Move values M of the initial position, M1 after Black 1, M2 after Black 1 - Black 2.

Mgote = (-4 - (-6)) / 2 = 1 = - 5 - (-6) = Msente (ambiguous)
M1sente = 1
M2gote = 2

If we perceive Black's ambiguous short sequence as a 1-move gote sequence, we have M = 1 of part I of the sequence Black 1 - Black 2 - White 3 and M1sente = 1 of part II (Black 2 - White 3) so the move values are constant.

If we perceive Black's ambiguous short sequence as a 2-move sente sequence, we have M = 1 of part I (Black 1 - White 2).

With neither perception, the move values increase from part I to part II. So your example is not a counter-example. Oh, wait. Did I allow constant for increasing-or-constant? It is, of course, enough hint to reveal the following counter-example:

Click Here To Show Diagram Code

[go]$$ Increasing move values
$$ -----------------------
$$ . X . . X . X X X . O .
$$ . X O O O O O O O O O .
$$ . X . . . . . . . . . .
$$ . . . . . . . . . . . .[/go]

M2gote = 3.5.

M1sente = 3 (Black's simple sente, serves also as follow-up move value for the initial position).

Tentatively, Mgote = (-7 - (-11)) / 2 = 2 and Msente = -10 - (-11) = 1 so
M1sente > Mgote > Msente <=> 3 > 2 > 1 so
M = Msente = 1 (Black's simple sente).

We have M < M1sente <=> 1 < 3. The increasing move values let this be a counter-example.

Note that we do not have traversal / reversion because there is no alternating 3-move sequence of moves with move value larger than 0.

Click Here To Show Diagram Code

[go]$$ Distant sente
$$ -----------------
$$ . X 1 3 4 X X O .
$$ . X O O O O O O .
$$ . X . . . . . . .
$$ . . . . . . . . .[/go]

elsewhere

gains 1 pt. gains 2 pts.

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

Visualize whirled peas.

Everything with love. Stay safe.

Click Here To Show Diagram Code

[go]$$ Increasing move values
$$ -----------------------
$$ . X . . X . X X X . O .
$$ . X O O O O O O O O O .
$$ . X . . . . . . . . . .
$$ . . . . . . . . . . . .[/go]

This is a 1 pt. sente. The threat is a 3 pt. sente.

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

Visualize whirled peas.

Everything with love. Stay safe.

The question was not whether (in your example) White 4 gains 4 points but whether Black 3 - White 4 as a unit (sente sequence) gains more than Black 1.

That was far from clear. How do you decide how to divide up a sequence of play into units? gains 1 pt., and each gain 2 pts. We could have had a sequence, which could occur with correct play in a real game, , , where gains 1 pt. while and each gain 2 pts.

Edit:

In your example we could have a sequence, , , where gains 3 pts. while and each gain 3.5 pts.

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

Visualize whirled peas.

Everything with love. Stay safe.

When we still start at the initial position, first, iteratively and backwards identify the traversal sequences. Second, iteratively and backwards identify the short gote / ambiguous / short sente sequences. This is how we determine counts and, at the starts of sequences parts, move values. Intermediate positions require calculations of the remaining move values, if needed.

Thermography has most of the properties you want, I think. You can even identify ambiguities with colored thermography. For instance, for the example I just gave, colored thermography reveals that it is ambiguous, with an average gain of 1 pt. For your example it reveals that it is a sente with a 3 pt. threat. However, in neither case does it tell us about the value of later moves, which you seem to be interested in.

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

Visualize whirled peas.

Everything with love. Stay safe.

I presume that thermography is an answer to everything, is it? :) However, currently I am interested in theory applicable by players (incl. myself). We are not plotters and most of us do not want to create and solve systems of linear equations during a game. Playing difference games is about the limit we might learn. Unless there are difficult kos, it is good enough to handle evaluation of long sequences, and often analysis is easier by applying our research (you will be surprised how useful it is!). Maybe I need to study thermography when I want to understand evaluation of difficult kos. Currently not. Sente and gote are entertaining enough even without them. Should I study thermography seriously some time, first of all I would want a translation applicable to players.

No, but it is a good start.

I gather from the remainder of your note that you are looking for something that is more practical. I share that sentiment.

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

Visualize whirled peas.

Everything with love. Stay safe.

Remarks

A, B are counts. PA, PB,... are positions.



Presuppositions

Let there be the combinatorial game PA := {PB|PE} := {PF|PC||PE} := {PF||x|PG|||PE}. PA, PB,..., PG are combinatorial games; PA, PB, PC are not numbers; x is a number. PG ≤ PE (1).

