SS TICK GRAPHS : EXAMPLES AND COUNTER - EXAMPLES
Irena Rusu
LS2N, University of Nantes, France
Abstract
Stick graphs are the intersection graphs of vertical and horizontal segments inthe plane whose bottom and respectively left endpoints belong to a line with negative slope.Such a representation is called a Stick representation of the graph. Very few results existon Stick graphs: only trivial classes of Stick graphs have been identified; recognizing Stickgraphs is an open problem; and even building examples of graphs that are not Stick graphsis quite tricky.In this paper, we first propose structural results on Stick graphs, connecting the neighbor-hoods of two vertices to the order of these two vertices in a possible Stick representation. Wededuce local obstructions for the existence of a Stick representation, which we use for theconstruction of graphs that are not Stick graphs. We next disapprove the idea, suggested bythese examples and the other examples from the literature, that a graph which is not a Stickgraph necessarily contains a hole. Finally, we show that bipartite complements of circular-arc graphs and of circle graphs are subclasses of Stick graphs.Keywords: grid intersection graphs, Stick graphs, circular arc graphs, circle graphs
Graph classes defined as intersection graphs of geometric objects have been extensively studied. Intersectiongraphs of intervals on the real line (called interval graphs ), of arcs on a circle (called circular arc graphs ),of trapezoids with bases on two fixed parallel lines (called trapezoid graphs ), of chords of a circle (called circle graphs ), of straight line segments in the plane (called segment graphs ) are only a few examples. Manyof them, together with references and relations between them, may be found in the book [2] as well as in [7].Among segment graphs, the class of grid intersection graphs [11] plays a particular role, since it concernsonly horizontal and vertical segments, yielding applications identified in [5]. In the same paper, severalsubclasses of grid intersection graphs are proposed and their intersections are studied. One of them is theclass of Stick graphs.A
Stick graph is the intersection graph of a set A of vertical segments and a set B of horizontal segmentsin the plane (respectively called A - and B -segments ), whose bottom and respectively left endpoints lie ona “ground” line with slope -1. Each of these endpoints is called the origin (which is then an A -origin or B -origin) of the corresponding segment. It is assumed that all the origins are distinct (otherwise, one mayappropriately move some of them close to their initial position, and modify some segment lengths, withoutaffecting the segment intersections).Stick graphs, as all the grid intersection graphs, are bipartite graphs. Given a Stick graph, a represen-tation of it with vertical and horizontal segments as mentioned above is called a Stick representation of thegraph. The problem of recognizing Stick graphs - denoted STICK - requires, given a bipartite graph G ,to decide whether G has a Stick representation or not. STICK is an open problem. Variants of STICK,assuming that the order of the A -origins and/or the order of the B -origins are known, have been consideredand solved in [13, 6], including in the case where the lengths of the segments are known. However, very1 a r X i v : . [ c s . D M ] A ug ittle is known about Stick graphs in general. Trivial examples of Stick graphs exist, namely chordless cy-cles and paths, as well as trees. Bipartite permutation graphs, defined as the intersection graphs of segmentswith one endpoint on each of two given parallel lines such that the resulting graph is bipartite, are also Stickgraphs (deduced from the definition). At the opposite, finding examples of graphs that are not Stick graphsneeds some work, since none of them is trivial.Unlike the previous approaches seeking to test whether a graph is a Stick graph, our aim is to investigateStick graphs - and therefore also non-Stick graphs - from the viewpoint of their structure. More precisely,we look for sufficient structural conditions ensuring that a given graph is a Stick graph or, on the contrary,that the graph is not a Stick graph. As a consequence, we identify forbidden subgraphs of Stick graphs andwell-known classes of graphs that are subclasses of Stick graphs.The paper is structured as follows. We give in Section 2 the main definitions and notations. In Sections 3and 4, we identify local properties of a Stick representation and deduce particular configurations of graphsthat act as obstructions to such a representation. Using them, we build several examples of graphs or familiesof graphs that are not Stick graphs, and are thus forbidden subgraphs of Stick graphs. All these examples,as well as the other examples presented in [5] and [13], have a common feature: they contain a hole, i.e. aninduced cycle with at least five vertices (in reality, only holes with an even number of vertices, thus at leastsix, will appear in bipartite graphs). However, in Section 5 we provide an example of a forbidden subgraphwithout any hole. At the opposite of these concerns related to non-Stick graphs, in Section 6 we focus onconditions that make that a graph is a Stick graph and we identify particular classes of Stick graphs, namelybipartite complements of circular arc graphs and bipartite complements of circle graphs. As a consequence,their subclasses of interval bigraphs and bipartite interval graphs are also Stick graphs. Section 7 is theconclusion. Related work.
Our results were obtained by investigating open questions, yet it appears that - whilestudying related issues - other authors have obtained, previously to the current paper, a result similar to ours.Their proof being more complex than ours, because the authors pursue other goals, we kept our result andmentioned the alternative proof (see Remark 5).
All the graphs we use are undirected and simple, and the notations are classical. The bipartite graph forwhich a Stick representation is searched is always denoted G = ( A ∪ B, E ) , with | A | = n and | B | = m .The vertices in A , also called A -vertices, as well as the A -segments representing them in any (attempt of)Stick representation, are denoted A i , ≤ i ≤ b , and similarly for the B -vertices and B -segments, denoted B j , ≤ j ≤ m . Moreover, denote T ( a i ) the point where the horizontal line going through the top endpointof the segment A i meets the ground line. Similarly, denote T ( b j ) the point where the vertical line goingthrough the rightmost endpoint of the segment B j meets the ground line. Each of these points is called the tip of its corresponding segment, and is an A - or B -tip according to the type of the segment. As initiallydone for the origins only, we may consider that all the origins and the tips are distinct.For any two points x, y of the ground line, denote x ≺ y if x is above y . The following observation iseasy (and similar to those made for other classes, e.g. for max-point tolerance graphs [3]): Observation.
