Thermography

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Bill Spight
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Re: Thermography

Post by Bill Spight »

Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:Let's look at Gérard's infinitesimal with some White corridors.
Oops now my name is on an infinitesimal ! I am becoming famous :D :D :D !

BTW my idea can be use in various ways:
Click Here To Show Diagram Code
[go]$$
$$ -----------------------
$$ | . X . . . . . . . . |
$$ | O X . . . . . . . . |
$$ | . X . . . . . . . . |
$$ | a X X X . . . . . . |
$$ | O O O X X . . . . . |
$$ | . X X b X . . . . . |
$$ | O O O X X . . . . . |
$$ | . . O O O . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------[/go]
Here black can choose "a" to reach ↑ in sente or "b" to reach tiny in sente (instead of * and ↑ in my first example). The goal is to allow black to choose between two fuzzy options allowing interesting games. White to move can play "a" and reach ↑ in gote.
The play at temperature 1 (t = 0 in chilled go)
Click Here To Show Diagram Code
[go]$$W White first
$$ -----------------------
$$ | . X . . . . . . . . |
$$ | O X . . . . . . . . |
$$ | 2 X . . . . . . . . |
$$ | 1 X X X . . . . . . |
$$ | O O O X X . . . . . |
$$ | 3 X X 4 X . . . . . |
$$ | O O O X X . . . . . |
$$ | . . O O O . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------[/go]
Result: +3
Click Here To Show Diagram Code
[go]$$B Black first
$$ -----------------------
$$ | . X . . . . . . . . |
$$ | O X . . . . . . . . |
$$ | . X . . . . . . . . |
$$ | 1 X X X . . . . . . |
$$ | O O O X X . . . . . |
$$ | 2 X X 3 X . . . . . |
$$ | O O O X X . . . . . |
$$ | . . O O O . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------[/go]
Result: +4 (+3 at chilled go)

The new Gérard infinitesimal is

{6|↑||↑}
Gérard TAILLE wrote: yes Bill, with the experience of your first analyse it is now far simplier
Click Here To Show Diagram Code
[go]$$
$$ -----------------------
$$ | . X . . . . . . . . |
$$ | O X . . . . . . . . |
$$ | . X . . . . . . . . |
$$ | 3 X X X . . . . . . |
$$ | O O O X X . . . . . |
$$ | 2 X X 1 X . . . . . |
$$ | O O O X X . . . . . |
$$ | . . O O O . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------[/go]
after :b1:, :w2: reverses and black must play :b3:
but by beginning with :b3: black can transpose and we can conclude that the option black :b1: can be deleted.

If it is white to play
Click Here To Show Diagram Code
[go]$$W
$$ -----------------------
$$ | . X . . . . . . . . |
$$ | O X . . . . . . . . |
$$ | . X . . . . . . . . |
$$ | 1 X X X . . . . . . |
$$ | O O O X X . . . . . |
$$ | a X X . X . . . . . |
$$ | O O O X X . . . . . |
$$ | . . O O O . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------[/go]
:w1: dominates a white move at "a"

Finally, provided some training, the result is {6|↑||↑} is not so hard to discover.
BTW, for Black the atari dominates the connection.
Click Here To Show Diagram Code
[go]$$W Difference game
$$ -----------------------
$$ | . X . X . . O . O . |
$$ | O X X X . . O O O X |
$$ | . X . X . . O . O 7 |
$$ | 2 X X X . . O O O 6 |
$$ | O O O X X O O X X X |
$$ | 3 X B 5 X O 1 O O 4 |
$$ | O O O X X O O X X X |
$$ | . . O O O X X X . . |
$$ | . . . . O X . . . . |
$$ | . . . . O X . . . . |
$$ -----------------------[/go]
:b8: at :bc:

If :w5: at 6, :b6: at 5.
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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Without ko complications or seki, regular territory values (sans komi) are 6*, or 7, or 8*, etc., with an equal number of plays by each player. So 6* becomes 7 at area scoring, 8* becomes 9, etc. If you have a seki with an odd number of dame when fully played out, those values become 6, 7*, 8, etc., and the area scores become even. :)
In this case it is quite obvious with a simpe miai situation:
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | . a X O . X X b X |
$$ | O . X O O O O O X |
$$ | X X X O . . . - X |
$$ | . X O O . . . - - |
$$ | X X O . . . . . . |
$$ | O O O . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------[/go]
Each player can choose to change or not the parity of dame.
G = {5|0} + {*|-5} = {{5*|0},{5*|*}||{0|-5},{*|-5}}

G <> 0 ; Black can move to {5*|*} and White can move to {*|-5}

G - * <> 0 ; Black can move to {5*|0} + * and White can move to {0|-5} + *

G - *2 <> 0 ; *2 = {0,*|0,*} Whichever option in G Black or White chooses, after the other replies, the first player has the winning option in *2.

