DeepAI

# Vertex removal in biclique graphs

A biclique is a maximal bipartite complete induced subgraph. The biclique graph of a graph H, denoted by KB(H), is the intersection graph of the family of all bicliques of H. In this work we address the following question: Given a biclique graph G=KB(H), is it possible to remove a vertex β of G, such that G - {β} is a biclique graph? And if possible, can we obtain a graph H' such that G - {β} = KB(H')? We show that the general question has a “no” for answer. However, we prove that if G has a vertex β such that d(β) = 2, then G-{β} is a biclique graph and we show how to obtain H'.

08/31/2017

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## 1 Introduction

Given a family of sets , the intersection graph of is a graph that has the members of as vertices and there is an edge between two sets when and have non-empty intersection. A graph is an intersection graph if there exists a family of sets such that is the intersection graph of . Intersection graphs of certain special subgraphs of a general graph have been studied extensively.

The clique graph of , denoted by , is the intersection graph of the family of all cliques of . Clique graphs were introduced in [HamelinkJCT1968] and characterized in [RobertsSpencerJCTSB1971]. It was proved in [Alc'onFariaFigueiredoGutierrez2006] that the clique graph recognition problem is Complete.

The biclique graph of a graph , denoted by , is the intersection graph of the family of all bicliques of . It was defined and characterized in [GroshausSzwarcfiterJGT2010]. However, no polynomial time algorithm is known for recognizing biclique graphs. Refer to [puppo2, puppo] for some results when restricted to particular graph classes.

In this work we study the following question: Given a biclique graph , is it possible to remove a vertex of , such that is a biclique graph? And if possible, can we obtain a graph such that ? We show that in general, the answer of this question is “no”. However, when the biclique graph has vertices of degree two, we answer positively to this question. In particular, our main theorem ensures that if a biclique graph has a vertex of degree two, then is a biclique graph and furthermore, we show how to obtain the graph such that .

This work is motivated by the open computational complexity of the recognition problem (that we suspect to be Complete), therefore we try to understand more the structure of biclique graphs. In particular, our result helps us, as well, to know about how a graph should be, given its biclique graph . Moreover, it is also the first result allowing to maintain the property of being a biclique graph after removing a vertex (as far as we know, there are no results of this kind even for the more studied subject of clique graphs). Since our result might be generalized (not easily) to higher degrees of vertices, one could use these kind of results (along with the proposed conjectures of last section) to restrict the input graph to a more “controlled” structure in order to solve the recognition problem of biclique graphs. We hope that these tools give some light to this matter.

Also, as a direct application of our main result, it can help to decide if a graph is not a biclique graph since we can remove iteratively (if possible) vertices of degree two and if we obtain a graph that we know it is not a biclique graph, then and all intermediate graphs are not biclique graphs.

This work is organized as follows. In Section the notation is given. In Section we discuss the main question of this work and we show why, in general, the answer is “no”, while in Section we give a positive answer when biclique graphs have vertices of degree two. In the last section we present some open and interesting related problems.

## 2 Preliminaries

Along the paper we restrict to undirected simple graphs. Let be a graph with vertex set and edge set , and let and . Say that is a complete graph when every possible edge belongs to and say that is bipartite complete when every possible edge between partitions belongs to . A complete graph of vertices is denoted and a bipartite complete graph on and vertices in each partition respectively, is denoted . A clique of is a maximal complete induced subgraph, while a biclique is a maximal bipartite complete induced subgraph of . We will denote bicliques as where and are the two partitions of the biclique. A set of vertices is an independent set when there is no edge between any pair of them. The open neighborhood of a vertex , denoted by , is the set of vertices adjacent to . The closed neighborhood of a vertex , denoted by , is the set . The degree of a vertex , denoted by , is defined as . Two vertices are called false-twins if . A graph is false-twin-free if it does not contain any false-twins vertices. Consider all maximal sets of false-twin vertices and let be the set of representative vertices such that . We call to the graph obtained by the deletion of all vertices of , for (definition from [marinayo]). Observe that has no false-twin vertices. A vertex is universal if it is adjacent to all other vertices in . A path of vertices is denoted by and a cycle of vertices is denoted by . We assume that all the graphs of this paper are connected.

A diamond is a complete graph with vertices minus an edge. A gem is an induced path with vertices plus an universal vertex.

Given a family of sets , the intersection graph of is a graph that has the members of as vertices and there is an edge between two sets when and have non-empty intersection. A graph is an intersection graph if there exists a family of sets such that is the intersection graph of .

## 3 Removing vertices in biclique graphs

We recall the theorem in [GroshausSzwarcfiterJGT2010] that gives a necessary condition for a graph to be a biclique graph. The contrapositive of this theorem is useful to prove that a graph is not a biclique graph.