Proposition 1 [Black's 3-move traversal]

PA - PB - PC reverses to x.

Remarks

Bill Spight has proven the analogue proposition for White's 3-move traversal. Proposition 1 follows by symmetry.

Presuppositions

Let there be the combinatorial game PA := {PB|PE} := {PF|PC||PE} := {PF||x|PG|||PE}. PA, PB,..., PG are combinatorial games; PA, PB, PC are not numbers; x is a number.

1) PA and PC have the gote counts A and C, respectively.

Proposition 2 [Comparing Counts for Black's 3-move traversal]

PC ≤ PA <=> C ≤ A.

Proof

2) By proposition 1, PA = {x|PE} so PA and PC = {x|PG} only differ with respect to their white followers. White's start in PC or PA results in the count G or E, respectively.

PC ≤ PA <=>(2) G ≤ E <=> (x + G) / 2 ≤ (x + E) / 2 <=>(1)(2) C ≤ A.

Remarks

Is this a proof or wishful thinking? The first equivalence transformation is the difficult part. Is it correct that both directions of implication apply? Should the (2) reasoning be worked out and how?

I think that proposition 2 is necessary for using counts and move values to distinguish short from long alternating sequences played successively and assigning the correct values for earlier positions. Is it?

Bill, you suggest using conditions involving counts and move values of followers to determine the values of the initial local endgame. Such a method relies on an implicit assumption that such fulfilled conditions allow game tree simplification so that afterwards we can calculate values of the initial local endgame as if it were a simple gote or sente. If we may simplify, everything is fine. However, the question remains: why may we simplify due to some condition like C <= C2 or M <= M2 for counts and move values before move 1 and after move 2?

The mean value theorem, even if I understood it better, does not provide enough because short games are presumed so kos excluded. My proposition 2 is even more limited because it presumes gote counts, which exclude local sente or reverse sente endgames, and settled followers, which exclude, e.g., remaining basic endgame kos.

Therefore, I wonder whether theory of thermography has generally enabled the method you suggest. Or is the current state of research that conditions about counts and move values provide naive commands for tree simplifications whilst nobody has proven yet that they are as valid as they are for CGT reversal due to comparing POSITIONS? Why exactly may we simplify due to conditions using counts or move values?

Or do we not simplify at all but are there proofs purely for counts or move values distinguishing value calculation for short versus long sequences? If your suggested method is correct, I think I understand most of its application. However, why is it correct?

One aspect of application I do not understand because there are examples for which two variants of application produce different evaluations. The difficulty affects local sente with a sequence of at least 4 moves.

Application I: From move 2 on, increasing or constant move values every second move identify a long gote sequence. Treat it like one move. Then, for the first move, verify the simple sente condition.

Application II: From move 1 on, for each 2-move sente sequence part, verify its simple sente condition. And verify increasing or constant move values every second move to link the parts.

Which application is correct? Are both correct but neither identifies all long sentes?

Click Here To Show Diagram Code

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

How to evaluate this (in the corner locale) and the follow-up positions after moves 1 and 2 using conditions with move values? The problem is that distinguishing simple gote from simple sente or long sequences depend on intermediate follow-up positions involved in a still unsettled ko.

Click Here To Show Diagram Code

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

Click Here To Show Diagram Code

[go]$$W after move 2, prisoner difference = -1
$$ ----------------
$$ | . X . 2 O . O .
$$ | X O O . X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

I think I can evaluate after move 3, which is dominating in almost all practical environments:

Click Here To Show Diagram Code

[go]$$W after move 3, prisoner difference = -1
$$ ----------------
$$ | . X . X O . O .
$$ | X O O 3 X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

EDIT 4: added one liberty avoiding liberty shortage.

Yes, this depends upon the ko threat situation.

Click Here To Show Diagram Code

[go]$$W after move 3, prisoner difference = -1
$$ ----------------
$$ | . X . X O . O .
$$ | X O O 3 X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

This ko has an average value of less than 3 pts. for Black, with each move gaining 5+ points.

Edit: I am afraid that I goofed the analysis.
See later note.

First, let Black be komaster.

Click Here To Show Diagram Code

[go]$$W after move 2, prisoner difference = -1
$$ ----------------
$$ | . X . 2 O 3 O .
$$ | X O O 4 X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

Then is sente. Even though Black is komaster, her gain from winning the ko is so small that White can normally play with sente before Black can afford to play the ko.

