GoogleCTF2023 Writeup

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Google Capture The Flag 2023 Writeup

LEAST COMMON GENOMINATOR?

Someone used this program to send me an encrypted message but I can't read it! It uses something called an LCG, do you know what it is? I dumped the first six consecutive values generated from it but what do I do with it?!

Attachment

generate.py 
from secret import config
from Crypto.PublicKey import RSA
from Crypto.Util.number import bytes_to_long, isPrime

class LCG:
    lcg_m = config.m
    lcg_c = config.c
    lcg_n = config.n

    def __init__(self, lcg_s):
        self.state = lcg_s

    def next(self):
        self.state = (self.state * self.lcg_m + self.lcg_c) % self.lcg_n
        return self.state

if __name__ == '__main__':

    assert 4096 % config.it == 0
    assert config.it == 8
    assert 4096 % config.bits == 0
    assert config.bits == 512

    # Find prime value of specified bits a specified amount of times
    seed = 211286818345627549183608678726370412218029639873054513839005340650674982169404937862395980568550063504804783328450267566224937880641772833325018028629959635
    lcg = LCG(seed)
    primes_arr = []

    dump = True
    items = 0
    dump_file = open("dump.txt", "w")

    primes_n = 1
    while True:
        for i in range(config.it):
            while True:
                prime_candidate = lcg.next()
                if dump:
                    dump_file.write(str(prime_candidate) + '\n')
                    items += 1
                    if items == 6:
                        dump = False
                        dump_file.close()
                if not isPrime(prime_candidate):
                    continue
                elif prime_candidate.bit_length() != config.bits:
                    continue
                else:
                    primes_n *= prime_candidate
                    primes_arr.append(prime_candidate)
                    break

        # Check bit length
        if primes_n.bit_length() > 4096:
            print("bit length", primes_n.bit_length())
            primes_arr.clear()
            primes_n = 1
            continue
        else:
            break

    # Create public key 'n'
    n = 1
    for j in primes_arr:
        n *= j
    print("[+] Public Key: ", n)
    print("[+] size: ", n.bit_length(), "bits")

    # Calculate totient 'Phi(n)'
    phi = 1
    for k in primes_arr:
        phi *= (k - 1)

    # Calculate private key 'd'
    d = pow(config.e, -1, phi)

    # Generate Flag
    assert config.flag.startswith(b"CTF{")
    assert config.flag.endswith(b"}")
    enc_flag = bytes_to_long(config.flag)
    assert enc_flag < n

    # Encrypt Flag
    _enc = pow(enc_flag, config.e, n)

    with open ("flag.txt", "wb") as flag_file:
        flag_file.write(_enc.to_bytes(n.bit_length(), "little"))

    # Export RSA Key
    rsa = RSA.construct((n, config.e))
    with open ("public.pem", "w") as pub_file:
        pub_file.write(rsa.exportKey().decode())

分析可知:

  • flag是使用RSA加密的,已知公🔑 文件,即n,e

  • 使用LCG线性同余生成器生成素数

  • 已知LCG的种子和前6个连续生成的数字

  • config.it = 8

  • config.bits = 256

LCG是伪随机数生成器和流密码的一种,递推公式是 𝑋𝑛+1=(𝑎𝑋𝑛+𝑐) 𝑚𝑜𝑑 𝑚

已知初值和随后LCG连续生成的6个值,未知增量、乘数和模数.

我们可以通过攻击得到这三个值,然后模拟原算法通过LCG得到8个素数后,进一步计算n的欧拉函数并求逆元得到d,解密即可.

