FixedOrientation Equilateral Triangle Matching of Point Sets ^{†}^{†}thanks: Anil Maheshwari and Michiel Smid are supported by NSERC. Jasine Babu is supported by DFAIT.
Abstract
Given a point set and a class of geometric objects, is a geometric graph with vertex set such that any two vertices and are adjacent if and only if there is some containing both and but no other points from . We study graphs where is the class of downward equilateral triangles (ie. equilateral triangles with one of their sides parallel to the axis and the corner opposite to this side below that side). For point sets in general position, these graphs have been shown to be equivalent to half graphs and TDDelaunay graphs.
The main result in our paper is that for point sets in general position, always contains a matching of size at least and this bound cannot be improved above .
We also give some structural properties of graphs, where is the class which contains both upward and downward equilateral triangles. We show that for point sets in general position, the block cut point graph of is simply a path. Through the equivalence of graphs with graphs, we also derive that any graph can have at most edges, for point sets in general position.
Keywords:
Geometric graphs, Delaunay graphs, Matchings1 Introduction
In this work, we study the structural properties of some special geometric graphs defined on a set of points on the plane. An equilateral triangle with one side parallel to the axis and the corner opposite to this side below (resp. above) that side as in (resp. ) will be called a down (resp. up)triangle. A point set is said to be in general position, if the line passing through any two points from does not make angles , or with the horizontal [3, 12]. In this paper, we consider only point sets that are in general position and our results assume this precondition.
Given a point set , (resp. ) is defined as the graph whose vertex set is and that has an edge between any two vertices and if and only if there is a down(resp. up)triangle containing both points and but no other points from (See Fig. 1.). We also define another graph as the graph whose vertex set is and that has an edge between any two vertices and if and only if there is a downtriangle or an uptriangle containing both points and but no other points from . In Section 3 we will see that, for any point set in general position, its graph is the same as the well known Triangle Distance Delaunay (TDDelaunay) graph of and the half graph of on socalled negative cones. Moreover, is the same as the graph of [3, 5].
Given a point set and a class of geometric objects, the maximum matching problem is to compute a subclass of of maximum cardinality such that no point from belongs to more than one element of and for each , there are exactly two points from which lie inside . Dillencourt [8] proved that every point set admits a perfect circlematching. Ábrego et al. [1] studied the isothetic square matching problem. Bereg et al. concentrated on matching points using axisaligned squares and rectangles [2].
A matching in a graph is a subset of the edge set of such that no two edges in share a common endpoint. A matching is called a maximum matching if its cardinality is the maximum among all possible matchings in . If all vertices of appear as endpoints of some edge in the matching, then it is called a perfect matching. It is not difficult to see that for a class of geometric objects, computing the maximum matching of a point set is equivalent to computing the maximum matching in the graph .
The maximum matching problem, which is the same as the maximum matching problem on , was previously studied by Panahi et al. [12]. It was claimed that, for any point set of points in general position, any maximum matching of (and ) will match at least vertices. But we found that their proof of Lemma 7, which is very crucial for their result, has gaps. By a completely different approach, we show that for any point set in general position, (and by symmetric arguments, ) will have a maximum matching of size at least ; i.e, at least vertices are matched. We also give examples where our bound is tight, in all cases except when is one less than a multiple of three.
We also prove some structural and geometric properties of the graphs (and by symmetric arguments, ) and . It will follow that for point sets in general position, graphs can have at most edges and their block cut point graph is a simple path.
2 Notations
Our notations are similar to those used in [3], with some minor modifications adopted for convenience. A cone is the region in the plane between two rays that emanate from the same point, its apex. Consider the rays obtained by a counterclockwise rotation of the positive axis by angles of with around a point . (See Fig. 2.) Each pair of successive rays, and , defines a cone, denoted by , whose apex is . For , when is odd, we denote using and the cone opposite to using . We call a positive cone around and a negative cone around . For each cone (resp. ), let (resp. ) be its bisector. If , then let denote the distance between and the orthogonal projection of onto . Similarly, if , then let denote the distance between and the orthogonal projection of onto . For , let and . For any two points and , the smallest downtriangle containing and is denoted by and the smallest uptriangle containing and is denoted by . If and are graphs on the same vertex set, (resp. ) denotes the graph on the same vertex set whose edge set is the intersection (resp. union) of the edge sets of and .
