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 Post subject: 77th Honinbo Game 1 puzzle
Post #1 Posted: Mon Jun 06, 2022 4:46 am 
Oza

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I don't actually know the rules of go. For example, I know I don't know what the Japanese rule on bent four is, but I do know there is a rule. But I've just found out that there is a rule about boundary-play kos (yose-ko) that not only do I not know but I didn't know existed. And it came up, and apparently mattered, in Game 1 of the 77th Honinbo.

This is all part of the messy aspects of go rulesets that I put in a carboard box and hide in the closet under the stairs. However, there were also some related aspects of procedural rules and go theory that were of some interest to me and, to understand those, ideally I should know the rules of go. But rather than succumb to such electric-shock therapy, I'd like to ask the rules mavens to explain it all to me (in English, not Rulish). In return, they may find the procedural parts are interestingly new to them.

A reminder: Iyama won by 0.5 point after 357 official moves, a record length for Honinbo title matches. Ichiriki could have played on with some more futile ko threats, to extend the moves record, but he apparently preferred not to sully the game record. The video does not show the final stages clearly, so it is hard to be certain what happened then (and it also seems there is a web version of the game where 356 differs from what was played in the game according to the Nihon Ki-in paper record).

First, when Iyama played 357, Takao made a remark expressing surprise that the game finished with Iyama making the last move (and winning). Apparently there is a "well-known fact" that I did not know, to the effect that "in a half-point game the side that fills in the last dame loses." I assume that dame here is being used somewhat loosely (as is common) for the last necessary move. Although the official game record is definitely being quoted by the Nihon Ki-in as 357 moves, there are 7 more dames to fill in for the actual counting to start. That would make 364 moves, meaning White has filled the last dame (in the other sense).

We do know that these extra dames were filled in, however, because when it came to move 362, Ichiriki played the last White stone in his bowl. Iyama played 363 and then Ichiriki put his hand in his empty bowl and wiggled it around. Which Iyama took to mean he was passing. Technically, that meant Iyama had illegally made two moves in succession, which is grounds for a loss. He got away with it here, because the board and stones had been supplied by the local organisers and had not been checked. Apparently - a trivia fact I did not know - the official number of stones is 181 Black stones and 180 White.

There's a bit of gristle in that fillet, but the meaty part for me is the "last dame player loses" heuristic. Does it have any sort of theoretical justification?

But the chocolate souffle for dessert is the piece de resistance. There is apparently a Japanese rule that, sometimes at least, a player does not have to play a repair move in a yose-ko left over at the end of the game. In other words, he gets an extra point profit. And in this case a win! What's that all about?

And why is it being left to me to bring these points up? Shouldn't rules mavens be maving more proactively? :)


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 Post subject: Re: 77th Honinbo Game 1 puzzle
Post #2 Posted: Mon Jun 06, 2022 6:16 am 
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John Fairbairn wrote:
The video does not show the final stages clearly, so it is hard to be certain what happened then (and it also seems there is a web version of the game where 356 differs from what was played in the game according to the Nihon Ki-in paper record).
That's true. At least the Go4Go record is wrong and so is whatever server they presumably took it from. Maybe "incomplete" is more like it. I checked the video stream and made a diagram below. It's worth noting that the video of the 2nd Day was beyond YouTube's 12 hour limit and so there is a 1 and a half hour supplement (linked to the time of the incident): https://youtu.be/HhXcNpcTaXc?t=3300
John Fairbairn wrote:
We do know that these extra dames were filled in, however, because when it came to move 362, Ichiriki played the last White stone in his bowl. Iyama played 363 and then Ichiriki put his hand in his empty bowl and wiggled it around. Which Iyama took to mean he was passing. Technically, that meant Iyama had illegally made two moves in succession, which is grounds for a loss.
Well, you can watch the video linked above at 55:08 where Ichiriki plays 362 and then holds up his bowl to show it's empty as Iyama is playing 363. Who knows what was said but Iyama holds up a black stone for 365, Ichiriki nods agreement, Iyama looks to the Witness and Recorders for approval then to Ichiriki, who nods again, then Iyama double checks with the officials and Witness Hane Naoki nods agreement. Only then does Iyama place the 2nd black stone. So no worries.

