# Download PDF by Penttonen M., Schmidt E.M.: Algorithm Theory - SWAT 2002

By Penttonen M., Schmidt E.M.

This ebook constitutes the refereed court cases of the eighth Scandinavian Workshop on set of rules thought, SWAT 2002, held in Turku, Finland, in July 2002.The forty three revised complete papers provided including invited contributions have been rigorously reviewed and chosen from 103 submissions. The papers are prepared in topical sections on scheduling, computational geometry, graph algorithms, robotics, approximation algorithms, information verbal exchange, computational biology, and information garage and manipulation.

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Example text

Let a = 2 1/ and δ = rmax /a. Since cost(σ) ≥ rmax , we have δ ≤ cost(σ)/2. Let P be the sum of all edge weights in M. Deﬁne b = 2(t + 1)a2 and ∆ = P/b. Since every edge must be traversed to serve all requests, cost(σ) ≥ P and therefore ∆ ≤ cost(σ)/(4(t + 1)m). We deﬁne a new metric N with a constant number of points, which we use to approximate M. A junction of M is deﬁned to be a vertex of degree three or more. Deﬁne a essential path of M to be path whose endpoints are either leaves or junctions.

There are analogous possible constraints on vehicle ending positions. The most common variant when an ending position is speciﬁed is that the origin and the ending position are the same. We denote this variant as RTO (return to origin). When no ending position is speciﬁed, the most common variant is that each vehicle has a ﬁxed origin. We denote this variant as FO (ﬁxed origin). – In the online variants of these problems, jobs are unknown before their release times, and even the number of jobs is a priori unspeciﬁed.

Xi i in order. If we ﬁx Xij for 1 ≤ j ≤ t, 0 ≤ i ≤ R − 1 then note that this determines XR and TR , since all requests not served in phases 0 . . R − 1 must be served 0 0 is m + 1. Once XR is ﬁxed, it during phase R. The number of choices for XR is easy to determine the remaining schedule in O(n) time, since this is just the Hamiltonian path problem on a tree. For 1 ≤ i < R there are mt+1 + 1 possible choices for Xi0 , , . . , Xit . Therefore, the total number of possible schedules is at most (m+1)(mt+1 +1)R−1 ≤ (m+1)(t+1)(R−1)+1 , which is constant with respect to n.