Regularity and Planarity of Token Graphs
Abstract
Let be a graph of order and let be an integer. The token graph of is the graph whose vertices are all the subsets of , two of which are adjacent whenever their symmetric difference is a pair of adjacent vertices in . In this paper we characterize precisely, for each value of , which graphs have a regular token graph and which connected graphs have a planar token graph.
Keywords: Token graph; Johnson graph; Regularity; Planarity.
AMS Subject Classification Numbers: 05C10; 05C69.
1 Introduction.
Throughout this paper, denotes a simple graph of vertices and is an integer with . The token graph of is the graph whose vertices are all the subsets of , where two such subsets are adjacent whenever their symmetric difference is a pair of adjacent vertices in . The token graph was introduced in [4] where some of their properties were studied. In that paper the authors noted that:
“Thus vertices of correspond to configurations of indistinguishable tokens placed at different vertices of , where two configurations are adjacent whenever one configuration can be reached from the other by moving one token along an edge from its current position to an unoccupied vertex. ”
As an example, the token graph of the cycle graph is shown in Figure 1. Clearly, ; we say that and are the trivial token graphs of .
The Johnson graph is the graph whose vertices are the subsets of an set, where two such subsets and are adjacent whenever . Thus, the Johnson graph is isomorphic to the token graph of the complete graph , i.e., . Therefore, results obtained for token graphs also apply for Johnson graphs; Johnson graphs are widely studied due to connections with coding theory, see, e.g., [3, 5, 6, 9, 11].
We write whenever and are adjacent vertices in . The edge joining these vertices is denoted by . For a nonempty set , and a vertex , denotes the set of neighbors that has in , i.e. ; the degree of in is denoted by . For a vertex , denotes the set of neighbors that has in , i.e., ; and denotes the closed neighborhood of the vertex , i.e., . We denote by the degree of a vertex in , and by the minimum and maximum degree of , respectively. The complement of a nonempty set is denoted by and the complement of by . The subgraph induced by is denoted by . As usual, denotes the indicator function of , i.e., if and otherwise.
In this paper we characterize precisely, for each value of , which graphs have a regular token graph and which connected graphs have a planar token graph; in particular we show the following.
Theorem 1.1.
Let be a graph of vertices and let be an integer. Then is regular if and only if one of the following four cases holds.

is isomorphic to the complete graph on vertices;

is isomorphic to ;

is isomorphic to complete bipartite graph and ;

is isomorphic to and .
Theorem 1.2.
Let be a connected graph of order and let be an integer. Then is planar if and only if or , and .
2 Regularity
In this section we prove Theorem 1.1. We split the proof in two cases: whether is regular or not. This are shown in Theorems 2.4 and 2.9, respectively
Let be the complement of with respect to the Johnson graph, i.e.,
The following statement follows easily from the definitions.
Proposition 2.1.
for every graph .
Since the Johnson graph is a regular graph, Proposition 2.1 has the following direct consequence.
Corollary 2.2.
Let be a graph such that is regular; then is also regular.
2.1 Regular token graph of a regular graph
In this section we answer the following question: when is the token graph of a regular graph also regular? We show in Theorem 2.4 that there are exactly two regular graphs which produce regular token graphs.
Lemma 2.3.
Let be a regular graph and such that is regular. Then there exist a constant , depending on , the degree of and the degree of , such that for every and every .
Proof.
Fix and . Let with and let . Let and be the degrees of and , respectively.
We first claim that . Note that
Analogously, we obtain
Since is regular, the claim follows.
For every , let ; by the claim we have that . Note that for every , and that . Thus, we have for every . Furthermore, we have vertices with and vertices with . Therefore,
The result follows with . ∎
It is a simple fact that is a regular graph for every admissible if is any empty graph or any complete graph . The following result shows that they are the unique regular graphs with regular token graph for .
Theorem 2.4.
Let be regular graph not isomorphic to either or . Then its token graph , with , is nonregular.
Proof.
Suppose to the contrary that is regular. Since is not isomorphic to nor to , there exists a vertex of such that the following holds. Vertex is of degree at least one and there exists another vertex not adjacent to . Let be a vertex of such that and . Consider any vertex and let . Hence, we have that which contradicts Lemma 2.3. ∎
2.2 Regular token graphs of nonregular graphs
In this section we show that there are exactly two nonregular graphs which produce regular token graphs. Throughout this subsection, is a fixed nonregular graph, and are vertices of such that , and . Also, we partition into four subsets:
Note that consists precisely of the vertices of which are nonadjacent to neither nor (see Figure 2).
In this context, we have the following statements.
Lemma 2.5.
If , then is nonregular for .
Proof.
We analyze three cases separately.

.
We choose a fixed such that . Now consider the vertices of defined as follows: and . By definition of we know that the degree of the vertex (respectively, ) in corresponds to the number of edges of with one end in (respectively, ) and the other in (respectively, ). Let be the number of edges of with one end in and the other in . Then and . Since , we have . Hence, is not regular.

