On Graphs with Minimal Eternal Vertex Cover Number

12/12/2018 ∙ by Jasine Babu, et al. ∙ ERNET India The Fleet Street 0

The eternal vertex cover problem is a variant of the classical vertex cover problem where a set of guards on the vertices have to be dynamically reconfigured from one vertex cover to another in every round of an attacker-defender game. The minimum number of guards required to protect a graph from an infinite sequence of attacks is the eternal vertex cover number (evc) of the graph. It is known that, given a graph G and an integer k, checking whether evc(G) < k is NP-Hard. However, for any graph G, mvc(G) < evc(G) < 2 mvc(G), where mvc(G) is the minimum vertex cover number of G. Precise value of eternal vertex cover number is known only for certain very basic graph classes like trees, cycles and grids. Though a characterization is known for graphs for which evc(G) = 2mvc(G), a characterization of graphs for which evc(G) = mvc(G) remained open. Here, we achieve such a characterization for a class of graphs that includes chordal graphs and internally triangulated planar graphs. For some graph classes including biconnected chordal graphs, our characterization leads to a polynomial time algorithm to precisely determine evc(G) and to determine a safe strategy of guard movement in each round of the game with evc(G) guards. It is also shown that deciding whether evc(G) < k is NP-Complete even for biconnected internally triangulated planar graphs.

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1 Introduction

A vertex cover of a graph is a subset such that for every edge in , at least one of its endpoints is in . A minimum vertex cover of is a vertex cover of of minimum cardinality and its cardinality is the minimum vertex cover number of , denoted by . Equivalently, if we imagine that a guard placed on a vertex can monitor all edges incident at , then is the minimum number of guards required to ensure that all edges of are monitored.

The eternal vertex cover problem is an extension of the above formulation in the context of a multi-round game, where mobile guards placed on a subset of vertices of are trying to protect the edges of from an attacker. This problem was first introduced by Klostermeyer and Mynhardt [1]. In this formulation, guards are initially placed by the defender on some vertices with at most one guard per vertex. In each round of the game, the attacker gets the first turn when he attacks an edge of the graph and then the defender is allowed to let each guard remain in its current vertex or move it to a neighboring vertex, ensuring that at least one guard moves through the edge that was attacked in the current round. Then the game proceeds to the next round of attack-defense. The movement of guards in a round is assumed to happen in parallel, but no two guards are allowed to be on the same vertex at any time. Clearly, if the vertices occupied by the guards do not form a vertex cover at the beginning of each round, there is an attack which cannot be defended, namely an attack on an edge that has no guards on its end points.

If is a family of vertex covers of of the same cardinality, such that the defender can choose any vertex cover from as the starting configuration and successfully keep on defending attacks forever by moving among configurations in itself, then is an eternal vertex cover class of and each vertex cover in is an eternal vertex cover of . If is an eternal vertex cover belonging to an eternal vertex cover class , we say that is a configuration in . Eternal vertex cover number of , denoted by , is the minimum cardinality of an eternal vertex cover of .

Klostermeyer and Mynhardt [1] showed that, for , a cycle on vertices with , and for any tree on vertices with

, eternal vertex cover number is one more than its number of internal vertices. In particular, for a path on an odd number of vertices, its eternal vertex cover number is twice its vertex cover number. They also showed that, for any graph

, . From the examples given above, it can be seen that both these bounds are tight even for bipartite graphs.

Fomin et al. [2] discusses the computational complexity and derives some algorithmic results for the eternal vertex cover problem. They use an eternal vertex cover problem model in which more than one guard can be placed on a single vertex. They showed that given a graph and an integer , it is NP-hard to decide . The paper also gave an exact algorithm with time complexity and exponential space complexity. They also gave an FPT algorithm with eternal vertex cover number as the parameter, to solve the eternal vertex cover problem. They also describe a simple polynomial time 2-factor approximation algorithm for the eternal vertex cover problem, using maximum matchings. The above results also carry forward (with minor modifications in proofs) for the original model which allows at most one guard per vertex. It is not yet known if the problem is in NP. It is also unknown whether the eternal vertex cover problem for bipartite graphs is NP-hard. Some related graph parameters based on multi-round attacker-defender games and their relationship with eternal vertex cover number were investigated by Anderson et al. [3] and Klostermeyer et al. [4].

