神奇的 Base64 隐写

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Tr0y 6月 14, 2017 16:23:29 本文共 1.1k 字
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俄罗斯有个叫 Olympic_ctf 的 CTF, 在 2014 年有道 misc 题是关于 Base64 的隐写题. 做完真是感觉无处不能用于隐写.

题目

大意就是给了你一段字符串, 让你找 flag.
这里放上字符串, 便于读者实验

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复习一下 Base64 吧

BASE64 是一种编码方式, 是一种可逆的编码方式.
编码后的数据是一个字符串, 包含的字符为: A-Za-z0-9+/
共 64 个字符:26 + 26 + 10 + 1 + 1 = 64
其实是 65 个字符, “=”是填充字符.

64 个字符需要 6 位二进制来表示, 表示成数值为 0~63.
Base64 表

这样, 长度为 3 个字节的数据经过 Base64 编码后就变为 4 个字节

编码

比如, 字符串”Tr0”经过 Base64 编码后变为”VHIw”
Tr0

上面说的字符串长度为 3 个字节的数据位数是 8x3=24, 可以精确地分成 6x4.
如果字节数不是 3 的倍数, 则位数就不是 6 的倍数, 那么就不能精确地划分成 6 位的块.
此时, 需在原数据二进制值后面添加零, 使其字节数是 6 的倍数.
然后, 在编码后的字符串后面添加 1 个或 2 个等号”=”, 表示所添加的零值字节数.
比如, 字符串”Tr0y”经过 Base64 编码后变为”VHIweQ==”

Tr0y

橙色底纹就是添加的 0.
这是 Base64 编码的方式.

解码

解码就是编码的逆过程.

  1. 把 Base64 字符串去掉等号, 转为二进制数(VHIweQ== -> VHIweQ -> 010101000111001000110000011110010000).
  2. 从左到右, 8 个位一组, 多余位的扔掉, 转为对应的 ASCII 码(01010100 01110010 00110000 01111001 0000 -> 扔掉最后 4 位 -> 01010100 01110010 00110000 01111001 -> Tr0y)

隐写原理

注意红色的 0, 我们在解码的时候将其丢弃了, 所以这里的值不会影响解码. 所以我们可以在这进行隐写.
为什么等号的那部分 0 不能用于隐写? 因为修改那里的二进制值会导致等号数量变化, 解码的第 1 步会受影响. 自然也就破坏了源字符串.
而红色部分的 0 是作为最后一个字符二进制的组成部分, 还原时只用到了最后一个字符二进制的前部分, 后面的部分就不会影响还原.
唯一的影响就是最后一个字符会变化. 如下图
隐写
如果你直接解密’VHIweQ==’与’VHIweR==’, 得到的结果都是’Tr0y’.

当然, 一行 base64 顶多能有 2 个等号, 也就是有 2*2 位的可隐写位. 所以我们得弄很多行, 才能隐藏一个字符串, 这也是为什么题目给了一大段 base64 的原因.
接下来, 把要隐藏的 flag 转为 8 位二进制, 塞进去就行了.

加密

# -*- coding: cp936 -*-
import base64

flag = 'Tr0y{Base64isF4n}' #flag
bin_str = ''.join([bin(ord(c)).replace('0b', '').zfill(8) for c in flag])

base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'

with open('0.txt', 'rb') as f0, open('1.txt', 'wb') as f1: #'0.txt'是明文, '1.txt'用于存放隐写后的 base64
    for line in f0.readlines():
        rowstr = base64.b64encode(line.replace('\n', ''))
        equalnum = rowstr.count('=')

        if equalnum and len(bin_str):
            offset = int('0b'+bin_str[:equalnum * 2], 2)
            char = rowstr[len(rowstr) - equalnum - 1]
            rowstr = rowstr.replace(char, base64chars[base64chars.index(char) + offset])
            bin_str = bin_str[equalnum*2:]

        f1.write(rowstr + '\n')

解密

# -*- coding: cp936 -*-

b64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'

with open('1.txt', 'rb') as f:
    bin_str = ''
    for line in f.readlines():
        stegb64 = ''.join(line.split())
        rowb64 =  ''.join(stegb64.decode('base64').encode('base64').split())

        offset = abs(b64chars.index(stegb64.replace('=','')[-1])-b64chars.index(rowb64.replace('=','')[-1]))
        equalnum = stegb64.count('=') #no equalnum no offset

        if equalnum:
            bin_str += bin(offset)[2:].zfill(equalnum * 2)

        print ''.join([chr(int(bin_str[i:i + 8], 2)) for i in xrange(0, len(bin_str), 8)]) #8 位一组

Emmmmmm

很多国外的 CTF 出题思路异常开阔, 值得好好学学.
最后附上这道题的答案吧
flag

话说貌似 ZJPCCTF 也出过这种题.

End

What do you think?

本文标题: 神奇的 Base64 隐写
原始链接: http://www.tr0y.wang/2017/06/14/Base64steg/
发布时间: 2017.06.14-16:23
最后更新: 2018.11.03-20:29
版权声明: 本站文章均采用CC BY-NC-SA 4.0协议进行许可。转载请注明出处!