Haskell`randoms`功能在我的库中表现不佳

Question

我正在尝试编写一个用于加密安全随机数的Haskell库.代码如下:

module URandom (URandom, initialize) where

import qualified Data.ByteString.Lazy as B
import System.Random
import Data.Word

newtype URandom = URandom [Word8]

instance RandomGen URandom where
  next (URandom (x : xs)) = (fromIntegral x, URandom xs)
  split (URandom l) = (URandom (evens l), URandom (odds l))
    where evens (x : _ : xs) = x : evens xs
          odds (_ : x : xs) = x : odds xs
  genRange _ = (fromIntegral (minBound :: Word8), fromIntegral (maxBound :: Word8))

initialize :: IO URandom
initialize = URandom . B.unpack <$> B.readFile "/dev/urandom"

不幸的是,它并不像我想要的那样.特别是表演

take 10 . randoms <$> initialize

产量(类似于)

[-4611651379516519433,-4611644973572935887,-31514321567846,9223361179177989878,-4611732094835278236,9223327886739677537,4611709625714976418,37194416358963,4611669560113361421,-4611645373004878170,-9223329383535098640,4611675323959360258,-27021785867556,9223330964083681227,4611705212636167666]

对我而言,未经训练的,未经训练的眼睛,看起来并不是很随机.很多46 ......和92 ......在那里.

怎么可能出错？为什么这不能产生分布均匀的数字？值得注意的是,即使我将Word8s 连接在一起形成Ints,分布也没有改善,我认为这里的代码不值得包含.

编辑:这里有一些证据没有正确分发.我写了一个名为histogram的函数:

histogram :: ? t . (Integral t, Bounded t)
          => [t] -> Int -> S.Seq Int
histogram [] buckets = S.replicate buckets 0
histogram (x : xs) buckets = S.adjust (+ 1) (whichBucket x) (histogram xs buckets)
  where whichBucket x = fromIntegral $ ((fromIntegral x * fromIntegral buckets) :: Integer) `div` fromIntegral (maxBound :: t)

当我跑

g <- initialize
histogram (take 1000000 $ randoms g :: [Word64]) 16

我回来了

fromList [128510,0,0,121294,129020,0,0,122090,127873,0,0,120919,128637,0,0,121657]

有些水桶是空的!

Answer 1

问题是在错误random-1.0.1.1已修复中random-1.1.更改日志指向此票证.特别是,参考旧版本:

它还假设所有RandomGen实现产生与StdGen相同的随机值范围.

这里,一次产生8位随机性,这导致观察到的行为.

random-1.1 解决了这个:

此实现也适用于任何RandomGen,即使是每次下一次调用只产生一位熵或具有非零的最小边界的RandomGen.