1 Introduction
Our motivation for studying the problem arises from practical applications in the frequency/channel assignment problem in wireless/cellular networks. In adhoc mobile networks, each host(station/tower) is equipped with a RadioFrequency (RF) transceiver to provide reliable transmission inside a circular range, represented by a disk, within some distance. Each wireless client is equipped with corresponding receivers. The clients themselves are represented by a set of points in a plane. The disks representing the range (which is presumably the same for all stations) of each potential host is represented by the set . In the spirit of reducing interference in broadcast and other energysaving measures, we aim to limit or reduce the number of different frequencies(channels) assigned to each, represented by coloring. Typically, (WiFi) networks are built with 3 independent channels BHLL10 , hence the motivation for a study on the 3CDUDC problem. In the same spirit, we generalize the 3CDUDC to the CDUDC problem, where is an integer. We further generalize the problem by considering line segments and a continuous rectangular region as representing potential wireless clients (resp. the CLSDC and CRRC problems), instead of points.
1.1 Related Work
The 3CDUDC problem, to the best of our knowledge, was first studied by Biedl et al., BBL19 . They gave a 2approximation algorithm in time for the 3CDUDC problem. Their approach first partitions the plane into horizontal strips, solves the problem for every strip optimally, then returns the union of solutions of all strips. To solve the problem for any strip they show that at most a constant number of disks of an optimal solution intersect any vertical line. Based on this, they define a directed acyclic graph such that there exists a path from source to a destination corresponding to this optimal solution. In this paper, we attempt to improve upon this impractical running time. Our approach, however, focuses on the specific geometric properties that arise from the dual conditionals of the problem statement. Although both of the approaches, initially, begin by dividing the plane, we recognize a unique bound that exists in our need to bound the colorability and provide a novel solution in the same regard.
A notion of conflictfree coloring (CFcoloring) was introduced by Even et al., ELRS03 . and Smorodinsky S03 . In the CFcoloring problem we are given a set of points (representing client locations) and a set of base stations, the objective is to assign colors (representing frequencies) to the base stations such that any client lying within the range of at least one base station is covered by the base station whose color is different from the colors of the other base stations covering the client, and the number of colors used should be as minimum as possible. Here, the range of base stations is modeled as regions e.g., disks or other geometric objects. Even et al., ELRS03 proved that colors are always sufficient to CFcolor a set of disks in the plane, and in the worst case, colors are required. Note that this CFcoloring of disks is different from our notion of colorable disk cover of points. In the former overlapping disks may be given the same color if they dont share a client, whereas in the CDUDC overlapping disks must be colored with distinct colors regardless of whether they cover a common client. A generalization of CFcoloring is called a faulttolerant CFcoloring. Cheilaris et al., CGRS14 presented a polynomialtime approximation algorithm for the faulttolerant CFcoloring in 1dimensional space. Horev et al., HKS10 proved that colors are sufficient for any set of disks in the plane. For dynamic CFcoloring and results on CFcoloring of other geometric objects, we refer to BM19 and references therein.
A related problem of the CDUDC problem in the literature is the Discrete Unit Disk Cover (DUDC) problem. In the DUDC problem, we are given a set of points and a set of unit disks, our goal is to select as the smallest number of disks from as possible such that the union of these selected disks covers all points in . As in the CDUDC, here also, the sets and can be considered as representing a set of wireless clients and a set of base stations or towers, respectively. The DUDC problem is NPhard and is a very well studied one. There is a polynomial time approximation scheme (PTAS) with impractical running time for this problem MR10 . The current best approximation algorithm with reasonable running time is for any BAD15 . However, a series of approximation algorithms have been proposed for this problem by various authors over the past two decades, and a complete survey on this can be found in FL12 . When a line segment is used to represent a potential wireless client, the DUDC problem becomes a Line Segment Disk Cover (LSDC) problem. In a similar line, there is another variant of the DUDC problem, a Rectangular Region Cover (RRC) problem, in which all the continuous set of points lying in a rectangular region represent wireless clients. All the available results for the DUDC problem also extend to the LSDC and RRC problems B18 , with slightly different running time. We also extend our results for the CDUDC problem to solve the colorable variants of the LSDC and RRC problems, namely, the CLSDC and CRRC problems.
