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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