A Proof of Theorem 4.2

# Independent Sets in Edge-Clique Graphs

## Abstract

We first show that the independent set problem on edge-clique graphs of cographs can be solved in time. We then show that the independent set problem on edge-clique graphs of graphs without odd wheels is NP-complete. We present a PTAS for planar graphs and show that the problem is polynomial for planar graphs without triangle separators. Lastly, we show that edge-clique graphs of cocktail party graphs have unbounded rankwidth.

1

## 1 Introduction

Let be an undirected graph with vertex set and edge set . A clique is a complete subgraph of .

###### Definition 1

The edge-clique graph of a graph has the edges of as its vertices and two vertices of are adjacent when the corresponding edges in are contained in a clique.

Edge-clique graphs were introduced and studied by Albertson and Collins [1]. Two characterizations of edge-clique graphs have been presented by [10, 12], but the problem whether there is any polynomial-time recognition algorithm for edge-clique graphs remains open. Some results of edge-clique graphs can be found in [1, 8, 9, 10, 11, 12, 26, 29, 36, 38, 39]. Notice that edge-clique graphs were implicitly first used by Kou et al [28]. In the following we introduce the independent set problem.

An independent set in a graph is a nonempty set of vertices such that there is no edge between any pair of vertices in . The independent set problem asks the maximal cardinality of independent sets, which is called independent set number and denoted by . In general graphs, this problem is known to be NP-hard and it remains NP-hard even for triangle-free graphs [35] and planar graphs of degree at most three [18]. In some classes of graphs, for example claw-free graphs [41] and perfect graphs [20], this problem can be solved in polynomial time.

A subset of edges in a graph is independent if it induces an independent set in , that is, no two edges of are contained in a clique of . We denote the maximal cardinality of independent sets of edges of by . Notice that, for any graph , the clique number of its edge-clique graph satisfies

 ω(Ke(G))=(ω(G)2).

For the independent set number there is no such relation. For example, when has no triangles then is an independent set and the independent set problem in triangle-free graphs is NP-complete.

There is another problem relating to independent sets in edge-clique graphs, which is called the edge-clique cover problem. An edge-clique covering of is a family of complete subgraphs such that each edge of is in at least one member of the family. The edge-clique cover problem asks the minimal cardinality of such a family, which is called the edge-clique covering number and denoted by .

The problem of deciding if , for a given natural number , is NP-complete [23, 28, 33]. The problem remains NP-complete when restricted to graphs with maximum degree at most six [24]. Hoover [24] gave a polynomial time algorithm for graphs with maximum degree at most five. For graphs with maximum degree less than five, this was already done by Pullman [37]. Also for linegraphs the problem can be solved in polynomial time [33, 37]. In [28] it was shown that approximating the clique covering number within a constant factor smaller than two remains NP-complete. The independence number of equals for graphs that are chordal. For interval graphs the edge-clique covering number equals the number of maximal cliques [40].

We organized this paper as follows. In Section 2 and Section 3 we show that the independent set problem on edge-clique graphs is NP-hard in general and remains so for edge-clique graphs of graphs without odd wheels. We show in Sections 2.1 and 2.2 that the problem can be solved in time for cographs and in polynomial time for distance-hereditary graphs. In Section 4, we present a PTAS for planar graphs and show that the problem is polynomial for planar graphs without triangle separator.

Notice that cographs have rankwidth one. If edge-clique graphs of cographs would have bounded rankwidth, then computing would be easy, because computing the independence number is polynomial for graphs of bounded rankwidth. However, we show in Section 5 that this is not the case, not even for edge-clique graphs of cocktail parties. Finally, we conclude in Section 6.

## 2 Algorithms for cographs

In this section we begin with a NP-hardness proof for the computation of for arbitrary graphs . Then we provide algorithms for the computation of independence number of edge-clique graphs of cographs and distance-hereditary graphs.

The following lemma shows that the independent set problem in can be reduced to the independent set problem in .