Attachment:
Game1.PNG
Game1.PNG [ 552.81 KiB | Viewed 1391 times ]


Correct diagram (I think but it's early) from the video: Moves 351-365, W364 passes. Well, I'm just assuming that White passed. It's possible that Iyama asked Hane if they could pretend he was Ichiriki holding a White stone and they all agreed.
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------------------------
$$ | . . . . . 4 O 0 X X O . O X . X O . O -. . . . . O O O X X O . O X . X O . O |
$$ | . O X . O O X X 9 O 8 O O O X X O O O -. O X . O O X X X O O O O O X X O O O |
$$ | . O X O O X X X O O O X O O X X . X O -. O X O O X X X O O O X O O X X . X O |
$$ | . O X O O O X O O X X X X O O X X X X -. O X O O O X O O X X X X O O X X X X |
$$ | . O X X O O X X X X X X O O . O O O O -. O X X O O X X X X X X O O . O O O O |
$$ | . O . X O O O X X O O X X O O . O X 6 -. O . X O O O X X O O X X O O . O . O |
$$ | O O O O X O X X . X O O X X O O X O O -O O O O X O X X . X O O X X O O X O O |
$$ | 3 O X X X O O X X O O . X O O X X X . -X O X X X O O X X O O . X O O X X X 2 |
$$ | 5 X . X X O O O X X O O X X X . . X X -X X . X X O O O X X O O X X X . . X X |
$$ | X . X X O X . . O X O . O X X X X O X -X . X X O X . . O X O . O X X X X O X |
$$ | X X X X O O O O O O X 7 O O O X O O O -X X X X O O O O O O X X O O O X O O O |
$$ | X O O O X 2 . O X O X . . X X O O O . -X O O O X O 5 O X O X . . X X O O O . |
$$ | O O X X X O X X X O X . . X O O . O O -O O X X X O X X X O X . . X O O . O O |
$$ | O . O O X 1 O X X O X . X X X O O X O -O . O O X X . X X O X . X X X O O X O |
$$ | . O O X . X O X O X X . X O X . O X X -. O O X . X . X O X X . X O X 1 O X X |
$$ | . . O X X X X O O X . X O O O O O X . -. . O X X X X O O X . X O O O O O X . |
$$ | . X O O O X O . O O X X X O X X X X X -. X O O O X O . O O X X X O X X X X X |
$$ | . O O . O X O O O X X . X O X X X O . -. O O 3 O X O O O X X . X O X X X O . |
$$ | . O X X X X X O . O X X X O O X O O . -. O X X X X X O . O X X X O O X O O . |
$$ ---------------------------------------[/go]

Incorrect online SGF:
Click Here To Show Diagram Code
[go]$$W Moves 350 to 357
$$ ---------------------------------------
$$ | . . . . . 5 O . X X O . O X . X O . O |
$$ | . O X . O O X X . O 7 O O O X X O O O |
$$ | . O X O O X X X O O O X O O X X . X O |
$$ | . O X O O O X O O X X X X O O X X X X |
$$ | . O X X O O X X X X X X O O . O O O O |
$$ | . O . X O O O X X O O X X O O . O X . |
$$ | O O O O X O X X . X O O X X O O X O O |
$$ | 4 O X X X O O X X O O . X O O X X X . |
$$ | 6 X . X X O O O X X O O X X X . . X X |
$$ | X . X X O X . . O X O . O X X X X O X |
$$ | X X X X O O O O O O X 8 O O O X O O O |
$$ | X O O O X 3 . O X O X . . X X O O O . |
$$ | O O X X X 1 X X X O X . . X O O . O O |
$$ | O . O O X 2 O X X O X . X X X O O X O |
$$ | . O O X . X O X O X X . X O X . O X X |
$$ | . . O X X X X O O X . X O O O O O X . |
$$ | . X O O O X O . O O X X X O X X X X X |
$$ | . O O . O X O O O X X . X O X X X O . |
$$ | . O X X X X X O . O X X X O O X O O . |
$$ ---------------------------------------[/go]

