This 'n' that

Bill Spight · **#921**

The meaning of the walls of the thermograph and why a simple gote of the form, {t|-t}, provides a good model for the environment at temperature, t.

Suppose, for instance, we have this position at territorial go.

Click Here To Show Diagram Code: [go]$$W $$ --------------------- $$ | a X O X . . . . . | $$ | X O O X . . . . . | $$ | O O X X . . . . . | $$ | b O X . . . . . . | $$ | X X X . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

White to play captures 2 stones for a result of -4 plus a dame. Black to play captures 6 stones for a result of +13.

The Black wall of the thermograph is the line, s = +13 - t up to t = 8½ and the White wall is the line, s = -4 + t up to t = 8½. The mast of the thermograph rises vertically from the point, (s,t) = (4½, 8½), where the two walls coincide.

What is the point on the Black wall when t = 5? The original thermography applied a tax of 5 to each play, so the point is (s,t) = (13 - 5,5) = (8,5). The point on the White wall is (1,5). s is the same as if Black played to 13 and then White played to -5, or if White played to -4 and then Black played to +5.

The rule, as shown above, is to play locally until the temperature drops below t. Earlier I said that rule applied to Berlekamp's komaster theory, which is true, but I didn't go far enough. it applies to all thermography.

----

What about the case mentioned above where we have the environment, g1 ≥ g2 ≥ g3 ≥ g4 ≥ g5 ≥ g6 ≥ ...? If we take t = g1, then to find the thermograph we must add as many {g1|-g1} games to the environment as necessary. If we take g2 = t, then we may have only one {g1|-g1}, where g1 > t, but it is not part of the environment, it is part of the game or ko ensemble. If we take g4 = t, the same reasoning applies, g1 ≥ g2 ≥ g3 > t and all of g1, g2, and g3 are not in the environment, but are in the game or ko ensemble.

Gérard TAILLE · **#922**

Bill Spight wrote:

My main concern these days is spreading and promoting thermography in go. Thermography was invented by Conway in the 1970s and published in On Numbers and Games. A friend lent me a copy and thought that thermography might be of interest in go. Then thermography was defined in terms of applying a tax to plays. It seemed to produce the same evaluations of go positions and plays as methods that go players already used — I was unaware of the problems with complex ko positions at that time —, so I did not see any benefit from it.

That changed for me when I attended a lecture by Berlekamp in 1994 or 5 in which he presented his komaster theory. Despite the fact that it left open the question of whether the conditions for komaster were met in actual play, it provided a considerably more tractable theory for evaluating complex ko positions than the ko theory I had developed. My theory included all of the environment in the ko ensemble. The problem with doing that is that to evaluate a ko you have to read out the whole board. But, OC, if you can do that you don't need any theory. :lol:

I joined a small group consisting of Berlekamp, some of his students and former students, and visiting scholars, and myself, which mainly studied komaster theory. At first I solved problems by opining correct play and then drawing the thermograph from that. This irked Berlekamp, who was around 3 kyu, because he contended that thermography provided a way of finding correct play. He was right, OC. Thermographic lines generated by incorrect play do not appear in the final thermograph.

At temperature 0, where the game is played out, thermography indicates correct play, but that is guaranteed only by exhaustive search, or by perhaps other means of proving correct play. (If there is an encore, thermography may apply below temperature 0, but in go this is highly dependent upon the rules.) What thermography does is to provide correct play at each temperature. A play or line of play may be incorrect at one temperature and correct at another. Any play, given otherwise correct play, that produces the best result at a given temperature for the player, will indicate a point on the thermograph at that temperature.

Any theory allowing to help a player to find the best move is very valuable. Thermography (without ko) is a very strong tool. By just defining an ideal environment depending of only one parameter called temperature you are able to guess the local best move and this guess is correct for a very large panel of real (non ideal) environments.
The difficulty was to define the ideal environment which is not so obvious for a non mathematical guy.

Click Here To Show Diagram Code: [go]$$W $$ --------------------- $$ | . O b O O O a O . | $$ | X X X X X X X O . | $$ | . . . . . . O O . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Evaluating such corridor means OC to evaluate a move at "a" but you have first to analyse what happens with "b" after a white move at "a". To solve such problem you have to imagine a quite strange environment: in one hand the environment may have an arbitrary high number of gote points at temperature t but in the other hand you may also expect that the temperature may drop in order to be able to deal with the remaining move at "b".
As far as I am concerned I only visualise a rich environment ε, 2ε, 3ε, 4ε ... and I make the following assumption: it exists a sufficiently small value ε0 such that the best moves in the environment ε0, 2ε0, 3ε0, 4ε0 ... are the same as the best moves in the environment ε, 2ε, 3ε, 4ε ... providing ε ≤ ε0.

Taking into account ko is quite ambitious. This ambition have to be defined by various questions and in particular:
1) do we take into account only "simple" direct kos or do we want also to take into account more complex kos?
2) do we consider kos only in the local area, only in the environment or in both local area and environment?
3) do we want to take into account ko threats in the environment (remember the position we are studying with only one black ko threat)
4) do we want to take into account ko threats with various values?

As soon as you have clarified your ambitions then you can look for a new ideal enviroment to help you analyse the local area.

Just a small example to illustrate the case were you allow ko only in the enviroment

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O . . . | $$ | . . . . O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O 1 . . | $$ | . . . 2 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

The exchange :b1:

is sente but generally not correct in a ko environment. First of all unless temperature is quite low black must prefer to play in the environment in order to keep :b1:

as a ko threat.

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O . . . | $$ | . . . 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Secondly, even if temperature is low it is generally better to play this cut in order to get at the end a local ko threat.
After this :b1:

cut what is the best move for white?

First possibility:

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . 6 2 . O 5 . . | $$ | . 4 3 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

with a black ko threat at :b1:

Second possibily

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . 4 3 6 O 5 . . | $$ | . . 2 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

with still a black ko threat at :b1:

What is best for white? The second one is best because the ko threat is smaller as shown by the following diagram:

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . O 3 O O X . . | $$ | . . O 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

do not answer th ko threat
and white may later recapture two stones.

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O . . . | $$ | . . 2 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

As you see a good go player can easily find the sequence :w1:

.
The question now is the following: is it possible to build a theory and an ideal environment able find and show clearly this sequence?

Bill Spight · **#923**

Gérard TAILLE wrote:

Any theory allowing to help a player to find the best move is very valuable. Thermography (without ko) is a very strong tool. By just defining an ideal environment depending of only one parameter called temperature you are able to guess the local best move and this guess is correct for a very large panel of real (non ideal) environments.
The difficulty was to define the ideal environment which is not so obvious for a non mathematical guy.

