Exactly Hittable Interval Graphs

Given a set system $\mathcal{X} = \{\mathcal{U},\mathcal{S}\}$, where $\mathcal{U}$ is a set of elements and $\mathcal{S}$ is a set of subsets of $\mathcal{U}$, an exact hitting set $\mathcal{U}'$ is a subset of $\mathcal{U}$ such that each subset in $\mathcal{S}$ contains exactly one element in $\mathcal{U}'$. We refer to a set system as exactly hittable if it has an exact hitting set. In this paper, we study interval graphs which have intersection models that are exactly hittable. We refer to these interval graphs as exactly hittable interval graphs (EHIG). We present a forbidden structure characterization for EHIG. We also show that the class of proper interval graphs is a strict subclass of EHIG. Finally, we give an algorithm that runs in polynomial time to recognize graphs belonging to the class of EHIG.


Introduction
We study classes of simple graphs which are intersection graphs of set systems that have exact hitting sets.In particular, we introduce a class of interval graphs which are intersection graphs of intervals that have exact hitting sets.We refer to this class as Exactly Hittable Interval Graphs (EHIG).We also present an infinite family of forbidden structures for EHIG.In the following, we introduce a setting of exact hitting sets and intersection graphs, before presenting our results.
Exact Hitting Sets: Set systems are synonymous with hypergraphs.A hitting set of a hypergraph H is a subset T of the vertex set of H such that T has at least one vertex from every hyperedge.If every hyperedge has exactly one element from T , then T is called an exact hitting set.The EXACT HITTING SET problem is a wellstudied decision problem that aims to find if a given hypergraph has an exact hitting set.It finds applications in combinatorial cryptosystems (Downey and Fellows (2013)) and computational biology among many others.The EXACT HITTING SET problem is the dual of the EXACT COVER problem which, in turn, seeks to find a set cover that covers all vertices of a hypergraph such that the number of occurrences each vertex has in the cover is exactly one.Some famous examples of the EXACT COVER problem are sudoku, tiling dominoes, and the n-queens problem.The EXACT COVER problem is a special case of the MINIMUM MEMBERSHIP SET COVER problem (MMSC) (Karp (1972)).While the classic SET COVER problem seeks to find a set cover of minimum cardinality, MMSC aims to find a set cover that minimizes the maximum number of occurrences each vertex has in the cover.MMSC is known to be NP-complete on arbitrary set systems (Kuhn et al. (2005)).However, for interval hypergraphs, MMSC was shown to be solvable in polynomial time by Dom et al. (2006).If a hypergraph H has an exact hitting set, we refer to H as an exactly hittable hypergraph.Dhannya and Narayanaswamy (2020) have shown that a conflict-free coloring of a set of intervals is exactly a partition into sets of intervals, such that each set has an exact hitting set.This motivates the question of characterizing those sets of intervals which have an exact hitting set.A natural characterization is obtained by writing the hitting set linear program with one constraint per interval.This system is totally unimodular and thus defines an integer polytope (Dom et al. (2006)).Thus, the intervals have an exact hitting set if and only if the polytope defined by the exact hitting set linear program is non-empty.Further, it is possible to find if the interval hypergraph is exactly hittable in 2 K.K. Nisha, N.S.Narayanaswamy, S.M. Dhannya polynomial time (Dom et al. (2006)).In this work, we consider a related graph theoretic version of this question -can we characterize the class of interval graphs that are the intersection graphs (defined in Section 1.1) of a set of intervals that have an exact hitting set?We refer to this class as the class of Exactly Hittable Interval Graphs (EHIG).
Intersection Graphs: The theory of graphs and hypergraphs are connected by a very well-studied notion of intersection graphs (Erdős et al. (1966)).It is well-known that every graph G is an intersection graph of some hypergraph H (Harary (1969)).H is referred to as an intersection model or a set representation of G (Golumbic (2004); Harary (1969)).Interestingly, certain special classes of graphs are characterized by the structure of their intersection models.For instance, Gavril (1974) has shown that the class of chordal graphs are the intersection graphs of subtrees of a tree .When the hyperedges are restricted to be paths on a tree, the resulting intersection graph class is that of path chordal graphs, which is a proper subclass of the class of chordal graphs (Chalopin and Gonc ¸alves (2009); Gavril (1978); Lévêque et al. (2009); Monma and Wei (1986)).These characterizations result in recognition algorithms which are very well studied.The book by Golumbic (2004) can be considered a pilgrimage for anyone interested in the characterization and recognition of different natural sub-classes of perfect graphs.The recognition problem in the class of perfect graphs itself remained a fascinating open problem with a long history of results (survey in the classic book by Grötschel et al. (2012)) till the Strong Perfect Graph Conjecture was proven by Chudnovsky et al. (2003).While many classes have efficient recognition algorithms, there are those for which the recognition problem is NP-complete.Tolerance graphs are a sub-class of interval graphs, and the recognition problem for this class has been shown to be NP-Complete by Mertzios et al. (2010).The thinness of a graph, on the other hand, is a width parameter that generalizes certain properties of interval graphs.Interval graphs are exactly the graphs of thinness one.In their work, Bonomo-Braberman and Brito (2023) have presented characterizations of 2-thin and proper 2-thin graphs as intersection graphs of rectangles in the plane, as vertex intersection graphs of paths on a grid, and by forbidden ordered patterns.Forbidden induced subgraph characterization for restricted cases of known graph classes are well-studied.For instance, even though a structural characterization by minimal forbidden induced subgraphs for the entire class of circle graphs is not known, Bonomo-Braberman et al. (2022) have given a characterization by minimal forbidden induced subgraphs of circle graphs, restricted to split graphs.Rectangle intersection graphs are the intersection graphs of axis-parallel rectangles in the plane.A graph is said to be a k-stabbable rectangle intersection graph (k-SRIG), if it has a rectangle intersection representation in which k horizontal lines can be placed such that each rectangle intersects at least one of them.Chakraborty et al. (2021) have introduced some natural subclasses of 2-SRIG, and have shown that one of these subclasses can be recognized in linear-time if the input graphs are restricted to be triangle-free.Earlier, Chakraborty and Francis (2020) had developed a forbidden structure characterization for block graphs that are 2-ESRIG (in the case when each rectangle intersects exactly one of the k horizontal lines) and trees that are 3-ESRIG, which lead to polynomial-time recognition algorithms for these two classes of graphs.These forbidden structures are natural generalizations of asteroidal triples.
A result which has the flavour of Exact Hitting Set is in a recent paper by Bhyravarapu et al. (2021).They consider the problem of coloring the vertex set of a graph with k non-zero colors and one zero colour such that for each vertex v, there is a vertex u in N (v) which has a non-zero colour different from all the other vertices in N (v).This is called the CF ON * colouring problem, and the goal is to find the minimum value of k for which the graph has a CF ON * colouring.For k = 1, this problem is the Exact Hitting Set problem of a set system in which the sets are the set of neighbours of each vertex.For unit disk graphs, they show that testing if there is a CF ON * coloring with one non-zero colour is NP-complete.
Forbidden Structure Characterizations: While a graph G may be identified as an intersection graph of a structured hypergraph, characterization of G based on forbidden structures has also been equally well-studied.For instance, the class of chordal graphs are characterized by the absence of induced cycles of size 4 or more (Golumbic (2004)).Similarly, by the celebrated theorem of Kuratowski (West (2000)), the class of planar graphs must not have subgraphs that are subdivisions of K 5 and K 3,3 .Interval graphs are known to be the class of chordal graphs without an asteroidal triple as induced subgraph Lekkerkerker and Boland (1962).Recall that an asteroidal triple of a graph G is a set of three independent vertices such that there is path between each pair of these vertices that does not contain any vertex of the neighborhood of the third.The class of proper interval graphs is a subclass of interval graphs that do not have a K 1,3 as an induced subgraph (Roberts (1978)).Refer to Table 1 for a summary of these examples.Clearly, characterization of simple graphs based on their intersection models and forbidden structures are extremely well-studied notions in defining graph classes.

