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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #21 Posted: Tue Oct 10, 2017 10:41 am 
Tengen

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QUESTION 7

How to round fractions of local endgames with star, up, down?

Click Here To Show Diagram Code
[go]$$B Example 1: count 1 1/2*
$$ . . . . . . . .
$$ . . . X X X X .
$$ O O O . . . X .
$$ . X . . X X X .
$$ . X X X X . . .
$$ . . . . . . . .[/go]


Click Here To Show Diagram Code
[go]$$B Black follower B = 2 1/2
$$ . . . . . . . .
$$ . . . X X X X .
$$ O O O 1 . . X .
$$ . X . . X X X .
$$ . X X X X . . .
$$ . . . . . . . .[/go]


Click Here To Show Diagram Code
[go]$$W White follower W = 1/2
$$ . . . . . . . .
$$ . . . X X X X .
$$ O O O 1 . . X .
$$ . X . . X X X .
$$ . X X X X . . .
$$ . . . . . . . .[/go]


The black follower has the count B = 2 1/2. The white follower has the count W = 1/2. Therefore, the combinatorial game in the initial position is {B|W} = {2 1/2|1/2} = 1 1/2 + {1|-1}. This chills to 1 1/2 + {0|0} = 1 1/2*.

1 1/2 belongs to this count of the local endgame, which is not an empty corridor but also carries a *.

For an empty corridor, we would round 1 1/2* like 1 1/2: if Black starts, round up to 2; if White starts, round down to 1.

This rounding also has a meaning here: if only this local endgame is on the board, the following sequences occur (for the sake of simplicity, I ignore the dame) with the results B' = 2 or W' = 1, as predicted by the aforementioned rounding:

Click Here To Show Diagram Code
[go]$$B Black starts, result B' = 2
$$ . . . . . . . .
$$ . . . X X X X .
$$ O O O 1 . . X .
$$ . X 2 . X X X .
$$ . X X X X . . .
$$ . . . . . . . .[/go]


Click Here To Show Diagram Code
[go]$$W White starts, result W' = 1
$$ . . . . . . . .
$$ . . . X X X X .
$$ O O O 1 2 . X .
$$ . X . . X X X .
$$ . X X X X . . .
$$ . . . . . . . .[/go]


However, if we only consider the first move, we get the intermediate positions with the count B = 2 1/2 or W = 1/2.

Recall the count 1 1/2* of the initial position. If we treat 1 1/2 and * separately, we can consider rounding of only the remainder * of the local combinatorial game. This is rounded up to 1 if Black starts or rounded down to -1 if White starts. Therefore, the starting Black achieves 1 1/2 + 1 = 2 1/2 or the starting White achieves 1 1/2 - 1 = 1/2. These are the counts of the intermediate positions.

Which rounding makes sense? Why? Are the relations between rounding and counts only accidental or which of the mentioned rounding techniques have a general scope of application?

Click Here To Show Diagram Code
[go]$$B Example 2: count -1 1/2^
$$ . . . . . . . .
$$ . . O O O . . .
$$ . O O . O O . .
$$ . O . . . O O .
$$ . O . X . . O .
$$ . O O X O O O .
$$ . . . X . . . .
$$ . . . . . . . .[/go]


Click Here To Show Diagram Code
[go]$$B Black follower B = -1/2
$$ . . . . . . . .
$$ . . O O O . . .
$$ . O O . O O . .
$$ . O . 1 . O O .
$$ . O . X e . O .
$$ . O O X O O O .
$$ . . . X . . . .
$$ . . . . . . . .[/go]


e = -1/2. f = -1*. Therefore the black follower has the count B = -1/2 and the white follower has the count W = e + f + (-1) = -2 1/2*.

