1 Introduction
An obstacle representation of an (undirected simple) graph is pair where maps vertices of to distinct points in and is a set of connected subsets of with the property that, for every , if and only if the line segment with endpoints and is disjoint from . The elements of are called obstacles. It is easy to see that every graph has an obstacle representation: obtain a straightline drawing of by taking any that does not map three vertices of onto a single line, and let be the set of the open faces in the resulting arrangement of line segments.
Since every graph has an obstacle representation, this defines a natural graph parameter called the obstacle number, . Since their introduction by Alpert et al. [2], obstacle numbers have been studied extensively with the main goal of bounding the obstacle numbers of various classes of graphs (see e.g. [3, 7, 11, 12, 15, 16] and the references therein).
For planar graphs, there is also a natural notion of a plane obstacle representation which is an obstacle representation in which defines a plane straightline embedding of . This leads to a plane obstacle number: . Using Euler’s formula, it is not hard to see that the plane obstacle number of any vertex planar graph is : let define any plane drawing of with no three vertices collinear and take to be the set of open faces in this drawing. Since an vertex planar graph has at most faces, this implies .
Recently, Biedl and Mehrabi [5] studied nonblocking grid obstacle representations of graphs, consisting of the pair as before in which maps the vertices of the graph to points in the plane and is a set of obstacles, but the adjacency in the graph is represented by replacing straightline segments with shortest paths in the plane. That is, for every , if and only if some shortest path from to is disjoint from ; see Figure 1 for an illustration of these obstacle representations.
Geodesic obstacle representation.
In this paper, we generalize the notions of obstacle representations [2], plane obstacle representations, and grid obstacle representations [5] by introducing geodesic obstacle representations of graphs. This natural generalization of obstacle representations comes from viewing line segments as shortest paths in the Euclidean plane. An obstacle representation has the property that if and only if the shortest path from to does not intersect . The Euclidean distance is a very special case because the shortest path between any two points and is unique. To accommodate other distance measures, we extend the definition of obstacle representation by saying that if and only if some shortest path from to does not intersect . In this way, we can obtain many generalizations of obstacle representations by changing the underlying distance measure. For example, with the distance measure, every monotone path is a shortest path. Therefore, if is an obstacle representation under , then if and only if there is some monotone path from to that avoids . Analogous to plane obstacle representations, we can define noncrossing geodesic obstacle representations in which defines a plane embedding of graph . Under the metric, this noncrossing version is equivalent to nonblocking grid obstacle representations as defined by Biedl and Mehrabi [5].
Considering the metric in the plane, one can view a geodesic obstacle representation of as a partition of the neighbours of each vertex into four sets based on which of the four quadrants relative to the neighbours of are in the representation. Consequently, if in such a way that is in the same quadrant of as is in the quadrant of in a representation, then we must have since there is an monotone path from to in the representation. As such, it is now not clear whether every graph has a geodesic obstacle representation. Indeed, the focus of this paper is to determine, for a class of graphs, whether or not every member of has a geodesic obstacle representation (under some metric space). Clearly, the existence of such representations is more likely if one extends the definition of monotonicity by considering equalangled cones around each vertex (instead of quadrants), where is an integer. This leads us to the general question of, informally speaking, what is the minimum integer for which every member of has a geodesic obstacle representation when shortest paths are defined by monotone paths relative to such equalangled cones around each vertex. In this paper, with this “parameter ”, we study geodesic obstacle representations and its noncrossing version under polyhedral distance functions in as well as shortest path distances in graphs. See Section 2 for the formal definition of this generalized notion of obstacle representations.
Related work.
It is known that every vertex graph has obstacle number [3] and some vertex graphs have obstacle number [7]. For planar graphs, there exists planar graphs with obstacle number 2 (the icosahedron is an example [4]), but the best upper bound on the obstacle number of an vertex planar graph is . Recall the upper bound on the plane obstacle number of any vertex planar graph by Euler’s formula. A lower bound of is also not difficult: any plane drawing of the grid has at least bounded faces. Each of these faces has at least four vertices and therefore requires at least one obstacle, so . Gimbel et al. [11] have nailed the leading constant by showing that every planar graph has plane obstacle number at most , the maximum being attained by planar bipartite graphs. See [2, 3, 7, 11] and the references therein for more details of results on obstacle number and its plane version.
While the obstacle numbers have been extensively studied under the Euclidean distance as shortest path, not much is known about obstacle representations under other shortest path metrics. In fact, we are only aware of the works of Bishnu et al. [6], Biedl and Mehrabi [5] both of which considered only a restricted version of obstacle representations. Bishnu et al. [6] showed that any vertex planar graph has an obstacle representation on an grid in the plane under metric, with the additional restriction that, for any , the shortest path from to also avoids for all (in addition to avoiding ). Biedl and Mehrabi [5] relaxed this “vertex blocking” constraint and were able to show that every vertex planar bipartite graph has a nonblocking grid obstacle representation on an grid. They left open the problem of finding other classes of graphs for which such nonblocking grid obstacle representations exist and, in particular, whether every planar graph has such a representation.
Our results.
In this paper, we prove the following results:

