The results of 1997/8 can help. Note that some email addresses are old.
Message-ID: <360CA419.51AC@berlin.snafu.de>
Date: Sat, 26 Sep 1998 10:21:45 +0200
From: Robert Jasiek <jasiek@berlin.snafu.de>
To: ML go-rules <go-rules@usgo.org>
CC: feldmann@bsi.fr
Subject: Result Changes due to the 3rd Pass Stone of Equivalence Scoring
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Denis Feldmann has pointed out that almost always the 3rd pass stone of equivalence scoring does not change the result. I give a proof for this proposition based on earlier proofs, which are included at the end.
Proposition:
In a game without handicap, with standard area komi, on an ODD board, ending with an EVEN SekiParity, without passes before the final passes, and with equivalence scoring the 3rd pass stone does NOT cause a result change.
Proof:
Due to the proof by RJ on 1997-07-10 the only case to be considered is black playing the last neutral intersection, the smallest winning margin of the territory score without any 3rd pass stone being B+0.5, the smallest winning margin of the area score being B+1.5. There, for white has to pass last, the smallest winning margin of the equivalence score is B+1.5. Even for the evaluation of the territory score this does not change the result of black winning.
Remarks:
The presuppositions are essential. However, they cover most games. This includes most professional games in Japan, where a standard area komi of 5.5 is used. According to Ing statistics only about 1 game in 10000 have a result change due to the 3rd pass stone. Generalizations of the above proposition will be possible.
--
robert jasiek
http://www.snafu.de/~jasiek/rules.html
http://www.snafu.de/~jasiek/int.html
Subject: Re: Changing Komi to 6.5 Points
Date: Thu, 10 Jul 1997 12:23:22 +0200
From: Robert Jasiek <jasiek@berlin.snafu.de>
Organization: Unlimited Surprise Systems, Berlin
To: "Dr. Kuo-Chi Lin" <klin@pegasus.cc.ucf.edu>
CC: Matthew@jklmn.demon.co.uk, go-rules@lists.io.com
Newsgroups: rec.games.go
Dr. Kuo-Chi Lin wrote:
>
> Matthew Macfadyen (Matthew@jklmn.demon.co.uk) wrote:
>
>[...]
> : But there are an odd number of points on the board, so a simple parity
> : argument says that when this occurs the margin will be an odd number, so
> : the game might be moved from a 4 point Black win to 5, but not from 5 to
> : 6, thus with 5.5 points komi a player used to Japanese counting does not
> : have to worry about losing due to the number of dame points being odd.
>
> Can someone confirm this theory? Personally, I do not think it is correct.
>
> Kurt
I am also surprised about this, but I am guilty of having supplied the mathematical background for this as included at the end of the message.
So here is a sketch of a proof for the above:
Presumptions are OddBoardSize and EvenSekiParity and standard area komi s(t) = 2t - 0.5; t positive natural numbers; e.g. 5.5 komi.
We are interested in smallest possible winning margins:
For Japanese rules: B+0.5 or W+0.5.
For area rules (like AGA): B+1.5 or W+0.5.
(Proof see inclusions.)
Now we want to observe the consequences of a last neutral point possibly changing the Japanese result into another win under area rules. A last neutral point conforms to 1 point, thus only smallest Japanese winning margins are of interest. Theoretically 4 cases occur: Each possible smallest Japanese winning margin combined with each possible colour for the last neutral point to be got. Details:
(1) Japanese B+0.5, B last neutral point:
With area rules this gives B+1.5. Hence B remains the winner.
(2) Japanese B+0.5, W last neutral point:
This is not possible, proof see inclusions.
(3) Japanese W+0.5, B last neutral point:
This is not possible, proof see inclusions.
(4) Japanese W+0.5, W last neutral point:
With area rules this would be W+1.5. However, this is not possible, proof see inclusions.
Hence the only possible case is (1). There the winner is the same. So indeed in the most frequently occuring parity case players used to Japanese scoring need not worry about last neutral points with area rules.
-:)]
Further reasons to use standard area komi only are contained in the inclusions below. Especially this means that 7.5 komi are much more preferable than 6.5.