###### Lemma 1

The computation of for arbitrary graphs is NP-hard.

###### Proof

Let be an arbitrary graph. Construct a graph as follows. At every edge in add two simplicial vertices2, both adjacent to the two endvertices of the edge. Add one extra vertex adjacent to all vertices of . Let be the graph constructed in this way.

Notice that a maximum set of independent edges does not contain any edge of since it would be better to replace such an edge by two edges incident with the two simplicial vertices at this edge. Also notice that a set of independent edges incident with corresponds with an independent set of vertices in . Hence

 α′(H)=2m+α(G),

where is the number of edges of . ∎

### 2.1 Cographs

A cograph is a graph without induced . It is well-known that a graph is a cograph if and only if every induced subgraph with at least two vertices is either a join or a union of two smaller cographs. It follows that a cograph has a decomposition tree where is a rooted binary tree and is a bijection from the vertices of to the leaves of . Each internal node of , including the root, is labeled as or . The -node joins the two subgraphs mapped to the left and right subtree. The unions the two subgraphs. When is a cograph then a decomposition tree as described above can be obtained in linear time [13].

###### Lemma 2

Let be a cograph. Assume that is the join of two smaller cographs and . Then any edge in is adjacent in to any edge in .

For a vertex , let be the independence number of the subgraph of induced by , that is,

 d′(x)=α(G[N(x)]). (1)
###### Lemma 3

Let be a cograph. Then

 α′(G)=max{∑x∈Wd′(x)|W is an % independent set in G}. (2)
###### Proof

Cographs are characterized by the fact that every induced subgraph has a twin. Let be a false twin of a vertex in . Let be a maximum independent set of edges in . Let and be the sets of edges in that are incident with and . Assume that . Let be the set of endvertices in of edges in . then we may replace the set with the set

 {{y,z}|z∈Ω(x)}.

The cardinality of the new set is at least as large as . Notice that, for any maximal independent set in , either or . By induction on the number of vertices in , Equation (2) is valid.

Let be a true twin of a vertex in . Let be a maximum independent set of edges in and let and be the sets of edges in that are incident with and . If then .

Now assume that . Endvertices in of edges in and are not adjacent nor do they coincide. Replace with

 {{x,z}|{x,z}∈A(x)or{y,z}∈A(y)}

and set . Then the new set of edges is independent and has the same cardinality as .

Let be an independent set in . At most one of and is in . By induction on the number of vertices in , the validity of Equation (2) is easily checked. ∎

###### Remark 1

Notice that the righthand side of Equation (2) is a lowerbound for for any graph . In this paper we find that there is an equality for some classes of graphs. Notice that, for example, equality does not hold for .

###### Theorem 2.1

When is a cograph then satisfies Equation (2). This value can be computed in time.

###### Proof

We proved that satisfies Equation (2) in Lemma 3. Here we show that can be computed in time.

The algorithm first computes a decomposition tree for in linear time. For each node of let be the subgraph induced by the set of vertices that are mapped to leaves in the subtree rooted at . Notice that the independence number of each can be computed in linear time.

Compute for as follows. Assume is the union of and . For , let for vertices in . Then

 d′(x)={d′1(x)if x∈V1d′2(x)if x∈V2. (3)

Assume that is the join of and . Then

 d′(x)={max{d′1(x),α(G2)}if x∈V1max{d′2(x),α(G1)}if x∈V2. (4)

Evaluate as follows. Assume that is the union of and . Then

 α′(G)=α′(G1)+α′(G2). (5)

Assume that is the join of and . For , let

 α′i=max{∑x∈Wd′(x)|W is an% independent set in Gi}. (6)

Then

 α′(G)=max{α′1,α′2}. (7)

This completes the proof. ∎

###### Remark 2

For any graph whose edge-clique graph is perfect the intersection number of equals the fractional intersection number of  [26, Theorem 4.1]. It is easy to see that for cographs , is not necessarily perfect. For example, when is the join of and , then contain a as an induced subgraph.3

### 2.2 Distance-hereditary graphs

A graph is distance-hereditary if the distance between any two nonadjacent vertices, in any connected induced subgraph of , is the same as their distance in the  [25]. Bandelt and Mulder obtained the following characterization of distance-hereditary graphs.

