维吉尼亚密码破解（Python完整详细源码）

维吉尼亚密码破解（Python完整详细源码）

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https://lunan0320.github.io/

1、写在前面

关于维吉尼亚密码破解

希望大家弄明白原理，不要只要代码

这里因为latex写的报告拷贝出现乱码，就把我们实验报告中的原理部分以图片形式分享出来，供大家学习。

2、维吉尼亚密码原理

3、维吉尼亚密码破解代码

import vigenerecipher
import string
import re
def gcd(a,b):
    if a<b:
        a,b=b,a
    if a%b==0:
        return b
    else:
        return gcd(b,a%b)

def findstr(Ctext,str_):
    interval_list=[]
    loc=0
    array_locs=[]
    while loc<len(Ctext):
        loc=Ctext.find(str_,loc)
        
        if loc==-1:
            break
        array_locs.append(loc)
        loc=loc+1
    index=0
    while index+1<len(array_locs):
        interval=array_locs[index+1]-array_locs[index]
        print("间隔：",interval)
        interval_list.append(interval)
        index=index+1
    interval_len=gcd(interval_list[0],interval_list[1])
    print("当重复值为",str_,"时，两两之间距离的最大公因数：",interval_len,"\n")
    return interval_len
    
def find_repeat(Ciphertext,repeat_list):
    while len(Ciphertext)>200:
        Ciphertext=list(Ciphertext)
        Ciphertext.pop(1)
        Ciphertext=''.join(Ciphertext)
        list1=re.findall(r'.{3}',Ciphertext)
    
        list2=[0]*len(list1)
        
        for i in range(len(list1)):
            for j in range(len(list1)):
                if(list1[i]==list1[j]):
                    list2[i]=list2[i]+1
                    
        max_len=max(list2)
        a=list2.index(max(list2))

        if list1[a] not in repeat_list:
            if max_len==3:
                repeat_list.append(list1[a])
                print("重复次数次数：",max_len)
                print("重复值：",list1[a])
                print("\n")

def check_len(Ciphertext,interval_len):
    ListCiphertext=list(Ciphertext)
    Keylength=1

    while Keylength<interval_len+1:
        #指数初始化为０
        CoincidenceIndex = 0

        #使用切片分组
        for i in range(Keylength):
            Numerator = 0
            PresentCipherList = ListCiphertext[i::Keylength]

            #使用集合去重，计算每一子密文组重合指数
            for Letter in set(PresentCipherList):
                Numerator += PresentCipherList.count(Letter) * (PresentCipherList.count(Letter)-1)
            CoincidenceIndex += Numerator/(len(PresentCipherList) * (len(PresentCipherList)-1))

        #求各子密文组的拟重合指数的平均值
        Average=CoincidenceIndex / Keylength
        Keylength += 1

        #均值＞0.6即可退出循环
        if Average > 0.06:
            break

    Keylength -= 1
    print("经重合指数验证后，密钥长度最可能为:",Keylength,"\n")
    return Keylength
def keyword(Ciphertext,keylength):
    ListCiphertext = list(Ciphertext)
    #标准数据来源于课本
    Standard = {'A':0.082,'B':0.015,'C':0.028,'D':0.043,'E':0.127,'F':0.022,'G':0.020,'H':0.061,'I':0.070,'J':0.002,'K':0.008,'L':0.040,'M':0.024,'N':0.067,'O':0.075,'P':0.019,'Q':0.001,'R':0.060,'S':0.063,'T':0.091,'U':0.028,'V':0.010,'W':0.023,'X':0.001,'Y':0.020,'Z':0.001}

    while True:
        KeyResult = []

        for i in range(keylength):
            # 使用切片分组
            PresentCipherList = ListCiphertext[i::keylength]

            #初始化重合指数最大值为０，检验移动位数对应字符以＊代替
            QuCoincidenceMax = 0
            KeyLetter = "*"

            #遍历移动的位数
            #m是密钥对应的英文字母
            for m in range(26):
                #初始化当前移动位数的重合互指数为０
                QuCoincidencePresent = 0

                #遍历计算重合指数：各个字符的频率＊对应英文字符出现的标准频率－－－的和
                for Letter in set(PresentCipherList):
                    #fi/n
                    LetterFrequency = PresentCipherList.count(Letter) / len(PresentCipherList)

                    # 标准频率
                    #ord(Letter) - 65是将letter对应的字母化为26内的数值，然后与m运算，得到的k是对应的明文字母
                    k = chr( ( ord(Letter) - 65 - m ) % 26 + 65 )
                    StandardFrequency = Standard[k]

                    #计算重合互指数，累加遍历26个英文字母
                    QuCoincidencePresent = QuCoincidencePresent + LetterFrequency * StandardFrequency

                #保存遍历过程中重合指数的最大值，同时保存对应应对的位数，即对应key的字符
                if QuCoincidencePresent > QuCoincidenceMax:
                    QuCoincidenceMax = QuCoincidencePresent
                    #m是26个英文对应的位置，从0开始，+65是因为A在ascii中是65
                    KeyLetter = chr( m +65 )
            print("第",i+1,"个密钥字母为:",KeyLetter,"对应的重合互指数为:",QuCoincidenceMax)
            #保存当前位置key的值，退出循环，进行下一组子密文移动位数的尝试
            KeyResult.append( KeyLetter )
        #列表转为字符串
        Key = "".join(KeyResult)
        break
    return Key

if __name__ == '__main__':

    Ciphertext = input("输入密文：").upper()
    repeat_list=[]
    find_repeat(Ciphertext,repeat_list)
    print("kasisiki测试法:\n重复列表",repeat_list)
    for m in repeat_list:
        interval_len=findstr(Ciphertext,m)
    
    key_len=check_len(Ciphertext,interval_len)

    KeyResult = keyword(Ciphertext,key_len)

    print("密钥最可能为：" , KeyResult,"\n")

    #已知秘钥可用python自带维吉尼亚解密
    ClearText = vigenerecipher.decode( Ciphertext,KeyResult )
    print("解密结果为：" , ClearText)