Thermography

Bill Spight · Post by **Bill Spight** » Tue Nov 17, 2020 3:26 pm

Gérard TAILLE wrote:
Bill Spight wrote:
Gérard TAILLE wrote:It's not a burden for me because I am often very fond of game theories.
Afer having looked at some articles on Hackenbush game analysed with CGT it appears to me that atomic weight is only defined for infinitesimals, I mean only for flowers connected to the ground by a green edge. For all other positions a value might be defined if the game is equal to a number but it is another thing. I did not find any article talking about atomic weight for no infinitesimals game.
Well, maybe not. But the calculation of the atomic weight of an infinitesimal (i.e, by the operational definition of atomic weight) may require finding the atomic weight of non-infinitesimals. For instance, to calculate the atomic weight of {4|*||*}, which is an infinitesimal, requires calculating the atomic weight of {4|*}, which is not. Such is life.

Gérard TAILLE wrote:BTW what are you expected when comparing {4|*} and ↑* ?
Black to play wins all game of the form {4|*} - G when G is an infinitesimal, even if the atomic weight of G is very high!
Yes, but to compare the two we have to consider the result when White plays first, as well.

{4|*} > ↑, but

{4|*} <> ↑*. In fact,

{4|*} <> *

This time I cannot agree Bill.
Assume the game H = {4|*} has an atomic weight n>=0
Let's now take the game G = (n+2)↑
The atomic weight of G is n+2. As a consequence we should have G > H because aw(G) = aw(H) + 2. That means that white cannot win the game G - H.

The question is not that of winning the game, but of getting the last play at temperature 0. Black will win the game, 4 + ↓, but White will get the last play at temperature 0.

The atomic weight theory applies only to infinitesimals, but to define atomic weight for infinitesimals, Conway, Berlekamp, and Guy had to define atomic weight for non-infinitesimals, as well.

Gérard TAILLE · Post by **Gérard TAILLE** » Tue Nov 17, 2020 4:02 pm

Bill Spight wrote: The question is not that of winning the game, but of getting the last play at temperature 0. Black will win the game, 4 + ↓, but White will get the last play at temperature 0.

Let's take the game G = 4 + ↓
G = {3|} + {*|0}
The only white move is in ↓ to reach {3|} then black plays in 3 to reach {2|} and white cannot take the last play can he?

Bill Spight · Post by **Bill Spight** » Tue Nov 17, 2020 4:15 pm

Gérard TAILLE wrote:
Bill Spight wrote: The question is not that of winning the game, but of getting the last play at temperature 0. Black will win the game, 4 + ↓, but White will get the last play at temperature 0.
Let's take the game G = 4 + ↓
G = {3|} + {*|0}
The only white move is in ↓ to reach {3|} then black plays in 3 to reach {2|} and white cannot take the last play can he?

White plays to 4. That is the last play at temperature 0.

Black's reply to 3 loses 1 point. It is at temperature -1.

Gérard TAILLE · Post by **Gérard TAILLE** » Wed Nov 18, 2020 6:41 am

Bill Spight wrote:
Gérard TAILLE wrote: Let's take the game G = 4 + ↓
G = {3|} + {*|0}
The only white move is in ↓ to reach {3|} then black plays in 3 to reach {2|} and white cannot take the last play can he?
White plays to 4. That is the last play at temperature 0.

Black's reply to 3 loses 1 point. It is at temperature -1.

Your are joking Bill don't you? Numbers are of greatest importance because it gives one player n moves while the opponent has none. In a fight to get the last play it is essential you can play in a number.

Take the following example:

Click Here To Show Diagram Code: [go]$$B Black to play $$ --------------------- $$ | O X C C X . . O X | $$ | . X X X X . . O X | $$ | . X . . . . . O . | $$ | . X . . . . . O . | $$ | O O . . . . . O . | $$ | . . . . . . . X X | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ --------------------[/go]

The two points marked are here to compensate white advantage; it is the number 2 for this game.