In fact, G is confused with every nimber (nim heap). Very nice. :) :bow:
Another position is the following
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | . O . . . . X . X |
$$ | O O O O X X X X X|
$$ | X X X X O O O O O |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------[/go]
Beside the sagari by each player you have the possibility
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | . O . 1 2 3 X . X |
$$ | O O O O X X X X X|
$$ | X X X X O O O O O |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------[/go]
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | . O . X a X X . X |
$$ | O O O O X X X X X|
$$ | X X X X O O O O O |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------[/go]
with two particularities in this resulting position:
1) the parity of dame has changed
2) a black move at "a" is now black priviledge
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Bill Spight
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:Another position is the following
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | . O . . . . X . X |
$$ | O O O O X X X X X |
$$ | X X X X O O O O O |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------[/go]
Beside the sagari by each player you have the possibility
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | . O . 1 2 3 X . X |
$$ | O O O O X X X X X |
$$ | X X X X O O O O O |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------[/go]
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | . O . X a X X . X |
$$ | O O O O X X X X X |
$$ | X X X X O O O O O |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------[/go]
with two particularities in this resulting position:
1) the parity of dame has changed
2) a black move at "a" is now black privilege
As an English go term, privilege is used for the ability to make a sente play before the opponent can play the reverse sente.

Neither Japanese nor Korean rules count a as territory, but my territory rules and Lasker-Maas rules do. :) So do at least one form of Ikeda's territory rules, I think. Counting a as territory is consistent with CGT. It is not an infinitesimal. A one way dame is a point for the player who can fill it.

Let's write out the game in slash notation, a la Lasker-Maas rules.

{{14|0},{14||2|-2}|||{0|-18},{2|-2||-18}}
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— Winona Adkins

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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:Let's look at Gérard's infinitesimal with some White corridors.
Oops now my name is on an infinitesimal ! I am becoming famous :D :D :D !

BTW my idea can be use in various ways:
Click Here To Show Diagram Code
[go]$$
$$ -----------------------
$$ | . X . . . . . . . . |
$$ | O X . . . . . . . . |
$$ | . X . . . . . . . . |
$$ | a X X X . . . . . . |
$$ | O O O X X . . . . . |
$$ | . X X b X . . . . . |
$$ | O O O X X . . . . . |
$$ | . . O O O . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------[/go]
Here black can choose "a" to reach ↑ in sente or "b" to reach tiny in sente (instead of * and ↑ in my first example). The goal is to allow black to choose between two fuzzy options allowing interesting games. White to move can play "a" and reach ↑ in gote.
The play at temperature 1 (t = 0 in chilled go)
Click Here To Show Diagram Code
[go]$$W White first
$$ -----------------------
$$ | . X . . . . . . . . |
$$ | O X . . . . . . . . |
$$ | 2 X . . . . . . . . |
$$ | 1 X X X . . . . . . |
$$ | O O O X X . . . . . |
$$ | 3 X X 4 X . . . . . |
$$ | O O O X X . . . . . |
$$ | . . O O O . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------[/go]
Result: +3
Click Here To Show Diagram Code
[go]$$B Black first
$$ -----------------------
$$ | . X . . . . . . . . |
$$ | O X . . . . . . . . |
$$ | . X . . . . . . . . |
$$ | 1 X X X . . . . . . |
$$ | O O O X X . . . . . |
$$ | 2 X X 3 X . . . . . |
$$ | O O O X X . . . . . |
$$ | . . O O O . . . . . |
$$ | . . . . . . . . . . |
$$ | . . . . . . . . . . |
$$ -----------------------[/go]
Result: +4 (+3 at chilled go)

The new Gérard infinitesimal is

{6|↑||↑}

It is obviously greater than 0. It is greater than * and confused with ↑.

