Now that we have the whole problem, adding white's tiger mouth(at 7-2) simplifies the calculations by a lot.
White gains 1 point (A=1) from being able to block if black plays first, and white gains slightly more than zero points (B=0+) from being able to hane and not worry about the connection. Hence, overall calculating from black's perspective, the move gain should have changed by (-A - (-B))/2 = (B-A)/2 or a bit more than -0.5.
Hence, assuming Robert's calculations are correct, I will guess that the answer to the problem is a gain of
2 1/12 - 0.5 + (0+) ~ 1 2/3
where (0+) means something a bit more than zero.
A little under 2 should be correct, agreeing with lichigo's teacher and JF's estimate.
____
I will give my own answers to JF's questions.
John Fairbairn wrote:
(1) In practical terms what is the difference between 1.67 and "a little under 2". Or, more specifically, What is it about this difference that makes a too-difficult-for-me and error-prone (error even by a yose maven - see above) calculation worth attempting? The miniscule number of games in which it can make a genuine difference is probably more than outweighed by the number of times the calculations are miscalculated anyway.
It is a theorem of miai counting (see RJ's Endgame 5 I think) (ignoring kos) that if you don't play the right move when the largest endgame move has gain T, then you lose (under optimal play) at most 2T.(edit: I wrongly wrote T here before)
A simple gote is a position settled by one move (or one gote forcing sequence) by either side.
If only simple gotes remain and you play a move of gain t instead of T (t<T), then you lose at most 2(T-t).
For example not playing the final 0.5 gain but passing (0 gain) instead will lose you 2(0.5-0)= 1 point.
(This result can be found from a discussion between Robert and I previously).
I do not know how this generalises to more complicated endgames, but simple gotes are often a good rule of thumb.
As for whether it is worth it, I think such calculations are often not worth it unless it is a very close game and you have some time to work it out. To do a full miai calculation, you need to calculate every single possible optimal variation (including tenuki) locally which is very tiresome. It can be used to double check and pre-calculate the values of moves, allowing memorisation before a match (I think you mentioned how Rob van Zeijst memorised the values of hundreds of yose positions). Although the theory works perfectly and is simplest for small and simple problems in the late endgame, the principles still work for the early endgame, where the sizes of moves are bigger and hence it is more important to play accurately. You can use rules of thumb or estimates based on principles you learn from doing the simpler problem.
From a mathematical point of view, I find the theory impressive in how it solves what seems like a very complicated problem (the endgame of Go) in a fairly clean way, almost to perfection. Only almost because it can't easily handle ko and independence tends to require "immortal stones" which isn't always realistic.
John Fairbairn wrote:
(2) The REAL question, which everyone (pros and amateurs) seems to avoid answering, is still: why use one technique over another, especially when we keep getting told you can convert from to the other by dividing or multiplying by 2? Even if there is some obscure reason for saying one is superior to the other that can be demonstrated, it's not enough to say vague things like it helps with strategic choices. It may - but just saying just that is just like saying this margarine tastes like butter (and then wondering why most people and top chefs still prefer butter).
For simple gotes, deiri counting = miai counting * 2.
For anything more complicated, miai counting handles it perfectly, whereas deiri is a simplification (I don't know how you define deiri counting for more complicated problems, but probably by the swing - the difference between optimal play if black plays first and optimal play if white plays first.) Frankly, most of the time the half the swing is a very good estimate for a move gain. (NB: I may have slightly misused the term "swing" here).
It is a go player's intuition that it is a good rule of thumb, while it is a mathematicians job to prove that, and the only way is to use the more general theory in miai counting. Miai counting will handle all the weird and wonderful cases, as well as create "magic tricks", manufacture problems which are counter to intuition or any one rule of thumb.
Why multiply/halve by 2? Well, deiri simply counts the points difference between two real board positions, so it is natural to complain and ask what the point of dividing by 2 is. Miai counting requires a bit more abstraction/imagination and tries to look at all possible variations at once, but it makes it very natural to think in terms of each move making a contribution to the score. The final score is initial count plus the sum of the gains of all your moves minus the sum of the gains of your opponent's moves. S=C+Σm-Σm
In this case, the gain is like half the deiri move value, since it requires two moves (one by you or one by your opponent) to get to those two board positions being compared.
