最近科研的需要，需要測試二進制序列的隨機性

找遍所有內網都沒有找到自己合適的代碼，網上很多都是講解自己怎么去下載和安裝sts-2.1.2的開發包，于是我也是一開始就入坑了，之前也是寫了一篇比較完整的工具包的下載和安裝的教程，點擊打開鏈接，但是如果你按照我的安裝步驟的話，成功安裝肯定是沒有問題的，但是你的使用就會出現各種各樣的問題：

比如我在使用的過程中遇到的問題有：首先，就是經常出現UNDERFLOW的情況，這種情況下，一般都是因為數據量不夠大的情況下造成的，所以如果你用過，你就會發現，想要測試15項的內容，大約你需要1GBit,這對于我們正常的情況下，是非常難的，根本達不到這個數據量，沒錯，肯定有的人會說，你可以不測試那么多的數據量啊 ，你可以測試部分的測試項啊，我也想啊，但是網上根本沒有發現怎么使用，單獨測試，而不是使用15項

我按照他的步驟，每次單獨測試的時候總是會出現下面的情況，（我不知道你們有沒有遇到）

看這個提示：如果你不想測試所有的項，請輸入0，否則輸入1.

然后我就輸入0

從這里開始，無論你輸入0還是1，都是沒有響應了。我一直也沒有找到任何解決的辦法，如果有人知道，請留言交流。

但是又因為迫于科研的壓力，還是要解決這個事情，這個是NIST的官方文檔

參考開發文檔，和國外大牛寫的代碼：整理并編寫python如下：

這里把15個測試項都單獨的編寫了一個可以運行的python代碼：

先看一下我的結果圖：有兩項不pass，13項success、

第一個是：

cumulative_sums_test import math #from scipy.special import gamma, gammainc, gammaincc from gamma_functions import * #import scipy.statsdef normcdf(n):return 0.5 * math.erfc(-n * math.sqrt(0.5))def p_value(n,z):sum_a = 0.0startk = 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 - esum_b = 0.0startk = 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_breturn pdef cumulative_sums_test(bits):n = len(bits)# Step 1x = list() # Convert to +1,-1for 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 = 0forward_max = 0for e in x:pos = pos+eif abs(pos) > forward_max:forward_max = abs(pos)pos = 0backward_max = 0for e in reversed(x):pos = pos+eif abs(pos) > backward_max:backward_max = abs(pos)# Step 4p_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) 第二個是：maurers_universal_test
import mathdef pattern2int(pattern):l = len(pattern)n = 0for bit in (pattern):n = (n << 1) + bitreturn n def maurers_universal_test(bits,patternlen=None, initblocks=None):n = len(bits)# Step 1. Choose the block sizeif patternlen != None:L = patternlen else: ns = [904960,2068480,4654080,10342400,22753280,49643520,107560960,231669760,496435200,1059061760]L = 6if 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 blocksnblocks = int(math.floor(n/L))if initblocks != None:Q = initblockselse:Q = 10*(2**L)K = nblocks - Q# Step 3 Construct Tablensymbols = (2**L)T=[0 for x in range(nsymbols)] # zero out the tablefor i in range(Q): # Mark final position ofpattern = bits[i*L:(i+1)*L] # each patternidx = pattern2int(pattern)T[idx]=i+1 # +1 to number indexes 1..(2**L)+1# instead of 0..2**L# Step 4 Iteratesum = 0.0for i in range(Q,nblocks):pattern = bits[i*L:(i+1)*L]j = pattern2int(pattern)dist = i+1-T[j]T[j] = i+1sum = sum + math.log(dist,2)print (" sum =", sum)# Step 5 Compute the test statisticfn = sum/Kprint (" 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.pdfev_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)

這個好沒效率，15項的代碼太多了，等我把資源上傳到csdn吧，等審核通過，我再來附鏈接。

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