Clar Structure and Fries Set of Fullerenes and ( 4 , 6 )-Fullerenes on Surfaces

Fowler and Pisanski showed that the Fries number for a fullerene on surface Σ is bounded above by |V|/3, and fullerenes which attain this bound are exactly the class of leapfrog fullerenes on surface Σ. We showed that the Clar number of a fullerene on surface Σ is bounded above by (|V|/6) − χ(Σ), where χ(Σ) stands for the Euler characteristic of Σ. By establishing a relation between the extremal fullerenes and the extremal (4,6)-fullerenes on the sphere, Hartung characterized the fullerenes on the sphere S 0 for which Clar numbers attain (|V|/6) − χ(S 0 ). We prove that, for a (4,6)-fullerene on surface Σ, its Clar number is bounded above by (|V|/6) + χ(Σ) and its Fries number is bounded above by (|V|/3) + χ(Σ), and we characterize the (4,6)-fullerenes on surface Σ attaining these two bounds in terms of perfect Clar structure. Moreover, we characterize the fullerenes on the projective planeN 1 for which Clar numbers attain (|V|/6) − χ(N 1 ) in Hartung’s method.

The concepts of the Fries number and the Clar number were named after Fries [12] and Clar [13], respectively.An embedding of a graph in a surface is called cellular if each face is homeomorphic to an open disk.A map on a surface Σ is a graph embedded cellularly on Σ.Let  be a 2connected map.A perfect matching (or Kekulé structure)  of  is a set of edges such that each vertex is incident with exactly one edge in .The faces that have exactly half of their bounding edges in a perfect matching  of  are called alternating faces with respect to .The Fries number  () of  is the maximum number of alternating faces over all perfect matchings; the Clar number () of  is the maximum number of independent alternating faces over all perfect matchings.Note that the original definitions of the Fries number and the Clar number are the maximum number of alternating hexagonal faces and the maximum number of independent alternating hexagonal faces over all perfect matchings, respectively.The original definitions and 2 Journal of Applied Mathematics the definitions we used in this paper coincide when the 2connected map has no other even faces except hexagonal faces.A Fries set is a set of alternating faces with the maximum number of alternating faces.A Clar set is a set of independent alternating faces with the maximum number of alternating faces.If H is a Clar set of  and  0 is a perfect matching of  − H, then we say that (H,  0 ) is a Clar structure of .For a face  of , we say that an edge  in  exits  if  shares exactly one vertex with  and that  lies on  if both vertices of  are incident with .
Both Clar number and Fries number have their chemical significance.The Fries number was proposed as a stability index for hydrocarbon in the early decades of the nineteenth century, whereas the Clar number was later considered as another stability index for hydrocarbon in the 1960s.For buckminsterfullerene C 60 , Austin et al. think that the source of the stability of it is attributed to the fact that uniquely amongst the 1812 isomers it has a Fries Kekulé structure where all hexagons are alternating and all pentagons do not contain any double bond [14].Ye and Zhang [15] showed that the maximum Clar numbers of spherical fullerenes are (||/6) − 2 and further characterized all 18 extremal spherical fullerenes among all fullerene isomers of C 60 .Later, Zhang et al. [16] proposed a combination of Clar number and Kekulé count of spherical fullerenes, as a stability predictor of spherical fullerenes, which turns out to distinguish uniquely the icosahedral C 60 from its all 1812 fullerene isomers.
Leapfrog transformation L() for a 2-connected map  on surface Σ is usually defined in two ways.One way is first taking a dual of , then truncating the resulting graph [17,18].The other equivalent way is first taking triangulation (or omnicapping) on , then taking a dual of the resulting graph [17,19,20].Triangulation of a map is achieved by adding a new vertex in the center of each face followed by connecting it with each boundary vertex of this face [17,21].Dualization of a map on surface is built as follows: locate a point in the center of each face and join two such points if their corresponding faces share a common edge [17,20].Truncation of a map  is obtained by replacing each vertex V of degree  with  new vertices, one for each edge incident to V. Pairs of vertices corresponding to the edges of  are adjacent, and  new vertices corresponding to a single vertex of  are joined in a cycle of length  [17,18,20].If L() is the leapfrog graph of , we say that  is a parent of L().A fullerene (resp., (4,6)fullerene)   is called a leapfrog fullerene (resp., leapfrog (4,6)fullerene) if it is a leapfrog graph of some fullerene (resp., (4,6)-fullerene) .In this case, we denote the parent fullerene (resp., (4,6) The chemical significance of leapfrog transformation is that leapfrog cluster with at least one face of size 3 + 1 or 3 + 2 has a closed-shell electronic structure, with bounding HOMOs and antibonding LUMOs in Hückel theory, regardless of the electronic configuration of the parent fullerene [22].The other important fact for leapfrog transformation is that leapfrog fullerenes have at least one Kekulé structure containing the maximum number ||/3 of alternating hexagonal faces and thus to be maximally stable in a localised valence bond picture [23].Furthermore, the leapfrog spherical fullerenes obey the isolated-pentagon rule, while the leapfrog spherical (4,6)-fullerenes obey the isolated-square rule.
The paper is organized as follows.In Section 2, we show that the Clar number for a fullerene on surface Σ is bounded above by (||/6) − (Σ) and give some other useful results.In Section 3, we prove that the Clar number and Fries number of a (4,6)-fullerene on surface Σ are bounded above by (||/6) + (Σ) and (||/3) + (Σ), respectively.We characterize the (4,6)-fullerenes on surface Σ attaining these two bounds in terms of perfect Clar structure.In Section 4, by using the results in Section 3, we obtain further result that for the toroidal and Klein-bottle fullerenes, the upper bound for Clar number and Fries number always can be attained simultaneously.In Section 5, by establishing a relation between the fullerenes on the projective plane whose Clar numbers attain (||/6) − (Σ) and the extremal (4,6)fullerenes on the projective plane whose Clar numbers attain (||/6) + (Σ), we also characterize the fullerenes on the projective plane whose Clar numbers attain (||/6) − (Σ).

