Python代码实现NIST随机性测试

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本文介绍了在科研中如何使用Python实现NIST随机性测试，针对数据量不足导致的UNDERFLOW问题，作者分享了遇到的困难以及官方文档的参考。最终，作者整理并编写了15个单独的Python测试代码，结果显示13项成功，2项未通过。资源已上传至优快云，待审核后提供下载链接。

最近科研的需要，需要测试二进制序列的随机性

找遍所有内网都没有找到自己合适的代码，网上很多都是讲解自己怎么去下载和安装sts-2.1.2的开发包，于是我也是一开始就入坑了，之前也是写了一篇比较完整的工具包的下载和安装的教程，点击打开链接，但是如果你按照我的安装步骤的话，成功安装肯定是没有问题的，但是你的使用就会出现各种各样的问题：

比如我在使用的过程中遇到的问题有：首先，就是经常出现UNDERFLOW的情况，这种情况下，一般都是因为数据量不够大的情况下造成的，所以如果你用过，你就会发现，想要测试15项的内容，大约你需要1GBit,这对于我们正常的情况下，是非常难的，根本达不到这个数据量，没错，肯定有的人会说，你可以不测试那么多的数据量啊，你可以测试部分的测试项啊，我也想啊，但是网上根本没有发现怎么使用，单独测试，而不是使用15项

我按照他的步骤，每次单独测试的时候总是会出现下面的情况，（我不知道你们有没有遇到）

看这个提示：如果你不想测试所有的项，请输入0，否则输入1.

然后我就输入0

从这里开始，无论你输入0还是1，都是没有响应了。我一直也没有找到任何解决的办法，如果有人知道，请留言交流。

但是又因为迫于科研的压力，还是要解决这个事情，这个是NIST的官方文档

参考开发文档，和国外大牛写的代码：整理并编写python如下：

这里把15个测试项都单独的编写了一个可以运行的python代码：

先看一下我的结果图：有两项不pass，13项success、

第一个是：

cumulative_sums_test

import math
#from scipy.special import gamma, gammainc, gammaincc
from gamma_functions import *
#import scipy.stats

def normcdf(n):
    return 0.5 * math.erfc(-n * math.sqrt(0.5))

def p_value(n,z):
    sum_a = 0.0
    startk = int(math.floor((((float(-n)/z)+1.0)/4.0)))
    endk   = int(math.floor((((float(n)/z)-1.0)/4.0)))
    for k in range(startk,endk+1):
        c = (((4.0*k)+1.0)*z)/math.sqrt(n)
        #d = scipy.stats.norm.cdf(c)
        d = normcdf(c)
        c = (((4.0*k)-1.0)*z)/math.sqrt(n)
        #e = scipy.stats.norm.cdf(c)
        e = normcdf(c)
        sum_a = sum_a + d - e

    sum_b = 0.0
    startk = int(math.floor((((float(-n)/z)-3.0)/4.0)))
    endk   = int(math.floor((((float(n)/z)-1.0)/4.0)))
    for k in range(startk,endk+1):
        c = (((4.0*k)+3.0)*z)/math.sqrt(n)
        #d = scipy.stats.norm.cdf(c)
        d = normcdf(c)
        c = (((4.0*k)+1.0)*z)/math.sqrt(n)
        #e = scipy.stats.norm.cdf(c)
        e = normcdf(c)
        sum_b = sum_b + d - e 

    p = 1.0 - sum_a + sum_b
    return p
    
def cumulative_sums_test(bits):
    n = len(bits)
    # Step 1
    x = list()             # Convert to +1,-1
    for bit in bits:
        #if bit == 0:
        x.append((bit*2)-1)
        
    # Steps 2 and 3 Combined
    # Compute the partial sum and records the largest excursion.
    pos = 0
    forward_max = 0
    for e in x:
        pos = pos+e
        if abs(pos) > forward_max:
            forward_max = abs(pos)
    pos = 0
    backward_max = 0
    for e in reversed(x):
        pos = pos+e
        if abs(pos) > backward_max:
            backward_max = abs(pos)
     
    # Step 4
    p_forward  = p_value(n, forward_max)
    p_backward = p_value(n,backward_max)
    
    success = ((p_forward >= 0.01) and (p_backward >= 0.01))
    plist = [p_forward, p_backward]

    if success:
        print ("PASS")
    else:    
        print ("FAIL: Data not random")
    return (success, None, plist)

bits=[1,1,0,1,0,0,1,1,0,1,0,1,0,1,0,1,1,0,0,1,0,1,1,1,1,1,
          0,1,1,1,1,1,0,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,0,0,0,
          0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,1,0,0,1,0,0,0,0,1,0,1,0,
          0,1,1,0,0,0,1,1,1,0,1,0,0,0,0,1,0,0,1,0,1,0,1,0,0,1,1,
          0,0,0,1,1,0,1,0,1,1,1,0,0,1,1,1,1,1,0,0,0] 
if __name__ == "__main__":
     
