Thermography

[Gérard TAILLE]

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 attachment GTG.png is no longer available

Your post is really interesting
First of all we note the calculation giving the sequence
0, -1, -4½, -8¾, -12⅞
then handling the possibility to create a sente situation you made the correction
-12⅞ -> 13
and then the sequence ended by
13, 18
in order to give the correct thermograph.

From a theoritical point of view your method to reach the thermograph is correct but I see two major drawbacks for pratice:
1) The sequence 0, -1, -4½, -8¾, -12⅞, 13, 18 is a very poor sequence in terms of convergence towards the final value, because the most significant change through this sequence appears at the end of the sequence and not at the beginning
2) When you look in details to the correction -12⅞ -> 13 you discover that a part of the calculation to build the sequence 0, -1, -4½, -8¾, -12⅞ may be not necessary

With this in mind I look for another procedure for building this thermograph. Here it is.

Firstly we can note that for such corridor the thermograph looks like the following

corridor.png (5.09 KiB) Viewed 27011 times

Let's take again the corridor defined by the sequence 6, 4, 5, 6, 2

The first figure of the sequence (6 in the example) is the most important figure and this figure is the width of the thermograph base.
In order to draw the correct thermograph we have only to calculate the value x shown above in the attachement file.
My goal is to calculate this value x by following the sequence in the most efficient order and by handling at each step a min value and a max value for x.

Taking the first value of the sequence (6) I initialise
min = 0 and max = 6

Now I get to each figure in the sequence a weight designing the influence of the figure on the final result
The first figure (6) has the weight 1
The second figure (4) has the weight 1/2
The third figure (5) has the weight 1/4
The fourth figure (6) has the weight 1/8
The fifth figure (2) has the weight 1/16

Now let's procede:

Let's take the second figure (4), I handle the contribution of this figure to find x as follows:
The weight of this figure being 1/2 the second figure 4 can only increase the min value by figure between 2 and 4. The new min-max value are then
min = 2 max = 4

Let's take the third figure (5)
The weight of this figure being 1/4 this third figure 5 can only increase the min value by figure between 1¼ and 2½. The new min-max value are then
min = 3¼ max = 4
Note that the max value as not changed because previous_min + 2½ = 2 + 2½ = 4½ > previous_max

Let's take the fourth figure (6)
The weight of this figure being 1/8 the fourth figure 6 can only increase the min value by figure between ¾ and 1½. The new min max value are then
min = 4 max = 4

Because min = max = 4 you stop there and draw the thermograph accordingly without looking to the remaining figures defining the corridor!

BTW, if you are a little lazy and you look only on a good estimation of the thermograph you can of course stop for example after the third figure. Here you have min = 3¼ max = 4 and you may use your feeling to estimate x between 3¼ and 4. Seeing the following figure 6, which is quite high your feeling will certainly choose a value very near from the correct value 4 !

[Bill Spight]

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.

GTG.png

Your post is really interesting
First of all we note the calculation giving the sequence
0, -1, -4½, -8¾, -12⅞
then handling the possibility to create a sente situation you made the correction
-12⅞ -> 13
and then the sequence ended by
13, 18
in order to give the correct thermograph.

From a theoritical point of view your method to reach the thermograph is correct but I see two major drawbacks for pratice:
1) The sequence 0, -1, -4½, -8¾, -12⅞, 13, 18 is a very poor sequence in terms of convergence towards the final value, because the most significant change through this sequence appears at the end of the sequence and not at the beginning
2) When you look in details to the correction -12⅞ -> 13 you discover that a part of the calculation to build the sequence 0, -1, -4½, -8¾, -12⅞ may be not necessary

Good points. Note that this is not a thermographic technique. It is one that a player might use at the table. Yours is a thermographic technique.

With this in mind I look for another procedure for building this thermograph. Here it is.

Firstly we can note that for such corridor the thermograph looks like the following

corridor.png

Let's take again the corridor defined by the sequence 6, 4, 5, 6, 2

The first figure of the sequence (6 in the example) is the most important figure and this figure is the width of the thermograph base.
In order to draw the correct thermograph we have only to calculate the value x shown above in the attachement file.
My goal is to calculate this value x by following the sequence in the most efficient order and by handling at each step a min value and a max value for x.

