By Hee-Kap Ahn, Chan-Su Shin

ISBN-10: 3319130749

ISBN-13: 9783319130743

ISBN-10: 3319130757

ISBN-13: 9783319130750

This e-book constitutes the refereed complaints of the twenty fifth foreign Symposium on Algorithms and Computation, ISAAC 2014, held in Jeonju, Korea, in December 2014.

The 60 revised complete papers awarded including 2 invited talks have been rigorously reviewed and chosen from 171 submissions for inclusion within the booklet. the focal point of the quantity in at the following issues: computational geometry, combinatorial optimization, graph algorithms: enumeration, matching and task, information buildings and algorithms, fixed-parameter tractable algorithms, scheduling algorithms, computational complexity, computational complexity, approximation algorithms, graph conception and algorithms, on-line and approximation algorithms, and community and scheduling algorithms.

**Read or Download Algorithms and Computation: 25th International Symposium, ISAAC 2014, Jeonju, Korea, December 15-17, 2014, Proceedings PDF**

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**Additional resources for Algorithms and Computation: 25th International Symposium, ISAAC 2014, Jeonju, Korea, December 15-17, 2014, Proceedings**

**Example text**

Let I(P, ) be the set of all facility location intervals of P . According to the above discussion, to determine whether is a feasible value, it is suﬃcient to compute a minimum number of points that can cover all intervals of I(P, ), which can be done in O(n) time after the endpoints of all intervals of I(P, ) are sorted [17]. The overall time for solving the decision problem is O(n log n) due to the sorting. Below, we give an O(n) time algorithm, without sorting. Similar to Lemma 1, if is a feasible value, then there exists a feasible solution in which each facility serves a set of consecutive points of P .

If |H| > 5, then let h1 , . . , hk be a cyclic order of H and consider the signature graph GU [H] . Note that we can compute the signature graph in polynomial time using only U [H]. In the digraph DU [H] , the edges (hi , hi+1 ) and (hi+1 , hi ) will both be labeled H \ {hi−1 , hi , hi+1 , hi+2 } and thus {hi , hi+1 } is green in GU [H] for all 1 ≤ i ≤ k. On the other hand, the edge (hi , hj ) will be labeled H \ {hi , hj−1 , hj , hj+1 }, whereas (hj , hi ) will be labeled H \{hi−1 , hi , hi+1 , hj } for |i−j| > 1.

Similarly, Δ2 must contain a or both c and d. Suppose that Δ1 contains both c and d and let e be the third vertex of Δ1 . By the argument in the previous paragraph, we must have e = b. But then the forbidden regions of abd and Δ1 = bcd together cover all of abc. This is a contradiction since |V | ≥ 5. Symmetrically, Δ2 cannot contain both c and d. Hence, Δ1 must contain b and Δ2 must contain a. Furthermore, neither triangle can intersect cd since c or d would be in the forbidden region otherwise.

### Algorithms and Computation: 25th International Symposium, ISAAC 2014, Jeonju, Korea, December 15-17, 2014, Proceedings by Hee-Kap Ahn, Chan-Su Shin

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