2 CDUDC Problem
In this section we consider the following problem.

Colorable Discrete Unit Disk Cover (CDUDC): Given a set of points, and a set of unit disks (of radius=1), both lying in the plane, and a parameter , the objective is to compute a set that covers all points in such that the set can be partitioned into , where for each the disks in are pairwise disjoint, i.e., the disks in can be colored with at most colors such that the overlapping disks receive distinct colors and every point in is covered by a disk in .
As it was pointed out in BBL19 that there is a related problem, namely, Unit Disk Chromatic Number (UDCN) problem, that aims to color all nodes in a given unit disk graph with at most colors. The UDCN problem is NPhard for any CCJ90 . Similar to Biedl et al. BBL19 , we can center a set of unit disks in the plane such that there are at least pairwise nondisjoint disks that have a common intersection region and a unit disk graph induced by is connected. Let us then place a set of points in this intersection region. Now observe that the set has a cover which is at most colorable, whereas the graph is at least colorable. Hence, the CDUDC problem is different from the UDCN problem. Biedl et al. BBL19 showed that the 3CDUDC problem is NPhard by carefully incorporating a set of points in the NPhard proof of the UDCN problem with in CCJ90 . This directly implies that the CDUDC is NPhard since the CDUDC is a generalization of 3CDUDC. It is also easy to see that the CDUDC problem belongs to the class NP, as follows: Here, the certificate for any Yes instance of CDUDC is a set of distinct colors identified by nonnegative integers , and a mapping , where . A polynomial time verifier checks if every point in is covered by a disk in and for every pair of disks if , whether it is the case that .
2.1 4Approximate Algorithm
Here, our algorithm is based on partitioning the plane containing points into a grid and then determining bound on the number of unit disks that can participate in any colorable covering of points lying within any square of the grid. We first define a grid of width units that partitions the plane into squared regions. Each of these squared regions is a grid cell with a size . For simplicity assume no point of lies on the boundary of these grid cells. Let us associate a unique ID to each grid cell as follows; let be a point in and be the grid width, then , (see Fig. 1). Note that each grid cell has a unique ID associated with it but multiple points can be associated with the same ID (if they lie within the corresponding grid cell). Let and be any two arbitrary grid cells with base points and respectively. We define the greater than operator for an ID as follows: . Note that our defintion of implies that the grid cells are indexed from bottomleft to topright, this operator simply indicates the order of iteration that is followed by our algorithm. We move left to right rowwise starting from the bottomleft corner cell. A predefined order is essential to our handingover logic at line 11 of Algorithm 1.
Given a grid cell , the following lemma provides a bound on the cardinality of the colorable unit disks covering all points lying within the grid cell . Let be the set of colorable unit disks covering all points of lying within the grid cell . The proof of the lemma is based on the observation that determining this bound is the same as determining a maximum number of disjoint unit disks that could potentially intersect .
Observation 1.
If is a grid cell of size , then the maximum number of pairwise disjoint unit disks that could potentially intersect is at most if is even, and is atmost if
is odd.
Proof.
We will provide an upper bound to the number of pairwise disjoint unit disks that can cover a square of side length . Let us prove this by considering the two cases: being even and odd.
Since we aim at bringing an upper bound to the number of disks that have a common intersection point with , we divde the region of into two parts; the inner part of the square and the union of its outer edges on which these common intersection points can lie. To maximize the number of disks, it is intuitive to keep them as far as possible to increase the spacing between disks and thereby trying to increase the number of disks.
with even side length: When is a multiple of 2, it is quite intutive that a symmetric pattern is likely to give the best results. So we attempt two types of symmetric pattern.