Click Here To Show Diagram Code

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

fills at

So then the original position is a White sente with an average value of 10 pts. for Black. The reverse sente gains 3 pts.

Next, let White be komaster.

Click Here To Show Diagram Code

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

Then the original position is gote, because after Black cannot afford to make the ko. The average value is 6 pts. and each move gains 7 pts.



Yes, it does seem difficult for Black to be komaster. But not that easy for White to be komaster, either.

In case neither player is komaster, this is a possibility.

Click Here To Show Diagram Code

[go]$$W
$$ ----------------
$$ | . X . 4 1 . O .
$$ | X O O 8 X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

After is played in gote, Black may later play in gote, and then later play in gote.

One approach to the case where neither player is komaster is to assume a neutral threat environment (NTE), where both players have a large number of equivalent ko threats. That sometimes approximates real games, but cannot be counted on to give "correct" evaluations. If I have calculated the NTE case correctly, then the original position is gote, with an average value of 8½ for Black, and with each play gaining 4½ pts.

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

Visualize whirled peas.

Everything with love. Stay safe.

So you attack evaluation of such a position by a case analysis:
- Black komaster
- White komaster
- neither komaster, assume neutral threat environment, this gives expected local sequences to stable followers (except for remaining basic endgame kos) for the subcases Black starts and White starts, then derive the initial values from them
- neither komaster, assume some environment that is not a neutral threat environment

I think I have had in mind the last case. For it, I still have no idea how to evaluate if to be done only in terms of endgame values. What do you mean by not necessarily giving correct evaluations? Would this be a case with undefined count and move values of initial position, after move 1, after move 2? I.e., would we rather develop some ko fight theory instead of trying to apply endgame values? Such as comparing cases of Black winning a particular ko whilst White gains from ko threats versus White winning the ko whilst Black gains from ko threats? Or would that be a hybrid of ko fight evaluation and endgame evaluation, because we can calculate endgame values at least for the ko threats in the environment? For simple kos, local and global move values can be compared. Is this so also for the last case above?

If you do not abstract the ko threat environment via komaster or NTE, or something, you have to include the ko threats in the analysis in some way.



Ordinary move values tell you correct play in certain ideal environments. But they also tell you correct play in real games the vast majority of the time. NTE values tell you correct play in certain ideal environments, but cannot be counted on to tell you correct play in real games.



NTE analysis also suggests that White will normally respond at when neither player is komaster. Whether that is nearly all practical environments is another question.

It also suggests the following possibility.

Click Here To Show Diagram Code

[go]$$W
$$ ----------------
$$ | . X . 4 1 . O .
$$ | X O O 8 X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

After is played in gote, Black may later play in gote, and then later play in gote.

This sequence of play is not on the radar with either komaster analysis. Nor, I suspect, will most pros consider it, since after it looks obvious for White to make the ko. But if is so obvious, then Black will not play and we are back to the assumption that White is komaster.

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

Visualize whirled peas.

Everything with love. Stay safe.

Click Here To Show Diagram Code

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

This is quite an interesting position.

Black to play connects for 13 pts. of territory.

Click Here To Show Diagram Code

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

This ko is an obvious possibility when White plays first. If Black takes and wins the ko the result is again +13. (All territory values are from Black's perspective.)

Click Here To Show Diagram Code

[go]$$W White wins the ko
$$ ----------------
$$ | . B W C O C O .
$$ | X O O O X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

If White wins the ko White gets 1 pt. for each captured stone plus ⅓ pt. for the stone in the remaining ko, for -2⅓. Each play in the ko gains on average 5 1/9 pts. ((13 + 2⅓)/3), and the position after in the previous diagram has an average territorial count of 2 7/9.

Now let's look at the position just before White makes the ko.

Click Here To Show Diagram Code

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

Suppose that Black is komaster. Then, OC, Black can take and win the ko for 12 pts.

Click Here To Show Diagram Code

[go]$$W 1 Black prisoner, White sente
$$ ----------------
$$ | . X . X O 1 O .
$$ | X O O 2 X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

White to play can fill the ko with sente, for 10 pts. If Black takes and wins the ko she gains only 2 pts. in 2 moves, for an average value of 1 pt. per move.

However, White has another arrow in her quiver.

Click Here To Show Diagram Code

[go]$$W 1 Black prisoner, White makes ko
$$ ----------------
$$ | . X . X O 2 O .
$$ | X O O 1 X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

If White makes the ko and Black takes, the resulting position has an average value of 13 - 5 1/9 = 7 7/9 pts., which is better for White than 10 pts.

