Ok, now it is clear (except that I need to study mast values in general).

My mistake, I think. If Black plays C-19, no seki, no ko. The right play to make seki or the corner ko, I now think, is the sagari, E-19. Here is an sgf that explains the variations. Maybe I made a mistake, but I don't think so. Points are marked with circles, captured stones and places where stones have been captured are marked with triangles.

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

Visualize whirled peas.

Everything with love. Stay safe.

Yes, this must be the right timimg.

Why is D19 an »obvious error«? It seems to me that both the ko and the seki are worse results for White. What am I overlooking?

_________________
A good system naturally covers all corner cases without further effort.

Because Kano assumes that the position is miai.

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

Visualize whirled peas.

Everything with love. Stay safe.

What is the algebraic and graphical representation of a thermograph of an integer in the temperature range from -1 to 0?

What is the algebraic and graphical representation of a thermograph of a dyadic fraction in the temperature range from -1 to 0?

I do not even dare to ask for infinitesimals.

Are the basic slopes of a thermograph 1 (for gote in Black's wall), -1 (for gote in White's wall) or oo (for mast or sente), or are they -1 (for gote in Black's wall), 1 (for gote in White's wall) or 0 (for mast or sente)? The former is as a thermograph is drawn, the latter is according to Siegel. Is it just convention of what slopes are?

If there are no kos, the thermograph of an integer is a mast starting at temperature -1. With kos and superkos it is in theory possible to have a thermograph of an integer with a temperature less than 1 but more than -1..



Let's take the simplest subzero thermograph, that of ½ = {0,1}. At temperature -½ we subtract -½ from Black's move to 0, getting 0 - (-½) = ½. Similarly, we add -½ to White's move to 1, also getting ½. Since these values are equal, the mast rises from temperature -½ at score ½. At temperature -1 we subtract -1 from Black's move to 0, getting 1, and we add -1 to White's move to 1, getting 0. We end up with a thermograph that looks like the thermograph of {1|0}, except that the temperature is 1 point less.

Now let's look at the thermograph of ¼ = {0|½}. It has temperature -¼, above which the mast rises. The left wall below temperature -¼ is a straight line to 1 at temperature -1. (0 - (-1) = 1.) The right wall is a straight line to 0 at temperature -½ and a straight line down to 0 below that. It looks like the thermograph of {1|-½} with a temperature 1 point less.

In fact, the easy way to draw the thermograph of a dyadic fraction is to add 1 to the left side and subtract 1 from the right side and draw the thermograph of that, then subtract 1 from the temperature. Remember that non-ko go scores are integers. While {½|-½} is a combinatorial game, it is not a non-ko go game. It is a game at chilled go, OC.

For instance, the easy way to draw the thermograph of ⅜ = {¼|½} is to draw the thermograph of {1¼|-½} and then subtract 1 from the temperature.

Edit: Hmmm, The thermograph of 1¼ is that of {2|½}. Maybe that's not so obvious. How about this?

The temperature of a dyadic fraction, s/d, is 1 - 1/d, and the fraction is a gote. ⅜ = {¼|½}. Let's draw the right side. ½ has a temperature of -½. So we draw a line from (⅜,-⅛) down to (0,-½) and then a vertical line down to (0,-1). Now for the left side. ¼ has a temperature of -¼. So we draw a line from (⅜,-⅛) down to (½,-¾) and then a vertical line down to (½,-½), and then a line down to (1,-1).

Or perhaps this. We know that the temperature of 1¼ as a go score is ¾. 1¼ + ¾ = 2, and 1¼ - ¾ = ½. So we find the thermograph of {2|½).

Anyway, if you ever need to calculate it for a go position, it's fairly obvious.



Infinitesimals have a thermograph with a mast starting at temperature 0.

For instance, the thermograph of * = {0|0} has a mast at 0 and a left wall down to 1 and a right wall down to -1. The easy way to draw it is to draw the thermograph of {1|-1} and then subtract 1 from the temperature.



