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BY 4.0 license Open Access Published by De Gruyter Open Access January 29, 2021

Optimization of a k-covering of a bounded set with circles of two given radii

Alexander V. Khorkov EMAIL logo and Shamil I. Galiev
From the journal Open Computer Science

Abstract

A numerical method for investigating k-coverings of a convex bounded set with circles of two given radii is proposed. Cases with constraints on the distances between the covering circle centers are considered. An algorithm for finding an approximate number of such circles and the arrangement of their centers is described. For certain specific cases, approximate lower bounds of the density of the k-covering of the given domain are found. We use either 0–1 linear programming or general integer linear programming models. Numerical results demonstrating the effectiveness of the proposed methods are presented.

1 Introduction

Let G be a bounded convex closed set with a nonempty interior on the plane P, and let closed circles with the radii r1 and r2 be given. A set of these circles forms a k-covering (k ≥ 1) of the domain G if each point s of G belongs to at least k circles. The density of a covering of G is defined as the ratio of the total area of the covering circles to the area of G.

Problems of covering a given domain with circles of either identical or different radii are important for applications, including emergency response system design, locating radio stations, and navigation facilities. Other many applications areas are given in [1,2,3,4,5,6]. There are a lot of applications investigating wireless sensor networks for monitoring a given domain when sensors have equal or different ranges. In a typical scenario, several different types of sensors are available which can be appropriately placed in the given field [7,8,9,10,11,12]. Designing such a system requires solving the problem of finding a k-covering (k ≥ 1) of the given domain with circles of equal or different radii.

There are a lot of publications devoted to coverings with equal circles (see [13,14,15,16] and references therein). In studies devoted to coverings with equal circles, covering of specific domains: a square, a rectangle, a circle, and some others are considered. Covering problems for sets that do not coincide with whole plane, e.g., strips, are studied in papers [17,18,19], see also references therein. There are also publications in which k-coverings, k ≥ 1, are studied (e.g., see [10, 12, 20]). Some results for determining the number of circles needed for a k-covering of certain given convex bounded closed domains are reported in [21].

The k-covering problem of a finite set with circles of different radii is studied in [22], where a nonlinear model of the problem is constructed that is then transformed into a linear problem with a large number of variables, auxiliary variables, and constraints. The variable radius covering problem is considered in [6]. Mathematical programming formulations are proposed for the discrete problem and a solution approach is suggested and tested. See [23] for covering the plane with two kinds of circles.

In this paper, we propose a method for investigating k-coverings of an arbitrary convex bounded closed set with a nonempty interior with circles of two given radii. Cases with constraints on the distances between the covering circle centers are considered. We use 0–1 linear programming or general integer linear programming models. An algorithm for finding an approximate number of such circles and the arrangement of their centers is described. For certain specific cases, approximate lower bounds of the density of the k-covering of the given domain are found. Numerical results demonstrating the effectiveness of the proposed methods are presented. In some cases the obtained value of k-covering density coincides with the lower bound of the density of the k-covering, therefore, the coverings in these cases are not improvable; in other cases the difference between obtained value and lower estimates of the covering densities characterizes the quality of the covering.

In this paper, we use the results of [24,25,26,27], and some of them are given for completeness and clarity of presentation.

2 Problem of Finding a k-covering of a Bounded Set with Circles of Two Radii

Let r1 > r2 be an assumption that holds throughout. We will choose Δ > 0 and build a rectangular grid with the step size Δx = Δy = Δ in G. The set of grid lines forming this set in G generates the square cells CΔ with the side Δ. If such a cell lies entirely in G, then each of its vertex is a grid node. If the square CΔ belongs to G only, then each vertex CΔ in G, the points at which the boundary of G enters CΔ, and the points at which the boundary of G exits CΔ are considered as grid nodes. In addition, we add on the boundary F of the domain GCΔ node points such that the distance between them along the line F does not exceed Δ. Let TΔ consist of all grid points described above: TΔ = {t1, . . . , tn}, where tiG and 1 ≤ in.

Instead of the rectangular grid on G, we can construct another grid, e.g., an oblique one. If G is an equilateral triangle, then we construct the oblique grid as follows. Select a step Δ such that it exactly divides the triangle side length. Draw lines parallel to each side of the triangle with the step Δ along the triangle sides. As a result, these grid lines generate equilateral triangles CΔ with the sides Δ. The nodes of this grid form the set TΔ.

