Yes OC they are immortal. BTW I propose a change in the position in order that black "a" will not win the semeai.
That way I hope the problem is clearer.

Click Here To Show Diagram Code

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

I thought they were immortal.

I'm not going to have time to analyze these positions until probably next week.

Joyeux Noël!

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

My two main guides in life:
My mother and my wife.

Everything with love. Stay safe.

OK Bill, Joyeux Noël for you and your family. In France it is not easy due to restriction in realtion with the coronavirus. Anyway I can always continue to study Go

My own analysis is the following.
First of all because locally black wins the ko and because black has no interest to lose a move in order to provoke the ko we can assume that white will play first at "a". The problem is to find at which temperature white will play at "a".
If the temperature is too high black will not answer to a white play at "a" and the play will continue by:

Click Here To Show Diagram Code

[go]$$W
$$ -------------------------
$$ | . . . O . 1 X 5 . . O |
$$ | X X O . O X 3 X X O O |
$$ | X . X X X O X X O O . |
$$ | X X O O O O O O O O O |
$$ | . X . . . . . . . O . |
$$ | X X . . . . . . . O O |
$$ | . X . . . . . . . . . |
$$ | X X . . . . . . . . . |
$$ | . . . . . . . . . . . |
$$ | . . . . . . . . . . . |
$$ | . . . . . . . . . . . |
$$ -----------------------[/go]

tenuki
tenuki
tenuki
and a local score -14,5
If now black choose to win the ko then it follows

Click Here To Show Diagram Code

[go]$$W
$$ -------------------------
$$ | . . . O 2 1 X . . . O |
$$ | X X O 4 O X . X X O O |
$$ | X . X X X O X X O O . |
$$ | X X O O O O O O O O O |
$$ | . X . . . . . . . O . |
$$ | X X . . . . . . . O O |
$$ | . X . . . . . . . . . |
$$ | X X . . . . . . . . . |
$$ | . . . . . . . . . . . |
$$ | . . . . . . . . . . . |
$$ | . . . . . . . . . . . |
$$ -----------------------[/go]

tenuki
and a local score +13,5
The deiri value of the area is 28 points for a tally equal to 4. The value of a local move is then equal to 7.
I conclude that if temperature is greater than 7 then neither side will play in this area.
What happens if the temperature is less or equal to 7? Black will never play first in the area and white is then the only player who can choose the right timing.
Because the temperature is not greater than 7 then we know that in any case black will choose to win the ko. In that case white will gain in exchange one tenuki. That means that white have to play in the area at the highest temperature which is 7 points.
The strategy of white is then the following:
if temperature is 7 then white begins the ko.
if temperature is less than 7 (say 5 or 6) then white have to look for increasing the temperature up to 7. If white cannot increase the temperature then white have to provoque the ko as soon as possible.

I have now to wait Bill's corrections!

Merry Christmas.

Click Here To Show Diagram Code

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

Here is another example difficult for me to analyse.
Assume no ko threat in the environment. White may choose to play "w" in order to use the ko to give life to his marked stone but the questions are the following:
1) at which temperature white will play at "w"?
2) at which temperature black will play at "b"?

Click Here To Show Diagram Code

[go]$$W Sacrifice
$$ -----------------
$$ | X b . . . W X X 1 |
$$ | X W W W W W X O 2 |
$$ | X X X X X O X 4 O |
$$ | . . . . X O X X X |
$$ | X X X X X X O O O |
$$ | . . . . . . O . . |
$$ | . . . . . . O O . |
$$ | . . . . . . . O . |
$$ | . . . . . . . O . |
$$ -----------------[/go]

elsewhere

The result after is 26 - t, where gains t points.

White can play when he threatens to win the ko.

Click Here To Show Diagram Code

[go]$$W White wins ko
$$ -----------------
$$ | X b . . . W X X 1 |
$$ | X W W W W W X O . |
$$ | X X X X X O X 3 O |
$$ | . . . . X O X X X |
$$ | X X X X X X O O O |
$$ | . . . . . . O . . |
$$ | . . . . . . O O . |
$$ | . . . . . . . O . |
$$ | . . . . . . . O . |
$$ -----------------[/go]

, elsewhere

The result after is -17⅛ + 2t, on average, counting -2⅛ for the corridor.

White threatens to win the ko when t ≤ 14⅜. So that is when White will throw in to make the ko, as a rule.

Black has no inclination to play at b, because that will reduce Black's potential result by 1 point.

But what happens at the end of the game? By Japanese rules Black must play first to keep this from becoming 0 points. Black will have to play twice to kill the White stones, so the final score will be 24.

That means that, as a rule, White will have no incentive to play elsewhere when doing so gains less than 2 points. So White will normally play the throw-in when 2 ≤ t ≤ 14⅜.