Segments A i and B j intersect in a Stick representation iff T ( a i ) ≺ b j ≺ a i ≺ T ( b j ) , i.e. iffintervals [ T ( a i ) , a i ] and [ b j , T ( b j )] of the ground line overlap in this order from top to bottom.We may then see Stick graphs as a variant of interval bigraphs [10] defined as the intersection graphs oftwo sets of intervals, A versus B , of the real line. 2 (cid:2) (cid:1) (cid:3) (cid:1) (cid:4) (cid:1) (cid:5) (cid:6) (cid:2) (cid:6) (cid:3) (cid:6) (cid:4) (cid:7) (cid:4) (cid:8) (cid:4) (cid:8) (cid:2) (cid:8) (cid:3) (cid:7) (cid:2) (cid:7) (cid:3) (cid:7) (cid:5) (a) (cid:1) (cid:2) (cid:3) (cid:4) (cid:3) (cid:5) (cid:3) (cid:2) (cid:3) (cid:6) (cid:1) (cid:4) (cid:1) (cid:5) (b) Figure 1: a) A Stick graph and its Stick representation, with continuous lines. The dotted lines indicate howto compute the tips. b) The resulting flat Stick representation, where the plain circles represent the tips.
Conventions.
Throughout the paper, with only one exception, we use this flat
Stick representation, wherethe ground line is assumed to be the real line (with “top” becoming “left”, and “bottom” becoming “right”).We also use the term overlap for two intervals A i , B j only when T ( a i ) ≺ b j ≺ a i ≺ T ( b j ) . Example 1.
Figure 1 shows a Stick graph G together with its standard and flat Stick representations. Notethat the graph G (cid:48) obtained from G by adding a vertex B adjacent to vertices A , A , A is not a Stick graph(see Section 4 and Figure 2(a)).For X ⊆ A ∪ B , the intervals in X are called X -intervals . The origins of intervals in X are called X -origins . Since each vertex is identified with its interval, we transfer some of the graph terminology andnotations to intervals. Two intervals A i , B j are adjacent if A i B j ∈ E . We also say they are neighbors . Theneighborhood of an interval I is the set of all its neighbors, denoted N ( I ) . In order to avoid supplementaryterminology, all these notions and notations (adjacency, neighborhood) are extended to the origins of theintervals, i.e. for instance N ( a i ) = { b j | A i B j ∈ E } .In the next section, assuming that G has a Stick representation, we renumber the A -intervals as A ,A , . . . , A n such that i < j iff T ( a i ) ≺ T ( a j ) . Let ≤ v < t ≤ n and denote B = N ( A t ) \ N ( A v ) , B = N ( A v ) ∩ N ( A t ) , B = N ( A v ) \ N ( A t ) . Furthermore, denote β i = { b | b is a B i -origin } and T ( S ) = { T ( s ) | s ∈ S } for each set S of A -origins and B -origins.Interval A s is a partner of interval A k if s ≤ k and they have at least one common neighbor. Then wealso say that the A -origin a s is a partner of the A -origin a k . Let A s be a partner of A v or A t , which impliesthat s ≤ t . Let w be a subset of { , , } denoted as a word ( e.g. for { , } ), and denote B w = ∪ j ∈ w B j .Interval A s is called a B w -partner of (both) A v and A t if it satisfies the property that N ( A s ) ∩ B j (cid:54) = ∅ if andonly if j ∈ w . Similarly, a s is a B w -partner of (both) a v and a t . By definition, exactly one w exists for each a s such that a s is a B w -partner of a v and a t . Moreover, a t (respectively a v ) is a (trivial) B − (respectively B ) partner of a v and a t . Finally, denote N w ( a s ) = N ( a s ) ∩ B w . In this section, we investigate necessary conditions for G to be a Stick graph.3 R R R (C1) T ( a v ) ≺ β ≺ T ( a t ) ≺ β ∪ β ≺ a t ≺ T ( β ) ∪ β ≺ a v ≺ T ( β ) ∪ T ( β ) (C2) T ( a v ) ≺ β ∪ β ≺ T ( a t ) ≺ β ∪ β ≺ a v ≺ β ∪ T ( β ) ∪ T ( β ) ≺ a t ≺ T ( β ) ∪ T ( β ) ∪ T ( β ) (C2’) T ( a v ) ≺ β ∪ β ≺ a v ≺ T ( a t ) ≺ β ∪ T ( β ) ≺ a t ≺ T ( β ) ∪ T ( β ) Table 1:
Possible configurations for a v , a t and their neighbors. All sets may be empty. We have β k = ∪ j =1 , , β jk ,whenever these subsets exist. Notation X ≺ Y indicates that the whole set X is before ( i.e. to the left of) the wholeset Y on the real line. Each R i , ≤ i ≤ , denotes a set of origins and/or tips called a region. Proposition 1.
Configurations (C1), (C2) and (C2’) in Table 1 are the only possible configurations for theorigins and tips of the intervals in { A v , A t } ∪ N ( A v ) ∪ N ( A t ) . Proof.