Taking the first value of the sequence (6) I initialise
min = 0 and max = 6

That, I believe, is equivalent to finding the minimum and maximum values for the game {z|-17||-23}, where z ≥ -17. When z = -17 we get the minimum of 0, and when z ≥ -5 we get the maximum of 6. OC, we know that the maximum for z in this corridor is 0.

Now I get to each figure in the sequence a weight designing the influence of the figure on the final result
The first figure (6) has the weight 1
The second figure (4) has the weight 1/2
The third figure (5) has the weight 1/4
The fourth figure (6) has the weight 1/8
The fifth figure (2) has the weight 1/16

Now let's procede:

Let's take the second figure (4), I handle the contribution of this figure to find x as follows:
The weight of this figure being 1/2 the second figure 4 can only increase the min value by figure between 2 and 4. The new min-max value are then
min = 2 max = 4

For the game {-13 | -13 || -17 ||| -23} we get the minimum of 2. For the game {0 | -13 || -17 ||| -23} we get the maximum of 4. (OC, we could use Big instead of 0 and not calculate the mean of -13 and 0.)

Let's take the third figure (5)
The weight of this figure being 1/4 this third figure 5 can only increase the min value by figure between 1¼ and 2½. The new min-max value are then
min = 3¼ max = 4
Note that the max value as not changed because previous_min + 2½ = 2 + 2½ = 4½ > previous_max

For the game {-8 | -8 || -13 ||| -17 |||| -23} we get the minimum of 3¼. For the game {0 | -8 || -13 ||| -17 |||| -23} we get the maximum of 4, because {0 | -8 || -13 ||| -17} is a Black sente.

Let's take the fourth figure (6)
The weight of this figure being 1/8 the fourth figure 6 can only increase the min value by figure between ¾ and 1½. The new min max value are then
min = 4 max = 4

For the game {-2 | -2 || -8 ||| -13 |||| -17 ||||| -23} we get the minimum of 4, because {-2 | -2 || -8 ||| -13 |||| -17} is ambiguous between sente and gote. We do not have to check {0 | -2 || -8 ||| -13 |||| -17}, as it will be a Black sente.

Because min = max = 4 you stop there and draw the thermograph accordingly without looking to the remaining figures defining the corridor!

BTW, if you are a little lazy and you look only on a good estimation of the thermograph you can of course stop for example after the third figure. Here you have min = 3¼ max = 4 and you may use your feeling to estimate x between 3¼ and 4. Seeing the following figure 6, which is quite high your feeling will certainly choose a value very near from the correct value 4 !

The method that I used, starting from the left, assumes that each game is gote unless we find that it is sente. This is the non-thermographic method that I usually teach, because it seems that people understand it fairly quickly. Also, on the go board gote are more common than sente. You can also use the method of assuming that each game is sente unless we find that it is gote. That also works. In practice, we can also guess which it is and proceed until and unless proven wrong.

Let's apply the sente first method to this corridor, assuming that each step into the corridor is sente.

{Big | -17 || -23} count = -17 gain = 6

{Big | -13 || -17 ||| -23} count = -18 gain = 5

{Big | -8 || -13 ||| -17 |||| -23} count = -18 gain = 5

(We can infer this from your sequence, 6 + 4 + 5 + 6 + 2, because 5 > 4. This is obvious from the thermograph.)

This is consistent with {Z | -13 || -17} being sente, where Z ≥ -13, but we have failed to disprove it. Let's try to do so by substituting the next value in the corridor for Big in {Big | -8}.

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

The mast of {-2 | -8} - t and -13 + t intersect at t = 4. That means that this game is ambiguous between sente and gote, but its mast value is still -13, which is the same as if it were sente. The game when we add the next value is, OC, sente.

Edited for clarity and correctness. (I missed the ambiguity. }

[Gérard TAILLE]

It seems to me that we have a here a common understanding haven't we?
At least we have now on the table three methods that can be used to find the thermograph of a corridor and every one can have her preference.
You know my preference Bill, what is yours? BTW I suspect you could have two different preferences; one as thermography theorician and the other as go player?

[Bill Spight]

It seems to me that we have a here a common understanding haven't we?
At least we have now on the table three methods that can be used to find the thermograph of a corridor and every one can have her preference.
You know my preference Bill, what is yours? BTW I suspect you could have two different preferences; one as thermography theorician and the other as go player?

