Numsgil wrote: Anyway, I think I'm uncomfortable with positions like this because I don't know how to bound my error.
Do you mean your error in playing, not your error in calculating? If so, Berlekamp has a strategy called Sentestrat, which does so when there are no kos. It does require accurate calculation. Simply making the largest play does not bound your error.
Code: Sentestrat
Start with a variable called the temperature, equal to the miai value of the largest play on the board. Once that play is made, the value of the temperature becomes the miai value of the largest play now on the board, as long as it is the same or smaller than the current temperature. If the largest play on the board has a higher miai value than the current temperature, the value of the temperature remains the same, until the miai value of the largest play drops below it.
A sente play is defined as a play that puts a play on the board with a miai value greater than the current temperature.
Sentestrat says to answer all of the opponent's sente plays. :)
Sentestrat is not very sophisticated, but it does bound your error. 
Quote: The normal endgame counting can fall apart if you just blithely assume you'll get all the sente plays you want.
I take it that you mean that sometimes your opponent will get to make reverse sente plays.
Quote: So I have an abstract example (I'd try to make an actual position for it, but it's pretty tricky to work backwards from counts to a position).
Let's say I have 3 unrelated endgame sequences to resolve, and then the game is over. Assume that the point values below are all black's gain if he gets the move, and white doesn't gain any points when he plays.
Do you mean score or count? E. g., a position with this game tree,
Code: A
/ \
4 0
where / indicates a Black play and \ indicates a White play, is what you mean by one where Black gains 4 points and White gains nothing?
Quote: Sequence A: 4 point gote play (A1) with a 1 point sente followup (A2).
Sequence B: 5 point gote play (B1) with a 2 point sente followup (B2).
Sequence C: 4 point gote play (C1) with a 3 point gote followup (C2), which itself has a 5 point sente followup (C3).
With normal endgame counting, the values would be: A:5, B:7, C:8.
If so, it sounds like the game tree for A looks like this:
Code: A
/ \
A1 0
/ \
A2 4
/ \
Big 5
And B looks like this:
Code: B
/ \
B1 0
/ \
B2 5
/ \
Big 7
And C looks like this:
Code: C
/ \
C1 0
/ \
C2 4
/ \
C3 8
/ \
Big 13
Is that what you mean? (I am really not sure about C.)
Quote: So how do we actually handle this sort of position in a way that assures us we can play optimally and find the actual final score? If we use the endgame counting heuristic, are we at least assured that the first move we play will be optimal?
No, we do not have that assurance. The largest play is not always the best play. And in fact, always making the largest play can have a theoretically unbounded error. Berlekamp's Sentestrat does bound the error.
Now, with the game trees that I have constructed, a play in A gains 2.5 points, a play in B gains 3.5 points, and a play in C gains 4 points. The original count is 10 points. If each player takes the largest play, then
Black plays C, for a count of 14.
White plays C1, for a count of 10.
Black plays B, for a count of 13.5.
White plays A, for a count of 11.
Black plays B1 with sente, for a count of 11.
(You can verify that Black gets 4 in C and 7 in B.)
Can either player do better?
|
). Add everything up, and you get 14 points in gote for black. This takes me (probably) about 20 seconds to determine, plus another 10-20 seconds to verify, so it terms of time management, it's very feasible to do this whilst in byo-yomi. I think that no other method is anywhere near as fast.
in diag. 1 and 
in diag. 2 are gote):