维格纳密码，如何处理超出字符值范围的序数值

Question

我是一个学生在做我的课程，我们必须作出一个维格纳密码程序。 我知道我的程序是复杂的，但我不知道有任何其他方式来做它和它来不及修复。 我似乎有一个问题，当我添加的序数值的消息和关键字的新的序数值是超出范围的正常字母值。所以它会打印出像这样奇怪的字母。你知道吗

这是我的密码：

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

Answer 1

我认为解决这个问题最常见的方法是把数字再“包装”一下。也就是说，如果除以要允许的最大有效序数，然后使用余数。你知道吗

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'

我真的不知道这会影响你以后的描述，我还没有调查过，但你可能会想一想。你知道吗