The slope is ∆t/∆s, where t is the temperature and s is the score. By convention CGT reverses the order of s, so that the lines look backwards to the usual convention, and the apparent slopes are the negative of the real slopes. With the exception of some ko and superko thermographs, the slopes of the Black wall are < 0 or vertical, and the slopes of the White wall are > 0 or vertical, as Siegel says. It is just that they look the opposite of the usual convention.

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

Visualize whirled peas.

Everything with love. Stay safe.

Let me start with this simplest topic.

Suppose we have a number A.

I think Black's scaffold is A - T and White's scaffold A + T.

Now, if I equate A - T = A + T <=> 0 = 2T <=> 0 = T, I find the move value 0 (as it should be). So I calculate the count A - 0 = A + 0 = A.

Therefore, at least so I think, Black's wall is

A if T >= 0,

A - T if T < 0

and White's wall is

A if T >= 0,

A + T if T < 0.

Let me study T = -1. I get

for Black's wall A - (-1) = A + 1 and

for White's wall A + (-1) = A - 1.

Neither indicates a mast downwards to T = -1.

I would only get there by introducing the tax 1 for only negative temperature. Presumably, I have not understood something.

Siegel draw the thermograph of a simple gote with 45° segments reaching into negative temperatures. Does he do so maybe because introducing a tax only for them makes a difference?



Somebody says you have an example for that. What?

If A = ½, then Black's scaffold is the line, t = -s, starting at (s,t) = (1,-1), and White's scaffold is the line, t = s - 1, starting at (0,-1). They intersect when s - 1 = -s, i.e., s = ½ and t = -½. The mast rises vertically from there.

Dyadic rationals have more complicated thermographs.



Suppose that A = 1. Then the Black wall is the deponent line between (1,-1) and (1,-1). The same is true for the White wall.



The idea of subterranean thermographs originated with Conway or Berlekamp. They follow the same rules as other thermographs, originally defined in terms of taxes.



Maybe, but I don't think so. {shrug}

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

Visualize whirled peas.

Everything with love. Stay safe.

Was tired. I did not mean to ask about numbers but only about integers.

"line between (1,-1) and (1,-1)" - uh!

Karen Ye: "We do not tax numbers."

Is the -T or +T adjustment to an unsettled position the same as a tax?

If she is right at least for integers, then of course there is a mast in the [-1;0] temperature range.

However, I fundamentally don't understand why according to definitions.

Except for the max, Black's scaffold is calculated: β_T({L|R}) = W_T(L) - T.
How do we apply this definition to an integer A?

Except for the min, White's scaffold is calculated: ω_T({L|R}) = B_T(R) + T.
How do we apply this definition to an integer A?


Oh, wait, now I recall that we DO NOT. For a settled position / integer x, the different definition of the walls and count forgoes scaffolds and is

B_T(x) = W_T(x) = C(x) = x for all T.

Have I now got it right?


This is your reported position allegedly with an interesting negative graph.

Click Here To Show Diagram Code

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

2 EDITs

She is not talking about subterranean thermography.



What adjustment are you talking about? Anyway, numbers are not considered to be unsettled, as a rule. However, sometimes you have to play in numbers when your opponent does not agree to stop play. And, as we know, kos and superkos can make numbers worth playing.



She is not talking about subterranean thermography.



I don't know what you are talking about, and I redefined thermography in 1998. {shrug}






Look. Subterranean thermography is not difficult. If there were no kos, CGT would simply apply to chilled go. But go players, from time immemorial, have not played chilled go. Nobody plays chilled go. We keep on playing and stop with scores that are integers. We may lose some information that way, but we don't care.

I am going to do some simple examples, now. I won't post any thermographs, because they take some doing, and I haven't always gotten this site to accept them. If you want to explain what you are talking about, that might help. If I don't get it, I'll just ignore it.

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

Visualize whirled peas.

Everything with love. Stay safe.

Currently, the go board example is not important for me.

My understanding has not reached ko thermography yet. I am trying to apply basic thermographs a la Siegel section II.5. And for that, I still do not understand why an integer is a mast down to T = -1.

You say it is but I do not understand why and how it relates to the definitions of: scaffolds, minimum temperature defining move value and count, and walls.