.
We choose a fixed nonempty subset of such that . Thus the sets and are vertices of . Let denote the number of edges of with one end in and the other in . Clearly, and (recall that ). Thus , because . Hence, is not regular.

.
We choose a fixed nonempty subset of such that . Thus the sets and are vertices of . Let denote the number of edges of with one end in and the other in . It is easy to see that and . Thus , because . Hence, is not regular.
∎
Lemma 2.6.
If is a regular graph for some , then is adjacent to every vertex in in .
Proof.
Since is isomorphic to , we may assume that . By Lemma 2.5 and the hypothesis we have that . Then because . Suppose that there exists a vertex in that is nonadjacent to . Note that must be an element of .
Let be a fixed subset of such that . Thus the sets and are vertices of . Let be the number of edges of with one end in and the other in . Then and . By the regularity of we have , or equivalently
(1) 
Similarly, let be a fixed subset of such that . Thus the sets and are vertices of . Let be the number of edges of with one end in and the other in . Then and Since , we have that , or equivalently
(2) 
Which implies that , a contradiction. ∎
The next statement is an immediate consequence of Lemma 2.6.
Corollary 2.7.
If is a regular graph, for some , then and are empty sets.
Lemma 2.8.
If is a regular graph for some , then .
Proof.
Again, we may assume that because is isomorphic to . Suppose to the contrary that . By Corollary 2.7, and are empty; therefore, .
First suppose that . Let be a fixed subset of such that . Then the sets and are vertices of F. Let be the number of edges of with one end in and the other in . Then and . In the last equation we are using that (Lemma 2.6) . Since , we have that , or equivalently, . This implies that , a contradiction.
Now suppose that . Let be a fixed subset of such that . Then the sets and are vertices of F. Let be the number of edges of with one end in and the other in . Then and . Since , we have that , or equivalently, . This implies, , a contradiction. ∎
We are ready to prove the main result of this section.
Theorem 2.9.
Let be a nonregular graph such that is regular for some . Then is isomorphic to or to , and in both cases .
Proof.
Again, we assume that . First, we show that there are only two possible degrees in . Suppose to the contrary that there exists three vertices and such that . Apply Lemma 2.8 twice: once with and , and a second time with and . Then and . We obtain a contradiction by applying Lemma 2.6 with and , since is not adjacent to .
Let , , with , be the only two possible degrees in . By Lemma 2.8 we have that or . Let and be vertices of of degree and , respectively.
Suppose that . We claim that is the only vertex of degree . Suppose that there exists a second vertex with degree . We arrive at a contradiction by applying Lemma 2.6 with and , as and are not adjacent. Therefore, all vertices of distinct from have degree . Moreover, Lemma 2.6 with and implies that . Therefore, is isomorphic to . Now we show that . Let be any subset of with . Then the sets and , where is any element in , are vertices of with and . As is a regular graph we have that which implies that .
Suppose that . Then, Lemma 2.6 with and implies and is adjacent to every vertex in . Therefore, there cannot be another vertex distinct from adjacent to every vertex in since . Thus, is isomorphic to . Finally we show that . Let be any subset of with . Let and , where is any element in . Then and are vertices of with and . As is a regular graph we have that which implies that .
∎
3 Planarity
In this section we fully characterize, in terms of , when the token graph of is planar. Since , we have that and are planar if and only if is planar; therefore, we only consider the cases when and .
As usual, we denote by the graph obtained from graph by contracting the edge of ; and also, we denote by (, respectively) the graph obtained from graph by deleting the edge (vertex , respectively) of . A graph is a minor of a graph if a graph isomorphic to can be obtained from by contracting some edges, deleting some edges, and deleting some isolated vertices. A graph is a subdivision of a graph if can be obtained from by subdividing some edges.
Kuratowski [8] proved that a graph is planar if and only if it does not contain a subdivision of the complete graph nor a subdivision of the bipartite graph . Wagner [8] proved that a graph is planar if and only if it does not contain the complete graph nor the complete bipartite graph as a minor.
First we show that if is a minor of then is a minor of . This result, which is of independent interest, is used to prove the main theorems of this section.
Lemma 3.1.
If is a minor of then is a minor of .
Proof.
First suppose that is obtained from from applying one minor operation on . That is by deleting a vertex, deleting an edge or contracting an edge; we show that is a minor of .

is obtained from by deleting a vertex .
Then is isomorphic to the graph obtained from by deleting all the vertices of which contain . Thus, is a minor of .

is obtained from by deleting an edge .
Then is isomorphic to the graph obtained from by deleting all the edges of such that . Thus, is a minor of .

