Farthest-Point Voronoi Diagrams in the Hilbert Metric
Abstract
The Hilbert metric, introduced by David Hilbert in 1895, is a projective metric defined on a bounded convex domain in a Euclidean space. For a convex polygon with vertices and point sites lying inside the polygon in the plane, it is shown that the nearest-point Voronoi diagram in the Hilbert metric has combinatorial complexity of [Gezalyan and Mount, SoCG 2023]. In this paper, we show that the farthest-point Voronoi diagram in the Hilbert metric has combinatorial complexity , which is independent of the number of sites. Also, we present an efficient algorithm to compute the farthest-point Voronoi diagram.
Keywords and phrases:
Farthest-point Voronoi diagram, Hilbert metric, Complexity, AlgorithmCopyright and License:
2012 ACM Subject Classification:
Theory of computation Computational geometryAcknowledgements:
For helpful comments we thank Auguste H. Gezalyan, David M. Mount and the anonymous reviewers.Funding:
This research was supported by the Institute of Information & communications Technology Planning & Evaluation (IITP) grant funded by the Korea government (MSIT) (No. RS-2019-II191906, Artificial Intelligence Graduate School Program (POSTECH)). This research was partly supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (RS-2023-00219980).Editors:
Pat Morin and Eunjin OhSeries and Publisher:
Leibniz International Proceedings in Informatics, Schloss Dagstuhl – Leibniz-Zentrum für Informatik
1 Introduction
The Hilbert metric, introduced by David Hilbert in 1895 [16], defines a distance function between points in the interior of any convex body in -dimensional space. Hilbert geometry provides a generalized framework that applies to any convex body, transcending the limitations of Euclidean space [21]. The significance of Hilbert metric is highlighted by a critical role in convex approximation, which is widely used in applications such as approximate nearest neighbor searches in the Euclidean metric and other metric spaces [1, 8], optimal construction of -kernels [6], approximate closest vectors [12, 13, 19, 23], and approximations of polytopes with low combinatorial complexity [3, 5, 7]. Various elements used in these approximations, such as Macbeath regions and Dikin ellipsoids, act similarly to Hilbert balls [2].
Despite its potential, not much is known about construction algorithms for structures in Hilbert geometry. Nielsen and Shao characterized balls in the Hilbert metric [20]. Bumpus et al. investigated the properties of balls and bisectors in the Hilbert metric [9]. A Voronoi diagram is one of the most fundamental structures in understanding the underlying geometry. Thus, it has been studied under different metrics [10, 17, 18, 22]. Gezalyan and Mount presented an -time algorithm for computing the nearest-point Voronoi diagram of point sites in the Hilbert metric defined by a convex -gon [15] and an -time algorithm for computing the Delaunay triangulation [14]. They showed that the diagram has complexity . No previous work, however, is known about the farthest-point Voronoi diagram in the Hilbert metric. The farthest-point Voronoi diagram may reveal further characteristics of Hilbert geometry and its potential applications.
1.1 Our results and outline
We denote by the farthest-point Voronoi diagram of point sites in the Hilbert metric defined by a convex -gon . Unlike Voronoi diagrams in the Euclidean metric, poses a few difficulties in the computation. The Hilbert distance between points is defined as the logarithm of the cross-ratio of them and the intersections of the line through the points with the domain boundary. It takes time to compute the distance between points. The bisector between any two sites is a piecewise conic curve consisting of segments. Thus, a naïve divide-and-conquer algorithm using bisectors may take time.
We show that has combinatorial complexity , which is independent of the number of sites. Moreover, we present an -time algorithm for computing the diagram. This is the first algorithm for computing .
In Section 2, we introduce the Hilbert metric, Hilbert balls, Hilbert bisectors, and . In the Euclidean metric, a site of has a nonempty cell in the farthest-point Voronoi diagram if and only if it is a vertex of the convex hull of [11]. This is, however, not always the case in the Hilbert metric. If a site has a nonempty cell in , it is a vertex of the convex hull. But not every vertex of the convex hull always has a nonempty cell in . In Section 3, we characterize this phenomenon and show that each cell in is connected and incident to the boundary of . We give an ordering lemma for the sites of and their Voronoi cells appearing along the boundary of .
