Thermography

Gérard TAILLE · Post by **Gérard TAILLE** » Tue Nov 03, 2020 3:26 pm

Bill Spight wrote: However, if Move1 has a miai value, t1 <= 1, and Move2 has a miai value, t2 < t1, then Move1 dominates Move2, with the ko caveat. This is shown in "Mathematical Go" via the concept of chilling (See https://senseis.xmp.net/?Chilling ). Chilling is derived there, but the derivation uses aspects of CGT with which I am not familiar.

A great result of "Mathematical Go". Any clue to prove this result which allows to trust and use chilled go?
I understood you are not familiar with this point of "Mathematical Go" but you might have the main idea of the proof (?)

Bill Spight · Post by **Bill Spight** » Tue Nov 03, 2020 6:58 pm

Gérard TAILLE wrote:
Bill Spight wrote: However, if Move1 has a miai value, t1 <= 1, and Move2 has a miai value, t2 < t1, then Move1 dominates Move2, with the ko caveat. This is shown in "Mathematical Go" via the concept of chilling (See https://senseis.xmp.net/?Chilling ). Chilling is derived there, but the derivation uses aspects of CGT with which I am not familiar.
A great result of "Mathematical Go". Any clue to prove this result which allows to trust and use chilled go?
I understood you are not familiar with this point of "Mathematical Go" but you might have the main idea of the proof (?)

Regular go players are used to ignoring dame in territory scoring. For instance, in your recent example, they might write the sum of games this way:

{2|0} + {1|0}

But for these two games, {2|0} > {1|0}, i.e., {2|0} - {1|0} > 0, i.e.,

{2|0} + {0|-1} > 0

1) Black first will play in the left hand game and then White will answer in the right hand game, for a score of 2 - 1 = 1 > 0. Black wins.

2) White first will also play in the left hand game and then Black will reply in the right hand game, for a score of 0 + 0 = 0, with White to play. Black wins this way also.

Without the dame {2|0} and {1|0} are not incomparable, which justifies the ordinary go player's practice of ignoring the dame.

In CGT we write * (star) for a dame. It turns out, as you may verify, {1|0} is impossible on the go board. You either have {1|*} or {1*|0}. (1* means 1 plus a dame.) Likewise, you either have {2|0} or {2*|*}. This fact allows us to restore the dame to {1|0}, getting {1*|0} or {1|*} depending upon the context of the whole board. So ignoring the dame does not lose any significant information. (The result by area scoring will remain the same when any ignored dame are restored.) Ignoring the dame justifies chilling, because the strict ordering of chilled go is the same as the ordering before chilling when the dame are ignored. IOW, correct play at chilled go is also correct at regular go, up to a dame, with the ko caveat.

Edit: Why a dame? Because an even number of dame add to zero.

I think that chilling would not work if regular go had fractional scores on the board. Ignoring the dame would lose significant information that could not be restored. (Edit: For instance, ignoring dame would change 3 points without capturing to 2½ points without capturing.

) Restoring the dame depends upon the parity of the empty points on the board plus prisoners.

When I first read Mathematical Go I got it that chilling depends upon ignoring dame, which I was doing, anyway. So what's the problem?

Bill Spight · Post by **Bill Spight** » Tue Nov 03, 2020 7:20 pm

Correction. I said that only moves dominate, not games. That is not correct CGT terminology. Strictly speaking, CGT does not compare moves directly, except in terms of temperature, which is heuristic. What I should have said is that options dominate. If Black to play can move to option A or option B, and A ≥ B, then A dominates B. Likewise, if White to play can move to option C or option D, and D ≤ C, then D dominates C.

Edit: One game cannot dominate another, per se. It is only as options for one player or the other that one game can dominate another.

Edit2: Besides, in von Neumann game theory you do speak of one strategy dominating another, and playing one move rather than another is a behavioral strategy. So I'll continue to talk about one play dominating another, s'il vous plaît.

Bill Spight · Post by **Bill Spight** » Wed Nov 04, 2020 2:34 am

Click Here To Show Diagram Code: [go]$$ White to play $$ ----------------- $$ | O a O . . . . | $$ | X X O . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | X X O . . . . | $$ | . b O . . . . | $$ -----------------[/go]

By convention, outside stones are immortal.

