In a game {A||B|C} or {A|B||C}, M|F stands for the numbers that are the (hopefully sente) move value M and follow-up move value F. There, B need not equal 0 while your sente followers are 0.

Given two simple endgames in an environment with no ko caveat, G0 = {2*f0 | 0 || -m0} + c0 and G1 = {2*f1 | 0 || - m1} + c1, where f0, f1, m0, m1 are numbers ≥ 0 and c0 and c1 are numbers: WOLOG, let f0 ≥ f1.

1) If f0 = f1, it is better for Black to play in the game with the greater of m0 or m1.

2) If f0 > f1:
a) if m0 ≥ m1, it is better for Black to play in G0;
b) if m1 > m0 and 2*f0 > 2*f1 + m1, it is better for Black to play in G0;
c) if m1 > m0 and 2*f1 + m1 ≥ f0, which game is better for Black to play in depends upon the environment.

These rules do not depend upon local sente or gote, but when G0 is local sente, it may be right to play in G0, no matter how small m0 is. Thus, the usual heuristic of making the hotter play may be wrong.

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I need to understand some basics of adding a number K (call it the komi, if you like) to a game.


{2F | 0 || -M} has the sente move value 0 - (-M) = M and follow-up move value (2F - 0) / 2 = F.

{2F + K | 0 + K || -M + K} has the sente move value (0 + K) - (-M + K) = M and follow-up move value ((2F + K) - (0 + K)) / 2 = F.

Is {2F | 0 || -M} + K = {2F + K | 0 + K || -M + K}?


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}.


Hence, may we also write the following?

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

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

I follow the convention of using lower case letters for numbers and upper case letters for games (which might also be a number). In what follows below I will take all of your letters to indicate numbers, to keep things simple.



If F, M ≥ 0, yes.

Note: It is important that {2F | 0 || -M} is not a number.



To compare moves in an environment without a ko caveat, we may set up a difference game.

Thus, given G0 = {2*f0 | 0 || - m0} + c0 and G1 = {2*f1 | 0 || - m1 + c1, environment E, and komi, k, we start by setting up this 0 game:

{2*f0 | 0 || - m0} + c0 + {2*f1 | 0 || - m1 + c1 + E + k +
{m0 || 0 | -2*f0} - c0 + (m1 || 0 | -2*f1} - c1 - E - k
=
{2*f0 | 0 || - m0} + {2*f1 | 0 || - m1 +
{m0 || 0 | -2*f0} + (m1 || 0 | -2*f1}

Note that c0, c1, E, and k all drop out.

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

Visualize whirled peas.

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(You have omitted three } brackets.)

***

F, M ≥ 0 because they are move values.

{2F | 0 || -M} must be interesting to play as an unsettled game so, for that purpose, I understand why it should not be a number. Do you have another reason in mind why it should not be a number?

***

I see all those extra numbers dropping out in your difference game but what do you want to tell me with it for "theorem 1 + K"?

***

Since {2F | 0 || -M} + K = {2F + K | 0 + K || -M + K}, let me discuss the following again:

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.

Simply speaking, the proof works like this: [Strategy yields] the local endgames {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1}, where we assume 0 at the sente followers as our calibration WLOG. Therefore, by theorem 1, theorem 2 is proved for its presuppositions. That is, the creator's two local endgames have arbitrary numbers of their sente followers so that the move values are M0|F0 and M1|F1, where M0 and M1 are sente move values indeed, even for local gote(s).

***

A closer reading of theorem 3 and its proof comes next to verify or refute _sente_ move values even for local gotes.

Given: G0 = {2*f0 | 0 || -m0} and G1 = {2*f1 | 0 || -m1} in a non kocaveat environment, where m0 ≥ m1 ≥ 0 and f0 ≥ f1 ≥ 0.

To prove: A Black play in G0 is at least as good as a play in G1.

First, set up this zero game:

{2*f0 | 0 || -m0} + {2*f1 | 0 || -m1} + {m0 || 0 | -2*f0} + {m1 || 0 | -2*f1}

Let Black play in G0 and White play in -G1, to produce this difference game.

D = {2*f0 | 0} + {2*f1 | 0 || -m1} + {0 | -2*f1} + {m0 || 0 | -2*f0}

Black's play in G0 is at least as good as a play in G1 iff D ≥ 0, that is, White to play in D cannot win.