--
robert jasiek
Subject: Re: Why 5.5 komi???
Date: Fri, 23 May 1997 15:31:59 +0200
From: Robert Jasiek <jasiek@berlin.snafu.de>
Organization: Unlimited Surprise Systems, Berlin
Newsgroups: rec.games.go
> As I know, in chinese counting system, giving 6 or 6.5 has the same effect as giving 5.5. Am I right?
> Xin Kang
No, you are wrong. Proof:
Let n be the number of board points. It is sufficient to consider n being odd. For even n things are inverted. For simplicity two representative types of area rule sets shall be considered: that of AGA 1991 and that of Ing 1991. Komi for AGA type use half points: k = 0.5, 1.5, 2.5,... Komi for Ing type use natural numbers with black winning ties: k = 0, 1, 2,... Let a be the area winning margin on the board for black. (Negative values are a win for white.) Let w be the winning margin for black. By definition: w = a - k.
It is necessary to consider cases depending on BoardSizeParity, SekiParity, Winner. The board can have an even or an odd number of empty grid points that are not monochromely surrounded. (They occur in odd "sekis".)
n odd, SekiParity even, black win:
----------------------------------
a = 1, 3, 5, 7,... is possible.
The interesting value for a is the smallest with a black win.
Then w is the smallest possible winning margin.
AGA: a = k + 0.5 + (k - 0.5) mod 2
Ing: a = k + (k + 1) mod 2
Example:
R k a w
AGA 5.5 7 1.5
AGA 6.5 7 0.5
Ing 6 7 1
Ing 7 7 0
n odd, SekiParity even, white win:
----------------------------------
a = 1, 3, 5, 7,... is possible.
The interesting value for a is the greatest with a black loss.
Then w is the smallest possible winning margin.
AGA: a = k - 0.5 - (k + 0.5) mod 2
Ing: a = k - 1 - k mod 2
Example:
R k a w
AGA 5.5 5 -0.5
AGA 6.5 5 -1.5
Ing 6 5 -1
Ing 7 5 -2
n odd, SekiParity odd, black win:
----------------------------------
a = 0, 2, 4,... is possible.
The interesting value for a is the smallest with a black win.
Then w is the smallest possible winning margin.
AGA: a = k + 0.5 + (k + 0.5) mod 2
Ing: a = k + k mod 2
Example:
R k a w
AGA 5.5 6 0.5
AGA 4.5 6 1.5
Ing 6 6 0
Ing 5 6 1
n odd, SekiParity odd, white win:
---------------------------------
a = 0, 2, 4,... is possible.
The interesting value for a is the greatest with a black loss.
Then w is the smallest possible winning margin.
AGA: a = k - 0.5 - (k - 0.5) mod 2
Ing: a = k - 1 - (k + 1) mod 2
Example:
R k a w
AGA 5.5 4 -1.5
AGA 4.5 4 -0.5
Ing 6 4 -2
Ing 5 4 -1
---------------------------------
Results:
- For each case and each area rule set two komi values are equivalent.
- Consistency independent of SekiParity is only given with standard komi s:
AGA: s(t) = 2t - 0.5; t positive natural numbers
Ing: s(t) = 2t; t natural numbers
- For AGA s = 1.5, 3.5, 5.5, 7.5, 9.5,..., for Ing s = 0, 2, 4,...
- Inconsistent komi change the smallest winning margins to a win of the other player in case of odd SekiParity.
- With area rules the nearest a values for constant SekiParity have a difference of 2. Thus with properly chosen k as to the rule set the value of a is constant for all k with |a-k| <= 2.
--robert jasiek
Subject:
Re: Use for dame
Date:
Sat, 21 Jun 1997 19:25:48 +0200
From:
Robert Jasiek <jasiek@berlin.snafu.de>
To:
ML go-rules <go-rules@lists.io.com>
> But alternate filling is starting to catch on in the domestic Japanese professional tournaments too. One reason may be the discovery that it provides a check on the result. With 5.5-point compensation, if the margin of victory is half a point, and if the neutral points have been filled alternately, then the winner is (almost always) the player who got the last neutral point.
