A player's final-string is capturable-2 if it is neither uncapturable nor capturable-1 and the opponent cannot - with the same hypothetical-strategy - force both capture of the string's stones and no local-2 *** insert >>> and no local-1 <<< insert *** permanent-stone of the player.

In conjunction with the definition of capturable-1 you can now distinguish between

• Common case I: the player can force at least one permanent stone on local-1.
• Common case II: the player cannot force any permanent stone on local-1, but can force at least one permanent stone on local-2.
• Abnormal case: the player can neither force any permanent stone on local-1 nor on local-2 (the aim local-1 exclusive or local-2 is given before the try); but if the opponent can force no permanent stone on local-1, the player can force at least one permanent stone on local-2, and vice-versa.

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

Case II is not that common after the endgame.

Isolated for themselves, your alternative local/capturable definitions are possible but now what does that mean for trying to redo Chris's proof for such other definitions? IIRC from memory, conflicting positions are not known. So maybe one could try a new proof for the alternative situation.

OK, Case II is a bit nearer to "common" than to "abnormal", when regarding the examples.

Chris' proof is no problem. Just leave it as it is or follow my suggestion. (2b1) does not do any harm, it may be only superfluous.

The branch for capturable-1 covers all Nakade and Uttegaeshi.
The branch for not-capturable-1 covers the only remaining position, in which one single stone is transformed to one of only two eye points of a two-eye-formation.

-------------------------------------------------
-------------------------------------------------


But let's try another view.

To not interfere with the definitions in J2003, let us connect

"capturable" to "the opponent can force the capture of the string"

and (thereafter)

"pin1" to "the player can force at least one permanent stone in local-1" and
"pin2'" to "the player can force at least one permanent stone in local-2 minus local-1" (local-2 minus local-1 = "local-2'").

You can assign these definitions to my latest version of Chris' proof respectively as "capturable-pin1" for "capturable-1", "capturable-pin2" for "capturable-2" and "local-2'" for "local-2".

What remains open is your fear that a position could have been "forgotten", in which it would be important to use local-2 instead of local-2'.

Your fear is without any good reason. And, in my opinion, your fear is an unavoidable implication of the multiple-step-multiple-aim algorithm you use in J2003.

I think it necessary to add some motivation for "player" and "opponent" to make the application of "force" within your J2003 well-defined:

• The opponent shall try to reach "capture without any successor" >>> "capture with a successor in local-2 (may be local-2')" >>> "capture with a successor in local-1", in this order.
• The player shall try to reach "uncapturable" >>> "captured with a successor in local-1" >>> "captured with a successor in local-2 (may be local-2')", in this order.

Then you will have no mismatch with different aims in different steps of the evaluation, because there will be only one set of hypothetical sequences only.

By the way: One simple way for the opponent to avoid capturable-1 or capturable-2 would be to capture during the explicit search for "uncapturable", but not to capture during both of the following explicit searches for "capturable-n".


-------------------------------------------------
-------------------------------------------------


Let's return to your fear. I hope that the following text will make my view understandable. But it will be not of the sort you would accept as "proof", I suppose.

#4 is not really an example of this type of position, it resembles a snap-shot mid in the evaluation sequence (originally in #4, nothing had been captured). I suppose it will be extremely difficult to find one position of this type. But anyway, let's go on. I've thought a lot, so I found one example.

Let's bring to mind again that the string under evaluation has been captured.

If the player cannot force a permanent stone neither on local-1 nor on local-2', there are two explanations only.

The first one is simple: It is impossible for the player to have a permanent stone on local-2.

The second one is more complicated: It is not the player, who can force a permanent stone either on local-1 or at local-2', but it is the choice of the opponent. Who shouts: "Ha, ha, if you want capturable-pin1, I will let you have capturable-pin2' and vice versa."

Click Here To Show Diagram Code

[go]$$
$$ -----------------
$$ | X X X . X X X |
$$ | X O O O O O X |
$$ | X O . O . O X |
$$ | X O O O O O X |
$$ | O X X X X X O |
$$ | O X . X . X O |
$$ | O X X X X X O |
$$ | O O O . O O O |
$$ -----------------[/go]

As can be seen with ease, White's 5-stone-chains are not uncapturable.

Click Here To Show Diagram Code

[go]$$
$$ -----------------
$$ | X X X . X X X |
$$ | X O O O O O X |
$$ | X O . O . O X |
$$ | X O O O O O X |
$$ | O X X X X X O |
$$ | O X . X . X O |
$$ | O X X X X X O |
$$ | O O O 1 O O O |
$$ -----------------[/go]

Black captures with 1.