An Upper Bound for the Clar Number and Fries Number of Fullerenes on Surfaces
The following lemma is an extension of the result on the sphere in [24] to other surfaces.The proof of it is essentially similar to that in [24].However, to the completeness of this paper, we present it here.

Lemma 1.
Let  be a fullerene on surface Σ, and let H be a set of alternating faces with respect to a perfect matching .
Then there is an even number of edges in  (possibly 0) exiting any hexagon; there is an odd number of edges in  exiting any pentagon.
Proof.Let  be a face in .If  is a face in H, then no edges in  exit .Suppose that  is not in H. Any face of H adjacent to  is incident with two adjacent vertices on , as is any edge in  that lies on .Thus if  is a hexagon, then there is an even number of vertices (possibly 0) remaining to be covered by the edges in  that exit ; if  is a pentagon, then there is an odd number of vertices remaining to be covered by the edges in  that exit .
Zhang and Ye [25] proved that (||/6) − 2 is the upper bound of Clar number of spherical fullerenes.A new proof was given by Hartung [24].Gao and Zhang [26] generalized this bound to fullerenes on any surface as the following theorem.Here we give a new proof for this theorem using Hartung's method.
Proof.Let (H, ) be a Clar structure of .Since every face in H contains six vertices, every edge in  contains two vertices, and every vertex of  is incident with exactly one element of the Clar structure, we have the equation By Lemma 1 at least one edge in  exits each of the pentagons.There are additional edges in  if one of these edges exits a hexagon.By Euler's formula, a fullerene has exactly 6(Σ) pentagons, and thus || ≥ 3(Σ), where equality holds exactly when all edges from  connect pairs of pentagons.So We say that a fullerene on surface Σ is extremal if the Clar number of the fullerene attains the bound (||/6) − (Σ).As a direct corollary of the above theorem, the following result, which is also a generalization of the result on the sphere in [24], gives a necessary condition for the extremality of a fullerene.