###### Theorem 3.1 ([GroshausSzwarcfiterJGT2010]).

Let be a biclique graph. Then every induced of is contained in an induced diamond or an induced gem of as shown in Figure 1.

To answer the question stated in the introduction, consider the following observation.

###### Observation 3.2.

The graph , for , has no vertex such that, after its removal, the resulting graph is still a biclique graph

The explanation of Observation 3.2 is that, after removing any vertex of , we obtain an induced that is not contained in any induced diamond or gem, thus for every vertex , is not a biclique graph by Theorem 3.1. We can conclude that the answer to the question in general is “no”. See Figure 2 for an example.

To answer the question in a positive way, we show in the next section that biclique graphs with vertices of degree two, are still biclique graphs after removing them. Moreover, we show how to obtain the graph such that for a vertex of degree two.

## 4 Biclique graphs with vertices of degree two

###### Theorem 4.1.

Let for some graph and let be a vertex of such that . Then is a biclique graph. In particular, we can construct a graph such that .

Before proving Theorem 4.1 we need several lemmas.

###### Lemma 4.2.

Let be a false-twin-free graph with . If there exists a biclique such that either or , then in where is the corresponding vertex of to the biclique of .

###### Proof.

The proof is by induction on . For the base case , there are only false-twin-free graphs. We generated them all using NAUTY C library [nauty] and verified the lemma for those graphs that satisfy the hypothesis, i.e., whenever there exists a biclique such that either or , we obtained that in for the corresponding vertex of to the biclique of . Therefore, the base case holds.

Now, let be a false-twin-free graph with and let be a biclique of such that either or . Let and let such that is connected. Now, after removing , some false-twin vertices might appear in the sets and . Clearly, if is false-twin-free, the biclique still exists. Therefore, as , by induction hypothesis, in and since is an induced subgraph of then is a subgraph of ([marinayo]), that is, in as desired. We assume now that has false-twins.

Suppose first that there is a vertex , such that has only false-twins in the set . Consider the graph . Clearly, it is false-twin-free and the biclique still exists. Therefore, if the result follows by induction as before, otherwise . In fact, since has no false-twins, , and we deleted . Now, there are only three graphs with vertices that are false-twin-free, have either a or biclique such that its corresponding vertex in the biclique graph has degree less than . These graphs are shown in Figure 3. It is easy to see that wherever we add back the vertex in order to maintain the property of being false-twin-free, we would have at least one more biclique that intersects , that is, the degree of its corresponding vertex in the biclique graph is at least as we wanted to prove.

Suppose next that has false-twins in the set . Moreover, we can assume that for all , has false-twin vertices in , since if has false-twins only in , we proceed exactly as in the previous case.

We show first that . Clearly, as , if , then and there is nothing to show. Now, if and , we have that either for , or . In the first case, has false-twin vertices, since only two vertices in are not enough to make vertices of not false-twins, therefore a contradiction. For the other two cases we have that thus as has no false-twins111We remark that there exist false-twin-free graphs on vertices ([nauty]) therefore this case could also be easily verified with the computer.. If , then the only way to make the vertices in the partition of not false-twins using two vertices of is shown in Figure 4. Now depending on the adjacencies between vertices and (at least one edge must exist as are not false-twins), we can easily see that we always obtain at least other bicliques that intersect , that is, in . Finally, if , let . Since is false-twin-free, each and must be adjacent to at least one and at most two vertices of each partition of , and moreover, their neighbors of each partition cannot be the same. Therefore each and are not adjacent to at least one and at most two vertices of each partition of and these sets of non-neighbors of each partition cannot be the same neither. Now, taking along with the maximal set of adjacent vertices to of one partition and the maximal set of non-adjacent vertices to on the other, we obtain one biclique. Doing the same but taking adjacent and non-adjacent vertices to of opposite partitions we obtain a second biclique. Finally, using the same argument for , we obtain two more bicliques, that is, we have in total at least bicliques that intersect , so in . Thus we assume in what follows that .

We show now that if , then they belong to two different bicliques that intersect . This would imply, as , that we have other bicliques in that intersect , that is, in . Recall now that and have false-twins in . Let be the false-twins of in , with adjacent to and let be the false-twins of in , with adjacent to . We have the following cases:

• False-twins belong to different partitions of (see Fig. 5). Now and belong to two different bicliques that intersect unless is adjacent to , not adjacent to and not adjacent to . In this case and would belong to the same biclique containing . However, we would also obtain that and are contained in two different bicliques that intersect as and are not adjacent ( and ).

• False-twins belong to the same partition of . We have more cases now depending on and .

• are all different (see Fig. 6). We have that and belong to two different bicliques that intersect unless is not adjacent to , adjacent to and adjacent to , in which case we obtain that and are contained in two different bicliques intersecting as and are adjacent ( and ).