Click Here To Show Diagram Code

[go]$$
$$ -----------------
$$ | X X X 2 X X X |
$$ | X O O O O O X |
$$ | X O . O . O X |
$$ | X O O O O O X |
$$ | b X X X X X a |
$$ | . X . X . X . |
$$ | . X X X X X . |
$$ | . . . 1 . . . |
$$ -----------------[/go]

White gives Atari with 2 and it is Black now who can choose:

• If he connects at "a", White will play at "b" and establish a permanent stone in what is local-2' for her right-hand string and local-1 for her left-hand one.
• If he connects at "b", White will play at "a" and establish a permanent stone in what is local-1 for her right-hand string and local-2' for her left-hand one.

If you remember my suggestion above, you will realise that both players will meet at "capture(d) with a successor in local-2 (may be local-2')" for each of White's strings.

White cannot "force" capturable-1 and Black must not abandon the "force" to "force" capturable-2, what would be against the spirit of the rules.


And, what is important also, this can only happen with strings that will not become part of a two-eye-formation, so are outside the focus of Chris' proof.

The opponent has to capture "symmetrical", the player has to answer with a "symmetrical" threat. This is impossible, if any of the strings in action could become part of a two-eye-formation on its own.

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

You follow lots of different topics in your post. I will have to inspect each of them. Let me start with one topic: your example position per se.

Beautifully elegant, many thanks!

Next topic: Do I accept your post as a proof? No, of course not! First of all, it remains unclear what might be the proposition to be possibly proven.

Since by all means you are not interested in claiming honour alone, I grab part of it by writing down things more carefully. Here I do not address your "another view" but your previous view before that.

**************************************************************

Remarks:

J2003 always refers to version 35a. We use definitions as in J2003 for hypothetical-sequence, hypothetical-strategy, force, uncapturable, permanent-stone, local-1, capturable-1.

Definition:

For a player's final-string, "local-2\1" is local-2 without local-1.

Remarks:

Capturable-2 is defined as in J2003, i.e. the definition relies on local-2. Two-eye-formation, two-eye-alive, J2003-alive, WAGC-alive-in-seki, WAGC-alive are defined as in WAGCmod, using the basic terms of J2003.

Conjecture:

Given the above definitions, the final-position and a fixed, arbitrary string,

the string is WAGC-alive equals the string is J2003-alive.

Draft of a Proof:

Let us denote

> In a position, a string is _WAGC-alive-in-seki_ if it is
> J2003-alive and not two-eye-alive.

as

WAGC-alive-in-seki == J2003-alive && (!two-eye-alive )

In the same notation we also have from

> In a position, a string is _WAGC-alive_ if it is either
> two-eye-alive or WAGC-alive-in-seki.

WAGC-alive == two-eye-alive ^^ WAGC-alive-in-seki.

Substituting the former into the latter expression, we find

WAGC-alive == two-eye-alive ^^ (J2003-alive && (!two-eye-alive)).

In propositional calculus this reduces to

WAGC-alive == J2003-alive || two-eye-alive.

If we now also have the implication two-eye-alive -> J2003-alive if follows
that

WAGC-alive == J2003-alive.

For the implication two-eye-alive ->J2003-alive, imagine that a string is two-eye-alive. The string can either be uncapturable or not uncapturable.
(1) The string is uncapturable -> It is J2003-alive
(2) It is not uncapturable -> The string is either capturable-1 or not capturable-1
(2a) It is capturable-1 -> It is J2003-alive
(2b) It is not capturable-1 -> Because the string is two-eye-alive there is in every hypothetical-strategy of its opponent a hypothetical-sequence in which we reach a two-eye-formation that includes one of its intersections. For every
hypothetical-strategy H of the opponent, we choose a hypothetical-sequence S(H) in it where the oponent reaches a two-eye-formation and subsequently only passes. Because the two-eye-formation cannot be capture by only moves of its opponent, it consists of permanent-stones. In S(H) the
two-eye-formation that is formed on the captured string does not have a stone on local-1 of the string. Therefore local-1 of the string consists of the one or both of the empty intersections of the two-eye-formation. Actually,
it consists of one of the intersections since if it would consist of both, these would have to be adjacent to each other which contradicts the definition of a two-eye-formation. So, local-1 of the string consists of one intersection and during S(H) it becomes one of the the empty points of a two-eye-formation. This implies that this two-eye formation includes strings that occupy the intersections adjacent to local-1. Because local-1 consists of one intersections these adjacent intersections where empty or
occupied by opposing stones. Hence, these intersections belong to local-2\1 of the string. We see that the two-eye-formation that is formed in S(H) has permanent-stones on local-2\1 of the string. Hence, if every hypothetical-strategy of the opponent of the string there is a hypothetical-sequence where a permanent-stone is played on
local-2\1. Hence, the opponent cannot force both capture of the string and no local-2\1 permanent-stone. Hence, the string is capturable-2. Hence, it is J2003-alive. Hence, under the assumption that the string is two-eye-alive, we find that it is J2003-alive. QED.

