Independence numbers of some double vertex graphs and pair graphs
Abstract
The combinatorial properties of double vertex graphs has been widely studied since the 90’s. However only very few results are know about the independence number of such graphs. In this paper we obtain the independence numbers of the double vertex graphs of fan graphs and wheel graphs. Also we obtain the independence numbers of the pair graphs, that is a generalization of the double vertex graphs, of some families of graphs.
Keywords: Double vertex graphs; pair graphs; independence number.
AMS Subject Classification Numbers: 05C10; 05C69.
1 Introduction.
Let be a graph of order . The double vertex graph of is defined as the graph with vertex set all subsets of , where two vertices are adjacent in whenever their symmetric difference is an edge of . This concept, and its generalization called token graphs, has been redefined several times and with different names. The double vertex graphs were defined and widely studied by Alavi et al. [1, 2, 3], but we can find them earlier in a thesis of G. Johns [17], with the name of the subgraph graph of . T. Rudolph [21] redefined the double vertex graphs with the name of symmetric powers of graphs and used this graphs to studied the graph isomorphism problem and to study some problems in quantum mechanics and has motivated several works of different authors, see, e.g., [6, 7, 8, 13] and the references therein. Later, R. FabilaMonroy, et. al. [14] reintroduce this concept but now with the name of token graphs, where the double vertex graphs are precisely the token graphs, and studied several combinatorial properties of this graphs such as: connectivity, diameter, cliques, chromatic number and Hamiltonian paths. After this work, there are a lot of results about different combinatorial parameters of token graphs, see for example [4, 9, 11, 12, 15, 20, 19].
In H. de Alba, et. al. [5] began the study of the independence number of token graphs and in particular for the double vertex graphs of some special graphs such as: paths, cycles, complete bipartite graphs, star graphs, etc. A subset of vertices of is an independent set if no two vertices in are adjacent. The independence number of is the number of vertices in a largest independent set in . it is know that to determine the independence number is an NPhard [18] problem in its generality.
In this work we obtain the independence number of the double vertex graphs of fan graphs and wheel graphs. The fan graph is defined as the join graph , where denote the path graph of order and the complete graph of order , and wheel graph is defined as the joint graph , where denote the cycle of order . Our main results about independence number of double vertex graphs are the following:
Theorem 1.1.
Let be an integer. Then
Theorem 1.2.
Let be and integer. Then
1.1 Pair graph of graphs
Let be a graph of order . The pair graph of is the graph whose vertex set consists of all multisets of where tho vertices and are adjacent if and only if and if , then and are adjacent in . The pair graphs, also called complete double vertex graph, of a graph, were implicitly introduced by Chartrand et al. [10] and defined explicitly by Jacob et. al. [16], were the first combinatorial properties were studied. The pair graphs are a generalization of the double vertex graphs and is always isomorphic to a subgraph of .
For the case of the independence number of complete double vertex graphs we have the following results.
Theorem 1.3.
If is an integer, then
Theorem 1.4.
Let be an integer. Then
Theorem 1.5.
Let be an integer. Then
Theorem 1.6.
Let be an integer. Then
In the rest of the papers we prove all these results in different sections.
2 Preliminary results
In the proofs of some of our results, we use the following known facts.
Lemma 2.1.
If is an induced subgraph of , then .
Let denote the cartesian product of graphs and .
Proposition 2.2.
Let and be positive integers. Then
Proposition 2.3.
If , where is a component of with , for every , then
where .
The following proposition appears in the proof of Lema 12 in [11].
Proposition 2.4.
Let be a subset of and . Then is isomorphic to the graph obtained from by deleting all vertices in such that have al least one element of .
In [5] was proved that , . This is sequence in The OnLine Encyclopedia of Integer Sequences (OEIS) [22].
Proposition 2.5.
Let , .

.

, , .

