User-friendly Reading of the Japanese 1989 Rules

[Cassandra]

some sort of "cooperation" is what you fear

No, it is not cooperation what I "fear".

Conjecture: "The set of capturable-2 strings is a subset of the set of capturable-2\1 strings."

Now my fear is the possibility of falsehood of this conjecture. Compare Proposition 5. If we could prove the conjecture, then we would know the equality of capturable-2 and capturable-2\1. The current state of the art though is: Neither is the conjecture proven nor has a counter-example been found.

No player's stone will be found on local-1.

Our wish is not a proof.

The set "capturable-2 strings" is identical to the set "capturable-2\1" strings.

And there is no wish, but a proof.

By including local-1 in local-2 you had been overcautious from the very beginning.

If "force" is "I'll do my very best and the opponent has no way to prevent the result", then

• "the string remains uncaptured" is forced by the player
• "the string has a successor on at least one of its primary points" is forced by both
• "there is a permanent stone in a certain area" is forced by both
• "there in no permanent stone in a certain area" is forced by the opponent

Should "there is a permanent stone in a certain area" become relevant, then both sets "the string remains uncaptured" and "the string has a successor on at least one of its primary points" must be empty.

It follows that the "certain area" has nothing in common with the strings primary points.

[RobertJasiek]

The set "capturable-2 strings" is identical to the set "capturable-2\1" strings.

And there is no wish, but a proof.

Since you claim there to be a proof, write it down! It must start from the assumption "The string is capturable-2." and end with the implication "The string is capturable-2\1." What you have written so far does not even attempt doing that.

[Cassandra]

The set "capturable-2 strings" is identical to the set "capturable-2\1" strings.

And there is no wish, but a proof.

Since you claim there to be a proof, write it down! It must start from the assumption "The string is capturable-2." and end with the implication "The string is capturable-2\1." What you have written so far does not even attempt doing that.

Here is a translation of the proof given already:

Let us assume that we have identified everything related to local-1, which the opponent cannot force.

So we know,

• (A) which strings are uncapturable, and
• (B) which strings are capturable-1.

Let us further assume that we are evaluating a string that has been identified as capturable-2. It follows that there must be at least one permanent stone at local-2.

Local-2\1 is local-2 minus local-1, so there are three cases, concerning permanent stones on local-1 and local-2\1.

• (C1) There is at least one permanent stone on local-2\1, but no permanent stone on local-1.
• (C2) There is at least one permanent stone on local-2\1, and at least one permanent stone on local-1.
• (C3) There is no permanent stone on local-2\1, but at least one permanent stone on local-1.

(C1) shows capturable-2\1 being identical to capturable-2.

(C2) has three implications:

• The permanent stone on local-1 is irrelevant, because there is a stone on local-2\1.
• The permanent stone on local-1 cannot be forced by the opponent, because this would result in a contradiction to (B).
• No (post-capturable-1-) sequence forced by the opponent can result in a permanent stone on local-1.

(C3) has two implications:

• The permanent stone on local-1 cannot be forced by the opponent, because this would result in a contradiction to (B).
• No (post-capturable-1-) sequence forced by the opponent can result in a permanent stone on local-1.

As concluding result (C1) survives. Local-2\1 is sufficient.

[RobertJasiek]

Although you offer interesting food for thought for what might become a proof structure, after a (too?) quick reading I have objections to your claim of yours being proof:

- The (C1), (C2), (C3) classification may not be made without further justification because "force" does not necessarily suggest independence of the cases from each other.
- Your claimed contradictions to (B) are fakes because, during considering "force" for "capturable-2", it is immaterial what the different "force" did for "capturable-1". Both "force"s applications are independent from each other.

So my guess is that we cannot easily circumvent explicit, representative usage of hypothetical-strategies H and hypothetical-sequences S(H).

[Cassandra]

Have ALL hypothetical strategies and sequences FIRST.

Then evaluate. Using the instructions I outlined before.

Only ONE installation of "force" will exist.


The trouble with your kind of thinking is that you make "force" dependant on the aim to achive. This cannot work.

[TMark]

Why don't you two take this to email? 171 posts, 162 made by only you two and I don't see any great interest in the subject by anybody else. Discuss it privately, and then produce your agreed version, which everybody else can then ignore.

Best wishes.

[RobertJasiek]

Research and research attempts shall be public so that others can learn from the process. Nobody is forced to read this thread; if you do not like it, don't read it. The greater part of this thread is for those interested in following research live.

