2
$\begingroup$

Can someone view this video and comment?

https://cipherloc.net/cipherloc-presents-details-of-encryption-attacks/

The concept is presented between 8 and 13 minutes, and then results are presented at roughly 20 minutes, but I think the results are limited to some type of linear cipher.

They claim at 12 minutes that with as little as 5 collisions, a message can be obtained directly. The problem is that I don't see how this is effective against AES. I calculate that the probability of even a single collision occurring for an AES-128 encrypted message--for 1 million blocks in CBC mode-- is roughly 10^-27. Pardon me is my arithmetic is off, but the point remains. In any case this is the basis for which they claim a successor for AES is required...

$\endgroup$
  • $\begingroup$ Unless I am gravely mistaken, they are just talking about CBC information leakage, which is a known result and not a problem for block ciphers with sufficient block sizes, like AES (unless you consider encrypting unrealistic amounts of data under the same key). $\endgroup$ – DiscobarMolokai Dec 11 '16 at 20:38
  • 4
    $\begingroup$ Your question should be self-supporting. We should not have to go watch a video somewhere. You need to add more detail. $\endgroup$ – mikeazo Dec 11 '16 at 20:43
  • 3
    $\begingroup$ Cipherloc tries to push their own product (PKPA) using mindbogglingly false claims about block ciphers and modes of operation. They're at best utterly clueless, at worst utterly dishonest. $\endgroup$ – Dennis Dec 12 '16 at 3:04
  • $\begingroup$ Not sure how I can re-phrase this without referring to the link, length would be excessive $\endgroup$ – John Dhamas Dec 12 '16 at 16:35
4
$\begingroup$

Without having listened to the lecture, I agree with you. This makes no sense. With AES-128, the probability of getting collisions is very small. There has been a lot of work computing the security of known modes of encryption, and you have to be encrypting an insane amount to have even a reasonable probability of a collision happening. If you are using 3DES then this is a completely different story and indeed 64-bit blocks isn't enough anymore.

$\endgroup$

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