___
Just from watching Chinese pro commentaries careful counting beyond rough estimates (+/- 1 point) is rare. But they sometimes talk about 1/6, 1/12 etc. (sometimes in a joking way, or saying how such analysis can be very professional), or about how a cut on the 2nd line is best defended by a tiger mouth than the descend or solid connection (i.e. CGT Tinies/Minies), but I haven't heard much more than that. Some do seem to be able to calculate the sizes of small endgame using miai counting very quickly, but as kvasir says, they use double the gain instead, mimicking deiri numbers. But I don't know how much they know about the theory or invest it understanding it/using it behind the scenes. In my imagination, there are experts at universities. But in any case, I get the impression Bill Spight's understanding of and ability to analyse endgame is much more advanced than most top pros.
As for practical theory, I would like to push a version of miai counting on the board that I am surprised not to have seen anywhere else. Perhaps JF has been hinting at it?
Let's call it
Iterative deiri countingThe idea is to start with deiri and then iteratively improve estimates. Perhaps this will help people who are good at deiri counting make a slight upgrade in accuracy.
Start with deiriThe deiri estimate compares
- Click Here To Show Diagram Code
[go]$$B Black plays first
$$ | X X X . . . . . .
$$ | . . X . . . . . .
$$ | . X X X X O O O O
$$ | C X O O O . O . .
$$ | C 3 1 2 . . . . .
$$ ------------------[/go]
This diagram is a gote forcing sequence(
,
are sente)
to
- Click Here To Show Diagram Code
[go]$$W White plays first
$$ | X X X . . . . . .
$$ | . . X . . . . . .
$$ | . X X X X O O O O
$$ | 2 X O O O . O . .
$$ | . 1 . C . . . . .
$$ ------------------[/go]
The difference is 2 black points and 1 white point or 3 points total.
Divide by 2 to get a miai count of 3/2= 1 1/2.
In general, we divide by 2^k where k is the number of moves it takes to get a different move in the 2 diagrams we are comparing. In the above 2 diagrams, move 1 is already difference, so k=1 and 2^k=2 as expected.
First iterationThe second diagram was not a gote forcing sequence, so we need to insert
: tenuki
- Click Here To Show Diagram Code
[go]$$W White plays first, Black tenukis, White continues
$$ | X X X . . . . . .
$$ | . . X . . . . . .
$$ | 6 X X X X O O O O
$$ | 3 X O O O . O . .
$$ | . 1 5 . . . . . .
$$ ------------------[/go]
Here,
is gote, but
and
are miai.
In this diagram, black has lost a point at
, but as
is in atari, black has an extra 1/3.
Hence, this diagram gains white 2/3 over the second diagram (a deiri value).
We need to check if this position can arise. i.e. If
is sente, this variation is impossible. To check, at the divergence point, white has played two more moves than in the 2nd diagram. (i.e.
,
is 2 more W moves vs
is even). At the divergence point, the 2nd diagram has one more white move than the 1st diagram. We divide the move value by the number of moves difference. So compare
(2/3)/2 to (1 1/2)/1
The former is much less so
is gote for sure. So we include this variation.
But as this position arises from a follow-up, and divergence is on the 2nd move, k=2.
We must divide by 2^k to get (2/3 )/4 = 1/6.
There are no other variations. Therefore, the gain of the first move is 1 1/2 + 1/6 = 1 2/3.
Woohoo! My guess at the top of this post was correct.
The general process is to add more variations at depth 1, then depth 2 and so on as required. You do not need to do the full miai calculation starting from the bottom of the tree. Instead, deiri from the top with iterative deepening is a better practical method.
You can perhaps understand how this idea is helpful for estimating the size of very complicated large endgames. I have little doubt that pros and amateurs have been using this sort of technique once they realise deiri counting isn't perfect. But it takes a little miai counting theory to prove that it works and to make sure you are doing the steps correctly - checking gote/sente as well as choosing the correct k to divide by 2^k.
To me this method also reminds me of a lecture by Bill on influence functions. Those familiar with the halving influence down CGT corridors may know what I mean. I have plans to develop this theory further.
BTW: I tried to develop this method on the spot in this video:
https://www.youtube.com/watch?v=rcz9b6k ... e&index=15 but made a mess of it. Hopefully my explanation above is more clear.