The citations are from this thread, May 05 or 06 in 2019.
"3) 5 gote
Play in r if (s > a) or ((s > a - d + e) and (t > a + e)) or ((s > a - b + c - d + e) and (t > a + c - d + e))."
Which is it?
[(s > a) or ((s > a - d + e) and (t > a + e))] or ((s > a - b + c - d + e)] and (t > a + c - d + e))
(s > a) or [((s > a - d + e) and (t > a + e)) or ((s > a - b + c - d + e) and (t > a + c - d + e))]
"Now we can predict the conditions for 7 gote:
Play in r if (s > a) or ((s > a - f + g) and (t > a + g)) or ((s > a - d + e - f + g) and (t > a + e - f + g)) or ((s > a - b + c - d + e - f + g) and (t > a + c - d + e - f + g))."
Which is it?
[(s > a) or ((s > a - f + g) and (t > a + g))] or ((s > a - d + e - f + g) and (t > a + e - f + g)) or ((s > a - b + c - d + e - f + g) and (t > a + c - d + e - f + g))
[(s > a) or ((s > a - f + g) and (t > a + g)) or ((s > a - d + e - f + g) and (t > a + e - f + g))] or ((s > a - b + c - d + e - f + g) and (t > a + c - d + e - f + g))
(s > a) or [((s > a - f + g) and (t > a + g)) or ((s > a - d + e - f + g) and (t > a + e - f + g))] or ((s > a - b + c - d + e - f + g) and (t > a + c - d + e - f + g))
(s > a) or [((s > a - f + g) and (t > a + g)) or [((s > a - d + e - f + g) and (t > a + e - f + g)) or ((s > a - b + c - d + e - f + g) and (t > a + c - d + e - f + g))]]
"We can predict the conditions for 6 gote.
Play in r if ((s > a - f) and (t > a)) or ((s > a - d + e - f) and (t > a + e - f)) or ((s > a - b + c - d + e - f) and (t > a + c - d + e - f))."
Which is it?
[((s > a - f) and (t > a)) or ((s > a - d + e - f) and (t > a + e - f))] or ((s > a - b + c - d + e - f) and (t > a + c - d + e - f))
((s > a - f) and (t > a)) or [((s > a - d + e - f) and (t > a + e - f)) or ((s > a - b + c - d + e - f) and (t > a + c - d + e - f))]
This 'n' that
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Harleqin
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Re: This 'n' that
Logical 'or' is associative, so those are all equivalent (where there is no mistake in brackets).
A good system naturally covers all corner cases without further effort.
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Bill Spight
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Re: This 'n' that
Thanks. 
Let me format it differently.
Let me format it differently.Claro?Bill Spight wrote:Let's take a look at our little theory so far of how for White to play in r = {t | s || 0} or in a = {a | 0}, when the only other plays are in the simple gote, b = {b | 0}, c = {c | 0}, . . . , where the gote are labeled alphabetically in descending order of size.
For an odd number of gote we have this:
1) 1 gote
Play in r if s > a. (Else play in a understood.)
2) 3 gote
Play in r if (s > a) or
((s > a - b + c) and (t > a + c)).
3) 5 gote
Play in r if (s > a) or
((s > a - d + e) and (t > a + e)) or
((s > a - b + c - d + e) and (t > a + c - d + e)).
Now we can predict the conditions for 7 gote:
Play in r if (s > a) or
((s > a - f + g) and (t > a + g)) or
((s > a - d + e - f + g) and (t > a + e - f + g)) or
((s > a - b + c - d + e - f + g) and (t > a + c - d + e - f + g)).
An so on for a larger odd number of gote.
For an even number of gote we have this:
1) 2 gote
Play in r if ((s > a - b) and (t > a)).
2) 4 gote
Play in r if ((s > a - d) and (t > a)) or
((s > a - b + c - d) and (t > a + c - d)).
We can predict the conditions for 6 gote.
Play in r if ((s > a - f) and (t > a)) or
((s > a - d + e - f) and (t > a + e - f)) or
((s > a - b + c - d + e - f) and (t > a + c - d + e - f)).
And so on.
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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Everything with love. Stay safe.
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Re: This 'n' that
I still do not understand difference games comparing two positions P and -Q.
Black starts and achieves the score 0.
White starts and achieves the score 0.
What does this mean for the difference game P - Q >= 0?
Is P - Q >= 0 characterised by "Black starts and wins or ties by achieving at least the score 0 AND White starts and does not win by achieving at least the score 0" or by what else?
What does this mean for the difference game P - Q > 0?
Is P - Q > 0 characterised by "Black starts and wins by achieving a score larger than 0 AND White starts and does not win by achieving at least the score 0" or by what else?
Black starts and achieves the score 0.
