Life In 19x19

RobertJasiek wrote:The theorem takes as input a) an arbitrary local endgame tree with one player's gote option and sente option, single play sequences, and arbitrary resulting counts, and b) an arbitrary environment of arbitrarily many simple gotes without follow-ups with arbitary move values. The theorem always says which first move is correct. Hence, it solves an infinite number of such example positions.

When I saw this condition to apply your theorems for late endgame I immediatly understood that I will never apply your theorems for a very simple reason:

Click Here To Show Diagram Code

[go]$$ $$ ------------------- $$ . . . O . a X . . . $$ . . . O O O X . . . $$ . . . . . . . . . . $$ . . . . . . . . . .[/go]

For me, after a black move at "a", a follow-up exists because I use AGA rules! Because in practice it exists almost always such area in the environment your conditions for a late endgame are not fullfilled in practice.

RobertJasiek wrote:

$$B local start with the sente option, correct choice on move 4, C3 = -22 $$ +---------------------------------+ $$ | O . X X . 5 1 3 X . X X X X X X | $$ | O O . O X . O 2 O X X X . X O 6 | $$ | . O O O . O O B O X . X , X O B | $$ | . O C O O O C B O X . X X X O O | $$ | . O B B B O B B O X . X X O O O | $$ | . O B B 4 O O O O X . X X O . O | $$ | O O O O X X X X X X X X X O O . | $$ +---------------------------------+

Click Here To Show Diagram Code

[go]$$B local start with the sente option, correct choice on move 4, C3 = -22 $$ +---------------------------------+ $$ | O . X X . 5 1 3 X . X X X X X X | $$ | O O . O X . O 2 O X X X . X O 6 | $$ | . O O O . O O B O X . X , X O B | $$ | . O C O O O C B O X . X X X O O | $$ | . O B B B O B B O X . X X O O O | $$ | . O B B 4 O O O O X . X X O . O | $$ | O O O O X X X X X X X X X O O . | $$ +---------------------------------+[/go]

I do not understand why you call this option a SENTE option.
After you play in the environment. That means this sequence is gote isn't it?
In that case how can you apply your theorem 128?

Gérard TAILLE wrote: Seeing you did not find any other position (or tree) I conclude that we cannot have simultaneously
1) T = MGOTE
2) One player has a gote and a sente option
3) F < T
That means that your theorem cannot be applied if T = MGOTE

Non-existence is proved by a proof of non-existence.

Non-existence is not proved by referring to my limited time.

Gérard TAILLE wrote:I will never apply your theorems for a very simple reason:

$$ $$ ------------------- $$ . . . O . a X . . . $$ . . . O O O X . . . $$ . . . . . . . . . . $$ . . . . . . . . . .

Click Here To Show Diagram Code

[go]$$ $$ ------------------- $$ . . . O . a X . . . $$ . . . O O O X . . . $$ . . . . . . . . . . $$ . . . . . . . . . .[/go]

For me, after a black move at "a", a follow-up exists because I use AGA rules! Because in practice it exists almost always such area in the environment your conditions for a late endgame are not fullfilled in practice.

Right, these theorems are for territory scoring while area scoring during the late endgame requires more sophisticated maths, such as infinitesimals of CGT. Have fun always applying infinitesimals and full-blown CGT! ;)

Gérard TAILLE wrote:I do not understand why you call this option a SENTE option.

Because, for an assessment as a local sente, my (and Bill's) related definition of an option is purely local and only considers it and the opponent's sequence.

Local means local. Local does not include anything outside the local region. Therefore, it is local. Local - not global!

(Bill has sometimes used a different assessement a la John Conway by considering arbitrarily many multiple copies of the local region, combined play in the union of all copies and forming the average count for the limit to infinity on the number of copies.)

(Global sente is a different concept. A local sente might, or might not, be a global sente (e.g., at different ambient temperatures).)


1 EDIT

Gérard TAILLE wrote:

RobertJasiek wrote:

$$B local start with the sente option, correct choice on move 4, C3 = -22 $$ +---------------------------------+ $$ | O . X X . 5 1 3 X . X X X X X X | $$ | O O . O X . O 2 O X X X . X O 6 | $$ | . O O O . O O B O X . X , X O B | $$ | . O C O O O C B O X . X X X O O | $$ | . O B B B O B B O X . X X O O O | $$ | . O B B 4 O O O O X . X X O . O | $$ | O O O O X X X X X X X X X O O . | $$ +---------------------------------+

Click Here To Show Diagram Code

[go]$$B local start with the sente option, correct choice on move 4, C3 = -22 $$ +---------------------------------+ $$ | O . X X . 5 1 3 X . X X X X X X | $$ | O O . O X . O 2 O X X X . X O 6 | $$ | . O O O . O O B O X . X , X O B | $$ | . O C O O O C B O X . X X X O O | $$ | . O B B B O B B O X . X X O O O | $$ | . O B B 4 O O O O X . X X O . O | $$ | O O O O X X X X X X X X X O O . | $$ +---------------------------------+[/go]

