On k-resonant fullerene graphs
On k-resonant fullerene graphs∗ arXiv:0801.1483v1 [math.CO] 9 Jan 2008
Dong Ye, Zhongbin Qi and Heping Zhang† School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, P. R. China E-mails: [email protected], [email protected], [email protected]
Abstract A fullerene graph F is a cubic 3-connected plane graph with exact 12 pentagons and other hexagons. Let M be a perfect matching of F . A cycle C of F is M -alternating if and only if the edges of C appear alternately in and off M . A set H of disjoint hexagons of F is called a resonant pattern if F has a perfect matching M such that all hexagons in H are M -alternating. A fullerene graph F is k-resonant if any i (0 ≤ i ≤ k) mutually disjoint hexagons of F form a resonant pattern. In this paper, we prove that every hexagon of a fullerene graph is resonant and all leapfrog fullerene graphs are 2-resonant. Finally, we show that a k-resonant (k ≥ 3) fullerene graph has at most 60 vertices and construct all nine 3-resonant fullerene graphs, which are also k-resonant (k > 3). Keywords: Fullerene graph; Resonant pattern; k-resonant; Perfect matching AMS 2000 subject classification: 05C70, 05C90
1
Introduction
A fullerene graph is a cubic 3-connected plane graph with exact 12 pentagonal faces and other hexagonal faces. Fullerene graphs has been studied in mathematics as trivalent polyhedra for a long time [9, 11], for example, the dodecahedron is the fullerene graph with 20 vertices. Fullerene graphs have been studied in chemistry as fullerene molecules which have extensive applications in physics, chemistry and material science [6]. Let G be a plane 2-connected graph. A perfect matching or 1-factor M of G is a set of independent edges such that every vertex of G is incident with exact one edge in M. A cycle of G is M-alternating if edges of C is alternately in and off M. For a fullerene graph F , every edge of F belongs to a perfect matching [4]. A hexagon h of a fullerene graph F is ∗ †
This work is supported by NSFC. Corresponding author.
1
resonant if F has a perfect matching M such that h is M-alternating. It was proved that every hexagon of a normal benzenoid system is resonant [22]. This result was generalized to normal coronoid systems [2] and plane elementary bipartite graphs [24]. However a fullerene graph is a non-bipartite graph. It is natural to ask if every hexagon of a fullerene graph is resonant. The present paper first uses Tutte’s Theorem to give a positive answer to this question. A set H of disjoint hexagons of a fullerene graph F is a resonant pattern (or sextet pattern) if F has a perfect matching M such that every hexagon in H is M-alternating; equivalently, if F − H has a perfect matching, where F − H denotes the subgraph obtained from F by deleting all vertices of H together with their incident edges. The maximum cardinality of resonant patterns of F is called Clar number of F [3, 17], and the maximum number of M-alternating hexagons over all perfect matchings M of F is called the Fries number of F [8]. For fullerene graphs, Graver [10] explored some connections among Clar number, face independence number and Fries number of a fullerene graph, and obtained a lower bound for the Clar number of leapfrog fullerene graphs with icosahedral group. Zhang and Ye [25] ⌋ showed that Clar number of a fullerene graph Fn with n vertices satisfies c(Fn ) ≤ ⌊ n−12 6 which is sharp for infinite many fullerene graphs, including C60 whose Clar number is 8 [1]. Shiu, Lam and Zhang [18] computed the Clar polynomial and sextet polynomial of C60 by showing that every face independent set of C60 is also a resonant pattern. A fullerene graph is k-resonant if any i (0 ≤ i ≤ k) disjoint hexagons form a resonant pattern. Hence a fullerene graph with each hexagon being resonant is 1-resonant. Zheng [28, 29] characterized general k-resonant benzenoid systems. In particular, he showed that every 3-resonant benzenoid system is also k-resonant (k ≥ 3). This result also holds for coronoid systems [14], open-ended nanotubes [23], toroidal polyhexes [19, 26] and Kleinbottle polyhexes [20]. For a recent survey on k-resonant benzenoid systems, refer to [12]. Next the paper mainly consider general k-resonant fullerene graphs. We show that all leapfrog fullerene graph are 2-resonant and a 3-resonant fullerene graph has at most 60 vertices. Finally, we construct all 3-resonant fullerene graphs, which are all k-resonant (k ≥ 3).
2
1-resonance of fullerene graphs
Let G be a plane graph admitting a perfect matching. For a face f of G, let V (f ) and E(f ) be the sets of vertices and edges of f , respectively. Use ∂G to denote the boundary of G. Let G1 and G2 be two subgraphs of G. We say G1 and G2 are adjacent if ∂G1 ∩ ∂G2 6= ∅ and (G1 − ∂G1 ) ∩ (G2 − ∂G2 ) = ∅. A subgraph H of G is nice if G − H has a perfect matching. So a resonant pattern of G is also a nice subgraph of G. A graph G is cyclically k-edge connected if deleting fewer than k edges of G can not separate G into two components such 2
that each of them contains a cycle. By Tutte’s Theorem on perfect matching of graphs ([15], Theorem 3.1.1), we have following result. Theorem 2.1. A subgraph H of a graph G is nice if and only if for any S ⊆ V (G − H), Co (G − H − S) ≤ |S|, where Co (G − H − S) is the number of odd components of G − H − S. Theorem 2.2. Let G be a cyclically 4-edge connected cubic graph with a 6-length cycle. Then for every 6-length cycle H of G, either H is nice or G − H is bipartite. Proof: Let H be a 6-length cycle in G. If G − H has a perfect matching, then the theorem holds. If not, then by Theorem 2.1 there exists an S ⊂ V (G −H) such that Co (G −H −S) ≥ |S| + 2 by parity, i.e. |S| ≤ Co (G − H − S) − 2. Since G is cubic, S sends out at most 3|S| ≤ 3Co (G − H − S) − 6 edges. Let G1 , G2 , ..., Gk be all odd components of G−H −S, where k = Co (G−H −S). Because G is cyclically 4-edge connected and cubic, it has no bridge. Every Gi (i = 1, 2, ..., k) sends odd number edges, hence at least three edges, to H ∪ S. So ∪ki=1 Gi sends out at least 3Co (G − H − S) edges to either S or H. Since H is a 6-length cycle, there are at most 6 edges between H and ∪ki=1 Gi . So ∪ki=1 Gi sends at least 3Co (G−H −S)−6 edges to S. Hence there are precisely 3Co (G − H − S) − 6 edges between S and ∪ki=1 Gi . So S is an independent set, and every Gi sends out exact 3 edges, and G − H − S has no even component. In addition, since G is cyclically 4-edge connected, every Gi is a tree. We claim that each Gi is a singular vertex. If not, then an odd component Gi has at least 2 vertices. So Gi has at least two leaves. Every leaf of Gi is adjacent to at least two vertices in S ∪ H. So Gi sends at least four edges out, contradicting the fact that every Gi sends precise three edges out. Therefore G − H is a bipartite graph with bipartition (S, G − H − S). This completes the proof of the theorem. Lemma 2.3. [5, 16] Every fullerene graph is cyclically 5-edge connected. By Lemma 2.3 and Theorem 2.2, we immediately have following result. Theorem 2.4. Every hexagon of a fullerene graph is resonant. Proof: Let F be a fullerene graph and H be a hexagon. It is obvious that F − H is not bipartite. By Theorem 2.2 and Lemma 2.3, H is nice. That means H is resonant.
