By Darja Krushevskaja, S. Muthukrishnan (auth.), Leizhen Cai, Siu-Wing Cheng, Tak-Wah Lam (eds.)

This publication constitutes the refereed court cases of the twenty fourth overseas Symposium on Algorithms and Computation, ISAAC 2013, held in Hong Kong, China in December 2013. The sixty seven revised complete papers offered including 2 invited talks have been conscientiously reviewed and chosen from 177 submissions for inclusion within the ebook. the point of interest of the amount in at the following issues: computation geometry, trend matching, computational complexity, web and social community algorithms, graph conception and algorithms, scheduling algorithms, fixed-parameter tractable algorithms, algorithms and information constructions, algorithmic video game conception, approximation algorithms and community algorithms.

**Read Online or Download Algorithms and Computation: 24th International Symposium, ISAAC 2013, Hong Kong, China, December 16-18, 2013, Proceedings PDF**

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**Extra info for Algorithms and Computation: 24th International Symposium, ISAAC 2013, Hong Kong, China, December 16-18, 2013, Proceedings**

**Example text**

1 Introduction A graph G is called Laman if |E(G)| = 2|V (G)|−3 and |E(H)| ≤ 2|V (H)|−3 for any subgraph H of G with E(H) = ∅. A Laman graph has a property of being minimally rigid in the plane if it is realized as a generic bar-joint framework [5,4]. , [4]). , the set of the coordinates is algebraically independent over the rational ﬁeld) is rigid if and only if G contains a spanning Laman subgraph [5]. Laman graphs appear in a wide range of applications, not only statics but also mechanical design such as linkages, design of CAD systems, analysis of protein ﬂexibility, and sensor network localization [11,10].

Lemma 2 in particular implies that, for Q ⊆ P with |Q| = 4, the longest edge in K(Q) does not belong to MLG(P ) since K4 violates the (2, 3)-sparsity condition. This fact will be frequently used. Lemma 3. Let P be a semi-generic point set in the plane, and let ab ∈ MLG(P). (i) Let R be one half of lens(a, b) divided by the edge ab. Then there exists at most one point in the interior of R. (ii) Suppose that there exists one point in each half of lens(a, b) (say, p and q). Then pq > ab holds, and pq ∈ / MLG(P ).

Also, it passes through b, and thus bisect(ac) passes the right of b. Therefore, d lies on the left side of bisect(ac). Thus, ad < cd holds. Now let us consider bisect(ad). Let d be the intersection point of ad and the boundary of lens(ab) and m be the midpoint of ab. Since bisect(ad ) always 38 S. Bereg et al. bisect(ac’) a c’ c d d’ a c’ c 120° m bisect(ad) x y bisect(ac) d’ d b Fig. 3. Illustration of the proof of Lemma 5 b Fig. 4. Illustration of the proof of Lemma 6(i) passes below m, so does bisect(ad).