White starts and achieves the score 0.
What does this mean for the difference game P - Q <= 0?
Is P - Q <= 0 characterised by "Black starts and does not win by achieving at most the score 0 AND White starts and wins or ties by achieving at most the score 0" or by what else?
What does this mean for the difference game P - Q < 0?
Is P - Q < 0 characterised by "Black starts and does not win by achieving at most the score 0 AND White starts and wins by achieving a score smaller than 0" or by what else?
Black starts and achieves the score 0.
White starts and achieves the score 0.
What does this mean for the difference game P - Q >= 0?
Is P - Q >= 0 characterised by "Black starts and wins or ties by achieving at least the score 0 AND White starts and does not win by achieving at least the score 0" or by what else?
What does this mean for the difference game P - Q > 0?
Is P - Q > 0 characterised by "Black starts and wins by achieving a score larger than 0 AND White starts and does not win by achieving at least the score 0" or by what else?
Black starts and achieves the score 0.
White starts and achieves the score 0.
What does this mean for the difference game P - Q <= 0?
Is P - Q <= 0 characterised by "Black starts and does not win by achieving at most the score 0 AND White starts and wins or ties by achieving at most the score 0" or by what else?
What does this mean for the difference game P - Q < 0?
Is P - Q < 0 characterised by "Black starts and does not win by achieving at most the score 0 AND White starts and wins by achieving a score smaller than 0" or by what else?
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Bill Spight
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Re: This 'n' that
I don't remember exactly where this came from, but it looks like the question is one of reversal.
It does so if the position after
(Q) is greater than or equal to the original position (P) from Black's point of view, i.e., if P - Q ≤ 0. I.E., if Black to play in P - Q cannot win with correct play (by definition). To answer that question we do not have to look at play when White plays first. To show that in addition P != Q we also may have to look at the result when White plays first.
(The CGT definition says that White wins if Black plays first, but CGT regards 0 as a second player win. In the context of go, which allows ties, we just say that the first player, Black, cannot win.)
It does so if the position after
(Q) is greater than or equal to the original position (P) from Black's point of view, i.e., if P - Q ≤ 0. I.E., if Black to play in P - Q cannot win with correct play (by definition). To answer that question we do not have to look at play when White plays first. To show that in addition P != Q we also may have to look at the result when White plays first.(The CGT definition says that White wins if Black plays first, but CGT regards 0 as a second player win. In the context of go, which allows ties, we just say that the first player, Black, cannot win.)
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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— Winona Adkins
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Everything with love. Stay safe.
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RobertJasiek
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Re: This 'n' that
As long as there are no ko fights, one can compare any two local endgame positions P and Q. In particular, one can ask how their difference compares to 0.
CGT-reversal is one application for comparing two positions in a difference game. As you guessed from my positions, it is the application I am currently studying.
The relations >= and <= seem to be easier for go scoring than the relations > and <. Therefore, let me stick to the former in this message. I redo my study according to your reply, using Black's value perspective.
We have P - Q ≤ 0 because the correct resulting score is 0 so Black starting cannot win.
As an application, White's alternating sequence from P to Q reverses:
We have P - Q ≥ 0 because the correct resulting score is 0 so White starting cannot win.
As an application, Black's alternating sequence from P to Q reverses:
Correct?
CGT-reversal is one application for comparing two positions in a difference game. As you guessed from my positions, it is the application I am currently studying.
The relations >= and <= seem to be easier for go scoring than the relations > and <. Therefore, let me stick to the former in this message. I redo my study according to your reply, using Black's value perspective.
We have P - Q ≤ 0 because the correct resulting score is 0 so Black starting cannot win.
As an application, White's alternating sequence from P to Q reverses:
We have P - Q ≥ 0 because the correct resulting score is 0 so White starting cannot win.
As an application, Black's alternating sequence from P to Q reverses:
Correct?
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Bill Spight
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Re: This 'n' that
Correct.RobertJasiek wrote:We have P - Q ≥ 0 because the correct resulting score is 0 so White starting cannot win.
As an application, Black's alternating sequence from P to Q reverses:
Correct?
And, OC, if P - Q ≤ 0 and P - Q ≥ 0, then P - Q = 0.
Edit: When considering possible reverses at go, it is conceivable that you could reach a position where the second player has no play. E.g., there is an open point, but to play there would be illegal. In such a case you can require the second player to pass but at the cost of 1 point by territory scoring.
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins
Visualize whirled peas.
Everything with love. Stay safe.
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: This 'n' that
That is not my understanding Robert.RobertJasiek wrote:I still do not understand difference games comparing two positions P and -Q.
Black starts and achieves the score 0.
White starts and achieves the score 0.
What does this mean for the difference game P - Q >= 0?
