python - 降低Python中字典暴力破解的复杂度

Question

我有一个程序如下，它基本上比较标准字典中所有可能单词的异或，并将异或结果与密文的异或结果进行比较。但我猜复杂度是 O(n2)。我不知道如何降低复杂性。

def find_collision():
    a = int("4ADD55BA941FE954",16) ^ int("5AC643BE8504E35E",16)
    with open("/usr/share/dict/words", "r") as f:
        alist = [line.rstrip() for line in f]
    b = len(alist)

    for i in range(0,b,1):
        for j in range(i,b,1):
        if(((int(alist[i].encode('hex'), 16))^ (int(alist[j].encode('hex'), 16)))==a):
            print("Plain Text1: "+alist[i]+'\n'+"Plain Text2: "+alist[j])
            #print "Yes"
            break

任何帮助将不胜感激。

Answer 1

首先，让我们尝试简化。

def find_collision():
    key = 0b1000000011011000101100000010000010001000110110000101000001010
    # that's 0x4ADD55BA941FE954^0x5AC643BE8504E35E

然后我们方便的 itertools 模块可以为大列表完成繁重的工作。这取代了嵌套的 for 循环，并且可能工作得更快。

from itertools import combinations
##def find_collision()
##    key = 0b1000000011011000101100000010000010001000110110000101000001010
with open("/usr/share/dict/words", "r") as f:
    full_wordlist = combinations( map(str.rstrip,f.readlines()), 2 )
    # Combinations( { ('word1','word2'),('word1','word3'),('word1','word4'),
                    ('word2','word3') ... } )

但我们并不真正关心整件事，不是吗？我们关心的只是碰撞，所以让我们进行碰撞吧？ 编辑:并且由于这里肯定会有单词，我们无法转换为十六进制，因此:

#instead of full_wordlist = combinations(...)

import re
with open("usr/share/dict/words","r") as f:
    words = (word for word in map(str.rstrip,f.readlines()) if not re.search(r"[^0-9a-fA-F]",word))
    # you can avoid the need for regex by doing:
    # words = (word for word in map(str.rstrip,f.readlines()) if
    #         not any(char not in "0123456789abcdefABCDEF" for char in word))
    collisions = [keypair for keypair in combinations(words,2)
                 if bin(int(keypair[0],base=16)^int(keypair[1],base=16)) == key]

然后用一些理智的东西来消除碰撞，例如:

for collision in collisions:
    print("Collision between {0[0]}^{0[1]} and key".format(collision))