This means that when the largest plays on the rest of the board gain less than 3 pts. on average, White should normally take sente, but when the largest plays elsewhere gain between 3 and 5 1/9 pts. on average, White should normally make the ko and exchange a play somewhere else for losing the ko. This may seem counterintuitive, but in that case White should make the ko in order to lose it.

Let's back up one move.

Click Here To Show Diagram Code

[go]$$W 1 Black prisoner, White plays first
$$ ----------------
$$ | . X . 1 O . O .
$$ | X O O . X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

In this position White to play can prevent the ko for a local score of -1. If Black plays at 1 and White makes the ko (in order to lose it) Black will get 13 pts. in 2 net moves. Then each move is worth on average 4⅔ pts. ((13 + 1)/3). When the ambient temperature is less than 4⅔ and more than 3, and Black plays first, that is the expected play.

Now let's back up to the original position.

Click Here To Show Diagram Code

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

Then when the ambient temperature is between 4⅔ and 3 we expect this line of play when White plays first, which is the main line. takes the ko, plays elsewhere, and wins the ko. (OC, there may be a ko fight to force Black to fill. ) When the temperature is exactly 4⅔ each play gains on average 4⅔, and the territorial count is 8⅓.

When the temperature is 3 or less, we expect the following line of play.

Click Here To Show Diagram Code

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

fills the ko
The local score is 10. The change in the territorial value from 8⅓ to 10 represents an average loss of 1⅔ pts. to White, which explains why White does not wait, but makes the ko in order to lose it.

Next I'll look at the White komaster situation.

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

Visualize whirled peas.

Everything with love. Stay safe.

When White is komaster, let's look at this position.

Click Here To Show Diagram Code

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

As we have seen, White to play can make and win the ko in 2 moves, for an average result of -2⅓ pts. Black to play can connect at , for an average result of 10⅓. Each play gains on average 4 2/9 pts. ((10⅓ + 2⅓)/3), and this position before a play has an average territorial count of 6 1/9 pts. (10⅓ - 4 2/9).

Let's back up one play.

Click Here To Show Diagram Code

[go]$$B 1 Black prisoner
$$ ----------------
$$ | . X . 1 O . O .
$$ | X O O 2 X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

OC, if White plays first at 1, the result is -1. But if Black makes the ko in order to lose it, the average territorial count after is 1 8/9 pts. (6 1/9 - 4 2/9). Note that this assumes that the ko is played out, since each play in the ko gains 5 1/9 pts., which is greater than 4 2/9 pts. gains only 4 2/9 pts. because Black's threat is only to connect at 2.
So the average territorial count before is 1 8/9, and each play gains 4 2/9 on average.

Back to the original position.

Click Here To Show Diagram Code

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

As we know, Black to play can connect for 13 pts. The position after has an average value of 1 8/9 pts. This is a gote. Each play gains on average 5 5/9 pts. ((13 - 1 8/9)/2), and the average territorial count is 7 4/9 pts. ((13 + 1 8/9)/2). After when the temperature drops below 4 2/9 we expect that Black will make the ko in order to lose it.

Interesting that normally someone will make the ko in order to lose it. Who it is depends upon who is komaster. I don't recall having seen such a position before.

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

Visualize whirled peas.

Everything with love. Stay safe.

And, finally, the NTE analysis.

NTE stands for neutral threat environment. That's a general term, but specifically it means that each player has the equal and opposite threats to the other player, of the form {Θ | 0 || } for Black and { || 0 | -Θ} for White, and there are many such pairs of threats, so that if a threat of a certain size is needed, it is available. For convenience we assume that the threats are unremovable. (With enough threats of the same size we do not need that assumption.) The NTE is totally unrealistic, OC, but it does give us a way to abstract the idea that neither player is komaster.

Let's start with the original position.

Click Here To Show Diagram Code

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

NTE analysis gives this an average territorial count of 8, which lies between 8⅓ when Black is komaster and 7 4/9 when White is komaster, just as we expect. This is gote, and each play gains 5 pts.

Click Here To Show Diagram Code

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

The position after White has captured the Black stone has an average territorial count of 3, and it is also gote. Each play gains 4 pts. White, OC, can play to a position worth -1.

Click Here To Show Diagram Code

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

The position after Black has played atari has an average territorial count of 7, and it is also gote. Each play gains 3⅓ pts. Black to play can connect for 10⅓ pts.