Click Here To Show Diagram Code
[go]$$W White follower W = -2 1/2*
$$ . . . . . . . .
$$ . . O O O . . .
$$ . O O . O O . .
$$ . O . 1 . O O .
$$ . O e X f . O .
$$ . O O X O O O .
$$ . . . X . . . .
$$ . . . . . . . .[/go]


The local endgame in the initial position of Example 2 is {B|W} = {-1/2|-2 1/2*} = -1 1/2 + {1|-1*} = -1 1/2 + {1 | -1 {0|0}}, which chills to -1 1/2 + {0 | 0 {0|0}} = -1 1/2 + {0 |{0|0}} = -1 1/2 + {0||0|0} = -1 1/2 + {0|*} = -1 1/2^. The calculation has already not been so easy and the UP is surprising in a room surrounded mostly by white stones. The real difficulty, however, is the question of correct rounding.

For an empty corridor, we would round -1 1/2^ like 1 1/2: if Black starts, round up to -1; if White starts, round down to -2.

This rounding also has a meaning here: if only this local endgame is on the board, the following sequences occur (for the sake of simplicity, I ignore the dames) with the results B' = -1 or W' = -2, as predicted by the aforementioned rounding:

Click Here To Show Diagram Code
[go]$$B Black starts, result B' = -1
$$ . . . . . . . .
$$ . . O O O . . .
$$ . O O . O O . .
$$ . O . 1 . O O .
$$ . O . X 2 . O .
$$ . O O X O O O .
$$ . . . X . . . .
$$ . . . . . . . .[/go]


Click Here To Show Diagram Code
[go]$$W White starts, result W' = -2
$$ . . . . . . . .
$$ . . O O O . . .
$$ . O O . O O . .
$$ . O . 1 . O O .
$$ . O 3 X 2 . O .
$$ . O O X O O O .
$$ . . . X . . . .
$$ . . . . . . . .[/go]


However, if we only consider the first move, we get the intermediate positions with the count B = -1/2 or W = -2 1/2*.

Recall the count -1 1/2^ of the initial position. If we treat -1 1/2 and ^ separately, we can consider rounding of only the remainder ^ of the local combinatorial game. This is rounded up to 1 if Black starts or rounded down to 0 if White starts. Therefore, the starting Black achieves -1 1/2 + 1 = -1/2 or the starting White achieves -1 1/2 + 0 = -1 1/2. For Black's start, this is the count B = -1/2 of the intermediate position. For White's start, the -1 1/2 is NOT the count W = -2 1/2* of the intermediate position.

Since the rounding behaviour with UP (and DOWN) differs from the behaviour with STAR, I have to ask my questions again for Example 2, but this time for UP and DOWN:

Which rounding makes sense? Why? Are the relations between rounding and counts only accidental or which of the mentioned rounding techniques have a general scope of application?

And you thought that rounding was easy... For infinitesimals, it is a matter of well-definition and correct relation to semantics.

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #22 Posted: Tue Oct 10, 2017 11:54 am 
Judan

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In CGT {0|1} = ½. Whoever plays first loses ½ pt. Also, {0|1||1} = {½|1} = ¾. Whoever plays first loses ¼ pt. Therefore neither player wishes to play in one of these games, and indeed, neither player wishes to play in a number, because playing in a number entails a loss. Rather than play in a number the players can stop play and score the game.

All of that holds true in chilled go, if there is no ko. However, we do not actually play chilled go, although we could. But if you think that the ko rules under territory scoring are a mess, just consider the ko rules for chilled go! ;) 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. :)

Now let's consider ¾*. ¾* = {¾|¾}. Whoever plays first does not change the count. ¾^ = {¾||¾|¾}. Again, the score remains the same regardless of who plays first. What about chilled go? In chilled go, with correct play infinitesimals, like * and ^, are played before numbers. They do not change the score, but may affect who wins the game. Infinitesimals in chilled go have temperature 1 in territory scoring. If the player who plays first at temperature 1 also plays last, he gains 1 pt.; if the other player plays last, the count remains the same (except possibly regarding ko). There is no rounding.

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #23 Posted: Tue Oct 10, 2017 12:16 pm 
Judan

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How does rounding affect territory scoring? (As always, assuming no kos.) If, at temperature 1, the count is a fraction, then with correct play in the chilled go fractions, we can ignore any infinitesimals, as long as we play them first, OC.

