[thfuran]

The last several comments were responding to

>2^61 is still a very large number of course, but much more feasible to reach than 2^122 when doing a collision attack. This is the reason that cryptographic hashes are typically 256 bits or more (to make the cost of collision attacks >= 2^128).

[8organicbits]

I'm not sure. 6ghz is around 2^61 CPU cycles in 12 years. I.E. basic CPU instructions; counting, not computing a cryptographic hash. Otherwise, where is the cluster that's bruteforcing ~122 bit cryptographic hash collisions in minutes?

[vlovich123]

For what it’s worth generating a random UUID for the purposes of collision isn’t generally much more complicated than a few arithmetic instructions which is why I used counting as an example. And as the other poster mentioned generating a UUID collision isn’t a security problem since the UUID tends to be generated within your infrastructure where you can’t really go full blast at generating UUIDs for all sorts of reasons anyway.

For cryptographic applications it is really small because the previous poster is correct that 2^64 is very small for that purpose - a small supercomputing cluster or two could decrypt such a cipher in a reasonable amount of time, which is why symmetric keys are all 256 bits and up to guarantee there’s no way to attack them.

[8organicbits]

I don't think that's quite right. A 128-bit UUIDv4 having a 50% chance of having any collision after 2^61 generations is very different from finding a specific 128-bit symmetric key. The best cryptanalysis of AES-128 is 2^126; nowhere near 2^64. Which is why standards bodies like NIST still recommend AES-128 as a baseline.

[twiss]

You're right that AES-128 is fine. Normally the birthday paradox only applies to cryptographic hashes.

The only way it would apply to symmetric keys is if you have a server that stores 2^64 encrypted messages, and can somehow find out which messages used the same symmetric key (normally not possible unless they also have the same IV and plaintext), and can somehow coerce the user who uploaded message #1 to decrypt message #2 for you (or vice versa). Obviously that isn't realistic.