1 Introduction
A rectangle intersection representation of a graph is a collection of axisparallel rectangles on the plane such that each rectangle in the collection represents a vertex of the graph and two rectangles intersect if and only if the vertices they represent are adjacent in the graph. The graphs that have rectangle intersection representation are called rectangle intersection graphs. The boxicity of a graph is the minimum such that is representable as a geometric intersection graph of dimensional (axisparallel) hyperrectangles. A graph is an interval graph if and is a rectangle intersection graph if .
A stabbed rectangle intersection representation is a rectangle intersection representation, along with a collection of horizontal lines called stab lines, such that every rectangle intersects at least one of the stab lines. A graph is a stabbable rectangle intersection graph (SRIG), if there exists a stabbed rectangle intersection representation of . The stab number of a rectangle intersection graph, denoted by , is the minimum integer such that there exists a stabbed rectangle intersection representation of . In other words is the minimum integer such that is SRIG. Clearly, if a graph has boxicity at most 2, then is finite. For graphs with boxicity at least three, we define .
A exactly stabbed rectangle intersection representation is a stabbed rectangle intersection representation in which every rectangle intersects exactly one of the stab lines. A graph is a exactly stabbable rectangle intersection graph, or ESRIG for short, if there exists a exactly stabbed rectangle intersection representation of . The exact stab number of a rectangle intersection graph, denoted by , is the minimum integer such that there exists a exactly stabbed rectangle intersection representation of . In other words, is the minimum integer such that is ESRIG. When a graph has no exactly stabbed rectangle intersection representation for any integer , we define . A graph with is said to be an exactly stabbable rectangle intersection graph. Note that for a graph , and that a graph is an interval graph if and only if , or in other words, the class of interval graphs, the class of 1SRIGs, and the class of 1ESRIGs are all the same.
For a subclass of rectangle intersection graphs, is the minimum integer such that any graph with vertices satisfies , and is the minimum integer such that for any graph with vertices satisfies . A unit height rectangle intersection graph is a graph that has a rectangle intersection representation in which all rectangles have equal height. It is wellknown that all unit height rectangle intersection graphs are exactly stabbable rectangle intersection graphs (for the sake of completion, we prove this in Theorem 5 in Section 3).
1.1 Motivation and related work
Boxicity of a graph has been an active field of research for many decades [17, 2, 8, 9, 10]. While recognizing graphs with boxicity at most is NPcomplete for all [19, 23], there are efficient algorithms to recognize interval graphs, i.e. graphs with boxicity at most 1 [21, 11]. There seems to be a “jump in the difficulty level” of problems as the boxicity of the input graph increases from 1 to 2. For example, the Maximum Independent Set and Chromatic Number problems, while being lineartime solvable for interval graphs, become NPcomplete for rectangle intersection graphs (even with the rectangle intersection representation given as input) [20, 18]. Our goal is to understand the reason of this jump by studying graph classes that lie “in between” interval graphs and rectangle intersection graphs. For this purpose, we introduce a parameter called stab number for rectangle intersection graphs. The concept of stab number is a generalization of the idea behind a class of graphs known as “2SIG”, which was introduced in an earlier paper [6]. Even though our definitions of 2SRIG and 2ESRIG are both slightly different from that of “2SIG”, all three classes of graphs turn out to be equivalent (Theorem 2 shows that the classes SRIG and ESRIG are equivalent for any ). A stabbed rectangle intersection representation of a graph involves rectangles and horizontal lines. Such combined arrangements of lines and rectangles have been popular topics of study in the geometric algorithms community. For example, such arrangements appear in the works of Agarwal et al. [3] and Chan [7], who gave approximation algorithms for the Maximum Independent Set problem in unit height rectangle intersection graphs, and also in a paper by Erlebach et al. [16], who proposed a PTAS for Minimum Weight Dominating Set for unit square intersection graphs. Correa et al. [12] have studied the problems of computing independent and hitting sets for families of rectangles intersecting a diagonal line.
