Wireless Sensor Networks and Computational

Wireless Sensor Networks and Computational Geometry Problems Adil Erzin1,2 , Natalia Shabelnikova2 , Lydia Osotova2 , and Yedilkhan Amirgaliyev3 1

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Sobolev Institute of Mathematics, Novosibirsk, Russia 2 Novosibirsk State University, Novosibirsk, Russia Institute of Information and Computational Technologies, Almaty, Kazakhstan

Abstract. This paper contains the previously known results, as well as new our results by objective of constructing the least dense covers of the plane regions with disks, ellipses and sectors. Such problems are considered in the context of design an energy efficient wireless sensor networks, which are an example of a distributed network of data collection and transmission. Arising in this connection computational geometry problems are difficult to solve, so basically approximate solutions are searched. We proposed several new coverage models with the least known density in their class.

Keywords: Wireless Sensor Networks, Coverage Density

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Introduction and Problem Formulation

Wireless sensor network (WSN) consists of devices that collect information and transmit it to the base station via radio. Each sensor is equipped with a battery replacement or charging of which is either impossible or impractical. WSN’s lifetime is a time period during which the network collects and transfers data from a certain region. Energy consumption is associated with sensor’s monitoring area, which is called the coverage area of the sensor. Energy consumption is proportional to the area covered by sensor. Therefore, the multiple coverage entails excessive energy loss, which leads to a reduction of network’s lifetime. Therefore, the problem of energy-efficient monitoring can be reduced to the problem of finding the least dense cover, where the density of coverage an area of S is the ratio of the sum of squares of elements in the cover to S. The lower density, the better cover. Since in the applications a coverage area of a sensor can has different shapes (disk, ellipse, sector) and different size (radius, semi-axes, angle and radius), then the following general computational geometry problem can be formulated. General problem. For a given plane region, every point of which must be covered by at least one figure, the list of types of figures and admitted regions of their parameters, it is required to define the set of figures in the cover, values of

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CITech-2015, Almaty, Kazakhstan, 2015 September 24-27

the parameters and to determine the placement of each figure and its orientation in order to minimize the density of the cover. Since the set of covers is continual, in practice, typically the regular covers are considered. In the regular cover the region is split into the equal polygons (tiles), and all the tiles are covered identically. As a result, for the calculation of the density of a cover of the whole region it is sufficient to estimate the coverage density of one tile. In the cover one can use different types of figures. Thus, the two figures have the different types if they differ not only by shape, but even if they have different sizes. For example, a disk and a sector it is different types of figures. But two disks are different too if their radii differ. On the other hand, the two ellipses have the same type if their half-axis coincide regardless of their orientation. We introduce the class COV (p, q) as a set of regular covers, in each of which the tile is covered with p figures of q different types. Obviously, the more types of figures used in the cover, the lower density of cover may be. In this paper we restrict ourselves to the covers in the classes COV (p, q), q ≤ 2 (with figures of at most two types). Consider a cover of one tile in the class COV (p, q). Renumber the figures covering a tile and denote by eti the i-th figure of type t, t = 1, . . . , q, i = 1, . . . , nt , where nt is a quantity of figures of type t covering one tile. Obviously, q ∑ nt = p. Denote by A(eti ) tile area, where it can be located centre (for disk t=1

or ellipse) or vertex (for sector) of the figure eti . If one set the number of each type of figures nt and the placing domain of each figure A(eti ), then define the coverage model. The choice of the sizes, the centre (from A(eti )) and a tilt angle of each figure eti , i = 1, . . . , nt , t = 1, . . . , q, defines a concrete cover. Problem formulation. In this paper, we consider the problem of the construction of min-density regular covers in the classes COV (p, q) when one tile is covered by p figures of q (q = 1, 2) different types.

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One Type of Figures

If equal disks are used in the cover, then in [13] is proved that regular cover D1 in class COV (3, 1) is optimal (of minimum density) if the centres of three pairwise intersecting disks form the equilateral triangle – a tile, and only one √ point belongs to all three disks (Fig. 1a). The density of this cover equals 2π/ 27 ≈ 1.2091. Some covers using ellipses can be constructed from the covers that use disks by applying the affine transformation (AT). An AT is a transformation that preserves straight lines and ratios of distances between points lying on a straight line while keeping the coverage density the same. Examples of ATs include translation, expansion, reflection, and rotation. An AT is equivalent to a linear transformation followed by a translation. In [6] we noticed that after applying AT to the cover D1, one can get the cover E1 with equal ellipses in different classes depending on the tile (Fig. 1b). If tile is a triangle EBC, then the cover is in the class COV (3, 1). If tile is a triangle ABC, or a square ABDC, then the cover

Wireless Sensor Networks and Computational Geometry Problems

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belongs to COV (2, 1). If tile is a square EBCF , then the cover is in the class COV (4, 1).

