2021 | DragonCTF | Miscellaneous

Compress The Flag

题目¶

Technically this isn't a good compression benchmark, but it's the only one we have.

nc compresstheflag.hackable.software 1337

server.py

#!/usr/bin/env python3
import threading
import socket
import random
import codecs
import lzma as lz

with open("flag.txt", "rb") as f:
    FLAG = f.read().strip()

def none(v):
    return len(v)

def zlib(v):
    return len(codecs.encode(v, "zlib"))

def bzip2(v):
    return len(codecs.encode(v, "bz2"))

def lzma(v):
    return len(lz.compress(v))

COMPRESSION_FUNCS = [
    none,
    zlib,
    bzip2,
    lzma
]

def handle_connection(s, addr):
    s.sendall(
        ("Please send: seed:string\\n\n"
        "I'll then show you the compression benchmark results!\n"
        "Note: Flag has format DrgnS{[A-Z]+}\n").encode())

    data = b''
    while True:
        idx = data.find(b'\n')
        if idx == -1:
            if len(data) > 128:
                s.shutdown(socket.SHUT_RDWR)
                s.close()
                return

            d = s.recv(1024)
            if not d:
                s.close()
                return
            data += d
            continue

        line = data[:idx]
        data = data[idx+1:]

        seed, string = line.split(b':', 1)

        flag = bytearray(FLAG)
        random.seed(int(seed))
        random.shuffle(flag)
        print(flag)
        test_string = string + bytes(flag)

        response = []
        for cfunc in COMPRESSION_FUNCS:
            res = cfunc(test_string)
            response.append(f"{cfunc.__name__:>8} {res:>4}")

        response.append('')
        response.append('')
        s.sendall('\n'.join(response).encode())

    s.shutdown(socket.SHUT_RDWR)
    s.close()


def main():
    with socket.socket(socket.AF_INET, socket.SOCK_STREAM) as s:
        s.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
        s.bind(('0.0.0.0', 1337))
        s.listen(256)

        while True:
            conn, addr = s.accept()
            print(f"Connection from: {addr}")

            th = threading.Thread(
                target=handle_connection,
                args=(conn, addr),
                daemon=True
            )
            th.start()

if __name__ == "__main__":
    main()

解题思路¶

提供一个随机数种子和字符串，服务器使用随机数种子打乱 Flag 字符的顺序，再拼接上提供的字符串，返回最后生成字符串的原长度以及不同压缩算法压缩后的长度

Please send: seed:string\n
I'll then show you the compression benchmark results!
Note: Flag has format DrgnS{[A-Z]+}
1:A
    none   26
    zlib   34
   bzip2   63
    lzma   84

none 即不进行压缩，除去输入的字符串和 Flag 头，需要求解的部分的长度为 \(26 - 1 - 7 = 18\)
指定随机数种子，也就是说可以知道 Flag 打乱之后每一个字符原先的位置

先在本地用 DrgnS{ABCDEFGHIJKLMNOPQR} 观察一下

当种子为 1 时，字母 E 开头

观察发现，当输入字符串为 4 个相同字符时，EEEE 和打乱的 Flag 字符串使用 zlib 和 bzip2 （均含 RLE 和 Huffman 编码）压缩的长度会小于其他字符

1:EEEE
    none   29
    zlib   35
   bzip2   69
    lzma   88
1:AAAA
    none   29
    zlib   37
   bzip2   70
    lzma   88
1:SSSS
    none   29
    zlib   37
   bzip2   70
    lzma   88

先找到使每个字母打乱后作为开头的随机数种子

import random

flag_pos = b'DrgnS{ABCTEFGHIJKLMNOPQR}' # 为方便查找，每个字符均不相同
d = {}
for i in range(10000):
    random.seed(i)
    tflag = bytearray(flag_pos)
    random.shuffle(tflag)
    if tflag[0] in b'ABCTEFGHIJKLMNOPQR' and tflag[0] not in d:
        d[tflag[0]] = i
    if len(d) == 18:
        break

随后爆破每一个字母（有一定的运气成分，如果随机数种子没选好，导致打乱的 Flag 开头是连续字母就难以直接根据长度判断了）

import pwn

conn = pwn.remote("compresstheflag.hackable.software", 1337)
conn.recvline_contains('Note')

flag_pos = 'DrgnS{ABCTEFGHIJKLMNOPQR}'
flag = list('DrgnS{xxxxxxxxxxxxxxxxxx}')
for k in d:
    min, mic = 0xffff, 'A'
    for c in range(ord('A'), ord('Z') + 1):
        conn.send(f'{d[k]}:{chr(c) * 4}\n')
        l = int(conn.recvline_contains('zlib').split(b' ')[-1])
        if l < min:
            min, mic = l, chr(c)
    flag[flag_pos.find(chr(k))] = mic

print(''.join(flag))
# DrgnS{THISISACRIMEIGUESS}

参考资料¶

最后更新: 2021年12月15日 14:47:30

Contributors: YanhuiJessica, YanhuiJessica

Compress The Flag

题目¶

解题思路¶

参考资料¶

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