Well, both at the table and as an analyst, I usually guess whether a play is sente or not and try to disprove that. At the table I am generally satisfied with approximations. After all, thermography only provides heuristics. With this position at the table I would probably stop after getting the same estimate twice.

[Bill Spight]

Using your sequence, let's make some approximations to the temperature.

6 + 4 + 5 + 6 + 2

6: max temp = 6, min = 6/2 = 3, Estimate = 4½, max error = 1½

6 + 4: max temp = 3 + 4/2 = 5, min temp = 3 + 4/4 = 4, Estimate = 4½, max error = ½ (Maybe good enough at the table. )

6 + 4 + 5: max temp = min (5, 4 + 5/4) = 5, min temp = 4⅝. Estimate = 4¹³/₁₆, max error = ³/₁₆ (Very likely good enought at the table.)

6 + 4 + 5 + 6: max temp = min(5, 4⅝ + 6/8) = 5, min temp = 4⅝ + 6/16 = 5, Estimate = 5, max error = 0

This is basically your approach.

[Gérard TAILLE]

Using your sequence, let's make some approximations to the temperature.

6 + 4 + 5 + 6 + 2

6: max temp = 6, min = 6/2 = 3, Estimate = 4½, max error = 1½

6 + 4: max temp = 3 + 4/2 = 5, min temp = 3 + 4/4 = 4, Estimate = 4½, max error = ½ (Maybe good enough at the table. )

6 + 4 + 5: max temp = min (5, 4 + 5/4) = 5, min temp = 4⅝. Estimate = 4¹³/₁₆, max error = ³/₁₆ (Very likely good enought at the table.)

6 + 4 + 5 + 6: max temp = min(5, 4⅝ + 6/8) = 5, min temp = 4⅝ + 6/16 = 5, Estimate = 5, max error = 0

This is basically your approach.

You have a perfect understanding of my approach Bill.
I like very much the quite efficient convergence towards the correct value with an error decreasing rapidly.

In addition taking for example the second step
6 + 4: max temp = 3 + 4/2 = 5, min temp = 3 + 4/4 = 4
in a real game, after having identifying 6 + 4, and without identifying the following 5 + 6 figures, you may see clearly that the corridor continues being quite large. In that case, with max temp = 5 and min temp = 4, your experience will tell you to take obviously max temp as a better approximation !
Yes Bill it is really a good method for a real go player and I would be not surprised to learn pro used it, without knowing the thermography background for the justification. Any information on pro estimation of a corridor?

[Bill Spight]

Yes Bill it is really a good method for a real go player and I would be not surprised to learn pro used it, without knowing the thermography background for the justification. Any information on pro estimation of a corridor?

AFAIK, pros have not written on any corridors besides the typical linear closed corridors that appear in textbooks. Since the Japanese edition of Mathematical Go was published in 1994, I expect that at least some pros use the techniques there. However, before publication, 9 dans flunked some of the corridor problems in the book, and I rather expect that some 9 dans today would flunk similar problems.

[Bill Spight]

Maybe I spoke too soon. O Meien, in his book, Yose: Zettai Keisan (Yose: Absolute Calculation), 2004, has some short corridors.

$$Bc Corridors
$$ ---------------------------------------
$$ | X O O . O X O . O . O . O O . O O O X |
$$ | . O X X O X X X X X O X X X X X X X X |
$$ | X X X . O . . . . . O . . . . . X . . |
$$ | . . . , . . . . . , . . . . . X O O X |
$$ | . . . . . . . . . . . . . . . X X X O |
$$ | . . . . . . . . . . . . . . . . . X O |
$$ | . . . . . . . . . . . . . . . . . X . |
$$ | . . . . . . . . . . . . . . . . O O O |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . , . . . . . , . . . . . , . . . |

Click Here To Show Diagram Code

[go]$$Bc Corridors
$$ ---------------------------------------
$$ | X O O . O X O . O . O . O O . O O O X |
$$ | . O X X O X X X X X O X X X X X X X X |
$$ | X X X . O . . . . . O . . . . . X . . |
$$ | . . . , . . . . . , . . . . . X O O X |
$$ | . . . . . . . . . . . . . . . X X X O |
$$ | . . . . . . . . . . . . . . . . . X O |
$$ | . . . . . . . . . . . . . . . . . X . |
$$ | . . . . . . . . . . . . . . . . O O O |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . , . . . . . , . . . . . , . . . |[/go]

All stones connected to the center are immortal.

