Approximation and Parameterized Algorithms for Covering with Disks of Two Types of Radii
Abstract
We study the Discrete Covering with Two Types of Radii problem motivated by its application in wireless networks. In this problem, the goal is to assign either small-range high frequency or large-range low frequency to each access point, maximizing the number of users in high-frequency regions while ensuring that each user is in the range of an access point. Unlike other weighted covering problems, our problem requires satisfying two simultaneous objectives, which calls for novel approaches that leverage the underlying geometry of the problem. In our work, we present two new algorithms: the first is a polynomial-time -approximation, and the second is an exact algorithm for sparse instances, which is fixed-parameter tractable (FPT) in the number of large-radius disks. We also prove that such an FPT algorithm is impossible for general instances lacking sparsity, assuming the Exponential Time Hypothesis. Before our work, the best-known polynomial-time approximation factor was 4 for the problem.
Our approximation algorithm results from a fine-grained classification of points that can contribute to the gain of a solution. Based on this classification, we design two sub-algorithms with interdependent guarantees to recover the respective class of points as gain. Our algorithm exploits further properties of Delaunay triangulations to achieve the improved bound. The FPT algorithm is based on branching that utilizes the sparsity of the instances to limit the overall search space.
Keywords and phrases:
Covering, Disks, Approximation, FPTFunding:
Sayan Bandyapadhyay: The work has been supported by the NSF grant 2311397.Copyright and License:
2012 ACM Subject Classification:
Theory of computation Computational geometry ; Mathematics of computing Approximation algorithmsEditors:
Pat Morin and Eunjin OhSeries and Publisher:
Leibniz International Proceedings in Informatics, Schloss Dagstuhl – Leibniz-Zentrum für Informatik
1 Introduction
Set cover is among the most popular optimization problems, where given a ground set of elements and a family of subsets of the ground set, the goal is to find a minimum-sized subfamily of those subsets whose union contains all elements of the ground set. The problem admits a polynomial-time (poly-time) -approximation, and this bound is known to be tight [16, 14]. Set cover has been studied in various restricted settings with the goal of achieving better approximation factors. One such interesting setting is geometric covering, where elements are points in a geometric space, and subsets are induced by geometric objects such as disks and squares. Such restricted covering problems can admit poly-time constant-approximations [11, 9] or even approximation schemes [17, 27, 10], as one can exploit the natural constraints imposed by geometric spaces and objects. Indeed, this turned out to be a very popular area of research, as many practical problems can be described in geometric terms, e.g., in areas such as sensor networks, computational biology, image processing, and VLSI design [29, 6, 15, 13, 17].
While several formulations of geometric set cover are now well-studied, their real applications are sometimes hindered by practical constraints. There exist well-established techniques like local search [27, 19, 7, 1] to compute near-optimal solutions to fundamental geometric problems, but real-life constraints can invalidate the assumptions needed by these approaches. In this work, we study one such problem motivated by applications in wireless and sensor networks, which was introduced by Maheshwari et al. [22]. Informally, given a set of access points and users, the goal is to assign either low-range high frequency (i.e., high speed) or large-range low frequency to each access point, maximizing the number of users in high-frequency range while ensuring that each user is in the range of at least one access point of either kind.
We use the following notation to define the problem. For any point and a real , let denote the closed disk with center and radius . In Discrete Covering with Two Types of Radii (DC-2), we are given two point sets (of “users”) and (of “access points”) in the plane, and two real numbers and , such that and . The goal is to select for each , a value such that and the gain, i.e., the size of the set and is maximized. Thus, we want to select for each , either the small disk (of radius ) centered at or the large disk (of radius ) centered at , such that the set is covered by the chosen disks and the number of points in each of which is contained in at least one chosen small disk is maximized.
The problem is known to be NP-hard [22] similar to most of the covering problems with disks. Bandyapadhyay et al. [2] designed a poly-time 4-approximation for DC-2. In their approach, they converted the problem to a weighted covering problem with one type of radii (large) by assigning unit weight for each user point contained in a small disk, to a representative large disk. To solve this covering problem, they compute the Delaunay triangulation of the large disk centers. In particular, this triangulation is a planar graph, and hence can be colored by 4 colors. Consequently, a vertex cover of weight at most of the total weight can be computed, and the corresponding large disks are shown to cover all user points. Thus, at least of the total weight of the large disks remains unused while constructing the cover, which leads to the 4-approximation. Obtaining an improved approximation factor remained an interesting open question, as many covering problems involving unit disks admit PTASes [27, 21, 10].