1) White could play in {2*f0 | 0}, but we know that a play in {m0 || 0 | -2*f0} is at least as good. If White plays in {m0 || 0 | -2*f0}, Black replies in {2*f1 | 0 || -m1}, for jigo.

2) If White plays in {2*f1 | 0 || -m1} then Black can reply in {m0 || 0 | -2*f0}, and then White plays in {2*f0 | 0} and Black replies in {0 | -2*f1}, for a result of m0 - m1, which is a win or tie for Black.

3) White could play in {0 | -2*f1}, but we know that a play in {2*f0 | 0} is at least as good.

Therefore D ≥ 0. QED.

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

Visualize whirled peas.

Everything with love. Stay safe.

Consider this seki, {8 | -8 || 10 | -6} = 0. Suppose that we add 2 to each leaf of the game tree, to get, {10 | -6 || 12 | -4} = 0. not 2.

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

Visualize whirled peas.

Everything with love. Stay safe.

You proved theorem 1 earlier, therefore, I called it a theorem - not a conjecture:)



Oh! So it is not classic arithmetics.

This and many other difference games had been solved long before I learned combinatorial game theory.



You can add a number to the leaves of a game in reduced form with no problem.

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

Visualize whirled peas.

Everything with love. Stay safe.

Presuppositions

Suppose the local endgames each with one follow-up M0|F0 and M1|F1 (the sente move values M0, M1 and follow-up move values F0, F1) of the creator with M0 ≥ M1 (*0) and M0 + 2F0 ≥ M1 + 2F1 (*1) 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.


Discussion

We can write the local endgames as {2*F0 + K | 0 + K || 2*M0 + K} and {2*F1 + L | 0 + L || 2*M1 + L}. We do not specify whether either local endgame is a local gote, ambiguous or local sente because we do not specify M0 ? F0 and M1 ? F1, that is, whether the move values of either local endgame decrease, are constant or increase. Since we can write the local endgames as before, M0, M1 are the sente move values. However, we should state the presuppositions (note to myself: and annotate the font) more carefully: M0, M1 are the TENTATIVE sente move values. Thereby, the theorem also applies to one or two local gotes, whose gote move values we need not know. It was my mistake to just pretend that local endgames' moves values would be sente move values. Instead, I must declare to use the TENTATIVE sente move values. Does this solve everything?


Proof (Extract)

[Strategy yields these comparisons.]

M0 - F1 + F1 ≥ M1 - F0 + F0 <=> M0 ≥ M1, which is (*0).

M0 - F1 - F1 ≥ M1 - F0 - F0 <=> M0 + 2*F0 ≥ M1 + 2*F1, which is (*1).

M0 + M1 ≥ M1 + M0 <=> 0 ≥ 0 is a truth.


Proof (Detailed)

S is the starting preventer's strategy for (1) to (3). The M0 is the first move. The other denoted values of a bracket can be preceded, interrupted or succeeded by moves elsewhere in the environment E.
S' for (4) to (6) is S except for the denoted values. (4), (5), (6) substitute (1), (2), (3), respectively. The first, second and possibly third denoted value of S is substituted by the respective denoted value of S'.
We have these kinds of sequences depending on S or S':
1) S(M0;-F1,F1)
2) S(M0;-F1,-F1)
3) S(M0;M1)
4) S'(M1;-F0,F0)
5) S'(M1;-F0,-F0)
6) S'(M1;M0)
By definition of S and S', when comparing (1), (2), (3) to (4), (5), (6), respectively, the net profit P(E) of the gains of the plays elsewhere in the environment E in (1), (2) or (3) equals the net profit P'(E) of the gains of the plays elsewhere in the environment E in (4), (5) or (6), respectively. Therefore, P(E) - P'(E) = 0 can be ignored. For asserting the proposition, we assert it for each of the following cases and the denoted values in the brackets.
Case (1) is replaced by (4):
(1) ≥ (4) <=> M0 - F1 + F1 ≥ M1 - F0 + F0 <=> M0 ≥(*0) M1.
Case (2) is replaced by (5):
(2) ≥ (5) <=> M0 - F1 - F1 ≥ M1 - F0 - F0 <=> M0 + 2*F0 ≥(*1) M1 + 2*F1.
Case (3) is replaced by (6):
(3) ≥ (6) <=> M0 + M1 ≥ M1 + M0 <=> 0 ≥ 0 is a truth.□

Some 50 years ago, years before On Numbers and Games came out, I realized that a complete theory of the endgame could be constructed using only local scores, as traditional evaluations and move values were only heuristics. To that end I came up with some idiosyncratic names for score differences.