Does a White play at a dominate a play at b?

Let's set up a difference game where White plays at a and Black plays at the mirror of b.

Click Here To Show Diagram Code: [go]$$B Difference game $$ -------------------------------- $$ | O W O . . . . | X . X . . . . | $$ | X X O . . . . | O O X . . . . | $$ | . . . . . . . | . . . . . . . | $$ | . . . . . . . | . . . . . . . | $$ | . . . . . . . | . . . . . . . | $$ | X X O . . . . | O O X . . . . | $$ | . . O . . . . | . B X . . . . | $$ --------------------------------[/go]

Black to play cannot win.

Click Here To Show Diagram Code: [go]$$B Black first $$ -------------------------------- $$ | O W O . . . . | X 2 X . . . . | $$ | X X O . . . . | O O X . . . . | $$ | . . . . . . . | . . . . . . . | $$ | . . . . . . . | . . . . . . . | $$ | . . . . . . . | . . . . . . . | $$ | X X O . . . . | O O X . . . . | $$ | . 1 O . . . . | 3 B X . . . . | $$ --------------------------------[/go]

Net result: -1

If

is at 2,

at 1 plays to 0. This means that the White connection dominates the intrusion into the short corridor.

----

Does a Black play at a dominate a play at b?

Let's set up a difference game where Black plays at a and White plays at the mirror of b.

Click Here To Show Diagram Code: [go]$$B Difference game $$ -------------------------------- $$ | . B O . . . . | X . X . . . . | $$ | X X O . . . . | O O X . . . . | $$ | . . . . . . . | . . . . . . . | $$ | . . . . . . . | . . . . . . . | $$ | . . . . . . . | . . . . . . . | $$ | X X O . . . . | O O X . . . . | $$ | . . O . . . . | . W X . . . . | $$ --------------------------------[/go]

One

prisoner.

White to play cannot win.

Click Here To Show Diagram Code: [go]$$W White first $$ -------------------------------- $$ | . B O . . . . | X 2 X . . . . | $$ | X X O . . . . | O O X . . . . | $$ | . . . . . . . | . . . . . . . | $$ | . . . . . . . | . . . . . . . | $$ | . . . . . . . | . . . . . . . | $$ | X X O . . . . | O O X . . . . | $$ | 3 1 O . . . . | . W X . . . . | $$ --------------------------------[/go]

One

prisoner

Net result: +1

If

is at 2,

at 1 plays to 0. This means that the Black capture dominates closing off the short corridor.

The play that gains 1 point dominates the play that gains ½ point, no surprise.

Gérard TAILLE · Post by **Gérard TAILLE** » Wed Nov 04, 2020 6:54 am

Click Here To Show Diagram Code: [go]$$ White to play $$ ----------------- $$ | O a O . . . . | $$ | X X O . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | X X O . . . . | $$ | . b O . . . . | $$ -----------------[/go]

Oops I do not want to compare a move at "a" against a move at "b" I just want to compare the two positions themselves!

Let's call G1 the game corresponding to the position at the top of the board. I guess we can write G1 = {2|0}
Let's call G2 the game corresponding to the position at the bottom of the board. I guess we can write G2 = {1|*}

For a mathematical point of view I now define the game G as
G = {G1,G2|} or {G1,G2|H} if you prefer for lowering the miai value of G.
Taking into account your correction (options vs moves), my point is to compare the two white options G1 and G2. They appear uncomparable though miaiValue(G1) = 1 and miaiValue(G2) = ½

Bill Spight · Post by **Bill Spight** » Wed Nov 04, 2020 7:49 am

Gérard TAILLE wrote:
$$ White to play
$$ -----------------
$$ | O a O . . . . |
$$ | X X O . . . . |
$$ | . . . . . . . |
$$ | . . . . . . . |
$$ | . . . . . . . |
$$ | X X O . . . . |
$$ | . b O . . . . |
$$ -----------------
Click Here To Show Diagram Code
[go]$$ White to play $$ ----------------- $$ | O a O . . . . | $$ | X X O . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | X X O . . . . | $$ | . b O . . . . | $$ -----------------[/go]
Oops I do not want to compare a move at "a" against a move at "b" I just want to compare the two positions themselves!

