This is not a good game as it is full of errors. But I thought you may find it interesting because it contains a triple ko which was not resolved until the last move of the game, which resulted in a one point win.

_________________
Still officially AGA 5d but I play so irregularly these days that I am probably only 3d or 4d over the board (but hopefully still 5d in terms of knowledge, theory and the ability to contribute).

I believe, according to Japanese rules, Black is dead in the triple ko. In hypothetical play White can take one ko, putting Black in atari, then if Black takes a second ko, White can take the third ko, putting Black in atari again, and Black must pass, then White can take the Black group. If Black tries to kill White it is a double ko which White can always take back.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

I thought that Japanese rules did not recognize superko and that it would be a NR if neither side conceded it. Anyway, DGS certainly does not implement superko. I was counting very carefully at the end to make sure I would win if I gave up the ko.

_________________
Still officially AGA 5d but I play so irregularly these days that I am probably only 3d or 4d over the board (but hopefully still 5d in terms of knowledge, theory and the ability to contribute).

Why must Black pass instead of just taking the other ko?

Necessary during "Confirmation of Life & Death", according to Japanese Rules (1989).

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

The Japanese rules do not have a superko rule, but the ko rules are different in hypothetical play after play has stopped. During regular play White cannot capture Black, because of the triple ko, but Black cannot capture White, either, because of the double ko. In hypothetical play after play has stopped, Black cannot continue the triple ko because he has not passed for either ko that White has taken. When he does pass, he is already in atari and White can capture. In hypothetical play Black still cannot capture White because when Black takes one of the double kos, White takes the other one and Black cannot take back without passing first for that ko. Then White passes for the other ko and Black cannot win the double ko. Therefore Black is dead and White is alive.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Doesn't that simply mean that Black will not pass? Even if White passes, can't Black keep cycling?

_________________
Dave Sigaty
"Short-lived are both the praiser and the praised, and rememberer and the remembered..."
- Marcus Aurelius; Meditations, VIII 21

You mean in the double ko? Yeah, that may be a flaw in the Japanese 89 rules. But in any event Black cannot capture the White stones.

But you live in Japan. Check this kind of ko out.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

This all seems very obscure and abstruse. I think the elegance of NR is mch more appealing.

_________________
Still officially AGA 5d but I play so irregularly these days that I am probably only 3d or 4d over the board (but hopefully still 5d in terms of knowledge, theory and the ability to contribute).

Actually I think I am wrong. If White had passed at 240, Black can capture the ko at 238. But after White recaptures at 237, Black has no further move. There are a few 'ko' threats. However, sooner or later Black can only pass.

_________________
Dave Sigaty
"Short-lived are both the praiser and the praised, and rememberer and the remembered..."
- Marcus Aurelius; Meditations, VIII 21

Ok, lets bring this down to its core and give it a bit of analysis!

Here's the core of the situation:

Click Here To Show Diagram Code

[go]$$B
$$ . . O O O X X X . .
$$ . O O X X O O X X .
$$ O O X X . X O O X X
$$ O X . X X O O . O X
$$ O O X X O . O O O X
$$ -------------------[/go]

Is this a triple ko?

Well, there are three kos and all of them them are relevant to the life and death of this group, right? Maybe. But lets compare it to some other triple kos:

Here's a very simple traditional one. Two groups without eyes sharing three kos.

Click Here To Show Diagram Code

[go]$$W White's turn
$$ O O O X X X
$$ O X X O O X
$$ O X O . O X
$$ O X X O O X
$$ O X . X O X
$$ O X X O O X
$$ O X . X O X
$$ -----------[/go]

And here's another one, which consists of both a direct ko on the right and a double-ko seki in the corner providing "infinite ko-threats" for both players:

Click Here To Show Diagram Code

[go]$$B Black's turn.
$$ | . . . . X X X O O O .
$$ | X X X X X O O X X O O
$$ | O O O O X O . O X X O
$$ | . O . O X O O X . X O
$$ | O X O O X X O X X X O
$$ | X X X O O X O O O X O
$$ | . X . X O X O . O X O
$$ +----------------------[/go]

Now one thing that is different about these two, is that they are constantly urgent. Which is why I've noted whose turn it is. At any time in these triple kos, if one player plays elsewhere, the other can immediately resolve the ko.

