Space Pirates
| Name: | Space Pirates |
|---|---|
| Hint: | Jones and his crew have started a long journey to discover the legendary treasure left by the guardians of time in the early beginnings of the universe. Mr Jones, though, is wanted by the government for his crimes as a pirate. Our agents entered his |
| Base Points: | Easy - Retired [0] |
| Rated Difficulty: |  |
 |
HTB-Bot  |
| Creator: | P3t4  |
This time we have two files and the hint:
chall.py:
from sympy import *
from hashlib import md5
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from random import randint, randbytes,seed
from Crypto.Util.number import bytes_to_long
FLAG = b'HTB{dummyflag}'
class Shamir:
def __init__(self, prime, k, n):
self.p = prime
self.secret = randint(1,self.p-1)
self.k = k
self.n = n
self.coeffs = [self.secret]
self.x_vals = []
self.y_vals = []
def next_coeff(self, val):
return int(md5(val.to_bytes(32, byteorder="big")).hexdigest(),16)
def calc_coeffs(self):
for i in range(1,self.n+1):
self.coeffs.append(self.next_coeff(self.coeffs[i-1]))
def calc_y(self, x):
y = 0
for i, coeff in enumerate(self.coeffs):
y +=coeff *x**i
return y%self.p
def create_pol(self):
self.calc_coeffs()
self.coeffs = self.coeffs[:self.k]
for i in range(self.n):
x = randint(1,self.p-1)
self.x_vals.append(x)
self.y_vals.append(self.calc_y(x))
def get_share(self):
return self.x_vals[0], self.y_vals[0]
def main():
sss = Shamir(92434467187580489687, 10, 18)
sss.create_pol()
share = sss.get_share()
seed(sss.secret)
key = randbytes(16)
cipher = AES.new(key, AES.MODE_ECB)
enc_FLAG = cipher.encrypt(pad(FLAG,16)).hex()
f = open('msg.enc', 'w')
f.write('share: ' + str(share) + '\n')
f.write('coefficient: ' + str(sss.coeffs[1]) + '\n')
f.write('secret message: ' + str(enc_FLAG) + '\n')
f.close()
if __name__ == "__main__":
main()
msg.enc:
share: (21202245407317581090, 11086299714260406068)
coefficient: 93526756371754197321930622219489764824
secret message: 1aaad05f3f187bcbb3fb6c9e233ea339082062fc10a59604d96bcc38d0af92cd842ad7301
b5b72bd5378265dae0bc1c1e9f09a90c97b35cfadbcfe259021ce495e9b91d29f563ae7d49b66296f15e7999c
9e547fac6f1a2ee682579143da511475ea791d24b5df6affb33147d57718eaa5b1b578230d97f395c458fc2c9
c36525db1ba7b1097ad8f5df079994b383b32695ed9a372ea9a0eb1c6c18b3d3d43bd2db598667ef4f8084542
4d6c75abc88b59ef7c119d505cd696ed01c65f374a0df3f331d7347052faab63f76f587400b6a6f8b718df1db
9cebe46a4ec6529bc226627d39baca7716a4c11be6f884c371b08d87c9e432af58c030382b737b9bb63045268
a18455b9f1c4011a984a818a5427231320ee7eca39bdfe175333341b7c
Hint:
Jones and his crew have started a long journey to discover the legendary treasure left by
the guardians of time in the early beginnings of the universe. Mr Jones, though, is wanted
by the government for his crimes as a pirate. Our agents entered his
We have a very important clue in the msg.enc file. The shares and coefficient lead directly to Shamir Secret Sharing (as does the SSS line clearly saying Shamir). This is going to be very similar to the "Please don't share" challenge. We can also see in the next_coeff function of the chall.py file that it is an MD5 conversion. We can also (the n=10 or first numeric argument in the SSS line) that we only need the next 9 coefficients. We should be able to script out a method of finding these coefficients and reversing the Shamir Secret Sharing to break this message.
# sss_reverse.py
#!/bin/python3
import base64
from Crypto import Random
from Crypto.Cipher import AES
from random import randint, randbytes,seed
from hashlib import md5
import math
def next_coeff(val):
return int(md5(val.to_bytes(32, byteorder="big")).hexdigest(),16)
def calc_coeffs(init_coeff, coeffs):
for i in range(8):
coeffs.append(next_coeff(coeffs[i]))
def rev_secret(init_coeff, init_x, init_y, p):
coeffs = [init_coeff]
calc_coeffs(init_coeff, coeffs)
# init_y = (secret + coeffs[1] * init_x + coeffs[2] * init_x^1 + ... + coeffs[9] *
init_x^9) % P
sum = 0
for i in range(len(coeffs)):
init_y -= coeffs[i] * (init_x ** (i + 1))
GCD = math.gcd(p, 1)
secret = (GCD * (init_y)) % p
return secret
secret = rev_secret(93526756371754197321930622219489764824, 21202245407317581090,
11086299714260406068, 92434467187580489687)
enc = '1aaad05f3f187bcbb3fb5c9e233ea339082062fc10a59604d96bcc38d0af92cd842ad7301b5b72bd
5378265dae0bc1c1e9f09a90c97b35cfadbcfe259021ce495e9b91d29f563ae7d49b66296f15e7999c9e547
fac6f1a2ee682579143da511475ea791d24b5df6affb33147d57718eaa5b1b578230d97f395c458fc2c9c36
525db1ba7b1097ad8f5df079994b383b32695ed9a372ea9a0eb1c6c18b3d3d43bd2db598667ef4f80845424
d6c75abc88b59ef7c119d505cd696ed01c65f374a0df3f331d7347052faab63f76f587400b6a6f8b718df1d
b9cebe46a4ec6529bc226627d39baca7716a4c11be6f884c371b08d87c9e432af58c030382b737b9bb63045
268a18455b9f1c4011a984a818a5427231320ee7eca39bdfe175333341b7c'
seed(secret)
key = randbytes(16)
cipher = AES.new(key, AES.MODE_ECB)
enc_FLAG = cipher.decrypt(bytes.fromhex(enc)).decode("ascii")
print(enc_FLAG)
There it is! HTB{1_d1dnt_kn0w_0n3_sh4r3_w45_3n0u9h!1337}