Homothetic triangle representations of planar graphs

We prove that every planar graph is the intersection graph of homothetic triangles in the plane.


Introduction
Here, an intersection representation is a collection of shapes in the plane. The intersection graph described by such a representation has one vertex per shape, and two vertices are adjacent if and only if the corresponding shapes intersect. In the following we only consider shapes that are homeomorphic to disks. In this context, if for an intersection representation the shapes are interior disjoint, we call such a representation a contact representation. In such a representation, a contact point is a point that is in the intersection of (at least) two shapes.
Research on contact representations of (planar) graphs with predefined shapes started with the work of Koebe in 1936, and was recently widely studied; see for example the literature for disks [2,3,15], triangles [8], homothetic triangles [10,13,14,21], rectangles [6,22], squares [16,20], pentagons [7], hexagons [9], convex bodies [19], or (non-convex) axis aligned polygons [1,12]. In the present article, we focus on homothetic triangles. It has been shown that many planar graphs admit a contact representation with homothetic triangles.  Note that one cannot drop the 4-connectedness requirement from Theorem 1. Indeed, in every contact representation of K 2,2,2 with homothetic triangles, there are three triangles intersecting in a point (see the right of Figure 1). This implies that the triangulation (not 4-connected) obtained from K 2,2,2 by adding a degree three vertex in every face does not admit a contact representation with homothetic triangles. Some questions related to this theorem remain open. For example, it is believed that if a triangulation T admits a contact representations with homothetic triangles, it is unique up to some choice for the triangles in the outer-boundary. However this statement is still not proved. Another line of research lies in giving another proof to Theorem 1 (a combinatorial one), or in providing a polynomial algorithm constructing such a representation [4,21]. Theorem 1 has a nice consequence. It allowed Felsner and Francis [5] to prove that every planar graph has a contact representation with cubes in R 3 . In the present paper we remain in the plane. Theorem 1 is the building block for proving our main result. An intersection representation is said simple if every point belongs to at most two shapes.

Theorem 2.
A graph is planar if and only if it has a simple intersection representation with homothetic triangles.
This answers a conjecture of Lehmann that planar graphs are max-tolerance graphs (as max-tolerance graphs have shown to be exactly the intersection graphs of homothetic triangles [14]). Müller et al. [17] proved that for some planar graphs, if the triangle corners have integer coordinates, then their intersection representation with homothetic triangles needs coordinates of order 2 Ω(n) , where n is the number of vertices. The following section is devoted to the proof of Theorem 2.

Intersection representations with homothetic triangles
It is well known that simple contact representations produce planar graphs. The following lemma is slightly stronger.
are disjoint non-empty connected regions. Let us draw G by first choosing a point p u inside c • (u), for representing each vertex u. Then for each neighbor v of u, draw a curve inside c • (u) from p u to the border of c(u) ∩ c(v ) (in the border of c • (u)) to represent the half-edge of uv incident to u. As the regions c • (u) are disjoint and connected, this can be done without crossings. Finally, for each edge uv it is easy to link its two half edges by drawing a curve inside c(u) ∩ c(v ). As the obtained drawing has no crossings, the lemma follows.
Note that for any two homothetic triangles ∆ and ∆ , the set ∆ \ ∆ is connected. Lemma 3 thus implies the sufficiency of Theorem 2. For proving Theorem 2 it thus suffices to construct an intersection representation with homothetic triangles for any planar graph G. In fact we restrict ourselves to (planar) triangulations because any such G is an induced subgraph of a triangulation T (an intersection representation of T thus contains a representation of G). The following Proposition 4 thus implies Theorem 2.
From now on we consider a particular triangle. Given a Cartesian coordinate system, let ∆ be the triangle with corners at coordinates (0, 0), (0, 1) and (1, 0) (see Figure 2.(a)). Thus the homothets of ∆ have corners of the form (x, y ), (x, y + h) and (x + h, y ) with h > 0, and we call (x, y ) their right corner and h their height.  We call (a') this relaxation of condition (a), and we call "bad points", the points at the intersection of 3 triangles.
Let us now reduce their number (to zero) as follows (and thus fulfill condition (a)).
Note that the corners of the outer triangles do not intersect inner triangles. This property will be preserved along the construction below.
Let p = (x p , y p ) be the highest (i.e. maximizing y p ) bad point. If there are several bad points at the same height, take among those the leftmost one (i.e. minimizing x p ). Then let t(u), t(v ) and t(w ) be the three triangles pairewise intersecting at p. Let us denote the coordinates of their right corners by (x u , y u ), (x v , y v ) and (x w , y w ), and their height by h u , h v and h w . Without loss of generality we let p = (x u + h u , y u ) = (x v , y v ) = (x w , y w + h w ) (see Figure 2.(b)). By definition of p it is clear that p is the only bad point around t(u). Note also that none of t(u), t(v ) and t(w ) is an outer triangle. Step 1: By definition of p and t(u), the corner q = (x u , y u +h u ) of t(u) is not a bad point. Now inflate t(u) in order to have its right angle in (x u − 1 , y u ) and height h u + 1 , for a sufficiently small 1 > 0 (see Figure 3.(a)). Here 1 is sufficiently small to avoid new pairs of intersecting triangles, new triples of intersecting triangles, or an intersection between t(u) and an outer triangle on a too big triangle (with height ≥ ). Since the new t(u) contains the old one, the triangles originally intersected by t(u) are still intersected. Hence, t(u) intersects the same set of triangles, and the new representation is still a representation of T . Since there was no bad point distinct from p around t(u), it is clear by the choice of 1 > 0 that the new representation still fulfills (a') and (b). After this step we have the following.
Claim 5. The top corner of t(u) is not a contact point.
Step 2: For every triangle t(z) that intersects t(u) on a single point of the open segment ]p, q[ do the following. Denote (x z , y z ) the right corner of t(z), and h z its height. Note that t(z) is an inner triangle of the representation and that by definition of p there is no bad point involving t(z). Now inflate t(z) in order to have its right corner at (x z , y z − 2 ), and height h z + 2 , for a sufficiently small 2 > 0 (see Figure 3.(b)). Here 2 is again sufficiently small to avoid new pairs or new triples of intersecting triangles, and to preserve (b). Since t(z) was not involved in a bad point, the new representation still fulfills (a'). Since the new t(z) contains the old one, the triangles originally intersected by t(z) are still intersected. Hence, t(z) intersects the same set of triangles, and the new representation is still a representation of T . After doing this to every t(z) we have the following.
Claim 6. There is no contact point on ]p, q].