Corollary 3.
Let  be an extremal fullerene on surface Σ, and let (H, ) be a Clar set of .Then there exists a pairing of the 6(Σ) pentagons such that each pair is connected by a single edge.The edges in  are only the 3(Σ) edges between pairs of pentagonal faces.
Fowler [23] proved that the Fries number for a spherical fullerene  is bounded above by ||/3 and characterized the spherical fullerenes that attain this bound, that is, the leapfrog spherical fullerenes.Then Fowler and Pisanski [27] generalized this result to trivalent polyhedrons on any surface.Theorem 4 (see [27]).Let  be a fullerene on surface Σ.Then  () ≤ ⌊||/3⌋.
Proof.For any edge in a perfect matching  of a fullerene , it is contained in at most 2 alternating faces.Since each alternating face has 3 edges in the perfect matching  and  has ||/2 edges of , we have 3 () ≤ 2|| = ||.It implies that  () ≤ ⌊||/3⌋ since  () is an integer.
A perfect Clar structure [27] (or face-only vertex covering [24]) of a map  is a set of vertex-disjoint faces that include each vertex of  once and only once.A perfect matching  is a Fries structure [27] if and only if for every edge, , that belongs to  the two adjacent faces are alternating faces with respect to .
Fowler and Pisanski [27] presented a sufficient condition for a map on a given surface to be a leapfrog graph.Later, in [24], Hartung showed that for planar graphs, the condition is also necessary.We are going to indicate that the sufficient condition is necessary for graphs on any surface.Proof.If a map  on surface Σ is a leapfrog graph, then each face of the triangulation of the parent graph is triangular.The fact that  is 3-regular follows from the fact that the dual of a triangulation graph is 3-regular.Furthermore, the faces of  corresponding to the faces of the parent graph form a perfect Clar structure of  (see Figure 1).
Conversely, if  is 3-regular and has a perfect Clar structure H, then −(H) is a perfect matching  of .For each alternating face of  with respect to , we add a new vertex at the center of this alternating face; two vertices are adjacent if the corresponding alternating faces are adjacent.Denote the resulting graph by   .It is easy to check that  is a leapfrog graph of   on surface Σ.
The following theorem is first presented by Fowler for spherical fullerenes [23], then generalized by Fowler and Pisanski [27] to trivalent polyhedron on any surface.
Theorem 6 (see [27]).Let  be a fullerene on surface.Then  attains the maximum Fries number ||/3 if and only if  is a leapfrog fullerene.
For a fullerene  on surface Σ, if  attains the maximum Fries number ||/3, then there exists a perfect matching  such that the number of alternating faces is ||/3, and further  is a Fries structure of .Since each vertex incidents with two alternating faces and a nonalternating face and the nonalternating face contain no edges of , all nonalternating faces form a perfect Clar structure of .
The following theorem is a natural generalization of a similar result on planar graphs in [24] and the proof is similar too.
Theorem 7. Given a bipartite map  on surface with partition  = (, ).Then the sets of faces of L() corresponding to , , and faces set of  form a face 3-coloring of .
Proof.Since the set of faces in  is corresponding to a perfect Clar structure H of L(), the remaining faces in L() correspond to vertices in .Denote the faces set of L() corresponding to  and  by U and W, respectively.Since  is bipartite, each face of L() is bounded by faces alternating from U and W. By Theorem 5, any leapfrog graph is 3regular.Thus each vertex in L() is incident with exactly one face from each of sets U, W, and H.
It is well known that Eulerian triangulation of the plane is 3-colorable.This implies that, for a 3-regular planar graph ,  is face 3-colorable is equivalent to  = L(  ) for some planar bipartite graph   [24].Using the same method in [24], we generalize this result to 3-regular maps on any surface as follows.
Theorem 8. Let  be a 3-regular map on surface Σ.Then the following three statements are equivalent.
Proof.(1) ⇒ (2) Since  is a 3-regular map, each vertex is incident with a face of each color and each color class is a perfect Clar structure of .By Theorem 5, there exists a graph   such that  is the leapfrog graph of   .The faces of one color class in  are one-to-one correspondence to all faces in   .Other faces in  are one-to-one correspondence to all vertices in   .  is a bipartite map since two vertices are adjacent in   if and only if corresponding faces are adjacent in .
(2) ⇒ (1) If  is the leapfrog graph of a bipartite map   on surface Σ, then the sets of faces of  corresponding to the bipartite partition of   together with the set of faces of  corresponding to the set of faces of   form a face 3-coloring of .
(3) ⇒ (1) For any face of  there is at most one way to adjoin new faces in order to build a perfect Clar structure.So each face belongs to at most one perfect Clar structure.If  has three different perfect Clar structures, then since  is 3regular, each vertex is incident with a face in each perfect Clar structure and three different perfect Clar structures cover each face of  exactly once.So these three different perfect Clar structures of  form a face 3-coloring of .
(2)  has a perfect Clar structure containing all quadrilateral faces.
Conversely, suppose that  has a perfect Clar structure H containing no quadrilaterals.Then |H| = ||/6. − (H) is clearly a perfect matching of .Denote this perfect matching by .We prove that each face other than the faces in H is alternating.To see this, we simply observe that each face  of  different from any hexagon in H has at least one edge  in , since  ∪ (H) = ().Two faces  1 and  2 connected by  are nonalternating faces with respect to .If  is a quadrilateral face, then the edge  1 lying on  that satisfies (,  1 ) = 1 also belongs to  since  and  1 are exterior of the same nonalternating face, simultaneously.If  is a hexagonal face, then the two edges  1 and  2 lying on  that satisfy (,   ) = 1 ( = 1, 2) also belong to  since  1 and  2 are exterior edges of two nonalternating faces  1 and  2 , respectively.So  is an alternating face for the perfect matching .Since the number of faces of  other than any hexagon in H is (||/3) + (Σ),  attains the maximum Fries number (||/3) + (Σ).