Below, if not otherwise specified, we assume that the grid is rectangular. Here and in what follows, d(s, t) is the Euclidean distance between the points s and t.

Let G, r1, r2, k(k ≥ 1), and Δ be given and TΔ be constructed. Consider the following problems.

P1. Find a k-covering of the domain G with circles of the given radii r1 and r2 centered in G so as to minimize the total area of the covering circles and determine the location of their centers in G.

P2. Find a k-covering of the domain G with circles of the given radii r1 and r2 so as to minimize the total area of the covering circles and determine the location of their centers under the condition that the circle centers are at the points of TΔ and there is not more than one circle center at each point of TΔ.

P3. Find a k-covering of the set TΔ with circles of the given radii r1 and r2 so as to minimize the total area of the covering circles and determine the location of their centers under the condition that the circle centers are at the points of TΔ and there is not more than one circle center at each point of TΔ.

P4. Find a k-covering of the domain G with circles of the given radii r1 and r2 so as to minimize the total area of the covering circles and determine the location of their centers under the condition that the circle centers are at the points of TΔ.

P5. Find a k-covering of the set TΔ with circles of the given radii r1 and r2 so as to minimize the total area of the covering circles and determine the location of their centers under the condition that the circle centers are at the points of TΔ.

First, consider P3

It is easy to verify that a sufficient condition for the solvability of P3 is as follows:

(1) ForeachpointtjfromthesetTΔthereexistatleastk1differentpointsti(tiisnotequaltotj)inTΔsuchthatthedistancefromtjtothesepointsdoesnotexceedr2.

Let us construct a mathematical model of P3. We assume that Δ has already been selected and the (finite) set TΔ has been constructed. Introduce the coefficients aij and ai,n+j where 1 ≤ i, jn:

(2) aij={1,ifd(ti,tj)r1,0,ifd(ti,tj)>r1,ai,n+j={1,ifd(ti,tj)r2,0,ifd(ti,tj)>r2.

Define the following variables: zi is the number of circle centers of radius r1 located at the point ti, 1 ≤ in; zn+j is the number of circle centers of radius r2 located at the point tj, 1 ≤ jn. In the case of k-covering P2 and P3, it is clear that, at any point of TΔ cannot contain more than one center of a circle of radius r1 or r2. Therefore, we have the constraints

(3) zi{0,1},1i2n.

In addition, we have the constraints

(4) zi+zi+n1,1in.

If the total area of the covering circles is minimized, then for k = 1 constraints (4) are redundant.

We formulate the problem

(5) (z1++zn)+(r2r1)2(zn+1++z2n)min

subject to the constraints

(6) a11z1+a12z2++a1,2nz2nk1,an1z1+an2z2++an,2nz2nkn,

(7) zi{0,1},1i2n.

Here kj is the prescribed multiplicity of covering of the point tj, 1 ≤ jn. Constraints (6) ensure that each point tj in TΔ is covered by at least kj circles. If kj = k for 1 ≤ jn, then we have a k-covering of the set TΔ. If the variable zi is equal to one, then the point ti of TΔ(1 ≤ in) is the center of a circle of radius r1. If zn+i is equal to one, then the point ti of TΔ(1 ≤ in) is the center of a circle of radius r2.

Problem presented in relation (5)–(7), (4) is the problem of finding a k-covering of a finite set of points TΔ in which the total area of the covering circles is minimal. As a result, we have a 0–1 linear program of size 2n. Denote this problem by W1. It is known that the general problem W1 is NP-hard (see [28,29,30,31,32,33]).

Condition (1) for the solvability of P3, i.e., of problem W1, is difficult to verify. We give easily verifiable sufficient solvability conditions of W1 for some specific types of the set G for the covering multiplicity not exceeding ten.

First, consider the case when G is a square, which we will denote by Q. In Q, we construct a rectangular grid with a step Δx = Δy = Δ. For simplicity,we assume that Δ exactly divides the square side length. Consider the covering of Q with circles of radii r1 and r2. For a point s, denote by R(s) the closed circle of radius r2 centered at s. Let s be a vertex of Q; then, s coincides with a point of TΔ. Then, in the case r2 = Δ the set QR(s) obviously contains three different points of TΔ; if r2=Δ2 , then the set QR(s) contains four different points of TΔ, etc. For any other point s in TΔ (that is not a vertex of the square), the number of different points of TΔ in QR(s) is no less than this number in the case when s is a vertex of the square. Therefore, if 1 ≤ k ≤ 3, then Δr2; if 1 ≤ k ≤ 4, then Δr2/2 ; if 1 ≤ k ≤ 6, then Δr2/2; if 1 ≤ k ≤ 9, then Δr2/22 ; and if 1 ≤ k ≤ 11, then Δr2/3.

