Harleqin wrote:I would be interested in how you would treat the life and death examples No. 4 and 16 from the Nihon Kiin rules.
#4
Let's start with determining the status of White's upper side chain (left side one is identical).
$$
$$ -------------------
$$ | . X 1 2 X X X W X
$$ | X X W W W W W W X
$$ | . O X X X X X X X
$$ | . O X , . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .
- Click Here To Show Diagram Code
[go]$$
$$ -------------------
$$ | . X 1 2 X X X W X
$$ | X X W W W W W W X
$$ | . O X X X X X X X
$$ | . O X , . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .[/go]
Black begins.
$$
$$ -------------------
$$ | . X X O . 3 . O X
$$ | X X O O O O O O X
$$ | . O X X X X X X X
$$ | 4 O X , . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .
- Click Here To Show Diagram Code
[go]$$
$$ -------------------
$$ | . X X O . 3 . O X
$$ | X X O O O O O O X
$$ | . O X X X X X X X
$$ | 4 O X , . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .[/go]
$$
$$ -------------------
$$ | . X X O 5 X 7 O X
$$ | X X O O O O O O X
$$ | 6 O X X X X X X X
$$ | O O X , . . . . .
$$ | . O X . . . . . .
$$ | . O X . . . . . .
$$ | . O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .
- Click Here To Show Diagram Code
[go]$$
$$ -------------------
$$ | . X X O 5 X 7 O X
$$ | X X O O O O O O X
$$ | 6 O X X X X X X X
$$ | O O X , . . . . .
$$ | . O X . . . . . .
$$ | . O X . . . . . .
$$ | . O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .[/go]
White's chain is eliminated.
Status of Black's upper side chain (left side one is identical).
$$
$$ -------------------
$$ | . X . 2 B B B O X
$$ | X X O O O O O O X
$$ | . O X X X X X X X
$$ | . O X , . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .
- Click Here To Show Diagram Code
[go]$$
$$ -------------------
$$ | . X . 2 B B B O X
$$ | X X O O O O O O X
$$ | . O X X X X X X X
$$ | . O X , . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .[/go]
White begins and captures the chain. If she does nothing, we return to the sequence before.
$$
$$ -------------------
$$ | . X . O . 3 . O X
$$ | X X O O O O O O X
$$ | . O X X X X X X X
$$ | 4 O X , . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .
- Click Here To Show Diagram Code
[go]$$
$$ -------------------
$$ | . X . O . 3 . O X
$$ | X X O O O O O O X
$$ | . O X X X X X X X
$$ | 4 O X , . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | X O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .[/go]
$$m5
$$ -------------------
$$ | . X 5 O 3 X 7 O X
$$ | X X O O O O O O X
$$ | . O X X X X X X X
$$ | O O X , . . . . .
$$ | . O X . . . . . .
$$ | 1 O X . . . . . .
$$ | . O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .
- Click Here To Show Diagram Code
[go]$$m5
$$ -------------------
$$ | . X 5 O 3 X 7 O X
$$ | X X O O O O O O X
$$ | . O X X X X X X X
$$ | O O X , . . . . .
$$ | . O X . . . . . .
$$ | 1 O X . . . . . .
$$ | . O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .[/go]
$$
$$ -------------------
$$ | . X X . B B B . X
$$ | X X . . . . . . X
$$ | . O X X X X X X X
$$ | O O X , . . . . .
$$ | . O X . . . . . .
$$ | X O X . . . . . .
$$ | . O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .
- Click Here To Show Diagram Code
[go]$$
$$ -------------------
$$ | . X X . B B B . X
$$ | X X . . . . . . X
$$ | . O X X X X X X X
$$ | O O X , . . . . .
$$ | . O X . . . . . .
$$ | X O X . . . . . .
$$ | . O X . . . . . .
$$ | O O X . . . . . .
$$ | X X X . . . . . .[/go]
All primary occupied points have become part of a Black 2-eyed group.
White cannot do anything with Black's chain in the corner, so this one is "2-eyed", too.
$$W
$$ -------------------
$$ | . 2 . . 2 2 2 P X
$$ | 2 2 P P P P P P X
$$ | . P X X X X X X X
$$ | . P X , . . . . .
$$ | 2 P X . . . . . .
$$ | 2 P X . . . . . .
$$ | 2 P X . . . . . .
$$ | P P X . . . . . .
$$ | X X X . . . . . .
- Click Here To Show Diagram Code
[go]$$W
$$ -------------------
$$ | . 2 . . 2 2 2 P X
$$ | 2 2 P P P P P P X
$$ | . P X X X X X X X
$$ | . P X , . . . . .
$$ | 2 P X . . . . . .
$$ | 2 P X . . . . . .
$$ | 2 P X . . . . . .
$$ | P P X . . . . . .
$$ | X X X . . . . . .[/go]
So we have White chains in status c) (let's call them "remouvable" - I have no better short term in the moment) inside Black chains in status "2-eyed".
White's chains are "dead", Black's chains are "alive".
As you can see, both of White's chains are seperated by "living" Black groups. So there is no side-effect from one chain to the other. What I said when explaining "local".
#16
White cannot do anything here, so we will start determining the status of the single White stone on the left.
$$
$$ -----------------------------
$$ | 1 O O . O . O X . X O . O X
$$ | W X X O O O O X X O O O O X
$$ | X X . X X X O X O . O X X X
$$ | . . . , X O X X X O O X . .
$$ | . . . . X O X . X O X X . .
$$ | . . . . X O X X X O X . . .
$$ | . . . . X O O O O X X . . .
