Thermography

Talk about improving your game, resources you like, games you played, etc.
Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
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. ;)
If I understand correctly your example is equivalent to the following problem:
Click Here To Show Diagram Code
[go]$$W
$$ -----------------------
$$ | . O . . . . . . . . |
$$ | X O . . . . . . . . |
$$ | . O . . . . . . . . |
$$ | a O . . . . . . . . |
$$ | X X . . . . . . . . |
$$ | X X . . . . . . . . |
$$ | b O . . . . . . . . |
$$ | . O . . . . . . . . |
$$ | X O . . . . . . . . |
$$ -----------------------[/go]
You prove by a difference game that a black move at "a" dominates a black move at "b".
OC your example is relevant because with the chilled game I cannot distinguish between the sequence beginning by "a" and the sequence beginning by "b".
But there are no contradiction. If the chilled games cannot distinguish between the two sequences you can always try a difference game OC.

Now let me add a * to the game
Click Here To Show Diagram Code
[go]$$W
$$ -----------------------
$$ | . O X . X . . . . . |
$$ | X O O O X . . . . . |
$$ | . O . . . . . . . . |
$$ | a O . . . . . . . . |
$$ | X X . . . . . . . . |
$$ | X X . . . . . . . . |
$$ | b O . . . . . . . . |
$$ | . O . . . . . . . . |
$$ | X O . . . . . . . . |
$$ -----------------------[/go]
Now you can see clearly that the chilled game beginning by black "a" is better than the chilled game beginning by black "b" and you will be able to play correctly at "a".
Two points here:
1) A chilled game is easier to visualize in practice than a difference game (at least for me). If the chilled game is able to distinguish between two sequences then it's fine and I am happy to avoid playing a difference game
2) If the chilled game is not able to distinguish between two sequences you have two solutions:
2.1) you play at random one or the other sequence. You know that you may not play the infinitesimals in the best way but you know that for this specific game it does not harm
2.2) you try a difference game to be sure to play the infinitesimal in the best possible way. From a theoritical point of view it is far better but in practice it is not quite easy.
Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
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]
:)
Fine Bill :clap:

Do you know also an *2 position for semeai where we use cooling by 2?
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Re: Thermography

Post by Bill Spight »

Bill Spight wrote: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. ;)
Gérard TAILLE wrote:If I understand correctly your example is equivalent to the following problem:
Click Here To Show Diagram Code
[go]$$W
$$ -----------------------
$$ | . O . . . . . . . . |
$$ | X O . . . . . . . . |
$$ | . O . . . . . . . . |
$$ | a O . . . . . . . . |
$$ | X X . . . . . . . . |
$$ | X X . . . . . . . . |
$$ | b O . . . . . . . . |
$$ | . O . . . . . . . . |
$$ | X O . . . . . . . . |
$$ -----------------------[/go]
You prove by a difference game that a black move at "a" dominates a black move at "b".
OC your example is relevant because with the chilled game I cannot distinguish between the sequence beginning by "a" and the sequence beginning by "b".
But there are no contradiction. If the chilled games cannot distinguish between the two sequences you can always try a difference game OC.
Well, usually difference games are homework. :) There are many go infinitesimals that occur again and again. These two plays, and plays like them, are common. If you know which plays are dominated, you can reduce your reading effort.
Gérard TAILLE wrote:Now let me add a * to the game
Click Here To Show Diagram Code
[go]$$W
$$ -----------------------
$$ | . O X . X . . . . . |
$$ | X O O O X . . . . . |
$$ | . O . . . . . . . . |
$$ | a O . . . . . . . . |
$$ | X X . . . . . . . . |
$$ | X X . . . . . . . . |
$$ | b O . . . . . . . . |
$$ | . O . . . . . . . . |
$$ | X O . . . . . . . . |
$$ -----------------------[/go]
Now you can see clearly that the chilled game beginning by black "a" is better than the chilled game beginning by black "b" and you will be able to play correctly at "a".
Two points here:
1) A chilled game is easier to visualize in practice than a difference game (at least for me). If the chilled game is able to distinguish between two sequences then it's fine and I am happy to avoid playing a difference game
2) If the chilled game is not able to distinguish between two sequences you have two solutions:
2.1) you play at random one or the other sequence. You know that you may not play the infinitesimals in the best way but you know that for this specific game it does not harm
2.2) you try a difference game to be sure to play the infinitesimal in the best possible way. From a theoritical point of view it is far better but in practice it is not quite easy.
If there are many go infinitesimals on the board and one of them is unfamiliar, it may be easier to visualize a difference game between a play in it and a play in one of its rivals than to read the game out. Especially if you have experience with difference games. For instance, before reducing your first new infinitesimal, {4|*||*}, it was clear to me that it was less than ↑↑. :) Like ↑, it is confused with *, but it is greater than ↑. From that we can deduce that White should prefer to play in it than in ↑. In fact, in the difference game, White plays to * in one game and Black plays to * in another, so the difference game reduces to a comparison of the two games. :) All of this takes a matter of seconds to do in your head, if you are used to doing difference games. :)
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At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote: If there are many go infinitesimals on the board and one of them is unfamiliar, it may be easier to visualize a difference game between a play in it and a play in one of its rivals than to read the game out. Especially if you have experience with difference games. For instance, before reducing your first new infinitesimal, {4|*||*}, it was clear to me that it was less than ↑↑. :) Like ↑, it is confused with *, but it is greater than ↑. From that we can deduce that White should prefer to play in it than in ↑. In fact, in the difference game, White plays to * in one game and Black plays to * in another, so the difference game reduces to a comparison of the two games. :) All of this takes a matter of seconds to do in your head, if you are used to doing difference games. :)
Interesting Bill. I tried to generalize this:
Let's suppose G > H and GR <= HR where GR and HR are right options of G and H, then for white GR dominates HR.
Is it correct Bill?
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote: If there are many go infinitesimals on the board and one of them is unfamiliar, it may be easier to visualize a difference game between a play in it and a play in one of its rivals than to read the game out. Especially if you have experience with difference games. For instance, before reducing your first new infinitesimal, {4|*||*}, it was clear to me that it was less than ↑↑. :) Like ↑, it is confused with *, but it is greater than ↑. From that we can deduce that White should prefer to play in it than in ↑. In fact, in the difference game, White plays to * in one game and Black plays to * in another, so the difference game reduces to a comparison of the two games. :) All of this takes a matter of seconds to do in your head, if you are used to doing difference games. :)
Interesting Bill. I tried to generalize this:
Let's suppose G > H and GR <= HR where GR and HR are right options of G and H, then for white GR dominates HR.
Is it correct Bill?
You are right. :)