For any integer , there is a graph with vertices that does not have a geodesic obstacle representation with parameter . On the other hand, every vertex graph has a geodesic obstacle representation with every .

For any integer and any integer , there exists a graph that does not have a geodesic obstacle representation in with parameter . On the other hand, every vertex graph has a geodesic obstacle representation in with .

A planar graph has a noncrossing geodesic obstacle representation with if and only if is bipartite.

Every planar graph of treewidth at most 2 (and hence every outerplanar graph) has a noncrossing geodesic obstacle representation with ; i.e., a nonblocking obstacle representation.

Not every planar 3tree has a noncrossing geodesic obstacle representation with , answering the question asked by Biedl and Mehrabi [5] negatively. Moreover, not every planar 4connected triangulation has a noncrossing geodesic obstacle representation with .

Every planar 3tree has a noncrossing geodesic obstacle representation with . Furthermore, every 3connected cubic planar graph has a noncrossing geodesic obstacle representation with .

Every planar graph with maximum degree has a noncrossing geodesic obstacle representation with , where is a computable function depending only on .

Every vertex graph admits a noncrossing geodesic obstacle representation when taking the cube graph as the underlying distance metric, where for some constant .
Organization.
We give definitions and notation in Section 2. Then, we show our results for (general) geodesic obstacle representations in Section 3 and for its noncrossing version in Section 4. We give our results for graph metrics in Section 5, and conclude the paper with a discussion on open problems in Section 6.
2 Notation and Preliminaries
Let be a metric space. A curve over is a function . We call and the endpoints of the curve and define the image of as . A curve is a geodesic if, for every , . A path space is a triple , where is a metric space and is a set of curves over that has the following closure property: if the curve is in then, for every , also contains the curves and . A path space is connected if, for every distinct pair , there is some path in with endpoints and . For a path space and a subset , we denote the subspace induced by as . The subspace that avoids is defined as . Moreover, any curve in is called an avoiding curve. With these definitions in hand, we are ready to define a generalization of obstacle representations. An obstacle representation of a graph is a pair where is a onetoone mapping and is a set of connected subspaces of with the property that, for every , if and only if contains a avoiding geodesic with endpoints and .
Notice that it is now not clear whether every graph has an obstacle representation. Indeed, the focus of this paper is to determine, for a class of graphs and a particular path space , whether or not every member of has an obstacle representation. This is closely related to certain types of embeddings of into . An embedding of a graph into consists of a onetoone mapping and a function such that, such that for each , the endpoints of correspond to and . The embedding is geodesic if is a geodesic for every . Moreover, the embedding is noncrossing if is disjoint from , for every with . Observe that given an obstacle representation of , for each , we can choose some avoiding geodesic with endpoints and . Then, the pair gives a geodesic embedding of into . If we can choose such that is also noncrossing, then we say that the representation is noncrossing.
Distance functions.
In this paper, we focus on the obstacle representations using polyhedral distance functions in . For a set
of vectors in
, we define the polyhedral distance functionwhere is the subspace spanned by the vectors in and is the projection of on . Every such distance function defines a centrally symmetric polyhedron . The facets of determine the geodesics. For a (closed) facet of , we denote the cone as the union of all rays originating at the origin and containing a point on (this is the affine hull of ). For a point , the sector of is . For a facet of of , we say that a curve is monotone in direction if, for all , . We say that a curve is monotone if it is monotone in direction for some facet of . Observe that a curve is a geodesic for if and only if is monotone. If and are curves that are each monotone in direction and contains at least one point , then and .
When , we let to denote the set of curves over . For the sake of compactness, when , we denote the obstacle representation by obstacle representation. For the plane case , we define, for each integer , the regular distance function , where . In this case, the associated polygon is a regular gon. Moreover, we use obstacle representation as shorthand for obstacle representation. Moreover, for a point in , we denote the sector of by , for .
In addition to polyhedral distance functions, we consider obstacle representations under graph distance. For a graph , we denote the set of neighbours of a vertex in by and the degree of by . Moreover, let denote the graph distance and let be the set of curves that define paths in . Then, we call a obstacle representation an obstacle representation. If we consider the infinite square grid (resp., the infinite triangular grid ), for instance, then it is not difficult to argue that a graph has a noncrossing obstacle representation (resp., noncrossing obstacle representation) if and only if has a noncrossing obstacle representation (resp., noncrossing obstacle representation). In general, for any integer , define the cube graph to be the graph with vertex set and that contains the edge if and only if and differ in exactly one coordinate.
