No, because Black cannot capture anything locally.

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

Click Here To Show Diagram Code

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

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | X X 6 X O O . |
$$ | X O X X X O O |
$$ | O O O O X X O |
$$ | O O O 5 O X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | . 7 O X O O . |
$$ | . O X X X O O |
$$ | O O O O X X O |
$$ | O O O X 8 X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

Click Here To Show Diagram Code

[go]$$Bm9
$$ -----------------
$$ | 4 X 2 X O O . |
$$ | . O X X X O O |
$$ | O O O O X X O |
$$ | O O O . O X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

pass
pass

Click Here To Show Diagram Code

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

had been captured in the very beginning, but was reborn.

+ + + + +

Variation:

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[go]$$B
$$ -----------------
$$ | 4 X . X . O . |
$$ | X O X X X O O |
$$ | O O O O X X O |
$$ | O O O 3 O X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | O X 5 X 6 O . |
$$ | 7 O X X X O O |
$$ | O O O O X X O |
$$ | O O O X 8 X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

Click Here To Show Diagram Code

[go]$$Bm9
$$ -----------------
$$ | 2 X X X O O . |
$$ | X O X X X O O |
$$ | O O O O X X O |
$$ | O O O . O X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

pass

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

had been captured in the very beginning, but was reborn.

+ + + + +

Variation:

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | O X 6 X . O . |
$$ | . O X X X O O |
$$ | O O O O X X O |
$$ | O O O X . X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

pass

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | O 8 O X . O . |
$$ | . O X X X O O |
$$ | O O O O X X O |
$$ | O O O X . X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

pass

Click Here To Show Diagram Code

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

had been captured in the very beginning, but was reborn.

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

Post it and your analysis regardless of whether I have time!

OK Robert here it is:

Click Here To Show Diagram Code

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

First of all you can verify that in "normal play" the game is finished. Neither Black nor white want to add a single move.
If I analyse this position with J89 rule it appears that all strings of stones are alive and I am happy with this. I expected this result which seems the result for all (?) sekis without ko.

What about J2003? Here the things are different due to local-2 context.

Click Here To Show Diagram Code

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

All groups are uncapturable except OC the two small marked black groups.
Now is the point : the big black group being uncapturable, when considering the status of any of the small black groups, then the big black group acts as a barrier for local-2 consideration.
As a consequence it looks that the small black group on the right is alive (capturable-1) while the left one is dead.
I can help you to prove that but firstly do you agree to say (if is true OC!) that this status of the left black stones (dead stones) is not what we would expect?

Quickly reading status in my head, I think the left black string is J2003-alive but there are many variations and I am unsure so far.

Yes Robert there are many variations and for that reason I told you it is a beast. It is not obvious but I am quite sure left black string is really dead in J2003. If you do not want to find the variation by yourself I will help you.
BTW did you manage to prove that neither white nor black want to play a move in normal play? (=> the game stops in that position)

Good news Cassandra now I agree entirely with you
In the post https://lifein19x19.com/viewtopic.php?p=266502#p266502 you claimed it was a seki and you claimed the game was unfinished. That was a surprise for me because it looks against my logic of go. Maybe I was wrong but now I understand we have the same logic.

This is the key diagram.
In J89 confirmation phase the move is prohibited but is not quite logic is it?

That time we were talking about J89, not about our understanding of what logically closed, contradiction-free rules might be.

In J89, your example is a seki, so White would have had to add a move during "play" to avoid this result of "status confirmation". Thus, the position is unfinished under J89.

As Robert already mentioned, the pass-for-a-specific-ko rule is not really necessary even in J89 -- in most cases.
However, when utilising J89, it is urgently needed to achieve several of the intended results of the examples.
Apparently together with some additional, but hidden, rules, like "recapturing in a double-ko is forbidden".

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

Independent of what the result of a J2003 status assessment might be:

As I already mentioned earlier, a valid border of "local" needs the property "two-eyed".

The condition "uncapturable OR capturable-1" alone is NOT sufficient, as these properties can be also obtained by groups that are part of a seki.
But a seki is a compound of interconnected groups that must not be seperated logically.

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

Yes Cassandra that is exactly my point and I agree with you. In my example local-2 definition is too restrictive.

