Iced TEA
Catégorie: Crypto Difficulté: easy Flag: HTB{th1s_1s_th3_t1ny_3ncryp710n_4lg0r1thm_____y0u_m1ght_h4v3_4lr34dy_s7umbl3d_up0n_1t_1f_y0u_d0_r3v3rs1ng}
Challenge
Description
Locked within a cabin crafted entirely from ice, you're enveloped in a chilling silence. Your eyes land upon an old notebook, its pages adorned with thousands of cryptic mathematical symbols. Tasked with deciphering these enigmatic glyphs to secure your escape, you set to work, your fingers tracing each intricate curve and line with determination. As you delve deeper into the mysterious symbols, you notice that patterns appear in several pages and a glimmer of hope begins to emerge. Time is flying and the temperature is dropping, will you make it before you become one with the cabin?
Analyse du code
Voici le code :
import os
from secret import FLAG
from Crypto.Util.Padding import pad
from Crypto.Util.number import bytes_to_long as b2l, long_to_bytes as l2b
from enum import Enum
class Mode(Enum):
ECB = 0x01
CBC = 0x02
class Cipher:
def __init__(self, key, iv=None):
self.BLOCK_SIZE = 64
self.KEY = [b2l(key[i:i+self.BLOCK_SIZE//16]) for i in range(0, len(key), self.BLOCK_SIZE//16)]
self.DELTA = 0x9e3779b9
self.IV = iv
if self.IV:
self.mode = Mode.CBC
else:
self.mode = Mode.ECB
def _xor(self, a, b):
return b''.join(bytes([_a ^ _b]) for _a, _b in zip(a, b))
def encrypt(self, msg):
msg = pad(msg, self.BLOCK_SIZE//8)
blocks = [msg[i:i+self.BLOCK_SIZE//8] for i in range(0, len(msg), self.BLOCK_SIZE//8)]
ct = b''
if self.mode == Mode.ECB:
for pt in blocks:
ct += self.encrypt_block(pt)
elif self.mode == Mode.CBC:
X = self.IV
for pt in blocks:
enc_block = self.encrypt_block(self._xor(X, pt))
ct += enc_block
X = enc_block
return ct
def encrypt_block(self, msg):
m0 = b2l(msg[:4])
m1 = b2l(msg[4:])
K = self.KEY
msk = (1 << (self.BLOCK_SIZE//2)) - 1
s = 0
for i in range(32):
s += self.DELTA
m0 += ((m1 << 4) + K[0]) ^ (m1 + s) ^ ((m1 >> 5) + K[1])
m0 &= msk
m1 += ((m0 << 4) + K[2]) ^ (m0 + s) ^ ((m0 >> 5) + K[3])
m1 &= msk
m = ((m0 << (self.BLOCK_SIZE//2)) + m1) & ((1 << self.BLOCK_SIZE) - 1) # m = m0 || m1
return l2b(m)
if __name__ == '__main__':
KEY = os.urandom(16)
cipher = Cipher(KEY)
ct = cipher.encrypt(FLAG)
with open('output.txt', 'w') as f:
f.write(f'Key : {KEY.hex()}\nCiphertext : {ct.hex()}')Ici on utilise le Cipher en mode ECB, voici ce qu’il fait :
Ajouter un padding au message pour que sa taille soit un multiple de 8 bytes
Découper en blocs de 8 bytes le message
Chiffrer ces blocs individuellement
Retourner le résultat du chiffrement
Concernant le chiffrement des blocs, il les sépare en 2 sous blocs de 4 bytes et appliquent des opérations simples et réversibles dessus, puis les réassemble
Il suffit donc d’inverser l’ordre des opérations :
def encrypt_block(self, msg):
m0 = b2l(msg[:4])
m1 = b2l(msg[4:])
K = self.KEY
msk = (1 << (self.BLOCK_SIZE//2)) - 1
s = 0
for i in range(32):
