Author: Marc Bourzutschky
Date: 16:30:28 05/16/04
Go up one level in this thread
On May 16, 2004 at 18:54:28, Russell Reagan wrote: >On May 16, 2004 at 16:02:54, Marc Bourzutschky wrote: > >>>After reading the article, I would have to agree that 8,294,400 is the correct >>>count. How did you arrive at your number? > >>If reflection on the vertical does not change the castling options for either >>player, the position is game theoretically equivalent to its mirrored position >>(because all piece movements except for castling are mirror symmetric). The >>8,294,400 count double counts those positions. > >Then you are asking (at least) two different questions, to which there are (at >least) two different answers. The "mathematical" answer is 8,294,400. If we >don't use the mathematical approach, it becomes a mess. We could look at the >problem from a data storage point of view, using >reflection/rotation/mirroring/etc. You are currently saying that KQRRBBNN is >equivalent to NNBBRRQK, i.e. they belong to the same group, i.e. if we knew the >optimal game theoretical result of each opening position, we could count both of >these as (say) a win for white (or whatever the real outcome is). > >Having said that, I could argue that there are three game theoretical >possibilities. Three groups of positions where... > >1. White has a forced win >2. Black has a forced win >3. There is a forced draw > >Yes, that is a stretch on my part, but you get the point. Maybe all of the >answers you posted are correct, because you haven't nailed down the question you >want an answer to. You need to concretely define the question you are asking. >8,294,400 is the right answer, but that answer may not belong to the question >you are asking. "Game theoretically different positions" as in my original post is a precise formulation of the question, since it is the number of starting positions that have to be completely analyzed to get a full description of the game. How many end up wins, draws, etc. after this analysis is not relevant. 8,294,400 is the answer to the simpler question of the number of distinct combinatorial possibilities. In fairness to Noam Elkies, he probably focused on the purely combinatorial question, but in my post I quite explicitly did not...
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.