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](https://github.com/Nightxade/ctf-writeups/blob/master/assets/CTFs/UofT-CTF-2024/flag.enc)  [gen.py](https://github.com/Nightxade/ctf-writeups/blob/master/assets/CTFs/UofT-CTF-2024/gen.py)  
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We're given the ciphertext and a source file for the encryption. Here's `gen.py`:  ```pyimport osimport 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:  
```pyfrom binascii import unhexlifyfrom 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}