题解: 
import math
from functools import reduce

import gmpy2

from Crypto.PublicKey import RSA
from Crypto.Util.number import bytes_to_long, isPrime, long_to_bytes

dump_file = open("dump.txt")
output_values = [int(x) for x in dump_file.readlines()]  # 已知的 LCG 输出值

def crack_unknown_increment(states, modulus, multiplier):
    """
    已知：a,m,s0,s1
    求c
    """
    increment = (states[1] - states[0] * multiplier) % modulus
    return modulus, multiplier, increment

def crack_unknown_multiplier(states, modulus):
    """
    已知：m,s0,s1,s2
    求a
    """
    multiplier = (
        (states[2] - states[1]) * gmpy2.invert(states[1] - states[0], modulus) % modulus
    )  # 注意这里求逆元
    return crack_unknown_increment(states, modulus, multiplier)

def crack_unknown_modulus(states):
    """
    已知：s0-s6
    求a,c,m
    """
    diffs = [s1 - s0 for s0, s1 in zip(states, states[1:])]
    zeroes = [t2 * t0 - t1 * t1 for t0, t1, t2 in zip(diffs, diffs[1:], diffs[2:])]
    modulus = abs(reduce(math.gcd, zeroes))
    return crack_unknown_multiplier(states, modulus)

class LCG:
    def __init__(self, lcg_m, lcg_c, lcg_n, lcg_s):
        self.state = lcg_s
        self.lcg_m = lcg_m
        self.lcg_c = lcg_c
        self.lcg_n = lcg_n

    def next(self):
        self.state = (self.state * self.lcg_m + self.lcg_c) % self.lcg_n
        return self.state


m, a, c = crack_unknown_modulus(output_values)
seed = 211286818345627549183608678726370412218029639873054513839005340650674982169404937862395980568550063504804783328450267566224937880641772833325018028629959635
lcg = LCG(a, c, m, seed)
print(a, c, m)
primes_n = 1
primes_arr = []
for i in range(8):
    while True:
        prime_candidate = lcg.next()
        if not isPrime(prime_candidate):
            continue
        elif prime_candidate.bit_length() != 512:
            continue
        else:
            primes_n *= prime_candidate
            primes_arr.append(prime_candidate)
            break

print(list(primes_arr))

phi = 1
for k in primes_arr:
    phi *= k - 1

key = RSA.importKey(open("public.pem", "r").read())
n = key.n
e = key.e
d = gmpy2.invert(e, phi)

enc = open("flag.txt", "rb").read()

flag = pow(int.from_bytes(enc, "little"), d, n)
print(long_to_bytes(flag))

flag: CTF{C0nGr@tz_RiV35t_5h4MiR_nD_Ad13MaN_W0ulD_b_h@pPy}

参考:

NPC

A friend handed me this map and told me that it will lead me to the flag.
It is confusing me and I don't know how to read it, can you help me out?

Attachment

使用Graphviz工具将hint.dot转换为图片,得到下图

dot -Tjpg  hint.dot -o hint.jpg
hint.jpg

分析代码:

  • 从USACONST.TXT中随机选择N个单词生成密码,使用passphrase.encrypt(secret, password)对flag加密

  • 对于password中的每个字母,创建一个带有唯一ID的节点,并添加到图中

  • 按照password的顺序,对密码中相邻的两个字符创建一条边

  • 向图中随机添加int(len(password) ** 1.3)条边

  • 随机打乱图中节点和边的顺序

  • 随机交换一些节点的起点和终点,由于random() % 2只有在random函数返回值为0时才为False,我们假定图中每条边的起点和终点都被交换了一遍

  • 将节点和边的信息写入到hint.dot文件中

我是思路是:

  1. 遍历words_list中的单词,从每个节点出发,使用DFS判断是否可以在图中找到,这样过滤后得到30个单词,即密码一定由该30个单词中的某几个单词组成

  2. 枚举密码中所用的单词个数N

  3. 使用组合数从这30个单词选择N个单词

  4. 判断所选的N个单词组成的字符集和hint.dot中的字符集是否一致

  5. 对N个单词进行全排列,并尝试解密 (为了提高效率,这里还用到了多进程)

当N=5时,得到passwordstandardwatersigngivenchosen和flag.

