Vignere密码解密函数未返回正确的解密python 2.7

我为Vignere密码创建了一个加密和解密程序。加密机工作了，它把鹰降落到了WHZ RCOOE PNU OAILRF。然而，当我试图用达芬奇解密WHZ RCOOE PNU OAILRF时，它并没有返回鹰已经着陆的消息。相反，它会返回一个随机的混乱。问题是在decryptVignere中还是在前面的函数中？我知道维格纳工作得很好

"""
This program will decrypt/encrypt a Vignere with the Keyword: DaVinci
"""

# helper functions
# letter to index
def letterToIndex(ch):
   alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ "
   idx = alphabet.find(ch)
   if idx < 0:
      print ("error: letter not in the alphabet", ch)
   return idx

# index to letter
def indexToLetter(idx):
   alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ "
   if idx > 25:
      print ('error: ', idx, ' is too large')
      letter = ' '
   elif idx < 0:
      print ('error: ', idx, ' is less than 0')
      letter = ' '
   else:
      letter = alphabet[idx]
   return letter

# looking up a letter in the Vignere square
def vignereIndex(keyLetter, plainTextLetter):
   keyIndex = letterToIndex(keyLetter)
   ptIndex = letterToIndex(plainTextLetter)
   newIdx = (ptIndex + keyIndex) % 26
   return indexToLetter(newIdx)

def decryptVignere(key, cipherText):
   plainText = ""
   keyLen = len(key)
   for i in range (len(cipherText)):
      ch = cipherText[i]
      if ch == ' ':
         plainText = plainText + ch
      else:
         plainText = plainText + vignereIndex(key[i%keyLen], ch)
   return plainText

#encrypting a message using the Vignere cipher
def encryptVignere(key, plainText):
   cipherText = ""
   keyLen = len(key)
   for i in range (len(plainText)):
      ch = plainText[i]
      if ch == ' ':
         cipherText = cipherText + ch
      else:
         cipherText = cipherText + vignereIndex(key[i%keyLen], ch)
   return cipherText

# decrypt
messageOne = "WHZ RCOOE PNU OAILRF" # raw_input("Enter your string: ")?
keyOne = "DAVINCI"

deStr = decryptVignere(keyOne, messageOne)
print deStr 

---

您的解密方法与加密方法完全相同！解密方法应该做相反的事情，即减去密钥而不是添加密钥

一种可能的解决方案是向vignereIndex方法添加一个参数，以便在加密和解密之间切换

def vignereIndex(keyLetter, plainTextLetter, encrypt):
    keyIndex = letterToIndex(keyLetter)
    ptIndex = letterToIndex(plainTextLetter)
    if not encrypt:
        keyIndex *= -1
    newIdx = (ptIndex + keyIndex) % 26
    return indexToLetter(newIdx)

---

代码注释：您可以使用字典将字母转换为数字，反之亦然，而不是每次都查看字符串。@Dschoni：更干净的解决方案是使用字母的ascii值。将ord和chr用于转换Python不使用C样式的三元组。@wvdz:Python也不使用！为了否定。你用什么解释器来测试你的代码？@DSM:也解决了这个问题。你可能已经猜到了，我正在使用内部大脑编译器IBC。