# Writeup Yunnyit from ASIS 2018 CTF

by: dorianr

First a disclaimer, we did not actually solve this challenge during the competition, but the servers were left running…

A server is provided: nc 37.139.22.174 22555

It greets us with the following text:

|-------------------------------------|
| Welcome to the Yunnyit crypto task! |
|-------------------------------------|
| Options:                            |
| [M]ixed encryption function of FLAG |
| [D]ecrypting cipher                 |
| [E]ncryption & decryption function  |
| [F]LAG encrypting...                |
| [Q]uit                              |
|-------------------------------------|
Submit a printable string X, such that sha256(X)[-6:] = 92730d


Option M outputs:

def EFoF(self, suffix, prefix):
assert len(FLAG) == 32
assert len(self.key) == 16
return AESCipher(self.key).encrypt(suffix + FLAG + prefix)


and E:

def encrypt(self, raw):
digest = hmac.new(self.key, iv + raw, sha1).digest()
cipher = AES.new(self.key, AES.MODE_CBC, iv)
return b64encode(iv + cipher.encrypt(pad(raw + digest)))

def decrypt(self, enc):
enc = b64decode(enc)
iv = enc[:BLOCK_SIZE]
cipher = AES.new(self.key, AES.MODE_CBC, iv)
raw, digest = plain[:-20], plain[-20:]
if hmac.new(self.key, iv + raw, sha1).digest() == digest:
return raw
else:
raise Exception


A quick websearch for AESCipher leads us to some code. Hence, we can presume that

pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS)
unpad = lambda s : s[0:-ord(s[-1])]


Using option F, we can enter a nonempty prefix and suffix and the server returns iv + enc(iv, k, suffix + flag + prefix + hmac + padding) [sic].

With D, we can let the server decrypt some ciphertext. The server responds with What?????, Great job :D or Catch FLAG if you can :P. We will use these outputs to construct an oracle. What????? is returned if the ciphertext is either not valid base64 or its length is not a multiple of 16. This is of no use to us.

One might think that Catch FLAG if you can :P is returned if the HMAC validation fails, but that is not the case here. In fact, it seems like the given decrypt function is not used at all. If it were used, we could exploit the naive unpad function for a POODLE-like attack (I only learned about POODLE because it is mentioned in the flag…). Decryption rather seems to be done with a function like this:

def decrypt2(self, enc):
enc = b64decode(enc)
iv = enc[:BLOCK_SIZE]
cipher = AES.new(self.key, AES.MODE_CBC, iv)
plain = cipher.decrypt(enc[BLOCK_SIZE:])
return "Great job :D" if FLAG in plain else "Catch FLAG if you can :P"


This function is also vulnerable. But first, let us write the code to interact with the server.

class YunnyIt:
def __init__(self):
self.p = remote("37.139.22.174", 22555)

def proof_of_work(self):
def check(target, i):
s = str(i)
if hashlib.sha256(s).hexdigest()[-6:] == target:
return s

line = self.p.readline_contains('Submit a printable string X, such that sha256')
target = re.search(' = ([0-9a-f]{6})', line).group(1)
s = search(target, check, 4)
self.p.sendline(s)

def encrypt(self, prefix, suffix):
self.p.sendline('F')
self.p.sendline(prefix)
self.p.sendline(suffix)
line = self.p.readline_startswith('Mixed encrypted FLAG =')
return base64.b64decode(re.search(' = (\S+)', line).group(1))

def decrypt(self, c):
c = base64.b64encode(c)
self.p.sendline('D')

y = YunnyIt()
y.proof_of_work()


The search function is from our little library bruteforce which helps parallelize such proof of work tasks:

import multiprocessing

def _search(check, target, id, proc_count, queue):
res = None
i = id
while res is None:
res = check(target, i)
i += proc_count
queue.put(res)

def search(target, check, proc=multiprocessing.cpu_count()):
q = multiprocessing.Queue()
procs = [multiprocessing.Process(target=_search, args=(check, target, i, proc, q)) for i in xrange(proc)]
for p in procs:
p.start()
res = q.get()
q.close()
for p in procs:
p.terminate()
return res


Now, let iv, c0, c1, c2 be the first four blocks of y.encrypt('a', 'a'). Notice that the plaintext for c2 only contains the last byte of the flag, which we know is }. For some ciphertext c let p = dec(k, c), i.e., its “real” plaintext (like in ECB mode). Then y.decrypt(iv + c0 + c1 + c) == True iff p[0] ^ c1[0] == '}'. Let us obtain for all bytes b a ciphertext iv, c0, c1, c2 such that c1[0] == i.

def find_ciphertexts(from_file=True):
"""for each byte b, find a valid ciphertext c with c[32] = b"""
if from_file:

cs = {}
while len(cs) < 256:
# encrypt plaintext: a | flag[0:15] || flag[15:31] || flag[31] | a | digest[0:14] || digest[14:20] | padding
c = y.encrypt('a', 'a')
b = ord(c[32])
cs[b] = c
print len(cs)

pickle.dump(cs, open('cs.p', 'wb'))
return cs

cs = find_ciphertexts()
last_flag_byte = ord('}')


Using our observations from the previous paragraph, we can find the first byte of the real plaintext of any ciphertext block:

def first_byte_of_real_plaintext(c):
"""returns first byte of real plaintext, i.e. dec(k, c)[0]"""
for i in xrange(256):
c2 = cs[i][0:48] + c
if y.decrypt(c2):
return i ^ last_flag_byte  # c[0] ^ i == last_flag_byte


Finally, we can get the flag:

def get_flag_byte(i):
"""returns flag[i]"""
c = y.encrypt('a', 'a' * (16 - i % 16))  # pad s.t. a plaintext block starts with flag[i]
block_index = 2 if i < 16 else 3
block = c[block_index * 16:(block_index + 1) * 16]
b = first_byte_of_real_plaintext(block)
return b ^ ord(c[(block_index - 1) * 16])  # xor with previous block

for i in xrange(32):
sys.stdout.write(chr(get_flag_byte(i)))
# ASIS{M1T!g473_TH3_p0od|E_nl;2\$E}