hack.lu CTF 2025 - Crypto: MÄNUAL by hubertf

hack.lu CTF 2025 - Crypto: MÄNUAL
Solution by hubertf & Claude.ai

(venv-hack.lu) neuland% cat 1gemini.py
import socket
import struct
import sys
from sympy import Matrix, GF

# --- Challenge Details ---
HOST = "manual.flu.xxx"
PORT = 1024
# --- Crypto Parameters ---
PIECE_COUNT = 32
PIECE_SIZE = 256

buffer = b''
SOCKET_TIMEOUT = 5

def recvline(s: socket.socket) -> bytes:
    """Liest Daten zeilenweise vom Socket, speichert überschüssige Daten im Puffer."""
    global buffer

    while b'\n' not in buffer:
        try:
            data = s.recv(4096)
        except socket.timeout:
            raise TimeoutError("Socket timeout during recvline.")

        if not data:
            raise EOFError("Connection closed unexpectedly.")
        buffer += data

    line, buffer_rest = buffer.split(b'\n', 1)
    buffer = buffer_rest
    return line.strip()

def connect_and_receive_initial_message():
    """Stellt die Verbindung her und liest die Willkommensnachricht."""
    try:
        s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
        s.settimeout(SOCKET_TIMEOUT)
        s.connect((HOST, PORT))
        print(f"[+] Connected to {HOST}:{PORT}")

        for _ in range(6): # Lese 6 Zeilen des Banners
            recvline(s)

        return s
    except Exception as e:
        print(f"[-] ERROR: Could not connect or read initial data: {e}")
        sys.exit(1)

def query_oracle(s: socket.socket, query_count: int, plaintext_bytes: bytes) -> bytes:
    """Sendet einen Hex-kodierten Klartext und gibt den Chiffretext als Bytes zurück."""
    try:
        # 1. Lese die Prompt-Zeile
        prompt_line = recvline(s)
        if not prompt_line.startswith(b'('):
            # Der Prompt kann in zwei Teile geteilt sein, versuche erneut zu lesen
            prompt_line += recvline(s)
            if not prompt_line.startswith(b'('):
                 raise ValueError(f"Failed to find expected prompt on query {query_count}: {prompt_line.decode(errors='ignore')}")

        # Sende den Hex-kodierten Klartext
        s.sendall(f"{plaintext_bytes.hex()}\n".encode())

        # 2. Lese die Ausgabe des Interns: "Here is what the intern put together:"
        recvline(s)
        # 3. Lese die nächste Zeile: der Chiffretext
        ciphertext_hex = recvline(s).decode()

        if not all(c in '0123456789abcdef' for c in ciphertext_hex.lower()):
            raise ValueError(f"Received non-hex string: {ciphertext_hex}")

        return bytes.fromhex(ciphertext_hex)
    except Exception as e:
        print(f"[-] ERROR during query {query_count}: {e}")
        s.close()
        sys.exit(1)

def solve_ctf():
    print("=" * 50)
    print("         CRYPTO CTF EXPLOIT (FLU.XXX)")
    print("=" * 50)

    s = connect_and_receive_initial_message()

    # --- 1. Bestimme K_total (Affiner Schlüssel) ---
    print("\n[*] 1. Determining K_total (sending all zeros)...")
    zero_plaintext = b'\x00' * PIECE_COUNT
    k_total_bytes = query_oracle(s, 0, zero_plaintext)
    k_total_int = int.from_bytes(k_total_bytes, 'big')
    print(f"[+] K_total (hex): {k_total_bytes.hex()}")

    # --- 2. Erstelle die Lineare Transformationsmatrix M ---
    print("\n[*] 2. Building the 256x256 linear transformation matrix M...")
    matrix_M_cols = []

    for i in range(PIECE_SIZE):
        p_i_int = 1 << i
        p_i_bytes = p_i_int.to_bytes(PIECE_COUNT, 'big')

        c_i_bytes = query_oracle(s, i + 1, p_i_bytes)
        c_i_int = int.from_bytes(c_i_bytes, 'big')

        m_i_int = c_i_int ^ k_total_int

        # Bits von 0 (kleinste) bis 255 (größte)
        column_vector = [(m_i_int >> j) & 1 for j in range(PIECE_SIZE)]
        matrix_M_cols.append(column_vector)

        if (i + 1) % 50 == 0:
            print(f"[*] {i+1}/{PIECE_SIZE} basis vectors queried...")

    print(f"[+] All {PIECE_SIZE} basis vectors queried successfully.")

    M_rows = [list(x) for x in zip(*matrix_M_cols)]

    # --- FEHLERBEHEBUNG HIER: Übergabe als Liste von Listen ---
    M = Matrix(M_rows, domain=GF(2)) # Removed PIECE_SIZE, PIECE_SIZE
    print("[+] Matrix M built successfully. Starting inversion...")

    # --- 3. Entschlüsselung der Flagge ---

    # Lese den letzten Prompt (256/300)
    recvline(s)
    s.sendall(b"finish\n")

    # Lese die Abschlussnachrichten
    recvline(s)
    c_flag_hex = recvline(s).decode()
    recvline(s)
    s.close()

    c_flag_int = int(c_flag_hex, 16)
    print(f"\n[+] Flag Ciphertext C_flag (hex): {c_flag_hex}")

    c_prime_int = c_flag_int ^ k_total_int
    c_prime_vector_list = [(c_prime_int >> j) & 1 for j in range(PIECE_SIZE)]
    C_prime_vector = Matrix(c_prime_vector_list, domain=GF(2))

    # Löse das lineare Gleichungssystem
    try:
        M_inv = M.inv()
        print("[+] Matrix M inverted.")
    except ValueError:
        print("[-] The matrix M is singular (not invertible).")
        return

    P_flag_vector = M_inv * C_prime_vector

    p_flag_int = 0
    for i in range(PIECE_SIZE):
        p_flag_int |= int(P_flag_vector[i]) << i

    p_flag_bytes = p_flag_int.to_bytes(PIECE_COUNT, 'big')

    print("-" * 50)
    print("[!!!] Decrypted Flag:")
    try:
        cleaned_bytes = p_flag_bytes.lstrip(b'\x00').rstrip(b'\x00')
        print(cleaned_bytes.decode('ascii'))
    except UnicodeDecodeError:
        print(f"Raw Hex: {p_flag_bytes.hex()}")
    print("-" * 50)

if __name__ == "__main__":
    solve_ctf()


Run it:

(venv-hack.lu) neuland% python 1gemini.py DEBUG=1
==================================================
         CRYPTO CTF EXPLOIT (FLU.XXX)
==================================================
[+] Connected to manual.flu.xxx:1024

[*] 1. Determining K_total (sending all zeros)...
[+] K_total (hex): 920ce6c15e68443b395b312bb6b00c05b3d8ab3ec35c5faec4b3ae1dfe77e045

[*] 2. Building the 256x256 linear transformation matrix M...
[*] 50/256 basis vectors queried...
[*] 100/256 basis vectors queried...
[*] 150/256 basis vectors queried...
[*] 200/256 basis vectors queried...
[*] 250/256 basis vectors queried...
[+] All 256 basis vectors queried successfully.
[+] Matrix M built successfully. Starting inversion...

[+] Flag Ciphertext C_flag (hex): 23c97b2c92a121fae9da8cc74f31d0b4be0876b33fc1ceca7c7e7fe017fa55dc
[+] Matrix M inverted.
--------------------------------------------------
[!!!] Decrypted Flag:
flag{crypt0_kn0wl3dg3_g4in3d_:3}
--------------------------------------------------
(venv-hack.lu) neuland%