Cyber Apocalypse
2024
IcedTEA
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?
Crypto
300 points
Walkthrough
We start by downloading the source files.
We are give a
source.py
file and an output.txt
file. It looks like the output is generated using the source code, so we start by analyzing the source 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()}')
This is the tiny encryption algortihm, which is a simple block cipher.
We can modify the script to decrypt the output:
import os
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)
def decrypt_block(self, msg):
m = b2l(msg)
m0 = m >> (self.BLOCK_SIZE//2)
m1 = m & ((1 << (self.BLOCK_SIZE//2)) - 1)
K = self.KEY
msk = (1 << (self.BLOCK_SIZE//2)) - 1
s = self.DELTA << 5
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
return l2b((m0 << 32) + m1)
def decrypt(self, ct):
blocks = [ct[i:i+self.BLOCK_SIZE//8] for i in range(0, len(ct), self.BLOCK_SIZE//8)]
pt = b''
if self.mode == Mode.ECB:
for block in blocks:
pt += self.decrypt_block(block)
elif self.mode == Mode.CBC:
X = self.IV
for block in blocks:
pt += self._xor(X, self.decrypt_block(block))
X = block
return pt
if __name__ == '__main__':
FLAG = b'HTB{TEST_FLAG}'
KEY = bytes.fromhex('850c1413787c389e0b34437a6828a1b2')
cipher = Cipher(KEY)
ct = cipher.encrypt(FLAG)
# print(ct.hex())
# output = ct.hex()
# output = bytes.fromhex(output)
# print(cipher.decrypt(output))
# real flag
output = bytes.fromhex('b36c62d96d9daaa90634242e1e6c76556d020de35f7a3b248ed71351cc3f3da97d4d8fd0ebc5c06a655eb57f2b250dcb2b39c8b2000297f635ce4a44110ec66596c50624d6ab582b2fd92228a21ad9eece4729e589aba644393f57736a0b870308ff00d778214f238056b8cf5721a843')
print(cipher.decrypt(output).decode())
Running this script gives us the flag.
python3 reverse.py
HTB{th1s_1s_th3_t1ny_3ncryp710n_4lg0r1thm_____y0u_m1ght_h4v3_4lr34dy_s7umbl3d_up0n_1t_1f_y0u_d0_r3v3rs1ng}
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