For instance, if the chilled value is 1½^, 1½*, or 1½, if Black plays first the correct result will be 2, and if White plays first it will be 1. Similarly, if the count is -2½, if Black plays first the correct result will be -2, and if White plays first it will be -3, regardless of any chilled go infinitesimal.

Edit: To any reader to whom that is not clear, if Black plays first and last at temperature 1, she will gain 1 pt. But then White will play first in the "fraction", and will round the score down to the next integer. For instance, if the count is 1½, Black will make it 2½, and then White will "round it down" to 2. But suppose that Black plays first at temperature 1 but White plays last. Then the count will stay the same but then Black will round up to the next integer. If the count is 1½, Black to play will leave it unchanged at 1½, and then Black will round it up to 2. The result will be the same in either case.

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Last edited by Bill Spight on Wed Oct 11, 2017 10:37 pm, edited 1 time in total.
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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #24 Posted: Wed Oct 11, 2017 9:53 pm 
Tengen

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QUESTION 8:

Can TINY_x - TINY_x-1 be simplified?

Can TINY_x-1 - TINY_x be simplified?

Can TINY_x - TINY_y be simplified with x <> y?

Can MINY_x - MINY_x-1 be simplified?

Can MINY_x-1 - MINY_x be simplified?

Can MINY_x - MINY_y be simplified with x <> y?

Can 0^N|TINY_x - 0^N-1|TINY_x be simplified?

Can 0^N-1|TINY_x - 0^N|TINY_x be simplified?

Can MINY_x|0^N - MINY_x|0^N-1 be simplified?

Can MINY_x|0^N-1 - MINY_x|0^N be simplified?

I ask because some of these expressions can occur in the values of incentives and Mathematical Go Endgames does not address this topic, except for pointing out that it can become complicated.

If simplifications are not available, how do we compare two incentives having such differences? If know that 0 < TINY_y << TINY_x for y > x and 0 > MINY_y >> MINY_x for y > x.

Also we have TINY_x = -MINY_x.

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #25 Posted: Wed Oct 11, 2017 11:11 pm 
Judan

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Probably.

For instance,

Tiny_1 - Tiny_2 = {{Tiny_1 + {2|0}} || Tiny_1, {0|-1}}

If I haven't made a mistake. ;)

We get the left follower because Black only plays in Miny_2. We get the right followers because White can play in either Tiny_1 or Miny_2. Black's reverse sente in Tiny_1 is dominated. Also, if White plays in Tiny_1, Black plays in Miny_2 with sente.

I'm not sure if we want to call the right side of that equation a simplification of the left side. ;)

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #26 Posted: Thu Oct 12, 2017 1:47 am 
Tengen

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What about approximations? Smaller / larger than some infinitesimal plus-minus some infinitesimal_epsilon?

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #27 Posted: Thu Oct 12, 2017 2:23 am 
Judan

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IIRC, Tiny_1 - Tiny_2 ≅ Tiny_1.

Etc.

Anyway, the key approximation is uppitiness.

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #28 Posted: Mon Oct 16, 2017 12:08 am 
Tengen

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QUESTIONS 9:

Click Here To Show Diagram Code
[go]$$B Position 1
$$| X X . . .
$$| O X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .
$$| . . . . .[/go]


The local endgame in position 1 is the game {2|0} with the count C = (2+0)/2 = 1 and move value M = (2-0)/2 = 1.

Extracting the count, we can write the game as {2|0} = 1 + {1|-1}, which chills to 1 + {0|0} = 1 + * = 1*.

Therefore, we have the chilled count c = 1* and chilled move value m = 1*. Rounding of c in favour of the starting player predicts the outcome. So far so clear.

Click Here To Show Diagram Code
[go]$$B Position 2
$$| X X . . .
$$| . X . . .
$$| O X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .[/go]


The local endgame in position 2 is the game {3|0} with the count C = (3+0)/2 = 1 1/2 and move value M = (3-0)/2 = 1 1/2.

