Triangle-Free Planar Graphs as Segments Intersection Graphs

Transcription

1 Triangle-ree Planar Graphs as Segments Intersection Graphs N. de Castro 1,.J.Cobos 1, J.C. Dana 1,A.Márqez 1, and M. Noy 2 1 Departamento de Matemática Aplicada I Universidad de Sevilla, Spain {natalia,cobos,dana,almar}@cica.es 2 Dpto. de Matemática Aplicada II Universitat Politècnica de Catalnya, Spain noy@grec.pc.es Abstract. We prove that every triangle-free planar graph is the graph of intersection of a set of segments in the plane. Moreover, the segments can be chosen in only three directions (horizontal, vertical and obliqe) and in sch a way that no two segments cross, i.e., intersect in a common interior point. 1 Introdction Given a set S of segments in the plane, its intersection graph has a vertex for every segment and two vertices are adjacent if the corresponding segments intersect. Intersection graphs of segments and other geometrical objects have been widely stdied in the past. or instance, if the segments are contained in a straight line then we have the interval graphs [4], a well-known family of perfect graphs. If the segments are chords of a circle then the intersection graph is called a circle graph, see for instance [6,8]. In the general case, there is no satisfactory characterization, bt some reslts are known for planar graphs. The most interesting one is de to de raysseix, Osona de Mendez and Pach [3], and independently to Ben-Arroyo Hartman, Newman and Ziv [2], which says that every planar bipartite graph is the intersection graph of a set of horizontal and vertical segments (on the other hand, it is known that the recognition of sch graphs is an NP-complete problem [5]). This reslt provides a partial answer to a qestion of Scheinerman [7]: is every planar graph the intersection graph of a set of segments in the plane? The main reslt in this paper, which is a significant extension of [3], is that every triangle-free planar graph is the intersection graph of a family of segments. Moreover, the segments can be drawn in only three directions and in sch a way that they do not cross. This particlar class of intersection graphs is also known as contact graphs. We call sch a representation a segment representation of the graph. Partially spported by DGES-MEC-PB C02 J. Kratochvíl (Ed.): GD 99, LNCS 1731, pp , c Springer-Verlag Berlin Heidelberg 1999

2 342 N. de Castro et al. A key point in or proof is Grötszch s Theorem [9], which garantees that every planar triangle-free graph is 3-colorable. The sketch of the proof is as follows. Given a triangle-free plane graph G, adding new vertices and (indced) paths between the vertices of G we can obtain a new triangle-free plane graph which is a sbdivision of a 3-connected graph. Starting with a 3-coloring, a segment representation in three directions is obtained for this new graph sing several technical lemmas. inally, removing the dmmy vertices and paths, we obtain a segment representation of the graph G. The three directions considered are horizontal, vertical and obliqe (parallel to the bisector of the second qadrant of the plane). o1 v1 h1 h1 v2 o2 v5 v3 v4 o2 v5 v1 v2 v3 v4 h3 h2 h5 h4 h3 h2 o1 h4 o3 v7 v6 h6 o3 h5 v6 v7 h6 ig. 1. A segment representation of a planar graph. 2 Convex aces with Three Directions Let I G be a segment representation of a plane graph G. Asegment path of length n is a seqence of segments P = {s 1,...,s n } sch that segment s i is adjacent to s i 1 and s i+1 for 1 <i<n. The representation I G divides the plane into regions that we call faces of the representation and which correspond to the faces of G. A face of a segment representation is convex if its bondary can be divided into for paths (clockwise) P 1, P 2, P 3 and P 4 sch that if s is a segment with extremes in the points (x 1,y 1 ) and (x 2,y 2 ) then: 1. if s is in P 1, then x 1 <x 2 if s is horizontal, y 1 >y 2 if s is vertical and x 1 <x 2 and y 1 >y 2 if s is obliqe; 2. if s is in P 2, then x 1 >x 2 if s is horizontal, y 1 >y 2 if s is vertical and x 1 >x 2 and y 1 <y 2 if s is obliqe; 3. if s is in P 3, then x 1 >x 2 if s is horizontal, y 1 <y 2 if s is vertical and x 1 >x 2 and y 1 <y 2 if s is obliqe; 4. if s is in P 4, then x 1 <x 2 if s is horizontal, y 1 <y 2 if s is vertical and x 1 <x 2 and y 1 >y 2 if s is obliqe.

