12 Oct 2020

# DamCTF: crypto/guess-secret

This challenge asks us to break a ‘super secure and efficient communication method’. It also gives us the server code, although this isn’t really necessary—the server itself tells us all we need to know.

The server explains in detail what it does. It accepts a string, and creates the plaintext by concatenating it to the flag. The plaintext is then compressed using the lossless deflate (zlib) algorithm. Lastly, the compressed data is encrypted using AES_CTR_128. The resulting ciphertext is returned to us.

There are two key insights here. The first is that the deflate algorithm eliminates duplicate strings. The second is that CTR mode reveals the exact length of the plaintext (or compressed data, in this case). This leaks how “compressible” the data is, since inputting AAAAAAAA should result in a shorter ciphertext than a uniformly random string of equivalent length.

The server also helpfully tells us the format of the flag, which is dam{[0-9a-f]+} with 32 hexadecimal chars. So all we have to do is start with dam{ and start guessing flag characters.

import ast
from pwn import *

io = remote('chals.damctf.xyz', 30308)
# initial text
io.recv()

chars = '0123456789abcdef'
flag = 'dam{'

for i in range(32):
min_l, min_c = 999, '.'
for c in chars:
# query text
io.recv()
# send partial flag guess
io.sendline(flag + c)
# compare ciphertext lengths
l = len(ast.literal_eval(io.recv().decode('ascii')))
if l < min_l:
min_l, min_c = l, c
flag += min_c

print(flag + '}')


It turns out that this works exactly as you would expect. With each correct guess, the length of the ciphertext remains constant at 65 bytes, while incorrect guesses are exactly one larger.

Running it gives us the flag within a few minutes:

\$ python sol.py
dam{9f64ee1d4a7d6d8fe9136c3e9a74fc76}