> a) compute xor of the hashes > b) compute sum of the hashes > c) compute product of the hashes (with 0 replaced by 1 to avoid 0*n=0) > Then combine the three values somehow. For instance: (a*17+b)*17+c
That is really good one. Better than pure XOR. I also found this in scala: https://github.com/scala/scala/blob/2.11.x/src/library/scala/util/hashing/MurmurHash3.scala#L88 On 2020/07/15 20:16:13, Vladimir Sitnikov <sitnikov.vladi...@gmail.com> wrote: > > things might be bad if we want do dedup RexNode children using > Set<RexNode> > > I guess the following might work better than XOR: > > a) compute xor of the hashes > b) compute sum of the hashes > c) compute product of the hashes (with 0 replaced by 1 to avoid 0*n=0) > Then combine the three values somehow. For instance: (a*17+b)*17+c > > That would result in a hash code that tolerates item reordering, it is > better than XOR, and it should be fast. > > Unfortunately, we can't use the improved function in Pair since Map.Entry > specifies the exact computation formula :-/ > > Vladimir >