Click Here To Show Diagram Code: [go]$$W $$ --------------------- $$ | . O b O O O a O . | $$ | X X X X X X X O . | $$ | . . . . . . O O . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Evaluating such corridor means OC to evaluate a move at "a" but you have first to analyse what happens with "b" after a white move at "a". To solve such problem you have to imagine a quite strange environment: in one hand the environment may have an arbitrary high number of gote points at temperature t but in the other hand you may also expect that the temperature may drop in order to be able to deal with the remaining move at "b".
As far as I am concerned I only visualise a rich environment ε, 2ε, 3ε, 4ε ... and I make the following assumption: it exists a sufficiently small value ε0 such that the best moves in the environment ε0, 2ε0, 3ε0, 4ε0 ... are the same as the best moves in the environment ε, 2ε, 3ε, 4ε ... providing ε ≤ ε0.

As a practical matter, this is what Berlekamp did by making ε = 0.01. But to have enough simple gote to accommodate kos, he also had 99 or 101 multiples of each gote in the environment.

Such an environment will sometimes yield the wrong thermograph, but any error would very likely be tiny. For a simple ko, for instance, the difference between 0.33 and ⅓ is small.

Gérard TAILLE wrote:

Taking into account ko is quite ambitious. This ambition have to be defined by various questions and in particular:
1) do we take into account only "simple" direct kos or do we want also to take into account more complex kos?

All ko and superko positions, although to resolve some questions you must rely upon the rules.

Gérard TAILLE wrote:

2) do we consider kos only in the local area, only in the environment or in both local area and environment?

The environment consists of simple gote only.

Gérard TAILLE wrote:

3) do we want to take into account ko threats in the environment (remember the position we are studying with only one black ko threat)

We may in some circumstances consider a pair of simple gote of the same size as a defensive ko threat, which prevents the komaster from becoming a komonster by gaining from the drop in temperature. But that is a feature, not a bug.

Gérard TAILLE wrote:

Just a small example to illustrate the case were you allow ko only in the enviroment

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O . . . | $$ | . . . . O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O 1 . . | $$ | . . . 2 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

The exchange :b1:

is sente but generally not correct in a ko environment. First of all unless temperature is quite low black must prefer to play in the environment in order to keep :b1:

as a ko threat.

This is not a simple gote, and therefore not part of a thermographic environment. OC, it can be a game or part of a game, and ko threats in the game may well matter.

Gérard TAILLE wrote:

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O . . . | $$ | . . . 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Secondly, even if temperature is low it is generally better to play this cut in order to get at the end a local ko threat.
After this :b1:

cut what is the best move for white?

First possibility:

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . 6 2 . O 5 . . | $$ | . 4 3 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

with a black ko threat at :b1:

Second possibily

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . 4 3 6 O 5 . . | $$ | . . 2 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

with still a black ko threat at :b1:

What is best for white? The second one is best because the ko threat is smaller as shown by the following diagram:

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . O 3 O O X . . | $$ | . . O 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

do not answer th ko threat
and white may later recapture two stones.

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O . . . | $$ | . . 2 1 O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

As you see a good go player can easily find the sequence :w1:

.
The question now is the following: is it possible to build a theory and an ideal environment able find and show clearly this sequence?

If this is part of a larger game (ko ensemble) that includes a ko or potential ko, then sure.

Gérard TAILLE · **#924**

Bill Spight wrote:

Gérard TAILLE wrote:

Any theory allowing to help a player to find the best move is very valuable. Thermography (without ko) is a very strong tool. By just defining an ideal environment depending of only one parameter called temperature you are able to guess the local best move and this guess is correct for a very large panel of real (non ideal) environments.
The difficulty was to define the ideal environment which is not so obvious for a non mathematical guy.

Click Here To Show Diagram Code: [go]$$W $$ --------------------- $$ | . O b O O O a O . | $$ | X X X X X X X O . | $$ | . . . . . . O O . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Evaluating such corridor means OC to evaluate a move at "a" but you have first to analyse what happens with "b" after a white move at "a". To solve such problem you have to imagine a quite strange environment: in one hand the environment may have an arbitrary high number of gote points at temperature t but in the other hand you may also expect that the temperature may drop in order to be able to deal with the remaining move at "b".
As far as I am concerned I only visualise a rich environment ε, 2ε, 3ε, 4ε ... and I make the following assumption: it exists a sufficiently small value ε0 such that the best moves in the environment ε0, 2ε0, 3ε0, 4ε0 ... are the same as the best moves in the environment ε, 2ε, 3ε, 4ε ... providing ε ≤ ε0.

As a practical matter, this is what Berlekamp did by making ε = 0.01. But to have enough simple gote to accommodate kos, he also had 99 or 101 multiples of each gote in the environment.

Such an environment will sometimes yield the wrong thermograph, but any error would very likely be tiny. For a simple ko, for instance, the difference between 0.33 and ⅓ is small.

This problem seems easy to solve: as soon as you give a value result, you have to eliminate all terms which appear mathematically negligeable comparing to others.

Let's take a simple ko as an example:
Black takes the ko => scoreBlackTakesTheKo = N - g1 - g2 + g3 ...
Black takes g1 and White connects the ko => scoreWhiteConnectsTheKo = g1 + g2 - g3 + ...
scoreBlackTakesTheKo - scoreWhiteConnectsTheKo = (N - g1 - g2 + g3 ...) - (g1 + g2 - g3 + ...) = N - 2g1 - 2g2 + 2g3 - 2g4 ...
Assuming 2g2 - 2g3 + 2g4 ... = g2 then
scoreBlackTakesTheKo - scoreWhiteConnectsTheKo = N - 2g1 - g2 = N - 3g1 + ε
scoreBlackTakesTheKo ≥ scoreWhiteConnectsTheKo <=> N - 3g1 + ε ≥ 0 <=> g1 <= (N + ε) / 3
and here, from a mathematical point if view, we have ε negligeable comparing to N thus the result becomes
scoreBlackTakesTheKo ≥ scoreWhiteConnectsTheKo <=> g1 <= N / 3

Bill Spight wrote:

Gérard TAILLE wrote:

2) do we consider kos only in the local area, only in the environment or in both local area and environment?

The environment consists of simple gote only.

with such answer you can forget my following example:

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O . . . | $$ | . . . . O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Assuming there is no ko in the environment you cannot find the best sequence in this local area. OC it is a pity but if you claim you have not the ambitious to consider ko in the environment we cannot do anything with this local area.

