The probability of a hash collision (kevingal.com)
from learnbyexample@programming.dev to programming@programming.dev on 05 Jul 03:03
https://programming.dev/post/33374522

#programming

threaded - newest

The2b@lemmy.vg on 05 Jul 04:54 next collapse

This is valid on a single unlisted assumption: The hash function has equal distribution. If your has function ends by multipliying the has value by 4, for example, your number of possible boxes is 1/4th the otherwise expected value based on the size of the hash output

Colloidal@programming.dev on 05 Jul 13:51 next collapse

The assumption is there though.

Wouldn’t multiplying the hash simply relabel the hash sites, as hashes non divisible by the factor simply be not accessible/not exist?

squaresinger@lemmy.world on 05 Jul 16:11 collapse

Of course. That’s one of the basic requirements for something to be an actual hash function.

[deleted] on 05 Jul 08:29 collapse

.