Click Here To Show Diagram Code

[go]$$W 1 Black prisoner, Ko
$$ ----------------
$$ | . X . X O . O .
$$ | X O O O X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

Suppose that White makes the ko. The key to NTE analysis is that among the pairs of equal and opposite threats for a simple ko there is one pair for each temperature of the ko fight such that it does not matter who wins the ko. (Or more pairs of threats in more complex positions.) In this case, at an ambient temperature of 3⅓, that threat pair is {9⅓ | 0 || } for Black and { || 0 | -9⅓} for White. The ko is hot, so we assume that it is played out.

First, let Black take the ko and ignore White's threat. After Black wins the ko and White carries out the threat, the result is 13 pts. for Black minus 9⅓ pts. for White, or 3⅔ pts. That equals 7 - 3⅓, as advertised.

Next, let Black take the ko and answer White's threat. Then let White ignore Black's threat. After White wins the ko and Black carries out the threat, the result is -2⅓ + 9⅓ = 7. The count has remained the same, as it should after an even number of plays.

----

Wow! I even goofed the NTE analysis, which was the easiest. I was really out of it the other morning. I had been sick (still am), but I thought that I was able to concentrate. I was a bit sleepy, too, but really!

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

Visualize whirled peas.

Everything with love. Stay safe.

I do not understand yet how you calculate counts and move values under NTE.

(Self-reminder: For your two previous messages, later I need to check whether your statements about when to play locally given an environment with temperature do fit our earlier theory, as presumed.)

For a simple ko it is easy, as only one pair of equal and opposite threats is required for each ambient temperature. There is a second potential ko fight here, but for temperatures above ⅓ we can ignore it.

Click Here To Show Diagram Code

[go]$$W 1 Black prisoner, Ko
$$ ----------------
$$ | . X . X O . O .
$$ | X O O O X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

In this case, if White wins the ko the result is -2⅓ pts. of territory, on average; if Black takes and wins the ko the result is 13 pts. At temperature 0, what is the result of the ko fight in an NTE if it does not matter who wins the ko? It is the average of 13 and -2⅓, or 5⅓ pts. This, OC, is a fictitious number, since go scores are integers, but we are abstracting. To be exact, we should figure the result at temperature ⅓, since below that temperature the small ko in the corner is active. But calculating at temperature 0 is easier, and gives us the right result at temperatures greater than ⅓.

Suppose that the ko fight goes this way. Black takes the ko, White plays a threat, Black wins the ko, White carries out the threat. Result: 13 - Θ. Or suppose that it goes this way. Black takes the ko, White plays a threat, Black answers, White takes the ko back, Black plays a threat, White wins the ko, Black carries out the threat, White plays a dame or passes. (To be comparable, both lines of play must have the same parity.) Result: -2⅓ + Θ. The results will be the same if Θ = 7⅔, namely 5⅓ pts. of territory.

Now we know that when the ambient temperature is 5 1/9 the result will be 2 7/9. At that temperature Θ = 10 2/9. If Black wins the ko fight the result will be 13 - 10 2/9. If White wins the ko fight the result will be -2⅓ + 10 2/9 - 5 1/9. The last number represents what White gets by playing elsewhere after Black carries out the threat. At temperature T, Θ = 7⅔ + T/2, when ⅓ <= T <= 5 1/9. And the result is 5⅓ - T/2.

Now let's consider this position.

Click Here To Show Diagram Code

[go]$$W 1 Black prisoner, Ko
$$ ----------------
$$ | . X . X O . O .
$$ | X O O . X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

Black to play moves to a position worth 10⅓ pts. on average. White to play makes the ko, which is worth 5⅓ - T/2. (When ⅓ <= T <= 5 1/9, OC). What is the intersection of the lines, M = 10⅓ - T and M = 5⅓ + T/2? They intersect at M = 7, T = 3⅓. Since ⅓ <= 3⅓ <= 5 1/9, then if White makes the ko it is played out, the result is given by the equation of the line, and we are done.

Now let's consider this previous position.

Click Here To Show Diagram Code

[go]$$W 1 Black prisoner, Ko
$$ ----------------
$$ | . X . . O . O .
$$ | X O O . X O O .
$$ | X . O . X O . .
$$ | X X . X X O . .
$$ | . X X X X O . .
$$ | . . . . . . . .[/go]

White to play moves to a position worth -1. (7 + 1)/2 = 4 > 3⅓, so Black's move at temperature 4 is gote, and the ko comes later. The average territorial count of this position is (7 - 1)/2 = 3.

The rest is easy.

_________________
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— Winona Adkins

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Everything with love. Stay safe.