Hamiltonian cycles in planar cubic graphs with facial 2‐factors, and a new partial solution of Barnette's Conjecture

Abstract We study the existence of hamiltonian cycles in plane cubic graphs G having a facial 2‐factor Q. Thus hamiltonicity in G is transformed into the existence of a (quasi) spanning tree of faces in the contraction G∕Q. In particular, we study the case where G is the leapfrog extension (called vertex envelope of a plane cubic graph G0. As a consequence we prove hamiltonicity in the leapfrog extension of planar cubic cyclically 4‐edge‐connected bipartite graphs. This and other results of this paper establish partial solutions of Barnette's Conjecture according to which every 3‐connected cubic planar bipartite graph is hamiltonian. These results go considerably beyond Goodey's result on this topic.


| INTRODUCTION AND PRELIMINARY DISCUSSION
Hamiltonian graph theory has its roots in the icosian game which was introduced by W.R. Hamilton in 1857. However, Kirkman presented his paper On the presentation of polyhedra [13], to the Royal Society already in 1855; and it was published in 1856.
The early development of hamiltonian graph theory focused to a large extent on planar cubic graphs; and there are good reasons for this course of development. For, in 1884, Tait conjectured that every cubic 3-connected planar graph is hamiltonian [16]. And Tait knew that the validity of his conjecture would yield a simple proof of the Four Color Conjecture. On the other hand, the Petersen graph is the smallest nonplanar 3-connected cubic graph which is not hamiltonian [15]. Tait's Conjecture was disproved by Tutte in 1946, who constructed a counterexample with 46 vertices [18]; other researchers later found even smaller counterexamples. However, none of these known counterexamples are bipartite. Tutte himself conjectured that every cubic 3-connected bipartite graph is hamiltonian [19], but this was shown to be false by the construction of a counterexample, the Horton graph [11]. Barnette proposed a combination of Tait's and Tutte's Conjectures that every counterexample to Tait's Conjecture is nonbipartite.
Barnette's Conjecture [1] Every 3-connected cubic planar bipartite graph is hamiltonian. This conjecture was verified for graphs with up to 64 vertices by Holton, Manvel, and McKay [10]. The conjecture also holds for the infinite family of graphs where all faces are either quadrilaterals or hexagons, as shown by Goodey [9]. Without the assumption of 3connectedness, it is NP-complete to decide whether a 2-connected cubic planar bipartite graph is hamiltonian, as shown by Takanori, Takao, and Nobuji [17].
For a more detailed account of the early development of hamiltonian graph theory we refer the interested reader to [2].
Given the fact that the existence of hamiltonian cycles is an NP-complete problem (in rather special classes of graphs), one has to develop ad hoc proof techniques depending on the class of graphs, whose members are being shown to be hamiltonian.
As for the terminology used in this paper we follow [3] unless stated explicitly otherwise. In particular, the subset E v ( ) of E G ( ) denotes the set of edges incident to ∈ v V G ( ). We note, however, that given graphs may contain multiple edges but no loops, whereas loops arising in the process of contracting cycles, will be deleted.
Next, we state some definitions and remarks.
Definition 1. A cubic graph G is k-cyclically edge connected if at least k edges must be removed to disconnect G either into two components each of which contains a cycle provided G contains two disjoint cycles, or else into two nontrivial components. The cyclic edge connectivity of G is the maximum k such that G is cyclically k-edgeconnected, denoted by κ G ( ) ′ c . Definition 2. Let C be a cycle in a plane graph H . The cycle C divides the plane into two disjoint open domains. The ( ) interior exterior of C is the bounded (unbounded) domain and is denoted by int C ext C ( ) ( ( )). Correspondingly, we say a cycle C′ is inside of then C is said to be a separating cycle in H. However, in the case of a plane embedding we distinguish between the unbounded or outer face F o (ie, ) and a bounded face F i (ie, Remark. 1. Two edges e xy = and e xy ′ = in a plane graph are called parallel edges if the digon D defined by e and e′ has no vertices inside. A maximal set of parallel edges is called a bundle; if its members are incident to u and v then we speak of a uv-bundle. For completeness sake, we call e uv = a uv-bundle if there is no parallel edge uv. We say two triangles T and T′ share a uv-bundle if both triangles contain an edge joining u and v, and these edges are parallel edges or they are identical. T and T′ with V T V T ( ) = ( ′) are called equivalent if their respective edges belong to the same bundles. Equivalent separating digons are defined likewise. Now, if two triangles T 1 and T 2 share a bundle, then they do not share another bundle unless they are equivalent, or there is 1 2 defines a separating digon. Clearly, the definition of bundles in a plane graph G defines an equivalence relation on the parallel edges of G and, correspondingly, the set of bundles in G defines uniquely a partition of E G ( ). Moreover, the set of triangles can be partitioned into equivalence classes of triangles. This partition and a partial order of its equivalence classes are discussed in Section 2 of this paper. 2. Given a 2-connected plane graph, sometimes we do not distinguish between faces and their face boundaries. Observe that in planar 3-connected graphs H , the face boundaries are independent from any actual embedding of H in the plane or sphere. is given by the counterclockwise cyclic ordering of the edges incident to v. We consider two cyclic orderings at v the same if one arises from the other by a cyclic permutation of the indices.