In Section 4, we analyze the combinatorial complexity of . We show that has edges and each Voronoi cell is connected. Using this, we show that the number of sites with nonempty cells is . By Euler’s formula, we show that has vertices. We conclude that the combinatorial complexity of is . Additionally, since the complexity of is at least the complexity of , a lower bound on the combinatorial complexity of is . As a result, we give a tight upper bound on the combinatorial complexity of , which is and independent of the number of the sites of .
In Section 5, we give an efficient algorithm of time for computing . Our algorithm consists of three stages. The subsections in Section 5 describe these stages of the algorithm. In the first stage, we compute the convex hull of and remove the points contained in the interior of the convex hull of . In the second stage, we construct restricted to the boundary of incrementally by adding the sites of one by one in clockwise orientation along the convex hull of . In doing so, we maintain a list of sites that have nonempty boundary cell. Initially, contains the first two consecutive sites of . When the next site is added, we check whether is the farthest from an intersection point of the boundary cells corresponding to the first and the last sites stored in . If is not the farthest one from the point, it has an empty cell so we are done and move on to the next iteration. If is the farthest one, it has a nonempty cell so we append to . We compute the boundary cell of along the boundary of from the intersection point. In this way, we compute all sites of whose Voronoi cell is nonempty in time.
In the third stage, we construct by subdividing into Voronoi cells using restricted to the boundary of . Our algorithm uses a divide-and-conquer technique similar to the one by Shamos and Hoey [24] that computes the nearest-site Voronoi diagram in the plane under the Euclidean metric. Roughly speaking, we partition a site sequence into two subsequences, and , of roughly equal size, and compute by merging and along their bisector recursively. But the merge step requires a novel algorithmic idea and an in-depth analysis as each bisector is a piecewise conic curve consisting of segments. In each recursive step, we compute approximate bisectors such that they become the exact bisectors in the last recursive depth. We show that this can be done in time in total. Therefore, can be computed in time in total.
2 Preliminaries
For any two points , we denote the line segment connecting and by , and denote the Euclidean distance between and by . A set of points is convex if for any two points , every point is contained in . The convex hull of is the smallest convex set that contains , which we denote by . We denote the boundary of by and the interior of by . For any two points , let denote the intersection of the line through two points and with .
2.1 The Hilbert metric and Hilbert balls
Definition 1 (Hilbert metric).
Given a convex body and two points , let and denote the endpoints of such that appear in order along . See Figure 1(a). The Hilbert distance between and is defined as
The quantity in the logarithm is the cross-ratio of . If or lies on , can be formally defined to be . This corresponds to a limiting case that a denominator approaches zero. The Hilbert distance is extended to all pairs of points by letting . It satisfies the axioms of a metric, and in particular, it is symmetric and the triangle inequality holds. Observe if , and are collinear in [21].
In the remainder of the paper, we abuse the notation and use to denote a convex polygon with vertices in the plane. For any two points , a simple binary search makes it possible to determine the two edges of , each containing an endpoint of . Thus, the Hilbert distance between two points can be computed in time.
For a point and , let denote the Hilbert ball of radius centered at . Nielsen and Shao characterized the shape of Hilbert balls and showed how to compute them [20]. Consider the set of ’s for vertices of , which we call the spokes of . See Figure 1(a). For each , there are exactly two points lying on whose Hilbert distance from is . Those points are all in convex position. Then is the convex polygon with vertices on those points. We say is the Hilbert ball centered at with radius . For a point , ; the equality holds for lying on the boundary of . For a point , . See Figure 1(b).
We extend the concept of Hilbert balls to ones centered at points in , which we call infinite Hilbert balls. Let be a point in , and let be a point in . For arbitrarily small , let be the point on with . As approaches 0, any sequence of Hilbert balls in the family converges to the infinite Hilbert ball centered at with radius , which we denote by . See Figure 1(c).
2.2 Hilbert bisectors
Gezalyan and Mount [15] introduced the Hilbert bisector of two points , denoted by , which is the set of points satisfying . For any point , and . For any point , and . For a point , let be two endpoints of , and let be two endpoints of . Let , , and be the lines containing the segments , , and , respectively. If , let and let be the line tangent to at . It is shown that two cross ratios, and , are the same if and only if these three lines meet at a common point. It follows , and thus is on the bisector. See Figure 2(a-b).