I think that it's atomic weight is 2.
Yes, Bill.
In addition this infinitesimals seems also also greater than ↑↑* and confused with ↑↑
BTW what is the definition of atomic weight?
Let's call UP(1) = ↑, UP(2) = ↑↑, UP(3) = ↑↑↑, ... and similarly DOWN(n)
My guessing is the following : the atomic weight of an infinitesimal G is the highest value n such that
G ≥ UP(n) or G ≥ UP(n) + * or G ≤ DOWN(n) or G ≤ DOWN(n) + *
but it is only my guessing not the real definition!
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Bill Spight
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote:The new Gérard infinitesimal is

{6|↑||↑}

It is obviously greater than 0. It is greater than * and confused with ↑.

I think that it's atomic weight is 2.
Yes, Bill.
In addition this infinitesimals seems also also greater than ↑↑* and confused with ↑↑
BTW what is the definition of atomic weight?
I have had occasion to use the definition of atomic weight less than a handful of times in 26 years. ;) It's a bit tricky and I have never derived it myself, so I can't claim to understand it. However, I can talk about it in terms of prototypes. :) I did an online search for the definition without finding one. {sigh}

A synonym for atomic weight is uppityness, so you might think that the prototypical game with atomic weight 1 is an ↑. But it isn't. The prototypical game with atomic weight 1 is ↑*. ;) Now, ↑* <> 0, so being ahead by 1 atomic weight is not enough to win if your opponent plays first. You have to be ahead by 2 to be sure.

The prototypical game with atomic weight 2 is ↑* + ↑* = ↑↑, which is greater than 0. The prototypical game with atomic weight 3 is ↑↑↑*, etc.

In chilled go the prototypical atomic weight game is then the sum of two closed corridors, each with the payoff of a stone at the end, and one of them being a *.

For instance:
Click Here To Show Diagram Code
[go]$$B Atomic weight = 4
$$ ------------------------
$$ | . . O . . . . . O X . .
$$ | . . O X X X X X X X . .
$$ | . . O . O X . . . . . .
$$ | . . O X X X . . . , . .
$$ | . . . . . . . . . . . .
$$ | . . . . . . . . . . . .[/go]
In chilled go this is ↑↑↑↑* + * = ↑↑↑↑ . Each play by White in the long corridor reduces the atomic weight by 1. That is, it reduces the value of the game by ↑*. In 4 plays White moves to 0 in chilled go, where we may ignore the actual score. ;)
The Adkins Principle:
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— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:The new Gérard infinitesimal is

{6|↑||↑}

It is obviously greater than 0. It is greater than * and confused with ↑.

I think that it's atomic weight is 2.
Yes, Bill.
In addition this infinitesimals seems also also greater than ↑↑* and confused with ↑↑
BTW what is the definition of atomic weight?
I have had occasion to use the definition of atomic weight less than a handful of times in 26 years. ;) It's a bit tricky and I have never derived it myself, so I can't claim to understand it. However, I can talk about it in terms of prototypes. :) I did an online search for the definition without finding one. {sigh}

A synonym for atomic weight is uppityness, so you might think that the prototypical game with atomic weight 1 is an ↑. But it isn't. The prototypical game with atomic weight 1 is ↑*. ;) Now, ↑* <> 0, so being ahead by 1 atomic weight is not enough to win if your opponent plays first. You have to be ahead by 2 to be sure.

The prototypical game with atomic weight 2 is ↑* + ↑* = ↑↑, which is greater than 0. The prototypical game with atomic weight 3 is ↑↑↑*, etc.

In chilled go the prototypical atomic weight game is then the sum of two closed corridors, each with the payoff of a stone at the end, and one of them being a *.

For instance:
Click Here To Show Diagram Code
[go]$$B Atomic weight = 4
$$ ------------------------
$$ | . . O . . . . . O X . .
$$ | . . O X X X X X X X . .
$$ | . . O . O X . . . . . .
$$ | . . O X X X . . . , . .
$$ | . . . . . . . . . . . .
$$ | . . . . . . . . . . . .[/go]
In chilled go this is ↑↑↑↑* + * = ↑↑↑↑ . Each play by White in the long corridor reduces the atomic weight by 1. That is, it reduces the value of the game by ↑*. In 4 plays White moves to 0 in chilled go, where we may ignore the actual score. ;)
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | O X . . X a . . X |
$$ | . X . X X O O O O |
$$ | b X X . X O . X O |
$$ | O O X X X O O O O |
$$ | . X . . X . X O . |
$$ | O O X X X O O O . |
$$ | . O O O O O . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------[/go]
in your post viewtopic.php?p=261486#p261486 you analysed as the fourth variation the above position and after having showed rather long sequences you conclude that black should prefer "a" rather than "b". Should you had used atomic weight you will have choose without any hesitation a move at "a" simply because you know for sure that "b" is a losing one giving the game (after the white forced answer) an atomic value equal to -2.