###### Lemma 4 ([5])

A graph is distance-hereditary if and only if every induced subgraph has an isolated vertex, a pendant vertex or a twin.

The papers [5] and [25] also contain characterizations of distance-hereditary graphs in terms of forbidden induced subgraphs.

Let be distance-hereditary. Then does not necessarily satisfy Equation (2). A counterexample is as follows. Take two 4-wheels and join the two centers by an edge. The maximum edge-independent set consists of the edges of the two 4-cycles and the edge joining the two centers. However, there is no independent set such that (2) gives this value.

## 3 NP-completeness for graphs without odd wheels

A wheel is a graph consisting of a cycle and one additional vertex adjacent to all vertices in the cycle. The universal vertex of is called the hub. It is unique unless . The edges incident with the hub are called the spokes of the wheel. The cycle is called the rim of the wheel. A wheel is odd if the number of vertices in the cycle is odd.

Lakshmanan, Bujtás and Tuza investigate the class of graphs without odd wheels in [30]. They prove that Tuza’s conjecture holds true for this class of graphs (see also [22]). Notice that a graph has no odd wheel if and only if every neighborhood in induces a bipartite graph. It follows that . Obviously, the class of graphs without odd wheels is closed under taking subgraphs. Notice that, when has no odd wheel then every neighborhood in is either empty or a matching. Furthermore, it is easy to see that contains no diamond (every edge is in exactly one triangle), no and no odd antihole.

For graphs without odd wheels coincides with the anti-Gallai graphs introduced by Le [32], since . For general anti-Gallai graphs the computation of the clique number and chromatic number are NP-complete.

Let us mention that the recognition of anti-Gallai graphs is NP-complete. Even when each edge in is in exactly one triangle, the problem to decide if is an anti-Gallai graph is NP-complete [3, Corollary 5.2]. The recognition of edge-clique graphs of graphs without odd wheels is, as far as we know, open. Let us also mention that the edge-clique graphs of graphs without odd wheels are clique graphs [9]. The recognition of clique graphs of general graphs is NP-complete [2].

###### Theorem 3.1

The computation of is NP-complete for graphs without odd wheels.

###### Proof

We reduce 3-SAT to the vertex cover problem in edge-clique graphs of graphs without odd wheels.

Let , ie, . See Fig. 2(A). Let be a 3-sun as depicted in Fig. 2(B). The graph is obtained from by adding three edges between pairs of vertices of degree two in .4 Call the three vertices of degree four in , the ‘inner triangle’ of and call the remaining three vertices of the ‘outer triangle.’

Notice that has 3 maximum independent sets of edges. Each maximum independent set of edges is an induced consisting of one edge from the inner triangle, one edge from the outer triangle, and two edges between the two triangles. The three independent sets partition the edges of .

The six edges of between the inner and outer triangle form a 6-cycle in . Let denote this set of edges in .

For each clause take one copy of . Take an independent set of three edges contained in and label these with , and .

For each variable take a triangle. Label one edge of the triangle with the literal and one other edge of the triangle with its negation . Then add a simplicial vertex to the unlabeled edge of the triangle.

Let be the graph constructed in this manner. Consider the triangles which are the edge-clique graphs of the triangles in containing the newly-added simplicial vertices. Notice that simplicial vertices of a graph may be removed without changing the complexity of the vertex cover problem and so the vertices of can be removed without changing the complexity. Let be the graph obtained from by removing the edges of these triangles.

Let be the number of variables, let be the number of clauses in the 3-SAT formula. Assume that there is a satisfying assignment. Then choose the vertices in corresponding to literals that are true in the vertex cover. The variable gadgets need vertices and the links need vertices in the vertex cover. Since this assignment is satisfying, we need at most vertices to cover the remaining edges in the clause structures, since the outgoing edge from each literal which is true is covered. Thus there is a vertex cover of with vertices.