Our results
1. We begin our set of results with a simple extension to a well-known theorem by Harary (1969) that every graph G is the intersection graph of some hypergraph H.
Observation 1 Every simple undirected graph is the intersection graph of an exactly hittable hypergraph.Further, if G is a chordal graph, then it is the intersection graph of an exactly hittable set of subtrees of a tree.
We present proof of this observation in Section 2. Further to this observation, we look at a subclass of chordal graphs, namely interval graphs, which are intersection graphs of subpaths on a path.We ask if there is an exactly hittable intersection model for every interval graph, where the intersection model consists of subpaths on a path.Interestingly, the answer is no.
2. We introduce the class of Exactly Hittable Interval Graphs (EHIG), which is the set of interval graphs that have an exactly hittable interval representation.A given set of intervals defines a unique interval graph, but an interval graph can have many interval representations.We say that an interval graph is an exactly hittable interval graph if and only if it has at least one exactly hittable interval representation.
Definition 1 (Exactly Hittable Interval Graphs) The class of exactly hittable interval graphs is the class of interval graphs which are intersection graphs of intervals that have exact hitting sets.
We present a forbidden structure characterization for EHIG.First, we define a family F of simple graphs as follows: Definition 2 For each k ≥ 1, F k denotes the set of connected interval graphs whose vertex set can be partitioned into an induced path P consisting of k vertices and the open neighbourhood of P (consisting of only those vertices which are not in P ) which is an independent set of size k + 3. Further, F is defined to be k≥1 F k .
Our main contribution in this paper is to prove that every graph in F is a forbidden structure for EHIG.See Fig. 1 for examples of forbidden structures.In Fig. 1(i), u is the induced path P consisting of one vertex with an independent set of four vertices a, b, c, d in its neighbourhood.Similarly, in Fig. 1(ii), a-b is the induced path P consisting of two vertices and {c, d, u, e, f } is an independent set of five vertices in the neighbourhood of P .It may, however, be noted that, Fig. 2(ii) contains K 1,4 as an induced subgraph, which itself is a forbidden structure in F 1 .
Theorem 1 An interval graph G is exactly hittable if and only if it does not contain any graph from the set F as an induced subgraph.
This theorem is proved in Section 3. We believe that this result is an interesting addition to the existing graph characterizations, primarily because we could not find such an equivalence elsewhere in the literature, including graph classes repositories like graphclasses.org.(Gilmore and Hoffman (2011);Golumbic (2004)).An interval representation is constructed from the ordering such that the intersection graph of this representation is isomorphic to G. By construction, there exists exactly one canonical interval representation for every interval graph.While the canonical representation may be of independent interest, this representation is crucial in proving Theorem 1 in this paper.

In
In Section 2, we prove the following theorem.
Theorem 2 Let G be an interval graph.Let H G be its canonical interval representation constructed as described in Section 2.Then, G is exactly hittable if and only if H G is exactly hittable.
4. Given an interval graph G and its canonical interval representation H G , we show that the algorithm by Dom et al. (2006) to solve the MMSC problem in interval hypergraphs can be used to recognize EHIG.
We present the details in Section 3.2.
5. We show that the class EHIG is positioned between the class of proper interval graphs and the class of interval graphs in the containment hierarchy of graph classes.
The proof of the second part of the above theorem follows from the definition of EHIG.We prove the first part of the containment relationship in Section 3.3.Interestingly, the smallest forbidden structure of EHIG is K 1,4 whereas that of the class of proper interval graphs is K 1,3 .

Preliminaries
Definition 3 (Intersection Graphs) Given a set system X = (U, S), the intersection graph G(X ) of sets in X is the simple graph obtained as follows.For every set S ∈ S, there exists a vertex v S ∈ G.An edge (v Si , v Sj ) occurs in G if and only if there exists two sets S i , S j ∈ F such that S i ∩ S j = ∅.The family S is called a set representation of the graph G.A set representation is also referred to as an intersection model (Golumbic (2004), Harary (1969)).
A hypergraph H = (V, E) is a graph theoretic representation of a set system X = (U, S), where the set V corresponds to U and the set E corresponds to S. The set V contains vertices of hypergraph H and the set E contains hyperedges.In the intersection graph G, for every hyperedge E ∈ E, there exists a vertex v E ∈ G.An edge (v Ei , v Ej ) occurs in G if and only if the hyperedges E i and E j have a non-empty intersection.
Definition 4 (Interval Graphs) A graph G = (V, E) is an interval graph if there exists an assignment of intervals on the real line to each vertex v ∈ V (G) such that for each edge (u, v) Each hyperedge in I is a set of consecutive integers, which we call an interval.In an interval I = {i, i + 1, . . ., j}, i and j are the left and right endpoints of I respectively, which we denote by l(I) and r(I), respectively.We use V(H) (or simply V) and I(H) (or simply I) to denote the vertex set and the hyperedge set, respectively, of an interval hypergraph H.An interval hypergraph is said to be proper if no interval is contained in another interval.If, for an interval graph G, there exists an interval representation in which no interval is properly contained inside another interval, then G is a proper interval graph.
An interval graph is characterized by the existence of a linear ordering of its maximal cliques.In Section 3, we use the following characterization to obtain an exactly hittable interval representation for an interval graph, if such a representation exists.
Theorem 4 (Gilmore and Hoffman ( 2011)) The maximal cliques of an interval graph G can be linearly ordered such that, for every vertex x of G, the maximal cliques containing x occur consecutively.
The class of interval graphs is a subfamily of the class of chordal graphs, which, in turn, is a subfamily of the class of perfect graphs.A chordal graph is a simple graph that does not contain any induced cycle of size ≥ 4 (Golumbic (2004)).Chordal graphs are known to be intersection graphs of subtrees of a tree (Gavril (1974)).A clique tree T of a graph G is a tree with the maximal cliques of G as nodes, such that for every vertex v of G, the maximal cliques containing v induce a subtree T (v) in T .In fact, chordal graphs are exactly the graphs that admit a clique tree (McKee and McMorris (1999)).A clique tree is also known as a tree decomposition of a graph.