BTW you cannot say that the environment consists of simple gote only if you accept to say that one player may be komaster. That means that you accept a number of ko threats in the environment but you do not accept a ko. Fine, it is your choice Bill, but we can also try other choices (not easy OC, is it?) to try and find really the best sequence in my example above.

Bill Spight · **#925**

Gérard TAILLE wrote:

Bill Spight wrote:

Gérard TAILLE wrote:

Any theory allowing to help a player to find the best move is very valuable. Thermography (without ko) is a very strong tool. By just defining an ideal environment depending of only one parameter called temperature you are able to guess the local best move and this guess is correct for a very large panel of real (non ideal) environments.
The difficulty was to define the ideal environment which is not so obvious for a non mathematical guy.

Click Here To Show Diagram Code: [go]$$W $$ --------------------- $$ | . O b O O O a O . | $$ | X X X X X X X O . | $$ | . . . . . . O O . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Evaluating such corridor means OC to evaluate a move at "a" but you have first to analyse what happens with "b" after a white move at "a". To solve such problem you have to imagine a quite strange environment: in one hand the environment may have an arbitrary high number of gote points at temperature t but in the other hand you may also expect that the temperature may drop in order to be able to deal with the remaining move at "b".
As far as I am concerned I only visualise a rich environment ε, 2ε, 3ε, 4ε ... and I make the following assumption: it exists a sufficiently small value ε0 such that the best moves in the environment ε0, 2ε0, 3ε0, 4ε0 ... are the same as the best moves in the environment ε, 2ε, 3ε, 4ε ... providing ε ≤ ε0.

As a practical matter, this is what Berlekamp did by making ε = 0.01. But to have enough simple gote to accommodate kos, he also had 99 or 101 multiples of each gote in the environment.

Such an environment will sometimes yield the wrong thermograph, but any error would very likely be tiny. For a simple ko, for instance, the difference between 0.33 and ⅓ is small.

This problem seems easy to solve: as soon as you give a value result, you have to eliminate all terms which appear mathematically negligeable comparing to others.

Let's take a simple ko as an example:
Black takes the ko => scoreBlackTakesTheKo = N - g1 - g2 + g3 ...
Black takes g1 and White connects the ko => scoreWhiteConnectsTheKo = g1 + g2 - g3 + ...
scoreBlackTakesTheKo - scoreWhiteConnectsTheKo = (N - g1 - g2 + g3 ...) - (g1 + g2 - g3 + ...) = N - 2g1 - 2g2 + 2g3 - 2g4 ...
Assuming 2g2 - 2g3 + 2g4 ... = g2 then
scoreBlackTakesTheKo - scoreWhiteConnectsTheKo = N - 2g1 - g2 = N - 3g1 + ε
scoreBlackTakesTheKo ≥ scoreWhiteConnectsTheKo <=> N - 3g1 + ε ≥ 0 <=> g1 <= (N + ε) / 3
and here, from a mathematical point if view, we have ε negligeable comparing to N thus the result becomes
scoreBlackTakesTheKo ≥ scoreWhiteConnectsTheKo <=> g1 <= N / 3

I appreciate your efforts, but this is a solved problem. It is possible to eliminate any errors, given a particular position (game), but that takes some work. Berlekamp's idea was to have an environment that he could use for any position, with only negligible errors, without having to do the work to find all the relevant positions and temperatures before drawing the thermograph. Thermographs only promise approximations to perfect play, anyway.

But suppose that we just want an ideal environment for both t = ½ and t = ⅓ (there is a ko or possible ko). An environment with a minimum positive temperature of 1/12 will do.

⅓ - ¼ + ⅙ - 1/12 = ⅙

½ - 5/12 + ⅙ = ¼

OC, to accommodate a ko fight we need a sufficiently large odd number of gote at the relevant temperatures.

We could start out with 11 and increase the number if necessary.

N.B. If, because of conditions on the board, there turn out to be an even number of plays at the maximum temperature, the environment is still ideal. E.g.,

½ - ½ + 5/12 - ⅓ + ¼ - ⅙ + 1/12 = ¼

OC, the environment is not ideal for t = 5/12, but that is a feature, not a bug.

Gérard TAILLE wrote:

Bill Spight wrote:

Gérard TAILLE wrote:

) do we consider kos only in the local area, only in the environment or in both local area and environment?

The environment consists of simple gote only.

with such answer you can forget my following example:

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | . . . . . O . . . | $$ | . . . . O O X X X | $$ | O O O O X X X . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Assuming there is no ko in the environment you cannot find the best sequence in this local area.

I don't know where you get that idea.

Gérard TAILLE wrote:

BTW you cannot say that the environment consists of simple gote only if you accept to say that one player may be komaster. That means that you accept a number of ko threats in the environment but you do not accept a ko.

If a thermographic environment consisting of simple gote cannot solve for komaster then Berlekamp's original komaster theory, based upon taxation, is wrong. That is not the case.

Gérard TAILLE · **#926**

Bill Spight wrote:

I don't know where you get that idea.

Gérard TAILLE wrote:

BTW you cannot say that the environment consists of simple gote only if you accept to say that one player may be komaster. That means that you accept a number of ko threats in the environment but you do not accept a ko.

If a thermographic environment consisting of simple gote cannot solve for komaster then Berlekamp's original komaster theory, based upon taxation, is wrong. That is not the case.

OK it seems I was not clear in my explanations. Let's take another example:

Click Here To Show Diagram Code: [go]$$W $$ --------------------- $$ | . . . b X . . . . | $$ | . a O O X . . . . | $$ | . . O X X . . . . | $$ | . O O X . . . . . | $$ | . O X X . . . . . | $$ | . X . . . . . . . | $$ | . X . . . . . . . | $$ | X X . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Let me simplify the reasoning by considering for black only the three moves a, b and tenuki (I mean a play in the environment) and assume I am asking a good go player the basic following question : "which move is best in this position?"

OC Bill can skip here some of the following paragraphs which adress those who are not mastering thermography.

If she is a good player she will clearly see that the best move depends on the exact configuration of the environment and she will try to have some information about this environment.

Now begins the problem :
1)if you give a lof information about the environment the answer would be quite reliable but the analyse will be extremely complex because you will analyse a great number of various environments with more or less complex area, with more or less complex ko, with various kind of ko threat etc. etc.
2)Assume now you give no information about the environment. In that case she can only imagine what could be the environment of such position. I think she will assume that we are certainly not at the end of the yose phase. For the local area I guess she will choose to play at "a" to save various possible options but how deciding between a play at "a" and a tenuki (a move in the environment)?

