I think some of the confusion comes from looking at the 5 stones first. I don't think this is how "Life & Death Confirmation" works. Considering the L&D status of the 5 stones by placing new uncapturable stones with an already independently living group is not "confirmation" affording to the definition of the word.

If you don't like dictionaries or diagrams, my main point is this:

Click Here To Show Diagram Code

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

This position that can result in determining the L&D status of the 5 White stones is the same position that can result when determining the L&D status of the 1 White ko-stone. The 1 White ko-stone is alive because even though it can be captured it can create the two stones. This same position does not somehow also "confirm" the living status of the 5 White stones. There is no more uncertainty about this position and nothing left to "confirm." If the players consider the 5 White stones together with the 1 stone stone, then of course the 6 White stones are alive. But replaying the same sequence adds nothing to the players understanding of White's L&D status.

Because the 1 White stone is independently alive without the 5 White stones and without their capture or not, then the same position does not also prove that the 5 White stones are alive. Repeating the same analysis is not what "confirmation" (kakunin 確認) is.

-----------

第九条－２
対局の停止後、双方が石の死活及び地を確認し、合意することにより対局は終了する

According to Kenkuysha's J-E dictionary:
Kakunin 確認 n. confirmation; affirmation; certification; corroboration; validation.
Jisho.org gives: confirmation; verification; validation; review; check; affirmation; identification.

As for the English definitions of these words, the OED indicates that they all require some uncertainty or informality in a determination or declaration that is now being solidified, firmed, or formalized.

Click Here To Show Diagram Code

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

The players performing 死活確認 L&D Kakunin need to find all the L&D that is uncertain and solidify and formalize the status. They will of course start with the definition of 活き石 Living/Alive Stones: 相手方の着手により取られない石、又は取られても新たに相手方に取られない石を生じうる石は「活き石」という。So first, identify stones that cannot be captured and confirm their living status. Then identify stones that can be captured, but even though they can be captured, they can create new uncapturable stones. It seems many people prefer diagrams so here I go.

STEP 1
The players confirming White's L&D status look for White's stones that cannot be captured. The do not look for White's stones that can be captured yet. They can confirm that White's 17 stones cannot be captured. Not even if Black plays all of because is an illegal move. Of course. This part is easy.

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | V . V . X . Y Q Q 1 Q Y
$$ | V V V X V X Q Q Q Q Q Y
$$ | X X X Q Y Q Q . Q Y Y Y
$$ | X . X Q Q Q Q Q Y . . .
$$ | . X X Y Y Y Y Y Y . . .
$$ | X X . . . . . . . . . .
$$ | . . . . . . . . . . . .
$$ | . . . . . . . . . . . .[/go]

At this point of 死活確認, the 1 White ko-stone is not considered because it can be captured:

Click Here To Show Diagram Code

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

The 5 White stones in the corner are also not considered because they can be captured:

Click Here To Show Diagram Code

[go]$$ pass
$$ ----------------------
$$ | P 7 P 5 X 2 3 O O . O
$$ | P P P X 6 X O O O O O
$$ | X X X O 1 O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

STEP 2
Now that the players have confirmed the uncapturable stones, they know that White's 17 stones are alive. But what about White's 1 ko-stone that can be captured? Is it alive? Can it create new uncapturable stones? Yes. is a new uncapturable stone that also makes uncapturable. Black does not bother with the 5 White stones because capturing them has nothing to do with whether the 1 stone can be captured and it also has nothing to do with whether the 1 capturable White stone can create a new uncapturable stone. Of course, Black also cannot just connect to ko, nor can Black play self-atari. Black captured the 1 White stone and White played and , which are then uncapturable. So the living status of the 1 White stone is confirmed.

The stones that are not part of the confirmation are half-shown.

Click Here To Show Diagram Code

[go]$$ pass, pass, pass
$$ ----------------------
$$ | V . V . X 6 . V V . V
$$ | V V V X 4 X V V V V V
$$ | X X X V 1 V V . V . .
$$ | X . X V V V V V . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

What if Black does try to capture the 5 White stones? White could throw-in or they could also play this way. Either way the 5 White stones are captured. But again, that has nothing to do with the 1 White stone which is the stone whose L&D status the players are confirming.

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | V 5 V 3 X 4 2 V V . V
$$ | V V V X 6 X V V V V V
$$ | X X X V 1 V V . V . .
$$ | X . X V V V V V . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

RECAP
Let's just recap what the current agreement of the players is as to L&D status of the White stones. The 17 White stones are alive because they cannot be captured and the 1 White stone is alive because although it can be captured, it can create at least one new uncapturable stone.