, .
3 Proof of Theorem 1.1
In , we consider that , and . We use some propositions in order to prove our main result in this section.
If is the set of all subsets of and , then is a partition of . Notice that the subgraph of induced by is isomorphic to and the subgraph induced by is isomorphic to . Sometimes we use and as set of vertices or as the corresponding induced subgraph.
For , we define the following subsets of vertices of
In fact, , for every (see Figure 1 for an example).
Proposition 3.1.
Let be an integer. Then , for all .
Proof.
By proposition 2.4, is isomorphic to the double vertex graph of , for every . If , then is isomorphic to and hence . If , then consists of two components: one isomorphic to and the other isomorphic to . Therefore
Finally, if , then consists of two components: and . Then, by Proposition 2.3 it follows that the double vertex graph of , that is isomorphic to , has three components. The first one is the double vertex graph of and is isomorphic to . The second component is the double vertex graph of and is isomorphic to . Finally, the third component is isomorphic to the grid graph . Therefore
and hence
∎
In Figure 2 (a) and (b) we show the graphs and , respectively.
Proposition 3.2.
Let be an integer. Let be a nonempty subset of such that in there does not exist consecutive integers. Then . Even more, if , then .
Proof.
If the result follows from Proposition 3.1. If then is an induced subgraph of , for every and by Lemma 2.1 it follows that . In the view of Lemma 2.1, it is enough to prove the second part of the affirmation for . The proof is by contradiction. Let , with . Suppose that . Let be an independent set in of cardinality . We have two cases.
Case 1. If , then , and hence . Also because , by hypothesis. Therefore the set will be another independent set of of cardinality greater than , a contradiction.
Caso 2. . Let and . As the vertices and belong to , then both vertices does not belong to . The open neighborhood of is a subset of . Of this vertices, only could be in . Therefore, the set is and independent set in (because ) of cardinality greater than , a contradiction. ∎
In Figure 2 (c) we show the graph .
It is clear that . We are ready to prove our result.
Theorem 1.1.
Let be an integer. Then
Proof.
The case can be checked by hand or by computer. We suppose that . As is isomorphic to it follows from Lemma 2.1 that . We will show that . We use the fact that the subgraph of induced by the vertex set is isomorphic to and .
Let be an independent set in . We have the following cases.
Case 1. If , then .
Case 2. If , then because .
Case 3. , with , and both nonempty. Let . As is an independent set, in there does not exists any two consecutive integers. We claim that . Indeed, suppose that . Then , for some . Without lost of generality we can suppose that , with . By definition of , vertex belong to . Now that is an edge in , and hence in . But this is a contradiction, since is an independent set. This shows that is a subset of and hence
(1) 
Now we show that . If , then and by Proposition 3.1
where the last inequality follows from Proposition 2.5(2). Finally, consider that . As , then . By Proposition 3.2 and Equation (1) we have that que
∎
4 Proof of Theorem 1.2
For the wheel graph we consider that , and . Let denote the subgraph of induced by all subsets of . Let denote the subgraph of induced by the vertex set . The graph is isomorphic to and is isomorphic to . We also use and as vertex sets. It is wellknown that and in [5] was proved that , .
It can be checked by computer that . We now prove our main result in this section.
Theorem 1.2.
Let be and integer. Then
Proof.
As is isomorphic to , then every independent set in satisfies . We will show that .
Let be an independent set in . If , then . If , then
Now suppose that , where , , and with both non empty sets. As is isomorphic to then . For , let defined as in previous section. Let
In a similar way that in the proof of Theorem 1.1, it can be showed that .
Therefore we have .
By Proposition 2.4, the graph is isomorphic to the double vertex graph of , for every . But as is isomorphic to then . We like to bound using that . First, consider that , for some , that is . By the previous paragraphs we have that
where the last inequality holds because .
Now, suppose that . First note that if , then the double vertex graph of is isomorphic to the double vertex graph of . Therefore, by Proposition 2.4 it follows that
(2) 
We like to obtain . By Equation (2), after we delete one set and the vertex from , for , we obtain an isomorphic copy of , which in turn contains isomorphic copies of the remaining sets , for . Then we are in Case (3) of the proof of Theorem 1.1 for and (with the corresponding relabeling given by the isomorphism between and ). Therefore, for any
Using Proposition 3.2 for and we have that
And hence
∎
5 Proof of Theorem 1.3
The proof of Theorem 1.3 follows directly from the following result.
Theorem 5.1.
For any non negative integer we have
Proof.
Without loss of generality we can suppose that for , , and for in , . Let be the function given by . It is an exercise to show that this function is a graph isomorphism between and . ∎
6 Proof of Theorem 1.4
The vertex set of can be partitioned in where (as induced graph) is isomorphic to (that is isomorphic to by Theorem 5.1) and
Notice that the subgraph of induced by is isomorphic to .
We define the following subset of vertices of .
The following proposition will be useful in the proof of Theorem 1.4
Proposition 6.1.
For , we have that , for any .
Proof.
Notice that for , the graph is isomorphic to the graph . We have several cases.
Case 1. If , then the graph is isomorphic to , that it is isomorphic to (by Theorem 5.1), and hence .
Case 2. If , then consists of three components as follows. One component that is isomorphic to . Such component is either the vertex , or the vertex if or , respectively. Another component consists either, in the subgraph generated by the vertices , when , or when . This component is isomorphic to . The last component is isomorphic to : when , will be the subgraph generated by the set of vertices , and when , will be the subgraph generated by the set of vertices . Then
where, for the last inequality, we use the fact that and part 2 of Proposition 2.5.
Case 3. If , then consists of three components that came from the double vertex graph of . The first component is isomorphic to , that in fact is isomorphic to . The other component is isomorphic to that is isomorphic to , which in turn is isomorphic to . The last component of is the subgraph of induced for the set of vertices of the form with and . This last component is isomorphic to the grid graph . Therefore
Therefore
∎
Corollary 6.2.
, for every , .
Proof.
In Figure 3 we show graph .
Theorem 1.4.
Let be an integer. Then
Proof.
Let be an independent set in of cardinality . As is an independent set in then .
We will show that . The case is easy so that we suppose that . Let be an independent set in . We have several cases.
Case 1. If , then .
Case 2. If , then , because and .
7 Proof of Theorem 1.5
We proof Theorem 1.5 by mean of Propositions 7.3 and 7.5. For , let
It is clear that , for every , and that is a partition of . The following proposition shows that most of the sets are independent sets in .
Proposition 7.1.
If is not an independent set in , then , where .
Proof.
Clearly and are independent sets and hence . As is not an independent set and , then there exists two adjacent vertices, say and , in . Notice that and . Therefore and . From these equations we obtain that , which implies that . ∎
Let be a graph and let and subsets of . We say that and are linked in , and is denoted by , if has an edge such that and .
Proposition 7.2.
Let . The subsets of previously defined are linked as follows:

, for .

, for .

All the links between the elements in are given by (1) and (2).
Proof.
(1) For , the sets and are linked because and is an edge in .
(2) By definition of it follows that and belongs to , and the vertices and belongs to . As and are edges in we obtain that .
(3) If is linked with a set , with , then, by the construction of , the unique possible vertices in that could be adjacent with vertices in are and . But this would implies that and we are in Case 2.
∎
Proposition 7.3.
Let be an integer. Then
Proof.
Now, as is a subgraph of with and , then . Using Corollary LABEL:indepcpn we obtain
∎
The following proposition will be useful for the case odd.
Proposition 7.4.
Let