Thermography

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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote: I suspect that the two parts of the White strategy are incompatible, at least for difference games.
OK Bill, let's try to separate as clearly as possible, all the potential environments.

Let's call E the set of all possible environments and let me divide E in three parts called E1, E2, E3 in the following way:

Firstly I define E1:
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 1 2 . . O |
$$ | X X 3 . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
by definition an environment is in E1 if, after the :b1:, :w2: exchange then :b3: appears (one of) the best move.

All other environments are put either in E2 or in E3

Secondly I define E2:
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . . 2 1 . O |
$$ | X X . . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
by definition an environment is in E2 if it is not in E1 and if after the move :b1:, the move :w2: appears (one of) the best move.

Finally all environments not in E1 or E2 are by definition in E3

Now we can take each environment one after the other:

Suppose the environment is in E1:
if black plays the kosumi she cannot avoid the sequence
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 1 2 . . O |
$$ | X X 3 4 . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
In this case you already proved keima dominates kosumi by the sequence:
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 3 1 2 . O |
$$ | X X . 4 . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
Suppose the environment is in E2:
if black plays the kosumi she cannot avoid the sequence
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 1 2 5 . O |
$$ | X X 4 3 6 . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
In this case, against the black monkey jump, because we know for sure that :w2: is by defintion of E2, (one the) best move, white cannot avoid the sequence
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . . 2 1 . O |
$$ | X X . 3 . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
and it looks that the kosumi and the monkey jump will give the same result

Eventually suppose the environment is in E3:
if black plays the kosumi she cannot avoid the sequence
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 1 . . . O |
$$ | X X . 2 . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
What about the black monkey jump with an enviroment in E3
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . . a 1 . O |
$$ | X X . . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
because we know (by definition of E3) that white cannot play at "a" white cannot avoid to play
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 3 . 1 . O |
$$ | X X . . 2 . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
note also that the following sequence looks not better for white
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 3 a 1 . O |
$$ | X X 2 . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
because, with the environment E3 white obviously still cannot play at "a".
We see clearly that the monkey jump dominates the kosumi for this type of environment.

This time I am quite convinced that it proves keima + monkey jump dominates kosumi.

But I would clearly repeat that in practise you cannot ignore the kosumi move for an obvious reason: in all the environments for which kosumi and keima are both one the best move, then I prefer the kosumi because this move is far better for future ko fights !

Do you agree Bill?
Bill Spight
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote: I suspect that the two parts of the White strategy are incompatible, at least for difference games.
OK Bill, let's try to separate as clearly as possible, all the potential environments.
It seems that you are not talking about difference games now. :)
Let's call E the set of all possible environments and let me divide E in three parts called E1, E2, E3 in the following way:

Firstly I define E1:
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 1 2 . . O |
$$ | X X 3 . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
by definition an environment is in E1 if, after the :b1:, :w2: exchange then :b3: appears (one of) the best move.
IOW, :b3: dominates all other Black plays on the board, i.e., including moves in the environment.
All other environments are put either in E2 or in E3

Secondly I define E2:
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . . 2 1 . O |
$$ | X X . . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
by definition an environment is in E2 if it is not in E1 and if after the move :b1:, the move :w2: appears (one of) the best move.
Dominating all other White moves on the board.

It seems that we need to distinguish between environments. Use subscripts? E1 for the first environment, E2 for this one?
Finally all environments not in E1 or E2 are by definition in E3

Now we can take each environment one after the other:

Suppose the environment is in E1:
if black plays the kosumi she cannot avoid the sequence
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 1 2 . . O |
$$ | X X 3 4 . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
Why not? Couldn't :w4: be in the environment? OC, :w4: could be a mistake, as could :w2:.
In this case you already proved keima dominates kosumi by the sequence:
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 3 1 2 . O |
$$ | X X . 4 . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
Really? Domination is not just local now, is it?

