A new rational algorithm for view updating in relational databases

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. In order to apply the rationality result of belief dynamics theory to various practical problems, it should be generalized in two respects: first it should allow a certain part of belief to be declared as immutable; and second, the belief state need not be deductively closed. Such a generalization of belief dynamics, referred to as base dynamics, is presented in this paper, along with the concept of a generalized revision algorithm for knowledge bases (Horn or Horn logic with stratified negation). We show that knowledge base dynamics has an interesting connection with kernel change via hitting set and abduction. In this paper, we show how techniques from disjunctive logic programming can be used for efficient (deductive) database updates. The key idea is to transform the given database together with the update request into a disjunctive (datalog) logic program and apply disjunctive techniques (such as minimal model reasoning) to solve the original update problem. The approach extends and integrates standard techniques for efficient query answering and integrity checking. The generation of a hitting set is carried out through a hyper tableaux calculus and magic set that is focused on the goal of minimality.

Appendix:

Proof of Theorem 1

Follows from Algorithm 1 and 2. □

Proof of Theorem 2

Follows from the result of [7] □

Proof of Lemma 1

Follows from the result of [5] □

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. 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 ${\Delta }^{\prime }$.

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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

Follows from the lemma 2,3 (minimal test) and 5,6 (materialized view) of [7] □

Proof of Theorem 3

Follows from Lemma 4 and Theorem 1. □

Proof of Theorem 4

Follows from Lemma 7 and Theorem 3. □