Consider first the possibilities to have a t ≺ a v . Sets β and β must be between T ( a t ) and a t since B ∪B = N ( A t ) , whereas T ( β ) (respectively T ( β ) ) must be placed between a t and a v (respectivelyafter a v ) in order to respect the sought adjacencies and non-adjacencies. Now, a v ≺ T ( β ) since B ⊆ N ( A v ) , thus the B -origins may be placed everywhere before a v except between T ( a t ) and a t (otherwise A t would have neighbors in B ). Configuration (C1) is then the only one possible when a t ≺ a v .The reasoning is similar when a v ≺ a t , with reversed roles for β and β , except that some B -originscould be placed between T ( a v ) and T ( a t ) . Their corresponding tips must then be placed after a v , that is,either between a v and a t , or after a t . Three types of B -intervals are then possible, yielding subsets β , β and β . This gives configuration (C2). However, when B = ∅ , a v may be placed between the two tips T ( a v ) and T ( a t ) , yielding configuration (C2’). Convention.
As configuration (C2’) is highly similar to configuration (C2), in the remaining of the paperwe explicitly consider only configurations (C1) and (C2), and assume (C2’) behaves as (C2) in case B = ∅ ,with T ( a t ) and a v reversed.With the aim of identifying the possible positions for the partners a s of a v and a t , we indicated in Table 1four regions. Each region R i is understood as a set of origins and/or tips, so that R i = T ( X ) ∪ Y for some X and Y . For instance, when R = β ∪ T ( β ) ∪ T ( β ) in (C2), we have Y = β and X = β ∪ β . Placing a s in R i means - for the moment - that the origins and tips currently in R i may be distributed as needed, and a s may be inserted before or after each element in R i , as long as the borders T ( a t ) and a t are not crossed.No difference exists between the placement of a s just before or just after T ( a t ) (the same holds for a v , or a t ), so that when possible we avoid this redundancy. For a given w , a B w -partner a s may be placed in someregion R i provided two types of necessary local conditions are satisfied: • the partnership conditions ensure that - making abstraction of the other partners that must be placed -there is a possible position for a s in R i such that A s can overlap B -intervals from each B j with j ∈ w .We may see these conditions as general, contextual, constraints. • the placement conditions ensure - in a region validated by the partnership conditions - the existenceof a precise position for a s such that the interval A s overlaps all its neighbors in the bipartite graph G . Remark 1.
Partners A s of A v and A t with s < v overlap each neighbor B q of A v or A t such that b q ≺ a s ≺ T ( b q ) since T ( a s ) ≺ T ( a v ) ≺ b q . Partners A s of A v and A t with v < s < t satisfy the sameproperty for all b q with b q ≺ a s ≺ T ( b q ) and b q (cid:54)∈ R . When b q ∈ R , one may have b q ≺ T ( a s ) implyingthat A s and B q do not overlap (recall that by convention we use the term overlap only for overlaps with T ( a s ) ≺ b q ≺ a s ≺ T ( b q ) ). 4 C1) (C2) w R Partnership conditions R Partnership conditions w = 13 R β (cid:54) = ∅ R β = ∅ R β = ∅ R β (cid:54) = ∅ w = 12 R β = ∅ or β ≺ T ( a s ) ≺ T ( a t ) R β = ∅ or β ≺ T ( a s ) ≺ T ( a t ) , and R β = ∅ or β ≺ T ( a s ) ≺ T ( a t ) , and N ( a s ) ⊇ β N ( a s ) ⊇ β R w = 23 R β (cid:54) = ∅ R R N ( a s ) ⊇ β R N ( a s ) ⊇ β R R β (cid:54) = ∅ w = 1 R β = ∅ or β ≺ T ( a s ) ≺ T ( a t ) R β = ∅ or β ≺ T ( a s ) ≺ T ( a t ) , and β = ∅ R β = ∅ or β ≺ T ( a s ) ≺ T ( a t ) , and β = ∅ R w = 2 R β = ∅ or β ≺ T ( a s ) ≺ T ( a t ) R β ∪ β = ∅ or β ∪ β ≺ T ( a s ) ≺ T ( a t ) R R w = 3 R β (cid:54) = ∅ R β ∪ β (cid:54) = ∅ R β (cid:54) = ∅ R R β = ∅ R β = ∅ R R β (cid:54) = ∅ w = 123 R β (cid:54) = ∅ R N ( a s ) ⊇ β R N ( a s ) ⊇ β R Table 2:
Possible positions of the B w -partners a s of A v , A t , depending on w , the configuration ( C ) and the neigh-borhood of a s in G . When a s can be placed in a region R i without any partnership condition, an empty line indicatesit. For each w , we assume that the condition β j (cid:54) = ∅ (equivalently B j (cid:54) = ∅ ) holds for all j ∈ w . Denote R g ( a s ) := { b q ∈ R | T ( a s ) ≺ b q } , the set of good B -origins in R with respect to a s , whichare the B -origins not concerned by the exception in the previous remark. Then R b ( a s ) = R − R g ( a s ) isthe set of bad B -origins of a s . Remark 2.
Concerning the placement conditions for a s , note first that R g ( a s ) ⊇ N w ( a s ) ∩ R must hold,and thus R b ( a s ) ∩ N w ( a s ) = ∅ . Consider now a region R i = T ( X ) ∪ Y ( i = 1 , , ), where each of X or Y (but not both) may be empty, X ∪ Y ⊆ B ∪ B ∪ B and X ∩ Y = ∅ . The position of a s in R i mustsatisfy the following properties: • the neighbors of a s in Y and the tips of its non-neighbors in X − R b ( a s ) are situated on the left sideof a s • the non-neighbors of a s in Y and the tips of its neighbors in X are situated on the right side of a s .The tips of the non-neighbors of a s in R b ( a s ) are not precisely placed with respect to a s , they may besituated either on its left or on its right side. In summary, the placement condition for a s is: ( Y ∩ N w ( a s )) ∪ ( T ( X ) − T ( R b ( a s )) − T ( N w ( a s ))) ≺ ( Y − N w ( a s )) ∪ ( T ( X ) ∩ T ( N w ( a s ))) . so that a s can be placed between the two sets above.In Table 2, we list all the possible positions of B w -partners a s , for all w , in configurations (C1) and (C2)as obtained from Proposition 2 below. In this table, the columns respectively indicate the set w , and for eachconfiguration (C1) and (C2): the regions where a s may be placed, according to the notations in Table 1, andthe partnership conditions that must be satisfied to allow the placement of a s in each region. Proposition 2.