Click Here To Show Diagram Code: [go]$$B $$ --------------------- $$ | O X . . X . . O X | $$ | . X X X X . . O X | $$ | b X . . . . . O a | $$ | 2 X . . . . . O 3 | $$ | O O . . . . . O 1 | $$ | . . . . . . . X X | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ --------------------[/go]

After

,

if you ignore that a number will give moves for one player then white plays at "b" and gets tedomari. If now you accept to play in number then white cannot get tedomari which makes sense in this example.

Whatever the temperature or the tax you cannot prohibit a play in a number can you?

Bill Spight · Post by **Bill Spight** » Wed Nov 18, 2020 3:13 pm

Gérard TAILLE wrote:
Bill Spight wrote:
Gérard TAILLE wrote: Let's take the game G = 4 + ↓
G = {3|} + {*|0}
The only white move is in ↓ to reach {3|} then black plays in 3 to reach {2|} and white cannot take the last play can he?
White plays to 4. That is the last play at temperature 0.

Black's reply to 3 loses 1 point. It is at temperature -1.
Your are joking Bill don't you? Numbers are of greatest importance because it gives one player n moves while the opponent has none. In a fight to get the last play it is essential you can play in a number.

I was not joking. We were talking about atomic weights. Atomic weights do not predict the result of playing in numbers. That is why every number has an atomic weight of 0.

Take the following example:
$$B Black to play
$$ ---------------------
$$ | O X C C X . . O X |
$$ | . X X X X . . O X |
$$ | . X . . . . . O . |
$$ | . X . . . . . O . |
$$ | O O . . . . . O . |
$$ | . . . . . . . X X |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------
Click Here To Show Diagram Code
[go]$$B Black to play $$ --------------------- $$ | O X C C X . . O X | $$ | . X X X X . . O X | $$ | . X . . . . . O . | $$ | . X . . . . . O . | $$ | O O . . . . . O . | $$ | . . . . . . . X X | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ --------------------[/go]
The two points marked are here to compensate white advantage; it is the number 2 for this game.

It only compensates for White's average advantage.

I take it we are going to play chilled go.

$$B
$$ ---------------------
$$ | O X . . X . . O X |
$$ | . X X X X . . O X |
$$ | b X . . . . . O a |
$$ | 2 X . . . . . O 3 |
$$ | O O . . . . . O 1 |
$$ | . . . . . . . X X |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------
Click Here To Show Diagram Code
[go]$$B $$ --------------------- $$ | O X . . X . . O X | $$ | . X X X X . . O X | $$ | b X . . . . . O a | $$ | 2 X . . . . . O 3 | $$ | O O . . . . . O 1 | $$ | . . . . . . . X X | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ --------------------[/go]
After , , if you ignore that a number will give moves for one player then white plays at "b" and gets tedomari. If now you accept to play in number then white cannot get tedomari which makes sense in this example.

In chilled go the position on the left has the value, 5 + ↑↑*, and the position on the right has the value -5 + {2|0||0|||0}. ↑↑* has an atomic weight of 2, {2|0||0|||0} has an atomic weight of -1. So Black has an advantage of atomic weight 1 at temperature 0, which is what atomic weights are about.

In fact, Black's advantage is enough that Black can get the last play at temperature 0 even if White plays first.

Click Here To Show Diagram Code: [go]$$W $$ --------------------- $$ | O X . . X . . O X | $$ | 6 X X X X . . O X | $$ | 3 X . . . . . O 5 | $$ | 1 X . . . . . O 4 | $$ | O O . . . . . O 2 | $$ | . . . . . . . X X | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ | . . . . . . . . . | $$ --------------------[/go]

Whatever the temperature or the tax you cannot prohibit a play in a number can you?

No, but playing in a number with gote entails an actual loss. Another way of putting that is to say that playing in a number with gote happens at a subzero temperature. CGT has what is called the number avoidance theorem, which indicates that optimal play is not in a number if a non-number option exists. If only numbers are left, then the sum of those numbers is the score of the game. Playing in a number is possible, but unnecessary.

That is why we do not normally consider play in numbers in chilled go, and are only concerned with plays at or above temperature 0.