John Fairbairn wrote:
There's a bit of gristle in that fillet, but the meaty part for me is the "last dame player loses" heuristic. Does it have any sort of theoretical justification?
I don't know. I haven't heard of this before. Maybe it is simply a consequence of having an odd number of intersections on the board and a half-point yose ko.
John Fairbairn wrote:
But the chocolate souffle for dessert is the piece de resistance. There is apparently a Japanese rule that, sometimes at least, a player does not have to play a repair move in a yose-ko left over at the end of the game. In other words, he gets an extra point profit. And in this case a win! What's that all about?
If it was an unfilled yose-ko it would have dame in the empty intersection, not an extra point of territory. Under the rules, that dame would cause the living stones nearby to be seki. So if the stones are supposed to be seki then it doesn't need to be filled (shouldn't be filled in cases I can think of) since it doesn't matter.

Update: I found an example from the Nihon Kiin rules. This diagram is "the same" (同) and the preceding diagram is "seki diagram" (セキの図)
Attachment:
Ko.PNG
Ko.PNG [ 7.38 KiB | Viewed 1383 times ]

Seems like the case here except this is a collapse.
Click Here To Show Diagram Code
[go]$$
$$ -----
$$ . O X T X O . O |
$$ O O O X X O O O |
$$ X O O X X . X O |
$$ X X O O X X X X |
$$ X O O . O O O O |[/go]


John Fairbairn wrote:
And why is it being left to me to bring these points up? Shouldn't rules mavens be maving more proactively? :)
Well... you know which rules are read and which aren't.


Last edited by CDavis7M on Mon Jun 06, 2022 7:17 am, edited 2 times in total.
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 Post subject: Re: 77th Honinbo Game 1 puzzle
Post #3 Posted: Mon Jun 06, 2022 7:06 am 
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Consider the case of 6.5 komi (I think that's what is currrently used, but I could be wrong).

Suppose black wins by 0.5 points, and white's territory was T, with white having taken P prisoners and black 0 prisoners. (One can assume 0 black prisoners without loss of generality: black and white return an equal number prisoners until the black total is zero, so P may be negative.) Then black must have T+P+7 points.

So the total territory is 2*T+7+P.

Now consider the number of stones on the board after the last dame point is filled, assuming no seki. Suppose there are B black stones. Then there are B+P white stones if white played the last dame; or B+P-1 if black played it.

Under the no seki assumption, the total number of stones on the board plus territory must equal 361.

Hence the total territory plus number of stones is 2*T+7+2*P+2*B if white played last, which is odd. So this could equal 361, and is possible.
If black played last, the total is 2*T+6+2*P+2*B, which is even, which is impossible as 361 is odd.

So with no seki and 6.5 komi, the heuristic is solid. With 5.5 or 7.5 komi, it is reversed.

Edit: The B+P or B+P-1 depends on there having been no passing, so if someone passes in a half point ko, this could fail. I guess if there are dame left, they could fill a dame and then it would still work. It also depends on the players alternating in the dame filling - I don't know if this usually done in Japanese professional games or not.


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 Post subject: Re: 77th Honinbo Game 1 puzzle
Post #4 Posted: Mon Jun 06, 2022 8:32 am 
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About the ko - I don't understand exactly what ko is being referred to. It looks like all the kos on the board involve groups that are dead (and so there's no need to play them - and the side that owns the territory doesn't want to play them). What am I missing?

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Post #5 Posted: Mon Jun 06, 2022 12:07 pm 
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rooklift wrote:
About the ko - I don't understand exactly what ko is being referred to. It looks like all the kos on the board involve groups that are dead (and so there's no need to play them - and the side that owns the territory doesn't want to play them). What am I missing?
I wondered too. I went back and skimmed the live commentary of Hane and Takao and I couldn't find the discussion about losing a point in scoring. But they were discussing a variation in the upper-right ko battle using a "horikomi" throw-in (different from the throw-in on the edge of the board). That horikomi variation led to a direct "hon ko" instead of the approach "yose ko" in the game, which loses a point since the yose ko eventually needs to be filled in (losing a point). But even if there were a different double ko elsewhere on the board (not in this game), the yose ko should be filled in and lose a point (otherwise it's dead because of the "pass" for retaking a ko rationale).

