this post was submitted on 13 Oct 2024
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[–] LodeMike@lemmy.today 60 points 1 month ago (28 children)

I love how it did not at all explain what they broke. It mentioned "rectangle"? Whats that? How does it have any relation to AES? Because AES is NOT vulnerable to quantum computing. Did they get the key by knowing the ciphertext and the original data?

[–] tal@lemmy.today 17 points 1 month ago* (last edited 1 month ago) (15 children)

Because AES is NOT vulnerable to quantum computing.

I have not been following the quantum computing attacks on cryptography, so I'm not current here at all.

I can believe that current AES in general use cannot be broken by existing quantum computers.

But if what you're saying is that AES cannot be broken by quantum computing at all, that doesn't seem to be what various pages out there say.

https://crypto.stackexchange.com/questions/6712/is-aes-256-a-post-quantum-secure-cipher-or-not

Is AES-256 a post-quantum secure cipher or not?

The best known theoretical attack is Grover's quantum search algorithm. As you pointed out, this allows us to search an unsorted database of n entries in n−−√ operations. As such, AES-256 is secure for a medium-term against a quantum attack, however, AES-128 can be broken, and AES-192 isn't looking that good.

With the advances in computational power (doubling every 18 months), and the development of quantum computers, no set keysize is safe indefinitely. The use of Grover is just one of the gigantic leaps.

I would still class AES as quantum resistant, so long as the best-known attack is still some form of an exhaustive search of the keyspace.

[–] LodeMike@lemmy.today 3 points 1 month ago (8 children)
[–] 4am@lemm.ee 1 points 1 month ago (2 children)

Because you cannot reverse a hash. Information is lost from the result.

[–] tal@lemmy.today 4 points 1 month ago* (last edited 1 month ago)

So, I haven't read up on this quantum attack stuff, and I don't know what Kairos is referring to, but setting aside quantum computing for the moment, breaking a cryptographic hash would simply require being able to find a hash collision, finding another input to a hash function that generates the same hash. It wouldn't require being able to reconstitute the original input that produced the hash. That collision-finding can be done -- given infinite conventional computational capacity, at any rate -- simply from the hash; you don't need additional information.

[–] LodeMike@lemmy.today 2 points 1 month ago

Nobody is wanting to make a magical algorithm that gets the input to the hash.

I mean, there's provably at least one person who does, but there are infinite inputs that lead to the same hash.

Breaking a hash is being able to easily create new input data that leads to the same hash (with or without the constraint of needing the original input data)

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