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1 algorithm directional resolution is consistent with

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Unformatted text preview: in bucket1 (if the bucket is empty, assign Q1 an arbitrary value); Step i. After assigning values to Q1; :::; Qi01, assign a value to Qi , so that together with the previous assignments it will satisfy all clauses in bucketi. Proof: assignments q1; :::; qi01 for the rst i 0 1 symbols that satis es all the clauses in the buckets Suppose the model generation process is not backtrack-free, i.e there exists a truth of Q1,..., Qi01 , but cannot be extended by any truth value of Qi without falsifying some clauses in bucketi. Let and be two clauses in the bucket of Qi that clash, i.e. cannot be satis ed simultaneously, given the assignment q1 ; :::; qi01. Clearly, and contain Qi with opposite signs; in one Qi appears negatively and in the other positively. Consequently, while being processed by directional-resolution, and should have been resolved upon, thus resulting in a clause that must appear now in a bucketj , j < i. Such a clause, if existed, would not have allowed the partial model q1; :::; qi, thus leading to a contradiction. 2 Corollary 1: [6] A theory has a non-empty directional extension i it is satis able. 2 Clearly, the e ectiveness of directional resolution both for satis ability and for subsequent query processing depends on the the size of its output theory Ed ('). 5 Theorem 2: (complexity ) Given a theory ' and an ordering d of its propositional symbols, the time complexity of algorithm directional resolution is O(n 1 jEd (')j2 ), where n is the number of propositional symbols in the language. Proof: There are at most n buckets, each containing no more clauses than the directional extension of the theory, and resolving pairs of clauses in a bucket is quadratic in its size. 2 The bound above, although it could be loose, demonstrates the dependence of the algorithm's complexity on the size of its resulting output. Once Ed (') is compiled, determining the entailment of a single literal involves checking the bucket of that literal rst. If the literal appears there as a unit clause, it is entailed; if not, the negation of that literal must be added to the theory and the algorithm must be restarted from that bucket. If the empty clause is generated, the literal is entailed. To determine the entailment of an arbitrary clause, the negation of each literal in that clause must be added to the corresponding bucket; then the processing must be restarted from the highest of those buckets. Therefore, in knowledge bases with queries involving a restricted subset of the alphabet the ordering for directional resolution should start with the symbols of that subset. In summary, Theorem 3: (entailment) Given a directional extension Ed (') and a constant c, the entailment of clauses involving only the rst c symbols in d is polynomial in the size of Ed ('). 2 4 Tractable classes Consider the following two examples demonstrating the e ect of ordering on Ed ('). Example 1: Let '1 = f(B; A) ,(C; :A); (D; A); (E; :A)g: For the ordering d1 = (E; B; C; D; A), all...
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This note was uploaded on 09/17/2013 for the course PMATH 330 taught by Professor W. alabama during the Spring '09 term at Waterloo.

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