Note:
We draw the reader's attention to the distinction between interval hypergraphs and interval graphs, and proper interval hypergraphs and proper interval graphs, as these are used extensively throughout the paper.Furthermore, recall that an interval graph is an Exactly Hittable Interval Graph if it has an intersection model, made of intervals, that has an exact hitting set.On the other hand, an Exactly Hittable Interval Hypergraph is one that has an exact hitting set.
Observation 2 Since our goal is to characterize interval graphs that have an exactly hittable interval representation, we assume without loss of generality that, in the graph G, for every sequence of consecutive maximal cliques in a linear ordering, there is at most one vertex which starts and ends in this sequence.
Indeed, if a given graph violates this property and there are two or more vertices that start at the same clique and end at the same clique in a sequence, then we retain only one of those vertices.The justification for this assertion is that if the resulting graph has an exactly hittable interval representation, so does the original graph.
Notations: All other definitions and notations on simple graphs, used throughout this paper, have been taken from West (2000).

A Canonical Interval Representation
In this section, we obtain a canonical interval representation H G of a given interval graph G.The canonical interval representation is nothing but a special intersection model of G. Consequently, the intersection graph of intervals in H G is isomorphic to G. The construction follows a well-defined set of steps with the result that every interval graph has a unique canonical interval representation.The canonical representation H G is obtained by stretching intervals so that all intervals have distinct left endpoints and distinct right endpoints.In other words, no pair of intervals start at the same point or end at the same point.The canonical interval representation is crucial to the proof of our main result in Section 3.

Outline:
The starting point of this construction is to use the well known linear ordering of maximal cliques associated with an interval graph (Golumbic (2004)) (refer Theorem 4).Fig. 3 gives an illustration of how to obtain the canonical interval representation of an interval graph.Let G = (V, E) be the given interval graph.
To construct the canonical interval representation, we associate a gadget D i with maximal clique Q i , for 1 ≤ i ≤ t.For every maximal clique Q i , we look at D i and stretch those intervals in I ′ that either start at i or end at i. Intuitively, we can think of I(v) as being stretched to the left if l(v) = i and as being stretched to the right if r(v) = i.Inside gadget D i , there is a point, which we denote by z i , with the following property: any interval for which l(v) = i, starts at z i or to the left of z i and any interval for which r(v) = i, ends at z i or to the right of z i .We refer to z i as the zero-point of gadget D i .The exact construction of stretched intervals is detailed in the subsequent paragraphs.
The gadgets D 1 , D 2 . . ., D t are arranged in the same order as that of the maximal cliques in O. Further, for each v ∈ V (G), the stretched interval associated with I(v) has D l(v) as its left-most gadget and D r(v) as its rightmost gadget.To complete the construction, between each pair of consecutive gadgets, we add an additional point, and we refer to these points as intermediate points.These points play a crucial role in our characterization of EHIGs in Section 3.1.The stretched interval of I(v) contains all these additional points between consecutive gadgets in the ordered set {D l(v) , D l(v)+1 , . . ., D r(v) }.Let H G = (V, I) denote the canonical interval hypergraph thus obtained.V is the set of all points internal to the gadgets (defined below) and the t − 1 additional points between consecutive gadgets (as described above).The intervals in I are the stretched intervals corresponding to each interval in I ′ .We now describe the gadget D i associated with maximal clique In other words, the ordered set considers the intervals whose left endpoint is i in descending order of their right endpoints.Then, for each 1 ≤ k ≤ m, the left endpoint of the interval of I(v k ) is stretched k − 1 points to the left.By this, l(v 1 ) is kept at z i itself, as no stretching is done on it (see Fig. 4).On the integer line, the left end point of I(v k ) ∈ D 1 , which is the most left stretched interval in D 1 , is taken as point 1.We next consider those intervals In other words, the ordered set considers the intervals whose right endpoint is i in ascending order of their left endpoints.Then, for each 1 ≤ k ≤ m, the right endpoint of the interval of I(v k ) is stretched k − 1 points to the right.On the integer line, the right endpoint of the stretched interval of I(v k ) would be z i + k − 1.This completes the description of the gadget D i .Note that for I(v) in I, the stretched interval is stretched to the left only in the leftmost gadget in which it is present, and it is stretched to the right in the rightmost gadget in which it is present.By construction, no two intervals share the same left endpoint and the same right endpoint.
Lemma 1 Let H G be the canonical interval representation of graph G as constructed using the above procedure.Then, G is isomorphic to the intersection graph of intervals in H G .
Proof: The gadgets D 1 , . . ., D t are arranged in the same order as the maximal cliques in the ordered set