- Click Here To Show Diagram Code
[go]$$
$$ -----------------------------
$$ | 1 O O . O . O X . X O . O X
$$ | W X X O O O O X X O O O O X
$$ | X X . X X X O X O . O X X X
$$ | . . . , X O X X X O O X . .
$$ | . . . . X O X . X O X X . .
$$ | . . . . X O X X X O X . . .
$$ | . . . . X O O O O X X . . .[/go]
Black captures.
$$
$$ -----------------------------
$$ | X O O . O . O X 2 X O . O X
$$ | 3 X X O O O O X X O O O O X
$$ | X X . X X X O X O . O X X X
$$ | . . . , X O X X X O O X . .
$$ | . . . . X O X . X O X X . .
$$ | . . . . X O X X X O X . . .
$$ | . . . . X O O O O X X . . .
- Click Here To Show Diagram Code
[go]$$
$$ -----------------------------
$$ | X O O . O . O X 2 X O . O X
$$ | 3 X X O O O O X X O O O O X
$$ | X X . X X X O X O . O X X X
$$ | . . . , X O X X X O O X . .
$$ | . . . . X O X . X O X X . .
$$ | . . . . X O X X X O X . . .
$$ | . . . . X O O O O X X . . .[/go]
Whatever happens on the right, Black will be able to connect.
If we want to determine the status of White's two-stone chain in the upper left, ...
$$
$$ -----------------------------
$$ | X W W 3 O . O X O . O . O X
$$ | . X X O O O O X X O O O O X
$$ | X X . X X X O X O . O X X X
$$ | . . . , X O X X X O O X . .
$$ | . . . . X O X . X O X X . .
$$ | . . . . X O X X X O X . . .
$$ | . . . . X O O O O X X . . .
- Click Here To Show Diagram Code
[go]$$
$$ -----------------------------
$$ | X W W 3 O . O X O . O . O X
$$ | . X X O O O O X X O O O O X
$$ | X X . X X X O X O . O X X X
$$ | . . . , X O X X X O O X . .
$$ | . . . . X O X . X O X X . .
$$ | . . . . X O X X X O X . . .
$$ | . . . . X O O O O X X . . .[/go]
... Black will capture this chain with 3.
$$
$$ -----------------------------
$$ | X . 5 X O . O X O . O . O X
$$ | . X X O O O O X X O O O O X
$$ | X X . X X X O X O . O X X X
$$ | . . . , X O X X X O O X . .
$$ | . . . . X O X 4 X O X X . .
$$ | . . . . X O X X X O X . . .
$$ | . . . . X O O O O X X . . .
- Click Here To Show Diagram Code
[go]$$
$$ -----------------------------
$$ | X . 5 X O . O X O . O . O X
$$ | . X X O O O O X X O O O O X
$$ | X X . X X X O X O . O X X X
$$ | . . . , X O X X X O O X . .
$$ | . . . . X O X 4 X O X X . .
$$ | . . . . X O X X X O X . . .
$$ | . . . . X O O O O X X . . .[/go]
White captures Black's chain in the middle, Black connects on the left. It does not help White to recapture at 5 with 4.
As will have become clear here, Black will not be able to capture White's left-hand 7-stone-group. This chain cannot be turned into "2-eyed" by White, but will remain on the board, so let us call this type of status "stable" in short (again I have not found a better term).
$$
$$ -----------------------------
$$ | . P P . @ . @ # . Z @ . @ X
$$ | P X X @ @ @ @ # # @ @ @ @ X
$$ | X X . X X X @ # P . @ X X X
$$ | . . . , X O # # # @ @ X . .
$$ | . . . . X O # . # @ X X . .
$$ | . . . . X O # # # @ X . . .
$$ | . . . . X O O O O X X . . .
- Click Here To Show Diagram Code
[go]$$
$$ -----------------------------
$$ | . P P . @ . @ # . Z @ . @ X
$$ | P X X @ @ @ @ # # @ @ @ @ X
$$ | X X . X X X @ # P . @ X X X
$$ | . . . , X O # # # @ @ X . .
$$ | . . . . X O # . # @ X X . .
$$ | . . . . X O # # # @ X . . .
$$ | . . . . X O O O O X X . . .[/go]
The big chains in the Double-Ko-Seki on the right are stable, too.
Both single stones in Atari cannot reach a "stable" status, so they are "removable".
Combining what we have found, we see chains, which are "removable", but are not situated between "2-eyed" opponent's chains. There are neither "alive" chains nor "dead" ones, so the whole position is a giant Seki.
This does not represent the 1989 Nihon Kiin rules result, I know. But I'm convinced that there must be a clear distinction between "Evalutate" and "Count". The method of taking "dead" stones off the board without capturing them is just a shortcut. Otherwise you would have to capture them in reality and get compensation by your opponent for your moves, either by playing into territory or by presenting a prisoner.
"Killing" White's 7-stone chain would only be possible after eleminating her single stone on the left. But this would mean changing the position at the end of "Play", which sets the foundation for "Evaluate".
Black would have had the one-sided option to turn this position into "neither victory nor defeat" during "Play", using the Triple-Ko.