[Harleqin]

Why don't you two take this to email? 171 posts, 162 made by only you two and I don't see any great interest in the subject by anybody else. Discuss it privately, and then produce your agreed version, which everybody else can then ignore.

No, why? I think it is very interesting to see this discussion. I believe that it is a major argument for rules that do not need this kind of discussion.

[RobertJasiek]

That the discussion can be used for yet better descriptions of Japanese style rules I consider just a byproduct. The discussion uses J2003 just because J2003 already has a fully worked out definition set. It could be for other rulesets instead (or even generalized for large sets of rulesets). The research is mainly for the sake of exploring go theory (currently related to life).

[RobertJasiek]

Have ALL hypothetical strategies and sequences FIRST.

This changes nothing.

Only ONE installation of "force" will exist.

There is one instance of force per string per object uncapturable, capturable-1, capturable-2, capturable-2\1. For the same string, the different "force"s can have different sets of hypothetical-strategies and hypothetical-sequences.

Besides of what I have written earlier,...

Let us assume that we have identified everything related to local-1, which the opponent cannot force.

Local-1 is not something to be forced indeed.

So we know,

* (A) which strings are uncapturable, and
* (B) which strings are capturable-1.

Let us further assume that we are evaluating a string that has been identified as capturable-2. It follows that there must be at least one permanent stone at local-2.

Rather it follows that the opponent cannot prevent at least one local-2 permanent-stone of the player.

* (C1) There is at least one permanent stone on local-2\1, but no permanent stone on local-1.
* (C2) There is at least one permanent stone on local-2\1, and at least one permanent stone on local-1.
* (C3) There is no permanent stone on local-2\1, but at least one permanent stone on local-1.

You should write down in case more carefully: E.g. (C1), When the opponent tries to but cannot prevent at least one local-2 permanent-stone of the player, the player can choose and chooses to create at least one permanent-stone on local-2\1, but no permanent-stone on local-1.

(C1) shows capturable-2\1 being identical to capturable-2.

No. It shows that in this case the capturable-2 string would also be capturable-2\1 if the opponent chould not have chosen any other case.

The permanent stone on local-1 is irrelevant, because there is a stone on local-2\1.

Explain "irrelevant"!

No (post-capturable-1-) sequence forced by the opponent can result in a permanent stone on local-1.

What do you mean by post-capturable-1? What do you mean by a sequence being forced?

[Cassandra]

Only ONE installation of "force" will exist.

There is one instance of force per string per object uncapturable, capturable-1, capturable-2, capturable-2\1. For the same string, the different "force"s can have different sets of hypothetical-strategies and hypothetical-sequences.

We will not come together as long as you insist on your "per ... per" combination.

Each of your "per ... per" combinations - based on what aim to achive - creates a new instance of "force".

But there can be only ONE force per string.

This force relies on following the instructions I have given for player and opponent.

[RobertJasiek]

Mathematical truth does not depend on whether our opinions converge.

If you want to bring your opinion in agreement with mathematical truth or falsehood, then start from assumption, then use step after step so that each step can be seen as a mathematical transformation.

[Cassandra]

Why don't you two take this to email? 171 posts, 162 made by only you two and I don't see any great interest in the subject by anybody else. Discuss it privately, and then produce your agreed version, which everybody else can then ignore.

Best wishes.

Dear TMark,

You are not forced to read.

Anyway, let us assume that your Go does not make any progress.

So you will need some advice by someone, e.g.

"This move has been superfluous."
"Here you could have done better."

What would be wrong with that, even if only you and the someone are interested in your progress ?

[Cassandra]

Mathematical truth does not depend on whether our opinions converge.

If you want to bring your opinion in agreement with mathematical truth or falsehood, then start from assumption, then use step after step so that each step can be seen as a mathematical transformation.

The attachment shows the process you are not willing to apply.

[RobertJasiek]

Concerning table 2:

- Why is cell of column B31 row B32 a contradiction?
- What shall the table do to assist your claimed proof draft?
- "force" in the table is not strictly the same "force" of J2003.
- You must specify that the opponent moves first, the player second.
- Each "force" would have to be checked independently of the other "force"s before creating the table.

Concerning Min-max:

- What is minimized / maximized?
- Please explain how the tables represent an algorithm!
- What (beyond basic Min-max) is that algorithm?
- What is that algorithm used for with respect to your claimed proof draft?