Part 10/12

White Starts in the Initial Position

White only has non-ko options. Dia. 23 shows White's best non-ko option. White's scaffold is derived from Black's wall of the remaining ko in the position after move 1 by adding the tax T for White 1.

White scaffold:

1/3 + T if T ≥ 1/3

1 - 2T + T = 1 - T if T ≤ 1/3


Dias. 31 - 33: We interpret White's scaffold.

Click Here To Show Diagram Code

[go]$$W Dia. 31: T ≥ 1/3, count = 1/3, result = 1/3 + T
$$ -----------
$$ | . W X X .
$$ | O X X . X
$$ | 1 O X X X
$$ | O O O O X
$$ | . O . O .[/go]

Dia. 31: This is the case T ≥ 1/3 of White's scaffold. After move 1, the players play elsewhere. The marked remaining basic endgame ko has the count 1/3. White's 1 excess play incurs the tax T. Therefore, the result is 1/3 + T.

Click Here To Show Diagram Code

[go]$$W Dia. 32: T ≤ 1/3
$$ -----------
$$ | 2 O X X .
$$ | O X X . X
$$ | 1 O X X X
$$ | O O O O X
$$ | . O . O .[/go]

Click Here To Show Diagram Code

[go]$$W Dia. 33: White 3 elsewhere, prisoners = 1, count = 1, result = 1 - T
$$ -----------
$$ | X 4 X X .
$$ | O X X . X
$$ | O O X X X
$$ | O O O O X
$$ | . O . O .[/go]

Dias. 32 + 33: This is the case T ≤ 1/3 of White's scaffold. The players play locally if they have a legal local play. Due to the prisoner difference 1, we have the count 1. Black's 1 excess play incurs the tax -T. Therefore, the result is 1 - T.

Part 11/12

Thermograph


We have

Black scaffold:

3 - 3T if T ≥ 7/9

2/3 if 1/3 ≤ T ≤ 7/9

1 - T if T ≤ 1/3

White scaffold:

1/3 + T if T ≥ 1/3

1 - T if T ≤ 1/3


Case T ≤ 1/3:

Black wall: 1 - T

White wall: 1 - T


Case 1/3 ≤ T ≤ 7/9:

Compare 2/3 ? 1/3 + T <=> 1/3 ? T

so 2/3 ≤ 1/3 + T <=> 1/3 ≤ T.

Black's scaffold is at most White's scaffold. We are in a cave.

Black wall: 2/3

White wall: 2/3


Case T ≥ 7/9:

Compare 3 - 3T ? 1/3 + T <=> 2 2/3 ? 4T <=> 2/3 ? T

so 3 - 3T < 1/3 + T <=> 2 2/3 < 4T <=> 2/3 < T

Black's scaffold is smaller than White's scaffold. The cave continues.

Set the walls as in the previous, lower case.

Black wall: 2/3

White wall: 2/3


Thermograph for All Cases:

Black wall:

2/3 if T ≥ 1/3. Mast.

1 - T if T ≤ 1/3.

White wall:

2/3 if T ≥ 1/3. Mast.

1 - T if T ≤ 1/3.


Move value and count:

We find equality of the walls for the smallest temperature of the mast. This occurs for the case 1/3 ≤ T ≤ 7/9.

Equality: 2/3 = 1/3 + T <=> 1/3 = T

Move value: M := T = 1/3

Count: 2/3 = 1/3 + 1/3 = 2/3

Part 12/12

Epilogue

Exercise: Correct all my mistakes! If there are none, prove it!

Exercise: Draw the scaffolds and thermographs!

In message #12,
https://www.lifein19x19.com/viewtopic.p ... 29#p262229
Bill Spight has proclaimed the thermograph that now I have calculated for the colour-reversed case.

I have hardly studied strategy yet. The cases T >= 7/9, T = 7/9, 1/3 <= T <= 7/9, T = 1/3, T <= 1/3 must be distingushed for Black's or White's start. Gerard Taille, Bill Spight and Schachus have made first suggestions but a rewriting and completion would be useful. However, maybe this task should be delayed until the other initial positions will have received their accurate calculations so we confirm their counts, can derive the gains and thereby better understand correct strategy.

How can you reach the above position taking into account the rules you explained in your pervious post ?

On a player's move, choose his best among the available of these options:
- best local non-ko play
- his ko option followed by the opponent's local play that does not recapture this ko
- his ko option followed by the opponent's play elsewhere and the player's local play

The three options do not come as rules but as definition of thermographic calculus for generalised thermography. On his turn, the player considers all up to three options - those available. Each option has T is a parameter. Therefore, for every T, the second-moving opponent, if White, minimises among options 2 and 3, then the player, if Black, maximises among this and option 1. For reversed colours, swap min-max.

