A Novel Geometric Diagram and its Applications in ... - Infocom 2004
A Novel Geometric Diagram and Its Applications in Wireless Networks Guangbin Fan* and Jingyuan Zhang† *
Department of Computer and Information Science, University of Mississippi University, MS 38677, Email: [email protected] † Department of Computer Science, University of Alabama Tuscaloosa, AL 35487, Email: [email protected]
Abstract—Wireless networks have a lot of geometric properties since the signal strength corresponds directly to the distance. Earlier research findings in geometry, such as Voronoi diagram, have found a lot of applications in wireless networks. In this paper, a new geometric diagram, referred to as the umbrella diagram, is proposed. The umbrella diagram is comparable to the Voronoi diagram. The Voronoi diagram deals with a set of points, whereas the umbrella diagram deals with a set of different networks. The umbrella diagram of a set of hexagonal networks is to divide the plane into regions such that the points in a region have a larger distance to one network than to the other networks. Unlike the Voronoi diagram, the regions in the umbrella diagram are not necessarily convex. The paper gives an efficient solution to compute the umbrella diagram of n networks with the same cell size and orientation. Like the Voronoi diagram, the umbrella diagram has potential to be used in many areas. To illustrate its usefulness, the paper further gives two applications in wireless networks: the optimal network deployment and the maximum base station reuse. The optimal network deployment intends to find the best position to fix the network such that the network is closest to all existing base stations. The maximum base station reuse problem is to maximize the number of base stations that can be reused under a given bound. Keywords-Voronoi Diagram; Umbrella Diagram; Wireless Networks; Computational Geometry
I. INTRODUCTION It is well known that there are two popular tessellations of a plane with regular polygons of the same kind: square and hexagonal [11]. In the square tessellation, each square has four (or eight) neighbors, and in the hexagonal tessellation, each hexagon has six neighbors. Fig. 1.1 illustrates both square and hexagonal tessellations. A tessellation can also be represented by the centers of its corresponding polygons, shown as the dots in Fig. 1.1. A lot of designs are based on those two popular tessellations. For example, mesh-connected computers are based on the square tessellation. The design of cellular networks is based on the hexagonal tessellation. In this paper we will consider the hexagonal tessellation, which closely approximates the circular radiation patterns and achieves the maximum coverage with a given number of nodes. Specifically, we will propose and compute the umbrella diagram of a set of different hexagonal tessellations, and show how to use the umbrella diagram to solve problems in the area
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of wireless networks. (The reason that the diagram is named as the umbrella diagram will be explained after its formal definition.)
Figure 1.1 Square and hexagonal tessellations.
To understand the umbrella diagram, we first introduce the well known closest-point and farthest-point Voronoi diagrams [9]. We assume a set S of n points in the plane, S = {P0, P1,…,Pn-1}. The closest-point Voronoi diagram (or simply Voronoi diagram) of S is to find, for each point Pi in S, the locus of points that are closer to Pi than to any other points of S. The farthest-point Voronoi diagram of S is to find, for each point Pi in S, the locus of points that are farther to Pi than to any other points of S. The closest-point diagram of S divides the plane into n convex regions, each of which is associated with a point in S. Although the farthest-point diagram of S also divides the plane into convex regions (unbounded), each region is associated with a point in S that is located on the convex hull of S. A point in S that is within the convex hull of S does not affect the diagram [9]. Fig. 1.2 illustrates the closest-point and farthest-point Voronoi diagrams of 8 planar points. The left figure corresponds to the closest-point diagram that consists of 8 regions. The right figure, corresponding to the farthest-point diagram, consists of only 5 regions. The points located within the convex hull, P5, P6 and P7, do not have an associated region. The Voronoi diagrams have found a lot of applications in areas ranging from archaeology to zoology [1, 4-8]. It can be used in graphics, image processing, wireless networks, robotics, motion planning, pattern recognition, etc. In [2], the author has presented a detailed survey of Voronoi diagrams and their applications. Voronoi diagrams are essential for
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wireless networks. In a cellular network, the area covered by a base station can be viewed as a Voronoi region of the base station. Voronoi diagram has also been used in mobile ad hoc networks and sensor networks. In [8], a localized algorithm is developed based on the Voronoi diagram to solve the exposure problems in sensor networks.
Figure 1.2 The closest-point and farthest-point Voronoi diagrams of a set of planar points.
The proposed umbrella diagram is comparable to the Voronoi diagram. The Voronoi diagram deals with a set S of n planar points, whereas the umbrella diagram deals with n networks (patterns, or tessellations), each of which is determined by a point in S. In this paper, we assume honeycomb networks. It is easy to see that a honeycomb network is determined by three factors: cell size, network orientation, and the position of one cell center. Therefore, if the cell size and orientation is fixed, one single point will determine a network. A set of n points will determine n networks (or n tessellations). The umbrella diagram of a set S of n planar points divides the plane into regions. Each region is associated with a point in S. The points in the region associated with the point p in S have a larger distance to the network determined by p than to the networks determined by the other points in S. Here the distance from a point to a network is the distance between the point and its closest cell center in the network. Please refer to Fig. 2.2 for an illustration of the umbrella diagram of 3 planar points. We will give an efficient solution to compute the umbrella diagram of n planar points with a time complexity of O(n3). The umbrella diagram can be used to solve problems in wireless networks. In this paper, we apply the diagram to solve two problems in wireless networks. One problem, referred to as the optimal network deployment, can be described as: given a set of existing base stations that are non-uniformly distributed, and a planned cellular network, find a deployment of the planned network such that the maximal distance between the existing base stations and their corresponding cell centers is minimized. The second application considers more constraints in network planning. In practice, most system designs permit a base station to be positioned up to one-fourth the cell radius away from the cell center to ensure an acceptable coverage [10]. Therefore the optimal network computed in the first application might not meet this constraint for all base stations. We assume the maximum allowed distance to be a constant d. So the problem, referred to as the maximum base station reuse, is to deploy the network such
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that number of nodes (old base stations) that can be located within distance d to their corresponding closest cell center is maximized. To the best of our knowledge, there is no solution to those problems reported in the literature. These problems can get solved utilizing the proposed umbrella diagram. One uses one application of the umbrella diagram of multiple nodes, and the other uses multiple applications of the umbrella diagram of two nodes. Although the solutions are given in cellular networks, the idea in general describes the important relationship between the maximum coverage (honeycomb) tessellation and the arbitrarily distributed nodes, which can be further used in the wireless ad hoc networks, image processing, etc. The rest of this paper is organized as follows. In Section 2, we define the umbrella diagram of a set of planar points. In Section 3, we show how to find the umbrella diagram. We further apply this diagram to solve two problems in wireless networks in Section 4. Finally we will summarize the results in Section 5. II. DEFINITION OF THE UMBRELLA DIAGRAM In this section, we will define the umbrella diagram of a set of planar points. The hexagonal plane tessellation divides the plane into hexagons as shown on the right side of Fig. 1.1. By using the hexagon geometry, the fewest number of cells can cover a geographic area, and the hexagon closely approximates a circular radiation pattern which would occur with an omni-directional antenna and free space propagation [10]. The hexagonal tessellation is also known as the honeycomb network [11]. In the following discussion, we will omit the word honeycomb when no confusion will arise. A network can be represented by a set of hexagons or a set of hexagon centers. A network is completely determined by three factors: the cell size, network orientation and the position of one cell center. If the cell size and orientation is fixed, the hexagon is solely determined by the position of one cell center. We define a network with a fixed cell size and orientation as movable. Definition 2.1 A network is movable if the cell size and orientation of the network is fixed but the position of the network is not fixed. In the following discussion, we will also refer to a movable network as a network pattern. Given a network pattern, if we fix the center of one cell at the position C, the network is completely fixed. We call the fixed network as the spreading network of C. Definition 2.2 Given a position C and a network pattern T, the spreading network of C with the pattern T, denoted as T(C), is the network with C being one of its cell centers. C is referred to as the starting point of T(C). Fig. 2.1 illustrates two spreading networks, T(P0) and T(P1), with a network pattern T. Next we will define the distance from a point to a network. It can be used to describe
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the error between a base station and its ideal position in a newly planned network.