I am going to stop here for the nonce. There are two different game theories that may apply. One is traditional (Western) game theory, where it matters who has the move and in the game tree results are backed up to evaluate current nodes. In this theory results and values are strictly ordered and there is no environment.

The other is combinatorial game theory, which accords with traditional go theory if there are no ko fights or potential ko fights. In this theory it does not matter who has the move, and the value of the current node is not simply a backed up value from the leaves of the tree. Furthermore, the values are not strictly ordered, but two results may be incomparable. However, if one option for a play dominates another, it also dominates it in any non-ko environment. The non-ko environment is another combinatorial game which is added to the game being considered. Difference games are part of combinatorial game theory, but not of von Neumann game theory.

This argument seems to basically belong to von Neumann game theory. Black has the move, and there is a best play for Black, instead of possibly incomparable plays. At the same time, you are incorporating the idea of environments into von Neumann game theory. Not that this is a bad idea, in fact it might be a great idea. :) But it does seem to raise questions about the idea of best play in a part of the board, when best play may be elsewhere. Dominance in von Neumann game theory applies to results in the whole game, not a part of it. It is global, not local. Combinatorial game theory arose out of the idea of local games which could be combined into a single game. One consequence of that idea is that values are not strictly ordered.

Now the local result after the keima is better for Black that the result after :w4: in the previous diagram, but I think that you may have to claim that that :w4: is globally correct to say that the keima is better globally. If :w4: is better elsewhere, then Black may be able to come back and do better locally.
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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
The other is combinatorial game theory, which accords with traditional go theory if there are no ko fights or potential ko fights. In this theory it does not matter who has the move, and the value of the current node is not simply a backed up value from the leaves of the tree. Furthermore, the values are not strictly ordered, but two results may be incomparable. However, if one option for a play dominates another, it also dominates it in any non-ko environment. The non-ko environment is another combinatorial game which is added to the game being considered. Difference games are part of combinatorial game theory, but not of von Neumann game theory.

This argument seems to basically belong to von Neumann game theory. Black has the move, and there is a best play for Black, instead of possibly incomparable plays. At the same time, you are incorporating the idea of environments into von Neumann game theory. Not that this is a bad idea, in fact it might be a great idea. :) But it does seem to raise questions about the idea of best play in a part of the board, when best play may be elsewhere. Dominance in von Neumann game theory applies to results in the whole game, not a part of it. It is global, not local. Combinatorial game theory arose out of the idea of local games which could be combined into a single game. One consequence of that idea is that values are not strictly ordered.
I understand Bill and, in my mind, my analyse is valid only under the non-ko environment defined by the combinatorial game theory.

It is not so easy but I will try to explain my view in more details.

But for that I have first to verify we have a common understanding of the combinatorial game theory.
My view is the following for this theory:

First of all it is very important to know:
1) what is meant when, in combinatorial game theory, we assume the independancy between the local area and the environment
2) why ko fights create a mess in the theory

Let's take one of the major result of the theory:

Let's consider two positions A et B surrounded by an environment E with "good" carateristics (independancy? non-ko? ...)
Assume for example that, black to play, the optimum result of the game beginning from A+E is a win for black by say 10 points
Assume also that, black to play, the optimum result of the game beginning from B+E is a win for black by say 7 points
Let's call B' the mirror position of position B and let's call E' the mirror position of position E
The amazing result of theory is the following: black to play will wins the game A+B' by 3 points !!!

Let's us try to prove this "theorem":
Take two boards, board1 and board2.
On board1 you put the position A+E and on board2 you put the position B'+E'.
Now the players will play a difference game on these two boards.
With the assumptions above I guess black has the advantage and I would like to prove that, black to move, black will win this difference game.
For proving the win for black, I have only to find a winning black strategy and that is quite easy:
Black begins by playing on board1, with the intention to win on this board by 10 points.
After this first move white will choose a board for her answer and will play on this board.
From this point till the end of the game the strategy of black is to always answer white move by a move on the board chosen for white last move.
With this strategy you can see black will win by 10 points on board1 and white will win by 7 points on board2.
So, black will win the difference game by 3 points
At that point comes the assumption that all areas (A, B, E) have good caracteristics allowing the following simplification:
The above difference game is made of the four areas A, B', E and E'. When you look at these four areas you see in particular the areas E and E' which look like perfect miai areas.
Here, we discover the basic assumption of all the theory:
Because E and E' are perfect miai areas, if black can win the game A+B'+E+E' then, providing good independance between the four areas, we can completly ignore the presence of the two areas E and E' => black wins the game A+B' by again 3 points.