Exactly Hittable Interval Graphs
Characterizing simple graphs as intersection graphs is a well-pursued line of study in graph theory.Harary (1969) had presented results on this problem in his book.We address the question of when a simple graph is the intersection graph of an exactly hittable hypergraph.We modify the proof given by Harary to answer this question.In addition, we present similar results for the class of chordal graphs (refer to Section 1.1 for definition).We recall and prove Observation 1 about arbitrary graphs and arbitrary chordal graphs.
Observation 1 Every simple undirected graph is the intersection graph of an exactly hittable hypergraph.Further, if G is a chordal graph, then it is the intersection graph of an exactly hittable set of subtrees of a tree.
Proof: The proof of the first statement is based on a slight modification to the intersection model constructed from G in Theorem 2.5 in the book by Harary (1969).Let H = (V, E) be the intersection model constructed as follows.The universe V of the hypergraph is , and E v contains all the edges incident on v and the element v. Clearly, the intersection graph of H is isomorphic to G and V (G) is an exact hitting set of H.The proof of the second statement, which is for a chordal graph G, is similar and is as follows.Since G is a chordal graph let it be isomorphic to the intersection graph of some subtrees of a tree T .In particular, let T be the clique tree of the chordal graph G (Golumbic (2004)).Let {T v | v ∈ V (G)} be the set of subtrees in T , where T v is the subtree associated with v and the tree nodes in T v correspond to those maximal cliques in G which contain the vertex v.We modify T to get T ′ by adding n = |V (G)| new nodes, each corresponding to a vertex in V (G).For each v ∈ V (G), the new node corresponding to v is made adjacent in T to some node in T v .The resulting tree is T ′ and T ′ v is the subtree of T ′ consisting of T v and the new node corresponding to v. Clearly, the newly added nodes form an exact hitting set of the set

and the intersection graph of the subtrees {T
Interestingly, not every interval graph has an exactly hittable interval representation.In this paper, we present a forbidden structure characterization for the class of interval graphs that have an exactly hittable interval representation.In this section, we prove that every graph in F (see Definition 1) is a forbidden structure for EHIG.First, we state and prove one direction of Theorem 1.
We use the following notations throughout the section.H ′ denotes an interval representation of G.We denote the open neighbourhood of vertex v by N (v).N (P ) denotes open neighbourhood of all vertices in path P , excluding the vertices in P .I(P ) denotes the set of intervals in H ′ corresponding to vertices in path P , X N (P ) denotes the set of independent vertices in N (P ) and I(X N (P ) ) denotes set of intervals in H ′ corresponding to X N (P ) .
Lemma 2 Let G be an interval graph.Let F ∈ F be any forbidden structure.If G contains F as an induced subgraph, then G is not an Exactly Hittable Interval Graph.
Proof: Our proof is by contradiction.Let H ′ be any exactly hittable interval representation of G and let G contain F ∈ F as an induced subgraph.Let F contain P , an induced path of length k in G that has an independent set of at least k + 3 vertices in its neighbourhood.Let S be an exact hitting set of H ′ .Recall that I(P ) denotes the set of intervals in H ′ corresponding to vertices in path P .By our assumption that G contains F , the number of intervals in I(X N (P ) ) is at least k + 3. Hence |I(X N (P ) ) ∩ S| ≥ k + 3. Since X N (P ) is an independent set, there can be at most two intervals in I(X N (P ) ) that have at least one endpoint each outside the union of intervals in I(P ) -one on either side of P .Therefore, even if these two intervals in I(X N (P ) ) are hit outside the intervals in I(P ) at either ends, the remaining k + 1 independent intervals have to be hit inside the union of intervals in I(P ).Hence |I(P ) ∩ S| ≥ k + 1.But there are only k intervals inside I(P ).Therefore, by the pigeonhole principle, at least one interval among the intervals in I(P ) has to be hit more than once.Thus S cannot be an exact hitting set of H ′ .We have arrived at a contradiction to the assumption that H ′ is exactly hittable.Since we started with an arbitrary exactly hittable representation and arrived at a contradiction, we conclude that G is not exactly hittable. ✷ Now, we prove the other direction of Theorem 1, i.e, an interval graph G which contains no graph from the set F as an induced subgraph is exactly hittable.Let O = {Q 1 , Q 2 . . .Q t } be a linear ordering of maximal cliques in G (refer Theorem 4 and Section 2).Let H G be the canonical interval representation of G obtained from O.
We use the following notations in this section.We denote a minimum clique cover of the neighbourhood of a vertex v, which is formed by the minimum number of maximal cliques in O, by C(N [v]).Recall that a clique cover for a vertex set S is a set of cliques such that each vertex in S appears in at least one clique.Note that such a clique cover exists.We prove a simple observation here.
Proof: We prove this by contradiction.Let us assume that , there exists a vertex u in Q j which is not in Q j−1 .It follows that u is not contained in any maximal cliques that occur before Q j−1 in O since the maximal cliques containing a vertex occur consecutively in the linear ordering of maximal cliques of an interval graph.Therefore, if ✷ From now on, when we refer to a minimum clique cover of the input graph, we mean a minimum clique cover formed by the minimum number of maximal cliques in O unless specified otherwise.Let |C(N [v])| denote the number of cliques in C(N [v]).Similarly, we denote a minimum clique cover of vertices in the maximal cliques Our proof is based on the structural properties of a path P in G, the construction of which is presented in Algorithm 1.The structural properties of path P are proved as lemmas later in the section.