I do not really understand why you are reluctant to use such tool Bill.
Any tool allowing to avoid reading the game is useful isn't it?
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Bill Spight
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Re: Thermography

Post by Bill Spight »

Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:The new Gérard infinitesimal is

{6|↑||↑}

It is obviously greater than 0. It is greater than * and confused with ↑.

I think that it's atomic weight is 2.
Yes, Bill.
In addition this infinitesimals seems also also greater than ↑↑* and confused with ↑↑
BTW what is the definition of atomic weight?
I have had occasion to use the definition of atomic weight less than a handful of times in 26 years. ;) It's a bit tricky and I have never derived it myself, so I can't claim to understand it. However, I can talk about it in terms of prototypes. :) I did an online search for the definition without finding one. {sigh}

A synonym for atomic weight is uppityness, so you might think that the prototypical game with atomic weight 1 is an ↑. But it isn't. The prototypical game with atomic weight 1 is ↑*. ;) Now, ↑* <> 0, so being ahead by 1 atomic weight is not enough to win if your opponent plays first. You have to be ahead by 2 to be sure.

The prototypical game with atomic weight 2 is ↑* + ↑* = ↑↑, which is greater than 0. The prototypical game with atomic weight 3 is ↑↑↑*, etc.

In chilled go the prototypical atomic weight game is then the sum of two closed corridors, each with the payoff of a stone at the end, and one of them being a *.

For instance:
Click Here To Show Diagram Code
[go]$$B Atomic weight = 4
$$ ------------------------
$$ | . . O . . . . . O X . .
$$ | . . O X X X X X X X . .
$$ | . . O . O X . . . . . .
$$ | . . O X X X . . . , . .
$$ | . . . . . . . . . . . .
$$ | . . . . . . . . . . . .[/go]
In chilled go this is ↑↑↑↑* + * = ↑↑↑↑ . Each play by White in the long corridor reduces the atomic weight by 1. That is, it reduces the value of the game by ↑*. In 4 plays White moves to 0 in chilled go, where we may ignore the actual score. ;)
Gérard TAILLE wrote:
Click Here To Show Diagram Code
[go]$$B
$$ ---------------------
$$ | O X . . X a . . X |
$$ | . X . X X O O O O |
$$ | b X X . X O . X O |
$$ | O O X X X O O O O |
$$ | . X . . X . X O . |
$$ | O O X X X O O O . |
$$ | . O O O O O . . . |
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ --------------------[/go]
in your post viewtopic.php?p=261486#p261486 you analysed as the fourth variation the above position and after having showed rather long sequences you conclude that black should prefer "a" rather than "b". Should you had used atomic weight you will have choose without any hesitation a move at "a" simply because you know for sure that "b" is a losing one giving the game (after the white forced answer) an atomic value equal to -2.

I do not really understand why you are reluctant to use such tool Bill.
Any tool allowing to avoid reading the game is useful isn't it?
I am not at all reluctant to use atomic weight. My purpose was pedagogical for our readers. :)
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
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Bill Spight
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Re: Thermography

Post by Bill Spight »

OK. What is the atomic weight of {4|*||*}?

Well, it looks like it is 1. Also, it is greater than 0, less than ↑↑, and confused with ↑*. That's a big clue. ;)

But let's apply the definition. Hidden out of courtesy. :)
First, we figure out a new game, G, using the atomic weights of the options of {4|*||*}.

We know that the atomic weight of * is 0, but what is the atomic weight of {4|*}? Let's figure out a new game, H, for it.

Again, the atomic weight of * is 0. And I can tell you that the atomic weight of any number is also 0. Makes sense, right?

So H = {0|0} :)

But we are not done with H. Now we have to start off like we are cooling H by 2 degrees. That gives us

H0 = {-2|2} = 0

So far, so good. But that does not mean that the atomic weight of {4|*} is 0. We now have to check whether it is an eccentric case.

It is an eccentric case iff H0 is an integer and H is greater than or less than a far star.

What, you may ask, is a far star? It is a sufficiently big nimber, *N, when N is large enough. When is N large enough? When it is larger than any nimber anywhere in the game tree of H, including H itself.