Click Here To Show Diagram Code

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

STEP 3
The only remaining uncertainty is the L&D status of the 5 White stones . What is their status?

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | Q . Q . X . . V V . V
$$ | Q Q Q X V X V V V V V
$$ | X X X V . V V . V . .
$$ | X . X V V V V V . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

Here, Black proves that the 5 White stones are dead.

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | Q 5 Q 3 X 4 2 V V . V
$$ | Q Q Q X 6 X V V V V V
$$ | X X X V 1 V V . V . .
$$ | X . X V V V V V . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

Wait, but what about this position? Don't the 2 new uncapturable stones confirm that the 5 White stones are alive?

Click Here To Show Diagram Code

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

No. This is the same position that was already considered when determining the L&D status of the 1 White stone. There is no more uncertainty about this position and nothing left to "confirm." If they players consider the 5 White stones together with the 1 stone stone then of course the 6 White stones are alive. But this adds nothing to their understanding of L&D status. The 5 White stones are still dead when considered separately.

The 5 dead White stones cannot be removed from the board because Black's 3 stones are dead.

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | P C P . X . . O O . O
$$ | P P P X O X O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

Point is surrounded by dead stones of one player. It is not surrounded by living stones of one player. Therefore, it is not an eye 目, it is dame 駄目. Only 目, not 駄目, can be territory 地. So White does not count as territory.

Black cannot capture the White stones during game-play but because Article 7-2 requires a pass before retaking a ko, the 5 White stones are capturable for purposes of Life and Death confirmation. Because the 5 White stones are capturable, they are dead because the only new uncapturable stones that can be played are stones that are only uncapturable by virtue of other stones that are independently alive and can already be deemed alive.

As I explained above, the White stones with 2 eyes are uncapturable and the 1 White ko-stone, even though it is capturable, can create two (or three) new uncapturable stones. The 2 new uncapturable stones only prove the living status of the 1 White ko-stone. They cannot be used to prove that the 5 White stones are alive because playing those two (or three) stones only re-confirms that the 1 White ko-stone is already independently alive -- those 2 stones uncapturable stones can be played in L&D confirmation without the 5 White stones being captured at all. There is nothing being "confirmed" by doing that. Even believers of the so-called "enable" rule in one English translation should admit that it is not the capture of the 5 white stones that "enables" the 2 new uncapturable stones, it is only the capture of the 1 white ko-stone that "enables" the 2 new uncapturable stones.

Click Here To Show Diagram Code

[go]$$ pass, pass, above 1,
$$ ----------------------
$$ | O . O . X 4 2 O O . O-O . O . X Q . O O . O
$$ | O O O X O X O O O O O-O O O X Q . O O O O O
$$ | X X X O 1 O O . O . .-X X X O . O O . O . .
$$ | X . X O O O O O . . .-X . X O O O O O . . .
$$ | . X X . . . . . . . .-. X X . . . . . . . .
$$ | X X . . . . . . . . .-X X . . . . . . . . .[/go]

It is the capture of the 1 White ko-stone that enables the 2 new uncapturable White stones , not any capture of the 5 White stones because the 5 White stones do not have to be captured to be able to play the 2 new uncapturable stones . Only the 1 White ko-stone needs to be captured.

A player cannot just add a stone to an already living group to pretend that other stones are alive. A player cannot just fill 1 point of would-be territory to pretend that a dead group is alive.

The 1 White stone is alive. The 6 White stones including the 1 White stone and the 5 White stones are alive. But the 5 White stones considered separately are dead. So the intersecting point between the 5 White stones is dame, not territory.

I try to understand your point but it is not that easy.
After step 1 you identified 17 stones uncapturable and two other capturable groups of stones. One group with only one stone and another group with 5 stones.
You begin step 2 by looking at the status of the group with only 1 stone. My question is the following : what is your analyse if you begin step 2 by looking first at the status of the group with 5 stones?

In my view it is quite simple: if that is the result of your interpretation of the japanese rules, then either the interpretation or the rules are wrong.

_________________
A good system naturally covers all corner cases without further effort.

Click Here To Show Diagram Code

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

I agree with you Harleqin, in this position I do not want white to add a move.
We all know that articles 7.1 and 7.2 are difficult to interpret correctly and they even seem in contradiction with the examples shown in the rule, especialy in case of double ko.