being a pair of consecutive edges in L implies j i d e gv = ± 1 (mod ( )), for every ∈ v V H ( ).
• As a consequence, in an A-trail in a 2-connected plane graph any two consecutive edges belong to a face boundary (here O v ( ) + derives from the embedding in the planecf. second part of Definition 3).
BAGHERI GH ET AL.
| 271 Definition 6. Let G be a 2-connected plane graph and let v be a vertex of G with , is given by the counterclockwise cyclic ordering of the edges incident to v.
in such a way that the result is a plane graph again. A plane graph obtained from G by truncating all vertices of G is called G truncation of and denoted by Tr G ( ) subject to the condition that Alternatively and more formally, the leapfrog extension Lf G . In the case of cubic G, it can be viewed as obtained from G by replacing every ∈ v V G ( ) by a hexagon C v ( ) 6 , with C v ( ) 6 and C w ( ) 6 sharing an edge if and only if ∈ vw E G ( ); and these hexagons are faces of Lf G ( ).
Next we quote some known results.  Theorem C (Payan and Sakarovitch [14]). Let G be a cyclically 4-edge-connected cubic graph of order ≡ n 2 (mod 4). Then G has an independent set S of order ∕ n ( + 2) 4 such that G V G S [ ( )\ ] is a tree.
Below we present a proof of Theorem D which relies exclusively on Theorem C and differs therefore from the proof in [6].
Proof. We draw Lf G ( ) in the plane and draw G so to speak inside of Lf G ( ) in such a way that ∈ v V G ( ) lies inside the corresponding hexagonal face ⊂ C v Lf G ( ) ( ) 6 and ∈ vw E G ( ) crosses the edge lying in ∩ C v C w ( ) ( ) 6 6 (see Definition 6(ii)). Now, since G is a cyclically 4-edge-connected cubic graph of order ≡ n 2 (mod 4), by Theorem C there exists an independent set ⊂ S V G ( ) such that T G V G S = [ ( )\ ] is a tree. Now, if we delete in Lf G ( ) those edges of C x ( ) 6 which do not belong to any other C y ( ) 6 , for every ∈ x S, then we obtain the plane graph G T ( ) covered by the hexagonal faces C q ( ) 6 , where ∈ q V G S V T ( )\ = ( ). Letting K be the set of these hexagonal faces, it follows that T I K = ( ), where I K ( ) is the intersection graph of K . Since for every pair ∈ C t C u K ( ), ( ) 6 6 we have ∩ ∅ C t C u ( ) ( ) = 6 6 or a single edge of Lf G ( ), and because I K ( ) is a tree and V G T V Lf G ( ( )) = ( ( )) by construction, it follows that G T ( ) has a (unique) hamiltonian cycle which is also a hamiltonian cycle of Lf G ( ). □ We note in passing that others speak of vertex envelope, or leap frog construction, or leap frog operation, or leap frog transformation (see, eg, [6,7,12,20]).
We need some considerations before formulating Definition 7 below. Let G be a plane cubic graph. It is a well known fact that G is hamiltonian if and only if its dual graph G* has a vertex decomposition V V { , } * * 1 2 such that the graph T* i induced by V * i is a tree, i = 1, 2. Correspondingly, there is a hamiltonian cycle C in G. We consider the outerplane graph G 1 consisting of C and the chords of C lying in int C ( ); without loss of generality T* 1 is the weak dual of G 1 (ie, T* 1 is the intersection graph of the boundaries of the bounded faces of G 1 ). We draw T* 1 inside of C whose vertices lie inside the corresponding faces and whose edges cross the corresponding edges of G 1 .
