This 'n' that

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Bill Spight
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Re: This 'n' that

Post by Bill Spight »

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]
:b5: = 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]

:w6:, :w8:, :w10: = 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]
:w2: tenuki
:w4: tenuki
:w6: 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]
:w2: 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]
:b7: tenuki
:b9: 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]
:w2:, :w6:, :w8: = t, :b7: 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]
:w2:, :b5:, :b7:, :b9: = t, :w6:, :w8: 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 :w10: 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 :b9:. Black could play there after :w6: or :w4:. 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 :b9: 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. :)
Last edited by Bill Spight on Tue Jun 15, 2021 5:38 am, edited 1 time in total.
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Re: This 'n' that

Post by Bill Spight »

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. :)
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— Winona Adkins

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Re: This 'n' that

Post by Gérard TAILLE »

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]
:w2:, :w6:, :w8: = t, :b7: 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]
:w2:, :b5:, :b7:, :b9: = t, :w6:, :w8: 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.
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Re: This 'n' that

Post by Bill Spight »

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]
:w2:, :w6:, :w8: = t, :b7: 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. :b5: 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]
:w2:, :b5:, :b7:, :b9: = t, :w6:, :w8: 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. :)
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— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
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Re: This 'n' that

Post by Gérard TAILLE »

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).
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Re: This 'n' that

Post by Bill Spight »

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. :)
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Re: This 'n' that

Post by Gérard TAILLE »

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?
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Re: This 'n' that

Post by Bill Spight »

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]
:b3:, :b5: = 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]
:b4:, :b6: = 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]
:w2:, :w4: = 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. :D

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. :)
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Re: This 'n' that

Post by Gérard TAILLE »

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: :b2: the result is quite different. If now white plays in the environment black can kill the corner. I conclude :w1: :b2: 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]
:b4: 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]
:b4: tenuki
:b6: 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]
:w1: tenuki
:b4: tenuki
:b6: tenuki
:b8: 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?
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Re: This 'n' that

Post by Bill Spight »

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: :b2: the result is quite different. If now white plays in the environment black can kill the corner. I conclude :w1: :b2: 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]
:b4: 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]
:b4: tenuki
:b6: 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]
:w1: tenuki
:b4: tenuki
:b6: tenuki
:b8: 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]
:b1: = u, :b3: = 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]
:b3: = u, :b5: = 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]
:w1: = -u, :b4:, :b6:, :b8: = 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]
:b4: = u, :b6: = 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.
Last edited by Bill Spight on Wed Jun 16, 2021 11:27 pm, edited 2 times in total.
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

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Re: This 'n' that

Post by Bill Spight »

OK. We know that Black first can not do better than to play the hane, with some possible but now unknown exceptions. What if White plays first and takes the ko?

Let us add this game to the corner to get the ko ensemble. ;)

{u||||2u|0||-u|||-2u}, where 6½ > u > t > 1

Variation 1.

White first takes ko. (Written territorial values below are averages.)
Click Here To Show Diagram Code
[go]$$W White first takes ko
$$ --------------
$$ | X 1 X O X . .
$$ | 3 X O O X . .
$$ | X 5 O X X . .
$$ | . O O X . . .
$$ | 2 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
:b4: = u :b6: = t

s = -5 + u + t

Now let :w1: take -u. There are two variations.

Variation 2.
Click Here To Show Diagram Code
[go]$$W White takes -u
$$ --------------
$$ | X 3 X O X . .
$$ | 5 X O O X . .
$$ | X 7 O X X . .
$$ | . O O X . . .
$$ | 2 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
:b4:, :b6:, :b8: = u

s = - 5 + 2u

Taking the ko is better for White than variation 2 by u - t. How does that happen?

:w1: plays to {2u|0||-u|||-2u}

After :b2: plays the hane, :w3: cannot, on average, afford to play to -2u, because :b4: will kill the corner for

s = 21 - 2u > 8.