Extracting the count, we can write the game as {3|0} = 1 1/2 + {1 1/2|-1 1/2}, which chills to 1 1/2 + {1/2|-1/2}. This does not contain *.

Does this mean that chilling does not add any infinitesimal to the count?

Rounding C in favour of the starting player predicts the outcomes 2 for Black's start or 1 for White's start. So with rounding, we do not predict the correct outcomes.

Using the count and move value, we can predict the outcomes: C + M = 1 1/2 + 1 1/2 = 3 if Black starts and C - M = 1 1/2 - 1 1/2 = 0 if White starts. (Dame ignored.)

Back to position 1, we can predict the outcomes: C + M = 1 + 1 = 2 if Black starts and C - M = 1 - 1 = 0 if White starts.

Why does chilling as an alternative for predicting the outcome work for position 1 but does not work for position 2? Or how to use chilling so that it also works for position 2?

Click Here To Show Diagram Code
[go]$$B Position 3
$$| X X . . .
$$| O X . . .
$$| O X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .[/go]


The local endgame in position 3 is the game {4|0} with the count C = (4+0)/2 = 2 and move value M = (4-0)/2 = 2.

Extracting the count, we can write the game as {4|0} = 2 + {2|-2}, which chills to 2 + {1|-1}. The right summand is not *.

If we extract 3, we can write the game as {4|0} = 3 + {1|-3}, which chills to 3 + {0|-2}. Again, the right summand is not *. Extracting a number different from the count cannot produce *, either.

Does this mean that chilling does not add any infinitesimal to the count?

Rounding C in favour of the starting player predicts the outcomes 2 for Black's start or 2 for White's start. So with rounding, we do not predict the correct outcomes.

Using the count and move value, we can predict the outcomes: C + M = 2 + 2 = 4 if Black starts and C - M = 2 - 2 = 0 if White starts.

Why does chilling as an alternative for predicting the outcome work for position 1 but does not work for position 3? Or how to use chilling so that it also works for position 3?

***

I study positions 2 and 3 because they occur as white followers for positions with tiny:

Click Here To Show Diagram Code
[go]$$B Position 4
$$| X X . . .
$$| . X . . .
$$| O X . . .
$$| . X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .[/go]


To calculate White's incentive in the local endgame in position 4, we need to know whether the chilled count of the local endgame in position 2 contains a *. Does it?

I think that the chilled count of the local endgame in position 4 is c = 3Tiny_1. Right?

Click Here To Show Diagram Code
[go]$$B Position 5
$$| X X . . .
$$| O X . . .
$$| O X . . .
$$| . X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .[/go]


To calculate White's incentive in the local endgame in position 5, we need to know whether the chilled count of the local endgame in position 3 contains a *. Does it?

I think that the chilled count of the local endgame in position 5 is c = 4Tiny_2. Right?

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #29 Posted: Mon Oct 16, 2017 2:16 am 
Judan

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RobertJasiek wrote:
QUESTIONS 9:

Click Here To Show Diagram Code
[go]$$B Position 1
$$| X X . . .
$$| O X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .
$$| . . . . .[/go]


The local endgame in position 1 is the game {2|0} with the count C = (2+0)/2 = 1 and move value M = (2-0)/2 = 1.


That is average move value, the same as the miai value.

Quote:
Extracting the count, we can write the game as {2|0} = 1 + {1|-1}, which chills to 1 + {0|0} = 1 + * = 1*.


OK. :)

Quote:
Therefore, we have the chilled count c = 1* and chilled move value m = 1*.


1* = 1 + * = {1|1}, which is a game with a count of 1. The average move value in that game is (1-1)/2 = 0.

The incentive for Black to move is the difference between the result of the Black move minus the original game, or

1 - 1* = *

The incentive for White to move is the difference between the original game minus the result after the White move, or

1* - 1 = *

Quote:
Rounding of c in favour of the starting player predicts the outcome. So far so clear.


What rounding?

