12 ms·
True. But in the case I'm evaluating, that isn't my worry. All I care about is the expected number of collisions over some extremely large number of hashes.
by cmurphycode 15y ago
True. But in the case I'm evaluating, that isn't my worry. All I care about is the expected number of collisions over some extremely large number of hashes.
- orijing 15y agoThat's not the definition of a cryptographic hash. One of the four requirements of a function being a cryptographic hash function is that "it is infeasible to find two different messages with the same hash."
- cmurphycode 15y agoI understand. I worded my question poorly- what I was looking for was what they found that invalidated it as a cryptographic hash, so I could figure out whether it would fit my (non-crypto) requirements.
- maaku 15y agoThen a cryptographic hash is not what you need.
- zobzu 15y agoExactly, it's be interesting to have the average number of hashes needed before you hit a collision. Crypto hashes are made so that collisions are as rare as possible. Sometimes, you need something very, very fast, with as little collisions are possible (aka it "never" happen, like crypto hashes) yet those are not hashes of passwords and the like, so if it would happen it wouldn't be a security issue. For that reason, this question is in fact interesting. Not that I have the answer to that :-)