Klostermeyer and Mynhardt [1] gave a characterization for graphs which have . The characterization follows a nontrivial constructive method starting from any tree which requires guards to protect it. They also give a few examples of graphs for which such as complete graph on vertices(), Petersen graph, , (where represents the box product) and grid, if or is even. However, they mention that an elegant characterization of graphs for which seems to be difficult.

Here, we achieve such a characterization for a class of graphs that includes chordal graphs and internally triangulated planar graphs. For some graph classes including biconnected chordal graphs, our characterization leads to a polynomial time algorithm to precisely determine and to determine a safe strategy of guard movement in each round of the game with guards. On the other hand, it is shown that deciding whether is NP-Complete for biconnected internally triangulated planar graphs.

Klostermeyer and Mynhardt [1] proved that a graph with two disjoint minimum vertex covers and each edge contained in some maximum matching has . They had posed a question whether it is necessary for every edge of to be present in some maximum matching, to satisfy . We also present an example which answers this question in negative.

2 A necessary condition for

For the rest of the paper, without loss of generality, we assume that the input graph is connected and has at least two vertices. In this section, we derive a necessary condition for a graph to have . For this, we look at a slightly more general question.

For any subset , let be the minimum cardinality of a vertex cover of that contains all vertices of and let be the minimum number of guards required to provide eternal protection to in such a way that in every configuration of the eternal vertex cover class, all vertices of are occupied by guards. Note that when , and .

Here, we derive a necessary condition for for a graph , where . Suppose . Let be an eternal vertex cover class of in which each configuration has size equal to and contains . Note that for every vertex , there must be a configuration in in which there are guards on all vertices of . Otherwise, an attack on an edge incident to cannot be defended, since after this attack, there must be a guard on , implying that we must be in a configuration that contains . Since every configuration in has size equal to , we have the following observation.

Observation 1

Let be a connected graph with at least two vertices and . If , then for every vertex , .

From the above observation, it is clear that for a graph to have , it is necessary that for every vertex , has a minimum vertex cover containing . It is interesting to check if this necessary condition is also sufficient for .

For a path on vertices, where is an even number, each vertex belong to some minimum vertex cover; but still . Therefore, the simple necessary condition is not sufficient for all graphs. In fact, among graphs which are not biconnected, it is easy to find several such examples.

Still, it is interesting to ask if the obvious necessary condition is sufficient for biconnected graphs. Here, we give a biconnected bipartite planar graph of maximum degree which answers this question in negative. Consider the bipartite graph with and shown in Fig. 1. This graph consists of two copies of on vertex sets and connected by two edges and . From the figure, it can be easily seen that and it has only one minimum vertex cover that contains . Therefore, for defending an attack on an edge incident on the vertex , the guards need to move to the configuration . In this configuration, when there is an attack on the edge , has to move to a configuration containing . The only minimum vertex covers of containing are , and . Since the edge does not belong to any maximum matching of , a transition from to is not legal. Configurations and both contain . Following the attack on in configuration , when the guard on moves to , no other guard can move to , because no neighbor of is occupied in . Thus, transitions to and are also not legal. Hence, the attack on cannot be handled and therefore .

Figure 1: (a) A biconnected bipartite planar graph with all vertices in some minimum vertex cover and (b) A planar drawing of the same graph.

3 Characterizing graphs with for some graph classes

In this section, we will show that for some graph classes, including chordal graphs and internally triangulated planar graphs, the obvious necessary condition discussed in the previous section is also sufficient to have .

For any subset , denotes the induced subgraph of on the vertex set . A vertex cover of a graph is called a connected vertex cover if is connected. The connected vertex cover number, , is the size of a minimum cardinality connected vertex cover of . In the lemma below we show that if every vertex cover of , with and , is connected, then the necessary condition mentioned in Observation 1 is also sufficient to get .

Lemma 1

Let be a connected graph with and let . Suppose that every vertex cover of of size that contains is connected. Then if and only if for every vertex , .

Proof

Let . Suppose every vertex cover of with and is connected.

If then by Observation 1, the forward direction of the lemma holds.

To prove the converse, assume that for every vertex , . We will show the existence of an eternal vertex cover class of with exactly guards such that in every configuration of , all vertices in are occupied. We may take any vertex cover of with and as the starting configuration. It is enough to show that from any vertex cover of with and , following an attack on an edge such that , we can safely defend the attack by moving to a vertex cover such that and .

