Homothetic triangle representations of planar graphs
HHomothetic triangle representations of planar graphs
Daniel Gon¸calves, Benjamin L´evˆeque, Alexandre Pinlou
Abstract
We prove that every planar graph is the intersection graph of homothetic triangles in the plane . Here, an intersection representation is a collection of shapes in the plane. The intersection graph described bysuch a representation has one vertex per shape, and two vertices are adjacent if and only if the correspondingshapes 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 contactrepresentation . In such a representation, a contact point is a point that is in the intersection of (at least) twoshapes.Research on contact representations of (planar) graphs with predefined shapes started with the work ofKoebe 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], convexbodies [19], or (non-convex) axis aligned polygons [1, 12]. In the present article, we focus on homothetictriangles. It has been shown that many planar graphs admit a contact representation with homothetic triangles.Figure 1: Contact representations with homothetic triangles.
Theorem 1. [13] Every 4-connected planar triangulation admits a contact representation with homothetictriangles.
Note that one cannot drop the 4-connectedness requirement from Theorem 1. Indeed, in every contactrepresentation of K , , with homothetic triangles, there are three triangles intersecting in a point (see the rightof Figure 1). This implies that the triangulation (not 4-connected) obtained from K , , by adding a degreethree vertex in every face does not admit a contact representation with homothetic triangles. Some questionsrelated to this theorem remain open. For example, it is believed that if a triangulation T admits a contact This result was already annouced in [13] a r X i v : . [ c s . D M ] A ug epresentations 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 hasa contact representation with cubes in R . In the present paper we remain in the plane. Theorem 1 is thebuilding block for proving our main result. An intersection representation is said simple if every point belongsto 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 graphshave shown to be exactly the intersection graphs of homothetic triangles [14]). M¨uller et al. [17] proved thatfor some planar graphs, if the triangle corners have integer coordinates, then their intersection representationwith homothetic triangles needs coordinates of order 2 Ω( n ) , where n is the number of vertices. The followingsection is devoted to the proof of Theorem 2. It is well known that simple contact representations produce planar graphs. The following lemma is slightlystronger.
Lemma 3.
Consider a graph G = ( V, E ) given with a simple intersection representation C = { c ( v ) : v ∈ V } . Ifthe shapes c ( v ) are homeomorphic to disks, and if for any couple ( u, v ) ∈ V the set c ( u ) \ c ( v ) is non-emptyand connected, then G is planar. Proof.
Observe that since C is simple, the sets c ◦ ( u ) = c ( u ) \ (cid:0) ∪ v ∈ V \{ u } c ( v ) (cid:1) are disjoint non-empty connectedregions. Let us draw G by first choosing a point p u inside c ◦ ( u ), for representing each vertex u . Then for eachneighbor 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 bedone 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. (cid:3) Note that for any two homothetic triangles ∆ and ∆ (cid:48) , the set ∆ \ ∆ (cid:48) is connected. Lemma 3 thus implies thesufficiency of Theorem 2. For proving Theorem 2 it thus suffices to construct an intersection representationwith homothetic triangles for any planar graph G . In fact we restrict ourselves to (planar) triangulations becauseany such G is an induced subgraph of a triangulation T (an intersection representation of T thus contains arepresentation 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 trianglewith corners at coordinates (0 , ,
1) and (1 ,
0) (see Figure 2.(a)). Thus the homothets of ∆ have cornersof 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 . Proposition 4.
For any triangulation T with outer vertices a , b and c , for any three triangles t ( a ) , t ( b ) , and t ( c ) homothetic to ∆ , that pairewise intersect but do not intersect (i.e. t ( a ) ∩ t ( b ) ∩ t ( c ) = ∅ ), and for any (cid:15) > , there exists an intersection representation T = { t ( v ) : v ∈ V ( T ) } of T with homothets of ∆ such that:(a) No three triangles intersect.(b) The representation is bounded by t ( a ) , t ( b ) , and t ( c ) and the inner triangles intersecting those outertriangles intersect them on a point or on a triangle of height less than (cid:15) . Proof.
