1 Introduction
A spanner (or hop spanner) of a connected graph is a subgraph , where , with the additional property that the distance between any two vertices in is at most times the distance in [20, 29]. (The distance between two vertices is the minimum number of edges on path between them.) The graph itself is a hop spanner. The minimum for which is a spanner of is referred to as the hop stretch factor of . An alternative characterization of spanners is given in the following lemma of [29].
Lemma 1 ([29]).
The subgraph is a spanner of the graph if and only if the distance between and in is at most for every edge .
If the subgraph has only edges, then is called a sparse spanner. In this paper we are concerned with constructing sparse spanners (with small ) for unit disk graphs in the plane. Given a set of points in the plane (or any metric space), the unit disk graph (UDG) is a geometric graph on the vertex set whose edges connect points that are at most unit distance apart. Recognizing UDGs was shown to be NPHard by Breu and Kirkpatrick [7].
Unit disk graphs are commonly used to model network topology in adhoc wireless networks and sensor networks. They are also used in multirobot systems for practical purposes such as planning, routing, power assignment, searchandrescue, information collection, and patrolling; refer to [2, 16, 19, 24, 28] for some applications of UDGs. For packet routing and other applications, a boundeddegree plane geometric spanner of the wireless network is often desired but not always feasible [5]. Since a UDG on points can have a quadratic number of edges, a common desideratum is finding sparse subgraphs that approximate the respective UDG with respect to various criteria. Plane spanners, in which no two edges cross, are desirable for applications where edge crossings may cause interference.
Obviously, for every , every graph on vertices has a spanner with edges. If is the complete graph, a star rooted at any vertex is a hop spanner with edges. However, the bound on the size of a hop spanner cannot be improved in general; a classic example [20] is that of a complete bipartite graph with vertices on each side. We are not aware of any analogous statement for . We show that for unit disk graphs we can do much better in terms of the number of edges for every .
Spanners in general and unit disk graph spanners in particular are used to reduce the size of a network and the amount of routing information. They are also used for maintaining network connectivity, improving throughput, and optimizing network lifetime [4, 18, 19, 23, 30].
Spanners for UDGs with hop stretch factors bounded by a constant were introduced by Catusse, Chepoi, and Vaxès in [9]. They constructed (i) hop spanners with at most edges for vertex UDGs; and (ii) plane hop spanners with less than edges. Recently, Biniaz [4] improved both these results, and showed that for every vertex unit disk graph, there exists a hop spanner with at most edges. He also showed how to construct a plane hop spanner for a vertex unit disk graph. It is straightforward to verify that the algorithms presented in [4, 9] run in time that is polynomial in .
Our results.
The following are shown for unit disk graphs.

[label=()]

For every , there exists an element point set such that every plane hop spanner on has hop stretch factor at least . If is sufficiently large, the lower bound can be raised to (Theorems 4 and 5 in Section 4). A trivial lower bound of can be easily obtained by placing four points at the four corners of a square of sidelength .
Notation and terminology.
For two points , we denote the Euclidean distance by or sometimes by . The distance between two sets, , is defined by .
Throughout the paper we use the notations and to denote a UDG and a subgraph of it, respectively. Given a graph, denotes the set of vertices adjacent to . For , let denote a shortest path in , i.e., a path containing the fewest edges; and denote the corresponding hop distance (number of edges). For brevity, a hop spanner for a point set is a hop spanner for the UDG on .
A geometric graph is plane if every two edges are either disjoint or share a common endpoint. Whenever we discuss plane graphs (plane spanners in particular), we assume that the points (vertices) are in general position, i.e., no three points are collinear.
A unit disk (resp., circle) is a disk (resp., circle) of unit radius. The complete bipartite graph with parts of size and is denoted by ; in particular, is a star on vertices. We use the shorthand notation for the set .
Related work.
Peleg and Schäfer [29] have shown that the problem of determining, for a given graph and a positive integer , whether there exists a spanner with at most edges is NPcomplete. They also showed that for every graph on vertices, a spanner with edges can be constructed in polynomial time. In particular, every graph on vertices has a spanner with edges. Their result was improved by Althöfer et al. [1], who showed that a spanner with edges can be constructed in polynomial time. This result was shown to be best possible. Kortsarz and Peleg [21] obtained approximation algorithms for the problem of finding a spanner in a given graph, with minimum maximum degree.
In the geometric setting, where the vertices are embedded in a metric space, spanners have been studied in [3, 8, 10, 11, 22, 24] and many other papers. In particular, plane geometric spanners were studied in [5, 6, 14, 15]. The reader is also referred to the surveys [6, 17, 25] and the monograph [27] dedicated to this subject.