Case 1: Considering the square to be symmetric along the vertical axis, we arrange the disks in two possible cases: either a disk is arranged with edge of as tangent such that the center of lies vertically above/below the disk (see Fig. (a)a), or the vertical partition is tangent to some of the disks (see Fig. (b)b). In the first case the maximum number of disks along horizontal part of outer edge would be , since the diameter of disk is 2. In the second case it can be shown that the maximum number of disks along the horizontal part would be .
However, as we see for the case =2 (shown in Fig (a)a and Fig (b)b), both the cases break down to the same case since if for a given pair of parallel edges of if one of the above case is true, then for the other pair the other case stands true. By induction we can prove that this stands true for all the even values of . Through this we get an upper limit in the number of disks along the edges. For the inner area of the maximum number of disjoint disks are . So the maximum number of disjoint disks are (/2) +2 /2 +2 (/2 + 1), which is 2+2+(/2).
Case 2: Considering to be diagonally symmetric, here we consider one of its diagonals as the symmetry line and place the disk such that their center is on the diagonal and intersects the square at its corner. As the disk covers a part of the edge of the square, say where (i.e., radius of disk) and , apart from the case where the disks are arranged at the diagonals again (see Fig. (a)a), the maximum number of disks on an edge is still equivalent to . The number of disks within the interiors of as stated above in the previous case will remain same. Therefore, the total number of maximum possible pairwise disjoint disks are , which is again equivalent to (see Fig. (a)a, Fig (b)b, Fig (a)a, Fig (b)b, and Fig (c)c for illustration of the case ).
with odd side length: When is not a multiple of 2, again it is quite intutive that a symmetric pattern is likely to give the best results. So we attempt two types of symmetric pattern.
Case 1: Like the case for even values of we consider the symmetric distribution along the horizontal and vertical axes. Again we have two possibilities either the center of disk lying along the axes or symmetric about the axes for both the pairs of edges. It can be shown that for both the pair of edges we would have only one amongst the two configurations at a time for getting the miximum number of disks. In the first case, disks can completely be accomodated on one edge and one disk as a common disk between two adjacent edges. So, there will be disks in the exterior part for this case. The number of disks in the interior part as in the even case would be . So, the total becomes (see Fig. (a)a and Fig. (b)b).
Case 2: When diagonally symmetric, the case is quite similar to the previous case with 4 disks at the corners and among the four edges of . The total again is calculating to .
∎
Lemma 1.
If is a grid cell of size and is a colorable solution for , then .
Proof.
From Observation 1 the cardinality of any set of pairwise disjoint unit disks intersecting with a grid cell is at most 7 as . Therefore, . Now consider another sets , , , , each of which can either be a replica of the same collection of disks in or a rotation or transformation of such that disks within each set remain pairwise disjoint and intersect . Hence, any colorable solution if the union of the disks in covers all the points in . Thus, the lemma follows. ∎
The outline of our algorithm (Algorithm 1) for computing a cover of the points is as follows. We first partition the rectangular region containing the objects in and into individual grid cells of size . By utilizing the bound obtained in Observation 1 (for e.g., for the actual bound is in Lemma 1) we compute a colorable cover of the points lying in each grid cell, in an exhaustive manner. To ensure that there is no conflict in the overall aggregate solution, we use a handingover logic. Only disks of any particular grid cell cover centered within the same grid cell are colored with the associated color set. If a disk is required to be a part of this grid cell cover, but is centered in another grid cell, it is handedover to that grid cell. Based on the grid width and the diameter of the disk, we then define a coloring scheme that assigns a color to each disk in the union of all the individual grid cell covers computed. Finally, we return the pair . Since the diameter of the disks is fixed to be two units, the approximation factor of the algorithm is implied by the choice of the value . If the value of is 2, then a unit disk can participate in the colorable covers of points lying in four adjacent grid cells. Hence, we prove that Algorithm 1 is a 4approximate algorithm (see Theorem 1). Later, we show that by varying the grid width , which results in a unit disk participating in more than four individual grid cell covers, we can obtain a family of algorithms with approximation factors corresponding to the choice of the value of (see Subsection 2.3).
We now define any coloring function that assigns colors to disks to be conflictfree if for any pair of nondisjoint disks (i.e., overlapping disks) the colors assigned to them are different.