Let's work it out. :) We have the sum, G + H, and we want to compare a Right (White) move in G to GR with one in H to HR.

If H ≤ G and

GR ≤ HR

Then H + GR ≤ G + HR

I.e., in G + H, White's option, GR + H, dominates the option, HR + G.
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Re: Thermography

Post by Gérard TAILLE »

Another theoritical question.
Let's take a game G and let's call L(G) and and R(G) the left stop and the right stop.
Is it possible to have L(G) < R(G) and G not a number?
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:Another theoritical question.
Let's take a game G and let's call L(G) and and R(G) the left stop and the right stop.
Is it possible to have L(G) < R(G) and G not a number?
Not if G is a combinatorial game.

For combinatorial games:

if L(G) < R(G) then G is a number
if L(G) = R(G) then G = L(G) or G = L(G) + an infinitesimal
if L(G) > R(G) then L(G) ≥ m(G) ≥ R(G) where m(G) is the mean value of G

:)
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:Another theoritical question.
Let's take a game G and let's call L(G) and and R(G) the left stop and the right stop.
Is it possible to have L(G) < R(G) and G not a number?
Not if G is a combinatorial game.

For combinatorial games:

if L(G) < R(G) then G is a number
if L(G) = R(G) then G = L(G) or G = L(G) + an infinitesimal
if L(G) > R(G) then L(G) ≥ m(G) ≥ R(G) where m(G) is the mean value of G

:)
OK Bill. Thank you.
BTW what is the best procedure to prove that a game is a number or not?
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:BTW what is the best procedure to prove that a game is a number or not?
First, if G has no Right option or if G has no Left option, G is a number, in fact, it is an integer.

Second, if all of the Left options of G are less than all of its Right options, then G is a number.

Third, even if the first two conditions do not hold, reduce G to its simplest form, and check if they hold.

Or, as a practical matter, you can guess a number, N, and see if it is equal to G by playing G - N.

-----

Example:

{*|*} does not satisfy either of the first two conditions, but since * is an infinitesimal, if {*|*} is any number, it is 0.

Let's play {*|*} - 0 = {*|*}

If Left plays to *, White plays to 0 and wins.
If Right plays to *, Left plays to 0 and wins.

So {*|*} = 0.