As mentioned in [2], DC-2 can be interpreted as a combination of two problems: Minimum Weight Unit Disk Cover (WUDC) and Maximum Coverage with Unit Disks (MCUD). In WUDC, the goal is to select a subset of unit disks that cover all input points such that the sum of the weights of the selected disks is minimized. In MCUD, for a given , the goal is to select a subset of unit disks that cover the maximum number of points possible. Both of these problems are well-studied in computational geometry.
Mustafa and Ray [27] developed a PTAS for WUDC when all weights are equal, i.e., for the unweighted version. Their approach, which relies on a local search scheme, is inadequate for addressing weighted instances. Finding the optimal cover even for weighted unit disks has long been known to be strongly NP-hard [18]. Regardless, approaches have been developed in the last two decades for computing approximate solutions for weighted covering problems for broad classes of objects, such as those with low union complexity [8, 11, 30, 4]. Chan et al. [9] designed constant approximations for covering with weighted disks. Subsequently, Li and Jin [21] designed a PTAS for WUDC, utilizing a shifting strategy [18] that partitions the plane into squares and addresses a restricted version of the problem within each square. We note that this approach cannot be used directly for DC-2 when users are covered by multiple small disks, because the assignment of user weights to small disks cannot be done arbitrarily, and so the weight of a small disk in the optimal solution is an unknown in DC-2.
Chaplick et al. [10] designed a local search PTAS for the MCUD problem. It follows that for some , there are small disks in any optimal solution to DC-2 such that they cover the points that contribute to the optimal gain. However, the best set of small disks that maximizes the coverage might not lead to a feasible solution, as the remaining large disks might not cover the points not in the union of the chosen small disks. Dealing with two different but dependent tasks is the main challenge in tackling DC-2. The need to satisfy one objective while optimizing another necessarily means that any decision made at the local scope can have global consequences. For example, selecting one access point for a small disk may force a neighbor to be selected as a large disk to maintain coverage, and such a dependency may further propagate to disks that are far away. As such, any solution for this kind of constrained problem must take into account its global properties while making any decision to achieve good approximation.
We also study DC-2 from the perspective of parameterized complexity. Designing parameterized algorithms for geometric problems is an active area of research that is becoming increasingly popular. A problem is called fixed-parameter tractable (FPT) in a parameter if it can be solved in time for a computable function , where is the input size. A known fact is that if a problem with parameter is W[1]-hard, then it cannot be FPT in [12]. Marx [23] considered the problem of covering with unit squares. Here the size of a cover is a natural parameter. He proved that the problem is W[1]-hard. One consequence of this is that the optimization version does not admit an Efficient PTAS or EPTAS [23], i.e., the -time PTAS due to Hochbaum and Maass [17] cannot be improved to an EPTAS. In a different work, Marx [24] studied the parameterized complexity of common optimization problems in geometric intersection graphs and obtained a mixed set of algorithmic and hardness results. In a follow-up work, Marx and Pilipczuk [26] designed a framework for designing sub-exponential () time algorithms for a wide range of geometric facility location problems. Applying this framework, they designed an -time algorithm for covering with unit disks. This result is also tight, assuming ETH [24] (also see this survey [25]). The framework is based on Voronoi diagram separators. We note that this framework is not useful for DC-2, again due to the mixed nature of the problem. Kowalska and Pilipczuk [20] recently studied parameterized approximation algorithms for covering with line segments. Lastly, Banik et al. [3] studied the parameterized complexity of a conflict-free version of geometric covering.
1.1 Our results and contributions
In our work, we design for any , an -time approximation algorithm for DC-2 with the improved approximation factor of . The algorithm is based on a fine-grained classification of user points that may contribute to the gain of a solution. As mentioned before, the best-known factor for DC-2 is based on a weight assignment scheme that assigns 1 weight for each user point contained in a small disk, to a large disk. Then using the Delaunay triangulation of the large disk centers, a cover by large disks can be computed where at least of the total weight remains unused. In our work, we identify a subset of user points for which of the total weight can be saved by using the same Delaunay triangulation based algorithm, but along with a new weight assignment scheme that exploits further properties of the triangulation. Specifically, this is the subset of user points that are contained in 2 or more small disks. Unfortunately, the weight saving in this way is not possible for the other user points contained in exactly 1 small disk and we still gain of the total weight for them. However, in a separate approach, we show that the points of the latter type that contribute to an optimal solution can be recovered almost exactly by treating them as the weights on the large disks and solving an instance of WUDC. Hence, our main algorithm runs two subalgorithms and returns the larger gain. Our analysis shows that this is at least times the optimal gain, which improves on the previous factor of .
