简单替换密码用一个字母代替另一个字母。由于字母A有 26 种可能的替换，B有 25 种可能的替换，C有 24 种可能的替换，等等，所以可能的键的总数是26 × 25 × 24 × 23 × ... × 1，即 403291461126605635584000000 个密钥！对于一台超级计算机来说，这对于暴力破解来说太多了，所以项目 7“凯撒破解”中使用的密码破解方法不能用于对抗简单的密码。不幸的是，狡猾的攻击者可以利用已知的弱点来破解代码。如果你想了解更多关于密码和密码破解的知识，你可以阅读我的书《Python 密码破解指南》（NoStarch 出版社，2018）。
当您运行simplesubcipher.py时，输出将如下所示：
Simple Substitution Cipher, by Al Sweigart
A simple substitution cipher has a one-to-one translation for each
symbol in the plaintext and each symbol in the ciphertext.
Do you want to (e)ncrypt or (d)ecrypt?
> e
Please specify the key to use.
Or enter RANDOM to have one generated for you.
> random
The key is WNOMTRCEHDXBFVSLKAGZIPYJQU. KEEP THIS SECRET!
Enter the message to encrypt.
> Meet me by the rose bushes tonight.
The encrypted message is:
Fttz ft nq zet asgt nigetg zsvhcez.
Full encrypted text copied to clipboard.
Simple Substitution Cipher, by Al Sweigart
A simple substitution cipher has a one-to-one translation for each
symbol in the plaintext and each symbol in the ciphertext.
Do you want to (e)ncrypt or (d)ecrypt?
> d
Please specify the key to use.
> WNOMTRCEHDXBFVSLKAGZIPYJQU
Enter the message to decrypt.
> Fttz ft nq zet asgt nigetg zsvhcez.
The decrypted message is:
Meet me by the rose bushes tonight.
Full decrypted text copied to clipboard.密钥的 26 个字母中的每一个的位置对应于字母表中相同位置的字母：
:字母表中的字母如何用一个以WNOM开头的密钥加密。若要解密，请将底部的字母替换为上面相应的字母。
用这个密钥，字母A加密到W （而W解密到A，字母B加密到N，以此类推。LETTERS和key变量被分配给charsA和charsB（或者在解密时反过来）。用charsB中的相应字符替换charsA中的任何消息字符，以产生最终翻译的消息。
"""Simple Substitution Cipher, by Al Sweigart email@protected
A simple substitution cipher has a one-to-one translation for each
symbol in the plaintext and each symbol in the ciphertext.
More info at: https://en.wikipedia.org/wiki/Substitution_cipher
This code is available at https://nostarch.com/big-book-small-python-programming
Tags: short, cryptography, math"""
import random
try:
import pyperclip # pyperclip copies text to the clipboard.
except ImportError:
pass # If pyperclip is not installed, do nothing. It's no big deal.
# Every possible symbol that can be encrypted/decrypted:
LETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
def main():
print('''Simple Substitution Cipher, by Al Sweigart
A simple substitution cipher has a one-to-one translation for each
symbol in the plaintext and each symbol in the ciphertext.''')
# Let the user specify if they are encrypting or decrypting:
while True: # Keep asking until the user enters e or d.
print('Do you want to (e)ncrypt or (d)ecrypt?')
response = input('> ').lower()
if response.startswith('e'):
myMode = 'encrypt'
break
elif response.startswith('d'):
myMode = 'decrypt'
break
print('Please enter the letter e or d.')
# Let the user specify the key to use:
while True: # Keep asking until the user enters a valid key.
print('Please specify the key to use.')
if myMode == 'encrypt':
print('Or enter RANDOM to have one generated for you.')
response = input('> ').upper()
if response == 'RANDOM':
myKey = generateRandomKey()
print('The key is {}. KEEP THIS SECRET!'.format(myKey))
break
else:
if checkKey(response):
myKey = response
break
# Let the user specify the message to encrypt/decrypt:
print('Enter the message to {}.'.format(myMode))
myMessage = input('> ')
# Perform the encryption/decryption:
if myMode == 'encrypt':
translated = encryptMessage(myMessage, myKey)
elif myMode == 'decrypt':
translated = decryptMessage(myMessage, myKey)
# Display the results:
print('The %sed message is:' % (myMode))
print(translated)
try:
pyperclip.copy(translated)
print('Full %sed text copied to clipboard.' % (myMode))
except:
pass # Do nothing if pyperclip wasn't installed.
def checkKey(key):
"""Return True if key is valid. Otherwise return False."""
keyList = list(key)
lettersList = list(LETTERS)
keyList.sort()
lettersList.sort()
if keyList != lettersList:
print('There is an error in the key or symbol set.')
return False
return True
def encryptMessage(message, key):
"""Encrypt the message using the key."""
return translateMessage(message, key, 'encrypt')
def decryptMessage(message, key):
"""Decrypt the message using the key."""
return translateMessage(message, key, 'decrypt')
def translateMessage(message, key, mode):
"""Encrypt or decrypt the message using the key."""
translated = ''
charsA = LETTERS
charsB = key
if mode == 'decrypt':
# For decrypting, we can use the same code as encrypting. We
# just need to swap where the key and LETTERS strings are used.
charsA, charsB = charsB, charsA
# Loop through each symbol in the message:
for symbol in message:
if symbol.upper() in charsA:
# Encrypt/decrypt the symbol:
symIndex = charsA.find(symbol.upper())
if symbol.isupper():
translated += charsB[symIndex].upper()
else:
translated += charsB[symIndex].lower()
else:
# The symbol is not in LETTERS, just add it unchanged.
translated += symbol
return translated
def generateRandomKey():
"""Generate and return a random encryption key."""
key = list(LETTERS) # Get a list from the LETTERS string.
random.shuffle(key) # Randomly shuffle the list.
return ''.join(key) # Get a string from the list.
# If this program was run (instead of imported), run the program:
if __name__ == '__main__':
main() 试着找出下列问题的答案。尝试对代码进行一些修改，然后重新运行程序，看看这些修改有什么影响。
- 如果删除或注释掉第 122 行的
random.shuffle(key)并输入密钥RANDOM会发生什么？ - 如果将第 16 行的
LETTERS字符串扩展成'ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890'会发生什么？