Remarks:

The sketch follows Chris Dams's text and idea. I have not made any attempt yet to read and verify or reject the contents of this draft. Therefore I do not call it "proof" and I speaks of conjecture instead of proposition. I fear the gap before the sentence "Hence, the string is capturable-2." though.

If you want a definition of plain "capturable", let me suggest:

Definition:

A string is "capturable" unless it is uncapturable.

Remarks:

Now let me grasp your "another view". Up to "local-1", definitions are as in J2003.

Definitions:

A string is "capturable" unless it is uncapturable.

A player's capturable final-string is "capturable-pin1" if the player moving second can force at least one permanent-stone of his on local-1.

For a player's final-string, "local-pin2" is local-1 and, recursively, any adjacent intersection without a stone of a string that is of the player and either uncapturable or capturable-pin1.

For a player's final-string, "local-2\1" is local-pin2 without local-1.

A player's neither uncapturable nor capturable-pin1 final-string is capturable-pin2 if the player moving second can force at least one permanent-stone of his on local-2\1.

Remarks:

Up to here, I am already not sure whether exactly this contents is your intention. Please clarify! Note that this usage of "force" is careless. A final version will have to revert to the J2003 style wording. This is a tricky technical detail only though.

Before answering to your previous posts, let me try to develop a "proof" for my statement that there is nothing to fear from any "simultaneity" of capturable-1 and capturable-2, concerning the relationship to two-eye-formation, which is the topic of Chris' proof.

The player's string under evaluation is neither uncapturable nor capturable-1.

Following your definition of local-1, local-2, and local-2\1, there can be no player's string in local-2\1, which is uncapturable or capturable-1.

Local-2\1 can be empty of player's strings or contain player's strings in status capturable-2 only.

If it is empty, the player has to establish (at least) one new permanent stone in it to reach status capturable-2 for his string.

This is possible only, if he captures some bordering unconnected opponent's stones, must be at least 2, and takes (at least) one of the now empty board points.

The opponent can prevent this by connecting his endangered strings during "Play" or may be even possible during "Evaluate" (before capturing the player's string). This type of "theoretical" construction will not turn the string under evaluation into a two-eye-formation; it remains unaffected from this capturing action far away. This is true, even if the opponent does not connect during "Play", there will remain the question of "capturable-2" for some of the player's strings only. So let us have this type out of the focus of consideration for the following.

If local-2\1 is not empty of further player's strings, we have to remember that it takes two moves in a row to turn a capturable-2 string into a capturable-1 one. This is, after the first move of the opponent, ...

• The player can force capturing something. (a)
• The opponent makes a Pass or plays Tenuki. (b)
• The player can force connection (of at least one stone) to an already uncapturable string of his own or two eyes inside his string. (c)

Be aware that (b) violates "force".

If the player has captured the "something" with a stone on local-1, this stone cannot be permanent; else the string would be capturable-1. Thereafter a permanent stone will be possible on local-2\1 only. Not forcing no player's stone on local-1 and following (b) will be against the spirit of evaluation, and more than infantile.

Else the player's stone(s), which has / have captured the "something" during (a), has / have been placed in local2\1 already. At the same time it follows that the primary player's string must contain more than 1 stone.

To establish a permanent stone on local-1, the player would have to (re)capture any of the opponent's strings that surrounded the player's primary string. As a matter of course, the opponent could prevent this, because the player's string is not capturable-1. Not preventing this and following (b) will be against the spirit of evaluation, and more than infantile.

Strings of more than 1 stone in status capturable-2 cannot become two-eye-formations.

This follows from the last part of Chris' proof, which refers to one single stone in status capturable-2 only. It is not valid for strings of more that one stone.


