我在编码/解码编程时遇到了麻烦。我只应该使用列表、字典和循环。编辑:我添加了解密我有。GetCharList()只获取一个包含字母表的列表。我不知道哪里出了问题,它使解密的输出不是原始消息。
def encryptVig(msg, keyword):
alphabet = getCharList() #Get char list is another function which creates a list containing a - z
key = keyword.upper()
keyIndex = 0
dicList = []
for symbol in msg:
num = alphabet.find(key[keyIndex])
if num != -1:
num += alphabet.find(key[keyIndex])
alphabet.find(key[keyIndex])
num%= len(alphabet)
if symbol.isupper():
dicList.append(alphabet[num])
elif symbol.islower():
dicList. append(alphabet[num].lower())
keyIndex += 1
if keyIndex == len(key):
keyIndex = 0
else:
dicList.append(symbol)
return " " .join(dicList)
def decryptVig(msg, keyword):
getCharList()
key = keyword.upper()
keyIndex = 0
dicList = []
for symbol in msg:
num = alphabet.find(key[keyIndex])
if num != -1:
num -= alphabet.find(key[keyIndex])
alphabet.find(key[keyIndex])
num%= len(alphabet)
if symbol.isupper():
dicList.append(alphabet[num])
elif symbol.islower():
dicList. append(alphabet[num].lower())
keyIndex -= 1
if keyIndex == len(key):
keyIndex = 0
else:
dicList.append(symbol)
return " " .join(dicList)
与其自己破解字母表,另一种方法是使用ord
和chr
来消除处理字母的复杂性。至少可以考虑使用itertools.cycle
和itertools.izip
来构造一个加密/解密对的列表。下面是我的解决方法:
def letters_to_numbers(str):
return (ord(c) - ord('A') for c in str)
def numbers_to_letters(num_list):
return (chr(x + ord('A')) for x in num_list)
def gen_pairs(msg, keyword):
msg = msg.upper().strip().replace(' ', '')
msg_sequence = letters_to_numbers(msg)
keyword_sequence = itertools.cycle(letters_to_numbers(keyword))
return itertools.izip(msg_sequence, keyword_sequence)
def encrypt_vig(msg, keyword):
out = []
for letter_num, shift_num in gen_pairs(msg, keyword):
shifted = (letter_num + shift_num) % 26
out.append(shifted)
return ' '.join(numbers_to_letters(out))
def decrypt_vig(msg, keyword):
out = []
for letter_num, shift_num in gen_pairs(msg, keyword):
shifted = (letter_num - shift_num) % 26
out.append(shifted)
return ' '.join(numbers_to_letters(out))
msg = 'ATTACK AT DAWN'
keyword = 'LEMON'
print(encrypt_vig(msg, keyword))
print(decrypt_vig(encrypt_vig(msg, keyword), keyword))
>>> L X F O P V E F R N H R
A T T A C K A T D A W N
我不知道Vigenere应该怎么工作。但是我很确定在
之后 num = alphabet.find(key[keyIndex])
if num != -1:
num -= alphabet.find(key[keyIndex])
num
为零