Python无密钥破解Vigenere密码示例
【这是密码学课上的一个作业题】
Vigenere密码即为传统多表代换密码，使用一密钥词对每个明文字母选择一个字母表来加密，如果到了密钥词的最后一个字母，则从头开始继续。
举例如下，比如使用密钥deceptive加密这样一段话：we are discovered save yourself
| Key | d | e | c | e | p | t | i | v | e | d | e | c | e | p | t | i | v | e | d | e | c | e | p | t | i | v | e |
| PlainText | w | e | a | r | e | d | i | s | c | o | v | e | r | e | d | s | a | v | e | y | o | u | r | s | e | l | f |
| CipherText | Z | I | C | V | T | W | Q | N | G | R | Z | G | V | T | W | A | V | Z | H | C | Q | Y | G | L | M | G | J |
加密之后密文即为：ZICVTWQNGRZGVTWAVZHCQYGLMGJ ，加密的方法即明文字母根据密钥字母向后平移不同的位数。与单字母表加密方法不同，单字母表每个明文字母向后平移相同的位数。所以单字母表加密保留了明文的英文字母的统计特性，由于英文字母的相对使用频率是固定的统计值，所以很容易被破译。
Vigenere密码因为需要猜测更多的字母表，并且频率分布特性也更加平坦，所以使得密码破译相对困难。
但是可以通过分析密文中的重复性来猜测密钥的长度，因为如果两个相同明文序列之间的距离是密钥词长度的整数倍，那么产生的密文序列也是相同的。再根据英文字母的频率分布特性，即可以实现破解。
这里即使用这种方法猜解Vigenere密码的密钥并实现破译。
密文如下，保存于vigenereCipherText文件中：
wubefiqlzurmvofehmymwtixcgtmpifkrzupmvoirqmmwozmpulmbnyvqqqmvmvjleymhfefnzpsdlppsdlpevqmwcxymdavqeefiqcaytqowcxymwmsemefcf
wyeyqetrliqycgmtwcwfbsmyfplrxtqyeexmruluksgwfptlrqaerluvpmvyqycxtwfqlmtelsfjpqehmozciwciwfpzslmaeziqvlqmzvppxawcsmzmorvgvvqszetrlqz
pbjazvqiyxewwoiccgdwhqmmvowsgntjpfppaybiybjutwrlqklllmdpyvacdcfqnzpifppksdvptidgxmqqvebmqalkezmgcvkuzkizbzliuammvz
编写程序如下：
#File vigenere.py
def decode(cipherText):
'''解密'''
length = findKeyLen(cipherText) #得到密钥长度
key = findKey(cipherText,length) #找到密钥
keyStr = ''
for k in key:
keyStr+=k
print('the Key is:',keyStr)
plainText = ''
index = 0
for ch in cipherText:
c = chr((ord(ch)-ord(key[index%length]))%26+97)
plainText += c
index+=1
return plainText
def openfile(fileName):
'''读取文件'''
file = open(fileName,'r')
text = file.read()
file.close();
text = text.replace('\n','')
return text
def findKeyLen(cipherText):
'''寻找密钥长度'''
length = 1
maxCount = 0
for step in range(1,11):#假定密钥长度在1到10之间
count = 0
for i in range(step,len(cipherText)-step):
if cipherText[i] == cipherText[i+step]:
count += 1
print(count)
if count>maxCount:
maxCount = count
length = step
return length
def findKey(text,length):
'''找出密钥'''
key = []
#定义字母表频率列表
alphaRate =[0.08167,0.01492,0.02782,0.04253,0.12705,0.02228,0.02015,0.06094,0.06996,0.00153,0.00772,0.04025,0.02406,0.06749,0.07507,0.01929,0.0009,0.05987,0.06327,0.09056,0.02758,0.00978,0.02360,0.0015,0.01974,0.00074]
matrix =textToList(text,length)
for i in range(length):
w = [row[i] for row in matrix] #取出每组密文
#print('w:',w)
li = countList(w) #统计w中a-z出现的频率
powLi = [] #算乘积
for j in range(26):
Sum = 0.0
for k in range(26):
Sum += alphaRate[k]*li[k]
powLi.append(Sum)
li = li[1:]+li[:1]#循环移位
Abs = 100
ch = ''
for j in range(len(powLi)):
if abs(powLi[j] -0.065546)<Abs:
Abs = abs(powLi[j] -0.065546)
ch = chr(j+97)
key.append(ch)
return key
def countList(lis):
'''统计列表中a-z出现的频率'''
li = []
alphabet = [chr(i) for i in range(97,123)]
for c in alphabet:
count = 0
for ch in lis:
if ch == c:
count+=1
li.append(count/len(lis))
return li
def textToList(text,length):
'''按密钥长度将text分组'''
textMatrix = []
row = []
index = 0
for ch in text:
row.append(ch)
index += 1
if index % length ==0:
textMatrix.append(row)
row = []
return textMatrix
#按行输出矩阵
def printTextMatrix(matrix):
for i in range(len(matrix)):
print(matrix[i])
if __name__ == '__main__':
cipherText = openfile('vigenereCipherText.txt')
print('cipherText:\n',cipherText)
print('====================')
plainText= decode(cipherText)
print('====================')
print('plainText:\n',plainText)
运行结果：
Key:
emily
plainText:
sittheedownandhavenoshamecheekbyjowlandkneebykneewhatcareiforanynamewhatfororderordegreeletmescrewtheeupapegletmeloosethytongu
ewithwinecallestthouthatthingalegwhichisthinnestthineorminethoushaltnotbesavedbyworksthouhastbeenasinnertooruinedtrunksonwitheredforksem
ptyscarecrowsiandyoufillthecupandfillthecanhavearousebeforethemorneverymomentdiesamaneverymomentoneisborn