Suppose now that G has a facial 2-factor . Let 1  denote the set of faces of  lying in int C ( ) and let c 1  be the faces not in  but also lying in C int( ). Let  be a set of bounded faces whose boundaries are pairwise edge-disjoint and such that every vertex of H is contained in some element of  . We define a subgraph H  of H by  is a tree, then we call H  a quasi spanning tree of faces of H , and the vertices in U V H U ( ( )\ ) are called ( ) proper quasi vertices. If U V H = ( ), then H  is called a spanning tree of faces. In other words, a spanning tree of faces is a spanning bridgeless cactus whose cycles are face boundaries.
We observe that if H is a plane eulerian graph with ≥ δ H ( ) 4 having an A-trail T ε , then T ε defines uniquely a quasi spanning tree of faces as follows (see [5, pp. VI V I . 71 − . 77]). Starting with a 2-face-coloring of H with colors 1 and 2, suppose the outer face of H is colored 1. Then T ε defines in every ∈ v V H ( ) a 1-splitting or a 2-splitting thus defining a vertex partition

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We point out, however, that the concept of (quasi) spanning tree of faces is a somewhat more general tool to deal with hamiltonian cycles in plane graphs, than the concept of A-trails. Below we shall prove the existence of (quasi) spanning trees of faces in plane graphs H derived from plane cubic graphs G having a facial 2-factor (rather than being bipartite-which implies the existence of three disjoint facial 2-factors), provided the cubic graphs satisfy some extra conditions. This, in turn yields hamiltonian cycles in G. In particular, we obtain new partial solutions of Barnette's Conjecture (cf. Theorem 8 and Corollary 10). In this context we also want to point out that every simple 4-connected eulerian triangulation of the plane has a quasi spanning tree of faces (see Corollary 7 below), whereas it is an unsolved problem (see [5,Conjecture VI. 86]), that every simple 4-connected eulerian triangulation of the plane admits an A-trail.
Finally observe that we did not include figures in proofs. Instead we elaborated arguments to such an extent that the reader himself/herself may draw such figures easily (and in a unique way) as he/she sees fit. We also wish to point out that this paper is in part the result of extracting certain results and their proofs of [4] (they have not been published yet in any refereed journal). On top of it, the first author of this paper succeeded in developing additional results and their proofs, basing his contribution on some of the work in [4]. Moreover, we relate some of the results of this paper to the theory of A-trails, as developed in [5].

| HAMILTONIAN CYCLE FROM QUASI SPANNING TREE OF FACES
In what follows G always denotes a 3-connected cubic planar graph having a facial 2-factor  (ie, a 2-factor whose cycles are face boundaries of G), together with a fixed imbedding in the Euclidean plane, such that  does not contain the boundary of the outer face. We denote the set of face boundaries of G not in  by c  . In general, when we say that a face F is an - face, we mean that ∈ F  . Let H always denote the reduced graph obtained from G by contracting the -faces to single vertices; ie, Suppose H has a quasi spanning tree of faces H  with proper vertex set U . Then the subgraph H  has a unique A-trail (obtained by traversing consecutive sections of the elements of  ). This A-trail can be transformed into a hamiltonian cycle C G of G such that the -faces correspond to the quasi vertices. Moreover, the face of G corresponding to the outer face of H ( ) G has no edge in common by hypothesis, C G can be transformed into an A-trail of H  . Now it is easy to see that H  is a quasi spanning tree of faces of H whose quasi vertices correspond to the -faces in ext C ( ) G . We summarize the preceding considerations together with the considerations preceding Definition 7 in the following result. Example 2.1. In Figure 1, a 3-connected cubic planar graph G 0 is given with a facial We return now to our general considerations. Suppose all -c  faces of G are either quadrilaterals or hexagons, while the -faces are arbitrary. Suppose the reduced graph H has a triangle T that contains at least one vertex in int T ( ), such that int T ( ) does not contain a separating digon.
We shall successively simplify the inside of the triangle T , while preserving the property that there is no separating digon inside of T , but allowing the presence of separating triangles inside of T, but with the following requirement. In what follows we delete loops (but not multiple edges) which may arise when contracting a triangle ⊂ T T ′ (such loop may arise when ∈ e E T ( ′) is a multiple edge). Also, when speaking of a digon or triangle T′ not being a face boundary, we mean that there is an equivalent digon or triangle T″ such that int T ( ″) contains at least one vertex.