B

D

B

A

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A

C

F

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Fig. 1. a) Optimal cover with equal disks. b) Optimal cover with equal ellipses.

Minimum density of the plane coverage √ with equal disks or ellipses does not depend on their size and is equal to 2π/ 27 ≈ 1.2091. A completely different situation with sectors. Denote the sector as a pair (R, α), where R is the radius, and α is the angle of a sector. Then the density of the cover S1 ∈ COV (1, 1) (Fig. 2a) depends on α and equals D(α) = α/ sin α, and D(α) → 1 when α → 0.

A

B

B

R 1

R

D A

D C

a)

C

D

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Fig. 2. a) Plane coverage model S1; b) Stripe coverage model M 1.

But, starting from a√certain angle, the values of the function D(α) = α/ sin α become more than 2π/ 27 ≈ 1.2091, and the optimal coverage with equal disks or ellipses is preferred. So, if the parameters (R, α) are fixed, it is necessary to find the best coverage model. Let consider now the stripe coverage with equal disks, ellipses and sectors.

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As √ we showed in [3, 5] the coverage density with equal disks or ellipses tends to 2π/ 27 ≈ 1.2091 when the number of disks covering one tile tends to infinity.

C

B

B

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D R

R ɴ

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Fig. 3. Stripe covering with equal sectors: a) Model M 2; b) Model M 3.

If equal sectors are used to cover a stripe, then we propose three efficient models M 1, M 2 and M 3 (Fig. 2b and Fig. 3). It was found that (but not published yet), depending on the parameters of a sector, the best (having a minimum density) may be any of these covers. Figure 4 shows the areas of preference of the coverage models, depending on the parameters of the sector. In the blue area model M 1 is the best, in the green area model M 2 has the lowest density, and in the red area model M 3 is preferable. Set, for example, α = 36◦ . If R = 0.96, then model M 1 has the lowest density among the models M 1, M 2, M 3; if R = 1.2, then the best model is M 3; if R = 1.6, again the best model is M 1; and if R > 1.9, then model M 2 has the lowest density among the models M 1, M 2, M 3. Of course, if there are no any bounds on the α and R, then one can set R sin α = 1, and the density α/ sin α of the cover (like S1) tends to 1 when α tends to 0 (in turn, R tends to infinity). However, in practice, the sector angle may not be less than some positive number. Recently we proved the Theorem 1. If α ∈ [π/180, π/2], then the minimum density of stripe covering with equal sector does not exceed 1.000051.

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Two Types of Figures

In [9] proposed a cover with two types of disks (Fig. 5a) which density tends to 1.0189 when the number of small disks tends to the infinity. It is a strong


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Fig. 4. The areas of preference.

a)

b)

Fig. 5. a) Cover in the class COV (∞, 2) which density tends to 1.0189 when the number √ of small disks tends to ∞; b) Cover in the class COV (4, 2), which density is 11π/ 972 ≈ 1.1084.

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result, but not of practical importance because of the unlimited number of circles involved in the covering of one tile. In [14] we proposed a√plane coverage model with two types of disks (Fig. 5b). Its density, equals 11π/ 972 ≈ 1.1084, is minimal in the class COV (4, 2) when a tile is an equilateral triangle. One can apply AT to get the covers with ellipses in the different classes, but having the same density [6].

a)

b)

Fig. 6. Cover in the class COV (6, 2): a) with ellipses, density is D ≈ 1.0786; b) with sectors, density is D ≈ 1.0321.