I do not understand why you call this option a SENTE option.
After you play in the environment. That means this sequence is gote isn't it?
In that case how can you apply your theorem 128?

imho
1) is sente in the local game but not in the whole game
2) if the whole game consist of N copies of the local game then would always be sente

RobertJasiek wrote:

Gérard TAILLE wrote: Seeing you did not find any other position (or tree) I conclude that we cannot have simultaneously
1) T = MGOTE
2) One player has a gote and a sente option
3) F < T
That means that your theorem cannot be applied if T = MGOTE

Non-existence is proved by a proof of non-existence.

Non-existence is not proved by referring to my limited time.

What about this one?

Click Here To Show Diagram Code

[go]$$ $$ +-------------------------+ $$ | O . X X . . . . X . . X | $$ | O O . O X . O . O X X X | $$ | . O O O . O O X O X . X | $$ | . O X O O O . X O X . X | $$ | . O X X X O X X O X . X | $$ | . O X X . O O O O X . X | $$ | O O O O X X X X X X X X | $$ +-------------------------+[/go]

dany wrote:

RobertJasiek wrote:

Gérard TAILLE wrote: Seeing you did not find any other position (or tree) I conclude that we cannot have simultaneously
1) T = MGOTE
2) One player has a gote and a sente option
3) F < T
That means that your theorem cannot be applied if T = MGOTE

Non-existence is proved by a proof of non-existence.

Non-existence is not proved by referring to my limited time.

What about this one?

$$ $$ +-------------------------+ $$ | O . X X . . . . X . . X | $$ | O O . O X . O . O X X X | $$ | . O O O . O O X O X . X | $$ | . O X O O O . X O X . X | $$ | . O X X X O X X O X . X | $$ | . O X X . O O O O X . X | $$ | O O O O X X X X X X X X | $$ +-------------------------+

Click Here To Show Diagram Code

[go]$$ $$ +-------------------------+ $$ | O . X X . . . . X . . X | $$ | O O . O X . O . O X X X | $$ | . O O O . O O X O X . X | $$ | . O X O O O . X O X . X | $$ | . O X X X O X X O X . X | $$ | . O X X . O O O O X . X | $$ | O O O O X X X X X X X X | $$ +-------------------------+[/go]

Thank you for this attempt Dany.
If I count correctly: MGOTE = MSENTE = F = 5 (=>ambiguous)
Because I assumed T = MGOTE then F < T is not fulfilled.

Distinguish MGOTE_GOTE_OPTION from MGOTE_SENTE_OPTION!

RobertJasiek wrote:Distinguish MGOTE_GOTE_OPTION from MGOTE_SENTE_OPTION!

What are your figures?

Gérard TAILLE wrote:

dany wrote:
What about this one?

$$ $$ +-------------------------+ $$ | O . X X . . . . X . . X | $$ | O O . O X . O . O X X X | $$ | . O O O . O O X O X . X | $$ | . O X O O O . X O X . X | $$ | . O X X X O X X O X . X | $$ | . O X X . O O O O X . X | $$ | O O O O X X X X X X X X | $$ +-------------------------+

Click Here To Show Diagram Code

[go]$$ $$ +-------------------------+ $$ | O . X X . . . . X . . X | $$ | O O . O X . O . O X X X | $$ | . O O O . O O X O X . X | $$ | . O X O O O . X O X . X | $$ | . O X X X O X X O X . X | $$ | . O X X . O O O O X . X | $$ | O O O O X X X X X X X X | $$ +-------------------------+[/go]

Thank you for this attempt Dany.
If I count correctly: MGOTE = MSENTE = F = 5 (=>ambiguous)
Because I assumed T = MGOTE then F < T is not fulfilled.

H = -11
MGOTE = (H - R) / 2 = (-11 -(-23))/2 = 6

S = -18
MSENTE = S - R = -18 - (-23) = 5

MGOTE_SENTE_OPTION = (-8 - (-18)) / 2 = 5 = MSENTE = F (ambiguous).

Theorem for the early endgame: F < T = MGOTE_GOTE_OPTION = 6 so either start in the environment or local with the gote option is correct.

Method of reading and counting with the follow-up having the count 0 delayed, as it must be during the early endgame:

Click Here To Show Diagram Code

[go]$$B start in the environment, -23 $$ +-------------------------+ $$ | O C B B C C C 2 X . . X | $$ | O O C O B C O C O X X X | $$ | . O O O C O O B O X . X | $$ | . O X O O O C B O X . X | $$ | . O X X X O B B O X . X | $$ | . O X X 1 O O O O X . X | $$ | O O O O X X X X X X X X | $$ +-------------------------+[/go]

Click Here To Show Diagram Code

[go]$$B start locally with the gote option, -23 $$ +-------------------------+ $$ | O C B B C C 2 3 X . C X | $$ | O O C O B C O 1 O X X X | $$ | . O O O C O O X O X . X | $$ | . O B O O O . X O X . X | $$ | . O B B B O X X O X . X | $$ | . O B B 4 O O O O X . X | $$ | O O O O X X X X X X X X | $$ +-------------------------+[/go]

3 EDITs: 2x early endgame; minor corrections.

RobertJasiek wrote:Distinguish MGOTE_GOTE_OPTION from MGOTE_SENTE_OPTION!