3 General Representations
In this section, we show our results for the general representations. We first consider the special case of and will then discuss our results for higher dimensions. We start by the following result. For any , there exists a graph with vertices such that has no obstacle representation for any .
Proof.
For some constant and all sufficiently large , there exists a graph with vertices and edges and that contains no as subgraph [1]. Let be a obstacle representation of and let be an embedding of obtained by taking, for each , to be some shortest avoiding path from to . From this point on we identify the vertices of with the points they are embedded to and the edges of with the curves the are embedded to.
By definition each edge is monotone. Since has at most facets and each edge is monotone in at least two of these directions, this means that it has some facet such that contains edges that are monotone in direction . Consider the graph consisting of only these edges and the embedding of . Observe that if two edges and of intersect at some point , then (after appropriate relabelling), this implies that there is a avoiding geodesic from to as well as from to . Therefore, .
Therefore, if contains an tuple of pairwise crossing edges, then contains a subgraph. Now, observe that the edges of are monotone in some direction and (after an appropriate rotation) we can assume that they are monotone. We call this an monotone embedding. Valtr [17] has shown that for every fixed , there exists a constant such that any monotone embedding of any vertex graph with more than edges contains a a set of pairwise crossing edges. In our case, this means that contains a subgraph if , which gives a contradiction when . The result then follows by choosing any . ∎
As , becomes the usual Euclidean distance function and obstacle representations are just the usual obstacle representations, which we know every graph has. Thus, for every , there is a threshold value such that every vertex graph has a obstacle representation. Theorem 3 shows that and the following theorem shows that . Every vertex graph has a obstacle representation for .
Proof.
Consider the regular gon with vertices at , for . It is well known that the pairs of vertices of this gon determine only distinct directions and that these directions are for .
Therefore, to obtain an obstacle representation of , place its vertices on this regular gon, join its vertices with straightline segments and take the obstacles to be the faces of the resulting arrangement of line segments. That this is a obstacle representation follows from the fact that no monotone path uses two different directions determined by pairs of vertices and no two edges in the same direction cross each other. ∎
Higher dimensions.
The proof of Theorem 3 makes critical use of the fact that obstacle representations live in the plane so that any sufficiently dense (sub)graph has a tuple of pairwise crossing edges. An obvious question, then, is whether every graph has a obstacle representation in (i.e., an obstacle representation), where is some polyhedral distance function. The following theorem shows that the answer to this question is no.
Let be a polyhedral distance function over whose corresponding polyhedron has facets, for . Then, there exists an vertex graph that has no obstacle representation.
Proof.
Let be an vertex graph with no clique and no independent set of size larger than . The existence of such graphs was shown by Erdős and Renyi [9]. Suppose, for the sake of contradiction, that has some obstacle representation . Let denote lexicographic order over points in .
We will colour the pairs of vertices of where the colours are facets of . A pair with is coloured with a facet of such that . If more than one such facet exists, we choose one arbitrarily. For each , let denote the partial order obtained by restricting the total order to the pairs of vertices in with colour . We claim that for at least one , contains a chain of size . To see why this is so, observe that, by Dilwerth’s Theorem, if does not contain a chain of length , then it contains an antichain of size . Now, proceed inductively on and , observing that every pair in is coloured with .
Next, consider the relation over in which if and only if and . Observe that is a partial order over . Therefore, by Dilwerth’s Theorem, it contains a chain of size at least or it contains an antichain of size at least . A chain corresponds to a clique in and an antichain corresponds to an independent set in . This contradicts our choice of when , which is true for all and all sufficiently large . ∎
Theorem 3 shows that, for some vertex graphs , any obstacle representation of must use a distance function with facets. Our next result shows that, even in , a polyhedral distance function with facets is indeed sufficient.
Let be any polyhedral distance function in for which the polyhedron has at least facets. Then, every vertex graph has a obstacle representation.
Proof.
