Let me work out a few details of your proofs of the following two propositions:
Proposition 1: {^|0} = *.
Proof:
G := {^|0}.
To prove {^|0} = *, we play the difference game {^|0} - * = 0 <=> {^|0} + * = 0 and below show that, regardless of the starting player, this is always fulfilled.
Code: Select all
G + * = 0
/ \ / \
^ 0 0 0
/ \
0 *
/ \
0 0
Black plays first.
1) Black plays from * to 0, then White plays from G to 0. The result is 0 + 0 = 0.
2a) Black plays from G to ^, then White plays from ^ to *. The result is * + * = 0.
2b) Black plays from G to ^, then White plays from * to 0, then Black plays from ^ to 0. The result is 0 + 0 = 0.
White plays first.
1) White plays from G to 0, then Black plays from * to 0. The result is 0 + 0 = 0.
2) White plays from * to 0, then Black plays G -> ^ -> * -> 0. The result is 0 + 0 = 0.
QED.
***
Proposition 2: {0|v} = *.
Proof: By proposition 1, {^|0} = * so {0|v} = -{^|0} = -* = *. QED.
***
Bill Spight wrote:
{^|0} = {0|0} = *. It seems wrong because ^ > 0. However, {^|0} = {0||0|0|||0} and Black's play to ^ reverses through * to
0. It does so because {^|0} = *
(I prefer to call CGT reversal "traversal".)
For traversal, equality is sufficient, right?
In the game
we consider the sequence G -> ^ -> * -> 0. In this sequence, we traverse from G to 0 iff * <= G. Therefore, to simplify the game to
we must have * <= G. By proposition 1, * = G so * <= G is fulfilled. Therefore, we may do the simplification and have G = G' <=> {^|0} = *. We already know this from proposition 1.
Hence, traversal seems useful here but actually is only useful when we already have proposition 1.
***
$$B
$$ . . . . . . . . .
$$ . . . X X X . . .
$$ . X X X . X X . .
$$ . X . . . . X . .
$$ . X . O . X X . .
$$ . X X O X X . . .
$$ . . . O . . . . .
$$ . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B
$$ . . . . . . . . .
$$ . . . X X X . . .
$$ . X X X . X X . .
$$ . X . . . . X . .
$$ . X . O . X X . .
$$ . X X O X X . . .
$$ . . . O . . . . .
$$ . . . . . . . . .[/go]
As I have discussed earlier, this is {2 1/2^|1/2} = 1 1/2 + {1^|-1}.
1 1/2 + {1^|-1} chills to 1 1/2 + {^|0} = 1 1/2* (by proposition 1).
Is my analysis of the position right now?
More generally, is the following procedure right?
To calculate the count and any infinitesimals of a local endgame,
- calculate the black follower B and white follower W (if necessary, iteratively),
- annotate them as the game {B|W} (if necessary, iteratively),
- extract a number N from the game (write the sum of N and the game modified by subtracting N from each leaf) to ease identification of any infinitesimals,
- chill (which does not affect the extracted number but applies the tax to each leaf of the game according to a player's excess plays from the root along the sequence to the leaf),
- if possible, represent any infinitesimals of the chilled game,
- if possible, simplify the infinitesimals,
- the game's count is the ordinary number calculated and the game's infinitesimals are those calculated.
***
Similarly, the colour-inverse chills to -1 1/2* (by proposition 2).
***
Bill Spight wrote:
RobertJasiek wrote:
$$B
$$ . . . . . . . . .
$$ . . . X X X . . .
$$ . X X X . X X . .
$$ . X . . . . X . .
$$ . X . O . X X . .
$$ . X X O X X . . .
$$ . . . O . . . . .
$$ . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B
$$ . . . . . . . . .
$$ . . . X X X . . .
$$ . X X X . X X . .
$$ . X . . . . X . .
$$ . X . O . X X . .
$$ . X X O X X . . .
$$ . . . O . . . . .
$$ . . . . . . . . .[/go]
I get a chilled value of 1½*.
If so, then two copies should equal 3, since * + * = 0. Let's see.
$$B Black first
$$ . . . . . . . . . . . . . . . . . .
$$ . . . X X X . . . . . . X X X . . .
$$ . X X X . X X . . . X X X . X X . .
$$ . X . 1 3 . X . . . X . 5 4 . X . .
$$ . X . O 2 X X . . . X . O 6 X X . .
$$ . X X O X X . . . . X X O X X . . .
$$ . . . O . . . . . . . . O . . . . .
$$ . . . . . . . . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$B Black first
$$ . . . . . . . . . . . . . . . . . .
$$ . . . X X X . . . . . . X X X . . .
$$ . X X X . X X . . . X X X . X X . .
$$ . X . 1 3 . X . . . X . 5 4 . X . .
$$ . X . O 2 X X . . . X . O 6 X X . .
$$ . X X O X X . . . . X X O X X . . .
$$ . . . O . . . . . . . . O . . . . .
$$ . . . . . . . . . . . . . . . . . .[/go]
2½ + ½ = 3. Check. :)
The play could also go :b1: - :w2:, :b5: - :w6:, :b3: - :w4:. Note that :w2: at 4 would be a mistake. Then :b3: could play at 2 and get 4 pts.
$$W White first
$$ . . . . . . . . . . . . . . . . . .
$$ . . . X X X . . . . . . X X X . . .
$$ . X X X . X X . . . X X X . X X . .
$$ . X . 2 1 . X . . . X . 4 6 . X . .
$$ . X . O 3 X X . . . X . O 5 X X . .
$$ . X X O X X . . . . X X O X X . . .
$$ . . . O . . . . . . . . O . . . . .
$$ . . . . . . . . . . . . . . . . . .