We also study the parameterized complexity of the problem, with being the number of large disks to be selected. This is a natural choice of parameter, as in practice, one would like to open as few large disks as possible to increase the overall quality of the coverage. In particular, the large disks correspond to low-speed antennas, and hence, the selection of too many such disks might slow down the data transfer, leading to a subpar experience for customers. We first show that the problem is W[1]-hard when is the number of large disks to be selected. In fact, we prove that assuming ETH, there is no time algorithm for the problem for any computable function . In stark contrast, we show that the problem is fixed-parameter tractable (FPT) in if each point is contained in, at most, a constant number of large disks. We define the notion of -sparse instances (see Section 3 for a formal definition) and show that the problem can be solved in time on any such instance, where . This result indicates that the W[1]-hardness arises solely from the absence of sparsity in the instances. We note that the problem remains NP-hard for -sparse instances [22].
While the -approximation works for any instance of DC-2, in practice, the FPT algorithm may yield vastly better run time when points are low-depth. DC-2 is directly motivated by wireless networks, where low-depth points frequently occur in certain environments like sparsely-populated cities. As such, we chose to describe both algorithms with the hope of improving the variety of results available for practical applications.
Organization.
2 An Approximation Algorithm for DC-2
Our main algorithm is an aggregate of two separate algorithms. The first is based on the Delaunay triangulation of large disk centers. The second algorithm is based on a reduction to the Minimum Weight Unit Disk Cover problem. Thereafter, we combine the two algorithms to obtain our main algorithm which has the desired approximation guarantee.
A set of disks is said to cover a set of points if the union of the disks in contains the points in . Let (resp. ) be the set of small (resp. large) disks centered at the access points in . Let such that each point in does not lie in any (small) disks of . Wlog, we assume that each point of is in at least two disks of . It must be in at least one disk, otherwise there is no feasible cover. Also, if it is in exactly one disk, this disk will always be part of the solution. Let such that each point in lies in at least one (small) disk of . Thus, is a partition of . Let such that each point in lies in exactly one disk of . Similarly, let such that each point in lies in at least two disks of . Thus, is a partition of . Let and . Fix an optimal solution of DC-2, i.e., the selected subset of disks of . Let OPT be the set of points in covered by the optimal solution. Let OPT OPT and OPT OPT. Also, fix an error parameter .
Observation 1.
The gain of any solution is at most .
As each point in is contained in a small disk and also in the corresponding large disk , the following observation holds.
Observation 2.
Consider any subset where for each , either or is in . Then is a cover for .
2.1 The first algorithm
Let be the Delaunay triangulation of the points in . In particular, is a connected planar graph where the vertices are the points in and each edge is a segment connecting two points of . We prove the following lemma.
Lemma 3.
For any disk in the plane that contains at least two points of , there exist two points in and a path in between and such that is contained in .
Proof.
Continuously decrease the radius of until it contains a point, say , of on its boundary. This process gives us a disk that contains all the points in . If there is no other point than on the boundary of , shrink along keeping the point fixed on its boundary until it contains another point on its boundary and denote it as . The resulting disk contains all the points in and has and both on its boundary (Figure 1). We prove that there is a path in between and such that is contained in , proving the lemma. The proof is similar to that of Theorem 1 [5], adapted in our context.
The proof is by induction on the number of points in . If there is no point of in the interior of , then is an edge of and we are done. Otherwise, suppose there is a point in the interior of . Shrink along , keeping the point fixed on its boundary until it contains on its boundary. This process gives us a disk that contains and on its boundary and does not contain . Similarly, shrink along , keeping the point fixed on its boundary until it contains on its boundary. This process gives us a disk that contains and on its boundary and does not contain . By induction, there is a path in between and such that is contained in and a path in between and such that is contained in . Merging the two paths and , we obtain the desired path .