Clearly we do need to give some information concerning the environment. What kind of information is the most relevant? Here I am sure everybody would appreciate what is called "temperature" which you can translate as the value of a move in the environment. OC the temperature does not give the exact configuration of the environment but it gives a very interesting information about it allowing to guess the best move with a high probability to be correct.

It remains a lot of incertainty concerning the exact configuration of the environment (which could be quite complex) but you cannot underestimate how good the guess of the best move could be correct with only this information. In addition you have to appreciate the simplicity of the result like:
above t = 9 you play tenuki; betwen t = 6 and t= 9 you play "a"; under t = 6 you play "a" or "b" (do not take any importance to the values here).

Now is my point Bill.

In order to analyse the position with only the temperature as information on the environment, the current theory imagine a set of pure gote points with various good properties. That sounds a good approach for simplifing the analyse but my feeling is that a good go player is able to analyse easily the position taking into account a far more complex environment.

As an exemple assume you know the temperature is t = 1. The simple analyse, with the ideal environment defined by the theory, will tell you to play "a" or "b" but the go player will imagine (without defining it precisely) a far more complex environment. She will say that a small (⅓ point or a little higher) ko may exist in the environment. In that case a move at "a" is a bad move because this move implies the loss of a local ko threat.
For a good go player it looks easy but, if I am right, this kind of analyse is not in the current theory because ko are not taken into account in the environment.

I do not know how we can improve the theory on this point. My feeing is that it is impossible to add in the environment various ko but because humans are able to take into account "potential ko" it must exist a way to define it and improve the analysis. BTW, in my previous post I mentioned that a go player is also able to choose a move that gives the best ko threat.

IOW I think that by keeping only temperature of the environment we can imporve the analysis by taking into account the "potential" kos that may exist in the environment. Sure we will find some good ideas in the near future, simply because humans manage to do that without major difficulties. OC,we certainly have to limit our ambition to simple (and small?) "potential" kos. We will see.

Bill Spight · **#927**

I know I did some work on this position some years ago, because I commented on it on SL, at https://senseis.xmp.net/?L2GroupWithDescent#toc4 . But it doesn't ring a bell. There are a number of kos in the game tree, but I am not well, and I am not interested in doing a full analysis, at least not now.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . . . . X . . $$ | . . O O X . . $$ | . . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

However, a few days ago I found some interesting things with the following first 4 plays.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . 2 . . X . . $$ | 4 1 O O X . . $$ | 3 . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

At this point White threatens to win the ko in 1 move. We assume that t > 1.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . 2 . . X . . $$ | 4 1 O O X . . $$ | 3 6 O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

= t

Since Black played first we can leave it at that.

Result: -5½ + t

Or Black can kill the corner in 3 net moves.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | 5 2 7 9 X . . $$ | 4 1 O O X . . $$ | 3 . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

,

= t

Result: 21 - 3t

The temperature of indifference occurs when

-5½ + t = 21 -3t, or when

t = 6⅝

We anticipate each play in the ko gaining 6½ points on average, not 6⅝. Verrry interesting.

More later.

Bill Spight · **#928**

Gérard TAILLE wrote:

Bill Spight wrote:

I don't know where you get that idea.

Gérard TAILLE wrote:

BTW you cannot say that the environment consists of simple gote only if you accept to say that one player may be komaster. That means that you accept a number of ko threats in the environment but you do not accept a ko.

If a thermographic environment consisting of simple gote cannot solve for komaster then Berlekamp's original komaster theory, based upon taxation, is wrong. That is not the case.

OK it seems I was not clear in my explanations. Let's take another example:

Click Here To Show Diagram Code: [go]$$W $$ --------------------- $$ | . . . b X . . . . | $$ | . a O O X . . . . | $$ | . . O X X . . . . | $$ | . O O X . . . . . | $$ | . O X X . . . . . | $$ | . X . . . . . . . | $$ | . X . . . . . . . | $$ | X X . . . . . . . | $$ | . . . . . . . . . | $$ ---------------------[/go]

Let me simplify the reasoning by considering for black only the three moves a, b and tenuki (I mean a play in the environment) and assume I am asking a good go player the basic following question : "which move is best in this position?"

OC Bill can skip here some of the following paragraphs which adress those who are not mastering thermography.

If she is a good player she will clearly see that the best move depends on the exact configuration of the environment and she will try to have some information about this environment.

Now begins the problem :
1)if you give a lof information about the environment the answer would be quite reliable but the analyse will be extremely complex because you will analyse a great number of various environments with more or less complex area, with more or less complex ko, with various kind of ko threat etc. etc.
2)Assume now you give no information about the environment. In that case she can only imagine what could be the environment of such position. I think she will assume that we are certainly not at the end of the yose phase. For the local area I guess she will choose to play at "a" to save various possible options but how deciding between a play at "a" and a tenuki (a move in the environment)?

Clearly we do need to give some information concerning the environment. What kind of information is the most relevant? Here I am sure everybody would appreciate what is called "temperature" which you can translate as the value of a move in the environment. OC the temperature does not give the exact configuration of the environment but it gives a very interesting information about it allowing to guess the best move with a high probability to be correct.

It remains a lot of incertainty concerning the exact configuration of the environment (which could be quite complex) but you cannot underestimate how good the guess of the best move could be correct with only this information. In addition you have to appreciate the simplicity of the result like:
above t = 9 you play tenuki; betwen t = 6 and t= 9 you play "a"; under t = 6 you play "a" or "b" (do not take any importance to the values here).

Now is my point Bill.

In order to analyse the position with only the temperature as information on the environment, the current theory imagine a set of pure gote points with various good properties. That sounds a good approach for simplifing the analyse but my feeling is that a good go player is able to analyse easily the position taking into account a far more complex environment.

As an exemple assume you know the temperature is t = 1. The simple analyse, with the ideal environment defined by the theory, will tell you to play "a" or "b" but the go player will imagine (without defining it precisely) a far more complex environment. She will say that a small (⅓ point or a little higher) ko may exist in the environment. In that case a move at "a" is a bad move because this move implies the loss of a local ko threat.
For a good go player it looks easy but, if I am right, this kind of analyse is not in the current theory because ko are not taken into account in the environment.

I do not know how we can improve the theory on this point. My feeing is that it is impossible to add in the environment various ko but because humans are able to take into account "potential ko" it must exist a way to define it and improve the analysis. BTW, in my previous post I mentioned that a go player is also able to choose a move that gives the best ko threat.

