How do Japanese rules handle this?

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Re: How do Japanese rules handle this?

Post by Cassandra »

CDavis7M wrote:
Cassandra wrote:
CDavis7M wrote:* snip *
Do you actually care what a "definition" is?
Yes. Did you get a copy of the Advanced Learners dictionary yet?
* snip *
Example for a definition:
String = a set of stones of the same colour, which are solidly connected to each other along the lines of the board.

Example for a result of the application of a given procedure:
This string is considered not fulfilling the definition of "alive".
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Re: How do Japanese rules handle this?

Post by CDavis7M »

Cassandra wrote:String = a set of stones of the same colour, which are solidly connected to each other along the lines of the board.
I don't know why you are defining words when we are talking about definitions of things other than words (ie board positions). A bit irrelevant but OK.
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Re: How do Japanese rules handle this?

Post by Cassandra »

CDavis7M wrote:
Cassandra wrote:String = a set of stones of the same colour, which are solidly connected to each other along the lines of the board.
I don't know why you are defining words when we are talking about definitions of things other than words (ie board positions). A bit irrelevant but OK.
In my Go world, "string" is a (part of a) "board position". Not in yours???

And I don't think that you want to TALK. You want to SPEAK, letting others know YOUR opinion.
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Re: How do Japanese rules handle this?

Post by CDavis7M »

Cassandra wrote:And I don't think that you want to TALK. You want to SPEAK, letting others know YOUR opinion.
It's not my opinion. It's Kano Yoshinori's opinion. It's Kudo Norio's opinion. And the opinions of Sakai Takeshi, Oeda Yusuke, Saijo Masataka, Goro Fujita, Shinoda Shigehito, and Shigerushi Hata.

They wrote that the Revision provides definitions. Not me. I just read it.
They never wrote anything to suggest that so-called hypothetical play must be used to determine L&D status. I recognize that fact.
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Re: How do Japanese rules handle this?

Post by CDavis7M »

Gérard TAILLE wrote:
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X . . X X X |
$$ | X X O O O O O |
$$ | . O X X X X X |
$$ | . O X Q Q Q X |
$$ | X O X Q Q . X |
$$ | X O X Q Q Q X |
$$ | X O X Q Q Q . |
$$ +---------------+[/go]
What is the status of the white marked stones?

CDavies7M, I did not see your interpretation concerning this position above. I believe you will conclude that the marked white stones are dead but I am not quite sure.
Could you show us your analysis?
I'll rely on Fairbairn's translation.
Stones which cannot be captured by moves of the opponent, or stones which, even if they can be captured, can newly form stones that cannot be captured by the opponent, are called 'live stones.' Stones other than live stones are called 'dead stones.'
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X 1 2 X X X -. X X O . 3 5 -. X X O 7 X X -. X X . X X X |
$$ | X X @ O O O O -X X @ O O O O -X X @ O O O O -X X . . . . . |
$$ | . O X X X X X -. O X X X X X -. O X X X X X -. O X X X X X |
$$ | . O X Q Q Q X -4 O X O O O X -O O X O O O X -O O X O O O X |
$$ | X O X Q Q . X -X O X O O . X -. O X O O . X -. O X O O B X |
$$ | X O X Q Q Q X -X O X O O O X -6 O X O O O X -O O X O O O X |
$$ | X O X Q Q Q . -X O X O O O . -. O X O O O . -. O X O O O B |
$$ +---------------+[/go]
The 5 White stones on the upper side are alive because even though they can be captured by Black, they can newly form stones :w4: and :w6: that cannot be captured by Black. Moves like :w4: are not uncapturable unless a move like :b1: reduces liberties in capturing. The 5 stones are capturable and it is those capturable stones that are the ones that can newly form stones that cannot be captured by the opponent because it is those stones ( :ws: ) that reduce Black's liberties such that the uncapturable stones can be formed. And without :ws: , Black can just connect.

Unlike other people, I'm not going to pretend that :w4: and :w6: used to show that the 5 stones are alive can somehow be used to prove that the different and separate group of 11 White stones alive. Those 11 stones do not contribute anything to the formation of the new uncapturable stones. The uncapturable stones can be formed without those stones being captured at all.

----------

I haven't spent much time looking at the other seki collapse position that was presented, but I believe I have a similar position there (also against the position of others).
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Re: How do Japanese rules handle this?