2 Sparse (possibly nonplane) hop spanners
In this section we construct hop spanners with a linear number of edges that provide various tradeoffs between the two parameters of interest: number of hops and number of edges.
2.1 Construction of 5hop spanners
We start with a short outline of the hop spanner constructed by Biniaz [4, Theorem 3]; it is based on a hexagonal tiling with cells of unit diameter. Inside every nonempty cell, a star is formed rooted at an arbitrarily chosen point inside the cell. Then, for every pair of cells, exactly one edge of the UDG is chosen, if such an edge exists. Biniaz showed that the resulting graph is a hop spanner with at most edges.
We next provide a more detailed description and an improved analysis of the construction. Consider a regular hexagonal tiling in the plane with cells of unit diameter; refer to Fig. 1 (left). We may assume that no point lies on a cell boundary. Every point in lies in the interior of some cell of (and so the distance between any two points inside a cell is less than ). Let be a point that lies in cell . The six cells adjacent to form the first layer around ; the twelve cells form the second layer around (in a BFS order in the dual graph of the tiling ).
For every pair of cells take an arbitrary edge , , , if such an edge exists; we call such an edge a bridge. Each cell can have bridges to at most other cells, namely those in the two layers around . A bridge is short if it connects points in adjacent cells and long otherwise.
Lemma 2.
Let be a point that lies in cell . The unit disk centered at intersects at most five cells from the second layer around .
Proof.
Let be the center of (shaded gray in Fig. 1 (right)). Subdivide into six regular triangles incident to . By symmetry, we can assume that , where .
Note that for , and is disjoint from the five cells , , , , and . Now, observe that . Hence, intersects at most one of these two pairs. Consequently, intersects at most cells from the second layer around . ∎
Obviously, any two points in a cell at within unit distance apart, Observe that the unit disk centered at always intersects the six cells . As such, Lemma 2 immediately yields the following.
Corollary 1.
For every point , all neighbors of lie in at most other cells around .
Theorem 1.
The (possibly nonplane) hop spanner constructed by Biniaz [4, Theorem 3] has at most edges.
Proof.
Let be a set of points. Let be the number of points in a hexagonal cell . The construction has inner edges that make a star and at most outer edges (bridges) connecting points in with points in other cells. We analyze the situation depending on .
If , there are no inner edges and at most outer edges by Corollary 1. As such, the degree of the (unique) point in is at most .
If , there is one inner edge and at most outer edges. Indeed, by Lemma 2, each point has neighbors in at most five cells from the second layer around (besides points in in the six cells in the first layer). Two points in can jointly have neighbors in at most other cells. As such, the average degree for points in is at most .
If , there are inner edges and at most outer edges. As such, the average degree for points in is at most
Summation over all cells implies that the average degree in is at most (the bottleneck is the first case), thus has at most edges. ∎
2.2 Construction of 3hop spanners
Here we show that every point set in the the plane has a hop spanners of linear size. This brings down the hopstretch factor of Biniaz’s construction from to at the expense of increasing in the number of edges (from to ).
Theorem 2.
Every vertex unit disk graph has a (possibly nonplane) hop spanner with at most edges.
Proof.
Our construction is based on a hexagonal tiling with cells of unit diameter (as in Subsection 2.1). Let be the 5hop spanner described in Section 2.1. We construct a new graph that consists of all bridges from and, for each nonempty cell , a spanning graph of the points in defined as follows. (In particular, if is connected, then is also connected.)
Let be a nonempty cell and be the number of points in . Let . For every cell in the two layers around , if and contains a bridge , where and , then we add the edge to . Observe that when is a short bridge, then is the center of a spanning star on . In addition, if no short bridge is incident to any point in , then we add a spanning star of (centered at the endpoint of a long bridge, if any) to .
It is easy to see that the hop distance between any two points within a cell is at most . Indeed, by construction, the points in each nonempty cell are connected by a spanning star. Consider now a pair of points , , , where . By construction, there is a bridge between the cells and . As such, is connected to by a hop path . Refer to Fig. 2 for an illustration.