Maybe between these two issues (hon ko vs yose ko; yose ko filling for life) that there was some rules discussion regarding Ichiriki's loss by half a point. It's also possible that Takao made a comment/variation on Yugen no Ma which is no longer available on the server. I Googled 第1局 第77期本因坊戦 "ヨセコウ" and had no luck.

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Post #6 Posted: Mon Jun 06, 2022 5:40 pm 
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John Fairbairn wrote:
There's a bit of gristle in that fillet, but the meaty part for me is the "last dame player loses" heuristic. Does it have any sort of theoretical justification?


It is almost elementary but if territory and stones played are to sum to an odd number, then assuming black plays one more or same number of moves as white and all dame is filled it has to be the case that the territory difference is an even number when black played last and an odd number when white played last. To see this for the case of 361 intersections on a Go board and jigo or B+1 result let N be the number of white moves and T the size of white's territory and see that 2 * N + 2 * T + 1 = 361 and that this is really only a matter of if the "+ 1" is a black move or a black territory because black either played N + 1 moves for T points of territory or N moves for T + 1 points of territory. You can adjust accordingly for handicap, komi and seki.

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Post #7 Posted: Tue Jun 07, 2022 2:52 am 
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Thanks for the explanations of what I called the heuristic, but that sort of thing makes me need to lie down in a darkened room. It's like the gardening. I know I CAN do it, but the thought of it appalls me, and so I prefer to switch the brain off. But why didn't it work here? Why was Takao surprised?

As to the "what am I missing" comments, I don't really know, and that's why I raised the query in the first place. But there is something. It is clearly not the case that all yose-kos are filled in at the end. The following sentence appeared in one of the Japanese commentaries I read: 一手ヨセコウ残りは日本囲碁規約のルール上、手入れを省いて1目得できる場合があります。 That is, "under the Japanese rules of go, there are cases where a fill-in move in a one-move boundary-play ko left over at the end can be omitted with 1 point profit." (And presumably 2-movers with 2 points profit, etc?)

Yes, but which cases? As far as I can see, in the case of this game it was to do with the total lack of ko threats for White (which Iyama had cleverly engineered?).

And is the pointed reference to Japanese rules significant? Would it have been an Iyama win under Chinese or Korean or Taiwanese or AGA rules, for instance?

In any event, it was an incredibly complex game and a great advertisement both for human play and the two-day system.

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Post #8 Posted: Tue Jun 07, 2022 3:14 am 
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My understanding was that this is was explicitly was denied in the (first? written?) Nihon Kiin rules in the 1930's and 1940's, there being two disputes involving Go Seigen around this (the first establishing that (mis?) understanding, the second whether the Nihon Kiin rules applied in that match). This was then also a guiding principle for the 1989 rules.

My japanese is not as firm as yours, but maybe there is a bit of ambiguity with »上«, which as far as I know also can mean something like »before«, so the comment would refer to rules »before« the Nihon Kiin rules, which would conform with my understanding.

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Post #9 Posted: Tue Jun 07, 2022 6:31 am 
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John Fairbairn wrote:
The following sentence appeared in one of the Japanese commentaries I read: 一手ヨセコウ残りは日本囲碁規約のルール上、手入れを省いて1目得できる場合があります。 That is, "under the Japanese rules of go, there are cases where a fill-in move in a one-move boundary-play ko left over at the end can be omitted with 1 point profit." (And presumably 2-movers with 2 points profit, etc?)
My first time seeing half-sized kana "in the wild." Books don't seem to use it. Maybe a newspaper thing?

