1到sys.maxsize范围内的随机数始终为1 mod 2 ^ 10

Question

我试图通过使用频率测试,运行测试和卡方检验来找到Python(2.7.10)中可用的PRNG的统计属性.

为了进行频率测试,我需要将生成的随机数转换为二进制表示,然后计算1's和0's 的分布.我正在试验python控制台上随机数的二进制表示,并观察到这种奇怪的行为:

>>> for n in random.sample(xrange(1, sys.maxsize), 50):
...     print '{0:b}'.format(n)
... 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111110000001110010001110111101110101010110001110000000000000001
100000101101000110101010010000101101000011111010001110000000001
101001011101100011001000011010010000000111110111100010000000001
10110101010000111010110111001111011000001111001100110000000001
10110111100100100011100101001100000000101110100100010000000001
10010111110001011101001110000111011010110100110111110000000001
111011110010110111011011101011001100001000111001010100000000001
101001010001010100010010010001100111101110101111000110000000001
101011111010000101010101000110001101001001011110000000000001
1010001010111101101010111110110110000001111101101110000000001
10111111111010001000110000101101010101011010101100000000001
101011101010110000001111010100100110000011111100100100000000001
111100001101111010100111010001010010000010110110010110000000001
100111111000100110100001110101000010111111010010010000000000001
100111100001011100011000000000101100111111000111100110000000001
110110100000110111011101110101101000101110111111010110000000001
>>> 

如您所见,所有数字都以数字结尾0000000001,即所有数字都是1 mod 2^10.为什么会这样？

此外,当范围是时,会观察到此行为1 to sys.maxsize.如果指定范围1 to 2^40,则不会观察到.我想知道这种行为的原因以及我的代码中是否有任何错误.

实现我正在使用的PRNG的随机库的文档在这里.

如果我应该提供更多信息,请告诉我.

Answer 1

@roeland暗示了原因:在Python 2中,重复sample()使用int(random.random() * n).查看源代码(在Python中Lib/random.py)以获取完整的详细信息.简而言之,random.random()返回不超过53个有效(非零)前导位; 然后int()用零填充其余的低位(你显然在机器上sys.maxsize == 2**63 - 1); 然后xrange(1, sys.maxsize)用一个偶数整数索引你的base(),其中"很多"的低位0位总是返回一个具有相同数目的低位0位的奇数(除了最后一位).

在Python 3中,没有发生这种情况 - random在Python 3中使用了更强大的算法,并且只random.random()在必要时才回退.例如,在Python 3.4.3下:

>>> hex(random.randrange(10**70))
'0x91fc11ed768be3a454bd66f593c218d8bbfa3b99f6285291e1d9f964a9'
>>> hex(random.randrange(10**70))
'0x7b07ff02b6676801e33094fca2fcca7f6e235481c479c521643b1acaf4'

编辑

这是一个更直接相关的例子,在64位框下的3.4.3下:

>>> import random, sys
>>> sys.maxsize == 2**63 - 1
True
>>> for i in random.sample(range(1, sys.maxsize), 6):
...    print(bin(i))
0b10001100101001001111110110011111000100110100111001100000010110
0b100111100110110100111101001100001100110001110010000101101000101
0b1100000001110000110100111101101010110001100110101111011100111
0b111110100001111100101001001001101101100100011001001010100001110
0b1100110100000011100010000011010010100100110111001111100110100
0b10011010000110101010101110001000101110111100100001111101110111

random.random()在这种情况下,Python 3根本不会调用,而是迭代地从底层的Mersenne Twister中抓取32位的块(32位无符号整数是来自MT实现的"自然"输出),将它们粘贴在一起构建一个合适的指数.因此,在Python 3中,平台浮动与它无关; 在Python 2中,浮动行为的怪癖与它有关.

Answer 2

这取决于很多事情,比如RNG的实现方式,它使用的状态位数以及sample函数的实现方式.

这是文档说的内容:

几乎所有模块函数都依赖于基本函数random(),它在半开放范围[0.0,1.0]内均匀生成随机浮点数.Python使用Mersenne Twister作为核心生成器.它产生53位精度浮点数,周期为2**19937-1.

因此,如果sample确实random()在引擎盖下使用,那么您应该只期望结果中有53位有意义的位.