使用Z3破解简单的XOR加密

使用Z3破解简单的XOR加密

翻译:无名侠

原文地址: https://yurichev.com/blog/XOR_Z3/

如果我们有一段用简单XOR加密过的文本,怎么寻找密钥呢?密钥的长度可能很长,所以暴力破解不是明智的选择。

明文、密文、密钥三者的关系可以用一些简单的方程组来描述,明文可能含有尽可能多的小写字母(a....z)。

import sys, hexdump
from z3 import *

def xor_strings(s,t):
    # https://en.wikipedia.org/wiki/XOR_cipher#Example_implementation
    """将两个字符串进行xor操作"""
    return "".join(chr(ord(a)^ord(b)) for a,b in zip(s,t))

def read_file(fname):
    file=open(fname, mode='rb')
    content=file.read()
    file.close()
    return content

def chunks(l, n):
    """将输入的数据分成n组"""
    n = max(1, n)
    return [l[i:i + n] for i in range(0, len(l), n)]

def print_model(m, KEY_LEN, key):
    # 从model中解析key(model()函数返回的是解
    test_key="".join(chr(int(obj_to_string(m[key[i]]))) for i in range(KEY_LEN))
    print "key="
    hexdump.hexdump(test_key)

    # 使用解出来的key解密数据:
    tmp=chunks(cipher_file, KEY_LEN)
    plain_attempt="".join(map(lambda x: xor_strings(x, test_key), tmp))
    print "plain="
    hexdump.hexdump(plain_attempt)

def try_len(KEY_LEN, cipher_file):
    cipher_len=len(cipher_file)
    print "len=", KEY_LEN
    s=Optimize()

    # key的每一个字节对应一个未知数变量:
    key=[BitVec('key_%d' % i, 8) for i in range (KEY_LEN)]
    # 输入密文的每一个字节对应一个未知数变量:
    cipher=[BitVec('cipher_%d' % i, 8) for i in range (cipher_len)]
    # 每一个明文字节对应的未知数变量:
    plain=[BitVec('plain_%d' % i, 8) for i in range (cipher_len)]
    # 纯文本的每个字节的变量:1如果字节在'a'…“z”范围:
    az_in_plain=[Int('az_in_plain_%d' % i) for i in range (cipher_len)]

    for i in range(cipher_len):
        # 根据输入的密文数据给每一个密文字节变量赋值
        s.add(cipher[i]==ord(cipher_file[i]))
        # plain text is cipher text XOR-ed with key:
        # 明文等于密文xor key
        s.add(plain[i]==cipher[i]^key[i % KEY_LEN])
        # 明文的每一个字节的应该是可以打印的字符,或CR 或 LF
        s.add(Or(And(plain[i]>=0x20, plain[i]<=0x7E),plain[i]==0xA,plain[i]==0xD))
        # 如果是小写字母则为1,否则为0.
        s.add(az_in_plain[i]==If(And(plain[i]>=ord('a'),plain[i]<=ord('z')), 1, 0))
   
    # 寻找小写字母个数最多的解。
    s.maximize(Sum(*az_in_plain))
    
    if s.check()==unsat:
        return
    m=s.model()
    print_model(m, KEY_LEN, key)

cipher_file=read_file (sys.argv[1])

for i in range(1,20):
    try_len(i, cipher_file)

#try_len(17, cipher_file)

( https://github.com/DennisYurichev/yurichev.com/blob/master/blog/XOR_Z3/1.py )

这里有一段需要被加密过的文件(350 字节):https://github.com/DennisYurichev/yurichev.com/blob/master/blog/XOR_Z3/cipher1.txt

运行尝试。

% python 1.py cipher1.txt
len= 1
len= 2
len= 3
len= 4
len= 5

...