Gérard TAILLE · Post by **Gérard TAILLE** » Thu Nov 19, 2020 10:41 am

Bill Spight wrote: No, but playing in a number with gote entails an actual loss. Another way of putting that is to say that playing in a number with gote happens at a subzero temperature. CGT has what is called the number avoidance theorem, which indicates that optimal play is not in a number if a non-number option exists. If only numbers are left, then the sum of those numbers is the score of the game. Playing in a number is possible, but unnecessary.

That is why we do not normally consider play in numbers in chilled go, and are only concerned with plays at or above temperature 0.

I agree with with you Bill but we are moving away from my point. Often it is more easier to understand the theory than to understand the difficulty of somebody else.

Bill Spight wrote: The question is not that of winning the game, but of getting the last play at temperature 0.

Let's take this obvious game

Click Here To Show Diagram Code: [go]$$W White to play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O . | $$ | . X . . . . . . O . | $$ | . X . . . . . . O . | $$ | . X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

Click Here To Show Diagram Code: [go]$$W White to play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O 6 | $$ | 5 X . . . . . . O 4 | $$ | 3 X . . . . . . O 2 | $$ | 1 X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

If your goal is only to get tedomari then black wins by the sequence above but it is not correct is it?
The result being not correct the rule of the game must be such that this way of playing gives the win to white.

Let's take a game for which it is time to play infinitesimals.
I agree it is not correct to play in numbers in chilled games but it is also not correct to look only for tedomari.
It seems to me that the correct interpretation stands somewhere beetween our two statements.
In chilled games, when a player passes, then it reamains only numbers.
My view is the following : the best sequence is the one which gives you the best score (the best number) and in case of equality the sequence which gives you tedomari.

With this rule, let's take again the following game

Click Here To Show Diagram Code: [go]$$W White to play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O 6 | $$ | 5 X . . . . . . O 4 | $$ | 3 X . . . . . . O 2 | $$ | 1 X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

The final score is 0 and black gets tedomari

Click Here To Show Diagram Code: [go]$$W White to play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O . | $$ | 4 X . . . . . . O 5 | $$ | 3 X . . . . . . O 2 | $$ | 1 X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

The final score in +1 and white gets tedomari

Due to the final score +1 this second sequence is better than first one. The score of the game has the proeminence over tedomari.

Bill Spight · Post by **Bill Spight** » Thu Nov 19, 2020 12:04 pm

Gérard TAILLE wrote:
Bill Spight wrote: No, but playing in a number with gote entails an actual loss. Another way of putting that is to say that playing in a number with gote happens at a subzero temperature. CGT has what is called the number avoidance theorem, which indicates that optimal play is not in a number if a non-number option exists. If only numbers are left, then the sum of those numbers is the score of the game. Playing in a number is possible, but unnecessary.

That is why we do not normally consider play in numbers in chilled go, and are only concerned with plays at or above temperature 0.
I agree with with you Bill but we are moving away from my point. Often it is more easier to understand the theory than to understand the difficulty of somebody else.

The general topic is atomic weights. I am not sure what your problem is. Or if you actually have one.

Gérard TAILLE wrote:
Bill Spight wrote: The question is not that of winning the game, but of getting the last play at temperature 0.
Let's take this obvious game
$$W White to play
$$ -----------------------
$$ | O X . . . . . . O X |
$$ | O X . . . . . . O . |
$$ | . X . . . . . . O . |
$$ | . X . . . . . . O . |
$$ | . X . . . . . . X X |
$$ | O O . . . . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------
Click Here To Show Diagram Code
[go]$$W White to play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O . | $$ | . X . . . . . . O . | $$ | . X . . . . . . O . | $$ | . X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

$$W White to play
$$ -----------------------
$$ | O X . . . . . . O X |
$$ | O X . . . . . . O 6 |
$$ | 5 X . . . . . . O 4 |
$$ | 3 X . . . . . . O 2 |
$$ | 1 X . . . . . . X X |
$$ | O O . . . . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------
Click Here To Show Diagram Code
[go]$$W White to play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O 6 | $$ | 5 X . . . . . . O 4 | $$ | 3 X . . . . . . O 2 | $$ | 1 X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]
If your goal is only to get tedomari then black wins by the sequence above but it is not correct is it?
The result being not correct the rule of the game must be such that this way of playing gives the win to white.

