The problem with Territory rules as i see them are: - When to pass? Pass as early as possible.
- Scoring After the end of the game you need something like the hypothetical-sequence as described in http://home.snafu.de/jasiek/j2003com.html I guess this can be done with a quick and dirty routine. (was thinking about the Java refbot but i am wondering if Javarefbot is nice with seki's, my first impression is that seki's are captured because he dooesn't allow passes before all liberties are filled A more technical problem - Territory scoring breaks simple transpositions. (just the boardposition and ko. The same position kan be lost draw or won depending on the number of stones captured before the position was reached . (off course then there need to be passes before the position was reached. And then rest the problems of some other problems like points in seki, bent 4 in the corner, fpositions with points without playing ect. On Thu, Apr 29, 2010 at 7:40 AM, Aja <[email protected]> wrote: > Hi Robert, > > Thanks a lot. The proof is very interesting. I will take a careful look. > > Aja > > ----- Original Message ----- From: "Robert Jasiek" <[email protected]> > To: <[email protected]> > Sent: Wednesday, April 28, 2010 11:57 PM > Subject: Re: [Computer-go] Many Faces and Japanese rules > > > On 2010-04-28 15:09, Aja wrote: >> >> I don't really understand what is the relation between what you tried to >> explain and the equivalence in normal cases of Chinese, Japanese scoring > > [...] >> >> if there will be strict mathematical proofs and conclusions from Berlekamp >> and you > > The local region territory thing was done much earlier. Berlekamp did a > sketch, I wrote down the details, see further below. > > Forget about CHN and JAP scoring - talk about Area vs. Territory. As we > know, e.g., from the AGA equivalence theorem and the AGA Rules' White > passes a pass stone as the last player, in normal shape games (no sekis > etc.) Black's area score tends to be 1 greater than the territory score. > Otherwise it is simple: 1 area stone not on the board because of being > "dead" / removed equals 1 prisoner for the opponent. I think you do not > also need my seki parity proofs. > > -- > robert jasiek > > > > > Robert Jasiek > 7 Mär. 2004 21:48 > Newsgroups: rec.games.go > Von: Robert Jasiek <[email protected]> > Datum: Sun, 07 Mar 2004 21:48:21 +0100 > Lokal: So 7 Mär. 2004 21:48 > Betreff: Playing in a Region > > Studying go mathematics or go rules mathematics is like learning go > for the second time. It has the advantage that certain aspects of go > are solved in general and thereby can teach everybody, who accepts to > learn from such research. > > PROPOSITION > > Under the Japanese 2003/29 Rules and during the alternating-sequence, > let > - P be a position, > - R be a not in-seki region in P, > - A be the player with the alive strings in R of P, > - B be the opponent of A, > - the players make all plays in R, > - PA be the number of plays of A, > - PB be the number of plays of B, > - S(X) be the score of position X. > Let Q be the final-position so that > - R is a not in-seki region in Q, > - A is the player with the alive strings in R of Q. > > Then > > S(Q) = S(P) + PB - PA. > > PROOF > > Each stone of an A-string scores 0 in favour A if it is on the > board and scores 1 in favour of A for the intersection plus 1 in > favour of B for the prisoner if it is removed. Therefore it does > not matter if stones of A-strings are removed. > > Each stone of a B-string scores 2 in favour of A if it is on the > board and scores 1 in favour of A for the intersection plus 1 in > favour of A for the prisoner if it is removed. Therefore it does > not matter if stones of B-strings are removed. > > Each of the PA stones played by A scores 1 in favour of B because > it occupies one intersections that otherwise is empty. Therefore > all PA stones score -PA in favor of A. > > Each of the PB stones played by B changes an empty intersection > that scores 1 in favour of A into an intersection occupied by a > B-stone and scoring 2 in favour of A. Therefore all PB stones > score PB in favor of A. > > The sum of all score changes from P to Q is 0 + 0 -PA + PB. > QED. > > EXAMPLE > > P > > # # O # > # O O # > . O # # > O O # # > # # # # > . . . . > > S(P) = 17. > > Alternating-sequence from P to Q with # first > > 3 . O # 1, 4, 5 = pass > . O O # > 2 O # # Note: prisoner-difference = -3 > O O # # > # # # # > . . . . > > A = #, B = O. > > PA = 1. PB = 1. > > Q > > # . O # > . O O # > O O # # > O O # # > # # # # > . . . . > > S(Q) = S(P) + PB - PA = 17 + 1 - 1 = 17. > > Capturing 3 stones is superfluous. > > -- > robert jasiek > _______________________________________________ > Computer-go mailing list > [email protected] > http://dvandva.org/cgi-bin/mailman/listinfo/computer-go > _______________________________________________ > Computer-go mailing list > [email protected] > http://dvandva.org/cgi-bin/mailman/listinfo/computer-go > _______________________________________________ Computer-go mailing list [email protected] http://dvandva.org/cgi-bin/mailman/listinfo/computer-go