Outline of Algorithm 1:
We construct an induced path P which contains a minimal set of vertices from graph G.The vertices in path P are selected such that every maximal clique in O has a non-empty intersection with path P .Further, we incrementally construct a clique cover of G by taking the clique cover of the closed neighbourhood of each of the individual vertices in P .
Let {v 1 , v 2 , . . ., v p } be the ordered set of vertices in the constructed path P with respect to the linear ordering O. Let v i , . . ., v j , 1 ≤ i ≤ j ≤ p be any subset of vertices in path P .We use ), is stored in K.We denote the maximal cliques which constitute K in the order in which they appear in Algorithm 1: Construction of path P and computation of clique cover Input: An interval graph G with a linear ordering of maximal cliques which has largest right endpoint; If there are more than one such vertex, then the one with the smallest left endpoint is chosen 11: Here the notation α ′ is used to indicate that K is a minimal clique cover of G rather than a minimum clique cover.
In any perfect graph, the size of a minimum clique cover equals the size of a maximum independent set.Based on this and the fact that interval graphs are perfect graphs, we state an observation which we use in proving some important properties of the constructed clique cover of the neighbourhood of vertices in path P .
Observation 4 In any perfect graph G ′ , for each maximal clique K in a minimum clique cover K of G ′ , there exists a vertex u ∈ K such that u does not belong to any other maximal clique in K.
) contains only the maximal cliques from the linear ordering O. From Observation 4, it follows that for each maximal clique Q ∈ C(N [v i ]), there exists a vertex w which is unique to Q. Since |C(N [v i ])| > 3, there exists at least 4 such vertices each belonging to different maximal cliques in C(N [v i ]).Let those vertices be denoted as w 1 , w 2 , w 3 , w 4 .We can easily see that the vertices w 1 , w 2 , w 3 , w 4 form an independent set, since each of them belong only to their respective maximal cliques.It follows that v i together with w 1 , w 2 , w 3 , w 4 form a forbidden structure K 1,4 (refer Fig. 1 (i)).This is a contradiction to our premise that G does not contain any forbidden structure.
Proof: The proof is by contradiction.Assume that there exists a vertex v i ∈ P, Fig. 6: Forbidden structure formation and v i+1 , being adjacent, form an edge in the path P .We consider the following cases based on the cardinality of and Q i r are the maximal cliques in the ordering O which contains the right endpoints of the intervals corresponding to v i−1 and v i respectively.For any vertex | is greater than 5, then those vertices in N [v i+1 ] which are not covered in Q i r will be covered by 3 maximal cliques, which would make The proof is by contradiction to our premise that G does not contain any forbidden structure.We first show that C(N Then, using Observation 4, we show that there exists a forbidden structure.By definition, Therefore, each of the three maximal cliques in C(N [v i ]) has atleast one unique vertex which does not belong to any other maximal clique.Since ) that are mutually disjoint and form an independent set of size five.The edge (v i , v i+1 ), together with V ′ form a forbidden structure (see Definition 2).Thus we have arrived at a contradiction.Therefore Proof: By construction of path P , that covers all the vertices in A,.In other words, From Lemma 4 and Observation 5, we present the following claim.
Lemma 5 In path P , there is at most one vertex v where Proof: The proof of this claim is again by contradiction.Assume that there is more than one vertex with size of minimum clique cover equal to 3 in path P .Let v i be the first such vertex in P in the increasing order of left endpoints.By Lemma 4, we know that the minimum clique cover of v i+1 is of size less than 3.By our assumption, ∃j > i + 1 such that minimum clique cover of v j is of size 3. Let the number of vertices in the subpath of P from v i to v j (including both v i and v j ) be l.It follows from Observation 5 that for each vertex It follows that each of the l − 2 vertices in {v i+1 , . . ., v j−1 } increments the size of the clique cover by 1.That is, We can see that the vertices from v i to v j form a path of length l which has an independent set of size l + 3 in its neighbourhood.The vertices from v i to v j , together with the independent set of size l + 3 in its neighbourhood forms a forbidden structure.We have arrived at a contradiction to our premise that G does not contain any forbidden structures.Therefore it is proved that in path P , there is at most one vertex which has a minimum clique cover of size 3. ✷

Computing the exact hitting set from K
We now complete the characterization of EHIGs.We first prove the characterization when p is at most 3 and then complete the proof by an inductive argument.From Algorithm 1, we use the minimal clique cover, K = {K 1 , K 2 . . .K α ′ }, which consists of maximal cliques in O and the path P consisting of vertices v 1 , . . ., v p from left to right, in the arguments below.Further, for each 1 ≤ i ≤ α ′ , we use D i to denote the gadget corresponding to K i .For each 1 ≤ i ≤ p, let L i and R i denote the leftmost and rightmost, respectively, maximal cliques in O which contains v i .DL i and DR i denote the gadgets in H G corresponding to L i and R i , respectively.Recall that H G denotes the canonical interval representation of G.It is useful to note that DL 1 and D 1 both denote the gadget associated with K 1 , and DR p and D α ′ both denote the gadget associated with K α ′ .The inductive argument below refers to two interval graphs G and G ′ , and the gadgets we refer to are in H G or H G ′ , and will be clear from the context.In all our arguments that follow, we use the maximal clique ordering O to reason about the position of a maximal clique with respect to another maximal clique.We thus use the words and phrases before, after, at or before, at or after with respect to O.These relationships are also represented by <, >, ≤, ≥, whenever it is more convenient to use these symbols.
We now prove the characterization based on the different cases for p.For p = 1, we explicitly present the hitting set, and when 2 ≤ p ≤ 3, we present the exact hitting sets which satisfy some additional properties.These additional properties are useful in the inductive argument for p ≥ 3.In all the following lemmas in this section, we assume that G is an interval graph which does not have an F ∈ F as an induced subgraph.Further, all the statements are based on the value of p which is the number of vertices in the path P computed by Algorithm 1, and K.
Lemma 6 Let v 1 be the only vertex in P .Then the canonical interval representation H G satisfies one of the following two statements.
Proof: In the first case, K 1 and K 2 are the first and last cliques of O, respectively.Let D 1 and D 2 be the gadgets corresponding to K 1 and K 2 , respectively, in G.By the definition of H G , the interval associated with v 1 has z 1 as the left endpoint and z 2 as the right endpoint.Consider the set {z 1 , z 2 + 1}.z 1 hits all the intervals whose left endpoint is in D 1 , and z 2 + 1 hits all the intervals whose left endpoint is to the right of D 1 , and right endpoint is to the right of z 2 in D 2 .Further, each interval is hit, and no interval is hit twice.By a symmetric argument, {z 1 − 1, z 2 } is also an exact hitting set.
In the second case, consider the cliques We show that the set {z 1 − w 1 , z 2 , z 3 + w 3 } is an exact hitting set of H G .First, we observe that the point z 2 hits all the intervals in D 2 .Further, from the construction of H G it follows that, if I 1 is an interval in D 1 \ D 2 and I 2 is an interval in D 1 ∩ D 2 , then in D 1 the left endpoint of I 1 is smaller than the left endpoint of I 2 .Thus, among the w 1 intervals in D 1 ∩ D 2 , the longest interval has the left endpoint in z 1 and the remaining w 1 − 1 intervals start at different points in {(z 1 − 1), . . ., (z 1 − (w 1 − 1))}.Therefore, the point z 1 − w 1 does not hit any of the w 1 intervals that belong to D 1 ∩ D 2 .Further, it hits all the intervals in D 1 \ D 2 .A symmetric argument shows that z 3 + w 3 hits all the intervals in D 3 \ D 2 and does not hit any interval in D 3 ∩ D 2 .The remaining intervals are all in D 2 , and they are not hit by either z 1 − w 1 or z 3 + w 3 , and they all contain z 2 which is the zero-point of gadget D 2 .Thus, this base case is proved.✷ Lemma 7 Let P consist of only v 1 and v 2 .Then, the following statements hold for H G .
Then H G has an exact hitting set which contains {z 1 − w 1 , z 4 + w 4 }.
Proof: If p = 2, then K either has 3 maximal cliques or 4 maximal cliques.Thus α ′ which is the index of the last clique in K is in {3, 4}.Further, in all the cases, z 1 − w 1 is in D 1 and for α ′ ∈ {3, 4}, In the first two cases, z 1 − w 1 hits the intervals associated with vertices in K 1 \ L 2 and does not hit any interval associated with vertices in K 1 ∩ L 2 .Consider the interval graph obtained by removing Consider the maximal cliques from K 2 to K 3 .v 2 is common to all of these cliques.We now consider the first case and the second case separately.
In the first case, the minimum clique cover of N [v 2 ] is of size 2. Let w 3 be the number of vertices in K 2 ∩K 3 .Consider the exact hitting set {z 2 , z 3 + w 3 }.Clearly, z 2 hits all the intervals in D 2 , and z 3 + w 3 hits all the intervals in D 3 \ D 2 and does not hit anything in D 2 ; this is because the right endpoints of the intervals in D 2 ∩ D 3 are smaller than z 3 + w 3 .Thus, for the first case {z 1 − w 1 , z 2 , z 3 + w 3 } is an exact hitting set of H G .
In the second case, the minimum clique cover of and consider H G ′ .L 2 is the first maximal clique in G ′ and in this case let z denote the zero-point of the gadget of DL 2 in H G .By using the second case of Lemma 6, consider the exact hitting From this set, we construct an exact hitting set of H G based on the following two subcases.The first subcase is that there are vertices in L 2 \ K 1 whose leftmost clique is L 2 , and whose rightmost clique is before K 3 .Then, the interval in H G ′ of such a vertex is hit by z − |L 2 ∩ K 3 | and this also hits all the intervals associated with vertices in L 2 \ K 3 .Thus, in this subcase, we get an exact hitting set for H G consisting of In the second subcase, for all the vertices whose leftmost clique is L 2 , the rightmost clique is at or after K 3 .Let h be the intermediate point just preceding DL 2 , which is the gadget associated with L 2 in H G .Consider the set {z 1 − w 1 , h, z 3 , z 4 + |K 3 ∩ K 4 |}.z 1 − w 1 hits all intervals associated with vertices in K 1 \ L 2 , h hits all intervals associated with vertices in L 2 \ (L 2 ∩ K 3 ), and this set includes K 1 ∩ L 2 .z 3 hits all intervals associated with vertices in (L 2 ∩ K 3 ) ∪ (K 3 ∩ K 4 ), and z 4 + |K 3 ∩ K 4 | hits all intervals associated with vertices in K 4 \ K 3 (that is, the intervals in D 4 \ D 3 ).Further, it is clear that it is an exact hitting set.Thus, the first two cases are proved.The third case is symmetric to the second case (i) , and thus it is also true.Hence the lemma.✷