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2-resonant fullerene graphs
Let F be a fullerene graph. The leapfrog operation on F is defined [7] as follows: for any face f ∈ F (F ), add a new vertex vf in f and join vf to all vertices in V (∂f ) to obtain a 3
new triangular graph F ′ ; Take the geometry dual of the graph F ′ and denote it by F ∗ (see Figure 1). Clearly, F ∗ is a fullerene graph since every vertex of F ′ is 6-degree excluding exact 12 5-degree vertices and every face of F ′ is a triangle. The edges of F ∗ cross the edges of F ⊂ F ′ in the geometry dual operation form a perfect matching M 0 of F ∗ . A fullerene graph is called leapfrog fullerene graph if it arises from a fullerene graph by the leapfrog operation. Fowler and Pisanski [7] explored the connection between leapfrog fullerenes and the Clar theory of aromatic sextet had been explored and showed that a fullerene graph is a leapfrog fullerene if and only if it has a Fries structure, i.e. a perfect matching which avoids edges in pentagons and is alternating on the maximal number n/3 hexagons.
Figure 1: The leapfrog operation on the dodecahedron F20 and the perfect matching M 0 of C60 (double edges).
Let F ∗ be a leapfrog fullerene graph arising from F . A face f of F ∗ is called a heritable face if it lies completely in some face of F , and a fresh face, otherwise. For example, C60 is the leapfrog fullerene graph of the dodecahedron and every pentagon is a heritable face and all hexagons are fresh faces. The perfect matching M 0 is corresponding to the Fries structure of C60 (see Figure 1). For a leapfrog fullerene graph, we have following result. Lemma 3.1. Let F be a leapfrog fullerene graph. Then every fresh face is M 0 -alternating and all heritable faces are independent. Let F be a leapfrog fullerene and f be a heritable face. A subgraph of F consisting of f together with all adjacent fresh faces is called the territory of f and denoted by T [f ]. For two heritable faces f1 and f2 , it is easily seen that there are at most 2 common fresh faces in their territories, which are adjacent. Theorem 3.2. Every leapfrog fullerene graph is 2-resonant. Proof: Let F be a leapfrog fullerene graph and f1 , f2 be any two disjoint hexagons. If both f1 and f2 are fresh faces, then clearly M 0 is alternating on both of them by Lemma 3.1. So 4
suppose that at least one of them is a heritable face, say f1 . If f2 is fresh, then f2 * T [f1 ] and it is adjacent to at most one face in T [f1 ]. Let us denote clockwise the 6 fresh hexagons in T [f1 ] by h0 , h1 , ..., h5 . Let M1 := M 0 ⊕ h1 ⊕ h3 ⊕ h5 if f2 adjoins none of h1 , h3 and h5 ; otherwise we choose h0 , h2 and h4 instead. Then M1 is a perfect matching and alternating on both f1 and f2 . So, in the following, we suppose both f1 and f2 are heritable. Let h′0 , h′1 , ..., h′5 be the six fresh hexagons of T [f2 ] in clockwise. First suppose T [f1 ] and T [f2 ] has a common hexagon; that is, there are i0 , j0 ∈ Z6 such that hi0 and h′j0 are the same hexagon. Let M2 := M 0 ⊕ hi0 ⊕ hi0 +2 ⊕ hi0 +4 ⊕ h′j0 +2 ⊕ h′j0 +4 . It is clear that M2 is a perfect matching alternating on both f1 and f2 . Now suppose T [f1 ] and T [f2 ] have no common hexagons. If no face in T [f2 ] adjoins one of h1 , h3 and h5 , let M3 := M 0 ⊕ h1 ⊕ h3 ⊕ h5 ⊕ h′1 ⊕ h′3 ⊕ h′5 ; otherwise, let M3 := M 0 ⊕ h0 ⊕ h2 ⊕ h4 ⊕ h′1 ⊕ h′3 ⊕ h′5 . Then M3 is also a perfect matching alternating on both f1 and f2 . So the theorem holds.
Figure 2: The dodecahedron F20 (left) and the Fullerene graph F24 with a perfect matching M (right).
There exist non-leapfrog 2-resonant fullerene graphs. The dodecahedron F20 is a trivial example. F24 (see Figure 2) is also 2-resonant since the two hexagons in it are simultaneously M-alternating with respect to the perfect matching M illustrated by double edges in Figure 2. Another non-trivial example is C70 . It has two perfect matchings M1 and M2 as shown in Figure 3. Let M3 := M1 ⊕ h2 ⊕ h10 and M4 := M1 ⊕ h2 ⊕ h4 ⊕ h6 ⊕ h8 ⊕ h10 . Let H := {h, h′ } be a set of any two disjoint hexagons in C70 . If h, h′ ∈ / {h1 , h3 , h5 , h7 , h9 }, then h and h′ are simultaneously M1 -alternating. So suppose h ∈ {h1 , h3 , h5 , h7 , h9 }. By symmetry, we may assume h = h1 . If h′ ∈ {h1 , h3 , h5 , h7 , h9 }, then both h and h′ are M4 -alternating. If h′ ∈ {h11 , h12 }, let h′ = h11 by the symmetry of h11 and h12 . Then both h and h′ are M2 -alternating. Finally, if h′ ∈ / {h3 , h5 , h7 , h9 , h11 , h12 }, then h and h′ are simultaneously M3 -alternating. By symmetry, we have C70 is 2-resonant. On the other hand, we can construct infinitely many fullerene graphs which are not 2resonant. Let H be a set of at most k disjoint facial cycles of a plane graph G. By Theorem 2.1, if there exists a vertex set S ⊆ V (G − H) such that Co (G − H − S) > |S|, then H is not a resonant pattern and G is not k-resonant. For convenience, the vertices in such S and the faces in such H are always illustrated in grey color and the vertices of odd components of G − H − S are illustrated in black color from now on. Let R5 and R6 be the graphs arising from deleting the outer pentagonal facial cycle from 5
Figure 3: C70 with two perfect matchings M1 (left) and M2 (right).
F20 and deleting the outer hexagonal cycle from F24 (see Figure 2), respectively.
Figure 4: R5 and R6 and the illustration for the proof of Theorem 3.3.
Theorem 3.3. Let F be a fullerene graph different from F20 and F24 . If F contains R5 or R6 as subgraphs, then F is not 2-resonant. Proof: First suppose R5 ⊂ F . Since F is different from F20 , there are at least two faces adjacent to R5 are hexagons and disjoint. Let H be the set of these two hexagons (grey hexagons in Figure 4). Then the four grey vertices of R5 in Figure 4 form a vertex set S such that such that F − H − S contains five isolated vertices (black vertices of R5 in Figure 4). So H is not a resonant pattern. Now suppose R6 ⊂ F . Since F is different from F24 , at least one of the faces of F adjacent to R6 is hexagonal. Let H be the set consisting of this hexagon together with the center hexagon of R6 . It is easy to see that H is not a resonant pattern (see Figure 4). Using R5 and R6 as caps, we can construct infinitely many fullerene graphs which are 1-resonant but not 2-resonant. It is interesting to characterize 2-resonant fullerene graphs. Here, we present a conjecture about 2-resonant fullerene graphs which is supported by Theorem 3.2. Conjecture 3.4. Every fullerene graph without adjacent pentagons is 2-resonant.
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4
k-resonant fullerene graphs (k ≥ 3)
Let f be a face of a fullerene graph F . A vertex v ∈ V (F − V (f )) is incident to f if N(v) ∩ V (f ) 6= ∅, where N(v) is the neighbor set of v. Let G∗ be a graph as shown in Figure 5. Then G∗ is a forbidden subgraph of k-resonant (k ≥ 3) fullerene graphs since deleting the three hexagons of G∗ isolates the vertex v and hence the three hexagons do not form a resonant pattern by Theorem 2.1.
Figure 5: A forbidden subgraph G∗ for 3-resonant fullerene graphs.