Post by Cassandra »

CDavis7M wrote:Unlike other people, I'm not going to pretend that :w4: and :w6: used to show that the 5 stones are alive can somehow be used to prove that the different and separate group of 11 White stones alive. Those 11 stones do not contribute anything to the formation of the new uncapturable stones. The uncapturable stones can be formed without those stones being captured at all.
Although I am a fan of "THEREAFTER" (rebirth is only possible AFTER death, not before), I also sympathise with jann's interpretation.
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | ? ? ? ? ? ? ? |
$$ | ? ? ? ? ? ? ? |
$$ | ? ? X X X X X |
$$ | ? ? X O O O X |
$$ | ? ? X O O . X |
$$ | ? ? X O O O X |
$$ | ? ? X O O O . |
$$ +---------------+[/go]
Neither does Black control the lower right part of the board (any move there is suicidal; consequently, he cannot remove White's stones), ...
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X . . X X X |
$$ | X X O O O O O |
$$ | . O X X X X X |
$$ | . O X ? ? ? ? |
$$ | X O X ? ? ? ? |
$$ | X O X ? ? ? ? |
$$ | X O X ? ? ? ? |
$$ +---------------+[/go]
... nor does he control the upper left one (as we know, he cannot remove ALL of White's stones).

White does not control these areas, too, thus the entire board is a large seki.
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X X . . X X |
$$ | X X O O O O O |
$$ | X O Y Y Y Y Y |
$$ | . O Y O O O Y |
$$ | . O Y O O . Y |
$$ | X O Y O O O Y |
$$ | X O Y O O O . |
$$ +---------------+[/go]
As it seems, the L&D-status of White's stones inside a temporary seki at one side of Black's :bt: string depends on the L&D-status of her stones outside that temporary seki on the other side of Black's :bt: string.
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Re: How do Japanese rules handle this?

Post by Gérard TAILLE »

CDavis7M wrote:
Gérard TAILLE wrote:
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X . . X X X |
$$ | X X O O O O O |
$$ | . O X X X X X |
$$ | . O X Q Q Q X |
$$ | X O X Q Q . X |
$$ | X O X Q Q Q X |
$$ | X O X Q Q Q . |
$$ +---------------+[/go]
What is the status of the white marked stones?

CDavies7M, I did not see your interpretation concerning this position above. I believe you will conclude that the marked white stones are dead but I am not quite sure.
Could you show us your analysis?
I'll rely on Fairbairn's translation.
Stones which cannot be captured by moves of the opponent, or stones which, even if they can be captured, can newly form stones that cannot be captured by the opponent, are called 'live stones.' Stones other than live stones are called 'dead stones.'
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X 1 2 X X X -. X X O . 3 5 -. X X O 7 X X -. X X . X X X |
$$ | X X @ O O O O -X X @ O O O O -X X @ O O O O -X X . . . . . |
$$ | . O X X X X X -. O X X X X X -. O X X X X X -. O X X X X X |
$$ | . O X Q Q Q X -4 O X O O O X -O O X O O O X -O O X O O O X |
$$ | X O X Q Q . X -X O X O O . X -. O X O O . X -. O X O O B X |
$$ | X O X Q Q Q X -X O X O O O X -6 O X O O O X -O O X O O O X |
$$ | X O X Q Q Q . -X O X O O O . -. O X O O O . -. O X O O O B |
$$ +---------------+[/go]
The 5 White stones on the upper side are alive because even though they can be captured by Black, they can newly form stones :w4: and :w6: that cannot be captured by Black. Moves like :w4: are not uncapturable unless a move like :b1: reduces liberties in capturing. The 5 stones are capturable and it is those capturable stones that are the ones that can newly form stones that cannot be captured by the opponent because it is those stones ( :ws: ) that reduce Black's liberties such that the uncapturable stones can be formed. And without :ws: , Black can just connect.

Unlike other people, I'm not going to pretend that :w4: and :w6: used to show that the 5 stones are alive can somehow be used to prove that the different and separate group of 11 White stones alive. Those 11 stones do not contribute anything to the formation of the new uncapturable stones. The uncapturable stones can be formed without those stones being captured at all.

----------

I haven't spent much time looking at the other seki collapse position that was presented, but I believe I have a similar position there (also against the position of others).
Seeing your argument "The uncapturable stones can be formed without those stones being captured at all" I can propose another example for which this argument does not exist:
Click Here To Show Diagram Code
[go]$$B
$$ +-------------------+
$$ | X X X . O . X . X |
$$ | O O O O O X X X X |
$$ | X X X X X Q Q Q . |
$$ | . . . . X Q . Q Q |
$$ | X X X X X Q Q Q . |
$$ | O O O O O X X X X |
$$ | X X X . O . X . X |
$$ +-------------------+[/go]
In this position black can choose to capture either the white stones in the upper left corner or the white stones in the bottom left corner (like in the previous position) but here (unlike the previous position) black must always BEGIN by capturing the white marked stones.
Do you conclude to the same conclusion i.e. white marked stone are dead (though you cannot capture them without allowing white to play new uncapturable stones) ?
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Re: How do Japanese rules handle this?

