嵇爾的吐槽

#没事画轮子的嵇尔不定期的(W)碎(E)碎(B)念(B)和(L)吐(O)槽(G)

Problem 59 XOR decryption 2017-04-20 09:32:00

Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.

A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.

For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption key in different locations, and without both “halves”, it is impossible to decrypt the message.

Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.

Your task has been made easy, as the encryption key consists of three lower case characters. Using cipher.txt (right click and ‘Save Link/Target As…’), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.

import math
import sys

avalible = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ,. ?!\'"()-1234567890:;'

source = [79,59,12,2,79,35,8,28,20,2,3,68,8,9,68,45,0,12,9,67,68,4,7,5,23,27,1,21,79,85,78,79,85,71,38,10,71,27,12,2,79,6,2,8,13,9,1,13,9,8,68,19,7,1,71,56,11,21,11,68,6,3,22,2,14,0,30,79,1,31,6,23,19,10,0,73,79,44,2,79,19,6,28,68,16,6,16,15,79,35,8,11,72,71,14,10,3,79,12,2,79,19,6,28,68,32,0,0,73,79,86,71,39,1,71,24,5,20,79,13,9,79,16,15,10,68,5,10,3,14,1,10,14,1,3,71,24,13,19,7,68,32,0,0,73,79,87,71,39,1,71,12,22,2,14,16,2,11,68,2,25,1,21,22,16,15,6,10,0,79,16,15,10,22,2,79,13,20,65,68,41,0,16,15,6,10,0,79,1,31,6,23,19,28,68,19,7,5,19,79,12,2,79,0,14,11,10,64,27,68,10,14,15,2,65,68,83,79,40,14,9,1,71,6,16,20,10,8,1,79,19,6,28,68,14,1,68,15,6,9,75,79,5,9,11,68,19,7,13,20,79,8,14,9,1,71,8,13,17,10,23,71,3,13,0,7,16,71,27,11,71,10,18,2,29,29,8,1,1,73,79,81,71,59,12,2,79,8,14,8,12,19,79,23,15,6,10,2,28,68,19,7,22,8,26,3,15,79,16,15,10,68,3,14,22,12,1,1,20,28,72,71,14,10,3,79,16,15,10,68,3,14,22,12,1,1,20,28,68,4,14,10,71,1,1,17,10,22,71,10,28,19,6,10,0,26,13,20,7,68,14,27,74,71,89,68,32,0,0,71,28,1,9,27,68,45,0,12,9,79,16,15,10,68,37,14,20,19,6,23,19,79,83,71,27,11,71,27,1,11,3,68,2,25,1,21,22,11,9,10,68,6,13,11,18,27,68,19,7,1,71,3,13,0,7,16,71,28,11,71,27,12,6,27,68,2,25,1,21,22,11,9,10,68,10,6,3,15,27,68,5,10,8,14,10,18,2,79,6,2,12,5,18,28,1,71,0,2,71,7,13,20,79,16,2,28,16,14,2,11,9,22,74,71,87,68,45,0,12,9,79,12,14,2,23,2,3,2,71,24,5,20,79,10,8,27,68,19,7,1,71,3,13,0,7,16,92,79,12,2,79,19,6,28,68,8,1,8,30,79,5,71,24,13,19,1,1,20,28,68,19,0,68,19,7,1,71,3,13,0,7,16,73,79,93,71,59,12,2,79,11,9,10,68,16,7,11,71,6,23,71,27,12,2,79,16,21,26,1,71,3,13,0,7,16,75,79,19,15,0,68,0,6,18,2,28,68,11,6,3,15,27,68,19,0,68,2,25,1,21,22,11,9,10,72,71,24,5,20,79,3,8,6,10,0,79,16,8,79,7,8,2,1,71,6,10,19,0,68,19,7,1,71,24,11,21,3,0,73,79,