Bill Spight wrote:
The other is combinatorial game theory, which accords with traditional go theory if there are no ko fights or potential ko fights. In this theory it does not matter who has the move, and the value of the current node is not simply a backed up value from the leaves of the tree. Furthermore, the values are not strictly ordered, but two results may be incomparable. However, if one option for a play dominates another, it also dominates it in any non-ko environment. The non-ko environment is another combinatorial game which is added to the game being considered. Difference games are part of combinatorial game theory, but not of von Neumann game theory.
This argument seems to basically belong to von Neumann game theory. Black has the move, and there is a best play for Black, instead of possibly incomparable plays. At the same time, you are incorporating the idea of environments into von Neumann game theory. Not that this is a bad idea, in fact it might be a great idea.
But it does seem to raise questions about the idea of best play in a part of the board, when best play may be elsewhere. Dominance in von Neumann game theory applies to results in the whole game, not a part of it. It is global, not local. Combinatorial game theory arose out of the idea of local games which could be combined into a single game. One consequence of that idea is that values are not strictly ordered.
I understand Bill and, in my mind, my analyse is valid only under the non-ko environment defined by the combinatorial game theory.
It is not so easy but I will try to explain my view in more details.
But for that I have first to verify we have a common understanding of the combinatorial game theory.
My view is the following for this theory:
First of all it is very important to know:
1) what is meant when, in combinatorial game theory, we assume the independancy between the local area and the environment
2) why ko fights create a mess in the theory
Let's take one of the major result of the theory:
Let's consider two positions A et B surrounded by an environment E with "good" carateristics (independancy? non-ko? ...)
Assume for example that, black to play, the optimum result of the game beginning from A+E is a win for black by say 10 points
Assume also that, black to play, the optimum result of the game beginning from B+E is a win for black by say 7 points
Let's call B' the mirror position of position B and let's call E' the mirror position of position E
The amazing result of theory is the following: black to play will wins the game A+B' by 3 points !!!
Let's us try to prove this "theorem":
Take two boards, board1 and board2.
On board1 you put the position A+E and on board2 you put the position B'+E'.
Now the players will play a difference game on these two boards.
With the assumptions above I guess black has the advantage and I would like to prove that, black to move, black will win this difference game.
For proving the win for black, I have only to find a winning black strategy and that is quite easy:
Black begins by playing on board1, with the intention to win on this board by 10 points.
After this first move white will choose a board for her answer and will play on this board.
From this point till the end of the game the strategy of black is to
always answer white move by a move on the board chosen for white last move.
With this strategy you can see black will win by 10 points on board1 and white will win by 7 points on board2.
So, black will win the difference game by 3 points
At that point comes the assumption that all areas (A, B, E) have good caracteristics allowing the following simplification:
The above difference game is made of the four areas A, B', E and E'. When you look at these four areas you see in particular the areas E and E' which look like perfect miai areas.
Here, we discover the basic assumption of all the theory:
Because E and E' are perfect miai areas, if black can win the game A+B'+E+E' then, providing good independance between the four areas, we can completly ignore the presence of the two areas E and E' => black wins the game A+B' by again 3 points.
Now just a small example to show why a ko is a mess for the theory. Let's take the following position:
Black to play
- Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | X a X X b X X X X |
$$ | X O O X O O O O O |
$$ | X O O X X O . O . |
$$ | X O . O X O O O O |
$$ | O O O O X X X X X |
$$ | . O X X . X . X O |
$$ | O O X X X X X O c |
$$ | X X X O O O O . O |
$$ | O O O O O O O O . |
$$ ---------------------[/go]
"a" and "b" are perfect miai points counted as very simple 4 points miai value.
In this game the correct sequence is black takes ko, white "a", black connects ko, white "b".
You can see clearly that the result would be very different if you ignore the miai points "a" and "b".
In other words the ko at "c" is really a mess because you cannot simplify the game by removing the miai areas.
Before continuing with our previous discussion I am waiting for your first comments on this very interesting theory.
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Bill Spight
is (one of) the best move in answer to 
and 
are forced. Then 
nails down the left and 
takes the sente on the right. The result is jigo. Therefore, Black cannot win and the keima is at least as good as the kosumi sequence.
captures two stones and wins by 1 point.
prisoner
connects and wins by 1 point.
prisoner
makes a mirror position, for jigo. OC, if 
is (one of) the best move for white then, against the kosumi, I choose the tsuke "a"
) you will be able to take into account whole white strategy!