Post by CDavis7M »

Gérard TAILLE wrote:
CDavis7M wrote:The 5 White stones on the upper side are alive because even though they can be captured by Black, they can newly form stones :w4: and :w6: that cannot be captured by Black. Moves like :w4: are not uncapturable unless a move like :b1: reduces liberties in capturing. The 5 stones are capturable and it is those capturable stones that are the ones that can newly form stones that cannot be captured by the opponent because it is those stones ( :ws: ) that reduce Black's liberties such that the uncapturable stones can be formed.
Seeing your argument "The uncapturable stones can be formed without those stones being captured at all" I can propose another example for which this argument does not exist...
Note my point that I underlined above.

Let's try to capture the 9 White stones. Indeed there is a new uncapturable stone :ws: . But does a new uncapturable stone mean anything by itself? I don't think so because in an actual game-play position, new uncapturable stones can always be played somewhere on the board (filling dame, territory, etc).
Click Here To Show Diagram Code
[go]$$B
$$ +-------------------+
$$ | X X X . O . X . X-X X X . O . X . X -X X X . O . X . X
$$ | O O O O O X X X X-O O O O O X X X X-O O O O O X X X X
$$ | X X X X X Q Q Q .-X X X X X Q Q Q 3-X X X X X . . . X
$$ | . . . . X Q . Q Q-. . . . X Q 5 Q Q-. . . . X . X . .
$$ | X X X X X Q Q Q 1-X X X X X Q Q Q X-X X X X X . . . X
$$ | O O O O O X X X X-O O O O O X X X X-O O O O O X X X X
$$ | X X X 2 O . X . X-. 4 . O O . X . X-. @ . O O . X . X
$$ +-------------------+[/go]
Backing up, the real question is whether it is these 9 White stones that can newly form stones that cannot be captured by the opponent -- is there some game-mechanic in the Game of Go that gives those 9 stones the ability to form stones that cannot be captured by the opponent. Or is it actually some other stones that already have that ability give the game mechanics?

Let's see what happens when Black tries to prevent the formation of the new uncapturable stones with :b3: . If Black tries, :wc: will reduce his liberties of :bc: and White can capture with :w6: . So yes, those 9 White stones are living stones because they contribute to the formation of the new uncapturable stones.
Click Here To Show Diagram Code
[go]$$B
$$ +-------------------+
$$ | X X X . O . X . X-X X X . O . X . X-X X X . O . X . X
$$ | O O O O O X X X X-O O O O O X X X X-O O O O O X X X X
$$ | X X X X X O O O .-X X X X X O O O .-X X X X X O O O .
$$ | . . . . X O . O O-. . . . X O . O W-. . . . X O . O O
$$ | X X X X X O O O 1-X X X X X W W W B-X X X X X O O O .
$$ | O O O O O X X X X-O O O O O B B B X-O O O O O . . . .
$$ | X X X 2 O . X . X-. 3 5 O O 4 X 6 X-. X X O O O . O .
$$ +-------------------+[/go]

This is easy to confirm. Consider if :wc: did not have the ability to reduce the liberty of :bc: . In this case, :w6: does not capture Black's stones because :bc: would have a liberty were it not for :wc: . :w6: cannot even be played, letting :b7: capture White, preventing the formation of new uncapturable stones.
Click Here To Show Diagram Code
[go]$$B
$$ +-------------------+
$$ | X X X . O . X . X-X X X . O . X . X-X X X . O . X . X
$$ | O O O O O X X X X-O O O O O X X X X-O O O O O X X X X
$$ | X X X X X O O O .-X X X X X O O O .-X X X X X O O O .
$$ | . . . . X O . O O-. . . . X O . O O-. . . . X O . O O
$$ | X X X X X W O O 1-X X X X X W O O X-X X X X X W O O X
$$ | O O O O O B X X X-O O O O O B X X X-. . . . . B X X X
$$ | X X X 2 O . X . X-7 3 5 O O 4 X 6 X-X X X . . . X . X
$$ +-------------------+[/go]
The 9 stones are live. This is similar to Example 4.