We claim that there exists a general position point set of size at least with the property that no geodesic contains 3 points of . Given such a point set, we obtain an embedding of by letting be any onetoone mapping of onto and letting, for each , be the line segment with endpoints and . In this way, no path of length 2 or more in becomes a geodesic, so is a obstacle representation of . Furthermore, since is in general position, no two edges of the embedding cross. Therefore, taking yields a obstacle representation of . All that remains is to show the existence of the set . In the following, we will ignore the general position requirement on , since it will be clear that the set we find can be slightly perturbed to ensure it is in general position.
Since is symmetric, the facets of come in opposing pairs; let contain one representative facet from each such pair, and let be a set of balls, where each ball is contained in the interior of . Finally, let be a set of sets of lines, where each contains all lines through the origin that intersect . Note that, since the balls are disjoint, so are the line sets .
It suffices to construct a point set , , such that, for any triple , there exists , , such that is parallel to some line in and is parallel to some line in . We construct such a set inductively. If , satisfies our requirements.
For , apply induction to obtain a set of points , such that, for any triple , there exists , , such that is parallel to some line in and is parallel to some line in . Now, choose two balls and such that, for every pair of points , , is parallel to some line in . Finally, scale and translate to obtain point set and another point set and take . Clearly, . By the inductive hypothesis, if or , then is parallel to some line and is parallel to some line in , with , . Otherwise, assume without loss of generality that and . Then, is parallel to some line in and is parallel to some line for some . ∎
If we take generic unit vectors in , then the polyhedral distance function determined by these vectors defines a polyhedron having vertices and triangular faces. Theorem 3 therefore implies that a polyhedral distance function determined by unit vectors is sufficient to allow an obstacle representation of any vertex graph.
In constant dimensions , there exists sets of vectors in defining polytopes with facets. Therefore, in , every vertex graph has a obstacle representation with vectors.
4 NonCrossing Representations
In this section, we consider noncrossing obstacle representations. The following lemma shows that these representations are equivalent to plane obstacle embeddings. A graph has a noncrossing obstacle representation if and only if has a noncrossing obstacle embedding.
Proof.
First, suppose has a plane obstacle embedding . Then, we claim that by taking , we obtain an obstacle representation .
If then is an avoiding monotone curve. On the other hand if then, since is plane, any avoiding monotone curve with endpoints and would determine a monotone path from to in . The definition of obstacle embedding does not allow this.
Next, we argue that a noncrossing obstacle representation implies the existence of a noncrossing obstacle embedding of . of . By the definition of noncrossing, we immediately obtain a function that maps edges of onto geodesic curves joining their endpoints, and any two of these curves are disjoint unless they share a common endpoint.
The resulting embedding is almost a plane obstacle embedding except that it may contains pairs of edges and such that and cross each other. In this case, we observe that both and are monotone in direction for the same value of . This makes it easy to eliminate crossings by repeatedly swapping segments of the curves and making local modifications around the crossings. Repeating this for every crossing pair of edges gives a plane obstacle embedding of . ∎
Lemma 4 allows us to focus our effort on studying the existence (or not) of plane obstacle embeddings.
4.1 Obstacle Representations
In this section, we show that the class of planar graphs that have plane obstacle embeddings are exactly planar bipartite graphs. Since a graph has a plane obstacle embedding if, and only if, it can be straightline embedded without monotone paths of length 2, we prove a slightly more general result about such embeddings.
A combinatorial embedding of a planar graph (the counterclockwise order of the neighbours of each vertex) has a straightline plane embedding with the same neighbour orders and no monotone paths of length 2 if, and only if, the graph is bipartite.
Proof.
If the graph is not bipartite, then it has an odd cycle, whose embedding must have at least one
monotone path of length 2. On the other hand, we show how to construct the desired straightline plane embedding if the graph is bipartite and has at least 3 vertices. The construction has three stages: input transformations to simplify the embedding; the embedding itself; and adaptation of the embedding to the original input.The first input transformation adds vertices and edges to the graph so a 2connected quadrilateralization results. The graph is first made connected by repeatedly adding edges between outer vertices of different connected components. To make it 2connected, we need to deal with cut vertices. If is a cut vertex, let be a neighbour of whose next vertex in the counterclockwise order around lies in a different 2connected component than and add a path of length 2 between and , as in Figure 2. This path addition merges the 2connected components of and , so eventually a 2connected graph remains.
