[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 . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]
Here black can choose "a" to reach ↑ in sente or "b" to reach tiny in sente (instead of * and ↑ in my first example). The goal is to allow black to choose between two fuzzy options allowing interesting games. White to move can play "a" and reach ↑ in gote.
[go]$$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]
at
If at 6, at 5.
_________________ The Adkins Principle: At some point, doesn't thinking have to go on? — Winona Adkins
My two main guides in life: My mother and my wife.
Posts: 328 Liked others: 4 Was liked: 18
Rank: 1er dan
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.
Quote:
In this case it is quite obvious with a simpe miai situation:
[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.
[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
[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}}
_________________ The Adkins Principle: At some point, doesn't thinking have to go on? — Winona Adkins
My two main guides in life: My mother and my wife.
[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 . . . . . | $$ | . . . . . . . . . . | $$ | . . . . . . . . . . | $$ -----------------------[/go]
Here black can choose "a" to reach ↑ in sente or "b" to reach tiny in sente (instead of * and ↑ in my first example). The goal is to allow black to choose between two fuzzy options allowing interesting games. White to move can play "a" and reach ↑ in gote.
[go]$$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!
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 *.
[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: At some point, doesn't thinking have to go on? — Winona Adkins
My two main guides in life: My mother and my wife.
Posts: 328 Liked others: 4 Was liked: 18
Rank: 1er dan
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 *.
[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.
[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 https://lifein19x19.com/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?
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 *.
[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.
[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 https://lifein19x19.com/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
My two main guides in life: My mother and my wife.
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
H_{0} = {-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 H_{0} 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 H_{0} = {-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 H_{0} 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 H_{0}.
OK, so the atomic weight of {4|*} is 1. Now we are ready to find the atomic weight of {4|*||*}.
G = {1|0}
G_{0} = {-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.
G_{0} = {-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: At some point, doesn't thinking have to go on? — Winona Adkins
My two main guides in life: My mother and my wife.
Posts: 328 Liked others: 4 Was liked: 18
Rank: 1er dan
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
H_{0} = {-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 H_{0} 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 H_{0} = {-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 H_{0} 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 H_{0}.
OK, so the atomic weight of {4|*} is 1. Now we are ready to find the atomic weight of {4|*||*}.
G = {1|0}
G_{0} = {-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.
G_{0} = {-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?
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
H_{0} = {-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 H_{0} 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 H_{0} = {-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 H_{0} 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 H_{0}.
OK, so the atomic weight of {4|*} is 1. Now we are ready to find the atomic weight of {4|*||*}.
G = {1|0}
G_{0} = {-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.
G_{0} = {-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:
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.
_________________ The Adkins Principle: At some point, doesn't thinking have to go on? — Winona Adkins
My two main guides in life: My mother and my wife.
Posts: 328 Liked others: 4 Was liked: 18
Rank: 1er dan
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
H_{0} = {-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 H_{0} 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 H_{0} = {-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 H_{0} 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 H_{0}.
OK, so the atomic weight of {4|*} is 1. Now we are ready to find the atomic weight of {4|*||*}.
G = {1|0}
G_{0} = {-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.
G_{0} = {-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:
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|*} ?
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.
Quote:
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
My two main guides in life: My mother and my wife.
Posts: 328 Liked others: 4 Was liked: 18
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Bill Spight wrote:
Gérard TAILLE wrote:
Quote:
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.
Quote:
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!
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.
Quote:
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|*} <> *
_________________ The Adkins Principle: At some point, doesn't thinking have to go on? — Winona Adkins
My two main guides in life: My mother and my wife.
Posts: 328 Liked others: 4 Was liked: 18
Rank: 1er dan
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.
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.
_________________ The Adkins Principle: At some point, doesn't thinking have to go on? — Winona Adkins
My two main guides in life: My mother and my wife.
Posts: 328 Liked others: 4 Was liked: 18
Rank: 1er dan
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?
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.
_________________ The Adkins Principle: At some point, doesn't thinking have to go on? — Winona Adkins
My two main guides in life: My mother and my wife.
Posts: 328 Liked others: 4 Was liked: 18
Rank: 1er dan
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.
[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.
[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?
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.
[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.
[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.
[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]
Quote:
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.
_________________ The Adkins Principle: At some point, doesn't thinking have to go on? — Winona Adkins
My two main guides in life: My mother and my wife.
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