Álvaro Begué wrote: > > On Dec 20, 2007 10:19 AM, Jason House <[EMAIL PROTECTED] > <mailto:[EMAIL PROTECTED]>> wrote: > > > > On Dec 20, 2007 10:15 AM, Arthur Cater <[EMAIL PROTECTED] > <mailto:[EMAIL PROTECTED]>> wrote: > > With 8 hashes per position, the chance of two different boards > producing a different set of hashes but > the same canonical hash is greater than 1/2^64, because there > will be > a bias in the choice of canonical > hashes - toward numerically lower numbers, for instance. > > I think. > > > More importantly, how does it differ from 8/2^64 = 1/2^61? > > > If you are going to compute all 8 hash keys, you can just add them up > at the end instead of picking the minimum. Wouldn't that be better? I think that's pretty workable. XOR is definitely wrong here. If you use xor, then the empty board would hash to the same value as the position after a stone (of either color) is placed on e5 as well as any other symmetry like this. I also think symetries like putting a black stone on 2 points across from each other (such as in diagonal corners) would create a zero hash because you have 2 sets of 4 hashes that cancel each other. I think addition as Álvaro suggests fixes this.
- Don > > > ------------------------------------------------------------------------ > > _______________________________________________ > computer-go mailing list > computer-go@computer-go.org > http://www.computer-go.org/mailman/listinfo/computer-go/ _______________________________________________ computer-go mailing list computer-go@computer-go.org http://www.computer-go.org/mailman/listinfo/computer-go/
