HTB Hack The Boo Practice: SPG

Challenge Information

Attribute	Details
Event	Hack The Boo Practice
Category	Cryptography
Challenge	SPG
Difficulty	Very Easy

After successfully joining the Applied Cryptography Academy of Ghosts, there is an eclass login system. The challenge provides a password generator script that needs to be reverse-engineered. By analyzing the password generation algorithm and extracting the master key, students can decrypt the encrypted flag using AES encryption.

Analysis

The password generator uses a custom algorithm that converts bits from a master key into characters from an alphabet. The algorithm works as follows:

def generate_password():
    master_key = int.from_bytes(MASTER_KEY, 'little')
    password = ''

    while master_key:
        bit = master_key & 1
        if bit:
            password += random.choice(ALPHABET[:len(ALPHABET)//2])
        else:
            password += random.choice(ALPHABET[len(ALPHABET)//2:])
        master_key >>= 1

    return password

Each bit of the master key determines which half of the alphabet to sample from. The alphabet is split in half:

Bits = 1: First half (uppercase/lowercase letters)
Bits = 0: Second half (digits and special characters)

The encrypted flag is encrypted using:

encryption_key = sha256(MASTER_KEY).digest()
cipher = AES.new(encryption_key, AES.MODE_ECB)

Challenge Output:

Your Password : gBv#3%DXMV*7oCN2M71Zfe0QY^dS3ji7DgHxx2bNRCSoRPlVRRX*bwLO5eM&0AIOa&#$@u
Encrypted Flag : tnP+MdNjHF1aMJVV/ciAYqQutsU8LyxVkJtVEf0J0T5j8Eu68AxcsKwd0NjY9CE+Be9e9FwSVF2xbK1GP53WSAaJuQaX/NC02D+v7S/yizQ=

Solution

Step 1: Reconstruct the MASTER_KEY from the password

The password can be reversed to reconstruct the bits of the master key:

import string

ALPHABET = string.ascii_letters + string.digits + '~!@#$%^&*'

def reverse_password_generation(password):
    reconstructed_master_key = 0

    for char in password:
        half_index = len(ALPHABET) // 2
        if char in ALPHABET[:half_index]:
            # First half = bit 1
            reconstructed_master_key = (reconstructed_master_key << 1) + 1
        else:
            # Second half = bit 0
            reconstructed_master_key = (reconstructed_master_key << 1)

    return reconstructed_master_key

password_from_output = "gBv#3%DXMV*7oCN2M71Zfe0QY^dS3ji7DgHxx2bNRCSoRPlVRRX*bwLO5eM&0AIOa&#$@u"
master_key = reverse_password_generation(password_from_output)

Step 2: Derive the encryption key and decrypt the flag

Once the master key is reconstructed:

from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
from base64 import b64decode
from hashlib import sha256

encrypted_flag = "tnP+MdNjHF1aMJVV/ciAYqQutsU8LyxVkJtVEf0J0T5j8Eu68AxcsKwd0NjY9CE+Be9e9FwSVF2xbK1GP53WSAaJuQaX/NC02D+v7S/yizQ="

# Convert reconstructed master key back to bytes
master_key_bytes = reconstructed_master_key.to_bytes(
    (reconstructed_master_key.bit_length() + 7) // 8, 'big'
)

# Derive encryption key
encryption_key = sha256(master_key_bytes).digest()

# Decrypt
encoded_flag = b64decode(encrypted_flag)
cipher = AES.new(encryption_key, AES.MODE_ECB)
decrypted_flag = unpad(cipher.decrypt(encoded_flag), AES.block_size)

print(f'Decrypted Flag: {decrypted_flag.decode()}')

Key Takeaways

Custom password generation algorithms can be reverse-engineered if the logic is known
Bit manipulation patterns leak information about the underlying key material
Never use random choice with a predictable key schedule
AES with ECB mode is deterministic - identical plaintexts produce identical ciphertexts
The security of AES depends on the unpredictability of the key, not the algorithm

Challenge Information

Summary

Analysis

Solution

Key Takeaways