Note that because the graph is 2connected and has at least 3 vertices, a counterclockwise traversal of any face will not repeat edges or vertices. Therefore, the neighbour ordering of the original graph is preserved if we preserve the faces of this 2connected combinatorial embedding, and this is what the construction will achieve. To conclude this first input transformation, we obtain a quadrilateralization by repeatedly inserting edges between vertices three edges apart in faces with more than four vertices, which is possible since faces all have an even number of vertices greater than two.
For the second input transformation, while there are two quadrilaterals sharing exactly two adjacent edges, we merge these quadrilaterals into one by erasing these two edges and the isolated vertex, as in Figure 3. Note that no cut vertices are introduced. Also, this merging reduces the number of quadrilaterals by one, so this process eventually ends.
Finally, for the third input transformation, while there is a vertex that is not part of the outer face, we first name the neighbours of in counterclockwise order as and the vertex opposing in the quadrilateral to the left of as . Then we remove and its incident edges and connect to , as in Figure 4. Note that no cut vertices are introduced here either and that the quadrilaterals originally incident to did not share any edges but the ones incident to due to our second input transformation. Furthermore, each of these steps removes a vertex, so this process also terminates.
The reason performing the embedding is now easy is that we are dealing with a 2connected outerplanar graph, whose dual graph is thus a tree. We can thus remove leaves from this tree until a single vertex remains. Therefore, we can start by embedding the quadrilateral corresponding to this vertex in any feasible way and keep embedding quadrilaterals that share a single (outer) edge with the current embedding, one at a time. These new quadrilaterals are embedded as concave quadrilaterals small enough so that they do not overlap with the rest of the embedding and so that their edges have the same slope as the shared edge, as in Figure 5. Because they have the same slope, no monotone paths of length 2 are created.
Now we have an embedding with no monotone paths of length 2, but we need to “undo” the transformations we did to the original graph in the reverse order they were made. To undo a transformation of the third type (Figure 4), note that must have been embedded all to the left or all to the right of . W.l.o.g., we assume the latter case. First we erase the edges from to and we embed the vertex close to , to the left of and to the right of . If is close enough to , there will be no crossings or change in slope when we reinsert the edges from to , as in Figure 6.
Upon close inspection, the second type of transformation (Figure 3) can be interpreted as a transformation of the third type where . The procedure just outlined may be then used to obtain a valid embedding prior to these transformations. As for transformations of the first type, they are simply vertex and edge insertions, so removing these vertices and edges clearly does not create monotone paths of length 2, providing us our desired embedding. ∎
A planar graph has a straightline obstacle embedding if, and only if, it is bipartite.
4.2 Obstacle Representations
In this section, we focus on plane obstacle embeddings. Recall that these are equivalent to the nonblocking planar grid obstacle representation studied by Biedl and Mehrabi [5]. We begin with the positive result that all graphs of treewidth at most 2 (i.e., partial 2trees) have plane obstacle embeddings.
Treewidth.
A tree is any graph that can be obtained in the following manner: we begin with a clique on vertices and then we repeatedly select a subset of the vertices that form a clique and add a new vertex adjacent to every element in . The class of trees is exactly the set of edgemaximal graphs of treewidth . A graph is called a partial tree if it is a subgraph of some tree. The class of partial trees is exactly the class of graphs of treewidth at most . We will make use of the following lemma, due to Dujmović and Wood [8] in proving Theorem 4.2 and later in Section 4.3. [Dujmović and Wood [8]] Every tree is either a clique on vertices or it contains a nonempty independent set and a vertex , such that (i) is a tree, (ii) , and (iii) every element in is adjacent to and elements of .
Every partial 2tree has a plane straightline obstacle embedding.
Proof.
Let be a partial 2tree. We can, without loss of generality, assume that is connected. If , then the result is trivial, so we can assume . We now proceed by induction on .
Let be a 2tree with vertex set and that contains . Apply Lemma 4.2 to find the vertex set and the vertex . Let and be the neighbours of in . Now, apply induction to find a plane straightline obstacle embedding of the graph whose vertex set is and whose edge set is . Denote by (resp., ) the neighbours of (resp., ) that belong to .
Now, observe that, since has degree 2 in and the edges and are in , this embedding does not contain any monotone path of the form or for any . Therefore, if we place the vertices in sufficiently close to , we will not create any monotone path of the form or for any and any . What remains is to show how to place the elements of in order to avoid unwanted monotone paths of the form , , or for any . There are three cases to consider:

and for some . W.l.o.g., assume that does not intersect the segment . Then, we can embed the elements of in without creating any new monotone paths; see Figure 7(a).

for some . There are two subcases: (i) At least one of or is in . Suppose . Then we embed in and embed in ; see Figure 7(b). The only monotone paths this creates are of the form with , which is acceptable since . (ii) Neither nor is in . In this case, we embed all of in (see Figure 7(c)). This does not create any new monotone paths.

and for some . We have three subcases to consider: (i) . In this case, assume . Then, we embed the vertices of in and we embed the vertices of in . See Figure 7(d). The only monotone paths this creates are of the form with , which is acceptable since . (ii) . In this case, we embed the vertices of in and we embed the vertices of in (see Figure 7(e)). The only monotone paths this creates are of the form with and with , which is acceptable since . (iii) . In this case, we embed all of into (see Figure 7(f)). This does not create any new monotone paths.
This completes the proof of the theorem. ∎
We next show that not every planar 3tree admits a plane obstacle embedding. To this end, we first need some preliminary results. The vertices of any triangle can be labelled such that for some .
Proof.
Consider the vertex and assume w.l.o.g. that , for some . Notice that . If or , then we are done. Otherwise, we must have or , which proves the lemma by a relabelling. ∎
A (1level) subdivision of a triangle is obtained by adding a vertex in the interior of and adding the edges , , . A level subdivision of is obtained by repeating this process recursively to a depth of . Let be a noncrossing obstacle embedding of some graph, and let be a threecycle in embedded with and . Then, does not contain a 3level subdivision in its interior.
Proof.
W.l.o.g., assume that and is above the edge . Consider the location of the vertex that subdivides . There are three cases to consider:

The vertex is placed in . In this case, there will be a monotone path from to the vertex that subdivides .

The vertex is placed in . In this case, there will be a monotone path from to the vertex that subdivides .