Anyway, there are a few examples in the Japanese rules to show when teire is required for life and when it is not required. Teire is required for a 本劫 hon-ko but not a one-step ko 一手ヨセ劫. The simple explanation is that the extra teire move is not required when the opponent's stones are dead. But with a direct ko the status-assessment flips and the "winning" side is actually dead because of the pass-to-retake rationale. One player might think they are "winning" the ko because they have other ko threats on the board but that doesn't work for life and death assessment. This same ruling holds for more complex situations, like a series of kos leading towards a bent-4 in the corner. Bent-4 is deemed dead so that's the end of that. Pretty mundane unfortunately.
Attachment:
itte yose ko.PNG
itte yose ko.PNG [ 58.08 KiB | Viewed 1197 times ]

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Post #10 Posted: Tue Jun 07, 2022 8:02 am 
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It is not clear what the confusion is. If it is about the eye vs. no-eye capturing race then rest assured that black need not fight over the ko.

For example...

Click Here To Show Diagram Code
[go]$$W White can never capture.
$$ ---------------------------------------
$$ |. . . . . O O O X X O . O X . X O . O |
$$ |. O X . O O X X X O O O O O X X O O O |
$$ |. O X O O X X X O O O X O O X X . X O |
$$ |. O X O O O X O O X X X X O O X X X X |
$$ |. O X X O O X X X X X X O O . O O O O |
$$ |. O . X O O O X X O O X X O O . O . O |
$$ |O O O O X O X X . X O O X X O O X O O |
$$ |X O X X X O O X X O O . X O O X X X O |
$$ |X X . X X O O O X X O O X X X 1 2 X X |
$$ |X . X X O X . . O X O 3 O X X X X O X |
$$ |X X X X O O O O O O X X O O O X O O O |
$$ |X O O O X O X O X O X . 4 X X O O O . |
$$ |O O X X X O X X X O X . . X O O . O O |
$$ |O . O O X X . X X O X . X X X O O X O |
$$ |. O O X . X . X O X X . X O X X O X X |
$$ |. . O X X X X O O X . X O O O O O X . |
$$ |. X O O O X O . O O X X X O X X X X X |
$$ |. O O X O X O O O X X . X O X X X O . |
$$ |. O X X X X X O . O X X X O O X O O . |
$$ ---------------------------------------[/go]


or

Click Here To Show Diagram Code
[go]$$W Taking the ko doesn't chance that white can't capture.
$$ ---------------------------------------
$$ |. . . . . O O O X X O . O X . X O . O |
$$ |. O X . O O X X X O O O O O X X O O O |
$$ |. O X O O X X X O O O X O O X X . X O |
$$ |. O X O O O X O O X X X X O O X X X X |
$$ |. O X X O O X X X X X X O O . O O O O |
$$ |. O . X O O O X X O O X X O O . O . O |
$$ |O O O O X O X X 3 X O O X X O O X O O |
$$ |X O X X X O O X X O O . X O O X X X O |
$$ |X X . X X O O O X X O O X X X 1 2 X X |
$$ |X . X X O X . . O X O 5 O X X X X O X |
$$ |X X X X O O O O O O X X O O O X O O O |
$$ |X O O O X O X O X O X . 4 X X O O O . |
$$ |O O X X X O X X X O X . . X O O . O O |
$$ |O . O O X X . X X O X . X X X O O X O |
$$ |. O O X . X . X O X X . X O X X O X X |
$$ |. . O X X X X O O X . X O O O O O X . |
$$ |. X O O O X O . O O X X X O X X X X X |
$$ |. O O X O X O O O X X . X O X X X O . |
$$ |. O X X X X X O . O X X X O O X O O . |
$$ ---------------------------------------[/go]


On the other hand if black didn't removed a liberty before black would no longer win the eye vs. no eye without defending in the ko.