Is P - Q >= 0 characterised by "Black starts and wins or ties by achieving at least the score 0 AND White starts and does not win by achieving at least the score 0" or by what else?
First of all I understand the notation P - Q >= 0 only in the context of CGT where the result of a game is a win or a loss but never a draw. As a consequence when you mentionned a "score 0" it is not a sufficient information to decide whether it is a win or a loss. You have to look at which side is taking the last move.
Taking your example the result are the following:
Black starts and loses
White starts and loses
In this context what about the status of P - Q ?
Here my understanding (for black point of view) is the following:
Black starts and loses means P - Q <= 0
White starts and loses means P - Q >= 0
and taking both results I conclude P - Q = 0
My understanding of P - Q > 0 is a little more complicated because here I need to know the resut when Black plays first and also when white plays first:RobertJasiek wrote: What does this mean for the difference game P - Q > 0?
P - Q > 0 means Black starts and wins AND White starts and loses
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Re: This 'n' that
The CGT definition uses last play or chilling but we need not be slaves of CGT when we are really interested in counts and scores. We must just be careful to translate the outcome classes and their conditions correctly.
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Gérard TAILLE
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Re: This 'n' that
OK Robert. In order for me to try and answer your questions, how do you suggest to translate the CGT sentence "Black starts and loses" in your go language made of counts and scores?RobertJasiek wrote:The CGT definition uses last play or chilling but we need not be slaves of CGT when we are really interested in counts and scores. We must just be careful to translate the outcome classes and their conditions correctly.
In your example it seems that "Black starts and achieves the score 0" means "Black starts and loses" but with other examples I doubt it is always true.
You can see the problem : if you use an ambiguous language you cannot expect an accurate answer to a question made in this language.
In CGT language a win or a loss is very clear and relations like P - Q >= 0 or P - Q > 0 can be defined with very good properties.
For the time being I am not aware of a translation of this CGT language into yours.
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Bill Spight
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Re: This 'n' that
Like I say, I am really quite busy these days, but let me offer a brief response.RobertJasiek wrote:The CGT definition uses last play or chilling but we need not be slaves of CGT when we are really interested in counts and scores. We must just be careful to translate the outcome classes and their conditions correctly.
As Gérard points out, and as we all know, CGT does not have ties. We all also know that ko fights do not fit into the CGT framework, but there are workarounds. We all also know that seki can have quite different results, depending on the rules. We all also know that the group tax is part of the CGT framework for both area and territory scoring, but we can work around the lack of a group tax.
The CGT rules of comparison with 0 are simple. Given a game, G, let us play it twice by correct minimax play, once with Black playing first, and once with White playing first.
1) if G is a second player win, G = 0;
2) if G is a first player win, G <> 0;
3) if G is a Black win, G > 0;
4) if G is a White win, G < 0.
Assume that we have taken care of the group tax and seki. All of these results are still possible. But we will also have jigo results. What is lost if we treat those results the same as CGT does? Will correct minimax play assigning wins and losses a la CGT not be correct when we allow jigo? Sure, sometimes the CGT classification of G will be a Black win instead of jigo, but best play a la CGT will still be best play with jigo. It's just that Black will get the last play.
Edit: It may seem that a half point komi confounds the CGT model by making the last play irrelevant. The half point komi can turn a Black win into a Black loss.
But all you have to do is implement the komi with this game, Komi = {-1 | 0}, and play G + Komi. Then if Black wins G by making last play to 0, White plays to 0 in the Komi and wins G + Komi.
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Visualize whirled peas.
Everything with love. Stay safe.
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Visualize whirled peas.
Everything with love. Stay safe.
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RobertJasiek
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Re: This 'n' that
I have seen a result function R, which we can reinterpret as 'count', in some CGT texts. Below is my naive attempt of defining the outcome classes.
Playing the difference game compares to 0 the result of play in a local position and some colour-reversed local position. In combinatorial game theory, two local endgames A and B are compared using a result function R as follows: A ? B :<=> R(A + E) ? R (B + E) for all finite, non-cyclic environments E. A result function assigns the count after both players' best play in the local endgames. We compare the result of the local difference game to 0, that is, R(A - B) ? 0. The environment E drops out when forming the difference.
Suppose we compare the two local positions P and Q. The colour-reversed of Q is its negative, that is, -Q. Accordingly, we can write and transform the comparison as follows: P ? Q <=> P - Q ? 0 <=> P + (-Q) ? 0. In other words, the difference game compares the imagined combined position to 0. We have
P > Q if White starts and Black achieves more than 0 as a count (White starting cannot prevent Black's win),
P ≥ Q if White starts and Black achieves at least 0 as a count (White starting cannot win),
P = Q if the count is 0 for Black's and White's starts (neither starting player can win),
P ≤ Q if Black starts and White achieves at most 0 as the count (Black starting cannot win),
P < Q if Black starts and White achieves less than 0 as the count (Black starting cannot prevent White's win),
P || Q if 1) Black starts and achieves a count larger than 0 (Black starting wins) and 2) White starts and achieves a count smaller than 0 (White starting wins).