-----

Is {↑*|↓*} a number? ↑* > ↓* , so we might think not. However, both ↑* and ↓* are confused with 0, so

{↑*|↓*} = 0.
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:BTW what is the best procedure to prove that a game is a number or not?
First, if G has no Right option or if G has no Left option, G is a number, in fact, it is an integer.

Second, if all of the Left options of G are less than all of its Right options, then G is a number.

Third, even if the first two conditions do not hold, reduce G to its simplest form, and check if they hold.

Or, as a practical matter, you can guess a number, N, and see if it is equal to G by playing G - N.

-----

Example:

{*|*} does not satisfy either of the first two conditions, but since * is an infinitesimal, if {*|*} is any number, it is 0.

Let's play {*|*} - 0 = {*|*}

If Left plays to *, White plays to 0 and wins.
If Right plays to *, Left plays to 0 and wins.

So {*|*} = 0.

-----

Is {↑*|↓*} a number? ↑* > ↓* , so we might think not. However, both ↑* and ↓* are confused with 0, so

{↑*|↓*} = 0.
Nice answer Bill. Thank you.
Guessing the number x = G seems to me a very good way of proving G is a number.
What about proving G is not a number?
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:What about proving G is not a number?
If L(G) > R(G) then G is not a number.

If L(G) = R(G) != G then G is not a number
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:What about proving G is not a number?
If L(G) > R(G) then G is not a number.

If L(G) = R(G) != G then G is not a number
Oops it does not help very much Bill.

Definition :Let G be a short game. We define numbers L(G) and R(G), the left and right stops of G, as follows.
If G is (equal to) a number, we set L(G) = R(G) = G. Otherwise,
L(G) = max(R(GL) and R(G) = min(L(GR).

With this definition, in order to calculate L(G) and R(G), I need first to know if G is a number! Obviously I cannot use L(G) and R(G) to know if G is a number, can I?
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote:
Gérard TAILLE wrote:What about proving G is not a number?
If L(G) > R(G) then G is not a number.

If L(G) = R(G) != G then G is not a number
Oops it does not help very much Bill.

Definition :Let G be a short game. We define numbers L(G) and R(G), the left and right stops of G, as follows.
If G is (equal to) a number, we set L(G) = R(G) = G. Otherwise,
L(G) = max(R(GL) and R(G) = min(L(GR).

With this definition, in order to calculate L(G) and R(G), I need first to know if G is a number! Obviously I cannot use L(G) and R(G) to know if G is a number, can I?
I understand how that definition came about. It assumes that you know what a number is. In that case, anything that does not fit the definition of a number is not a number. Easy. :)

But what if we do not know whether G is a number or not. How then can we use the first if clause? So let's use the second clause. A game, G, is not a number iff

max(R(GL)) > min(L(GR)) or

max(R(GL)) = min(L(GR)) != G
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote: I understand how that definition came about. It assumes that you know what a number is. In that case, anything that does not fit the definition of a number is not a number. Easy. :)

But what if we do not know whether G is a number or not. How then can we use the first if clause? So let's use the second clause. A game, G, is not a number iff

max(R(GL)) > min(L(GR)) or

max(R(GL)) = min(L(GR)) != G
I am sure you know it is not quite satisfactory because the problem occurs at each step of the induction : to calculate R(GL) and L(GR) I still need to know if GL or GR are numbers etc.

I am wondering if I can simply play the games "black to move" and "white to move" and compare the two results (which are here only integers and not any number). If the score "black to move" is greater than the score "white to move" is it true that it could not be a number ?
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote: I understand how that definition came about. It assumes that you know what a number is. In that case, anything that does not fit the definition of a number is not a number. Easy. :)

But what if we do not know whether G is a number or not. How then can we use the first if clause? So let's use the second clause. A game, G, is not a number iff

max(R(GL)) > min(L(GR)) or

max(R(GL)) = min(L(GR)) != G
I am sure you know it is not quite satisfactory because the problem occurs at each step of the induction : to calculate R(GL) and L(GR) I still need to know if GL or GR are numbers etc.

I am wondering if I can simply play the games "black to move" and "white to move" and compare the two results (which are here only integers and not any number). If the score "black to move" is greater than the score "white to move" is it true that it could not be a number ?
Yes, I thought about that, and I'm not sure. OC, in go it is not a problem. And if G is reduced to its simplest terms, it's not a problem. Certainly the test for whether G is a number may require reducing G to its simplest terms. I don't think we can get away from the possibility that that may be necessary.
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Visualize whirled peas.

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