IOW I think that by keeping only temperature of the environment we can imporve the analysis by taking into account the "potential" kos that may exist in the environment. Sure we will find some good ideas in the near future, simply because humans manage to do that without major difficulties. OC,we certainly have to limit our ambition to simple (and small?) "potential" kos. We will see.

It seems to me that you are using the term, environment, here to mean the rest of the board asid from the local position. You then seem to be complaining that knowing only the temperature of the environment is not enough to say what is best play. But that is a claim that nobody is making. {shrug}

Gérard TAILLE · **#929**

Bill Spight wrote:

I know I did some work on this position some years ago, because I commented on it on SL, at https://senseis.xmp.net/?L2GroupWithDescent#toc4 . But it doesn't ring a bell. There are a number of kos in the game tree, but I am not well, and I am not interested in doing a full analysis, at least not now.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . . . . X . . $$ | . . O O X . . $$ | . . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

However, a few days ago I found some interesting things with the following first 4 plays.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . 2 . . X . . $$ | 4 1 O O X . . $$ | 3 . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

At this point White threatens to win the ko in 1 move. We assume that t > 1.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . 2 . . X . . $$ | 4 1 O O X . . $$ | 3 6 O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

= t

Since Black played first we can leave it at that.

Result: -5½ + t

Or Black can kill the corner in 3 net moves.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | 5 2 7 9 X . . $$ | 4 1 O O X . . $$ | 3 . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

,

= t

Result: 21 - 3t

The temperature of indifference occurs when

-5½ + t = 21 -3t, or when

t = 6⅝

We anticipate each play in the ko gaining 6½ points on average, not 6⅝. Verrry interesting.

More later.

OK Bill my understanding is the following

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . O . . X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

if t > 6⅝ both players play in the environment
if 6½ < t < 6⅝ white will not play in the environment; black will play in the corner and white will not defend the corner:

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 5 X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

tenuki

tenuki
Finally if t < 6½ both player will play in the corner. If the case black plays in the corner then white will defend with

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 4 X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | X 6 X O X . . $$ | 8 X O O X . . $$ | X 0 O X X . . $$ | . O O X . . . $$ | 5 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

tenuki

Gérard TAILLE · **#930**

Bill Spight wrote:

It seems to me that you are using the term, environment, here to mean the rest of the board asid from the local position. You then seem to be complaining that knowing only the temperature of the environment is not enough to say what is best play. But that is a claim that nobody is making. {shrug}

Yes I use the term environment to mean the rest of the board asid from the local position.
But I am not complaining that knowing only the temperature of the environment is not enough to say what is best play. On contrary I appreciate to try and find the best move with only this information. What I am saying is that, only with this temperature, we can go further in the analyse of the local position in order to try and find best move, taking into account the building of ko threats. The point being that you do not know if a ko appears in the environment but it could be and the probability is certainly high in practice.

Bill Spight · **#931**

Gérard TAILLE wrote:

Bill Spight wrote:

I know I did some work on this position some years ago, because I commented on it on SL, at https://senseis.xmp.net/?L2GroupWithDescent#toc4 . But it doesn't ring a bell. There are a number of kos in the game tree, but I am not well, and I am not interested in doing a full analysis, at least not now.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . . . . X . . $$ | . . O O X . . $$ | . . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

However, a few days ago I found some interesting things with the following first 4 plays.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . 2 . . X . . $$ | 4 1 O O X . . $$ | 3 . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

At this point White threatens to win the ko in 1 move. We assume that t > 1.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . 2 . . X . . $$ | 4 1 O O X . . $$ | 3 6 O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

= t

Since Black played first we can leave it at that.

Result: -5½ + t

Or Black can kill the corner in 3 net moves.

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | 5 2 7 9 X . . $$ | 4 1 O O X . . $$ | 3 . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

,

= t

Result: 21 - 3t

The temperature of indifference occurs when

-5½ + t = 21 -3t, or when

t = 6⅝

We anticipate each play in the ko gaining 6½ points on average, not 6⅝. Verrry interesting.

More later.

OK Bill my understanding is the following

Click Here To Show Diagram Code: [go]$$ Corner ko $$ -------------- $$ | . O . . X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

if t > 6⅝ both players play in the environment
if 6½ < t < 6⅝ white will not play in the environment; black will play in the corner and white will not defend the corner:

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 5 X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

tenuki

tenuki
Finally if t < 6½ both player will play in the corner. If the case black plays in the corner then white will defend with

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 4 X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | X 6 X O X . . $$ | 8 X O O X . . $$ | X 0 O X X . . $$ | . O O X . . . $$ | 5 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

tenuki

Right. White cannot typically afford to play :w4:

in the corner until the temperature drops to 6½.

But note the peculiarity after :w4:

.

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 4 X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 5 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

,

= t,

fills the ko

If White allows Black to do so, Black can kill the corner in 3 net plays.

Result: 21 - 3t

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 4 X . . $$ | O X O O X . . $$ | X 0 O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

,

,

= t,

,

take ko

Result: -6 + 2t

The temperature of indifference occurs when

-6 + 2t = 21 - 3t, or when

t = 5.4

Really? Do we really have a 5 stage iterated ko? They exist, OC, but is this really one? Does White have to wait until t = 5.4 before playing :w4:

?

No. Take a look at the position after :w8:

.

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | . O . O X . . $$ | O X O O X . . $$ | X 0 O X X . . $$ | . O O X . . . $$ | 9 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

At this point :b9:

is essentially a 1 point sente.

Now

gains only 1 point and the result is -5 + t, so the temperature of indifference occurs when

-5 + t = 21 - 3t , or when

t = 6½

The thing is, White does not play :w10:

until Black has played :b9:

.

Now, Black does not have to wait until after :w8:

to play

. Black could play there after :w6:

or

. The point is, though, that it is played at temperature 1. And since we are assuming t > 1, it need not be played at all in this sequence. White stops after :w8:

. (We may show :b9:

for clarity, OC.)

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 9 O . O X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | a O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

Note that

gains, on average, 6½ points. It is not an inferior play, although futile. In practice, White will wait for Black to play at a before taking the ko to start with.

More later.

