Abstract
Jarnicki, Myrvold, Saltzman, and Wagon conjectured that if is Hamilton-connected and not , then its Mycielski graph is Hamilton-connected. In this paper, we confirm that the conjecture is true for three families of graphs: the graphs with , generalized Petersen graphs and , and the cubes . In addition, if is pancyclic, then is pancyclic.
1. Introduction
All graphs considered in this paper are simple and finite. For notations and terminologies not defined here, we refer to Bondy and Murty [1]. A spanning cycle (path) of a graph is called Hamilton cycle (Hamilton path). A graph which contains a Hamiltonian path between every two vertices of is called Hamilton-connected (HC). Mycielski [2] proved that the chromatic numbers of triangle-free graphs can be arbitrarily large by introducing a graph transformation as follows. For a graph on vertices , its Mycielski graph, denoted by , is the graph on vertices with edges for all and edges , , and for all edges in . In recent years, a number of papers are devoted to various properties of Mycielski graphs, such as Hamilton-connectivity, Hamiltonicity [3–7], total chromatic number [8, 9], circular chromatic number [10–14], and connectivity [15, 16]. Fisher et al. [4] obtained the following results.
Theorem 1 (see Fisher et al. [4]). The following results hold for a graph :(1)If is Hamiltonian, then is Hamiltonian(2)If is not connected, then is not Hamiltonian(3)If has at least two pendant vertices, then is not HamiltonianCheng, Wang, and Liu studied Hamiltonicity and Hamilton-connectedness in Mycielski graphs of bipartite graphs.
Theorem 2 (see Cheng et al. [3]). For a bipartite graph , the following are true:(1)If is Hamiltonian, then is balanced(2)If is Hamiltonian, then has a Hamilton pathIn 2017, Jarnicki et al. [17] established the following results for being Hamilton-connected or not.
Theorem 3 (see Jarnicki et al. [17]). The following results hold for a graph :(1)If is an odd cycle, then is Hamilton-connected(2)If is a Hamilton-connected graph with order odd, then is Hamilton-connected(3)If is an even cycle, then is not Hamilton-connectedThey posed the following conjecture.
Conjecture 1 (see Jarnicki et al. [17]). If is Hamilton-connected and not , then is Hamilton-connected.
In this paper, we confirm that the conjecture is true for three families of graphs: the graphs with , generalized Petersen graphs and , and the cubes . In addition, if is pancyclic, then is pancyclic.
2. Mycielski Factor
Let be a connected graph of order even, and . We call a connected spanning subgraph of to be a Mycielski factor starting at if it consists of an even number of odd cycles (possibly ) and an even cycle with the chord (possibly empty), joined by edges , where for each such that for each , and the chord joins and a vertex at distance even on .
Lemma 1. Assume that a graph is Hamilton-connected. If, for any , there exists a Mycielski factor starting at , then is Hamilton-connected.
Proof. As in the assumption, let be HC. Trivially, has a Hamilton cycle. By Theorem 3 (2), is HC if the order of is odd. So, it remains to tackle the case when the order is even. Let , where . Recall that . Take any two vertices . We consider five cases in terms of the location of and in and .
Case 1. and .
Without loss of generality, let and . Since is HC, there exists a Hamilton path connecting and in . We shall find a Hamilton path of depending on as follows. Zigzag up from until is reached. Then, jump via z to , and zigzag right until is reached. Formally, it isas shown in Figure 1.
Case 2. and .
Without loss of generality, let and . Since is HC, there exists a Hamilton path connecting and in . Thus, there exists a neighbor, say , of . Zigzag up from to and then back to , zigzag up to and then up to , and zigzag left to and then up to and , as shown in Figure 2. Formally,
Case 3. .
Without loss of generality, let and . Since is HC, there exists a Hamilton path connecting and in . We are able to find a Hamilton path joining and : zigzag up from to and then up to , and zigzag left to , and then reach and , as shown in Figure 3. Formally, it is
Case 4. .
Without loss of generality, let . Since is HC, has a Hamilton cycle . Label the vertices of as . We are able to find a Hamilton path joining and : zigzag from to and then go to , and then zigzag right to , and finish at , as shown in Figure 4. Formally, it is
Case 5. .
Without loss of generality, let . Let be a Mycielski factor of starting at , which consists of an even number of odd cycles (possibly ) and an even cycle with the chord (possibly empty), joined by edges , where for each such that for each , and the chord joins and a vertex at distance even on .
If , for every integer , label the vertices of in the clockwise order . One can find a Hamilton path of as follows:where and .
Let be an even cycle with chord in . One can find a Hamilton path of as follows:where and .
Thus, is a Hamilton path of from joining and .
3. Hamiltonian Connectedness
Theorem 4. Assume that is a Hamilton-connected graph of order . If , then is Hamilton-connected.
Proof. Let be a vertex of . We consider a Hamilton cycle of . Let be a neighbor of on . Since , it has a neighbor at distance even on . By Lemma 1, is HC.
The power of a graph , denoted by , is a graph with the same vertex set as in which two vertices are adjacent if and only if their distance in is at most . Thus, . We need the following result due to Karaganis [18].
Theorem 5 (see Karaganis [18]). The cube of every connected graph of order is Hamilton-connected.
Theorem 6. For any connected graph of order , is Hamilton-connected.
Proof. By Theorem 5, is HC for . Since is a spanning subgraph of for any spanning graph of , to show is HC, it suffices to show that is HC for any tree of order . Since is HC, by Theorem 3 (1), we may assume that is even. By Lemma 1, it remains to show that has a Mycielski factor starting from each vertex .
