Sente, gote and endgame plays

[Bill Spight]

You rely on the CGT definitions of local temperatures and mean. Can you explain the relation between the definitions and their application, please?

I'm not sure what you are asking.

Your answer "the method of multiples finds mean values, and temperatures can be derived from them" goes in the intended direction of my question.

You need a strategy at each decision point. So you end up with several comparisons.
[...] The method of multiples is inefficient. [...]
Each decision point requires a different strategy, so the eventual strategy must say what to do at each node. Finding the best strategy for an arbitrary tree can take a lot of work.
[...] you must have compared every possible strategy to find out which is correct [...]
in the arbitrary case the correct strategy can be arbitrarily complex [...]
why generalize, why indeed? Thermography is more efficient.

OIC. Looking at your examples, I thought that it was a 1 ply decision-making at the top level. You explain that it is iterative strategic decision-making bottom-up. If not thermography, we may use methods for value comparisons to distinguish gote from sente for short or long sequences worth playing successively. I also suspect that this would be more efficient than methods involving multiples, at least when we would need 8+ multiples.

Can you write down your general method as a procedure applicable to all examples of a class?

Not sure what you are asking. It's a general method for comparing strategies and classifying positions, but more work needs to be done to verify means and temperatures.

If it is a general method, state its procedure explicitly - otherwise, it is not general:)

For the class of all examples with one player's simple follow-up, stating the procedure is straightforward. Can you also state it so that it applies to iterative follow-ups? (If you have the time for doing so and think it is worthwhile.)

The overall strategy must say what each player plays at each choice point. Each option that is part of the strategy must be equally represented in the combined instances. These simple examples have only two choice points. The sente strategy gives only one option at each choice point, and each player makes the same number of plays, so all it requires is a single instance. The gote strategy gives an option for each player at each choice point, and requires four instances. So to compare the results of the two strategies requires four instances.

Suppose that we had another follower, with only one obvious choice for each player. Then we would have these possible strategies with White playing first: White sente, first play by White gote and the second play by White sente, each play by White gote. The comparisons would require 8 instances.

I am attaching another SGF file with comments that might make things clearer.

[RobertJasiek]

Code: Select all

          A
         / \
        /   \
       /     \
      B       C
     / \     / \
    D   E   F   G

1) Does a local endgame position exist with (B-C)/2 = E-C = B-F > 0? That would be a doubly ambiguous local endgame with short sequences and without other options having long sequences.

2) Does a local endgame position exist with B-F = (B-C)/2 > E-C > 0? That would be Black's sente and ambiguous on White's sente sequence despite the only short sequences. If you have the opposing colour preference, it would be 0 < B-F < (B-C)/2 = E-C.

[Bill Spight]

Code: Select all

          A
         / \
        /   \
       /     \
      B       C
     / \     / \
    D   E   F   G

1) Does a local endgame position exist with (B-C)/2 = E-C = B-F > 0? That would be a doubly ambiguous local endgame with short sequences and without other options having long sequences.

(B-C)/2 = E-C = B-F

(B+C)/2 = E = F = A

This is miai or double ko threat.

2) Does a local endgame position exist with B-F = (B-C)/2 > E-C > 0? That would be Black's sente and ambiguous on White's sente sequence despite the only short sequences. If you have the opposing colour preference, it would be 0 < B-F < (B-C)/2 = E-C.

F = (B+C)/2 > E

If F > 0 > E then A = 0; this is seki.

If F > E ≥ 0 then A = E.

If 0 ≥ F > E then A = F.

[RobertJasiek]

(B-C)/2 = E-C = B-F

(B+C)/2 = E = F = A

This is miai or double ko threat.

Do you say that the first is miai and the second a mutual ko threat?

How does a connected local endgame that is a miai look like? Note the required short sequences: we shall have numbers after the first two plays. Do such positions exist on the board?

A local endgame that is a mutual ko threat has the move value 0 and therefore does not fulfil my > 0 assumption.

F = (B+C)/2 > E

If F > 0 > E then A = 0; this is seki.

A seki has the move value 0 and therefore does not fulfil my > 0 assumption.

If F > E ≥ 0 then A = E.

If 0 ≥ F > E then A = F.

My problem is: how do such actual connection local endgame positions on the board look like with only short sequences before reaching numbers? Do they exist as positions on the board?

If we allow long sequences, there are such positions. The trouble is finding some for short sequences or proving their inexistence.

[Bill Spight]

(B-C)/2 = E-C = B-F

(B+C)/2 = E = F = A

This is miai or double ko threat.

Do you say that the first is miai and the second a mutual ko threat?

Except for the equation indicating equality with A, the two are equivalent.

E-C = (B-C)/2
E = (B-C+2C)/2 = (B+C)/2

(B-C)/2 = B-F
F = (2B-B+C)/2 = (B+C)/2

How does a connected local endgame that is a miai look like? Note the required short sequences: we shall have numbers after the first two plays. Do such positions exist on the board?

One example. Many similar positions go positions exist.

Code: Select all

          A
         / \
        /   \
       /     \
      B       C
     / \     / \
    9   5   5   1

A local endgame that is a mutual ko threat has the move value 0 and therefore does not fulfil my > 0 assumption.

This one satisfies your requirement. (B-C)/2 = B-F = E-C = 4 > 0.

F = (B+C)/2 > E

If F > 0 > E then A = 0; this is seki.

A seki has the move value 0 and therefore does not fulfil my > 0 assumption.

Seki example.

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          A
         / \
        /   \
       /     \
     16      -6
     / \     / \
   37  -5   5  -17

Your required relation: B-F = (B-C)/2 > E-C > 0

16-5 = 11 = (16-(-6))/2 > -5-(-6) = 1 > 0

If F > E ≥ 0 then A = E.

If 0 ≥ F > E then A = F.

My problem is: how do such actual connection local endgame positions on the board look like with only short sequences before reaching numbers? Do they exist as positions on the board?

If we allow long sequences, there are such positions. The trouble is finding some for short sequences or proving their inexistence.

I am unaware of any example on the go board with F > E which satisfies all of your requirements. However, if such examples exist, they will in theory have the values given. (As we know from the latest Japanese rules, the scores may be different from the theoretical values. )