 # Planar graphs as L-intersection or L-contact graphs

The L-intersection graphs are the graphs that have a representation as intersection graphs of axis parallel shapes in the plane. A subfamily of these graphs are L, |, ---contact graphs which are the contact graphs of axis parallel L, |, and -- shapes in the plane. We prove here two results that were conjectured by Chaplick and Ueckerdt in 2013. We show that planar graphs are L-intersection graphs, and that triangle-free planar graphs are L, |, ---contact graphs. These results are obtained by a new and simple decomposition technique for 4-connected triangulations. Our results also provide a much simpler proof of the known fact that planar graphs are segment intersection graphs.

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## 1 Introduction

The representation of graphs by contact or intersection of predefined shapes in the plane is a broad subject of research since the work of Koebe on the representation of planar graphs by contacts of circles [Koebe36]. In particular, the class of planar graphs has been widely studied in this context.

More formally, assigning a shape of the plane for each vertex of a graph , we say that is a -intersection graph if there is a representation of such that every vertex is assigned to a shape , and two shapes , intersect if and only if the vertices they are assigned to are adjacent in . In the case where the shape is homeomorphic to segments (resp. discs), a -contact system is a collection of shapes such that if an intersection occurs between two shapes, then it occurs at one of their endpoints (resp. on their border). We say that a graph is a -contact graph if it is the intersection graph of a -contact system. This definition can be easily generalized if the representation of each vertex is chosen among a family of shapes.

The case of shapes that are homeomorphic to a disc has been widely studied; see for example the literature for triangles [FOR94, GLP12], homothetic triangles [KKLS06, Swg17], axis parallel rectangles [T84], squares [Kenyon04, Schramm93], hexagons [GHKK12], or convex bodies [S90]. We here focus on the representation of planar graphs as contact or intersection graphs, where the assigned shapes are segments or polylines in the plane. The simplest definition of representation of graphs by intersection of curves is the so-called string-representation: each vertex is represented by a curve, and two curves intersect if and only if the vertices they represent are adjacent in the graph. It is known that every planar graph has a string-representation [ehrlich1976intersection]. However, this representation may contain pairs of curves that cross any number of times. One may thus take an additional parameter into account, namely the maximal number of crossings of any two of the curves: a 1-string representation of a graph is a string representation where every two curves intersect at most once. The question of finding a 1-string representation of planar graphs has been solved by Chalopin et al. in the positive [chalopin2010planar], and additional parameters are now studied, like order-preserving representations [BiedlSOFSEM2017].

Segment intersection graphs are in turn a specialization of the class of 1-string graphs. It is known that bipartite planar graphs are -contact graphs [Ben91, Fraysseix94] (i.e. segment contact graphs with vertical or horizontal segments). De Castro et al. [Castro02] showed that triangle-free planar graphs are segment contact graphs with only three different slopes. De Fraysseix and Ossona de Mendez [Fraysseix07] then proved that a larger class of planar graphs are segment intersection graphs. Finally, Chalopin and the first author extended this result to general planar graphs [chalopin2009every], which was conjectured by Scheinerman in his PhD thesis [schein84].

A graph is said to be a VPG-graph (Vertex-Path-Grid) if it has a contact or intersection representation in which each vertex is a path of vertical and horizontal segments (see [AFjgaa15, cohen2016posets]). Asinowski et al. [asinowski2012vertex] showed that the class of VPG-graphs is equivalent to the class of graphs admitting a string-representation. They also defined the class -VPG, which contains all VPG-graphs for which each vertex is represented by a path with at most bends (see [Fdam16] for the determination of the value of for some classes of graphs). It is known that -VPG -VPG, and that the recognition of graphs of -VPG is an NP-complete problem [chaplick2012bend]. These classes have interesting algorithmic properties (see for example [Marxiv17] for approximation algorithms for independence and domination problems in -VPG graphs), but most of the literature studies their combinatorial properties.

Chaplick et al. [chaplick2013planar] proved that planar graphs are -VPG graphs. This result was recently improved by Biedl and Derka [BDjocg16], as they showed that planar graphs have a 1-string -VPG representation.