The largest nimber in {4|*} is * = *1. So *2 is a far star for it. Then we have the question,

{4|*} + {0,*|0,*} <?> 0

OC, Black to play can win by playing to 4 + *2

White to play cannot win by playing in *2, so let him play to * + *2. Then Black replies to * + * = 0 and wins.

That means that H > *2 and we have an eccentric case.

Now we look at the White options of H0 = {-2|2}. Now we find the largest integer, I, for which I <| 2. The largest integer which is less than or confused with 2 is 1. If there were more White options of H0 we would look for the largest integer which is less than or confused with all of the White options. If H were less than *2 we would do the opposite for the Black options of H0.

OK, so the atomic weight of {4|*} is 1. Now we are ready to find the atomic weight of {4|*||*}.

G = {1|0}

G0 = {-1|2} = 0.

Here we are again. ;)

Well, we know that the far star is *2. We ask,

{4|*||*} + {0,*|0,*} <?> 0

Black, OC, wins by playing to {4|*} + *2, which we know is positive.

If White plays in {4|*||*} to * + *2 Black wins, so let White play in *2. OC, if White plays to {4|*||*}, it is positive, so Black wins. Finally, let White play to * + {4|*||*}; then Black replies in * to {4|*||*} and wins.

So {4|*||*} > *2.

G0 = {-1|2}

We know that the largest integer, I <|2 is 1.

As we suspected, the atomic weight of {4|*||*} is 1.

Whew!
The Adkins Principle:
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— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:OK. What is the atomic weight of {4|*||*}?

Well, it looks like it is 1. Also, it is greater than 0, less than ↑↑, and confused with ↑*. That's a big clue. ;)

But let's apply the definition. Hidden out of courtesy. :)
First, we figure out a new game, G, using the atomic weights of the options of {4|*||*}.

We know that the atomic weight of * is 0, but what is the atomic weight of {4|*}? Let's figure out a new game, H, for it.

Again, the atomic weight of * is 0. And I can tell you that the atomic weight of any number is also 0. Makes sense, right?

So H = {0|0} :)

But we are not done with H. Now we have to start off like we are cooling H by 2 degrees. That gives us

H0 = {-2|2} = 0

So far, so good. But that does not mean that the atomic weight of {4|*} is 0. We now have to check whether it is an eccentric case.

It is an eccentric case iff H0 is an integer and H is greater than or less than a far star.

What, you may ask, is a far star? It is a sufficiently big nimber, *N, when N is large enough. When is N large enough? When it is larger than any nimber anywhere in the game tree of H, including H itself.

The largest nimber in {4|*} is * = *1. So *2 is a far star for it. Then we have the question,

{4|*} + {0,*|0,*} <?> 0

OC, Black to play can win by playing to 4 + *2

White to play cannot win by playing in *2, so let him play to * + *2. Then Black replies to * + * = 0 and wins.

That means that H > *2 and we have an eccentric case.

Now we look at the White options of H0 = {-2|2}. Now we find the largest integer, I, for which I <| 2. The largest integer which is less than or confused with 2 is 1. If there were more White options of H0 we would look for the largest integer which is less than or confused with all of the White options. If H were less than *2 we would do the opposite for the Black options of H0.

OK, so the atomic weight of {4|*} is 1. Now we are ready to find the atomic weight of {4|*||*}.

G = {1|0}

G0 = {-1|2} = 0.

Here we are again. ;)

Well, we know that the far star is *2. We ask,

{4|*||*} + {0,*|0,*} <?> 0

Black, OC, wins by playing to {4|*} + *2, which we know is positive.

If White plays in {4|*||*} to * + *2 Black wins, so let White play in *2. OC, if White plays to {4|*||*}, it is positive, so Black wins. Finally, let White play to * + {4|*||*}; then Black replies in * to {4|*||*} and wins.

So {4|*||*} > *2.

G0 = {-1|2}

We know that the largest integer, I <|2 is 1.

As we suspected, the atomic weight of {4|*||*} is 1.

Whew!
Oops now I am lost Bill.
I understood that any infinitesimal is approximatively equal to a certain number of uppityness (↑*). Well, it could not be a definition but it makes sense from an intuitive point of view and we can try to compare any infinitesimal to n(↑*).
Where I am lost? When you try to find the atomic weight of {4|*}. For me {4|*} is not a infinitesimal and I cannot compare this game with n(↑*).
Can you clarify what you mean by atomic weight of {4|*} ?
BTW a number not equal to 0 is not an infinitesimal is it?
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Bill Spight
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
Bill Spight wrote:OK. What is the atomic weight of {4|*||*}?