Assume that has a vertex cover with vertices. At least vertices in are covering the edges in the variable gadgets and at least vertices in are covering the edges in the links. The other vertices of are covering the edges in the clause gadgets. Take the literals of the variable gadgets that are in the vertex cover as an assignment for the formula. Each clause gadget must have one literal vertex of which the outgoing edge is covered. Therefore, the assignment is satisfying.

This proves the theorem. ∎

## 4 Algorithms for planar graphs

In this section we obtain a PTAS for a maximum independent set of edges in planar graphs.

###### Lemma 5

Let . There exists a linear-time algorithm which computes a maximum independent set of edges for graphs of treewidth at most .

###### Proof

Notice that the problem can be formulated in monadic second-order logic. The claim follows from Courcelle’s theorem [14]. ∎

Consider a plane graph . If all vertices lie on the outerface, then is 1-outerplanar. Otherwise, is -outerplanar if the removal of all vertices of the outerface results in a -outerplanar graph. A graph is -outerplanar if it has a -outerplanar embedding in the plane [4].

###### Theorem 4.1 ([7])

The treewidth of a -outerplanar graph is at most .

###### Theorem 4.2

There exists a polynomial-time approximation scheme for a maximum independent set of edges in planar graphs.

We use Baker’s technique [4] in the proof of this theorem and we moved the proof to Appendix A.

###### Remark 3

It is easy to see that the treewidth algorithm, for graphs of treewidth , can be implemented to run in time. Since the treewidth of planar graphs is bounded by , this gives an exact algorithm for a maximum independent set of edges which runs in time.

###### Remark 4

Blanchette, Kim and Vetta prove in [6] that there is a PTAS for edge-clique cover of planar graphs.

### 4.1 Planar graphs without triangle separator

A triangle separator (of a connected graph) is a triangle the removal of which disconnect the graph. We compute for planar graphs without triangle separator in the following theorem.

###### Theorem 4.3

There exists an algorithm that computes a maximum independent set of edges in planar graphs without triangle separator.

###### Proof

Let be an embedding of a planar graph without triangle separator. We may assume that . In all the faces that have length more than three add a new vertex and make it adjacent to all vertices in the face. Let be this graph. Give the edges of a weight 1 and give the new edges a weight 0.

Let be the dual of . In the weight of an edge is the weight of the edge in that it crosses. We claim that a solution is obtained by computing a maximum weight matching in  [16].

Let be a matching in . We may assume that contains no edges of weight 0. Then every edge of crosses an edge of . We claim that this is an independent set of edges in . Since is a matching no two triangular faces of incident with different edges of coincide. Assume that two edges of lie in a triangle of . Then cannot be a face of since is a matching. But then is a separator which contradicts our assumption.

Let be an independent set of edges in and let be the corresponding edges of . Since is an independent set of edges no two edges of lie in a triangle. Assume that is not a matching, and let be a common face of of two edges in . Then is contained in a face of which is not a triangle. But then the edges of are the same. ∎

## 5 Rankwidth of edge-clique graphs of cocktail parties

In this section we show that the related problem to compute is probably much harder than the independence number. We show that relates to well-known open problems for cocktail party graphs . Cocktail parties are a proper subclass of cographs.

###### Definition 2

The cocktail party graph is the complement of a matching with vertices.

Gyárfás [21] showed an interesting lowerbound in the following theorem. Here, two vertices and are equivalent if they are adjacent and have the same closed neighborhood.

###### Theorem 5.1

If a graph has vertices and contains neither isolated nor equivalent vertices then .

Notice that a cocktail party graph has no equivalent vertices. Thus, by Theorem 5.1,

 θe(cp(n))≥log2(2n+1).