----------

Here we can see that :wt: have no ability to form new uncapturable stones. There's no game mechanic that let's them contribute to the formation. It is the other White stones that already have that ability. They are already alive when considered separately.
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X . . X X X |
$$ | X X O O O O O |
$$ | . O X X X X X |
$$ | . O X Q Q Q X |
$$ | X O X Q Q . X |
$$ | X O X Q Q Q X |
$$ | X O X Q Q Q . |
$$ +---------------+[/go]
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Re: How do Japanese rules handle this?

Post by CDavis7M »

We can do the same confirmation with the other position. Is it the 5 White stones that can newly form stones that cannot be captured by the opponent? Or is it the 1 White stone that can do that.

What if Black connects the ko in an attempt to prevent White from forming an uncapturable stone by retaking the ko. Normally this just ends in Black being captured.
Click Here To Show Diagram Code
[go]$$ :w4: pass, :b7: pass
$$ ----------------------
$$ | O . O . X 2 3 O O . O-O . O 8 X 6 X O O . O-O . O O . O . O O . O
$$ | O O O X O X O O O O O-O O O X 5 X O O O O O-O O O . . . O O O O O
$$ | X X X O 1 O O . O . .-X X X O X O O . O . .-X X X O . O O . O . .
$$ | X . X O O O O O . . .-X . X O O O O O . . .-X . X O O O O O . . .
$$ | . X X . . . . . . . .-. X X . . . . . . . .-. X X . . . . . . . .
$$ | X X . . . . . . . . .-X X . . . . . . . . .-X X . . . . . . . . .
$$ | . . . . . . . . . . .-. . . . . . . . . . .-. . . . . . . . . . .
$$ | . . . . . . . . . . .-. . . . . . . . . . .-. . . . . . . . . . .[/go]
But what if the stones :wt: do not contribute to reducing the liberties of :bt: ? Then :b7: is not suicide and :b9: can capture. But that is not the point. The question is: can Black can prevent :w8: from being uncapturable even if :wt: do not reduce liberties? No (this should be the case because their L&D status is independent). Even in a situation where :wt: contribute nothing, White can still form an uncapturable stone :w8: even in the situation where Black connects. Because White can form uncapturable stones even if the 5 White stone's contribute nothing, then it is not the 5 White stones in the corner that can newly form stones that cannot be captured by the opponent. Those 5 White stones are dead.
Click Here To Show Diagram Code
[go]$$ :w4: pass
$$ ----------------------
$$ | Q . Q . Y 2 3 O O . O-Q . Q 7 Y 6 X O O . O-. 9 . X Y O 8 O O . O
$$ | Q Q Q Y O Y O O O O O-Q Q Q Y 5 Y O O O O O-. . . Y X Y O O O O O
$$ | X X X O 1 O O . O . .-X X X O X O O . O . .-X X X O X O O . O . .
$$ | X . X O O O O O . . .-X . X O O O O O . . .-X . X O O O O O . . .
$$ | . X X . . . . . . . .-. X X . . . . . . . .-. X X . . . . . . . .
$$ | X X . . . . . . . . .-X X . . . . . . . . .-X X . . . . . . . . .
$$ | . . . . . . . . . . .-. . . . . . . . . . .-. . . . . . . . . . .
$$ | . . . . . . . . . . .-. . . . . . . . . . .-. . . . . . . . . . .[/go]
----------

Contrast this to the other position, where Black actually can prevent uncapturable stones if the marked stones did not contribute to liberty reduction. This different shows which stones actually can newly form stones that cannot be captured by the opponent and which cannot.
CDavis7M wrote:Consider if :wc: did not have the ability to reduce the liberty of :bc: . In this case, :w6: does not capture Black's stones because :bc: would have a liberty were it not for :wc: . :w6: cannot even be played, letting :b7: capture White, preventing the formation of new uncapturable stones.
Click Here To Show Diagram Code
[go]$$B
$$ +-------------------+
$$ | X X X . O . X . X-X X X . O . X . X-X X X . O . X . X
$$ | O O O O O X X X X-O O O O O X X X X-O O O O O X X X X
$$ | X X X X X O O O .-X X X X X O O O .-X X X X X O O O .
$$ | . . . . X O . O O-. . . . X O . O O-. . . . X O . O O
$$ | X X X X X W O O 1-X X X X X W O O X-X X X X X W O O X
$$ | O O O O O B X X X-O O O O O B X X X-. . . . . B X X X
$$ | X X X 2 O . X . X-7 3 5 O O 4 X 6 X-X X X . . . X . X
$$ +-------------------+[/go]
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Re: How do Japanese rules handle this?

