Nash Williams Conjecture and the Dominating Cycle Conjecture
Abstract
The disproved Nash Williams conjecture states that every 4regular 4connected graph has a hamiltonian cycle. We show that a modification of this conjecture is equivalent to the Dominating Cycle Conjecture.
Nash Williams Conjecture and the Dominating Cycle Conjecture
Arthur HoffmannOstenhof
Technical University of Vienna, Austria
Keywords: dominating cycle, hamiltonian cycle, 3regular, 4regular,
4connected, cyclic 4edge connected.
1 Basic definitions and main result
For used terminology which is not defined here we refer to [1, 2]. A dominating cycle (DC) of a graph is a cycle which contains at least one endvertex of every edge of . Let then denotes the set of edges incident with . A closed trail is a closed walk in which all the edges are distinct. All graphs here are considered to be loopless and finite.
The following two conjectures are well known in graph theory. The first one was disproved by Meredith, see [5].
Nash Williams Conjecture (NWC): Every regular connected graph has a hamiltonian cycle.
Dominating Cycle Conjecture (DCC): Every cyclically edge connected cubic graph has a dominating cycle.
The DCC is open and so far there is neither a promising approach known to prove it nor to disprove it. For a survey on this conjecture, we refer to [2].
We need the following definitions for introducing the modified NWC.
Definition 1.1
Let be a regular graph with a transition system , i.e. where is a partition of
the four edges incident with into two sets of size ; each of these two sets is called a transition of , of and of .
A trail is said to follow a transition if the two edges of the transition are consecutive edges of the trail.
Moreover, is said to be Thamiltonian if contains a Ttrail, that is a spanning closed trail of such that for each one of the following two conditions is fulfilled:
a) (in this case may follow no transition of )
b) follows both transitions of (in this case ).
For an example see Figure 1. Observe that is hamiltonian if is hamiltonian. Hence, hamiltonicity generalizes the concept of hamiltonian graphs. Now, we introduce the modification of the NWC.
NWC*: Let be a regular connected graph with a transition system , then is hamiltonian.
We state the main result.
Theorem 1.2
The DCC is equivalent to the NWC*.
Remark 1.3
Consider the NWC* as false. Then by the previous observation a counterexample to the NWC* is also a counterexample to the NWC. Hence in order to disprove the NWC* one needs a counterexample to the NWC with additional properties. Therefore and by Theorem 1.2, the DCC is harder to disprove than originally the NWC.
Theorem 1.2 implies another result. For stating it we use the following definition.
Definition 1.4
We say that a cycle dominates a matching of a graph if the cycle contains at least one endvertex of each edge of . Let denote the graph which results from by contracting each edge of to a distinct vertex.
Conjecture 1.5
Let be a cubic graph with a perfect matching such that is connected. Then contains a cycle which dominates .
2 Proof of the main result
A graph is called vertex connected if and is connected for every with . We abbreviate vertex connected by connected. A graph is called edge connected if and is connected for every with . A set of a connected graph is called an edge cut of if is disconnected. Moreover, if at least two components of are not a tree then is also called a cyclic edge cut of . If contains two vertex disjoint cycles, then is the minimum size over all cyclic edge cuts of . The line graph of a graph is denoted by . A cycle of length is called a triangle.
Proposition 2.1
If the NWC* is true, then the DCC is true.
Proof. Let be a cubic graph satisfying . We show that has a DC. It is well known and not difficult to see that is regular and connected. Note that every vertex corresponds to a unique triangle of . Moreover, is a partition of .
Define the following transitions which imply a transition system of : each pair of edges which is incident with the same vertex of and belongs to the same triangle of forms a transition, see Figure 2.
Since the NWC* holds, is hamilitonian. If has a hamiltonian cycle, then has a DC by Th. 5 in [2] (see also [3]) and we are finished.
Hence we may assume that has a trail which contains a positive minimum number of valent vertices. contains at most one valent vertex in every triangle of . Otherwise follows by Def.1.1 all transitions in at least two vertices of which implies that also the remaining third vertex of is a valent vertex of . However this is not possible since obviously cannot follow now both transitions of the third vertex contradicting Def.1.1. Let be a valent vertex of and let with be one triangle of which contains . Define the new trail of which results from by replacing the two edges of which are contained in and incident with by the remaining edge of . Then contains fewer valent vertices than which contradicts the definition of and thus finishes the proof.
For the formulation and proof of the next results we use the following definition.
Definition 2.2
Let be a regular graph with a transition system . Let denote the cubic graph which results from by firstly splitting every vertex of into two vertices , of degree such that the corresponding edges of each transition of remain adjacent and by secondly adding the new edge . Set which is a perfect matching matching of satisfying .
The following lemma can be verified straightforwardly.
Lemma 2.3
Let be a regular graph with a transition system . Then has a cycle which dominates if and only if has a trail.
We apply the subsequent three lemmata for the proof of the next proposition.
Lemma 2.4
Let be a matching of a cubic graph such that consists of two components and . Moreover, let every edge of have precisely one endvertex in , . Then is even for every perfect matching of .
Proof: straightforward.
Lemma 2.5
Let be a regular connected graph with a transition system , then is edge connected. Moreover, if has a cyclic edge cut , then one of the two components of is a triangle.