Figure 4: (a) Step 3 (b) Condition (c)
Step 3: Now translate t(u) downwards in order to have its right corner in (x u , y u − 3 ), and inflate t(v ) in order to have its right angle in (x v − 3 , y v ), and height h v + 3 , for a sufficiently small 3 > 0 (see Figure 4.(a)). Here 3 is again sufficiently small to avoid new pairs or triples of intersecting triangles, and to preserve (b) but it is also sufficiently small to preserve the existing pairs of intersecting triangles. This last requirement can be fulfilled because the only intersections that t(u) could loose would be contact points on ]p, q], which do not exist. After these three steps, it is clear that the new representation has one bad point less and induces the same graph. This proves the proposition for 4-connected triangulations. The conditions (a) and (b) imply the following property.
(c) For every inner face xy z of T , there exists a triangle t(xy z), negatively homothetic to ∆, which interior is disjoint to any triangle t(v ) but which 3 sides are respectively contained in the sides of t(x), t(y ) and t(z). Furthermore, there exists an > 0 such that any triangle t homothetic to ∆ of height with a side in t(x) ∩ t(xy z) does not intersect any triangle t(v ) with v = x, and similarly for y and z (see Figure 4.(b) where the grey regions represent the union of all these triangles).
We are now ready to prove the proposition for any triangulation T . We prove this by induction on the number of separating triangles. We just proved the initial case of that induction, when T has no separating triangle (i.e. when T is 4-connected). For the inductive step we consider a separating triangle (u, v , w ) and we call T in (resp. T out ) the triangulation induced by the edges on or inside (resp. on or outside) the cycle (u, v , w ). By induction hypothesis T out has a representation fulfilling (a), (b), and (c). Here we choose arbitrarily the outer triangles and . Since uv w is an inner face of T out there exists a triangle t(uv w ) and an > 0 (with respect to the inner face uv w ) as described in (c). Then it suffices to apply the induction hypothesis for T in (which outer vertices are u, v and w ), with the already existing triangles t(u), t(v ), and t(w ) , and for = min( , ). Then one can easily check that the obtained representation fulfills (a), (b), and (c). This completes the proof of the proposition.

Conclusion
Given a graph G its incidence poset is defined on V (G) ∪ E(G) and it is such that x is greater than y if and only if x is an edge with an end at y . A triangle poset is a poset which elements correspond to homothetic triangles, and such that x is greater than y if and only if x is contained inside y . It has been shown that a graph is planar if and only if its incidence poset is a triangle poset [18] 2 . Theorem 2 improves on this result. Indeed, in the obtained representation the triangles t(u) corresponding to vertices intersect only if those vertices are adjacent, and the triangles corresponding to edges uv , t(u) ∩ t(v ), are disjoint.
In R 3 , one can define tetrahedral posets as those which elements correspond to homothetic tetrahedrons in R 3 , and such that x is greater than y if and only if x is contained inside y . Unfortunately, graphs whose incidence poset is tetrahedral do not always admit an intersection representation in R 3 with homothetic tetrahedrons. This is the case for the complete bipartite graph K n,n , for a sufficiently big n. It is easy to show that its incidence poset is tetrahedral. In an intersection representation with homothetic tetrahedrons, let us prove that the smallest tetrahedron t has a limited number of neighbors that induce a stable set. Let t be the tetrahedron centered at t and with three times its size. Note that every other tetrahedron intersecting t, intersects t on a tetrahedron at least as big as t. The limited space in t implies that one cannot avoid intersections among the neighbors of t, if they are too many. The interested reader will see in [11] that these graphs defined by tetrahedral incidence posets also escape a characterization as TD-Delaunay graphs.