Post by Gérard TAILLE »

CDavis7M wrote:
Gérard TAILLE wrote:
CDavis7M wrote:The 5 White stones on the upper side are alive because even though they can be captured by Black, they can newly form stones :w4: and :w6: that cannot be captured by Black. Moves like :w4: are not uncapturable unless a move like :b1: reduces liberties in capturing. The 5 stones are capturable and it is those capturable stones that are the ones that can newly form stones that cannot be captured by the opponent because it is those stones ( :ws: ) that reduce Black's liberties such that the uncapturable stones can be formed.
Seeing your argument "The uncapturable stones can be formed without those stones being captured at all" I can propose another example for which this argument does not exist...
Note my point that I underlined above.

Let's try to capture the 9 White stones. Indeed there is a new uncapturable stone :ws: . But does a new uncapturable stone mean anything by itself? I don't think so because in an actual game-play position, new uncapturable stones can always be played somewhere on the board (filling dame, territory, etc).
Click Here To Show Diagram Code
[go]$$B
$$ +-------------------+
$$ | X X X . O . X . X-X X X . O . X . X -X X X . O . X . X
$$ | O O O O O X X X X-O O O O O X X X X-O O O O O X X X X
$$ | X X X X X Q Q Q .-X X X X X Q Q Q 3-X X X X X . . . X
$$ | . . . . X Q . Q Q-. . . . X Q 5 Q Q-. . . . X . X . .
$$ | X X X X X Q Q Q 1-X X X X X Q Q Q X-X X X X X . . . X
$$ | O O O O O X X X X-O O O O O X X X X-O O O O O X X X X
$$ | X X X 2 O . X . X-. 4 . O O . X . X-. @ . O O . X . X
$$ +-------------------+[/go]
Backing up, the real question is whether it is these 9 White stones that can newly form stones that cannot be captured by the opponent -- is there some game-mechanic in the Game of Go that gives those 9 stones the ability to form stones that cannot be captured by the opponent. Or is it actually some other stones that already have that ability give the game mechanics?

Let's see what happens when Black tries to prevent the formation of the new uncapturable stones with :b3: . If Black tries, :wc: will reduce his liberties of :bc: and White can capture with :w6: . So yes, those 9 White stones are living stones because they contribute to the formation of the new uncapturable stones.
Click Here To Show Diagram Code
[go]$$B
$$ +-------------------+
$$ | X X X . O . X . X-X X X . O . X . X-X X X . O . X . X
$$ | O O O O O X X X X-O O O O O X X X X-O O O O O X X X X
$$ | X X X X X O O O .-X X X X X O O O .-X X X X X O O O .
$$ | . . . . X O . O O-. . . . X O . O W-. . . . X O . O O
$$ | X X X X X O O O 1-X X X X X W W W B-X X X X X O O O .
$$ | O O O O O X X X X-O O O O O B B B X-O O O O O . . . .
$$ | X X X 2 O . X . X-. 3 5 O O 4 X 6 X-. X X O O O . O .
$$ +-------------------+[/go]

This is easy to confirm. Consider if :wc: did not have the ability to reduce the liberty of :bc: . In this case, :w6: does not capture Black's stones because :bc: would have a liberty were it not for :wc: . :w6: cannot even be played, letting :b7: capture White, preventing the formation of new uncapturable stones.
Click Here To Show Diagram Code
[go]$$B
$$ +-------------------+
$$ | X X X . O . X . X-X X X . O . X . X-X X X . O . X . X
$$ | O O O O O X X X X-O O O O O X X X X-O O O O O X X X X
$$ | X X X X X O O O .-X X X X X O O O .-X X X X X O O O .
$$ | . . . . X O . O O-. . . . X O . O O-. . . . X O . O O
$$ | X X X X X W O O 1-X X X X X W O O X-X X X X X W O O X
$$ | O O O O O B X X X-O O O O O B X X X-. . . . . B X X X
$$ | X X X 2 O . X . X-7 3 5 O O 4 X 6 X-X X X . . . X . X
$$ +-------------------+[/go]
The 9 stones are live. This is similar to Example 4.

