I'll show you mine if you show me, er, mine
hal at finney.org
Wed Feb 23 12:14:04 PST 2005
Markus Jakobsson is a really smart guy who's done some cool stuff, so I
think this is probably better than it sounds in the article. His web
site is http://www.informatics.indiana.edu/markus/ but I don't see any
papers there that sound like what the article describes. I tried to
reverse engineer the protocol from the article, and the results are below.
But first let me put this into context.
The security property seems to be that you send something to the server,
and it sends you back something that proves that it knows your password.
But neither a passive eavesdropper nor a MITM can learn anything about
your password from observing or influencing the exchange. The best an
attacker can do is to try to brute force your password by guessing it
repeatedly and trying each guess out at the server. And this can be
easily prevented by having the server refuse to answer more than a few
bad password attempts.
Note that this is different from simple PK based authentication,
because the secret is human memorizable. And it's different from,
say, having the server respond with a keyed hash of your passphrase,
because an eavesdropper could then do an offline brute force search.
The key feature is that the only attack is online brute forcing.
There are already a lot of protocols in the literature which do this,
often performing key agreement at the same time. The original one
and most famous was SPEKE. There is a long list of such protocols at
http://grouper.ieee.org/groups/1363/passwdPK/submissions.html. I don't
know what properties this new protocol has that the old ones don't.
Maybe it does have some and I am missing the point. Or there might be
some patent issues that it is trying to work around.
Anyway, here's my attempt at mimicking the protocol, based on the
description of envelopes and carbon paper.
You have a password, and so does the site you will login to. (Or,
maybe the site has a salted hash of your password; you could use that
instead.) You set up a homomorphic encryption system. This is one where
you can send an encrypted value to someone else, and he can do certain
operations on the encrypted value, like multiplying it by a constant.
In this case I think we only need to encrypt the value 1, and let the
other guy multiply by his constant, which makes it simpler.
I think ElGamal could work: you encrypt 1 as (g^k, y^k), where you'd
make up a key y = g^x on the spot. You send this to the other guy who
picks a random power j and raises both elements to that power, then
multiplies the 2nd one by c: (g^(k*j), y^(k*j) * c), and sends it back
to you. This is now a valid ElGamal encryption of c. But an observer
can't tell what c is.
For a first cut at this protocol, you take each bit of the password (or
salted hash) and create two encryptions of m = 1. It would look like
E(1) E(1) E(1) E(1) E(1) ...
E(1) E(1) E(1) E(1) E(1) ...
You send all these to the server. The server knows your password (or
salted hash) and, for each pair of encrypted values, multiplies the
one corresponding to password bit b_i by some constant c_i. The other
one of the pair, corresponding to !b_i, it multiplies by a random r_i.
The server sets it up so that the sum of all the c_i is zero. Then it
sends all of them back to you. If your passphrase started 01101...
it would be:
E(c_1) E(r_2) E(r_3) E(c_4) E(r_5) ...
E(r_1) E(c_2) E(c_3) E(r_4) E(c_5) ...
Now, you decrypt just the ones corresponding to the bits b_i and add up
the decrypted plaintexts, giving you sum of c_i. If the result is zero,
you know the server knew your password (or salted hash).
Actually this is not quite right, because the article says that you are
not supposed to be able to decrypt both ciphertext values in the pair
that corresponds to a password bit. Otherwise an imposter might be able
to figure out your passphrase by doing one interaction with the server,
then finding an element from each pair such that they all sum to zero.
This is kind of knapsacky and it might not be that hard, I'm not sure.
So I think what you could do is to send a valid ElGamal encryption of
1, and a bogus value which is not an ElGamal encryption of anything.
But the remote party wants to be sure that you can't decrypt them both.
One way to achieve this is to arrange that the first members of each pair,
g^k in the good encryption, multiply to some fixed value F for which the
discrete log is not known. Maybe it's the hash of "I don't know if this
will work." You can't know the DL of that hash, so you can't find two
g^k values which multiply to that hash. That means that if you have a
pair of ElGamal ciphertexts which have this property, only one is a real,
valid ElGamal ciphertext and so only one is decryptable (I think!). So
you would send, in the example above:
(g^k0, y^k0) (F/g^k1, junk) (F/g^k2, junk) (g^k3, y^k3) ...
(F/g^k0, junk) (g^k1, y^k1) (g^k2, y^k2) (F/g^k3, junk) ...
When the server did its multiplications as above you'd still get the
correct encryptions of c_i, but the other pair would be junk and you
wouldn't learn the r_i values:
E(c_1) junk junk E(c_4) junk ...
junk E(c_2) E(c_3) junk E(c_5) ...
Now you can still decrypt it and verify your password. But for someone
who is impersonating you and doesn't know your password, they're going
to get a mix of c_i and r_i values that won't add up to zero, and that
won't give them any clue about what the real password is, other than
that they guessed wrong.
I'm not 100% sure this will work, that the attacker can't create a bogus
pair (F/g^k, junk) which will allow him to determine what value the server
multiplied by. At a minimum I see that if junk = F/g^k then it will be
obvious what the constant was, so the server would have to check for that.
This is why it's good to have provable security!
This way of doing things would also be quite inefficient; there are
two ElGamal encryptions going back and forth (typically 2048 bits each)
for every bit of your password. I'll bet the actual paper has a much
more clever scheme which improves the efficiency and has a nice proof
of security. I'm looking forward to seeing it.
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