1 Introduction
Graph classes defined as intersection graphs of geometric objects have been extensively studied. Intersection graphs 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. Many of 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 concerns only horizontal and vertical segments, yielding applications identified in [5]. In the same paper, several subclasses of grid intersection graphs are proposed and their intersections are studied. One of them is the class of Stick graphs.
A Stick graph is the intersection graph of a set of vertical segments and a set of horizontal segments in the plane (respectively called  and segments), whose bottom and respectively left endpoints lie on a “ground” line with slope 1. Each of these endpoints is called the origin (which is then an origin or origin) of the corresponding segment. It is assumed that all the origins are distinct (otherwise, one may appropriately move some of them close to their initial position, and modify some segment lengths, without affecting the segment intersections).
Stick graphs, as all the grid intersection graphs, are bipartite graphs. Given a Stick graph, a representation of it with vertical and horizontal segments as mentioned above is called a Stick representation of the graph. The problem of recognizing Stick graphs  denoted STICK  requires, given a bipartite graph , to decide whether has a Stick representation or not. STICK is an open problem. Variants of STICK, assuming that the order of the origins and/or the order of the origins are known, have been considered and solved in [13, 6], including in the case where the lengths of the segments are known. However, very little is known about Stick graphs in general. Trivial examples of Stick graphs exist, namely chordless cycles and paths, as well as trees. Bipartite permutation graphs, defined as the intersection graphs of segments with one endpoint on each of two given parallel lines such that the resulting graph is bipartite, are also Stick graphs (deduced from the definition). At the opposite, finding examples of graphs that are not Stick graphs needs 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 investigate Stick graphs  and therefore also nonStick 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 and wellknown 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 3 and 4, we identify local properties of a Stick representation and deduce particular configurations of graphs that act as obstructions to such a representation. Using them, we build several examples of graphs or families of 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. an induced cycle with at least five vertices (in reality, only holes with an even number of vertices, thus at least six, will appear in bipartite graphs). However, in Section 5 we provide an example of a forbidden subgraph without any hole. At the opposite of these concerns related to nonStick graphs, in Section 6 we focus on conditions that make that a graph is a Stick graph and we identify particular classes of Stick graphs, namely bipartite 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 the conclusion.
Related work. Our results were obtained by investigating open questions, yet it appears that  while studying 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 and mentioned the alternative proof (see Remark 5).
2 Definitions and notations
All the graphs we use are undirected and simple, and the notations are classical. The bipartite graph for which a Stick representation is searched is always denoted , with and . The vertices in , also called vertices, as well as the segments representing them in any (attempt of) Stick representation, are denoted , , and similarly for the vertices and segments, denoted , . Moreover, denote the point where the horizontal line going through the top endpoint of the segment meets the ground line. Similarly, denote the point where the vertical line going through the rightmost endpoint of the segment meets the ground line. Each of these points is called the tip of its corresponding segment, and is an  or tip according to the type of the segment. As initially done for the origins only, we may consider that all the origins and the tips are distinct.
For any two points of the ground line, denote if is above . The following observation is easy (and similar to those made for other classes, e.g. for maxpoint tolerance graphs [3]):
Observation. Segments and intersect in a Stick representation iff , i.e. iff intervals and 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 of two sets of intervals, versus , of the real line.
Conventions. Throughout the paper, with only one exception, we use this flat Stick representation, where the 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 only when .
Example 1.
For , the intervals in are called intervals. The origins of intervals in are called origins. Since each vertex is identified with its interval, we transfer some of the graph terminology and notations to intervals. Two intervals are adjacent if . We also say they are neighbors. The neighborhood of an interval is the set of all its neighbors, denoted . In order to avoid supplementary terminology, all these notions and notations (adjacency, neighborhood) are extended to the origins of the intervals, i.e. for instance .
In the next section, assuming that has a Stick representation, we renumber the intervals as such that iff . Let and denote , , . Furthermore, denote and for each set of origins and origins.
Interval is a partner of interval if and they have at least one common neighbor. Then we also say that the origin is a partner of the origin . Let be a partner of or , which implies that . Let be a subset of denoted as a word (e.g. for ), and denote . Interval is called a partner of (both) and if it satisfies the property that if and only if . Similarly, is a partner of (both) and . By definition, exactly one exists for each such that is a partner of and . Moreover, (respectively ) is a (trivial) (respectively ) partner of and . Finally, denote .
3 General properties
In this section, we investigate necessary conditions for to be a Stick graph.
Proposition 1.
Configurations (C1), (C2) and (C2’) in Table 1 are the only possible configurations for the origins and tips of the intervals in .
Proof. Consider first the possibilities to have . Sets and must be between and since , whereas (respectively ) must be placed between and (respectively after ) in order to respect the sought adjacencies and nonadjacencies. Now, since , thus the origins may be placed everywhere before except between and (otherwise would have neighbors in ). Configuration (C1) is then the only one possible when .
The reasoning is similar when , with reversed roles for and , except that some origins could be placed between and . Their corresponding tips must then be placed after , that is, either between and , or after . Three types of intervals are then possible, yielding subsets and . This gives configuration (C2). However, when , may be placed between the two tips and , yielding configuration (C2’).
(C1)  

