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From: Chris Dams <chr...@wn5.nospamplease.nl>
Newsgroups: rec.games.go
Subject: Re: Model for the World Amateur Go Championship Rules
Date: Wed, 12 Apr 2006 20:56:06 +0000 (UTC)
Message-ID: <e1jph6$13q$1@wnnews.sci.kun.nl>
References: <p67q32p0mo82bm1c2h5gpolhs93hncpu52@4ax.com> <e1j9jn$pla$1@wnnews.sci.kun.nl> <27gq32p0vhdvn6j0e5o1fneud7erkaas4a@4ax.com>


Dear Robert,

Robert Jasiek <jas...@snafu.de> writes:

>On Wed, 12 Apr 2006 16:24:23 +0000 (UTC), Chris Dams
><chr...@wn5.nospamplease.nl> wrote:
>>>     In a position, a string of a player is _two-eye-alive_ if the
>>>opponent cannot force no intersection of the string with a
>>>two-eye-formation on. 
>>
>>>     _J2003-alive_ is defined like in J2003 as either uncapturable,
>>>capturable-1, or capturable-2. 
>>
>>>     In a position, a string is _WAGC-alive-in-seki_ if it is
>>>J2003-alive and not two-eye-alive. 
>>
>>>     In a position, a string is _WAGC-alive_ if it is either
>>>two-eye-alive or WAGC-alive-in-seki.  
>>
>>From these definitions it follows that WAGC-alive is identical to
>>J2003-alive.

>I doubt this. If you claim it, then please present a formal proof!

I have to admit that I, at first, interpreted "either" as "or" in the
definition of WAGC-alive. However, I think the identity of J2003-alive
and WAGC-alive is still provable. Proof is given below.

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 has either a
         stone on local-1 of the string or it does not have a stone on local-1
         of the string
         (2b1) If it has a stone on local-1 of the string, it is also on
            local-2.
         (2b2) If it does not have a stone on local-1 of the string,
             then 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 of the string.
         In both (2b1) and (2b2) we see that the two-eye-formation that is
         formed in S(H) has permanent stones on local-2 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. Hence, the opponent cannot force both caputre of the string
         and no local-2 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.

Best,
Chris