The same difference games that suggest playing the gote instead of the reverse sente apply as a heuristic is both cases. The second case is perhaps more striking because the gote gains less by itself (3½ points) than the reverse sente (4 points).



When I first joined Professor Berlekamp's study group he was a bit irritated with me because I figured out optimal play and used that to draw the thermograph. His point was that you can use thermography to find optimal play.



Commonly this area is described as a 4 point sente for Black. Better to say that it is a 4 point reverse sente for White, since it is the reverse sente that gains 4 points. I don't know how many players I have met who thought that Black gained 4 points with a 4 point sente.

Good players will also say that this area (the top two lines of the board) is worth 3 points for White. The thermographer adds the qualification, on average.



I think your 2 is a typo for 4. Above temperature 4 the average value (count) of the top is -3.



To be clear, that's the minimax value at that temperature, depending upon who plays first. In this case, below temperature 4 there are two minimax values at each temperature, which the sides of the thermograph indicate.



I don't know which area you are talking about. The bottom right corner? It is gote. Then I don't know why you are talking about a temperature drop.



For an ideal environment with a sufficiently high temperature.

And not every good player has the idea of an ideal environment.



Indeed. The practical player can usually dispense with that information and spend her time on other things.



g = 3, OC.



Do you mean that the reverse sente gains 4 points, on average?



An ideal environment for every possible game has an infinite nember of plays. However, given a specific game or go position, there is at least one finite ideal environment. Representing a simple gote that gains g points for either side as ±g, the environment, ±4, ±3½, ±3, ±2½, ±2, ±1½, ±1, ±½ should be ideal for the top side position.



OK. Any significant temperature drop can indicate a kind of tedomari. I.e., getting the last move at a specific temperature can be important.



The thermographer will be alert to the gote at temperature 2, because the left wall of the thermograph for the top position has an inflection point at temperature 2.

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

Visualize whirled peas.

Everything with love. Stay safe.

OK OC for the typo you corrected!
In addition instead of each player will recognize that the area is a good 3 points gote point you should read 4 points gote. Sorry for that.

Concerning the area on the bottom right. I know that on the diagram the gote value is g=3 but I know you have undertood that I vary this value to analyse various environment.
When I wrote The temperature of the environment is equal to 4 and the value of our local area (against an ideal environment) is 4 points in reverse sente. I do not mean that the reverse sente gains 4 points, on average.
If mean black would be advised to play in sente before temperature drops under 4 because, as soon as the temperature drops under 4 white, by playing the reverse sente at temperature t, will gain 4-t.



No real disagreement indeed it maybe only a problem of vocabulary. The basic reasons I said a real environment can never be ideal are the following:
- firstly when you calculate the real score with a real environment the resulting score is always a round number and cannot be something like n + t/2 for a total of say 2⅔, can it?
- secondly the environment, ±4, ±3½, ±3, ±2½, ±2, ±1½, ±1, ±½ looks quite ideal especially because when you play in this environment you gain exactly 2 points i.e. t/2 witch is fine. But as soon as the first gote point has been played, the new best gote point is ±3½, and this resulting environment cannot be also ideal with a gain of t/2.

Example of standard calculation without (?) thermography:

Click Here To Show Diagram Code

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

Imagine a good player analysing the local area in the assumption that the best gote move in the environment has a value > 4.

Because she is pretty good player she will imagine the following sequence:

Black to play:
Black begins by the sente sequence

Click Here To Show Diagram Code

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

then white continues in sente by

Click Here To Show Diagram Code

[go]$$W Black to play
$$ -----------------
$$ | . 2 B 1 B . O |
$$ | X X W . W . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ -----------------[/go]

and finally you add the following exchange:

Click Here To Show Diagram Code

[go]$$B Black to play
$$ -----------------
$$ | . B B W B 2 O |
$$ | X X W 1 W . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ -----------------[/go]

to reach the position

Click Here To Show Diagram Code

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

and now by a simple average calculation she counts the position as -3.

White to play is far more simple:

Click Here To Show Diagram Code

[go]$$W white to play
$$ -----------------
$$ | . . . . . . O |
$$ | X X 1 . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ -----------------[/go]

and here again by a simple average calculation she counts the position as -7.

In her mind the conclusion is simply a 4 point reverse sente for white.

But here is the point: without being aware of thermography the player has indeed made all the thermography analysis and all the calculation above can be visualised by the followng thermograph in which you can see in the bottom of the thermograph the two vertical lines corresponding to the average scores calculated effectively by the player.

Note that the wordings reverse sente and average used by the player implies indirectly the use of an ideal environment.