For instance, although I did not use slash notation, for the game, G = {a|b||c|d}, where a > b > c > d, I called b - c the double sente value (which was the traditional name), but I called a - d the double gote value (definitely not the traditional name). I called a - c the forward gote value and b - d the backward gote value. For the game, H = {e|f||g}, where e > f > g, I called f - g the sente value, regardless of whether H was a local sente or not. The concept of local sente is only a heuristic. I called e - g the gote value, since there was no forward gote value or backward gote value unless there was an odd number of dame along with g, and in that case they were equal, anyway. For the game, I = {h|i}, where h > i, I called h - i the gote value, which was the traditional name. I did consider values such as b - (c+d)/2 for G, but only as approximations.

I am not sure, but it seems to me that you are taking the heuristics as basic. The heuristics are quite useful. Professor Berlekamp recommended starting the analysis of a game by deriving its thermograph. But it's the final scores that are basic.

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

Visualize whirled peas.

Everything with love. Stay safe.

Scores are basic indeed.

Otherwise, I am not dogmatic about what further basics to use. Different systems of basics coexist and all have their advantages and disadvantages. The systems study the same so create intersections where they agree. The system I am currently developing upon some of your earlier ideas uses both tentative values and confirmed values, and does so successfully for the basic types of local endgames possibly with a basic environment.

You like to express things in terms of resulting counts by relating them explicitly. Thereby, you can avoid distinguishing tentative from confirmed. I prefer to reduce the number of values by combining them into what we call counts, move values and gains because fewer remaining values increase applicability. During evaluation, I need tentative values until we confirm values and identify the type of a local endgame. Nevertheless, tentative values (or their explicit representation) can provide more information.

Bill, I try to understand your theorem and proof below.


Outcome Classes in CGT by Siegel

o(G) is the outcome of G.

G = 0 :<=> o(G) = the second-moving player can force a win,
G ≥ 0 :<=> o(G) ≥ the second-moving player can force a win,
G ≤ 0 :<=> o(G) ≤ the second-moving player can force a win.

What does it mean that o(G) is at least / at most the stated outcome? I understand it for the outcome classes in go below. However, what other outcomes are larger / smaller than the outcome "the second-moving player can force a win"? In what
sense larger / smaller?


Some Outcome Classes in Go

G ≥ 0 if White starts and Black achieves at least 0 as a result,
G = 0 if the result is 0 for Black's and White's starts,
G ≤ 0 if Black starts and White achieves at most 0 as the result.


Presupposition

In a combinatorial game A, which can include any environment without kos now or later, the starting player can move to B or C.


Theorem [Black playing the difference game of two moves]

"Black starts the difference game B - C, White continues and does not win it." <=> B ≥ C.


Proof

1) "Playing the difference game B - C" is the comparison B - C ? 0.

2) By theorem 1.20 "G ≥ H if and only if no G_R ≤ H and G ≤ no H_L" and the definition of the outcome classes in the book Combinatorial Game Theory, Aaron N. Siegel, 2013, p. 58.

"Black starts the difference game B - C, White continues and does not win it." <=>(1)(2) B - C ≥ 0 <=> B ≥ C.


Explanation of (1)

- We have B - C and consider it together as one game (B - C).
- Since the game B - C is a difference, we call it a difference game.
- Playing a game relies on the implicit assumption that one player starts and the two players alternate.
- Therefore, playing the difference game B - C considers two cases: the player's start results in an outcome and the opponent's start results in the same or a different outcome.
- Possible outcome classes are defined as comparison to 0 by combining the outcomes of both cases.
- We play the specific game B - C, determine its outcomes and then apply the definition of outcome classes to determine to which outcome class the game belongs.

