Lucky Thirteen: Breaking the TLS and DTLS Record Protocols

http://www.isg.rhul.ac.uk/tls/

by agl • 13 years ago

77 19 13 years ago

isg.rhul.ac.uk

agl6• 13 years ago

Post on implementing the (rather complex) fix: http://www.imperialviolet.org/2013/02/04/luckythirteen.html

ajross2• 13 years ago

Can anyone with expertise provide details? From the description, it's a timing attack based on the content of the stream. But my understanding (and presumably that of the protocol designers) was that TLS post-key-negotiation was mostly deterministic and not subject to this kind of thing. What am I missing?

agl6• 13 years ago

The attacker lets the handshake complete normally and then intercepts the transmission of the encrypted record that contains the secret that they wish to decrypt. (Assuming that the parties negotiated a CBC ciphersuite.)

The attacker can then alter the encrypted record and send it onwards to the server. Because the record has been altered, the server will reject it and close the connection. However, the amount of time that it takes to reject the record reveals something about the plaintext padding. If the client will repeatedly send the same plaintext secret (e.g. HTTP cookie) over many connections (e.g. a web browser repeatedly requesting a resource) then the attacker can learn a little from each connection and, after thousands of connections, start to decrypt the secret.

ajross2• 13 years ago

> However, the amount of time that it takes to reject the record reveals something about the plaintext padding.

This is the part I don't get. Symmetric cyphers in CBC mode don't traditionally behave that way. You encrypt the block and get your answer. Is there a variable-length HMAC in there somewhere or something?

api• 13 years ago

If designing new crypto stuff, use a mode like GCM (Galois Counter Mode) or EAX. Avoid the old non-authenticated modes.

agl6• 13 years ago

Both AES and GHASH have their own timing side-channel issues.

If you need to encrypt something, I recommend NaCl: http://nacl.cr.yp.to/

ctz• 13 years ago

Are you saying that there is an equivalent attack for the AES-GCM TLS 1.2 ciphersuites? If so, paper please :)

moonboots• 13 years ago

Besides the difficulties of patching openssl and nss, what are other technical and political challenges preventing the adoption of djb's crypto primivatives in tls?

moxie2• 13 years ago

@agl, it looks to me as if the timing differences they are measuring would be a complete wash in any realistic scenario?

The original "good enough" timing fix (in response to Vaudenay) was based on the assumption that an attacker would not realistically be able to measure the small remaining timing differences.

This paper has demonstrated that it is possible to measure the small remaining timing differences, but it seems as if that requires:

1) The server not to be doing anything else or processing any other TLS sessions.

2) The attacker to be in a hyper-controlled networking environment with extremely low latency to the server and zero interference.

Do you see this attack ever being possible in a practical situation?

agl6• 13 years ago

Against DTLS it appears to practical, both because the connection isn't destroyed and because the difference can be amplified by the researcher's Heartbeat trick from http://www.isg.rhul.ac.uk/~kp/dtls.pdf.

Against TLS it's tougher. The paper is suggesting a difference of ~10000 cycles, which may be ~5us. So, yes, you need to have a good connection to the server to be able to measure that difference, but that's not unthinkable to me. I know that you could get such a connection to a Google server if you were willing to pay some money to get a machine hosted in the right place.

But it also appears that it's fairly easy to get a process instantiated on the same EC2 host as a target application. Or maybe your server is running on a small ARM device and so the difference is much larger.

Don't get me wrong - this is a trivial issue compared to, say, the Java vulnerabilities that we've seen recently. But it's huge compared to the level of security that plain TLS should be providing. Saying that it's secure as long as you don't allow the attacker to be close by, or you don't run it on a slow machine isn't good enough for me.

But I admit that it's unlikely to anyone will ever actually perform this attack in anger. Although I would have said the same about MD5 collisions at the beginning of 2012 :)

moxie2• 13 years ago

Interesting. And beyond network position, do you think it's possible to measure these micro timing differences effectively against a server that experiences the type of load patterns Google's servers do?

(And yeah, I have little faith in DTLS generally).

JoachimSchipper3• 13 years ago

Assuming that jitter will hide the signal seems optimistic to me - "attacks always get better", and e.g. Figure 5 seems to suggest that just throwing more measurements at the problem is fairly effective. Of course, more measurements means that the attack takes longer, but note that the bruteforcing of the initial two bytes dominates the time the attack takes. If those bytes are partially or fully known, e.g. because you know some headers and can manipulate the length of certain fields, the attack becomes comparatively fast.

Of course, you do need a victim which automatically reconnects or uses DTLS, both of which are somewhat questionable.

caf2• 13 years ago

Isn't it about time that a new TLS version switches to Encrypt-then-MAC?

Lucky Thirteen: Breaking the TLS and DTLS Record Protocols

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