Clar Number and Fries Number of Klein-Bottle Fullerenes and Toroidal Fullerenes
Since a Klein-bottle fullerene (toroidal fullerene) coincides with a Klein-bottle (4,6)-fullerene (toroidal (4,6)-fullerene, resp.), we may use the results of the previous section to obtain more information about fullerenes on the Klein bottle and torus whose Clar numbers attain the upper bound in Theorem 9 and fullerenes on the Klein bottle and torus whose Fries numbers attain the upper bound in Theorem 11.
Theorem 13.For a toroidal or a Klein-bottle fullerene , the following four properties are equivalent.
The following theorem reveals an interesting appearance of leapfrog transformation on Klein-bottle fullerenes.

Theorem 14. The leapfrog fullerene of a bipartite Kleinbottle fullerene is nonbipartite, and the leapfrog fullerene of a nonbipartite Klein-bottle fullerene is bipartite.
Proof.We prove the first part of this theorem by presenting an odd cycle in the leapfrog fullerene of a bipartite Klein-bottle fullerene.Given a bipartite Klein-bottle fullerene (, , ), let  1 be a noncontractible cycle along the vertical direction of (, , ).Then  1 is an even cycle since (, , ) is bipartite.The vertices of L((, , )), on either side of  1 , which are proximal to cycle  1 induce a noncontractible cycle  2 of L((, , )) (see Figure 3).Clearly, the length of  2 is equal to the length of  1 plus double the times that  1 passes through the edges of (, , ) and minus one.So  2 is an odd cycle in L((, , )).
We prove the second part of this theorem by presenting a bipartition in the leapfrog fullerene of a nonbipartite Kleinbottle fullerene (, , ).For each face of the leapfrog fullerene of a nonbipartite Klein-bottle fullerene, which is contained entirely in some face of (, , ), we collect the vertices, along the boundary of this face, into  and  in the same manner, alternately (see Figure 4).Clearly  ∪  contains all vertices of the leapfrog graph of (, , ), and both  and  are independent.Theorem 14, together with Theorem 8, implies that none of the bipartite Klein-bottle fullerenes (, , ) are face 3-colorable and a nonbipartite Klein-bottle fullerene is a leapfrog fullerene if and only if it is face 3-colorable.
Combining Theorem 13 and remarks after Theorem 14 with Theorems 15, 16, and 17, respectively, we obtain the following three corollaries.
Corollary 18.For a toroidal fullerene (, , ), the following six properties are equivalent.
Corollary 19.For a Klein-bottle fullerene (, , ), the following five properties are equivalent.
Corollary 20.For a Klein-bottle fullerene (, , ), the following six properties are equivalent.

