marcy
Site Admin
Joined: 23 Oct 2005
Posts: 209
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 Posted: Jun Mon 19, 2006 2:20 pm Post subject: Q: Can we "combine" candidate a.s. and a.s. defini |
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[Thanks to Nam Tran]
Q: Can the definitions of candidate answer sets and answer sets be combined as follows?
View <M1,R1> is better than view <M2,V2> iff:
1. R1 \subset R2; or
2. neither R1 \subset R2 nor R2 \subset R1, but R1 dominates R2.
M1 is an answer set if there exists a view <M1,R1> of P such that, for every view <M2,R2>, <M2,R2> is not better than <M1,R1>.
A: NO, the proposed definition is not equivalent to the original ones. Consider the following program, P:
| Code: |
r1: p +-.
r2: q +-.
r3: s +-.
prefer(r3,r1) :- not q.
:- s,p.
:- s,q.
:- not a.
a :- p.
a :- s.
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The views of P are:
V1=<{a,p,prefer(r3,r1)},{r1}>
V2=<{a,p,q},{r1,r2}> <--- no prefer atom
V3=<{a,s,prefer(r3,r1)},{r3}>
If we apply the original definitions, we obtain:
1. V3 dominates V1. Candidate answer sets: V2 and V3
Hence, the answer sets of P are V2 and V3.
If we apply the proposed definition, we get:
* rules(V1) \subset rules(V2). Hence, V1 is better than V2.
* V3 dominates V1. Hence, V3 is better than V1.
Therefore, according to the proposed definition, V3 is the only answer set of P. |
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