Various classes of graphs have been showed to have 1-string -VPG representations, such as planar partial 3-trees [BDarxiv15] and Halin graphs [francis2016vpg]. Interestingly, it has been showed that the class of segment contact graphs is equivalent to the one of -VPG contact graphs [kobourov2013combi]. This implies in particular that triangle-free planar graphs are -VPG contact graphs. This has been improved by Chaplick et al. [chaplick2013planar] as they showed that triangle-free planar graphs are in fact -contact graphs (that is without using the shapes and ).

The restriction of -VPG to -intersection or -contact graphs has been much studied (see for example [Fdam16]) and it has been shown that they are in relation with other structures such as Schnyder realizers, canonical orders or edge labelings [chaplick2013equilateral]. The same authors also proved that the recognition of -contact graphs can be done in quadratic time, and that this class is equivalent to the one restricted to equilateral shapes. Finally, the monotone (or linear) -contact graphs have been recently studied further, for example in relation with MPT (Max-Point Tolerance) graphs [2017arXiv170301544R, Cdam17max].

### Our contributions

The two main results of this paper are the following:

###### Theorem 1

Every triangle-free planar graph is a -contact graph.

###### Theorem 2

Every planar graph is a -intersection graph.

Both results were conjectured in  [chaplick2013planar]. Theorem 1 is optimal in the sense that a -contact graph with vertices has at most edges, while triangle-free planar graphs may have up to edges. However, up to our knowledge, the question of whether every triangle-free planar graph is a -contact graph is open 111In fact, it has been proven in the Masters thesis (in German) of Björn Kapelle in 2015 [Kapelle15, Sec. 3.3], but never published.. Theorem 2 implies that planar graphs are in -VPG, improving the results of Biedl and Derka [BDjocg16] stating that planar graphs are in -VPG. Since a -intersection representation can be turned into a segment intersection representation [middendorf1992max], this also directly provides a rather simple proof of the fact that planar graphs are segment intersection graphs [chalopin2009every]. Note that a simple modification of our method can be used to prove that 4-connected planar graphs have a -EPG representation [BP17], where vertices are represented by paths on a rectangular grid with at most 3 bends, and adjacency is shown by sharing an edge of the grid.

The common ingredient of the two results is what we call 2-sided near-triangulations. In Section 2, we present the 2-sided near-triangulations, allowing us to provide a new decomposition of planar 4-connected triangulations (see [BDjgaa16] and [Whitney31] for other decompositions of 4-connected triangulations). This decomposition is simpler than the one provided by Whitney [Whitney31] that is used in [chalopin2009every]. In Section 3, we define thick -contact systems, (i.e., -contact representations in which the shapes have some thickness ) with specific properties. We then show that every 2-sided near-triangulation can be represented by such a system. This result is used in Section 4 to prove Theorem 1. Then in Section 5 we use 2-sided near-triangulations to prove Theorem 2.

## 2 2-sided near-triangulations

In this paper we consider plane graphs without loops nor multiple edges. In a plane graph there is an infinite face, called the outer face, and the other faces are called inner faces. A near-triangulation is a plane graph such that every inner face is a triangle. In a plane graph , a chord is an edge not incident to the outer face but that links two vertices of the outer face. A separating triangle of is a cycle of length three such that both regions delimited by this cycle (the inner and the outer region) contain some vertices. It is well known that a triangulation is 4-connected if and only if it contains no separating triangle. Given a vertex on the outer face, the inner-neighbors of are the neighbors of that are not on the outer face. We define here -sided near-triangulations (see Figure 1) whose structure will be useful in the inductions of the proofs of Theorem 2, and Theorem 7.

###### Definition 3

A 2-sided near-triangulation is a 2-connected near-triangulation without separating triangle and such that going clockwise on its outer face, the vertices are denoted , with and , and such that there is no chord or (that is an edge or such that ).