After :w3: takes the ko, :b4: plays to {2u|0||-u}. :w5: must still take the ko.

After :w5: takes the ko, :b6: plays to {2u|0}. :w7: must win the ko.

Now :b8: plays to 2u, for s = -5 + 2u.

Variation 3.

What if :b2: does not play the hane now but follows Berlekamp's Sentestrat strategy by playing to {2u|0||-u}, instead? Now let :w3: play to -u.
Click Here To Show Diagram Code
[go]$$W White takes -u
$$ --------------
$$ | X 5 X O X . .
$$ | 7 X O O X . .
$$ | X 9 O X X . .
$$ | . O O X . . .
$$ | 4 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
:w1: = -u, :b2: = u, :w3: = -u, :b6:, :b8:. :b10: = t

s = -5 - u + 3t

This result is worse for White than taking the ko because

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

t < u , which is presumed.

But that simply means that :b2: should not play Sentestrat. :)

Edit: I think that Berlekamp would object to my calling :b2: in this case an example of Sentestrat, because :w1: does not actually raise the local temperature, it keeps it the same. But it is like Sentestrat in that it replies to your opponent's play instead of making your own play.
Last edited by Bill Spight on Thu Jun 17, 2021 8:21 am, edited 1 time 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.
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Re: This 'n' that

Post by Gérard TAILLE »

Bill Spight wrote:OK. We know that Black first can not do better than to play the hane, with some possible but now unknown exceptions. What if White plays first and takes the ko?

Let us add this game to the corner to get the ko ensemble. ;)

{u||||2u|0||-u|||-2u}, where 6½ > u > t > 1

Variation 1.

White first takes ko. (Written territorial values below are averages.)
Click Here To Show Diagram Code
[go]$$W White first takes ko
$$ --------------
$$ | X 1 X O X . .
$$ | 3 X O O X . .
$$ | X 5 O X X . .
$$ | . O O X . . .
$$ | 2 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
:b4: = u :b6: = t

s = -5 + u + t

Now let :w1: take -u. There are two variations.

Variation 2.
Click Here To Show Diagram Code
[go]$$W White takes -u
$$ --------------
$$ | X 3 X O X . .
$$ | 5 X O O X . .
$$ | X 7 O X X . .
$$ | . O O X . . .
$$ | 2 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
:b4:, :b6:, :b8: = u

s = - 5 + 2u

Taking the ko is better for White than variation 2 by u - t. How does that happen?

:w1: plays to {2u|0||-u|||-2u}

After :b2: plays the hane, :w3: cannot, on average, afford to play to -2u, because :b4: will kill the corner for

s = 21 - 2u > 8.

After :w3: takes the ko, :b4: plays to {2u|0||-u}. :w5: must still take the ko.

After :w5: takes the ko, :b6: plays to {2u|0}. :w7: must win the ko.

Now :b8: plays to 2u, for s = -5 + 2u.

Variation 3.

What if :b2: does not play the hane now but follows Berlekamp's Sentestrat strategy by playing to {2u|0||-u}, instead? Now let :w3: play to -u.
Click Here To Show Diagram Code
[go]$$W White takes -u
$$ --------------
$$ | X 5 X O X . .
$$ | 7 X O O X . .
$$ | X 9 O X X . .
$$ | . O O X . . .
$$ | 4 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
:w1: = -u, :b2: = u, :w3: = -u, :b6:, :b8:. :b10: = t

s = -5 - u + 3t

This result is worse for White than taking the ko because

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

t < u , which is presumed.

But that simply means that :b2: should not play Sentestrat. :)
Oops, surely you are teasing me Bill.
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]
I showed the white move :w1: is bad in a pure gote environment {g1|-g1}, {g2|-g2}, {g3|-g3} ....
and now you find an environment {u||||2u|0||-u|||-2u}, where 6½ > u > t > 1 and where :w1: is better than playing -u. Yes OC Bill it is possible to find a special environment (even without ko!) in which :w1: could be better.