To describe certain structural properties we return to our discussion in Remark 1 and consider the equivalence classes of triangles. Let  be the set of all equivalence classes of separating triangles in H . Define a relation ≼ on  in the following way. For every ∈ , . This relation is a partial order, and it is well defined indeed: for, if We observe that by a suitable choice of representatives this partial order carries over to a set of representatives. In fact, it suffices for every . Thus we may also speak of (direct) successors of triangles.
More explicitly, considering different equivalence classes ∈ , And finally, two direct successors T 2 and T 3 of a separating triangle T 1 are distinct if they belong to distinct equivalence classes ∈ , 2 3    containing T T , 2 3 , respectively, in H ; and , 2 3   in turn are direct successors of ∈ Note that by planarity, if the graph has no separating digons, then no separating triangle can be a direct successor of two inequivalent triangles.
At all steps in the simplification of the inside of the triangle T , we shall require that no triangle , has three distinct direct successors T 2 , T′ 2 , and T″ 2 . We define the invariant property for T to be such that T and every triangle inside of T has at most two distinct direct successors and there is no separating digon inside of T. In particular, any bounded facial triangle has the invariant property. Note that if the triangle T has the invariant property, then every triangle inside of T also has the invariant property (this is sort of a "relative hereditary property"). We say that a graph H has the invariant property if every triangle in H (and the outer face of H if it is a triangle) satisfies the invariant property.
The following theorem is of a more technical nature and is key to the subsequent results.
, and after contracting T′ to a single vertex T will still satisfy the invariant property.
Proof. Suppose T satisfies the invariant property. Let  be the set of all separating triangles T′ inside of T such that no triangle inside of T′ is separating. That is, T′ has no direct successors. Moreover, int T ( ′) does not contain a separating digon because T satisfies the invariant property by supposition. Observe that the equivalence class ′  containing ∈ T′  corresponds to a sink in the Hasse diagram of ≼ ( , )  . We have two cases.
Case 1. There exists a triangle ∈ T 1  whose interior contains at least two vertices. Set In this case, v 1 has at least two distinct neighbors v 4 and v 5 inside of T 1 . For if v 1 has no such neighbors, then v 1 belongs to a triangle inside of T 1 that has an edge v v 2 3 which forms a separating digon together with the corresponding edge of T 1 , contrary to the assumption that there is no separating digon inside of T (since T satisfies the invariant property). And if v 1 has precisely one such neighbor v 4 inside of T 1 , then for Claim 1. By contracting T 2 , the triangle T still satisfies the invariant property.

Set
∕ H H T ′ = 2 . We note that in H′, the triangle T 1 will not contain any separating digon inside since such a digon would derive from a separating triangle inside T 1 , contrary to the choice of ∈ T 1 . Next we show that after the contraction of T 2 , the triangle T 1 has at most two direct successors in H′.
In H′ however, there may appear separating triangles in int T ( ) 1 . Such triangles derive from possibly three different types of quadrilaterals:  or v v 5 9 inside; otherwise, there was a separating triangle in H inside of T 1 , again a contradiction to the choice of ∈ T 1 . This implies that for all such quadrilaterals Q * 2 and Q ** 2 containing v 4 we have either ⊂ Q Q * ** | 277 been said above. Let Q ′ 1 be the quadrilateral of the same type as Q 1 containing v 5 and with all quadrilaterals of the same type as Q 1 contained in int Q ( ) ′ 1 . The analogous properties also hold for the quadrilaterals of the same type as Q 3 , but these are again of only one kind, namely the interior of Q 3 does not contain v 1 , since they are contained in the triangle T 1 . Let Q ′ 3 be the quadrilateral not containing v 1 but with all quadrilaterals of the same type as Q 3 2 (defining the partial order ≼ in H′ as we did in H ). Subsequently we shall conclude that at most one of Q ′ 1 and Q″ 2 exist.
. Therefore, we have two possibilities for v′ 8 .
2 , we have two possibilities for v″ 8 as we had with respect to v′ 8 above.
; this contradicts the choice of ∈ T 1 .
Thus, at most one of Q ′ 1 and Q″ 2 exists. Therefore, • the above relations 2 and the fact that at most one of Q′ 1 and Q″ 2 exists preclude that T 1 has three or more direct successors in H′; • if T 1 has exactly one direct successor T Q in H′, then 3 ; • if T 1 has two direct successors in H′, then they are either T Q′ 2 and T Q ″ 2 , or T Q′ 2 and T Q′ 3 .
Note that every triangle deriving from a quadrilateral of the same type as Q 2 not containing v 4 , or every triangle deriving from a quadrilateral of the same type as Q 3 has at most one direct successor deriving from a quadrilateral of its respective type.
