[DCTF 2017] hack-tac-toe Writeup
Description:
Pair up! Let’s play a game of tic tac toe :)
https://hacktactoe.dctf-f1nals-2017.def.camp/
Author:
Anatol, Lucian Nitescu
Stats:
360 points / 10 solvers
Solution:
The challenge started with this simple website:
If we change the name of Player 1 to “test”, we get:
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Encrypted_Game_Session=ITSSn%2FICisO1i6CBSOBGkzM0kJWzU%2FS02%2B%2B92B6zVJt3N5aRvRTSxbbP9tlP4VKdcziQlroU0pa7nqPZS%2BMCyXBol8PoENSSt5rwjx7lB5gkOZfD6EDRlb%2BZ8JALvBeSdDSFy%2Bgc1dW02vzVE6U0xCF3wtepQJLM%2FJPz1wvpBcUlPIWe%2BkuIza35qoxQ4haIciyc1rNEitX8yqmfTKVfwXo%2BnNazR4rV%2FMqpn0y1RpMpNJee9A%3D%3D
If we change the name of Player 2 to “test2”, we get:
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Encrypted_Game_Session=ITSSn%2FICisO1i6CBSOBGkzM0kJWzU%2FS02%2B%2B92B6zVJt3N5aRvRTSxbbP9tlP4VKdcziQlroU0pa7nqPZS%2BMCyXBol8PoENSSt5rwjx7lB5gkOZfD6EDRlb%2BZ8JALvBeSdDSFy%2Bgc1dW02vzeG71GqEAOp6WJcbD3j6nG7SmHZKhADqeliXGw94%2Bpxu0ph2SKe32dkLNT3pbizPTPEvRenHosw9boFZLM%2FJPz1wv0B8cya4We4EuCzPyT8NcL9AfHMmuVh7IYise01A%3D%3D
Inside the Encrypted_Game_Session we can observe some statically stored data (because of the small variation).
If we request https://hacktactoe.dctf-f1nals-2017.def.camp/action.php?name=%00%00%00%00%00%00%00%00%00%00%00%00%00%00 we get:
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Encrypted_Game_Session=ITSSn%2FICisO1i6CBSOBGkzM0kJWzU%2FS02%2B%2B92B6zVJt3N5aRvRTSxbbP9tlP4VKdcziQlroU0pa7nqPZS%2BMCyXBol8PoENSSt5rwjx7lB5gkOZfD6EDRlb%2BZ8JALvBeSdDSFy%2Bgc1dW02vzeG71GqEAOp6WJcbD3j6nG7SmHZKhADqeliXGw94%2Bpxu0ph2SKe32dkLNT3pbizPTPEvRenHosw9boFZLM%2FJPz1wv0B8cya4We4EuCzPyT8NcL9AfHMmuVh7IYise01A%3D%3D
We now use a URL decoder:
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urldecoded(Encrypted_Game_Session):
ITSSn/ICisO1i6CBSOBGkzM0kJWzU/S02++92B6zVJt3N5aRvRTSxbbP9tlP4VKdcziQlroU0pa7nqPZS+MCyXBol8PoENSSt5rwjx7lB5gkOZfD6EDRlb+Z8JALvBeSdDSFy+gc1dW02vzeG71GqEAOp6WJcbD3j6nG7SmHZKhADqeliXGw94+pxu0ph2SKe32dkLNT3pbizPTPEvRenHosw9boFZLM/JPz1wv0B8cya4We4EuCzPyT8NcL9AfHMmuVh7IYise01A==
We now use a base64-to-hex decoder:
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decode_base64_to_hex(urldecoded(Encrypted_Game_Session):
21 34 92 9f f2 02 8a c3 b5 8b a0 81 48 e0 46 93 33 34 90 95 b3 53 f4 b4 db ef bd d8 1e b3 54 9b 77 37 96 91 bd 14 d2 c5 b6 cf f6 d9 4f e1 52 9d 73 38 90 96 ba 14 d2 96 bb 9e a3 d9 4b e3 02 c9 70 68 97 c3 e8 10 d4 92 b7 9a f0 8f 1e e5 07 98 24 39 97 c3 e8 40 d1 95 bf 99 f0 90 0b bc 17 92 74 34 85 cb e8 1c d5 d5 b4 da fc de 1b bd 46 a8 40 0e a7 a5 89 71 b0 f7 8f a9 c6 ed 29 87 64 a8 40 0e a7 a5 89 71 b0 f7 8f a9 c6 ed 29 87 64 8a 7b 7d 9d 90 b3 53 de 96 e2 cc f4 cf 12 f4 5e 9c 7a 2c c3 d6 e8 15 92 cc fc 93 f3 d7 0b f4 07 c7 32 6b 85 9e e0 4b 82 cc fc 93 f0 d7 0b f4 07 c7 32 6b 95 87 b2 18 8a c7 b4 d4