1.2 Contributions and organization of the paper
In this paper, we introduce the notion of “stab number” of a rectangle intersection graph and study this parameter for various subclasses of rectangle intersection graphs. In Section 2, we give some definitions and notation that will be used throughout the paper. We prove some basic results about SRIGs and ESRIGs in Section 3. We first show a simple necessary and sufficient condition for a graph to be a ESRIG and also show why the classes SRIG and ESRIG are equivalent when (Theorem 2). Then we prove that the class of unit height rectangle intersection graphs is a proper subset of the class of rectangle intersection graphs with finite exact stab number (Theorem 3), which in turn is a proper subset of rectangle intersection graphs (Theorem 5). This leads us to the natural question of finding exactly stabbable graphs whose exact stab number is strictly greater than the stab number. We show that for each , there exist trees which are SRIG but not ESRIG (Theorem 36). Therefore, even for graphs that are exactly stabbable, like trees (Theorem 10), the stab number and the exact stab number may differ. We prove this result only in Section 6.3, after the machinery required for the proof is developed in Section 6. In Section 4, we show a lower bound on the stab number of rectangle intersection graphs in terms of the clique number and the pathwidth, and then study upper bounds on the stab number of rectangle intersection graphs that are also (a) split graphs, or (b) block graphs. In particular, we show (a) that all rectangle intersection graphs that are also split graphs have exact stab number at most 3 and that this bound is tight, and (b) an upper bound of on the exact stab number of block graphs with blocks (this bound is shown to be asymptotically tight in Section 6.1). Then in Section 5, we describe a forbidden structure for SRIG and ESRIG, which we call “asteroidal(non()SRIG)” subgraphs and “asteroidal(non()ESRIG)” subgraphs respectively. These obstructions are a natural generalization of the wellknown “asteroidaltriples” of Lekkerkerker and Boland [21], which are obstructions for interval graphs. In Section 5.2, we discuss some general properties possessed by the blocktrees of graphs without these kinds of obstructions. In Section 6, we show that the absence of these forbidden structures is enough to characterize block graphs that are 2ESRIG (Theorem 19) and trees that are 3ESRIG (Theorem 20). These results lead to polynomialtime algorithms to recognize block graphs that are 2SRIG and trees that are 3SRIG. In Section 6.2, we develop a geometric argument that allows us to show that this kind of forbidden structure is not sufficient to characterize block graphs that are 3SRIG (Theorem 23) or trees that are SRIG, for any (Theorem 24). We conclude by listing some open problems and suggesting some possible directions for further research on this topic.
2 Preliminaries
We present some definitions in this section. Let be a graph with vertex set and edge set . Let and denote the open neighbourhood and the closed neighbourhood of a vertex , respectively. For , we denote by the subgraph induced in by the vertices in , and by the graph obtained by removing the vertices in from . For an edge , we denote by the graph on vertex set having edge set .
Let be a rectangle intersection graph with rectangle intersection representation . A rectangle in corresponding to the vertex is denoted as . All rectangles considered in this article are closed rectangles. Denote by , the coordinate of the right (left) bottom corner of . Also is the coordinate of the left top (bottom) corner of . In other words, . The span of a vertex , denoted as , is the projection of on the axis, i.e. . For two intervals and , we write to indicate that . Clearly, if and only if or . For an edge , we define .
Let be a SRIG with a stabbed rectangle intersection representation in which the stab lines are , , , , where . The top (resp. bottom) stab line of is the stab line (resp. ). For , we say that is the stab line “just above” the stab line and that is the stab line “just below” the stab line . We also say that the stab lines and are “consecutive”. A vertex is said to be “on” a stab line if intersects that stab line. Two vertices of “have a common stab” if there is some stab line that intersects both and . Similarly, a set of vertices is said to have a common stab if there is one stab line that intersects the rectangles corresponding to each of them. It is easy to see that if , then there must be either a stab line such that and are on it or two consecutive stab lines such that is on one of them and is on the other. Whenever the stabbed rectangle intersection representation of a graph under consideration is clear from the context, the terms , , , , , for every vertex and usages such as “on a stab line”, “have a common stab”, “span” etc. are considered to be defined with respect to this representation. Clearly, both the classes SRIG and ESRIG are closed under taking induced subgraphs. We say that a graph is a nonSRIG (resp. nonESRIG) if it is not a SRIG (resp. ESRIG). Similarly, we say that a graph is a noninterval graph if it is not an interval graph.