Let the domain G be an equilateral triangle Gt. Then, we construct an oblique grid on Gt as was described above. Let s be a vertex of the triangle Gt; therefore, s coincides with a point of TΔ. Then, in the case r2 = Δ the set GtR(s) contains three different points of TΔ; for r2=3Δ , GtR(s), contains four different points of TΔ, etc. For any other point s in TΔ (that is not a vertex of the triangle), the number of different points of TΔ in GtR(s), is no less than this number in the case when s is a vertex of the triangle. Therefore, if 1 ≤ k ≤ 3, then Δr2; if 1 ≤ k ≤ 4, then Δr2/3 ; if 1 ≤ k ≤ 6, then Δr2/2; if 1 ≤ k ≤ 8, then Δ2r2/33 ; and if 1 ≤ k ≤ 10, then Δr2/3.

As a result, we have relations between the covering multiplicity, the grid step size, the radius of the covering circles, and the domain G. If the sufficient conditions formulated above are satisfied, then condition (1) is also satisfied; therefore, P3 is solvable.

Consider P2

Let the grid step Δ be chosen, and the rectangular grid and the set TΔ be constructed. By solving problem W1,we obtain a k-covering of the set TΔ for which the total area of the circles covering TΔ is minimal. If TΔ is covered with circles of the given radii r1 and r2, it is not guaranteed that the given set G is completely covered. The grid with the step Δ on the set G generates squares with the diagonal equal to Δ2 . Let r2>Δ2 . If we decrease the radii r1 and r2 by the half of this diagonal when solving the P3, then the circles of the original radii r1 and r2 for the arrangement of circles obtained in this problem certainly cover the entire set G. Therefore, an approximate solution of P2 can be obtained by solving P3 for TΔ with the circle radii decreased by α=Δ2/2 , and then again increasing the radii of the found circles to r1 and r2. The sufficient solvability conditions for P2 are the same as for the solvability of P3 with r2 replaced by r2α. Thus, to check the solvability of P2 and P3, we may check conditions (1) or obtain the sufficient conditions as described above.

Below, we use the notation m1/m2 to indicate covering with m1 circles of radius r1 and m2 circles of radius r2.

Numerous computational experiments show that the less the grid step size Δ is, the closer the result to the optimal covering. For example, when the unit square Q is covered with circles of radii 0.55 and 0.2 for Δ equal to 0.1, 0.05, 0.025, 0.0125, and 0.01, the density of the 1-covering of Q is, respectively, 2.654, 1.829, 1.760, 1.704, and 1.454. The covering densities were obtained for the circle numbers equal, respectively, to 2/6, 1/7, 0/14, 1/6 and 1/4. Thus the choice of the value Δ is very important to obtain an acceptable solution of the covering problem.

The approximate solution of P2 can be considered as an approximate solution of P1.

Let us consider the problem P5 then P4

The solution of problem P5 is reduced to the solution of problem (5)–(6) with the following constraint:

(8) ziZ+,1i2n,whereZ+={0,1,2,3,}.

As a result, we have an integer linear programming (ILP) problem (5), (6) and (8) of size 2n. Denote this problem by W2.

Let TΔ, TΔ ≠ ∅, be already constructed. Then one can easily see that system W2 is solvable and, consequently, so is P5.

For ensuring the solvability of P4, we require that r2 > 2α, α=Δ2/2 . For solving P4, we introduce new circle radii r1α and r2α and solve P5 with them. We treat locations of centers of the mentioned circles in the solution of P5 as an approximate solution to P4, assuming that circle radii again are equal to initial values r1 and r2, correspondingly.

Thus we see that for solving P4 it is necessary to solve P5. We treat the solution to P4 as an approximate solution to P1. Therefore, one can solve P1 by solving problem W1, where zi ∈ {0, 1}, 1 ≤ i ≤ 2n, or by solving problem W2, where ziZ+, 1 ≤ i ≤ 2n. Later in Section 5 we establish that these solutions can differ.