Do your 1998 redefinitions have an explicit definition that the graph of an integer is a mast from -1 to oo?

Go board examples help to make sense of subterranean thermography.



I trust that you understand that an integer has a vertical mast when t ≥ 0 under territory scoring. All territory scores are integers, even though maybe they should not be. Therefore no non-ko scaffolds with territory scores can intersect at an integer score when 1 > t > 0. Regular go thermographs are subterranean thermographs for chilled go. All non-ko values at t = -1 are integers. So all non-ko thermographs with scaffolds that intersect when 0 > t > -1 have fractional masts. Suppose that the Black scaffold of a subterranean non-ko thermograph is a vertical line, s = 1. There are only three possible non-ko subterranean White scaffolds that intersect it or coincide with it. One is, obviously, the line, s = 1. Obviously the thermograph is the vertical mast starting at t = -1. Another is the line, s = t + 2. Again, the mast of the thermograph starts at t = -1. The third is the line, s = t + 1. Its mast starts at t = 0. But it is not the thermograph of a number, but of an infinitesimal. You can do the cases where the Black scaffold is the line, s = -t. Integers do not always have subterranean thermographs with masts intersecting or coinciding at t = -1. For instance, 0 = { | }, and has neither a Black scaffold or a White scaffold. But when integers do have non-ko subterranean thermographs with masts that start at 0 > t ≥ -1, they start at t = -1.



No. My redefinition was aimed at handling multiple kos and superkos, which defied Berlekamp's komaster analysis. Nonetheless, Berlekamp and I agreed about the mast values and temperatures of multiple kos and superkos. There had to be a way of formalizing our mutual understanding.

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

Visualize whirled peas.

Everything with love. Stay safe.

Let’s start with the simplest playable position under territory scoring, a dame.

Click Here To Show Diagram Code

[go]$$ There is nothing like a dame
$$ ---------------------------------------
$$ | . . . . . . . X C O O . . . . . . . . |
$$ | . . . . . . . X X X O . . . . . . . . |
$$ | . . . . . . . . . . O . . . . . . . . |
$$ | . . . , . . . . . , . . . . . , . . . |[/go]

Per convention, we assume that all stones framing the position are immortal.

What is the thermograph of this dame? We can write the dame as {0|0}. In (s,t) format the Black scaffold is the line, s = -t, and the White scaffold is the line, s = t. Obviously they intersect at (0,0). The mast, s = 0, rises vertically from there. Easy. ,

But in chilled go the dame is {-1|1}. Now, this is not the simplest form of 0, which is { | }, but let’s use it, anyway. The Black scaffold is the line, s = -1 - t, and the White scaffold is s = t + 1. These scaffolds do not intersect when t ≥ 0. and there are rules that allow us to figure out what number {-1|1} is, but let’s use subterranean thermography. When we do that, we find that the scaffolds intersect at (0,-1), and we draw the vertical mast, s = 0, from there. Wonderful! {-1|1} = 0.

Click Here To Show Diagram Code

[go]$$ 1 + 1 = 0
$$ ---------------------------------------
$$ | . . . . . . . X C O O O C X . . . . . |
$$ | . . . . . . . X X X O X X X . . . . . |
$$ | . . . . . . . . . . O . . . . . . . . |
$$ | . . . , . . . . . , . . . . . , . . . |[/go]

At regular go, two dames equal 0, an integer. Consider {0|0||0|0}. Whoever plays first, the other player can return to 0. Each player can guarantee at least 0 for herself, no matter who plays first. So {0|0||0|0} = 0. But even if we do not figure that out, we can draw the thermograph, which has a Black scaffold of s = -t and a White scaffold of s = t, which intersect at (0,0).

N. B. If you try to do the thermograph directly in chilled go, you have (-1|1||-1|1}, which is confusing, to say the least. You have to remember that {-1|1}= 0, or that it has a mast, s = 0, that starts at t = -1. The easy thing to do is to do the thermographs in regular go and then subtract 1 point from the temperature to get the subterranean thermographs for chilled go.

Now let’s do some closed corridors, which provide prototypical fractions in chilled go.