Click Here To Show Diagram Code

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

BTW, though the global ko-pass defined in J2003 is also a big progress, it is a pity to see that the status of white stones is still dead stones (=> seki => white should add a move)

Click Here To Show Diagram Code

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

Click Here To Show Diagram Code

[go]$$W
$$ -----------------
$$ | O . X . X 5 O |
$$ | O 3 X X X O O |
$$ | O O X O O O O |
$$ | X O X O O O O |
$$ | X O X O O O O |
$$ | X O X O 2 4 O |
$$ | . O X O 7 6 O |
$$ -----------------[/go]

Click Here To Show Diagram Code

[go]$$Bm8
$$ -----------------
$$ | O . X . X O O |
$$ | O O X X X O O |
$$ | O O X O O O O |
$$ | X O X O O O O |
$$ | X O X O O O O |
$$ | X O X O . 1 O |
$$ | 2 O X O O . O |
$$ -----------------[/go]

Click Here To Show Diagram Code

[go]$$Bm8
$$ -----------------
$$ | O . X . X O O |
$$ | O O X X X O O |
$$ | O O X O O O O |
$$ | . O X O O O O |
$$ | 4 O X O O O O |
$$ | . O X O 3 X O |
$$ | O O X O O 5 O |
$$ -----------------[/go]

Click Here To Show Diagram Code

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

had been captured, but was reborn
had been captured, and disappeared forever

? ? ?

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

almost perfect Cassandra.

Click Here To Show Diagram Code

[go]$$Bm8
$$ -----------------
$$ | O 6 X . X O O |
$$ | O O X X X O O |
$$ | O O X O O O O |
$$ | . O X O O O O |
$$ | 4 O X O O O O |
$$ | . O X O 3 X O |
$$ | O O X O O 5 O |
$$ -----------------[/go]

You must only add above because this point is part of the local-2 string of black left group!

After a minor modification...

Click Here To Show Diagram Code

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

... the result is a position that is structurally similar to Life-and-Death Example 4 and identical in terms of possible game outcomes or analysis results. Shouldn't similar considerations apply here as for Gérard's position? By this I mean especially the thought "the big black group acts as a barrier for local-2 consideration" (though I must admit I never fully understood J2003).

The answer to your question in No. By the way in the example 4 as it stands this barrier exists already and you do not need to change the position.
Now I will try to explain why this barrier has no effect in your example (or in example 4 because it is the same).
The barrier is made of black stones => The barrier can have an effect only when looking for the status of black strings of stones. Here is the point : with no barrier, an "apparently" dead group of black stones (like the left small black group in my example) can become an "alive" group because new black alive stones can appear elsewhere on the board. With a barrier it could be different : the "apparently" dead group may stay dead because the new black alive stones are on the other side of the barrier. This happen in my example where the "apparently" left black group stay dead because of the barrier.
In your example all the black groups are all alive. Because there are no "apparently" dead group, the existing barrier has no effect.

Oops I am not sure I have been very clear

I think it was Robert's intention to develop a logically closed and contradiction-free set of rules, which was able to achieve the desired results of the J89 life-and-death examples.
As far as I understand J2003, this particular ko-pass-rule is both appropriate and necessary to fulfill the above-mentioned task.

By the way:

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | P X . X . O . |
$$ | 1 O X X X O O |
$$ | O 2 O O X X O |
$$ | O O O . O X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

It does not matter that White would have to add a move to get rid of the seki in J89 status corfirmation.
During actual play, Black is able to capture with , forcing White to connect at , also gaining that specific point, due to the just won prisoner.
However, this raises the question, why Black should benefit in the status confirmation from having made a mistake during actual play?

+ + + + + + + + + +

By the way #2:
An insight that I have gained from working with various sets of (territorial) rules:

BEFORE designing such a rule set, you will have to decide about at least two decisive questions:




Do you want to behave a final position differently, depending on whether a specific DOUBLE-Ko is already part of the final position

Click Here To Show Diagram Code

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

or not, but arises during the course of the status confirmation?

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | O X . X . O . |
$$ | 1 O X X X O O |
$$ | O 2 O O X X O |
$$ | O O O . O X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

Do you want to behave a final position differently, whether it contains a solidly connected TRIPLE-Ko that is shared by only two groups

Click Here To Show Diagram Code

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

or not, but has the three ko-shapes farther away from each other?

Click Here To Show Diagram Code

[go]$$B
$$ +---------------------------------------+
$$ | 1 O O . O . O X 2 X O . O X . . . . . |
$$ | O X X O O O O X X O O O O X . . . . . |
$$ | X X . X X X O X O 3 O X X X . . . . . |
$$ | . . . , X O X X X O O X . . . , . . . |
$$ | . . . . X O X . X O X X . . . . . . . |
$$ | . . . . X O X X O X X . . . . . . . . |
$$ | . . . . X O O O O . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |[/go]

J89 answers both quesions with "Yes!", but I doubt that this is mandatory.

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

In what context are you Cassandra? In all rules I know, if the exchange occurs in normal play then all the black stones at the top are declared dead while all white stones are alive. I do not see when this exchange can give black an adding point: it is true that black gains a prinoner but at the same time white gets a point of territorry for the same intersection doesn't she? To be complete black stones becomes a prisoner but loses a point of territory.

