Question

https://en.wikipedia.org/wiki/Caesar_cipher

嘿所有，我这样做是为了Uni的任务。我已经接近完成，但我仍然坚持到最后一节。我确定这是基本的，但我花了将近四个小时试图使用print语句找到我的错误。

基本上，我们的想法是通过循环遍历所有-26个偏移来强行解密用户的加密消息。我的问题是我可以让代码循环26次，但它根本不会为每个偏移量解密消息..如果这是有道理的。

正确的样品输出：

Please enter string to decrypt: ykixkz&yw{oxxkr

Offset: -1 = Decrypted string: xjhwjy%xvznwwjq
Offset: -2 = Decrypted string: wigvix$wuymvvip
Offset: -3 = Decrypted string: vhfuhw#vtxluuho
Offset: -4 = Decrypted string: ugetgv"uswkttgn
Offset: -5 = Decrypted string: tfdsfu!trvjssfm
Offset: -6 = Decrypted string: secret squirrel
Offset: -7 = Decrypted string: rdbqds}rpthqqdk
Offset: -8 = Decrypted string: qcapcr|qosgppcj
Offset: -9 = Decrypted string: pb`obq{pnrfoobi
Offset: -10 = Decrypted string: oa_napzomqennah
Offset: -11 = Decrypted string: n`^m`oynlpdmm`g
Offset: -12 = Decrypted string: m_]l_nxmkocll_f
Offset: -13 = Decrypted string: l^\k^mwljnbkk^e
Offset: -14 = Decrypted string: k][j]lvkimajj]d
Offset: -15 = Decrypted string: j\Zi\kujhl`ii\c
Offset: -16 = Decrypted string: i[Yh[jtigk_hh[b
Offset: -17 = Decrypted string: hZXgZishfj^ggZa
Offset: -18 = Decrypted string: gYWfYhrgei]ffY`
Offset: -19 = Decrypted string: fXVeXgqfdh\eeX_
Offset: -20 = Decrypted string: eWUdWfpecg[ddW^
Offset: -21 = Decrypted string: dVTcVeodbfZccV]
Offset: -22 = Decrypted string: cUSbUdncaeYbbU\
Offset: -23 = Decrypted string: bTRaTcmb`dXaaT[
Offset: -24 = Decrypted string: aSQ`Sbla_cW``SZ
Offset: -25 = Decrypted string: `RP_Rak`^bV__RY
Offset: -26 = Decrypted string: _QO^Q`j_]aU^^QX

我的输出：

Please enter string to decrypt: ff

Offset: -1 = Decrypted string: ff
Offset: -2 = Decrypted string: ff
Offset: -3 = Decrypted string: ff
Offset: -4 = Decrypted string: ff
Offset: -5 = Decrypted string: ff
Offset: -6 = Decrypted string: ff
Offset: -7 = Decrypted string: ff
Offset: -8 = Decrypted string: ff
Offset: -9 = Decrypted string: ff
Offset: -10 = Decrypted string: ff
Offset: -11 = Decrypted string: ff
Offset: -12 = Decrypted string: ff
Offset: -13 = Decrypted string: ff
Offset: -14 = Decrypted string: ff
Offset: -15 = Decrypted string: ff
Offset: -16 = Decrypted string: ff
Offset: -17 = Decrypted string: ff
Offset: -18 = Decrypted string: ff
Offset: -19 = Decrypted string: ff
Offset: -20 = Decrypted string: ff
Offset: -21 = Decrypted string: ff
Offset: -22 = Decrypted string: ff
Offset: -23 = Decrypted string: ff
Offset: -24 = Decrypted string: ff
Offset: -25 = Decrypted string: ff
Offset: -26 = Decrypted string: ff

我的代码（我把大部分程序都删掉了）

choice = 0

print ('*** Menu ***\n')

print ('1. Encrypt string')
print ('2. Decrypt string')
print ('3. Brute force decryption')
print ('4. Quit\n')

    elif choice == 3:
        print ('In command 3 - Brute force')
        userString = input('\nPlease enter string to decrypt: ')
        userList = list(userString)
        offsetValue = 0
        decryptIndex = 0
        while offsetValue != -26 : # Once the count reaches -26 stop, hammer time
            while decryptIndex < len(userList):
                decryptChr = chr(ord(userList[decryptIndex]) + offsetValue)
                userList[decryptIndex] = decryptChr
                decryptIndex += 1
                offsetValue -= 1
            userString = ''.join(userList)
            print ('Offset',offsetValue,' = Decrypted string:' ,userString)
        print ('\n*** Menu ***\n')
        print ('1. Encrypt string')
        print ('2. Decrypt string')
        print ('3. Brute force decryption')
        print ('4. Quit\n')
        choice = int(input('What would you like to do? [1,2,3,4]? '))
        while choice != 1 and choice != 2 and choice != 3 and choice != 4:
            choice = int(input('\nPlease re-enter either [1,2,3,4] '))

print ('\nGoodbye.')