Consider an attack on the edge such that . Let is a vertex cover of with and . We will show that it is possible to safely defend the attack on by moving from to , where is an arbitrary minimum vertex cover such that the cardinality of its symmetric difference with is minimized. Let , and . Since is a vertex cover of that is disjoint from , we can see that is an independent set. Similarly, is also an independent set. Hence, is a bipartite graph. Further, since we also have .

Claim 1

has a perfect matching.

Proof

Note that . Consider any . Since is a vertex cover of , we have . If , then is a vertex cover of size smaller than with , violating the fact that . Therefore, , and by Hall’s theorem[5] has a perfect matching.

Since , we have .

Claim 2

, the bipartite graph has a perfect matching.

Proof

If is empty, then the claim holds trivially. Consider any non-empty subset . By Claim 1, . If , then is a vertex cover of with and . This contradicts the choice of , since the symmetric difference of and has lesser cardinality than that of and . Therefore, and . Hence, for all subsets , and by Hall’s theorem, has a perfect matching.

We will now describe how the attack on the edge can be defended by moving guards.

  • Case 1. :
    By Claim 2, there exists a perfect matching in . In order to defend the attack, move the guard on to and also all the guards on to along the edges of the matching .

  • Case 2. :
    Recall that . By our assumption, the vertex cover is connected. Let be a shortest path from to in . By the minimality of , it has exactly one vertex from and will be an endpoint of . Suppose where , for . By Claim 2, there exists a perfect matching in . In order to defend the attack, move the guard on to , to and to , . In addition, move all the guards on to along the edges of the matching .

In both cases, the attack can be defended by moving the guards as mentioned and the new configuration is . ∎

We now show a necessary condition for a graph to have .

Lemma 2

Let be any connected graph. Let be the set of cut vertices of . If , then .

Proof

Suppose and be a minimum eternal vertex cover class of . If , the result holds trivially. If , we will show that in any minimum eternal vertex cover class of , all cut vertices of have to be occupied with guards in all configurations.

Let be any cut vertex of . Let be a connected component of , and . Note that and are edge-disjoint subgraphs of with being their only common vertex. Let and . It is easy to see that . Since , there must be a vertex cover configuration in the eternal vertex cover class such that . Either or or both. If both and , then and has no minimum vertex covers without . This would immediately imply that in every configuration of , is occupied by a guard.

Therefore, without loss of generality, we need to consider the only case when has no minimum vertex cover containing . If is not occupied by a guard in a configuration , we must have and . In this configuration, consider an attack on an edge in . A guard must move to from and no guard from can move to at this point. This is impossible because has only guards on it and for occupying , at least guards are required on . Hence, in this case also, is occupied by a guard in every configuration of .

Since was an arbitrary chosen cut vertex, this implies that all vertices of must be occupied in all configurations of the eternal vertex cover class and hence . ∎

The following is a corollary of Lemma 2 and Observation 1.

Corollary 1

For any connected graph with at least three vertices and minimum degree one, .

The corollary holds because a degree one vertex and its neighbor (which is a cut vertex, if the graph itself is not just an edge) cannot be simultaneously present in a minimum vertex cover of .

The following theorem, which follows from Lemma 1 and Lemma 2, gives a necessary and sufficient condition for a graph to satisfy , if every minimum vertex cover of that contains all cut vertices is connected.

Theorem 3.1

Let be a connected graph with and be the set of cut vertices of . Suppose every minimum vertex cover of with is connected. Then if and only if for every vertex , there exists a minimum vertex cover of such that .

Proof

Suppose every minimum vertex cover of with is connected. Let .

If , by Lemma 2, . Hence, by Lemma 1, for every vertex , there exists a minimum vertex cover of such that .

For the converse, assume that for every vertex , there exists a minimum vertex cover of such that . This implies that and from our assumption, every vertex cover of with and is connected. Hence, by Lemma 1, . ∎

Theorem 3.1 gives a method to determine , if is biconnected and all its minimum vertex covers are connected.

Theorem 3.2

Let be a biconnected graph for which every minimum vertex cover is connected. If for every vertex , there exists a minimum vertex cover of such that , then . Otherwise, .