The same is not true for the ko of the game. In the first diagram above, I have not marked whose turn it is, because it doesn't matter. Neither player, in the position presented, can resolve the ko. Either of them could play there and make the situation urgent, but after a few captures, the situation will get back to a stable configuration where playing is not urgent. That's why DrStraw was able to leave it until the end of the game and only then decide whether he could afford to lose those stones.

Now, let's say he couldn't. Suppose the result of the game hinged on the result of the ko. How would that play out?

First, lets consider Chinese style rules. Nothing fancy, just superko and area counting (CHinese, AGA, New Zealand or whatever)

Suppose black wants to win the ko. How would he go about that?

Click Here To Show Diagram Code

[go]$$B Only option for black
$$ . . O O O X X X . .
$$ . O O X X O O X X .
$$ O O X X 2 X O O X X
$$ O X . X X O O . O X
$$ O O X X O 1 O O O X
$$ -------------------[/go]

Black only has one possible move inside, and white only has one sane reply. But how now? After black has no move inside. So he has to play elsewhere (a.k.a. a ko-threat). Suppose white responds to the ko threat, then what? (that's and to keep count)

Click Here To Show Diagram Code

[go]$$B Only one option for black again
$$ . . O O O X X X . .
$$ . O O X X O O X X .
$$ O O X X O 5 O O X X
$$ O X . X X O O . O X
$$ O O X X 6 X O O O X
$$ -------------------[/go]

No other option than to play like this, right? But after black is again out of options.

This continues until black runs out of legal moves. At no point will white ever have to play a ko-threat, so at no point will white ever get in trouble locally for responding to black's moves.

Conclusion: Black cannot kill, not even with ko.

How about white?

Click Here To Show Diagram Code

[go]$$W at , at
$$ . . O O O X X X . .
$$ . O O X X O O X X .
$$ O O X X 1 B O O X X
$$ O # 3 X X O O . O X
$$ O O X X W 2 O O O X
$$ -------------------[/go]

After black would like to recapture at but can't because that would break superko (in triple ko, you repeat after 6 moves). So black has to find a ko-threat. Suppose he finds one that is big enough, and black responds? (that was and to keep count).

Click Here To Show Diagram Code

[go]$$W Return to original position
$$ . . O O O X X X . .
$$ . O O X X O O X X .
$$ O O X X . X O O X X
$$ O 8 . X X O O . O X
$$ O O X X O . O O O X
$$ -------------------[/go]

Now with we have returned to the original position. What can white do? Well it doesn't really matter. White can just play elsewhere, because as we concluded earlier black has no way to kill.

And after playing elsewhere, white has the option of starting up the exact same sequence above again, and again black will have to find a ko-threat. So white needs no ko-threat, while black needs an unlimited amount of them.

Effectively, under Chinese style rules black is dead.

How about Japanese rules? Well, now we're getting into more of a bit of "rules arcana"

First off, for the more obsure effects of Japanese rules to take effect, play has to stop first. That is important, and it is one of the reasons I pointed out the distinction between this position and the other triple ko positions earlier.

Traditional triple-ko positions are "No Result" because neither player can stop playing! You pass, you lose. The whole "determination of life and death" phase never comes into play at all.

That is not so in this case. This position has a stable configuration where both players can pass, and then they can determine life and death. At that point the outcome is no longer ever "No Result". It is one of "both alive", "black alive, white dead", "white alive, black dead" or "both dead", for the local position. After which we can take the dead stones off and count.

So, let's again see what happens.

So first a bit of context. Japanese life and death resolution is often described simplified as "there are no ko-threats after both players pass". That rules is achieved by saying: For any ko, after it has been captured, you may not retake it no matter what moves have been played after, except if you "pass for that ko", i.e. you say pass and point at the specific ko you now wish to be able to take back.

First, let's say black tries to prove white is dead.