One of the main problem is the meaning of "capturable". This word appears in article 7.1 while the pass-for-ko rule appears only in article 7.2. As a consequence it is not clear if the status "capturable" has to be analysed with or without the pass-for-ko rule.
In my view if a group of stones cannot de captured in "normal" play then I consider it is alive, even if this group is capturable by using the pass-for-ko rule.
IOW I use the pass-for-ko rule only for analysing more deeply a group of capturable stones.

In the example above the groupe of 5 white stones is not capturable in "normal" play but capturable with the pass-for-ko rule. Here is the issue.

I pointed out in another thread that white is alive because the rule is not that stone must not be capturable but in fact the following:

Stones are said to be "alive" if they cannot be captured by the opponent, or if capturing them would enable a new stone to be played that the opponent could not capture. Stones which are not alive are said to be "dead."

This doesn't really help with J89 in general but white does have a clear argument for example because is a new stone that can't be captured.

Click Here To Show Diagram Code

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

Click Here To Show Diagram Code

[go]$$Bcm5
$$ ----------------------
$$ | O 3 O 1 X 2 X O O . O
$$ | O O O X 4 X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

Black has no way to avoid being played while capturing because connecting the ko doesn't let black capture.

Click Here To Show Diagram Code

[go]$$Bcm5 pass for ko
$$ ----------------------
$$ | O . O 3 X 1 X O O . O
$$ | O O O X 4 X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

However a definition of what exactly a new "enabled" stone is is not really available but J89 does use this concept to good effect to determine statuses in snapbacks. Those snapback back determinations are really key to what we know as Japanese rules. I think this position can really be thought of as a snapback shape with a ko, if black tries to take he is capturable in a ko and therefore it is not surprising that white doesn't have to add a move by the principle of "no ko in in status confirmation".

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | O . O . X a . O b . O
$$ | O O O X O X O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

In L&D confirmation phase it is true that black can capture the 5 stones and it is true that white will be able to put an uncapturable stone in "a". But it is not true to say that the capture of the 5 white stones by black "enables" white to put an uncapturable stone in "a". In any case black cannot prevent white to put an uncapturable stone at "a". That means that "a" point as the same status as "b" point : black cannot prevent white to put an uncapturable stone at "a" or "b".

It doesn't matter what order they are analyzed. If the players first attempt to determine whether the 5 stones are dead, they will quickly realize that they are determining the status of 6 stones, not 5, because the 1 ko-stone is involved. The 6 stones are alive, but is that by virtue of the 1 stone ko-stone or all 6 stones? The players will confirm the status of the 1 stone separately.

The players will realize that it is the capture of the 1 stone that actually "enables" the new uncapturable stones and that the 5 stones do not actually need to be captured at all. If the 5 stones do not actually have to be captured then they cannot be said to "enable" the new capturable stones. Of course it is just the the 1 stone alone that actually "enables" the new uncapturable stones.

When determining the status of the 1 White stone, Black does not need to attempt to capture the 5 stones because it is not their status that is being confirmed and their capture does not help black prove the status of the 1 stone.

Click Here To Show Diagram Code

[go]$$ pass, pass, above 1,
$$ ----------------------
$$ | O . O . X 4 2 O O . O-O . O . X Q . O O . O
$$ | O O O X O X O O O O O-O O O X Q . O O O O O
$$ | X X X O 1 O O . O . .-X X X O . O O . O . .
$$ | X . X O O O O O . . .-X . X O O O O O . . .
$$ | . X X . . . . . . . .-. X X . . . . . . . .
$$ | X X . . . . . . . . .-X X . . . . . . . . .[/go]

Yup. We are all talking about stones that are capturable and that can so-called "enable" a new uncapturable stone. The debate is not whether there is the possibility of new uncapturable stones, but which stones are the stones that "enable" the new uncapturable stones.

The definition of "enabled" is not available from the Japanese Rules because the Japanese Rules do not use this term. It is an unofficial interpretation of the Japanese Rules.

Stones are said to be "alive" if they cannot be captured by the opponent, or if capturing them would enable a new stone to be played that the opponent could not capture. Stones which are not alive are said to be "dead."

But if we take this guy's interpretation with "enable" and we look at the definition of "enable", it's easy to determine that it is not capturing the 5 stones that enables new uncapturable stones, it is capturing the 1 White stone that enables the new uncapturable stones. It is easy to see because the capturing of the 1 White stone enables new uncapturable stones without capturing the 5 stones. So it is the 1 white ko-stone alone that "enables" the new uncapturable stones. So the 1 stone is alive.