Well, in chilled go the atomic weight of the position on the right is -2, and the atomic weight on the left is +1. Why is the atomic weight +1?

I am not going to derive it, 1) because it is tedious, and 2) because you have objected to the assumptions of the derivation.

Click Here To Show Diagram Code: [go]$$W White to play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O . | $$ | . X . . . . . . O . | $$ | . X . . . . . . O . | $$ | 1 X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

After

the atomic weight on the left is 0. I don't know whether you agree with that or not. After all, it is a positive infinitesimal, so Black gets the last play locally, assuming correct play. The thing is, speaking informally, Black can ignore

at temperature 0 in chilled go, but not White's next play there.

Click Here To Show Diagram Code: [go]$$W Sente $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O . | $$ | 4 X . . . . . . O . | $$ | 3 X . . . . . . O . | $$ | 1 X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

In similar manner to evaluating mean values of hot positions, atomic weight theory assumes that sente are answered, as a rule. There are execptions, OC, but that's the default.

The position on the right has atomic weight -2. That means that at temperature 0 in chilled go, White can ignore two Black plays, as a rule.

Click Here To Show Diagram Code: [go]$$B Black first $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O . | $$ | . X . . . . . . O 3 | $$ | . X . . . . . . O 1 | $$ | . X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

is not sente, it is ambiguous, as it does not raise the local temperature. White can ignore it. In fact, after

the position on the right has atomic weight 0. It is gote, but the atomic weight is 0, even though White can reply locally.

Not that you do not know this, OC.

The atomic weight of the sum is -2 + 1 = -1. So White has the advantage in the fight for the last play at temperature 0. Getting the last play at temperature 0 is the point of atomic weight theory. That's what it is about.

Now, an advantage of 1 atomic weight is not enough to guarantee getting the last play at temperature 0, given correct play. CGT assumes correct or optimal play, unless otherwise stated. (Edit: OC, in proofs everything is stated.

But the default assumption is correct or optimal play.)

Click Here To Show Diagram Code: [go]$$W Incorrect play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O 6 | $$ | 5 X . . . . . . O 4 | $$ | 3 X . . . . . . O 2 | $$ | 1 X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

is a mistake, so this sequence lies outside the scope of atomic weight theory.

Click Here To Show Diagram Code: [go]$$W Correct play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O . | $$ | 4 X . . . . . . O 5 | $$ | 3 X . . . . . . O 2 | $$ | 1 X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]

is correct, as assumed by the theory as the default. And White does get the last play at temperature 0.

Gérard TAILLE · Post by **Gérard TAILLE** » Thu Nov 19, 2020 3:10 pm

Bill Spight wrote: The general topic is atomic weights. I am not sure what your problem is. Or if you actually have one.

My problem appear when you wrote:
The question is not that of winning the game, but of getting the last play at temperature 0

This phrasing creates some confusion when we reach the time to play infinitesimals because from this point we do not see clearly what is the relevant notion to take into account : the score to win the game, or tedomari to get the last infinitesimal?

Well I resolved my problem by creating my own solution taking into account both score and tedomari but I guess some reader may be not quite clear on this point.

The problem is really to define how to play a chilled game which is not that easy.
At the beginning it is quite simple: in order to have the right to move you must first pay a tax equal to one point. The game ends when both players passes.
Fine, but who will win the game when the two players passe?
Here the problem becomes far more difficult. Let's then assume that everybody understand(!) that all remaining areas on the board are numbers and the score of the game can be calculated just by adding these numbers.
Now I can tell who will win the game:
If the score is > 0 black wins
If the score is < 0 white wins
If the score is = 0 the player who gets tedomari wins.