• and (or and ). Clearly this case cannot happen.

• and . Since and are not adjacent (otherwise and would be adjacent as well which is a contradiction), we obtain that and belong to two different bicliques that intersect .

• and . Observe that should be adjacent. In this case, and belong to two different bicliques that intersect .

Since we covered all cases, we conclude that in for the corresponding vertex of to the biclique of , as desired. ∎

###### Lemma 4.3.

Let for some graph without false-twin vertices and . Let be a vertex of such that and let be its corresponding biclique in . Then we have one of the following:

1. and is the following graph where is an independent set.

2. and is the following graph where and .

###### Proof.

Suppose first that . Clearly we have that . By contradiction, suppose that have more than one common neighbor. Let be two neighbors of and . If and are adjacent then we have that , , and are contained in four different bicliques that intersect . This is a contradiction since in . Now if and are not adjacent, since has no false-twins, there exists a vertex adjacent to and not adjacent to . Now we have that , , and are contained in four different bicliques that intersect , again a contradiction. Therefore and have exactly one common neighbor, say , that is, . Now as , there are more vertices in the graph. Suppose that has two adjacent vertices, say . We obtain that , , and are contained in four different bicliques intersecting , a contradiction. We conclude then that in an independent set, that is, (1) holds.

Suppose last that is bigger than . Now if or , by Lemma 4.2, we have that in , that is, a contradiction. Therefore . Let as shown in the statement of the lemma. We will show that and by contradiction.

Case (A) Suppose first that has another neighbor . Now, as , we have the following cases depending on the adjacencies of .

Case (A1) is adjacent to and : as has no false-twins, there exists a vertex adjacent to and not adjacent to (equivalent for a vertex adjacent to and not to ). Therefore we have that the sets and are contained in two different bicliques that intersect . Finally, if is not adjacent to then is contained in a third biclique that intersects , otherwise, is contained in a third biclique that intersects (see Fig. 7). In both cases we obtain a contradiction, therefore cannot be adjacent to both and .

Case (A2) is adjacent to and not adjacent to (symmetric for adjacent to and not adjacent to ): We have more cases here.

Case (A2a) Suppose now that has another neighbor . As , we have the following cases:

Case (A2a1) is adjacent to and not adjacent to : Here we have that , and are contained in three different bicliques respectively, each of them intersecting , that is, a contradiction.

Case (A2a2) is adjacent to and not adjacent to : Observe first that if is adjacent to , we obtain that induces a , that is, there are different bicliques that intersect which is a contradiction. Then is not adjacent to and since has no false-twins, there exists a vertex adjacent to and not adjacent to (see Fig. 8). Now and are contained in two different bicliques respectively, intersecting . Finally, if is not adjacent to , then is contained in a third biclique intersecting , otherwise is contained in a third biclique intersecting . In both cases we obtain a contradiction.

We can conclude that does not have other neighbors than and .

Case (A2b) Suppose now that has another neighbor . Clearly, is not adjacent to (Case (A2a)) and is not adjacent to , as . In this case we obtain that the sets , and are contained in three different bicliques respectively, intersecting , a contradiction. We conclude then that does not have other neighbors than and .

Case (A2c) Suppose now that has another neighbor . As shown before, cannot be adjacent to either or . Now, if is not adjacent to , we obtain that , and are contained in three different bicliques respectively, intersecting , which is a contradiction. Therefore, is adjacent to . So far we have that and are contained in two different bicliques intersecting (see Fig. 9). Now, as , the graph contains more vertices:

Case (A2c1) Suppose has another neighbor . Clearly, is not adjacent to and . If is not adjacent to , we obtain that , and are contained in three different bicliques respectively, intersecting , a contradiction. Therefore, is adjacent to . We have the following cases now:

Case (A2c1a) is adjacent to : In this case we have that , , and are contained in four different bicliques respectively, intersecting , which is a contradiction.

Case (A2c1b) is not adjacent to : Now, as has no false-twin vertices, there exists a vertex adjacent to and not adjacent to (equivalent for a adjacent to and not adjacent to ). We have that and are contained in two different bicliques respectively, intersecting (see Fig 10). Finally, if is not adjacent to , , is contained in a third biclique that intersects , otherwise is contained in a third biclique. In both cases, that is a contradiction.

We conclude then that does not have other neighbors than and

Case (A2c2) Suppose has another neighbor . If is not adjacent to , refer to Case (A2c) (when is not adjacent to ). Otherwise, refer to Case (A2c1) (when is adjacent to ). We can conclude that has no other neighbors than and .

Case (A2c3) Suppose has another neighbor . Observe that cannot be adjacent to any of and , as otherwise we would be in a previous case. Therefore , and are contained in three different bicliques respectively, each of them intersecting , that is, a contradiction. Thus, cannot have other neighbors than and .