Figure 2.2 The umbrella diagram of P0, P1, and P2. Figure 2.1 Two spreading networks, T(P0) and T(P1), with a network pattern T.
The umbrella diagram can be also defined from another point of view as follows:
Definition 2.3 Given a point p and a network N, the distance between the point p and the network N, denoted by d(p,N), is defined as the distance between the point p and its closest cell center of N. d(p,N) is also called the distance from p to N.
Definition 2.6 Given a set S of n planar points, S = {P0, P1,…,Pn-1}, and a network pattern T, the umbrella diagram of S divides the plane into regions, each of which is associated with a point Pi in S, such that the spreading network starting from a point in the umbrella region of Pi is farther from Pi than any other points in S.
It is easy to verify that if the point p is located within a cell with its center being c, the network distance between p and N, d(p,N), is the distance between p and c since c is the closest cell center to p. Finally, we will define the umbrella diagram of a set of planar points. We will define the diagram of two points, followed by the diagram of multiple points. Definition 2.4 Given two planar points P0 and P1, and a network pattern T, the umbrella diagram of P0 and P1 consists of two regions, denoted as AT(P0, P1) and AT(P1, P0), where AT(P0, P1) is the locus of points x such that d(x, T(P0)) ≥ d(x, T(P1)), and AT(P1, P0) is the locus of points x such that d(x, T(P1)) ≥ d(x, T(P0)). The diagram is named as the umbrella diagram because of the umbrella shape of the 3d surface with equation z = x2 + y2, cut by a hexagon moving along the z axis. Definition 2.5 Given a set S of n planar points, S = {P0, P1,…,Pn-1}, and a network pattern T, the umbrella diagram of S is a set of n regions, {AT(Pi, S) | 0 ≤ i ≤ n-1 }, where AT(Pi, S) is the locus of points x such that d(x, T(Pi)) ≥ d(x, T(Pj)) for all j ≠ i. AT(Pi, S) is called the umbrella region of Pi . Fig. 2.2 illustrates the umbrella diagrams of 3 points. For clarity, we use U(Pi) instead of AT(Pi, S) in the figure. The polygons of the same shape belong to the umbrella region of a point. We mark only a few polygons of each umbrella region in the figure.
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III. COMPUTING THE UMBRELLA DIAGRAM In this section, we will show how to compute the umbrella diagram of a set of planar points with a network pattern T. We first show how to find the umbrella diagram of two points. Then we extend our idea to find the umbrella diagram of a set of points. Finally we will present the algorithm and analyze its time complexity. A. Umbrella Diagram of Two Points Given two points, P0 and P1, and a network pattern T, we will show how to compute AT(P0, P1) and AT(P1, P0). AT(P0, P1) consists of all points that have a larger distance to T(P0) than to T(P1). Since the network T(Pi), i = 0 or 1, is the repeating of the hexagon centered at Pi, the regions AT(P0, P1) and AT(P1, P0) are also repeatable. We will show how to compute AT(P0, P1). Without loss of generality, we will consider the points within the hexagon centered at P0, referred to as H(P0). Assume P is a point located within H(P0). It is easy to verify that d(P, T(P0)) = |PP0|. Next we will compute d(P, T(P1)). We assume H(P0) is not identical to any hexagon of the network T(P1). If they are identical, it is obvious that d(P, T(P1)) is always the same as d(P, T(P0)). Under that assumption, H(P0) will intersect with either 3 or 4 hexagons of T(P1) as shown in Fig. 3.1.
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most five edges. Therefore, AT(P0, P1) within H(P0) consists of 4 convex polygons, each of which has at most five edges.
Figure 3.1 Two kinds of intersections between H(P0) and T(P1).
We will consider the 4-intersection case with the 3intersection case being similar. Assume the 4 hexagons of T(P1) that are intersected with H(P0) are centered at Q0, Q1, Q2, and Q3 as shown in Fig. 3.2. Since H(Q0), H(Q1), H(Q2), and H(Q3) completely cover H(P0), we only need to consider H(Q0), H(Q1), H(Q2), and H(Q3) when computing d(P, T(P1)). Specifically we have the following. Lemma 3.1 Given two points, P0 and P1, and a network pattern T, assume Q0, Q1, Q2, and Q3 are the centers of the hexagons in T(P1) that intersect with H(P0). For any position P ∈ H(P0), d(P, T(P0)) = |PP0| and d(P, T(P1)) = |PQi| if P ∈ Ii = H(P0) I H(Qi). Proof. If P ∈ Ii= H(P0) I H(Qi), then P ∈ H(P0) and P ∈ H(Qi). Therefore d(P, T(P0)) = |PP0|, and d(P, T(P1)) = |PQi|.♦
However, in order to compute the diagram of multiple points based on the diagram of two points, we will compute AT(P0, P1) and AT(P1, P0) a little bit differently. We will find the intersections of the bisector line segments (or their extensions). The bisector line segments li and li+1 (or their extensions) intersects on the boundary of cells H(Qi) and H(Qi+1). This is because the intersection is the circumcenter of the triangle P0QiQi+1. If H(P0) intersects with 4 hexagons in T(P1), the polygon formed by the perpendicular bisectors (or their extensions) will be a quadrilateral; if H(P0) intersects with 3 hexagons in T(P1), the polygon formed will be a triangle as shown in Fig. 3.1. Denote the formed polygon by L(P0,P1) and the area outside L(P0,P1) by ~L(P0,P1). In addition, the contour of L(P0,P1) is visible from P0. We summarize our finding in the following theorem. Theorem 3.2 Given two points, P0 and P1, and a network pattern T, AT(P0,P1) = ~L(P0,P1) I H(P0) and AT(P1,P0) = L(P0,P1)
I H(P0) within H(P0). ♦
Equivalently, if we compute AT(P0, P1) and AT(P1, P0) within H(P1), then AT(P0, P1) = L(P1,P0) I H(P1) and
I H(P1). If we put together the I H(P0), the resulting shape will be exactly the same as AT(P0, P1) I H(P1). It is easy to see that
AT(P1,P0) = ~L(P1,P0) polygons of AT(P0,P1)
AT(P0, P1) and AT(P1, P0) can be computed in constant time. Fig. 3.3 illustrates the umbrella diagrams of P0 and P1. For clarity, we use U(P0) as AT(P0, P1) and U(P1) as AT(P1, P0) in the figure. The polygons of the same shape belong to the umbrella region of the same point. We mark only a few polygons of each umbrella region in the figure.
Figure 3.2 The construction of L(P0, P1).
Ii = H(P0) I H(Qi) is a convex polygon because both H(P0) and H(Qi) are convex polygons. It is easy to verify that Ii consists of at most six edges. To compute the umbrella diagram of P0 and P1, we construct the perpendicular bisector li of P0Qi in Ii= H(P0) I H(Qi) as shown in Fig. 3.2. By
Lemma 3.1, for any point P ∈ Ii = H(P0) I H(Qi), if P is on the line segment li, d(P, T(P0)) = |PP0| = |PQi| = d(P, T(P1)); if P is on the side that is closer to P0, d(P, T(P0)) = |PP0| < |PQi| = d(P, T(P1)); if P is on the side that is farther to P0, d(P, T(P0)) = |PP0| > |PQi| = d(P, T(P1)). It is also easy to see that Ii is divided by li into two convex polygons, each of which has at
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Figure 3.3 The umbrella diagram of P0 and P1.