In part 2 of my messages, only a preliminary study is done to find out whether the initial position is 'simple' (a simple loopy game). In the preliminary study, the choices due to thermographic calculus are not considered yet. Only the basic rules of the game and, I think, the basic ko rule apply. So let Black ko capture, White elsewhere, Black ko capture, White elsewhere, Black connect and you have this subposition, which meets the 'simple' definition's criteria.

Click Here To Show Diagram Code

[go]$$B tenuki tenki
$$ -----------
$$ | 1 5 X X .
$$ | O X X . X
$$ | 3 O X X X
$$ | O O O O X
$$ | . O . O .[/go]

Surely I must have missed a point.
How can you accept to consider a stupid move like black move in the diagram above?
If so then don't you have also to consider all the positions in the following very provocative stupid sequence:

White to play from the initial position:

Click Here To Show Diagram Code

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

Click Here To Show Diagram Code

[go]$$W
$$ -----------
$$ | 3 4 X X .
$$ | 5 X X . X
$$ | X O X X X
$$ | O O O O X
$$ | . O . O .[/go]

Click Here To Show Diagram Code

[go]$$W
$$ -----------
$$ | 7 X X X .
$$ | . X X . X
$$ | 6 O X X X
$$ | O O O O X
$$ | . O . O .[/go]

My view is that when building the tree needed to analyse a position you MUST simply delete all obvious stupid moves in order to reduce drastically the nodes you have to evaluate.

as a consquence the position

Click Here To Show Diagram Code

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

cannot appear in the tree.

Do you really want to put this position in the tree as a follower of the initial position?

The definition of simple (loopy game) does not depend on any classification of strong or weak moves.

Therefore, the position must also be considered.

You have correctly identified my mistake of (besides settled followers) not listing all those positions occurring after what go players call weak (stupid, if you prefer) moves. Also they must be listed and checked for the definition in principle.

In practice, you are right. A strict application of the (almost) formal definitions is impractical in most cases and we must (or want to) simplify analysis by skipping the obviously immaterial, like we do in tactical reading.

Similarly, in Dia. 14, I made the lazy mistake of not considering the non-ko play besides the ko option and letting Black maximise among both. You would say, it has been well done by me because the non-ko play is obviously inferior to the ko option:)

With some experience, one might look for pragmatic simplifications, such as combining the evalation of Dias. 13-15 or not dissecting every remaining basic endgame ko. However, I have wanted to stick application as close to the definitions as I could. First, one needs to understand what one is doing before one max simplify.

Simplification is often tempting but not always valid, especially not in accurate endgame evaluation or CGT. We must always be careful.

The more frightening thing about your White 1 is the possibility of a long cycle without basic ko recapture or suicide. So, strictly speaking, the initial position is not 'simple'. Only after pruning the obviously inferior moves and variations, the remaining game is 'simple'. It will become an exercise to evaluate corner stage ko with advanced ko thermography...;(

You are right Robert, strictly speaking the two stage ko you proposed is not 'simple'. If the goal for you is that you can take into account such position you have to modify something: Firstly by changing some defintions and maybe secondly by adding a procedure to avoid disturbing weak moves. Not that easy to formalize is it?

Obviously all the process you defined in your post allows to draw a thermograph but I do not see clearly what work do you do at each node.
Especially what about the count? In the diagram above you have shown how you derived the count to reach the value count = 2/3. Does that mean that you make this derivation for each node?
If yes can you telll us what is the count of the following position

Click Here To Show Diagram Code

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

Berlekamp invented this and Siegel tried his best to express it consistently in more accessible mathematical annotation but did not quite succeed. However, what we might want, easier practical application by pruning, does not fit easily into such mathematical theory but would require demanding extensions for it. Currently, we are better off by doing such pruning informally.

Ah, I see. The problem is that I have used "count" in two different manners:

1) A diagram's on-board count incl. the prisoners. Especially in a settled position.

2) A count derived during an algebraic calculation from the scaffolds. This presumes you are familiar with basic thermography. I have not explained this algebraic procedure here but presumed that the reader knows it. If not, you know where you can find it:) However, you might try to learn it by looking at my calculations.

Good joke. This will require about the same amount of work as I have applied to the other initial position! The count pops out while calculating the intial positon's thermograh after having analysed the followers and options towards them.

Why, of course. Thermography is a doubly recursive, iterative evaluation on the move options! Similar to tactical reading, except that emphasis is on value calculations.