85,87,79,38,18,27,68,6,3,16,15,0,17,0,7,68,19,7,1,71,24,11,21,3,0,71,24,5,20,79,9,6,11,1,71,27,12,21,0,17,0,7,68,15,6,9,75,79,16,15,10,68,16,0,22,11,11,68,3,6,0,9,72,16,71,29,1,4,0,3,9,6,30,2,79,12,14,2,68,16,7,1,9,79,12,2,79,7,6,2,1,73,79,85,86,79,33,17,10,10,71,6,10,71,7,13,20,79,11,16,1,68,11,14,10,3,79,5,9,11,68,6,2,11,9,8,68,15,6,23,71,0,19,9,79,20,2,0,20,11,10,72,71,7,1,71,24,5,20,79,10,8,27,68,6,12,7,2,31,16,2,11,74,71,94,86,71,45,17,19,79,16,8,79,5,11,3,68,16,7,11,71,13,1,11,6,1,17,10,0,71,7,13,10,79,5,9,11,68,6,12,7,2,31,16,2,11,68,15,6,9,75,79,12,2,79,3,6,25,1,71,27,12,2,79,22,14,8,12,19,79,16,8,79,6,2,12,11,10,10,68,4,7,13,11,11,22,2,1,68,8,9,68,32,0,0,73,79,85,84,79,48,15,10,29,71,14,22,2,79,22,2,13,11,21,1,69,71,59,12,14,28,68,14,28,68,9,0,16,71,14,68,23,7,29,20,6,7,6,3,68,5,6,22,19,7,68,21,10,23,18,3,16,14,1,3,71,9,22,8,2,68,15,26,9,6,1,68,23,14,23,20,6,11,9,79,11,21,79,20,11,14,10,75,79,16,15,6,23,71,29,1,5,6,22,19,7,68,4,0,9,2,28,68,1,29,11,10,79,35,8,11,74,86,91,68,52,0,68,19,7,1,71,56,11,21,11,68,5,10,7,6,2,1,71,7,17,10,14,10,71,14,10,3,79,8,14,25,1,3,79,12,2,29,1,71,0,10,71,10,5,21,27,12,71,14,9,8,1,3,71,26,23,73,79,44,2,79,19,6,28,68,1,26,8,11,79,11,1,79,17,9,9,5,14,3,13,9,8,68,11,0,18,2,79,5,9,11,68,1,14,13,19,7,2,18,3,10,2,28,23,73,79,37,9,11,68,16,10,68,15,14,18,2,79,23,2,10,10,71,7,13,20,79,3,11,0,22,30,67,68,19,7,1,71,8,8,8,29,29,71,0,2,71,27,12,2,79,11,9,3,29,71,60,11,9,79,11,1,79,16,15,10,68,33,14,16,15,10,22,73]


i = 0
source_group = [[], [], []]
for s in source:
    source_group[i].append(s)
    i += 1
    if i == 3:
        i = 0

result_char = [[], [], []]
for group_id in range(0, 3):
    for secret in range(ord("a"), ord("z") + 1):
        result = ""
        flag = True
        for s in source_group[group_id]:
            r = chr(s ^ secret)
            if r in avalible:
                result += r
            else:
                flag = False
                break
        if flag:
            result_char[group_id] = result
            break

ascii_sum = 0
for index_id in range(0, len(source)/3 + 1):
    for group_id in range(0, 3):
        if index_id < len(result_char[group_id]):
            sys.stdout.write(result_char[group_id][index_id])
            ascii_sum += ord(result_char[group_id][index_id])

print "\n\n" + str(ascii_sum)

在这题上走了相当多的弯路,其实不难。

首先,我自己又开始写轮子。自己写了个按位异或的函数,然后就被同事嘲笑了,才发现直接^就能搞定

然后,最开始的时候用排列组合列出了所有的可能秘钥,然后对着秘钥012循环解密,最后一起测试avaliable,结果怎么都出不来(我也不知道为什么),然后改变了循环方式,先把密文切成三块,用单一秘钥居然一个个都出来了,费解。

评论共