TetCTF——Crypto

Shares

Description: Yet another secret sharing scheme.
Service: nc 139.162.61.222 13371
Binary: https://drive.google.com/file/d/1sP0zKSKoF6TJZ-gtaUf7rv4dQ-bGr-cp/view?usp=sharing

用nc并沒能拿到題目代碼，不知道問題出哪了🐷一直顯示鏈接超時
所以比完賽才拿到題目

""" This is an (incomplete) implement for a new (and experimental) secret/password sharing scheme. The idea is simple. Basically, a secret or password is turned into a set of finite field elements and each share is just a linear combination of these elements. On the other hand, when enough shares are collected, the finite field elements are determined, allowing the original secret or password to be recovered. """ from typing import List from secrets import randbelow import stringALLOWED_CHARS = string.ascii_lowercase + string.digits + "_" P = len(ALLOWED_CHARS) INT_TO_CHAR = {} CHAR_TO_INT = {} for _i, _c in enumerate(ALLOWED_CHARS):INT_TO_CHAR[_i] = _cCHAR_TO_INT[_c] = _idef get_shares(password: str, n: int, t: int) -> List[str]:"""Get password shares.Args:password: the password to be shared.n: the number of shares returned.t: the minimum number of shares needed to recover the password.Returns:the shares."""assert len(password) <= tassert n > 0ffes = [CHAR_TO_INT[c] for c in password]ffes += [randbelow(P) for _ in range(t - len(password))]result = []for _ in range(n):coeffs = [randbelow(P) for _ in range(len(ffes))]s = sum([x * y for x, y in zip(coeffs, ffes)]) % Pcoeffs.append(s)result.append("".join(INT_TO_CHAR[i] for i in coeffs))return resultdef combine_shares(shares: List[str]) -> str:raise Exception("unimplemented")def main():pw_len = 16password = "".join(INT_TO_CHAR[randbelow(P)] for _ in range(pw_len))# how about n < t :Dn = 16t = 32for _ in range(2022):line = input()if line == password:from secret import FLAGprint(FLAG)returnelse:print(get_shares(password, n, t))if __name__ == '__main__':main()

考點依舊在格

對整個程序的描述是：
這是一個新的(實驗性的)秘密/密碼共享方案的(不完整的)實現。
這個想法很簡單。基本上，一個秘密或密碼被轉換成一組有限域元素，每個共享只是這些元素的線性組合。另一方面，當收集到足夠多的共享時，確定有限字段元素，允許恢復原始的秘密或密碼

對get_shares的描述是：
得到密碼的shares.
參數:
password:共享用戶的密碼。
N:返回的shares數量。
T:找回密碼所需的最小shares數量。
返回：shares。

讀程序
從這里開始，char表示允許的$37$個字符中的一個，而int表示$0$到$36$之間的整數。
正如我們在源代碼中看到的，一個1616個字符的密碼$c_1{c}_2\dots c_{16}$
?
一開始就確定了，我們的目標是猜測它。還可以獲得20212021股get_shares(pwd)，其中:
向密碼追加$16$個隨機字符
$P=c_1{c}_2\dots c_{16}{r}_1{r}_2\dots r_{16}$
生成$(\mathbb{Z}/37\mathbb{Z}{)}^{32}v$中$16$個32位的隨機向量$v_1,v_2,ldots,v_{16}$

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參考代碼

from pwn import * from tqdm import tqdm# local runs in 30 seconds, remote takes 20 minutes r = remote('139.162.61.222', 13371) # r = process('pypy3 shares.py', shell=True)P = 37 CHARS = "abcdefghijklmnopqrstuvwxyz0123456789_1" def INT_TO_CHAR(i): return CHARS[i] def CHAR_TO_INT(c): return CHARS.index(c)lhs, rhs = [], [] tq = tqdm(range(2000)) cnt = 0 for _ in tq:r.sendline(b'a')rv = eval(r.recvline())# Notice I keep the c1, c2, ..., c16 coefficients as vectorsrv = [[CHAR_TO_INT(c) for c in r] for r in rv]cs = Matrix(GF(P), [r[16:32] for r in rv])# Singular?if cs.rank() < 16:rs = Matrix(GF(P), [r[:16] + [r[-1]] for r in rv])res = cs.augment(rs).rref()[-1]# first 16 columns are the v_{i, j}lhs.append(res[16:32])rhs.append(res[32])if len(lhs) == 16:breakcnt += 1tq.set_description(f"{len(lhs)} / {cnt}")# solve for password ref = Matrix(GF(P), lhs).solve_right(vector(GF(P), rhs)) pwd = ''.join((INT_TO_CHAR(r) for r in ref)) r.sendline(pwd.encode()) print("Flag:", r.recvline().decode())

看人家的代碼看好幾天，雖然沒都懂，而且時間過了差不多也不能訪問了，但是還是有點收獲的，比如： 我太菜了😥

總結

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