Vigenere密码，如何处理超出字符值范围的序数值

Posted 2023-02-16

【中文标题】Vigenere密码，如何处理超出字符值范围的序数值

【英文标题】：Vigenere cipher, how to deal with ordinal values out of range of character values

【发布时间】：2015-12-10 10:15:41

【问题描述】：

我是一名学生，正在做我的课程，我们必须制作一个 Vigenere 密码程序。 我知道我的程序很复杂，但我不知道有任何其他方法可以解决它，而且修复它为时已晚。 当我添加消息的序数值和关键字时，我似乎有一个问题，新的序数值超出了正常字母值的范围。所以它会打印出像这样的奇怪字母 ÐÑÐÑÐÑÐÑÐÑÐÑÐÑÐÑ。

这是我的代码：

newmessage1 = []
stringposition = 0
number1 = 0
mesletter = 0
keyletter = 0
output =[]
keylist = []
stringofnumbs = []
question = input('Would you like to encrypt, decrypt a message or quit?')
encdec=[]
if question == 'encrypt'or'encrypt a string'or'ENCRYPT'or'encrypt_a_message'or'Encrypt':
    message1 = input('Please input a message to encrypt')
    keyword1 = input('Please input a keyword ')
    if message1.isalpha and keyword1.isalpha:#this check if the message and the keyword only contains letters
        messagelength=len(message1)#this sets the variable 'messagelength' to the length of message1
        newkeyword=''#this is an empty variable used to insert the length of the new keyword
        while len(keyword1)<=len(message1):#this will loop checks that the length of keyword1 is smaller thasn the length of message1.
            keyword1+=keyword1#this will repeat the keyword to fit the length of the message
            newkeyword=keyword1[:len(message1)]
            #this set the 'newkeyword' variable to the length of the new keyword
            #(the new keyword is the keyword repeated to fit the length of the message.) 

            for mesletter in message1:
                mesnumber = ord(mesletter)#what it does is it takes every letter in message1 and puts it into its unicode form.
                stringofnumbs.append(mesnumber)#once one letter has been put in its unicode value it will append that unicdoe from of that
                #letter in the variable 'stringofnumbs',it will do this process for every letter in the message. 

            for keyletter in keyword1:
                keynumber = ord(keyletter)#what it does is it takes every letter in keyword1 and puts it into its unicode form.
                keylist.append(keynumber)#once one letter has been put in its unicode value it will
                #append that unicdoe from of that letter in the variable 'stringofnumbs',it will do this process for every letter in the message.

                temp1 = int(stringofnumbs[stringposition])#temp1 is a variable that holds the ordinal values of letters in message1
                temp2 = int(keylist[stringposition-1])#and temp2 is a variable that holds the ordinal values of letters in the keyword1
                temp3 = temp2+temp1#temp3 then adds these ordinal values togther 
                encdec.append(temp3)#the ordinal values added togther are appended to the encdec variable

            for number1 in encdec:
                newletter1=chr(number1)#this for loop takes every letter in encdec and puts the ordinal values back into charcters
                newmessage1.append(newletter1)#then every letter that has been changed to a charcater value is appended to the varibale newmessage1
            print (' ')#this leaves a bit of space to make the encoded text clear    
            print ('the endcoded text is:')#this is just a bit of text to make the encoded text clear to the user
            print (''.join(newmessage1))#here the encoded message is printed
    else:
        ('The message or keyword is invalid please use only numbers')#this will print only if the keyword or message doesnt only contain letters and spaces

【参考方案1】：

我认为最常见的解决您问题的方法是再次“包装”数字。也就是说，如果你除以你想要允许的最大有效序数，然后使用余数。

例如，使用chr 作为ord 的倒数：

In [0]: chr(122)    # I guess this is the last valid character you want?
Out[0]: 'z'

In [1]: chr(123)    # Because now we go into symbols
Out[1]: ''

In [2]: 234%122    # We can take the remainder using modulo division
Out[2]: 112

In [3]: chr(234)    # Not what you want?
Out[3]: 'ê'

In [4]: chr(234%122)    # What you want?
Out[4]: 'p'

In [5]: chr(0) # On further inspection, you also need to add a constant when you divide
Out[5]: '\x00'

In [6]: chr(49)    # Try adding 49 after modulo dividing, and that should keep you alphanumeric.
Out[6]: '1'

我实际上不知道这是否会影响你之后的解密，我没有调查过，但你可能想考虑一下。

【参考方案2】：

您使用序数不必要地使事情复杂化。实际上，您使用的是包含 256 个符号的“字母”，而您只对 26 个符号感兴趣。

理想情况下，您只希望符号的数字范围在 0 到 25 之间，一旦达到 26，您就会再次扭曲到 0。您可以使用序数来实现这一点，但您必须使用正确的偏移量。

现在，假设您只使用大写字母。它们的序数从 65 到 90 不等。因此，一旦你达到 91，你就想扭曲到 65。

if temp3 > 90:
    temp3 -= 26

另一个问题是您只需添加temp1 和temp2。 temp1 可以保留为 65 到 90 之间的数字，但 temp2 应该是键偏移量，范围从 0 到 25。因此，当你计算 temp2 时，你真的应该这样做

temp2 = int(keylist[stringposition-1]) - 65

这在概念上可以解决您的问题，但您的算法仍然不正确。 while 循环的目的是将关键字的长度扩展为至少与消息的长度一样长。其他所有内容都必须向左缩进 4 个空格，这样它们就不是循环的一部分。明确地，

while len(keyword1)<=len(message1):
    keyword1+=keyword1
newkeyword=keyword1[:len(message1)]

for mesletter in message1:
    keynumber = ord(keyletter)
    stringofnumbs.append(mesnumber)

for keyletter in newkeyword:
    keynumber = ord(keyletter)
    keylist.append(keynumber)

    temp1 = int(stringofnumbs[stringposition])
    temp2 = int(keylist[stringposition]) - 65
    temp3 = temp2+temp1
    if temp3 > 90:
        temp3 -= 26
    encdec.append(temp3)

    stringposition += 1

etc

在上面的代码块中，我想指出一些我更正的错误。您应该遍历 newkeyword，而不是 keyword1，并且您还必须在您通过的每个字母时增加 stringposition 计数器。

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要获得更高效且不太复杂的算法，请考虑将您的代码发布到 Code Review，以便进行审核，并从您获得的反馈中学习。

虽然不是最好的编码方式，但在保持代码不变的情况下，您仍然可以做一些基本的整理工作。基本上，一旦您的消息和关键字的长度相同，您就可以遍历这对字母并即时计算加密的字母，以便将其存储在一个列表中。没有理由存储中间步骤，例如将消息和关键字转换为数字列表。

if message1.isalpha() and keyword1.isalpha():
    message1 = message1.upper()
    keyword1 = keyword1.upper()

    if len(keyword1) < len(message1):
        keyword1 = keyword1 * (len(message1) / len(keyword1)) + keyword1
    keyword1 = keyword1[:len(message1)]

    for m, k in zip(message1, keyword1):
        encrypted = ord(m) + (ord(k) - 65)
        if encrypted > 90:
            encrypted -= 26
        encdec.append(chr(encrypted))

    print ('\nThe endcoded text is:')
    print (''.join(encdec))