Life In 19x19

We can distinguish two views:

1) Usual practical playing advice.

2) Theoretical evaluation covering all cases.

***

You are concerned with (1), which is fine. I have approached (2), which is also fine (I have omitted the komonster cases though).

For (1), the stage ko fight is rare (or maybe even non-existent; don't know yet). For (1), your observation "make white believe it was good to play there once the ambient temperature is 0,5. This would be a mistake, as black could later potentially take the ko first" is useful.

***

For (2), we must consider the basic ko fight and the stage fight.

Possible reason why this stage go fight might never be played: White chooses the wrong option on move 2, due to which the initial count would be 1. White prefers move 2 in Dia. 2 so that the smaller, more favourable initial count is 2/3.

Possible reason why this stage go fight might be fought nevertheless: construct an environment so that the stage ko fight is correct for both players. Are other kos elsewhere necessary for that? Maybe the colour-reversed copy?

Compare the thermographs of both options from White's perspective. At all temperatures, does the basic ko fight dominate the stage ko fight, that is, does White achieve the smaller global count?

Can ko fights elsewhere and ko bans influence this?


Edited.

Here is the proof of the existence of a possible stage ko fight in perfect play. The file is meant to be on a 6x6 board. Download the file to view! Therefore, evaluation of the stage ko fight is also necessary.

RobertJasiek wrote:"miai" is only an informal term.

While I may be right about this,...

[...] Move value = (-1/3 - (-3)) / 4 = 2/3. [...] Count = -1/3 - 1 * 2/3 = -3 + 3 * 2/3 = -1. [...]

...I may be wrong about this.

Apparently, hyperactive kos must not be simply evaluated referring to swing divided by tally. Instead, some more general theory must be applied: general ko thermography in its algebraic or graphical representations. I need to learn it and will make another attempt later.

Like for stage kos in the middle of the board, I have already noticed that consistency demands possible transfers of initial positions of a corner stage ko, whose counts are related by adjusting virtual prisoners.

Click Here To Show Diagram Code

[go]$$B Count -2/3
$$ ---------------
$$ | . X . X . X .
$$ | X O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .[/go]

Click Here To Show Diagram Code

[go]$$B Count -4/9
$$ ---------------
$$ | O X . X . X .
$$ | . O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .[/go]

These two counts are related. When imagining transferring it from the first initial position with its postulated count -2/3, add the prisoner difference -1 to the second initial position so its modified count is -1 4/9. Now, if we find, as Bill has suggested, that a play gains 7/9, we have a consistent relation of the values:

-2/3 - 7/9 = -1 4/9.

We must understand thermography so that hopefully its application generates these counts and the gain for at least some (komaster) case.


Edited.

Bill used to say: "Draw the thermographs to determine the values!" The difficulty, however, is to know how to do so at all and correctly. After a week of studying mathematical theory about thermography, I am becoming increasingly aware of what I have always suspected: it is very difficult! Drawing is one means - algebraic calculation is another means. Either way, one must rely on the right numbers.

Standard ko thermography (too euphemistically advertised by its name 'generalised thermography') makes in particular these assumptions:
- Each play is compensated by 1 point in favour of its player so one can play out filling territory.
- The imagined environment of simple gotes or coupons has the temperature t with arbitrarily many such moves of the same value.
- Cyclic positions are 'simple' if for the initial position and each follow-up position only basic kos occur, there are no long cycles of alternating [board-]plays, Black has at most one basic ko capture, White has at most one basic ko capture.

As it turns out, a 2-stage ko in the corner is not too bad as it is a 'simple' cyclic position. Ko thermography is difficult and thermographs start at -1 (instead of 0) but we do not need those extra difficult additions for multiple kos or long cycles.

RobertJasiek wrote:INTRODUCTION

What are the move value, gains and count?

$$B Initial position
$$ ---------------
$$ | . X . X . X .
$$ | X O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .

Click Here To Show Diagram Code

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

Transformation to a follow-up:

$$W Transformation
$$ ---------------
$$ | 1 X 2 X . X .
$$ | X O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .

Click Here To Show Diagram Code

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

$$B Prisoners = -1, count = -2/3
$$ ---------------
$$ | O X X X . X .
$$ | . O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .

Click Here To Show Diagram Code

[go]$$B Prisoners = -1, count = -2/3
$$ ---------------
$$ | O X X X . X .
$$ | . O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .[/go]

The prisoner difference is -1 and the ko has the count 1/3 so the follow-up position has the count -1 + 1/3 = -2/3.