len= 16
len= 17
key=
00000000: 90 A0 21 52 48 84 FB FF  86 83 CF 50 46 12 7A F9  ..!RH......PF.z.
00000010: 36                                                6
plain=
00000000: 4D 72 2E 22 54 63 65 72  6F 6F 63 6D 27 48 6F 6C  Mr."Tceroocm'Hol
00000010: 6A 65 73 2C 22 70 63 6F  20 74 61 73 26 72 73 75  jes,"pco tas&rsu
00000020: 61 6B 6C 79 20 74 62 79  79 20 6F 61 74 63 27 69  akly tbyy oatc'i
00000030: 6E 20 73 68 65 20 6F 68  79 6E 69 6D 67 73 2A 27  n she ohynimgs*'
00000040: 73 61 76 62 0D 0A 75 72  68 65 20 74 6B 6F 73 63  savb..urhe tkosc
00000050: 27 6E 6F 74 27 69 6E 66  70 62 7A 75 65 6D 74 20  'not'infpbzuemt
00000060: 69 64 63 61 73 6E 6F 6E  73 22 70 63 65 6E 23 68  idcasnons"pcen#h
00000070: 65 26 70 61 73 20 72 70  20 61 6E 6B 2B 6E 69 64  e&pas rp ank+nid
00000080: 68 74 2A 27 77 61 73 27  73 65 61 76 62 6F 0D 0A  ht*'was'seavbo..
00000090: 62 74 20 72 6F 65 20 62  75 65 61 6B 64 66 78 74  bt roe bueakdfxt
000000A0: 20 77 61 62 6A 62 2E 20  49 27 73 74 6F 6D 63 2B   wabjb. I'stomc+
000000B0: 75 70 6C 6E 20 72 6F 65  20 68 62 61 72 74 6A 2A  upln roe hbartj*
000000C0: 79 75 67 23 61 6E 62 27  70 69 63 6C 65 64 20 77  yug#anb'picled w
000000D0: 77 2B 74 68 66 0D 0A 75  73 69 63 6B 27 77 68 69  w+thf..usick'whi
000000E0: 61 6F 2B 6F 75 71 20 76  6F 74 69 74 6F 75 20 68  ao+ouq votitou h
000000F0: 61 66 27 67 65 66 77 20  62 63 6F 69 6E 64 27 68  af'gefw bcoind'h
00000100: 69 6D 22 73 63 65 20 6D  69 67 6E 73 20 62 65 61  im"sce migns bea
00000110: 6F 72 65 2C 27 42 74 20  74 61 73 26 66 0D 0A 66  ore,'Bt tas&f..f
00000120: 6E 6E 65 2C 22 73 63 69  63 68 20 70 6F 62 63 65  nne,"scich pobce
00000130: 20 68 66 20 77 6D 68 6F  2C 20 61 75 6C 64 68 75   hf wmho, auldhu
00000140: 73 2D 6F 65 61 64 67 63  27 20 6F 65 20 74 6E 62  s-oeadgc' oe tnb
00000150: 20 73 6F 75 74 20 77 6A  6E 68 68 20 6A 73         sout wjnhh js
len= 18
len= 19

这段文本并不能流畅地阅读,有趣的是,只存在17字节的key。

我们对英语文本了解多少? 数字在文本中很稀少,有一种叫做“二合字母”的东西,比如:“th”、“he”、“in”、“er” 等等,我们可以通过计算它们在文本中出现的频率,然后寻找出现次数最多的那一组解。

...
	# 定义变量
    # 对于每一个明文的字节变量: 1 如果这个字节是数字符号:
    digits_in_plain=[Int('digits_in_plain_%d' % i) for i in range (cipher_len)]
    # 对于每一个明文的字节变量: 1 如果当前字节 + 下一字节恰好等于‘th’:
    th_in_plain=[Int('th_in_plain_%d' % i) for i in range (cipher_len-1)]
    # ... 同理 ‘he’:
    he_in_plain=[Int('he_in_plain_%d' % i) for i in range (cipher_len-1)]
    in_in_plain=[Int('in_in_plain_%d' % i) for i in range (cipher_len-1)]
    er_in_plain=[Int('er_in_plain_%d' % i) for i in range (cipher_len-1)]

...