Seeing the wordings MGOTE, MGOTE_GOTE_OPTION, MGOTE_SENTE_OPTION it seems a good place here, in this thread dealing with gote and sente moves to define clearly what is meant by this notion.

[img][/img]

Let's take this basic tree in the attachement, in which you can see a black gote option AB and another black option AC, and similarly a white gote option AE and another white option AD.
OC, at the beginning of the analysis you do not know if options AC and AD are gote or sente;

Let's call a, b, c ... the counts of positions A, B, C ...
You can calculate immediately c = (f+g)/2 and d = (h+i)/2

Now how do you proceed in order to know if AC is gote or sente? I understood you will use a MGOTE_SENTE_OPTION value and a MSENTE value but how will you calculate these values? I guess you will use c and g values but what about d and e, especially when you do not know yet if AD is sente or gote?

To calculate values of a player's option, temporarily prune the tree by removing the currently not considered alternative option. This leaves two options for the opponent's start.

Your tree has options for both starting players and this makes it more complicated than any of my theory uses explicitly. Nevertheless, your questions are valid for a possibly broader understanding.

With two options for the opponent's start remaining in the pruned tree, CGT techniques must be considered. In particular, we try to detect if any of the opponent's options is dominated or reversible so can also be pruned.

If thereby the tree is without alternative options, calculate the tentative gote move value, Black's tentative sente move value (for the left sente sequence) and White's tentative sente move value (for the right sente sequence) like we always calculate gote or sente move values, respectively.

If, however, two options of the opponent remain and other reasoning cannot choose clearly, then tentative move values are undefined (so far).

If all tentative move values, tentative counts and Black's and White's follow-up move values could be calculated for a particular option, we derive the assessments of the type of the initial position for that option.

"Definitions 17 [types]
For such a local endgame, we define these types:
local gote :<=> MGOTE < MB,SENTE, MW,SENTE,
Black's local sente :<=> MW,SENTE ≥ MGOTE > MB,SENTE,
White's local sente :<=> MB,SENTE ≥ MGOTE > MW,SENTE,
Black's ambiguous :<=> MW,SENTE > MGOTE = MB,SENTE,
White's ambiguous :<=> MB,SENTE > MGOTE = MW,SENTE,
doubly ambiguous :<=> MGOTE = MB,SENTE = MW,SENTE." [22]

Again, I do NOT define the types of individual moves. I define the types of positions.

https://www.lifein19x19.com/viewtopic.p ... 45#p143245

RobertJasiek wrote:To calculate values of a player's option, temporarily prune the tree by removing the currently not considered alternative option. This leaves two options for the opponent's start.

Your tree has options for both starting players and this makes it more complicated than any of my theory uses explicitly. Nevertheless, your questions are valid for a possibly broader understanding.

With two options for the opponent's start remaining in the pruned tree, CGT techniques must be considered. In particular, we try to detect if any of the opponent's options is dominated or reversible so can also be pruned.

If thereby the tree is without alternative options, calculate the tentative gote move value, Black's tentative sente move value (for the left sente sequence) and White's tentative sente move value (for the right sente sequence) like we always calculate gote or sente move values, respectively.

If, however, two options of the opponent remain and other reasoning cannot choose clearly, then tentative move values are undefined (so far).

If all tentative move values, tentative counts and Black's and White's follow-up move values could be calculated for a particular option, we derive the assessments of the type of the initial position for that option.

"Definitions 17 [types]
For such a local endgame, we define these types:
local gote :<=> MGOTE < MB,SENTE, MW,SENTE,
Black's local sente :<=> MW,SENTE ≥ MGOTE > MB,SENTE,
White's local sente :<=> MB,SENTE ≥ MGOTE > MW,SENTE,
Black's ambiguous :<=> MW,SENTE > MGOTE = MB,SENTE,
White's ambiguous :<=> MB,SENTE > MGOTE = MW,SENTE,
doubly ambiguous :<=> MGOTE = MB,SENTE = MW,SENTE." [22]

Again, I do NOT define the types of individual moves. I define the types of positions.

https://www.lifein19x19.com/viewtopic.p ... 45#p143245

[img][/img]
OK lets'take the following example: b = +3, e = -1, f = +28, g = h = +2, i = -24
What are the values of MGOTE, MB,SENTE and MW,SENTE ?