----------

Here we can see that :wt: have no ability to form new uncapturable stones. There's no game mechanic that let's them contribute to the formation. It is the other White stones that already have that ability. They are already alive when considered separately.
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X . . X X X |
$$ | X X O O O O O |
$$ | . O X X X X X |
$$ | . O X Q Q Q X |
$$ | X O X Q Q . X |
$$ | X O X Q Q Q X |
$$ | X O X Q Q Q . |
$$ +---------------+[/go]

Click Here To Show Diagram Code
[go]$$B Diag 1
$$ +---------------+
$$ | . X . . X X X |
$$ | X X O O O O O |
$$ | . O X X X X X |
$$ | . O X Q Q Q X |
$$ | X O X Q Q . X |
$$ | X O X Q Q Q X |
$$ | X O X Q Q Q . |
$$ +---------------+[/go]
Click Here To Show Diagram Code
[go]$$B Diag 2
$$ +-------------------+
$$ | X X X . O . X . X |
$$ | O O O O O X X X X |
$$ | X X X X X Q Q Q . |
$$ | . . . . X Q . Q Q |
$$ | X X X X X Q Q Q . |
$$ | O O O O O X X X X |
$$ | X X X . O . X . X |
$$ +-------------------+[/go]
Let's compare theses two positions.
In Diag1 you consider that the marked stones are dead because they have "no ability to form new uncapturable stones".
But in Diag2 you consider that the marked stones are alive because they have this ability to form new uncapturable stones.
This result is strange for me and I have to work harder in order to really understand.

My view is the following : for a go player looking for the status of a group, the sequence itself is not relevant. What is relevant is the final position reached after the chosen sequence. In this FINAL position the player can look at the captured stones as well as the new uncapturable stones created and decide whether a group of stones is alive or not.

Let's call A and B the two unmarked white groups of stones in Diag 1 or 2 and let's call C the white marked group in Diag 1 or 2
In both diagram:
1) the three A, B, C groups are capturable
2) the two A, B groups cannot be simultaneously captured
3) Black can capture A and C but in this case uncapturable stones are created towards B
4) Black can capture B and C but in this case uncapturable stones are created towards A
What difference you see between these two diagrams for a player looking at the status of these groups?

BTW what is the status of the two unmarked white groups in Diag 2 ?
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Re: How do Japanese rules handle this?

Post by jann »

Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X . . X X X |
$$ | X X O O O O O |
$$ | . O X X X X X |
$$ | . O X Q Q Q X |
$$ | X O X Q Q . X |
$$ | X O X Q Q Q X |
$$ | X O X Q Q Q . |
$$ +---------------+[/go]
Click Here To Show Diagram Code
[go]$$
$$ ---------------------------------
$$ | . O X X X O O O O X O O X Q Q |
$$ | X X O O X X X . O X . O X . Q |
$$ | X X O . O O X . O X . O X . Q |
$$ | O O O O . O X X O X X O X X Q |
$$ ---------------------------------[/go]
IMO there is zero doubt the bottom/right is also alive both cases. This is not like lightvector's position or torazu3 where there are at least alternatives (in lightvector's case the current ruling is only caused by careless wording, referring to enabled STONES instead of control as mentioned above). Similar variant for example 2:
Click Here To Show Diagram Code
[go]$$B example 2 variant - no prisoners, komi -2
$$ +-----------------------------------+
$$ | X X X X X X X . X O O . O X X O O |
$$ | X X O X X O X . X . O O O O X . O |
$$ | X X O O O O O X X X X X X O X . O |
$$ | X X O . O . O O O O O O O X X O O |
$$ +-----------------------------------+[/go]
Again the entire right is alive (both W strings). And B won't even dare to capture in real game - this is permanent seki. J89 (and even J2003) have no problem here: attempting to capture either W string would enable or give rise to new uncapturable stones (compared to the original position OC, for these reasons). It doesn't matter if he captures in one move or in two parts (nor have the rule any additional condition on new stones).

And this is not even specific to J89 or the enable rule: example 2 was considered alive and seki even in J1949 (and maybe even before that). J89 changed L/D for a few special cases, but for the most part it only introduced a logical explanation. Alive includes uncapturable stones, capturable but replayable stones, and stones that are not really capturable only exchangeable. Exchanges need to be played out, scoring cannot grant them. Especially cannot score a string as dead while the other side of the exchange (the B string that would be lost) as alive.
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Re: How do Japanese rules handle this?

Post by Gérard TAILLE »

