PCC Quals 2024 - Counter

A AES based crypto challenge

This blog has the solution for a crypto challenge that I solved at the Pakistan Cyber Security Challenges Qualification Round 2024.

Counter:

Challenge Description: its so frustrating to generate key again and again.

We were provided with the following files.

.
├── challenge.py
├── docker-compose.yml
├── Dockerfile.dist
├── flag.txt
└── new.py

The challenge.py had the following script.

#!/usr/local/bin/python3

import random
from Crypto.Util.number import long_to_bytes
from Crypto.Cipher import AES
from Crypto.Util import Counter
from Crypto.Util.Padding import pad

def get_flag():
    try:
        with open("/flag.txt", "rb") as f:
            FLAG = f.read().strip()
        return FLAG
    except:
        print("[ERROR] - Please contact an Administrator.")

key = random.randbytes(16)

def encrypt(data: bytes) -> bytes:
    cipher = AES.new(key, AES.MODE_CTR, counter=Counter.new(128))
    return cipher.encrypt(data)

# you'll have to find this, i guess.
f = open("a_quote_from_a_leader.txt", "rb")
data = f.read()
f.close()

encrypted = encrypt(data)
print(f"Encrypted Data: {encrypted.hex()}")

flag = get_flag()

encrypted_flag = encrypt(flag)
print(f"Encrypted Flag: {encrypted_flag.hex()}")

Script Breakdown:

This Python script encrypts both a quote from a file and a flag using AES encryption in CTR mode. First, the get_flag() function attempts to read the flag from a file /flag.txt, returning its content if successful, or printing an error message if the file cannot be accessed. The script generates a random 16-byte key using random.randbytes(16). The encryption is performed by the encrypt() function, which initializes an AES cipher in CTR mode with a 128-bit counter. The content of the file a_quote_from_a_leader.txt is read, encrypted using this cipher, and then printed in hexadecimal format. Afterward, the flag is fetched using the get_flag() function, encrypted similarly, and also printed in hexadecimal format. The user must discover the flag by reversing the process, but the random key and dynamic counter make decryption challenging without access to the original key and counter values.

Solution:

Hint: The quote is of Quaid-e-azam

So by looking at the hint I searched for quotes of Quaid-e-Azam and added them in my script one after the other and the one quote that gave me the flag was

I do not believe in taking the right decision, I take a decision and make it right.

from Crypto.Util.number import long_to_bytes

def xor_bytes(a: bytes, b: bytes) -> bytes:
    return bytes(x ^ y for x, y in zip(a, b))

def get_flag(known_guess: str, encrypted_quote_hex: str, encrypted_flag_hex: str):
    encrypted_quote = bytes.fromhex(encrypted_quote_hex)
    encrypted_flag = bytes.fromhex(encrypted_flag_hex)
    guess_bytes = known_guess.encode()
    keystream_guess = xor_bytes(guess_bytes, encrypted_quote[:len(guess_bytes)])
    decrypted_flag_guess = xor_bytes(encrypted_flag[:len(keystream_guess)], keystream_guess)
    return decrypted_flag_guess.decode(errors='ignore')

encrypted_quote_hex = "5667c3646a39e6bf6b1d098aadd003671854a7eab5ceaa3ad1ba21a9754cded388dc883114f7b1ee9aafb360c2d7ca48e8974903636a3f39b623784e344404bd8c29e178d08b033534848bf408e55f8292586f"
encrypted_flag_hex = "4f04e4700925b8a9141b1e87a3d2446c5f62f8b99eccf163d38272a96b7198e9d8e8883242e29dfeb2a1"

original_quote = "I do not believe in taking the right decision, I take a decision and make it right."
flag = get_flag(original_quote, encrypted_quote_hex, encrypted_flag_hex)
print(flag)

The challenge was an instance based challenge and you get the values encrypted_quote_hex and encrypted_flag_hex from the instance.

Running the script gives you the complete flag:

n0tabdu11ah@MNM:~/Downloads/crypto-count$ python3 exploit.py 
PCC{Cr1b_dragg1ng_1s_c00l_stvXfH9ShwvqIsA}

Flag: PCC{Cr1b_dragg1ng_1s_c00l_stvXfH9ShwvqIsA}

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THANKS

MUHAMMAD ABDULLAH