By Terry Dartnall (auth.), Grigoris Antoniou, John Slaney (eds.)

This e-book provides the completely refereed post-conference court cases of the eleventh Australian Joint convention on synthetic Intelligence, AI'98, held in Brisbane, Australia in July 1998.
The 28 revised complete papers awarded within the ebook have been rigorously reviewed and chosen from two times as many papers permitted for presentation on the convention.
Among the subjects lined are philosophical matters, fuzzy good judgment, agent structures, AI logics, making plans, wisdom illustration, computerized deduction, clever brokers, studying, constraint fixing, and neural networks.

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Extra resources for Advanced Topics in Artificial Intelligence: 11th Australian Joint Conference on Artificial Intelligence, AI’98 Brisbane, Australia, July 13–17, 1998 Selected Papers

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Consequently, to generate strong cuts, in practice each time a prime implicant is added to the BDD, a nogood is extracted from the new reduced BDD and added to the boolean formula. 3 QB DD (DPLL) Approach The algorithm 2 represents the QBF solver obtained by a combination of DPLL procedure with BDD. As we can see, the algorithm search for all models until the BDD representation (characterized by the global variable bdd initialized to the 0-terminal node) is reduced to a 1-terminal node, in that case the algorithm terminates and answers the validity of the QBF formula.

231–239. [2] H. Kautz and B. Selman. Pushing the Envelope: Planning, Propositional Logic, and Stochastic Search. Proc. of AAAI’96 (1996). [3] E. A. Hirsch, and A. Wolper. Algorithms for SAT based on Search in Hamming Balls. Proc. of STACS 2004, LNCS 2996 (2004), pp. 141–151. J. C. Stavropoulos. An algorithm for generating all maximal models of Boolean expressions. Proc. of the 4th Panhellenic Logic Symposium (2003), pp. 125-129. [5] R. Khardon and D. Roth. Reasoning with Models. Artificial Intelligence 87 (1996), pp.

If h(α(q/T )) = 1, then h(α(q/α(q/T ))) = h(α(q/T )) = 1. - If h(α(q/T )) = 0, then h(q) = 0 (otherwise h(α(q/T )) = h(α) = 1). So h(α(q/T )) = h(q) and h(α(q/α(q/T ))) = h(α) = 1. The proof of (ii) is similar. - If h(¬α(q/F )) = 0, then h(α(q/¬α(q/F ))) = h(α(q/F )) = 1 − h(¬α(q/F )) = 1. - If h(¬α(q/F )) = 1, then h(q) = 1 (otherwise h(¬α(q/F )) = h(¬α) = 0). So h(¬α(q/F )) = h(q) and h(α(q/¬α(q/F ))) = h(α) = 1. L EMMA 2. Let S1 , . . , Sn be the substitution string for a formula α with respect to an enumeration p and a map τ .

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