Let w be a subset of { , , } and (C) be a configuration among (C1) and (C2). The followingproperties hold: i ) Each B w -partner a s of a v and a t , except a v and a t themselves, must be placed in at least one of theregions identified in Table 2 for w and ( C ) ; in this case, the conditions identified in the table for eachregion must be satisfied. ( ii ) For every pair a s , a r of partners of a v and a t that are placed in the same region R i = T ( X ) ∪ Y ( i ∈ , , ), either: N w ( a s ) ∩ Y ⊆ N w ( a r ) ∩ Y and ( N w ( a s ) ∩ X ) − R b ( a r ) ⊇ ( N w ( a r ) ∩ X ) − R b ( a s ) , or N w ( a s ) ∩ Y ⊇ N w ( a r ) ∩ Y and ( N w ( a s ) ∩ X ) − R b ( a r ) ⊆ ( N w ( a r ) ∩ X ) − R b ( a s ) . Proof. ( i ) Recall that s < t , thus T ( a s ) ≺ T ( a t ) . The proof of each case follows the same principles:(1) identify the regions that may possibly (under conditions) contain points belonging to at least one intervalfrom each B i , for all i ∈ w ; and (2) for each such region, find the conditions that must be satisfied outsidethat region ( i.e. the partnership conditions) in order to ensure that a s belongs to B w for the value of w understudy.As an example, we give the proof for w = 13 and (C1). Regions R and R proposed in Table 1 arethe only possible ones, since placing a s before β or after T ( β ) would prevent any adjacency between A s and an interval in B . When a s is placed in region R , β is the only source of neighbors in B for a s , sowe must have β (cid:54) = ∅ . Moreover, β ≺ a s ≺ T ( β ) , thus A s overlaps all the intervals with good B -originswith respect to a s (by Remark 2 the origins of the intervals that A s must overlap are necessarily good B -origins). When a s is placed in region R , since (cid:54)∈ w we must have B = ∅ (otherwise A s overlaps all the B -intervals in B , a contradiction with w = 13 ). The neighbors of a s in B belong to β ∪ β , thus to β which is not empty by definition (otherwise no B -partner exists). The reasoning is similar for the othercases. ( ii ) If N w ( a s ) = N w ( a r ) , then we are done. Otherwise, assume w.l.o.g. that a s and a r are in this orderfrom left to right (and both are included in R i , by hypothesis). By Remark 2 applied to each of a s , a r , wethen have, using only the left sides of the placement conditions and the property that all the origins in Y areprecisely placed on one side or the other of a s (respectively a r ): ( Y ∩ N w ( a s )) ∪ ( T ( X ) − T ( R b ( a s )) − T ( N w ( a s ))) ⊆ ( Y ∩ N w ( a r )) ∪ ( T ( X ) − T ( R b ( a s )) − T ( N w ( a r ))) , which implies N w ( a s ) ∩ Y ⊆ N w ( a r ) ∩ Y . Since some origins in T ( X ) are not precisely placed withrespect to a s and a r , the similar deduction is not possible for T ( X ) . Using the right side of the placementcondition for a s , to the right of a s we have ( T ( X ) ∩ T ( N w ( a s ))) ∪ T ( R b → ( a s )) , where R b → ( a s ) is thesubset of origins in R b ( a s ) situated on the right side of a s . Similarly, to the right of a r we have the set oforigins ( T ( X ) ∩ T ( N w ( a r ))) ∪ T ( R b → ( a r )) . Then ( T ( X ) ∩ T ( N w ( a s ))) ∪ T ( R b → ( a s )) ⊇ ( T ( X ) ∩ T ( N w ( a r ))) ∪ T ( R b → ( a r )) . Now, the unions in the previous equations are disjoint unions, since R b ( a s ) ∩ N w ( a s ) = ∅ and R b ( a r ) ∩ N w ( a r ) = ∅ by Remark 2. Thus by removing the subset T ( R b → ( a s )) ∪ T ( R b → ( a r )) from both sides weobtain: ( T ( X ) ∩ T ( N w ( a s ))) − T ( R b → ( a r )) ⊇ ( T ( X ) ∩ T ( N w ( a r ))) − T ( R b → ( a s )) . By the definition of R b → ( a s ) and R b → ( a s ) and recalling that the tips of the neighbors of a s (respectively a r ) are situated to the right of a s ( a r ), we further deduce that: ( T ( X ) ∩ T ( N w ( a s ))) − T ( R b → ( a r )) = ( T ( X ) ∩ T ( N w ( a s ))) − T ( R b ( a r )) , and6 T ( X ) ∩ T ( N w ( a r ))) ∪ T ( R b → ( a r )) = ( T ( X ) ∩ T ( N w ( a r ))) − T ( R b → ( a s )) so that: ( T ( X ) ∩ T ( N w ( a s ))) − T ( R b ( a r )) ⊇ ( T ( X ) ∩ T ( N w ( a r ))) − T ( R b ( a s )) . When a s in placed on the right side of a r , the opposite inclusions hold. A B w -partner A p and a B w -partner A q of A v , A t are called B i -overlapping for some i ∈ w ∩ w if ( N ( A p ) − N ( A q )) ∩ B i (cid:54) = ∅ and ( N ( A q ) − N ( A p )) ∩ B i (cid:54) = ∅ . Proposition 3. If B (cid:54) = ∅ , then no Stick representation of G can represent A v , A t , with T ( a v ) ≺ T ( a t ) , anda pair of partners given by one of the following cases ( w = w is accepted): ( a ) a B w -partner and a B w -partner with w , w ∈ { , } that are B -overlapping. ( b ) a B w -partner and a B w -partner with w , w ∈ { , } that are B -overlapping. Proof.