Proof

Klostermeyer et al. [1] showed that is at most one more than the size of a connected vertex cover of . Hence, from our assumption that all minimum vertex covers of are connected, we have . Now, the theorem follows by Theorem 3.1, because is biconnected and therefore has no cut vertices. ∎

To illustrate the usefulness of Theorem 3.1 and Theorem 3.2, we will look at some graph classes for which every vertex cover that contains all cut vertices is a connected vertex cover.

A graph is locally connected if for every vertex of , its open neighborhood induces a connected subgraph in . Some well-known graph classes including biconnected chordal graphs and biconnected internally triangulated planar graphs are examples of locally connected graphs. Chartrand and Pippert [6] proved that if is a graph of order such that for every pair of vertices , , , then is locally connected. Some other sufficient conditions for a graph to be locally connected were given by Vanderjagt [7]. Threshold phenomenon for local connectivity of a random graph was given by Erdös Palmer and Robinson [8].

A block in a connected graph is either a maximal biconnected component or a bridge of . The following is an easy observation.

Observation 2

Let be a connected graph and be the set of cut vertices of . If every block of is locally connected, then every vertex cover of with is connected.

Proof

The restriction of a vertex cover of to a block will give a vertex cover of the block. Hence, to prove the observation, it is enough to show that all vertex covers of a biconnected locally connected graph are connected.

For contradiction, suppose is a biconnected locally connected graph and is a vertex cover of such that is not connected. Then, there exists a vertex and two components and of such that is adjacent to vertices and . Since is a vertex cover that does not contain , we have . Since is locally connected, we know that is connected and therefore, and must belong to the same component of , which is a contradiction. Hence, is connected. ∎

From Theorem 3.1, Theorem 3.2 and Observation 2, we have the following result.

Corollary 2

Let be a connected graph with in which every block is locally connected. Let be the set of cut vertices of .

  • if and only if for every vertex , there exists a minimum vertex cover of such that .

  • if is biconnected, then and
    .

Note that, the characterization mentioned in Corollary 2 does not immediately give a polynomial time algorithm for deciding whether a locally connected graph satisfies . In the remaining parts of this section, we study some graph classes where such a polynomial time algorithm can be derived.

A class of graphs is called hereditary, if deletion of vertices from any graph in would always yield another graph in .

Theorem 3.3

Let be a hereditary graph class such that :

  • for graphs in , their minimum vertex cover number computation can be done in polynomial time and

  • for every biconnected graph in , all vertex covers of are connected.

Then,

  1. for any connected graph in , in polynomial time we can decide whether

  2. for any connected graph in with , starting from any minimum vertex cover of that contains all cut vertices of as the first configuration, it is possible to determine a safe strategy of guard movements in each round of the game that uses exactly guards and requiring only polynomial time to compute the strategy in any round

  3. for any biconnected graph in , in polynomial time we can compute . Further, it is possible to determine a safe strategy of guard movement in each round of the game that uses exactly guards and requiring only polynomial time to compute the strategy in any round.

Proof
  1. For any connected graph in , in polynomial time we can compute . Identifying the set of cut vertices of can also be done in polynomial time. By Theorem 3.1, to decide whether , it is enough to check for every vertex whether has a minimum vertex cover . Checking whether has a minimum vertex cover containing is equivalent to checking whether , where . Since , we can compute and perform this checking in polynomial time.

  2. Consider a connected graph in with and be the set of cut vertices of . By Lemma 2, . By our assumption, for any block of , all its vertex covers are connected and hence, every vertex cover of with is a connected vertex cover. Therefore, by Observation 1, for every vertex , . We complete the proof by extending the basic ideas used in the proof of Lemma 1.

    Take any minimum vertex cover of with as the starting configuration. It is enough to show that from any minimum vertex cover of with , following an attack on an edge such that , we can safely defend the attack by moving to a minimum vertex cover such that . Consider an attack on the edge such that . To start with, choose an arbitrary minimum vertex cover of with as a candidate for being the next configuration.

    1. Suppose , and . By similar arguments as in the proof of Lemma 1, is a non-empty bipartite graph with a perfect matching. If , the bipartite graph has a perfect matching, then we can choose to be the new configuration and move guards as explained in the proof of Lemma 1. Otherwise, we describe a method to choose another configuration instead of .

    2. If the bipartite graph does not have a perfect matching for some , in polynomial time we can identify a subset for which , using a standard procedure described below. First find a max-matching in and identify an unmatched vertex . Let be the set of vertices in reachable via -alternating paths from in , together with vertex . If , it would result in an -augmenting path from , contradicting the maximality of . Thus, . (In fact, since has a perfect matching, and this would mean .) Now, let .