3 Triangle-ree Planar Graphs as Segments Intersection Graphs 343 It can be seen that the partition of a convex face into P 1, P 2, P 3 and P 4 is not niqe, for instance in the example of igre 2 the first segment of P 1 cold be in P 4. If the segments s 1,...,s n are the intersection between two convex faces and G, we will assign those segments to only one path in each face, following the newt rle: if this segments belongs to P 1 in, then they belongs to P 3 in G and reciprocally. Analogosly if they belongs to P 2 in, then they are in P 4 of G and reciprocally. This rle alow s to fix the partition of a convex face into the for paths depending on the adjacent faces if the partition is not niqe. P 4 P 1 P 3 P 2 Convex ace Non Convex ace ig. 2. The for paths of a convex face. Lemma 1. Let I G be a segment representation of a plane graph G sch that all its faces are convex, let k be a positive real nmber and let s be a segment of length l in a face of I G. Then, I G can be transformed into another segment representation I G satisfying: 1. I G and I G have the same faces and face bondaries as G, and the faces of I G are convex; 2. the length of s in I G is k + l; 3. at most two segments of (other than s) have different length in I G and I G. Proof. Let P 1, P 2, P 3 and P 4 denote the paths which make convex. Call the pper part (the lower part) of the face the nion of the paths P 4 and P 1 (P 2 and P 3, respectively). Similarly, the right part (the left part) is the nion of the paths P 1 and P 2 (P 3 and P 4, respectively). The proof is by indction on the nmber of faces of the graph G. Sppose that I G is jst the cycle and let s be the segment of we want to enlarge. Withot loss of generality we can sppose that s is a horizontal segment. If s is in the pper part of, we seek another horizontal segment s in the lower part of. If there exists sch a segment, we enlarge s and s by the same amont and we make a translation of all the segments between s and s, transforming into other convex face (see igre 3). Otherwise, there mst exist a vertical segment s and an obliqe segment s in the lower part of, becase the bondary of is a closed path. In this case, we increase the length of s, s and s and we make a translation of the rest of the segments (see igre 3).

4 344 N. de Castro et al. According to the above remark, we sppose in the rest of the proof that there exists a segment parallel to s in the opposite part of. Sppose now that I G has two convex faces and sharing the segments s 1,...,s n. irst of all, by the convexity of and, observe that if s 1,...,s n are segments of P 1 (P 2 ), then s 1,...,s n form part of P 3 (P 4, respectively). Ths, sppose that s 1,...,s n belong to P 1 in and to P 3 in.ifsis in the pper part of, we seek another horizontal segment s in the lower part of and we proceed as before, increasing the length of s and s. (a) (b) ig. 3. How to enlarge a segment of a convex face by any amont, (a) sing parallel segments or (b) sing three segments. Obviosly, if s is not one of the s i, the transformation in does not affect the segments of. Bt, if s is one of the s i then, on accont of the first case, we consider a horizontal segment s in the lower part of and a horizontal segment s in the pper part of. Increasing the length of s, s and s, and making a translation of all the segments of between s and s, and all the segments of between s and s, we obtain a new segment representation with the same convex faces as I G. Sppose now that I G has n convex faces, 1,..., n, and let s be a segment of a face n, where n lies in the oter face of I G, i.e., the vertices v 1,...,v k of G, corresponding to the segments s 1,...,s k of n, lie on the oter face, and v 2,...,v k 1 are of degree two. Let s consider the graph G 1 = G {v 2,...,v k 1 }. Removing the segments of I G corresponding to v 2,...,v k 1, we obtain a segment representation I G1 of G 1. Since I G1 has n 1 faces, we can transform its segment representation into another representation which has the same convex faces, enlarging in any amont some of the segments. To obtain this new representation, we have to enlarge other segments in I G1 besides s, bt we have preserved the strctre of the segments in the representation, so we are able to represent again the segments corresponding to v 2,...,v k 1 enlarging the length of one (or two) of them if necessary, obtaining a segment representation of G.