For the cocktail party graph an exact formula for appears in [19]. In that paper Gregory and Pullman (see also [17, 27]) prove that

 limn→∞θe(cp(n))log2(n)=1.

For a graph we denote the vertex-clique cover number of by . Notice that, for a graph ,

 θe(G)=κ(Ke(G)).

Albertson and Collins mention the following result (due to Shearer) [1] for the graphs , defined recursively by .

 α(Kre(cp(n)))≤3⋅(2r)!

Thus, for , . However, the following is easily checked (see Lemma 3).

###### Lemma 6

For

 α(Ke(cp(n)))=4.
###### Definition 3

A class of graphs is -bounded if there exists a function such that for every graph ,

 χ(G)≤f(ω(G)).

Dvořák and Král’ prove that the class of graphs with rankwidth at most is -bounded [15].

###### Theorem 5.2

The class of edge-clique graphs of cocktail parties has unbounded rankwidth.

###### Proof

It is easy to see that the rankwidth of any graph is at most one more than the rankwidth of its complement [34]. Assume that there is a constant such that the rankwidth of is at most whenever is a cocktail party graph. Let

 K={¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯Ke(G)|G≃cp(n),n∈N}.

Then the rankwidth of graphs in is uniformly bounded by . By the result of Dvořák and Král’, there exists a function such that

 log2(2n+1)≤θe(G)=κ(Ke(G))≤f(α(Ke(G)))=4f

for every cocktail party graph . This contradicts Lemma 6 and Theorem 5.1. ∎

## 6 Conclusion and discussions

In this paper we show polynomial-time algorithms and NP-completeness proofs for the independence number of edge-clique graphs of some classes of graphs. In particular we show that for cographs and distance-hereditary graphs the independence number of edge-clique graphs satisfies Formula (2). The results lead us ask new questions that for which other classes of graphs the formula also holds true and under what conditions the formula is true. For edge-clique cover problem we make a conjecture as follows.

###### Conjecture 1

The edge-clique cover problem is NP-complete for cographs.

We move the motivation of this conjecture to Appendix B.

Finally, trivially perfect graphs is the subclass of cographs that are chordal. A graph is trivially perfect if it does not contain nor as an induced subgraph. For this class of graphs we have equality:

###### Theorem 6.1

If a graph is connected and trivially perfect then .

###### Proof

If contains only one vertex then . Assume that contains more than one vertex. Since is connected then . When a graph is trivially perfect then the independence number is equal to the number of maximal cliques in . Thus . Finally, we conclude . ∎

## Appendix A Proof of Theorem 4.2

###### Theorem A.1

There exists a polynomial-time approximation scheme for a maximum independent set of edges in planar graphs.

###### Proof

We use Baker’s technique [4]. Consider a plane embedding of . Let be the set of vertices in the outerface. Remove the vertices of from and let be the new outerface. Continuing this process partitions the vertices into layers.

Fix an integer . For and , let be the subgraph of induced by the vertices in layers where

 t∈{jk+i+1,jk+i+2,…,jk+i+(k−1)}   and   t≥0.

Notice that the outerplanarity of each is at most . By lemma 5 there exists a linear-time algorithm which computes a maximum independent set of edges in .

Every edge in connects two vertices in either the same layer or in two adjacent layers. For fixed , the consecutive graphs skip two layers. So is an independent set of edges in .

Let be a maximum independent set of edges in . Sum over all graphs the edges of that are contained in . Then every edge of is counted times. Also, since is an exact solution for this is at least as big as the number of edges induced by in . Therefore, the sum of is at least times the optimum. If we take the maximum over , we find an approximation of size at least times the optimum.

This proves the theorem. ∎

## Appendix B Conjecture on edge-clique cover problem for cographs

According to Theorem 5.1 Gyárfás [21] result implies that the edge-clique cover problem is fixed-parameter tractable (see also [46]).

