Thermography

Talk about improving your game, resources you like, games you played, etc.
Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote: 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.
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.
Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

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.
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote: 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.
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.
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.
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.
Right. Did you think I missed that?
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.
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

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Everything with love. Stay safe.
Bill Spight
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
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.
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.
Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote: Right. Did you think I missed that?
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. :)
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote: Right. Did you think I missed that?
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.
Rest assured I am quite capable of missing and misunderstanding things. ;)
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At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

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

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:
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.
Your expected gain in a real game is closer to 0.05.
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?
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote: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.
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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

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?
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Re: Thermography

Post by Bill Spight »

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.
gerardTG.png
gerardTG.png (3.78 KiB) Viewed 30198 times
Last edited by Bill Spight on Tue Oct 13, 2020 12:51 pm, edited 2 times in total.
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
Bill Spight
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote: 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 ...
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.
Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
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.
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?
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote:
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.
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.
gerardTG00.png
gerardTG00.png (5.08 KiB) Viewed 30201 times
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.
Gérard TAILLE
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Re: Thermography

Post by 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 ;-)
Bill Spight
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote: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
GTG.png (4.03 KiB) Viewed 30179 times
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
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