is obtained from by contracting an edge .
Consider the following subsets of :
Clearly, are disjoint, and . Since we have that for each there is exactly one with in , in fact . Hence, by contracting these edges from , we obtain a subgraph isomorphic to . Thus, is a minor of .
Now suppose that is obtained by applying two or more minor operations on . Since the minor relation is a transitive relation, the result follows by induction on the number of these operations. ∎
Lemma 3.1 implies the following.
Theorem 3.2.
Let be a graph and let be a minor of such that every nontrivial token graph of is nonplanar. Then is nonplanar for .
Notice that the order of in Theorem 3.2 is greater than 4 because is a planar graph. The circumference of a graph is the supremum of the lengths of its cycles, if is a tree we define . The following theorem implies the nonplanarity of many token graphs.
Theorem 3.3.
Let be a graph and . If or are greater than or equal to then is nonplanar.
Proof.
Since contains as a minor a star graph or a cycle graph and by Theorem 3.2, it suffices to check that , and are nonplanar. As shown in Figures 1 and 3, and both contain a subdivision of and thus are not planar. For the case of , note that by contracting the edges joining vertices of the same color in Figure 3 we obtain the complete graph . Thus, is nonplanar and the proof is completed. ∎
The following results shows that others token graphs not included in the previous theorems are nonplanar, too. In particular paths do not verify the hypotheses of previous theorems; however, many of their token graphs are nonplanar.
Proposition 3.4.
Let be a graph containing a path on three vertices as a subgraph and let . If or have maximum degree greater than , then is nonplanar.
Proposition 3.5.
Let be a graph and . If contains two disjoint subgraphs isomorphic to and to respectively, then is nonplanar.
We now show that the nontrivial token graphs of all trees (with the exception ) of more than vertices are nonplanar.
Theorem 3.6.
Let be a tree of order nonisomorphic to . Then for every , the graph is nonplanar.
Proof.
Let . Since we have . The case when has maximum degree follows by Theorem 3.3. We consider the cases and separately.
Suppose that . Consider a vertex with . Note that if there exists a vertex with then we have a subgraph , and so is nonplanar by Proposition 3.5. Now, if , for every , and since , then there are vertices in such that the distance for with . Therefore, Proposition 3.5 implies that is nonplanar.
Suppose that . Consider a vertex with and let . Suppose first that no other vertex of has degree . Since , there is a path in and so Proposition 3.5 gives that is nonplanar. Suppose now that there are at least two vertices in with degree . Note that if there is a vertex with , then there is a path in , and so, Proposition 3.5 gives that is nonplanar. Thus we can assume that there is no with . Note that if then Proposition 3.5 gives that is nonplanar taking a in . Thus, we can assume without loss of generality that and for . Since , there exists a in , and so, Proposition 3.5 gives that is nonplanar. ∎
Note that the token graph of every path graph with vertices is planar, see [4, Figure 1]. However, the following result shows that the token graph of is nonplanar for and .
Theorem 3.7.
Let be a graph and . If contains a path with vertices, then is nonplanar.
Proof.
Let be a path of seven vertices in . We first show that contains as a subgraph. This follows immediately if . Thus, assume that . Fix tokens at vertices in and let be the subgraph of that results from moving the remaining four tokens freely. Then, contains as a subgraph, but . Thus is a subgraph of as claimed. Now, by Proposition 3.4 and the fact that maximum degree of is three, is nonplanar. The result follows. ∎
Theorem 2.
Let be a connected graph of order and let be an integer. Then is planar if and only if or , and .
Proof.
3.1 Graphs of small order
The only one nontrivial token graph of is , it is isomorphic to the octahedral graph, which is planar. Therefore, all token graphs of graphs with at most four vertices are planar. Theorem 1.2 implies that all the nontrivial token graphs of a connected graph not isomorphic of more than ten vertices are nonplanar. Thus, the graphs of more than four and at most ten vertices, whose nontrivial token graphs are planar, remain to be found. Note that if is a subgraph of then is a subgraph of . Therefore, it is sufficient to search for the connected graphs edgemaximal with the property that their token graphs are planar.
For each value of , we did an exhaustive search for these graphs as follows. Since , we considered only those graphs of order at least . Using nauty [10] we generated every connected graph on vertices and edges. We started our search at ; afterwards, we increased by one. We stopped as soon as all graphs of vertices and edges have nonplanar token graphs. For a given graph we tested whether its token graph is planar and whether the addition of any new edge to produces a graph whose token graph is nonplanar. To check for planarity we used sage [12], which in turn uses Boyer’s implementation of [1]. We found connected graphs edgemaximal with the property that their token graphs are planar; these are shown in Figure 4. For we found the two graphs of six vertices shown on Figure 5. All token graphs of connected graphs with seven or more vertices are nonplanar. For , all connected graphs of or more vertices have nonplanar token graphs.
Acknowledgments
The authors would like to thank the anonymous referee for his/her useful suggestions. W. C. was partially supported by the Spanish Ministry of Economy and Competitiveness through project MTM201346374P. J. L. was partially supported by CONACyT Mexico grant 179867 and by European Union and Republic of Slovenia through the grant ”Internationalization as the pillar of development of University of Maribor”. L. M. R. was partially supported by PROMEP grant UAZCA169 and PIFI (Mexico).
Footnotes
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