[Gérard TAILLE]

A special case of corridor has to be analysed in a different way : it is the common problem of answering a double sente move with another double sente move.

Let's suppose white plays a common "double sente" move which threatens to begin entering a corridor 6 + 6 + 5 + 7 and now suppose black answers with another "double sente" move which threatens another corridor 8 + 5 + 8.
Because we are talking about "double sente" move I assume the temperature of the environment is lower than 5.
What will now white play? Entering her corridor or blocking the opponent one? It seems (?) that the decision will be different if temperature is above or under t=3. I suspect a lot of mistakes can be made by real go players when facing such common situations.
What is your feeling Bill?

[Bill Spight]

A special case of corridor has to be analysed in a different way : it is the common problem of answering a double sente move with another double sente move.

Let's suppose white plays a common "double sente" move which threatens to begin entering a corridor 6 + 6 + 5 + 7 and now suppose black answers with another "double sente" move which threatens another corridor 8 + 5 + 8.
Because we are talking about "double sente" move I assume the temperature of the environment is lower than 5.
What will now white play? Entering her corridor or blocking the opponent one? It seems (?) that the decision will be different if temperature is above or under t=3. I suspect a lot of mistakes can be made by real go players when facing such common situations.
What is your feeling Bill?

We have a game, {24||||18|||12||7|0} + {0|-8||-13|||-21}, is that right? The question is, which is better for White, to move to {18|||12||7|0} + {0|-8||-13|||-21} or to {3||||-3|||-9||-14|-21}. If we played the difference game, I suspect that we would find that they are incomparable. I.e.,
{18|||12||7|0} + {0|-8||-13|||-21} <> {3||||-3|||-9||-14|-21}
Since you are asking the question.

What about the thermograph of the game? Let's start by raising the ambient temperature to 3½. Now we no longer see {7|0}, but only its count, 3½. That yields

{24|||18||12|3½} + {0|-8||-13|||-21}

Now let's raise the temperature to 4. We can no longer see {0|-8}, but only its count, -4. That yields

{24|||18||12|3½} + {-4|-13||-21}

Now let's raise the temperature to 4¼. We can no longer see {12|3½}, but only its count, 7¾. That yields

{24||18|7¾} + {-4|-13||-21}

Now let's raise the temperature to 4½. We can no longer see {-4|-13}, but only its count, -8½. That yields

{24||18|7¾} + {-8½|-21}

Now let's raise the temperature to 5⅛. We can no longer see {18|7¾}, but only its count, 12⅞. That yields

{24|12⅞} + {-8½|-21}

Now let's raise the temperature to 5 9/16. That yields

18 7/16 + {-8½|-21} = {9 15/16|-2 9/16}

Now let's raise the temperature to 6¼. That yields

3 11/16

Which is the count, and the right wall of the thermograph at that temperature. The question, I take it, is to find the minimax value of play in an ideal environment at each temperature. The reduces to finding the right wall of the thermograph.

BTW, having two such hot plays remaining on the board at an ambient temperature of 3 is not at all common.

At t = 6¼ it is 3 11/16, OC.

At t = 5 9/16 it is -2 9/16 + 5 9/16 = 3.
Note that Black gets the last play. We could also find this value by subtracting 11/16 from 3 11/16.

At t = 5⅛ it is -21 + 24 = 3.
That's no surprise, since the right wall is vertical between t = t = 5⅛ and t = 5 9/16.

At t = 4½ White closes off the corridor and so does Black, for the same value of 3.
If White enters the corridor, then so does Black, for a value of -8½ + 7¾ + 4½ = 3¾, which is plainly worse.

At t = 4¼ the value is still 3.
If White enters the corridor the result will be 7¾ - 4 = 3¾.

At t = 4 the value is still 3.
If White enters the corridor the result will be 3½ - 4 + 4 = 3½.

At t = 3½ the value is still 3.
If White enters the corridor the result will be 3½ + 0 = 3½.

At t = 3 the value is still 3.
If White enters the corridor the result will be 0 + 0 + 3 = 3.

If t < 3 the value will be t. White will enter the corridor.