Now just a small example to show why a ko is a mess for the theory. Let's take the following position:

Black to play
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | X a X X b X X X X |
$$ | X O O X O O O O O |
$$ | X O O X X O . O . |
$$ | X O . O X O O O O |
$$ | O O O O X X X X X |
$$ | . O X X . X . X O |
$$ | O O X X X X X O c |
$$ | X X X O O O O . O |
$$ | O O O O O O O O . |
$$ ---------------------[/go]
"a" and "b" are perfect miai points counted as very simple 4 points miai value.
In this game the correct sequence is black takes ko, white "a", black connects ko, white "b".
You can see clearly that the result would be very different if you ignore the miai points "a" and "b".
In other words the ko at "c" is really a mess because you cannot simplify the game by removing the miai areas.

Before continuing with our previous discussion I am waiting for your first comments on this very interesting theory.
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote:
The other is combinatorial game theory, which accords with traditional go theory if there are no ko fights or potential ko fights. In this theory it does not matter who has the move, and the value of the current node is not simply a backed up value from the leaves of the tree. Furthermore, the values are not strictly ordered, but two results may be incomparable. However, if one option for a play dominates another, it also dominates it in any non-ko environment. The non-ko environment is another combinatorial game which is added to the game being considered. Difference games are part of combinatorial game theory, but not of von Neumann game theory.

This argument seems to basically belong to von Neumann game theory. Black has the move, and there is a best play for Black, instead of possibly incomparable plays. At the same time, you are incorporating the idea of environments into von Neumann game theory. Not that this is a bad idea, in fact it might be a great idea. :) But it does seem to raise questions about the idea of best play in a part of the board, when best play may be elsewhere. Dominance in von Neumann game theory applies to results in the whole game, not a part of it. It is global, not local. Combinatorial game theory arose out of the idea of local games which could be combined into a single game. One consequence of that idea is that values are not strictly ordered.
I understand Bill and, in my mind, my analyse is valid only under the non-ko environment defined by the combinatorial game theory.

It is not so easy but I will try to explain my view in more details.

But for that I have first to verify we have a common understanding of the combinatorial game theory.
My view is the following for this theory:

First of all it is very important to know:
1) what is meant when, in combinatorial game theory, we assume the independancy between the local area and the environment
2) why ko fights create a mess in the theory

Let's take one of the major result of the theory:

Let's consider two positions A et B surrounded by an environment E with "good" carateristics (independancy? non-ko? ...)
Assume for example that, black to play, the optimum result of the game beginning from A+E is a win for black by say 10 points
Assume also that, black to play, the optimum result of the game beginning from B+E is a win for black by say 7 points
Let's call B' the mirror position of position B and let's call E' the mirror position of position E
The amazing result of theory is the following: black to play will wins the game A+B' by 3 points !!!

Let's us try to prove this "theorem":
Take two boards, board1 and board2.
On board1 you put the position A+E and on board2 you put the position B'+E'.
Now the players will play a difference game on these two boards.
With the assumptions above I guess black has the advantage and I would like to prove that, black to move, black will win this difference game.
For proving the win for black, I have only to find a winning black strategy and that is quite easy:
Black begins by playing on board1, with the intention to win on this board by 10 points.
After this first move white will choose a board for her answer and will play on this board.
From this point till the end of the game the strategy of black is to always answer white move by a move on the board chosen for white last move.
With this strategy you can see black will win by 10 points on board1 and white will win by 7 points on board2.
(Emphasis mine.)