    3. It is easy to see that is a minimum vertex cover of with and the symmetric difference of and is smaller than the symmetric difference of and . Now we replace with and iterate the steps above by redefining the sets , and and the graph .

    We will repeat these steps until we reach a point when the (re-defined) bipartite graph has a perfect matching, . This process will terminate in less than iterations, because in each iteration, the symmetric difference of the candidate configuration with is decreasing.

    The basic computational steps involved in this process are computing minimum vertex covers containing , finding maximum matching in some bipartite graphs and computing some alternating paths. All these computations can be performed in polynomial time [9].

  3. Let be a biconnected graph in . By Theorem 3.2, , . Therefore, by using part 1 of this theorem, can be decided exactly, in polynomial time. If , using part 2 of this theorem, we can complete the proof. If , we will make use of the fact that every minimum vertex cover of is connected. We will fix a minimum vertex cover and initially place guards on vertices of and one additional vertex. Using the method given by Klostermeyer et al. [1] to show that , we will be able to keep defending attacks while maintaining guards on all vertices of after end of each round of the game.

A graph is chordal if it contains no induced cycle of length four or more. It is well-known that chordal graphs form a hereditary graph class and computation of a minimum vertex cover of a chordal graph can be done in polynomial time [10]. It can also be easily seen that biconnected chordal graphs are locally connected. Hence, by Corollary 2, we have the following result.

Corollary 3
  1. For any connected chordal graph , if and only if for every vertex of that is not a cut-vertex, there is a minimum vertex cover of that contains and all the cut-vertices.

  2. For any chordal graph , we can decide in polynomial-time whether . Also, if , a polynomial-time strategy for guard movements as mentioned in Theorem 3.3 using guards exists.

  3. If is a biconnected chordal graph, then we can determine in polynomial-time. Moreover, a polynomial-time strategy for guard movements as mentioned in Theorem 3.3 using guards exists.

4 Complexity of computing eternal vertex cover number of locally connected graphs

Fomin et al. [2] showed that, given a graph and an integer , deciding whether is NP-hard. However, it is not known whether this problem is in NP or not. Further, the graph obtained by their reduction is not locally connected. In this section, we study the complexity of this decision problem for locally connected graphs.

Proposition 1

Given a locally connected graph and an integer , it is NP-complete to decide if . Moreover, it is NP-hard to approximate of locally connected graphs within any factor smaller than unless P=NP.

Proof

By Corollary 2, for a locally connected graph , is the minimum such that for every vertex , has a vertex cover of size containing . Hence, the decision problem of whether for a locally connected graph is equivalent to deciding if for every vertex , has a vertex cover of size at most containing . This is clearly in NP.

A famous result by Dinur et al. [11] states that given a connected graph , it is NP-hard to approximate the minimum vertex cover number of connected graphs within any factor smaller than . For a given connected graph and integer , we can construct a locally connected graph by adding a new vertex to and connecting it to all the existing vertices of . It can be seen easily that . Therefore, even for locally connected graphs, the minimum vertex cover number is NP-hard to approximate within any factor smaller than . By Corollary 2, for any locally connected graph , . Hence, the result follows. ∎

Now, we turn our attention to a subclass of locally connected graphs. A graph is an internally triangulated planar graph if it has a planar embedding in which all internal faces are triangles. It can be easily seen that biconnected internally triangulated planar graphs are locally connected. Hence, given a biconnected internally triangulated planar graph and an integer , deciding whether is in NP. We will show that this decision problem is NP-hard using a sequence of simple reductions. First we show that the classical vertex cover problem is NP-hard for biconnected internally triangulated planar graphs. Then we will show that an additive one approximation to vertex cover is also NP-hard for the same class and use it to derive the required conclusion.

Figure 2: Triangulating an internal face of with , by adding new vertices and edges.
Proposition 2

Given a biconnected internally triangulated planar graph and an integer , it is NP-complete to decide if .