The Hilbert distances and of a point are determined by the boundary edges of incident to and , respectively. Consider the subdivision of by all spokes of and all spokes of . Observe that for every point in a subregion, the set of edges of incident to and remains the same. Thus, and for points in a subregion can be formulated in explicit functions, and thus we can obtain the Hilbert bisector of and restricted to the subregion from the functions. It has been shown that in a subregion is a conic curve. It follows that is a piecewise conic curve whose joints lie on subregion boundaries [9]. Thus, we say that consists of bisector segments. See Figure 2(c).
Lemma 2 (Lemma 2 of [14]).
For any three points in not lying on the same line, there is at most one Hilbert ball whose boundary contains them.
Lemma 2 naturally extends for infinite Hilbert balls. The lemma implies that is at most one point for any three points not lying on the same line.
Lemma 3.
Two distinct (infinite) Hilbert balls intersect each other along their boundaries in at most two connected components.
2.3 Farthest-point Voronoi diagram
Let be a set of points lying in which we call sites. The farthest-point Voronoi diagram of in the Hilbert distance on , denoted by , is a subdivision of into cells in which the same point of is the farthest point in the Hilbert distance. The farthest-point Voronoi cell of a site , denoted by , is the set of points whose farthest site is in the Hilbert distance. Consider a point contained in . If , for all sites . If , there is a point sequence contained in such that the sequence converges to and for all sites and for all . Not every point of has a nonempty cell in . We will investigate this in Section 3.
General Position Assumption.
We assume that no three sites in are collinear, and in particular, the line passing through any pair of sites of and the lines extending any two edges of are not coincident at a common point, including all three being parallel. If this assumption does not hold, the bisectors separating Voronoi cells can widen into 2-dimensional regions. These general position conditions were also assumed by Gezalyan and Mount [15].
3 Conditions for valid sites
We study the conditions for a site to have a nonempty cell in . We say a site is valid for if . As for the farthest-point Voronoi diagram in the Euclidean metric, a site contained in the interior of cannot have a nonempty cell in .
Lemma 4.
If a site is contained in , is not valid for .
Proof.
Suppose that is valid for . Then there is a point such that for all . This implies that contains all sites of . Since is convex, is contained in . Then , contradicting that lies on the boundary of . In the Euclidean metric, every site has a nonempty cell in the farthest-point Voronoi diagram if it appears as a vertex of the convex hull of the sites. This is not necessarily true for in the Hilbert metric. For two distinct points , let and be two infinite Hilbert balls, each of which contains both and on its boundary. Let . Gezalyan et al. [14] showed that no Hilbert ball contains , and a point on its boundary. Using Lemma 2, they showed that every Hilbert ball centered at a point in and containing both and on its boundary contains . See Figure 3(a).
Lemma 5.
If a site is contained in for sites and , it is not valid for .
Proof.
Let be a site contained in for some sites and . Then no Hilbert ball contains , , and on its boundary [14]. This implies that does not intersect . Without loss of generality, we assume that separates and from . Let be a point contained in . Since the Hilbert ball centered at and containing and on its boundary contains , . This means that separates and from , and separates and from . Since does not intersect and , separates from . Then the side of containing does not intersect the side of containing . Since , we have , and thus is not valid for . See Figure 3(a) for an illustration.
By Lemma 5, a site may be not valid for even for .
3.1 restricted to the domain boundary
To find all valid sites of for and compute efficiently, we compute restricted to the boundary of . We denote it by . The Voronoi cell in of a site , denoted by , is the intersection of and . We call a site valid for if .
Lemma 6.
Let be a point in and be the endpoint of closer to than to . Then .
Proof.
Suppose there is a point but . Let be a site of with . By definition, and . Combining these we have , or equivalently as are collinear [21]. But this violates the triangle inequality, yielding a contradiction.
Since is the restriction of on the boundary of , a site valid for is also valid for . Lemma 6 implies that a site valid for is also valid for . Thus, we can identify all valid sites of for from .
Lemma 7.
is connected for a site .
Proof.
If is not valid for , and we are done. Assume for the contradiction that is valid for and is not connected for a site . Let and be two connected components of . Then there are two sites , not necessarily distinct, with connected components of and of such that appear in order along in clockwise orientation, as shown in Figure 3(b). Observe that and are separated by , and and are separated by . Since both the side of containing and the side of containing contain , and intersect at least two times. Since and are not collinear, this contradicts Lemma 2.