Remark:
The approach of obtaining an exact hitting set for H G from an exact hitting set for H G ′ , by using a preceding or following intermediate point, is a template that repeats in the following proofs.However, this template is used in different contexts, and it seems that the repeated case-by-case usage of this template is unavoidable.
Lemma 8 Let P consist of only v 1 , v 2 , v 3 , and let N [v 2 ] have a minimum clique cover of size 3. Further, let Then H G has an exact hitting set containing {z 1 −w 1 , z α ′ +w α ′ }.
Proof: Consider the interval graph G ′ obtained by removing K 1 \ L 2 .Let H G ′ be the canonical interval representation.The path obtained from Algorithm 1 on G ′ , using the cliques from L 2 to K α ′ is the path consisting of v 2 and v 3 only.The first maximal clique in H G ′ is L 2 which is the leftmost clique containing v 2 .Further, the rightmost clique containing v 2 is K α ′ −1 , and the minimum clique cover of N [v 2 ] is 3 and the minimum clique cover of N [v 3 ] is 2 (from Lemma 5).Let S be an exact hitting set of H G ′ obtained from using Lemma 7. Let z denote the zero-point of the gadget DL 2 .For some w > 0, let z − w be in S. Also, in the construction of S it is ensured that the interval associated with v 2 is not hit by z − w.From the structure of S defined in Lemma 7, {z − w, z α ′ −1 , z α ′ + w α ′ } ⊆ S. We get an exact hitting set for H G from S based on two cases.
The first case is that there are vertices in L 2 \K 1 for which the leftmost clique is L 2 and the interval associated with them are hit by z − w in H G ′ .Such vertices will also occur in the rightmost clique containing v 1 , since N [v 1 ] has a clique cover of size 2 in G. Thus in this case, S ∪ {z 1 − w 1 } is an exact hitting set for H G .The main reason is that the intervals associated with the vertices in L 2 \ K 1 whose leftmost clique is at or before L 2 will all be hit by z − w.Further, z 1 − w 1 hits all the intervals associated with vertices in K 1 \ L 2 .
In the second case, all the intervals in L 2 \ K 1 whose leftmost clique is L 2 are not hit by z − w; they are hit by another element of S. Thus, z − w is in the hitting set of H G ′ only to hit intervals whose left endpoint is to the left of DL 2 in H G .Let h be the intermediate point just to preceding the gadget associated with DL 2 in H G .Consider the set S \ {z − w} ∪ {z 1 − w 1 , h}.This is an exact hitting set of H G since S is an exact hitting (i) A detailed description of the symmetry is presented in the inductive argument in Lemma 9 set of H G ′ , the intervals associated with the vertices in L 2 \ K 1 whose leftmost clique is before L 2 will all be hit by h, and z 1 − w 1 hits all the intervals associated with vertices in K 1 \ L 2 .Hence the lemma.✷ Lemma 9 Let G be an interval graph which does not have an F ∈ F as an induced subgraph.Then the canonical interval representation H G has an exact hitting set.Further, if p ≥ 2, there is an exact hitting satisfying the following two additional properties: • The intervals corresponding to the vertices in Proof: The proof when p = 1 follows from Lemma 6.The proof is by induction on p ≥ 2. The base case is for p = 2, and from Lemma 7 we know H G has a hitting set which satisfies the additional properties.We now prove the claim for all p ≥ 3. First, we consider the case in which minimum clique cover of either ] is of size 3 and the minimum clique cover of either N [v p−1 ] or N [v p ] is of size 3. From Lemma 5, we know that there is at most one vertex in P which has a minimum clique cover of size 3. Therefore, in this case, it follows that p = 3, and N [v 2 ] has a minimum clique cover of size 3. From Lemma 8, we know that H G has an exact hitting set which satisfies the additional properties.
Next we consider the case in which either the minimum clique cover of N [v 1 ] and N [v 2 ] are both of size 2 and if not, the minimum clique cover of N [v p−1 ] and N [v p ] are both of size 2. The induction hypothesis is that the claim is true for all natural numbers in the set {2, . . ., p − 1}.Given this, we prove that the claim is true for p.
Let us first consider the case when N [v 1 ] and N [v 2 ] both have a minimum clique cover of size 2. By construction, in Algorithm 1, we know that K 2 is the rightmost clique in O which contains v 1 .From Observation 3, we know that K 2 is in a minimum clique cover of N [v 1 ].Any vertex in N [v 1 ] which is not present in K 2 will be present in K 1 , since minimum clique cover of N [v 1 ] is 2. Let L 2 be the leftmost clique in O which contains v 2 .We know that K 1 < L 2 ≤ K 2 .That is, L 2 is a clique to the right of K 1 and is not later than K 2 in O.
Consider the partition of K 1 into K 1 \ L 2 and K 1 ∩ L 2 .Let w 1 be the number of vertices in By the nature of the construction of H G , this point is to the left of the starting point of the intervals associated with vertices in K 1 ∩ L 2 .Consider the interval graph G ′ obtained by removing B 1 from G. Since the elements in N [v 1 ] are in the cliques K 1 and K 2 , the first clique in the maximal clique ordering of G ′ is L 2 .Further, among all the vertices in L 2 , v 2 is the vertex in L 2 for which the rightmost clique containing it has the largest index in O. Thus, Algorithm 1 on G ′ will compute the path P \ v 1 which has the p − 1 vertices v 2 , v 3 , . . .v p .Let L 3 denote the leftmost maximal clique in G ′ which contains v 3 .The additional properties satisfied by G ′ are as follows: • For each vertex v in G ′ for which the leftmost clique containing it (in O) is after K 1 and before L 2 , the rightmost clique containing v is before K 3 .Otherwise, the choice for v 2 by the algorithm is violated.
• For each vertex v in K 1 ∩ L 2 , the rightmost clique (in O) containing v is at or before K 2 .
• For each vertex u in N [v 2 ] for which the leftmost clique containing it is to the right of L 2 , u is also present in K 3 .Otherwise, let there exist such a vertex u for which the rightmost clique containing it is before K 3 .Then u along with one vertex each in L 2 and K 3 form an independent set of size 3. Thus, minimum clique cover of N [v 2 ] is 3.This contradicts our premise that for both N [v 1 ] and N [v 2 ] have a minimum clique cover of size 2.
• L 3 is after K 2 and not later to K 3 in O.We first show that h * hits the intervals in H G ′ corresponding to the vertices in K 1 ∩ L 2 .This is because, by the construction of P , for each vertex in K 1 , the rightmost clique containing it is at or before K 2 , and L 3 is after K 2 .Thus, the point in H G ′ that hits any interval corresponding to a vertex in K 1 ∩ L 2 has to be in a gadget to the left of D 2 .If this point is different from h * , then it would also hit v 2 .This contradicts the fact that in the exact hitting set S, v 2 is hit by a point to the left of the zero-point in the gadget DL 3 or to the intermediate point just before DL 3 .Clearly, both these points are different from h * which is a point in the gadget DL 2 and we know that DL 2 is different from DL 3 .
To construct the exact hitting set in H G from S, we define a point h in H G as follows based on two cases: • The first case is when there is a vertex such that the leftmost clique containing it is L 2 and the rightmost clique containing it is before K 3 (in O).Among all such vertices, let u be the vertex such that the rightmost clique containing it has the largest index in O.By construction of H G the left endpoint of the interval associated with u would have been the largest among all such vertices.We take h to be the left endpoint of the interval assigned to u in H G .
• In the second case there is no vertex such that the leftmost clique containing it is L 2 and the rightmost clique containing it is before K 3 (in O).In this case, h is the intermediate point between DL 2 and the preceding gadget in H G .Such a point exists, since L 2 is different from the first clique K 1 , and by construction of H G , the gadget for every clique in O except K 1 has a point to its left.
Then S \ {h * } ∪ {z 1 − w 1 , h} is a hitting set of H G , where z 1 − w 1 hits all the intervals in B 1 .h hits all the intervals associated with vertices in L 2 whose rightmost endpoint is before K 3 , and these are not hit by any other point in S, due to the induction hypothesis, and they are not hit by z 1 − w 1 .Further, the vertices in B 1 are not elements of L 2 , and thus are hit only by z 1 − w 1 .Finally, the intervals associated with all the other vertices are hit exactly once by S in gadgets different from the gadgets associated with L 2 and K 1 , and these gadgets have the same structure in H G ′ as in H G .Therefore, S \ {h * } ∪ {z 1 − w 1 , h} is an exact hitting set of H G , and it also satisfies the two additional properties.Next, we consider the case when N [v 1 ] or N [v 2 ] has a minimum clique cover of size 3.Then, since p ≥ 4, we know that both N [v p−1 ] and N [v p ] has a minimum clique cover of two cliques.Further, v p−1 and v p are distinct from v 1 and v 2 .From Observation 3, we know that K α ′ is in the minimum clique cover of N [v p ]. Consider v p−1 in P and we know that D α ′ −1 denotes the rightmost clique in O which contains v p−1 .To prove the claim in this case, we consider the reversed maximal clique ordering O, and the path obtained by executing Algorithm 1 on this ordering.Due to the symmetry of the vertices chosen during the algorithm to be added to the path, it is clear that the path computed will be the reversal of P .Also, we know that L p−1 ≤ K α ′ −2 < L p ≤ K α ′ −1 < K α ′ .We consider the argument for the previous case by using the reversal of O, the canonical representation constructed using the reversal of O, and the reversal of P .We compare the execution of Algorithm 1 using the reversal of O and its execution using O: K α ′ will be in the place of K 1 , K α ′ −1 in the place of L 2 , L p in the place of K 2 , K α ′ −2 in the place of L 3 , and L p−1 in place of K 3 .By an argument symmetric with the argument using the induction hypothesis for the previous case, using the canonical representation constructed using the reversal of O, it follows that S \ {h * } ∪ {z α ′ + w α ′ , h} is an exact hitting set of H G , and it satisfies the two additional properties.Hence the lemma.✷ We complete the proof of the forbidden structure characterization for EHIGs.
Theorem 1 An interval graph G is exactly hittable if and only if it does not contain any graph from the set F as an induced subgraph.
Proof: From Section 2, we know that every interval graph G has a unique canonical interval representation, which we denote by H G .Furthermore, if G does not have an F ∈ F as an induced subgraph, then by Lemma 9, H G has an exact hitting set.We have shown in Lemma 2 that if G has an exactly hittable interval representation, then G does not have any F ∈ F as an induced subgraph.This proves the theorem.✷ Using the above theorem, we prove Theorem 2.
Theorem 2 Let G be an interval graph.Let H G be its canonical interval representation constructed as described in Section 2.Then, G is exactly hittable if and only if H G is exactly hittable.
Proof: Let the hypergraph H G constructed from interval graph G has an exact hitting set.By Lemma 1, the intersection graph G ′ of H G is isomorphic to G. It follows that if G ′ has an exactly hittable interval representation, then G also has one.Thus, G is exactly hittable.
To show the other direction, let G be an EHIG.By Theorem 1, if G is exactly hittable, then G does not have any forbidden structure.Then, it follows from Lemma 9 that the canonical representation H G of G has an exact hitting set.✷

