Maximizing Voronoi Regions of a Set of Points Enclosed in a Circle with Applications to Facility Location
Summary (2 min read)
1 Introduction
- The main objective in any facility location problem is to judiciously place a set of facilities, serving a set of users (or demand points), such that certain optimality criteria are satisfied.
- In the discrete case, the problem of placing a new facility amidst existing ones, such that the number of users served by it is maximized, has been addressed very recently by Cabello et al. [8].
- In Figure 1(a), the authors demonstrate a situation, where a few mobile service stations are plying along a path R (on roads or on rescue ships) surrounding the affected region, supplying provisions to the distressed people.
- This problem can also be considered as an extension of the competitive facility location problems related to Voronoi games [1, 10].
- These two problems can be mathematically modeled as follows.
2 Properties of a Voronoi Zone
- Consider the rectangular coordinate system, with the origin at the point o and the horizontal axis aligned along the diameter of the circle ψ passing through the point p.
- Next, the authors consider the situation where more than one point is placed inside the circle ψ.
3 Problem P1
- The authors begin by showing that under the equidistant assumption Area{⋃ni=1 V R(pi, S∪{ψ})} is maximized when all the n points are placed on the regular n-gon with circumcenter at o.
- Observe that if n arcs of equal length are to be chosen on some circle, then a regular distribution of these arcs about the circle maximizes the length of their union.
- This establishes the concavity of the function F(x) on [0, π] ), Table 1 shows that as n increases the optimum ratio e(n) increases and Figure 4(b)) and at the limit reaches 1 ).
- Table 1 also demonstrates that the optimum area of the combined service zone of all the members in S increases to πr2 asymptotically.
4.1 Exact Solution for n = 1
- Let us choose a coordinate system where the point o is the origin, and the line joining o and the point p is the x-axis.
- Throughout the proof of this lemma, the authors shall refer to Figure 8(a).
4.2 Approximate Solution of Problem P2
- The techniques used in this section emulates the methods of Cheong et al. [9] for approximating the area of a Voronoi region a of new point, given a set of fixed points.
- Using this definition the authors now prove the following lower bound on OPTArea.
- Now, the authors need to define E(x) and describe a method to find the point xQ, for each grid cell Q.
- The authors now make use of the following simple lemma.
- Since the Voronoi diagram V (S ∪ {ψ}) and the largest reach ` can be computed in O(n log n) time [17], the next theorem follows from arguments exactly similar to those in Theorem 3.3 of Cheong et al. [9].
5 Conclusions
- These are motivated from various applications in facility location and disaster management problems, where both stationary and mobile service stations are deployed.
- The interior of the circle is partitioned into the Voronoi region of the points and the Voronoi region of the circle itself.
- Therefore, the special case where the stationary facilities are assumed to be equidistant from the center of the circle, is likely to provide the optimum solution for the general case as well.
- The author wishes to thank Professors Probal Chaudhuri, Sandip Das, and Subhas C. Nandy of the Indian Statistical Institute, Kolkata, and Professor Rolf Klein of the Institut für Informatik I, Universität Bonn, Germany, for their insightful suggestions, also known as Acknowledgement.
- The author is also grateful to the two anonymous referees for their critical comments, which have greatly improved the quality of the paper.
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Cites background from "Maximizing Voronoi Regions of a Set..."
...…to have the same speed, starting move in the same moment and not considering the direction, the K-regions can be approximate to the Voronoi theorem (Bhattacharya, 2010; Dashti, Kamali, & Aghaeepour, 2007; Dehne, Klein, & Seidel, 2005) that identifies all the field’s zone that are closer to the…...
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...It represents the locus of the points that a player could achieve before the others: In the easiest case, where the players are considered all to have the same speed, starting move in the same moment and not considering the direction, the K-regions can be approximate to the Voronoi theorem (Bhattacharya, 2010; Dashti, Kamali, & Aghaeepour, 2007; Dehne, Klein, & Seidel, 2005) that identifies all the field’s zone that are closer to the considered player; the K-regions (see Figure 1) work according to the time, and thus, they assign to the player all the points that he could achieve before the others in terms of time....