Theorem 4.1. Let F be a k-resonant (k ≥ 3) fullerene graph. Then |V (F )| ≤ 60. Proof: Since F is k-resonant (k ≥ 3), G∗ * F . So any v ∈ V (F ) is incident to at least one pentagon of F . On the other hand, for any pentagon f of F , there are at most 5 vertices in V (F − f ) incident to it. Hence |V (F )| ≤ 12 × 5 = 60 since F has exact 12 pentagons. So the theorem holds. Since C60 has the Fries Kekul´e structure M 0 which is alternating on every hexagon of C60 (see Figure 1), it is k-resonant (k ≥ 1). By Theorem 4.1, C60 is the unique k-resonant leapfrog fullerene graph for k ≥ 3 since it is the unique leapfrog fullerene graph with no more than 60 vertices. We restate this fact as follows: Corollary 4.2. C60 is the unique k-resonant (k ≥ 3) leapfrog fullerene graph. A fragment B of a fullerene graph F is a subgraph of F consisting of a cycle together with its interior. A face of F adjacent to B but not in B is called a neighboring face of B. The territory T [B] of B is the subgraph of F consisting of B together with its all neighboring faces. The following useful lemma is obtained by Ye and Zhang [21].
K
2
K 1, 3
T0
Figure 6: Trees: K2 , K1,3 and T0 .
Lemma 4.3. ([21], Lemma 2.1) Let B be a fragment of F and W be the vertex set consisting of all 2-degree vertices on ∂B. If 0 < |W | ≤ 4, then T = F − (V (B) \ W ) is a forest and 7
(1) T is K2 if |W | = 2; (2) T is K1,3 if |W | = 3; (3) T is the union of two K2 s, or a 3-length path, or T0 , if |W | = 4. For a fragment B of a fullerene graph F , let FB := F − (B − ∂B). Let B1 , B2 , B3 and B4 be the four fragments shown in Figure 7. Lemma 4.4. Let B be a fragment of a fullerene graph F such that there are six 2-degree vertices and six 3-degree vertices appearing alternately on ∂B. Then either all neighboring faces of B are hexagons or FB is isomorphic to R6 or one of Bi for i = 1, 2, 3, 4.
Figure 7: Five possible isomorphic graphs of FB in Lemma 4.4.
Proof: Let x be the number of the hexagonal neighboring faces of B and T [B] be the territory of B. Then T [B] is a fragment of F with x 2-degree vertices where 0 ≤ x ≤ 6 and x 6= 1 since there are six 2-degree vertices and six 3-degree vertices appearing alternately on ∂B. If x = 6, then the lemma is true. So suppose that x < 6. If x = 0, all neighboring faces of B are pentagons and therefore ∂T [B] is a 6-cycle without 2-degree vertex. So FB is isomorphic to R6 . If x = 2, up to isomorphism, T [B] has three possible cases which are showed in Figure 8. By Lemma 4.3, the two 2-degree vertices u1 and u2 on ∂T [B] are adjacent in F . So only the graph (c) is possible since F does not has a 3-length cycle or 4-length cycle. Hence FB is isomorphic to B1 . A similar analysis for x = 3 and x = 4 will bring us to the conclusion that FB is isomorphic to B2 and FB is isomorphic to B3 , respectively. Here we omit the detailed analysis. u2
u1
B
u1
u1
B
u2
B u2
Figure 8: Three possible cases of T [B] for x = 2.
8
Now suppose x = 5. Let u1 , u2 , u3, u4 and u5 be the five 2-degree vertices on ∂T [B] (see Figure 9 (a)). It is clear that u1 and u5 are not adjacent in F . If they have a common neighbor vertex w, then we obtain a fragment with 4 2-degree vertices (see Figure 9 (b)). So by Lemma 4.3, one pair of vertices u2 ,u3 and vertices u3 ,u4 will be adjacent or have a common neighbor vertex in F − T [B], contradicting that F has no 3-length cycle and 4length cycle. So u1 and u5 have no common neighbor vertex. That is the neighboring face f of T [B] with u1 and u5 on its boundary must be a hexagon (see Figure 9 (c)). Let B ′ be the territory of T [B] and y be the number of 2-degree vertices on ∂B ′ . Then 0 ≤ y ≤ 4 and y 6= 1. If y > 0, by Lemma 4.3, there exist at least a pair of 2-degree vertices on ∂B ′ such that either they are adjacent in F or they have a common neighbor vertex in F − V (B ′ ), also contracting that F has no 3-length cycle and 4-length cycle. So y = 0 and FB is isomorphic to B4 . w
u5
u1
u1
u4 u3
u5
B u4
u2 u3
u5
u1
B
B u2
f
u4
u2 u3
Figure 9: The illustration for the proof of Lemma 4.4 for x = 5. A fragment B of a fullerene graph F is maximal if all its neighboring faces are hexagonal. Lemma 4.5. Let F be a k-resonant (k ≥ 3) fullerene graph. Then F contains none of R6 and B1 as maximal fragments and F does not contain B2 and B4 . Proof: Theorem 3.3 implies that F can not contain R6 as its maximal fragment. If F contains B1 as a maximal fragment or B2 , then F contains the subgraph G∗ as subgraph which is forbidden in a k-resonant k ≥ 3 fullerene graphs, a contradiction (see Figure 10 (a) and (b)). If F contains B4 , then the three independent grey hexagonal faces of B4 in Figure 10 (c) is not a resonant pattern since deleting them will result in a component isomorphic to K1,3 which has no perfect matching. Hence F is not 3-resonant, also a contradiction. Hence the lemma is true. A fragment B of a fullerene graph F is said to be a pentagon fragment if its every inner face is a pentagon. Let γ(B) be the minimum number of pentagons adjoining the same pentagon in B. Let B ∗ be the inner dual of B. Then γ(B) is the the minimum degree of B ∗ . For example, γ(R5 ) = 3. Lemma 4.6. ([21], Lemma 3.6) Let B be a pentagon fragment of a fullerene graph F . Then: (1) R5 ⊆ B if γ(B) ≥ 3; (2) B has a pentagon adjoining exactly two adjacent pentagons of B if γ(B) = 2. 9
Figure 10: Illustration for the proof of Lemma 4.5.
A turtle is a pentagon fragment consisting of six pentagons as illustrated in Figure 11. Clearly, γ(B) = 1 if B is a turtle. The fragment B3 in Figure 7 contains a turtle. Following is a theorem characterizing maximal pentagon fragments of k-resonant (k ≥ 3) fullerene graphs.
Figure 11: Pentagon fragment turtle.
Theorem 4.7. Let F be a k-resonant (k ≥ 3) fullerene graph different from F20 and B be a maximal pentagon fragment of F . Then B is isomorphic to either a pentagon or a turtle. Proof: Let B be a maximal pentagon fragment of F . By Theorem 3.3, we may assume that B dose not contain R5 . Lemma 4.6 implies γ(B) ≤ 2. If γ(B) = 0, then B is a pentagon. So suppose that γ(B) > 0. Let H be a set of disjoint of hexagons. Since F is k-resonant (k ≥ 3), F − H has a perfect matching; that is, F contains no H such that F − H has no perfect matchings.
(∗)
Case 1. γ(B) = 1. That means B has a pentagonal face f0 which has only one neighboring pentagonal face f1 ⊂ B. Let h1 , h2 , h3 and h4 be the four neighboring hexagonal faces of B such that hi is adjacent to hi+1 (1 ≤ i ≤ 3) and both h1 and h4 are also adjacent to f1 . Further, let f2 and f3 be the other two neighboring faces of f1 as illustrated in Figure 12 (a). If one of f2 and f3 is hexagonal, say f2 , then H := {h2 , h4 , f2 } contradicts (∗) since F −H has an isolate vertex. So suppose both f2 and f3 are pentagons of B. Let f4 6= f1 be the face adjacent to both f2 and f3 . If f4 is a hexagon, then H := {h1 , h4 , f4 } contradicts (∗) since F − H has an isolate vertex. So f4 must be a pentagonal face of B. Let h5 and h6 be the two faces adjacent to a pair pentagons f2 , f4 and a pair pentagons f3 , f4 , respectively (see Figure 12 (b)). Let f5 be the face adjacent to h5 , f4 and h6 . If f5 is a hexagon, then H := {h1 , h4 , f5 } 10
also contradicts (∗) since one component of F − H is K1,3 . So f5 is a pentagonal face of B. Thus G := ∪5i=0 fi ⊆ B is isomorphic to a turtle. It sufficient to show that B = G. That is all neighboring faces of G are hexagons. Beside h1 , . . . , h6 , let h7 and h8 be other are two neighboring faces of G as illustrated in Figure 12 (b). Let G′ := G ∪ (∪8i=1 hi ). Because h1 , h2 , h3 and h4 are already hexagonal, we need only to show that h5 , h6 , h7 and h8 are hexagonal. Since R5 * B, one of h5 and h6 must be hexagonal, say h5 by symmetry. Then h6 is either a pentagon or a hexagon.
