{ ASIS CTF 2017 } - Crypto - Unbreakable - 200

ASIS CTF 2017 Quals Writeup: Unbreakable

Unbreakable

193

We think this challenge is unbreakable, even for you? Try it now!

unbreakable

flag.enc.orig

For this challenge we are given an executable and an encoded flag. When we run an executable it gives an error:

$ ./unbreakable
error, input not found 

Let’s run it with strace:

$ strace ./unbreakable
...
open("key", O_RDONLY)                   = -1 ENOENT (No such file or directory)
fstat(1, {st_mode=S_IFCHR|0620, st_rdev=makedev(136, 0), ...}) = 0
write(1, "error, input not found\n", 23error, input not found
) = 23
exit_group(1)                           = ?
+++ exited with 1 +++

So it needs a key file. When we supply one it encodes it and puts the output into flag.enc - so the key is not an encryption key, it’s the plaintext that gets encoded. Our job is to reverse the process and decode flag.enc we are supplied.

Reversing the executable in Snowman gives some scary looking encryption code. But before we jump into reversing let’s experiment with encoding a little bit. Supplying key composed of different character combinations produces some interesting results:

key	flag.enc
A	4D 16
AA	4D 16 16 4D
AAA	4D 16 16 4D 4D 16
B	2E 4D
BB	2E 4D 4D 2E
BBB	2E 4D 4D 2E 2E 4D
AABB	4D 16 16 4D 2E 4D 4D 2E

Things are not that scary any more. We can see that:

• Every character is encoded as a 2-byte sequence
• Every character is encoded independently of others
• The 2-byte sequence is reversed depending on whether the position is odd or even

So what we can do now is simply build a key file with a full range of bytes, encode it, and use the result as the translation table to reverse the original flag.enc:

from subprocess import call

tab = b""

# build a translation table for full range of bytes, 
# with each byte used twice (to account for odd and even positions)
for x in range(0x100):
 tab += chr(x)*2
 
open("key","wb").write(tab)

call(["./unbreakable"])

conv = open("flag.enc","rb").read()

# build a dictionary out of encoded values
dict = {}
for x in range(0x100):
 dict[conv[x*4:x*4+2]] = x
 dict[conv[x*4+2:x*4+4]] = x
 
puzzle = open("flag.enc.orig","rb").read()

# use the dictionary to decode the flag
out = b""
for x in range(len(puzzle)/2):
 out += chr(dict[puzzle[x*2:x*2+2]])
 
open("flag.dec","wb").write(out)

Once the flag file is recovered it turns out that it is a PNG image. The image contains the flag - ASIS{Ju5t_C0py_And_paSte_on_Sc0rebo4rd!!}.