Well, it looks like it is 1. Also, it is greater than 0, less than ↑↑, and confused with ↑*. That's a big clue. ;)

But let's apply the definition. Hidden out of courtesy. :)
First, we figure out a new game, G, using the atomic weights of the options of {4|*||*}.

We know that the atomic weight of * is 0, but what is the atomic weight of {4|*}? Let's figure out a new game, H, for it.

Again, the atomic weight of * is 0. And I can tell you that the atomic weight of any number is also 0. Makes sense, right?

So H = {0|0} :)

But we are not done with H. Now we have to start off like we are cooling H by 2 degrees. That gives us

H0 = {-2|2} = 0

So far, so good. But that does not mean that the atomic weight of {4|*} is 0. We now have to check whether it is an eccentric case.

It is an eccentric case iff H0 is an integer and H is greater than or less than a far star.

What, you may ask, is a far star? It is a sufficiently big nimber, *N, when N is large enough. When is N large enough? When it is larger than any nimber anywhere in the game tree of H, including H itself.

The largest nimber in {4|*} is * = *1. So *2 is a far star for it. Then we have the question,

{4|*} + {0,*|0,*} <?> 0

OC, Black to play can win by playing to 4 + *2

White to play cannot win by playing in *2, so let him play to * + *2. Then Black replies to * + * = 0 and wins.

That means that H > *2 and we have an eccentric case.

Now we look at the White options of H0 = {-2|2}. Now we find the largest integer, I, for which I <| 2. The largest integer which is less than or confused with 2 is 1. If there were more White options of H0 we would look for the largest integer which is less than or confused with all of the White options. If H were less than *2 we would do the opposite for the Black options of H0.

OK, so the atomic weight of {4|*} is 1. Now we are ready to find the atomic weight of {4|*||*}.

G = {1|0}

G0 = {-1|2} = 0.

Here we are again. ;)

Well, we know that the far star is *2. We ask,

{4|*||*} + {0,*|0,*} <?> 0

Black, OC, wins by playing to {4|*} + *2, which we know is positive.

If White plays in {4|*||*} to * + *2 Black wins, so let White play in *2. OC, if White plays to {4|*||*}, it is positive, so Black wins. Finally, let White play to * + {4|*||*}; then Black replies in * to {4|*||*} and wins.

So {4|*||*} > *2.

G0 = {-1|2}

We know that the largest integer, I <|2 is 1.

As we suspected, the atomic weight of {4|*||*} is 1.

Whew!
Oops now I am lost Bill.
I understood that any infinitesimal is approximatively equal to a certain number of uppityness (↑*). Well, it could not be a definition but it makes sense from an intuitive point of view and we can try to compare any infinitesimal to n(↑*).
Where I am lost? When you try to find the atomic weight of {4|*}. For me {4|*} is not a infinitesimal and I cannot compare this game with n(↑*).
Can you clarify what you mean by atomic weight of {4|*} ?
BTW a number not equal to 0 is not an infinitesimal is it?
This is from the calculation method (operational definition) of atomic weight in Winning Ways, by Conway, Berlekamp, and Guy. I suppose that to define atomic weight for infinitesimals, they had to define it for every finite game. :)

They first introduce the idea of atomic weight for a game called Hackenbush Hotchpotch, which is played on certain pictures using curves or lines colored blue, red, or green. In the original definitions of games the two players are Left and Right. In go Black is Left and White is Right. In Hackenbush Hotchpotch Blue is Left and Red is Right. (Edit: The color green indicates an option for either Blue or Red.)

They tell us that all Blue flowers in that game have an atomic weight of 1 and all Red flowers have an atomic weight of -1. In the game Blue flowers have green stalks topped by at least one blue closed curve called a petal. So a single Blue flower can be any of an infinite number of games.

The simplest Blue flower is {0,*|0} = ↑*. Other Blue flowers include these, each of which has atomic weight 1:

{0,↑*||0}
{0,{0,↑*||0}||0}
{0,{0,{0,↑*||0}||0}||0}
{0,*,*2|0,*}
{0,*,{0,*,*2|0,*}||0,*}
{0,*,{0,*,{0,*,*2|0,*}||0,*}||0,*}
{0,*,*2,*3|0,*,*2}
{0,*,*2,{0,*,*2,*3||0,*,*2}||0,*,*2}
{0,*,*2,*3,*4|0,*,*2,*3}

OC, {0|*} = ↑ has atomic weight 1, although it is not a Blue flower. ;) And {1|*} also has atomic weight 1, even though it is not an infinitesimal. Ditto for {4|*}. Them's the facts. {shrug}

Edit: {4|*} <> ↑*. Let's do the difference game.