    for i in range(cipher_len-1):
        # 约束条件
      	# ... 对于每一个明文的字节变量: 1 如果当前字节 + 下一字节恰好等于‘th’:
        s.add(th_in_plain[i]==(If(And(plain[i]==ord('t'),plain[i+1]==ord('h')), 1, 0)))
        # ... etc:
        s.add(he_in_plain[i]==(If(And(plain[i]==ord('h'),plain[i+1]==ord('e')), 1, 0)))
        s.add(in_in_plain[i]==(If(And(plain[i]==ord('i'),plain[i+1]==ord('n')), 1, 0)))
        s.add(er_in_plain[i]==(If(And(plain[i]==ord('e'),plain[i+1]==ord('r')), 1, 0)))

    # 寻找小写字母个数最多的解。
    s.maximize(Sum(*az_in_plain))
    # ... 且数字出现次数最少的解
    s.minimize(Sum(*digits_in_plain))

    # "th", "he", "in" , "er" 出现次数最多的解
    s.maximize(Sum(*th_in_plain))
    s.maximize(Sum(*he_in_plain))
    s.maximize(Sum(*in_in_plain))
    s.maximize(Sum(*er_in_plain))

...

( https://github.com/DennisYurichev/yurichev.com/blob/master/blog/XOR_Z3/2.py )

现在我们再从次执行脚本,好像更接近原文了

len= 17
key=
00000000: 90 A0 22 50 4F 8F FB FF  85 83 CF 56 41 12 7A FE  .."PO......VA.z.
00000010: 31                                                1
plain=
00000000: 4D 72 2D 20 53 68 65 72  6C 6F 63 6B 20 48 6F 6B  Mr- Sherlock Hok
00000010: 6D 65 73 2F 20 77 68 6F  20 77 61 73 20 75 73 75  mes/ who was usu
00000020: 66 6C 6C 79 23 76 65 72  79 20 6C 61 74 65 20 69  flly#very late i
00000030: 6E 27 74 68 65 23 6D 6F  72 6E 69 6E 67 73 2C 20  n'the#mornings,
00000040: 73 61 71 65 0D 0A 76 70  6F 6E 20 74 68 6F 73 65  saqe..vpon those
00000050: 20 6E 6F 73 20 69 6E 65  72 65 71 75 65 6E 74 20   nos inerequent
00000060: 6F 63 63 61 74 69 6F 6E  70 20 77 68 65 6E 20 68  occationp when h
00000070: 65 20 77 61 73 27 75 70  20 62 6C 6C 20 6E 69 67  e was'up bll nig
00000080: 68 74 2C 20 77 61 74 20  73 65 62 74 65 64 0D 0A  ht, wat sebted..
00000090: 61 74 20 74 68 65 20 65  72 65 61 68 66 61 73 74  at the ereahfast
000000A0: 20 74 61 62 6C 65 2E 20  4E 20 73 74 6C 6F 64 20   table. N stlod
000000B0: 75 70 6F 6E 20 74 68 65  20 6F 65 61 72 77 68 2D  upon the oearwh-
000000C0: 72 75 67 20 61 6E 64 20  70 69 64 6B 65 64 23 75  rug and pidked#u
000000D0: 70 20 74 68 65 0D 0A 73  74 69 63 6C 20 77 68 6A  p the..sticl whj
000000E0: 63 68 20 6F 75 72 20 76  69 73 69 74 68 72 20 68  ch our visithr h
000000F0: 62 64 20 6C 65 66 74 20  62 65 68 69 6E 63 20 68  bd left behinc h
00000100: 69 6E 20 74 68 65 20 6E  69 67 68 74 20 62 62 66  in the night bbf
00000110: 6F 72 66 2E 20 49 74 20  77 61 73 20 61 0D 0A 61  orf. It was a..a
00000120: 69 6E 65 2F 20 74 68 69  63 6B 20 70 69 65 63 65  ine/ thick piece
00000130: 27 6F 66 20 74 6F 6F 64  2C 20 62 75 6C 62 6F 75  'of tood, bulbou
00000140: 73 2A 68 65 61 67 65 64  2C 20 6F 66 20 74 68 65  s*heaged, of the

有几个字符还是有错误的,但是我们可以手动修复它们,因此添加限制条件。(不直接改原文的原因是我们的目标是寻找key,所以通过增加确定性限制条件重新求解key)

...
    # 3 known characters of plain text:
    s.add(plain[0xf]==ord('l'))
    s.add(plain[0x20]==ord('a'))
    s.add(plain[0x57]==ord('f'))
...