B. Umbrella Diagram of a Set of Points In this subsection, we assume a set S of n planar points, S = {P0, P1,…,Pn-1}. For each point Pi, we will find AT(Pi, S), the locus of points that have a larger distance to the spreading network of Pi than to the spreading network of any other
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points in S. In the previous subsection, we have shown how to compute the umbrella diagram of two points. Specifically for any two points Pi and Pj with j ≠ i, we know how to compute AT(Pi, Pj). By definition, AT(Pi, S) =
IA
T
( Pi , Pj ) . We will
j ≠i
show how to compute AT(Pi, S) within H(Pi). By Theorem 3.2, for j ≠ i, AT(Pi,Pj) = ~L(Pi,Pj) I H(Pi) within H(Pi), Therefore, within H(Pi), AT(Pi, S) =
IA
T
( Pi , Pj ) =
j ≠i
=
j
i
The algorithm to compute the umbrella diagram of n planar points can be described as follows.
j ≠i
H ( Pi ) I (I ~ L( Pi , Pj ))
Algorithm: Umbrella Diagram
j ≠i
=
H ( Pi ) I (~ U L( Pi , Pj ) ). j ≠i
Let
L(Pi,
S)
=
U L( P , P ) , i
j
then
AT(Pi,
S)
=
j ≠i
H ( Pi ) I ~L(Pi, S). We summarize this in the following theorem. Theorem 3.3 Given a set S of n planar points, S = {P0, P1,…,Pn-1}, and a network pattern T, AT(Pi, S) within H(Pi) is equal to
Once the umbrella regions of all Pi ∈ S are computed, the umbrella diagram can be formed. As illustrated in Fig. 3.5, the umbrella regions are likely to be non-convex. Furthermore, the region of a point, e.g. U(P0) in the figure, can consists of multiple polygons of different shape. C. The Algorithm and Its Time Complexity
I (~ L( P , P ) I H ( P )) i
The contour of L(P0, S) is shown in a darker color in the figure. The umbrella region of P0 within H(P0), which is represented by the hatched part in the figure, consists of all points that are outside L(P0, S) yet inside H(P0).
H ( Pi ) I ~ L(Pi, S) with L(Pi, S) =
U L( P , P ) .♦ i
j
j ≠i
Input: a set S of n planar points, S={P0, P1,…,Pn-1}, and a network pattern T; Output: the umbrella diagram of S, that is, AT(Pi, S) within H(Pi) for 0 ≤ i ≤ n-1. BEGIN FOR every point Pi ∈ S FOR every point Pj ∈ S ( j ≠ i ) Find the cell centers Q0,…,Qm-1 (m = 3 or 4) of T(Pj), the spreading network of Pj, such that H(Qk) intersects with H(Pi); Compute L(Pi,Pj) formed by the perpendicular bisectors of Pi Qk; END FOR; Compute L(Pi, S) = L( Pi , Pj ) ;
U j ≠i
Compute AT(Pi, S) = ~ L(Pi, S)
I H ( Pi ) ;
END FOR; END.
Figure 3.4 Computing AT(P0, S) within H(P0) with S = {P0, P1, P2, P3}.
Theorem 3.3 tells us how to compute the umbrella region of Pi, AT(Pi, S), within H(Pi). To compute AT(Pi, S) within H(Pi), we first find L(Pi,Pj) for every j ≠ i. Recall that L(Pi,Pj) is a triangle or a quadrilateral whose contour is visible from Pi. Then we compute L(Pi, S) = L( Pi , Pj ) . Finally find the part
U j ≠i
of H(Pi) that is located outside of L(Pi, S). Fig. 3.4 illustrates how to compute AT(P0, S) within H(P0) with S = {P0, P1, P2, P3}. L(P0, S) is the union of L(P0,P1), L(P0,P2), and L(P0,P3).
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Figure 3.5 Umbrella diagram of S = {P0, P1, P2, P3}.
Now we analyze the time complexity of the algorithm. Given two points Pi and Pj (j ≠ i), polygon L(Pi,Pj) is either a triangle or a quadrilateral which can be computed in constant
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time. Therefore, for a point Pi, it takes O(n) time to compute all the polygons L(Pi,Pj) for 0 ≤ j ≤ n-1 and j ≠ i. Next we will show how long it takes to compute L(Pi, S) =
U L( P , P ) . Lemma 3.4 shows that L(P , S) is a star-shaped i
i
j
j ≠i
polygon with O(n2) edges. Lemma 3.4 The outer contour of L(Pi, S) =
U L( P , P ) is a i
j
j ≠i
star-shaped polygon visible from Pi with at most 4(n-1) vertices.
2
Proof. For simplicity, we call L(Pi,Pj) as Lj. We will prove L(Pi, S) is a star-shaped polygon visible from Pi by induction. Suppose C(k) is the union of k Ls. We will compute C(k+1) that is the union of C(k) and Lk+1. We assume that p is an intersection of C(k) with Lk+1. Because p is visible from Pi in both C(k) and Lk+1, p is visible from the Pi in C(k+1). Therefore, all end-points in the union C(k+1) are visible from Pi, that is, C(k+1) is a star-shaped polygon. That concludes L(Pi, S) is a star-shaped polygon visible from Pi. Now we compute the number of vertices on L(Pi, S) in the worst case. An end-point on the outer contour of L(Pi, S) is either the intersection of any two Ls, or an original vertex of L. There are three cases concerning the intersections for a pair of Ls, L(Pi,Pa) and L(Pi,Pb), as shown below. L(Pi,Pa) ∩ L(Pi,Pb)
get the outer contour of the original n Ls. We first merge the
angle list θ 1, θ 2,…, θ u and the angle list θ ’1, θ ’2,…, θ ’v. Note that C(A) and C(B) can have at most one intersection between any two consecutive angles in the resulting list. Without lost of generality, we only need to consider 6 cases for two consecutive angles α and β as shown in Figure 3.6. (Some symmetric cases have not been listed.) Case a: Keep the end points at α and
β
.
Case b: Discard the end points at α and β . Case c: Find the intersection, and let γ be the angle for the intersection. Discard the end point at α , keep the end point at β , and add a new point at γ . Case d: Discard the end point at α , and keep the end point at β. Case e: Find the intersection, and let γ be the angle for the
intersection. Keep the end points at α and β , add a new point at γ . Case f: Find the intersection, and let γ be the angle for the intersection. Discard the end points at and add a new point at γ .
α
and
β
,
Number of intersections
triangle ∩ quadrilateral
≤6 ≤6 ≤8
triangle ∩ triangle quadrilateral ∩ quadrilateral
For each case, the number of intersections is no more than 8. There are (n-1)(n-2)/2 pairs of Ls. Therefore, the total number of end points on the outer contour of L(Pi, S) is at most 8(n1)(n-2)/2 + 4(n-1) = 4(n-1)2. ♦ Next we show that, given a point Pi, it takes O(n2) time to compute the outer contour of L(Pi, S) = U L( P , P ) . We apply i
j
j ≠i
the divide and conquer approach. Again we denote L(Pi,Pj) as Lj, and assume Pi is at the origin o. To find the outer contour of n Ls, we divide the n Ls into two groups, A and B, each of n/2 Ls. Assume the outer contours of A and B are C(A) and C(B) respectively. Realizing each end point p can be represented by the distance ρ between p and the origin o and the angle
θ
made by op with the x axis, the outer contour of k
ρ , 2 θ 2>,…, < ρ k, θ k> with θ 1 < θ 2 , < ρ 2, θ 2>,…, < ρ u, θ u> and C(B) = < ρ 1, θ ’1>, 2 ρ ρ < 2, θ ’2>,…, < v, θ ’v>. By Lemma 3.4, u ≤ ( n − 2) vertices can be represented by a list
,
d(P1, T(x)). So for any network with a starting point located outside L(P0,P1) but within H(P0), if P0 can be reused, P1 can be reused too because the network distance of P0 is greater than that of P1. Now we show how to incorporate the bound d into the contour. An old base station can be reused if and only if it is within the bound of a nearby cell center. If we draw a circle around the old base station P0 with the radius being the bound d as shown in Fig. 4.4, the circled area stands for the reuse region of P0. For any network with a starting point within the circle, it is obvious that P0 can be reused. Therefore P1 can be reused too because the network distance of P0 is greater than that of P1. Lemma 4.2 summarizes the above discussion. Lemma 4.2 Given two base stations Pi and Pj, the area that is outside L(Pi,Pj) but inside the circle that centered at Pi is the locus to deploy the cellular network such that both Pi and Pj can be reused subject to the condition that Pi has an equal or larger network difference than Pj. ♦
For a given bound d, it is possible that no matter how we design the network, not all the old base stations can be located within the bound. Then the problem is how to find the
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difference. The following theorem summarizes our findings concerning our sub-optimal solutions.
Figure 4.4 Bounded L(P0, P1) for P0 and P1.