Let w , w ∈ { , } . In either configuration (C1) or (C2), Table 2 ensures that the same region, R in (C1) or R in (C2), contains the A -origins of both B w - and B w -partners, since B (cid:54) = ∅ . Proposi-tion 2 ( ii ) then implies their neighborhoods are included in each other, so they cannot be B -overlapping.Let now w , w ∈ { , } . Again by Proposition 2 ( ii ) , in order to avoid comparable B -neighbor-hoods, the A -origins of the two partners must belong to different regions. In configuration (C1), one of thepartners, say A s , must then have its A -origin a s in T ( β ) ∪ β . But then the B -neighborhood of A s is B (according to the necessary conditions, both for w = 123 and w = 12 , N w ( a s ) ⊇ β ), and thus thisneighborhood cannot overlap the B -neighborhood of the other partner, a contradiction. The reasoning issimilar for (C2).In the next propositions, we consider the graph G = ( A ∪ B, E ) for which we seek a Stick representation.Then, the order of the tips is unknown. Given two vertices A v , A t of G we keep the previous notations,but replace the term partner (that was assuming a tip T ( a s ) on the left side of T ( a t ) ) by the term mate (which assumes nothing about the order of the tips). Thus we denote again B = N ( A t ) \ N ( A v ) , B = N ( A v ) ∩ N ( A t ) , B = N ( A v ) \ N ( A t ) . A B w - mate A s of A v , A t is an A -interval such that N ( A s ) ∩B i (cid:54) = ∅ iff i ∈ w . A B w -mate A p and a B w -mate A q of A v , A t are B i -overlapping for some i ∈ w ∩ w if ( N ( A p ) − N ( A q )) ∩ B i (cid:54) = ∅ and ( N ( A q ) − N ( A p )) ∩ B i (cid:54) = ∅ . Proposition 4.
Let G = ( A ∪ B, E ) be a bipartite graph. Assume that, for each vertex A t ∈ A , thereexists a vertex A v ∈ A such that the pair ( A t , A v ) satisfies B (cid:54) = ∅ and owns two mates verifying one of thefollowing conditions ( w = w is accepted): ( i ) a B w -mate and a B w -mate with w , w ∈ { , } that are B -overlapping. ( ii ) a B w -mate and a B w -mate with w , w ∈ { , } that are B -overlapping.Then G is not a Stick graph. Proof.
Assume by contradiction that G is a Stick graph, and assume as before that the vertices arenumbered A , . . . , A n in increasing order of their tips. Then T ( A n ) is the rightmost tip. By hypothesis,there exists a vertex A v such that A n , A v possess mates satisfying one of the properties ( i ) and ( ii ) . Thenall these mates are partners of A n , A v , since T ( a n ) is the rightmost tip and thus all these neighbors havetheir tips on its left side. But this contradicts Proposition 3.7 (cid:2) (cid:1) (cid:3) (cid:1)(cid:1)(cid:1) (cid:4) (cid:1) (cid:1) (cid:5) (a) (cid:1) (cid:2) (cid:1) (cid:3) (cid:1) (cid:4) (cid:1) (cid:5) (cid:1) (cid:6) (cid:1) (cid:7) (cid:1) (cid:8) (cid:1) (cid:9) (cid:1) (cid:10) (b) (cid:1)(cid:1) (cid:2) (cid:1) (cid:3) (cid:1) (cid:4) (cid:1) (cid:5) (cid:1) (cid:6) (cid:1) (cid:2)(cid:7)(cid:3) (cid:1) (cid:2)(cid:7)(cid:4) (cid:1) (cid:2)(cid:7)(cid:5) (cid:8) (cid:3) (cid:8) (cid:4) (cid:8) (cid:5) (cid:8) (cid:6) (cid:8) (cid:2)(cid:7)(cid:4) (cid:8) (cid:2)(cid:7)(cid:3) (cid:8) (cid:2) (cid:1) (cid:4)(cid:2) (cid:8) (cid:2)(cid:9)(cid:3)(cid:2)(cid:9)(cid:3) (cid:8) (cid:2)(cid:9)(cid:4) (cid:1) (cid:2)(cid:9)(cid:4) (cid:8) (cid:4)(cid:2) (cid:1) (cid:4)(cid:2)(cid:7)(cid:3) (cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1) (c) Graph J k , for odd k only. Figure 2: Forbidden subgraphs of Stick Graphs (I).The obstructions identified in the previous proposition allow to build examples of graphs that are notStick graphs.
Example 2.
The graph in Figure 2(a) was proposed in [13] and has the property that each pair of vertices A v , A t admits two B -mates that are B -overlapping. Proposition 4 thus applies with case ( ii ) for allvertices A t ∈ A . Example 3.