Click Here To Show Diagram Code

[go]$$ pass, pass, above 1,
$$ ----------------------
$$ | O . O . X 4 2 O O . O-O . O . X Q . O O . O
$$ | O O O X O X O O O O O-O O O X Q . O O O O O
$$ | X X X O 1 O O . O . .-X X X O . O O . O . .
$$ | X . X O O O O O . . .-X . X O O O O O . . .
$$ | . X X . . . . . . . .-. X X . . . . . . . .
$$ | X X . . . . . . . . .-X X . . . . . . . . .[/go]

The 1 stone can be captured and enable new uncapturable stones even if the 5 stones are not captured. The 5 stones do not enable anything. The 5 stones are dead and so they have dame.

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | O . O . X a . O b . O
$$ | O O O X O X O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

'b' would be part of the surrounding 'stones' also know as a group in English. Black can show that the corner can be captured, 'a' is therefore a point in an area that will after removal of stones from the board be (partially) surrounded by capturable stones and that is the difference between 'a' and 'b'.

Let's explain this further. In the following diagram the marked black stones are capturable and dead. These stones are removed from the board if and only if the surrounding stones are alive, if any of the surrounding stones are dead then these stone remain on the board (anti-seki). The marked white stones are capturable but unless they are also dead marked black stones will remain on the board. Therefore 'a' will be surrounded by either the marked black or marked white stones, in either case 'a' is (partially) surrounded by capturable stones. It is different from 'b' which is surrounded only by alive, uncapturable, white stones.

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | Q . Q . Y a . O b . O
$$ | Q Q Q Y O Y O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

I hope that clears up that 'a' is not same as 'b'



Black plays first and white is not entitle to play 'a'. Black must capture the white marked stones and defend against new stones. It doesn't matter if white could play the same stones in a different scenario. Sure there is an issue of how to understand 'enable' because it is not stated clearly in the rules text but I think this position is clear.

Claiming black need not capture the marked stones is something else. Status confirmation is not an issue if black refuses to capture the stones. It is only a question of if white can play 'new stones', stones not being new if they are part of existing 'stones'. If these rules were originally in English they would use Go terms differently and this point might be clearer, but there is no guarantee.

Anyway. 'a' is not same as 'b' because 'b' is part of existing 'stones' but 'a' would be a new 'stone' because the surrounding stones are not the same 'stones'. The surrounding stones do not have the same color or capturable status; 'a' is breaking new ground.

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | O . O . X a . O b . O
$$ | O O O X O X O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

A agree that, on number of points, the point "a" and "b" are OC different.
I was talking only on a specific point : can black prevent white to put an uncapturable stone on "a"?
It seems the answer is no. In that case can you claim that the five white in the corner are alive because if black kills them then white will be enabled to put a "new" uncapturable stone at "a"? It is only in this sense that "a" and "b" are comparable. BTW you can easily replace point "b" by a point surrounded by alive white stones AND alive black stones (in seki).

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | O . O . X a . O b . O
$$ | O O O X O X O O O O O
$$ | X X X O . O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

I'll start with your section position because it contradicts your previous position (appearing first in your original post). If you understand that 'a' is not the same as 'b' because 'b' is part of existing stones that are independently alive, then why do you not recognize that the new uncapturable stones are only alive because of the 1 single ko-stone alone, which can be deemed as alive without capturing the 5 white stones.

Arguing that the 5 stones are alive because of 1 stone that is independently alive is just like arguing that the 5 stones are alive if White can play at 'b'. Both of these arguments suffer from the same misconception.

That's where your wrong. It DOES matter if White can play the same stones without capturing the 5 white stones and it matters because of the definition of "enable." The OED defines "enable" as give (someone) the ability or means to do something; make possible. As shown, it is the capture of the 1 White stone that makes the uncapturable stones possible to play in L&D confirmation of the White stones. We can't just pretend that words have no meaning.

There is no doubt that the 6 White stones are alive. You say "status confirmation" but not of which stones. The Japanese Rules explanations are clear that stones are considered separately and together. The status of the 6 stones must be considered but also of the 1 stone alone and the 5 stones alone. This is clear from example 24. Even though all of Whites stones are alive together, the one is dead when considered alone. So it has dame.