Playing the difference game compares to 0 the result of play in a local position and some colour-reversed local position. In combinatorial game theory, two local endgames A and B are compared using a result function R as follows: A ? B :<=> R(A + E) ? R (B + E) for all finite, non-cyclic environments E. A result function assigns the count after both players' best play in the local endgames. We compare the result of the local difference game to 0, that is, R(A - B) ? 0. The environment E drops out when forming the difference.
Suppose we compare the two local positions P and Q. The colour-reversed of Q is its negative, that is, -Q. Accordingly, we can write and transform the comparison as follows: P ? Q <=> P - Q ? 0 <=> P + (-Q) ? 0. In other words, the difference game compares the imagined combined position to 0. We have
P > Q if White starts and Black achieves more than 0 as a count (White starting cannot prevent Black's win),
P ≥ Q if White starts and Black achieves at least 0 as a count (White starting cannot win),
P = Q if the count is 0 for Black's and White's starts (neither starting player can win),
P ≤ Q if Black starts and White achieves at most 0 as the count (Black starting cannot win),
P < Q if Black starts and White achieves less than 0 as the count (Black starting cannot prevent White's win),
P || Q if 1) Black starts and achieves a count larger than 0 (Black starting wins) and 2) White starts and achieves a count smaller than 0 (White starting wins).
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Bill Spight
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Re: This 'n' that
OK, but as Berlekamp defined count, all scores are counts but not all counts are scores.RobertJasiek wrote:I have seen a result function R, which we can reinterpret as 'count', in some CGT texts.
Let G be a finite combinatorial game, G, with a count, R.
G <> 0 or
(G > 0 and R > 0) or
{G < 0 and R < 0) or
(G = 0 and R = 0)
N.B. I am using <> to mean is confused with. || is ambiguous with slash notation.
E - E = 0.RobertJasiek wrote:Below is my naive attempt of defining the outcome classes.
Playing the difference game compares to 0 the result of play in a local position and some colour-reversed local position. In combinatorial game theory, two local endgames A and B are compared using a result function R as follows: A ? B :<=> R(A + E) ? R (B + E) for all finite, non-cyclic environments E. A result function assigns the count after both players' best play in the local endgames. We compare the result of the local difference game to 0, that is, R(A - B) ? 0. The environment E drops out when forming the difference.
{-1 | 1} = 0RobertJasiek wrote:Suppose we compare the two local positions P and Q. The colour-reversed of Q is its negative, that is, -Q. Accordingly, we can write and transform the comparison as follows: P ? Q <=> P - Q ? 0 <=> P + (-Q) ? 0. In other words, the difference game compares the imagined combined position to 0. We have
P > Q if White starts and Black achieves more than 0 as a count (White starting cannot prevent Black's win),
What you should be interested in is scores, not counts. If you do not play when the position has a score, then in {-1 | 1} White does not play to 1, but stops at 0; i.e., does not play at all. That solves that problem.
The particular scores you are interested in are stops. A stop is the first score reached with optimal minimax play. For game, G, let S(B) be the stop when Black plays first and S(W) be the stop when White plays first. OC, S(B) ≥ S(W).
1) If S(B) < 0 then G < 0.
2) If S(W) > 0 then G > 0.
3) If S(B) > 0 > S(W) then G <> 0.
4) If S(B) = 0 and S(W) < 0 then G <| 0.
5) If S(B) > 0 and S(W) = 0 then G |> 0.
6) If S(B) = S(W) = 0, then G = 0 or G is an infinitesimal.
Now, we may not always realize when we have reached a score without playing the game out. For example, {3|3||0|||0||-5|-7} = 0.
Also, for difference games who gets the last play in a jigo may well matter, because sente usually matters.
Last edited by Bill Spight on Mon Feb 01, 2021 8:15 am, edited 1 time in total.
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Visualize whirled peas.
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RobertJasiek
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Re: This 'n' that
Thanks, it seems to make sense!
(For those who don't know: 'confused with' is sometimes called 'incomparable'. <| means 'smaller or incomparable', |> means 'larger or incomparable'.)
EDIT: But where in your scheme do P ≥ Q and P ≤ Q fit?
(For those who don't know: 'confused with' is sometimes called 'incomparable'. <| means 'smaller or incomparable', |> means 'larger or incomparable'.)
EDIT: But where in your scheme do P ≥ Q and P ≤ Q fit?