I read that brute force attacks against a plaintext encrypted in OCB with unknown key and IV has approximately the same complexity as an attack where only the key is unknown.

  1. Why is that?
  2. Is there a description of this attack? If not, could you give me a hint on how this attack might work?

1 Answer 1


Ok, here is one possible sketch of a brute-force style attack against OCB, which doesn't assume you know the nonce. It takes 512 trial decryptions for every key tested, so it's a bit more expensive than, say, a brute-force attack on CBC, but only by a constant factor.

Let us assume that:

  • you have an encrypted message that is $16n+15$ bytes long
  • that you know the last 15 bytes of the plaintext
  • and there's another block $P_n$ that you have enough information that you can recognize it if we decrypt it properly.

If we look at the OCB processing of the last 15 bytes, we have:

$C_{last} = P_{last} \oplus Trunc( E(K, \Delta_{last} ))$

where $Trunc$ is a function that chops off the last byte. So, we know the first 15 bytes of $E(K, \Delta_{last})$, namely $C_{last} \oplus P_{last}$.

To test a value of $K$, we iterate through all 256 possible values of the last byte $B$, and compute

$\Delta_{last} = D(K, (C_{last} \oplus P_{last}) || B)$

From each $\Delta_{last}$ value, we compute the corresponding $\Delta_{n}$ value (which is a $GF(2^{128})$ multiplication, so that's easy), and then compute:

$P_n = \Delta_{n} \oplus D(K, C_n \oplus \Delta_n)$

and check if that is a plausible plaintext value.

Once we have a plausible values for $K$ and $\Delta_i$, we can check the authentication tag to validate the decryption.

Once we have all that, we can also recover the nonce that was used to encrypt the message; this may be useful if related nonces were used to encrypt other messages with the same key.

Now, obviously this attack can be improved if we can get a plaintext message which is $128n+127$ bits long; that is unlikely in practice.


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