For formulation W1, a heuristic algorithm described in [26] solves P3 and P2, and, consequently, P1, even when the problem dimension is as large as 14 000. Note that without the use of this algorithm, we can solve the mentioned problems with no more than 5 000 variables (moreover, in certain cases, the maximum number of variables guaranteeing the problem solvability is only 2 000). The heuristic algorithm essentially uses the fact that problem variables are Boolean. Unfortunately, the authors have failed to construct a similar algorithm for solving problem W2 with ziZ+, because the reduction of a general integer linear programming problem to a Boolean linear programming problem is a rather cumbersome task.

3 Constraints on the Distances Between the Circle Centers

In some cases, it is important that the centers of the covering circles are all different. For example, when navigation systems are designed, each point may contain no more than one device. Let us introduce the restrictions on the minimum distance between the centers of covering circles in the same way as it was done in [25, 26]. Note that for the covering problems, various restrictions were introduced besides coverage conditions. For example, in [34,35,36].

Suppose that it is required that the distance between the centers of the circles of radius r1 must be at least q1. For each point ti in TΔ = {t1, t2, . . . , tn}, let pi be the number of points tj in TΔ(ji) for which d(ti, tj) < q1. Define the coefficients:

bij={1ifd(ti,tj)<q1,0ifd(ti,tj)q1,ij,1i,jn;bii=pi,1in.

Now the condition that the distances between the centers of the circles of radius r1 must be at least q1 is written as (see [25, 26]):

(9) b11z1+b12z2++b1nznp1,bn1z1+bn2z2++bnnznpn.

If the problem W1 is solved subject to the additional constraints (9), then we obtain the desired covering in which the centers of the covering circles of radius r1 are at the distance at least q1.

Let it be required that the distance between the centers of the circles of radius r2 is not less than a given number q2. Let fi and cij be determined in the same way as pi and bij(1 ≤ i, jn) only using q2 instead of q1. Then, the conditions that the distance between the centers of the circles of radius r2 is not less than the given number q2 take the form:

(10) c11zn+1+c12zn+2++c1nz2nf1,cn1zn+1+cn2zn+2++cnnz2nfn.

Furthermore, suppose that it is required that the distance between the centers of the circles of radii r1 and r2 must be not less than a given number q3. Let gi and eij(1 ≤ i, jn) be determined in the same way as pi and bij but using q3 instead of q1. We also define hi,n+i = gi and hij = 0 for 1 ≤ in, n + 1 ≤ j ≤ 2n, and ji. Then, the conditions that the distance between the centers of the circles of radii r1 and r2 is not less than the given number q3 have the form (see [25]):

(11) (e11e1nhi,n+10e21e2n00en1enn0hi,n+n)×(z1z2z2n)(g1g2gn)

If q1 = q2 = q3, then we set pi = fi = gi for 1 ≤ in and bij = cij = eij for 1 ≤ i, jn in constraints (10) and (11). Constraints (9), (10) and (11) may be imposed simultaneously or one may impose only those of them that are needed in a specific problem. In each case, we obtain a ILP.

If constraints on the distance between the circle centers are imposed, the problem can be unsolvable. For example, for the problem of finding a 3-covering of an equilateral triangle with the side c, it is senseless to require that the distance between the circle centers is greater than c because in this case the circle centers will be outside the triangle.

4 Finding Lower Bounds on the Density of Covering for Specific Cases

In [26] one obtains approximate lower bounds for the density of the k-covering of the set G with circles of radii r1 and r2, provided that the minimum distance between centers of covering circles is not less than λ, λ ≥ 3Δ, and the covering multiplicity satisfies the condition 1 ≤ k ≤ 4.

In this paper, we obtain approximate lower bounds for densities of coverings with any multiplicity k ≥ 1, imposing no constraint on the minimum distance between centers of covering circles.

Problem P1 allows a k-covering such that the total area of covering circles is minimal (an optimal k-covering). Let an optimal covering contain m circles of given radii. Denote centers of these circles by c1, c2, . . . , cm. We use subscripts for circle centers even when their locations coincide. Since the set G is convex, the center of each covering circle belongs to G. The definition of a k-covering implies that for any point sG there exists at least k centers cj among c1, c2, . . . , cm such that d(s, cj) ≤ max{r1, r2}.