3 Preliminaries
In this section, we describe some basic properties of the geometric graphs described earlier and their equivalence with other geometric graphs which are well known in the literature.
The class of downtriangles (and uptriangles) admits a shrinkability property [1]: each triangle object in this class that contains two points and , can be shrunk such that and lie on its boundary. It is also clear that we can continue the shrinking process—from the edge that does not contain neither or —until at least one of the points, or , becomes a triangle vertex and the other point lies on the edge opposite to this vertex. After this, if we shrink the triangle further, it cannot contain and together. Therefore, for any pair of points and , () has one of the points or at a vertex of () and the other point lies on the edge opposite to this vertex. In Fig. 1, triangles are shown after shrinking.
By the shrinkability property, for the matching problem, it is enough to consider the smallest downtriangle for every pair of points from . Thus, is equivalent to the graph whose vertex set is and that has an edge between any two vertices and if and only if contains no other points from . Notice that if has as one of its vertices, then . The following two properties are simple, but useful.
Property 1
Let and be two points in the plane. Let . The point is in the cone if and only if the point is in the cone . Moreover, if is in the cone , then .
Proof
The first part of the claim is obvious. Now, without loss of generality, assume that and . (See Fig. 3.) Since is the bisector of and is the bisector of , and are parallel lines. Hence, is the perpendicular distance of to the line , which makes an angle with the horizontal and passes though . Similarly, is the perpendicular distance of to the line , which makes an angle with the horizontal and passes though . Hence both and are equal to the perpendicular distance between the lines and . ∎
Property 2
Let be a point set, and . If is nonempty, then, in , the vertex corresponding to the point in with the minimum value of is the unique neighbour of vertex in .
Proof
Assume . For any point in , it is easy to see that contains no points outside the cone . Let be the point with the minimum value of . The minimality ensures that does not contain any other point other than and from . Therefore, and are neighbours in .
In order to prove uniqueness, consider any point in other than and . It can be seen that contains the point and therefore, and are not adjacent in . Thus is the only neighbour of in . ∎
Consider a point set and let be two distinct points. By Property 1, such that or ; by the general position assumption, both conditions cannot hold simultaneously. Since has either or as a vertex, Property 2 implies that we can construct as follows. For every point , and for each of the three cones, , for , add an edge from to the point in with the minimum value of , if . This definition of is the same as the definition of the halfgraph on negative cones (), given by Bonichon et al. [3]. We can similarly define the graph using the cones instead of , for , and show that it is equivalent to the half graph on positive cones (), given by Bonichon et al. [3]. In Bonichon et al. [3], it was shown that for point sets in general position, the halfgraph, the triangular distanceDelaunay graph (TDDel) [5], which are 2spanners, and the geodesic embedding of , are all equivalent.
The graphs discovered by Clarkson [6] and Keil [9] in the late 80’s, are also used as spanners [10]. In these graphs, adjacency is defined as follows: the space around each point is decomposed into regular cones, each with apex , and a point of a given cone is linked to if, from , the orthogonal projection of onto ’s bisector is the nearest point in . In Bonichon et al. [3], it was shown that every graph is the union of two halfgraphs, defined by and cones. In our notation this is same as the graph , which by definition, is equivalent to . Thus, for a point set in general position, .
Now, we will prove some more properties of which will be used in the later sections of the paper.
Property 3
Let with , , for distinct . Then, in the graph , has at least one neighbour in .
