RichardKennaway comments on Prisoner's dilemma tournament results - Less Wrong

32 Post author: AlexMennen 09 July 2013 08:50PM

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Comment author: RichardKennaway 13 July 2013 09:19:05PM 0 points [-]

Thinking about this further, given enough assertions, there's no need to have the programs at all. Let the agents be just the assertions that they make about themselves. Each agent would consist of a set of axioms, perhaps written in Prolog, about who they cooperate or defect against, and running a contest between two agents would just be a matter of taking the union of the two sets of axioms and attempting to deduce (by the standard Prolog proof-search) what choice each one makes.

Comment author: skepsci 14 July 2013 05:10:51AM 2 points [-]

What happens if the two sets of axioms are individually consistent, but together are inconsistent?

Comment author: RichardKennaway 14 July 2013 07:40:34AM 0 points [-]

Deem both of the agents to have not terminated?

Comment author: skepsci 14 July 2013 02:21:55PM 1 point [-]

How will you know? The set of consistent axiom systems is undecidable. (Though the set of inconsistent axioms systems is computably enumerable.)

Comment author: RichardKennaway 14 July 2013 05:50:39PM 0 points [-]

You just run the Prolog (or whatever logic system implements all this), and it either terminates with a failure or does not terminate within the time allowed by the competition. The time limit renders everything practically decidable.

Comment author: skepsci 15 July 2013 05:14:34AM 2 points [-]

Perhaps it terminates in the time required proving that A defects and B cooperates, even though the axioms were inconsistent, and one could also have proved that A cooperates and B defects.

Comment author: RichardKennaway 15 July 2013 06:13:29AM 1 point [-]

The competitors know the deterministic proof-search algorithm (e.g. that of Prolog), and the first answer that produces within the time limit, that is the answer.