Abstract
A 4-tuple in a graph is a 3-arc if each of and is a path. The 3-arc graph of is the graph with vertex set all arcs of and edge set containing all edges joining and whenever is a 3-arc of . A Hamilton cycle is a closed path meeting each vertex of a graph. A graph including a Hamilton cycle is called Hamiltonian and has a Hamiltonian decomposition provided its edge set admits a partition into disjoint Hamilton cycles (possibly with a single perfect matching). The current paper proves that every connected 3-arc graph consists of more than one Hamilton cycle. Since the 3-arc graph of a cubic graph is 4-regular, it further proves that each 3-arc graph of a cubic graph in a certain family has a Hamiltonian decomposition.
1. Introduction
A path (cycle) in a graph passing each vertex is said to be a Hamilton path (cycle). A graph is hamiltonian if it consists of a Hamilton cycle and is Hamilton-connected if every pair of vertices are joined by a Hamilton path. A partition of the edge set into Hamilton cycles (or plus a perfect matching when is -regular with odd) is called a Hamiltonian decomposition of . The Hamiltonian problems and related topics are classical in combinatorics and have been received extensive treatment. See [1–3] for more results on this area of investigation.
The current paper concentrates on the class of 3-arc graphs. This class of graphs is produced from a graph operation named 3-arc graph operation, which is similar to a line graph operation. This operation was initially introduced in [4] in exploring certain families of imprimitive symmetric graphs. It was also exploited in investigating several classes of symmetric graphs. See [4, 5] and references therein.
We consider only a graph with finite order and undirected without loops. The term multigraph is used if there are multiple edges connecting the same pair of vertices. An arc of is a directed edge. If are adjoined in , we write to denote the directed edge outgoing from to , the directed edge from to , and the edge connecting and . A 4-tuple in a graph is a 3-arc if each of and is isomorphic to a path on three vertices. It may happen that . Suppose is a multigraph. A walk in of length is as follows:
With terms interchangely contained in vertex set and edge set , where and are connected by an edge , . A walk is closed if its originating and ending vertices coincide and is a trail if its all edges are pairwise different. A trail passing all edges of is an Eulerian trail, and an Eulerian tour is a closed Eulerian trail.
We refer to [1] for terms and notation not mentioned. The degree of in a graph is written as , and the minimum degree of is denoted by . We use to denote the set of all arcs outgoing from , and to represent the set of all arcs of .
In the following we give the formal definition of the 3-arc graph of a graph .
Definition 1. Let be a graph. The 3-arc graph of , is the graph with vertex set and all edges connecting and whenever is a 3-arc of .
We illustrate the definition of this construction by depicting the 3-arc graph of complete graph on four vertices and (see Figure 1). For instance, both and are 3-arcs in . The first one defines the adjacent relation between arcs and , and the latter gives rise to the adjacency between and .
It is not hard to see that is not directed with vertices and edges. It is shown that can actually be obtained from of by the next procedures: divide every vertex of into two, i.e., and ; for every pair of vertices of that are at a distance 2 away in , say, and are adjoined in , connect and . For more recent work in this direction, see [6]. Specifically, in [7], it is established that all connected graphs contain a Hamilton cycle.
In 1975, Sheehan proposed the following famous conjecture:
Conjecture 1 (Sheehan [8]). Each Hamiltonian graph that is regular of degree 4 consists of no less than two Hamilton cycles.
Motivated by Sheehan’s Conjecture and the work [7], we study the Hamiltonian decomposition problem for 3-arc graphs. We show every connected 3-arc graph contains a second Hamilton cycle, and the 3-arc graphs of a family of cubic graphs have Hamiltonian decompositions. As a corollary, we demonstrate that the 3-arc graph of a bipartite cubic graph includes one Hamiltonian decomposition.
To confirm Sheehan’s conjecture for 3-arc graphs, we actually prove a stronger result which asserts that any connected 3-arc graph contains no less than two Hamilton cycles. As the 3-arc graph of a cubic graph is 4-regular. We show further that each 3-arc graph of a cubic graph in a certain family has a Hamiltonian decomposition.
The primary results are presented as follows.
Theorem 1. Let be with a maximum degree and have a connected 3-arc graph . Then the number of Hamilton cycles of is no less than the following:
Remark 1. Since , is increasing and
Theorem 2. Suppose is a connected cubic graph. If contains an even IS-C decomposition, then has a Hamiltonian decomposition.
2. Proof of Theorems 1 and 2
We need the following notation and definitions.
A cactus is a graph in which any two cycles have at most one vertex in common. In particular, a tree is a special cactus. A cycle is said to be an odd (resp., even) cycle if it consists of an odd (resp., even) number of edges.
Definition 2. Let be a partition (i.e., and ) of . is said to be an independent set-cactus (IS-C) decomposition of if in the set induces an empty graph and induces a (connected) cactus.
Note that not every graph has an IS-C decomposition. For example, the complete graph on five or more vertices has no IS-C decomposition.
Since trees are acyclic, as a convention, we define the length of each cycle (which does not exist) of a tree to be 0. A cactus is called even if the length of each cycle is even. We treat a tree as an even cactus. An IS-C decomposition is called even if every cycle in the cactus is even.
Lemma 1. Each connected cubic graph has an IS-C decomposition.
Proof. Suppose is a connected cubic graph. If every two cycles of share one or less common vertex, then itself is a cactus. Clearly, an IS-C decomposition, , say, exists.
Suppose that has two distinct cycles sharing two or more common vertices. We construct an IS-C decomposition for as follows: choose any two cycles that share at least two common vertices. Denote the maximal common path occurring on both and by ; that is, any vertex not on occurs on at most one of the two cycles and . Then delete from and put it into the set ( takes the role of the independent set). Note that each neighbour of now has degree strictly less than 3, and hence occurs on at most one cycle in . So, the resultant graph is connected. To simplify notation, denote again by the graph resulting after deleting . If still has two cycles that share at least two common vertices. Denote the maximal common path occurring on both and by . Note that . Delete and put it into . Continue this process until the resultant graph becomes a cactus.
Since, at each stage, the set remains to be independent and the resulting graph stays to be connected, eventually becomes a desired IS-C decomposition.
Though every cubic graph has an IS-C decomposition, not every cubic graph has an even IS-C decomposition. For instance, has no even IS-C decomposition.
A 2-trail with the middle term is said to be a visit to. If , then and are viewed as different visits to . When we are not concerned about the directions of and , or its orientation is unknown, we denote . The following operation regarding two parallel edges will be needed.
Definition 3. Let be parallel edges joining two adjacent of , let be covered by a closed trail of length at least 4, . (It may happen that ). Then one of and , say , contains two visits with one covering and the other covering . Denote these two visits by and , where are two neighbors of other than , and, are edges between and respectively. Split the two 2-trails at and reconnect with , respectively. We call this the edge-shift operation of with respect to , and denote the resultant trail(s) by , or simply if .
Remark 2. A few comments on Definition 3 are ready:(1)If , remains to be a single closed trail if and only if the orientations of and are reverse to each other in ; in this case, this operation is, in fact, the edge version of the bow-tie operation (see, Definition 5 in [7]).(2)If , is a set of two closed trails if and only if the orientations and are the same between and in .(3)All edges covered by , are covered by .(4)After the edge-shift operation above, the three terms involved in are swapped. That is, is transformed into , and is transformed into . All other visits induced are kept or with orientation inverted.
Proof of Theorem 1. Suppose is a graph with connected. Clearly, . Denote the set of degree-2 vertices of by , the multigraph gained from by doubling its every edge by .
Suppose that . Since is connected, by [7], is Hamiltonian, and every vertex in has degree 2 or 3, is independent, and is connected. Further, it can be observed that contains at least 2 vertices. Suppose are two adjacent vertices in , is a neighbor of different from , and are two distinct neighbors of not equal (it may occur that one of equals ). Denote and .
Let , be two Eulerian tours of obtained without violating the condition (1) in Section 3 of [7], from extending to cover each edge of , respectively. Apply the bow-tie operation (Definition 5 in [7]) when necessary to process in such a way that the bipartite graph (Definition 3 in [7]) with respect to the resultant Eulerian tour has a perfect matching and the path is a segment of , . Note that both and exist. Consider . If has no perfect matching, then contains twin visits by Lemma 1 in [7], that is, . Apply the bow-tie operation to with regard to and anyone of the twin visits that is not occurring on , we get a new Eulerian tour , and one can observe that is a perfect matching of 3 independent edges. The bow-tie operation can be performed similarly when necessary at to produce a new Eulerian tour such that has a perfect matching and the path is maintained unchanged. It is analogous to showing that exists.
Let be two Hamilton cycles of derived from , , respectively. Denote the 3rd neighbor of other than by . Then one can observe that is connected to in , while connected to in , where are not necessarily distinct. Since , and are different Hamilton cycles of .
Suppose that . Suppose is the maximum-degree vertex, an Eulerian tour of with the property that contains a perfect matching for each (such a can be achieved by applying the bow-tie operations successively when needed). Denote the family of heavy visits to (that is, each element of is of the form , where is a neighbor of ). By performing a series of bow-tie operations, we will transform each visit of into a new visit containing three pairwise distinct terms.
To simplify notation we will still use to denote the new Eulerian tour produced by applying an operation. If , choose a pair of visits of , and apply bow-tie operation to with regarding to . Then in the new Eulerian tour , the pair is transformed into two visits with each containing three pairwise distinct terms. Thus, each time of applying the bow-tie operation to with regard to two elements of deducts the number of by two. Apply this operation until . If , we are done. Suppose . Let be an arbitrary visit outside . Apply the same operation to with regard to and . Again, can be eliminated without producing a new heavy visit. Eventually, we get an Eulerian tour, denoted again by , of such that each element of contains three pairwise distinct terms. Then in , each visit of is adjacent to arcs of , and each arc of is adjacent to visits of . That is, is a -regular bipartite graph.
If , is 2-regular, hence contains two distinct perfect matchings. Suppose that , and denote by the set of all perfect matchings of . Schrijver [9] states that every bipartite graph regular of degree with vertices contains at least perfect matchings. Thus, we have the following:For every vertex in , fix a perfect matching of . As in the Proof of Theorem 1 in [7], a Hamilton cycle of can be derived from by using the fixed perfect matchings of all bipartite graphs with , together with every single perfect matching of , . Note that corresponding to distinct , of , the derived Hamilton cycles of are also distinct. Therefore, has different Hamilton cycles and the result follows.
Suppose is a graph containing no isolates. If is connected and 4-regular, then clearly has a maximum degree of no less than 3. Hence by Theorem 1, we have the next result, which verifies Sheehan’s conjecture for 3-arc graphs.
Corollary 1. Every 4-regular 3-arc graph has no less than two Hamilton cycles.
Proof of Theorem 2. Suppose is a connected cubic graph and an even IS-C decomposition of . To simply notation, we employ to represent the cactus induced by .
Suppose is the multigraph achieved from by doubling its all edges, the multigraph achieved from by deleting all edges joining some vertex of . Equivalently, is the multigraph obtained from by doubling each of its edges.
Then is Eulerian. Assume is an Eulerian tour of , so that contains no heavy visit for each with 2 or 3 neighbors in . Note that such a can be obtained from any Eulerian tour of by applying a succession of bow-tie operations when necessary.
We next extend to an Eulerian tour of as follows: for each vertex with degree 1 in , let be the only neighbor of in , and the other two distinct neighbors of in . Then . We extend at to cover the four edges between and such that the new visit decomposition at is . In other words, we insert the trail into via the midvertex of the visit . For each vertex with degree 2 in , let be the unique neighbor of in , and the other two distinct neighbors of in . Then by the assumption on . We extend at to cover the two edges between and such that the new visit decomposition is . Applying this extension to every vertex of that has degree 1 or 2, we obtain an Eulerian tour of , denoted .
By the way is extended, one may observe that consists of no heavy visit if , and contains only heavy visits if . In particular, the bipartite graph is a set of three independent edges if , and if . Thus, in both cases has a perfect matching, and, a Hamilton cycle of , denoted by , can be derived from .
Since is cubic, for each pair of adjacent of , by Definition 1, the subgraph induced by vertices in in is isomorphic to .
For any two vertices adjacent in , let be two distinct neighbours of other than , and be two distinct neighbours of other than . Since each of and contains no heavy visit, we have and . Let be parallel between and in . Then each of is contained in a visit of and a visit of . Assume w.l.o.g. that contain and contain . Then each of and is a segment of length 3 of . We may define each of these two segments as a visit to the edge of , and denote them as , , respectively. From the definition of , each edge, and hence , of is visited twice by of . Represent by the set of two visits of to the edge , i.e., .
Then, each element of is a trail of length 3 which corresponds exactly to one edge of covered by . And, the two visits of are in fact corresponding to two independent edges of , namely, , , which are covered by . This means that covers exactly one perfect matching (two independent edges) of and leaves the other perfect matching uncovered by noting that contains two perfect matchings.
Apply the edge-shift operation (Definition 3) to with respect to the 2 parallel edges connecting and , denote the resulted trail(s) by . Then is transformed into . If stays as an Eulerian tour of , one can observe that any Hamilton cycle , derived from , will cover the pair of independent edges of which are not covered by .
The following claim shows that we can apply the edge-shift operations to every pair of parallel edges of once and still get an Eulerian tour of .
Claim. There are a series of edge-shift operations such that to every pair of parallel edges of , exactly one operation is performed and the resultant trail is an Eulerian tour of .
Proof of the Claim. First consider an edge belonging to none of the cycles of , then is a bridge of . Let be the two parallel edges between and in . Then the orientations of are opposite to each other in , by (1) of Remark 2, stays as an Eulerian tour of . Thus, we can apply the edge-shift operation to each pair of parallel edges of joining two end-vertices of a bridge of . To simplify notation, denote again by the Eulerian tour of after processing all such pairs of parallel edges.
Next we process edges on cycles in the cactus . Since , any two distinct cycles in are vertex- and edge-disjoint. We process these cycles one by one.
Let be an arbitrary cycle of , where is even. Since every vertex has degree 3 in , each has a neighbor, denoted by , where . Note that each edge is a bridge of . Denote by the component containing after deleting the edge . If proceeds along some visit into , must return back to the component containing the cycle via the visit immediately after finishing its all traverses in . We can shrink the traverse of in as a simple visit to as follows: suppose proceeds its traverses in using the following trail:We can first shrink this trail into the visit to , and this visit could be restored back to its original form above after the processing of the edges of . After shrinking all such traverses of in , where , we get an Eulerian tour of the multigraph ( is the multigraph obtained from the cycle by doubling each of its edges). Equivalently, is a single walk that covers each edge of the simple cycle twice. Then, for each edge of , denoted by , the two parallel edges between and .
Apply the edge-shift operation to with respect to . Since the orientations of in are the same, is a set of two closed trails, , , say. One can observe that the trail covering and that covering are distinct, where , are two parallel edges between and . After this operation, we immediately apply the edge-shift operation to , with respect to , . Then, this operation will concatenate and together, that is, (, ) will become again an Eulerian tour of .
Continue applying this operation to the pairs ; we finally obtain an Eulerian tour of , denoted again by . Then we restore back to an Eulerian tour of .
Using this procedure to process all cycles of , we finally obtain an Eulerian tour of after performing exactly one operation to every pair of parallel edges of . The claim follows.
Denote by , the Eulerian tour of produced after applying the edge-shift operation exactly once to each pair of parallel edges connecting two adjacent vertices of .
Consider every edge with one end-vertex (then is not in ). Then all neighbours of lie in . Denote , and , where . Then, by the construction of ,In , such a heavy visit is exactly corresponding to two adjacent edges of . For example, in this instance, if enters via and leaves via to complete the visit ; then, it 1-1 corresponds to the two edges , which is a 2-path in (also in the Hamiltonian cycle ), where is either or . Note that the other two edges in uncovered by will be available in the second Hamilton cycle .
We do not apply any operation on edges with one end-vertex in . We can just use the two uncovered edges in the new Hamilton cycle when constructing it. Equivalently, for each , the perfect matching employed in deriving is the one that shares no common edge with .
Then, is a Hamilton cycle sharing no edges with , the proof is completed.
Corollary 2. Suppose is cubic. If has a stable set such that whose removal leaves a tree, then has a Hamiltonian decomposition.
Since every cubic graph has an IS-C decomposition, every bipartite cubic graph has an even IS-C decomposition. We have the following.
Corollary 3. The 3-arc graph of any bipartite cubic graph has a Hamiltonian decomposition.
Let be the Petersen graph. It is not difficult to find a stable set of consisting of three vertices whose removal leaves a tree. Let be an -prism, . Then, is isomorphic to the Cartesian product . Denote , where , , and, each of and induces an -cycle. The edge set of is . Set . If is even, then is an even IS-C decomposition of with inducing a unicyclic graph; if is odd, then is an even IS-C decomposition of with inducing a tree. By applying 2 and 3, we obtain the following example:
Example 1. (1) The 3-arc graph of Petersen graph has a Hamiltonian decomposition; (2) The 3-arc graph of the -prism has a Hamiltonian decomposition.
Remark 3. The condition in Theorem 2 is sufficient but not necessary. For example, does not have an even IS-C decomposition but is still Hamiltonian decomposable. In , see Figure 1, each of the two sets of bold edges (in color black) and thin edges (in color blue) forms a Hamilton cycle [10].
Data Availability
The findings of this study are supported by the rigorous proofs which are included within the paper.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work was supported by the National Natural Science Foundation of China (11661084, 11801494, and 11971196), Science and Technology Foundation of Guizhou Province (Qian Ke He Ji Chu [2017]1201), Growth Project of Young Scientific and Technological Talents in Colleges and Universities of Guizhou Province (Qianjiaohe KY Zi [2020] 093), and Natural Science Foundation of Shandong Province (No. ZR2021MA012).