Bill Spight · **#932**

Gérard TAILLE wrote:

Bill Spight wrote:

It seems to me that you are using the term, environment, here to mean the rest of the board asid from the local position. You then seem to be complaining that knowing only the temperature of the environment is not enough to say what is best play. But that is a claim that nobody is making. {shrug}

Yes I use the term environment to mean the rest of the board asid from the local position.
But I am not complaining that knowing only the temperature of the environment is not enough to say what is best play. On contrary I appreciate to try and find the best move with only this information. What I am saying is that, only with this temperature, we can go further in the analyse of the local position in order to try and find best move, taking into account the building of ko threats. The point being that you do not know if a ko appears in the environment but it could be and the probability is certainly high in practice.

Sorry, I fail to see what's new. {shrug}

This does not detract from the brilliance of your original sequence.

Gérard TAILLE · **#933**

Bill Spight wrote:

Right. White cannot typically afford to play :w4:

in the corner until the temperature drops to 6½.

But note the peculiarity after :w4:

.

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 4 X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 5 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

,

= t,

fills the ko

Oops I do not understand this sequence Bill.

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | X 6 X 4 X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 5 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

Because we assume white plays :w4:

with t ≤ 6½ then after the hane :b5:

white cannot afford to play tenuki. I think he has to take the ko by :w6:

for a -5 + t result.

Bill Spight wrote:

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 4 X . . $$ | O X O O X . . $$ | X 0 O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

,

,

= t,

,

take ko

Result: -6 + 2t

Here again, unless temperature is very low, the move :w6:

takes ko is not good. White has to prefer to play tenuki hoping a drop of the temperature and, as a consequece, a drop of the result -5 + t.

Bill Spight · **#934**

Gérard TAILLE wrote:

Bill Spight wrote:

Right. White cannot typically afford to play :w4:

in the corner until the temperature drops to 6½.

But note the peculiarity after :w4:

.

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 4 X . . $$ | O X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 5 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

,

= t,

fills the ko

Oops I do not understand this sequence Bill.

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | X 6 X 4 X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 5 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

Because we assume white plays :w4:

with t ≤ 6½ then after the hane :b5:

white cannot afford to play tenuki. I think he has to take the ko by :w6:

for a -5 + t result.

Yes.

is a 1 point sente, raising the local temperature. White must reply.

Gérard TAILLE wrote:

Bill Spight wrote:

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | 1 O 3 4 X . . $$ | O X O O X . . $$ | X 0 O X X . . $$ | . O O X . . . $$ | a O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

,

,

= t,

,

take ko

Result: -6 + 2t

Here again, unless temperature is very low, the move :w6:

takes ko is not good. White has to prefer to play tenuki hoping a drop of the temperature and, as a consequece, a drop of the result -5 + t.

I may have misstated things in my previous note. Until Black plays the hane each play in the ko gains the temperature, t, on average. There is typically no loss in making a play that gains as much as the temperature. In fact, it is a good play.

The only thing that White must be careful about is not to win the ko before Black plays the hane at a.

What about the Black sente? Suppose that White has played :w4:

. The temperature is t and the local territorial value, s = -5 + t. Now Black plays the hane at a, threatening to kill the corner in one more play, for 21 points, gaining 26 - t points, and then White gains t, for a result of 21 - 2t. However, White takes and wins the ko in 3 plays, for a result of -5 + 2t. The hane has raised the local temperature to 6½. White must take and win the ko, tout de suite.

Gérard TAILLE · **#935**

Bill Spight wrote:

I may have misstated things in my previous note. Until Black plays the hane each play in the ko gains the temperature, t, on average. There is typically no loss in making a play that gains as much as the temperature. In fact, it is a good play.

The only thing that White must be careful about is not to win the ko before Black plays the hane at a.

What about the Black sente? Suppose that White has played :w4:

. The temperature is t and the local territorial value, s = -5 + t. Now Black plays the hane at a, threatening to kill the corner in one more play, for 21 points, gaining 26 - t points, and then White gains t, for a result of 21 - 2t. However, White takes and wins the ko in 3 plays, for a result of -5 + 2t. The hane has raised the local temperature to 6½. White must take and win the ko, tout de suite.

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | X . X 4 X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | a O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

Providing the temperature is lower than 6½ and not very low (I mean greater than t = 1) then the only concern for white is to avoid losing the corner. That is the purpose of :w4:

and white expect the result -5 + 2t.
In the above position white has no major reason to play in the corner. Instead white must prefer to play in the environment hoping for a drop of the temperature and thus a drop of the result -5 + 2t.
For black point of view, unless black is able to increase the temperature of the environment, black should play hane as soon as possible to get the best 5 + 2t result.
In this context what is the temperature of the local area? if the temperature of the environment is lower than 6½ then white will not play and black should play as soon as possible. Can we say that the temperature is near from t = 1 for white and is t = 6½ for black?
If yes then, effectively, you can say that the black hane increases the temperature of the corner but only for white point of view (not for black point of view).

Bill Spight · **#936**

Gérard TAILLE wrote:

Bill Spight wrote:

I may have misstated things in my previous note. Until Black plays the hane each play in the ko gains the temperature, t, on average. There is typically no loss in making a play that gains as much as the temperature. In fact, it is a good play.

The only thing that White must be careful about is not to win the ko before Black plays the hane at a.

What about the Black sente? Suppose that White has played :w4:

. The temperature is t and the local territorial value, s = -5 + t. Now Black plays the hane at a, threatening to kill the corner in one more play, for 21 points, gaining 26 - t points, and then White gains t, for a result of 21 - 2t. However, White takes and wins the ko in 3 plays, for a result of -5 + 2t. The hane has raised the local temperature to 6½. White must take and win the ko, tout de suite.

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | X . X 4 X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | a O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

Providing the temperature is lower than 6½ and not very low (I mean greater than t = 1) then the only concern for white is to avoid losing the corner. That is the purpose of :w4:

and white expect the result -5 + 2t.
In the above position white has no major reason to play in the corner. Instead white must prefer to play in the environment hoping for a drop of the temperature and thus a drop of the result -5 + 2t.

For a number of reasons we want t > 1. Since Black cannot win the ko and kill the corner in this position, Black should, as a rule, lose the ko as soon as possible. In this case she can do so by raising the temperature to 6½ in the corner, so that White must play there. Assuming, thermographically, that the ko is played out at the same temperature, in each ko exchange, where White makes one play in the ko while Black makes one play elsewhere, the net result is t - 6½, which is a net gain for White, That is why Black a is a kind of sente, not just for one turn, but for each ko exchange.

Gérard TAILLE wrote:

For black point of view, unless black is able to increase the temperature of the environment, black should play hane as soon as possible to get the best 5 + 2t result.
In this context what is the temperature of the local area?

