If white does not play into his or her territory, black can play as follows before ending the game:
- Click Here To Show Diagram Code
[go]$$Bc 2, 4 pass
$$ ---------------------------------------
$$ | X O O X X X X . O . X . O X . . . . . |
$$ | X X O O O O X X X X . X O X . . . . . |
$$ | X . X O . O O X O O X O O X . X . . . |
$$ | X X X O O O X O . O O . O O X X . . . |
$$ | O O X X X O X O O . O O X X . . . . . |
$$ | . O O X 1 O X X O O X X . . . . . . . |
$$ | O . O X O O O X X X O X X X . . . X . |
$$ | . O X X X X O 5 X O O O X O X X . . X |
$$ | O O O X 3 O O X . X O . O O O X X X O |
$$ | O O X X X X O . X X O O . . . O X O O |
$$ | . X . X O X O O O O X O . . O O O O . |
$$ | X X . X O O O X X X X X O O X . . X X |
$$ | O X X X X X X X . X X O X O X X X X . |
$$ | O O O X . X O O O X O O X X . X O O . |
$$ | . X O . O X . O X X O O . X . X X X . |
$$ | . O . O O X O O O O O X X X X O X . . |
$$ | . O O O X X O . . . O X X O O O O X X |
$$ | O X O X X X O . . . . O O . . O O O X |
$$ | . X X . X O O . . . . . . . . . O X X |
$$ ---------------------------------------[/go]
To me, this is Seki because white cannot capture black without playing self-atari and black can't capture white without self-atari.
I read Robert Jasiek's summary of the 1989 rules and 2003 rules and here are some quotes:
Quote:
A player's final-string is uncapturable if the opponent cannot force capture of its stones.
Quote:
A player can force something if there is at least one hypothetical-strategy of his so that each compatible hypothetical-sequence fulfils that something.
Quote:
For each hypothetical-strategy of the opponent, the player can find several answers but the player must find AT LEAST ONE answer to succeed.
If we consider the black stones, the opponent is white and any feasible hypothetical strategy of white's that aims to capture the black string will involve self-atari. Blacks hypothetical answer will be to capture white. That counts as at least one, in my mind, and is sufficient to prove that black is uncapturable.
Similarly, white is uncapturable.
Howerver, in truth, that is all irrelevant because:
Quote:
In the final-position, an intersection of a black-eye-string / white-eye-string is a black-eye-point / white-eye-point if the black-eye-string / white-eye-string is adjacent and only adjacent to intersections with stones of one or more than one alive black-strings / white-strings.
In the final-position, an eye-point is either a black-eye-point or a white-eye-point.
In the final-position, a dame is an empty intersection that is not an eye-point.
In the final-position, a black-region / white-region is an intersection with a stone of an alive black-string / white-string and, recursively, any adjacent intersection that has a stone of an alive black-string / white-string or is a black-eye-point / white-eye-point.
In the final-position, a black-region / white-region is in-seki if at least one of its intersections is adjacent to a dame.
So H10 is clearly "dame" and the result is clearly "seki".
If black were to play 1 and 3 and not play 5 and the final position looked as follows, things would be different.
- Click Here To Show Diagram Code
[go]$$Bc 2 pass
$$ ---------------------------------------
$$ | X O O X X X X . O . X . O X . . . . . |
$$ | X X O O O O X X X X . X O X . . . . . |
$$ | X . X O . O O X O O X O O X . X . . . |
$$ | X X X O O O X O . O O . O O X X . . . |
$$ | O O X X X O X O O . O O X X . . . . . |
$$ | . O O X 1 O X X O O X X . . . . . . . |
$$ | O . O X O O O X X X O X X X . . . X . |
$$ | . O X X X X O . X O O O X O X X . . X |
$$ | O O O X 3 O O X . X O . O O O X X X O |
$$ | O O X X X X O . X X O O . . . O X O O |
$$ | . X . X O X O O O O X O . . O O O O . |
$$ | X X . X O O O X X X X X O O X . . X X |
$$ | O X X X X X X X . X X O X O X X X X . |
$$ | O O O X . X O O O X O O X X . X O O . |
$$ | . X O . O X . O X X O O . X . X X X . |
$$ | . O . O O X O O O O O X X X X O X . . |
$$ | . O O O X X O . . . O X X O O O O X X |
$$ | O X O X X X O . . . . O O . . O O O X |
$$ | . X X . X O O . . . . . . . . . O X X |
$$ ---------------------------------------[/go]
At this point, the white stones would be uncapturable according to my former argument but black would be dead because, when considering whether black's groups are uncapturable, white gets to play the first move in the hypothetical sequence and white could play either H10 or H12.
Assuming that black is not an idiot, after 3, white should expect black 5 as in my first diagram because black would prefer the seki to outright death. To prevent black 5, surely white should play H10 or H12, sacrificing a point?
There is something, here, that I just don't understand.