5 Triangle-ree Planar Graphs as Segments Intersection Graphs Triangle-ree Graphs Using the reslts of Barnette in [1], it can be dedced easily that every planar triangle-free graph which is a sbdivision of a 3-connected graph can be redced, by deleting edges and paths with internal vertices of degree two, to a sbdivision of the complete graph K 4 in sch a way that in each step we have a sbdivision of a 3-connected graph. Moreover, if we fix a sbgraph of G which is a sbdivision of K 4, G can be redced, by deleting edges and paths, to this sbgraph. This reslt will allow s to bild a segment representation in three directions of sbdivisions of 3-connected graphs as follows. irstly we bild a segment representation of a sbdivision of K 4 and then insert, in reverse order, the edges and paths that were removed to obtain from G the sbdivision of K 4. In order to do this, we need two basic operations: (1) insert a path between two segments of the same face and (2) join two segments of the same face. The second operation is, at least, as difficlt as the first one, so we will concentrate only in the second operation. We can see, sing Lemma 1, that by enlarging some segments, a segment representation with convex faces can be transformed into another one preserving the topology of the embedding. Bt this is not sfficient to carry on the second operation, so we need another kind of transformation. ixing an index i and changing the drawing of the segments of the path P i of a face, it is possible to draw a new convex face where the segments that were in P i belong now to P i+1 or P i 1. The rest of the paths of only change in the length of some of the segments (see igre 4). P4 P1 P4' P1' a b c ' c b a P3 P2 P3' P2' ig. 4. How to restrctre a face throgh the path P 1. In the segment representation, the change of the face to prodces some changes in the faces adjacent to along the path P i, and we obtain another segment representation with convex faces as in the proof of Lemma 1 (see igre 5). When we change the drawing in the manner described above, we say that the face is restrctred throgh the path P i. We need at this point some new definitions. Given a 3-colored plane graph G with colors {h, o, v}, a path P = { 1,..., n } of G is rare if it verifies one of the following conditions:

6 346 N. de Castro et al. ig. 5. How to restrctre the faces adjacents to the restrctred face. Case 1. 1 is colored as h, 2 as o, and n as v. Case 2. 1 is colored as v, n 1 as h, and n as o. Case 3. 1 is colored as h, 2 as o, n 1 as h, and n as o. Case 4. 1 is colored as o, 2 as v, n 1 as v, and n as h. A face of a graph is called rare if there exists a path P i, with i =1,...,4, containing a rare sbpath. We say that a rare face is repaired if we sbdivide its rare path with a new vertex colored as v between 1 and 2 in the first and the third case, and between n 1 and n in the second case, or a new vertex colored as h between 1 and 2 in the forth case. On the other hand, notice that the vertical segments (except the last segment of P 1 and the last of P 3 ) can be extended to the interior of a convex face, the horizontal segments too if they are in P 1 or P 3 (except the first of P 1 and the first of P 3 ), and the obliqe segments if they are in P 2 or P 4 (except the first one of P 1 and P 3, and the last one of P 1 and P 3 ). Moreover, the horizontal segments in P 2 or P 4 can be extended to the interior of if there is a vertical segment to the right or an obliqe segment to the left of them; and the obliqe segments in P 1 or P 3 can be extended to the interior of if there is a horizontal segment to the right or a vertical to the left of them (see igre 2). So, restrctring the paths and applying Lemma 1, we can transform the face in sch a way that these transformations allow s to extend two segments inside to prodce a new adjacency between them. In the following lemma, when we say that two segments s and t can be joined by a path length k, it means a path s, s 1,s 2,...,s k,t. Lemma 2. Let I G be a segment representation of a plane graph G and let s 1 and s 2 be two distinct segments in a convex non-rare face. Then I G can be transformed into another segment representation I G satisfying 1. I G and I G have the same faces and face bondaries as G, and the faces of I G are convex; 2. s 1 can be joined to s 2 by a segment path of length k, for all k > 0, or directly if s 1 and s 2 have not the same direction inside the convex face, corresponding to.