Lemma 2.
The coloring defined by Algorithm 1 is conflictfree.
Proof.
For the sake of contradiction, let us assume that there are two disks such that , and , where along with is the output of Algorithm 1. Since , the distance between the centers of and is at most 2. Let the centers of and be lying in the grid cells and , respectively. Observe that and are either linearly or diagonally adjacent. If and are chosen to cover points lying only in the respective grid cells, then and are assigned colors from different color sets because the row and column numbers in their ID’s are not the same for both (see forloop at Line 17) (contradicting that ). Therefore, the only possibility for colorconflict to arise between and is that when both and are centered in the same grid cell , where covers a point lying in the cell above and covers a point lying in the cell below and each disk is initially chosen by the respective grid cell by means of the algorithm (Note that a similar case can be studied for horizontally and diagonally opposite grid cells). As per our color scheme (Line 1628), these grid cells are the nearest to have the same color set (say ) associated with them (see Fig. 1). Step 11 in the algorithm solves the conflict that arises in this case as follows. By means of the grid cell ID condition, disk is handed over to grid cell as its ID is greater and will receive the color set associated with that cell ( in this case, see Fig. 1). Disk however will not be handed over, but retains a color from the color set (contradicting that ). Thus, the lemma follows. ∎
Theorem 1.
Algorithm 1 is a 4approximation algorithm that runs in time for the CDUDC problem.
Proof.
In line 2 of Algorithm 1, the rectangular region is partitioned into grid cells. For each grid cell , let denote the number of points of lying in , i.e., . For each grid cell we then enumerate all subsets of such that each such subset contains at most 7 pairwise disjoint disks intersecting with the cell . In each iteration of the forloop at line 3, among collections of subsets together containing at most disks for some constant , we compute a set that covers all points lying in . In order to do this, we first compute the voronoi diagram on the center points of these disks. We then do point location queries for each of these points on to determine the closest center point. We will then test whether the corresponding disk centered at the closest center point covers this point. This step will take time. Therefore, we invest time to compute colorable cover of the points lying in the cell . The total time the forloop takes over all the nonempty grid cells is =. The remaining steps of the algorithm (including the preprocessing in steps 1 and 2) will take no more than time. Under the assumption that is much larger than the number of colors, the time complexity of the algorithm in total is . In order to assign the colors at a later step of the algorithm, with each grid cell we also associate and store the corresponding cover computed in the forloop at line 3. Thus, the additional space the algorithm requires is .
The handover behaviour at line 11 of the algorithm and Lemma 2 ensure that any colorconflict is resolved in an elegant way. Since we use four disjoint sets of distinct colors and for each grid cell we compute colorable unit disk cover (from Lemma 1), the approximation factor of the algorithm is 4. Thus, the theorem follows.
∎
The following corollary says that Theorem 1 yields a faster algorithm for CDUDC than that of Biedl et al.BBL19 , but at the cost of increase in approximation factor.
Corollary 1.
There exists a 4approximate algorithm that runs in time for the CDUDC problem.
Proof.
Follows from Theorem 1 by applying .∎
2.2 Improved Algorithm for a Grid Cell
In this subsection, we try to improve on the brute force approach of selecting 7 or fewer disks (7 being the max when is 2). We suggest various possible ways of improvement.
We divide our grid cell of size into 4 equal subgrid cells (see Fig. 6(a)). Therefore, the edge of each subgrid cell would be , i.e., it would be between to 1, since . We consider a subset of disks that can cover points in 2 diagonally opposite pair of subgrid cells such that the center of these disks lies in the respective quadrant of the subgrid cells, assuming that is placed such that the center of is at the origin of the coordinate system. Suppose that and are these subgrid cells chosen and that would contain only those disks that have centers above and towards the right for , and below and towards left for the dividing boundaries for (see Fig. 6).
Observation 2.
The maximum number of pairwise disjoint disks in is at most 5.
Proof.