As in these three cases we obtained a contradiction. We can conclude that has no other neighbors than and .

Case (A2d) Suppose now that has another neighbor . If is adjacent to , then refer to Case (A2c) (when is adjacent to ), otherwise we have that , and are contained in three different bicliques respectively, that intersect , which is a contradiction. Thus, cannot have other neighbors than .

To conclude Case (A2), if is adjacent to and not adjacent to , we obtain a contradiction.

To resume Case (A), whenever we supposed that had another neighbor , we have obtained a contradiction. So, we conclude that has no other neighbors than and , that is, .

Case (B) Suppose finally that has a neighbor and has a neighbor . Clearly as and has only and as neighbors (Case (A)). Using the same argument, we obtain that is not adjacent to neither to . Observe now that if is adjacent to , we obtain that induces a , that is, there are four different bicliques that intersect which is a contradiction. Suppose now that has another neighbor (equivalent if has another neighbor). Like before, is not adjacent to and . Now if is adjacent to , we obtain that , and are contained in three different bicliques respectively, each of them intersecting , that is, a contradiction. Therefore, is not adjacent to . Now as is false-twin-free, there exists a vertex adjacent to and not adjacent to . Clearly , otherwise we have an induced so four different bicliques intersecting . Therefore, we obtain that , and are contained in three different bicliques respectively, intersecting , a contradiction. We can conclude that either or has no other neighbors than . If we suppose that does not have other neighbors, we obtain that .

To finish the proof, combining Cases (A) and (B), we obtain that and as desired, that is, (2) holds. ∎

Now we can give the proof of Theorem 4.1.

Proof of Theorem 4.1. Let and let be a vertex of such that . We want to show that is a biclique graph and in particular, that we can construct a graph such that . We can assume that is false-twin-free as ([marinayo]). We can also assume that since it is easy to check the theorem if is false-twin-free and (there are only graphs with vertices, with and with  [nauty]. See Appendix of [marinaYoArxiv] for a list of all biclique graphs up to vertices).

Now by Lemma 4.3 we have that if is the corresponding biclique in to vertex of , then either and is the first graph shown in the statement of Lemma 4.3 where is an independent set, or and is the second graph shown in the statement of Lemma 4.3 where and .

In the first case observe that we have the biclique that intersects only the following two bicliques: and . Consider the graph obtained as follows: take , remove and , add a copy of vertex (with the same neighborhood), say , and finally add a vertex adjacent to and a vertex adjacent to (see Fig. 11). Note that has exactly one biclique less than (biclique ), moreover, we can associate bicliques of to the bicliques of , respectively. Clearly, as , and intersect, and since this set is equal to of , and have the same intersections in than and of (minus the biclique ). We can conclude then, that .

Finally, if and is the second graph shown in the statement of Lemma 4.3, take . It is clear that the only biclique that is lost in is , while the rest of the structure remains the same. Therefore where in this case, is a subgraph of , as we wanted to prove. ∎

In Figure 12, we can see an example of application of Theorem 4.1 using its contrapositive. We want to check if is not a biclique graph, therefore we remove one by one the vertices of degree two, (now with degree two) and , and we obtain the graph that we know it is not a biclique graph [marinaYoArxiv]. We can conclude then that and all intermediate graphs, are not biclique graphs as well. Remark that in all these graphs every induced is contained in an induced diamond or gem, therefore we cannot apply the contrapositive of Theorem 3.1.

## 5 Open problems

Our first conjecture states that the family of biclique graphs described in Observation 3.2 is actually unique in respect to the property of not having any vertex such that after its removal, the resulting graph is a biclique graph.

###### Conjecture 5.1.

Let be a biclique graph such that , for . Then, there exists a vertex such that is a biclique graph.

Note that does not imply that . Consider the graph obtained by taking a , , joining each vertex of the cycle to (zero or more) different new vertices and finally adding (zero or more) false-twin vertices to the vertices of the cycle. It easy to see that (recall that [marinayo]). We propose therefore the following conjecture that characterizes the class of graphs having as biclique graphs.

###### Conjecture 5.2.

Let be a biclique graph. Then , for , if and only if consists of an induced such that each vertex of the cycle is adjacent to at most one vertex (outside the cycle) that has degree one.

The if part is trivial as it was explained right before the statement. If the only if part is true, given a graph , one can verify if in linear time using the modular decomposition given in [habib] to obtain (then it is easy to check if is an induced with each vertex having at most one neighbor outside the cycle with degree one).

To finish, the last conjecture implies the first one when the biclique graphs have false-twin vertices.

###### Conjecture 5.3.

Let be a biclique graph and let be a false-twin vertex of . Then, is a biclique graph. In particular, is a biclique graph.