Corollary 4.
For any point , there exist two points such that the edge is in and is in both large disks and .
Proof.
Consider the disk . As is in at least two large disks, there must be at least two points of in this disk. Then by Lemma 3, there are two points in and a path in between and such that is contained in . Consider any edge on . Then as are in , is in both large disks and .
Corollary 5.
For any point , there exist two points such that the edge is in and is in both small disks and .
Proof.
Consider the disk . As is in at least two small disks, there must be at least two points of in this disk. Then by Lemma 3, there are two points in and a path in between and such that is contained in . Consider any edge on . Then as are in , is in both small disks and .
Now, we describe the pseudocode of the algorithm, which is self-explanatory.
Input: The sets , small radius and large radius .
Output: A set of large and small disks covering .
The four coloring of a planar graph can be computed in polynomial time, for example using the algorithm in [28]. Hence, the algorithm runs in polynomial time.
Lemma 6.
The solution computed by Weighted-Extraction covers all points of .
Proof.
Note that we pick either or in the solution for each access point . Thus, by Observation 2, the points in are covered by the selected disks. Now, consider any point . We prove that is covered in the solution. By Corollary 4, we know that there exist two points such that the edge is in and is in both large disks and . Note that the points in form an independent set of the graph by all sharing the same color. Hence, the set of points is a vertex cover of . It follows that at least one of and is in . As we pick large disks for all access points in , must be covered by or in the solution.
We need the following lemma to analyze the gain of the solution.
Lemma 7.
Let be the total weight of the points in . Then the set of small disks centered at the access points in contains at least points of .
Proof.
First, note that the points in form an independent set of the graph . Now, consider any point . We claim that it contributes 1 unit of weight to at most one point of . If is in , by our construction, it contributes either 1 or 0 depending on whether the center of the unique small disk that contains is in or not. If is in , then by our construction it contributes 1 unit each to two access points and . However, the edge is contained in . As is an independent set, it can contain at most one of and . Hence, in this case, contributes at most 1 unit as well. It follows that each 1 unit of weight of comes from a unique point of . Now, by our construction, 1 unit of weight for a point of is assigned to an access point only if is in the small disk . Hence, for each unit of weight of the points in , there is a unique point in that is contained in a small disk centered at a point of . So, the number of points contained in the set of small disks centered at the access points in is at least .
We note that in the above, the number of points covered by small disks may be more than , as the weight of each point contained in a small disk centered at a point of might not be assigned to the points of . Hence, is only a lower bound.
Corollary 8.
The solution computed by Weighted-Extraction has a gain of at least OPT OPT.
Proof.
First, note that by our construction, the sum of the weights of all points in is . Now, as we pick to be the maximum weight color class, the total weight of the points in must be at least the average weight, which is . By Lemma 7, the gain of the solution, that is the number of points of covered by the small disks with centers at points in , is at least . As OPT and OPT, the gain is at least OPT + OPT.
2.2 The second algorithm
This algorithm has three steps. In the first step, we construct an instance of the Minimum Weight Unit Disk Cover problem from the given instance of DC-2.
In Minimum Weight Unit Disk Cover (WUDC), given a set of unit disks along with a weight function and a set of points in the plane, the goal is to find a subset , such that covers and the sum of the weights of the disks in is minimized. We refer to this sum of weights as the cost of the solution .
The construction of the instance of WUDC is as follows. is set to be . is the set of all large disks in . The weight of each large disk is the number of points of in the corresponding small disk , i.e., .
In the second step of the algorithm, a -approximate solution to WUDC on is computed using the PTAS in [21], which runs in time . In the third step, a solution to DC-2 is computed as follows. First, add all the large disks in to the solution set. Next, for each large disk in , add the corresponding small disk to the solution set.
The above algorithm runs in time. Next, we analyze this algorithm. Let OPT′ be the optimal cost of .
Observation 9.
The solution computed by the above algorithm covers all points of .
Proof.
We pick either or in the solution for each access point . Thus, by Observation 2, the points in are covered by the selected disks. We also see that the solution to WUDC covers the points in . As we include the (large) disks of in the DC-2 solution and , the points of are also covered.
Lemma 10.
OPT′ is at most OPT1.
Proof.