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #30 Posted: Mon Oct 16, 2017 2:50 am 
Judan

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RobertJasiek wrote:
Click Here To Show Diagram Code
[go]$$B Position 2
$$| X X . . .
$$| . X . . .
$$| O X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .[/go]


The local endgame in position 2 is the game {3|0} with the count C = (3+0)/2 = 1 1/2 and move value M = (3-0)/2 = 1 1/2.


The local endgame is {3 || 0 | 0}. Don't forget the dame. The mean value of {0|0} = 0, and the mean value of {3|0} = 1½. The average move value = 1½.

Quote:
Extracting the count, we can write the game as {3|0} = 1 1/2 + {1 1/2|-1 1/2}, which chills to 1 1/2 + {1/2|-1/2}. This does not contain *.


{3||0|0} = {3|*}, which chills to {2|1}, or 1½ + {½|-½}.

Quote:
Does this mean that chilling does not add any infinitesimal to the count?


The chilled game does not equal the count plus an infinitesimal.

Quote:
Rounding C in favour of the starting player predicts the outcomes 2 for Black's start or 1 for White's start. So with rounding, we do not predict the correct outcomes.


2 and 1 are the correct outcomes in the chilled game. :)

Edit: To be clear, Berlekamp and Wolfe apply rounding to unchilled games with local temperature less than 1. The chilled game, {2|1}, has a temperature less than 1, so you could call this rounding, but they apply rounding to the unchilled game. and the temperature of {3|*} is greater than 1.

Quote:
Using the count and move value, we can predict the outcomes: C + M = 1 1/2 + 1 1/2 = 3 if Black starts and C - M = 1 1/2 - 1 1/2 = 0 if White starts. (Dame ignored.)


We can predict the mean values of the outcomes. That means, as you say, that we ignore the dame, or *.

Quote:
Back to position 1, we can predict the outcomes: C + M = 1 + 1 = 2 if Black starts and C - M = 1 - 1 = 0 if White starts.


Right.

Quote:
Why does chilling as an alternative for predicting the outcome work for position 1 but does not work for position 2? Or how to use chilling so that it also works for position 2?


Chilling is not an alternative for predicting the outcome. Rounding does work for position 2. It predicts the outcome in the chilled game, {2|1}. (Edit: As noted above, Berlekamp and Wolfe do not bother with rounding chilled games.) There is no rounding for position 1. The miai value for the chilled game, 1*, is 0. Using that miai value predicts the mean result of the chilled game, which is 1, regardless of who plays first. You used the miai value for the unchilled game to get the result of the unchilled game. That's why you thought that chilling "worked" for position 1.

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Last edited by Bill Spight on Mon Oct 16, 2017 3:47 am, edited 1 time in total.
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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #31 Posted: Mon Oct 16, 2017 3:04 am 
Judan

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RobertJasiek wrote:
Click Here To Show Diagram Code
[go]$$B Position 3
$$| X X . . .
$$| O X . . .
$$| O X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .[/go]


The local endgame in position 3 is the game {4|0} with the count C = (4+0)/2 = 2 and move value M = (4-0)/2 = 2.

Extracting the count, we can write the game as {4|0} = 2 + {2|-2}, which chills to 2 + {1|-1}. The right summand is not *.

If we extract 3, we can write the game as {4|0} = 3 + {1|-3}, which chills to 3 + {0|-2}. Again, the right summand is not *. Extracting a number different from the count cannot produce *, either.

Does this mean that chilling does not add any infinitesimal to the count?


The local game is {4|0}, which chills to {3|1} = 2 + {1|-1}. There is no infinitesimal.

Quote:
Rounding C in favour of the starting player predicts the outcomes 2 for Black's start or 2 for White's start. So with rounding, we do not predict the correct outcomes.


There is no rounding.

Quote:
Using the count and move value, we can predict the outcomes: C + M = 2 + 2 = 4 if Black starts and C - M = 2 - 2 = 0 if White starts.


Yes.

Quote:
Why does chilling as an alternative for predicting the outcome work for position 1 but does not work for position 3? Or how to use chilling so that it also works for position 3?