I understood from your post viewtopic.php?p=265525#p265525 (where you used the acronym "KISS") you prefer to keep things simple by avoiding adding even simple gote. Yes your are teasing me Bill don't you?

Though taking the ko by :w1: in the diagram above might be good in certain circumtances (maybe also if black has ko threats available but it is the another context), I consider that this move is in general a bad move.
Did you find an environment with only simple gote {u|-u} in which :w1: is better than playiing in the environment?
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Re: This 'n' that

Post by Gérard TAILLE »

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]
:b5: = 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]

:w6:, :w8:, :w10: = t

Result: 21 - 3t

The temperature of indifference occurs when

-5½ + t = 21 -3t, or when

t = 6⅝
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]
After this sequence you studied :b5: tenuki in the environment
I am wondering if this could really by a good sequence (I mean in a simple environment, either ideal or made of only simple {u|-u} gote points.
My feelig is that both :b1: :w2: and :b3: :w4: reverses. If it is true either black has to continue in the corner or black should prefer tenuki directly instead of :b1:
What is your feeling?
Bill Spight
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Re: This 'n' that

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote:OK. We know that Black first can not do better than to play the hane, with some possible but now unknown exceptions. What if White plays first and takes the ko?

Let us add this game to the corner to get the ko ensemble. ;)

{u||||2u|0||-u|||-2u}, where 6½ > u > t > 1

Variation 1.

White first takes ko. (Written territorial values below are averages.)
Click Here To Show Diagram Code
[go]$$W White first takes ko
$$ --------------
$$ | X 1 X O X . .
$$ | 3 X O O X . .
$$ | X 5 O X X . .
$$ | . O O X . . .
$$ | 2 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
:b4: = u :b6: = t

s = -5 + u + t

Now let :w1: take -u. There are two variations.

Variation 2.
Click Here To Show Diagram Code
[go]$$W White takes -u
$$ --------------
$$ | X 3 X O X . .
$$ | 5 X O O X . .
$$ | X 7 O X X . .
$$ | . O O X . . .
$$ | 2 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
:b4:, :b6:, :b8: = u

s = - 5 + 2u

Taking the ko is better for White than variation 2 by u - t. How does that happen?

:w1: plays to {2u|0||-u|||-2u}

After :b2: plays the hane, :w3: cannot, on average, afford to play to -2u, because :b4: will kill the corner for

s = 21 - 2u > 8.

After :w3: takes the ko, :b4: plays to {2u|0||-u}. :w5: must still take the ko.

After :w5: takes the ko, :b6: plays to {2u|0}. :w7: must win the ko.

Now :b8: plays to 2u, for s = -5 + 2u.

Variation 3.

What if :b2: does not play the hane now but follows Berlekamp's Sentestrat strategy by playing to {2u|0||-u}, instead? Now let :w3: play to -u.
Click Here To Show Diagram Code
[go]$$W White takes -u
$$ --------------
$$ | X 5 X O X . .
$$ | 7 X O O X . .
$$ | X 9 O X X . .
$$ | . O O X . . .
$$ | 4 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
:w1: = -u, :b2: = u, :w3: = -u, :b6:, :b8:. :b10: = t

s = -5 - u + 3t

This result is worse for White than taking the ko because

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

t < u , which is presumed.

But that simply means that :b2: should not play Sentestrat. :)
Oops, surely you are teasing me Bill.
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]
I showed the white move :w1: is bad in a pure gote environment {g1|-g1}, {g2|-g2}, {g3|-g3} ....
and now you find an environment {u||||2u|0||-u|||-2u}, where 6½ > u > t > 1 and where :w1: is better than playing -u. Yes OC Bill it is possible to find a special environment (even without ko!) in which :w1: could be better.
It's not just better, it's thermographically better. :) t remains the same while the ko ensemble is played out.