3 Basic Results
Given a collection of intervals, a hitting set of is a subset of such that each interval in contains at least one element of . The set is an exact hitting set of if each interval in contains exactly one element of . An interval graph is said to have an exact hitting set of size if there exists an interval representation of that has an exact hitting set of cardinality . Note that some collections of intervals may not have an exact hitting set of any cardinality. Also, there are interval graphs (for example, ) that have no exact hitting set.
Theorem 1.
A graph is a ESRIG if and only if there exists two interval graphs and such that and and at least one of has an exact hitting set of size .
Proof.
First we prove that if has a ESRIG representation, then there exist two interval graphs and such that and and at least one of them has an exact hitting set of size . Let be a exactly stabbed rectangle intersection representation of and be the set of stab lines in . Let be the interval graphs formed by taking the projections of the rectangles in on the and axes, respectively. In other words, is the interval graph given by the interval representation and is the interval graph given by the interval representation . It is clear that and . Furthermore, the set is an exact hitting set of the interval representation of . Hence, has an exact hitting set of size .
Now assume that there exist two interval graphs and such that and and at least one of them, say , has an exact hitting set of size . Let be an exact hitting set of an interval representation of . Also, let be an interval representation of . For each , define . It is easy to see that is a rectangle intersection representation of . Further, the lines , , , are horizontal lines such that each rectangle in intersects exactly one of them. Hence, , together with these lines, is a exactly stabbed rectangle intersection representation of and therefore, is a ESRIG. This completes the proof. ∎
Theorem 2.
When , the classes SRIG and ESRIG are equivalent.
Proof.
If a graph is ESRIG for some , then is also SRIG. Therefore it suffices to prove that if a graph has a stabbed rectangle intersection representation for some , then also has a exactly stabbed rectangle intersection representation. If , then there is nothing to prove. So we shall assume that . Let be a stabbed rectangle intersection representation of a graph with with stab lines , , , . We can assume without loss of generality that for any two distinct vertices , we have and that for any vertex , we have (note that if this is not the case, then the rectangles in can be perturbed slightly so that these conditions are satisfied). Let and be a positive real number such that . Let intersects the stab line . For each vertex , define , where and . Let be the rectangle intersection representation given by the collection of rectangles . It is now easy to verify that is a exactly stabbed rectangle intersection representation of . Indeed, is obtained from by the vertical shortening of some of the rectangles intersecting the stab line , and we only need to show that every rectangle that is so shortened still intersects with all the rectangles with which it originally has an intersection. The definition of guarantees that in , the bottom edge of any rectangle is no higher than and the top edge of any rectangle is no lower than . So when a rectangle is shortened in the manner described above, it does not become disjoint from a rectangle with which it previously had a nonempty intersection. Therefore is a valid rectangle intersection representation of . It is clear that any rectangle that intersects the stab line in intersects only the stab line in . This implies that is a exactly stabbed rectangle intersection representation of . ∎
In the following theorem, we show that for , the classes SRIG and ESRIG differ.
Theorem 3.
There is a graph such that and .
Proof.
We let , i.e. the complete bipartite graph in which each partite set contains four vertices each. Clearly, is a rectangle intersection graph with (see Figure 1(a)). We shall prove that , or in other words, is not an exactly stabbable rectangle intersection graph. First we prove the following claim.
Claim. Let be a cycle of length four and . There is no exactly stabbed rectangle intersection representation of , for any integer , in which have a common stab and have a common stab.
Assume for the sake of contradiction that there is a exactly stabbed rectangle intersection representation of , for some integer , in which have a common stab and have a common stab. Clearly, cannot all be on one stab line (as is not an interval graph). Since every vertex is on exactly one stab line and because , we can assume without loss of generality that are on the stab line just below the stab line on which are. Since and are nonadjacent in , again without loss of generality we can assume that . Since , we can infer that . Similarly, we can show that . But this implies that . Since are on the same stab line, this means that . As , this contradicts the fact that is a rectangle intersection representation of . This proves the claim.
Now suppose that has a exactly stabbed rectangle intersection representation for some . Let be the two partite sets of (recall that is isomorphic to ) and be a vertex on some stab line . Since each vertex is on exactly one stab line, and all vertices of are adjacent to , we know that each vertex of must be on the stab line , on the stab line just above , or on the stab line just below . By Pigeon Hole Principle, there exists such that and are both on one of these stab lines, say . Now, for the same reason as before, each vertex of must be on the stab line , on the stab line just above , or on the stab line just below . Again by Pigeon Hole Principle, there are two vertices such that and are both on one of these stab lines. Now, consider the cycle of length four with , that is an induced subgraph of . It can be seen that the rectangles in corresponding to the vertices of form a exactly stabbed rectangle intersection representation of in which have a common stab and have a common stab. This contradicts the claim proved above. Therefore, cannot have a exactly stabbed rectangle intersection representation for any . ∎
Corollary 4.