Proof
Without loss of generality, assume that and . It is easy to observe that, for any point , (See Fig. 2). Since and , for any point , . To find a vertex in which is a neighbour of in , we just need to find a point such that contains no point from other than .
We can choose any point to start with. If contains no point from other than , we are done. If not, replace with some other point inside and repeat the process. Since triangle sizes are going down in each step, eventually we will end up with a vertex in such that contains no point from other than . ∎
Property 4
Let with and for some . Then the vertex corresponding to has degree at least two in .
Proof
Without loss of generality, assume that and . If or , then by Property 2, has a neighbour in . On the other hand, if and , then, by Property 3, has at least one neighbour in .
By Property 2, we know that has a unique neighbour in too. Thus, the degree of is at least two. ∎
Property 5
Let be such that the vertex corresponding to is of degree one in . Suppose , such that and . Let be the element in . Then, .
Proof
Assume that and , for distinct . If or , then, by Property 4, the degree of is at least two in , which is a contradiction. Therefore, .
If , then by Property 3, has at least one neighbour each in and . If this is the case, the degree of is at least two, which is a contradiction. Therefore, . ∎
4 Some properties of
4.1 Planarity
Chew defined [5] TDDelaunay graph to be a planar graph and its equivalence with graph implies that is planar. This also follows from the general result that Delaunay graph of any convex distance function is a planar graph [4]. For the sake of completeness, we include a direct proof here.
Lemma 1
For a point set , its is a plane graph, where its edges are straight line segments between the corresponding endpoints.
Proof
Whenever there is an edge between and in , we draw it as a straight line segment from to . Notice that this segment always lies within . We will show that this gives a planar embedding of .
Consider two edges and of . If the interiors of and have no point in common, the line segments and can not cross each other. Suppose the interiors of and share some common area. The case that (or vice versa) is not possible, because in this case contains and (or contains and ), which contradicts its emptiness. Since and have parallel sides, this implies that one corner of infiltrates into or vice versa (see Fig. 4). Thus their boundaries cross at two distinct points, and . Since , the points and must be on that portion of the boundary of that does not lie inside . So the line through separates from . ∎
Throughout this paper, we use to represent both the abstract graph and its planar embedding described in Lemma 1. The meaning will be clear from the context.
4.2 Connectivity
In this section, we prove that for a point set , its is connected. As stated in the following lemma, between every pair of vertices, there exist a path with a special structure.
Lemma 2
Let be a point set with . Then, in , there is a path between and which lies fully in and hence is connected.
Proof
We will prove this using induction on the area of . For any pair of distinct points , if the interior of does not contain any point from , by definition, there is an edge from to in . By induction, assume that for pairs of points such that the area of is less than the area of , in the graph in , there is a path which lies fully in between and .
If the interior of does not contain any point from , there is an edge from to in . Otherwise, there is a point which is in the interior of . This implies and . Since the area of and the area of are both less than the area of , by the induction hypothesis, there is a path that lies in between and and there is a path that lies in between and . By concatenating these two paths, we get a path which lies in between and . ∎
4.3 Number of degreeone vertices
In this section, we prove for a point set , its has at most three vertices of degree one. This fact is important for our proof of the lower bound of the cardinality of a maximum matching in .
Lemma 3
For a point set , its has at most three vertices of degree one.
Proof
We will give a proof by contradiction. Let and be four points such that the vertices corresponding to them are of degree one in . Since the points are in general position, without loss of generality, we can assume that these points are given in the bottom to top order of their coordinates. We analyse different relative positionings of and with respect to and prove that in none of these cases, we can properly place all the four points consistently. Since is below and , the relative positioning of and should be one of the following :

Case 1 : .

Case 2 : but .

Case 3 : or .

Case 4 : , or , .
Case 1. Since , we have . Since is of degree one, by Property 4, . Since and are above , and is above , we have only the following subcases to consider. (See Fig. 5.)

Case 1a. and , where .