Chilling is not an alternative for predicting the outcome. It did not "work" for position 1, as you think.

More later. :)

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #32 Posted: Mon Oct 16, 2017 4:00 am 
Tengen

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Bill, many thanks for your, as usual, very helpful explanations!

Now, I realise that I must separate analysis and values of unchilled and chilled game carefully. MGE chilling is thrice as difficult as I feared: 1) understanding the unchilled game, 2) the chilled game and 3) how the findings for the chilled game provide information for the unchilled game.

Although I am awaiting your remaining answers, here is yet another question: you wrote that temperature theory was easier than CGT chilling but can linear algebra (no, not thermographs, which I perceive yet another layer of complication) representing temperature equations solve everything that MGE solves with chilling?

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Post #33 Posted: Mon Oct 16, 2017 8:15 am 
Judan

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RobertJasiek wrote:
Bill, many thanks for your, as usual, very helpful explanations!


De nada. :)

Quote:
Now, I realise that I must separate analysis and values of unchilled and chilled game carefully. MGE chilling is thrice as difficult as I feared: 1) understanding the unchilled game, 2) the chilled game and 3) how the findings for the chilled game provide information for the unchilled game.


Chilled go is a simplification of territory scoring, just as some forms of territory scoring, such as Spight or Lasker-Maas or Ikeda scoring may be regarded as simplifications of area scoring. And, with some exceptions, correct play in chilled go is also correct play by territory scoring, just as, with some exceptions, correct play by territory scoring is also correct play by area scoring. The exceptions, OC, have to do with ko.

So chilled go without kos is a simpler game than territory scoring, and that simplicity can help with analysis and with finding correct play. However, there are plays that are incorrect in chilled go but are still correct by territory scoring. Examples abound, even in pro play. Why not? They have read the game out, anyway. Chilled go also has something that territory go does not have, and that is a variety of infinitesimals. A dame is an infinitesimal, and you can construct others, but in modern territory go infinitesimals can be ignored. In chilled go they can be crucial to winning.

If you know CGT, you already know a good bit about infinitesimals, which you can immediately apply to chilled go. But most go players do not know CGT, and so chilling does not have that advantage for them. Go players do know that getting the last play can be important, but are generally clueless about how to go about doing that, except by reading the game out.

Now, in chilled go play stops once we reach a number, which may be a fraction. This may be useful to go players, because it means that, except in ko situations, once the global temperature drops below 1, they can generally stop reading and round the unchilled result up or down to the nearest integer, depending on who has the play. It is true that some positions with temperature below 1 are tricky, but that bridge can usually be crossed later. :)

Quote:
Although I am awaiting your remaining answers, here is yet another question: you wrote that temperature theory was easier than CGT chilling but can linear algebra (no, not thermographs, which I perceive yet another layer of complication) representing temperature equations solve everything that MGE solves with chilling?


Well, naive temperature theory, which says to make the hottest play, is easier than working out unfamiliar infinitesimals, but is more prone to error. Any time you rely upon mean values you introduce the possibility of error. Finding the mean value and temperature of a game is, as I have noted, a form of defuzzification, with the temperature as the maximum error when there are no kos. So using algebra with mean values and temperatures may improve on the naive theory, but will still produce errors. Chilling makes getting exact solutions easier. :)

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #34 Posted: Mon Oct 16, 2017 8:33 am 
Judan

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RobertJasiek wrote:
I study positions 2 and 3 because they occur as white followers for positions with tiny:

Click Here To Show Diagram Code
[go]$$B Position 4
$$| X X . . .
$$| . X . . .
$$| O X . . .
$$| . X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .[/go]


To calculate White's incentive in the local endgame in position 4, we need to know whether the chilled count of the local endgame in position 2 contains a *. Does it?


To get the miai value you need count, which does not contain any infinitesimal. To get the incentive you need the game. The chilled game in position 2 does not contain a *.

Quote:
I think that the chilled count of the local endgame in position 4 is c = 3Tiny_1. Right?


