2
$\begingroup$

Say there's a cryptosystem where encryption and authentication of data is done using a key derived from a user-given password using something like PBKDF2. Is it a good or bad idea to include the PBKDF2 salt in the MAC?

In other words, which of these two systems is better?

$$Packet = Salt + IV + Enc(Data)$$ $$Output = Packet + MAC(Packet)$$

Or

$$Packet = IV + Enc(Data)$$ $$Output = Salt + Packet + MAC(Packet)$$

Assuming $Output$ is the value being transmitted.

A real system like this would obviously be more complicated. This is just to give a rough idea of what I mean.

One argument for this is that it prevents the salt from being modified without notice. But you already get this from the salt deriving the MAC key.

One argument against this is that it allows for a piece of the MAC'd data to have influence on the derived key to calculate said MAC. But this shouldn't be feasible to attack given good KDF and MAC functions.

Which is better? Is there a difference?

$\endgroup$
  • $\begingroup$ I don't understand your argument against. How could an attacker influence the derived key if you include it in the MACed data? $\endgroup$ – Elias Jun 25 '17 at 7:56
  • $\begingroup$ @Elias Oversight on my part. As written, that's an argument for, not against. What I meant was that a part of the MAC'd data influences the key of the MAC itself, which feels wrong in a way I don't know how to prove or disprove. $\endgroup$ – Daffy Jun 25 '17 at 8:02
1
$\begingroup$

It seems unwise to give an attacker enough information to run an off-line attack against a password even if you're using a good PBKDF. PAKE would probably be better.

Assuming you have a strong key your question still holds. I interpret it as:

Given a key $k_1$ and a salt $r$ you derive $k_2 = KDF(k_1|r)$. You then send $m, r$ and $MAC_{k_2}(m|r)$. Is it safe to do this given that a change to $r$ will change both $k_2$ and $m|r$ or is there a chance that $MAC_{k_2'}(m|r') = MAC_{k_2}(m|r)$?

And my answer is, there is no chance of that happening because it would constitute a forgery. Given a good KDF $k_2'$ will be indistinguishable from random so a valid MAC for anything under that key including $MAC_{k_2'}(m|r')$ would be a forgery.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.