By M. Tamer Özsu (auth.), Qing Li, Ling Feng, Jian Pei, Sean X. Wang, Xiaofang Zhou, Qiao-Ming Zhu (eds.)
This publication constitutes the lawsuits of the joint foreign convention APWeb/WAIM 2009 which used to be held in Suzhou, China, in the course of April 1-4, 2009.
The forty two complete papers awarded including 26 brief papers and the abstracts of two keynote speeches have been conscientiously reviewed and chosen for inclusion within the booklet. the subjects lined are question processing, topic-based thoughts, internet information processing, multidimensional info research, circulate information processing, info mining and its purposes, and information administration aid to complex applications.
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Extra info for Advances in Data and Web Management: Joint International Conferences, APWeb/WAIM 2009 Suzhou, China, April 2-4, 2009 Proceedings
Example text
According to Lemma 3, if algorithm SNRA doesn’t halt , it will proceed to access the next level at least in one list until the stopping rule is satisfied. Assume that algorithm SNRA halts after dj sorted access to Lj (j = 1, 2, . . , m) and the objects output by algorithm SNRA are R1 , R2 , . . , Rk . Let R be an object not among R1 , R2 , . . , Rk . We must show that t(R) ≤ t(Ri ) for each i = 1, 2, . . , k. Let d = [d1 , d2 , . . , dm ]. Since algorithm SNRA halts at depth d, the best com(d) petitor ’s best value is less than Mk , then, B (d) (R) ≤ B (d) (best competitor) ≤ (d) (d) Mk and t(R) ≤ B (d) (R).
Suppose that the last best competitor has only one missing field when algorithm NRA is running. In this case, we still have to sorted access the best competitor ’s all other known fields. This will incur lots of useless sorted accesses if the last best competitor continue for a long time before the top-k objects are obtained. These observations motivate us to propose Selective-NRA algorithm. 2 Selective-NRA Algorithm (SNRA) We will show our algorithm in pseudo-code form. 1, algorithm NRA performs some sorted accesses which will definitely not reduce the best competitor ’s best value.
M} with values xi1 , xi2 , . . , xil , compute W (d) (R) = WS (R) = xi if i ∈ S t(w1 , w2 , . . , wm ) where wi = and B (d) (R) = BS (R) = 0 else xi if i ∈ S . t(b1 , b2 , . . , bm ) where bi = (d) xi else (d) • Let Tk , the current top k list, contain the k objects with the largest W (d) values seen so far (and their grades); if two objects have the same W (d) value, then ties are broken using the B (d) values, such that the object with the highest B (d) value wins (and arbitrarily among objects that tie for the highest B (d) (d) (d) value).