We can bound the average degree of the points in as follows. For each point , the unit disk centered at intersects at most other hexagonal cells. If is not incident to any bridge, we add at most edges between and other points in ; these edges increase the sum of degrees in by . Else assume that is incident bridges, for some . Then we add edges from to at most other points in . The bridges each have only one endpoint in . Overall, these edges contribute to the sum of degrees in . Finally, the spanning star, if needed, contributes to the same sum. Overall, the sum of degrees in is bounded from above by
Thus, the average vertex degree is at most in all . Consequently, the hop spanner has at most edges. ∎
3 Construction of 2hop spanners
In this section, we construct a 2hop spanner with edges for a set of points in the plane. We begin with a construction in the bipartite setting, and then extend it to the general setup. For two disjoint point sets, and , denote by the bipartite graph with parts and that contains an edge if and only if . Clearly, is a subgraph of the unit disk graph on .
Lemma 3.
Let be a set of points in the plane such that , , and . Then there is a graph on the vertex set with at most edges, such that for every edge of , contains a path of length at most between and .
Proof.
Our proof is constructive. We construct the graph incrementally, as a union of stars. Each iteration creates one star, and deletes some of the vertices from or . Due to the deletion, the number of vertices in and monotonically decrease, and the degree of every surviving vertex also decreases. Note that the properties and are invariants.
For every vertex , let be the set of neighbors of in ; and let
that is, the set of vertices at distance from in . As monotonically decreases in the course of our algorithm, both and monotonically decrease. Note that is a bipartite graph; in particular, implies and .
Our algorithm proceeds as follows. While , do the following. Let be the maximum degree in . Let be a vertex of maximum degree in , i.e., . Create a star centered at on the vertex set , and then delete from . When the while loop terminates, let be the union of the stars .
To prove correctness, we show that for every edge of the initial , (with , ), contains a path of length between and . Recall that each iteration of the algorithm removes vertices from either or , until or . We may assume, without loss of generality, that the algorithm removes vertex before . Consider the iteration in which the algorithm removes from . At his time, we have and . Since and , all edges between and are in the unit disk graph. In particular, the star centered at on the vertex set is a subgraph of the unit disk graph. Now the star , and hence the graph , contains the path of length between and , as required.
It remains to derive an upper bound on the number of edges in . One iteration decreases the number of vertices in by , and creates a star with edges. Recall that and observe that . It follows that
Since and we trivially have . Consequently, the ratio between the number of edges added and the number of vertices deleted is bounded as follows:
(1) 
Assume that the algorithm terminates in iterations, deleting the vertex sets . The total number of deleted vertices is bounded from above by . Using (1), the overall number of edges in the stars is
(2) 
as claimed. ∎
We now consider the general case.
Theorem 3.
Every vertex unit disk graph has a (possibly nonplane) hop spanner with edges.
Proof.
Let be a set of points in the plane. Consider a tiling of the plane with regular hexagons of unit diameter; and assume that no point in lies on the boundary of any hexagon. Let be the set of nonempty hexagons. Then is partitioned into sets . As noted above, for every , there are other cells within unit distance; see Fig. 1 (left).
For each cell , chose an arbitrary vertex , and create a star centered at on the vertex set . The overall number of edges in all stars , , is
For every pair of cells , where , consider the bipartite graph . By Lemma 3, there is a graph of size
Since every vertex appears in at most such bipartite graphs, the total number of edges in these graphs is at most
We show that the union of the stars , , and the graph is a hop spanner. Let be an edge of the unit disk graph. If both and are in the same cell, say , then is an edge in the star or the star contains the path . Otherwise, and lie in two distinct cells, say , such that . By Lemma 3, contains a path of length at most between and , as required. ∎
4 Lower bounds for plane hop spanners
A trivial lower bound of for the hop stretch factor of plane graphs can be easily obtained by taking the four corners of a square of sidelength . In this case, the UDG is a complete graph but exactly one diagonal of the square can be used to satisfy planarity. We show that a lower bound of holds if is sufficiently large; our main result in this section is Theorem 5. We begin with a lower bound of that holds already for .
Theorem 4.
For every , there exists an element point set such that every plane hop spanner on has hop stretch factor at least .
Proof.
Let be a set of successive points on a circle of radius , so that is a horizontal chord, , , , and . see Fig. 3 (left). Note that ; and that is vertically symmetric about the bisector of . Since is in convex position we may assume that for . We show that the hop stretch factor of is at least ; assume for contradiction that it is . We distinguish seven cases depending on the (Euclidean) length of the longest edge in ; trivially such an edge cannot connect two consecutive vertices.
Case 1: the longest edge is . (This case is symmetric with that when the longest edge is .) Since , we have . Then has at least hops, a contradiction.
Case 2: the longest edge is . (This case is symmetric with that when the longest edge is .) Since , we have . Since , we have . Since , we have . Then has at least hops, a contradiction.