This can be estimated by considering our earlier observation that for a grid cell of edge length less than 1, at most 4 disjoint disks are required to cover it maximally. The 4 disks required can be seen as 1 disk for covering each edge. Effectively when we consider disks with centers in a particular quadrant, we are covering only 3 sides for each subgrid cell. The sub grid cells being diagonally opposite have a common disk, so effectively we are looking for
nonintersecting disks to cover both the diagonally opposite subgrid cells maximally (see Fig. 6(b)). Also this being an upper bound for case where is 2 and reduces as the value of reduces. ∎Given the points lying in and being covered by , it is a fair possibility that we might require less than 5 disks to cover the points in diagonally opposite subgrid cells and . So we calculate combinations of disks of such that they cover the points in diagonally opposite subcells and are pairwise interiordisjoint. For each of this combination calculations we require possibilities respectively making the complexity to be . For each of these possibilities, we remove from the given disk set all those disks that might overlap with the chosen disks and then arrange the remaining disk centers lying in the quadrants containing the remaining subgrid cells in increasing values of and coordinate. We choose the disk for the upper leftmost subgrid cell (call it ) with its center having minimum and coordinates (i.e., a disk whose center is lexicographically closest to ) covering all the required points. Likewise, we choose the disk for the bottom rightmost grid cell (call it ) with its center having a minimum and coordinate covering the points. Overall the complexity for sorting turns out to be . Hence the overall complexity for selecting the disks reduces from (ignoring and ) to for disks for a given color.
We also repeat the above procedure by considering all possible combinations for the diagonally opposite subgrid cells and and for each subset of at most 5 nonintersecting disks from each of such combinations by finding two disks (pairwise disjoint from each of these 5 disks), lexicographically closest to subgrid cells and respectively, as above. The overall time is still excluding the factors and .
However, by presorting the disks in by both the  and coordinates of their centers and by maintaining these two orderings separately, we can improve the running time from to because given the cell , extracting the sorted lists of disks for the remaining subgrid cells and choosing the disks from these lists to cover the required remaining points all can be done in time.
Hence, for a grid cell we can generate all candidate subsets of at most 7 pairwise disjoint disks corresponding to each of colors, in time. Thus, the running time of our exaustive search for a colorable cover of all points lying in a cell can be improved by a factor of . The correctness of this procedure follows again due to the packing constraints (Lemma 1). Therefore, we have the following results.
Theorem 2.
We have a 4approximation algorithm that runs in time for the CDUDC problem.
Corollary 2.
There exists a 4approximate algorithm that runs in time for the CDUDC problem.
2.3 Generalization
In this subsection, we generalize the results from the preceding subsection to a general case observations for possible values of the width of each cell in the grid partitioning approach. We begin by attempting to generalize the potential coloring schemes for each grid cell similar to the observations presented in Lemma 1. We first define a parameter that represents the factor indicating the number of additional color sets needed to satisfy a union of independent solution sets. As a result, we obtain a family of approximation algorithms for the CDUDC problem, depending upon a different possible values for and .
Lemma 3.
The union of all independently optimal colorable solution sets for points lying in each grid cell of size is colorable, where
Proof.
We prove this by verifying the number of grid cells a unit disk can maximally intersect while considering each case. This determines the number of disjoint color sets, each consisting of at most distinct colors. Clearly, this number is the same as . Note that we are interested in finding the upper bound of such intersections.
Case 1. : Since the diameter of each disk is 2 and the width of grid cell is also ()2, certainly, no disk can span more than 2 linearly adjacent disks (see Fig. 7). Thus, maximal intersection count is achieved by placing the disk in any of the grid intersection corners. Here the disk will certainly intersect grid cells regardless of the width.
Case 2. : If the width of the grid cell is less than 2, surely, a unit disk can span 3 linearly adjacent grid cells. However, if the width is , the disk cannot span more than 2 diagonally adjacent grid cells (see Fig. 7). Hence, if the width is greater than or equal to , then no matter where the disk is centered (in Fig. 7), it can not intersect more than 6 grid cells simultaneously.