Choose a grid step Δ, Δ > 0, and construct the set TΔ = {t1, . . . , tn}, tiG, 1 ≤ in. By construction of the set TΔ, the distance between any point sG and the nearest to its point in TΔ does not exceed α=Δ2/2 .

If a circle center ci coincides with no point in TΔ, then we move this center to the nearest point tjTΔ. We move all circle centers in the mentioned way.

In addition to point shifts, we increase the radii of the covering circles by α. The set of circles of radii r1 and r2 centered in c1, c2, . . . , cm forms the desired optimal covering of G. Suppose that this set consists of n10 circles of radius r1 and n20 circles of radius r2. Then, the density of the optimal covering is P0(r1,r2)=(n10r12+n20r22)τ , where τ = π/S(G) and S(G) is the area of the set G.

After the circle centers have been shifted to the points of TΔ and their radii have been increased by α they form a k-covering of G and, therefore, a k-covering of TΔ. Define Pα#=(n10(r1+α)2+n20(r2+α)2)τ . Solve P5 of k-covering of TΔ with circles of radii r1 + α and r2 + α. As a result, we obtain n1 circles of radius r1 + α and n2 circles of radius r2 + α. It is important that n1 and n2 are known in contrast to the unknown quantities n10 and n20 . Define Pα,opt#=[n1(r1+α)2+n2(r2+α)2]τ . It is clear that Pα#Pα,opt#0 .

Let m1=n10n1 and m2=n20n2 . Then the inequality Pα#Pα,opt#0 implies

(12) m1(r1+α)2+m2(r2+α)20

Define the new quantity P*#(r1,r2)=(n1r12+n2r22)τ , where n1 and n2 are found by solving P5. For brevity, set PN=P*#(r1,r2) . We will examine the difference B=P0(r1,r2)P*#(r1,r2)=(m1r12+m2r22)τ for various possible values of m1 and m2.

  1. If m1 = m2 = 0, then n10=n1 , n20=n2 and B = 0.

  2. If m1 = 0 or m2 = 0, then (12) implies that either m2 ≥ 0 or m1 ≥ 0, respectively, and B ≥ 0; therefore,

    (13) P*#(r1,r2)P0(r1,r2)

  3. Consider the case m1 > 0, when the number of large circles decreases by a number p(p > 0) after solving P5. The number of small circles changes by a q,n2=n20+q(q0orq>0) . In this case, (12) implies [p(r1+α)2q(r2+α)2] ≥ 0. Then B=[pr12qr22]τ . If q ≤ 0, then it is clear that B ≥ 0, and inequalities (13) hold. Let q > 0. Then qp(r1+αr2+α)2 . If we assume B < 0 in this case, then pr12qr22<0 and q>pr12r22 . Inequalities qp(r1+αr2+α)2 and q>pr12r22 cannot simultaneously be true for any α ≥ 0. Hence, the assumption B < 0 does not hold, we have B ≥ 0, and inequality (13) holds.

  4. Consider the case m1 < 0, when the number of large circles increases by a number p > 0 after solving P5. If the number of large circles did not change, then m2 = 0, and (b) implies that inequality (13) holds. The number of circles of radius r1 and the number of circles of radius r2 cannot increase simultaneously. Suppose that when the number of large circles increased, the number of small circles decreased by q > 0. Then, inequality (12) implies

    (14) (n10(n10+p))(r1+α)2+(n20(n20q))(r2+α)2=p(r1+α)2+q(r2+α)20.

    Let B < 0. Then pr12+qr22<0 . Taking into account (14), we obtain two inequalities

    (15) p>q(r2r1)2,pq(r2+αr1+α)2

    where p and q are integers greater than zero. In the case under consideration n1=n10+p ; then p=n1n10 and, therefore, pn1. If n1 = 0, then p > 0 does not exist and the assumption B < 0 does not hold; therefore, B ≥ 0 and we have (13). Now consider the case n1 > 0, p > 0. Assume that both inequalities in (15) hold; i.e., for a certain q, p is within the interval Iq=(q(r2r1)2,q(r2+αr1+α)2] . In this case B < 0, and no lower bound for P0(r1, r2) is found. If such p and q do not exist, then the assumption B < 0 does not hold, and we have (13).

