1
$\begingroup$

I'm writing a paper on Authenticated Key Exchange Protocols. I've read Bellare and Rogway's seminal paper on the subject and I think I understand BR Model and I'm now reading Cenetti and Krawczyk's paper which aims to improve on it. I'm confused as to how the CK model is an improvement of the BR. As mentioned in the appendix of the CK paper, the BR paper phrases their analysis in terms of oracles. They mention a security flaw in the BR model but I need it thoroughly dumbed down.

In the definitions of (BR), the adversary points to an unexposed session of its choice, and receives a value $k_b$, where $k_0$ is the real session key of this session, $k_1$ is an independently chosen random value, and $b$ is a randomly chosen bit that is unknown the the adversary. The security requirement is that the adversary is unable to predict $b$ with non-neglible advantage over one half. The original version of these definitions requires that the adversary outputs its guess for $b$ immediately after it obtains the test value

This is largely fine and I include it to provide context for the part I don't get.

Consider your favourite secure key-exchange protocol $\pi$. Now add to the specifications of the protocol the following instruction for the party that completes the session establishment according to the protocol $\pi$: if at any point the party receives a message with the value $\mathrm{MAC}_\kappa(0)$, where $\mathrm{MAC}$ is a secure message authentication function and $\kappa$ is the established session key, then the party publicizes $\kappa$. However, the protocol never instructs any party to carry out such an instruction. As a result the protocol can be showed to pass the weakened definition. On the other hand, it is clear that such a protocol cannot be composed securely with an authentication application that uses the session key for MAC-ing information.

Doesn't the protocol directly tell the part to publish $\kappa$ if it receives $\mathrm{MAC}_\kappa(0)$? So, how is this protocol secure under the BR definition? How does CK eliminate this possibility?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.