By Eric Aaron, Juan Pablo Mendoza (auth.), Cory Butz, Pawan Lingras (eds.)

This e-book constitutes the refereed complaints of the twenty fourth convention on synthetic Intelligence, Canadian AI 2011, held in St. John’s, Canada, in could 2011. The 23 revised complete papers offered including 22 revised brief papers and five papers from the graduate pupil symposium have been rigorously reviewed and chosen from eighty one submissions. The papers hide a wide diversity of themes proposing unique paintings in all components of man-made intelligence, both theoretical or applied.

**Read or Download Advances in Artificial Intelligence: 24th Canadian Conference on Artificial Intelligence, Canadian AI 2011, St. John’s, Canada, May 25-27, 2011. Proceedings PDF**

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**Additional info for Advances in Artificial Intelligence: 24th Canadian Conference on Artificial Intelligence, Canadian AI 2011, St. John’s, Canada, May 25-27, 2011. Proceedings**

**Sample text**

The task is to find an expansion B of A that satisfies φ: A A A A B (L ∪ K; P , W , WlA , HCap , LA Cap , Q ) |= φ. B Interpretations of the expansion vocabulary ε = {Q}, for structures B that satisfy φ, is a mapping from items to knapsacks that satisfies the problem properties. The grounding task is to produce a ground formula ψ = Gnd(φ, A), such that models of ψ correspond to solutions for instance A. Formally, to ground we bring domain elements into the syntax by expanding the vocabulary with a new constant symbol for each element of the domain.

Example 4. (Continue From Example 3) βRi corresponds to the answer to variables li (i ∈ {1, 2}). So, R1 (R2 ) should have one free variable, namely l1 (l2 ). Having an answer to ti , an answer to ti , (αRi , βRi ), can be computed. By proposition 3, we have βR (2) = βR1 (7) βR2 (5) ∪ βR1 (5) βR2 (3). In other word, the answer to t is 2 if either t1 = 7 ∧ t2 = 5 or t1 = 5 ∧ t2 = 3. 1 Base Case for Complex Terms To extend our grounding algorithm to handle terms which cannot be evaluated out, we add the following base cases to the algorithm.

So, for a given mapping from Dx¯ to αR1 , we need to conjunct the conditions obtained from δR1 to find the necessary and sufficient condition for one of the cases where the output sum is exactly n. And the outside disjunction, finds the complete condition. Although what is described in (7) can be used directly to find an answer for the Sum aggregates, in practice many of the entries in R will be eliminated during grounding as they are joined with false formula or unioned with a true formula. So, to reduce the grounding time, we use a place holder, SUM place holder, in the form SU M (R1 , S, n, γ), as the formula corresponding to δR (γ, n).