Re: User-friendly Reading of the Japanese 1989 Rules
Posted: Thu Jun 24, 2010 10:15 am
Given a final-position, a string in it, and the definitions of local-2\1, capturable-2\1, J2003-alive-2\1, WAGC-alive-in-seki-2\1, WAGC-alive-2/1.
Proposition:
The string is WAGC-alive-2\1 equals the string is J2003-alive-2\1.
Proof:
Part 1 of the proof is analogue to part 1 of Chris Dams's proof.
Part 2 of the proof is as follows:
For the implication two-eye-alive -> J2003-alive-2\1, imagine that a string is two-eye-alive. Either the opponent cannot force capture of the string or he can force capture of the string.
(1) The opponent cannot force capture of the string -> It is uncapturable. -> It is J2003-alive-2\1.
(2) The opponent can force capture of the string. -> Because the string is two-eye-alive there is in every hypothetical-strategy of its opponent a hypothetical-sequence in which we reach the player's 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 player reaches such a two-eye-formation and subsequently only passes. Because the two-eye-formation cannot be captured by only moves of its opponent, it consists of permanent-stones. -> In S(H) the two-eye-formation that is formed on at least one intersection of 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.
(2a) The two-eye-formation has a stone on local-1 of the string -> the string is capturable-1 -> It is J2003-alive-2\1.
(2b) The two-eye-formation has no stone on local-1 of the string -> Local-1 of the string consists of 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 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 intersection these adjacent intersections were 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, in 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\1. Hence, it is J2003-alive-2\1.
Hence, under the assumption that the string is two-eye-alive, we find that it is J2003-alive-2\1. QED.
Proposition:
The string is WAGC-alive-2\1 equals the string is J2003-alive-2\1.
Proof:
Part 1 of the proof is analogue to part 1 of Chris Dams's proof.
Part 2 of the proof is as follows:
For the implication two-eye-alive -> J2003-alive-2\1, imagine that a string is two-eye-alive. Either the opponent cannot force capture of the string or he can force capture of the string.
(1) The opponent cannot force capture of the string -> It is uncapturable. -> It is J2003-alive-2\1.
(2) The opponent can force capture of the string. -> Because the string is two-eye-alive there is in every hypothetical-strategy of its opponent a hypothetical-sequence in which we reach the player's 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 player reaches such a two-eye-formation and subsequently only passes. Because the two-eye-formation cannot be captured by only moves of its opponent, it consists of permanent-stones. -> In S(H) the two-eye-formation that is formed on at least one intersection of 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.
(2a) The two-eye-formation has a stone on local-1 of the string -> the string is capturable-1 -> It is J2003-alive-2\1.
(2b) The two-eye-formation has no stone on local-1 of the string -> Local-1 of the string consists of 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 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 intersection these adjacent intersections were 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, in 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\1. Hence, it is J2003-alive-2\1.
Hence, under the assumption that the string is two-eye-alive, we find that it is J2003-alive-2\1. QED.