Provided we do not use certain superko rules.



Proof?



Proof?



Not precisely.



Please write down that definition!



Proof???

If the number of groups on the board is not too small, even super-Ko should be no obstacle when there is cooperation. And J2003 does not use super-Ko.


Opponent and player meet at (B31).
The opponent could not realize (B32). The player could not realize (A), (B2).
The opponent was not forced to allow a stone on local-1 (= B2). The player was not able to force a stone on local-1 (= B2).


See above: There will be no stone on local-1.


Why not ?


It is sufficient what I have written already. It is not necessary to define "certain area" for the purpose of showing that including local-1 in local-2 is overdoing things.


See above: There will be no stone on local-1.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

Just because of the usual, trivial wording and text embedding details.



This is not a mathematical proof. In particular because your (A) to (B32) phrases presume the players cooperation while what would have to be proven depends on force.



Since your mathematical proof is missing, the consequences may not be made (thus far).

The string will be captured and have a successor on at least one of its primary points. What more is needed for "capturable-1" ?


This is not correct, Robert.

Each of my (A) to (B32) is possible when player and opponent cooperate and some sort of "cooperation" is what you fear for not including local-1 in local-2. (A) to (B32) is nothing less than the set of all theoretically possible evaluation sequences (spoken in J2003-language "all hypothetical sequences of all hypothetical strategies").

But there will remain NO room at all for "cooperation" because of the instructions for player and opponent. These instructions clarify "force" for both. Following these instructions eliminates all "superfluous" sequences not relevant for the string under evaluation.


Let sequences survive the application of the instructions, which belong to (B31). No player's stone will be found on local-1.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

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.



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.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

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.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

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).

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.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

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.

_________________
No aji, keshi, kifu or kikashi has been harmed in the compiling of this post.
http://www.gogod.co.uk

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.

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.

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A good system naturally covers all corner cases without further effort.

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).

This changes nothing.



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,...



Local-1 is not something to be forced indeed.



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



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.



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.



Explain "irrelevant"!



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

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.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

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.

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 ?

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

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

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

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?