The structure of the 2-sided near-triangulations allows us to describe the following decomposition:

###### Lemma 4

Given a 2-sided near-triangulation with at least vertices, one can always perform one of the following operations:

• This operation applies if , if has no neighbor with , and if none of the inner-neighbors of has a neighbor with . This operation consists in removing from , and in denoting in anti-clockwise order the new vertices on the outer face, if any. This yields a 2-sided near-triangulation (see Figure 1(a)).

• This operation applies if , if has no neighbor with , and if none of the inner-neighbors of has a neighbor with . This operation consists in removing from , and in denoting in clockwise order the new vertices on the outer face, if any. This yields a 2-sided near-triangulation . This operation is strictly symmetric to the previous one.

• This operation applies if and , and if the unique common neighbor of and , denoted , has a neighbor with , and a neighbor with . If has several such neighbors, and correspond to the smaller possible values. This operation consists in cutting into three 2-sided near-triangulations , and (see Figure 1(b)):

• is the 2-sided near-triangulation contained in the cycle formed by vertices , and the vertex is renamed .

• (resp. ) is the 2-sided near-triangulation contained in the cycle (resp. ), where the vertex is denoted (resp. ).

Proof. Suppose that has no neighbor with and none of the inner-neighbors of has a neighbor with . We denote the inner-neighbors of in anti-clockwise order such that is connected to for every . Let be the graph obtained by removing and its adjacent edges from . It is clear that is a near-triangulation, and that it has no separating triangle (otherwise would have one too). Furthermore, as there is no chord incident to , and as has at least three vertices its outer face is bounded by a cycle, and is thus 2-connected. As is a -sided near-triangulation, has no chord , with , or with . By hypothesis, the inner-neighbors of have no neighbors with , thus there is no chord with and . There is no chord in with . Otherwise the vertices , , and would form a triangle with at least one vertex inside, , and at least one vertex outside, : it would be a separating triangle, a contradiction. Therefore is a -sided near-triangulation.

The proof for the -removal operation is analogous to the previous case.

Suppose that we are not in the first case nor in the second one. Let us first show that and . Towards a contradiction, consider that . Then as is 2-connected, it has at least three vertices on the outer face and . In such a case one can always perform the -removal operation, a contradiction.

Let us now show that is not adjacent to a vertex with . Towards a contradiction, consider that is adjacent to a vertex with . Then by planarity, (with ) has no neighbor with , and has no inner-neighbor adjacent to a vertex with . In such a case one can always perform the -removal operation, a contradiction. Symmetrically, we deduce that is not adjacent to a vertex with .

Vertices and have one common neighbor such that is an inner face. Note that as there is no chord incident to or , then is not on the outer face. They have no other common neighbor , otherwise there would be a separating triangle (separating from both vertices and ).

As we are not in the first case nor in the second case, we have that (resp. ) has (at least) one inner-neighbor adjacent to a vertex with (resp. with ). By planarity, is the only inner-neighbor of (resp. ) adjacent to a vertex with (resp. with ). We can thus apply the cutting operation.

We now show that , and are 2-sided near-triangulations. Consider first . It is clear that it is a near-triangulation without separating triangles. It remains to show that there are no chords or . By definition of , the only chord possible would have as an endpoint, but the existence of an edge with would contradict the minimality of . Thus is a -sided near-triangulation.

By definition, is also a near-triangulation containing no separating triangles. Moreover, there is no chord with as there are no such chords in . Therefore is a -sided near-triangulation. We show in the same way that is a -sided near-triangulation.

## 3 Thick {└}-contact system

A thick is a shape where the two segments are turned into -thick rectangles (see Figure 2(a)). Going clockwise around a thick from the bottom-right corner, we call its sides bottom, left, top, vertical interior, horizontal interior, and right.

A thick  is described by four coordinates such that and . It is thus the union of two boxes: . If not specified, the corner of a thick  denotes its bottom-left corner (with coordinate ). In the rest of the paper, all the thick  shapes have the same thickness .

###### Definition 5

Given a thick -contact system, a thick is said left if its horizontal interior is free (i.e., does not touch another ) and its left side is not contained in the side of another thick (see Figure 2(b)). Similarly, a thick is said bottom if its vertical interior is free and its bottom side is not contained in the side of another thick (see Figure 2(c)).