Bill Spight wrote:
$$W Incorrect play
$$ -----------------------
$$ | O X . . . . . . O X |
$$ | O X . . . . . . O 6 |
$$ | 5 X . . . . . . O 4 |
$$ | 3 X . . . . . . O 2 |
$$ | 1 X . . . . . . X X |
$$ | O O . . . . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------
Click Here To Show Diagram Code
[go]$$W Incorrect play $$ ----------------------- $$ | O X . . . . . . O X | $$ | O X . . . . . . O 6 | $$ | 5 X . . . . . . O 4 | $$ | 3 X . . . . . . O 2 | $$ | 1 X . . . . . . X X | $$ | O O . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]
is a mistake, so this sequence lies outside the scope of atomic weight theory.

You exclude

as being a mistake. I agree with you OC but to be rigourous you must firstly define what means winning a chilled game. That way you can show clearly why

is a mistake.
That is really my point : to show

is a mistake you must take into account the score and not just tedomari.

Bill Spight · Post by **Bill Spight** » Thu Nov 19, 2020 8:24 pm

Gérard TAILLE wrote:
Bill Spight wrote: The general topic is atomic weights. I am not sure what your problem is. Or if you actually have one.
My problem appear when you wrote:
The question is not that of winning the game, but of getting the last play at temperature 0

This phrasing creates some confusion when we reach the time to play infinitesimals because from this point we do not see clearly what is the relevant notion to take into account : the score to win the game, or tedomari to get the last infinitesimal?

I think a better way to have said that is that winning the game is outside the scope of the atomic weight theory. Atomic weight theory is only concerned with the question of getting the last play at temperature 0, given correct play otherwise. It is too much to ask atomic weight theory to address mistakes outside of its scope.

That is really my point : to show is a mistake you must take into account the score and not just tedomari.

That's my point, as well.

Gérard TAILLE · Post by **Gérard TAILLE** » Fri Nov 20, 2020 3:34 am

Bill Spight wrote: I think a better way to have said that is that winning the game is outside the scope of the atomic weight theory. Atomic weight theory is only concerned with the question of getting the last play at temperature 0, given correct play otherwise.

Well it appears a far better way because you added "given correct play otherwise".
Now, "getting the last play at temperature 0, given correct play otherwise" isn't it equivalent to "winning the chilled game at temperature 0" as I defined it in my previous post?

Concerning "atomic weight" I built my own definition which seems to work very well. I guess it is not the definition given by the theory but I do not know what is this last one.

BTW do you know if a position equal to *4 has been discovered in go game?

Bill Spight · Post by **Bill Spight** » Fri Nov 20, 2020 4:03 am

Gérard TAILLE wrote:
Bill Spight wrote: I think a better way to have said that is that winning the game is outside the scope of the atomic weight theory. Atomic weight theory is only concerned with the question of getting the last play at temperature 0, given correct play otherwise.
Well it appears a far better way because you added "given correct play otherwise".
Now, "getting the last play at temperature 0, given correct play otherwise" isn't it equivalent to "winning the chilled game at temperature 0" as I defined it in my previous post?

Without checking, I don't think so, because you can win many chilled games without playing the infinitesimals correctly.

Concerning "atomic weight" I built my own definition which seems to work very well. I guess it is not the definition given by the theory but I do not know what is this last one.

As I said, I have consulted the official method of calculation less often than once a year.

BTW do you know if a position equal to *4 has been discovered in go game?

Nakamura found a *2 in chilled go and I found another one later on. OC, you can construct *3 by adding * and *2, but no other *3 or larger *N has been discovered, AFAIK.

Gérard TAILLE · Post by **Gérard TAILLE** » Fri Nov 20, 2020 6:35 am

Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote: I think a better way to have said that is that winning the game is outside the scope of the atomic weight theory. Atomic weight theory is only concerned with the question of getting the last play at temperature 0, given correct play otherwise.
Well it appears a far better way because you added "given correct play otherwise".
Now, "getting the last play at temperature 0, given correct play otherwise" isn't it equivalent to "winning the chilled game at temperature 0" as I defined it in my previous post?
Without checking, I don't think so, because you can win many chilled games without playing the infinitesimals correctly.