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27 citations
19 citations
Cites background from "Maximizing Voronoi Regions of a Set..."
...Variations of this problem, involving maximization of the area of Voronoi regions of a set of points placed inside a circle, have been recently considered by Bhattacharya [7]....
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...Contents lists available at SciVerse ScienceDirect European Journal of Operational Research journal homepage: www.elsevier .com/locate /e jor Discrete Optimization New variations of the maximum coverage facility location problem q Bhaswar B. Bhattacharya a,⇑, Subhas C. Nandy b a Department of Statistics, Stanford University, CA, USA b Advanced Computing and Microelectronics Unit, Indian Statistical Institute, Kolkata 700 108, India a r t i c l e i n f o Article history: Received 23 December 2011 Accepted 12 August 2012 Available online 29 August 2012 Keywords: Reverse nearest neighbor Competitive location Computational geometry Facility location 0377-2217/$ - see front matter 2012 Elsevier B.V....
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...22nd Canadian ba, Canada, 2010, 241–244. m (B.B. Bhattacharya), nan-...
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Frequently Asked Questions (10)
Q2. What are the future works in "Maximizing voronoi regions of a set of points enclosed in a circle with applications to facility location" ?
It would be interesting to study similar maximization problems when the outer circle is replaced by other bounding shapes, for example a convex polygon. Acknowledgement: The author wishes to thank Professors Probal Chaudhuri, Sandip Das, and Subhas C. Nandy of the Indian Statistical Institute, Kolkata, and Professor Rolf Klein of the Institut für Informatik I, Universität Bonn, Germany, for their insightful suggestions.
Q3. What is the problem of maximizing Areani=1 V R(pi,?
The problem of maximizing Area{⋃ni=1 V R(pi, S ∪ {ψ})} is equivalent to choosing angles φ1, . . . , φn such that Area{ ⋃n i=1 Eφi} is maximized, where Eφ denotes a copy of E rotated by an angle φ about the center o in the clockwise direction.
Q4. What is the nearest service station for a user located in Z?
Since [β1, β2] is the perpendicular bisector of the line segment [p1, p2], the nearest service station for a user locatedin Z(p1) = V R(p1, {p1, p2, ψ}) is p1.
Q5. What is the optimum value of e for different values of n?
The optimum values of e for different values of n, to be denoted by e(n), are obtained by solving the equation ddeK(e, n) = 0 numerically using Mathematica 4.0.
Q6. Why is the area of the curve ni=1 Ei a?
This is because, under a regular distribution of the arcs, the length of the curve ⋃n i=1 Eφi ∩Cv is either the length of the circumference of Cv, or the sum of the lengths of Eφi ∩Cv, for every 0 ≤ v ≤ r.
Q7. What is the problem of finding the point xa in the plane?
Theorem 5. Given a set S of n points in the plane and a parameter ε > 0, one can deterministically compute, in time O(n/ε4 +n log n), a point xa such that Area(V R(xa, S∪ {xa, ψ})) ≥ (1− ε)OPTArea.
Q8. What is the limiting case of the circle?
In the other limiting case, when the point p lies on the circle ψ (i.e. b = r) then E is the radius of the circle through p, which can be interpreted as a degenerate ellipse with length of minor axis equal to zero.
Q9. how many points of Q are in the definition of V R(x, S ?
all points of S participating in the definition of V R(x, S ∪ {ψ}) lie in Q and the 24 grid cells at distance at most 2` from it.
Q10. What is the equation of the perpendicular bisector of line segment?
The equation of the perpendicular bisector of line segment [p, q′] (denoted by uv in Figure 8(a)) is y − b sin θ2 = − b cos θ+ab sin θ ( x− b cos θ−a2 ) .