The count is 3, in both chilled go and territory scoring.

Quote:
Click Here To Show Diagram Code
[go]$$B Position 5
$$| X X . . .
$$| O X . . .
$$| O X . . .
$$| . X . . .
$$| . X . . .
$$| O O . . .
$$| . . . . .
$$| . . . . .[/go]


To calculate White's incentive in the local endgame in position 5, we need to know whether the chilled count of the local endgame in position 3 contains a *. Does it?


The chilled game in position 2 does not contain a *.

Quote:
I think that the chilled count of the local endgame in position 5 is c = 4Tiny_2. Right?


The count is 4.

More on rounding later. :)

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #35 Posted: Mon Oct 16, 2017 9:51 am 
Judan

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Why Berlekamp and Wolfe do not apply rounding to chilled go games, but to regular go games under territory scoring, and only to the whole board

The game, {2|1}, has a mean value of 1½ and a temperature of ½. It may be written as 1½ + {½ | -½}. If Black plays first the result is 2, if White plays first the result is 1. Rounding up or down to the nearest integer works.

However, rounding does not work for the game, {¾ | -¼}. It may be written as ¼ + {½ | -½}. If Black plays first the result is ¾, not 1, and if White plays first the result is -¼, not 0. {¾ | -¼} could be a chilled go game, which is why rounding does not always work in chilled go.

But {¾ | -¼} is not a game under territory scoring, because territory scores are integers. (Yes, it is possible to construct positions which, in theory, should be fractions by territory scoring, but they involve kos and are not, therefore, combinatorial games.)

Rounding works for territory scoring because the scores are integers. Rounding requires these additional conditions:

1) The game is a CGT game, i.e., without any ko.
2) The temperature, t, of the game is such that 0 < t < 1.
3) The mean value of the game lies between two integers.

Rounding is useful when chilling because we stop play in chilled go when the game is a number, and if that number is fractional, play will continue in regular go under territory scoring. Rounding then tells us what the result will be with correct play under territory scoring. We do not have to read anything out, unless we don't know correct play. We only apply rounding to the whole board, because reaching a number in a local position does not necessarily stop play. There may be other places left to play in.

For example, suppose that we end play in chilled go with a result of ¾. In regular go that is the game, {2|1||0}, ignoring dame. White to play moves to 0, Black to play moves to {2|1} and then White continues to 1. We do not have to work this out. All we have to do is round down to 0 if White has the move and round up to 1 if Black has the move. Easy. :)

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Last edited by Bill Spight on Mon Oct 16, 2017 2:50 pm, edited 1 time in total.
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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #36 Posted: Mon Oct 16, 2017 12:41 pm 
Tengen

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Extraordinarily clear explanation!

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #37 Posted: Mon Oct 16, 2017 2:50 pm 
Judan

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Glad you like. :D

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #38 Posted: Tue Oct 17, 2017 11:48 am 
Tengen

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QUESTION 10:

Suppose the game {2|0} chilling to {1|1}. From the chilled game, we extract a count, 1, to identify the infinitesimal * in the sum 1 + *.

Which count do we extract? That of the game or that of the chilled game?

Is the count of the game equal to the count of the chilled game if a) there are no kos and b) the count of the chilled game is a number possibly plus infinitesimals? Exactly when in general are the two counts equal?

If there are no kos, is it possible (how) that the count of the chilled game is not a number possibly plus infinitesimals?

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #39 Posted: Tue Oct 17, 2017 2:39 pm 
Judan

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Chilling does not affect the count. :)

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 Post subject: Re: Studying Microendgame and Infinitesimals
Post #40 Posted: Tue Oct 17, 2017 8:48 pm 
Tengen

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Ok.

Is there a theorem for this, maybe in Aaron Siegel's CGT book?

So you are saying that even local kos and global kos do not affect the count when chilling?

***

How about the move values? Is the move value of a game always 1 larger than the move value of the chilled-by-1 game? Is there a theorem for this? Independent of any kos?

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