Every triangle deriving from a quadrilateral of the same type as Q 2 containing v 4 has at most one direct successor deriving from either a quadrilateral of its type or a quadrilateral of the same type as Q 3 .
Every triangle deriving from a quadrilateral of the same type as Q 1 containing v 5 has either at most one direct successor (deriving from either a quadrilateral of its type or from a quadrilateral of the same type as Q 2 not containing v 4 , or from a quadrilateral of the same type as Q 3 ) or at most two direct successors deriving from two quadrilaterals, one of the same types as Q 2 not containing v 4 and one of the same type as Q 3 .
Thus, T still satisfies the invariant property in H′. This finishes the proof of Claim 1 and thus finishes the consideration of Case 1.
Case 2. The interior of every member of  contains precisely one vertex.
In this case, there is a triangle T 1 (possibly T T = 1 ) satisfying the invariant property and such that either it has one direct successor ∈ T 2  or it has two direct successors ∈ T T , 2 3 . That is, there exist at most two separating triangles-T 2 or T 2 and T 3 -in . By contracting any triangle inside of T 2 , we will not create a separating digon since such a digon would derive from a separating triangle T 0 in ; so by the assumption of Subcase 2.1, T 0 is not equivalent to T 1 or T 3 . Thus, T 1 would contain a distinct direct successor other than T 2 and T 3 in H , contradicting the choice of T 1 .
Note that if a quadrilateral ⊂ Q H shares two bundles with T 2 , then the contraction of any triangle inside of T 2 will not transform Q into a separating triangle or a separating digon (otherwise, either T 1 would contain in H a distinct separating triangle other than T 2 and T 3 in H or T 2 would not be a direct successor of T 1 in Hwhich contradicts the choice of T 1 or T 2 ). (**) Let v 0 be the single vertex inside of T v v v v = 2 1 2 3 1 . We must again consider quadrilaterals Supposing that Claim 2 fails and starting with a fixed Q 1 , we consider all possibilities for Q 2 vis-a-vis Q 1 . Suppose , and T 2 is a direct successor of T 1 by the choice of T 1 and ⊂ Q T i 1 , for ≤ ≤ i 1 3, we conclude that T* and T 1 are equivalent. Thus, v 1 is a vertex of T 1 and, consequently, 3 , which is a contradiction.
Suppose v v = 7 4 . Then, an analogous reasoning yields T v v v v = 1 1 4 5 1 and the same conclusion as above holds is a separating digon in T 1 which contradicts the invariant property of T 1 .
Suppose  Claim 2 now follows. Therefore, we may assume without loss of generality that either there is no quadrilateral Q 1 containing v 3 in its interior, or there is no quadrilateral Q ′ 1 not containing v 3 in its interior such that after identifying v 1 and v 2 a new separating triangle arises.
The contraction of the triangle v v v v 0 1 2 0 creates only a sequence of triangles with pairwise containment involving the new vertex ≡ v v 1 2 , apart from the triangle T 3 , thus preserving the property that no triangle has three direct successors. Having shown at the beginning of this subcase that the contraction of v v v v 0 1 2 0 does not create a separating digon, we now conclude that the invariant property is being preserved. Proposition 3. Suppose G is a 3-connected cubic planar graph with a facial 2-factor . Assume that the faces not in are either quadrilaterals or hexagons, while the faces in are arbitrary. Suppose the reduced graph H G = / satisfies the invariant property, and that the outer face of H is a triangle. If H has an odd number of vertices, then H has a spanning tree of faces that are triangles, and so G is hamiltonian.