There could be three cases of intersection between L(Pi,Pj) and the circle centered at Pi: (a). L(Pi,Pj) is totally outside the circle; (b). L(Pi,Pj) is totally inside the circle; (c). L(Pi,Pj) is intersecting with the circle. If L(Pi,Pj) is totally outside the circle, it means Pi cannot be reused in any network that has a larger distance to Pi than to Pj. If the L(Pi,Pj) is totally inside the circle, it means that both Pi and Pj can reused in a network that has a larger distance to Pi than to Pj. In this case, the area that is outside of L(Pi,Pj) but inside the circle is the locus to deploy a cellular network in which both old base stations can be reused subject to the condition that Pi have a equal or larger network difference than Pj. Finally, if L(Pi,Pj) is intersecting with the circle as shown in Fig. 4.4, it means that the area that is outside L(Pi,Pj) but inside the circle is the locus to deploy a cellular network in which both old base stations can be reused subject to the condition that Pi has a equal or larger network difference than Pj. Next, we will present our solution to the problem of maximum base station reuse. For each old base station Pi, we will show how to get the circular arcs that will be used to solve the sub-maximum base station reuse problem that is subject to the condition that Pi has the largest network difference. The maximum number of base stations that can be reused can be found by combining sub-optimal solutions. Suppose we have computed all the L(P0,Pj) for all j ≠ 0 , and draw them around P0 as shown in Fig. 4.5. Next we add a circle around P0 with the radius being the bound. Now if we walk from a point x on the circle straight towards the center P0, the number of lines we encounter is one less than the number of base stations that can be reused if the new network is deployed at x. If the encountered line is a part of L(P0,Pj), Pj can be reused. Note P0 can always be reused. By walking from all points on the circle, we can find the maximum number of base stations that can be reused. Of course all statements here are subject to the condition that P0 has the largest network
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Figure 4.5 Inspecting maximum reused base stations.
Theorem 4.3 Given a set S of n base stations, S={P0, P1,…,Pn1}, if the maximum number of contours L(P0,Pj) with j ≠ 0 that intersect a radius starting from P0 is k, the maximum number of base stations that can be reused is k+1 under the condition that P0 has the largest network difference. Proof. Subject to the condition that P0 has the largest network difference, we know that the network starting point must be located outside the contour L(P0,Pj). In terms of P0 if the starting point is inside the bounding circle centered at P0, P0 can be reused. Otherwise, P0 cannot be reused. For a base station Pj other than P0, if the starting point is outside the contour L(P0,Pj), Pj can be reused as long as P0 can be reused because, in this case, nd(P0,T) is greater than nd(Pj,T). Therefore, if we connect a point on the circle to the center P0, the number of intersecting contours corresponds to the number of base stations that can be reused subject to the condition that P0 is farther from the network than other base stations. It is easy to see that if the maximum number of contours that can be intersected with a radius is k, the maximum number of base stations that can be reused for the given bound and subject to the condition that P0 has the largest network difference is k+1. ♦
For example, in Fig. 4.5, if we walk from point A towards the center P0, only L(P0, P1) is encountered. Therefore, if the network is deployed at A, only old base stations P0 and P1 can be reused. However, if we walk from point B towards the center, L(P0, P1), L(P0, P2) and L(P0, P3) will be encountered. Therefore, in this case, the base stations, P0, P1, P2 and P3, can be reused when the network is deployed at B. However, it is impossible for us to inspect every radius to find the maximum number of base stations that can be reused. Next, we will use circular arcs to solve the maximum base station reuse problem. As shown in Fig. 4.5, we consider the intersections of the circle and L(P0,Pj). Each intersection chord corresponds to a circular arc [12]. L(P0,Pj) introduces at most 4 circular arcs. L(P0,P1), L(P0,P2),…, L(P0,Pn-1) will introduce
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a total of 4(n-1) circular arcs. To find the maximum number of circular arcs that intersect with a radius, we can first sort all the arc end points. Then we use a sweep line to get the maximum number of the arcs that intersect with the sweep line. The maximum number of overlapping arcs for n circular arcs can be found in O(nlogn) time. For each old base station Pi, we can compute the maximum number of old base stations that can be reused under the condition that Pi has the largest network difference. Next we will try to solve the maximum base station reuse problem without any condition. Assume Tmax is the deployed network in which the maximum number of base stations can be reused, and Pk is the old base station that has the largest network difference to Tmax. Then the maximum number of old base stations can be reused is the same as the maximum number of old base stations can be reused under the condition that Pk is farthest from the network. The following theorem summarizes how to find the solution to the maximum reuse problem. Theorem 4.4 Given a set S of n base stations, S={P0, P1,…,Pn1}, if ki (i=0..n-1) is the maximum number of old base stations that can be reused under the condition that Pi has a larger network difference than all other old base stations, and km is the maximum among all ki (i=0..n-1), km is the maximum number of base stations that can be reused within the given bound. ♦
Next we consider the time complexity of the algorithm. As we know, given two points Pi and Pj (j ≠ i), polygon L(Pi,Pj) is either a triangle or quadrilateral which can be computed in constant time. For each Pi, we have a circular arc graph of a maximum of 4(n-1) arcs. The maximum number of arcs that can intersect with a radius can be found in O(nlogn). There are n circular arc graphs, each of which corresponds to an old base station. Therefore it takes O(n2logn) time to compute the maximum number of old base stations that can be reused. V. SUMMARY
Computational geometry has found a lot of applications in the area of wireless networks. In a cellular network, the area served by a base station can be viewed as a Voronoi region of that base station in the Voronoi diagram. In this paper, we have proposed a new geometric diagram, referred to as the umbrella diagram. The umbrella diagram is comparable to the famous Voronoi diagram. Whereas the Voronoi diagram deals with a set of points, the umbrella diagram deals with a set of different networks, each of which is derived from a point. The umbrella diagram of a set of hexagonal networks is to divide the plane into regions such that the points in a region have a larger distance to one network than to the other networks. The distance from a point to a network is the distance between the point and its closest cell center of the network. Unlike the Voronoi diagram, the regions in the umbrella diagram are not necessarily convex. The paper shows that the umbrella diagram of n networks with the same cell size and orientation (or the umbrella diagram of n points under a network pattern) can be computed in O(n3) time.
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The umbrella diagram characterizes the relationship between uniform patterns and arbitrarily distributed nodes and it can be used to solve many problems in wireless networks. In this paper we show how to use the diagram to solve two problems: the problem of the optimal deployment of a planned network, and the problem of maximum base station reuse with a given bound. The optimal deployment of a planed network is to minimize the distance between the existing base stations and the deployed network if all existing base stations are to be reused. The maximum base station reuse problem is to find the maximum number of old base stations that can be reused under a given bound. These problems can be solved via the proposed umbrella diagram elegantly. Although two applications in wireless networks are given in this paper, the ideas have potential to be used in other areas. REFERENCE [1]
E. U. Acar, H. Choset and P. N. Atkar, “Complete sensor-based coverage with extended-range detectors: a hierarchical decomposition in terms of critical points and Voronoi diagrams,” Proceedings of 2001 IEEE/RSJ International Conference on Intelligent Robots and Systems, 2001, pp.1305 -1311.
[2]
F. Aurenhammer, “Voronoi diagrams - a survey of a fundamental geometric data structure,” ACM Computing Surveys, vol. 23, no. 3, Sept. 1991, pp. 345-405.
[3]
U. Black, Mobile and Wireless Networks, Prentice Hall, 1996.
[4]
B. Debaque, S. Gobert, G. Ruckebusch and G. Stamon, “Strong-fromweak model sensor estimation using Voronoi diagrams,” Proceedings of International Conference on Image Analysis and Processing, 1999, pp. 1067-1072.
[5]
C. Gentile and M. Sznaier, “An improved Voronoi-diagram-based neural net for pattern classification,” IEEE Transactions on Neural Networks, vol. 12, no. 5, September 2001, pp. 227-1234.
[6]
T. Jeon and J. Moon, “A systematic approach to signal space detection,” IEEE Transactions on Magnetics, vol. 33, Sept. 1997, pp. 2737-2739.
[7]
Q. Li and C. N. Georghiades, “On a geometric view of multiuser detection for synchronous DS/CDM channels,” IEEE Transactions on Information Theory, vol. 46, no. 7, Nov. 2000, pp. 2723 -2731.