Figure 12: Illustration for the proof of Subcase 1.1 of Lemma 4.7.
Subcase 1.1. h6 is a pentagon. If h7 is a hexagon, then H := {h1 , h4 , h7 } contradicts (∗) since F − H has an odd component with five vertices (see Figure 12 (b)). So h7 must be a pentagon. If h8 is a pentagon, then G′ is a fragment with only two 2-degree vertices w1 and w2 on h2 (see Figure 12 (c)). By Lemma 4.3, w1 and w2 are adjacent in F − (G′ − {w1 , w2 }), resulting in a 2-length cycle of F , a contradiction. So h8 must be a hexagon. Then the fragment G′ contains four 2-degree vertices (see Figure 12 (d)). If G′ ⊂ F , it is easy to check that there will be at least one face of F with size of at most 4 by Lemma 4.3, also a contradiction. The contradiction implies that h6 is not a pentagon.
Figure 13: Illustration for the proof of Subcase 1.2 of Lemma 4.7.
Subcase 1.2. h6 is a hexagon. Let h9 , h10 be other two neighboring faces of G′ as shown 11
in Figure 13 (a) and let G′′ := G′ ∪ h9 ∪ h10 . We claim that both h7 and h8 are hexagons. Suppose to the contrary that at least one of them is a pentagon, say h7 . If at least one of h8 , h9 and h10 is a pentagon, then G′′ is a fragment with at most four 2-degree vertices. By Lemma 4.3, it is readily checked that G′′ can not be a subgraph of F . So all of h8 , h9 and h10 are hexagons. Then H := {h3 , h8 , h9 } contradicts (∗) since F − H has an odd component with 15 vertices (see Figure 13 (b)). Hence both h7 and h8 are hexagons. So all neighboring faces of G are hexagonal and hence B = G. Case 2. γ(B) = 2. Lemma 4.6 implies that B contains a pentagonal face f0 which has two adjacent pentagonal neighboring faces f1 and f2 . Let h1 , h2 and h3 be the other three hexagonal neighboring faces of f0 as shown in Figure 14 (a).
Figure 14: Illustration for the proof of Case 2 of Lemma 4.7.
Let f3 6= f0 be the face adjacent to both f1 and f2 . If f3 is a hexagon, then H := {h1 , h3 , f3 } contradicts (∗) since F − H has an isolate vertex. Therefore f3 must be a pentagon in B. Let h4 , f4 , h5 be the other three neighboring faces of f3 as shown in Figure 14 (a). If f4 is hexagonal, then H := {h1 , h3 , f4 } also contradicts (∗) since one component of F − H is K1,3 . Since R5 * B, at least one of h4 and h5 is a hexagon, say h4 by symmetry. If h5 is also a hexagon, then H := {h2 , h4 , h5 } contradicts (∗) since one component of F − H is K1,3 . So h5 is a pentagon. Let h6 and h7 be other two neighboring faces of f4 as shown in Figrue 14. If h6 is a hexagon, then H := {h1 , h3 , h6 } contradicts (∗) since F − H has an odd component with size seven (see Figure 14 (b)). Hence h6 is a pentagon. Similarly, h7 is also a pentagon (see Figure 14 (c)). Now, we have a fragment G := (∪4i=0 fi ) ∪ (∪7j=1 hj ) with four 2-degree vertices w1 , w2 , w3 , w4 (see Figure 14 (d)). By Lemma 4.3, G * F . This completes the proof of the theorem.
5
Construction of k-resonant (k ≥ 3) fullerene graphs
Let {fi |i ∈ Zl , l ≥ 5} be a set of l faces of a fullerene graph F such that fi ∩ fi+1 (i ∈ Zl ) is an edge, denoted by ei . The subgraph R := ∪i∈Zl fi is called a ring if {ei |i ∈ Zl } is an 12
independent edge set and l is called the size of the ring R, denoted by l(R). A ring is called a pentagon ring if every fi (i ∈ Zl ) is a pentagon. The R5 and R6 in Figure 4 are two pentagon rings with size five and six, respectively. Note that R5 is the unique pentagon ring with size five. C60 is the unique fullerene graph with no more than 60 vertices and without adjacent pentagons. By Theorem 4.1 and Theorem 4.7, a k-resonant (k ≥ 3) fullerene graph different from C60 contains either maximal pentagon fragments each of which is isomorphic to a turtle or pentagon rings. Lemma 5.1. Let F be a k-resonant (k ≥ 3) fullerene graph containing no pentagon rings. 1 Then F is isomorphic to C60 or the graph F36 shown in Figure 15.
1 Figure 15: The k-resonant (k ≥ 3) fullerene graph F36 with a perfect matching M
Proof: C60 is clearly k-resonant (k ≥ 3) since the Fries Kekul´e structure M 0 is alternating 1 on every hexagon (see Figure 1). For F36 showed in Figure 15, the perfect matching M 1 illustrated in Figure 15 is also alternating on its every hexagon. So F36 is also k-resonant (k ≥ 3). Let F be a k-resonant (k ≥ 3) fullerene graph without pentagon rings as subgraphs. By Theorem 4.7, every maximal pentagon fragment of F is isomorphic to a pentagon or a turtle. If F does not contain a maximal pentagon fragment isomorphic to a turtle, then F is isomorphic to C60 by Theorem 4.1. So suppose that F contains a maximal pentagon fragment B isomorphic to a turtle. Denote clockwise the hexagonal neighboring faces of B by h1 , h2 , ..., h8 as shown in Figure 16. Let G0 := B ∪ h3 ∪ h4 ∪ h7 ∪ h8 . Then G0 is is a fragment isomorphic to B3 in Figure 7. Clearly, G0 satisfies the conditions of Lemma 4.4. So either FG0 := F − (G0 − ∂G0 ) is isomorphic to one of R6 and Bi (i = 1, 2, 3, 4) in Figure 7 or all neighboring faces of G0 is hexagonal. By Theorem 3.3 and Lemma 4.5, either FG0 is isomorphic to B3 or all neighboring faces of G0 are hexagonal. 1 If FG0 is isomorphic to B3 , then F is isomorphic to F36 since B is maximal. Now suppose that all neighboring faces of G0 are hexagonal. Let G1 be the fragment consisting of G0 together with its all hexagonal neighboring faces. A similar analysis for G1 yields that either 13
Figure 16: The fragments G0 , G1 and G2 in the proof of Lemma 5.1.
FG1 := F − G0 is isomorphic to B3 or all neighboring faces of G1 are hexagonal. If the former holds, then there are two fullerene graphs with 48 vertices containing two subgraphs isomorphic to B3 as shown in Figure 17. Both of the two fullerene graphs are not 3-resonant since each of them has three disjoint hexagons whose deleting results in a odd component (grey hexagons in Figure 17).