Let's call G1 the game corresponding to the position at the top of the board. I guess we can write G1 = {2|0}
Let's call G2 the game corresponding to the position at the bottom of the board. I guess we can write G2 = {1|*}

For a mathematical point of view I now define the game G as
G = {G1,G2|} or {G1,G2|H} if you prefer for lowering the miai value of G.
Taking into account your correction (options vs moves), my point is to compare the two white options G1 and G2. They appear uncomparable though miaiValue(G1) = 1 and miaiValue(G2) = ½

Well, you already proved that G1 and G2 are incomparable. The question now is how to construct

G = {{2|0},{1|*}|H}

on the go board. Chilling is useful for go because of the properties of go positions, including the prisoner count. That does not mean that it is useful for the game, G.

As for G = {{2|0),{1|*}| }, I think it is equal to 1, which is equal to {0| }. I.e.,

{{2|0},{1|*}| } + { |0} = 0

1) Let Black play to {2|0} in G and then White replies to 0 in G. Black has no move, so White wins.
2) Let Black play to {1|*} in G and then White replies to * in G, then Black plays to 0 from *. Now White plays to 0 in { |0}. Black has no play and White wins.
3) Let White play to 0 in { |0}. Now Black plays to {1|*} and wins.

The sum is a second player win and is thus equal to 0.

So {{2|0),{1|*}| } = 1

Gérard TAILLE · Post by **Gérard TAILLE** » Wed Nov 04, 2020 8:11 am

Bill, in MathematicalGo/discussion https://senseis.xmp.net/?MathematicalGo%2Fdiscussion, you wrote:
Bill: Here, in the main, is what Mathematical Go is about. There is a game, chilled go. Unless ko is a consideration, if you win chilled go you win the regular game of go that you get if you continue playing. Chilled go has values that do not appear in regular go, such as fractional scores and infinitesimals (other than dame). Infinitesimals are well understood in combinatorial game theory. This knowledge about infinitesimals can be applied to chilled go, and, since winning chilled go usually allows you to win regular go, it can be applied to regular go, as well.

What means "usually" in the phrase "winning chilled go usually allows you to win regular go"?
Because we consider only non-ko situation, does that mean it exists exceptional non-ko situation for which it is not true?

Bill Spight · Post by **Bill Spight** » Wed Nov 04, 2020 8:36 am

Gérard TAILLE wrote:Bill, in MathematicalGo/discussion https://senseis.xmp.net/?MathematicalGo%2Fdiscussion, you wrote:
Bill: Here, in the main, is what Mathematical Go is about. There is a game, chilled go. Unless ko is a consideration, if you win chilled go you win the regular game of go that you get if you continue playing. Chilled go has values that do not appear in regular go, such as fractional scores and infinitesimals (other than dame). Infinitesimals are well understood in combinatorial game theory. This knowledge about infinitesimals can be applied to chilled go, and, since winning chilled go usually allows you to win regular go, it can be applied to regular go, as well.

What means "usually" in the phrase "winning chilled go usually allows you to win regular go"?

I meant that the ko caveat, mentioned earlier in that paragraph, usually does not make optimal play in chilled go non-optimal in regular go.

Bill Spight · Post by **Bill Spight** » Wed Nov 04, 2020 8:53 am

For fun, let's consider the game,

G = {{2|0},{1|*}|-1}

It has a count of 0 and White's play gains 1 point on average.

Orthodox play for Black above ambient temperature ½ up to ambient temperature 1 is to play to {2|0}. Between ambient temperature of 0 and ½, either option could be optimal. At temperature -1 the {1|*} option leads to a local score of +1.

Gérard TAILLE · Post by **Gérard TAILLE** » Wed Nov 04, 2020 10:32 am

Bill Spight wrote:
Gérard TAILLE wrote:Bill, in MathematicalGo/discussion https://senseis.xmp.net/?MathematicalGo%2Fdiscussion, you wrote:
Bill: Here, in the main, is what Mathematical Go is about. There is a game, chilled go. Unless ko is a consideration, if you win chilled go you win the regular game of go that you get if you continue playing. Chilled go has values that do not appear in regular go, such as fractional scores and infinitesimals (other than dame). Infinitesimals are well understood in combinatorial game theory. This knowledge about infinitesimals can be applied to chilled go, and, since winning chilled go usually allows you to win regular go, it can be applied to regular go, as well.