By Lemmas 6 and 7, is nonempty and connected for every valid site of . Aronov et al. [4] gave an Ordering Lemma for the farthest-point geodesic Voronoi diagram for point sites in a simple polygon. We extend the ordering lemma for with respect to .
Lemma 8.
The order of valid sites along their convex hull is the same as the order of their Voronoi cells along the boundary of .
Proof.
For ease of description, we relabel the sites such that is the sequence of sites valid for in clockwise orientation along their convex hull. We use modulo for indices. First, we show that and are adjacent to each other for every . Suppose that and are not adjacent to each other for some . Let be a site such that and are adjacent to each other and appears right after in clockwise orientation along . Since is an endpoint of , and are the same. Then . See Figure 3(c).
Observe that separates from since all sites appear in clockwise orientation along their convex hull. Without loss of generality, assume that is contained in the side of not containing . Let be the endpoint of other than . Thus, is contained in the side of of containing . Then and are the same, and thus contains both and on its boundary. By Lemma 3, and intersect each other along their boundaries at most twice. Thus, is contained in . Since , it follows from Lemma 5 that is not valid for . This is a contradiction that is valid. Thus, and are adjacent to each other.
There are two cases: appear in order in clockwise orientation along or they appear in order in counterclockwise orientation along . Suppose the latter case. Then for some , the side of containing contains a point in . Let and be the endpoints of such that , and appear in clockwise orientation along . By definition, is contained in . Observe that is contained in by the argument in the previous paragraph. Thus, is contained in . By Lemma 5, is not valid for , a contradiction. Thus, appear in order in clockwise orientation along .
4 Complexity
We analyze the combinatorial complexity of by counting the numbers of vertices, bisector segments, and edges of . Let denote the set of sites valid for such that appear in clockwise orientation along their convex hull. We use modulo for the indices of the sites in . Obviously, and are the same.
For two sites , we call each connected portion of that appears in a Voronoi edge defined by and . Thus, a Voronoi edge is a curve consisting of bisector segments of , with the exception that the first and last segments might be truncated. Each endpoint of a Voronoi edge is a Voronoi vertex, which has degree 3 (if contained in the interior of ) or 1 (if contained in ). Each endpoint of a bisector segment is either a Voronoi vertex or a vertex with degree 2. We show that has Voronoi edges, and that consists of Voronoi cells and Voronoi vertices. This implies that has complexity . Since the complexity of is at least the complexity of , has complexity . Thus, the combinatorial complexity of is .
4.1 Complexity of
Gezalyan and Mount [15] showed that the combinatorial complexity of the nearest-point Voronoi diagram is . We show that the combinatorial complexity of the farthest-point Voronoi diagram is . Recall that every bisector is a piecewise conic with joints lying on spokes. To analyze the combinatorial complexity of , we count the number of Voronoi vertices and the total number of bisector segments in . Let denote the set of farthest sites for a point in . If consists of a single element, we use to denote the element. The following lemma shows an upper bound on the number of bisector segments.
Lemma 9.
There are bisector segments in .
Proof.
We count the number of bisector segments in by counting the spokes of that cross the boundary of for each . Consider a bisector segment of in for two sites and . Since is a piecewise conic with joints on spokes of and , each endpoint of lies on a spoke of or unless it is a Voronoi vertex. Let be a vertex of such that an endpoint of lies on . Then intersects . By Lemma 6, intersects at most two bisector segments in . See Figure 4(a). This implies that the number of bisector segments in can be obtained by counting the total number of spokes of that intersect for each .
Consider for a site and a vertex of such that . Then either or . See Figure 4(b). If , is incident to at most two Voronoi cells, and thus the total number of such spokes is .
Consider the case that . Assume for the contradiction that there is another site () such that but . Let be the intersection of and , and let be the intersection of and . See Figure 4(c). If , , and are collinear, and , implying .
So assume that , , and are not collinear. Since separates and from and , crosses each of , , and in an odd number of times. Thus, there is a point on that is not contained in . By definition, . Since , and are all contained in , we have or , implying that and thus .
Hence, the total number of such spokes is . Since there are spokes that cross cell boundaries in , the total number of bisector segments in is . It remains to count the number of Voronoi vertices, which can be done by Euler’s formula.