代码 
import concurrent.futures
import itertools
import re
from collections import Counter

from pyrage import passphrase

from encrypt import get_word_list


class Node:
    def __init__(self, letter):
        self.letter = letter
        self.adjacent = []

    def __str__(self) -> str:
        return f"{self.letter} -> {[x.letter for x in self.adjacent]}"

    __repr__ = __str__

def build_nodes():
    pattern = r"\s+(\d+)\s+\[label=(\w+)\];"
    pattern2 = r"\s+(\d+)\s+--\s+(\d+);$"
    nodes = dict()
    with open("hint.dot", "r") as f:
        for line in f:
            if "label" in line
                match = re.match(pattern, line)
                node_id = match.group(1)
                letter = match.group(2)
                nodes[node_id] = Node(letter)
            elif "--" in line:
                match = re.match(pattern2, line)
                start = match.group(1)
                end = match.group(2)
                # nodes[start].adjacent.append(nodes[end])
                nodes[end].adjacent.append(nodes[start])
    return nodes

visited = set()

def dfs(node, index, word):
    if index == len(word):
        return True

    if node.letter != word[index]:
        return False

    visited.add(node)

    for adj in node.adjacent:
        if adj not in visited:
            if dfs(adj, index + 1, word):
                return True

    visited.remove(node)
    return False

def check(password):
    with open("secret.age", "rb") as f:
        enc = f.read()
    try:
        print("[FLAG] is ", passphrase.decrypt(enc, password))
        print("Password is ", password)
    except Exception as e:
        pass


def filter_words(nodes):
    ans = set()
    for word in get_word_list():
        visited.clear()
        for node in nodes.values():
            if dfs(node, 0, word):
                ans.add(word)
                break

    print("ans:", len(ans), ans)
    return ans


def main():
    nodes = build_nodes()

    letters = sorted([node.letter for node in nodes.values()])

    ans = filter_words(nodes)
    for num in range(1, len(ans) + 1):
        print(f"Num is {num}")
        for comb in itertools.combinations(ans, num):  # 组合
            if sorted("".join(comb)) == letters:
                passwords = [
                    "".join(perm) for perm in itertools.permutations(comb, len(comb))
                ]

                with concurrent.futures.ProcessPoolExecutor(max_workers=8) as executor:
                    executor.map(check, passwords)


if __name__ == "__main__":
    main()

flag: CTF{S3vEn_bR1dg35_0f_K0eN1g5BeRg}

参考:

UNDER-CONSTRUCTION

We were building a web app but the new CEO wants it remade in php.

Attachment
https://under-construction-web.2023.ctfcompetition.com
https://under-construction-php-web.2023.ctfcompetition.com

题目提供了Flask和PHP两个站点,用户可以在Flask站点进行注册,注册的账号可以同时用于登录Flask和PHP两个站点.

分析代码:

Flask会将HTTP请求原始查询参数转发到PHP应用程序中完成用户注册.

# File: /flask/authorized_routes.py
@authorized.route('/signup', methods=['POST'])
def signup_post():
    raw_request = request.get_data()
    ...
    requests.post(f"http://{PHP_HOST}:1337/account_migrator.php", 
        headers={"token": TOKEN, "content-type": request.headers.get("content-type")}, data=raw_request)
    return redirect(url_for('authorized.login'))

只有gold级别的用户,在PHP站点登录后才可以看到FLAG

# File: /php/index.php
function getResponse()
{
    ...
    $response = "Login successful. Welcome " . htmlspecialchars($username) . ".";

    if ($tier === "gold") {
        $response .= " " . getenv("FLAG");
    }

    return $response;
}

Flask会对查询参数进行校验,防止创建高权限的用户.

# File: /flask/authorized_routes.py
@authorized.route('/signup', methods=['POST'])
def signup_post():
    ...
    tier = models.Tier(request.form.get('tier'))
    if(tier == models.Tier.GOLD):
        flash('GOLD tier only allowed for the CEO')
        return redirect(url_for('authorized.signup'))
    ...

HTTP查询参数中存在重复的key时,在Flask和PHP有不同的行为,flask会取第一个值,而PHP会取最后一个值.

因此我们可以构造如下命令,以此绕过Flask对查询参数的校验,并在PHP中注册高权限用户.

curl

curl -X POST https://under-construction-web.2023.ctfcompetition.com/signup -d "username=admin&password=admin&tier=blue&tier=gold"

然后用上述的用户名和密码去PHP站点登录即可.

flag: CTF{ff79e2741f21abd77dc48f17bab64c3d}

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