I say that because you seem to be resisting the idea of including moves in the environment in the ko ensemble. I have sympathy for your approach, because it is one that I took for some 25 years.
Gérard TAILLE wrote:Though taking the ko by :w1: in the diagram above might be good in certain circumtances (maybe also if black has ko threats available but it is the another context), I consider that this move is in general a bad move.
Bad move is a strong judgement, which, IMHO, is unwarranted. You haven't even shown that there is a loss at all from playing in the ko. (Not that it is a good idea. It is obvious in general that White should wait for Black to play the hane.)
Gérard TAILLE wrote:Did you find an environment with only simple gote {u|-u} in which :w1: is better than playiing in the environment?
In the thermographic environment, which is composed of simple gote, the gain from White taking the ko is equal to the gain from playing in the environment. That is unusual, since we are at a lower temperature than that of the mast value, above which neither player is likely to play locally. It is a result of the inclined mast.

So we have two moves that are thermographically equal in value. Since thermography plays the averages, that does not mean that are exactly equal. We do expect that normally any difference between them will be slight, so that even if one is wrong, it may not make any difference in the final score. That is not the stuff of bad moves. It also means that we cannot say, a priori, which move is better. Now we know, as a practical matter, that it is in general better for White not to take the ko, but thermography tells us that there are potentially cases where taking the ko is better, and that often it does not matter which play is made.
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Re: This 'n' that

Post by Gérard TAILLE »

Bill Spight wrote: In the thermographic environment, which is composed of simple gote, the gain from White taking the ko is equal to the gain from playing in the environment. That is unusual, since we are at a lower temperature than that of the mast value, above which neither player is likely to play locally. It is a result of the inclined mast.

So we have two moves that are thermographically equal in value. Since thermography plays the averages, that does not mean that are exactly equal. We do expect that normally any difference between them will be slight, so that even if one is wrong, it may not make any difference in the final score. That is not the stuff of bad moves. It also means that we cannot say, a priori, which move is better. Now we know, as a practical matter, that it is in general better for White not to take the ko, but thermography tells us that there are potentially cases where taking the ko is better, and that often it does not matter which play is made.
No doubt that by adding the area {u||||2u|0||-u|||-2u}, where 6½ > u > t > 1, to the ko ensemble taking the ko gains u -t.
But the area {u||||2u|0||-u|||-2u} is in fact very subtil in order to reach this goal.
If instead you take only the simple gote {u|-u} always with 6½ > u > t > 1, then taking the ko will lose 2(u - t) and here also playing in the environment for white is not just better it's thermographically better!

That's my point : unless you add to the ko ensemble a very subtle area, playing in the environement for white is thermographically better than taking the ko.

Look at my previous calculation with an environment g1 ≥ g2 ≥ g3 ≥ g4 ...
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

Now I like to use all the power of thermography tool. My understanding is that an ideal environment is in fact a strange mathematical object with proporties that looks contradictory.
In one hand you can consider g1 = g2 = g3 = g4 ... but at the same time you can also wait for a drop of the temperature!
IOW you may consider 2(g1 - g3) = 0 and at the same time you can assume 2(g1 - g3) > 0 but very small.
That the reason why I claim that playing in the environment is in general better than taking the ko.
BTW by taking g1 = u and g2 = t then you have 2(g1 - g3) = 2(u - t).

It is the same thing in the following position
Click Here To Show Diagram Code
[go]$$B
$$ --------------
$$ | X . X O X . .
$$ | . X O O X . .
$$ | X . O X X . .
$$ | . O O X . . .
$$ | 1 O X X . . .
$$ | . X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
$$ | . . . . . . .
$$ | . . . , . . .[/go]
In one hand you can delay without harm the black hane :b1: but in the other hand you must play hane immediatly.

It just a matter of using this famous ideal environment taking into account all its power.

As far as I am concern I like to say in an ideal environment we have g1 = g2 + ε with ε equal to infinity small but ε > 0

OC, with specific environment anything may happen.
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