有什么想法吗？！

Answer 1

更容易使用str.translate

chars = "abcdefghijklmnopqrstuvwxyz"
rot1 = str.maketrans(chars, chars[1:]+chars[0])

message = input()

for i in chars:
    print(message)
    message = message.translate(rot1)

Answer 2

在Python中解码凯撒密码的一种更简洁的方法是使用字符串切片。

取字母表，即：

src = 'abcdefghijklmnopqrstuvwxyz'

现在，对于每种可能性，使用shift n = 0, ..., 25，我们可以为cypher构建一个新的字符集：

for n in range(26):
    dest = src[n:] + src[:n]

然后，您可以通过查找源集和目标集中每个字符的位置来解码字符串s：

    decoded = ''.join(dest[src.index(c)] for c in s)

Answer 3

原始程序的问题在于

行

offsetValue -= 1

不正确地缩进。

您不会重新初始化循环计数器，而且由于ord(userList[decryptIndex]) + offsetValue是无效字符，您可能会导致chr抛出错误。

转换为列表并返回是笨拙且不必要的。使用字符串连接更清晰。虽然对于一个非常干净的解决方案，其他解决方案显示了解决这个问题的pythonic方法。

所以你的程序应该是：

choice = 0

print ('*** Menu ***\n')

print ('1. Encrypt string')
print ('2. Decrypt string')
print ('3. Brute force decryption')
print ('4. Quit\n')

    elif choice == 3:
        print ('In command 3 - Brute force')
        userString = input('\nPlease enter string to decrypt: ')
        offsetValue = 0

        while offsetValue != -26 : # Once the count reaches -26 stop, hammer time
            decryptIndex = 0
            decryptedString = ""
            while decryptIndex < len(userString):
                c = ord(userString[decryptIndex]) + offsetValue 
                if c < 0: c += 128
                decryptChr = chr(c)
                decryptedString += decryptChr
                decryptIndex += 1
            offsetValue -= 1
            print ('Offset',offsetValue,' = Decrypted string:' ,decryptedString)
        print ('\n*** Menu ***\n')
        print ('1. Encrypt string')
        print ('2. Decrypt string')
        print ('3. Brute force decryption')
        print ('4. Quit\n')
        choice = int(input('What would you like to do? [1,2,3,4]? '))
        while choice != 1 and choice != 2 and choice != 3 and choice != 4:
            choice = int(input('\nPlease re-enter either [1,2,3,4] '))

print ('\nGoodbye.')

Answer 4

你可以按照以下方式做点什么：

from string import ascii_lowercase as alphabet 
import string

def caesar(plaintext, shift):
    shifted_alphabet = alphabet[shift:] + alphabet[:shift]
    table = string.maketrans(alphabet, shifted_alphabet)
    return plaintext.translate(table)

plain='hello there'  
shift=3
coded=caesar(plain,shift)  
decoded=caesar(coded,-shift)    
print plain
print coded
print decoded

打印：

hello there
khoor wkhuh
hello there

Answer 5

我会从gnibbler那里无耻地偷走，因为他是迄今为止最好的答案。

我有点担心你的菜单处理。您打印出菜单，然后在选项内再次打印。也许这样一个“老派”的CLI菜单可以更优雅地处理。

我没有为gnibbler（肯定得到我的upvote）的实际答案做出贡献，但我试图解决另一个问题，这可能并希望可以帮助你获得“更好”的代码。

这是我的想法：

#! /usr/bin/python3.2

import sys
from collections import defaultdict

menu = '''
*** Menu ***
[1] Encrypt
[2] Decrypt
[3] Brute force
[4] Quit
'''

def caesar (s, offset):
    chars = 'abcdefghijklmnopqrstuvwxyz'
    return s.translate (str.maketrans (chars, chars [offset:] + chars [:offset] ) )

def decrypt ():
    s = input ('Enter string to decrypt: ')
    offset = int (input ('Enter offset: ') )
    print (caesar (s, (-offset) % 26) )

def encrypt ():
    s = input ('Enter string to encrypt: ')
    offset = int (input ('Enter offset: ') )
    print (caesar (s, offset) )

def bruteForce ():
    s = input ('Enter string to brute force: ')
    for offset in range (26):
        print ('Offset {}:\t{}'.format (offset, caesar (s, (-offset) % 26) ) )

def unrecognized ():
    print ('unrecognized command')

options = {'1': encrypt,
           '2': decrypt,
           '3': bruteForce,
           '4': sys.exit}

options = defaultdict (lambda: unrecognized, options)

while ...: options [input (menu) ] ()

（inb4“奇怪的whitespacing”）

蛮力凯撒密码Python

5 个答案: