I don't remember exactly where this came from, but it looks like the question is one of reversal.

Click Here To Show Diagram Code

[go]$$W Does - reverse?
$$ -----------------
$$ | . . 2 1 . . O .
$$ | X X X X O O O .
$$ | . . . . . . . .[/go]

It does so if the position after (Q) is greater than or equal to the original position (P) from Black's point of view, i.e., if P - Q ≤ 0. I.E., if Black to play in P - Q cannot win with correct play (by definition). To answer that question we do not have to look at play when White plays first. To show that in addition P != Q we also may have to look at the result when White plays first.

(The CGT definition says that White wins if Black plays first, but CGT regards 0 as a second player win. In the context of go, which allows ties, we just say that the first player, Black, cannot win.)

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

As long as there are no ko fights, one can compare any two local endgame positions P and Q. In particular, one can ask how their difference compares to 0.

CGT-reversal is one application for comparing two positions in a difference game. As you guessed from my positions, it is the application I am currently studying.

The relations >= and <= seem to be easier for go scoring than the relations > and <. Therefore, let me stick to the former in this message. I redo my study according to your reply, using Black's value perspective.

Click Here To Show Diagram Code

[go]$$B Example 1: P + (-Q)
$$ ---------------------------------
$$ | . . . . . . O . X . . X O . . |
$$ | X X X X O O O . X X X O O O O |
$$ | . . . . . . . . . . . . . . . |[/go]

Click Here To Show Diagram Code

[go]$$B Black starts I, score 0
$$ ---------------------------------
$$ | . . . . 1 . O . X . 2 X O . . |
$$ | X X X X O O O . X X X O O O O |
$$ | . . . . . . . . . . . . . . . |[/go]

Click Here To Show Diagram Code

[go]$$B Black starts II, score 0
$$ ---------------------------------
$$ | . . 3 2 4 . O . X . 1 X O . . |
$$ | X X X X O O O . X X X O O O O |
$$ | . . . . . . . . . . . . . . . |[/go]

We have P - Q ≤ 0 because the correct resulting score is 0 so Black starting cannot win.

As an application, White's alternating sequence from P to Q reverses:

Click Here To Show Diagram Code

[go]$$W reversal
$$ -----------------
$$ | . . 2 1 . . O .
$$ | X X X X O O O .
$$ | . . . . . . . .[/go]

Click Here To Show Diagram Code

[go]$$B Example 2: P + (-Q)
$$ ---------------------------------
$$ | . . . . . . X . O . . O X . . |
$$ | O O O O X X X . O O O X X X X |
$$ | . . . . . . . . . . . . . . . |[/go]

Click Here To Show Diagram Code

[go]$$W White starts I, score 0
$$ ---------------------------------
$$ | . . . . 1 . X . O . 2 O X . . |
$$ | O O O O X X X . O O O X X X X |
$$ | . . . . . . . . . . . . . . . |[/go]

Click Here To Show Diagram Code

[go]$$W White starts II, score 0
$$ ---------------------------------
$$ | . . 3 2 4 . X . O . 1 O X . . |
$$ | O O O O X X X . O O O X X X X |
$$ | . . . . . . . . . . . . . . . |[/go]

We have P - Q ≥ 0 because the correct resulting score is 0 so White starting cannot win.

As an application, Black's alternating sequence from P to Q reverses:

Click Here To Show Diagram Code

[go]$$B reversal
$$ -----------------
$$ | . . 2 1 . . X .
$$ | O O O O X X X .
$$ | . . . . . . . .[/go]

Correct?

Correct.

And, OC, if P - Q ≤ 0 and P - Q ≥ 0, then P - Q = 0.

Edit: When considering possible reverses at go, it is conceivable that you could reach a position where the second player has no play. E.g., there is an open point, but to play there would be illegal. In such a case you can require the second player to pass but at the cost of 1 point by territory scoring.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

That is not my understanding Robert.

First of all I understand the notation P - Q >= 0 only in the context of CGT where the result of a game is a win or a loss but never a draw. As a consequence when you mentionned a "score 0" it is not a sufficient information to decide whether it is a win or a loss. You have to look at which side is taking the last move.
Taking your example the result are the following:
Black starts and loses
White starts and loses

In this context what about the status of P - Q ?
Here my understanding (for black point of view) is the following:
Black starts and loses means P - Q <= 0
White starts and loses means P - Q >= 0
and taking both results I conclude P - Q = 0



My understanding of P - Q > 0 is a little more complicated because here I need to know the resut when Black plays first and also when white plays first:
P - Q > 0 means Black starts and wins AND White starts and loses

The CGT definition uses last play or chilling but we need not be slaves of CGT when we are really interested in counts and scores. We must just be careful to translate the outcome classes and their conditions correctly.

OK Robert. In order for me to try and answer your questions, how do you suggest to translate the CGT sentence "Black starts and loses" in your go language made of counts and scores?
In your example it seems that "Black starts and achieves the score 0" means "Black starts and loses" but with other examples I doubt it is always true.
You can see the problem : if you use an ambiguous language you cannot expect an accurate answer to a question made in this language.
In CGT language a win or a loss is very clear and relations like P - Q >= 0 or P - Q > 0 can be defined with very good properties.
For the time being I am not aware of a translation of this CGT language into yours.

Like I say, I am really quite busy these days, but let me offer a brief response.

As Gérard points out, and as we all know, CGT does not have ties. We all also know that ko fights do not fit into the CGT framework, but there are workarounds. We all also know that seki can have quite different results, depending on the rules. We all also know that the group tax is part of the CGT framework for both area and territory scoring, but we can work around the lack of a group tax.

The CGT rules of comparison with 0 are simple. Given a game, G, let us play it twice by correct minimax play, once with Black playing first, and once with White playing first.

1) if G is a second player win, G = 0;
2) if G is a first player win, G <> 0;
3) if G is a Black win, G > 0;
4) if G is a White win, G < 0.

Assume that we have taken care of the group tax and seki. All of these results are still possible. But we will also have jigo results. What is lost if we treat those results the same as CGT does? Will correct minimax play assigning wins and losses a la CGT not be correct when we allow jigo? Sure, sometimes the CGT classification of G will be a Black win instead of jigo, but best play a la CGT will still be best play with jigo. It's just that Black will get the last play.

Edit: It may seem that a half point komi confounds the CGT model by making the last play irrelevant. The half point komi can turn a Black win into a Black loss.

But all you have to do is implement the komi with this game, Komi = {-1 | 0}, and play G + Komi. Then if Black wins G by making last play to 0, White plays to 0 in the Komi and wins G + Komi.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

I have seen a result function R, which we can reinterpret as 'count', in some CGT texts. Below is my naive attempt of defining the outcome classes.

Playing the difference game compares to 0 the result of play in a local position and some colour-reversed local position. In combinatorial game theory, two local endgames A and B are compared using a result function R as follows: A ? B :<=> R(A + E) ? R (B + E) for all finite, non-cyclic environments E. A result function assigns the count after both players' best play in the local endgames. We compare the result of the local difference game to 0, that is, R(A - B) ? 0. The environment E drops out when forming the difference.

Suppose we compare the two local positions P and Q. The colour-reversed of Q is its negative, that is, -Q. Accordingly, we can write and transform the comparison as follows: P ? Q <=> P - Q ? 0 <=> P + (-Q) ? 0. In other words, the difference game compares the imagined combined position to 0. We have

P > Q if White starts and Black achieves more than 0 as a count (White starting cannot prevent Black's win),

P ≥ Q if White starts and Black achieves at least 0 as a count (White starting cannot win),

P = Q if the count is 0 for Black's and White's starts (neither starting player can win),

P ≤ Q if Black starts and White achieves at most 0 as the count (Black starting cannot win),

P < Q if Black starts and White achieves less than 0 as the count (Black starting cannot prevent White's win),

P || Q if 1) Black starts and achieves a count larger than 0 (Black starting wins) and 2) White starts and achieves a count smaller than 0 (White starting wins).

OK, but as Berlekamp defined count, all scores are counts but not all counts are scores.

Let G be a finite combinatorial game, G, with a count, R.

G <> 0 or

(G > 0 and R > 0) or

{G < 0 and R < 0) or

(G = 0 and R = 0)

N.B. I am using <> to mean is confused with. || is ambiguous with slash notation.



E - E = 0.



{-1 | 1} = 0

What you should be interested in is scores, not counts. If you do not play when the position has a score, then in {-1 | 1} White does not play to 1, but stops at 0; i.e., does not play at all. That solves that problem.

The particular scores you are interested in are stops. A stop is the first score reached with optimal minimax play. For game, G, let S(B) be the stop when Black plays first and S(W) be the stop when White plays first. OC, S(B) ≥ S(W).

1) If S(B) < 0 then G < 0.

2) If S(W) > 0 then G > 0.

3) If S(B) > 0 > S(W) then G <> 0.

4) If S(B) = 0 and S(W) < 0 then G <| 0.

5) If S(B) > 0 and S(W) = 0 then G |> 0.

6) If S(B) = S(W) = 0, then G = 0 or G is an infinitesimal.

Now, we may not always realize when we have reached a score without playing the game out. For example, {3|3||0|||0||-5|-7} = 0.

Also, for difference games who gets the last play in a jigo may well matter, because sente usually matters.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Thanks, it seems to make sense!

(For those who don't know: 'confused with' is sometimes called 'incomparable'. <| means 'smaller or incomparable', |> means 'larger or incomparable'.)

EDIT: But where in your scheme do P ≥ Q and P ≤ Q fit?

It's not my scheme, it's Conway, Berlekamp, and Guy's scheme.

Let Black play first in G, with correct play. The results are the opposite if White plays first.

1) If Black wins, then G |> 0.

2) If White wins, then G ≤ 0.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

I think you mean

2) If White wins or ties (that is, Black cannot win), then G ≤ 0.

No, I meant in the CGT sense. Either player can win a tie.

I came up with the "cannot win" formulation for certain difference games, such as those for comparing two options or for testing for reversal. The point being that in the difference game P - Q for those questions, P - Q has the same number of stones of each color and an even number of empty points. If you get a jigo result it should come after a even number of plays and thus indicate a second player win in CGT. That has been my experience. But that is not the case in general.

I did not want to teach unfamiliar CGT concepts for people to be able to use difference games for those purposes. However, it is certainly possible for jigo to occur in such difference games with an odd number of plays and an odd number of dame, but, IMX, that would be a rare occurrence. I think I was wrong not to address that issue. You can do so in terms of sente and gote, I think, without having to invoke CGT. And, IMX, it is very unlikely to come up, anyway.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

In in the great majority of cases when S(B) = S(W) = 0 the optimal sequences are gote for who plays first.
Do you have an example in which one of the optimal sequence (black plays first or white plays first) is sente with S(B) = S(W) = 0 ?

Not IMX. The positive and negative infinitesimals are sente for one side, gote for the other. Then you have infinitesimals that are confused with 0, which are gote for both sides, and the cases where G = 0, which are sente for both sides, unless a play by either side is non-existent or a loss.



Well, Robert's example, where G = 0, is one.

Click Here To Show Diagram Code

[go]$$B Example 1: P + (-Q)
$$ ---------------------------------
$$ | . . . . . . O . X . . X O . . |
$$ | X X X X O O O . X X X O O O O |
$$ | . . . . . . . . . . . . . . . |[/go]

P - Q has the same number of Black stones and White stones, and an even number of empty points. A jigo result is thus very likely to be sente for both sides.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Oops I wrote the contrary of what I had in mind. Sorry for that.
In the majority of cases, when S(B) = S(W) = 0 in a difference game, the two areas P and -Q are miai and the sequence is sente for who plays first (=> in CGT context it is a loss for the player who plays first)
Do you know situations where one of the sequence is gote and S(B) = S(W) = 0?

How silly of me! Consider {0,*|0}. If we were to use a difference game to choose between the Black options, we could get DG = * - 0 = *, which is confused with 0.

OC, such examples are not uncommon in chilled go, but I don't remember finding one in regular go when I was asking that question. But all you need is a choice between two seki with different parity.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

In the video https://www.youtube.com/watch?v=3kj3YQb9P74 Michael Redmond propose the following postion :

Click Here To Show Diagram Code

[go]$$B White to play
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . O . . . O . O . . . . |
$$ | . . . . . . . . . . . . . . O X X . . |
$$ | . . . . . . . . . . . . . . . O . . . |
$$ | . . . . . . . . . . . . . . . O O . . |
$$ | . . . . . . . . . . . . . . X O X . . |
$$ | . . . . . . . . . . . . . . . X X . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ -----------------------------------------[/go]

and shows the failure of white in the sequence:

Click Here To Show Diagram Code

[go]$$B White to play
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . O . . . O . O . . . . |
$$ | . . . . . . . . . . . . . . O X X . . |
$$ | . . . . . . . . . . . . . . 7 O 3 2 6 |
$$ | . . . . . . . . . . . . . . . O O 1 5 |
$$ | . . . . . . . . . . . . . . X O X 4 8 |
$$ | . . . . . . . . . . . . . . . X X . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ -----------------------------------------[/go]

at
at

Click Here To Show Diagram Code

[go]$$Bcm11 White to play
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . O . . . O . O . . . . |
$$ | . . . . . . . . . . . . . . O X X 3 . |
$$ | . . . . . . . . . . . . . . X O X O O |
$$ | . . . . . . . . . . . . . . 1 O O 2 O |
$$ | . . . . . . . . . . . . . . X O X O O |
$$ | . . . . . . . . . . . . . . . X X 5 6 |
$$ | . . . . . . . . . . . . . . . . . 4 . |
$$ | . . . . . . . . . . . . . . . . 7 . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ -----------------------------------------[/go]

My question concerns only the last move
Is it the best move to capture the white stones?

Isn't it better to play

Click Here To Show Diagram Code

[go]$$Bcm11 White to play
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . O . . . O . O . . . . |
$$ | . . . . . . . . . . . . . . O X X X . |
$$ | . . . . . . . . . . . . . . X O X O O |
$$ | . . . . . . . . . . . . . . X O O O O |
$$ | . . . . . . . . . . . . . . X O X O O |
$$ | . . . . . . . . . . . . . . . X X X O |
$$ | . . . . . . . . . . . . . . . . 7 O . |
$$ | . . . . . . . . . . . . . . . . . 8 . |
$$ | . . . . . . . . . . . . . . . . 9 . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . a . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ -----------------------------------------[/go]

in order to avoid white sente move around "a"?

Remark
We have studied the following theorems.


Definitions
The following local endgames M0|F0 and M1|F1 each with one simple follow-up have the sente(?) move values M0, M1 and follow-up move values F0, F1.


Presuppositions
Suppose the local endgames {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} with M0 ≥ M1 and F0 ≥ F1 and a constant number C.

Theorem 1
Playing to {2*F0 | 0} + {2*F1 | 0 || -M1} is at least as good as playing to {2*F1 | 0} + {2*F0 | 0 || -M0}.


Definitions
If M0 ≥ M1 and F0 ≥ F1, we write M0|F0 for the larger local endgame and M1|F1 for the smaller local endgame.


Presuppositions
Suppose the local endgames with simple follow-ups M0|F0 and M1|F1 of the creator with M0 ≥ M1 and F0 ≥ F1 in a (possibly empty) environment E without kos now or later.

Theorem 2
The creator's start in M0|F0 is at least as good as in M1|F1.


Presuppositions
Suppose the local endgames each with one follow-up M0|F0 and M1|F1 of the creator with M0 ≥ M1 and M0 + 2*F0 ≥ M1 + 2*F1 in a (possibly empty) environment E without kos now or later.

Theorem 3
The preventer's start in M0|F0 is at least as good as in M1|F1.


Presuppositions
Suppose the local endgames each with one follow-up A|B and C|D of the creator with A ≥ C and A + 2B ≥ C + 2D.

Theorem 4
A correct start of the creator is playing at A.


Remarks and Questions
- For theorems 1 + 2, you have said that the local endgames can be local gotes, ambiguous or local sentes.
- For theorem 1, it is obvious that M0 and M1 are sente move values.
- For theorem 2, we have seen in the proof that {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} appear so we also know that M0 and M1 are sente move values.
- Is it correct for theorem 3 that M0 and M1 are sente move values even if one or both local endgames are local gotes?
- Is it correct for theorem 4 that A and C are sente move values even if one or both local endgames are local gotes?

IIRC, {2F0 | 0 || -M0} + C0 in CGT notation is equivalent to M0|F0 in your notation, where C0 is a number, but F0 and M0 need not be. If F0 and M0 are numbers, then we can use a difference game to prove things about M0|F0 and M1|F1. In that case, whether we call the positions sente or gote is irrelevant to the proof.

These days I have a lot to do, and I am ailing. Not with any serious ailment, but enough to make it difficult to get things done.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.