(6) 𝑁(𝑝, 𝑞, 𝑡) is face 3-colorable.
A map on a surface is called fully-benzenoid if it has a perfect Clar structure consisting only of hexagons.Gutman and Babić [30] gave a characterization of fully-benzenoid hydrocarbons earlier in 1991.Recently Gutman and Salem [31] showed that any fully-benzenoid hydrocarbon has a unique Clar set.Corollary 18, together with Corollaries 19 and 20, implies that fully-benzenoid (, , ) also has this property; however, fully-benzenoid fullerenes (, , ) and (, , ) each have exactly three Clar sets, while by Theorem 2, none of the spherical fullerenes and projective fullerenes is fully-benzenoid.
A map on the projective plane is the antipodal quotient of a centrosymmetric spherical map, that is, its vertices, edges, and faces are obtained by identifying antipodal vertices, edges, and faces of the centrosymmetric spherical map.Maps on the projective plane are usually drawn inside a circular frame where antipodal boundary points are identified [9].
For an extremal fullerene  on surface Σ, let (H, ) be a Clar structure of .Then the expansion of  is defined as follows: widen each of 3(Σ) edges in  between pentagonal pairs into a quadrilateral.Under this expansion, we can see that each vertex covered by  becomes an edge, and each pentagon becomes a hexagon (see Figure 5).Denote the resulting (4,6)-fullerene by E(H, ) and the set of quadrilateral faces by M.