Click Here To Show Diagram Code

[go]$$ Closed corridor, length 2
$$ ---------------------------------------
$$ | . . . . . . . X . . O . . . . . . . . |
$$ | . . . . . . . X X X O . . . . . . . . |
$$ | . . . . . . . . . . O . . . . . . . . |
$$ | . . . , . . . . . , . . . . . , . . . |[/go]

As we all know, this corridor has an average territorial value, or count, of ½. Black to play closes the corridor for 1 point; White to plays intrudes into the corridor for a dame. We write this position as {1||0|0}. The Black scaffold is the line, s = 1 - t, and the White scaffold is the line, s = t. The scaffolds intersect at (½,½). The mast is a vertical line, s = ½, starting at t = ½. To get the subterranean thermograph for chilled go, subtract 1 point, so that the mast starts at t = -½.

Click Here To Show Diagram Code

[go]$$ ½ + ½ = 1
$$ ---------------------------------------
$$ | . . . . . . . X . . O . . X . . . . . |
$$ | . . . . . . . X X X O X X X . . . . . |
$$ | . . . . . . . . . . O . . . . . . . . |
$$ | . . . , . . . . . , . . . . . , . . . |[/go]

This is worth 1 point at chilled go, 1 point plus a dame at regular go. Whoever plays first, the opponent can return to 1 point plus a dame, so that is what it is worth. We can write this position as {2||1|1|||1|1||0}. We know from inspection that it equals {1|1}, but let’s do the thermograph.

The Black scaffold of the thermograph is the right wall of {2||1|1} - t. That is, s = min(1, 1½ - t), and White scaffold is the left wall of {1|1||0} + t. That is, s = max(1, ½ + t). The two scaffolds coincide when 0 ≤ t ≤ ½. The mast is the vertical line, s = 1, starting at t = 0. In chilled go, it starts at t = -1. So this is an integer at chilled go.

Enough for now.

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

Visualize whirled peas.

Everything with love. Stay safe.

You try to tell me as much as possible about negative thermographs without answering my question:)



Why "for instance"? What integers other than 0 = { | } do not have subterranean thermographs with masts intersecting or coinciding at t = -1?

Click Here To Show Diagram Code

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

The position simplifies to 0 = { | }.



Why, ACCORDING TO THE DEFINITIONS, does 0 = { | } have neither a Black scaffold nor a White scaffold? Let me try.


Siegel (slightly modified):

II.3.16: "Let G be a short game. The Left stop L(G) and the Right stop R(G) are defined recursively by

L(G) = G if G is equal to a number;
max_G_L (R(G_L)) otherwise;

R(G) = G if G is equal to a number;
min_G_R (L(G_R)) otherwise."

II.5.1: "Let t >= -1. We define G cooled by t, denoted by Gt, as follows. If G is equal to an integer n, then simply Gt = n. Otherwise, put G't = { G_L,t - t | G_R,t + t }. Then Gt = G't, unless there is some t' < t such that G't', is infinitesimally close to a number x. In that case, fix the smallest such t' and put Gt = x."

II.5.2: "The temperature of G, denoted by t(G), is the smallest t >= -1 such that Gt is infinitesimally close to a number."

II.5.3: "The Left and Right scores of G, denoted by Lt(G) and Rt(G), are defined by Lt(G) = L(Gt) and Rt(G) = R(Gt) [...] The thermograph of G is the ordered pair (Lt (G), Rt (G))"

II.5.8: "Let G be a game. We define trajectories β_t(G) and ω_t(G) (the walls of G), for t >= -1, as follows. If G is equal to an integer n, then β_t (G) = ω_t (G) = n for all t. Otherwise, first define the scaffolds β't (G) and ω't (G) by β't (G) = max_G_L (ω_t (G_L) - t) and ω't (G) = min_G_R (β_t (G_R) + t)."


0 = { | } is equal to an integer so, by II.5.1, the cooled game is Gt = 0 for t >= -1. (*)

By II.5.2, the temperature is t(0) = -1.

