古典密码学 凯撒密码 Python 实现 Caesar Cipher in Python

原理

将26个英文字母分别对应一个数字，循环移位。移位的个数是密钥。

• 如密钥是 B，则移位 1 位；
• 如密钥是 Ｃ，则移位 ２ 位，以此类推。

密钥是C（移位2位）对应图如图所示:

代码

class CaesarCipher:
    map1 = { "A":0 ,"B":1 ,"C":2 ,"D":3 ,"E":4 ,"F":5 ,"G":6 ,"H":7 ,"I":8 ,"J":9 ,"K":10 ,"L":11 ,"M":12 ,"N":13 ,"O":14 ,"P":15 ,"Q":16 ,"R":17 ,"S":18 ,"T":19 ,"U":20 ,"V":21 ,"W":22 ,"X":23 ,"Y":24 ,"Z":25 
    ,0:"A" ,1:"B" ,2:"C" ,3:"D" ,4:"E" ,5:"F" ,6:"G" ,7:"H" ,8:"I" ,9:"J" ,10:"K" ,11:"L" ,12:"M" ,13:"N" ,14:"O" ,15:"P" ,16:"Q" ,17:"R" ,18:"S" ,19:"T" ,20:"U" ,21:"V" ,22:"W" ,23:"X" ,24:"Y" ,25:"Z" }
    def __init__(self):
        pass
    def enc(self, plaintext:str, key:str)->str:
        """加密函数
        
        Attribute:
        - plaintext: 明文
        - key: 密钥
        """
        plaintext = plaintext.upper()
        assert sum([i not in map1 for i in set(list(plaintext))]) =  = 0,"Exist charactor not in plaintext or cyphertext space."
        return "".join([map1[(map1[i]+map1[t_key])%26] for i in plaintext])
    def dec(self,cyphertext:str, key:str)->str: 
        """解密函数
        
        Attribute:
        - cyphertext: 密文
        - key: 明文"""
        cyphertext = cyphertext.upper()
        assert sum([i not in map1 for i in set(list(cyphertext))]) =  = 0,"Exist charactor not in plaintext or cyphertext space."
        return "".join([map1[(map1[i]-map1[t_key])%26] for i in cyphertext])

实验

输入在明文域内的明文:

t_plain = "HappyBirthday"
t_key = "C"
t = CaesarCipher()
print("明文是：",t_plain)
t_cypher = t.enc(t_plain, t_key)
print("密文是：",t_cypher)
t_result = t.dec(t_cypher, t_key)
print("解密结果是：",t_result)

结果是:

明文是： HappyBirthday
密文是： JCRRADKTVJFCA
解密结果是： HAPPYBIRTHDAY

输入不在明文域内的明文:

t_plain = "HappyBirthday!!"
t_key = "B"
print("明文是：",t_plain)
t_cypher = t.enc(t_plain, t_key)

结果是:

AssertionError: Exist charactor not in plaintext or cyphertext space.