(C2)  
(C2’) 
Convention. As configuration (C2’) is highly similar to configuration (C2), in the remaining of the paper we explicitly consider only configurations (C1) and (C2), and assume (C2’) behaves as (C2) in case , with and reversed.
With the aim of identifying the possible positions for the partners of and , we indicated in Table 1 four regions. Each region is understood as a set of origins and/or tips, so that for some and . For instance, when in (C2), we have and . Placing in
means  for the moment  that the origins and tips currently in
may be distributed as needed, and may be inserted before or after each element in , as long as the borders and are not crossed. No difference exists between the placement of just before or just after (the same holds for , or ), so that when possible we avoid this redundancy. For a given , a partner may be placed in some region 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 in such that can overlap intervals from each with . We may see these conditions as general, contextual, constraints.

the placement conditions ensure  in a region validated by the partnership conditions  the existence of a precise position for such that the interval overlaps all its neighbors in the bipartite graph .
Remark 1.
Partners of and with overlap each neighbor of or such that since . Partners of and with satisfy the same property for all with and . When , one may have implying that and do not overlap (recall that by convention we use the term overlap only for overlaps with ).
Denote , the set of good origins in with respect to , which are the origins not concerned by the exception in the previous remark. Then is the set of bad origins of .
Remark 2.
Concerning the placement conditions for , note first that must hold, and thus . Consider now a region (), where each of or (but not both) may be empty, and . The position of in must satisfy the following properties:

the neighbors of in and the tips of its nonneighbors in are situated on the left side of

the nonneighbors of in and the tips of its neighbors in are situated on the right side of .
The tips of the nonneighbors of in are not precisely placed with respect to , they may be situated either on its left or on its right side. In summary, the placement condition for is:
so that can be placed between the two sets above.
In Table 2, we list all the possible positions of partners , for all , in configurations (C1) and (C2) as obtained from Proposition 2 below. In this table, the columns respectively indicate the set , and for each configuration (C1) and (C2): the regions where may be placed, according to the notations in Table 1, and the partnership conditions that must be satisfied to allow the placement of in each region.
(C1)  (C2)  

R  Partnership conditions  R  Partnership conditions  
or  or , and  
or , and  
or  or , and  
or , and  
or  or  
Proposition 2.
Let be a subset of and (C) be a configuration among (C1) and (C2). The following properties hold:

Each partner of and , except and themselves, must be placed in at least one of the regions identified in Table 2 for and ; in this case, the conditions identified in the table for each region must be satisfied.

For every pair of partners of and that are placed in the same region ), either:
and , or
and .
Proof. Recall that , thus . 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 interval from each , for all ; and (2) for each such region, find the conditions that must be satisfied outside that region (i.e. the partnership conditions) in order to ensure that belongs to for the value of under study.
As an example, we give the proof for and (C1). Regions and proposed in Table 1 are the only possible ones, since placing before or after would prevent any adjacency between and an interval in . When is placed in region , is the only source of neighbors in for , so we must have . Moreover, , thus overlaps all the intervals with good origins with respect to (by Remark 2 the origins of the intervals that must overlap are necessarily good origins). When is placed in region , since we must have (otherwise overlaps all the intervals in , a contradiction with ). The neighbors of in belong to , thus to which is not empty by definition (otherwise no partner exists). The reasoning is similar for the other cases.
If , then we are done. Otherwise, assume w.l.o.g. that and are in this order from left to right (and both are included in , by hypothesis). By Remark 2 applied to each of , we then have, using only the left sides of the placement conditions and the property that all the origins in are precisely placed on one side or the other of (respectively ):
which implies . Since some origins in are not precisely placed with respect to and , the similar deduction is not possible for . Using the right side of the placement condition for , to the right of we have , where is the subset of origins in situated on the right side of . Similarly, to the right of we have the set of origins . Then
Now, the unions in the previous equations are disjoint unions, since and by Remark 2. Thus by removing the subset from both sides we obtain:
By the definition of and and recalling that the tips of the neighbors of (respectively ) are situated to the right of (), we further deduce that:
so that:
When in placed on the right side of , the opposite inclusions hold.
4 Obstructions and first examples of forbidden subgraphs
A partner and a partner of are called overlapping for some if and .
Proposition 3.
If , then no Stick representation of can represent , with , and a pair of partners given by one of the following cases ( is accepted):

a partner and a partner with that are overlapping.