Now, Bill has used this count of the follow-up to derive the alleged count -2/3 of the initial position.

$$B Miai?
$$ ---------------
$$ | W X B X . X .
$$ | X O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .

Click Here To Show Diagram Code

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

His argument has been that the circled stones could be exchanged in either order in a sente-like sequence so would be miai and therefore the initial position would inherit the count -2/3 from its sente-like follower.

However, "miai" is only an informal term. In the transformation above, Black need not play at 2 but he has the alternative option of playing elsewhere and fighting the stage ko. There is no a priori justification why the initial count would have to be determined by the basic ko fight due to the connection Black 2.

My point of view is different.
I do not consider the circled stones could be exchanged in either order, and I do not consider a stage ko could take place, at least in a theoritical approach.
My view is the following:

Click Here To Show Diagram Code

[go]$$B
$$ ---------------
$$ | . X 1 X . X .
$$ | X O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .[/go]

the black move above is a very small move and this move cannot be played if temperature is greater than 1/3.

Click Here To Show Diagram Code

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

On the other hand after the white move above black shoud connect if temperature is less than 7/9.

The conclusion is simple: white waits until temperature drops between 7/9 and 1/3 and white plays the sente exchange
In this case the count of the initial position is -2/3

Yours is a useful strategic advice. However, first we must establish and justify the 7/9.

RobertJasiek wrote:Yours is a useful strategic advice. However, first we must establish and justify the 7/9.

OC in pracice you don't need to calculate the exact value 7/9. You have just to consider that is sente as soon as temperature is a little above 1/3.
Anyway how do I calculate the value 7/9 ?

Click Here To Show Diagram Code

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

white to play

Click Here To Show Diagram Code

[go]$$W :b2: tenuki, :b4: tenuki
$$ ---------------
$$ | O 3 1 X . X .
$$ | . O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .[/go]

score : S1 = 2t - 2

Black to play

Click Here To Show Diagram Code

[go]$$B :w2: tenuki
$$ ---------------
$$ | O X 1 X . X .
$$ | . O X X X X X
$$ | O O O O . . .
$$ | . O . O . . .[/go]

score : S2 = -t + 1/3

S1 = S2 <=> 2t - 2 = -t + 1/3 <=> t = 7/9

The white tenuki is essy to justify. The hard part is to justify using the intermediate ko position for evaluating the initial position with two black stones. Apart from these assumptions, your calculation for the intermediate position is correct without komaster.

RobertJasiek wrote:The white tenuki is essy to justify. The hard part is to justify using the intermediate ko position for evaluating the initial position with two black stones. Apart from these assumptions, your calculation for the intermediate position is correct without komaster.

Yes, basically a ko introduces always a difficulty but for a theoritical point of view using the intermediate poisition to find a sente move is only natural:

Click Here To Show Diagram Code

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

Click Here To Show Diagram Code

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

Using the position after white in order to see that ambiant temperature has increased is common approach to discover a sente move isn't it?

Ko thermography is not as straightforward as sente because balloon trajectories need not proceed vertically but can have vertical parts.

RobertJasiek wrote:Ko thermography is not as straightforward as sente because balloon trajectories need not proceed vertically but can have vertical parts.

Yes I know. My answer was only related to your "hard part":
"The hard part is to justify using the intermediate ko position for evaluating the initial position with two black stones."

OC I agree ko thermography is far more difficult.

Let me propose an other position which is very simple and very very common but however not that easy to evaluate due to the existence of a future ko:

Click Here To Show Diagram Code

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

When will you play at "a" depending of the temperature?
Is the answer the same for white and black?

See https://www.lifein19x19.com/viewtopic.p ... 37#p276737 for a hint why this might not be "very simple".

RobertJasiek wrote:See https://www.lifein19x19.com/viewtopic.p ... 37#p276737 for a hint why this might not be "very simple".

I agree with you to say that the evaluation of the position is not easy.
Concerning the position itself (and not it's evaluation) I consider this position very simple because the sequence to play locally (I mean if ambiant temperature is t = 0) is absolutly obvious.
The difficulty arises as soon as you take into account an ambiant temperature t > 0.
BTW, without giving all the calculation, could you just give us the final result just to motivate the readers to try and find the correct justification?

I am busy with stage kos... For your position, thermography is an overkill - the method of making a hypothesis will do. To verify the second-most naive assumption of Black's long sente (with White's 3-move sequence), explore the tree, calculate the counts of the followers, derive the gains and compare each to the assumed move value. I do not know if the method of comparing the opponent's branches applies; if it does, it might be faster.