( https://github.com/DennisYurichev/yurichev.com/blob/master/blog/XOR_Z3/3.py )

这一次key正确了:

len= 17
key=
00000000: 90 A0 21 50 4F 8F FB FF  85 83 CF 56 41 12 7A F9  ..!PO......VA.z.
00000010: 31                                                1
plain=
00000000: 4D 72 2E 20 53 68 65 72  6C 6F 63 6B 20 48 6F 6C  Mr. Sherlock Hol
00000010: 6D 65 73 2C 20 77 68 6F  20 77 61 73 20 75 73 75  mes, who was usu
00000020: 61 6C 6C 79 20 76 65 72  79 20 6C 61 74 65 20 69  ally very late i
00000030: 6E 20 74 68 65 20 6D 6F  72 6E 69 6E 67 73 2C 20  n the mornings,
00000040: 73 61 76 65 0D 0A 75 70  6F 6E 20 74 68 6F 73 65  save..upon those
00000050: 20 6E 6F 74 20 69 6E 66  72 65 71 75 65 6E 74 20   not infrequent
00000060: 6F 63 63 61 73 69 6F 6E  73 20 77 68 65 6E 20 68  occasions when h
00000070: 65 20 77 61 73 20 75 70  20 61 6C 6C 20 6E 69 67  e was up all nig
00000080: 68 74 2C 20 77 61 73 20  73 65 61 74 65 64 0D 0A  ht, was seated..
00000090: 61 74 20 74 68 65 20 62  72 65 61 6B 66 61 73 74  at the breakfast
000000A0: 20 74 61 62 6C 65 2E 20  49 20 73 74 6F 6F 64 20   table. I stood
000000B0: 75 70 6F 6E 20 74 68 65  20 68 65 61 72 74 68 2D  upon the hearth-
000000C0: 72 75 67 20 61 6E 64 20  70 69 63 6B 65 64 20 75  rug and picked u
000000D0: 70 20 74 68 65 0D 0A 73  74 69 63 6B 20 77 68 69  p the..stick whi
000000E0: 63 68 20 6F 75 72 20 76  69 73 69 74 6F 72 20 68  ch our visitor h
000000F0: 61 64 20 6C 65 66 74 20  62 65 68 69 6E 64 20 68  ad left behind h
00000100: 69 6D 20 74 68 65 20 6E  69 67 68 74 20 62 65 66  im the night bef
00000110: 6F 72 65 2E 20 49 74 20  77 61 73 20 61 0D 0A 66  ore. It was a..f
00000120: 69 6E 65 2C 20 74 68 69  63 6B 20 70 69 65 63 65  ine, thick piece
00000130: 20 6F 66 20 77 6F 6F 64  2C 20 62 75 6C 62 6F 75   of wood, bulbou
00000140: 73 2D 68 65 61 64 65 64  2C 20 6F 66 20 74 68 65  s-headed, of the
00000150: 20 73 6F 72 74 20 77 68  69 63 68 20 69 73         sort which is

因此,这就是正确的key。

不用说,输入的数据越大越好。这个350字节的文件实际上是另一个更大加密文件的前戏(cipher2.txt, 12903 bytes) ,这个大文件的key可以不用添加那些些启发式条件也能求得。

SMT Solver 可以秒杀这些东西,我曾经很天真的解决了这些问题,这里有一个更快的版本: https://yurichev.com/blog/XOR_mask_2/尽管如此,这仍然是另一个优化问题的演示。

posted @ 2018-02-09 15:42  无名侠  阅读(1016)  评论(0编辑  收藏  举报