What means "usually" in the phrase "winning chilled go usually allows you to win regular go"?
I meant that the ko caveat, mentioned earlier in that paragraph, usually does not make optimal play in chilled go non-optimal in regular go.

For the time being I don't see clearly how you can say who is winning in chilled-go. Playing the game until we reach the first infintesimal seems clear but then what will happen? Are all the infinitesimals played in chilled go? I guess the answer is yes but I am not sure.
Let's now suppose the last infinitesimal has been played. That means it remains only areas which miai value is, in a non chilled game, ≤1. How do you continue in order to design the winner?
Can we add all remaining scores and round it to the correct ordinal number depending which side plays first?
Who wins if the final score is equal?
Should we continue the game without chilling in order to know who will play the last move?

Bill Spight · Post by **Bill Spight** » Wed Nov 04, 2020 11:27 am

Gérard TAILLE wrote:For the time being I don't see clearly how you can say who is winning in chilled-go. Playing the game until we reach the first infintesimal seems clear but then what will happen? Are all the infinitesimals played in chilled go? I guess the answer is yes but I am not sure.

Edit: OC, we are assuming no kos.

In theory, you may be able to count the score without playing out all the chilled infinitesimals. In practice, you nearly always play them out. Some rules require you to do so.

Back during the Japanese rules discussions in the early 20th century, several players objected to any idea of requiring the dame to be filled during play, because leaving them unfilled was part of the beauty of Japanese go. OC, it is possible to score the game without filling the dame.

Click Here To Show Diagram Code: [go]$$ Miai $$ ----------------- $$ | O . O . . . . | $$ | X X O . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | X X O . . . . | $$ | O . O . . . . | $$ -----------------[/go]

In chilled go, you could leave these infinitesimals unplayed and count the combination as 2 points for Black. In practice, you play them out.

Let's now suppose the last infinitesimal has been played. That means it remains only areas which miai value is, in a non chilled game, ≤1. How do you continue in order to design the winner?

Continue normally.

Edit: I.e., biggest play first is optimal.

But you have reminded me of cases where winning the chilled go game does not necessarily mean winning the regular game. For instance, suppose that the chilled go game ends with Black ahead by ½ point with White to play. Black wins the chilled go game, but the regular go game will be jigo. But if Black has played optimally in the chilled go game she will have also played optimally in the regular go game. Similarly, suppose that with komi White wins the chilled go game by ¼ point, with Black to play. Black can "round up" to the next integer, gaining ¾ points in the process and winning by ½ point. Again, if White's play was optimal in chilled go, it was optimal in regular go. I had forgotten about such positions.

Can we add all remaining scores and round it to the correct ordinal number depending which side plays first?

Yes, as indicated.

Who wins if the final score is equal?

Well, regular go allows ties, or maybe White wins jigo. Black wins jigo is possible, OC, but I am unaware of any such rule.

Should we continue the game without chilling in order to know who will play the last move?

In regular territorial go with no ko who plays the last move does not affect the score, right? In some forms of go it does. For instance, in the Capture Game.

Bill Spight · Post by **Bill Spight** » Wed Nov 04, 2020 1:01 pm

The value of chilled go to me

1) Learning go infinitesimals.

Many of them are easy enough that, IMO, almost every amateur dan player can in practice play them correctly without breaking a sweat. This knowledge enabled me to see quickly occasional game losing pro errors in playing them, including the error in the ancient game for the golden lidded bowls.

2) Reassuring me that, except for ko situations, playing the biggest play that gains less than 1 point is optimal.

3) A slight saving of time and effort from not reading small plays to the end.

4) Making it easier to compose problems by thinking of them as chilled go problems. I also feel that the problems are more interesting that way.