Thus, in some cases we have not found a lower bound for P0(r1, r2) (both inequalities in (15) hold). If inequalities (15) do not hold, then the assumption B < 0 is not true and inequality (13) holds.

Consider some particular cases of the covering. Notice that unlike [26] where the P3 is studied, here for solving P5 it is not necessary to calculate the minimum distance between centers of covering circles that was mentioned above.

Let Q be a unit square and it is required to cover it with circles of radii r1 = 0.55 and r2 = 0.30. Choose α = 0.01 and Δ=0.012 . By solving P5, we obtain n1 = 0 and n2 = 6. Then it is clear that m1 ≥ 0. According to (b), for m1 = 0 we obtain (13); for m1 > 0, (c) implies (13). As a result, the lower bound for P0(0.55, 0.30) is P*#(0.55,0.30)=1.696 .

Consider one more particular case. Suppose that we want to find a 2-covering of Q with circles of radii r1 = 0.55 and r2 = 0.25. By setting α = 0.0125, Δ=0.01252 , and solving P5, we obtain n1 = 0 and n2 = 16. Next, we obtain the lower bound for the density of the 2-covering equal to 3.141. If we set α = 0.01 and Δ=0.012 instead of α = 0.0125 and Δ=0.01252 , then by solving P5 we obtain n1 = 1 and n2 = 12. Next, using the interval I1 = (0.20661, 0.21556], we find, as in the second particular case, that inequality (13) holds, and the lower bound for the density of the 2-covering is 3.306. This bound is greater than the earlier bound 3.141; therefore, we improved the bound by decreasing α and Δ.

5 Numerical Results

Consider a covering of a rectangular (1.22 × 0.82) set R. Evidently, its area is close to 1 (more precisely, 1.22×0.82 = 1.0004). Let the radii of covering circles r1 = 0.58 and r2 = 0.46 and let the covering multiplicity be 2. Using problem W1with Δ = 0.02,we conclude that the set R can be covered with three circles of the radius r1 and one circle of the radius r2, while the density of the 2-covering equals 3.835. Using problem W2 with the same grid step Δ, we conclude that the set R can be twice covered with two circles of the radius r1 with coinciding centers and two circles of the radius r2 whose centers also coincide. We found that the density of this covering is 3.443. Therefore, problems W1 and W2 give different results with the same initial data. See Figure 1(a) and Figure 1(b) for the mentioned coverings of the set R obtained with the help of problems W1 and W2.

Figure 1 Covering of a rectangle (1.22 × 0.82) with the circles of radii r1 = 0.58 and r2 = 0.46 a) 2-covering is obtained by problem W1, b) 2-covering is obtained by problem W2, c) 2-covering is obtained by problem W1 with restriction min[d(ci, cj) ≥ λ : 1 ≤ i, j ≤ m, i  ≠ j], where λ = 0.6r2.
Figure 1

Covering of a rectangle (1.22 × 0.82) with the circles of radii r1 = 0.58 and r2 = 0.46 a) 2-covering is obtained by problem W1, b) 2-covering is obtained by problem W2, c) 2-covering is obtained by problem W1 with restriction min[d(ci, cj) ≥ λ : 1 ≤ i, jm, i  ≠ j], where λ = 0.6r2.

Using problem W2 with the grid step Δ = 0.02, we conclude that the set R can be twice covered with two circles of the radius r1 with coinciding centers and two circles of the radius r2 whose centers also coincide, consequently n1 = 2 and n2 = 2. For the case m1 ≥ 0 according (a)–(c)n from section 4 we obtain (13). It can be shown that for the case under consideration m1 < 0 is excluded. As a result, the lower bound for P0(0.58, 046) is P*#(0.58,0.46)=3.443 .

We performed corresponding computations for covering R with various r1 and r2, the results are presented in Table 1. The problem P5 for obtaining the lower bounds of the covering density was solved exactly without using heuristics. As has been mentioned above the lower bound on the covering density is not guaranteed to be found. However, no such cases were encountered in our computations.

