Crypto - XorD

I just discovered bitwise operators, so I guess 1 XOR 1 = 1?

The flag is encrypted byte-by-byte using XOR with a random key. The Python random module is used with a fixed seed (1337), making the keystream predictable.

XOR is reversible. random.seed(1337) makes the random numbers deterministic The same random keystream can be regenerated to decrypt the flag

We have xord.py and output.txt from the challenge.

xord.py

import os
import random

def xor(a, b):
    return bytes([a ^ b])

flag = os.getenv('FLAG', 'pascalCTF{REDACTED}')
encripted_flag = b''
random.seed(1337)

for i in range(len(flag)):
    random_key = random.randint(0, 255)
    encripted_flag += xor(ord(flag[i]), random_key)

with open('output.txt', 'w') as f:
    f.write(encripted_flag.hex())

output.txt

cb35d9a7d9f18b3cfc4ce8b852edfaa2e83dcd4fb44a35909ff3395a2656e1756f3b505bf53b949335ceec1b70e0

Solve.py

import random

cipher_hex = "cb35d9a7d9f18b3cfc4ce8b852edfaa2e83dcd4fb44a35909ff3395a2656e1756f3b505bf53b949335ceec1b70e0"
cipher = bytes.fromhex(cipher_hex)

random.seed(1337)

flag = ""
for b in cipher:
    flag += chr(b ^ random.randint(0, 255))

print(flag)

FLAG: pascalCTF{1ts_4lw4ys_4b0ut_x0r1ng_4nd_s33d1ng}