Corollary 22.
Let  be an extremal fullerene on the sphere or the projective plane.Then the expansion E(H, ) of  has a perfect Clar structure containing all quadrilateral faces.
Theorem 10 and Corollary 22 imply that expansion of each extremal fullerene on surface Σ is always a leapfrog (4,6)-fullerene.A natural question is that given a leapfrog (4,6)-fullerene L(), can we always construct an extremal fullerene  such that L() is the expansion of ?
For each quadrilateral face of L(), we choose a pair of two opposite hexagons adjacent to the quadrilateral, denoted by P. The reverse expansion procedure on the pair (L(), P), E −1 (L(), P) is defined as follows: for each quadrilateral face of L(), shrink each of the two opposite edges that the quadrilateral shares with paired hexagons into a vertex.We then obtain the graph E −1 (L(), P), the reverse expansion of L() with respect to P (see Figure 5).
The following three theorems are the natural generalizations of the similar results on the sphere in [24].The proofs are analogous to the corresponding proofs and are omitted here.
Theorem 23.Suppose L() is the leapfrog fullerene of a (4,6)fullerene  on the sphere or the projective plane Σ.Let P be the set of hexagons paired on opposite sides of quadrilateral faces of L().The reverse expansion of this pair,  = E −1 (L(B), P), is a fullerene if and only if P is a set of 6(Σ) distinct hexagons.When  is a fullerene,  is extremal.
A diagonalization of a (4,6)-fullerene  is defined as a choice of diagonal vertices for each quadrilateral face so that no vertex is chosen twice.It follows that L() can be contracted into an extremal fullerene exactly when a diagonalization of  is possible.A perfect diagonalization of a bipartite (4,6)-fullerene  is a diagonalization in which all vertices chosen are in the same class of the bipartition.
Perfect diagonalization of a bipartite (4,6)-fullerene  on the sphere or the projective plane has been proposed to characterize the extremal fullerenes which has a Fries set of size ||/3.The hypothesis condition that  is bipartite always holds for (4,6)-fullerene on the sphere, since it is well known that a 3-regular planar graph is bipartite if and only if it has only faces of even degree.But it is indispensable for (4,6)-fullerenes on the projective plane.In fact, as we will see later, there exist infinitely many nonbipartite (4,6)-fullerene on the projective plane.The following theorem is somewhat interesting.
Theorem 26.The leapfrog fullerene of a bipartite (4,6)fullerene on the projective plane is nonbipartite, and the leapfrog fullerene of a nonbipartite (4,6)-fullerene on the projective plane is bipartite.
Proof.Suppose  is a bipartite (4,6)-fullerene on the projective plane, we prove the first part of this theorem by presenting an odd cycle in L().Let  be any noncontractible cycle of .Then the length of  is even.The vertices of L(), on either side of , which are proximal to cycle  induce a cycle  1 of L() (see Figure 6(a)).The length of  1 is equal to the length of  plus double the times that  passes through edges of  and minus one.So the cycle  1 is an odd cycle.
Suppose  is a nonbipartite (4,6)-fullerene on the projective plane.We prove the second part of this theorem by proving that there are no odd cycles in L().
For any cycle  in L(), if  is a contractible cycle, then  is an even cycle since the vertices of  and the interior of  induce a planar bipartite graph.So let us assume that  is a noncontractible cycle.
Let  1 be any noncontractible cycle in .We will show that  1 is an odd cycle.Suppose to the contrary that  1 is an even cycle.Since  is nonbipartite, there is an odd cycle  2 in .It is clear that  2 is also a noncontractible cycle of , and the intersection of  1 and  2 is nonempty.Without loss of generality, we may assume that V 1 , V 2 , . . ., V  are intersection vertices of  1 and  2 such that only one edge incident with V  on  1 does not lie on  2 (see Figure 6(b)).Since the lengths of  1 and  2 have opposite parity, at least one of the boundaries Clearly,  1 is a planar graph and is a subgraph of .Each face of  1 except the outer face has length 4 or 6, thus the length of boundary V 2  1 V 3  2 V 2 is even, a contradiction.So  1 is an odd cycle.
The vertices of , on either side of , which are proximal to  induce a noncontractible cycle  3 of  (see Figure 6(c)).
The length of  is equal to the length of  3 plus double the times that  passes through edges of  minus 1, thus the lengths of  and  3 have the opposite parity.Since  3 is an odd cycle,  is an even cycle.
The following theorem presented by Mohar in [32] will be useful for determining the face chromatic number of a (4,6)fullerene on the projective plane.