7 Triangle-ree Planar Graphs as Segments Intersection Graphs 347 Proof. The proof falls natrally into two cases: the segments have the same direction, or they do not. In both cases, we have divided the proof in a seqence of sbcases. Case 1: s 1 and s 2 have not the same direction. In this case, we can sppose that s 1 is vertical and s 2 is horizontal (similar argments apply to the other configrations on the directions of the segments s 1 and s 2 ). The sbcases are the following: 1. s 1 and s 2 are in the same path P i. a) s 1 and s 2 are in P 2 (respectively in P 4 ). i. s 1 appears (clockwise) before s 2. If an obliqe segment precedes s 2, by Lemma 1 the segments can be joined directly. Otherwise, the face mst be restrctred transforming P 2 into P 3 (respectively, P 4 into P 1). ii. s 2 appears (clockwise) before s 1. Since the face is not rare, s 2 precedes a vertical segment. Then, by Lemma 1, the segments can be joined directly. b) s 1 and s 2 are in P 1 (respectively in P 3 ). If s 1 appears (clockwise) before s 2, by Lemma 1 the segments can be joined directly. If s 2 appears (clockwise) before s 1, the face mst be restrctred transforming P 1 into P 2 (respectively, P 3 into P 4). 2. s 1 and s 2 are in different paths. If s 2 is in P 2 (respectively in P 4 ) and it precedes an obliqe segment and s 1 is in P 3 (respectively P 1 ), the face mst be restrctred transforming P 2 into P 1 (respectively, P 4 into P 3). Otherwise, by Lemma 1 the segments can be joined directly. Case 2: s 1 and s 2 have the same direction. In this case, it sffices to join the segments by a path of length 1, becase this path can be sbstitted by any other of length greater than 1. Again there are several sbcases: 1. Both are vertical segments. In this case, the segments can be joined by a path of length 1 directly sing Lemma Both are horizontal segments. If the segments are in P 2 (respectively in P 4 ) and the first one precedes an obliqe segment, and a vertical segment precedes the second one, then the face mst be restrctred transforming P 2 into P 3 (respectively, P 4 into P 1). Otherwise, sing Lemma 1 the segments can be joined directly. 3. Both are obliqe segments. If the segments are in P 1 (respectively in P 3 ), the first one precedes a vertical segment, and a horizontal segment precedes the second one, then the face mst be restrctred transforming P 1 into P 2 (respectively, P 3 into P 4). Otherwise, they can be joined directly by Lemma 1. Now we prove: Lemma 3. Any triangle-free plane graph which is a sbdivision of a 3-connected graph can be represented by segments in three directions.

8 348 N. de Castro et al. Proof. As G is a triangle-free plane graph, in virte of Grötzsch s Theorem [9], G admits a 3-coloring with the colors h, v and o (horizontal, vertical and obliqe, respectively). Since G is a sbdivision of a 3-connected graph, we can find a vertex, not in the oter face of G, connected by three disjoint paths to three vertices in the oter face. These paths and the oter face determine a graph K, which is a sbdivision of K 4. Using Barnette s reslts in [1] it is possible to bild a seqence of graphs G 1,...,G n and a seqence of paths Q 1,...,Q n sch that G 1 = G, G i is a sbdivision of a 3-connected plane graph, G i is obtained from G i 1 by deleting the path Q i 1, and the graph obtained from G n deleting Q n is K. When the path Q i is deleted, a new face appears in G i+1 as the nion of two faces in G i. The bondary of can be divided into two paths; on the one hand the path beginning in the first vertex of Q i and ending in the last one and, on the other hand, the rest of the bondary. If one of them is rare, we mst repair sbdividing an edge with a new vertex (that we label as a repaired vertex), to make the face non-rare. So, we can constrct another seqence G 1,...,G n where G i is G i with a new repaired vertex, if necessary. ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ig. 6. How to remove the repaired vertex in the for cases. It is easy to give a segment representation of K with all its faces convex. It sffices to represent convexly the oter face of K and proceed according to the