Proof. Set . Obviously, is connected. Suppose has a bridge , then by Lemma 2.4 (with ), . Hence, corresponds to a cut vertex of which contradicts that is connected.
Suppose is a edge cut of . Then is a matching, otherwise has a bridge. By Lemma 2.4 we have two cases.
Case 1. .
is a simple graph otherwise is not connected.
Thus, is also a simple graph. Therefore both components of contain more than two vertices. Hence, corresponds to a vertex cut of which contradicts the definition of .
Case 2. .
Then corresponds to a edge cut of . Since is connected and since the edge connectivity is greater or equal the vertex connectivity, this is impossible.
Hence, is edge connected which finishes the first part of the proof.
Let be a cyclic edge cut of . Then is a matching and consists of two components, otherwise would contain a edge cut for some . Moreover, at most one component of is a triangle since otherwise and is not connected.
Suppose by contradiction that no component of is a triangle. Hence both components have more than three vertices. Since every graph contains an even number of vertices of odd degree, every component has at least five vertices. Denote one of the two components of by . By Lemma 2.4, we need to consider two cases: Case A and Case B. Set .
Case A. .
Let us suppose first that .
Set . Obviously, (depending on whether one edge of covers one endvertex of and one of ).
Hence corresponds to a vertex cut of for some which contradicts that is connected.
Thus, we may assume that . Hence, . Set . Since is a simple graph, and have the following properties:
(1) and are not contained together in a cycle of length .
(2) Neither nor is contained in a triangle.
Denote by the unique valent vertex of which is neither matched by nor by . Denote by and the remaining two valent vertices in . If , then ; otherwise the two vertices in corresponding to and would form a vertex cut of . Hence we can set and such that .
By (2), is adjacent to precisely one endvertex of and to one of . Thus, we have three cases:
Case 1. .
Then must contain the edge twice which is impossible since is a simple cubic graph.
Case 2. .
Then contains the triangle consisting of the vertices , , which contradicts (2).
Case 3. .
Then contains either double edges or a cycle of length consisting of the vertices: which contradicts (1).
Hence, Case A cannot occur.
Case B. . Since both components of have at least five vertices and since , corresponds to to a vertex cut of which contradicts the definition of and thus finishes the proof.
Definition 2.6
Let be a regular simple graph with a transition system . Denote by the graph which results from by contracting every triangle of to a distinct vertex.
Note that each pair of triangles of (in Def.2.6) is vertex disjoint. Hence, is well defined.
Lemma 2.7
Let be a regular connected graph with a transition system , then either or .
Proof. Since , and thus . Suppose first that does not contain two disjoint cycles. Since and by Theorem 1.2 in [4], it follows that .
Now, assume that has two disjoint cycles. Suppose by contradiction that is a cyclic edge cut of for some . Then corresponds to a cyclic edge cut of . Lemma 2.5 implies that and that one of the two components of is a triangle. Since this triangle is contracted to a vertex in , one of the two components of is a vertex. Hence, is not a cyclic edge cut which is a contradiction and thus finishes the proof.
Proposition 2.8
If the DCC is true, then the NWC* is true.
Proof. Let be a regular connected graph with a transition system . Since is connected, . Set . Then and we have the following two cases.
Case 1. . Then, by assumption has a DC which thus dominates (Def. 2.2). By Lemma 2.3, is hamiltonian.
Case 2. . Consider . Every edge of corresponds to an edge of . Thus, every subgraph , say, of induces by its corresponding edge set in , a subgraph of which we denote by .
Note that and have a dominating cycle. Therefore, and by Lemma 2.7 and since the DCC holds, has a dominating cycle . The corresponding subgraph is not a cycle if and only if there is a vertex which has been obtained by contracting a triangle in . We denote this triangle by and call such a vertex , a bad vertex of . We define the cycle depending on : for each bad vertex of , we extend to by adding the unique path of length which is contained in and which connects two endvertices of two edges of ; if has no bad vertex, then is already a cycle and we set . We show that dominates .
Since is connected, is simple. Thus, no triangle of contains an edge of . Hence, it suffices to show that the edge is dominated by for every . If , then and thus is dominated by . If , then two cases are possible.
Case A. is a chord of .
Then, by the construction of both endvertices of are contained in .
Case B. Precisely one endvertex of is contained in .
Then, one endvertex of is contained in .
Hence, dominates . By Lemma 2.3, is hamiltonian which finishes the proof.
Theorem 1.2 The DCC is equivalent to the NWC*.
Corollary 2.9
Conjecture 1.5 is equivalent to the DCC.
Proof. Suppose that Conjecture 1.5 holds. We show that in this case the NWC* holds which implies by Th.1.2 the truth of the DCC. Let be a regular connected graph with a transition system . Since Conjecture 1.5 holds, the cubic graph has a cycle dominating and thus by Lemma 2.3, the first part of the proof is finished.
Suppose that the DCC and thus by Th.1.2 also the NWC* holds. Let and be defined as in Conjecture 1.5. We want to find a cycle which dominates . Set and let be the transition system of such that . Since the NWC* holds, is hamiltonian. By Lemma 2.3, has a cycle dominating where equals which finishes the proof.
Acknowledgment: This work was funded by the Austrian Science Fund (FWFProject 26686).
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