The graph in Figure 2(b) is not a Stick graph either. For each vertex A t ( t = 1 , , ) andeach vertex A v such that N ( A t ) ∩ N ( A v ) (cid:54) = ∅ there exist two disjoint paths with four edges joining A t and A v . The two A -origins in these paths, which are not endpoints, are B -neighbors of A v , A t which are B -overlapping. Proposition 4 thus applies with case ( i ) for all vertices A t ∈ A . Note however that thisgraph is not a minimal non-Stick graph, since the induced subgraph obtained by removing the vertices ofthe outer cycle C and their incident edges is not a Stick graph either (the proof is done by a case study). Example 4.
Consider the family of graphs J k , for odd k only, in Figure 2(c). For each vertex A t of J k ,the pair made of A t and the vertex A v following A t in the clockwise direction satisfies B = { B v } (cid:54) = ∅ and B = { B t , B t + k } . Thus, with the notation A p for the A -vertex following A t in the counterclockwisedirection, the B -partners A p and A p + k of A v , A t are B -overlapping. By Proposition 4 ( i ) , J k is not a Stickgraph. The abovementioned local obstructions allow to build forbidden subgraphs of Stick graphs, as proved above.However, the graphs proposed in the previous section, as well as those proposed in [5] - invariably containan induced cycle on six vertices (a C ). The following questions arise then (a hole is an induced cycle with h ≥ vertices): Q1.
Are there any C -free forbidden subgraphs of Stick graphs? Q2.
More generally, are there any hole-free forbidden subgraphs of Stick graphs?Both answers are positive. We prove it only for the more general question Q2.
Proposition 5.
Question Q2 has a positive answer, as shown by the graph
P y in Figure 3(a), which isbipartite, hole-free and not a Stick graph. (cid:2) (cid:3) (cid:2) (cid:3)(cid:4) (cid:5) (cid:3)(cid:4) (cid:6) (cid:3)(cid:4) (cid:7) (cid:3)(cid:4) (cid:8) (cid:3) (cid:5) (cid:3) (cid:6) (cid:3) (cid:7) (cid:3) (cid:8) (cid:1) (cid:5) (cid:1) (cid:6) (cid:1) (cid:7) (cid:1) (cid:8) (cid:1) (cid:9) (a) 2-pyramid P y (b ) (a' ) b a a a a (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (a' ) (b ) (b ) (b ) (b ) (b )
22 4 (b ) (b ) (b) Possible Stick representations (without A and B ) Figure 3: Forbidden subgraphs of Stick graphs (II).
Proof.
For this proof we use the classical 2-dimensional Stick representation. The graph
P y is bipar-tite, since each edge goes from vertex set A to vertex set B . It is also hole-free, since the graph inducedby all the vertices except A , B is hole-free, and B (respectively A ) is adjacent to all the vertices in A (respectively in B ).If, by contradiction, a Stick representation exists for P y , then it is symmetrical with respect to A , A ,A , A and we may assume w.l.o.g. that b ≺ a ≺ a ≺ a ≺ a . Then the segments B and A (resp. A , resp. A , resp A ) intersect as in Figure 3(b) (thick lines). The segment B has two possible positions,denoted as ( b ) and ( b ) in the figure, and the same holds for B , B , B in this order. Note that theorder of the B -origins b , ( b ) , ( b ) , ( b ) , ( b ) cannot be modified since if ( b j ) goes below ( b k ) with k < j (respectively below b ) then B j (placed in position ( b j ) ) intersects A k (respectively A ), a contradiction(except if ( b ) goes below b but in this case ( b ) and ( b ) represent the same case for the position of B ).For each possible position of B i , ≤ i ≤ , the possible positions of A (cid:48) i are shown in Figure 3(b) and aredenoted ( a (cid:48) hi ) , ≤ h ≤ . Note that position ( a (cid:48) i ) may be chosen for A (cid:48) i only if B i is in position ( b i ) andnone of B i +1 , . . . , B is respectively in position ( b i +1 ) , . . . , ( b ) , otherwise B i intersects an A j , j ≥ i + 1 ,that it should not intersect.In order to be able to place B (respectively A ) in the Stick representation, a horizontal segment mustexist that intersects all the A -segments (respectively B -segments). We show that we cannot choose thepositions of B i and A (cid:48) i , ≤ i ≤ , so that both B and A may be placed.9 ase 1. If position ( b ) is chosen for B , then either position ( a (cid:48) ) is chosen for A (cid:48) , or B is in position ( b ) , so that position ( a (cid:48) ) is chosen for A (cid:48) . The latter case implies that the horizontal segments B and B are separated by the vertical segment A , and thus A cannot intersect both of them. We continue withthe former case. Since a horizontal segment must intersect A (cid:48) and A (cid:48) , we deduce that the position of ( A (cid:48) ) is ( a (cid:48) ) (otherwise B separates A (cid:48) and A (cid:48) ) and the position ( b ) cannot be chosen for B (since then B separates A (cid:48) and A (cid:48) ). Thus B is position ( b ) and therefore A (cid:48) is in position ( a (cid:48) ) or ( a (cid:48) ) . But then B separates A (cid:48) and A (cid:48) implying that no horizontal segment can intersect both A (cid:48) and A (cid:48) , thus B cannot beplaced. Case 2.
If position ( b ) is chosen for B , we have again two cases. If B is in position ( b ) , then A separates B (independently of its position) and B , so A cannot be placed. If B is in position ( b ) , thesegment B must be in position ( b ) and must be as long as needed to avoid the same separation of B and B by A (otherwise A separates again B and B ). But then if A (cid:48) is in position ( a (cid:48) ) , B separates A (cid:48) and A (cid:48) (which must be in position ( a (cid:48) ) or ( a (cid:48) ) ). As a consequence, A (cid:48) can be only in position ( a (cid:48) ) ,implying that B is in position ( b ) (otherwise, with B is in position ( b ) , the segment B intersects A before intersecting A (cid:48) ). But then B and B are separated by A , thus A cannot be placed. Remark 3.
