UofT CTF 2024
Repeat [100 pts]
I’m a known repeat offender when it comes to bad encryption habits. But the secrets module is secure, so you’ll never be able to guess my key!
Author: SteakEnthusiast
flag.enc
gen.py
We’re given the ciphertext and a source file for the encryption. Here’s gen.py:
import os
import secrets
flag = "REDACATED"
xor_key = secrets.token_bytes(8)
def xor(message, key):
return bytes([message[i] ^ key[i % len(key)] for i in range(len(message))])
encrypted_flag = xor(flag.encode(), xor_key).hex()
with open("flag.enc", "w") as f:
f.write("Flag: "+encrypted_flag)
Seems like a very standard XOR encryption.
Conveniently, the key length is 8 bytes. This is crucial because we actually know the first 8 bytes of the flag, i.e. uoftctf{. Therefore, since XOR is a reversible function, we can relatively easily get the flag. See below:
\(First\; 8\; bytes\; of\; ciphertext\; = ct[:8] = key \oplus \text{"uoftctf\{"}\)
\(ct[:8] \oplus \text{"uoftctf\{"} = key \oplus \text{"uoftctf\{"} \oplus \text{"uoftctf\{"}\)
\(ct[:8] \oplus \text{"uoftctf\{"} = key\)
Therefore, we can easily get the key and decrypt the flag! Here’s the short implementation:
from binascii import unhexlify
from Crypto.Util.strxor import strxor
ct = unhexlify("982a9290d6d4bf88957586bbdcda8681de33c796c691bb9fde1a83d582c886988375838aead0e8c7dc2bc3d7cd97a4")
key = strxor(ct[:8], b'uoftctf{')
def xor(message, key):
return bytes([message[i] ^ key[i % len(key)] for i in range(len(message))])
flag = xor(ct, key)
print(flag)
Run the script to get the flag!
uoftctf{x0r_iz_r3v3rs1bl3_w17h_kn0wn_p141n73x7}