Edit: Hmmm. I may well be wrong about the Japanese rules, as White has no chance to live because Black wins the ko. In which case Black does not have to spend any play to kill White and White will normally make the ko when t ≤ 14⅜,

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

My two main guides in life:
My mother and my wife.

Everything with love. Stay safe.

I do not understand why black has not to play at "b" if the temperature is low. Hasn't black to avoid the sequence

Click Here To Show Diagram Code

[go]$$W
$$ -----------------
$$ | X 3 . 5 . O X X 1 |
$$ | X O O O O O X O 2 |
$$ | X X X X X O X 4 O |
$$ | . . . . X O X X X |
$$ | X X X X X X O O O |
$$ | . . . . . . O . . |
$$ | . . . . . . O O . |
$$ | . . . . . . . O . |
$$ | . . . . . . . O . |
$$ -----------------[/go]

Thanks. I replied too quickly. I thought I could just assume that Black was komaster. Given the source, I should have known better.

Click Here To Show Diagram Code

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

After White can play first and lose the ko for a result of 25 - t when t ≤ 13¾.

That means that the left scaffold for the thermograph is 25 - 2t. (Edit: At that temperature range, OC. ) I'll go to work on the right scaffold now, considering the local ko threat.

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

My two main guides in life:
My mother and my wife.

Everything with love. Stay safe.

yes Bill due to this sequence it is not so obvious to analyse the situation.
OC, as I said, I assume there are no ko threat in the environment but localy it exists one ko threat (not as big as 14⅜ but it exists).
I understand your count : (26 + 17⅛) / 3 = 14⅜ but this count assume that black can kill all white stone by winning the ko but this is not true because white is able to choose to save some of his stones instead of playing in the environment.
Due to this white possibility my feeling is that a white play at "w" is not as big as 14⅜ and here is my difficulty when looking for the temperature at which white has to play at "w".

OK. Let's try again.

Click Here To Show Diagram Code

[go]$$Bc Black first
$$ -----------------
$$ | X 1 . . . W X X 2 |
$$ | X W W W W W X O 3 |
$$ | X X X X X O X 5 O |
$$ | . . . . X O X X X |
$$ | X X X X X X O O O |
$$ | . . . . . . O . . |
$$ | . . . . . . O O . |
$$ | . . . . . . . O . |
$$ | . . . . . . . O . |
$$ -----------------[/go]

, elsewhere

Result: 25 - 2t

Click Here To Show Diagram Code

[go]$$Wc White first
$$ -----------------
$$ | X 3 . 5 . W X X 1 |
$$ | X W W W W W X O 2 |
$$ | X X X X X O X 4 O |
$$ | . . . . X O X X X |
$$ | X X X X X X O O O |
$$ | . . . . . . O . . |
$$ | . . . . . . O O . |
$$ | . . . . . . . O . |
$$ | . . . . . . . O . |
$$ -----------------[/go]

elsewhere

Result: 4 + t

Assuming that these are the sequences of play when t is small enough, we find that t by solving the equation,

25 - 2t = 4 + t

t = 21/3 = 7

And we can verify that when Black plays first, White throws in to make the ko when t ≤ 13¾ , and when White plays first and Black takes the ko, White plays the ko threat when t ≤ 11. So these are the normal sequences of play when t ≤ 7.

The mast value of the position is 11.

If I haven't goofed, that is.

Edit: We need to check the case when 26 - t < 11. I.e., when t > 15. White does not threaten to make and win the ko when t > 14⅜, so we're OK.

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

My two main guides in life:
My mother and my wife.

Everything with love. Stay safe.

I agree that black will play first if the temperature has droped to 7.
But I do not understand why the mast value of the position is 11 instead of 14⅜.
If for example the temperature is equal to 13 then white should immediatly provoque the ko without waiting for temperature droping to 11 shouldn't he?
BTW how will you eventually draw the thermograph Bill? It looks quite unusual doesn't it?

The mast value is equal to 4 + 7 or 25 - 2*7. 14⅜ is the temperature below which White threatens to win the ko.



If White throws in at temperature 13, then:

1) if White lets Black win the ko the result is 26 - 13 = 13, which is worse for White than 11;

2) if White plays the threat then we reach this position with Black to play.

Click Here To Show Diagram Code

[go]$$Wc White first
$$ -----------------
$$ | X W . . . W X X . |
$$ | X W W W W W X O B |
$$ | X X X X X O X . O |
$$ | . . . . X O X X X |
$$ | X X X X X X O O O |
$$ | . . . . . . O . . |
$$ | . . . . . . O O . |
$$ | . . . . . . . O . |
$$ | . . . . . . . O . |
$$ -----------------[/go]

One prisoner

If plays elsewhere White can take and win the ko for a result of -18 + 3t (from the original position), which is equal to 39 - 18 = 21. That's even worse for White than above, so Black will play elsewhere and so will White.