s += self.DELTA
m0 += ((m1 << 4) + K[0]) ^ (m1 + s) ^ ((m1 >> 5) + K[1])
m0 &= msk
m1 += ((m0 << 4) + K[2]) ^ (m0 + s) ^ ((m0 >> 5) + K[3])
m1 &= msk
m = ((m0 << (self.BLOCK_SIZE//2)) + m1) & ((1 << self.BLOCK_SIZE) - 1) # m = m0 || m1
return l2b(m)
def decrypt_block(self, ct: bytes) -> bytes:
m0 = b2l(ct[:4])
m1 = b2l(ct[4:])
K = self.KEY
msk = (1 << (self.BLOCK_SIZE // 2)) - 1
# Comme dans le chiffrement on ajoute DELTA à s à chaque tour de boucle
# Ici s commence à sa valeur max et on soustrait DELTA
s = 32 * self.DELTA
for i in range(32):
# Littéralment juste inverser l'ordre des opérations
# Il faut cependant bien laisser le ET logique avec le mask après, ça ne fait pas parti des opérations, c'est pour garder le résultat sur 4 4bytes
m1 -= ((m0 << 4) + K[2]) ^ (m0 + s) ^ ((m0 >> 5) + K[3])
m1 &= msk
m0 -= ((m1 << 4) + K[0]) ^ (m1 + s) ^ ((m1 >> 5) + K[1])
m0 &= msk
s -= self.DELTA
# on reassemble les blocs
ct = ((m0 << (self.BLOCK_SIZE // 2)) + m1) & ((1 << self.BLOCK_SIZE) - 1)
return l2b(ct)Script de résolution
import os
from Crypto.Util.Padding import pad, unpad
from Crypto.Util.number import bytes_to_long as b2l, long_to_bytes as l2b
from enum import Enum
class Mode(Enum):
ECB = 0x01
CBC = 0x02
class Cipher:
def __init__(self, key, iv=None):
self.BLOCK_SIZE = 64
self.KEY = [b2l(key[i:i + self.BLOCK_SIZE // 16]) for i in range(0, len(key), self.BLOCK_SIZE // 16)]
self.DELTA = 0x9e3779b9
self.IV = iv
if self.IV:
self.mode = Mode.CBC
else:
self.mode = Mode.ECB
def _xor(self, a, b):
return b''.join(bytes([_a ^ _b]) for _a, _b in zip(a, b))
def decrypt(self, ct: bytes) -> bytes:
blocks = [ct[i:i + self.BLOCK_SIZE // 8] for i in range(0, len(ct), self.BLOCK_SIZE // 8)]
msg = b''
if self.mode == Mode.ECB:
for pt in blocks:
msg += self.decrypt_block(pt)
elif self.mode == Mode.CBC:
X = self.IV
for pt in blocks:
enc_block = self.decrypt_block(self._xor(X, pt))
ct += enc_block
X = enc_block
return unpad(msg, self.BLOCK_SIZE//8)
def decrypt_block(self, ct: bytes) -> bytes:
m0 = b2l(ct[:4])
m1 = b2l(ct[4:])
K = self.KEY
msk = (1 << (self.BLOCK_SIZE // 2)) - 1
s = 32 * self.DELTA
for i in range(32):
m1 -= ((m0 << 4) + K[2]) ^ (m0 + s) ^ ((m0 >> 5) + K[3])
m1 &= msk
m0 -= ((m1 << 4) + K[0]) ^ (m1 + s) ^ ((m1 >> 5) + K[1])
m0 &= msk
s -= self.DELTA
ct = ((m0 << (self.BLOCK_SIZE // 2)) + m1) & ((1 << self.BLOCK_SIZE) - 1)
return l2b(ct)
def solve():
KEY = bytes.fromhex("850c1413787c389e0b34437a6828a1b2")
CIPHERTEXT = bytes.fromhex("b36c62d96d9daaa90634242e1e6c76556d020de35f7a3b248ed71351cc3f3da97d4d8fd0ebc5c06a655eb57f2b250dcb2b39c8b2000297f635ce4a44110ec66596c50624d6ab582b2fd92228a21ad9eece4729e589aba644393f57736a0b870308ff00d778214f238056b8cf5721a843")
cipher = Cipher(KEY)
flag = cipher.decrypt(CIPHERTEXT).decode()
print(f"Flag: {flag}")
if __name__ == '__main__':
solve()Dernière mise à jour
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