Let be a neighbor of in , and let and be the components of containing and , respectively. Let and be the order of and , respectively. Let be a neighbor of in , and let be a neighbor of in . Case 1: both and are at least 3. Subcase 1.1: both and are odd. By Theorem 5, both and are HC. Let and be Hamilton cycles of and , respectively. One can see that is a Mycielski factor of starting at . Subcase 1.2: both and are even. By the induction hypothesis, has a Hamilton path joining and , and has a Hamilton path joining and . One can see that is a Mycielski factor of starting at . Case 2: . Subcase 2.1: . Subcase 2.1.1: . Since is an even number at least 3, is an odd number at least 3. By Theorem 5, let be a Hamilton cycle of containing . It is easy to see that is a Mycielski factor of starting at . Subcase 2.1.2: . If , then . Trivially, has a Mycielski factor starting at . If , then is an even number at least 4. By Theorem 5, let be a Hamilton path of . It can be seen that is a Mycielski factor starting at . Subcase 2.2: .If , then has a Mycielski factor starting at . Next, we assume that . We can choose a neighbor of such that . Combining with our assumption that , we have . By Theorem 5, let be a Hamilton path of . It can be checked that is a Mycielski factor starting at .
In 1969, Watkins [19] introduced the notion of the generalized Petersen graph , , as follows. The vertex set is , and the edge set is , where the subscript arithmetic performs modulo using the residues . In 1971, Frucht et al. [20] showed that is vertex-transitive if and only if or . Next, we consider the Hamilton-connectedness of the Mycielski graph of the generalized Petersen graphs and .
Theorem 7 (see Alspach and Liu [21]). The generalized Petersen graph with is Hamilton-connected if and only if .
Theorem 8. If is Hamilton-connected for , then is Hamilton-connected.
Proof. In view of Lemma 1, it suffices to show that has a Mycielski factor starting at any . We consider two cases: Case 1: . Since is odd, is vertex-transitive, and we may assume that , without loss of generality. Let and be the outer cycle and inner cycle of . Let be a vertex of . It is clear that is a Mycielski factor of starting at . Case 2: .By the symmetry, it suffices to tackle two possibilities according to the location of in : lies on the outer cycle or inner cycle of . Without loss of generality, let or .
First, for the case when , we can find a Mycielski factor of as follows:whereFor , by inserting 12 new vertices to of , we get as illustrated in Figures 5–7 for the case that lies in the outer cycle and for the case that lies in the inner cycle as illustrated in Figures 6, 8, and 9. For the case when , by inserting 12 new vertices to with type insertion, we obtain a Mycielski factor of starting at .
Theorem 9 (see Alspach and Liu [21]). The generalized Petersen graph with is Hamilton-connected if and only if is odd.
Theorem 10. If is Hamilton-connected, then is Hamilton-connected.
Proof. Since is Hamilton-connected, by Theorem 9, is an odd number at least 7. Let be a vertex of . In view of Lemma 1, it suffices to show that has a Mycielski factor starting at . We consider two cases: Case 1: . Since is odd, is vertex-transitive; by the symmetry, we may assume that , without loss of generality. Let and be the outer cycle and inner cycle of . It is clear that is a Mycielski factor of starting at . Case 2: .By the symmetry, it suffices to tackle two possibilities according to the location of in : lies on the outer cycle or inner cycle of . Without loss of generality, let or .
First, for the case when , we can find a Mycielski factor of starting at as follows:where .
For , by inserting 12 new vertices to of , we get as illustrated in Figures 10–12. For the case when , by inserting 12 new vertices to with type insertion, we obtain a Mycielski factor of starting at .
4. Pancyclicity
In this section, we show that if a graph is pancyclic, then is also pancyclic.
Theorem 11. If is pancyclic, then is pancyclic.
Proof. Let be a pancyclic graph of order . Since contains as its subgraph, contains a cycle of length for each .
Now, we find a cycle of length in . Take a cycle of length in . Without loss of generality, let be a path resulting from deleting an edge. It can be seen that is a cycle of length , as illustrated in Figure 13. In a similar way, one can find a cycle of length in in terms of a cycle of length in .
Next, we will find a cycle of length in for each . Take a Hamilton cycle . Without loss of generality, let in . We consider two cases according to the parity of : Case 1: is odd. One can find a cycle of length n+k in μ(G), as shown in Figure 14. Formally, it is Case 2: is even.Zigzag up from to and left to , then zigzag left to , , , and , and go right to and back to , as shown in Figure 15. Formally, it is
5. Conclusion
In this paper, we introduce the notion of the Mycielski factor of a graph. If a graph has a Mycielski factor starting at for any , then is Hamilton-connected. Applying this result, we are able to show that if a graph belongs to three (well-defined) families of graphs, then is Hamilton-connected. However, the full conjecture of Jarnicki, Myrvold, Saltzman, and Wagon is not yet solved. We also prove that if is pancyclic, then is pancyclic.
One of the reviewers proposed the following two interesting problems.
Zhong et al. [7] showed that the line graph of the generalized Petersen graph is always Hamilton-connected. Is it easy to show that the Mycielski graph of is Hamilton-connected?
It is known that the line graph of a Hamilton-connected graph is also Hamilton-connected. Is Hamilton-connected if is Hamilton-connected? [22].
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This research was supported by the Key Laboratory Project of Xinjiang (2018D04017), NSFC.(No. 12061073, 11801487), and XJEDU2019I001.