Wolv CTF 2024
Blocked 1 [100 pts]
The WOLPHV group (yes, this is an actual article) group encrypted our files, but then blocked us for some reason. I think they might have lost the key. Let’s log into their accounts and find it…
nc blocked1.wolvctf.io 1337
We’re provided a Python source file:
"""
----------------------------------------------------------------------------
NOTE: any websites linked in this challenge are linked **purely for fun**
They do not contain real flags for WolvCTF.
----------------------------------------------------------------------------
"""
import random
import secrets
import sys
import time
from Crypto.Cipher import AES
MASTER_KEY = secrets.token_bytes(16)
def generate(username):
iv = secrets.token_bytes(16)
msg = f'password reset: {username}'.encode()
if len(msg) % 16 != 0:
msg += b'\0' * (16 - len(msg) % 16)
cipher = AES.new(MASTER_KEY, AES.MODE_CBC, iv=iv)
return iv + cipher.encrypt(msg)
def verify(token):
iv = token[0:16]
msg = token[16:]
cipher = AES.new(MASTER_KEY, AES.MODE_CBC, iv=iv)
pt = cipher.decrypt(msg)
username = pt[16:].decode(errors='ignore')
return username.rstrip('\x00')
def main():
username = f'guest_{random.randint(100000, 999999)}'
print(""" __ __
_ ______ / /___ / /_ _ __
| | /| / / __ \\/ / __ \\/ __ \\ | / /
| |/ |/ / /_/ / / /_/ / / / / |/ /
|__/|__/\\____/_/ .___/_/ /_/|___/
/_/""")
print("[ password reset portal ]")
print("you are logged in as:", username)
print("")
while True:
print(" to enter a password reset token, please press 1")
print(" if you forgot your password, please press 2")
print(" to speak to our agents, please press 3")
s = input(" > ")
if s == '1':
token = input(" token > ")
if verify(bytes.fromhex(token)) == 'doubledelete':
print(open('flag.txt').read())
sys.exit(0)
else:
print(f'hello, {username}')
elif s == '2':
print(generate(username).hex())
elif s == '3':
print('please hold...')
time.sleep(2)
# thanks chatgpt
print("Thank you for reaching out to WOLPHV customer support. We appreciate your call. Currently, all our agents are assisting other customers. We apologize for any inconvenience this may cause. Your satisfaction is important to us, and we want to ensure that you receive the attention you deserve. Please leave your name, contact number, and a brief message, and one of our representatives will get back to you as soon as possible. Alternatively, you may also visit our website at https://wolphv.chal.wolvsec.org/ for self-service options. Thank you for your understanding, and we look forward to assisting you shortly.")
print("<beep>")
main()
Basically, the idea is that we need to modify a ciphertext encrypted with AES CBC mode such that, when decrypted, it will return something different.
This is very similar to this challenge that I made a writeup about Iris CTF 2024’s Integral Communication challenge. For an explanation, read this writeup. Only, ignore the part about errors and fixing the IV, since WolvCTF’s challenge conveniently includes this:
username = pt[16:].decode(errors='ignore')
The explanation should be comprehensive for WolvCTF’s challenge here, so here is the final implementation:
from pwn import *
import binascii
from Crypto.Util.strxor import *
p = remote('blocked1.wolvctf.io', 1337)
p.recvuntil(b'as: ')
username = p.recvline().decode('ascii')[:-1]
p.sendlineafter(b'> ', b'2')
enc = binascii.unhexlify(p.recvline().decode('ascii')[:-1])
injection = strxor(strxor(enc[16:16+12], username.encode()), b'doubledelete')
payload = enc[:16] + injection + enc[16+12:]
payload = binascii.hexlify(payload)
p.sendlineafter(b'> ', b'1')
p.sendlineafter(b'> ', payload)
print(p.recvrepeat(1).decode('ascii'))
Run the script to get the flag!
wctf{th3y_l0st_th3_f1rst_16_byt35_0f_th3_m3ss4g3_t00}