Click Here To Show Diagram Code

[go]$$B Only option for black
$$ . . O O O X X X . .
$$ . O O X X O O X X .
$$ O O X X 2 X O O X X
$$ O X . X X O O . O X
$$ O O X X O 1 O O O X
$$ -------------------[/go]

After black has no move. Playing a ko-threat elsewhere on the board is not possible, it doesn't work. His only option is to pass and point at saying he wants the option to retake that ko. But white will pass and point at , and then after both players have retaken the ko they pointed at we're right back in the same stalemate. From black's point of view, this is a simple double-ko and he cannot kill white.

How about white?

Click Here To Show Diagram Code

[go]$$W
$$ . . O O O X X X . .
$$ . O O X X O O X X .
$$ O O X X 1 X O O X X
$$ O . 3 X X O O . O X
$$ O O X X O 2 O O O X
$$ -------------------[/go]

After there are no legal local moves left (remember, you cannot take any ko, no matter what has been played elsewhere, without first pasing and pointing!)

So this is simple. Black passes and points at and then white just captures everything because black is in atari.

So white can prove he can kill black under Japanese rules for determining life and death.

End result: Black is dead, white is alive.

So, to summarize: because the position has a stable configuration, it will not end in NR, as it can reach the "determination of life and death" phase, at which point Japanese rules result in black being dead.

BONUS:

A bit earlier, I mentioned four possible outcomes of life & death determination. Three of those "both alive" (seki), "black alive, white dead", "white alive, black dead" should be rather familiar to everyone here. But what about "both dead"? That seems counterintuitive, to say the least, right?

Well, that's because liofe & death determination is not actually about fights, it's about individual groups. So there are only two possible outcomes: "alive" or "dead". That's it.

So lets have another look at a position I posted earlier:

Click Here To Show Diagram Code

[go]$$W
$$ O O O X X X
$$ O X X O O X
$$ O X O . O X
$$ O X X O O X
$$ O X . X O X
$$ O X X O O X
$$ O X . X O X
$$ -----------[/go]

Simple right? Triple ko, no result. Sure, yes, except if one player decides to pass. A valid option. Perhaps, like DrStraw, he has counted that he can lose the triple ko and still win, so he doesn't want a "No Result".

So say white passes in this position.

Black, tired after the gruelling hour-long duel, thinks "Oh, right, it's over then" and also passes. Totally forgetting to take the stones. What now?

Well, now suddenly life and death determination rules can com into play!

Can black prove white is dead? Sure, he can take those stones with his first move.

How about white?

Click Here To Show Diagram Code

[go]$$W White's turn
$$ O O O X X X
$$ O X X O O X
$$ O X O 2 O X
$$ O X X O O X
$$ O X 1 X O X
$$ O X X O O X
$$ O X 3 X O X
$$ -----------[/go]

After there are no more legal moves (remember, no taking back kos) so black can do nothing and white will take the black stones off the board.

Both dead.

Very helpful post, HermanHiddema.


So, in the determination-of-life-and-death phase, both players can ask for a chance to go first? That seems odd, and the both-dead result looks unnatural. I thought only the player who didn't pass last can go first...

On a different note, I don't understand why some computer programs (like Zenith Go) use Japanese scoring when clearly they can handle none of this stuff!

No. Because it's all about "stones", not about "semeai".

Article 7.1 of the J1989 rules is: "Stones are said to be 'alive" if they cannot be captured by the opponent ..."

This implies that -- for determining the status of your stones -- your opponent plays first.
For determining the status of your opponent's stones, you have the first move.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

Maybe this is the time to ask this stupid question:
I never understood how this rule translates to the rule we know.

if there is a dead white shape with a black nakade then usually the nakade stones CAN be "caputed", but somehow it doesnt make a difference, because there in white territory anyway after taking black stones. How is that inside the rules?.

So lets say I have this position at game end:

Click Here To Show Diagram Code

[go]$$W
$$ . . . . . .
$$ . . . . . .
$$ . . X X X X
$$ . . X O O O
$$ . X . O . X
$$ . X X O X X
$$ . . X O X X
$$ -----------[/go]

and white for some reason claims my stones were dead, becuase he can capture it. What would be the right way to answer this by the rules. Do I need to say "yes they are, but so is your group?"