Lemma 10.
For a site , is connected.
Proof.
Assume for the contradiction that is not connected for a site . Let and be two connected components of . For each , let be a point lying inside , and let be the point closer to than among two endpoints of . By Lemma 6, . However, and are not connected in , which contradicts Lemma 7.
Lemma 11.
There are sites in .
Proof.
Theorem 12.
The farthest-point Voronoi diagram of point sites lying inside a convex -gon in the Hilbert metric has combinatorial complexity .
Proof.
5 Algorithm
In this section, we present an algorithm to compute . Our algorithm computes and identifies all valid sites. Then it computes in a divide-and-conquer manner.
5.1 Computing restricted to the domain boundary
By Lemma 4, every site of contained in is not valid for . Hence, we compute and remove all sites contained in its interior from . This can be done in time. Thus, all remaining sites in are in convex position. We relabel them such that is the sequence of sites in clockwise orientation along their convex hull.
Some sites of , however, may not be valid for . We will identify all valid sites of by computing . We do this incrementally by inserting sites one by one in order, checking if the newly inserted one is valid for the Voronoi diagram of the sites inserted so far, and handling the cases accordingly. We maintain a few structures. Let be the subset of consisting of the first sites of for . Let denote the Voronoi cell of in . Let be the list that maintains all valid sites for .
Initially, let and . We compute by computing the endpoints of . In the -th iteration with , we compute and maintain as follows. Let and be the first element and the last element in , respectively. Then there is a point shared by and . By Lemma 8, is valid if and only if . If , is not valid for , and thus it is not valid for . We set and move on to the next iteration. If , is valid for . We set . Then we update from by computing the endpoints of . This can be done by searching them along the boundary of from , once in clockwise orientation and once in counterclockwise orientation. In doing so, we update accordingly, and then append to .
Consider the case that we search for an endpoint of along the boundary of in counterclockwise orientation from . Let be the endpoint of other than . Observe that for site previous to in . If , one endpoint of lies on . We compute the endpoint of on . Otherwise, is not valid for , and thus it is not valid for . We remove from and move on to . We repeat this until one endpoint of is found. See Figure 5.
We find the other endpoint of in clockwise orientation similarly. After computing both endpoints of , we append to .
Lemma 13 (Lemma 7 of [14]).
Given an -sided convex polygon and any two points , the endpoints of on can be computed in time.
We can compute and in time. If , we are done. If , we identify all sites not valid for among sites valid for , and remove them from . This takes time in total over all insertion steps. Then we compute the endpoints of . Observe that one endpoint of is an endpoint of and the other endpoint of is an endpoint of , where and are the first and the last elements in . We can compute the endpoints of in time by Lemma 13. Since the number of sites is , it takes time in total to compute .
Lemma 14.
can be computed in time.
5.2 Computing
We have preprocessed of size and all sites valid for and by Lemma 14. We give an algorithm for computing in time by subdividing into Voronoi cells using . Our algorithm is based on a divide-and-conquer technique similar to the one by Shamos and Hoey [24] that computes the nearest-site Voronoi diagram in the plane under the Euclidean metric. But the merge step requires a novel algorithmic idea and an in-depth analysis to reduce the running time as each bisector is a piecewise conic curve consisting of segments.
Let denote the sequence of sites valid for such that appear in clockwise orientation along . Observe . We use modulo for the indices. For a site , let denote the region in bounded by , , , where and are two endpoints of . By Lemma 6, . Observe that every spoke of intersecting intersects . Let denote the number of the spokes of intersecting . The total number of for all is by Lemma 9.
We compute , which is the same as . Roughly speaking, we partition a site sequence into two subsequences, and , of roughly equal size, and compute by merging and along recursively. The bisector satisfies the property that any point of lying on one side of is farthest from some site in and any point of lying on the other side of is farthest from some site in . Observe that is connected in with two distinct points on . If was not connected or it had more than two points on , it partitions into four or more pieces, each belonging to either or . Then the cells of or are not consecutive along , contradicting Lemma 8. Also, has no closed loop since no Voronoi cell in the loop is incident to .
Consider the -th recursive step of the algorithm. Let and be two input sequences of sites in the step. We have and . For a site , let denote the cell of in , and let denote the cell of in for and the cell of in for . Let and . Let and .