The vertex is placed in . In this case, consider the vertex that subdivides . The preceding two arguments prevent from being placed in or . However, placing in creates a monotone path from to .
∎
Let be a noncrossing obstacle embedding of some graph, and let be a threecycle in with for some . Then, does not contains a 4level subdivision in its interior.
Proof.
There exists a planar 3tree that does not have a noncrossing obstacle embedding.
Proof.
Notice that the graph in Theorem 4.2 has treewidth 3. Figure 8 shows that the infinite triangular grid has a noncrossing obstacle embedding. This means that while not all planar graphs of treewidth 3 have a noncrossing obstacle embedding, there are planar graphs of treewidth that admit such an embedding.
We next prove that even 4connectivity does not help to guarantee the existence of noncrossing obstacle embeddings. The idea is to show that a 4connected triangulation having a plane obstacle representation must have a constrained 4colouring in the sense that, for the neighbours of a vertex, which colours and in what order are they allowed to be assigned to them. We then find a 4connected triangulation that does not have such a constrained 4colouring. We next give the details. Let be a noncrossing obstacle representation of a 4connected triangulation; we call a vertex of an outer vertex if it is incident to the outerface of . For every internal vertex of , there is an such that has exactly one neighbour in , exactly one neighbour in and all remaining neighbours in .
Proof.
Suppose for a contradiction that this is not the case. Then, since is 4connected, would have two nonadjacent neighbours and such that and , for some . This means that the path is monotone, but and are not adjacent in — a contradiction. ∎
Lemma 8
, classifies the internal vertices of
into four types 0, 1, 2, and 3. For each internal vertex , we define as the type of the vertex . For an internal vertex with , we can assume by Lemma 8 that has of its neighbours in and has no neighbours in .For an internal vertex of with , for some , the neighbour (resp., ) of in (resp., ) is either an outer vertex or (resp., ).
Proof.
Since and , the vertices and are two neighbours of that are both in . This means that , unless is an outer vertex. An analogous argument applies to the vertex . ∎
Lemma 4.2 suggests the following property for an internal vertex of . If and all neighbours of are internal (i.e., has no neighbour on the outerface), then has two consecutive neighbours and such that (i) and , and (ii) is before in the counterclockwise ordering of the neighbours of . See Figure 9. We call this as the twoneighbour property of .
The partial function is a proper colouring of the internal vertices of .
Proof.
Let be an internal vertex with . Then, by Lemma 4.2, the neighbour (resp., ) of in (resp., in ) is either an outer vertex or (resp., ). Moreover, any vertex has at least one neighbour (namely, ) in and so is either an outer vertex or . ∎
Now, consider the graph shown in Figure 10(a) in which the labels of the vertices denote the colour of the vertices.
Let be a planar graph that contains such that all the vertices of are internal in . If has a plane obstacle embedding, then (referring to the graph shown in Figure 10(a)) , , and up to rotation.
Proof.
Since is planar and all the vertices of are internal in , the twoneighbour property must hold for every vertex of by Lemma 4.2. W.l.o.g., assume that and so by Lemma 4.2. We next consider these three cases.
Case I: and . Then, we consider the two cases for . (i) If , then because otherwise cannot satisfy the twoneighbour property (i.e., cannot have two consecutive neighbours and such that , , and appears before in the counterclockwise ordering around ). By Lemma 4.2, and so we then have . If , then we must have in order to satisfy the twoneighbour property for . This will then imply that , but then the twoneighbour property does not hold for . If , then and by Lemma 4.2. But then, does not have the twoneighbour property. (ii) If , then we must have in order to satisfy the twoneighbour property for . This implies that by proper colouring, but now the twoneighbour property does not hold for .
Case II: and . We consider the two cases for . (i) If , then . Notice that cannot be 3 because then by proper colouring, but then the twoneighbour property does not hold for . Since and the colours 1 and 3 must appear at two consecutive neighbours of in counterclockwise, we must have and . This gives by proper colouring, but then the twoneighbour property does not hold for (notice that and , but they do not appear in counterclockwise around ). (ii) If , then . If , then the remaining vertices are forced to be and (by the twoneighbour property for ), and then and (by proper colouring) in this order. Similarly, if , then (by the twoneighbour property for ), and (by the twoneighbour property for ), and by proper colouring. Observe that in either case , , and up to rotation, satisfying the ordering stated in the lemma.
Case III: and . We again consider the two cases for . (i) If , then we must have and (in order to satisfy the twoneighbour property for ). This will then imply that we have and (in order to satisfy the twoneighbour property for ). But, then we cannot satisfy this property for . (ii) If , then we must have and (again, in order to satisfy the twoneighbour property for ). But, then we cannot satisfy this property for ; notice that although and , they do not appear in the counterclockwise around as required.
Therefore, the only way can have a plane obstacle embedding is to have and in which case we get , , and up to rotation. ∎
There exists a 4connected triangulation with maximum degree 7 that has no plane obstacle embedding.
Proof.
Graph is shown in Figure 10(b). First, it is easy to see that is planar, 4connected and has maximum degree 7. Moreover, contains two copies of graph “attached” to each other in its interior. By Lemma 4.2, the four vertices on the outerface of each of these copies of must have the ordering type given in Lemma 4.2. However, since these two copies of share two of the outerface vertices, it is not possible to satisfy such ordering type for the outerface vertices at the same time. As such, does not admit a plane