exfiltrated entropy
by sealldev
🚩 CTFs HackTheBox University CTF 2024 rev
Suggested:
#network-forensics
exfiltrated entropy / HackTheBox University CTF 2024
Original Writeup on seall.dev
We are given a server.py, client.py, params.py and a PCAP.
We are given some initial parameters:
a = 0xa1d41ebef9c575ac113fcfd5ac8dbda9
b = 0x8dcf3cf766e0b6c30e753416a70e2367
m = 0x100000000000000000000000000000000
It’s an LCG encryption system:
class LCG:
def __init__(self):
self.state = SEED
self.a = a
self.b = b
self.m = m
def _next(self):
self.state = (self.a * self.state + self.b) % self.m
return self.state
def generate_key(self, l):
return bytes([self._next() & 0xff for _ in range(l)])
def generate_packet_uuid(self):
return hex(self._next())
def encrypt(self, msg):
key = self.generate_key(len(msg))
return xor(msg, key)
def decrypt(self, msg):
return self.encrypt(msg)
The issue is with every _next it sets the state again, so we can predict it with information with the PCAP and decrypt the message.
The blue is the server, so the id we are seeing is the first self._next() for the string as its sent before the output is encrypted, and we can calculate all future ones for the encrypted string from that, meaning we can decrypt outputs.
The state encryption is made with 3 known params (a, b, and m) and the state from the previous execution.
We can start by making a state calculation function:
def calculate_next_state(S_n,a,b,m):
return (a * S_n + b) % m
The function for key generation is using a new state with a & 0xff for each character in the length of the message, so the key covers the full message. The message is then XOR’ed with this key.
def decode_and_xor(encoded_message, key):
decoded_message = base64.b64decode(encoded_message)
state = int(key,16)
key_bytes = []
for i in range(len(decoded_message)):
state = calculate_next_state(state, a, b, m)
val = state & 0xff
key_bytes.append(val)
key_bytes = bytes(key_bytes)
return xor(decoded_message, key_bytes)
The message is first base64 decoded, then we convert the hexadecimal state to an integer. For each character in the message the state is calculated and then the key is created in key_bytes. The key is then used with the xor function from the pwn library.
Now that the script works, we can use the id from the 2nd server message and the response from the client’s 2nd message to decode the output:
import base64
from itertools import product
from pwn import xor
from params import *
def calculate_next_state(S_n,a,b,m):
return (a * S_n + b) % m
def decode_and_xor(encoded_message, key):
decoded_message = base64.b64decode(encoded_message)
state = int(key,16)
key_bytes = []
for i in range(len(decoded_message)):
state = calculate_next_state(state, a, b, m)
val = state & 0xff
key_bytes.append(val)
key_bytes = bytes(key_bytes)
return xor(decoded_message, key_bytes)
key = 'eba14c429a64b2251717da016e096091'
encoded_message = 'Ve7i4H2jQpnQaq2QVgLqprnTCzzM8xLx3TzrV/17HYPvkpZOkcLiOWhXSybh+QMNAie+CTVC7lZ928epBo/yMoQ1KfAlfwBckLny2pSb86i8RcHlz/aG9kEjfNy8ek/VKciP0V+Duq1xT9c5cH/Cl5mzM0I1z3bP4B/CMJMf/2eJKzdt+jucTAz7OwONh3twYB/a/R0nzzBG5iKUZe/SE2wSA8lDHCbh8dOP5DIj2cLR+XiIrooI'
result = decode_and_xor(encoded_message, key)
print(f'Key: {key} -> Decoded XOR Result: \n{result.decode(errors='ignore')}')
$ python3 solve.py
Key: eba14c429a64b2251717da016e096091 -> Decoded XOR Result:
uid=1000(user) gid=1000(user) groups=1000(user),20(dialout),24(cdrom),25(floppy),27(sudo),29(audio),30(dip),44(video),46(plugdev),106(netdev),110(lpadmin),114(bluetooth),116(scanner)
Yay!
Now decoding the huge output we get the flag.
$ python3 solve.py
Key: 8829affbed23334d34016095627ba065 -> Decoded XOR Result:
0█████000█████0███████████0███████████000000███0000000000█████00000███00████00████000000000000000000000000000000000█████00000000000000000000000000000000000000000000000000000000███00000000000000000000000000000000000█████0000000000
░░███000░░███0░█░░░███░░░█░░███░░░░░███0000██░0000000000░░███00000░░░00░░███0░░███00000000000000000000000000000000░░███00000000000000000000000000000000000000000000000000000000░░░00000000000000000000000000000000000░░███00000000000
0░███0000░███0░000░███00░00░███0000░███000██00000█████00███████000████00░███00░███000000000000████████0000██████00███████00000000000000██████000██████00████████000█████0█████0████00████████0000██████000██████000███████00000000000
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Flag: HTB{still_not_convinced_about_LCG_security?}