What is the mast value of this position? From here White can take and win the ko for a result of -18 + 2t.

If instead Black wins the ko we have this position.

Click Here To Show Diagram Code

[go]$$Wc White first
$$ -----------------
$$ | X W . . . W X X . |
$$ | X W W W W W X . B |
$$ | X X X X X O X B . |
$$ | . . . . X O X X X |
$$ | X X X X X X O O O |
$$ | . . . . . . O . . |
$$ | . . . . . . O O . |
$$ | . . . . . . . O . |
$$ | . . . . . . . O . |
$$ -----------------[/go]

Three prisoners

If White lives the result is 4. If Black kills the result is 26. The mean value is 15 and at or below temperature 11 White will live.

To find the temperature for this position at or below which the players will play locally, let's first assume that White lives. Then we solve for

-18 + 2t = 4

t = 22/2 = 11

Fine. We are done. The mast value of this position is 4.

Assuming that at temperature 13 White makes the ko and then plays the threat, the result will be 4 + 13 = 17 (from the original position), which is worse for White than 11, so White does best to wait until t = 7.



Actually, it looks like a simple ko thermograph. Above temperature 7 the mast rises at a territory value of 11. Below that the right wall is 4 + t and the left wall is 25 - 2t. I'll add the thermograph below, probably tomorrow.

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

My two main guides in life:
My mother and my wife.

Everything with love. Stay safe.

At temperature 13 the result 26 - 13 = 13 looks to me the best result for white. How do you manage to reach the score 11?

Let m be the mast value.

The scaffolds have been derived above.

Left scaffold up to temperature 7:

25 - 2t

Right scaffold up to temperature 7:

4 + t

4 + t = 25 - 2t

3t = 21

t = 7

m = 4 + 7 = 25 - 14 = 11

Above temperature 7 all White has to do to guarantee the mast value is to play elsewhere.

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

My two main guides in life:
My mother and my wife.

Everything with love. Stay safe.

OK BIll, I understand.
Because this result is not the result I expected let me propose a modification

Click Here To Show Diagram Code

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

Why I add the 3 black marked stone?
The only difference is that the figure 14⅜ is now replaced by the figure 16⅜. As a consequence I can consider a white move at temperature 16 and now 26 - t can be less than 11. That is the point I wanted to highlight.
My analyse is then the following:
Above temperature 16⅜ black and white play tenuki
Between temperature 15 and 16⅜ white provoques the ko in the corner to reach the score 26 - t
Between temperature 7 and 15 black and white wait and play tenuki
Under temperature 7 both white and black want to play in the area to reach a score 4 + t or 25 - 2t
Is it correct Bill?
If yes, how do you draw the termograph?

That's what I get, as well.

This is a phenomenon that Professor Berlekamp discovered over 30 years ago. I bow to his memory.



The right scaffold of the thermograph is 4 + t up to t = 11, then it is 26 - t up to t = 16⅜, then it is -6¾ + t upwards from there.

The left scaffold is 25 - 2t up to t = 16¾, then -5½ - t upwards from there.

As already noted, the scaffolds intersect at (m,v) = (11,7). Then a black mast rises at m = 11 up to (11,15), at which point the right scaffold intersects the mast. Then an inclined red mast is v = 26 - t up to (9⅝, 16⅜). Then a black mast rises upwards from there at m = 9⅜. The mast value of the thermograph is 9⅝.

The inclined mast was already known and understood by some players, but, OC, not by that name. But I doubt if the zigzag mast and other strange masts were understood or even suspected before Berlekamp's discoveries.

I am having trouble posting a thermograph now. I'll post them when the problem is cleared up.

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

My two main guides in life:
My mother and my wife.

Everything with love. Stay safe.

Here is an example of semeai leading to the studied position:

Click Here To Show Diagram Code

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

Click Here To Show Diagram Code

[go]$$W white to play
$$ -----------------
$$ | . . . . . 5 6 2 . |
$$ | . . X O O O X 1 . |
$$ | . . X X . O X . 3 |
$$ | . . . . X O X X 4 |
$$ | . . . . X X O O 7 |
$$ | . . . . . . O . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ -----------------[/go]

The handling of such position appears quite difficult.
Assume white is able to use several ko threats (of different values) in the environment then it is not easy to answer the following questions:
When will white provoque the ko?
Should black use one or several moves to destroy some white ko threats at the expense of few points?

Do somebody know if such position already appeared in pro game?