Proof

The vertex cover problem on biconnected planar graph is known to be NP-hard [12]. We show a reduction from the vertex cover problem on biconnected planar graph to the vertex cover problem on biconnected internally triangulated planar graph. Suppose we are given a biconnected planar graph and an integer . We construct such that is an induced subgraph in . First, compute a planar embedding of in polynomial time [13]. We know that, in any planar embedding, each face of a biconnected planar graph is bounded by a cycle [14]. To construct , each internal face of with more than three vertices on its boundary is triangulated by adding four new vertices and some edges (see Fig. 2). Let be a cycle bounding an internal face of , with . Let and be two distinct indices from . Add three vertices , and inside . Now, add edges (, ), (, ) (, ), (, ), (, ) (, ), (, ), (, ), (, ) and (, ) in such a way that the graph being constructed does not loses its planarity. Add a new vertex inside the triangle formed by , and . Now, make adjacent to , and by adding edges (), () and (). Repeat this construction procedure for all faces of bounded by more than vertices. As per the construction, it is clear that the resultant graph is biconnected, internally triangulated and planar. It can be seen easily that the biconnected triangulated planar graph has a vertex cover of size at most if and only if the biconnected internally triangulated planar graph has a vertex cover of size at most where is the number of internal faces of bounded by more than vertices. ∎

In the proof of Proposition 1, we used the APX-hardness of vertex cover problem of locally connected graphs to derive the APX-hardness of eternal vertex cover problem of locally connected graphs. However, a polynomial time approximation scheme is known for computing the minimum vertex cover number of planar graphs [15]. Hence, we need a different approach to show the NP-hardness of eternal vertex cover problem of planar graphs. We will show that if minimum vertex cover number of biconnected internally triangulated planar graphs can be approximated within an additive one error, then it can be used to precisely compute the minimum vertex cover number of graphs of the same class.

Figure 3: NP-hardness reduction for additive one approximation of vertex cover number of biconnected internally triangulated planar graphs.
Proposition 3

Getting an additive -approximation for computing the minimum vertex cover number of biconnected internally triangulated planar graphs is NP-hard.

Proof

Let be the given biconnected internally triangulated planar graph. Consider a fixed planar internally triangulated embedding of . The reduction algorithm constructs a new graph as follows. Make two copies of namely, and . For each vertex , let and denote its corresponding vertices in and respectively. Choose any arbitrary edge on the outer face of . Add new edges and maintaining the planarity. Now, the new graph is biconnected and planar; but the face with boundary needs to be triangulated. For this, we follow the same procedure we used in the proof of Proposition 2 which adds four new vertices and some new edges inside this face (see Fig. 3). The resultant graph is biconnected, internally triangulated and planar.

Consider a minimum vertex cover of such that . It is clear that either or is in . It is easy to see that is a vertex cover of with size . Similarly, at least vertices from and and at least vertices among {, , , } has to be chosen for a minimum vertex cover of . This shows that .

Suppose there exist a polynomial time additive -approximation algorithm for computing the minimum vertex cover number of biconnected internally triangulated graphs. Let be the approximate value of minimum vertex cover of , computed by this algorithm. Then, . This implies that , giving a polynomial time algorithm to compute . Hence, by Proposition 2, getting an additive -approximation for computing the minimum vertex cover for biconnected internally triangulated planar graphs is NP-hard. ∎

By Corollary 2, for a biconnected internally triangulated graph, . Therefore, a polynomial time algorithm to compute would give a polynomial time additive -approximation for . Hence, by Proposition 3, we have the following result.

Proposition 4

Given a biconnected internally triangulated planar graph and an integer , it is NP-complete to decide if .

Note that, using the PTAS designed by Baker et al. [15] for computing the minimum vertex cover number of planar graphs, it is possible to derive a polynomial time approximation scheme for computing the eternal vertex cover number of biconnected internally triangulated planar graphs. A summary of the complexity results presented here are given in Fig. 4.

Figure 4: Complexity of deciding whether .
11footnotetext: All locally connected graphs are biconnected, with the exception of .

5 A graph with an edge not contained in any maximum matching but

Klostermeyer et al. [1] proved that if a graph has two disjoint minimum vertex covers and each edge is contained in a maximum matching then . They had asked if , is it necessary that for every edge of there is a maximum matching of that contains . Here, we give a biconnected chordal graph for which the answer is negative. The graph shown in Fig. 5 has , a maximum matching of size 4 and the edge not contained in any maximum matching. It can be shown that because has an evc class with two configurations, and .

Figure 5: and size of maximum matching is 4. Edge is not contained in any maximum matching.

Hence, even for a graph class such that for all , , there could be a graph with and an edge not present in any maximum matching of .

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