[8]
S. Meguerdichian, S. Slijepcevic, V. Karayan and M. Potkonjak, “Localized Algorithms in Wireless Ad-Hoc Networks: Location Discovery and Sensor Exposure,” 2001 ACM Symposium on Mobile Ad Hoc Networking & Computing, Oct. 2001, pp. 106-116.
[9]
F. P. Preparata and M. I. Shamos, Computational Geometry: An Introduction, Springer-Verlag, 1985.
[10] T. S. Rappaport, Wireless Communications-Principles and Practice, Prentice Hall, 1996. [11] I. Stojmenovic, “Honeycomb networks: Topological properties and communication algorithms,” IEEE Transactions on Parallel and Distributed Systems, vol. 8, no. 10, October 1997, pp. 1036-1042. [12] U. I. Gupta, D. T. Lee and J. Y. T. Leung, “Efficient algorithms for interval circular-arc graphs,” Networks, vol. 12, 1982, pp. 459-467.
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Department of Computer and Information Science, University of Mississippi University, MS 38677, Email: [email protected] † Department of Computer Science, University of Alabama Tuscaloosa, AL 35487, Email: [email protected]
Abstract—Wireless networks have a lot of geometric properties since the signal strength corresponds directly to the distance. Earlier research findings in geometry, such as Voronoi diagram, have found a lot of applications in wireless networks. In this paper, a new geometric diagram, referred to as the umbrella diagram, is proposed. The umbrella diagram is comparable to the Voronoi diagram. The Voronoi diagram deals with a set of points, whereas the umbrella diagram deals with a set of different networks. The umbrella diagram of a set of hexagonal networks is to divide the plane into regions such that the points in a region have a larger distance to one network than to the other networks. Unlike the Voronoi diagram, the regions in the umbrella diagram are not necessarily convex. The paper gives an efficient solution to compute the umbrella diagram of n networks with the same cell size and orientation. Like the Voronoi diagram, the umbrella diagram has potential to be used in many areas. To illustrate its usefulness, the paper further gives two applications in wireless networks: the optimal network deployment and the maximum base station reuse. The optimal network deployment intends to find the best position to fix the network such that the network is closest to all existing base stations. The maximum base station reuse problem is to maximize the number of base stations that can be reused under a given bound. Keywords-Voronoi Diagram; Umbrella Diagram; Wireless Networks; Computational Geometry
I. INTRODUCTION It is well known that there are two popular tessellations of a plane with regular polygons of the same kind: square and hexagonal [11]. In the square tessellation, each square has four (or eight) neighbors, and in the hexagonal tessellation, each hexagon has six neighbors. Fig. 1.1 illustrates both square and hexagonal tessellations. A tessellation can also be represented by the centers of its corresponding polygons, shown as the dots in Fig. 1.1. A lot of designs are based on those two popular tessellations. For example, mesh-connected computers are based on the square tessellation. The design of cellular networks is based on the hexagonal tessellation. In this paper we will consider the hexagonal tessellation, which closely approximates the circular radiation patterns and achieves the maximum coverage with a given number of nodes. Specifically, we will propose and compute the umbrella diagram of a set of different hexagonal tessellations, and show how to use the umbrella diagram to solve problems in the area
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of wireless networks. (The reason that the diagram is named as the umbrella diagram will be explained after its formal definition.)
Figure 1.1 Square and hexagonal tessellations.
To understand the umbrella diagram, we first introduce the well known closest-point and farthest-point Voronoi diagrams [9]. We assume a set S of n points in the plane, S = {P0, P1,…,Pn-1}. The closest-point Voronoi diagram (or simply Voronoi diagram) of S is to find, for each point Pi in S, the locus of points that are closer to Pi than to any other points of S. The farthest-point Voronoi diagram of S is to find, for each point Pi in S, the locus of points that are farther to Pi than to any other points of S. The closest-point diagram of S divides the plane into n convex regions, each of which is associated with a point in S. Although the farthest-point diagram of S also divides the plane into convex regions (unbounded), each region is associated with a point in S that is located on the convex hull of S. A point in S that is within the convex hull of S does not affect the diagram [9]. Fig. 1.2 illustrates the closest-point and farthest-point Voronoi diagrams of 8 planar points. The left figure corresponds to the closest-point diagram that consists of 8 regions. The right figure, corresponding to the farthest-point diagram, consists of only 5 regions. The points located within the convex hull, P5, P6 and P7, do not have an associated region. The Voronoi diagrams have found a lot of applications in areas ranging from archaeology to zoology [1, 4-8]. It can be used in graphics, image processing, wireless networks, robotics, motion planning, pattern recognition, etc. In [2], the author has presented a detailed survey of Voronoi diagrams and their applications. Voronoi diagrams are essential for
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wireless networks. In a cellular network, the area covered by a base station can be viewed as a Voronoi region of the base station. Voronoi diagram has also been used in mobile ad hoc networks and sensor networks. In [8], a localized algorithm is developed based on the Voronoi diagram to solve the exposure problems in sensor networks.
Figure 1.2 The closest-point and farthest-point Voronoi diagrams of a set of planar points.
The proposed umbrella diagram is comparable to the Voronoi diagram. The Voronoi diagram deals with a set S of n planar points, whereas the umbrella diagram deals with n networks (patterns, or tessellations), each of which is determined by a point in S. In this paper, we assume honeycomb networks. It is easy to see that a honeycomb network is determined by three factors: cell size, network orientation, and the position of one cell center. Therefore, if the cell size and orientation is fixed, one single point will determine a network. A set of n points will determine n networks (or n tessellations). The umbrella diagram of a set S of n planar points divides the plane into regions. Each region is associated with a point in S. The points in the region associated with the point p in S have a larger distance to the network determined by p than to the networks determined by the other points in S. Here the distance from a point to a network is the distance between the point and its closest cell center in the network. Please refer to Fig. 2.2 for an illustration of the umbrella diagram of 3 planar points. We will give an efficient solution to compute the umbrella diagram of n planar points with a time complexity of O(n3). The umbrella diagram can be used to solve problems in wireless networks. In this paper, we apply the diagram to solve two problems in wireless networks. One problem, referred to as the optimal network deployment, can be described as: given a set of existing base stations that are non-uniformly distributed, and a planned cellular network, find a deployment of the planned network such that the maximal distance between the existing base stations and their corresponding cell centers is minimized. The second application considers more constraints in network planning. In practice, most system designs permit a base station to be positioned up to one-fourth the cell radius away from the cell center to ensure an acceptable coverage [10]. Therefore the optimal network computed in the first application might not meet this constraint for all base stations. We assume the maximum allowed distance to be a constant d. So the problem, referred to as the maximum base station reuse, is to deploy the network such
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that number of nodes (old base stations) that can be located within distance d to their corresponding closest cell center is maximized. To the best of our knowledge, there is no solution to those problems reported in the literature. These problems can get solved utilizing the proposed umbrella diagram. One uses one application of the umbrella diagram of multiple nodes, and the other uses multiple applications of the umbrella diagram of two nodes. Although the solutions are given in cellular networks, the idea in general describes the important relationship between the maximum coverage (honeycomb) tessellation and the arbitrarily distributed nodes, which can be further used in the wireless ad hoc networks, image processing, etc. The rest of this paper is organized as follows. In Section 2, we define the umbrella diagram of a set of planar points. In Section 3, we show how to find the umbrella diagram. We further apply this diagram to solve two problems in wireless networks in Section 4. Finally we will summarize the results in Section 5. II. DEFINITION OF THE UMBRELLA DIAGRAM In this section, we will define the umbrella diagram of a set of planar points. The hexagonal plane tessellation divides the plane into hexagons as shown on the right side of Fig. 1.1. By using the hexagon geometry, the fewest number of cells can cover a geographic area, and the hexagon closely approximates a circular radiation pattern which would occur with an omni-directional antenna and free space propagation [10]. The hexagonal tessellation is also known as the honeycomb network [11]. In the following discussion, we will omit the word honeycomb when no confusion will arise. A network can be represented by a set of hexagons or a set of hexagon centers. A network is completely determined by three factors: the cell size, network orientation and the position of one cell center. If the cell size and orientation is fixed, the hexagon is solely determined by the position of one cell center. We define a network with a fixed cell size and orientation as movable. Definition 2.1 A network is movable if the cell size and orientation of the network is fixed but the position of the network is not fixed. In the following discussion, we will also refer to a movable network as a network pattern. Given a network pattern, if we fix the center of one cell at the position C, the network is completely fixed. We call the fixed network as the spreading network of C. Definition 2.2 Given a position C and a network pattern T, the spreading network of C with the pattern T, denoted as T(C), is the network with C being one of its cell centers. C is referred to as the starting point of T(C). Fig. 2.1 illustrates two spreading networks, T(P0) and T(P1), with a network pattern T. Next we will define the distance from a point to a network. It can be used to describe
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the error between a base station and its ideal position in a newly planned network.