Figure 17: Two fullerene graphs with 48 vertices and two subgraphs isomorphic to B3 . Now suppose that all neighboring faces of G1 are hexagonal. Let G2 be the fragment consisting of G1 together with its all hexagonal neighboring faces. Let FG2 := F − G1 . Since |V (G1 )| = 36 and |V (F )| ≤ 60, |V (FG2 )| ≤ 24. By Theorem 3.3, Lemma 4.4 and Lemma 4.5, FG2 is isomorphic to B3 since |V (B3 )| = 24. Then we have two fullerene graphs with 60 vertices as shown in Figure 18. Both of the two fullerene graphs are not 3-resonant since each of them has three disjoint hexagons whose deletion results in a single vertex (the grey hexagons in Figure 18).
Figure 18: Two fullerene graphs with 60 vertices and two subgraphs isomorphic to B3 . 1 Summarizing above discussions, F is isomorphic to C60 or F36 .
14
1 Let F be a k-resonant (k ≥ 3) fullerene graph different from C60 and F36 . By Lemma 5.1, F should contain a pentagon ring. Let R ⊂ F be a pentagon ring with size l(R). Clearly, R has two faces different from the pentagons of it. Let C and C ′ be the boundaries of the two faces, respectively. Suppose that C and C ′ have s(R) and s′ (R) 2-degree vertices, respectively. Assume s(R) ≤ s′ (R). We always embed R in the plane such that C ′ = ∂R. The cycles C and C ′ are called inner cycle and outer cycle of R, respectively. Since s′ (R) + s(R) = l(R) and s(R) ≤ s′ (R), s(R) ≤ ⌊ l(R) ⌋. Clearly, s(R) 6= 1 and s′ (R) 6= 1 2 since F is 3-connected. Let ψl (F ) := min{s(R)| all pentagon rings of F with size l}. For example, ψ5 (F20 ) = 0 and ψ6 (F24 ) = 0. Let r(R), n6 (R) and n5 (R) be the numbers of vertices, hexagons and pentagons within C, respectively. By Euler formula,
n5 (R) = 6 + s(R) − l(R)
(1)
and
1 (2) n6 (R) = l(R) + (r(R) − s(R)) − 5. 2 Equation (2) implies that r(R) ≡ s(R) (mod 2). For a fullerene graph F , let τ (F ) := min{l(R)|R is a pentagon ring of F }. For example, τ (F20 ) = 5 and τ (F24 ) = 6. Lemma 5.2. Let F be a k-resonant (k ≥ 3) fullerene with τ (F ) = 5 or 6. Then F is isomorphic to F20 or F24 . Proof: Since both F20 and F24 are 2-resonant and contains no more than three pentagons, they are also k-resonant (k ≥ 3). Now let F be a k-resonant (k ≥ 3) fullerene graph. If τ (F ) = 5, Theorem 3.3 implies that F is isomorphic to F20 since R5 is the unique pentagon ring with size five. Now suppose τ (F ) = 6. Let R be a pentagon ring with size l(R) = τ (R) = 6 and let C and C ′ be the inner cycle and outer cycle of R, respectively. Let f0 , f1 , ..., f5 be the 6 pentagons of R in clockwise. Then s(R) ≤ ⌊ 62 ⌋ = 3 and s(R) 6= 1.
Figure 19: Illustration for the proof of Lemma 5.2
If s(R) = 3, there are three 2-degree vertices on C and also three 2-degree vertices on C ′ . By Lemma 4.3, the three 2-degree vertices on C have a common neighbor vertex within C 15
(see Figure 19 (a)). Hence F contains a R5 . So τ (F ) = 5, a contradiction. If s(R) = 2, there are two 2-degree vertices v1 , v2 on C and four 2-degree vertices on C ′ . By Lemma 4.3, v1 and v2 are adjacent in F . Since every face of F is a pentagon or a hexagon, v1 and v2 lie on fi and fi+3 (i ∈ Z6 ), respectively (see Figure 19 (b)). Thus F contains a R5 , contradicting τ (F ) = 6. So s(R) = 0. Then R is isomorphic to R6 . By Theorem 3.3, F is isomorphic to F24 . Lemma 5.3. There is no fullerene graphs with τ (F ) = 7. Proof: Suppose to the contrary that pentagon ring of F with size l(R) = Lemma 4.3, whether s(R) = 2 or 3, yellow). So τ (F ) = 6, a contradiction
F is a fullerene graph with τ (F ) = 7. Let R be the 7. Then s(R) ≤ ⌊ l(R) ⌋ = 3. So s(R) = 2 or 3. By 2 F contains a R6 (see Figure 20, R6 is illustrated in to τ (F ) = 7.
Figure 20: Illustration for the proof of Lemma 5.3
Lemma 5.4. Let F be a k-resonant (k ≥ 3) fullerene graph with τ (F ) = 8. Then F is isomorphic to F28 shown in Figure 21. Proof: Let R be a pentagon ring of F with size l(R) = τ (F ) = 8 and s(R) = ψ8 (F ) and let C and C ′ be the inner cycle and outer cycle of R, respectively. Let f0 , f1 , · · · , f7 be the 8 pentagons of R in clockwise ordering. Obviously, 2 ≤ ψ8 (F ) = s(R) ≤ ⌊ l(R) ⌋ = 4. 2
Figure 21: The illustration for Case 1 of the proof of Lemma 5.4.
Case 1. ψ8 (F ) = 2. Then C contains two 2-degree vertices v1 and v2 . By Lemma 4.3, v1 and v2 are adjacent. Since each face of F is either a pentagon or a hexagon, v1 and v2 lie on two pentagons fi and fi+4 (i ∈ Z8 ) (f2 , f6 in Figure 21 (a)), respectively. So there are 16
exact two adjacent hexagons h′ and h′′ within C. Let h1 and h2 be the two faces out of C ′ such that h1 is adjacent to faces fi−1 , fi and fi+1 , while h2 is adjacent to faces fi+3 , fi+4 and fi+5 (see Figure 21 (a)). If both h1 and h2 are hexagons, let S := {v} and H := {h1 , h2 , h′ }. Then F − H − S has two isolated vertices v1 and v2 (see Figure 21 (a)). By Theorem 2.1, H is not a resonant pattern, contradicting that F is k-resonant (k ≥ 3). So at least one of h1 and h2 is a hexagon, say h2 . If h1 is a pentagon, let h3 and h4 be other two neighboring faces of h1 as shown in Figure 21 (b). By Lemma 4.3, it is easy to check that both h3 and h4 are hexagons. Hence F is isomorphic to the fullerene graph F30 shown in Figure 21 (b). Clearly, H := {h1 , h4 , h′′ } is not a resonant pattern since F − V (H) has a singular vertex (the vertex v in Figure 21 (b)), also contradicting that F is not k-resonant (k ≥ 3). So both h1 and h2 are pentagons. By Lemma 4.3, F is isomorphic to the fullerene graph F28 shown in Figure 21. The perfect matching of F28 consisting all double edges illustrated in Figure 21 alternates on and off its all hexagons. So F is k-resonant.
Figure 22: The illustration for Case 2 of the proof of Lemma 5.4.
Case 2. ψ8 (F ) = 3. By Lemma 4.3 and the equations (1) and (2), we have r(R) = 1, n5 (R) = 1 and n6 (R) = 2. Immediately, we have a subgraph of F as shown in Figure 22. Hence F contains a pentagon ring R′ with size 8 and s(R′ ) = 2 (see Figure 22, R′ is illustrated in yellow). So 3 = ψ8 (F ) ≤ s(R′ )=2, a contradiction. The contradiction implies that there is no k-resonant (k ≥ 3) fullerene graph F with ψ8 (F ) = 3.
Figure 23: The illustration for Case 3 of the proof of Lemma 5.4.
