL.ByteString -> ListS Bool
bitStream' = concat' . map' splitByte' . fromBS

To find the longest run we'll keep track of the previous value, the length of the current run, and the length of the longest run. We make the fields strict so that the strictness of the fold prevents chains of thunks from being accumulated in memory. Making a strict data type for a state is an easy way to get control over both its memory representation and when its fields are evaluated.

data LongestRun = LongestRun !Bool !Int !Int

{-# INLINE extendRun #-}
extendRun (LongestRun previous run longest) x = LongestRun x current (max current longest)
  where
    current = if x == previous then run + 1 else 1

{-# INLINE longestRun #-}
longestRun :: ListS Bool -> Int
longestRun l = longest
 where
   (LongestRun _ _ longest) = build l extendRun (LongestRun False 0 0)

And we're done

main :: IO ()
main = do
    bs <- BSL.getContents
    print $ longestRun $ bitStream' bs

This is much faster, but not quite the performance of c.

longest-seq-c       0m00.12s    (C)
longest-seq         0m08.65s    (Haskell ByteString)
longest-seq-fuse    0m00.81s    (Haskell ByteString fused)

The program allocates about 1 Mb to read 1000000 bytes from input.

total alloc =   1,173,104 bytes  (excludes profiling overheads)

Updated github code

回答2:

I found another solution that is on par with C. The Data.Vector.Fusion.Stream.Monadic has a stream implementation based on this Coutts, Leshchinskiy, Stewart 2007 paper. The idea behind it is to use a destroy/unfoldr stream fusion.

Recall that list's unfoldr :: (b -> Maybe (a, b)) -> b -> [a] creates a list by repeatedly applying (unfolding) a step-forward function, starting with an initial value. A Stream is just an unfoldr function with starting state. (The Data.Vector.Fusion.Stream.Monadic library uses GADTs to create constructors for Step that can be pattern-matched conveniently. It could just as well be done without GADTs, I think.)

The central piece of the solution is the mkBitstream :: BSL.ByteString -> Stream Bool function that turns a BytesString into a stream of Bool. Basically, we keep track of the current ByteString, the current byte, and how much of the current byte is still unconsumed. Whenever a byte is used up another byte is chopped off ByteString. When Nothing is left, the stream is Done.

The longestRun function is taken straight from @Cirdec's solution.

Here's the etude:

{-# LANGUAGE CPP #-}
#define PHASE_FUSED [1]
#define PHASE_INNER [0]
#define INLINE_FUSED INLINE PHASE_FUSED
#define INLINE_INNER INLINE PHASE_INNER
module Main where

import           Control.Monad.Identity            (Identity)
import           Data.Bits                         (shiftR, (.&.))
import qualified Data.ByteString.Lazy              as BSL
import           Data.Functor.Identity             (runIdentity)
import qualified Data.Vector.Fusion.Stream.Monadic as S
import           Data.Word8                        (Word8)

type Stream a = S.Stream Identity a   -- no need for any monad, really

data Step = Step BSL.ByteString !Word8 !Word8   -- could use tuples, but this is faster

mkBitstream :: BSL.ByteString -> Stream Bool
mkBitstream bs' = S.Stream step (Step bs' 0 0) where
    {-# INLINE_INNER step #-}
    step (Step bs w n) | n==0 = case (BSL.uncons bs) of
                            Nothing        -> return S.Done
                            Just (w', bs') -> return $ 
                                S.Yield (w' .&. 1 == 1) (Step bs' (w' `shiftR` 1) 7)
                       | otherwise = return $ 
                                S.Yield (w .&. 1 == 1) (Step bs (w `shiftR` 1) (n-1))


data LongestRun = LongestRun !Bool !Int !Int

{-# INLINE extendRun #-}
extendRun :: LongestRun -> Bool -> LongestRun
extendRun (LongestRun previous run longest) x  = LongestRun x current (max current longest)
    where current = if x == previous then run + 1 else 1

{-# INLINE longestRun #-}
longestRun :: Stream Bool -> Int
longestRun s = runIdentity $ do
    (LongestRun _ _ longest) <- S.foldl' extendRun (LongestRun False 0 0) s
    return longest

main :: IO ()
main = do
    bs <- BSL.getContents
    print $ longestRun (mkBitstream bs)