In an actual game, when the temperature is below 3½ or so, White will explore the possibility of getting the last play before playing in the environment.

[Gérard TAILLE]

Looking at our previous discussion I had no idea on the way you will answer: taking the corridors from the end (pure analyst approach) or trying to take the corridor by the beginning (go player approach ?).

I fully understand your quite precise analyst approach. Let's now try to take the corridors by the beginning.
the white corridor is 6 + 6 + 5 + 7
the black corridor is 8 + 5 + 8
Because I assume the temperature of the environment lower than 5, then you can see that each player can always block the opponent corrodor in sente because the opponent should of course block immediatetly the other corridor.
Having this is mind the problem is far simplier: we imagine each player playing in their corridor and we look the opportunity for one player to block in sente the two corridors and take the environment.
Remember my only question is whether or not white has to play her first move in her corridor or for blocking black ones.

Let's proceed:
At the beginning, it is white to play and white have to choose between taking the environment (I mean blocking first the two corridors) or playing in her corridor.
1) Let'suppose white plays in her corridor => local score -6 (for black point of view)
2) if black blocks the corridor it is a success for white because black loses 6 points (the local score) while the temperature is lower (we assumed t < 5) => no choice in the context of the goal (I mean finding the best first white move) : black plays in her corridor => the local score becomes +2 (-6 and +8)
3) Because the score for black is >0 white has no choice : if the first white move in the corridor is correct then, at this stage, white must continue in her corridor for a new local score -4 (-6 -6 and +8)
4) We reach an interesting point : with a local score of -4 we see that black can choose to block the two corridors and black will gain something if the temperature of the environment is greater than 4 => first result : it t > 4 then white must choose, at her very first move to block the two corridors. Let's continue assuming t < 4. In that case black should continue her corridor for a local score of +1 (-6 -6 and +8 +5)
5) Here again, like at satge 3) above, because the score for black is >0 white has no choice => white plays in the corridor for a local score -4 (-6 -6 -5 and +8 +5)
6) The temperature of the environment being t<4 black continue in her corridor for a local score +4 (-6 -6 -5 and +8 +5 +8)
7) Here again, like at stage 3) or 5) above, because the score for black is >0 white has no choice => white plays in the corridor for a local score -3 (-6 -6 -5 -7 and +8 +5 +8). Like in point 4) above we see that all this sequence can be good for white only if t <= 3

That concludes the analysis white must block the two corridors if t > 3 and white must play in her corridor if t < 3.

[Bill Spight]

Looking at our previous discussion I had no idea on the way you will answer: taking the corridors from the end (pure analyst approach) or trying to take the corridor by the beginning (go player approach ?).

I fully understand your quite precise analyst approach. Let's now try to take the corridors by the beginning.
the white corridor is 6 + 6 + 5 + 7
the black corridor is 8 + 5 + 8
Because I assume the temperature of the environment lower than 5, then you can see that each player can always block the opponent corrodor in sente because the opponent should of course block immediatetly the other corridor.
Having this is mind the problem is far simplier: we imagine each player playing in their corridor and we look the opportunity for one player to block in sente the two corridors and take the environment.
Remember my only question is whether or not white has to play her first move in her corridor or for blocking black ones.

The thermograph of a sum of games (positions) has not been featured here, so I took the task of at least finding the right wall of the thermograph. Maybe later I will do the left wall.

In a real game White can presumably tell that all of the gains in each corridor are at least 3½. It is also plain that if each player plays to the end of each corridor the right wall will be t at temperature t. It is also plain that if White blocks one corridor below temperature 5 (taking a quick approximation) Black will block the other one and the right wall will be at most 3 points for Black. Putting these two facts together suggests that at or below temperature 3 White should play the corridors to the end. Either player might be able to do better during the play in the corridors at a higher temperature, which would invalidate that hypothesis, so White checks that possibility.

[Gérard TAILLE]

The thermograph of a sum of games (positions) has not been featured here, so I took the task of at least finding the right wall of the thermograph. Maybe later I will do the left wall.

In a real game White can presumably tell that all of the gains in each corridor are at least 3½. It is also plain that if each player plays to the end of each corridor the right wall will be t at temperature t. It is also plain that if White blocks one corridor below temperature 5 (taking a quick approximation) Black will block the other one and the right wall will be at most 3 points for Black. Putting these two facts together suggests that at or below temperature 3 White should play the corridors to the end. Either player might be able to do better during the play in the corridors at a higher temperature, which would invalidate that hypothesis, so White checks that possibility.