Case 3. : If the width of the grid cell is , surely, a unit disk can span 3 linearly adjacent grid cells and the middle cell from the next adjacent column or row of the linearly adjacent grid cells (see Fig. 7). Thus, we use the position shown in Fig. 7 to indicate maximal count possible.
Case 4. : For width , a unit disk could potentially intersect 3 linearly and diagonally adjacent grid cells (see Fig. 7).
We do not consider the grid partitioning with grid cells of width as it is inconsequential to our study and provides no substantial results (as the colorability increases substantially with no real improvement to the number of intersecting disks).∎
In the same spirit, we provide a generalization of the observation presented in Lemma 3 for the maximum number of pairwise disjoint unit disks that can intersect a square of size . We define a parameter that is the count of mutually nonintersecting unit disks that can intersect a grid cell of width .
Lemma 4.
The bound on the cardinality of a set of pairwise disjoint unit disks that can intersect a grid cell of size is , for an integer , where
Proof.
Consider a grid cell of size . Imagine placing four unit disks, each centered farthest apart from one another, but outside the cell , and touching one of the four corners of the cell . Since the cell size is , we can not place any more disk that intersects the cell, but at the same time disjoint from each of these four disks. Hence, the bound for the case of width . Proof for each of the other cases can be done similar to the proofs provided for the cases given above, and given in Lemma 1 (see, for example, the cases and being illustrated in Fig. 8).
We do not consider grid cells of width as it provides no improvement to the running time while the approximation factor stays the same after (as seen in Lemma 3). On the other hand, is also not considered because the approximation factor becomes arbitrarily very high for . One can observe here that the upper bound for obtained in Observation 1 is tight in case of being even, whereas in case of odd, it is a loosely bound.∎
Theorem 3.
There exists a approximation algorithm to solve the CDUDC problem, that has a runing time of for a given grid width , where
,
Proof.
The input of Algorithm 1, in addition to a set of points, a set of unit disks, and an integer , also consists of a grid partitioning parameter . From Lemmata 3 and 4, it is clear that any reasonable value for the parameter will imply the values of and . Hence, we have a approximation algorithm in time. ∎
2.4 Further Generalization
In this subsection, as in Theorem 3 we further attempt to generalize the packing constraints that define the values of and our stated algorithm might take for a given value of . We attempt to bring an upper bound to the values of and and hence an approximate algorithm with complexity in terms of .
The density of a packing of two or more objects in the interior of any region is the ratio between the area of the union of these objects and the total area of . Then, we have the following packing lemmas.
Lemma 5 (Fejes Tóth, Ft53 ).
Every packing of two or more congruent disks in a convex region has density at most .
Lemma 6.
Consider a unit ball in the space , and consider any collection of pairwise disjoint hydercubes of side lendths at least that overlap , then .
Theorem 4.
There exists a approximation algorithm to solve the CDUDC problem, that has a runing time of for a given grid width , where and .
Proof.
The region that encloses all pairwise disjoint unit disks that participate in colorable cover of points lying in a grid cell of size is the Minkowski sum of and a disk of radius 2. The total area of this region is . It is obvious that this region is convex. Now, using Lemma 5 we can bound the the maximum number of pairwise disjoint unit disks that can cover points in as follows. The density of packing unit disks in the interior of is . Therefore, as referred to in the previous sections can be said to be bounded by . For getting a bound on we try to enclose as many squares as possible along the diameter vertically and horizontally. As many as + 1 squares can be arranged along with one of the diagonals. Therefore arranging the squares as grid by setting an upper value for number of squares in rows and columns, at most which is + +1 (as in Lemma 6). Since would be the dominating term here for values of , hence it can be said that is bounded by . Substituting the bounds of and in Theorem 3, the algorithm is a approximation algorithm with a running time of , where .
Now we argue that should be chosen such that . To provide an appropriate argument for our chosen values of we consider the following 2 cases
Case 1: Consider cases where . The Observation 1 holds true only for integer values of . So for values of it can be observed that there would be at most 4 disjoint disks, by keeping each of the disks at the 4 vertices. Therefore the value of , in this case, is bounded by 4. Hence the running time remains . The approximation factor increases arbitrarily. It turns out that a large number of color sets get used without any improvement in running time. So values of do not provide us with much of an advantage.