Here a sketch for a proof is given.
def) Definitions:
n: # board points
k: komi
wt: winning margin for territory score
tB: # B points for score
tW: # W points for score
e: # empty board points
eB: # empty board points surrounded by B
eW: # empty board points surrounded by W
eN: # empty board points that are not surrounded
fB: # surrounded empty board points in B seki
fW: # surrounded empty board points in W seki
cB: # prisoners of B color
cW: # prisoners of W color
ms: # stones played during game
mp: # passes during game
sB: # B stones played during game
sW: # W stones played during game
pB: # B passes during game
pW: # W passes during game
D := eB - eW + cW - cB
Presuppositions:
0) Rules: Japanese 1989
1) n = 361 : ODD
2) k = 5.5
3) wt = 0.5
4) eN : EVEN at game end
5) fB = fW
6) pB = 1
7) pW = 1
Implications:
10) def => wt = | tB - tW |
11) def => tB = eB - fB + cW
12) def => tW = eW - fW + cB + k
13) def => e = eB + eW + eN
15) def => ms = sB + sW
16) def => mp = pB + pW
17) setup with empty board => eN : ODD at game start
18) (1) and setup with empty board => e = n = 361 : ODD at game start
19) (6), (7), (16) => mp = 1 + 1 = 2 : EVEN. This also means: no move after any pass.
Now we show:
A) D is even in case of a B win.
B) D is odd in case of a W win.
C) D is even in case of the last played stone being B.
D) D is odd in case of the last played stone being W.
Then it follows: A <=> C AND B <=> D.
The details:
(A)
30) (3), (10) => wt = 0.5 = | tB - tW |
31) (30) and B win => 0.5 = tB - tW
32) (30) and W win => 0.5 = tW - tB
33) (11), (12), (30) =>
0.5 = | tB - tW | = | eB - fB + cW - eW + fW - cB - k |
34) (5), (33) => 0.5 = | tB - tW | = | eB - eW + cW - cB - k |
35) (31), (34) => 0.5 = eB - eW + cW - cB - k in case of B win
36) def, (2), (35) => 6 = eB - eW + cW - cB = D is EVEN in
case of B win.
(B)
40) (32), (34) => 0.5 = eW - eB + cB - cW + k in case of W win
41) (2), (40) => -5 = eW - eB + cB - cW is ODD in case of W win.
42) (41) => 5 = eB - eW + cW - cB = D is ODD in case of W win.
(C)
Presupposition:
50) The last stone of the game is played by B.
Implications:
51) (50) => sB = sW + 1.
52) (15), (51) => ms = 2*sW + 1
53) (52) => ms is ODD.
Now the parity changes during the game are analysed:
60) (17), (18) => At the game start eB = eW = cB = cW = 0,
so D is EVEN. Besides e = eN = 361 are ODD.
61) After the first move eB = 360 is EVEN, eW = cB = cW = 0.
So D is still EVEN.
62) Changes due to a B move: placing the stone: e -> e - 1;
then if capture of i W stones: eB -> eB + i for the i board
points of capture AND cW -> cW + i.
63) Net effect of (62) capture: D -> D + 2*i keeps its parity.
64) Changes due to a W move: placing the stone: e -> e - 1;
then if capture of j B stones: eW -> eW + j for the j board
points of capture AND cB -> cB + j.
65) Net effect of (64) capture: D -> D + 2*j keeps its parity.
66) (62), (64) => each succession of a B and a W move changes
without consideration of captures e -> e - 2 keeping parity.
67) (53), (60), (66) => e EVEN at game end
68) (4), (13), (67) => eB, eW either both EVEN or both ODD
at game end
69) (68) => The parity of eB - eW is EVEN and independent of
eN value at game end.
70) (53), (61), (63), (65), (69) => D EVEN at game end.
71) Removals after the game end behave like captures during
the game and do not affect the parity of D.
(D)
Prove as in (C) but now
72) last stone by W => ms EVEN => e ODD at game end =>
eB - eW ODD at game end => D odd at game end
-:)]
Feel free to present a generalisation for aribitrary komi, scoring systems, board sizes, winning margins, SekiParities, single pass occurances, monochromely surrounded empty points in sekis, handicaps...
--robert jasiek
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