Consider the optimal solution of DC-2. The solution covers all points of and in particular the points of . As each point of is not in any small disk of , it must be covered by a selected large disk. Thus, the set of selected large disks, say , is a valid cover for the points of , and so is a feasible solution to . We show that the cost of this solution is OPT1, proving the lemma.
Now, the gain of the DC-2 solution from the points in is OPT1. Thus the union of small disks that are not selected (corresponding to the large disks in ) contains OPT1 points of . By construction of , these points are the only points that are assigned to the large disks in . Hence, the cost of , that is the sum of the weights of the large disks in is OPT1.
By the above lemma and the fact that is a -approximate solution, we have the following observation.
Observation 11.
The cost of is at most OPT OPT.
Lemma 12.
The gain of the DC-2 solution computed by the second algorithm is at least .
Proof.
Note that the gain is at least the number of points of in the small disks that are not selected in the DC-2 solution. Equivalently, the gain is at least the sum of the weights of the large disks that are in . By Observation 11, it follows that the gain is at least
The inequality follows, as by Corollary 8, . Hence, we obtain the desired lemma.
2.3 Combining the two algorithms
Our main algorithm is a combination of the two above-mentioned algorithms. We invoke both on the DC-2 instance and return the solution with the larger gain.
Lemma 13.
The gain of the larger-gain solution is at least .
Proof.
Let be such that OPT OPT. By Lemma 8 and 12, the gain of the larger solution is at least
This function of is minimized when
| Or, | |||
| Or, |
Thus, the gain is at least . By scaling by a constant, we obtain the desired bound. We state our main result of this section in the following theorem.
Theorem 14.
There is an -time -approximation algorithm for DC-2.
3 An FPT Algorithm for Sparse Instances
The parameterized version of DC-2 includes a fixed parameter , where in addition to the original goal, we are assured that for at most access points . We refer to this problem as Par-DC-2. Here, we reuse the notation from the previous section. Let OPT be the gain of any optimal solution. Recall the sets of small disks and large disks , and the sets of vulnerable points and non-vulnerable points . An instance of Par-DC-2 is called -sparse for an integer if each point of is in at most large disks of . We note that in such an instance, a point of can lie in more than large (or small) disks. We prove that Par-DC-2 is fixed-parameter tractable in .
For a set , let the gain of be the number of points of contained in the disks of . For a subset of large disks , let sm be the set of corresponding small disks.
Next, we describe our algorithm. Our algorithm is based on a recursive branching strategy. Algorithm 2 shows the pseudo-code of a recursive subroutine that is the main procedure used by our algorithm. The subroutine Recursive-Branching takes as input a subset of points that are yet to be covered, a subset of disks already chosen to cover, and an integer that denotes the depth of recursion. The subroutine returns a True/False flag. If the flag is True, it also outputs a set of large disks such that covers , and the gain corresponding to the solution , i.e., the gain of sm. Inside the subroutine, we first check if is empty, and if that is the case, itself is returned as the solution along with its gain. If the depth , but , the False flag is returned, indicating there is no feasible solution along this branch, so it should be pruned here. Otherwise, and . In that case, we consider any arbitrary point . There are at most large disks in that cover . We branch on each such large disk . In each branch, we remove the points of that are covered by and recurse on the set of remaining points with being the new set of selected disks and as the new depth. We also keep track of the largest gain solution returned by these calls and return it if the flag counter is set to True. If is False, we return the False flag.
The main algorithm makes a call to Recursive-Branching. If the call returns True, it outputs the solution where the large disks of are selected. Since there are no remaining points in to cover, the remaining small disks sm may be selected to cover a number of points in equal to the gain of . Otherwise, it concludes that there is no feasible solution with large disks.
Input: A set of points , a subset , and an integer .
Output: A flag True/False, and if the flag is True, a set of large disks such that covers and the gain corresponding to ; otherwise, and 0.
Next, we analyze the algorithm. Consider the recursion tree corresponding to Recursive-Branching. We define the level of the nodes in as follows. The level of the root is 1. The level of any other node is 1 plus the level of its parent. Let denote the large disks selected in the optimal solution. In our analysis, we assume that does not contain any redundant disk, i.e., removing any disk from leaves uncovered. If it contains such a redundant disk, then we can try our algorithm with a smaller value .
We prove the correctness of Recursive-Branching with the following lemma.
