[Computer-go] Many Faces and Japanese rules
Robert Jasiek
jasiek at snafu.de
Wed Apr 28 08:57:43 PDT 2010
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 <jas... at snafu.de>
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
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