Proof. Let T be the outer face of H. Apply Theorem 2 repeatedly to contract triangular bounded faces inside of T to single vertices while preserving the invariant property. Each step reduces the number of vertices by two, so at each step the order of the resulting graph remains odd until we are left with just the outer face T and parallel edges but no vertex in int T ( ). We claim that the triangle corresponding to the innermost face T 0 inside of T involving all three vertices together with the triangles contracted in this process forms a set of faces  of H and defines a spanning tree of faces H  . Now, guarantees that  covers all of V H ( ), and H  is connected by construction. If H  is not a spanning tree of faces, then there exists a set of triangles otherwise. Assume T i 0 is the last contracted triangle in the contraction process of the T i 's, ≤ ≤ i k 1 . Thus after the contraction of T i for all ≤ ≠ ≤ i i k T 1 , i 0 0 is being transformed into a digon. This contradicts the selection of T i 0 by Theorem 2. Proposition 3 now follows. □ We note in passing that by using Lemma 9 below and Theorem A(ii), Theorem D can be shown to be a special case of Proposition 3. Moreover, let G be a 3-connected cubic graph and as described in Theorem B. Then H G = / is a triangulation of the plane. Thus, by Proposition 3 and Theorems A(ii) and B, we have the following corollary. Note that G 0 has an odd number of faces if ≡ n 2 (mod 4) where n is the order of G 0 and thus for G Lf G = ( ) 0 we have that H G = / is of odd order. Satisfying the invariant property is an essential condition in Corollary 4. As shown in Figure 2, for a simple 2-connected cubic planar graph G, in H Lf G = ( )/ the triangle v v v v 1 2 3 1 has seven direct successors, but as we show below, H has no spanning tree of faces nor a quasi spanning tree of faces. Proof. Since H is 4-connected and v 0 is a vertex of degree 4 inside of T, therefore v 0 belongs to 4 triangles at most one of which shares a bundle with T . Therefore, by cyclically rotating the labels in O v ( ) + 0 if necessary, there exist triangles v v v v 0 4 5 0 and v v v v 0 6 7 0 which share no bundle with each other nor with T . Thus H′ is well defined. By 4connectedness of H , it has no separating digon nor a separating triangle. Note that H′ has no separating digon; otherwise, H has a separating triangle, which is a contradiction. We show that every triangle in H′ has at most two direct successors. We first observe that there cannot exist simultaneously two quadrilaterals , each containing a vertex inside, x and x′, respectively, other than v v v v , , , 4 5 6 7 , and . Otherwise, v v = * * 8 9 , in which case there is a separating triangle containing x or x′ in H , contradicting that H is 4-connected. Thus, without loss of generality, suppose Q′ does not exist. Note that the quadrilateral   . Since every simple 4-connected eulerian triangulation of the plane has at least six vertices of degree 4, the following is an immediate corollary of Propositions 3 and 6.
Corollary 7. Every simple 4-connected eulerian triangulation of the plane has a quasi spanning tree of faces.
Example 2.2. We claim that the 3-connected triangulation of the plane of Figure 2 above has no quasi spanning tree of faces.
Proceeding by contradiction, we first assume that there is a set  of faces and H  is a spanning tree of faces in H . For every degree three vertex v 0 and ≤ ≤ v i , 7 12 i , there exists precisely one triangle in  containing v 0 or v i . Without loss of generality and because of symmetry,  Therefore,  (otherwise, there is a cycle of faces in H  ). Thus, there is no face in  containing v 11 , which is a contradiction. By a similar argument one can show that H has no quasi spanning tree of faces, observing that quasi vertices must have even degree and thus without loss of generality, v 5 would be a quasi vertex and Corollary 7 implies a result on hamiltonicity in planar cubic bipartite graphs.
Theorem 8. Let G be a bipartite cubic planar graph with the following properties: (i) In the natural 3-face coloring of G with colors 1, 2, 3, two of the color classes (without loss of generality, color classes 1  and 2  ) contain hexagons only.
(ii) The contraction of the faces in color class 3  is 4-connected.
Then G is hamiltonian.
Lemma 9. Let G be a simple cubic planar graph and let  be the set of faces in Lf G ( ) corresponding to the faces of G. Then, ∕ κ G κ Lf G ( ) = ( ( ) ). ′ c  Proof. Let H Lf G = ( )/. Note that by Definition 6 (ii), the reduced graph H is a triangulation of the plane and every edge of H corresponds to a unique edge of G, and vice versa; and every vertex of H corresponds to a unique face of G, and vice versa. Note that G and H can be drawn in the plane in such a way that ∈ f V H ( ) lies in int F ( ) contracting T′ to a single vertex, T 1 will satisfy the invariant property. Now, let H′ be the graph obtained from H by contracting T′.
It is easy to see that H′ satisfies all hypotheses of Theorem 11 and its order is n − 2. (Note that every separating digon in H′ would derive from a separating triangle inside T 1 in H , contrary to the choice of T 1 ). Thus by induction, H′ has a spanning tree of faces that are triangles with face set ′  . Let  be the union of the set of the corresponding faces of ′  in H and the set T { ′}. It can be easily seen that H  is a spanning tree of faces in H . This completes the proof of Theorem 11. □ We note finally that in [6], hamiltonicity in the leapfrog extension of a plane cubic graph was studied from a different point of view.