    #bits = [1,1,0,0,1,0,0,1,0,0,0,0,1,1,1,1,1,1,0,1,
           # 1,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,0,1,
            #0,1,1,0,1,0,0,0,1,1,0,0,0,0,1,0,0,0,1,1,
            #0,1,0,0,1,1,0,0,0,1,0,0,1,1,0,0,0,1,1,0,
            #0,1,1,0,0,0,1,0,1,0,0,0,1,0,1,1,1,0,0,0]
    
    success, _, plist = cumulative_sums_test(bits)

    print ("success =",success)
    print ("plist = ",plist)

第二个是：maurers_universal_test

import math

def pattern2int(pattern):
    l = len(pattern)
    n = 0
    for bit in (pattern):
        n = (n << 1) + bit
    return n          
         
def maurers_universal_test(bits,patternlen=None, initblocks=None):
    n = len(bits)

    # Step 1. Choose the block size
    if patternlen != None:
        L = patternlen  
    else: 
        ns = [904960,2068480,4654080,10342400,
              22753280,49643520,107560960,
              231669760,496435200,1059061760]
        L = 6
        if n < 387840:
            print ("Error. Need at least 387840 bits. Got %d." % n)
            exit()
        for threshold in ns:
            if n >= threshold:
                L += 1 

    # Step 2 Split the data into Q and K blocks
    nblocks = int(math.floor(n/L))
    if initblocks != None:
        Q = initblocks
    else:
        Q = 10*(2**L)
    K = nblocks - Q
    
    # Step 3 Construct Table
    nsymbols = (2**L)
    T=[0 for x in range(nsymbols)] # zero out the table
    for i in range(Q):             # Mark final position of
        pattern = bits[i*L:(i+1)*L] # each pattern
        idx = pattern2int(pattern)
        T[idx]=i+1      # +1 to number indexes 1..(2**L)+1
                        # instead of 0..2**L
    # Step 4 Iterate
    sum = 0.0
    for i in range(Q,nblocks):
        pattern = bits[i*L:(i+1)*L]
        j = pattern2int(pattern)
        dist = i+1-T[j]
        T[j] = i+1
        sum = sum + math.log(dist,2)
    print ("  sum =", sum)
    
    # Step 5 Compute the test statistic
    fn = sum/K
    print ("  fn =",fn)
       
    # Step 6 Compute the P Value
    # Tables from https://static.aminer.org/pdf/PDF/000/120/333/
    # a_universal_statistical_test_for_random_bit_generators.pdf
    ev_table =  [0,0.73264948,1.5374383,2.40160681,3.31122472,
                 4.25342659,5.2177052,6.1962507,7.1836656,
                 8.1764248,9.1723243,10.170032,11.168765,
                 12.168070,13.167693,14.167488,15.167379]
    var_table = [0,0.690,1.338,1.901,2.358,2.705,2.954,3.125,
                 3.238,3.311,3.356,3.384,3.401,3.410,3.416,
                 3.419,3.421]
                 
    # sigma = math.sqrt(var_table[L])
    mag = abs((fn - ev_table[L])/((math.sqrt(var_table[L]))*math.sqrt(2)))
    P = math.erfc(mag)

    success = (P >= 0.01)
    return (success, P, None)
    

if __name__ == "__main__":
    #bits = [0,1,0,1,1,0,1,0,0,1,1,1,0,1,0,1,0,1,1,1]
    bits=[1,1,0,1,0,0,1,1,0,1,0,1,0,1,0,1,1,0,0,1,0,1,1,1,1,1,
          0,1,1,1,1,1,0,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,0,0,0,
          0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,1,0,0,1,0,0,0,0,1,0,1,0,
          0,1,1,0,0,0,1,1,1,0,1,0,0,0,0,1,0,0,1,0,1,0,1,0,0,1,1,
          0,0,0,1,1,0,1,0,1,1,1,0,0,1,1,1,1,1,0,0,0] 
    success, p, _ = maurers_universal_test(bits, patternlen=2, initblocks=4)
    
    print ("success =",success)
    print ("p       = ",p)

这个好没效率，15项的代码太多了，等我把资源上传到csdn吧，等审核通过，我再来附链接。

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