The class of exactly stabbable rectangle intersection graphs is a proper subset of the class of rectangle intersection graphs.
The above theorem shows that there are graphs whose stab number is a constant but their exact stab number is infinite. Later on, in Theorem 36, we shall show that there are even trees whose stab number and exact number differ, even though both these parameters are finite for trees.
(a)  (b) 
Theorem 5.
The class of unit height rectangle intersection graphs is a proper subset of the class of exactly stabbable rectangle intersection graphs.
Proof.
We shall first give a proof for the wellknown fact that every unit height rectangle intersection graph is an exactly stabbable rectangle intersection graph. We shall prove the following stronger claim.
Claim. Given a unit height rectangle intersection representation for a graph , there exists a set of horizontal lines , , , (for some integer ), where , such that each rectangle in intersects exactly one of them and .
Let and let and . Now consider the unit height rectangle intersection representation of . By the induction hypothesis, there exists a set of horizontal lines , , , , for some integer , where , such that each rectangle in intersects exactly one of them and . Since every rectangle in lies completely above the horizontal line , we have that . Therefore, we have . Since , this means that for , no rectangle of intersects both the horizontal lines and . Since every rectangle in intersects the horizontal line , and every rectangle in intersects exactly one of the horizontal lines , , , , it follows that each rectangle of intersects exactly one of the horizontal lines , , , , . This proves the claim.
We shall now show the existence of an exactly stabbable rectangle intersection graph that is not a unit height rectangle intersection graph. Consider the graph , i.e. the complete bipartite graph in which each partite set contains three vertices each. Clearly, is an exactly stabbable rectangle intersection graph (see Figure 1(b)). We shall prove that is not a unit height rectangle intersection graph.
A rectangle intersection representation of a graph is crossingfree if for any two rectangles and in , the regions and are both arcconnected. Note that a unit height rectangle intersection representation of a graph is crossingfree. We shall show that if a trianglefree graph has a crossingfree rectangle intersection representation, then must be a planar graph. It then follows directly that is not a unit height rectangle intersection graph.
Let be a crossingfree rectangle intersection representation of a trianglefree graph and let be the set of vertices of having degree one. Let . Clearly, is planar if and only if is planar. Let be obtained from by removing all the rectangles corresponding to the vertices in . Note that is a trianglefree graph and is crossingfree.
Claim. There is no rectangle in which is contained in some other rectangle of .
Assume for the sake of contradiction that for vertices we have in . Since is a vertex of , we know that must have degree at least two in . Let be a neighbour of other than in . Then in , we have . Since , this implies that . But now form a triangle in , contradicting the fact that is trianglefree. This proves the claim.
Since is trianglefree, we have that in , for any vertex and any two vertices in , . This, together with the fact that is crossing free, implies that the region is arcconnected and nonempty. (To see this, observe that if is nonempty, but is not arcconnected, then there exists two points and a simple curve such that and are in different arcconnected components of . Since for any two vertices in , we have , we know that there exists some such that . But this means that and are in different arcconnected components of , contradicting the fact that is crossingfree. If is empty, then . Again, since for any two vertices in , we have , it must be the case that there exists some such that . But this contradicts the claim proved above.) Now choose for every vertex , a point in . In other words, is a point in which is not contained in any rectangle other than . For every edge , choose a point that is contained in the rectangular region . Further, for each edge , choose a simple curve between and that is completely contained in and a simple curve between and that is completely contained in such that for any curve in the collection , none of its interior points are contained in any other curve in the collection. Now the set of simple curves corresponds to the edges of and gives a planar embedding of (please see Figure 2 for an example). Hence, is a planar graph.
∎
4 Bounds on the stab number for some graph classes
In this section, we study the stab number of some subclasses of rectangle intersection graphs. We show a lower bound on for any rectangle intersection graph , which is used to derive an asymptotically tight lower bound for the stab number of grids. We also derive upper bounds on when is a split graph or a block graph.