In [6] we proposed a new cover in the class COV (6, 2) with two types of ellipses having density D ≈ 1.0786 (Fig. 6a). If instead of ellipses use sectors, then we get another cover in the same class which density is D ≈ 1.0321. In the regular cover, a tile typically has the shape of a regular polygon (triangle, square or hexagon). Finally, we consider a regular cover using equilateral triangle with two types of sectors and estimate its density depending on the number of small sectors. Without loss of generality, set the length of the side of the triangular tile equal 2. The proposed cover ST2 consists of three equal sectors of radius 1 with vertices at the nodes of the triangle. A uncovered curvilinear triangle in the center of the tile can be split into six equal curvilinear triangles. Let us consider one of them – curvilinear triangle ABC which is covered with equal sectors (r, β), r = 2 sin β, β = π/12n (Fig. 7a), as follows. Triangle AEF is covered by 1 sector (r, β), triangle EGH – by 3 nonoverlapping sectors, triangle GBI – by 5 non-overlapping sectors, and so on (Fig. 7a). Just to cover all curvilinear triangles enough Qt =

n ∑ k=1

(2k − 1) = n2

(1)

Wireless Sensor Networks and Computational Geometry Problems C

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3 S 3 sin 2 12n 1

3 S 2 3 sin 2 12n

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Fig. 7. a) Example of the cover ST2 when n = 3; b) Coverage of triangle BCD.

sectors (r, β). Then if we take 2 sectors directed in opposite directions with a common side, they cover the rectangle EHJF . The rectangle GIDJ can be covered by 8 sectors. And so on. The total number of sectors covering all rectangles is equal to n n−1 ∑ ∑ (n − 1)n(2n − 1) Qs = 2 (k − 1)2 = 2 k2 = . (2) 3 k=2

k=1

Note that the form of the triangle BCD does not depend on sectors (r, β). The 2n sectors with the common vertex form a sector (r, nβ) = (r, π/6). We cover the triangle √ BCD with rectangles whose height is r/2 and length is reduced from 1 − 3/2 to r, and each rectangle, in turn, cover with the sectors (r, π/6) (Fig. 7b). As a result, the number of rectangles is ⌉ √ ⌈ ⌈ ⌉ √ √ 1/ 3 − 1/2 2− 3 2− 3 √ √ L= ≤ + 1, = π π r/2 2 3 sin 12n 2 3 sin 12n and the number of sectors (r, β) in l-th rectangle (counting from the bottom) is (⌉ ⌈ ) ( ) √ √ √ √ π π 1 − 3/2 − l 3 sin 12n 1 − 3/2 − l 3 sin 12n ql = 2n + 1 ≤ 2n +3 . π r sin 12n Then the total number of sectors (r, β) covering the triangle BCD is ( ) √ L−1 ∑ √ 2− 3 q= ql + 2n ≤ 2n + 2n(L − 1) 3 + π − L 3/2 . 2 sin 12n

(3)

l=1

Theorem 2. The minimum coverage density of the equilateral triangle with two π π , 12n ) types of sectors tends to 1.00857 when the number n of sectors (2 sin 12n tends to infinity.

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π π Proof. The total number of sectors (2 sin 12n , 12n ) sufficient to cover the curvilinear triangle ABC is Q = Qt , +Qs + q. Using (1)-(3), we get ( √ √ ) √ ) ( π (12 − 2 3) sin 12n + 2 − 3 (2 − 3) √ Q ≤ n n + 2 + (n − 1)(2n − 1)/3 + . π 4 3 sin2 12n

The area of one sector equals s = r2 β/2 =

π )2 π ( 2 sin . 24n 12n

√ The density of the cover is D(n) = (π/12 + Qs)/S, where S = 1/ 12. Then √ √ π 3 π 3 (n − 1)(2n − 1) π { D(n) ≤ + sin2 n+2+ + 6 3 12n 3 √ √ ) √ ( π (12 − 2 3) sin 12n + 2 − 3 (2 − 3) } √ . (4) π 4 3 sin2 12n So, it is easy to show that lim D(n) ≤ 1.00857.

n→+∞

The proof is over. We found the upper bound for the density depending on the number of sectors in the cover n (4). In particular, lim D(n) ≤ 1.00857 which is much less than n→+∞

the coverage density with disks of two radii (which tends to 1.0189 when the number of disks tends to infinity) [9]. Moreover, D(n) ≤ 1.00859 < 1.0189 when the number of sectors is finite, for example, n ≥ 10000. Acknowledgments. This research was supported jointly by the Russian Foundation for Basic Research (grant 13-07-00139) and the Ministry of Education and Science of the Republic of Kazakhstan (grant 0115PK00550).

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