Figure 2.2 The umbrella diagram of P0, P1, and P2. Figure 2.1 Two spreading networks, T(P0) and T(P1), with a network pattern T.
The umbrella diagram can be also defined from another point of view as follows:
Definition 2.3 Given a point p and a network N, the distance between the point p and the network N, denoted by d(p,N), is defined as the distance between the point p and its closest cell center of N. d(p,N) is also called the distance from p to N.
Definition 2.6 Given a set S of n planar points, S = {P0, P1,…,Pn-1}, and a network pattern T, the umbrella diagram of S divides the plane into regions, each of which is associated with a point Pi in S, such that the spreading network starting from a point in the umbrella region of Pi is farther from Pi than any other points in S.
It is easy to verify that if the point p is located within a cell with its center being c, the network distance between p and N, d(p,N), is the distance between p and c since c is the closest cell center to p. Finally, we will define the umbrella diagram of a set of planar points. We will define the diagram of two points, followed by the diagram of multiple points. Definition 2.4 Given two planar points P0 and P1, and a network pattern T, the umbrella diagram of P0 and P1 consists of two regions, denoted as AT(P0, P1) and AT(P1, P0), where AT(P0, P1) is the locus of points x such that d(x, T(P0)) ≥ d(x, T(P1)), and AT(P1, P0) is the locus of points x such that d(x, T(P1)) ≥ d(x, T(P0)). The diagram is named as the umbrella diagram because of the umbrella shape of the 3d surface with equation z = x2 + y2, cut by a hexagon moving along the z axis. Definition 2.5 Given a set S of n planar points, S = {P0, P1,…,Pn-1}, and a network pattern T, the umbrella diagram of S is a set of n regions, {AT(Pi, S) | 0 ≤ i ≤ n-1 }, where AT(Pi, S) is the locus of points x such that d(x, T(Pi)) ≥ d(x, T(Pj)) for all j ≠ i. AT(Pi, S) is called the umbrella region of Pi . Fig. 2.2 illustrates the umbrella diagrams of 3 points. For clarity, we use U(Pi) instead of AT(Pi, S) in the figure. The polygons of the same shape belong to the umbrella region of a point. We mark only a few polygons of each umbrella region in the figure.
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III. COMPUTING THE UMBRELLA DIAGRAM In this section, we will show how to compute the umbrella diagram of a set of planar points with a network pattern T. We first show how to find the umbrella diagram of two points. Then we extend our idea to find the umbrella diagram of a set of points. Finally we will present the algorithm and analyze its time complexity. A. Umbrella Diagram of Two Points Given two points, P0 and P1, and a network pattern T, we will show how to compute AT(P0, P1) and AT(P1, P0). AT(P0, P1) consists of all points that have a larger distance to T(P0) than to T(P1). Since the network T(Pi), i = 0 or 1, is the repeating of the hexagon centered at Pi, the regions AT(P0, P1) and AT(P1, P0) are also repeatable. We will show how to compute AT(P0, P1). Without loss of generality, we will consider the points within the hexagon centered at P0, referred to as H(P0). Assume P is a point located within H(P0). It is easy to verify that d(P, T(P0)) = |PP0|. Next we will compute d(P, T(P1)). We assume H(P0) is not identical to any hexagon of the network T(P1). If they are identical, it is obvious that d(P, T(P1)) is always the same as d(P, T(P0)). Under that assumption, H(P0) will intersect with either 3 or 4 hexagons of T(P1) as shown in Fig. 3.1.
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most five edges. Therefore, AT(P0, P1) within H(P0) consists of 4 convex polygons, each of which has at most five edges.
Figure 3.1 Two kinds of intersections between H(P0) and T(P1).
We will consider the 4-intersection case with the 3intersection case being similar. Assume the 4 hexagons of T(P1) that are intersected with H(P0) are centered at Q0, Q1, Q2, and Q3 as shown in Fig. 3.2. Since H(Q0), H(Q1), H(Q2), and H(Q3) completely cover H(P0), we only need to consider H(Q0), H(Q1), H(Q2), and H(Q3) when computing d(P, T(P1)). Specifically we have the following. Lemma 3.1 Given two points, P0 and P1, and a network pattern T, assume Q0, Q1, Q2, and Q3 are the centers of the hexagons in T(P1) that intersect with H(P0). For any position P ∈ H(P0), d(P, T(P0)) = |PP0| and d(P, T(P1)) = |PQi| if P ∈ Ii = H(P0) I H(Qi). Proof. If P ∈ Ii= H(P0) I H(Qi), then P ∈ H(P0) and P ∈ H(Qi). Therefore d(P, T(P0)) = |PP0|, and d(P, T(P1)) = |PQi|.♦
However, in order to compute the diagram of multiple points based on the diagram of two points, we will compute AT(P0, P1) and AT(P1, P0) a little bit differently. We will find the intersections of the bisector line segments (or their extensions). The bisector line segments li and li+1 (or their extensions) intersects on the boundary of cells H(Qi) and H(Qi+1). This is because the intersection is the circumcenter of the triangle P0QiQi+1. If H(P0) intersects with 4 hexagons in T(P1), the polygon formed by the perpendicular bisectors (or their extensions) will be a quadrilateral; if H(P0) intersects with 3 hexagons in T(P1), the polygon formed will be a triangle as shown in Fig. 3.1. Denote the formed polygon by L(P0,P1) and the area outside L(P0,P1) by ~L(P0,P1). In addition, the contour of L(P0,P1) is visible from P0. We summarize our finding in the following theorem. Theorem 3.2 Given two points, P0 and P1, and a network pattern T, AT(P0,P1) = ~L(P0,P1) I H(P0) and AT(P1,P0) = L(P0,P1)
I H(P0) within H(P0). ♦
Equivalently, if we compute AT(P0, P1) and AT(P1, P0) within H(P1), then AT(P0, P1) = L(P1,P0) I H(P1) and
I H(P1). If we put together the I H(P0), the resulting shape will be exactly the same as AT(P0, P1) I H(P1). It is easy to see that
AT(P1,P0) = ~L(P1,P0) polygons of AT(P0,P1)
AT(P0, P1) and AT(P1, P0) can be computed in constant time. Fig. 3.3 illustrates the umbrella diagrams of P0 and P1. For clarity, we use U(P0) as AT(P0, P1) and U(P1) as AT(P1, P0) in the figure. The polygons of the same shape belong to the umbrella region of the same point. We mark only a few polygons of each umbrella region in the figure.
Figure 3.2 The construction of L(P0, P1).
Ii = H(P0) I H(Qi) is a convex polygon because both H(P0) and H(Qi) are convex polygons. It is easy to verify that Ii consists of at most six edges. To compute the umbrella diagram of P0 and P1, we construct the perpendicular bisector li of P0Qi in Ii= H(P0) I H(Qi) as shown in Fig. 3.2. By
Lemma 3.1, for any point P ∈ Ii = H(P0) I H(Qi), if P is on the line segment li, d(P, T(P0)) = |PP0| = |PQi| = d(P, T(P1)); if P is on the side that is closer to P0, d(P, T(P0)) = |PP0| < |PQi| = d(P, T(P1)); if P is on the side that is farther to P0, d(P, T(P0)) = |PP0| > |PQi| = d(P, T(P1)). It is also easy to see that Ii is divided by li into two convex polygons, each of which has at
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Figure 3.3 The umbrella diagram of P0 and P1.