4.1 Lower bounds
It is clear that given a stabbed rectangle intersection representation of a graph , a set of colours can be used to properly colour the vertices whose rectangles have a common stab (since the subgraph induced in by these vertices is an interval graph). This means that if is exactly stabbable, we can use two sets of colours each to colour the vertices on alternate stab lines of a exactly stabbed representation of (for some ) to obtain a proper colouring of . Thus, if is an exactly stabbable rectangle intersection graph, then . For general rectangle intersection graphs, we can adapt the same colouring strategy to get the following observation.
Observation A.
For a rectangle intersection graph , we have , or in other words, .
Remarks. Even though for a 3SRIG , the above observation gives only , we can use Theorem 2 to infer that is actually ESRIG, and therefore . Note that for any rectangle intersection graph , [4]. The question of whether there exists an upper bound on for rectangle intersection graphs that is linear in is open.
We now strengthen the above observation and show that the in the lower bound can be replaced by , where is the “pathwidth” of . A path decomposition of a graph is a collection of subsets of , where is some positive integer, such that for each edge , there exists such that and for each vertex , if , where , then for . The width of a path decomposition of is defined to be . The pathwidth of a graph , denoted by , is the width of a path decomposition of of minimum width.
We adapt a proof by Suderman [22] to show that if a graph is SRIG then has pathwidth at most .
Theorem 6.
Let be a rectangle intersection graph. Then , or in other words, .
Proof.
Let be a rectangle intersection graph with . We shall show that . Let be a stabbed rectangle intersection representation of . Let such that . For , let us define the subset . We claim that is a path decomposition of . To see this, note that for any edge , where , . Also, if some vertex , where , then contains both and , implying that it also contains , for . Therefore, , for . To complete the proof, we only need to show that . Suppose that for some , there exists such that and all the vertices of have a common stab. Since and the rectangles corresponding to the vertices of all intersect a common stab line, we have that the vertices of form a clique in , which is a contradiction to the fact that is the clique number of . Therefore, for any , there exists at most vertices in that have a common stab. Since there are only stab lines in , we now have that for each . ∎
The grid is the undirected graph with and .
Corollary 7.
Let be the grid. Then .
Proof.
It is clear that and from a result of [15] we know that the pathwidth of the grid is . From these facts and Theorem 6, we can infer that, . It is easy to see that the grid graph has a exactly stabbed rectangle intersection representation as shown in Figure 3, and therefore . The statement of the corollary now follows from the fact that . ∎
(a)  (b) 
The above corollary shows that Grids. This also shows that there are trianglefree rectangle intersection graphs on vertices whose stab number can be . Moreover, these trianglefree rectangle intersection graphs are exactly stabbable.
4.2 Split graphs
A split graph is a graph whose vertex set can be partitioned into a clique and an independent set. It is known that split graphs can have arbitrarily high boxicity [13]. So it is natural to ask whether the split graphs within rectangle intersection graphs are all exactly stabbable rectangle intersection graphs. We show that any split graph with boxicity at most 2 is 3ESRIG (Theorem 8) and that there exists a split graph with boxicity at most 2 which is not 2ESRIG (Theorem 9). From Theorem 2, it then follows that the stab number and exact stab number are equal for any split graph that has boxicity at most 2. Adiga et al. showed that deciding whether a split graph has boxicity at most 3 is NPcomplete [1]. But as far as we know, the problem of deciding whether the boxicity of a split graph is at most 2 is not known to be polynomialtime solvable or NPcomplete. By our observations below, it follows that this problem is equivalent to deciding whether a given split graph is 3ESRIG (or equivalently, 3SRIG).
Theorem 8.
A split graph is a rectangle intersection graph if and only if is a 3ESRIG.
Proof.
As is a split graph, there exists a partition of into sets and such that is a clique and is an independent set. If is a 3ESRIG then is a rectangle intersection graph. Now let be a split graph having a rectangle intersection representation such that for any two vertices , (note that such a rectangle intersection representation exists for any rectangle intersection graph). We shall assume without loss of generality that in this representation, the origin is contained in . For every vertex , define the region . It is easy to see that . It follows that for vertices such that and , . Also, is a rectangle (by the Helly property of rectangles) with nonzero height and width. This means that we can choose a point in that is not on the axis for each vertex , while satisfying the additional property that no two points in have the same coordinate. Consider . Since the degenerate rectangle given by the point intersects all the rectangles in , we can replace the rectangle with the degenerate rectangle given by the point to obtain a new rectangle intersection representation of . Let be the rectangle intersection representation of obtained in this fasion, i.e. (see Figure 4(a)).