In the proof above, the contradictions are obtained using only the hypothesis that each pair ofvertices in { A (cid:48) , A (cid:48) , A (cid:48) , A (cid:48) } as well as each pair of vertices in { B , B , B , B } has a common neighbor.Thus replacing A , B by another set of vertices, and appropriate incident edges, ensuring that the aboveproperties hold allows us to build a family of non-Stick graphs (which may however be non-minimal withthis property, and usually contain holes). Remark 4.
Proposition 5 is equivalent to the affirmation that not all chordal bipartite graphs, defined ashole-free bipartite graphs, are Stick graphs. This result may also be deduced from [4], whose constructionof chordal bipartite graphs of arbitrarily high boxicity implies the existence of chordal bipartite graphswhich are not grid intersection graphs, and thus not Stick graphs. Note however that the graph
P y is botha chordal bipartite and a grid intersection graph, so that it cannot be obtained by the construction in [4]. Corollary 1.
The class of chordal grid intersection graphs and the class of Stick graphs are incomparable.
A class of graphs included in the class of Stick graphs should therefore contain none of the forbiddensubgraphs, and families of forbidden subgraphs, we proposed above.Interval graphs, circular arc and circle graphs are classical classes of intersection graphs (defined in theIntroduction). Interval bigraphs, defined in Section 3, are sometimes mistaken for bipartite interval graphs,which are a strict subclass of interval bigraphs (the graph C shows the difference). We first show that: Theorem 1.
The graphs that are both bipartite and complements of circular arc graphs are Stick graphs.
Proof.
Let G = ( A ∪ B, E ) be a graph satisfying the hypothesis. Then its complement G is a circular arcgraph whose vertices are covered by two disjoint cliques, induced by A and B . As remarked in [9] based ona result from [15], since G is a circular arc graph with clique cover number two then it has a representationas follows. The circle C has top point p , rightmost point r , bottom point q and leftmost point s . The familyof arcs of the circle representing G is K = { S v | v ∈ A } ∪ { T w | w ∈ B } such that: each arc S v contains s and q but neither p nor r ; each arc T w contains p and r but neither s nor q ; arcs S v and T w intersect iff vw is an edge of G . The endpoints of the arcs may be considered as distinct.10enote C and C respectively the top half of the circle (containing p ) and the bottom half of the circle(containing q ). For each arc ( xy ) := S v such that x ∈ C and y ∈ C , let y (cid:48) be the point of the circle that issituated symmetrically to y with respect to the horizontal line sr . For each arc ( zt ) := T w such that z ∈ C and t ∈ C , let t (cid:48) be the point of the circle that is situated symmetrically to t with respect to the horizontalline sr . Then x, z, y (cid:48) , t (cid:48) ∈ C . We show that the set of intervals A = { [ x, y (cid:48) ] | ( xy ) ∈ S v , x ∈ C , y ∈ C } and B = { [ z, t (cid:48) ] | ( zt ) ∈ T w , z ∈ C , t ∈ C } are a flat Stick representation of G (the half-circle C is thenassumed to represent the horizontal line).Let ( xy ) ∈ S v and ( zt ) ∈ T w be two arcs representing the vertices v and w of G . Case 1. ( xy ) and ( zt ) do not intersect. Then s, x, z, p, r, t, y, q appear in this order in the clockwisedirection along C , so that in C the same direction but using now y (cid:48) and t (cid:48) gives the order s, x, z, p, y (cid:48) , t (cid:48) , r .The intervals [ x, y (cid:48) ] and [ z, t (cid:48) ] overlap with [ x, y (cid:48) ] to the left of [ z, t (cid:48) ] . Case 2. ( xy ) and ( zt ) intersect only between s and p . Then s, z, x, p, r, t, y, q appear in this order in theclockwise direction along C , and we deduce that s, z, x, p, y (cid:48) , t (cid:48) , r appear in this order in C in the clockwisedirection. The interval [ x, y (cid:48) ] is contained in the interval [ z, t (cid:48) ] . Case 3. ( xy ) and ( zt ) intersect only between q and r . Then s, x, z, p, r, y, t, q appear in this order in theclockwise direction along C , so that the order using y (cid:48) and t (cid:48) is now s, x, z, p, t (cid:48) , y (cid:48) , r . The interval [ z, t (cid:48) ] iscontained in the interval [ x, y (cid:48) ] . Case 4. ( xy ) and ( zt ) intersect both between s and p , and between q and r . Then s, z, x, p, r, y, t, q appear in this order in the clockwise direction along C . We deduce that the order using y (cid:48) and t (cid:48) is s, z, x, p, t (cid:48) , y (cid:48) , r . The interval [ z, t (cid:48) ] intersects the interval [ x, y (cid:48) ] , with [ z, t (cid:48) ] to the left of [ x, y (cid:48) ] .In conclusion, the order x, z, y (cid:48) , t (cid:48) occurs iff ( xy ) and ( zt ) do not intersect (in case 1), which happensiff in G the vertices v and w corresponding respectively to ( xy ) and ( zt ) are not adjacent. But that means [ x, y (cid:48) ] and [ z, t (cid:48) ] overlap (in this order from left to right) iff v, w are adjacent in G , thus we have a flat Stickrepresentation of G . Corollary 2.