Discussion of (2)

- The comparison G ≥ H is derived recursively.
- "no G_R ≤ H" means "there is no Right option G_R of G so that G_R ≤ H".
- "G ≤ no H_L" means "there is no Left option H_L of H so that G ≤ H_L".
- Recursion terminates due to leaves with an outcome due to which player loses by not having any next move.
- We only need the outcome classes listed above.
- What is the meaning of an inequation in "no G_R ≤ H" or "G ≤ no H_L"?

Discussion of the Proof's Core

- Do we need (1) for the first <=> transformation?
- Do we need (2) for the second <=> transformation?
- Why does Black's start and White's failure to win mean ≥ in the comparison?
- We need the definition of outcome classes, as mentioned by (2).
- Why do we need the implied definition of G ≥ H in (2) at all? Why, in CGT, may we not naively transform B - C ≥ 0 <=> B ≥ C as we would do in linear algebra?

For finite combinatorial games, Left (Black in go) wins the game iff the final result of alternating play is a number greater than 0 or 0 where Black plays last or Right (White in go) to play has no move, and conversely for Right. This is different from games where 0 is a tie. In that case a tie may be a win for either player in CGT.

Given that:

G = 0 iff G is a second player win.

G ≥ 0 iff Black wins with White to play.

G ≤ 0 iff White wins with Black to play.



We do. However, 0 is well defined, so to show that B ≥ C we show that B - C ≥ 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 may be stirring things up, but I wanted to ask something regarding difference games.

In Siegel's "Combinatorial Game Theory", section II.6, the concept of >= (mod Inf) is introduced. In this setting a game G is >=0 (modulo infinitesimal) if and only if Right playing first in G can only get a score >=0. More or less what Bill just said about "0 being either player's win".

In particular II.6.4 is very interesting. The lemma shows that this condition only loses an infinitesimal with respect to the more accurate difference game taking into account the last move. Furthermore, this difference game is conceptually easier to understand for go players ("why would I care about the last move?" they could say).

In the context of go, that last half point may be easily ignored for the sake of simplicity, right? Everything else is the same, except the weaker condition gives us reduced canonical forms, which have the same thermographs anyway.

I thought that section was very interesting, but got a bit lost when even-tempered and odd-tempered. I hope that isn't important!

When using difference games for comparing moves in go when ko fights are not an issue, we do not need to rely upon CGT per se. The commonsense notion that a game is at least as good for one player versus the other if the other player plays first but cannot win is all we need.

Example:



When we set up the original 0 game, there will always be a even number of board points in play.

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

Visualize whirled peas.

Everything with love. Stay safe.

miai values in area or territory counting.

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | a O . O . O . |
$$ | X O O O O O b |
$$ | X X c X d X X |
$$ | . X O X O X . |
$$ | . X X X . X . |
$$ | . . . X X X . |
$$ | . . . . . . . |
$$ -----------------[/go]

In territory counting the values of the moves in a, b, c, d are respectively 0, ½, 1, 1½
In area counting the values of the moves in a, b, c, d are respectively 1, 1½, 2, 2½
We see that for gote move the difference is equal to 1.
In area counting a pass move gains 0 points and we have seen above the existence of the values 1, 1½, 2, 2½.
Here I have a quite naïve question : in area counting does it exist a gote move with the value ½ ? Isn't it strange to see such missing value?

Click Here To Show Diagram Code

[go]$$ Evaluate this board. (Area scoring)
$$ -------------------------
$$ | . . X O O . O O . O X . |
$$ | O X X O . O X O O O X X |
$$ | O O X O O X X X X X O . |
$$ | O . X O X X . X X O O O |
$$ | X X X O X . X . X X O . |
$$ | O O O O X X X X X O O O |
$$ -------------------------[/go]

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

Visualize whirled peas.

Everything with love. Stay safe.

Beautiful idea Bill, congratualation. A white move at "a" build a seki and a black move at "a" lead to an exchange with the ko threat at "b". The final result is quite clear : a move at "a" is gote with the value ½. Thank you Bill

your conclusion is the following:
"if there is no ko fight, in the original position B is at least as good for Black as A."

That leads to another issue: assume a ko may arise in the environment. In that case we know that the result of the difference game is not reliable but here is my question : can you build an example of ko environment in which A appears strictly better than B ?

If not I will say B dominates A in a strong manner, otherwise I will only say B dominates A in a weak manner.