###### Definition 6

A convenient thick -contact system (CTLCS) is a contact system with thick  shapes (which implies that the thick  shapes interiors are disjoint) with a few properties:

• Two thick  shapes intersect either on exactly one segment or on a point (Figure 4 lists the allowed ways two shapes can intersect). If the intersection is a segment, then it must be exactly one side of a thick . If the intersection is a point, then it is the bottom right corner of one thick  and the top left corner of the other one.

• Every thick is bottom or left.

Remark that the removal of any thick  still leads to a CTLCS. This definition implies that in a CTLCS there is no three shapes intersecting as in Figure 5.

We now make a link between CTLCS and 2-sided near-triangulations (See Figure 6 for an illustration).

###### Theorem 7

Every 2-sided near-triangulation can be represented by a CTLCS with the following properties:

• every thick  is included in the quadrant ,

• has the rightmost corner and has the up-most corner,

• every vertex is represented by a bottom thick  whose corner has coordinates , with , and

• every vertex is represented by a left thick  whose corner has coordinates , with .

Proof. We proceed by induction on the number of vertices. The theorem clearly holds for the 2-sided near-triangulation with three vertices. Let be a 2-sided near-triangulation; it can thus be decomposed using one of the three operations described in Lemma 4. We go through the three operations successively.

(-removal) Let be the 2-sided near-triangulation resulting from an -removal operation on . By the induction hypothesis, has a CTLCS with the required properties (see Figure 6(a)). We can now modify this CTLCS slightly in order to obtain a CTLCS of (thus adding a thick  corresponding to vertex ). Move the corners of the thick  corresponding to vertices slightly to the right. Since these are left thick  shapes, one can do this without modifying the rest of the system. Then one can add the thick  of such that it touches the thick  of vertices and (as depicted in Figure 6(b)). One can easily check that the obtained system is a CTLCS of and satisfies all the requirements.

(-removal) This case is strictly symmetric to the previous one.

(cutting) Let , and be the three 2-sided near-triangulations resulting from the cutting operation described in Lemma 4. By induction hypothesis, each of them has a CTLCS satisfying the requirements of Theorem 7. Consider the CTLCS of (see Figure 8). Move the corner of slightly upward. Since is a bottom vertex, one can do this without modifying the rest of the system. Then one can add the CTLCS of below vertex and the one of on its left (see Figure 8, bottom). One can easily check that the obtained system satisfies all the requirements.

## 4 {└,|,−}-contact systems for triangle-free planar graphs

We can now use the CTLCS systems to prove Theorem 1. Recall that a -contact system is a contact system with some , some vertical segments , and some horizontal segments , such that if an intersection occurs between two of these objects, then the intersection is an endpoint of one of the two objects. We need the following lemma as a tool (it is proved in appendix).

###### Lemma 8

For any plane triangle-free graph , there exists a 4-connected triangulation containing as an induced subgraph.

We can now prove Theorem 1, which asserts that every triangle-free planar graph has a -contact system.

Proof. Consider a triangle-free planar graph . According to Lemma 8, there exists a 4-connected triangulation containing as an induced subgraph. As the exterior face of is a triangle, is a -sided near-triangulation (denoting the three exterior vertices in clockwise order). By Theorem 7, has a CTLCS and removing every thick  corresponding to a vertex of leads to a CTLCS of .

If a thick   has its bottom side included in the horizontal interior side of another thick  , then is bottom, and so does not intersect anyone on its horizontal interior side. Furthermore, does not intersect anyone on its right side nor on its bottom right corner. Indeed, if there was such an intersection with a thick  , then and would also intersect, contradicting the fact that is triangle-free (see Figure 9). One can thus replace the thick  of by a thick . Figure 9: If a thick └ x has its bottom side included in the horizontal interior of a thick └ y, then x has no intersection with a thick └ z on its right side and on its bottom right corner.

Similarly, if a thick   has its left side included in the vertical interior side of a thick  , we can replace the thick  of by a thick .

