维吉尼亚解密密钥

目录1.脚本破解前置安装numpy安装wordninja2.网站解密写在前面：参考大佬博客，加自己的一点想法，仅供学习参考1.脚本破解前置运行环境:python3 numpy 、wordninja两个库安装numpy命令pip install numpy或者python -m pip install numpy安装是否成功检验打开cmd输入python输入import numpy as nu如上图

_Jasmine_

1700人浏览 · 2021-11-08 18:00:12

_Jasmine_ · 2021-11-08 18:00:12 发布

写在前面：参考大佬博客，加自己的一点想法，仅供学习参考

1.脚本破解

前置

运行环境:python3 numpy 、wordninja两个库

安装numpy

命令

pip install numpy

或者

python -m pip install numpy

安装是否成功检验

打开cmd

输入python

输入

import numpy as nu

如上图所示，则安装成功

安装wordninja

命令

pip install wordninja

或者

python -m pip install wordninja

安装成功检验

输入

import wordninja

维吉尼亚密码密钥解密脚本

import numpy as np
import wordninja


def alpha(cipher):  # 预处理,去掉空格以及回车
    c = ''
    for i in range(len(cipher)):
        if (cipher[i].isalpha()):
            c += cipher[i]
    return c


def count_IC(cipher):  # 给定字符串计算其重合指数
    count = [0 for i in range(26)]
    L = len(cipher)
    IC = 0.0
    for i in range(len(cipher)):
        if (cipher[i].isupper()):
            count[ord(cipher[i]) - ord('A')] += 1
        elif (cipher[i].islower()):
            count[ord(cipher[i]) - ord('a')] += 1
    for i in range(26):
        IC += (count[i] * (count[i] - 1)) / (L * (L - 1))
    return IC


def count_key_len(cipher, key_len):  # 对字符串按输入个数进行分组，计算每一组的IC值返回平均值
    N = ['' for i in range(key_len)]
    IC = [0 for i in range(key_len)]
    for i in range(len(cipher)):
        m = i % key_len
        N[m] += cipher[i]
    for i in range(key_len):
        IC[i] = count_IC(N[i])
    # print(IC)
    print("长度为%d时,平均重合指数为%.5f" % (key_len, np.mean(IC)))
    return np.mean(IC)


def length(cipher):  # 遍历确定最有可能的密钥长度返回密钥长度
    key_len = 0
    mins = 100
    aver = 0.0
    for i in range(1, 10):
        k = count_key_len(cipher, i)
        if (abs(k - 0.065) < mins):
            mins = abs(k - 0.065)
            key_len = i
            aver = k
    print("密钥长度为%d,此时重合指数每组的平均值为%.5f" % (key_len, aver))
    return key_len


def count_MIC(c1, c2, n):  # n=k1-k2为偏移量,计算c1,c2互重合指数MIC
    count_1 = [0 for i in range(26)]
    count_2 = [0 for i in range(26)]
    L_1 = len(c1)
    L_2 = len(c2)
    MIC = 0
    for i in range(L_1):
        if (c1[i].isupper()):
            count_1[ord(c1[i]) - ord('A')] += 1
        elif (c1[i].islower()):
            count_1[ord(c1[i]) - ord('a')] += 1
    for i in range(L_2):
        if (c2[i].isupper()):
            count_2[(ord(c2[i]) - ord('A') + n + 26) % 26] += 1
        elif (c2[i].islower()):
            count_2[(ord(c2[i]) - ord('a') + n + 26) % 26] += 1
    for i in range(26):
        MIC += count_1[i] * count_2[i] / (L_1 * L_2)
    return MIC


def count_n(c1, c2):  # 确定两个子串最优的相对偏移量n=k1-k2
    n = 0
    mins = 100
    k = [0.0 for i in range(26)]
    for i in range(26):
        k[i] = count_MIC(c1, c2, i)
        # print(i,k[i])
        if (abs(k[i] - 0.065) < mins):
            mins = abs(k[i] - 0.065)
            n = i
    return n


def group_k(cipher, key_len):  # 完成分组操作并计算每一组与第一组的最优相对偏移量并返回
    N = ['' for i in range(key_len)]
    MIC = [0 for i in range(key_len)]
    s = [0 for i in range(key_len)]
    for i in range(len(cipher)):  # 对密文进行分组
        m = i % key_len
        N[m] += cipher[i]
    for i in range(1, key_len):  # 计算与第一组之间的相对偏移量
        s[i] = count_n(N[0], N[i])  # s[i] = k1-k(i+1)
        MIC[i] = count_MIC(N[0], N[i], s[i])  # MIC[i] = MIC(1,i+1)
        print("第1组和第%d组之间偏移为%d时，互重合指数为%.5f" % (i + 1, s[i], MIC[i]))
    return s


def miyao(key_len, s, k):  # k为第一个子串的移位，输出密钥并返回密钥所有字母的下标
    mi = ['' for i in range(key_len)]
    for i in range(key_len):
        s[i] = -s[i] + k  # k2=k1-n
        mi[i] = chr((s[i] + 26) % 26 + ord('a'))
    print("第一个偏移量为%d,密钥为%s时" % (k, mi))
    return s


def the_end(cipher, key_len, s):  # 输入密文密钥返回明文结果
    plain = ''
    i = 0
    while (i < len(cipher)):
        for j in range(key_len):
            if (cipher[i].isupper()):
                plain += chr((ord(cipher[i]) - ord('A') - s[j] + 26) % 26 + ord('A'))
            else:
                plain += chr((ord(cipher[i]) - ord('a') - s[j] + 26) % 26 + ord('a'))
            i += 1
            if (i == len(cipher)):
                break
                # print(plain)
    return plain
5

if __name__ == "__main__":
    fp = open("C:/Users/Jasmine/Desktop/1.txt", "r")
    cipher = ''
    for i in fp.readlines():
        cipher = cipher + i
    fp.close()
    cipher = alpha(cipher)
    key_len = length(cipher)
    s = group_k(cipher, key_len)
    m = s.copy()
    for k in range(26):
        s = m.copy()
        s = miyao(key_len, s, k)
        plain = the_end(cipher, key_len, s)
        print(plain[0:20])  # 输出部分明文确定偏移量k1
    print("参考输出，请输入第一个子串的偏移量:", end='')
    k = int(input())
    m = miyao(key_len, m, k)
    plain = the_end(cipher, key_len, m)

    '''对英文文本进行分词'''
    word = wordninja.split(plain)
    plain = ''
    for i in range(len(word)):
        plain += word[i]
        plain += ' '
    print("明文为\n" + plain)

代码需要优化（菜鸡之后写bug）,不是每一个都能很准确的跑出来，但还是很好用的，感谢大佬！