Interval bigraphs and bipartite interval graphs are Stick graphs.
Proof.
By definition, interval bigraphs are bipartite graphs. Moreover, their complements are circulararcs graphs [12], so the affirmation results from Theorem 1. Furthermore, it follows easily by definition thatbipartite interval graphs are interval bigraphs.
Remark 5.
In [14], the authors study the 2-directional orthogonal ray graphs (2DOR graphs) defined as theintersection graphs of horizontal and vertical rays, which are half-lines in the plane that extend rightwardand respectively upward to infinity. They show - using properties of the adjacency matrices - that 2DORgraphs are exactly the bipartite complements of circular arc graphs. On the other side, 2DOR graphs areStick graphs [5], which implies an alternative - but indirect - proof for Theorem 1 and its corollary.
Theorem 2.
The graphs that are both bipartite and complements of circle graphs are Stick graphs.
Proof.
In [8], the authors give an alternative proof of a result by Bouchet [1]. This result states thatevery graph that is a bipartite complement of a circle graph is also a circle graph. Given a graph G =( A ∪ B, E ) which is a bipartite complement of a circle graph, the proof of Theorem 1 in [8] shows that G isthe intersection graph of a set of chords with pairwise distinct endpoints in a circle C and such that all thechords intersect a line l . The line l may be assumed horizontal such that p is the leftmost point of the circleand q its rightmost point. Assume G is connected (otherwise make the same reasoning for each connectedcomponent) and let S A , S B be the sets of chords corresponding to the vertices in A and B . Then each chord [ xy ] in S A ( [ zt ] in S B ) has its endpoint x (respectively z ) between p and q in the top half of the circle C , and11ts endpoint y (respectively t ) in the bottom half of the circle C . W.l.o.g. we also assume that the endpoint y or t closest to p is the endpoint y of a chord [ x y ] .We show that: ( P ) Each pair of intersecting chords [ xy ] ∈ S A and [ zt ] ∈ S B has the property that p, z, x, q, t, y, p are inthis order along C in the clockwise direction.Note that [ x y ] and each chord intersecting it satisfy this property. If, by contradiction, the property isfalse, then there exists a pair of intersecting chords [ x (cid:48) y (cid:48) ] ∈ S A and [ z (cid:48) t (cid:48) ] ∈ S B such that p, x (cid:48) , z (cid:48) , q, y (cid:48) , t (cid:48) , p are in this order along C in the clockwise direction. Since G is connected, there exists: • either a sequence [ x y ] , [ z t ] , [ x y ] , [ z t ] , . . . , [ x u y u ] , [ z u t u ] of chords with [ x u y u ] = [ x (cid:48) y (cid:48) ] and [ z u t u ] = [ z (cid:48) t (cid:48) ] • or a sequence [ x y ] , [ z t ] , [ x y ] , [ z t ] , . . . , [ x u y u ] , [ z u t u ] , [ x u +1 y u +1 ] of chords with [ x u +1 y u +1 ] =[ x (cid:48) y (cid:48) ] and [ z u t u ] = [ z (cid:48) t (cid:48) ] .In both cases, we may assume that the sequence is as short as possible between [ x , y ] and a pair ofchords [ x (cid:48) y (cid:48) ] and [ z (cid:48) t (cid:48) ] with a bad order of endpoints along C . Now, in the former case we have that p, z u − , x u , q, t u − , y u , p (on the one hand) and p, x u , z u , q, y u , t u , p (on the other hand) are in this orderalong C , which implies that p, z u − , x u , z u , q, t u − , y u , t u , p are in this order along C and thus [ z u , t u ] and [ z u − , t u − ] intersect, a contradiction. In the latter case, we similarly have that p, z u , x u , q, t u , y u , p (on theone hand) and p, x u +1 , z u , q, y u +1 , t u , p (on the other hand) are in this order along C , which implies that p, x u +1 , z u , x u , q, y u +1 , t u , y u , p are in this order along C and thus [ x u , y u ] and [ x u +1 , y u +1 ] intersect, acontradiction.Now, let q l and q r be two points of C situated immediately before and immediately after q in clockwisedirection, so that no endpoint of a chord exists between q l and q r . Cut the circle at the point q and unfoldit such that it becomes a straight horizontal line with left endpoint q l and right endpoint q r . Each chord in S A (respectively S B ) becomes an interval in A (respectively B ) on this line (which represents a segment ofthe real line). Then, using Property ( P ) , each pair of intersecting intervals [ x, y ] ∈ A and [ z, t ] ∈ B has theproperty that q l , x, z, p, y, t, q r are in this order along the real line. Indeed, the intersecting chords becameintersecting intervals, and non-intersecting chords became intervals included in each other. Therefore theintervals in A and B provide a flat Stick representation for G . Stick graphs are a challenging class of graphs, for which holes play a crucial role in the construction offorbidden subgraphs. In this paper, we have proposed several examples of graphs and families of graphssupporting this idea, but we have also found a forbidden subgraph with no hole. We have also shown that- when restricted to bipartite graphs, since Stick graphs are bipartite - several classical classes of graphs ortheir complements are included in the class of Stick graphs.These progresses allow us to much better understand the frontiers of the class of Stick graphs. However,the list of minimal forbidden subgraphs is not complete (nor characterized in some way) and other subclassesof Stick graphs may exist. Characterizing Stick graphs, either by forbidden subgraphs or in alternate waysinvolving structural properties, remains an open problem, which may be the first step towards finding thecomplexity of the STICK problem, which is open too.12 eferences [1] Andr´e Bouchet. Bipartite graphs that are not circle graphs. In
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