Case 3: the longest edge is . (This case is symmetric with that when the longest edge is .) Since , we have . Then has at least hops, a contradiction.
Case 4: the longest edge is . Since , we have . Then has at least hops, a contradiction (note that by the assumption).
Case 5: the longest edge is . (This case is symmetric with that when the longest edge is .) Then has at least hops, a contradiction (note that by the assumption).
Case 6: the longest edge is . (This case is symmetric with that when the longest edge is .) Since , we have . Then has at least hops, a contradiction (note that by the assumption).
Case 7: the longest edge is . (This case is symmetric with that when the longest edge is .) Since , we have . Then has at least hops, a contradiction (note that by the assumption).
Since there are no other cases, the analysis is complete. Thus, we have shown that every plane hop spanner on has hop stretch factor of at least . Using we can construct an element point set for every such that every plane hop spanner on has hop stretch factor of at least . ∎
We next derive a better bound assuming that is sufficiently large.
Theorem 5.
For every sufficiently large , there exists an element point set such that every plane hop spanner on has hop stretch factor at least .
Proof.
Consider a set of points that form the vertices of regular gon inscribed in a circle , where the circle is just a bit larger than the circumscribed circle of an equilateral triangle of unit edge length. Formally, for a given , put and choose the radius of such that every sequence of consecutive points from makes a subset of diameter at most ; and any larger sequence makes a subset of diameter larger than . Note that . (We may set , which yields .)
The short circular arc between two consecutive vertices of is referred to as an elementary arc. (Its center angle is .) If is a set of elementary arcs, denotes its set of endpoints; obviously , with equality when covers the entire circle .
Suppose, for the sake of contradiction, that the unit disk graph has a plane subgraph with hop number at most . First, arbitrarily augment (if possible) by adding edges from one by one without introducing crossings as long as possible. Adding edges does not increase the hop number of , which remains at most .
Define maximal edges in as follows. Associate every edge of (or ) with the shorter circular arc between its endpoints. Note that containment between arcs is a partial order (poset). An edge of is maximal, if the associated arc is maximal in this poset. Then one can show that the maximal edges form a convex cycle, i.e., a convex polygon . Refer to Fig. 4. Each edge of the chain determines a set of points called block (endpoints of the edge are included). By the choice of , we have . Since the length of each edge of is at most , the restriction of to the vertices in a block is a triangulation.
Let be the sets of elementary arcs in counterclockwise order covering the th block. and are separated by a common vertex , where the triangle is the (unique) triangle adjacent to the chord in the triangulation of the th block. ( is the last endpoint of an elementary arc in and the first endpoint of an elementary arc in .) As such, we have
(3) 
By definition we have
(4) 
By maximality of the blocks in we have
(5) 
By the maximality of , we also have , since otherwise an averaging argument gives us two adjacent blocks, say, and , that can be merged by adding one chord of length at most and so that the merged sequence of points has size at most
which is a contradiction. We claim that
(6) 
where indices are considered with wraparound. Assume for contradiction that holds for some . Consider the elementary arcs preceding the arcs in and the elementary arcs following the arcs in , in counterclockwise order. Denote these sets of arcs by and , respectively (). Recall that and thus .
We claim that there exist and such that and . Indeed, since contains at most
consecutive points. This proves the first part of the claim. For the second part, we can take as one the two vertices preceding that is not and similarly, we can take as one the two vertices following that is not . For this choice we have and and passes through . Consequently,
We have reached a contradiction and so inequality (6) holds. Adding this inequality for all vertices in and using (3) and the inequality yields
This last contradiction completes the proof. ∎
An upper bound for points on a circle.
For many problems dealing with finite point configurations in the plane, it is sometimes enlightening to study points on a circle or even vertex sets of regular polygons; for such cases it may be possible to obtain tighter bounds; see, e.g., [12, 13, 26, 31]. It is certainly also natural for us, since the lower bound construction consists of points on a circle. We prove that the lower bound of is tight in this case.
Theorem 6.
For every point set on a circle , there exists a plane hop spanner.
Proof.
We may assume that is connected. Let and denote the center and radius of , respectively. If , then , we set , i.e., a star centered at an arbitrary point. This yields for every . We therefore subsequently assume that ; this implies that no edge of is incident to .
Let be a shortest arc of covering the points in , where the points are labeled counterclockwise on . We claim that , for . Indeed, let be the smallest index such that . Then and therefore and are disconnected in , a contradiction.
