Question

我正在尝试使用此代码将密文＆＃34; htrgti＆＃34; 解密为纯文本。我一直在＆＃34; wigvix＆＃34; 这不是我应该得到的信息。

鉴于纯文本（应该是一个常用的单词或短语）和密钥，对于每个说ciphertext的地点，我用plaintext替换它，并且每个地点都显示plaintext 1}}：

def caesar(ciphertext, shift):  
    alphabet=["a","b","c","d","e","f","g","h","i",
              "j","k","l","m","n","o","p","q","r",
              "s","t","u","v","w","x","y","z"]  
    plaintext = ""

    for i in range(len(ciphertext)):                        
        letter = ciphertext[i]                  

        # Find the number position of the ith letter
        num_in_alphabet = alphabet.index(letter)        

        # Find the number position of the cipher by adding the shift    
        plain_num = (num_in_alphabet + shift) % len(alphabet)   

        # Find the plain letter for the cipher number you computed
        plain_letter = alphabet[plain_num]          

        # Add the cipher letter to the plaintext
        plaintext = plaintext + plain_letter            

    return plaintext

Answer 1

如果你正确地调用它，你的功能是有效的，正如@L.Llama暗示的那样，右移caesar("htrgti", 11)是secret。您的功能本身运行正常。这是代码的更好版本，但它仍然存在每个.index查找为O（alphalen）的问题。所以，作为奖励，我添加了@ Two-Bit Alchemist建议的实现，以使用str.translate。

import string
alphabet = string.ascii_lowercase
alphalen = len(alphabet)
shift = 11

def caesar(ciphertext, shift):  
    plaintext = ""
    for letter in ciphertext:
        num_in_alphabet = alphabet.index(letter)        
        plain_num = (num_in_alphabet + shift)  %  alphalen
        plain_letter = alphabet[plain_num]          
        plaintext = plaintext + plain_letter            
    return plaintext

print(caesar("htrgti", shift))
# secret

table = str.maketrans(alphabet, alphabet[shift:] + alphabet[:shift])
print("htrgti".translate(table))
# secret

使用凯撒密码解密密文

1 个答案: