Abstract
The circumference of a graph is the length of a longest cycle in , denoted by . For any even number , let = min { is a 3-connected cubic triangle-free plane graph with vertices}. In this paper, we show that an upper bound of is for .
1. Introduction
Graph theory has applications in various research areas. In chemistry and biology, it gives an illustration of molecular structures in [1, 2], and it identifies interactions between humans and nature in [3]. Graph theory can be combined together with other branches of mathematics such as algebraic graph theory, matroid theory, graph labeling, and topological graph theory. Recently, it is combined with fixed point theory in [4, 5].
We consider only simple undirected graph. Let = (, ) be a graph, where is the vertex set of and is the edge set of . The degree of a vertex is the number of edges incident to . A graph is cubic if all vertices have degree exactly three. A plane graph is a graph embedded in a plane without edge crossing. An -path of a graph is a sequence of distinct vertices of such that , for all . Similar to -path, a cycle of a graph is defined as a sequence of distinct vertices , of such that , , for all . The triangle is a cycle of length 3, and a graph is triangle-free if it contains no triangle. A graph is connected if there is a -path for all distinct . A graph is -connected if , and for any of size , a graph is connected. A -leg fragment is a connected plane graph such that there are exactly vertices of with degree two, and the remaining vertices have degree three. A Hamilton cycle of a graph is a cycle containing all vertices of .
Finding Hamilton cycles in graphs is a famous topic in graph theory. In 1880, Tait [6] conjectured that a 3-connected cubic plane graph has a Hamilton cycle. Tutte [7] constructed the Tutte fragment, a triangle-free 3-leg fragment with 15 vertices, and showed the following statement.
Lemma 1 (see [7]). Let be the Tutte fragment in Figure 1. Then,(i)There is an -path (respectively, an -path) containing all vertices of , and(ii)For each -path , , i.e., misses at least one vertex from .
Using three Tutte fragments and one vertex, the Tutte graph is the first counterexample of Tait’s conjecture with 46 vertices. Later, Holton and McKay [8] showed that a smallest counterexample of this conjecture is the Barnette–Horák–Lederburg graph, a 38-vertex triangle-free graph containing two Tutte fragments. Moreover, Grinberg graph [9] and Faulkner–Younger graph [10] are other counterexamples of this conjecture without any Tutte fragment.
For 4-connected graphs embedded in surfaces, Tutte [11] proved that every 4-connected plane graph has a Hamilton cycle. Extending Tutte’s result, Thomas and Yu [12] proved that every 4-connected projective plane graph has a Hamilton cycle. In the case of toroidal graph, Grünbaum [13] and Nash-Williams [14] independently conjectured that every 4-connected toroidal graph has a Hamilton cycle. For the closest result of this conjecture, Thomas and Yu [15] proved that every 5-connected toroidal graph has a Hamilton cycle. A closed knight’s tour on the chessboard is an example of Hamilton cycle problem. Some results of a closed knight’s tour are shown in [16, 17].
Define the circumference of a graph as the length of a longest cycle in a graph , denoted by . Since Tait’s conjecture is false, the generalized question is what is the minimum circumference of any 3-connected cubic (triangle-free) plane graph with vertices? Then, let { is a 3-connected cubic plane graph with vertices} and { is a 3-connected cubic triangle-free plane graph with vertices}.
Note that , and must be even because the sum of degrees of all vertices is even. For the exact value of , Holton and McKay [8] showed that if , and . Note that is still unknown for .
For a lower bound of , Jackson [18] showed that an -vertex 3-connected cubic graph has circumference for sufficiently large . Bilinski et al. [19] improved this to , and then, Liu et al. [20] improved to . This implies for sufficiently large .
For an upper bound of for all even numbers . Let be the sequence of graphs such that is the the Barnette–Horák–Lederburg graph, and is obtained from by replacing all vertices of by triangle. Then, is a -vertex graph with circumference . Note that a longest cycle of misses vertices. Let be a graph obtained by by replacing vertices of by triangle, where . Then, . This gives if .
For a triangle-free graph, Lu [21] constructed a -vertex graph, where and , with circumference . If , then this graph implies for infinitely many integers . Next, Lo and Schmidt [22] constructed a -vertex graph with circumference at most for some positive integer . This implies for some which are divisible by 122.
In this paper, we show that an upper bound of is 133 in Theorem 1, and the graph , a -vertex graph, has the circumference in Lemma 4. In Theorem 2, we improve an upper bound of to , for all .
2. Main Results
First, we construct the 4-leg fragment with 136 vertices as in Figure 2(a). Note that is triangle-free and contains nine Tutte fragments. Next, for any positive integer , we let be a copy of such that , , , and correspond to , , , and in , respectively. Finally, we construct a graph from the union of graphs , by adding edges , , , and , for all , as in Figure 3. Note that is a 3-connected cubic triangle-free plane graph with vertices.
(a)
(b)
By the construction of , a longest cycle of , where , must pass through some fragments . Since is a 4-leg fragment, are either an -path, where , , or a union of two disjoint paths where all end vertices of these paths are in . In order to find the circumference of , it is necessary to show the maximum size of each path of in Lemma 2 and of each union of two disjoint paths of in Lemma 3.
Lemma 2. Let be the 4-leg fragment in Figure 2(a). Then,(i)Each -path in contains at most 130 vertices(ii)Each -path (respectively, -path) in contains at most 132 vertices(iii)Each -path (respectively, -path) in contains at most 133 vertices(iv)Each -path in contains at most 119 vertices
Proof. Let be the graph obtained from by contracting all Tutte fragments into vertices , , , , , , , , and as in Figure 2(b). Note that and, the vertices , , , and of belong to the fragments represented by , , , and , respectively. Let be a longest -path, where , and let be a path of corresponding to . So, is the -path, where . Note that if passes through all vertices in , then , i.e., may miss the vertex and one vertex from each Tutte fragments. Otherwise, . Then, should be the maximum in . Thus, we consider four cases as follows. Case 1: is an -path. Then, is an -path. If , then either or . Suppose . Since there is only one -path missing only vertex , the sequence of vertices of is , , , , , , , , and . This implies passes through each fragment in this order. By Lemma 1, misses vertex and other five vertices from the fragments , , , , and . Then, . Case 2: is an -path, or a -path. By the symmetry of , we consider only -path. Since there is only one -path containing all vertices of , the sequence of vertices of is , , , , , , , , , and . By Lemma 1, misses four vertices from the fragments , , , and . Then, . Case 3: is an -path, or a -path. By the symmetry of , we consider only -path. Since there is only one -path containing all vertices of , the sequence of vertices of is , , , , , , , , , and . By Lemma 1, misses three vertices from the fragments , , and . Then, . Case 4: is a -path. Then, is a -path. If , then either or . Suppose . Since there is only one -path missing only vertex , the sequence of vertices of is , , , , , , , , and . Note that misses the vertex so misses 15 vertices from fragment . By Lemma 1, misses other two vertices from the fragments and . Then, .
For a union of two disjoint paths, we define a -dual path as the disjoint union of a -path and a -path. Note that the intersection of an -path and a -path in is not empty. By the symmetry of , there are exactly two cases of -dual path, where in as follows.
Lemma 3. Let be the 4-leg fragment in Figure 2(a). Then,(i)Each -dual path in contains at most 133 vertices(ii)Each -dual path in contains at most 134 vertices
Proof. Let be the graph defined from in Figure 2(b). Let be the largest -dual path in such that . Let be the dual path corresponding to . Without loss of generality, we consider two cases as follows: Case 1: is an -dual path. Then, is an -dual path. Since there is exactly one dual path containing all vertices of , the sequences of these two paths of are , , , , , , , and , , , respectively. By Lemma 1, misses three vertices from the fragments , , and . Then, . Case 2: is an -dual path. Then, is an -dual path. Without the loss of generality, there is only one dual path containing all vertices of . Then, the sequences of these two paths of are , , , , , , and , , , , respectively. By Lemma 1, misses two vertices from the fragments and . Then, .
Since we obtain Lemma 2 and Lemma 3, we have by showing the following theorem.
Theorem 1. There is a 136-vertex 3-connected cubic triangle-free plane graph with circumference 133.
Proof. Let be a graph obtained from the fragment by adding edges and (see Figure 4(a)). Note that is a 3-connected cubic triangle-free plane graph with 136 vertices. Let be the graph obtained from by contracting all Tutte fragments into vertices , , , , , , , , and as in Figure 4(b).
Let be a longest cycle of satisfying the condition in each following case.
(a)
(b)
Case 1. Neither nor is in . Let be the cycle of corresponding to . Then, neither nor is in . If , then either or . Suppose . Since there is only one cycle containing all vertices of , the sequence of vertices of is , , , , , , , , , , and . By Lemma 1, misses five vertices from the fragments , , , , and . Then, .
Case 2. Only one edge from is in . Assume without loss of the generality that , but . Then, is an -path in . By Lemma 2 (ii), .
Case 3. Both and are in . Then, is an -dual path in . By Lemma 3(i), .
Comparing the length of in each case, we conclude that .
Recall that a graph has vertices. Then, we give the circumference of as follows.
Lemma 4. The circumference of is for all positive integers .
Proof. For , we use the similar proof to Theorem 1 to show that = 134. Suppose . Let be a longest cycle of passing through all fragments . Without loss of generality, we generate each case by the possible cases of as follows. Case 1: is an -path, where and . Then, is an -path, where and , for any (see Figure 5(a)). By Lemma 2, . This implies . Case 2: is either a -path or an -path. Assume without loss of generality that is a -path. Then, is an -path, and is an -dual path, for any (see Figure 5(b)). By Lemma 2(ii) and Lemma 3(i), . Case 3: is an -dual path. Then, is an -dual path, for any . Note that , , , , for all (see Figure 5(c)). By Lemma 3, .For the case that a longest cycle does not intersect some fragments , we can assume, without loss of generality, that if , otherwise (see Figure 5(d)). This case is similar to Case 2, so we can show that = .
Then, we conclude that a longest cycle in is a cycle defined in Case 3 which implies that .
(a)
(b)
(c)
(d)
The ladder, denoted by , is a graph with and . Then, is a triangle-free 4-leg fragment with vertices. It is easy to show the following statements:(i)There is a -path (respectively, -path) containing all vertices of (ii)If is odd, then there are a longest -path (respectively, -path) containing all vertices of , and a longest -path (respectively, -path) missing exactly one vertex from (iii)If is even, then there are a longest -path (respectively, -path) containing all vertices of , and a longest -path (respectively, -path) missing exactly one vertex from (iv)For any such that except for the case , there is a -dual path containing all vertices of
Combining a graph and a ladder , we have the main theorem as follows.
Theorem 2. For any even number , there is an -vertex 3-connected cubic triangle-free plane graph with circumference .
Proof. For the case that is the multiple of 136, it is shown in Lemma 4. Assume that is not the multiple of 136. Let and . Let be an -vertex graph with and (see Figure 6). By this construction, it is easy to show that is a 3-connected cubic triangle-free plane graph. Let be a longest cycle of . Then, is a cycle defined in Case 3 of Lemma 4 which misses vertices from . Since is a subgraph of , a longest cycle of should miss at least vertices, i.e., .
Define paths : , , , and : , , . Let . Then, is a cycle in containing vertices. Hence, .
3. Discussion and Open Problems
It can be implied from Theorem 1 and Theorem 2 that , and if . This improves most of upper bound of from [21, 22] except for some cases in [21]. Then, we will compare ours to that result as follows.
Theorem 3 (see [21]). For any integer and , there is a 3-connected cubic triangle-free plane graph with vertices and circumference .
Putting appropriate and , this theorem gives a better upper bound of than ours in the following cases: if , if , if , if {270, 342, 414, 486, 558, 630}, if , if , if , if , if , if , if , if , if , if , if , if , if , and if .
For an open problem, we define (respectively, ) as the minimum of , where is a 3-connected cubic triangle-free projective plane graph (respectively, toroidal graph) with vertices. The Petersen graph, a 10-vertex graph, is a smallest 3-connected cubic triangle-free nonhamiltonian graph which can be embedded in both projective plane and torus. This implies that if , and . Since a plane graph is also a projective plane graph and a toroidal graph, we have the following inequality: and . Then, we give the following questions.
Question 1. What is an upper bound of ?
Question 2. What is an upper bound of ?
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 work was supported by Ramkhamhaeng University.