Note that now the intersections are on small segments, or on a point, between the bottom right corner of a thick  or , and the top left corner of a thick  or . Then, we replace each thick , , and by thin ones as depicted in Figure 10. It is clear that we obtain a -contact system whose contact graph is . This concludes the proof. An example of the process is shown in Figure 11. Figure 10: Replacing thick └, |, and − by thin ones. Figure 11: Given a CTLCS of a triangle-free graph G, we first replace some thick └ by thick | and thick −, and then replace every thick shape by a thin one according to Figure 10.

## 5 The └-intersection systems

An -intersection system (LIS) is an intersection system of shapes where every two shapes intersect on at most one point. Using Theorem 7, one could prove that that every 4-connected triangulation has a LIS. To allow us to work on every triangulation (not only the 4-connected ones) we need to enrich our LISs with the following notion that was introduced in [Fdam16] under the name of private region.

An anchor can be seen as a union of three segments, or as the union of two . It has two corners, which correspond to the shapes corners. There are two types of anchors. A horizontal anchor is a set where and (see Figure 12a). The middle corner of such a horizontal anchor is defined as the point . A vertical anchor is a set where and (see Figure 12b). The middle corner of such a vertical anchor is defined as the point . Consider a near-triangulation , and any inner face of . Given a LIS of , an anchor of is an anchor crossing the shapes of and and no other , and such that the middle corner is in the square described by , and as depicted in Figure 12.

###### Definition 9

A full -intersection system (FLIS) of a near-triangulation is a LIS of with an anchor for every (triangular) inner face of , such that the anchors are pairwise non-intersecting.

Let us now prove that every 2-sided near-triangulation admits a FLIS.

###### Proposition 10

Every 2-sided near-triangulation has a FLIS such that among the corners of the shapes and the anchors:

• from left to right, the first corners are those of vertices and the last one is the corner of vertex , and

• from bottom to top, the first corners are those of vertices and the last one is the corner of vertex .

As the of and (resp. and ) intersect, the FLIS is rather constrained. This is illustrated in Figure 14, where the grey region contains the corners of the inner vertices, and the corners of the anchors. Figure 14: Illustration of Proposition 10 when p>1 and q>1, when p=1 and q>1, and when p>1 and q=1.

Proof. We proceed by induction on the number of vertices.

The result clearly holds for the 2-sided near-triangulation with three vertices, whatever and , or and . Let be a 2-sided near-triangulation with at least four vertices. By Lemma 4 we consider one of the following operations on :

(-removal) Consider the FLIS of obtained by induction and see in Figure 15 how one can add a for and an anchor for each inner face with and for the inner face . One can easily check that the obtained system verifies all the requirements of Proposition 10.

(-removal) This case is symmetric to the previous one.

(cutting) Consider the FLISs of , and . Figure 16 depicts how to combine them, and how to add an anchor for , in order to get the FLIS of . One can easily check that the obtained system verifies all the requirements of Proposition 10.

We now prove Theorem 2 which asserts that every planar graph is a -intersection graph. It is well known that every planar graph is an induced subgraph of some triangulation (see [chalopin2010planar] for a proof similar to the one of Lemma 8). Thus, given a planar graph , one can build a triangulation whose is an induced subgraph. If one can create a FLIS of , then it remains to remove the shapes assigned to vertices of along with the anchors in order to get a -representation of . In order to prove Theorem 2, we thus only need to show that every triangulation admits a FLIS.

###### Proposition 11

Every triangulation with outer-vertices has a FLIS such that among the corners of the shapes and the anchors:

• the corner of is the upmost and leftmost,

• the corner of is the second leftmost, and

• the corner of is the bottom-most and rightmost. Figure 17: Illustration of Proposition 11, and the FLIS obtained after reflection with respect to a line of slope 1.

Note that in this proposition there is no constraint on , so by renaming the outer vertices, other FLISs can be obtained. Another way to obtain more FLISs is by applying a reflection with respect to a line of slope . In such FLIS (see Figure 17) among the corners of the shapes and the anchors:

• the corner of is the bottom-most and rightmost,

• the corner of is the second bottom-most, and

• the corner of is the upmost and leftmost.

