I'll give a simpler example to explain what I mean. Linked on the SL page for tedomari, I see the following problem:
- Click Here To Show Diagram Code
[go]$$B Tedomari problem 1
$$ -------------------
$$ | . . X X a X O . . |
$$ | X X . X O . O . . |
$$ | X O X X O O O O O |
$$ | X O O O . O X X b |
$$ | O O . O O O O O X |
$$ | . O O X O X O X X |
$$ | . O X X X X X . X |
$$ | O O c O X . . X . |
$$ | . X . X X . . . . |
$$ -------------------[/go]
The problem is small enough that *maybe* I can read out each combination of play to see who will win in the end.
But the solution shows this approach:
1.) Calculate miai values of each location - seems easy enough. Maybe this is simple to do following the way of calculating miai values.
2.) Get overall count - kind of makes sense how this is done, but I don't see what the point is in terms of finding optimal play.
3.) Get tedomari and win. The solution shows how using the greedy approach (the one I would usually use unless I can read it out exhaustively) fails, and how getting the last play gives you a score that wins the game.
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OK, easy example, and I can see what is said is true, for the most part.
But this still leaves me confused:
1.) Why do you need overall count to play optimally?
2.) In this example, trying to "get" tedomari seemed to work to give you the right answer. But:
a. Is tedomari always optimal?
b. In general, is this the approach: calculate miai values, try to get tedomari? Let's say there are 10 spots on the board. So I calculate the miai value of each of those 10 spots and then I try to play to get the last play?
Basically, I wonder if this is imprecise, and when it fails, condition the method fails, and so on.
My current understanding:
* You can use miai values as a heuristic to estimate the order of play. Naïve way to go about this is to then take greedy approach, taking largest miai value as you get it.
* Tedomari is another heuristic that (usually?) works that lets you say, "OK. I'm not going to take the biggest moves in order, but I'm going to give up a big move so I can get the last play, and probably have more points..."
If my current understanding is correct, it would seem that my endgame approach should be as follows:
1.) If possible, read out entire game tree with all combinations to see best sequence (many search possibilities, but straightforward).
2.) If #1 is not feasible, calculate miai values for each spot. Then iterate all combinations of miai value plays until I see which gives greatest result. Is this approach optimal?
3.) If #1 and #2 are not feasible, try to figure out ordering that gives me tedomari. This approach is not always optimal...?
4.) If #1, #2, and #3 are all not feasible, order miai values largest to smallest, and play in that order.
5.) If #1, #2, #3, and #4 are not feasible... Try to keep sente.
6.) If #1, #2, #3, #4, and #5 are not feasible... Try to play big spots.
7.) If #1, #2, #3, #4, #5, and #6 are not feasible... Resign...?
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Is this the best approach?
Basically, to me it sounds like we have a bunch of heuristics and stuff, and I want to define precisely when an approach is optimal, and if we don't have time for the optimal approach, the next best approach to getting the solution.
Am I making any sense, or am I just typing?
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Bill Spight
rather than the one on the left edge, because it's better to leave a plain gote move rather than an asymmetric move that favors black.
gets rid of the asymmetric black-favoring position at the top, transforming it into a 1 point gote. Black tries to do the same to white with 
.
to convert it to an ordinary 1 point gote. So when the remaining 1 point gote moves trade off, white gets the last one with 
and wins tedomari.
. I tried a few variations and they all seem to come out the same. Either way, white takes the other one, and we end up with the same result.
does not. 