Let us first prove the proposition for 4-connected triangulations. Theorem 1 tells us that 4-connectedtriangulations have such a representation if we relax condition (a) by allowing 3 triangles t ( u ), t ( v ) and t ( w ) tointersect if they pairewise intersect in the same single point p (i.e. t ( u ) ∩ t ( v ) = t ( u ) ∩ t ( w ) = t ( v ) ∩ t ( w ) = p ).2 pt(u) t(v)t(w) (a) (b)Figure 2: (a) The triangle ∆ (b) The triangles t ( u ), t ( v ) and t ( w ).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 preservedalong 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 sameheight, take among those the leftmost one (i.e. minimizing x p ). Then let t ( u ), t ( v ) and t ( w ) be the threetriangles 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 ). Notealso that none of t ( u ), t ( v ) and t ( w ) is an outer triangle. q pt(u) t(v)t(w) pq t(z)t(u) t(v)t(w) (a) (b)Figure 3: (a) Step 1 (b) Step 2 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 ) inorder to have its right angle in ( x u − (cid:15) , y u ) and height h u + (cid:15) , for a sufficiently small (cid:15) > (cid:15) is sufficiently small to avoid new pairs of intersecting triangles, new triples of intersecting triangles, oran intersection between t ( u ) and an outer triangle on a too big triangle (with height ≥ (cid:15) ). Since the new t ( u )contains the old one, the triangles originally intersected by t ( u ) are still intersected. Hence, t ( u ) intersects thesame set of triangles, and the new representation is still a representation of T . Since there was no bad pointdistinct from p around t ( u ), it is clear by the choice of (cid:15) > Claim 5.
The top corner of t ( u ) is not a contact point. tep 2: For every triangle t ( z ) that intersects t ( u ) on a single point of the open segment ] p, q [ do thefollowing. 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 therepresentation and that by definition of p there is no bad point involving t ( z ). Now inflate t ( z ) in order to haveits right corner at ( x z , y z − (cid:15) ), and height h z + (cid:15) , for a sufficiently small (cid:15) > (cid:15) 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 theold one, the triangles originally intersected by t ( z ) are still intersected. Hence, t ( z ) intersects the same set oftriangles, and the new representation is still a representation of T . After doing this to every t ( z ) we have thefollowing. Claim 6.
There is no contact point on ] p, q ] . pq t(u) t(v)t(w) (a) (b)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 − (cid:15) ), and inflate t ( v ) inorder to have its right angle in ( x v − (cid:15) , y v ), and height h v + (cid:15) , for a sufficiently small (cid:15) > (cid:15) is again sufficiently small to avoid new pairs or triples of intersecting triangles, and to preserve (b) butit is also sufficiently small to preserve the existing pairs of intersecting triangles. This last requirement can befulfilled because the only intersections that t ( u ) could loose would be contact points on ] p, q ], which do notexist.After these three steps, it is clear that the new representation has one bad point less and induces thesame graph. This proves the proposition for 4-connected triangulations. The conditions (a) and (b) imply thefollowing property.(c) For every inner face xy z of T , there exists a triangle t ( xy z ), negatively homothetic to ∆, which interioris 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 (cid:15) (cid:48) > t homothetic to ∆ of height (cid:15) (cid:48) witha side in t ( x ) ∩ t ( xy z ) does not intersect any triangle t ( v ) with v (cid:54) = x , and similarly for y and z (seeFigure 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 numberof 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 ). Byinduction hypothesis T out has a representation fulfilling (a), (b), and (c). Here we choose arbitrarily the outertriangles and (cid:15) . Since uv w is an inner face of T out there exists a triangle t ( uv w ) and an (cid:15) (cid:48) > uv w ) as described in (c). Then it suffices to apply the induction hypothesis for T in (whichouter vertices are u , v and w ), with the already existing triangles t ( u ), t ( v ), and t ( w ) , and for (cid:15) (cid:48)(cid:48) = min( (cid:15), (cid:15) (cid:48) ).Then one can easily check that the obtained representation fulfills (a), (b), and (c). This completes the proofof the proposition. (cid:3) 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 andonly if x is an edge with an end at y . A triangle poset is a poset which elements correspond to homothetictriangles, and such that x is greater than y if and only if x is contained inside y . It has been shown that a graphis planar if and only if its incidence poset is a triangle poset [18] . Theorem 2 improves on this result. Indeed,in the obtained representation the triangles t ( u ) corresponding to vertices intersect only if those vertices areadjacent, and the triangles corresponding to edges uv , t ( u ) ∩ t ( v ), are disjoint.In R , one can define tetrahedral posets as those which elements correspond to homothetic tetrahedrons in R , and such that x is greater than y if and only if x is contained inside y . Unfortunately, graphs whose incidenceposet is tetrahedral do not always admit an intersection representation in R 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 itsincidence poset is tetrahedral. In an intersection representation with homothetic tetrahedrons, let us prove thatthe smallest tetrahedron t has a limited number of neighbors that induce a stable set. Let t (cid:48) be the tetrahedroncentered at t and with three times its size. Note that every other tetrahedron intersecting t , intersects t (cid:48) ona tetrahedron at least as big as t . The limited space in t (cid:48) implies that one cannot avoid intersections amongthe neighbors of t , if they are too many. The interested reader will see in [11] that these graphs defined bytetrahedral incidence posets also escape a characterization as TD-Delaunay graphs. References [1] M.J. Alam, T. Biedl, S. Felsner, M. Kaufmann, S.G. Kobourov, and T. Ueckerdt, Computing cartogramswith optimal complexity,
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