Click Here To Show Diagram Code
[go]$$W Black in atari.
$$ ---------------------------------------
$$ |. . . . . O O O X X O . O X . X O . O |
$$ |. O X . O O X X X O O O O O X X O O O |
$$ |. O X O O X X X O O O X O O X X . X O |
$$ |. O X O O O X O O X X X X O O X X X X |
$$ |. O X X O O X X X X X X O O . O O O O |
$$ |. O . X O O O X X O O X X O O . O . O |
$$ |O O O O X O X X 3 X O O X X O O X O O |
$$ |X O X X X O O X X O O 5 X O O X X X O |
$$ |X X . X X O O O X X O O X X X 1 2 X X |
$$ |X . X X O X . . O X O . O X X X X O X |
$$ |X X X X O O O O O O X 4 O O O X O O O |
$$ |X O O O X O X O X O X . . X X O O O . |
$$ |O O X X X O X X X O X . . X O O . O O |
$$ |O . O O X X . X X O X . X X X O O X O |
$$ |. O O X . X . X O X X . X O X X O X X |
$$ |. . O X X X X O O X . X O O O O O X . |
$$ |. X O O O X O . O O X X X O X X X X X |
$$ |. O O X O X O O O X X . X O X X X O . |
$$ |. O X X X X X O . O X X X O O X O O . |
$$ ---------------------------------------[/go]


Instead of removing a liberty directly during the game it was also possible to just connect the ko.

Click Here To Show Diagram Code
[go]$$W Black is just in time in the eye vs. no eye.
$$ ---------------------------------------
$$ |. . . . . O O O X X O . O X . X O . O |
$$ |. O X . O O X X X O O O O O X X O O O |
$$ |. O X O O X X X O O O X O O X X . X O |
$$ |. O X O O O X O O X X X X O O X X X X |
$$ |. O X X O O X X X X X X O O . O O O O |
$$ |. O . X O O O X X O O X X O O . O . O |
$$ |O O O O X O X X B X O O X X O O X O O |
$$ |X O X X X O O X X O O . X O O X X X O |
$$ |X X . X X O O O X X O O X X X 1 2 X X |
$$ |X . X X O X . . O X O 5 O X X X X O X |
$$ |X X X X O O O O O O X . O O O X O O O |
$$ |X O O O X O X O X O X 6 3 X X O O O . |
$$ |O O X X X O X X X O X . 4 X O O . O O |
$$ |O . O O X X . X X O X . X X X O O X O |
$$ |. O O X . X . X O X X . X O X X O X X |
$$ |. . O X X X X O O X . X O O O O O X . |
$$ |. X O O O X O . O O X X X O X X X X X |
$$ |. O O X O X O O O X X . X O X X X O . |
$$ |. O X X X X X O . O X X X O O X O O . |
$$ ---------------------------------------[/go]


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Post #11 Posted: Tue Jun 07, 2022 11:26 pm 
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John Fairbairn wrote:
Apparently there is a "well-known fact" that I did not know, to the effect that "in a half-point game the side that fills in the last dame loses." I assume that dame here is being used somewhat loosely (as is common) for the last necessary move.


See https://www.lifein19x19.com/viewtopic.p ... 13#p273713 for history and my mathematical theorem.


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 Post subject: Re: 77th Honinbo Game 1 puzzle
Post #12 Posted: Fri Jun 10, 2022 6:04 am 
Judan

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drmwc wrote:
Consider the case of 6.5 komi (I think that's what is currrently used, but I could be wrong).

Suppose black wins by 0.5 points, and white's territory was T, with white having taken P prisoners and black 0 prisoners. (One can assume 0 black prisoners without loss of generality: black and white return an equal number prisoners until the black total is zero, so P may be negative.) Then black must have T+P+7 points.

So the total territory is 2*T+7+P.

Now consider the number of stones on the board after the last dame point is filled, assuming no seki. Suppose there are B black stones. Then there are B+P white stones if white played the last dame; or B+P-1 if black played it.

Under the no seki assumption, the total number of stones on the board plus territory must equal 361.

Hence the total territory plus number of stones is 2*T+7+2*P+2*B if white played last, which is odd. So this could equal 361, and is possible.
If black played last, the total is 2*T+6+2*P+2*B, which is even, which is impossible as 361 is odd.

So with no seki and 6.5 komi, the heuristic is solid. With 5.5 or 7.5 komi, it is reversed.

Edit: The B+P or B+P-1 depends on there having been no passing, so if someone passes in a half point ko, this could fail. I guess if there are dame left, they could fill a dame and then it would still work. It also depends on the players alternating in the dame filling - I don't know if this usually done in Japanese professional games or not.