Proof. We proceed by induction on the number of vertices in . Let be a triangulation with outer vertices .

If is 4-connected, then it is also a 2-sided near-triangulation. By Proposition 10 and by renaming the outer-vertices to , to and to , has a FLIS with the required properties.

If is not 4-connected, then it has a separating triangle formed by vertices , and . We note and the triangulations obtained from by removing the vertices outside and inside respectively.

By the induction hypothesis, has a FLIS verifying Proposition 11 (considering the outer vertices to be in the same order). Without loss of generality we can suppose that the shapes of , and appear in the following order: the upmost and leftmost is , the second leftmost is and the bottom-most is . There are two cases according to the type of the anchor of the inner face .

If the anchor of in the FLIS of is vertical (see Figure 17(a)), then applying the induction hypothesis on with as outer vertices considered in that order, has a FLIS as depicted on the Figure 17(b). Figure 17(c) depicts how to include the FLIS of in the close neighborhood of the anchor of . As is not a face of , the close neighborhood of its anchor is indeed available for this operation. (a) The vertical anchor of abc in the FLIS of Tout

Now suppose that the anchor of in the FLIS of is horizontal (see Figure 18(a)). By application of the induction hypothesis on with as outer vertices considered in that order, then has a FLIS as depicted on the Figure 18(b). By a reflection of slope , has a FLIS such that is the up-most and left-most, is the second left-most and is bottom-most (see Figure 18(c)). Similarly to the previous case, we include this last FLIS of in the one from (see Figure 18(d)). As and cover , and intersect only on the triangle , and as every inner face of is an inner face in or in , these constructions clearly verify Proposition 11. This concludes the proof of the proposition. (a) The horizontal anchor of abc in the FLIS of Tout

## Appendix A From triangle-free planar graphs to 4-connected triangulations

We here prove Lemma 8. Figure 20: A planar triangle-free graph G (in black) and a 4-connected near-triangulation containing it as an induced subgraph (adding red vertices and edges). The boundary lists of the two inner faces of G are respectively {1,(1,2),2,(2,3),3,(3,4),4,(4,7),7,(7,1)}, {1,(1,7),7,(7,4),4,(4,5),5,(5,6),6,(6,5),5,(5,1)} The outer face is {1,(1,2),2,(2,3),3,(3,4),4,(4,5),5,(5,1)}.

Proof. The main idea of the construction of is to insert vertices and edges in every face of (even for the exterior face).

For the sake of clarity, vertices of are said black and vertices of are said red. The new graph contains as an induced subgraph, along with other vertices and edges. More precisely, for every face of , let be the list of vertices and edges along the face boundary (see Figure 20), where is the edge between vertices and ; there can be repetitions of vertices or edges. For each face of , given the list , the graph contains a vertex for each vertex , a vertex for each edge , and an additionnal vertex . Each vertex is connected to and (with subscripts addition done modulo the size of the face), each vertex is connected to , and , and the vertex is connected to all vertices and (see Figures 20 and 21 for examples).

The new graph is a triangulation, and we now show that it is 4-connected, i.e., has no separating triangle. Suppose that there is a separating triangle in the new graph. There are four cases depending on the colors of the edges of this triangle:

• The separating triangle contains three black edges. It is impossible since is triangle-free.

• The separating triangle contains exactly one red edge. One of its endpoints must be a red vertex. But a red vertex is adjacent to only red edges, a contradiction.

• The separating triangle contains exactly two red edges. Then their common endpoint is a red vertex, and the triangle is made of two vertices and , together with the vertex . All these triangles are faces, a contradiction.

• The separating triangle contains three red edges. Since for each face, the red vertices (vertices , and ) induce a wheel graph centered on , with at least peripheral vertices (vertices and ), this separating triangle has at least one black vertex. As two adjacent black vertices are linked by a black edge, this separating triangle has exactly one black vertex. As the two red vertices are two adjacent or vertices, we have that those are and , for some and for or for . Such a triangle is not separating, a contradiction.

This concludes the proof of the lemma.