Algorithm to recognize exactly hittable interval graphs
In this section, we present an algorithm to recognize an exactly hittable interval graph.This algorithm makes use of the canonical interval representation in Section 2 and the result by Dom et al. (2006) for MMSC problem (described in Section 1) on interval hypergraphs.In their paper, Dom et al. showed that an integer linear programming (ILP) formulation, say L, for MMSC problem on interval hypergraphs can be solved in polynomial time.The coefficients of inequalities in L results in a totally unimodular matrix and the polyhedron corresponding to L is an integer polyhedron.If the input instance to ILP is an exactly hittable instance, then the solution returned is 1.We use this algorithm below to test if a given interval hypergraph instance is exactly hittable.
Algorithm isEHIG: Given an interval graph G, construct the canonical interval representation as described in Section 2. Let H G be the resulting interval representation.Run MMSC algorithm by Dom et al. (2006) on H G as input.If the algorithm returns value 1, then return yes.Else return no.
Lemma 10 Algorithm isEHIG(G) outputs yes if and only if G is exactly hittable in polynomial time.
Proof: The proof follows from Lemma 1, Theorem 2 and the correctness of algorithm for MMSC problem on interval hypergraphs.✷ It is also clear that the inductive argument in Lemma 9 can be converted into a polynomial time combinatorial algorithm to check if H G has an exact hitting set.This leverages the fact that a minimum clique cover of a perfect graph can be computed in polynomial time.