Without the hane at a the local temperature is the same as the temperature of the whole board. As the temperature of the whole board drops, so does the temperature of the corner. That is why Black should not wait for the temperature of the whole board to drop, because that makes Black's share of the corner drop as well. Black's share is -5 + t, where t is the temperature of the whole board. (OC, thermography only promises approximations.)

Gérard TAILLE wrote:

if the temperature of the environment is lower than 6½ then white will not play and black should play as soon as possible.

It is almost certainly urgent for Black to hane at a. Before Black does so, it is a real question whether White should take the ko or play elsewhere.

Gérard TAILLE wrote:

Can we say that the temperature is near from t = 1 for white and is t = 6½ for black?

The global temperature is what it is. We assume that it lies between 6½ and 1. After Black plays the hane, the local temperature is 6½ for White. The ko rule prevents it from being that high for Black, because Black cannot take the ko back.

Gérard TAILLE wrote:

If yes then, effectively, you can say that the black hane increases the temperature of the corner but only for white point of view (not for black point of view).

Because of the ko ban.

Gérard TAILLE · **#937**

Bill Spight wrote:

Gérard TAILLE wrote:

Click Here To Show Diagram Code: [go]$$B Corner ko $$ -------------- $$ | X . X 4 X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | a O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

Providing the temperature is lower than 6½ and not very low (I mean greater than t = 1) then the only concern for white is to avoid losing the corner. That is the purpose of :w4:

and white expect the result -5 + 2t.
In the above position white has no major reason to play in the corner. Instead white must prefer to play in the environment hoping for a drop of the temperature and thus a drop of the result -5 + 2t.

For a number of reasons we want t > 1. Since Black cannot win the ko and kill the corner in this position, Black should, as a rule, lose the ko as soon as possible. In this case she can do so by raising the temperature to 6½ in the corner, so that White must play there. Assuming, thermographically, that the ko is played out at the same temperature, in each ko exchange, where White makes one play in the ko while Black makes one play elsewhere, the net result is t - 6½, which is a net gain for White, That is why Black a is a kind of sente, not just for one turn, but for each ko exchange.

Gérard TAILLE wrote:

For black point of view, unless black is able to increase the temperature of the environment, black should play hane as soon as possible to get the best 5 + 2t result.
In this context what is the temperature of the local area?

Without the hane at a the local temperature is the same as the temperature of the whole board. As the temperature of the whole board drops, so does the temperature of the corner. That is why Black should not wait for the temperature of the whole board to drop, because that makes Black's share of the corner drop as well. Black's share is -5 + t, where t is the temperature of the whole board. (OC, thermography only promises approximations.)

your answer "Without the hane at a the local temperature is the same as the temperature of the whole board" upset me a little. I understood that, in thermography, the local temperature correspond to the temperature of the low part of the mast going vertically to infinity.

Click Here To Show Diagram Code: [go]$$B $$ -------------- $$ | X . X O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

If I try to draw the termograph of this position I will draw the line -5 + 2t starting at the bottom from the point -5 at t= 0 till the point +8 at t = 6½ and then, from that point (+8, 6½) I will draw a vertical mast to intinity.
Such thermograpph is not quite usual but by definition the local temperature seems to me t = 6½ isn't it?
BTW you use the value -5 + t where I use the value -5 + 2t. How do you draw the thermograph Bill?

Bill Spight · **#938**

Gérard TAILLE wrote:

your answer "Without the hane at a the local temperature is the same as the temperature of the whole board" upset me a little. I understood that, in thermography, the local temperature correspond to the temperature of the low part of the mast going vertically to infinity.

Click Here To Show Diagram Code: [go]$$B $$ -------------- $$ | X . X O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

If I try to draw the termograph of this position I will draw the line -5 + 2t starting at the bottom from the point -5 at t= 0 till the point +8 at t = 6½ and then, from that point (+8, 6½) I will draw a vertical mast to intinity.
Such thermograpph is not quite usual but by definition the local temperature seems to me t = 6½ isn't it?
BTW you use the value -5 + t where I use the value -5 + 2t. How do you draw the thermograph Bill?

For the Black wall, let's find out the territorial count, s, when t > 1.

Click Here To Show Diagram Code: [go]$$B Black first $$ -------------- $$ | X 2 X O X . . $$ | 4 X O O X . . $$ | X 6 O X X . . $$ | . O O X . . . $$ | 1 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

= t

The final local score is -5.

s = -5 + 2t

Now for the White wall.

Click Here To Show Diagram Code: [go]$$W White wall $$ -------------- $$ | X 1 X O X . . $$ | 3 X O O X . . $$ | X 5 O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

= t

The result is the same. Big surprise. :lol:

But to find the vertical mast, we have to find out where the two non-vertical walls diverge. That can happen when White is indifferent to Black's threat to kill the corner.

Click Here To Show Diagram Code: [go]$$B Black first $$ -------------- $$ | X 3 X O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 1 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

,

= t

s = 21 - 2t

The walls meet when

-5 + 2t = 21 - 2t , that is, when

t = 6½ , and

s = 8.

As advertised, because we know that we do not have a 5 move iterated ko where each move gains 5.4 points.

So when t ≥ 6½, the mast rises from (s,t) = (8,6½). When 6½ ≥ t ≥ 1, the mast is inclined along the line, s = -5 + 2t, and when 1 ≥ t ≥ 0, the mast is a vertical line at s = -3.

Let's confirm the value when 1 ≥ t ≥ 0.

Click Here To Show Diagram Code: [go]$$B Black first $$ -------------- $$ | X 2 X O X . . $$ | 4 X O O X . . $$ | X 6 O X X . . $$ | 5 O O X . . . $$ | 1 O X X . . . $$ | 3 X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

Final local score, s = -3.

Ditto when White plays first.

----

I probably gave the impression that the thermograph was inclined along the line, s = -5 + t when 6½ ≥ t ≥ 1, because that was the Black wall in that range for the other thermograph. I apologize for that.

But the thing is, the mast of this thermograph is inclined at, say. t = 4. The mast does not determine the local temperature, it's the other way around. The temperature determines the mast, when the two walls coincide. The Black and White walls are the same at every temperature. :shock:

So, even though the mast rises vertically at s = 8, that is not the average territorial value at t = 4. That value is -5 + 2t = 3.

----

Click Here To Show Diagram Code: [go]$$B $$ -------------- $$ | X O . O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

The thermograph for this position, after White has captured one Black stone has the inclined mast when 6½ ≥ t ≥ 1, along the line, s = -5 + t. At t = 4, s = -1, which is 4 points better for White than s = +3. That is why capturing the stone gains, on average, t.

