Independent Sets in Edge-Clique Graphs
We show that the edge-clique graphs of cocktail party graphs have unbounded rankwidth. This, and other observations lead us to conjecture that the edge-clique cover problem is NP-complete for cographs. We show that the independent set problem on edge-clique graphs of cographs and of distance-hereditary graphs can be solved in time. We show that the independent set problem on edge-clique graphs of graphs without odd wheels remains NP-complete.
Let be an undirected graph with vertex set and edge set . A clique is a complete subgraph of .
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 minimal cardinality of such a family is the edge-clique covering number, and we denote it by .
The problem of deciding if , for a given natural number , is NP-complete [38, 51, 33]. The problem remains NP-complete when restricted to graphs with maximum degree at most six . Hoover  gives 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 . Also for linegraphs the problem can be solved in polynomial time [51, 54].
In  it is shown that approximating the clique covering number within a constant factor smaller than two remains NP-complete.
Gyárfás  showed the following interesting lowerbound. Two vertices and are equivalent if they are adjacent and have the same closed neighborhood.
If a graph has vertices and contains neither isolated nor equivalent vertices then .
Gyárfás result implies that the edge-clique cover problem is fixed-parameter tractable (see also ). Cygan et al showed 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 . (Mujuni and Rosamond also mention that the edge-clique cover problem probably has no polynomial kernel.)
2 Rankwidth of edge-clique graphs of cocktail parties
The cocktail party graph is the complement of a matching with vertices.
Notice that a cocktail party graph has no equivalent vertices. Thus, by Theorem 1.1,
For the cocktail party graph an exact formula for appears in . In that paper Gregory and Pullman prove that
Let be a graph. The edge-clique graph has as its vertices the edges of and two vertices of are adjacent when the corresponding edges in are contained in a clique.
Albertson and Collins prove that there is a 1-1 correspondence between the maximal cliques in and . The same holds true for the intersections of maximal cliques in and in .
For a graph we denote the vertex-clique cover number of by . Thus
Notice that, for a graph ,
Albertson and Collins mention the following result (due to Shearer)  for the graphs , defined inductively by .
Thus, for , . However, the following is easily checked.
Let be the complement of a matching , for . Let . Obviously, every pair of edges in the matching induces an independent set with four vertices in .
Consider an edge in . The only edges in that are not adjacent to in , must have endpoints in or in . Consider an edge for some . The only other edge incident with , which is not adjacent in to nor to is .
The only edge incident with which is not adjacent to nor is . This proves the lemma. ∎
A class of graphs is -bounded if there exists a function such that for every graph ,
Dvořák and Král proved that the class of graphs with rankwidth at most is -bounded .
We now easily obtain our result.
The class of edge-clique graphs of cocktail parties has unbounded rankwidth.
It is easy to see that the rankwidth of any graph is at most one more than the rankwidth of its complement . Assume that there is a constant such that the rankwidth of is at most whenever is a cocktail party graph. Let
Then the rankwidth of graphs in is uniformly bounded by . By the result of Dvořák and Král, there exists a function such that
It is easy to see that for cographs , is not necessarily perfect. For example, when is the join of and then contains as an induced subgraph.
3 Independent set in edge-clique graphs of cographs
Notice that, for any graph , the clique number of its edge-clique graph satisfies
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. We write
We say that a subset of edges in a graph is independent if it induces an independent set in . In other words, a set of edges in is independent if no two edges of are contained in a clique of .
A graph is trivially perfect if it does not contain nor as an induced subgraph.
If a graph is connected and trivially perfect then .
When a graph is trivially perfect then the independence number is equal to the number of maximal cliques in . Therefore, and since is connected . ∎
The following lemma shows that the independent set problem in can be reduced to the independent set problem in .
The computation of for arbitrary graphs is NP-hard.
Let be an arbitrary graph. Construct a graph as follows. At every edge in add two simplicial vertices, 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
where is the number of edges of . ∎
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 .
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 .
Let and be edges in and , respectively. Then the four endpoints induce a clique in . ∎
For a vertex , let be the independence number of the subgraph of induced by , that is,
When is a cograph then
There exists an algorithm that computes the independence number of for cographs . Here is the number of vertices of .
Let be a cograph. 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 as follows.
Let be an internal node and let and be the two children of . For , write instead of . Let be labeled with and let be the join of and . Then
Assume that is labeled with . Then let be the union of and . In that case
Consider two disjoint, nonempty subsets of vertices, and , such that the graph is either a join or a union of and . Let be the maximal cardinality of an independent set of edges in such that no element has both endpoints in . Assume that is the union of and . Then
Assume that is the join of and . Then consider the following three cases. First assume that is the union of two smaller cographs and . In that case
Consider the case where is the join of two smaller cographs and . In that case
Finally, assume that . In that case
The independence number of equals for graphs that are chordal. For interval graphs the edge-clique cover number equals the number of maximal cliques .
3.1 Distance-hereditary graphs
In this section we briefly look at the independence number of edge-clique graphs of 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 . Bandelt and Mulder obtained the following characterization of distance-hereditary graphs.
Lemma 4 ()
A graph is distance hereditary if and only if every induced subgraph has an isolated vertex, a pendant vertex or a twin.
Let be distance hereditary. Then satisfies Equation (2). This value can be computed in polynomial time.
Consider an isolated vertex in . Then is a maximum independent set of edges in if and only if is a maximum independent set of edges in the graph . By induction, Equation (2) is valid for .
Let be a pendant vertex and let be the unique neighbor of in . Since is not in any triangle, the edge is in any maximal independent set of edges in . Therefore,
Let be an independent set which maximizes Equation (2) for . If then goes up by one when adding the vertex . If , then is an independent set in and .
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
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
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 . The validity of Equation (2) is easily checked. ∎
4 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 . They prove that Tuza’s conjecture holds true for this class of graphs.
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 . 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 of general graphs is NP-complete . 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 . The recognition of clique graphs of general graphs is NP-complete .
The computation of is NP-complete for graphs without odd wheels.
We reduce 3-SAT to the vertex cover problem in edge-clique graphs of graphs without odd wheels.
Let , ie, the complement of .
Let be a 3-sun. The graph is obtained from by adding
three edges between pairs of vertices of
degree two in .
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 .
Construct links between variable gadgets and clause gadgets as follows. Let be a clause. Identify one endpoint of the edge in the clause gadget with an endpoint of the edge labeled in the variable gadget. Add an edge between the other two endpoints. Construct links for the other two literals in the clause in the same manner.
Let be the graph constructed in this manner.
Notice that contains some simplicial vertices;
namely the unlabeled edges in each variable gadget and the
unlabeled edges in the links. Notice that these simplicial vertices
can be removed without changing the complexity of the vertex cover
Let be the number of variables and 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 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. 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. ∎
5 Concluding remark
As far as we know, the recognition of edge-clique graphs is an open problem.
Let denote the complete multipartite graph with partite sets each having vertices. Obviously, is a cograph with vertices.
Theorem 5.1 ()
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 maximum number of pairwise 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  show that
A lowerbound for the speed at which grows was obtained by Wilson, who showed that when is sufficiently large . Better bounds for , for some specific values of , were obtained by various authors (see eg ).
See eg  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. Despite many efforts, the existence of three pairwise orthogonal Latin squares of order 10 is, as far as we know, still unclear.
The edge-clique cover problem is NP-complete for cographs.
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