(a)  (b) 
Let (respectively ) be the set of vertices is above (respectively, below) the axis . Let and . For each vertex , we define to be the degenerate rectangle given by the vertical line segment whose bottom end point is and top end point has coordinate . Similarly, for each vertex , we define to be the degenerate rectangle given by the vertical line segment whose top end point is and bottom end point has coordinate . As each rectangle in corresponding to a vertex in contains the origin, we have that for any such that and , the rectangle intersects if and only if contains . Therefore, the collection of rectangles given by is a rectangle intersection representation of . It is easy to see that this rectangle intersection representation, together with the horizontal lines , , and , forms a 3ESRIG representation of (see Figure 4(b)). ∎
Theorem 9.
There is a split graph which is a rectangle intersection graph but not a 2ESRIG.
Proof.
Let be the split graph whose vertex set is partitioned into a clique on four vertices and an in independent set of 14 vertices, and whose edges are defined as follows. Let be the set of all subsets of with . For every , there is exactly one vertex such that . See Figure 5(a) for a drawing of the graph . Clearly, has a rectangle intersection representation as shown in Figure 5(b).
(a)  (b) 
Now assume for the sake of contradiction that has a 2ESRIG representation . We can assume that the stab lines are and . We shall further assume that all the rectangles are contained in the strip of the plane between the two stab lines, i.e. for each , we have and (it is easy to see that every 2ESRIG representation can be converted to such a 2ESRIG representation by “trimming” the parts of the rectangles that lie above the top stab line and below the bottom stab line).
Observe that for each , the rectangle intersects all the rectangles in and is disjoint from each rectangle in . Now choose a point . Clearly, and .
Let (not necessarily distinct) such that and . Let be two distinct vertices in . By our choice of and , we have and , or in other words .
Claim. The vertices and have a common stab.
Suppose for the sake of contradiction that and do not have a common stab. Then, since and , it follows that the rectangle . We thus have . But this contradicts the fact that there exists a point such that and . This proves the claim.
By the above claim, we shall assume without loss of generality that and are on the stab line and that . This implies that (recall that ). Note that , implying that . But this contradicts the fact that there exists a point such that and . ∎
4.3 Block graphs
A graph is a block graph if every block (i.e 2connected component) of is a clique. Note that all trees are block graphs. It is not hard to see that all trees, and indeed all block graphs, are rectangle intersection graphs. We show that all block graphs are exactly stabbable rectangle intersection graphs and give an upper bound of for the exact stab number of block graphs with blocks, where . Note that this implies an upper bound of for the exact stab number of trees on vertices. We shall show in Section 6.1 that this bound is asymptotically tight, by constructing trees whose stab number is .
Let be a block graph. Given a exactly stabbed rectangle intersection representation of , we say that a set of vertices , where is a block in , is accessible if all vertices in are on the bottom stab line of and for any vertex either is not on the bottom stab line or for every vertex .
Theorem 10.
For any block graph with blocks, .
Proof.
Note that we only need the statement of the theorem to be proved for connected graphs. In fact, we shall prove the following stronger claim for connected graphs.
Claim. Let be any connected block graph with blocks and let . Then for any block of , any subset of , any such that , and any such that , there is a exactly stabbed rectangle intersection representation of with stab lines , , , , such that:

is accessible,

for every vertex , ,

for every vertex that is on the bottom stab line, we have , and

for every vertex that is not on the bottom stab line, we have .
Proof. We prove the claim by induction on . When , is an interval graph. It is not hard to see that the statement of the claim is true in this case. From here onwards, we shall assume that , and that the statement of the claim is true when the number of blocks in the graph is lesser than .
Let be the set of components of . It is easy to see that each graph is a block graph and at most one of them can have greater than blocks. We shall denote the graph in that has greater than blocks, if it exists, as . For a vertex , let . Note that for such that , . Also, since is connected, is a partition of . If exists, let be the vertex such that .
Let
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