{4|*} + * + {*|0}

Black plays in ↓ to *, yielding {4|*} > 0, and wins.
White plays in {4|*} to *, yielding ↓ < 0, and wins.
:)
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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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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:OK. What is the atomic weight of {4|*||*}?

Well, it looks like it is 1. Also, it is greater than 0, less than ↑↑, and confused with ↑*. That's a big clue. ;)

But let's apply the definition. Hidden out of courtesy. :)
First, we figure out a new game, G, using the atomic weights of the options of {4|*||*}.

We know that the atomic weight of * is 0, but what is the atomic weight of {4|*}? Let's figure out a new game, H, for it.

Again, the atomic weight of * is 0. And I can tell you that the atomic weight of any number is also 0. Makes sense, right?

So H = {0|0} :)

But we are not done with H. Now we have to start off like we are cooling H by 2 degrees. That gives us

H0 = {-2|2} = 0

So far, so good. But that does not mean that the atomic weight of {4|*} is 0. We now have to check whether it is an eccentric case.

It is an eccentric case iff H0 is an integer and H is greater than or less than a far star.

What, you may ask, is a far star? It is a sufficiently big nimber, *N, when N is large enough. When is N large enough? When it is larger than any nimber anywhere in the game tree of H, including H itself.

The largest nimber in {4|*} is * = *1. So *2 is a far star for it. Then we have the question,

{4|*} + {0,*|0,*} <?> 0

OC, Black to play can win by playing to 4 + *2

White to play cannot win by playing in *2, so let him play to * + *2. Then Black replies to * + * = 0 and wins.

That means that H > *2 and we have an eccentric case.

Now we look at the White options of H0 = {-2|2}. Now we find the largest integer, I, for which I <| 2. The largest integer which is less than or confused with 2 is 1. If there were more White options of H0 we would look for the largest integer which is less than or confused with all of the White options. If H were less than *2 we would do the opposite for the Black options of H0.

OK, so the atomic weight of {4|*} is 1. Now we are ready to find the atomic weight of {4|*||*}.

G = {1|0}

G0 = {-1|2} = 0.

Here we are again. ;)

Well, we know that the far star is *2. We ask,

{4|*||*} + {0,*|0,*} <?> 0

Black, OC, wins by playing to {4|*} + *2, which we know is positive.

If White plays in {4|*||*} to * + *2 Black wins, so let White play in *2. OC, if White plays to {4|*||*}, it is positive, so Black wins. Finally, let White play to * + {4|*||*}; then Black replies in * to {4|*||*} and wins.

So {4|*||*} > *2.

G0 = {-1|2}

We know that the largest integer, I <|2 is 1.

As we suspected, the atomic weight of {4|*||*} is 1.

Whew!
Oops now I am lost Bill.
I understood that any infinitesimal is approximatively equal to a certain number of uppityness (↑*). Well, it could not be a definition but it makes sense from an intuitive point of view and we can try to compare any infinitesimal to n(↑*).
Where I am lost? When you try to find the atomic weight of {4|*}. For me {4|*} is not a infinitesimal and I cannot compare this game with n(↑*).
Can you clarify what you mean by atomic weight of {4|*} ?
BTW a number not equal to 0 is not an infinitesimal is it?
This is from the calculation method (operational definition) of atomic weight in Winning Ways, by Conway, Berlekamp, and Guy. I suppose that to define atomic weight for infinitesimals, they had to define it for every finite game. :)

They first introduce the idea of atomic weight for a game called Hackenbush Hotchpotch, which is played on certain pictures using curves or lines colored blue, red, or green. In the original definitions of games the two players are Left and Right. In go Black is Left and White is Right. In Hackenbush Hotchpotch Blue is Left and Red is Right. (Edit: The color green indicates an option for either Blue or Red.)

They tell us that all Blue flowers in that game have an atomic weight of 1 and all Red flowers have an atomic weight of -1. In the game Blue flowers have green stalks topped by at least one blue closed curve called a petal. So a single Blue flower can be any of an infinite number of games.

