使用频率分析/密码分析技术破译密码文本

Question

本网站(https://www.guballa.de/substitution-solver)他做到了。

我必须通过频率分析来做到这一点(https://en.wikipedia.org/wiki/Frequency_analysis)

我面临的主要问题是，当我替换时，检查单词是否看起来像英语单词。

请指导我如何处理这个问题

谢谢哈基德

Answer 1

这可能是一个迟来的答案，但这段代码可以作为你的开始。

from operator import itemgetter

letterFrequency = [
                   [12.00, 'E'], [9.10, 'T'],
                   [8.12, 'A'], [7.68, 'O'],
                   [7.31, 'I'], [6.95, 'N'],
                   [6.28, 'S'], [6.02, 'R'],
                   [5.92, 'H'], [4.32, 'D'],
                   [3.98, 'L'], [2.88, 'U'],
                   [2.71, 'C'], [2.61, 'M'],
                   [2.30, 'F'], [2.11, 'Y'],
                   [2.09, 'W'], [2.03, 'G'],
                   [1.82, 'P'], [1.49, 'B'],
                   [1.11, 'V'], [0.69, 'K'],
                   [0.17, 'X'], [0.11, 'Q'],
                   [0.10, 'J'], [0.07, 'Z']]


plain_to_cipher = {
       "a": "l", "b": "f",
       "c": "w", "d": "o",
       "e": "a", "f": "y",
       "g": "u", "h": "i",
       "i": "s", "j": "v",
       "k": "z", "l": "m",
       "m": "n", "n": "x",
       "o": "p", "p": "b",
       "q": "d", "r": "c",
       "s": "r", "t": "j",
       "u": "t", "v": "q",
       "w": "e", "x": "g",
       "y": "h", "z": "k",
       }
cipher_to_plain = {v: k for k, v in plain_to_cipher.items()}
alphabet = "qwertyuioplkjhgfdsazxcvbnm"


message = input("Enter message to encrypt: ")
message = message.lower()
ciphertext = ""


for c in message:
    if c not in alphabet:
        ciphertext += c
    else:
        ciphertext += plain_to_cipher[c]
print("\nRandom substitution Encryption is: \n\t{}".format(ciphertext))

# .......................................................................
# calculate letter frequency of ciphertext

letter_list = []
cipher_len = 0
for c in ciphertext:
    if c in alphabet:
        cipher_len += 1
        if c not in letter_list:
            letter_list.append(c)

letter_freq = []
for c in letter_list:
    letter_freq.append([round(ciphertext.count(c) / cipher_len * 100, 2), c])

# ....................................................................................
# Now sort list and decrypt each instance of ciphertext according to letter frequency

letter_freq = sorted(letter_freq, key=itemgetter(0), reverse=True)
decrypted_plaintext = ciphertext

index = 0
for f, c in letter_freq:
    print("Replacing {} of freq {} with {}.".format(c, f, letterFrequency[index][1]))
    decrypted_plaintext = decrypted_plaintext.replace(c, letterFrequency[index][1])
    index += 1
print("\nThe Plaintext after decryption using frequency analysis: \n\t{}".format(decrypted_plaintext))

旁注：这个程序可以在大部分时间成功解密最常用的字母，如e， t， a， o，但是不能成功映射较少使用的字母（因为频率差异开始减少，使得结果更难预测）。这个问题可以通过分析英语中最常用的字母（如th）并使用结果做出更准确的预测来克服一点。你可以利用的另一个注意事项是，字母a很容易被打破，这样打破字母i就不那么痛苦了，因为任何一个句子之间有一个密文字符都可能对应于a（例如：一本书）或i（例如：我去了）（我们已经推导出a，所以任何其他单个密文字符都可能是i）