Local-2\1 is sufficient to identify all members of uncapturable, capturable-1 and capturable-2, which are also members of two-eye-formation.

Using local-2 instead has become a matter of not argueing about "To what am I forced ?"; it is no question of "correct" status evaluation any more.

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

Text is OK.

And there is nothing to fear, see my posting above.

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

There are 2 possibilities.

(A)
A player's capturable final-string is "capturable-1" if the player moving second can force at least one permanent-stone of his on local-1.
For a player's final-string, "local-2" is local-1 and, recursively, any adjacent intersection without a stone of a string that is of the player and either uncapturable or capturable-1.
For a player's final-string, "local-2\1" is local-2 without local-1.
A player's neither uncapturable nor capturable-1 final-string is capturable-2 if the player moving second can force at least one permanent-stone of his on local-2\1.

(B)
A player's capturable final-string is "capturable-1" if the player moving second can force at least one permanent-stone of his on local-1.
For a player's final-string, "local-2" is, recursively, any local-1 adjacent intersection without a stone of a string that is of the player and either uncapturable or capturable-1.
A player's neither uncapturable nor capturable-1 final-string is capturable-2 if the player moving second can force at least one permanent-stone of his on local-2.


I would suggest you to prefer the logic of (A) - together with your "the opponent cannot force no" terminology -, because it would prevent mismatch with the "traditional" use of your terms in all of your papers so far.
In my personal view on J2003, capturable-1 is bound to "at least one successor", capturable-2 to "no successor". This has nothing to do how far away the permanent stone securing capturable-2 might be played (bordered by any or only the own uncapturable and capturable-1 strings).

If you want to have everything well-defined, use "capturable-2\1" instead of "capturable-2", despite in my opinion this will not be mandatory.

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

I know that you would want me to prefer (A). There are different application aims though: If one want to model the official J1989 commentary results, then (B) is needed. If one wants to use a life-equivalence proof, then (B) is needed. For (A), such a proof is not available (yet); your informal talk does not convince me at all. Personally I do not prefer (A) or (B) for a greater elegance - both have some kind of elegance.



Exactly.



Always. And on the semi-formal level, e.g. J2003 is well-defined.

As so often, you post contains too little explicit references and too much informal talk. I would need 4 to 6 hours to translate it to a more formal context and decide which parts might make some sense. I do not know yet whether I will have the time. I do not want to do your work.

Wishful thinking.



It is well-defined but maybe not obvious from a high-level Go player's strategic view.



This is a nice aim for rules design but incompatible to the J2003 approach. It requires an independent, new design, which does not care for any precedents.

Suppose we wanted to redesign force, two-eye-formation, uncapturable, C1, C2 (or only one capturable type) afresh, then we would have greater freedom for the capturable-x definitions. One thing I want to maintain though: A life-equivalence a la Dams.



You suppose correctly.

Why do only (these) two explanations exist?

See attachment.

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

See attachment for a proof in short.

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

In your attachment Mappe 1, I am having difficulties to follow the start of your sketch. In table row E, implicitly you claim this


Conjecture:

The player cannot force a local-1 permanent-stone
AND
The player cannot force a local-2\1 permanent-stone

<=>

The player cannot force a local-2 permanent-stone.


Remarks:

By definition, we know that local-1 + local-2\1 = local-2. But now how do you prove the conjecture? A naive logic operation is insufficient because the "force" for local-2 works independently of the "force"s for local-1 and local-2\1. It is not just a textual addition that you dream of.

Only the "<=" direction of the conjecture is given trivially. The other direction you need to prove explicitly!

Do not call everything you write or paint a proof!

In Mappe2, you would need to prove every non-trivial entry!

Mappe2 contains at least one mistake; by far I have not checked all entries. The existence of the mistake means though that you should verify already the draft of Mappe2 much more carefully. The mistake: Example 0000 fits into column 9 row D, where you have written a "not possible".

You should be more carefully with your usage of "mistake".

Column 9 has "NO" for "Permanent stone after capture in local-1" and "NO" for "Permanent stone after capture in local-2\1".

Local-2 is by definition the compound of local-1 and local-2\1. It follows that a permanent stone in local-2 is impossible, because none is in any of its parts.

As consequence of your seeing a "mistake" in a table where there is none, I would you suggest to rely Chris' proof on "local-1", "local-2", and "local-2\1" as primary conditions for the clauses.

"capturable-1" and "capturable-2" should be used as secondary parameters only, when needed to refer to "live", because they can be deducted from the primary ones.

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