- Click Here To Show Diagram Code
[go]$$W White first
$$ . . . . . . . . . . . . . . . . . .
$$ . . . X X X . . . . . . X X X . . .
$$ . X X X . X X . . . X X X . X X . .
$$ . X . 2 1 . X . . . X . 4 6 . X . .
$$ . X . O 3 X X . . . X . O 5 X X . .
$$ . X X O X X . . . . X X O X X . . .
$$ . . . O . . . . . . . . O . . . . .
$$ . . . . . . . . . . . . . . . . . .[/go]
½ + 2½ = 3. Check. :)
First, I need to understand the structure of your analysis. I think what you trying to do is using the method of multiples.
I am not familiar with the underlying theory. Why is the method of multiples well-defined, that is, why may we conclude from M copies of a game that the count and infinitesmals of the game are: the count and infinitesmals of M copies of the game divided by M?
(When) do we need the assumption "with no kos now or later"?
***
Second, you get the count 3 and no infinitesimals for M = 2 copies of the game. I understand that, for 2 copies, 3 = G + G = 1 1/2 + 1 1/2. So you confirm the count. However, where does the infinitesimal * come from? 3 = G + G = 1 1/2 + 1 1/2 but also 3 = G + G = 1 1/2* + 1 1/2*. With your analysis, you do not, I think, determine 1 1/2* (yet). Instead, so far, you can only determine that 1 1/2 and 1 1/2* belong to the set of candidate solutions. With your analysis, you still need to prove why 1 1/2 is not a solution but 1 1/2* is the only solution. How do you do this to complete your kind of analysis?
***
Bill Spight wrote:
Counts are numbers.
We have a disagreement of terminology. Let's discuss it.
Traditionally (uhm, you introduced it, did you?), a count (in go theory) is a number. Infinitesimals are games.
However, we also work with infinitesimals like we work with ordinary (real, or in go counts, rational) numbers: we do arithmetics with infinitesimals, we compare infinitesimals or express incomparability, we compare infinitesimals with ordinary numbers.
There are purposes of application when we want to perceive infinitesimals as games and other purposes when we want to perceive infinitesimals as numbers.
Infinitesimals can occur when considering an unchilled or a chilled game. So they can be related to counts as well as chilled counts.
My pragmatic suggestion is: be tolerant with the meaning of count (and chilled count). Allow both meanings: 1) count meaning ordinary number and 2) count meaning to consist of its ordinary number and infinitesimal components.
What does Mathematical Go Endgames say? In tables for corridor positions, it says "area" when meaning the infinitesimal component of a count. But area is a misnomer; we use area for the count or score under area scoring.
In the preface, it describes "count" as the "traditional go player's notion of the count" (uhm, but the traditional go player did not have a clear notion in the sense of a term but was just counting numbers for territorial values of regions); uses "the value" as if it were something well understand, does not define it but characterises it by mentioning the count, "a considerable amount of additional information about the local situation" and to depend on only a local analysis of the relevant partial board position.
In chapter 2.1, it speaks of "values" with a meaning related to counts. In chapter 2.3, it speaks of "values" when meaning (possibly chilled) move values.
In chapter 2.4, it uses "values" to refer to [ordinary, which it calls "conventional"] numbers and infinitesimals. Then it speaks of "final scores" being infinitesimals or [ordinary] numbers.
Its most consistent use is the phrase "value(s)" (or, where applicable, "chilled value(s)" to describe terms consisting of an [ordinary] number and (possibly) infinitesimals.
However, "values" is too unspecific. The word can refer to counts, move values, incentives and unrelated other values. We have values consisting of [ordinary] numbers and infinitesimals expressing counts, values consisting of [ordinary] numbers and infinitesimals expressing move values, and values consisting of [ordinary] numbers and infinitesimals expressing incentives.
Although the book is ambiguous about its use of such terms, it agrees with me that "count" (and, if applicable, "chilled count") is a value that can consist of [ordinary] numbers and infinitesimals.
So why not use "count" with this freedom of meaning? Where disambiguation is needed, it can be given.
If you want to restrict counts to numbers, then what do you call the infinitesimal components of values associated with counts? Don't tell me "infinitesimal components of values associated with counts". I prefer to say "the count 3*" rather than "the count 3 associated with the infinitesimal component *".
- 
, 
- 
, 
- 
. Note that 
But, even though we may find it convenient to think in terms of chilled go for non-ko positions, we are not actually playing chilled go. A chilled go value of ½ corresponds to {1|0} by territory scoring. It has a mean value of ½, but each player gains ½ pt. instead of losing ½ pt. A chilled value of ¾ corresponds to {2|1||0}. Each player gains ¾ pt. instead of losing ¼ pt. Note that both ½ and ¾ by chilled go are bounded by 0 and 1 by territory scoring. Thus, with correct play in a chilled go fraction that lies between N and N+1, the result by territory scoring will be N if White plays first and N+1 if Black plays first. That is what we mean by rounding a chilled go fraction.