B. Umbrella Diagram of a Set of Points In this subsection, we assume a set S of n planar points, S = {P0, P1,…,Pn-1}. For each point Pi, we will find AT(Pi, S), the locus of points that have a larger distance to the spreading network of Pi than to the spreading network of any other
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points in S. In the previous subsection, we have shown how to compute the umbrella diagram of two points. Specifically for any two points Pi and Pj with j ≠ i, we know how to compute AT(Pi, Pj). By definition, AT(Pi, S) =
IA
T
( Pi , Pj ) . We will
j ≠i
show how to compute AT(Pi, S) within H(Pi). By Theorem 3.2, for j ≠ i, AT(Pi,Pj) = ~L(Pi,Pj) I H(Pi) within H(Pi), Therefore, within H(Pi), AT(Pi, S) =
IA
T
( Pi , Pj ) =
j ≠i
=
j
i
The algorithm to compute the umbrella diagram of n planar points can be described as follows.
j ≠i
H ( Pi ) I (I ~ L( Pi , Pj ))
Algorithm: Umbrella Diagram
j ≠i
=
H ( Pi ) I (~ U L( Pi , Pj ) ). j ≠i
Let
L(Pi,
S)
=
U L( P , P ) , i
j
then
AT(Pi,
S)
=
j ≠i
H ( Pi ) I ~L(Pi, S). We summarize this in the following theorem. Theorem 3.3 Given a set S of n planar points, S = {P0, P1,…,Pn-1}, and a network pattern T, AT(Pi, S) within H(Pi) is equal to
Once the umbrella regions of all Pi ∈ S are computed, the umbrella diagram can be formed. As illustrated in Fig. 3.5, the umbrella regions are likely to be non-convex. Furthermore, the region of a point, e.g. U(P0) in the figure, can consists of multiple polygons of different shape. C. The Algorithm and Its Time Complexity
I (~ L( P , P ) I H ( P )) i
The contour of L(P0, S) is shown in a darker color in the figure. The umbrella region of P0 within H(P0), which is represented by the hatched part in the figure, consists of all points that are outside L(P0, S) yet inside H(P0).
H ( Pi ) I ~ L(Pi, S) with L(Pi, S) =
U L( P , P ) .♦ i
j
j ≠i
Input: a set S of n planar points, S={P0, P1,…,Pn-1}, and a network pattern T; Output: the umbrella diagram of S, that is, AT(Pi, S) within H(Pi) for 0 ≤ i ≤ n-1. BEGIN FOR every point Pi ∈ S FOR every point Pj ∈ S ( j ≠ i ) Find the cell centers Q0,…,Qm-1 (m = 3 or 4) of T(Pj), the spreading network of Pj, such that H(Qk) intersects with H(Pi); Compute L(Pi,Pj) formed by the perpendicular bisectors of Pi Qk; END FOR; Compute L(Pi, S) = L( Pi , Pj ) ;
U j ≠i
Compute AT(Pi, S) = ~ L(Pi, S)
I H ( Pi ) ;
END FOR; END.
Figure 3.4 Computing AT(P0, S) within H(P0) with S = {P0, P1, P2, P3}.
Theorem 3.3 tells us how to compute the umbrella region of Pi, AT(Pi, S), within H(Pi). To compute AT(Pi, S) within H(Pi), we first find L(Pi,Pj) for every j ≠ i. Recall that L(Pi,Pj) is a triangle or a quadrilateral whose contour is visible from Pi. Then we compute L(Pi, S) = L( Pi , Pj ) . Finally find the part
U j ≠i
of H(Pi) that is located outside of L(Pi, S). Fig. 3.4 illustrates how to compute AT(P0, S) within H(P0) with S = {P0, P1, P2, P3}. L(P0, S) is the union of L(P0,P1), L(P0,P2), and L(P0,P3).
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Figure 3.5 Umbrella diagram of S = {P0, P1, P2, P3}.
Now we analyze the time complexity of the algorithm. Given two points Pi and Pj (j ≠ i), polygon L(Pi,Pj) is either a triangle or a quadrilateral which can be computed in constant
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time. Therefore, for a point Pi, it takes O(n) time to compute all the polygons L(Pi,Pj) for 0 ≤ j ≤ n-1 and j ≠ i. Next we will show how long it takes to compute L(Pi, S) =
U L( P , P ) . Lemma 3.4 shows that L(P , S) is a star-shaped i
i
j
j ≠i
polygon with O(n2) edges. Lemma 3.4 The outer contour of L(Pi, S) =
U L( P , P ) is a i
j
j ≠i
star-shaped polygon visible from Pi with at most 4(n-1) vertices.
2
Proof. For simplicity, we call L(Pi,Pj) as Lj. We will prove L(Pi, S) is a star-shaped polygon visible from Pi by induction. Suppose C(k) is the union of k Ls. We will compute C(k+1) that is the union of C(k) and Lk+1. We assume that p is an intersection of C(k) with Lk+1. Because p is visible from Pi in both C(k) and Lk+1, p is visible from the Pi in C(k+1). Therefore, all end-points in the union C(k+1) are visible from Pi, that is, C(k+1) is a star-shaped polygon. That concludes L(Pi, S) is a star-shaped polygon visible from Pi. Now we compute the number of vertices on L(Pi, S) in the worst case. An end-point on the outer contour of L(Pi, S) is either the intersection of any two Ls, or an original vertex of L. There are three cases concerning the intersections for a pair of Ls, L(Pi,Pa) and L(Pi,Pb), as shown below. L(Pi,Pa) ∩ L(Pi,Pb)
get the outer contour of the original n Ls. We first merge the
angle list θ 1, θ 2,…, θ u and the angle list θ ’1, θ ’2,…, θ ’v. Note that C(A) and C(B) can have at most one intersection between any two consecutive angles in the resulting list. Without lost of generality, we only need to consider 6 cases for two consecutive angles α and β as shown in Figure 3.6. (Some symmetric cases have not been listed.) Case a: Keep the end points at α and
β
.
Case b: Discard the end points at α and β . Case c: Find the intersection, and let γ be the angle for the intersection. Discard the end point at α , keep the end point at β , and add a new point at γ . Case d: Discard the end point at α , and keep the end point at β. Case e: Find the intersection, and let γ be the angle for the
intersection. Keep the end points at α and β , add a new point at γ . Case f: Find the intersection, and let γ be the angle for the intersection. Discard the end points at and add a new point at γ .
α
and
β
,
Number of intersections
triangle ∩ quadrilateral
≤6 ≤6 ≤8
triangle ∩ triangle quadrilateral ∩ quadrilateral
For each case, the number of intersections is no more than 8. There are (n-1)(n-2)/2 pairs of Ls. Therefore, the total number of end points on the outer contour of L(Pi, S) is at most 8(n1)(n-2)/2 + 4(n-1) = 4(n-1)2. ♦ Next we show that, given a point Pi, it takes O(n2) time to compute the outer contour of L(Pi, S) = U L( P , P ) . We apply i
j
j ≠i
the divide and conquer approach. Again we denote L(Pi,Pj) as Lj, and assume Pi is at the origin o. To find the outer contour of n Ls, we divide the n Ls into two groups, A and B, each of n/2 Ls. Assume the outer contours of A and B are C(A) and C(B) respectively. Realizing each end point p can be represented by the distance ρ between p and the origin o and the angle
θ
made by op with the x axis, the outer contour of k
ρ , 2 θ 2>,…, < ρ k, θ k> with θ 1 < θ 2 , < ρ 2, θ 2>,…, < ρ u, θ u> and C(B) = < ρ 1, θ ’1>, 2 ρ ρ < 2, θ ’2>,…, < v, θ ’v>. By Lemma 3.4, u ≤ ( n − 2) vertices can be represented by a list
,
d(P1, T(x)). So for any network with a starting point located outside L(P0,P1) but within H(P0), if P0 can be reused, P1 can be reused too because the network distance of P0 is greater than that of P1. Now we show how to incorporate the bound d into the contour. An old base station can be reused if and only if it is within the bound of a nearby cell center. If we draw a circle around the old base station P0 with the radius being the bound d as shown in Fig. 4.4, the circled area stands for the reuse region of P0. For any network with a starting point within the circle, it is obvious that P0 can be reused. Therefore P1 can be reused too because the network distance of P0 is greater than that of P1. Lemma 4.2 summarizes the above discussion. Lemma 4.2 Given two base stations Pi and Pj, the area that is outside L(Pi,Pj) but inside the circle that centered at Pi is the locus to deploy the cellular network such that both Pi and Pj can be reused subject to the condition that Pi has an equal or larger network difference than Pj. ♦
For a given bound d, it is possible that no matter how we design the network, not all the old base stations can be located within the bound. Then the problem is how to find the
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difference. The following theorem summarizes our findings concerning our sub-optimal solutions.
