Remarks on Barnette's Conjecture

Let $P$ be a cubic $3$-connected bipartite plane graph which has a $2$-factor which consists only of facial $4$-cycles, and suppose that $P^{*}$ is the dual graph. We show that $P$ has at least $3^{\frac{2|P^{*}|}{\Delta^{2}{(P^{*})}}}$ different Hamilton cycles.


Introduction
All graphs considered in this paper are finite and simple. We use [1] as reference for undefined terms.
Let P denote the family of cubic 3-connected bipartite plane graphs. Barnette, in 1969 ( [8], Problem 5), conjectured that every graph in P has a Hamilton cycle. In [4], Goodey proved that if a graph in P has only faces with 4 or 6 sides, then it is hamiltonian. Feder and Subi [3] generalized Goodey's result by showing that when the faces of a graph in P are properly 3-colored and if two of the three color classes contain only faces with either 4 or 6 sides then the conjecture holds. Holton, Manvel and McKay [5] used computer search to confirm Barnette's conjecture for graphs up to 64 vertices, and also to confirm the following property (called H ± property) for graphs up to 40 vertices: If any two edges are chosen on the same face, then there is a Hamilton cycle through one and avoiding the other. Kelmans [6] proved that Barnette's conjecture holds if and only if every graph in P has H ± property.
Florek in [2] studied the class P(4) of those graphs in P which possess a 2-factor consisting only of facial 4-cycles. He proved in particular the following: Theorem 1.1. Let G ∈ P(4) and suppose that F is a 2-factor of G which consists only of facial 4-cycles. In this paper we prove the following theorem which complements the above results.
Theorem 1.2. Let G ∈ P(4) and suppose that F is a 2-factor of G which consists only of facial 4-cycles. If two faces have a common edge which is not in F , there is a Hamilton cycle containing all edges of the first face, except this edge, and avoiding all edges of the second face which are not in F . On the other hand, for every n 3, there is a graph G n ∈ P for which the above corollary is not satisfied (see Figure 1). Notice that G n possesses a 2-factor which consists of n facial 4-cycles and one facial 6-cycle.
We will use a result that expresses hamiltonicity in the terms of the dual graph. Let E be the dual family to P. Hence, E is the family of all Eulerian plane triangulations. Let G ∈ E. We say that a subgraph of G dominates all faces of G if every face of G is incident with a vertex of this subgraph. Stein [7] proved that G * (the dual graph of G) is hamiltonian if and only if G contains two disjoint induced tree-subgraphs which together cover all vertices of G (see Figure 2). Hence follows that G * is hamiltonian if and only if G contains an induced tree-subgraph, say T , dominating all faces of G, and that G − T is another induced tree-subgraph dominating all faces of G.
Let E(4) be the dual family to P(4). Hence, E(4) is the family of all Eulerian plane triangulations with an independent set of vertices of degree 4 dominating all faces of the graph. Thus, our aim is to prove the following (Theorem 2.1): Let G ∈ E(4) and suppose that U ⊂ V (G) is an independent set of vertices of degree 4 dominating all faces of G. For every edge ae, a, e / ∈ U, there is an induced tree-subgraph of G, say T , dominating all faces of G, which avoids all neighbours of a different than e (a is a leaf of T ) and contains all neighbours of e not belonging to U. Hence follows that all neighbours of e not belonging to U and of degree 4 are leaves of T (see Figure 2).

Main results
Let G ∈ E(4) with the vertex set V (G) and the edge set E(G). Florek [2] proved that every graph in E(4) is 4-connected. Hence, every 3-cycle is a face boundary. Therefore the set of neighbours of any vertex in G induces a cycle of G. For W ⊂ V (G), the neighbours in V (G) \ W of vertices in W are called neighbours of W . If the induced graph G[W ] is connected, then G/W is a graph obtained from G by contracting the set W to a single vertex, which becomes adjacent to all the former neighbours of W (see [1], p. 19). A cycle C in G is called separating if the subgraph G − C is disconnected. Proof For |G| = 6 the assertion holds; we now assume that |G| > 6. We apply induction on |G|. Fix an edge ae, a, e / ∈ U. We first consider the case d(e) = 4. Let abcda be a cycle induced by the set of neighbours of e. Hence, b, d ∈ U and c / ∈ U. Certainly, there is no separating 4-cycle containing the path bed, because G is 4-connected and |G| > 6. By contracting the set {b, e, d} to a vertex b 0 we obtain a graph G 0 = G/{b, e, d} ∈ E(4) such that (U\{b, d}) ∪ {b 0 } is an independent set of vertices of degree 4 dominating all faces of G 0 . By the dual version of Theorem 1.1(a), G 0 has an induced tree-subgraph, say T 0 , dominating all faces of G 0 , which avoids all neighbours of a different than b 0 and contains the path ab 0 c. Hence, V (T 0 −b 0 )∪{e} induces the tree-subgraph of G dominating all faces of G, which avoids all neighbours of a different than e and contains the path aec. Now let d(e) 6. Suppose that abc, is a path induced by three consecutive neighbors of e. Let us consider the following two cases: (i) there is a separating 4-cycle containing the path abc, (ii) there is no separating 4-cycle containing the path abc.
Case (i). By (i), there is a separating 4-cycle C = aecf a such that f is not a neighbourhood of e and d(f ) 6, because G is 4-connected and d(e) 6. Since C is induced, G − C consists of two components of G, say C 1 and C 2 , and each of them has at least two verices. Notice that vertices of C are of odd degree in G − C i , for i = 1, 2, because they not belong to U. By adding a vertex b i and edges b i a, b i e, b i c, b i f to G − C i we obtain a graph G i ∈ E(4) such that the set U i = (U\V (C i )) ∪ {b i } is an independent set of vertices of degree 4 dominating all faces of G i . Certainly, |G i | < |G|.
By induction, G i , for i = 1, 2, has an induced tree-subgraph, say T i , dominating all faces of G i , which avoids all neighbours of a (in G i ) different than e and contains all neighbours of e (in G i ) not belonging to U i . Since b i / ∈ T i , T i is the induced tree-subgraph of G such that T 1 ∩ T 2 = aec. Hence, T 1 ∪ T 2 is the induced tree-subgraph of G dominating all faces of G, which has the desired properties.
Case (ii). By contracting the set {a, b, c} to a vertex a ′ we obtain a graph G ′ = G/{a, b, c} ∈ E(4) such that U ′ = U\{b} is an independent set of vertices of degree 4 dominating all faces of G ′ .
By induction, G ′ has an induced tree-subgraph, say T ′ , dominating all faces of G ′ , which avoids all neighbours of a ′ different than e and contains all neighbours of e not belonging to U ′ . Since a ′ is a leaf of T ′ , the set V (T ′ − a ′ ) ∪ {a, c} induces the tree-subgraph of G dominating all faces of G, which has the desired properties.