Case 1b. , where , and .

Case 1c. and , where and .

Case 1d. , , where .
Without loss of generality, assume that and .
Case 1a : We have , implying that and .
Since and , by Property 4, the degree of is at least two.
This is a contradiction.
Case 1b : We have . This implies that and .
Since and , by Property 2, the degree of is at least two. This
is a contradiction.
Case 1c : We have . This implies that and .
Since , we have . Since we already had ,
by Property 2, the degree of is at least two, which is a contradiction.
Case 1d : Since and , by Property 5, we get . This contradicts the property , that we made at the beginning of the analysis of Case 1.
Case 2. Without loss of generality, assume that . Thus, . Since , if we have , by Property 4, the degree of is at least two. On the other hand, if , by Property 2, the degree of is at least two. Since both cases lead to contradictions, we have and . Since and are above , this implies that . This gives us and (See Fig. 6.).
Since and is of degree one, by Property 4, we get and by Property 2, we get . Since is above , this implies that and hence . Since we already had , by Property 2, the degree of is at least two, which is a contradiction.
Case 3. We need only consider the situation . The other situation is symmetric to this. Since , we get . Since , and is of degree one, by Properties 2 and 4, we get and .
Since is above , this means that , which gives . Since by assumption, we also have (See Fig. 6.). These two observations give us and . Applying Property 2, it follows that the degree of is at least two, which is a contradiction.
Case 4. We need only consider the situation and . The other situation is symmetric. Since , we have . Since is of degree one, and , by Properties 2 and 4, we get and .
Since is above , this means that , which gives .
Similarly, since , we have . Since is of degree one, and , by Properties 2 and 4, we get and . Since is above , this means that , which gives . Since we already had , using Property 2, it follows that the degree of is at least two, which is a contradiction (See Fig. 6.).
Thus in each of the four possible placements of and , we concluded that the configuration is impossible. This completes the proof. ∎
4.4 Internal triangulation
In this section, we will prove that for a point set , the plane graph is internally triangulated. This property will be used in Section 5 to derive the lower bound for the cardinality of maximum matchings in .
Lemma 4
For a point set , all the internal faces of are triangles.
Proof
Consider an internal face of . We need to show that is a triangle. Let be the vertex with the highest coordinate among the vertices on the boundary of . Since is an internal face, has at least two neighbours on the boundary of . Let and be the neighbours of on the boundary of such that is to the right of the line passing through and making an angle of with the horizontal and any other neighbour of on the boundary of is to the right of the line passing through and making an angle with the horizontal. Because of the general position assumption, and can be uniquely determined.
We will prove that is also an edge on the boundary of and there is no point from in the interior of the triangle whose vertices are and . This will imply that the face is the triangle whose vertices are and .
We know that . By Property 2, it cannot happen that both , for any . Other possibilities are shown in Fig. 7, where is assumed to be above . An analogous argument can be made when is above as well.
Since and are edges in , we know that and .
Notice that, the area bounded by the lines (1) the horizontal line passing through , (2) the line passing through and making an angle of with the horizontal, and (3) the line passing through and making an angle of with the horizontal, will define an equilateral down triangle with , and on its boundary. Let us denote this triangle by .
Claim
.
Proof
For contradiction, let us assume that there exists a point . Because of the general position assumption, cannot be on the boundary of . Therefore, does not contain and . By Lemma 2, in , there exists a path between and which lies inside . Let this path be . Since , and , we know that all vertices in the path lie inside the region .
Let be the cone with apex bounded by the rays and . Observe that for any point , the line segment lies inside the cone . Since and is an edge (in the path from to ), the line segment corresponding to the edge lies inside in .