Well, the area is worth 3 points to White, on average. The 4 points refers to how much the reverse sente gains, again, on average, which we can drop if you like. It is important to distinguish between the count of an area and the gain of a move. For the sake of our readers.



I think it is a question of definition. For a specific game an ideal environment is one that produces the thermograph for the game. That's the purpose of the ideal environment.



Well, of course that last requirement cannot be met with a finite environment.



Well, for one thing there are plays that the player has not tried. E.g.,

Click Here To Show Diagram Code

[go]$$Bc Black to play
$$ -----------------
$$ | . . 3 . 1 4 O |
$$ | X X . . 2 . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ -----------------[/go]

Which she may have seen in a textbook.

It is true that my feeling was that this was too passive for White, and I came up with this.

Click Here To Show Diagram Code

[go]$$Bc Black to play
$$ -----------------
$$ | . . 5 2 1 . O |
$$ | 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]

One reason I stopped here was that I had calculated the thermograph for the textbook result years ago and could therefore tell that it was worse for White on average than this result. But if I had not done so I would have had to do some calculation, or in a game I might have just gone with my gut feeling.

I know from long experience that pros miscalculate yose in textbooks. Any sizable text, with some recent exceptions, will probably have more than 100 calculation errors. Probably because they go with their gut feelings.

Edit:


Simply knowing the rules for finding the territory values for sente and gote does not imply any idea of an ideal environment. The rules can be derived in other ways. In fact the original thermography did not refer to an environment at all. Berlekamp came up with the idea of an ideal environment, although he used different terminology. I had independently come up with the idea of an environment, by which I was able to derive those rules before learning thermography. And later I was able to define thermography in terms of an ideal environment in order to apply thermography to multiple ko and superko positions.

Now, pros certainly have an idea about the environment. Sometimes even strong amateurs have trouble with the idea that correct play in a position without an environment (as we would say, at temperature 0) may be different from the play used to calculate the territory of that position. "This is correct play in this position. Why don't we use correct play to find the territory?"

But if pros in general had the idea of an ideal environment, there would be no confusion about double sente. Thermography indicates double sente at temperatures where both walls of the thermograph are vertical. A simple and clear explanation.

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

Visualize whirled peas.

Everything with love. Stay safe.

I quite undertand what the ideal environment is but I see also that a non mathematician go player will have difficulty to imagine what means a limit calculation with an infinite gote points.

On contrary the environment ±4, ±3½, ±3, ±2½, ±2, ±1½, ±1, ±½ you proposed yourself is the kind of environment which sounds quite obvious for any player. With this in mind I looked for some ideas to give such environment good carateristics and in particular the basic result:
the gain expected by playing first in the environment is equal to t/2.

Finally my idea is the following:
Imagine that for each gote value (½, 1, 1½, 2, 2½, ...) you put a random number of gote points. The point is the following : because the number of gote points on say 1½ value, is a random number I claim that for 50% this number is even and for 50% it is odd.
Let's show you that for such environment the gain expected for the player with the move is always t/2.
Firstly this formula is correct at temperature ½ beacuse
if the number of ½ gote points is even the gain is 0
and if the number of ½ gote points is odd the gain is ½
On average the gain of this environment at temperature ½ is ¼.
Now suppose the formula is correct at temperature 3 for example and let's consider the environment at temperature 3½.
if the number of 3½ gote points is even then the gain on level 3½ is 0 but the sequence is sente and the gain is thus equal to the gain at level 3, and this gain is 1½ by hypothesis
if the number of 3½ gote points is odd then the gain on level 3½ is 3½ but the gain for remaining gote point (1½) will be taken by the opponent and finally the total gain is 3½ - 1½ = 2.
On average (without knowing if the number of 3½ gote points is even or odd) the gain for playing first in the environment is (1½ + 2)/2 = 1¾ and you see that 1¾ is equal to t/2 with t = 3½.

I do not know is this approach may be useful but at least it is quite understandable that the formula saying the gain the envionment is t/2 makes really sense doesn'it?

Indeed, it does.

As you may recall, early on I suggested an environment where the number of plays of each size was unknown, thus its parity was unknown. I did not appeal to a random number generator, but the effect is basically the same. We may estimate the value of making the top play in the environment at ambient temperature t as t/2. Also, my original environment where each successive temperature is less than or equal to its previous temperature has the same property.

One drawback to all of these is that there may be sizable drops in temperature in them. That is why Berlekamp preferred what he called a rich environment, where each temperature has an odd number of plays, but the temperature drops are small. He even suggested an environment where each temperature drop is 0.01 point. He did not envision players actually making successive plays in the environment, but making sealed bids of the temperature at which they are willing to make a play in the game.

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

Visualize whirled peas.

Everything with love. Stay safe.