From this, and after a few retries, we can determine that “40 0e a7 a5 89 71 b0 f7 8f a9 c6 ed 29 87 64 a8” is static and is also the key for our XOR “encryption”. After running the following Python script we get:
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import sys
def xor_strings(xs, ys):
return "".join(chr(ord(x) ^ ord(y)) for x, y in zip(xs, ys))
buf = "2134929ff2028ac3b58ba08148e0469333349095b353f4b4dbefbdd81eb3549b77379691bd14d2c5b6cff6d94fe1529d73389096ba14d296bb9ea3d94be302c9706897c3e810d492b79af08f1ee50798243997c3e840d195bf99f0900bbc1792743485cbe81cd5d5b4dafcde1bbd46a8400ea7a58971b0f78fa9c6ed298764a8400ea7a58971b0f78fa9c6ed2987648a7b7d9d90b353de96e2ccf4cf12f45e9c7a2cc3d6e81592ccfc93f3d70bf407c7326b859ee04b82ccfc93f0d70bf407c7326b9587b2188ac7b4d4"
key = "400ea7a58971b0f78fa9c6ed298764a8"
binary_a = buf.decode("hex")
binary_b = key.decode("hex")
xored = xor_strings(binary_a, binary_b).encode("hex")
print xored
Result: 613a353a7b733a343a22666c6167223b
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lucian@nitescu:~$ python
Python 2.7.12 (default, Nov 19 2016, 06:48:10)
[GCC 5.4.0 20160609] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> buf = "613a353a7b733a343a22666c6167223b"
>>> buf.decode("hex")
'a:5:{s:4:"flag";'
>>>
At this point we are sure about our XOR key, and we just have to improve our Python script a bit, like this:
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import sys
def xor_strings(xs, ys):
return "".join(chr(ord(x) ^ ord(y)) for x, y in zip(xs, ys))
buf = "2134929ff2028ac3b58ba08148e0469333349095b353f4b4dbefbdd81eb3549b77379691bd14d2c5b6cff6d94fe1529d73389096ba14d296bb9ea3d94be302c9706897c3e810d492b79af08f1ee50798243997c3e840d195bf99f0900bbc1792743485cbe81cd5d5b4dafcd513a534c42177c2d7a94092ccfc93f3d70be905c5253c859efa4b88cdadf9aa8c50e21688722c9cd6b3448ad5fccaa99f4ca55fc17a3e9cd6b3478ad5fccaa99f4cb546932934979ef4"
key = "400ea7a58971b0f78fa9c6ed298764a8"
while len(buf) > 32:
binary_a = buf.decode("hex")
binary_b = key.decode("hex")
xored = xor_strings(binary_a, binary_b).encode("hex")
xored = xored.decode("hex")
print xored
buf = buf[len(key):]
On execution we get the flag:
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lucian@nitescu:~$ python 1.py
a:5:{s:4:"flag";
s:70:"DCTF{57403
79144eb29f04ff65
36733eba47e4bdfa
0f0faade836b7bc0
d70fa1ab006}";s:
4:"name";s:8:"Pl
ayer 1";s:5:"nam
e2";s:8:"Player
2";s:5:"score";i
:0;s:6:"score2";
DCTF{5740379144eb29f04ff6536733eba47e4bdfa0f0faade836b7bc0d70fa1ab006}