Case 3. ψ8 (F ) = 4. By Lemma 4.3, the subgraph G of F induced by R together with all vertices within C is isomorphic to one of the four graphs illustrated in Figure 23. If G is isomorphic to the graph (a) or the graph (b), then F contains a pentagon ring with size six, contradicting τ (F ) = 8. If G is isomorphic to the graph (c) or the graph (d), then F contains a pentagon ring R′ with size eight and s(R′ ) = 2, contradicting
17
ψ8 (F ) = 4. The contradictions imply that there is no k-resonant (k ≥ 3) fullerene graph F with ψ8 (F ) = 4. Lemma 5.5. Let F be a k-resonant (k ≥ 3) fullerene graph with τ (F ) = 9. Then F is isomorphic to the fullerene graph F32 as shown in Figure 24. Proof: Let R be a pentagon ring of F with size l(R) = 9 and s(R) = ψ9 (F ) and let C and C ′ be the inner cycle and outer cycle of R, respectively. If s(R) ≤ 2, then F has a face with size more than six within C by Lemma 4.3, contradicting that F is a fullerene graph. So ⌋ = 4. Let f0 , f1 , · · · , f8 be the nine pentagons of R in clockwise. 3 ≤ ψ9 (F ) = s(R) ≤ ⌊ l(R) 2
Figure 24: Illustration for Case 1 of the proof of Lemma 5.5.
Case 1. ψ9 (F ) = 3. By Lemma 4.3 and the equations (1) and (2), we have r(R) = 1, n5 (R) = 0 and n6 (R) = 3. Denote the three hexagons within C by h1 , h2 and h3 . The three 2-degree vertices on C may lie on the pentagons fj , fj+3 and fj+6 (j, j + 3, j + 6 ∈ Z9 ) (see Figure 24 (a) where j = 1). Let h′1 , h′2 , · · · , h′6 be the six faces adjacent to R clockwise along C ′ such that h′1 is adjacent to fj−1, fj and fj+1 (see Figure 24 (a)). If at least two of h′1 , h′3 and h′5 are hexagons, say h′1 and h′5 , then H := {h′1 , h′5 , h1 } is not a resonant pattern since F − V (H) has an isolated vertex v (see Figure 24 (a)), contradicting the assumption that F is k-resonant (k ≥ 3). So suppose that only one of h′1 , h′3 and h′5 is a hexagon. Then by Lemma 4.3, F should has a face with size either four or seven, a contradiction. So all of h′1 , h′3 and h′5 are pentagons. By Lemma 4.3 again, F is isomorphic to the graph (b) in Figure 24, also the fullerene graph F32 in Figure 24. Conversely, we need to show that F32 is k-resonant k ≥ 3. Since there are no more than two disjoint hexagons in F32 , it suffices to show that any two disjoint hexagons of F32 are simultaneously resonant. By symmetry, we need only to show that {h1 , h′4 } and {h1 , h′6 } are resonant patterns. Let M be the perfect matching of F32 consisting of the double edges illustrated in Figure 24. Clearly, all of h1 , h′4 and h′6 are M-alternating. Hence both {h1 , h′4 } and {h1 , h′6 } are resonant patterns of F32 . So F is isomorphic to F32 . Case 2. ψ9 (F ) = 4. Let G be the subgraph of R induced by the vertices of R together with the vertices with in C. By Lemma 4.3 and the equations (1) and (2), either r(R) = 0, 18
n5 (R) = 1 and n6 (R) = 2, or r(R) = 2, n5 (R) = 1 and n6 (R) = 3. By Lemma 4.3, G is isomorphic to the graph (a) in Figure 25 if the former holds, and G is isomorphic to the graph (b) or the graph (c) in Figure 25 if the latter holds.
Figure 25: The illustration for Case 2 of the proof Lemma 5.5.
If G is isomorphic to the graph (a), then F contains a pentagon ring with size eight, contradicting τ (F ) = 9. If G is isomorphic to the graph (b), then F contains a pentagon ring R′ with size 9 and s(R′ ) = 3. Further, s(R′ ) = 3 < 4 = ψ9 (F ) ≤ s(R′ ), a contradiction. So suppose G is isomorphic to the graph (c). Let f be the face adjacent to R along a 4-length path as shown in Figure 25 (d). If f is a pentagon, then F contains a pentagon ring with size 8 which consists of 7 pentagons of R and f , contradicting τ (F ) = 9. So f is a hexagon. Let H := {f, h1 , h2 } where h1 , h2 are two disjoint hexagons within C (see Figure 25 (d)). Then F − V (H) has an isolated vertex v (see Figure 25 (d)). So H is not a resonant pattern and F is k-resonant (k ≥ 3), a contradiction again. So there is no k-resonant (k ≥ 3) fullerene graph F with ψ9 (F ) = 4. The following lemma is due to Kutnar and Maruˇsiˇc [13]: Lemma 5.6. [13] Let F be a fullerene graph containing a ring R of five faces, and let C and C ′ be the inner cycle and the outer cycle of R, respectively. Then either (1) C or C ′ is the boundary of a face, or (2) both C and C ′ are of length 10, and the five faces of R are all hexagonal. Lemma 5.7. Let F be a k-resonant (k ≥ 3) fullerene graph with τ (F ) = 10. Then F is 2 isomorphic to F36 in Figure 26 or F40 in Figure 27. Proof: Let R be a pentagon ring of F with size l(R) = τ (F ) = 10 and s(R) = ψ10 (F ) and let C be the inner cycle of R. Let f0 , · · · , f9 be the 10 pentagons of R10 in clockwise. Clearly, ψ10 (F ) ≥ 2. If ψ10 (F ) = 2, then the two 2-degree vertices are adjacent by Lemma 4.3. Then there are two faces h0 , h1 of F within C and |h0 | + |h1 | = l(R) + s(R) + 2 = 14 since the edge within C is counted twice in |h0 | + |h1 |. So at least one of h0 and h1 has size more than seven, a contradiction. If ψ10 (F ) = 3, then the three 2-degree vertices of R on C together with vertices within C induce a K1,3 by Lemma 4.3. Hence there are three faces h0 , h1 , h2 19
P of F within C and i∈Z3 |hi | = l(R) + s(R) + 6 = 19. So one of h0 , h1 , h2 with size at least ⌋ = 5. seven, a contradiction. So 4 ≤ ψ10 (F ) = s(R) ≤ ⌊ l(R) 2
Figure 26: The illustration for Case 1 of the proof of Lemma 5.7.