If three consecutive sites are contained in for some , and are the same. Thus, by Lemma 6, implying that is .
Lemma 15.
The followings hold.
-
1.
is connected.
-
2.
does not appear in .
-
3.
is connected for any .
Proof.
Let . Consider Claim 1. Since has an endpoint in , it suffices to show that is connected and is connected. Assume for the contradiction that is not connected. Then there is a curve contained in which has an endpoint on for some site . Consider the case . Then there is a point such that the farthest site in of is , contradicting .
Thus, or . By definition, consists of a line segment and a part of ’s for some . If has an endpoint on , there is a point such that the farthest site of is . This contradicts . Thus, intersects at least three times, contradicting Lemma 6. Therefore, is connected. We can show that is connected by exchanging the roles of and . Thus, Claim 1 holds.
Consider Claim 2. Let be the part of for and that appears in . Then and . Since for every , . Thus, . By contraposition, Claim 2 holds.
Consider Claim 3. Let be a site in . Let be the maximal portion of contained in . By Claim 1, it suffices to show that is connected. Assume for the contradiction that is not connected. Then there is a region bounded by and satisfying . Since is connected, does not intersect . By Lemma 10, there is a site with . This contradicts that each Voronoi cell intersects . Similarly, we can show the case for .
5.2.1 Divide-and-conquer algorithm
Now we are ready to describe our algorithm for computing from . In the -th recursive step, we maintain a region which contains for . As the base case, we initialize to . For the -th recursive step, we compute from for inductively. In the last recursive step, we have for every .
Consider the recursive step with and . We compute the point . Then we trace starting from until the bisector meets or . Consider the case that meets at first. We shoot a ray from towards and find the point of hit by the ray. Let be the curve consisting of the part of from to , and . Then partitions and into two subregions. We set to the subregion of containing for each . See Figure 6. We handle the case that meets first analogously.
Consider the -th recursive step. Let and be the two input sequences of sites. Without loss of generality, let be the last site of and be the first site of . Let and . We find the point . Then we trace until it leaves (1) , (2) , (3) , or (4) .
Consider the case that (1) holds but (3) does not hold. Let be the point where leaves and enters for . If has been visited before, we trace a ray from towards satisfying , until the first intersection with . If has not been visited before, we repeat tracing from inwards as before until it leaves one of those four regions. See Figure 7(a). We handle the case that (2) holds but (4) does not hold analogously.
Consider the case that (3) holds. Let be the point where leaves . We trace a ray from towards satisfying until the first intersection point with . See Figure 7(a). We handle the case that (4) holds analogously.
Let be the resulting curve from the tracing above. Then partitions into two subregions for some . We set to the subregion of containing .
Observe that for all by the construction of the algorithm. Since , the following lemma holds.
Lemma 16.
Let and be two input sequences in the -th recursive step. Then and are the same in .
We show how to maintain for all in a map . Each region in has at most four sides. In the base case, we store the subdivision of by the spokes of in for each . We construct from by shaving off the edges in along from an endpoint . Each edge of belongs to one of three types: (1) part of a spoke of , (2) an edge connecting a Voronoi vertex and , and (3) part of . See Figure 7(b). Observe that is connected and it is part of the boundary of . When crosses an edge of type (1), we shave off the part of the edge not contained in . The resulting edge is incident to the intersection point. See Figure 7(c-d). Then we continue tracing . When crosses an edge of type (2), we remove the edge and continue tracing . When crosses an edge of type (3), we insert a Voronoi vertex at the intersection and insert an edge of type (3). If leaves , we finish the construction of by removing the boundary curve of from to .
5.2.2 Analysis
We analyze the running time for the -th recursive step. Since we already have , we can find an endpoint of in time. By Lemmas 15.3 and 16, in has complexity . Thus, has complexity . The same is visited times. We can check whether meets the same for the second time in additional time. The ray shooting and partitioning take time linear to the complexity of each cell.
Now we analyze the running time of the algorithm. The recursion depth is and by Lemma 9. The total complexity of for sites over all recursive calls at a fixed level of recursion is . Thus, the algorithm runs in time, after computing in time.
Theorem 17.
The farthest-point Voronoi diagram of point sites lying inside a convex -gon in the Hilbert metric can be computed in time.
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