9 Triangle-ree Planar Graphs as Segments Intersection Graphs 349 above remark. Using Lemma 2 it is possible to add the path Q i to the segment representation of G i+1 to obtain a convex face representation of G i. In order to obtain a segment representation of the graph G, we mst remove the repaired vertices. These vertices cannot be removed directly, so it is necessary to modify the representation. The for possible cases are illstrated in igre 6. Notice that we cold not remove the repaired vertices if the rare path wold have exactly three segments 1, 2 and 3, bt this case it is not possible becase joining 1 with 3 it will be form a triangle, and G was a triangle-free graph. We can now formlate or main reslt as follows: Theorem 1. Every triangle-free planar graph is the intersection graph of a set of segments in three directions. Proof. Let G be a triangle-free plane graph. Since G has no triangles, we can obtain a 3-coloring of G sing Grötszch s Theorem [9]. The colors will be labeled as h, v and o (horizontal, vertical and obliqe, respectively). We can bild a new triangle-free plane graph G 1, sbdivision of a 3-connected graph, which contains G as a sbgraph, adding new vertices and edges joining the blocks of G, possibly sbdividing some edges of the blocks of G sing new vertices. If this is the case, these vertices are labeled as dmmy vertices. When an added edge prodces a triangle, we sbdivide it, and when a new edge joins two vertices with the same color, we sbdivide it too. We call these new vertices and edges virtal. All the new vertices (dmmy and virtal) can be colored so that the 3-coloring is preserved. By Lemma 3, G 1 admits a segment representation. Ot of this segment representation we mst remove all the vertices and edges added. A virtal edge (or a path bilt with virtal edges and vertices) in the segment representation of G 1 is an adjacency between two segments (or a virtal path joining two segments). It sffices to break this adjacency and to shorten these segments (note that the segments do not cross, they only contact). The dmmy vertices are removed as the repaired vertices in Lemma 3. 4 Conclding Remarks The hypothesis that the graph has no triangles can be relaxed in some cases. No dobt there exist planar graphs with triangles that admits segment representations (see igre 1). The problem lies in the fact that we have followed a constrctive proof to obtain a segment representation sing the convex faces representation of a sbdivision of a 3-connected graph. This constrction wold not be possible in general if the graph contains triangles, becase if we fix the three directions of the segments we can observe that the graph in igre 7 cannot be represented by non-crossing segments. In this example we see that a segment representation contains faces that do not correspond to faces of the embedding of the graph. So, the topology of the embedding of the plane graph cannot be preserved. This problem admits a

10 350 N. de Castro et al. o1 v2 v1 h1 o1 v1 h1 o2 h2 o2 h2 v2 ig. 7. Two segments mst cross. partial soltion as follows. A 3-coloring of a plane graph G with colors {h, v, o} is good if all the triangles of G are colored, in clockwise order, as h v o. The proof of Theorem 1 can be adapted yielding the following reslt: Theorem 2. Let G be a 3-colored plane graph. The coloring is good if and only if exists a segment representation I G verifying that the faces of G correspond to faces of I G, and the bondaries of the faces are preserved. References 1. D.W. Barnette, On Steinitz s theorem concerning convex 3-polytopes and on some properties of planar graphs, The many facets of graph theory. Lectres Notes in Mathematics Vol. 110, Springer, Berlin, pp , I. Ben-Arroyo Hartman, I. Newman and R. Ziv, On grid intersection graphs, Discrete Math. Vol. 87, pp , H. de raysseix, P.Osona de Mendez and J. Pach, Representation of planar graphs by segments, Colloqia Mathematica Societatis János Bolyai, Intitive Geometry, Szeged (Hngary), M.C. Golmbic, Algorithmic Graph Theory and Perfect Graphs, Academic Press, J. Kratochvíl, A special planar satisfiability problem and a conseqence of its NP-completeness, Discrete Applied Math., Vol. 52, pp , W. Naji, Reconnaisance des graphes de cordes, Discrete Math. Vol. 54, pp , E.R. Scheinerman, Intersection classes and mltiple intersection parameters of graphs, Ph D. thesis,, Princeton University, J. Spinrad, Recognition of circle graphs, J. of Algorithms, Vol. 16, pp , C. Thomassen, Grötzsch s 3-color theorem and its conterparts for the tors and the projective plane, J. Comb. Theory B Vol. 62, pp , 1994.