The simplest Blue flower is {0,*|0} = ↑*. Other Blue flowers include these, each of which has atomic weight 1:

{0,↑*||0}
{0,{0,↑*||0}||0}
{0,{0,{0,↑*||0}||0}||0}
{0,*,*2|0,*}
{0,*,{0,*,*2|0,*}||0,*}
{0,*,{0,*,{0,*,*2|0,*}||0,*}||0,*}
{0,*,*2,*3|0,*,*2}
{0,*,*2,{0,*,*2,*3||0,*,*2}||0,*,*2}
{0,*,*2,*3,*4|0,*,*2,*3}

OC, {0|*} = ↑ has atomic weight 1, although it is not a Blue flower. ;) And {1|*} also has atomic weight 1, even though it is not an infinitesimal. Ditto for {4|*}. Them's the facts. {shrug}

Edit: {4|*} <> ↑*. Let's do the difference game.

{4|*} + * + {*|0}

Black plays in ↓ to *, yielding {4|*} > 0, and wins.
White plays in {4|*} to *, yielding ↓ < 0, and wins.
:)
I need some time to analyse all this stuff Bill.
I am not quite familier with Hackenbush game so sorry in advance if my question is a little stupid: how do you draw in Hackenbush game the game {4|*} ?
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Bill Spight
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Re: Thermography

Post by Bill Spight »

Gérard TAILLE wrote:
OC, {0|*} = ↑ has atomic weight 1, although it is not a Blue flower. ;) And {1|*} also has atomic weight 1, even though it is not an infinitesimal. Ditto for {4|*}. Them's the facts. {shrug}

Edit: {4|*} <> ↑*. Let's do the difference game.

{4|*} + * + {*|0}

Black plays in ↓ to *, yielding {4|*} > 0, and wins.
White plays in {4|*} to *, yielding ↓ < 0, and wins.
:)
I need some time to analyse all this stuff Bill.
Sorry, I didn't mean to overburden you. :( But you can see why I'm usually satisfied with estimating the atomic weight of a chilled go position without going through the calculations.
I am not quite familier with Hackenbush game so sorry in advance if my question is a little stupid: how do you draw in Hackenbush game the game {4|*} ?
I have only a passing acquaintance with Hackenbush, myself. I do not know how to draw such a position.
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

Bill Spight wrote:
Gérard TAILLE wrote:
OC, {0|*} = ↑ has atomic weight 1, although it is not a Blue flower. ;) And {1|*} also has atomic weight 1, even though it is not an infinitesimal. Ditto for {4|*}. Them's the facts. {shrug}

Edit: {4|*} <> ↑*. Let's do the difference game.

{4|*} + * + {*|0}

Black plays in ↓ to *, yielding {4|*} > 0, and wins.
White plays in {4|*} to *, yielding ↓ < 0, and wins.
:)
I need some time to analyse all this stuff Bill.
Sorry, I didn't mean to overburden you. :( But you can see why I'm usually satisfied with estimating the atomic weight of a chilled go position without going through the calculations.
I am not quite familier with Hackenbush game so sorry in advance if my question is a little stupid: how do you draw in Hackenbush game the game {4|*} ?
I have only a passing acquaintance with Hackenbush, myself. I do not know how to draw such a position.
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.

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!
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Re: Thermography

Post by Bill Spight »

Bill Spight wrote:
Gérard TAILLE wrote:
OC, {0|*} = ↑ has atomic weight 1, although it is not a Blue flower. ;) And {1|*} also has atomic weight 1, even though it is not an infinitesimal. Ditto for {4|*}. Them's the facts. {shrug}

Edit: {4|*} <> ↑*. Let's do the difference game.

{4|*} + * + {*|0}

Black plays in ↓ to *, yielding {4|*} > 0, and wins.
White plays in {4|*} to *, yielding ↓ < 0, and wins.
:)
I need some time to analyse all this stuff Bill.
Sorry, I didn't mean to overburden you. :( But you can see why I'm usually satisfied with estimating the atomic weight of a chilled go position without going through the calculations.
I am not quite familier with Hackenbush game so sorry in advance if my question is a little stupid: how do you draw in Hackenbush game the game {4|*} ?
I have only a passing acquaintance with Hackenbush, myself. I do not know how to draw such a position.
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|*} <> *

:)
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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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Gérard TAILLE
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Re: Thermography

Post by Gérard TAILLE »

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.
G - H = (n+2)↑ + {*|-4}
but you can see that if white plays in H white will win easily and that contradicts G - H > 0
That proves that you cannot assign an atomic weight to H and that confirms that atomic weight concerns only infinitesimals.
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