LINECTF 2021: babycrypto1 (Crypto)

nc 35.200.115.41 16001

babycrypto1.py

In this challenge we are provided with the ip and port of a network service and a python script which seemingly contains the code of the network service. Lets have a look at the code shall we?

flag = open("flag", "rb").read().strip()
COMMAND = [b'test',b'show']


The code reads in the flag from a file, and defines 2 commands: test and show.

if __name__ == '__main__':
server = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
server.bind(('0.0.0.0', 16001))
server.listen(1)

while True:

aes_key = get_random_bytes(AES.block_size)
token = b64encode(get_random_bytes(AES.block_size*10))[:AES.block_size*10]

process = multiprocessing.Process(target=run_server, args=(client, aes_key, token))
process.daemon = True
process.start()


As seen above, a key and token is randomly generated (per session) upon a client connecting to the server. The client is then serviced by the run_server() function.

class AESCipher:
def __init__(self, key):
self.key = key

def encrypt(self, data):
iv = get_random_bytes(AES.block_size)
self.cipher = AES.new(self.key, AES.MODE_CBC, iv)

def encrypt_iv(self, data, iv):
self.cipher = AES.new(self.key, AES.MODE_CBC, iv)

def decrypt(self, data):
raw = b64decode(data)
self.cipher = AES.new(self.key, AES.MODE_CBC, raw[:AES.block_size])

def run_server(client, aes_key, token):
client.send(b'test Command: ' + AESCipher(aes_key).encrypt(token+COMMAND[0]) + b'\n')
client.send(b'**Cipher oracle**\n')
client.send(b'IV...: ')
iv = b64decode(client.recv(1024).decode().strip())
client.send(b'Message...: ')
msg = b64decode(client.recv(1024).decode().strip())
client.send(b'Ciphertext:' + AESCipher(aes_key).encrypt_iv(msg,iv) + b'\n\n')
while(True):
tt = client.recv(1024).strip()
tt2 = AESCipher(aes_key).decrypt(tt)
client.send(tt2 + b'\n')
if tt2 == token+COMMAND[1]:
client.send(b'The flag is: ' + flag)
client.close()
break


In summary, the logic of run_server() is as follows:

1. Print an encrypted test command AESCipher(aes_key).encrypt(token + b'test')
2. Provide us with a “Cipher oracle”
• lets us supply an IV and message
• performs encryption AESCipher(aes_key).encrypt_iv(msg, iv) and returns the ciphertext
3. Continuously accepts (encrypted) commands from us, decrypts it, sends us the decrypted command, and sends us the flag if the decrypted command is token + b'show'

Additionally we note that more specifically (with reference to the AESCipher class) the commands are being encrypted with AES-128-CBC. We know that the block size is 16 bytes (128 bits) because AES.block_size yields 16.

At this point, the question we should be asking ourselves is:

How the heck do we generate an encrypted command (ciphertext) that decrypts to token + b'show' if we dont have access to the IV and key?

It’s possible, but requires some ingenuity on our part. The first step to enlightment is contingent on our understanding of how AES works in CBC mode.

Lets first try to understand how decryption works in CBC mode. We can generalize the above diagram as follows:

Where $$n$$ is the block number, $$C$$ is the Ciphertext, $$P$$ is the Plaintext, and $$D(x)$$ is the decryption function, and $$K$$ is the key,

$P_n = \begin{cases} D(C_{n},~K)~\oplus~\text{IV} & n = 1 \\ D(C_{n},~K)~\oplus~C_{n-1} & n > 1 \\ \end{cases}$

Lets approach this problem by zooming in on the part we are most interested in, the command.

token = b64encode(get_random_bytes(AES.block_size*10))[:AES.block_size*10]


As seen from the snippet of code above, the size of our token is going to be aligned to the block size. Therefore, we can assume that the last block contains only the command (plus padding) since encryption is performed on token + command. So lets focus solely on what actually matters.

With reference to the simplified diagram above, we can now reframe our problem as follows:

What should the ciphertext in the last block be such that it decrypts to ‘show’?

Or more formally,

$D(?,~K) \oplus~C_{n-1}~=~\text{show}$

How are we going to figure out what the ciphertext should be? If you still haven’t figured it out by now, I would encourage you to mentally (or physically?) flip the above diagram upside down and compare it to the diagram below.

Do you see the resemblance? No? Well let me help you out a little.

That’s right, all we have to do is to encrypt ‘show’ and set the IV to $$C_{n-1}$$ to figure out what the ciphertext should be! Conveniently, the challenge provides us with a “Cipher Oracle” which lets us encrypt an arbitrary message using an IV which we can also supply.

So lets recap our game plan:

1. Read in the test command and extract out the second last block i.e. $$C_{n-1}$$
2. Use the “Cipher Oracle” to encrypt ‘show’ with $$C_{n-1}$$ from step 1 as the IV. Read in the ciphertext.
3. Append the ciphertext from step 2 to the test command from step 1 (less the last block since we are replacing it).
4. Get flag. Profit!

We now craft the following script to implement the steps above.

from pwn import *
from base64 import b64decode, b64encode

HOST = '35.200.115.41'
PORT = 16001

TESTCMD_PROMPT = 'test Command: '
IV_PROMPT = 'IV...: '
MSG_PROMPT = 'Message...: '
CIPHERTEXT_PROMPT = 'Ciphertext:'
COMMAND_PROMPT = 'Enter your command: '
FLAG_MARKER = 'The flag is: '

COMMAND = [b'test', b'show']

def chunks(l, n):
return [l[i:i + n] for i in range(0, len(l), n)]

io = remote(HOST, PORT)

io.recvuntil(TESTCMD_PROMPT)

testcmd_encrypted = b64decode(io.recvline())
testcmd_blocks = chunks(testcmd_encrypted, 16)

iv = testcmd_blocks[-2]
pt = COMMAND[1]

io.sendlineafter(IV_PROMPT, b64encode(iv))
io.sendlineafter(MSG_PROMPT, b64encode(pt))
io.recvuntil(CIPHERTEXT_PROMPT)

poison_block = b64decode(io.recvline().rstrip())

poison_ciphertext = b''.join(testcmd_blocks[:-2] + [poison_block, ])

io.sendlineafter(COMMAND_PROMPT, b64encode(poison_ciphertext))

io.recvuntil(FLAG_MARKER)

log.success(io.recvuntil('}').decode())

io.close()


Flag: LINECTF{warming_up_crypto_YEAH}