Proper Interval Graphs is a subclass of EHIG
We now recall and complete the proof of Theorem 3.
Proof: Let G be a proper interval graph and let it be the intersection graph of the interval hypergraph H = (V, I) in which no interval properly contains another.Since H is a proper interval hypergraph, no two intervals in I can have the same left endpoint.Hence order intervals in I according to increasing order of their left endpoints.Let this ordering be I 1 < I 2 < . . .< I m .Add r(I 1 ) (which is the smallest right endpoint among all intervals) to set S. Remove all intervals hit by r(I 1 ).Recurse on the remaining set of intervals until all the intervals are hit by S. Clearly, S is an exact hitting set.
To show the strict containment, we show that the graph K 1,3 which is a forbidden structure (Roberts (1978)) for Proper Interval Graphs has an exactly hittable interval representation.Let the vertices of the K 1,3 be {u, a, b, c} and edges be {

Discussion
Our results indicate that there is an interesting hierarchy among the class of interval graphs based on the number of times an interval is hit by a hitting set.We have shown that proper interval graphs have an exactly hittable interval representation.Further it is a strict subclass of the set of EHIGs.The natural question is to characterize interval graphs which have a representation such that each interval is hit at most k times by a hitting set.We also believe that the recognition problem for such graphs is fundamental and interesting.

Fig. 1 :Fig. 2 :
Fig. 1: Two simple instances of forbidden structures in G, the associated intervals I(u) and I(v) have a non-empty intersection.The set of intervals {I(v)} v∈V (G) is an interval representation or intersection model of G. Open and Closed neighborhoods: For a vertex v in a graph G = (V, E), the open neighborhood of v in G, denoted by N (v), is the set {u ∈ V | {u, v} ∈ E} and the closed neighborhood of v in G, denoted by N [v], is the set N (v) ∪ {v}.Definition 5 Cheilaris and Smorodinsky (2012) An interval hypergraph is any hypergraph H = ([n], I), where [n] = {1, . . ., n} and I

Fig. 5 :
Fig. 5: Construction of path P and by our premise, | C(N [v i ]) | = 3. Therefore those vertices of N [v i+1 ] which are not covered by Q i r have to be covered by exactly one more clique.It follows that |

Fig. 7 :
Fig. 7: Vertices vi and vj belong to three consecutive cliques in the clique cover are hit by a point in D 1 to the left of z 1 andD 1 ∩ N [v 2 ]is hit by a point in DL 2 which is to the left of the zero-point of DL 2 or the intermediate point just before DL 2 .•The intervals corresponding to vertices inK α ′ \ N [v p−1 ] are hit by a point in D α ′ to the right of z α ′ and D α ′ ∩ N [v p−1 ] is hit by a point in D α ′ −1 to the right of the zero-point of D α ′ −1 or the intermediate point just after D α ′ −1 .

Fig. 8 :
Fig.8: Exactly hittable interval representation of K1,3 (here, {1,3,5} is an exact hitting set) Section 2, we introduce, what we refer to as, a canonical interval representation for an interval graph.Given an interval graph G, a canonical interval representation H G is an interval hypergraph given by H G = ([n], I), where [n] = {1, . . ., n} and I ⊆ {{i, i + 1, . . ., j} | i ≤ j, i, j ∈ [n]}, and all intervals have distinct left endpoints and distinct right endpoints.Further, for each v ∈ G, I v ∈ I denotes the corresponding interval.For construction of H G , we start with the well known linear ordering of maximal cliques associated with an interval graph By the induction hypothesis applied to G ′ , H G ′ has an exact hitting set such that the intervals corresponding to the vertices in L 2 \ N [v 3 ] are hit at a point to the left of the zero-point of the gadget DL 2 in H G ′ and the intervals corresponding to vertices in L 2 ∩N [v 3 ] are hit by a point to the left of the zero-point of the gadget DL 3 or by the intermediate point just before DL 3 .In particular, v 2 which is in L 2 ∩ N [v 3 ] is hit by a point to the left of the zero-point in the gadget DL 3 or by the intermediate point just before DL 3 .Note that the intermediate point just before DL 3 exists since DL 3 is to the right of D 2 .Let this hitting set be denoted by S. Let h * ∈ S be the point in the gadget DL 2 in H G ′ .Note that DL 2 is the leftmost gadget in H G ′ .