If the point is outside the face , edge will cross the boundary of , which is contradicting the planarity of . Since cannot be outside the face , the edge belongs to the boundary of . Since lies inside the cone and , this means that is a neighbour of on the boundary of such that is to the left of the the line passing through and making an angle of with the horizontal. This is a contradiction to our assumption that is the only neighbour of on the boundary of , lying to the left of the the line passing through and making an angle of with the horizontal. ∎
Let us continue with the proof of Lemma 4. Since the triangle with vertices and is inside the triangle , from the above claim, it is clear that there is no point from , other than the points and , inside the triangle whose vertices are and . Since the edges and belong to the boundary of , to show that is a triangle, it is now enough to prove that is also an edge in . This fact also follows from the above claim as explained below.
Since , by the claim above, cannot contain any point from other than and . Moreover, since lies above and , we know that . Therefore, . Therefore, is an edge in .
Thus, has to be a triangle bounded by the edges , and . ∎
Corollary 1
For a point set , all the cut vertices of lie on its outer face.
Proof
Consider any vertex of which is not on its outer face. Since is internally triangulated, each neighbour of in lies on a cycle in the graph . Since is connected, remains connected. Thus, cannot be a cut vertex. ∎
5 Maximum matching in
In this section, we show that for any point set of points, contains a matching of size ; i.e, at least vertices are matched. Consider a point set containing points. If we have only two points in , then the graph contains a perfect matching. Hence, we assume that .
We construct a graph such that it is a connected planar graph of minimum degree at least and then make use of the following theorem of Nishizeki [11] to get a lower bound on the size of a maximum matching of . Using this, we will then derive a lower bound on the size of a maximum matching of .
Theorem 5.1 ([11])
Let be a connected planar graph with vertices having minimum degree at least and let be a maximum matching in . Then,
Initialize to be the same as . Consider a simple closed curve in the plane such that (1) the entire graph (all vertices and edges) lies inside the bounded region enclosed by , (2) the vertices of which lie on are precisely the degreeone vertices of , (3) except for the end points, every edge of lies in the interior of the bounded region enclosed by .
Let the degreeone vertices of be denoted by . In the previous section, we proved that .
If , the region of the outer face of bounded by the curve can be divided into regions where is the region bounded by the edge at , the edge at , the boundary of the outer face of and the curve . See Fig. 8. (Here onwards, in this subsection we assume that indices of vertices and regions are taken modulo .) Notice that every vertex on the outerface of lies on at least one of these regions and lies on the regions and , for . We insert new vertices into . (To visualize the abstract graph , vertex may be assumed to lie on the boundary of the region , a point distinct from and .) New edges are added between and , for . We also insert new edges into between each and all the vertices of which lie on the region , for . This transformation maintains planarity. (Edges between new vertices and old vertices can be drawn inside the corresponding region . The edges among the new vertices can be drawn outside these regions, except at their end points.)
Each degreeone vertex , , of lies on two regions and , in it gets two new edges; one to and the other to . Thus the degree of becomes . All other vertices on the outer face of were of degree at least two. Since they belong to at least one of the regions , they get at least one new edge in and their degree is at least three in . Since is an internally triangulated planar graph, we know that all vertices except those on the outer face had degree at least . These vertices maintain the same degrees in as in . The degree of , , is also at least in , because it is adjacent to and at least one more vertex on the outer face of . Thus, has minimum degree at least three.
If or , the modification of is similar but rather simpler. We insert a new vertex in the outer face of and add edges between and all other vertices in the outer face of . This transformation maintains planarity. As earlier, all vertices in except the vertex (present only when ) have degree at least three now.
If , the degree of has become two in at this stage. In this case, let be a face of the current graph which contains and . Modify by inserting a new vertex inside and adding edges from this new vertex to all other vertices belonging to . As earlier, this transformation maintains planarity. Now, the degree of becomes .
Claim
The graph is connected.
Proof
It is easy to observe that none of the newly inserted vertices can be a cut vertex of . For any vertex which was not on the outer face of , the induced subgraph on its neighbours form a cycle in as it was in . They cannot be cut vertices.
Consider any vertex which was on the outer face of