Gérard TAILLE · **#939**

Click Here To Show Diagram Code: [go]$$B $$ -------------- $$ | X . X O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

Obviously we are now in line for the analyse and the thermograph of the above position. That's fine because it was not that obvious.

Now you propose the following position:

Bill Spight wrote:

Click Here To Show Diagram Code: [go]$$B $$ -------------- $$ | X O . O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

The thermograph for this position, after White has captured one Black stone has the inclined mast when 6½ ≥ t ≥ 1, along the line, s = -5 + t. At t = 4, s = -1, which is 4 points better for White than s = +3. That is why capturing the stone gains, on average, t.

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 1 X O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

This white move :w1:

(takes ko) looks strange and I hardly see a gain with this move.
The point is the following : instead of the :w1:

white can afford to play in the environment and he will be still able to defend the corner.

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 1 X O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

After the exchange :w1:

the result is quite different. If now white plays in the environment black can kill the corner. I conclude :w1:

reverses and white must continue in the corner then

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 1 X O X . . $$ | 3 X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki
After :w3:

in the corner and :b4:

tenuki white still cannot play in the environment which is a quite bad news for white.
The all squence is then

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 1 X O X . . $$ | 3 X O O X . . $$ | X 5 O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

tenuki

Let's us compare with the sequence after :w1:

in the environment

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 3 X O X . . $$ | 5 X O O X . . $$ | X 7 O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

tenuki

tenuki

tenuki

When comparing the two sequences we can see that the corner is the same but the moves in the environment are different:
ScoreWhiteTakesKo : g1 + g2 - g3 + g4 - g5 + g6 ....
ScoreWhiteTenuki : -g1 + g2 + g3 + g4 - g5 + g6 ....
White should prefer taking the ko if:
ScoreWhiteTakesKo < ScoreWhiteTenuki <=> 2(g1 - g3) < 0 which is never true.
Taking the ko in the corner is not a good idea isn'it? OC in a real game with potential black ko threats then you change all the assumptions.

Where is the gain you founnd by taking the ko?

Bill Spight · **#940**

Gérard TAILLE wrote:

Click Here To Show Diagram Code: [go]$$B $$ -------------- $$ | X . X O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

Obviously we are now in line for the analyse and the thermograph of the above position. That's fine because it was not that obvious.

Now you propose the following position:

Bill Spight wrote:

Click Here To Show Diagram Code: [go]$$B $$ -------------- $$ | X O . O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

The thermograph for this position, after White has captured one Black stone has the inclined mast when 6½ ≥ t ≥ 1, along the line, s = -5 + t. At t = 4, s = -1, which is 4 points better for White than s = +3. That is why capturing the stone gains, on average, t.

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 1 X O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | . O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

This white move :w1:

(takes ko) looks strange and I hardly see a gain with this move.
The point is the following : instead of the :w1:

white can afford to play in the environment and he will be still able to defend the corner.

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 1 X O X . . $$ | . X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

After the exchange :w1:

the result is quite different. If now white plays in the environment black can kill the corner. I conclude :w1:

reverses and white must continue in the corner then

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 1 X O X . . $$ | 3 X O O X . . $$ | X . O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki
After :w3:

in the corner and :b4:

tenuki white still cannot play in the environment which is a quite bad news for white.
The all squence is then

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 1 X O X . . $$ | 3 X O O X . . $$ | X 5 O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

tenuki

Let's us compare with the sequence after :w1:

in the environment

Click Here To Show Diagram Code: [go]$$W $$ -------------- $$ | X 3 X O X . . $$ | 5 X O O X . . $$ | X 7 O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

tenuki

tenuki

tenuki

tenuki

When comparing the two sequences we can see that the corner is the same but the moves in the environment are different:
ScoreWhiteTakesKo : g1 + g2 - g3 + g4 - g5 + g6 ....
ScoreWhiteTenuki : -g1 + g2 + g3 + g4 - g5 + g6 ....
White should prefer taking the ko if:
ScoreWhiteTakesKo < ScoreWhiteTenuki <=> 2(g1 - g3) < 0 which is never true.
Taking the ko in the corner is not a good idea isn'it? OC in a real game with potential black ko threats then you change all the assumptions.

Where is the gain you founnd by taking the ko?

If you want to find the gain from taking the ko, start by taking the ko.

If you want to find the gain from playing in a non-thermographic environment, make that play part of the game. That is, part of the ko ensemble.

For instance, if you add the simple gote, {u|-u}, 6½ ≥ t > 1, to the corner position. we can try to find the thermograph of that ko ensemble. Let's find the walls when 6½ ≥ u ≥ t.

Click Here To Show Diagram Code: [go]$$B Black takes u first $$ -------------- $$ | X 2 X O X . . $$ | 4 X O O X . . $$ | X 6 O X X . . $$ | . O O X . . . $$ | 5 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

= u,

= t

s = -5 + u + t

Now let's try playing the hane first.

Click Here To Show Diagram Code: [go]$$B Black plays the hane first $$ -------------- $$ | X 2 X O X . . $$ | 4 X O O X . . $$ | X 6 O X X . . $$ | . O O X . . . $$ | 1 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

= u,

= t

s = -5 + u + t

All same same.

Now let's try White first.

Click Here To Show Diagram Code: [go]$$W White takes -u first $$ -------------- $$ | X 3 X O X . . $$ | 5 X O O X . . $$ | X 7 O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

= -u,

,

,

= t

s = -5 - u + 3t

Click Here To Show Diagram Code: [go]$$W White takes ko first $$ -------------- $$ | X 1 X O X . . $$ | 3 X O O X . . $$ | X 5 O X X . . $$ | . O O X . . . $$ | 2 O X X . . . $$ | . X . . . . . $$ | . X . . . . . $$ | X X . . . . . $$ | . . . . . . . $$ | . . . , . . .[/go]

= u,

= t

s = -5 + u + t

Taking -u first is better. Big duh.

The walls meet when

s = -5 - u + 3t = -5 + u + t , that is, when

t = u

When u > t > 1 , then the White wall follows the line, s = -5 - u + 3t and the Black wall follows the line, s = -5 + u + t.

To get this far. we assumed that u > t. But when 6½ ≥ t ≥ u the thermograph is the same as above, when u was out of the picture. The thermograph has an inclined mast following the line s = -5 + 2t starting at (s,t) = (-5 + 2u, u) and a vertical mast at rising from (s,t) = (8,6½).

----

When u > t, then White does better to take -u. But when t ≥ u, White does at least as well to take the ko.

This 'n' that

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