Figure 4.4 Bounded L(P0, P1) for P0 and P1.
There could be three cases of intersection between L(Pi,Pj) and the circle centered at Pi: (a). L(Pi,Pj) is totally outside the circle; (b). L(Pi,Pj) is totally inside the circle; (c). L(Pi,Pj) is intersecting with the circle. If L(Pi,Pj) is totally outside the circle, it means Pi cannot be reused in any network that has a larger distance to Pi than to Pj. If the L(Pi,Pj) is totally inside the circle, it means that both Pi and Pj can reused in a network that has a larger distance to Pi than to Pj. In this case, the area that is outside of L(Pi,Pj) but inside the circle is the locus to deploy a cellular network in which both old base stations can be reused subject to the condition that Pi have a equal or larger network difference than Pj. Finally, if L(Pi,Pj) is intersecting with the circle as shown in Fig. 4.4, it means that the area that is outside L(Pi,Pj) but inside the circle is the locus to deploy a cellular network in which both old base stations can be reused subject to the condition that Pi has a equal or larger network difference than Pj. Next, we will present our solution to the problem of maximum base station reuse. For each old base station Pi, we will show how to get the circular arcs that will be used to solve the sub-maximum base station reuse problem that is subject to the condition that Pi has the largest network difference. The maximum number of base stations that can be reused can be found by combining sub-optimal solutions. Suppose we have computed all the L(P0,Pj) for all j ≠ 0 , and draw them around P0 as shown in Fig. 4.5. Next we add a circle around P0 with the radius being the bound. Now if we walk from a point x on the circle straight towards the center P0, the number of lines we encounter is one less than the number of base stations that can be reused if the new network is deployed at x. If the encountered line is a part of L(P0,Pj), Pj can be reused. Note P0 can always be reused. By walking from all points on the circle, we can find the maximum number of base stations that can be reused. Of course all statements here are subject to the condition that P0 has the largest network
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Figure 4.5 Inspecting maximum reused base stations.
Theorem 4.3 Given a set S of n base stations, S={P0, P1,…,Pn1}, if the maximum number of contours L(P0,Pj) with j ≠ 0 that intersect a radius starting from P0 is k, the maximum number of base stations that can be reused is k+1 under the condition that P0 has the largest network difference. Proof. Subject to the condition that P0 has the largest network difference, we know that the network starting point must be located outside the contour L(P0,Pj). In terms of P0 if the starting point is inside the bounding circle centered at P0, P0 can be reused. Otherwise, P0 cannot be reused. For a base station Pj other than P0, if the starting point is outside the contour L(P0,Pj), Pj can be reused as long as P0 can be reused because, in this case, nd(P0,T) is greater than nd(Pj,T). Therefore, if we connect a point on the circle to the center P0, the number of intersecting contours corresponds to the number of base stations that can be reused subject to the condition that P0 is farther from the network than other base stations. It is easy to see that if the maximum number of contours that can be intersected with a radius is k, the maximum number of base stations that can be reused for the given bound and subject to the condition that P0 has the largest network difference is k+1. ♦
For example, in Fig. 4.5, if we walk from point A towards the center P0, only L(P0, P1) is encountered. Therefore, if the network is deployed at A, only old base stations P0 and P1 can be reused. However, if we walk from point B towards the center, L(P0, P1), L(P0, P2) and L(P0, P3) will be encountered. Therefore, in this case, the base stations, P0, P1, P2 and P3, can be reused when the network is deployed at B. However, it is impossible for us to inspect every radius to find the maximum number of base stations that can be reused. Next, we will use circular arcs to solve the maximum base station reuse problem. As shown in Fig. 4.5, we consider the intersections of the circle and L(P0,Pj). Each intersection chord corresponds to a circular arc [12]. L(P0,Pj) introduces at most 4 circular arcs. L(P0,P1), L(P0,P2),…, L(P0,Pn-1) will introduce
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a total of 4(n-1) circular arcs. To find the maximum number of circular arcs that intersect with a radius, we can first sort all the arc end points. Then we use a sweep line to get the maximum number of the arcs that intersect with the sweep line. The maximum number of overlapping arcs for n circular arcs can be found in O(nlogn) time. For each old base station Pi, we can compute the maximum number of old base stations that can be reused under the condition that Pi has the largest network difference. Next we will try to solve the maximum base station reuse problem without any condition. Assume Tmax is the deployed network in which the maximum number of base stations can be reused, and Pk is the old base station that has the largest network difference to Tmax. Then the maximum number of old base stations can be reused is the same as the maximum number of old base stations can be reused under the condition that Pk is farthest from the network. The following theorem summarizes how to find the solution to the maximum reuse problem. Theorem 4.4 Given a set S of n base stations, S={P0, P1,…,Pn1}, if ki (i=0..n-1) is the maximum number of old base stations that can be reused under the condition that Pi has a larger network difference than all other old base stations, and km is the maximum among all ki (i=0..n-1), km is the maximum number of base stations that can be reused within the given bound. ♦
Next we consider the time complexity of the algorithm. As we know, given two points Pi and Pj (j ≠ i), polygon L(Pi,Pj) is either a triangle or quadrilateral which can be computed in constant time. For each Pi, we have a circular arc graph of a maximum of 4(n-1) arcs. The maximum number of arcs that can intersect with a radius can be found in O(nlogn). There are n circular arc graphs, each of which corresponds to an old base station. Therefore it takes O(n2logn) time to compute the maximum number of old base stations that can be reused. V. SUMMARY
Computational geometry has found a lot of applications in the area of wireless networks. In a cellular network, the area served by a base station can be viewed as a Voronoi region of that base station in the Voronoi diagram. In this paper, we have proposed a new geometric diagram, referred to as the umbrella diagram. The umbrella diagram is comparable to the famous Voronoi diagram. Whereas the Voronoi diagram deals with a set of points, the umbrella diagram deals with a set of different networks, each of which is derived from a point. The umbrella diagram of a set of hexagonal networks is to divide the plane into regions such that the points in a region have a larger distance to one network than to the other networks. The distance from a point to a network is the distance between the point and its closest cell center of the network. Unlike the Voronoi diagram, the regions in the umbrella diagram are not necessarily convex. The paper shows that the umbrella diagram of n networks with the same cell size and orientation (or the umbrella diagram of n points under a network pattern) can be computed in O(n3) time.
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The umbrella diagram characterizes the relationship between uniform patterns and arbitrarily distributed nodes and it can be used to solve many problems in wireless networks. In this paper we show how to use the diagram to solve two problems: the problem of the optimal deployment of a planned network, and the problem of maximum base station reuse with a given bound. The optimal deployment of a planed network is to minimize the distance between the existing base stations and the deployed network if all existing base stations are to be reused. The maximum base station reuse problem is to find the maximum number of old base stations that can be reused under a given bound. These problems can be solved via the proposed umbrella diagram elegantly. Although two applications in wireless networks are given in this paper, the ideas have potential to be used in other areas. REFERENCE [1]
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[10] T. S. Rappaport, Wireless Communications-Principles and Practice, Prentice Hall, 1996. [11] I. Stojmenovic, “Honeycomb networks: Topological properties and communication algorithms,” IEEE Transactions on Parallel and Distributed Systems, vol. 8, no. 10, October 1997, pp. 1036-1042. [12] U. I. Gupta, D. T. Lee and J. Y. T. Leung, “Efficient algorithms for interval circular-arc graphs,” Networks, vol. 12, 1982, pp. 459-467.
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