Re: User-friendly Reading of the Japanese 1989 Rules
Posted: Thu Jun 17, 2010 10:30 am
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.
**************************************************************
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.