By (*) and II.3.16, the Black stop is L(Gt) = L(0) = 0 and the White stop is R(Gt) = R(0) = 0. (**)

By II.5.3 and (**), the Black score is Lt(0) = L(Gt) = 0 and the White score is Rt(0) = R(Gt) = 0.

By II.5.8 for t >= -1, 0 = { | } is an integer so the walls are β_t (0) = ω_t (0) = 0 and there are no scaffolds.

Click Here To Show Diagram Code

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

The position simplifies to 1.

1 is equal to an integer so, by II.5.1, the cooled game is Gt = 1 for t >= -1. (*)

By II.5.2, the temperature is t(1) = -1.

By (*) and II.3.16, the Black stop is L(Gt) = L(1) = 1 and the White stop is R(Gt) = R(1) = 1. (**)

By II.5.3 and (**), the Black score is Lt(1) = L(Gt) = 1 and the White score is Rt(1) = R(Gt) = 1.

By II.5.8 for t >= -1, 1 is an integer so the walls are β_t (1) = ω_t (1) = 1 and there are no scaffolds.

Click Here To Show Diagram Code

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

The position simplifies to -1.

-1 is equal to an integer so, by II.5.1, the cooled game is Gt = -1 for t >= -1. (*)

By II.5.2, the temperature is t(-1) = -1.

By (*) and II.3.16, the Black stop is L(Gt) = L(-1) = -1 and the White stop is R(Gt) = R(-1) = -1. (**)

By II.5.3 and (**), the Black score is Lt(-1) = L(Gt) = -1 and the White score is Rt(-1) = R(Gt) = -1.

By II.5.8 for t >= -1, -1 is an integer so the walls are β_t (-1) = ω_t (-1) = -1 and there are no scaffolds.

Is it clear to you that the thermograph of an integer is a vertical mast when t ≥ 0? That means that if an integer has a thermograph when 0 ≥ t ≥ -1 the mast of the thermograph is an integer. For non-ko thermographs there are only three ways that can happen, given that the only values at t = -1 are integers. 1) The thermograph is a vertical mast when t ≥ 1. 2) One wall of the thermograph is a vertical line and the other is a line from the next integer. In that case the lines intersect at t = 0 and the game is an infinitesimal. 3) Both walls of the thermograph are inclined, one from the next integer greater than the mast value, and one from the next integer less than the mast value. These lines also intersect at t = 0. Thus, the only subterranean non-ko thermograph that has an integer mast and is not of an infinitesimal has temperature -1. Does that not answer your question? (Yes, I left a some work for you to do, but you, like me, like to work things out for yourself. )



The simplest forms of all integers do not have thermographs with two walls.

0 = { | }
1 = {{ | }|| }
2 = {{{ | }|| }||| }

etc.

There is no White wall for any non-negative integer in its simplest form.



Doesn't this indicate that the temperature of an integer is -1?

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

Visualize whirled peas.

Everything with love. Stay safe.

I am so busy with other things that I cannot work out this now. Your explanations are constructive while currently I need to understand mainly the defining part. When something is important to me and it is not convincingly worked out, then I may work out it, as you have seen elsewhere:)

I study a bit of thermography in the non-negative range and just want to have a preliminary understanding of the negative part as fall-back.

For what purpose? Making a hypothesis and its limitations:)

As I indicated, I tried to cater to that.



Non-ko thermography of regular ko when 1 > t ≥ 0 will tell you everything about subterranean non-ko thermography when 0 > t ≥ -1, because regular go is subterranean to chilled go.

Integers are basic to non-ko thermography, not the other way around. You don't define integers by their thermographs. Abstractly speaking, that would be possible, but then it would be turtles all the way down. As for ko, superko, and multiple ko thermography, we have to rely upon the rules to tell us how to end play and tell us how to score end positions with unfilled kos and dame. CGT does not tell us. I ran into that problem with my 1998 paper, which is why my discussion of molasses ko got cut in the editing.

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

Visualize whirled peas.

Everything with love. Stay safe.

Has Elwyn Berlekamp invented sentestrat?

Yes, long ago.

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

Visualize whirled peas.

Everything with love. Stay safe.