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Merge pull request #223 from ZeusWPI/csc_radium

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Fix explanation
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David Vandorpe 2018-03-09 12:09:13 +01:00 committed by GitHub
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@ -48,9 +48,7 @@ So, how do we get our flag? Let's list some ideas:
Let's see what these ideas lead to.
We found a [writeup](https://shrikantadhikarla.wordpress.com/2016/03/08/des-ofb-writeup-boston-key-party-ctf/) from another CTF that cracks DES encryption in OFB mode. The key weakness here is using a weak DES key (making encryption symmetric), in combination with OFB. But wait a second, isn't AES a symmetric algorithm? So basically we would expect the second block of ciphertext to just be the plaintext XOR'd with the IV. However, for reasons unknown to me this turned out not to be the case. **Update:** Thanks to Toon Willems for pointing out that AES modifies the key after every round. For those interested, see [act 3](https://www.moserware.com/2009/09/stick-figure-guide-to-advanced.html).
![AES Key Expansion](https://www.moserware.com/assets/stick-figure-guide-to-advanced/aes_act_3_scene_06_key_expansion_part_1_1100.png)
We found a [writeup](https://shrikantadhikarla.wordpress.com/2016/03/08/des-ofb-writeup-boston-key-party-ctf/) from another CTF that cracks DES encryption in OFB mode. The key weakness here is using a weak DES key (guaranteeing `x == DES(DES(x))`), in combination with OFB. But wait a second, the program uses the same function to encrypt and decrypt as well! So basically we would expect the second block of ciphertext to just be the plaintext XOR'd with the IV. Some testing proved that our logic was flawed: using the same function worked because of the symmetry of the XOR operation in OFB mode, and not because `IV == AES(AES(IV))` which would break AES OFB.
Setting the encryption key to NULL on the server seemed easy enough, but sadly this would also mean that the server couldn't decrypt the messages from the client (as this key was used for both encryption and decryption by both sides).