undefined

points

[-]

ok damn, I did not know this, obviously. Thanks.

I guess I've never actually had this problem because was always hashing things that were static, or specialty cases like password hashes where the salt obviously guarantees uniqueness.

by 3 hours ago|

parent|

[-]

deleted

by kro4 hours ago|

parent|

prev|

[-]

It's very very unlikely to get collisions there, but still not impossible. Whenever you map data of arbitrary length (infinite possibilities) to a limited length collisions are possible.

by eru4 hours ago|

parent|

[-]

Doesn't even have to be arbitrary length.

Whenever you map into a smaller space, you get collisions. The bigger space doesn't have to be infinite.

by bryanrasmussen9 minutes ago|

parent|

[-]

with a password you may be mapping into a smaller space or a bigger space, because what you want is to get them all same length, but yeah you may in some cases be mapping into a smaller space, hadn't thought of that, although I sort of also think it is unlikely.

But there it doesn't matter anyway because the password is put together with the email to identify the user so in practicality passwords will never collide even if they could in theory.

by eru4 minutes ago|

parent|

[-]

For passwords: the input _space_ is bigger. That doesn't say anything about the length of any particular password.

> But there it doesn't matter anyway because the password is put together with the email to identify the user so in practicality passwords will never collide even if they could in theory.

For passowrds, you are not worried so much about two users accidentally getting the same hash, you are worried about people finding a pre-image that hashes to the same output as your user's password.

by ogogmad2 hours ago|

parent|

prev|

[-]

Let's consider a hash table with an allocation of 1MB, which is about 2^20 bytes. Assume also that each entry occupies a byte. Assuming that the hash function's values are distributed randomly, the probability of there being a collision with only 1000 entries is approximately 38% = 1-(2^20)!/(2^20 - 1000)!/(2^20)^1000. See the "Birthday Problem".