The 5 white stones are only alive if you consider the 6 stones together. Why is Black capturing the 1 stone if trying to decide whether the separate 5 White stones are alive? Because the life of the 5 stones depends on the life of the 1 stone. But the life of the 1 stone does not depend on the life of the 5 stones. The 1 stone is independently alive because it enables new uncapturable stones. They 5 stones are not "enabling" for L&D confirmation.

You controversially discuss two conceivable sequences of status confirmation, but BOTH end with the SAME result.


Sequence #1:

Click Here To Show Diagram Code

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

captures White's single stone.

Click Here To Show Diagram Code

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

THEREAFTER, is established on the board as a stone that cannot be captured. Thus, White's captured stone is considered "alive".
connects.
White's five stones in the corner remain uncaptured, due to Black's damezumari. Thus, these stones are also considered "alive".



Sequence #2:

Click Here To Show Diagram Code

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

captures White's single stone. THEREAFTER, ...

Click Here To Show Diagram Code

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

White suspects that this time (after the obvious failure in sequence #1) Black wants to attack her stones in the corner, and so gives atari with , causing damezumari temporarily.
captures, so resolving his damezumari.

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[go]$$Bm5 pass for the upper ko
$$ ----------------------
$$ | O 3 O 1 X 2 X O O . O
$$ | O O O X . X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

gives atari in the corner.
recaptures the ko at the upper edge.
captures White's five stones in the corner.

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[go]$$Wm8
$$ ----------------------
$$ | . X . X X O . O O . O
$$ | . . . X 1 X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

THEREAFTER, is established on the board (you will easily realise the "snap-back feature" that has been highlighted by kvasir) ...

Click Here To Show Diagram Code

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

... as a stone that cannot be captured. Thus, ALL of White's captured stones are considered "alive".

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

If something is already possible, or if a previous action already makes something possible, then a later action cannot be said to "enable" that thing to be possible.

This is wrong. White's captured stone is not deemed "alive" because the stone is not actually "enabled" (made possible) by the captured stone (previously at ). The capture does not "enable" because White could play after a Black move that does not capture the 1 White stone. Also recognize that is alive because the group with 2 eyes is uncapturable, not because the 1 White stone is capturable. Therefore, a move at does not "confirm" (determine something unknown) the L&D status of any stones.


I found the point you misunderstand. If the players are confirming the L&D of the 1 White stone then both White and Black have no reason to care whether the stones in the corner are attacked or captured because their capture does NOT affect the status of the 1 White stone considered by itself.

If the players were attempting to confirm the L&D of the 5 white stones, then why is Black capturing a 6th stone? So then the players are assessing 6 stones, not 5 stones. I agree, the 6 stones considered together are alive. But the Japanese Rules show that the L&D status of stones is also considered separately.

Assessing the L&D status of the 1 White stone separate from the 5 White stones in the corner, the players will see that the 1 White stone is alive. That is, the capture of the 1 White stone enables White to place new uncapturable stones even without the 5 stones being captured. Since the new uncaptured stones are already possible (already enabled) by the capture of the 1 stone, the capture of the 5 White stones is not "enabling."

Wrong. The 6 White stones together are alive. The 1 White stone by itself is alive. But the 5 White stones in the corner are dead because their capture does not "enable" any uncapturable stones that were not already enabled by the capture of the 1 White stone by itself.

White has dame and loses.

----------------

The situation we are discussing is similar to Example 24: https://www.nihonkiin.or.jp/match/kiyak ... 22_25.html


Considered as a whole, the 13 White stones are alive because at least 11 of the White stones have 2 eyes.
Considering separately, it is alive because if Black captures then White can connect at 'c' and is uncapturable.
Considering separately, it is dead because it can be captured and it's capture does not "enable" any new uncapturable stones that were not already "enabled." Playing a stone at 'c' is not "enabled" by being captured. Therefore, is dead and point 'b' is dame.

Similarly, is alive and are dead. After is captured, playing stones at c and are not enabled by the 5 stones being captured, they are already enabled by being captured. So at least point 'b' is dame (as are others). In that area near with the 3 black stones, only point 'd' is territory.

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | Q b Q . X c . O O . O
$$ | Q Q Q X @ X O O O O O
$$ | X X X O d O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

I think that ...



There is simply NO "enable" in the current Japanese original of J89.

You continue to conclude on the basis of a translation into English of which we know neither the original Japanese text nor the extent to which it faithfully reflects the content intended at the time.

The current legal text tells us that stones, which are captured in the course of status confirmation, are nevertheless considered "alive", if a "new stone" is feasible (as a matter of course after that capture), which will not become captured.
And quite naturally, the reference area is limited to the disputed area.