In the first phase, we construct a polygonal chain , on a subset of elements of with the vertices chosen counterclockwise by a greedy algorithm starting with ( is determined by the algorithm). is part of the plane spanner; the following properties will be satisfied.

, for ,

, .
In the current step, assume that has already been selected; where precedes . The algorithm checks subsequent points counterclockwise on , say As noted previously, since is connected, we have . The algorithm selects , where is the largest index such that for , i.e., for all successive points until ; or , if the last point is reached. If precedes , the algorithm updates and continues with the next iteration; if , we set . When this process terminates, is set.
If , the edge is added to close the chain, i.e., is a convex polygon whose edges belong to , in particular, ; note that there may be points of on the arc ¿ . It is possible that , in which case is an open chain with edges. In this case there are no other points of on the arc ¿ . Each edge of the chain determines a set of points called block (endpoints of the edge are included). Depending on whether the chain is open or closed, there are either blocks or blocks.
In the second phase, for every edge (with wrap around), we connect with all other points (if any) in that block (i.e., make a star whose apex is ); refer to Fig. 5 for an example. This completes the plane spanner construction.
It remains to analyze its hop factor. Let be any edge of ; we may assume that is horizontal and lies below the center . Refer to Fig. 6 (right). We show that can have at most one other edge of below it. Assume that is an edge of the polygon that lies strictly below . We claim that , i.e., and so this edge is unique if this occurs. Note that if , then the chain is closed.
Assume that . Since is below the horizontal diameter of , we have , and thus the greedy algorithm would have chosen or another vertex beyond counterclockwise, instead of as the other endpoint of the edge incident to , a contradiction. This proves the claim.
By the claim, the endpoints of every edge lie either in the same block, in two adjacent blocks, or in two blocks that are separated by exactly one other block. Consequently, can be connected by a hop path, for some . Fig. 6 (left) shows the case when the endpoints belong to two blocks that are separated by exactly one other block: the connecting path is . Fig. 6 (right) shows the case when the endpoints belong to two adjacent blocks: the connecting path is . When belong to a single block of the chain, they are connected either directly or by a path of length via the center of the corresponding star. ∎
5 The maximum degree of hop spanners cannot be bounded
A standard counting argument shows that dense (abstract) graphs do not admit constant bounded degree hop spanners (irrespective of planarity). We start by illustrating the case of a complete unit disk graph . (In general, for our results, we need to be also concerned about the geometric realizability as UDGs.)
Theorem 7.
For every pair of integers and , there exists a set of points such that the unit disk graph on has no spanner whose maximum degree is at most .
Proof.
Consider a set of points so that the unit disk graph on is the complete graph (e.g., points in a disk of unit diameter). Choose a point . Let . Let denotes the set of vertices adjacent to in . Since the degree of in is at most , . Let . The points in have edges to a set of at most points in . In general, the set contains at most points in . Consider the sets . Now it is easy to check that
Let denote the above expression in , be a point in and . Observe that , whereas , and so is not a spanner for . ∎
An alternative argument is included below—in a form that we use later in this section. We arrange points so that the unit disk graph on is the complete graph (e.g., points in a disk of unit diameter). Assume that the points are labeled from to ; and assume there exists a subgraph whose maximum degree is at most that is a spanner of . For each vertex , label the elements in by (the maximum label is ), in some arbitrary fashion. For every edge , , there is a connecting path of at most edges in . Such a path can be uniquely encoded by a string of length over the alphabet : by specifying the start vertex followed by an encoding of the edges in the path. There are at most choices for the first edge in the path; and at most choices for any subsequent edge and zero for indicating the end of a path whose length is shorter than ; the encoding of a path whose length is shorter than has trailing zeros at the end. As such, there are at most encodings. If
i.e., if , some edge has no encoding, which is a contradiction, and this completes the proof. ∎
The above result can be extended as follows.
Theorem 8.
Let , , be an integer function that tends to with . For every pair of integers and , there exist arbitrarily large sets of points (i.e., ) such that

the unit disk graph on has edges, and

has no spanner whose maximum degree is at most .
Proof.
Observe that . Let be large enough so that (we can choose infinitely many with this property). Write . For a given , arrange points into groups of size and a remaining group (if any) of size . Arrange each group make a complete subgraph and the groups be linearly connected in a path (every two consecutive groups are connected by an edge of unit length). Assume that there exists a subgraph whose maximum degree is at most that is a spanner of . Encode paths in as in the proof of Theorem 7. The number of edges in is bounded from above and from below as follows:
(7)  
(8)  
(9) 
We distinguish two cases: is large or is small as specified below.
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