Cygan et al [44] show that, under the assumption of the exponential time hypothesis, there is no polynomial-time algorithm which reduces the parameterized problem to a kernel of size bounded by . In their proof the authors make use of the fact that is a [sic] “hard instance for the edge-clique cover problem, at least from a point of view of the currently known algorithms.” Note that, in contrast, the parameterized edge-clique partition problem can be reduced to a kernel with at most vertices [48]. (Mujuni and Rosamond also mention that the edge-clique cover problem probably has no polynomial kernel.) These observations lead us to investigate edge-clique graphs of cocktail parties in Section 5.

Let denote the complete multipartite graph with partite sets each having vertices. Obviously, is a cograph with vertices.

###### Theorem B.1 ([49])

Assume that

 3≤m≤n+1.

Then if and only if there exists a collection of at least pairwise orthogonal Latin squares of order .

Notice that, if there exists an edge-clique cover of with cliques, then these cliques are mutually edge-disjoint. Finding the maximal number of mutually orthogonal Latin squares of order is a renowned open problem. The problem has a wide field of applications, eg in combinatorics, designs of experiments, group theory and quantum informatics.

Unless is a prime power, the maximal number of MOLS is known for only a few orders. We briefly mention a few results. Let denote the maximal number of MOLS of order . The well-known ‘Euler-spoiler’ shows that only for and . Also, for all , and Chowla, Erdös and Straus [43] show that

 limn→∞f(n)=∞.

Define

 nr=max{n|f(n)

A lowerbound for the speed at which grows was obtained by Wilson, who showed that when is sufficiently large [50]. Better bounds for , for some specific values of , were obtained by various authors (see eg [42]).

See eg  [45] for some recent computational attempts to find orthogonal Latin squares. The problem seems extremely hard, both from a combinatorial and from a computational point of view [27]. Despite many efforts, the existence of three pairwise orthogonal Latin squares of order 10 is, as far as we know, still unclear.

Finally, the observations mentioned above lead us to conjecture that the edge-clique cover problem is NP-complete for cographs.

## Appendix C Planar graphs of bounded treewidth

###### Theorem C.1

Let be a planar graph with treewidth at most . Then can be computed in time, where is the number of vertices of .

###### Proof

Let be a nice tree decomposition of , where is a family of subsets of and is a tree whose nodes are the subsets  [47]. Since is nice, is a rooted tree and there are four types of node . We choose an arbitrary node in as the root of .

For each node , is a subgraph of formed by all nodes in sets , with or a descendant of . Let denote a set of the edges induced by . For node in tree decomposition and a subset , we define as the maximum cardinality of independent set of edges of with . Note that if does not exist.

In the following, we consider the four types of nodes in and calculate .

Leaf: Let be a leaf node with . Then and .

Join: Let be a node with children , . Then

 α′(i,F)=α′(j1,F)+α′(j2,F)−|F|. (8)

Introduce: Let be a node with child node such that , for some vertex . Let . Then . Let be a subset of the edges in that are incident with . Then if is independent,

 α′(i,F∪F′)=α′(j,F)+|F′|. (9)

Otherwise, if is not independent then .

Forget: Let be a node with child node such that , for some vertex . Let and let be a subset of the edges in that are incident with . Then

 α′(i,F)=max{α′(j,F),α′(j,F∪F′)}. (10)

For node in tree decomposition we compute a table with all values via dynamic programming. Since is planar, each node contains edges. Thus for any type of nodes, a table contains entries and can be calculated in time. Therefore, can be obtained in time.

This completes the proof. ∎

## References

### Footnotes

1. institutetext: Department of Computer Science
National Tsing Hua University, Taiwan
chinghao.liu@gmail.com    spoon@cs.nthu.edu.tw
2. An additional argument would show that adding one simplicial per edge is sufficient to prove NP-hardness.
3. Let and . There is a with vertices
4. In [31, Theorem 14] the authors prove that every maximal clique in contains a simplicial vertex if and only if does not contain, as an induced subgraph, nor a 3-sun with 0, 1, 2 or 3 edges connecting the vertices of degree two.

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