jann wrote:
Click Here To Show Diagram Code
[go]$$B
$$ +---------------+
$$ | . X . . X X X |
$$ | X X O O O O O |
$$ | . O X X X X X |
$$ | . O X Q Q Q X |
$$ | X O X Q Q . X |
$$ | X O X Q Q Q X |
$$ | X O X Q Q Q . |
$$ +---------------+[/go]
Click Here To Show Diagram Code
[go]$$
$$ ---------------------------------
$$ | . O X X X O O O O X O O X Q Q |
$$ | X X O O X X X . O X . O X . Q |
$$ | X X O . O O X . O X . O X . Q |
$$ | O O O O . O X X O X X O X X Q |
$$ ---------------------------------[/go]
IMO there is zero doubt that the bottom/right is also alive in both cases. This is not like lightvector's position or torazu3 where there are at least alternatives (in lightvector's case the current ruling is only caused by careless wording, referring to enabled STONES instead of control as mentioned above). A similar variant can be made for example 2:
Click Here To Show Diagram Code
[go]$$B example 2 variant - no prisoners, komi -2
$$ +-----------------------------------+
$$ | X X X X X X X . X O O . O X X O O |
$$ | X X O X X O X . X . O O O O X . O |
$$ | X X O O O O O X X X X X X O X . O |
$$ | X X O . O . O O O O O O O X X O O |
$$ +-----------------------------------+[/go]
Again the entire right is alive (both W strings). And B won't even dare to capture in real game - this is permanent seki. J89 (and even J2003) has no problems here: attempting to capture either W string would enable or give rise to new stones (compared to the original position OC, for these reasons). It doesn't matter if he captures in one move or in two parts (nor does the rule have any additional condition on new stones).

And this is not even specific to J89 or the enable rule: example 2 was considered alive and seki even in J1949 (and maybe even before that, in traditional go). J89 changed a few special cases wrt L/D, but for the most part it only introduced a logical explanation. Alive includes uncapturable stones, capturable but replayable stones, and stones that are not really capturable only exchangeable. Exchanges need to be played out, scoring cannot grant them. And especially cannot score a string as dead while the other side of the exchange (the B string that would be lost) as alive.
I agree with you Jann, I also consider all groups alive in these diagrams.

BTW I would like to add an argument not related to the rule as it is written but related to my logic as a go player:
Click Here To Show Diagram Code
[go]$$
$$ ---------------------------------
$$ | . O X X X Q Q Q Q X O O X O O |
$$ | X X O O X X X . Q X . O X . O |
$$ | X X O . O O X . Q X . O X . O |
$$ | O O O O . O X X Q X X O X X O |
$$ ---------------------------------[/go]
Let's assume I am quite hesitant for the status of the marked stones in the diagramm above. Though I do not know for sure this status I am sure of the two following points:
1) If the marked stones are captured then the two other white groups at the right are also captured
2) If the marked stones are not cpatured then the two other white groups at the right are also not captured
IOW the marked stones and the two other white groups on the right are strongly tied => for my logic of go player these three groups must have the same status. If not I would have serious doubt concerning the consistency of the rule.

Same remark with your example 2.
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Re: How do Japanese rules handle this?

Post by CDavis7M »

Gérard TAILLE wrote:In Diag1 you consider that the marked stones are dead because they have "no ability to form new uncapturable stones".
But in Diag2 you consider that the marked stones are alive because they have this ability to form new uncapturable stones.
This result is strange for me and I have to work harder in order to really understand.

My view is the following : for a go player looking for the status of a group, the sequence itself is not relevant. What is relevant is the final position reached after the chosen sequence. In this FINAL position the player can look at the captured stones as well as the new uncapturable stones created and decide whether a group of stones is alive or not.
Well, because we are talking about capturable stones then the sequence is relevant. And even if you ignore the sequence, the game mechanics can still recognized from the final board position alone. It should be clear whether which stones were the ones that could create the uncapturable stone. In some diagrams a group of stones is completely separate. In other diagrams they are not. The relation between the stones and the new uncapturable stone can be seen just by looking at the intersections.
Gérard TAILLE wrote: Let's call A and B the two unmarked white groups of stones in Diag 1 or 2 and let's call C the white marked group in Diag 1 or 2
In both diagram:
1) the three A, B, C groups are capturable
2) the two A, B groups cannot be simultaneously captured
3) Black can capture A and C but in this case uncapturable stones are created towards B
4) Black can capture B and C but in this case uncapturable stones are created towards A
What difference you see between these two diagrams for a player looking at the status of these groups?

BTW what is the status of the two unmarked white groups in Diag 2 ?
In diagram 1, the capture of C happens by extraneous moves. Playing moves to capture C is similar to as if the player determined L&D in the upper corner and then started to play an atari in a nakade shape in the lower corner. There is no need to do that because L&D status in the upper and lower corners can already be determined. This is the same issue with the 5 stones in diagram previously discussed. Capture of the 5 stones is extraneous. The L&D of the 1 stone is not interdependent on the 5 stones.

In diagram 2, the capture of C is not extraneous. The life of the groups of stones are interdependent. A (or B) cannot be captured because C is reducing the liberties of Black's stones. Contrast to Diagram 1 where the liberties of the seki in the lower right make no difference to the other separate groups of stones.
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Re: How do Japanese rules handle this?

