Increasing Collisions in Password Hashing

With this post I will try to show how, in a theoretical scenario, deliberately weakening the collision resistance of a password hash function can increase security. This is by no means a formal proof.

Suppose Alice has an account on two computer systems X and Y. She uses the same password on both systems.

Suppose both systems implement the following password hashing scheme: Let t be some positive integer. Given a password P, compute the entropy of the password H(P). Create and store a salted B-bit hash of P where B = H(P) - lg(t). To verify a password, wait for t units of time then check the hash in the usual manner. Here, the entropy of the password is the logarithm base two of the number of guesses it will take an adversary to guess the exact password. Also suppose that it takes one unit of time to compute a hash.

I claim that using this scheme, Alice's accounts are no less secure than if we used 256-bit hashes (if H(P) < 256), and that if System X gets compromised, then her account on System Y is more secure than if we used 256-bit hashes.

To see why, note two things:

  1. The probability of any random string S having the same hash as Alice's password P is 1/(2^B). So if Mallory tries to guess Alice's password using an online attack, it will take about 2^B guesses, so about 2^B * t time. So the effective security against an online attack is B + lg(t). By the construction, B + lg(t) = H(P) so this is no easier than guessing P.
  2. Suppose Mallory roots System X and gets the B-bit hash of Alice's password. It takes Mallory about 2^B guesses to find a preimage. But since B = H(P) - lg(t), Mallory will find about t collisions when she performs the same search she would if it was instead a 256-bit hash (this is not at all a formal argument, but hopefully you see the idea behind it).

By supposition, Mallory has rooted System X, so there is no point trying to secure Alice's account on X anymore. All that remains is the security of Alice's account on system Y. This breaks down if Mallory can get read-only access to the hash database without rooting... but suppose the root password is stored in plain text in the hash database ;) .

Because System Y is using different salts, it is unlikely that any of the t collisions Mallory found, except P itself, can be used to authenticate as Alice to System Y (reader: double check this). So Mallory's best option is to try authenticating to System Y as Alice using the t collisions as passwords guesses, since she knows P is one of those t collisions, and by construction lg(t) < H(P). This will take t * t time, for an effective security of 2 * lg(t) bits.

If, instead, System X was using standard 256-bit hashes, Mallory would have found the exact P and could get into System Y on the first try. Using this scheme it takes about t tries. So by weakening the hash, Alice's account on System Y is more secure than it would have been had the hash not been weakened.

Of course, this system is impractical, since it assumes knowledge of how Mallory would guess P, and thus the entropy of P. It also assumes that if Alice's hash on System X gets compromised, then the security of her account on X no longer matters at all, which is obviously not always the case. It is nevertheless interesting to see how the altruism of the compromised system can increase the security of Alice's account on the other system.

Another problem is that the user must wait t units of time when logging in, so in practice there would be tradeoff between security and log-in wait time. Locking the accounts after significantly less than t attempts might be a better approach.

It doesn't look like it can be made any better than this, since if you decrease B any further, finding collisions with an online attack becomes easier than guessing P, but it is still an online (thus detectable, stoppable) attack so it might still be harder. I would guess that the wait time t would be better spent computing iterations of PBKDF2 or similar.