Basically I see we are now often in line concerning the understanding of the various available methods. OC, we may have different plilosophical point of view on these various methods but at least no misunderstanding. Because that was not the case at the beginning of our discussions I am quite happy because it sounds like I made some progress! Thank you Bill.

[Bill Spight]

I wrote:

We have a game, {24||||18|||12||7|0} + {0|-8||-13|||-21}, . . .

What about the thermograph of the game? Let's start by raising the ambient temperature to 3½. Now we no longer see {7|0}, but only its count, 3½. That yields

{24|||18||12|3½} + {0|-8||-13|||-21}

Now let's raise the temperature to 4. We can no longer see {0|-8}, but only its count, -4. That yields

{24|||18||12|3½} + {-4|-13||-21}

Now let's raise the temperature to 4¼. We can no longer see {12|3½}, but only its count, 7¾. That yields

{24||18|7¾} + {-4|-13||-21}

Now let's raise the temperature to 4½. We can no longer see {-4|-13}, but only its count, -8½. That yields

{24||18|7¾} + {-8½|-21}

Now let's raise the temperature to 5⅛. We can no longer see {18|7¾}, but only its count, 12⅞. That yields

{24|12⅞} + {-8½|-21}

Now let's raise the temperature to 5 9/16. That yields

18 7/16 + {-8½|-21} = {9 15/16|-2 9/16}

Now let's raise the temperature to 6¼. That yields

3 11/16

Which is the count, and the right wall of the thermograph at that temperature. The question, I take it, is to find the minimax value of play in an ideal environment at each temperature. . . . {I find the right wall of the thermograph, which indicates the minimax result of White playing first at each temperature.}

At t = 6¼ it is 3 11/16, OC.

At t = 5 9/16 it is -2 9/16 + 5 9/16 = 3.
Note that Black gets the last play. We could also find this value by subtracting 11/16 from 3 11/16.

At t = 5⅛ it is -21 + 24 = 3.
That's no surprise, since the right wall is vertical between t = t = 5⅛ and t = 5 9/16.

At t = 4½ White closes off the corridor and so does Black, for the same value of 3.
If White enters the corridor, then so does Black, for a value of -8½ + 7¾ + 4½ = 3¾, which is plainly worse.

At t = 4¼ the value is still 3.
If White enters the corridor the result will be 7¾ - 4 = 3¾.

At t = 4 the value is still 3.
If White enters the corridor the result will be 3½ - 4 + 4 = 3½.

At t = 3½ the value is still 3.
If White enters the corridor the result will be 3½ + 0 = 3½.

At t = 3 the value is still 3.
If White enters the corridor the result will be 0 + 0 + 3 = 3.

If t < 3 the value will be t. White will enter the corridor.

Let me now find the left wall of the thermograph, using previous calculations.

At t = 6¼ it is 3 11/16, OC.

At t = 5 9/16 it is 3 11/16 + 11/16 = 4⅜.

At t = 5⅛ it is 12⅞ - 8½ = 4⅜, no surprise.
Note that between t = 5 9/16 and t = 5⅛ this combination is double sente, since both walls of the thermograph are vertical.

At t = 4½ it is 18 - 8½ - 4½ = 5.
Black enters one corridor and then answers when White enters the other. Then White gains 4½ points by playing in the environment.

At t = 4¼ it is 18 - 13 = 5.

At t = 4 it is still 5.
If Black continues in the corridor instead, the result is 12 - 4 - 4 = 4, which is worse.

At t = 3½ it is still 5.
If Black continues to the end of the corridor the result is 0 + 3½ = 3½, which is even worse.

Below t < 3½ Black can answer White's first entry into that corridor and guarantee 5 points, or play to the end of the corridor she enters and reply to White's last play in the other corridor for a value of 7 - t. Below t = 2 that will be a good strategy.

We have found another double sente range of temperature, between t = 3 and t = 4½, where the White wall is worth 3 and the Left wall is worth 5. These double sente ranges are apparent in the thermograph.

dbl sente TG.png (4.14 KiB) Viewed 26856 times

[Gérard TAILLE]

Concerning thermography how is handled yose ko?

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