"New", according to Oxford Advanced Learner's Dictionary, has the meaning of "not existing before", "recently made, invented, introduced, etc."

It has NOTHING to do with something like "could come into existence in another szenario".

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

I feel like only Cassandra understood the argument.

To the other points raised I can only try convey an opinion. The status of the triangle marked stones is alive unless black can both capture them and defend against new 'stones'. We can argue over which of the circle marked points qualify as new stones when white is able to play there, if they are all the same or different. At least I make an argument that being able to play on some of them is different from playing on unrelated intersections.

The some of the circle marked points would be on the boundary of the triangle stones after stone removal (if the triangle stones are alive) and in this case all of them allow white to live if they were played. I think the boundary argument is good, one can argue if one considers all of them to be the same (J2003 does this, or does it exclude 2 of them?) or if one emphasis that being allowed to play some of them is similar to how snapbacks are resolved is a different matter. I am trying to explain why this makes sense, not trying to make a formal system, so I choose not to be reductive.

Click Here To Show Diagram Code

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

To the argument that white can play there anyway or in a different way. I can only offer almost tautologies: "It is not white's turn to play." "Black is supposed to show a sequence that captures the triangle stones, not something else."

I don't know Japanese but I know that it leads to misconceptions when you try to read deeper meanings into everyday words. Most word are meant to be read in the flow of the overall text, not examined individually. In this case "would enable" is subjunctive mood and "a stone to be played" is passive voice, we already know that it should better be "stones" not "a stone" but never mind that. I think the emphasis here has to be on the condition "if capturing them" and the action in the consequence "a stone to be played" not on the subjunctive "would enable" but I think it is clear why this kind of investigation leads nowhere. An example: "if buying an airplane ticket would enable a person to travel", do we claim that "enable" has non-superfluous meaning? I think not, it is just there to comply with grammar. The subjunctive is also called the conjunctive because it connects independent and dependent sub-sentences, I think that is all there is to this sentence from the English translation. It is hard to argue with someone that says "would enable" is a meaningful phrase, but I still think that is incorrect and the connection between "if capturing" and "a stone to be played" is not being defined but instead glossed over.

Surely I can understand that the following english article may not be good a good translation

1. Stones are said to be "alive" if they cannot be captured by the opponent, or if capturing them would enable a new stone to be played that the opponent could not capture. Stones which are not alive are said to be "dead."

I do not know japanese; can anybody tell us what could be a better translation?

Click Here To Show Diagram Code

[go]$$
$$ ----------------------
$$ | Q . Q . X C C C C C C C C C O . O . O |
$$ | Q Q Q X O X O O O O O O O O O O O O O |
$$ | X X X O C O O . . . . . . . . . . . . |
$$ | X . X O O O . . . . . . . . . . . . . |
$$ | . X X . . . . . . . . . . . . . . . . |
$$ | X X . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |[/go]

In this diagram which marked intersections are related or unrelated to the status of five white marked stones in the corner?

The points at the upper edge except for a maximum of one are irrelevant:

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

captures.

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[go]$$B pass for ko
$$ -----------------------
$$ | O . O 3 X . . . . . .
$$ | O O O X . X O O O O O
$$ | X X X O X O O . O . .
$$ | X . X O O O O O . . .
$$ | . X X . . . . . . . .
$$ | X X . . . . . . . . .
$$ | . . . . . . . . . . .
$$ | . . . . . . . . . . .[/go]

gives atari at White's stones in the corner.

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

re-captures the ko / gives atari at Black's three stones at the top.
captures White's stones in the corner.
THEREAFTER ...

Click Here To Show Diagram Code

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

... connects, so establishing a permanent "new" stone.
As a matter of course, White could have captured at A, instead, to the same effect.

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

If black captures the corner then white can play new stones on all of these points and also these two.

Click Here To Show Diagram Code

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

Does it really matter? These points are a superset of the points in the first position. White can still play the same points if black captures the corner as before. I don't think it matters if it is all of them or the same ones as in the first position.

In a previous post you said



I agree with you that in this last diagram "a" and "b" look quite different.
The first diagram is intending to show that the frontier of the surrounding stones in the corner is not very easy to define. As a feeling I can say that the fisrt intersection or maybe the first two intersections are relevant but as far as the wording of a rule is concerned I do not know how to express this feeling for all types of positions.
That's the reason why I ask for a better translation of the original japanese 7.1 article (I understood that the word "enable" may not be a good translation of this japanese text).