J89, but apparently I was mistaken

+ + + + + + + + + + + + + +

Click Here To Show Diagram Code

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

If Black does nothing, White has to add a move during "play", in order to avoid "seki" by status confirmation.

White has ten points occupied with dead Black stones => 20 points.
White has six points of unoccupied territory => 6 points.

Black has two points of territory => 2 points.

=> White wins the game by 24 points.

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | O X . X . O . |
$$ | 1 O X X X O O |
$$ | O 2 O O X X O |
$$ | O O O . O X O |
$$ | X X O O X X O |
$$ | . X X O O O . |
$$ | X . X X X O O |
$$ -----------------[/go]

During "play", Black plays his kikashi in the upper left.

Click Here To Show Diagram Code

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

White has eleven points occupied with dead Black stones => 22 points.
White has six points of unoccupied territory => 6 points.

Black has two points of territory => 2 points.
Black got one prisoner => 1 point.

=> White wins the game by 25 points.

It seems that I did not visualise before that capturing one White stone empties the respective point, turning it into White territory.

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

Click Here To Show Diagram Code

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

OK Cassandra, I think we agree now that black, in normal play, cannot have any interest to take the ko in the corner instead of passing to stop the game.
It remains only the question : in normal play, should white add a move before passing?
In J89 and J2003 we proved white should add this move due to the you use of the pass-ko rule.
In https://lifein19x19.com/viewtopic.php?p=266571#p266571 you claimed that with your own understanding of japonese rule (I mean not the current J89 or J2003 as they are written) white should not add a move and I agree with you because in normal play black cannot save her stones even by playing first! For that reason I think the pass-ko rule, without minimizing the number of advantages it has, has though here a flaw which was certainly not expected.
Do you agree Cassandra?

Click Here To Show Diagram Code

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

OK Cassandra let's try to have a common understanding of tripel ko or double.

Let's begin by the solidly connected TRIPLE-Ko which seems the easiest one.

Click Here To Show Diagram Code

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

In J89 and J2003 (where I am talking about J89 or J203 I am refering strictly to the test and not to the examples because they could be not coherent with the text), my understanding is the following:

In J89 the marked stones are dead because they can be captured and no new stone are enable to be played that the opponent could not capture. All other stones are alive => the left part of the board is seki

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | . Z Z Z P P P |
$$ | O O Z Z P . . |
$$ | O X . Z P . . |
$$ | O O Z Z P . . |
$$ | O . O Z P . . |
$$ | O O Z Z P P P |
$$ | O . O Z P . . |
$$ -----------------[/go]

In J2008 the three single stones are also dead due to the double barrier with black and white marked stones. All other stones are alive => the left part of the board is seki

Is seki the expected result? I do not know but the losing player made a mistake by passing. She would have continue the game by playing the triple ko to reach a NO RESULT game. IOW the result of the confirmation phase is not relevant. In any case the potential losing player can always continue the game to reach a NO RESULT.

In principle, yes.

However, in my opinion the only "advantage" of such a special-ko-ban rule is to ensure the intended results of several of the life-and-death examples of J89. Otherwise, it would not be needed, in order to create a self-contained, contradiction free rule set.
I consider it extremely likely that for EVERY ruleset "beast" could be created, which status-confirmation result "common sense" did not expect.

But let's return to J89.
I suppose that the inventors of the special-ko-ban rule wanted to avoid "no result" in several TRIPLE-KO cases.
But it seems that they overlooked the effects of their rule on "NESTED" ko shapes (as in the board position you created). The status-assessment of the position reached after two moves (Black captured in the upper left corner, White connected the resulting atari below) -- if THIS position was the final position of the game -- would match "common sense".
As you corrected me, Black's initional move would COST him one point -- if played BEFORE the game stopped.
So, what it the reasoning that he shall GAIN one point without capturing during play -- through status-confirmation?

+ + + + + + + + + + + + + + + +

By the way:

The life-and-death examples in J89 -- treating a compound of bent-four and double-ko -- do NOT cover ALL possible variations of shared double-ko formations. J89 suppresses an example like the following one -- with NESTED ko shapes.

Click Here To Show Diagram Code

[go]$$
$$ +----------------------
$$ | . O . X O X O . . . .
$$ | O X X X O X O O . . .
$$ | . X O O O X X O O . .
$$ | X X O O . O X X O , .
$$ | O O O . O X . X O . .
$$ | X X O O X . X X O . .
$$ | . X X O O X X O O . .
$$ | . . X X X O O O . . .
$$ | . . . . X . . . . . .
$$ | . . . , . . . . . , .
$$ | . . . . . . . . . . .[/go]

To be honest, I have forgotten ("misplaced") the results of my J89-application trials with it. But for sure, at that time I did not dream of some restricting side conditions that prevent any triple-ko cycles.

However, I know that a status-assessment without any special ko-ban-rule will give interesting results (as well).

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