Gérard TAILLE · Post by **Gérard TAILLE** » Wed Nov 04, 2020 1:54 pm

I begin to understand the interest of chilling-go but I need to know another result. At the end of the chilled-go, after each infinitesimal has been played, each of the remaining areas has a miai value ≤1.
Let's call G1, G2, G3 ... all these remaining areas and let's call s1, s2, s3 .. the score of these areas.
The remaining game is G = G1 + G2 + G3 + ....
Is it true that, at the end of chilling-go we have:
1) G score = s1 + s2 + s3 + ....
2) G miai value ≤1

Bill Spight · Post by **Bill Spight** » Wed Nov 04, 2020 2:32 pm

Gérard TAILLE wrote:I begin to understand the interest of chilling-go but I need to know another result. At the end of the chilled-go, after each infinitesimal has been played, each of the remaining areas has a miai value ≤1.

In terms of regular go, chilled go infinitesimals have a miai value of 1. So the remaining plays have a miai value strictly less than 1. (Miai value is a go term, not a CGT term. It applies to plays, not areas. In a way, we have too much terminology.

)

One important lesson of Mathematical Go is that the fight for the last play of the game in go is actually about the fight for the last play at temperature 1; i.e., for the last play in the chilled go game.

Hence the subtitle: Chilling gets the last point. That discovery was new to go theorists, and many of them may still be unaware of it, unless they have read Mathematical Go. AFAICT, the bots haven't learned that, either. It is not exactly obvious.

Let's call G1, G2, G3 ... all these remaining areas and let's call s1, s2, s3 .. the score of these areas.

In chilled go they are scores, in regular go they are counts or territorial values. (Too much terminology.

)

The remaining game is G = G1 + G2 + G3 + ....
Is it true that, at the end of chilling-go we have:
1) G score = s1 + s2 + s3 + ....
2) G miai value ≤1

Well, the chilled score for G will be the sum of the scores of its components. Which will be their combined count in regular go.

In regular go the effective miai value of the sum of games will be less than 1. For instance, although the miai value of each play in the following diagram is 1, the effective miai value of their sum is -1.

Click Here To Show Diagram Code: [go]$$ Miai $$ ----------------- $$ | O . O . . . . | $$ | X X O . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | X X O . . . . | $$ | O . O . . . . | $$ -----------------[/go]

Gérard TAILLE · Post by **Gérard TAILLE** » Wed Nov 04, 2020 2:41 pm

Bill Spight wrote:
Gérard TAILLE wrote:I begin to understand the interest of chilling-go but I need to know another result. At the end of the chilled-go, after each infinitesimal has been played, each of the remaining areas has a miai value ≤1.
In terms of regular go, chilled go infinitesimals have a miai value of 1. So the remaining plays have a miai value strictly less than 1. (Miai value is a go term, not a CGT term. It applies to plays, not areas. In a way, we have too much terminology. )

One important lesson of Mathematical Go is that the fight for the last play of the game in go is actually about the fight for the last play at temperature 1; i.e., for the last play in the chilled go game. Hence the subtitle: Chilling gets the last point. That discovery was new to go theorists, and many of them may still be unaware of it, unless they have read Mathematical Go. AFAICT, the bots haven't learned that, either. It is not exactly obvious.

Let's call G1, G2, G3 ... all these remaining areas and let's call s1, s2, s3 .. the score of these areas.
In chilled go they are scores, in regular go they are counts or territorial values. (Too much terminology. )

The remaining game is G = G1 + G2 + G3 + ....
Is it true that, at the end of chilling-go we have:
1) G score = s1 + s2 + s3 + ....
2) G miai value ≤1
Well, the chilled score for G will be the sum of the scores of its components. Which will be their combined count in regular go.

In regular go the effective miai value of the sum of games will be less than 1. For instance, although the miai value of each play in the following diagram is 1, the effective miai value of their sum is -1.

$$ Miai
$$ -----------------
$$ | O . O . . . . |
$$ | X X O . . . . |
$$ | . . . . . . . |
$$ | . . . . . . . |
$$ | . . . . . . . |
$$ | X X O . . . . |
$$ | O . O . . . . |
$$ -----------------
Click Here To Show Diagram Code
[go]$$ Miai $$ ----------------- $$ | O . O . . . . | $$ | X X O . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | . . . . . . . | $$ | X X O . . . . | $$ | O . O . . . . | $$ -----------------[/go]

In case of miai, the miai value of the sum is 0, isn't it?

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