It is not clear because in the post https://lifein19x19.com/viewtopic.php?p=276760#p276760 you analyse the position

Click Here To Show Diagram Code

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

as a follower and you only derive the black scaffold: 3 - 3T
With my (poor) knowledge of thermography I cannot deduce the count with only this information.
Then it seems you do not need to derive the count for this position and you confirm that in your previous post https://lifein19x19.com/viewtopic.php?p=276787#p276787 where you claimed it would require a great deal of new work

My basic question is the following : do you use count only on the leaves or also through the nodes of the tree by iteration?

If a position occurs as subposition in an evaluation of another initial position, it does not mean that the evaluation as a subposition would be the same as its evaluation as an initial position! The root position, prisoners and ko bans matter!

To evaluate the other initial position, what we do need from this subposition is exactly and just the black scaffold.

<edit> Since we only need the subposition's black scaffold, we do not need its count as if it were the initial position. For the black scaffold, we also do not need an auxiliary count in this subposition. We need the on-board and prisoners count 3, which is the 3 in the term 3 - 3T, in the settled subposition, from which then we inherit the black scaffold of this unsettled subposition. </edit>

The leaves need their on-board and prisoners count.

When eventually we determine the initial position's thermograph, we can also infer its count (more precisely: its mast value, which for God's sake I call "count"; these additional CGT terms would only confuse us; Berlekamp and Spight have also used the word "count"; recall that the count depends on the assumption of no / a komaster and used model of the environment).

For intermediate subpositions, we need their scaffolds and / or walls. The walls might be equal or unequal to the scaffolds for different temperature ranges and the related trajectory segments. Where they are unequal, move value (aka local temperature) and count (aka mast value) are used to derive the walls from the scaffolds. Like the leaf counts, these counts are auxiliary and only used during the analysis of the initial position.

To determine the intrinsic count of a subposition, consider it as an initial position and evaluate it as such. No prisoners or new ko bans during analysis sequences yet.


Edited.

Mast is defined as long as the walls coincide so this whole ko thermograph is called the mast, whose lower part inclines.

Now it seems correct to me.
As you know my approach is different : we start both with a tree and an evaluation of leaves but then you use black and white scafolds through the nodes while I work only on the nodes of the tree by handling a count and a value at each node. The advantage of your approach is that you use a pure recursive approach while my approach is basically recursive EXCEPT when I detect that a move increases the temperature (=> expected sente move).
Anyway I completly agree with the fact that a count for a node depends on the initial position. It is perfectly clear through my approach.
Are the following three points a good understanding of your approach ?
1) the count makes sense only on the leaves and the initial position
2) the value of moves make sense only on initial position
3) for intermediate nodes only black and white scafolds make sense.

Beside the "count" issue I see also an issue with the word "position". For intermediate nodes in the tree I prefer to simply use the word "node" instead of "position" because a node contains in my view some historical characteristics and in particular
1) the identity of the last move in case of a ko option
2) the information saying if the last two moves were an option ko followed by a tenuki
3) the number of prisoners
note 1 : concerning the root node I have no problem to call it a "position"
note 2 : concerning a leave node it seems acceptable to use the word "position" providing you put in the "position" wording the number of prisonners.

In your first post (https://lifein19x19.com/viewtopic.php?p=276756#p276756) you define basic terms :
"In this definition, 'game' means 'position', 'subposition' means 'the position itself or a follow-up position', 'ko' means 'basic ko' or 'local position with alternating 2-play cycle', 'loop' seems to have the intended meaning 'positional cycle of plays', 'Left' means 'Black', 'Right' means 'White', 'option' means 'next move', 'ko option' means 'basic ko capture'"
but the basic word "position" is not defined while constantly used. Maybe you can improve your definitions on this point to avoid ambiguity and misunderstanding.

Compare the stated black and white walls - they are equal!

It is not my method but Berlekamp's method according to Siegel's description. I am just one who currently applies it.

You mention your method but have you stated it somewhere? If alternative methods exist, fine. However, what is it and what is its (proven) scope? Generalised thermography is proven and its scope is simple loopy games (incl. non-ko games).

For intermediate subpositions, move value and count are often needed but you can call them auxiliary move value and count or, in CGT parliance, local temperature and mast value (for ko thermography, mean value would be the wrong term).

Node is a valid term in graph theory or informatics. Siegel uses the term subposition when he studies CGT. CGT explores many games, not just Go, so most CGT terms do not conform to go terms. Left = Black, Right = White etc. Go players use the term follow-up position and (as I have clarified) follower for an unsettled follow-up position. A leaf is a settled position and related to a CGT number. One can assign states to a node aka subposition aka follow-up position.

It is an exaggeration to say that my first post defined terms; rather, I have given some semi-formal descriptions of terms occurring in Siegel's attempted definition. I have not wanted to repeat Siegel's maths annotation but rather tried to explain it briefly for everybody having some thermographic background.