• Category: Crypto
• Points: 100
• Solves: 13
• Description:

Here’s some kind of semi-secure shell, however we don’t have the key. Can you obtain the secret nevertheless?

turbo.py

## Write-up

This was a fun challenge. Again the service is written in python. This services seems to involve two secrets we don’t know. We can send a password and a command, if the password is correct it will be executed. The flag and the output are then encrypted with AES-CBC and an unkown key.


#!/usr/bin/env python3
from Crypto.Cipher import AES
import binascii, gzip, subprocess

try:
pw, cmd = input().split(' ', 1)
except ValueError:
exit(1)

assert len(goodpw) >= 32

assert len(key) in AES.key_size
assert len(iv) == AES.block_size
aes = AES.new(key, AES.MODE_CBC, iv)

r = b''
r += 'Today\'s special: {}.\n'.format(flag).encode()
if pw == goodpw:
r += subprocess.check_output(cmd, shell = True)
else:



The password is at least 32 bytes long, so too long to guess. But if we take a closer look at the code, we see that the data is compressed before it is encrypted. Furthermore we control parts of the output, because the password is added to the encrypted data. We can use the service as an oracle, by abusing the compression. Remember CRIME and BREACH?

So the way this attack works is that we will guess the flag byte by byte and supply our guess of the flag as the password. The flag is already in the data that is compressed. If we guessed the flag correctly the compression will remove this redundant data and we will see no increase in length. If we guessed wrong we will see longer data. The tricky part here is that we won’t see a change in length if the bytes we are guessing are not at the “boundary” of a cipher block. But this is relatively easy to achieve by gradually filling the supplied data with random characters, that probably won’t be compressed, until we see a length change.

One thing that puzzeled me while playing around is that there is a bug in the padding code. If you have zero bytes of padding you will get an additional ciphertext block, filled with 0x01 bytes. Fortunately this doesn’t matter much for the attack, we just have to make sure that there is always 2 bytes padding in the last block.

In this case we know that the flags start with hxp{. So we can start by checking whether this increases the length of ciphertext. Of course we have to account for the bytes that the backreference that is added by gzip needs. As expected it didn’t increase the length, so we are on a good track. Now we have to guess the following bytes, of which we can assume they are printable. So we start by supplying hxp{a, hxp{b etc. until we receive a ciphertext that has the same length as only hxp{. Then we have guessed correct and we can continue with the next byte: hxp{1a, hxp{1b etc.

I wrote the following script, which will guess the flag byte by byte using the oracle. Unfortunately I didn’t manage to complete the script in time and it finished guessing only a few minutes after the CTF ended. The service was still up so I got the flag hxp{1_r34LLy_L1k3_0r4cL3s__n0T_7h3_c0mp4nY}


import sys
from pwn import remote, log, context, process
import string
import zlib

#context.log_level = 'debug'
BLKSIZE = 16
#doremote = True
doremote = False

def get_length(inp):
rem = None
inp = inp + " a"
if doremote:
rem.sendline(inp)
rem.shutdown("write")
ret = rem.recvline().strip()
else:
rem = process("./turbo.py")
out, err = rem.communicate(inp + "\n")
ret = out.strip()
log.info("\n".join(ret.split("\n")[:-1]))
ret = ret.split("\n")[-1].strip()
if err:
log.info(err)
rem.close()
del rem
if not ret:
return len(ret) / 2 / BLKSIZE

printables = string.digits + "_" + string.ascii_letters
printables = printables[::-1]

# We can find out how long the flag is by gradually increasing the length of
# the password we supply. If we get another block of encrypted data, we know we
# filled up the padding. Therefore we know how much padding there is given an
# empty password. We can then calculate the length of the flag by subtracting
# the length of the padding and the other known plaintext. The only thing
# that's left must be the length of the flag. At least that's the idea. For
# some reason (I guess gzip) this isn't working so well...
flaglen = 0
minlen = get_length("")
datalen = len(zlib.compress('Today\'s special: .\nWrong password: .\n', 9))
pl = minlen
for i in range(1, BLKSIZE):
log.info("trying {} chars".format(i))
l = get_length(printables[:i])
log.info("got length {}".format(l))
if l != pl:
break
flaglen = ((minlen - 1) * 16) - datalen - padlen
flaglen -= 2  # to account for padding bug
log.info("flag must be {} bytes long".format(flaglen))

#sys.exit(0)

#### some testing

# the password will be the guessed flag.
# "hxp{" + guess
log.info("Trying hxp{ flag prefix")
log.info(s)
pl = get_length(s)
assert pl == minlen
s = printables[:(padlen - 4)] + ""
x = s + "hxp{"
log.info(x)
l = get_length(x)
assert l == minlen

if not doremote:
# these strings will be compressed and replaced by a ref to lz dictionary
for teststr in ("hxp{1", "hxp{1_r3", testflag):
x = s + teststr
log.info(x)
l = get_length(x)
assert l == minlen, "teststr not compressed: " + repr(teststr)

x = s + "hxp{X"
log.info(x)
l = get_length(x)
assert l > minlen

#### The actual flag bruteforcing,
# byte by byte using the compression oracle

# block boundary, # bytes that can be added before next block
# the padding length from above - some guesssed value (because of gzip? I don't
# know)
# the guess results
previous = []
pl = 0
for i in range(len(previous), flaglen):
log.info("guessing character number {}".format(i))
test = "hxp{" + "".join(previous)
test = printables[:BB] + test
log.info("trying prefix: " + test)
pl = get_length(test)
log.info("got length = {}".format(pl))
foundit = False
for c in printables + "}":
test = "hxp{" + "".join(previous)
test = printables[:BB] + test + c
log.info("trying prefix: " + test)
l = get_length(test)
log.info("l = {}, pl = {}".format(l, pl))
if l == pl:
log.info("deduced next char: " + c)
previous.append(c)
foundit = True
break
else:
log.info("guess was wrong")
if foundit:
log.info("Current flag prefix guess is \'hxp{}{}\'"
.format("{", "".join(previous)))
else:
log.error("Couldn't guess char {}".format(i))
if previous[-1] == "}":
break

flag = "hxp{" + "".join(previous)
if flag[-1] != "}":
flag += "}"
log.info("final flag should be " + flag)

with open("./real_flag.txt", "w") as f:
f.write(flag)