Surely I was not clear ennough because you above take my point.
Let's s the two following proposals for building on environment:
1) you take as an environment at temperature t, say t=4, the environment made of the 8 gote points 4, 3½, 3, 2½, 2, 1½, 1, ½. It looks fine because the expected gain by playing first in this environment is :
gain(8 gote points) = 4 - 3½ + 3 - 2½ + 2 - 1½ + 1 - ½ = 2.
But as soon as you have played the highest got move it remains 7 gote points 3½, 3, 2½, 2, 1½, 1, ½ and this time the gain when playing first in this remaining environment at temperature 3½ is:
gain(7 gote points) = 3½ - 3 + 2½ - 2 + 1½ - 1 + ½ = 2 which if different from t/2 = 3½ / 2 = 1¾
2) you take as an environment at say temperature t=4, a set of N gote points each with a value of 4 points. Because you take N at random you cannot say if N is even or odd and as a consequence the gain expected from this environment is clearly t/2. That is fine but this environment cannot be ideal because the temperature can only drop from 4 to 0 without any intermediate temperature.

My idea is mixing the ideas of this two environments. To create my environment I need only a random generator (a piece of money?) which gives as a result the value "true" or the value "false".
As an example let's create an environment at temperature t=4.
First of all I put in this environment a 4 points gote point.
Then I initialise a variable v to v=t, this variable being the current value of the gote point I will add to my environment
and now I continue the building of the environment by the following loop:
loop
...I ask for the result of the random generator :
...if the result is "true" I add to the environment a gote point with value v
...if the result is "false" I do not add a gote point in the environment but instead I decrease the v value to v -> v-½
...You continue the loop until v reach 0
end of loop

The purpose of my previous post was to proof the gain expected from such environment at temperature t is equal to t/2.
Note the following other caracteristic of this environment:
For any value v < t the mean number of gote points of value v is equal to 1. More precisely, the number of gote points on this level v is 0 with probability ½, 1 with probability ¼, 2 with probability ⅛ ...
You can now see why this environment is a kind of mixture of the two previous environments I mentionned above but this time with good caracteristics:
gain in the environment equal to t/2, temperature droping slowly from t to 0, small number of gote points (about 2t gote points).

OC, if you may prefer, instead of decreasing v by the value ½, you can decrease this value by 0.01 but it is another point.

Click Here To Show Diagram Code

[go]$$W white to play
$$ -----------------
$$ | . . . . . . O |
$$ | X X a . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ -----------------[/go]

In the diagram above, white to play, a move at "a" is analysed as a 4 point reverse snete.
Assuming that the environment is at temperature t=4 we conclude that it is equivalent for white to play at "a" or to play first in the environment.
My question is the following : looking only at the highest gote points of the environment can white improve her average score?
The answer seems yes:
You count the number of 4 points gote in the environment. If this number is even then you play at "a" leaving the opponent, on average, a slightly miai environment. If this number is odd then you play in the environment in order to play first in a slightly tedomari environment. That way you can probably expect a gain in average equal to ¼.

From the perspective of thermography, it is not ideal because Black to play can move to a 2 point gote, but there is no play in the environment that gains 2 points.



Right. Did you think I missed that?



Berlekamp was willing to accept a slight deviation from t/2 in exchange for limiting the potential error.

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

Visualize whirled peas.

Everything with love. Stay safe.

Your expected gain in a real game is closer to 0.05.

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

Visualize whirled peas.

Everything with love. Stay safe.

Oops you know perfectly that I will not dare to think you can miss something like that Bill but I know that, perhaps due to my bad english, I can be sometimes unclear, especially for the other readers.
Be sure I consider you are the best expert on thermography, at least on this forum.

Rest assured I am quite capable of missing and misunderstanding things.

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

Visualize whirled peas.

Everything with love. Stay safe.

I do not understand how you calculate the figure 0.05
My analyse is the following:
if the number of 4 point gote is even then these gote are miai. If I assume that in a real game the first following not miai level is at level 3½ then the gain from the environment is only 1¾ and not 2 => by chosing the reverse sente I expect to gain an extra ¼ point.
In the other hand if the number of 4 point gote is odd then these gote at level 4 are tedomari. The gain from the environment is now
4 - 1¾ = 2¼ => by chosing the environment rather than the reverse sente I expect to gain here also an extra ¼ point.

How did you find your 0.05 figure? By assuming that, on average, a real game gote environment looks like: 4, 3.9, 3.8, 3.7 .... 0.2, 0.1 ?
I lack statistics on real games but my feeling (it is only my feeling!) is that, on average, when temperature drops following a given move, it drops on average for ½ points.
Do you have such statistics figures on real game?