a partner and a partner with that are overlapping.
Proof. Let . In either configuration (C1) or (C2), Table 2 ensures that the same region, in (C1) or in (C2), contains the origins of both  and partners, since . Proposition 2 then implies their neighborhoods are included in each other, so they cannot be overlapping.
Let now . Again by Proposition 2, in order to avoid comparable neighborhoods, the origins of the two partners must belong to different regions. In configuration (C1), one of the partners, say , must then have its origin in . But then the neighborhood of is (according to the necessary conditions, both for and , ), and thus this neighborhood cannot overlap the neighborhood of the other partner, a contradiction. The reasoning is similar for (C2).
In the next propositions, we consider the graph for which we seek a Stick representation. Then, the order of the tips is unknown. Given two vertices of we keep the previous notations, but replace the term partner (that was assuming a tip on the left side of ) by the term mate (which assumes nothing about the order of the tips). Thus we denote again , , . A mate of is an interval such that iff . A mate and a mate of are overlapping for some if and .
Proposition 4.
Let be a bipartite graph. Assume that, for each vertex , there exists a vertex such that the pair satisfies and owns two mates verifying one of the following conditions ( is accepted):

a mate and a mate with that are overlapping.

a mate and a mate with that are overlapping.
Then is not a Stick graph.
Proof. Assume by contradiction that is a Stick graph, and assume as before that the vertices are numbered in increasing order of their tips. Then is the rightmost tip. By hypothesis, there exists a vertex such that possess mates satisfying one of the properties and . Then all these mates are partners of , , since is the rightmost tip and thus all these neighbors have their tips on its left side. But this contradicts Proposition 3.
The obstructions identified in the previous proposition allow to build examples of graphs that are not Stick graphs.
Example 2.
Example 3.
The graph in Figure 2(b) is not a Stick graph either. For each vertex () and each vertex such that there exist two disjoint paths with four edges joining and . The two origins in these paths, which are not endpoints, are neighbors of which are overlapping. Proposition 4 thus applies with case for all vertices . Note however that this graph is not a minimal nonStick graph, since the induced subgraph obtained by removing the vertices of the outer cycle 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 , for odd only, in Figure 2(c). For each vertex of , the pair made of and the vertex following in the clockwise direction satisfies and . Thus, with the notation for the vertex following in the counterclockwise direction, the partners and of are overlapping. By Proposition 4, is not a Stick graph.
5 Stick graphs and holes
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 contain an induced cycle on six vertices (a ). The following questions arise then (a hole is an induced cycle with vertices):
Q1. Are there any free forbidden subgraphs of Stick graphs?
Q2. More generally, are there any holefree 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 in Figure 3(a), which is bipartite, holefree and not a Stick graph.
Proof. For this proof we use the classical 2dimensional Stick representation. The graph is bipartite, since each edge goes from vertex set to vertex set . It is also holefree, since the graph induced by all the vertices except is holefree, and (respectively ) is adjacent to all the vertices in (respectively in ).
If, by contradiction, a Stick representation exists for , then it is symmetrical with respect to and we may assume w.l.o.g. that . Then the segments and (resp. , resp. , resp ) intersect as in Figure 3(b) (thick lines). The segment has two possible positions, denoted as and in the figure, and the same holds for , , in this order. Note that the order of the origins cannot be modified since if goes below with (respectively below ) then (placed in position ) intersects (respectively ), a contradiction (except if goes below but in this case and represent the same case for the position of ). For each possible position of , , the possible positions of are shown in Figure 3(b) and are denoted , . Note that position may be chosen for only if is in position and none of is respectively in position , otherwise intersects an , , that it should not intersect.
In order to be able to place (respectively ) in the Stick representation, a horizontal segment must exist that intersects all the segments (respectively segments). We show that we cannot choose the positions of and , , so that both and may be placed.
Case 1. If position is chosen for , then either position is chosen for , or is in position , so that position is chosen for . The latter case implies that the horizontal segments and are separated by the vertical segment , and thus cannot intersect both of them. We continue with the former case. Since a horizontal segment must intersect and , we deduce that the position of is (otherwise separates and ) and the position cannot be chosen for (since then separates and ). Thus is position and therefore is in position or . But then separates and implying that no horizontal segment can intersect both and , thus cannot be placed.
Case 2. If position is chosen for , we have again two cases. If is in position , then separates (independently of its position) and , so cannot be placed. If is in position , the segment must be in position and must be as long as needed to avoid the same separation of and by (otherwise separates again and ). But then if is in position , separates and (which must be in position or ). As a consequence, can be only in position , implying that is in position (otherwise, with is in position , the segment intersects before intersecting ). But then and are separated by , thus cannot be placed.
Remark 3.
In the proof above, the contradictions are obtained using only the hypothesis that each pair of vertices in
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