Theorem 27 (see [32]).Let  be an Eulerian triangulation of the projective plane.Then () ≤ 5 and  has a color factor.Moreover, if  is any color factor of , then  Theorem 27 implies that the unique projective (4,6)fullerene with chromatic number 4 is  4 .All other projective (4,6)-fullerenes are 3-colorable.This fact further implies that there is a polynomial time algorithm to determine the chromatic number of a projective (4,6)-fullerene.
Theorem 28.The face chromatic number of a nonbipartite (4,6)-fullerene on the projective plane is 3, the face chromatic number of a bipartite (4,6)-fullerene on the projective plane is 4 except L( 3 ), and the face chromatic number of the bipartite (4,6)-fullerene L( 3 ) on the projective plane is 5.
Proof.The theorem holds directly from Theorem 27, the definition of the leapfrog transformation, and the fact that the chromatic number of a map is equal to the face chromatic number of the dual map.
Theorem 23, together with Theorems 25 and 28, generates a method of constructing extremal fullerenes on the sphere and the projective plane from (4,6)-fullerene on these surfaces using the leapfrog transformation and the reverse expansion procedure.Given a (4,6)-fullerene  0 on the sphere or the projective plane, we define iterated leapfrog fullerenes   = L  ( 0 ) = L( −1 ), ( = 1, 2, . ..).Then all   ( ≥ 1) have a diagonalization, so the reverse expansion  +1 with respect to 3(Σ) pairs of hexagons corresponding to a diagonalization of   generates an extremal fullerene   on Σ.By being more specific on the projective plane, if  0 is nonbipartite, then each   ( ≥ 1) has a diagonalization and  2−1 is bipartite and isolated, thus having a perfect diagonalization.If  0 is bipartite, then each   ( ≥ 1) has a diagonalization and  2 ( ≥ 1) is bipartite and isolated, thus having a perfect diagonalization (see Figure 7).The reverse expansion with respect to 3(Σ) pairs of hexagons corresponding to a perfect diagonalization of   generates an extremal fullerene   on Σ which also has a Fries set of size ||/3.All hexagons, which are contained entirely in some hexagon of   , form a Clar set of resulting graph, whereas all faces corresponding to a color class of a bipartite   form a Fries set of resulting graph.A projective fullerene generated from reverse expansion L 2 ( 4 ) (see Figure 8(b)) with respect to 3(Σ) pairs of hexagons corresponding to a diagonalization (resp., perfect diagonalization) of L( 4 ) is illustrated by Figure 8(a) (resp., Figure 8(c)).Furthermore, two projective fullerenes, which are illustrated in Figures 8(a) and 8(c), are the only two projective fullerenes on 30 vertices.It is noticeable that the corresponding centrosymmetric spherical fullerene of the projective fullerene in Figure 8(c) is the famous buckminsterfullerene C 60 .
Theorem 29.A map on the projective plane is a leapfrog graph if and only if the corresponding centrosymmetric spherical map is a leapfrog graph.
Proof.Suppose a map  on the projective plane is a leapfrog graph.Then by Theorem 5,  is 3-regular and has a perfect Clar structure H. Since each face of H is obtained by identifying two antipodal faces of the corresponding centrosymmetric spherical map   , all faces in   corresponding to faces of H in  form a perfect Clar structure H  of   .Since each vertex of  is obtained by identifying two antipodal vertices of the corresponding centrosymmetric spherical map   ,   is also 3-regular.By Theorem 5, the corresponding centrosymmetric spherical map   of  is a leapfrog graph.Conversely, suppose the corresponding centrosymmetric spherical map   of  is a leapfrog graph.Then by Theorem 5,   is 3-regular and has a perfect Clar structure H  .Since the parent graph L −1 (  ) is also a centrosymmetric spherical map and each face of H  is corresponding to a face of L −1 (  ), the antipodal face of each face in H  is also in H  .The set of faces obtained by identifying all antipodal faces of H  forms a perfect Clar structure H of .Clearly,  is 3regular.By Theorem 5,  is a leapfrog graph.
As a corollary of Theorems 10 and 29, we have the following result.

Figure 1 :
Figure 1: Illustration for the proof of Theorem 5.

Figure 3 :
Figure 3: Illustration for the first part of proof in Theorem 14.

Figure 4 :
Figure 4: Illustration for the second part of proof in Theorem 14.

Figure 6 :
Figure 6: Illustration for the proof of Theorem 26.

( 1 )
() = 3 if and only if  −  is bipartite; (2) () = 4 if and only if  −  is not bipartite and does not contain a quadrangulation of the projective plane; (3) () = 5 if and only if  −  is not bipartite and contains a quadrangulation of the projective plane.Such a quadrangulation is necessarily nonbipartite.
For any edge in a perfect matching  of a (4,6)fullerene , it is contained in at most 2 alternating faces.Suppose that I is the set of alternating faces in  with respect to a perfect matching  and  is the number of quadrilaterals in I. Since each alternating hexagon has 3 edges, and each alternating quadrilateral has 2 edges in the perfect matching , we have 3(|I| − ) + 2 ≤ 2|| = ||.