Case 1. ψ10 (F ) = 4. By Lemma 4.3 and the equation (1) and (2), n5 (R) = 0 and r(R) = 0, n6 (R) = 3 or r(R) = 2, n6 (R) = 4. If r(R) = 0 and n6 (R) = 3, the four 2-degree vertices on C belong to the pentagons fj , fj+1, fj+5 and fj+6 for some j ∈ Z10 , say j = 3 (see Figure 26 (a)). Let h1 , h2 and h3 be the three hexagons within C such that h1 ∩ f2 6= ∅, h2 ∩ f3 6= ∅ and h3 ∩ f4 6= ∅. Let f be a common neighboring face of f2 , f3 , f4 and f5 (see Figure 26 (a)). If f is a pentagon, then F contains a pentagon ring (R − {f3 , f4 }) ∪ f with size 9, contradicting τ (F ) = 10. So suppose f is a hexagon. Then H := {f, h1 , h3 } is not a resonant pattern since F − V (H) has an isolated vertex (the black vertex in Figure 26 (a)), contradicting that F is k-resonant (k ≥ 3). So suppose r(R) = 2 and n6 (R) = 4. By Lemma 4.3, the four 2-degree vertices on C together with all vertices within C induce a T0 . Hence, the four 2-degree vertices belong to the pentagons fj , fj+3 , fj+5 and fj+8 for some j ∈ Z10 , say j = 1 (see Figure 26 (b)). Let h1 , h2 , h3 and h4 be the four hexagons within C in clockwise and h1 ∩f0 6= ∅. Let h′1 , h′2 , · · · h′6 be the six faces adjacent to R clockwise along its boundary such that h′1 ∩ f1 6= ∅ (see Figure 26 (b)). If h′4 and h′6 are both hexagons, then H := {h4 , h′4 , h′6 } is not a resonant pattern since F − V (H) has an isolated vertex v, contradicting that F is k-resonant (k ≥ 3). So one of h′4 and h′6 is a pentagon, say h′6 is a pentagon by symmetry. By the symmetry again, one of h′1 and h′3 is a pentagon. If h′1 is a pentagon, then F will has a face h′2 = h′5 with size 8 by Lemma 4.3, a contradiction. So h′3 is a pentagon, and both h′1 and h′4 are hexagons. 2 Immediately, F is isomorphic to the graph (c) in Figure 26, and also F36 in Figure 26. 2 In the following, it suffices to prove that F36 is k-resonant (k ≥ 3). Let M be a perfect 2 matching of F36 consisting of double edges illustrated in Figure 26. Let M1 := M ⊕ h2 ⊕ h4 , ′ M2 = M ⊕ h1 ⊕ h′4 and M3 := M ⊕ h2 ⊕ h4 ⊕ h′1 ⊕ h′4 . Let H a resonant pattern of F . If h1 , h′5 ∈ / H, then every hexagon in H is M-alternating. If h1 ∈ H but h′5 ∈ / H, then / H, then every hexagon in H is every hexagon of H is M1 -resonant. If h′5 ∈ H but h1 ∈ ′ ′ M2 -alternating. If h1 , h5 ∈ H, then H = {h1 , h5 } and hence they are H is M3 -alternating. 20
2 So F36 is k-resonant (k ≥ 3). Case 2. ψ10 (F ) = 5. By the equations (1) and (2), n5 (R) = 1 and n6 (R) = 5− 12 (5−r(R)). Subcase 2.1. There exist two vertices of R on C having a common neighbor vertex within C. Let G be the subgraph induced by R together with all vertices within C. By Lemma 4.3, G is isomorphic to the graph (a) or the graph (b) in Figure 27. If G is isomorphic to the graph (a), then F contains a pentagon ring R′ with size 10 and s(R) = 4. Hence s(R′ ) = 4 < ψ10 (F ) ≤ s(R′ ), a contradiction. So suppose G is isomorphic to the graph (b). Let f be a face adjacent to R along a 4-length path (see Figure 27, the common neighboring face f of f2 , ..., f5 ). If f is a pentagon, then F contains a pentagon ring with size 9 (the pentagon ring (R − {f3 , f4 }) ∪ f in Figure 27), contradicting τ (F ) = 10. So suppose f is a hexagon. Then F has a hexagon set H consisting of three mutually disjoint hexagons such that f ∈ H and F − H has an isolated vertex (the grey hexagons and the black vertex in Figure 27 (b)), contradicting that F is k-resonant (k ≥ 3). So there is no k-resonant fullerene graph satisfying the condition of Subcase 2.1.
Figure 27: The illustration for Subcase 2.1 of the proof of Lemma 5.7.
Subcase 2.2. Any two 2-degree vertices on C have no common neighbor vertex within C. Then the five faces adjacent to R within C form a ring R′ of F with C as its outer cycle. Since |C| = 15, the inner cycle of R′ bounds a face f ′ of F by Lemma 5.6. Note that r(R) ≡ s(R) = ψ10 (F ) (mod 2). So f ′ is a pentagon. Therefore the subgraph G induced by R together with all inner vertices of C is isomorphic to the graph (c) in Figure 27. A similar discussion yields that the five faces adjacent G along its boundary are hexagons and F − G is a pentagon. So F is isomorphic to F40 as shown in Figure 27. Now, it suffices to prove that F40 is k-resonant (k ≥ 3). Let M be the perfect matching of F40 consisting of all double edges illustrated in Figure 27. Clearly, all hexagons of F40 are M-alternating. Hence F40 is k-resonant (k ≥ 3). Summarizing Case 1 and Case 2, a k-resonant (k ≥ 3) fullerene graph F with τ (F ) = 10 2 is isomorphic to F36 in Figure 26 or F40 in Figure 27. Lemma 5.8. A fullerene graph F with τ (F ) = 11 is not k-resonant (k ≥ 3). Proof: Let R be a pentagon ring of F with size l(R) = τ (F ) = 11 and s(R) = ψ11 (F ) and let C be the inner cycle of R. By equation (1), n5 (R) = s(R)−5 and further ψ11 (F ) = s(R) ≥ 5. 21
⌋ = 5. So ψ11 (F ) = s(R) = 5 and n5 = 0; that is, On the other hand, ψ11 (F ) = s(R) ≤ ⌊ l(R) 2 there is no pentagons within C. Let v1 , v2 , v3 , v4 and v5 be the five 2-degree vertices clockwise on C. If two of these five vertices are adjacent, let v1 v2 ∈ E(F ). By Lemma 4.3, v3 , v4 and v5 have a common neighbor vertex within C, denoted by w. Let h be the face containing v1 , v2 , v3 , w and v5 . Note that any two vertices of v1 , v2 , ..., v5 are not adjacent on C. So |h| ≥ 7, a contradiction. If any two vertices of v1 , · · · v5 have no a common vertex, then the five faces adjacent to R along C form a ring R′ with size five and C as its out cycle. Since |C| = 16, the inner cycle of R′ bounds a face f ′ of F by Lemma 5.6. Note s(R) ≡ r(R) (mod 2). So f ′ is a pentagon, contradicting n5 (R) = 0.
Figure 28: The illustration of Lemma 5.8
So there exist two vertices of v1 , ..., v5 with a common neighbor vertex within C, say v1 and v2 . By Lemma 4.3 and n5 (R) = 0, the subgraph of F induced by R together with all vertices within C is isomorphic to the graph (a) in Figure 28. Let f be a face adjacent to R along a 4-length path on the boundary of R (see Figure 28 (b)). If f is a pentagon, then F contains a pentagon ring R′ with size l(R′ ) = 10. Then 11 = τ (F ) ≤ l(R′ ) = 10, a contradiction. So suppose f is a hexagon. Then F contains a set H consisting of three mutually disjoint hexagons such that F − V (H) has an isolated vertex (the grey hexagons and the black vertex in Figure 28 (b)), a contradiction to the assumption that F is k-resonant (k ≥ 3). So no fullerene graph F with τ (F ) = 11 is k-resonant (k ≥ 3).
Figure 29: The fullerene graph F48 with a perfect matching M0 .
Lemma 5.9. Let F be a k-resonant (k ≥ 3) fullerene graph with τ (F ) = 12. Then F is isomorphic to F48 as shown in Figure 29. 22
Proof: Let R be the unique pentagon ring of F with size l(R) = 12. Let C be the inner cycle of R. Since F has 12 pentagons only, there is no pentagons within C and hence n5 (R) = 0. By the equation (1), s(R) = 6. Let v0 , v1 , · · · , v5 be the six 2-degree vertices clockwise on C. First suppose that r(R) = 0; that is, there is no vertex within C. Then the subgraph of F induced by R together with all vertices within C is isomorphic to the graph G1 (see Figure 30 (left)) and F is isomorphic to the fullerene graph shown in Figure 30 (right). Let H be the set consisting of the three mutually disjoint grey hexagons of F . Then F − V (H) has an isolated vertex and hence F is not k-resonant (k ≥ 3), a contradiction.