Case 2: Consider cases where . As it can be observed from Lemma 3, the approximation factor remains 4 for values of greater than 2, however, the runtime increases a lot since is bounded by . Clearly, there is a huge overhead of running time with no improvement in the approximation factor.
We can infer that values of such that are preferred. ∎
3 Line Segment and Rectangular Region Cover
In the same spirit as the CDUDC problem is considered due to its practical application in frequency/channel assignment in wireless networks, we also define two problems, that generalize the locations of potential wireless clients from discrete set of points to line segments and from discrete set of points to a continuous rectangular region, namely, the Colorable Line Segment Disk Cover (CLSDC) and Colorable Rectangular Region Cover (CRRC) problems, respectively.
We begin our approach using a fundamental combinatorial result involving unit disks that helps us to transform the above problems into our original CDUDC problem. Given a set of unit disks in the plane, a sector is the smallest region bordered by the boundary lines of disks and is covered by the same set of disks in . Thus, the arrangement of all disks of subdivides the plane into many sectors. It is not hard to show that the worstcase complexity of the arrangement of any set of unit disks is quadratic, as stated below.
Observation 3 (Funke et al. Fkkls07 ).
The number of sectors created by intersection of unit disks in is .
To develop approximation algorithms for the CLSDC problem, we transform every instance of CLSDC problem into an instance of CDUDC problem as follows. In an instance of CLSDC problem, we have a set of unit disks covering a finite union of line segments of arbitrary length with arbitrary orientation, and an integer , The objective, here, is to compute a colorable cover of all the line segments. We split each of these line segments into slices such that each such slice lies within some sector. Now, for each subset of slices lying within a single sector, we add one point into the same sector and remove all the slices. This collection of points is referred to as . Hence, from Observation 3 we have that . This can be taken as an instance of the CDUDC problem, where is taken as the input set of points .
Similarly, we can do a similar transformation for the CRRC problem. Here, we have a set of unit disks covering a continuous rectangular region . Our objective is to compute a colorable set of units disks such that is covered by the union of these disks. As above, we split into sectors, as induced by the union of disks in , and add one point into each sector.
For the CLSDC problem, the construction of the set of points can be done as follows. We first preprocess the given set of unit disks into any reasonable data structure, e.g., a doubly connected edge list (DCEL), in time VSDO00 . We can build the Voronoi diagram on the center points of disks in in time. We store cross pointers between Voronoi cells (or disks) in and the faces (or sectors) of DCEL that are contributed by the corresponding disks. For each of the given line segments, we do point location query on for the left endpoint of the line segment to determine the disk in which it lies. We then follow the cross pointer to access the sector that contains it. Subsequently, we traverse the adjacent sectors of this sector in DCEL. As we do, we add points into those sectors ( also, into ) that covers a portion of the line segment and mark the corresponding faces (or sectors) as processed in DCEL. This step will take time. In the case of the CRRC problem, the construction of takes time as we have to test whether each sector is intersected by .
Therefore, we have the following results for the CLSDC and CRRC problems.
Theorem 5.
We have a approximation algorithm to solve the CLSDC problem, that has a runing time of for a given grid width , where and are defined as in Theorem 4.
Proof.
Theorem 6.
We have a approximation algorithm to solve the CRRC problem, that has a runing time of for a given grid width , where and are defined as in Theorem 4.
Proof.
Follows from Theorem 4 by applying .∎
4 Conclusion
In this paper, we have proposed constantfactor approximation algorithms for computing colorable unit disk covering of points, line segments, and a rectangular region. The algorithm is based on exhaustively searching an optimal colorable cover of points lying within a grid cell, which is a square of constant size. We believe that there seems to be no room for further improvement of approximation factor to smaller than 4 using the square grid approach. However, as future work, one could improve the almost bruteforce algorithm’s running time for a grid cell by exploiting various packingconstrained covering properties about grid square and unit disks. This will improve the running time of the overall algorithm.
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