For people not having read my theorems and proofs of 1997 - 2021, you make a plausibility argument but do not prove. You speak of the heuristic but, since my first proof in 1997, it is much more than that: it is a theorem and truth.

You are right that changing the komi parity reverses the winner.

You make or rely on these assumptions:
- 361 intersections,
- 6.5 komi,
- Black wins by 0.5,
- a moment after the last dame is filled,
- the last play is the last dame,
- no seki,
- so far no passing,
- a single pass might occur due to a play in a basic endgame ko (your phrase "half point ko" adds unnecessary ambiguity),
- alternate dame filling.

These your assumptions are unnecessarily restrictive:
- The last play is the last dame: the last play might also be another play if only then there are no more two-sided dame and teire.
- A moment after the last dame is filled: this is related to your restrictive "no seki" assumption so that then you need not distinguish two-sided dame.
- So far no passing: the game ending successive two passes can be allowed.
- A single pass might occur due to a play in a basic endgame ko: this is related to your restrictive "no seki" assumption so that you do not consider single passes due to one-sided dames.
- Alternate dame filling: not the alternation of dame filling is essential but the alternation of moves and (according to research so far) no single passes.

In my theorem 144, these assumptions are more general:
- odd parity of the number of intersections (an even board parity would reverse the winner),
- any komi with a 0.5 fraction,
- the territory score is 0.5 in favour of Black or White,
- final position with an even seki parity (an odd seki parity would reverse the winner) but I have not avoided the less restrictive assumptions of 'without capturable stones in sekis' and 'without asymmetric sekis' (in which the players have different numbers of eye points),
- no single passes,
- territory or area scoring are allowed.

My assumptions include while your assumptions overlook:
- no teire,
- no handicap stones,
- unequivocality of territory scoring,
- no suicide (allowed suicide requires future research).

Methodically, you presume a final position just after the last play as given but what is the difference between that final position and the preceding last play and inhowfar is it relevant that that play is the last one? My proofs start at the game start from the empty position, proceed move by move and thereby reach that last play. Thereby, the last play becomes meaningful.

You are right to study the cases of Black or White making the last play. This case study is needed for territory scoring, but furthermore is needed for the relation between area score and territory score and for verifying the same winner in 0.5 points (territory scored) games.

While I understand your basic reasoning, I do not understand every detail of your annotation. With White's territory T, why is Black's territory also T?! If White has taken P prisoners, then why does Black have P points for the dead black prisoner stones?! Where have you specified that dead stones have already been removed and added to the prisoners?! At the very least, you need a territory difference and a prisoner difference, but these are not magically even numbers by themselves; it is not the number 2 of players that would make them even. So please explain each '2' you insert in your terms! Rather I think that parity must be transformed from the beginning of the game to the moment after the last play (or move), as I have done in my proofs. That is part of the reason why they do not fill just one line but many pages.

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 Post subject: Re: 77th Honinbo Game 1 puzzle
Post #13 Posted: Sun Jun 12, 2022 2:45 am 
Judan

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drmwc wrote:
Consider the case of 6.5 komi


Except for missing or wrong assumptions, let us study the mistakes you make.

For the discussed game, the komi is right though.

Quote:
Suppose black wins by 0.5 points,


This is one of the cases to be studied.

Mistake 1: You do not discuss the other most interesting case that White wins by 0.5.

Quote:
and white's territory was T,


It is possible to call it T.

Mistake 2: You do not specify whether there may still be dead stones in White's territory or whether you they have already become prisoners.

Quote:
with white having taken P prisoners and black 0 prisoners. (One can assume 0 black prisoners without loss of generality: black and white return an equal number prisoners until the black total is zero, so P may be negative.)


This construction is possible but I am not delighted as it creates a necessity for additional explanation of how to deal with this "without loss of generality" in practice.

Quote:
Then black must have T+P+7 points.


Note: Your 7 seems to be the sum of 6.5 komi wrongly accounted for Black instead of White plus 0.5 points of Black's winning margin. Compare Mistake 5.