Table 1

Covering of a rectangle (1.22 × 0.82) for given radii and multiplicity of the covering

Radii of circles r1/r2 Multiplicity of covering k
1 2
Approximate lower bounds (PN) of covering density and corresponding number of circles Obtained covering density and corresponding number of circles Approximate lower bounds (PN) of covering density and corresponding number of circles Obtained covering density and corresponding number of circles
0.55/0.30 1.516 (1/2) 1.696 (0/6) 3.031 (2/4) 3.392 (0/12)
0.55/0.25 1.539 (1/3) 1.767 (0/9) 2.945 (0/15) 3.338 (0/17)
0.55/0.20 1.382 (0/11) 1.703 (1/6) 2.765 (0/22) 3.266 (0/26)
0.45/0.30 1.696 (0/6) 1.696 (0/6) 3.252 (2/7) 3.392 (0/12)
0.45/0.25 1.571 (0/8) 1.767 (0/9) 2.945 (0/15) 3.338 (0/17)
0.45/0.20 1.382 (0/11) 1.641 (1/8) 2.765 (0/22) 3.267 (0/26)
0.35/0.30 1.696 (0/6) 1.696 (0/6) 3.032 (2/8) 3.236 (4/6)
0.35/0.25 1.539 (4/0) 1.742 (3/3) 2.906 (5/5) 3.109 (4/8)

In the Figure 1(c) we see the 2-covering of R with circles of radii r1 = 0.58 and r2 = 0.46 obtained under the additional requirement that the minimum distance between centers of covering circles (that was mentioned above) should be not less than 0.6r2. This covering is performed with six circles of the radius r2 so that the covering density equals 3.989.

Let us now consider other covering of the rectangle R with circles of radii r1 and r2. Its results are shown in Table 1. See [26] for analogous results for the square covering problem and to compare with some well-known publications on this topic. Results given in Table 1 are new, so we can compare them with no other ones. We can only estimate these results, based on approximate lower bounds for the covering density and the calculated values of the density. In certain cases, lower bounds and obtained values coincide (see Table 1), so the coverings are unimprovable. In the rest of cases, the difference between obtained values and approximate lower bounds for the covering density characterizes the acceptability (quality) of the covering.

The results from Table 1 were obtained using a rectangular grid. For studying the dependence of the density on grid properties, we have covered the rectangle with rectangular and oblique grid. In these calculations the same value step size Δ were selected. We obtained the number of circles for covering with circles of radii 0.55 and 0.30 using an oblique grid. We got the number of circles and therefore the covering densities coinciding with those given in Table 1. Corresponding coincidences were obtained for the remaining values of r1 and r2 of Table 1 when using oblique and rectangular grids. Thus, in the considered cases no influence of the grid type was revealed. As noted in section 2 (and the example is given) the choice of the value Δ is important.

If we are not satisfied with the covering quality, we try to improve the obtained value and (or) lower bound of the covering density. This can be done by repeatedly lessening the value Δ and solving problems P4 and P5, provided that the computer capabilities allow one to do this.

The computations were performed on an Intel Core i7-3537U computer with 6 Gb of memory and clock rate of 2.5 GHz under Windows 10. The computation time for each of the results discussed above was 1–3 minutes. Since the covering problems in many applications are solved in advance rather than in the online mode, such a computation time is quite acceptable.

For the investigation of the ILPs problems constructed in this work, subgradient and other methods can be used. It is important that the approach proposed in this paper is simple and effective. The numerical results confirm the effectiveness of the method for various versions of k-covering problems.

6 Conclusion

A numerical method for investigating k-coverings of a convex bounded set with circles of two given radii is proposed. Cases with constraints on the distances between the covering circle centers are considered. An algorithm for finding an approximate number of such circles and the arrangement of their centers is described. For certain specific cases, approximate lower bounds of the density of the k-covering of the given domain are found. We use 0–1 linear programming or general integer linear programming models. Numerical results demonstrating the effectiveness of the proposed methods are presented.

The proposed method with constructing a grid on G can be extended for covering problems in spaces of three and more dimensions. While solving the problem of covering with circles of equal radii, we obtain an ILPs of dimension n × 2n, where n is the number of nodes in the grid on G. In the case of covering G with balls of s different (fixed) radii, the dimension of the ILP is (s × n) × 2n without taking into account constraints on the distances between the ball centers. The solution of such problems requires either longer computation time or other methods for solving large-scale ILPs.

Acknowledgement

Authors express their gratitude to the anonymous referees for the remarks and useful recommendations which allowed to improve this article.

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Published Online: 2021-01-29

© 2021 Alexander V. Khorkov et al., published by De Gruyter

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