One thing that combinatorial game theory (CGT) shares with von Neumann game theory in this case is that both want to know the minimax result, not just the result of some winning strategy.

When comparing two plays from a given position, we construct the difference game from a zero position, and any winning strategy for the player who goes first is good enough to show that a game is greater than or incomparable with zero. But for the game A - B we need more precision.
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote: Now just a small example to show why a ko is a mess for the theory. Let's take the following position:

Black to play
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | X a X X b X X X X |
$$ | X O O X O O O O O |
$$ | X O O X X O . O . |
$$ | X O . O X O O O O |
$$ | O O O O X X X X X |
$$ | . O X X . X . X O |
$$ | O O X X X X X O c |
$$ | X X X O O O O . O |
$$ | O O O O O O O O . |
$$ ---------------------[/go]
"a" and "b" are perfect miai points counted as very simple 4 points miai value.
In this game the correct sequence is black takes ko, white "a", black connects ko, white "b".
You can see clearly that the result would be very different if you ignore the miai points "a" and "b".
In other words the ko at "c" is really a mess because you cannot simplify the game by removing the miai areas.
Congratulations! You have independently discovered what I dubbed a virtual ko threat. (See https://senseis.xmp.net/?VirtualKoThreat ) :clap: :clap: :clap:
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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Fine Bill, at least I am sure we have the same understanding of the theory.

One more point concerning ko. We know that ko is a great problem for the difference game theory.
BTW, if you modify the rule of the ko and you take the rule defined in the japonese rule for the confirmation of life and death phase then the ko problem disappears completly! Sure you are able to prove that, otherwise I can show you my own analysis.
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:Let's take one of the major result of the theory:

Let's consider two positions A et B surrounded by an environment E with "good" carateristics (independancy? non-ko? ...)
Assume for example that, black to play, the optimum result of the game beginning from A+E is a win for black by say 10 points
Assume also that, black to play, the optimum result of the game beginning from B+E is a win for black by say 7 points
Let's call B' the mirror position of position B and let's call E' the mirror position of position E
The amazing result of theory is the following: black to play will wins the game A+B' by 3 points !!!


(A + E) - (B + E) = A - B

However, with Black to play the minimax result in A + E minus the minimax result in B + E does not equal the minimax result in A - B. :)
The Adkins Principle:
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— Winona Adkins

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Everything with love. Stay safe.
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:Fine Bill, at least I am sure we have the same understanding of the theory.
Not really. See above. :)
One more point concerning ko. We know that ko is a great problem for the difference game theory.
BTW, if you modify the rule of the ko and you take the rule defined in the japonese rule for the confirmation of life and death phase then the ko problem disappears completly! Sure you are able to prove that, otherwise I can show you my own analysis.
The Japanese rules take some pains to make any ko in the confirmation stage independent of the rest of the board.
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— Winona Adkins

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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:Let's take one of the major result of the theory:

Let's consider two positions A et B surrounded by an environment E with "good" carateristics (independancy? non-ko? ...)
Assume for example that, black to play, the optimum result of the game beginning from A+E is a win for black by say 10 points
Assume also that, black to play, the optimum result of the game beginning from B+E is a win for black by say 7 points
Let's call B' the mirror position of position B and let's call E' the mirror position of position E
The amazing result of theory is the following: black to play will wins the game A+B' by 3 points !!!


(A + E) - (B + E) = A - B

However, with Black to play the minimax result in A + E minus the minimax result in B + E does not equal the minimax result in A - B. :)
I don't understand your point, Bill



Assuming the position at the top of the board is A + E, the position at the bottom of the board is B + E, then the position on the left of the board is effectively A - B.
From that point I drawed on the right the position (A + E) - (B + E).
Now if I compare A - B to (A + E) - (B + E) they seem to me equivalent.
Can you clarify your point, Bill?
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Re: Thermography

Post by Bill Spight »



They are equivalent. :)

However, the minimax result with Black to play for (A + E) - (B + E) is +2, not 2 - 2 = 0.
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:

They are equivalent. :)

However, the minimax result with Black to play for (A + E) - (B + E) is +2, not 2 - 2 = 0.
Oops I have some difficulties with your comments.
The result
MinMax((A + E) - (B + E)) = MinMax(A - B)
is indeed a very interesting result and we know this is correct only if the areas A, B, E have good carateristics.