Well, as Cassandra already said, in this phase it is all about determining status of groups of stones. We're not playing anymore.

And yes, the end result certainly feels unnatural, but that's really just because the players passed in an unfinished position. Black should have captured.

Here's another, simpler, example of "both dead":

Click Here To Show Diagram Code

[go]$$B Both players passed
$$ . O O O X X . . .
$$ . O X X O X . . .
$$ . O X . O X . . .
$$ -----------------[/go]

Really either player should have played here, but if they both pass, and you follow the rules, both groups are dead.

Note that the Japanese rules allow either player to request a resumption of the game, but they must let their opponent play first if they do, so neither player would want to, in a position like this.



Because people want Japanese scoring, and this kind of stuff is rare enough that it can be safely ignored.

Sure, once every 10,000 games you'd have a rules "bug", but I'm pretty sure most people would be ecstatic if the programs they use daily only had a bug once every 10,000 times they used it

The "..." in my posting above -- referring to Article 7.1 -- reads:

", or if capturing them would enable a new stone to be played that the opponent could not capture."

Click Here To Show Diagram Code

[go]$$W
$$ . . . . . . |
$$ . . X X X X |
$$ . . X O O O |
$$ . X . O 1 X |
$$ . X X O X X |
$$ . . X O X X |
$$ -------------[/go]

Click Here To Show Diagram Code

[go]$$W
$$ . . . . . . |
$$ . . X X X X |
$$ . . X O O O |
$$ . X . O O . |
$$ . X X O . 2 |
$$ . . X O . . |
$$ -------------[/go]

Click Here To Show Diagram Code

[go]$$W
$$ . . . . . . |
$$ . . X X X X |
$$ . . X O O O |
$$ . X 4 O O 3 |
$$ . X X O 5 X |
$$ . . X O 6 . |
$$ -------------[/go]

This implies that Black's five stones in the corner are "alive".

As you will know for sure, White's group can be captured by Black, so her stones are "dead".

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

For details of the Japanese 1989 Rules, see
http://home.snafu.de/jasiek/j1989c.html

Good question!

The more complete rule is: Stones are dead if they can be captured, and the owner cannot then get living stones on the same intersections by continuing play.

A simple example is snapback:

Click Here To Show Diagram Code

[go]$$B
$$ X X X X
$$ X O O X
$$ O B . X
$$ O O X X[/go]

Even though white can capture the marked black stoned, this is ruled as: the white stones are dead, the marked black stone is not.

Why? Because black can get a living stone on the same intersection, which white cannot capture.

How?

Click Here To Show Diagram Code

[go]$$W
$$ X X X X | X X X X | X X X X | X X X X | X X X X |
$$ X O O X | X O O X | X . . X | X . . X | X . 4 X |
$$ O B . X | O . 1 X | O 2 . X | O X 3 X | O X . X |
$$ O O X X | O O X X | O O X X | O O X X | O O X X |[/go]

In the end, ended up on the same intersection as and could not be captured. Hence the original stone was alive.

The same is true for your example. Although white can capture the five black stones, black will ultimately capture white and can thus place living stones on (some of) the original five intersections. The same is not true for white. If white is captured here, any additional stones he triers to play inside will also be captured.

Yes, the both dead result is unnatural, IMO. But the Japanese 1989 rules have lasted for quite a while. {shrug}

However, there are two different points of view about scoring the final position. One is, as you say, that who has the move is important. That was the view of, among others, Honinbo Shusai and Go Seigen. (But at that time there were no passes, so in Herman's example if White did not play after Black took the ko, the White stones were dead.) Another view is that the score does not depend upon who has the move. That is the view of the Japanese 1949 rules and the Japanese 1989 rules, as well as the Ing rules. So under the Japanese rules to resolve a question of life and death after play has stopped it is not that each player has the chance to go first, it is that each player must go first. That is why the play is hypothetical; no stone is placed on the board.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

It is true that this is an example of both dead. However, it may also be an example of rule 13: Both players lose because they have left a gainful play on the board. AFAIK, that rule has never been invoked in professional play. But I doubt if it has ever been invoked in amateur play, where it could well occur.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.