Post by CDavis7M »

jann wrote:
Click Here To Show Diagram Code
[go]$$ ---------------------------------
$$ | . @ X X X O O O O X O O X Q Q |
$$ | X X O O X X X . O X . O X . Q |
$$ | X X O . O O X . O X . O X . Q |
$$ | O O O O . O X X O X X O X X Q |
$$ ---------------------------------[/go]
IMO there is zero doubt the bottom/right is also alive both cases.
Which portion of the rules provides a basis for this interpretation?
To me, this position looks like a simple combination of Example 1 and Examples 15-18. By the reasoning in Example 1, :ws: is alive. By the reasoning in Examples 15-18 :wt: is dead. The fact that :ws: can form new uncapturable stones only shows that :ws: is alive. It does not show that :wt: is alive.
It is the ability of :ws: to capture the capturing Black stones that allows formation of new stones. The :wt: are completely out of the picture. They don't contribute anything to :ws: 's formation of the uncapturable stones.
Click Here To Show Diagram Code
[go]$$ ---------------------------------
$$ | 1 @ X X X O O O O X O O X Q Q-X 2 X X X O O O O X O O X Q Q-5 O X X X O O O O X O O X Q Q |
$$ | X X O O X X X . O X . O X . Q-X X O O X X X . O X . O X . Q-4 3 O O X X X . O X . O X . Q |
$$ | X X O . O O X . O X . O X . Q-X X O . O O X . O X . O X . Q-6 . O . O O X . O X . O X . Q |
$$ | O O O O . O X X O X X O X X Q-O O O O . O X X O X X O X X Q-O O O O . O X X O X X O X X Q |
$$ ---------------------------------[/go]
----------
jann wrote:
Click Here To Show Diagram Code
[go]$$B example 2 variant - no prisoners, komi -2
$$ +-----------------------------------+
$$ | X X X X X X X . X O O . O X X O O |
$$ | X X O X X O X . X . @ O O O X . Q |
$$ | X X O O O O O X X X X X X O X . O |
$$ | X X O . O . O O O O O O O X X O O |
$$ +-----------------------------------+[/go]
Again the entire right is alive (both W strings). And B won't even dare to capture in real game - this is permanent seki. J89 (and even J2003) have no problem here: attempting to capture either W string would enable or give rise to new uncapturable stones (compared to the original position OC, for these reasons). It doesn't matter if he captures in one move or in two parts (nor have the rule any additional condition on new stones).
Again, I disagree. This position is like a combination of Example 2 and Example 14. As in Example 2, :ws: is alive because even though it can be captured it can newly form uncapturable stones. However, as in Example 14, the other White stones are dead without White teire. It doesn't matter whether :ws: can capture other black stones to form uncapturable stones. What matter is whether :wt: can form uncapturable stones itself. As in Example 2, :ws: doesn't need teire. But as in Example 14, :wt: does. I just follow the Examples.

----------
jann wrote: And this is not even specific to J89 or the enable rule: example 2 was considered alive and seki even in J1949 (and maybe even before that).
I don't see a position even close to this one mentioned in the 1949 Japanese Go Rules. I only see a different rendition of what is now Example 2. What section are you looking at? Or are you just guessing?
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Re: How do Japanese rules handle this?

Post by Gérard TAILLE »

CDavis7M wrote:
jann wrote:
Click Here To Show Diagram Code
[go]$$B example 2 variant - no prisoners, komi -2
$$ +-----------------------------------+
$$ | X X X X X X X . X O O . O X X O O |
$$ | X X O X X O X . X . @ O O O X . Q |
$$ | X X O O O O O X X X X X X O X . O |
$$ | X X O . O . O O O O O O O X X O O |
$$ +-----------------------------------+[/go]
Again the entire right is alive (both W strings). And B won't even dare to capture in real game - this is permanent seki. J89 (and even J2003) have no problem here: attempting to capture either W string would enable or give rise to new uncapturable stones (compared to the original position OC, for these reasons). It doesn't matter if he captures in one move or in two parts (nor have the rule any additional condition on new stones).
Again, I disagree. This position is like a combination of Example 2 and Example 14. As in Example 2, :ws: is alive because even though it can be captured it can newly form uncapturable stones. However, as in Example 14, the other White stones are dead without White teire. It doesn't matter whether :ws: can capture other black stones to form uncapturable stones. What matter is whether :wt: can form uncapturable stones itself. As in Example 2, :ws: doesn't need teire. But as in Example 14, :wt: does. I just follow the Examples.
I understand you consider the white stones at the right are dead. Does that mean that black has 14 points (territory + prisonners) on the right side ?
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