But why do you talk about an equation of the form
minMax(A) - minMax(B) = minMax(A-B)
which is completly different and which, in practise, is quite never true?
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:But why do you talk about an equation of the form
minMax(A) - minMax(B) = minMax(A-B)
which is completly different and which, in practise, is quite never true?
As a reply to your claim here.
Gérard TAILLE wrote:Let's take one of the major result of the theory:

Let's consider two positions A et B surrounded by an environment E with "good" carateristics (independancy? non-ko? ...)
Assume for example that, black to play, the optimum result of the game beginning from A+E is a win for black by say 10 points
Assume also that, black to play, the optimum result of the game beginning from B+E is a win for black by say 7 points
Let's call B' the mirror position of position B and let's call E' the mirror position of position E
The amazing result of theory is the following: black to play will wins the game A+B' by 3 points !!!
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote: Let's call E the set of all possible environments and let me divide E in three parts called E1, E2, E3 in the following way:

Firstly I define E1:
Click Here To Show Diagram Code
[go]$$B
$$ -----------------
$$ | . . 1 2 . . O |
$$ | X X 3 . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ ----------------[/go]
by definition an environment is in E1 if, after the :b1:, :w2: exchange then :b3: appears (one of) the best move.
Bill Spight wrote: IOW, :b3: dominates all other Black plays on the board, i.e., including moves in the environment.
I understand your point Bill and I need some time to really take into account this comment.

By proposing as an alternative the kosumi move you indirectly proposed a quite difficult problem:

Taking the three moves kosumi, keima and monkey jump is it possible to build a good (non ko) environment for which kosumi is the only correct move?
This is true for the keima and the monkey jump but what about the kosumi?

Two possibilities to resolve the problem:
1) You find an example of such environment!
2) You prove such environment does not exist

I tried the first possibilty but, for the time being, I failed to find such environment.
As a consequence I tried the very difficult second solution and seeing your comments it appears really not easy to be rigourous!

Anyway let's try to be optimist hoping to eventually resolve the problem through discussion.
BTW, do you have the feeling that such environment exist or do you have some doubts?
Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:But why do you talk about an equation of the form
minMax(A) - minMax(B) = minMax(A-B)
which is completly different and which, in practise, is quite never true?
As a reply to your claim here.
Gérard TAILLE wrote:Let's take one of the major result of the theory:

Let's consider two positions A et B surrounded by an environment E with "good" carateristics (independancy? non-ko? ...)
Assume for example that, black to play, the optimum result of the game beginning from A+E is a win for black by say 10 points
Assume also that, black to play, the optimum result of the game beginning from B+E is a win for black by say 7 points
Let's call B' the mirror position of position B and let's call E' the mirror position of position E
The amazing result of theory is the following: black to play will wins the game A+B' by 3 points !!!
it looks only as a misunderstanding:
When I talk about (A+E) result I am referring to the game on board1 and when I talk about (B+E) I am referring to the game on board2. In that case I can easily sum the results can't I?

Isn't it what you assume when you write:
(A+E) - (B+E) = A-B
The left side of sign "=" is really refering to two games on two different boards though on the right side your are on only on one board. That is the reason why the equation may be correct.
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Re: Thermography

Post by Gérard TAILLE »

Oops the notation I used is not correct because it is not clear when I am reasonning one two boards and when I switch to one board. As a a consequence I see that what I wrote is quite unclear.
Let me take some time to rephrase all my analysis before rediscussing.
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