Evaluation hodgepodge

[RobertJasiek]

Now, you are on track:)

Why does talking about mean values already imply m(B) ≥ m(G)?

For longer traversal, m(A) - m(G) ≥ m(A) - m(B) need not be fulfilled. It is sufficient for each gain to be at least t.

Kano's example is not exactly miai because, even without ko threat play, playing at C19 can sometimes be correct and did win FJ Dickhut a German Championship game.

[Bill Spight]

Kano's example is not exactly miai because, even without ko threat play, playing at C19 can sometimes be correct and did win FJ Dickhut a German Championship game.

Yes, Kano also assumes that Black cannot win the ko in the corner.

Why does talking about mean values already imply m(B) ≥ m(G)?

You have to ask? If m(G) > m(B), Black plays to C.

[RobertJasiek]

If you only use counts, everything is clear. However, if you use mean values, I have to ask because I am not firm with the CGT theory of mean values.

[Bill Spight]

If you only use counts, everything is clear. However, if you use mean values, I have to ask because I am not firm with the CGT theory of mean values.

Well, mean values are basic. Berlekamp came up with the term, count, to cover both mean values of finite combinatorial games and mast values of kos and superkos. If you stick to finite combinatorial games, go players were calculating mean values well over a century ago.

[Bill Spight]

Also to be clear. If we are considering a game, G, in an ideal environment, G may well consist of more than one local positions (games). A play in any of these positions is local to G. In particular, it may be correct to play suboptimally in one position in order to get the last play in another before a temperature drop.

[RobertJasiek]

Berlekamp came up with the term, count, to cover both mean values of finite combinatorial games and mast values of kos and superkos.

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

[Bill Spight]

Kano's example is not exactly miai because, even without ko threat play, playing at C19 can sometimes be correct and did win FJ Dickhut a German Championship game.

Yes, Kano also assumes that Black cannot win the ko in the corner.

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.

[RobertJasiek]

Yes, this must be the right timimg.

[Harleqin]

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?

[Bill Spight]

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?

Because Kano assumes that the position is miai.

[RobertJasiek]

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?

[Bill Spight]

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

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..

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

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.

I do not even dare to ask for infinitesimals.

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.

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?

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.

[RobertJasiek]

the thermograph of an integer is a mast starting at temperature -1.

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?

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..

Somebody says you have an example for that. What?

[Bill Spight]

the thermograph of an integer is a mast starting at temperature -1.

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.

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.

Neither indicates a mast downwards to T = -1.

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.

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?

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

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..

Somebody says you have an example for that. What?

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

[RobertJasiek]

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.

$$B
$$ ---------------------------
$$ | . . X O O . O O . O X . |
$$ | 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 |
$$ ---------------------------

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