Mistake 3: If White, from White's perspective, has T points (of empty intersections?) of territory, then, in general, Black, from Black's perspective, need not also have T points of territory. Instead, White can have Tw and Black can have Tb points. Furthermore, note that Tw or Tb might be odd numbers so that a sum or difference of Tw and Tb need not be an even number.

Mistake 4: Previously, you have set P as White's prisoners, from White's perspective, and, WLOG, 0 as Black's prisoners. Therefore, you may not now change Black's prisoners from 0 to P, from Black's perspective.

Mistake 5: The komi 6.5 favours White. If you account the komi for Black, you may therefore not add 6.5, but you must subtract 6.5.

Quote:
So the total territory is 2*T+7+P.


Note: You do not mean territory but the territory score.

Mistake 6: You apply previous values despite your previous mistakes.

Mistake 7: When considering both players' territories, we must assume either player's value perspective and form a difference of his territory minus the opponent's territory, such as Tw - Tb from White's perspective. However, your 2T = T + T forms a sum instead of a difference. Therefore, the total territory score does not contain a 2T. It would only make sense to use a sum when considering both players' total number of territory intersectiions. This you do not do; you refer to the territory score, i.e., an expression that might contain a difference such as Tw - Tb.

Mistake 8: In genral, 2T is wrong because the players need not have the same amount of territory or same numbers of territory intersections.

Quote:
Now consider the number of stones on the board after the last dame point is filled,


Note: that the last play occupies the last dame is a consequence of your too restricted related assumption.

Quote:
assuming no seki.


For the basic case, this assumption may be made. However, more cases need to be considered because seki is not that rare.

Quote:
Suppose there are B black stones.


Note: you mean black stones on the board. Of course, the value may be called B.

Quote:
Then there are B+P white stones if white played the last dame; or B+P-1 if black played it.


Mistake 9: The live (non-seki) white stones on the board do not magically disappear. They are also not accounted in P because live stones on the board do not become prisoners. The without loss of generality construction does not let them disappear, either, because it only accounts prisoners but does not account live stones. You do also need the number, say, W of life white stones on the board, or consider the difference B - W or, opposite perspective, W - B or define B (then better called S) to be such a difference.

Note: Due to your incomplete or wrong accounting of live stones on the board, I cannot judge yet how many, if any, additional mistakes your stone numbers of the last play cases might contain.

Quote:
Under the no seki assumption, the total number of stones on the board plus territory must equal 361.


Note: This presumes correcting your assumptions, such as all teire already being filled.

Quote:
Hence the total territory plus number of stones is 2*T+7+2*P+2*B if white played last,


Notes: Due to your mistakes, I cannot be sure what you mean by "total territory" here. I also cannot be sure whether the addition of 2P and the addition of 2B are the right operations or either should be subtraction. Besides 2P or 2B might have to be subtituted by differences, see above.

Mistake 10: You started with White's prisoners P, then made the mistake to derive Black's prisoners P. Your new mistake is to use this sum P + P = 2P in your term.

Mistake 11: Likewise, your new mistake is to use the sum B + B = 2B in your term, firstly because the players might have different numbers of live stones on the board and secondly because it must be clarified whether a difference W - B or B - W needs to be used here.

Quote:
which is odd.


Mistake 12: Given your earlier mistakes, this conclusion is premature.

Quote:
So this could equal 361, and is possible.


Note: Since now you compare your term to 361 (the number of board intersections), it remains necessary to convey the meaning of "the total territory plus number of stones", which also includes the '7'.

Quote:
If black played last, the total is 2*T+6+2*P+2*B, which is even, which is impossible as 361 is odd.


Mistake 13: as before.

Note: as before, but now you account 6 = 7 - 1, where the -1 deserves explanation.

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 Post subject: Re: 77th Honinbo Game 1 puzzle
Post #14 Posted: Fri Jun 17, 2022 4:26 am 
Lives in gote
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There are many mistakes in your post. I am not sure I can be bothered to respond to them all.

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Post #15 Posted: Fri Jun 17, 2022 4:58 am 
Judan

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If you think to have made fewer mistakes than indicated, express your statements more clearly so that there is less, or ideally no, ambiguity.

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