It is little worrying Bill. That means that I probably miss something.
Take my definition of what is a win in a chilling game. When you compare two sequences then
1)if the score of the two sequences is different the best sequence for a player is the one which gives her the best score
2)if the score of the two sequences is equal then the best sequence for a player is the one which gives her tedomari
In my view the first case is there to avoid incorrect play and the second case is there to play correctly the infinitesimals.

IOW I hoped that this definition allows me to detect if infinetisamls are playing correctly.

Now you say that it may not be the case (?). What is your criterias to know if infinitesimals are played correctly or not?

Gérard TAILLE · Post by **Gérard TAILLE** » Fri Nov 20, 2020 8:11 am

Bill Spight wrote: Nakamura found a *2 in chilled go and I found another one later on. OC, you can construct *3 by adding * and *2, but no other *3 or larger *N has been discovered, AFAIK.

Could you show us your *2 position you found Bill?

Bill Spight · Post by **Bill Spight** » Fri Nov 20, 2020 9:33 am

Gérard TAILLE wrote:
Bill Spight wrote:you can win many chilled games without playing the infinitesimals correctly.
It is little worrying Bill. That means that I probably miss something.
Take my definition of what is a win in a chilling game. When you compare two sequences then
1)if the score of the two sequences is different the best sequence for a player is the one which gives her the best score

Even if she does not play the infinitesimals correctly.

Gérard TAILLE wrote:2)if the score of the two sequences is equal then the best sequence for a player is the one which gives her tedomari
In my view the first case is there to avoid incorrect play and the second case is there to play correctly the infinitesimals.

IOW I hoped that this definition allows me to detect if infinetisamls are playing correctly.

Now you say that it may not be the case (?). What is your criterias to know if infinitesimals are played correctly or not?

Well, since infinitesimals occur frequently, even if they don't alway matter, I think that it is important to learn correct play that applies in general, if it exists. You learn that through difference games.

For instance, is it best for Black to play in ↓ or in {1|0||0}? Let's play the difference game. We set up the 0 game where we subtract these two from each other. Then let Black play in {1|0||0} to {1|0} and White play in {0||0|0} to {0|0}, leaving

{1|0} + {0||0|-1} + {0|0||0} + {0|0}

Black to play can win by playing to 1 in {1|0}, leaving 1 plus some infinitesimals.

If White plays in {1|0} to 0, then Black can win by playing to {0|0} in {0|0||0}, leaving {0||0|-1}, which is positive.
If White plays in {0||0|-1} to {0|-1}, then Black can win by playing to {0|0} in {0|0||0}, for a sum equal to 0.
If White plays in either of the other two games Black can win by playing to 1 in {1|0}.

So for Black to play in {1|0||0} instead of ↓ is always correct, even though ↓ has atomic weight -1.

What about the other way around? Is it better for Black to play in {0||0|-1} or in ↑? For the difference game we set up the same 0 game, but this time let Black play in {0||0|-1} and White play in {0|0||0}, leaving

{0||0|0} + {1|0||0}

Obviously, Black can win by playing to {1|0} in {1|0||0}.

White's best chance is to play to {0|0} in {0||0|0}, but then Black can win by playing to {1|0} in {1|0||0}.

So it is always correct for Black to play in {0||0|-1} instead of ↑.

OC, the ko caveat applies.

Bill Spight · Post by **Bill Spight** » Fri Nov 20, 2020 9:39 am

Gérard TAILLE wrote:
Bill Spight wrote: Nakamura found a *2 in chilled go and I found another one later on. OC, you can construct *3 by adding * and *2, but no other *3 or larger *N has been discovered, AFAIK.
Could you show us your *2 position you found Bill?

Click Here To Show Diagram Code: [go]$$ *2 $$ ------------------- $$ . . X O . . X O . . $$ . . X X X O O O . . $$ . . . . . . .. . .[/go]

Life In 19x19

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