Comparative study of view update algorithms in rational choice theory

Abstract

The dynamics of belief and knowledge is one of the major components of any autonomous system that should be able to incorporate new pieces of information. We show that knowledge base dynamics has interesting connection with kernel change via hitting set and abduction. The approach extends and integrates standard techniques for efficient query answering and integrity checking. The generation of hitting set is carried out through a hyper tableaux calculus and magic set that is focussed on the goal of minimality. Many different view update algorithms have been proposed in the literature to address this problem. The present paper provides a comparative study of view update algorithms in rational approach.

Appendix :

Proof of Theorem 1

Sound and Completeness are trivially to shown from the definition. □

Proof of Lemma 1

• 1. The fact that $\nvdash \neg \alpha$ and there exists a KB - closed locally minimal abductive explanation for α wrt K B _I , it is clear that there exists at least one α- kernel of KB. Suppose Δ (${\Delta }\in {\Delta }^{+} \cup {\Delta }^{-}$) is empty (i.e., $KB_{I}\vdash \neg \alpha$), then the required result follows immediately. If not, since Δ is a locally minimal abductive explanation, there exists a minimal subset $KB_{I}^{\prime }\subseteq KB_{I}$, s.t. Δ is minimal abductive explanation of α wrt $KB_{I}^{\prime }$. Since, Δ is KB-closed, it is not difficult to see that $KB_{I}^{\prime }\cup {\Delta }^{+} \cup {\Delta }^{-}$ is a α - kernel of KB.

• 2. Since X is a α - kernel of KB and Δ is the set of all abducibles in X, it follows that ${\Delta }^{+} \cup {\Delta }^{-}$ is a minimal abductive explanation of Δ wrt $X\backslash {\Delta }^{-} \cup {\Delta }^{+}$. It is obvious that ${\Delta }^{+} \cup {\Delta }^{-}$ is KB- closed, and so Δ is a KB-closed locally minimal abductive explanation for α wrt K B _I .

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Proof of Theorem 2

Sound and Completeness are trivially to shown from the Algorithm 1. □

Proof of Lemma 2 and 5

• 1. Consider a ${\Delta } ({\Delta }\in {\Delta }_{i} \cup {\Delta }_{j})\in S$. We need to show that Δ is generated by algorithm 3 at step 2. From lemma 1, it is clear that there exists a A-kernel X of D D B _G s.t. $X \cap EDB = {\Delta }_{j}$ and $X \cup EDB = {\Delta }_{i}$. Since $X \vdash A$, there must exist a successful derivation for A using only the elements of X as input clauses and similarly $X \nvdash A$. Consequently Δ must have been constructed at step 2.

• 2. Consider a ${\Delta }^{\prime }(({\Delta }^{\prime }\in {\Delta }_{i} \cup {\Delta }_{j})\in S^{\prime }$. Let ${\Delta }^{\prime }$ be constructed from a successful(unsuccessful) branch i via Δ _i (Δ _j ). Let X be the set of all input clauses used in the refutation i. Clearly $X\vdash A$($X\nvdash A$). Further, there exists a minimal (wrt set-inclusion) subset Y of X that derives A (i.e. no proper subset of Y derives A). Let ${\Delta } = Y \cap EDB$ ($Y \cup EDB$). Since IDB does not(does) have any unit clauses, Y must contain some EDB facts, and so Δ is not empty (empty) and obviously ${\Delta }\subseteq {\Delta }^{\prime }$. But, Y need not (need) be a A-kernel for I D B _G since Y is not ground in general. But it stands for several A-kernels with the same (different) EDB facts Δ in them. Thus, from lemma 1, Δ is a DDB-closed locally minimal abductive explanation for A wrt I D B _G and is contained in ${\Delta }^{\prime }$.

• 3. Since this proof easy to see materialized view update with minimal.

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Proof of Lemma 3 and 6

• 1. (Only if part) Suppose H is a minimal hitting set for S. Since $S \subseteq S^{\prime }$, it follows that $H \subseteq \bigcup S^{\prime }$. Further, H hits every element of $S^{\prime }$, which is evident from the fact that every element of $S^{\prime }$ contains an element of S. Hence H is a hitting set for $S^{\prime }$. By the same arguments, it is not difficult to see that H is minimal for $S^{\prime }$ too.

  (If part) Given that H is a minimal hitting set for $S^{\prime }$, we have to show that it is a minimal hitting set for S too. Assume that there is an element E∈H that is not in $\bigcup S$. This means that E is selected from some $Y \in S^{\prime }\backslash S$. But Y contains an element of S, say X. Since X is also a member of $S^{\prime }$, one member of X must appear in H. This implies that two elements have been selected from Y and hence H is not minimal. This is a contradiction and hence $H \subseteq \bigcup S$. Since $S \subseteq S^{\prime }$, it is clear that H hits every element in S, and so H is a hitting set for S. It remains to be shown that H is minimal. Assume the contrary, that a proper subset $H^{\prime }$ of H is a hitting set for S. Then from the proof of the only if part, it follows that $H^{\prime }$ is a hitting set for $S^{\prime }$ too, and contradicts the fact that H is a minimal hitting set for $S^{\prime }$ . Hence, H must be a minimal hitting set for S.

• 2. (If part) Given that H is a hitting set for $S^{\prime }$, we have to show that it is a hitting set for S too. First of all, observe that $\bigcup S = \bigcup S^{\prime }$, and so $H \subseteq \bigcup S$. Moreover, by definition, for every non-empty member X of $S^{\prime }$, $H \cap X$ is not empty. Since $S \subseteq S^{\prime }$, it follows that H is a hitting set for S too.

  (Only if part) Suppose H is a hitting set for S. As observed above, $H \subseteq \bigcup S^{\prime }$. By definition, for every non-empty member X∈S, $X \cap H$ is not empty. Since every member of $S^{\prime }$ contains a member of S, it is clear that H hits every member of $S^{\prime }$, and hence a hitting set for $S^{\prime }$.

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Proof of Lemma 4 and 7

Proof of Lemma 4 and 7. Follows from the lemma 3, 4 (minimal test) and 6, 7 (materialized view) of [11] □

Proof of Theorem 3

Follows from Lemma 3 and Theorem 2. □

Proof of Theorem 4

Follows from Lemma 6 and Theorem 2. □