Lemma 15.
For each , there is a node in of level with the property that a call is made at to Recursive-Branching such that is the subset of points of covered by the disks of , and is a subset of with disks.
Proof.
We use induction on the level . In the base case, consider . The root node at level 1 selects a point and recurses for each disk . Now, must be covered by a large disk that is in . Consider the call Recursive-Branching made at the root. By definition, is the set of points in that are covered by . Hence, the statement is true in the base case. Now, suppose the statement is true for any level . We prove that the statement is true for level as well.
By induction argument, there is a node of level that makes a call to Recursive-Branching such that is the subset of points of covered by the disks of , and is a subset of with disks. Denote the child of in by that corresponds to this call and thus receives the set of large disks . Now, does not cover , as otherwise, contains a redundant disk. Thus, we have that . Suppose is the point of arbitrarily chosen during the execution of the call corresponding to . Then a recursive call is made for each . Now, by our assumption, is not covered by the disks in . Hence, there must be a disk that covers and is contained in . Consider the call Recursive-Branching made at . By definition, is the subset of points of contained in . As any point in is in a disk of , is the exact subset of points of covered by the disks of . Also, . Hence, the statement is also true for . The next lemma shows that the algorithm returns a solution with optimal gain.
Lemma 16.
Recursive-Branching always returns the True flag and the corresponding solution has the gain value OPT.
Proof.
Consider the node in of level that makes a call to Recursive-Branching where . The existence of such a node follows due to Lemma 15. Now, note that covers , and thus . Hence, this call returns True, OPT to its parent node. Now, inside each call, we keep track of the maximum gain solution returned by all recursive calls and return it to the parent. Hence, the root node always returns True and the gain of the solution returned must be at least OPT.
We will show that the total number of nodes of is bounded by . As the algorithm spends only a polynomial time at each node, the desired result follows. Now, note that at each node, the algorithm can make at most calls, as each point of is in at most disks of . Moreover, as we prune any branch at level , the maximum level is . Thus, the total number of nodes is . We state our result in the following theorem.
Theorem 17.
Par-DC-2 can be solved in time on any -sparse instance.
4 Parameterized Hardness of DC-2
We give a reduction from Dominating set for unit disks. In this problem, we are given a set of unit disks (radius 1) in the plane and a parameter , and the goal is to select a subset of disks such that for each disk in , or there is a disk in such that .
Marx [24] proved that Dominating set for unit disks is W[1]-hard. In fact, he proved that assuming ETH, the problem cannot be solved in time for any computable function . Next, we describe our reduction.
Let be the given instance of Dominating set for unit disks. Let be the set of centers of the disks in . The distance between a point and a set is defined as . For any point , let be the minimum distance between and the disks in (see Figure 2). Let . Also, let with . Denote . Also, let . We set , , , and . Also, we use the same parameter . Let be the constructed instance of Par-DC-2. The construction takes time.
Observation 18.
For any , the small disk .
Proof.
Let . Consider any . First, consider the case when and , then . Thus, . Next, consider the case when or . This implies that there is another point such that was constructed from , i.e., and . Now, suppose . Then, . In particular, . But, by the definition of , this is a contradiction. So, in this case, and so . As the two cases are exhaustive, the observation follows. The above observation implies that the gain of any solution is 0. Thus, the hardness comes from the ability to cover the points of using large disks.
Lemma 19.
has a dominating set of size if and only if there is a feasible solution for .
Proof.
First, suppose has a dominating set of size . Let be the set of centers of the disks in the dominating set. We claim that the large disks at the centers in cover the points in . Consider a point and let be such that and . We know that there is such that intersects . Thus . So, . Hence, the large disk at covers .
Next, suppose there is a feasible solution for . Thus, there is a size subset such that the large disks at cover the points in . We claim that the set of disks at the centers in form a dominating set. Consider any disk . Suppose . Consider the point such that , and . Then there is such that covers . We claim that . Suppose that is not true. Then as per the definition of , is at least distance away from . Thus, . However, . We obtain a contradiction, and hence . Equivalently, , and so dominates . Hence, is a dominating set of size . Based on the above discussion, we have the following theorem.
Theorem 20.
Par-DC-2 is W[1]-hard. Moreover, assuming ETH, Par-DC-2 cannot be solved in time for any computable function .
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