The first move gains around 14 points. If game lasts to the end, 200+ moves later the last move typically gains 0 points. Not all of those moves are gote, OC, but games that are not resigned usually last longer than 200 moves.

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

Visualize whirled peas.

Everything with love. Stay safe.

I begin to look at corridor evaluation but it seems not so obvious.
For the common simpliest case:

Click Here To Show Diagram Code

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

the evaluation is quite easy. Depending on the length of the corridor you will find 1½ or 1¾ or 1⅞ ...
but what about a larger corridor like the following:

Click Here To Show Diagram Code

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

or a most difficult one like

Click Here To Show Diagram Code

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

Note that the situation above may be very common when white has just play somewhere a double sente move and black chooses a possible exchange by answering with the hane here with the marked black stone, followed by a move at "a" (which is not necessarily sente depending of the temperature of the environment).

Let's take an example where the moves to play are obvious.

Click Here To Show Diagram Code

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

The corridor begins at "a".
Now each time black plays in the area the score for black grows following 6 + 4 + 5 + 3 + 2 + 4
How do you calculate the miai value of a black move at "a" if the temperature of the environment is high (eg t = 8).
I find a way to calculate this value without running throudh all the corridor but I am not sure it is completly correct.
I find here the value 4⅞ but what is the correct value in this example?

Click Here To Show Diagram Code

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

Using slash notation here is this game.

{0|-4||-6|||-9||||-14|||||-18||||||-24}

Working from the bottom up, above temperature 2 we can replace 0|-4 with -2.

{-2|-6||-9|||-14||||-18|||||-24}

And we can replace -2|-6 with -4.

{-4|-9||-14|||-18||||-24}

Above temperature 2½ we can replace -4|-9 with -6½.

{-6½|-14||-18|||-24}

Above temperature 3¾ we can replace -6½|-14 with -10¼.

{-10¼|-18||-24}

Above temperature 3⅞ we can replace -10¼|-18 with -14⅛.

{-14⅛||-24}

Above temperature 4 ¹⁵/₁₆ we can replace {-14⅛||-24} with

-19 ¹/₁₆

Which is the count.

And the average gain per move is 4 ¹⁵/₁₆.

Edit: Here is the thermograph.

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

Visualize whirled peas.

Everything with love. Stay safe.

The counts are 0, -½, -1¼, -2⅛, . . . In this case it is -6 ¹/₁₂₈ and the average gain is ¹²⁷/₁₂₈.

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

Visualize whirled peas.

Everything with love. Stay safe.

I can follow your calculation but for the time being I am not quite sure of my understanding

Click Here To Show Diagram Code

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

With only the first 6 deiri value, the miai value of a move at "a" is 3.

Now take the following

Click Here To Show Diagram Code

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

Taking your calculation the value of a move at "a" is 3 + 4 = 7 but here is my point:
The best possible result black can expect from a move at "a" is that the move is sente and white answers immediately at "b".
That means that the miai value for a move at "at cannot be greater than 6.
Am I wrong?

Click Here To Show Diagram Code

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

Here is the game in slash notation.

{0|-16||-22}

Above temperature 8 we can replace 0|-16 with -8.

{-8||-22}

Above temperature 7 we could replace that with -15, but 7 < 8, which means that White's play is sente. So above temperature 6 we can replace {0|-16||-22} with -16.

The thermograph makes that clear.



The right scaffold of the thermograph intersects the left scaffold at temperature 6. If we draw the thermograph we do not even have to consider the possibility of temperature 7. But working non-graphically, it may take us some calculation to figure that out.

Note that the left wall, colored blue, extends up to temperature 8, which indicates that below that temperature Black can normally play with sente. The blue wall between temperature 6 and 8 indicates what go players call the privilege of Black to play the sente.

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

Visualize whirled peas.

Everything with love. Stay safe.

OK Bill, now I see clearly what is your calculation and due to a miscalculation my example is not the one I would have taken.
Instead of the sequence 6 + 4 + 5 + 3 + 2 + 4 can we take the simplier sequence 6 + 4 + 5 + 6 + 2
and the question is : at what temperature will white blocks the corridor by playing at "a"?
This time I found t=5 but I am not sure at 100% of such result

Let's see.

{0|-2||-8|||-13||||-17|||||-23}

Above temperature 1 we have

{-1|-8||-13|||-17||||-23}

Above temperature 3½ we have

{-4½|-13||-17|||-23}

Above temperature 4¼ we have

{-8¾|-17||-23}

Above temperature 4⅛ we might have

{-12⅞|-23} but 4⅛ < 4¼ so above temperature 4 we have

{-13|-23}

Above temperature 5 we have

-18.

The thermograph is this.

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

Visualize whirled peas.

Everything with love. Stay safe.