Figure 30: The subgraph G1 (left) and the fullerene graph containing G1 (right). So suppose r(R) ≥ 2 by r(R) ≡ s(R) (mod 2). Let G be the subgraph induced by the vertices within C and let n and m be the vertex number and edge number of G, respectively. Then n = r(R) ≥ 2. Since C has six 2-degree vertices, 3n − 2m = 6. If G is not connected, then G is a forest since F is 3-connected and cyclically 5-connected. Then m = n − w where w ≥ 2 is the number of components of G. So 3n − 2(n − w) = 6, and further n = 6 − 2w ≤ 2. So n = 2 and hence G is two isolated vertices, denoted by u, v. We may assume that N(u) = {v0 , v1 , v2 } and N(v) = {v3 , v4 , v5 }. Let f be the face within C containing vertices v0 , u, v2, v3 , v and v5 . Note that the six 2-degree vertices are not adjacent on C. So v2 v3 ∈ / E(f ) and v5 v0 ∈ / E(f ). Therefore |f | ≥ 8, a contradiction. So suppose that G is connected. Let ∂G be the boundary of G. Note that ∂G may be not a cycle and an edge which belongs only to the infinite face will contribute 2 to |∂G|. Let x be the number of inner faces of G. By Euler’s formula, m = n − 1 + x. So 3n − 2(n − 1 + x) = 6, and further n = 4 + 2x. On the other hand, since every inner face of G is also a face of F , every inner face of G is a hexagon. So |∂G| + 6x = 2m = 2n − 2 + 2x. Hence |∂G| = 6.
Figure 31: The subgraphs G2 and G3 in the proof of Lemma 5.9. If x = 0, then G is a tree. Since |∂G| = 6, G has three edges. So G is isomorphic to 23
a K1,3 or a 3-length path. Hence the subgraph of F induced by V (R ∪ G) is isomorphic to either G2 or G3 shown in Figure 31. Whether G2 ⊂ F or G3 ⊂ F , F has three mutually disjoint hexagons whose deletion results in an isolated vertex (the grey hexagons in Figure 31). So F is not k-resonant for k ≥ 3, a contradiction. So suppose x 6= 0. Since the length of every cycle of G is at least 6 and |∂G| = 6, ∂G is a 6-length cycle. Since F is cubic, there are six edges connecting 2-degree vertices v0 , v1 , · · · , v5 to vertices of G. So G is a hexagon. By symmetry of R, F is isomorphic to F48 shown in Figure 29. Now, it suffices to prove F48 is k-resonant (k ≥ 3). Let M0 be the perfect matching of F48 consisting of all double edges illustrated in Figure 29. Let f1 , f2 , f3 and f1′ , f2′ , f3′ be the faces marked in F48 and let M1 := M0 ⊕ f1 ⊕ f2 ⊕ f3 , M2 := M0 ⊕ f1′ ⊕ f2′ ⊕ f3′ and M3 := M0 ⊕ f1 ⊕ f2 ⊕ f3 ⊕ f1′ ⊕ f2′ ⊕ f3′ . For any set H of mutually disjoint hexagons, every hexagon of H is Mj -alternating for some j ∈ Z4 . So F48 is k-resonant (k ≥ 3). Summarizing above results, we have the following main theorem, Theorem 5.10. A fullerene graph F is k-resonant (k ≥ 3) if and only if F is isomorphic 1 2 to one of F20 , F24 , F28 , F32 , F36 , F36 , F40 , F48 and C60 . According to Theorem 5.10, a fullerene graph F is k-resonant (k ≥ 3) if and only if it is 3-resonant. This result is coincident with these for benzenoid systems [29], coronoid systems [2, 14], open-end nanotubes [23], toroidal polyhexes [26] and Klein-bottle polyhexes [20].
References [1] S. El-Basil, Clar sextet theory of buckminsterfullerene (C60 ), J. Mol. Struct. (Theochem) 531 (2000) 9-21. [2] R. Chen and X. Guo, k-coverable coronoid systems, J. Math. Chem. 12 (1993) 147- 162. [3] E. Clar, The Aromatic Sextet (Wiley, London, 1972). [4] T. Doˇsli´c, On lower bounds of number of perfect matchings in fullerene graphs, J. Math. Chem. 24 (1998) 359-364. [5] T. Doˇsli´c, Cyclical edge-connectivity of fullerene graphs and (k,6)-cages, J. Math. Chem. 33 (2003) 103-112. [6] P.W. Fowler and D.E. Manolopoulos, An Atlas of Fullerenes (Oxford Univ. Press, Oxford, 1995). [7] P. Fowler and T. Pisanski, Leapfrog transformation and polyhedra of Clar type, J. Chem. Soc. Faraday Trans. 90(19) (1994) 2865-2871.
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[8] K. Fries, Bicyclic compounds and their comparison with naphthalene III, Justus Liebigs Ann. Chem. 454 (1972) 121-324. [9] M. Goldberg, A class of multi-symmetric polyhedra, Tˆohoku Math. J. 43 (1937) 104-108. [10] J. Graver, Kekul´e structures and the face independence number of a fullerene, Europ. J. Combin. 28 (2007) 1115-1130. [11] B. Gr¨ unbaum, Convex Polytopes (Wiley, New York, 1967). [12] X. Guo, k-resonant benzenoid systems and k-cycle resonant graphs, MATCH Commun. Math. Comput. Chem. 57 (2006) 153-168. [13] K. Kutnar and D. Maruˇsiˇc, On cyclic edge-connectivity of fullerenes, Discrete Appl. Math. (2007), doi: 10.1016/j.dam.2007.08.046. [14] K. Lin and R. Chen, k-coverable polyhex graphs, Ars Combin. 43 (1996) 33-48. [15] L. lov´asz and M. D. Plummer, Matching Theory, Ann. Discrete Math., Vol. 29 (NorthHolland, Amsterdam, 1986). [16] Z. Qi and H. Zhang, A note on the cyclical edge-connectivity of fullerene graphs, J. Math. Chem. 43 (2008) 134-140. [17] M. Randi´c, Aromaticity of polycyclic conjugated hydrocarbons, Chem. Reviews 103 (9) (2003) 3449-3605. [18] W.C. Shiu, P.C.B. Lam and H. Zhang, Clar and sextet polynomials of buckminsterfullerene, J. Mol. Struct. (Theochem) 662 (2003) 239-248. [19] W.C. Shiu, P.C.B. Lam and H. Zhang, k-resonance in toroidal polyhexes, J. Math. Chem. 38 (4) (2005) 451-466. [20] W.C. Shiu and H. Zhang, A complete charaterization for k-resonant Klein-bottle polyhexes, J. Math. Chem. 43 (2008) 45-59. [21] D. Ye and H. Zhang, Extremal fullerene graphs with the maximum Clar number, submitted. [22] F. Zhang and R. Chen, When each hexagon of a hexagonal system covers it, Discrete Appl. Math. 30 (1991) 63-75. [23] F. Zhang and L. Wang, k-resonance of open-ended carbon nanotubes, J. Math. Chem. 35(2) (2004), 87-103.
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[24] F. Zhang and M. Zheng, Generalized hexagonal systems with each hexagon being resonant, Discrete Appl. Math. 36 (1992) 67-73. [25] H. Zhang and D. Ye, An upper bound for the Clar number of fullerene graphs, J. Math. Chem. 41 (2007) 123-133. [26] H. Zhang and D. Ye, k-resonant toroidal polyhexes, J. Math. Chem. (2007), doi: 10.1007/s10910-007-9310-2. [27] H. Zhang and F. Zhang, Plane elementary bipartite graphs, Discrete Appl. Math. 105 (2000) 291-311. [28] M. Zheng, k-resonant benzenoid systems, J. Mol. Struct. (Theochem) 231 (1991) 321334. [29] M. Zheng, Construction of 3-resonant benzenoid systems, J. Mol. Struct. (Theochem) 277 (1992) 1-14.
26
Dong Ye, Zhongbin Qi and Heping Zhang† School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, P. R. China E-mails: [email protected], [email protected], [email protected]
Abstract A fullerene graph F is a cubic 3-connected plane graph with exact 12 pentagons and other hexagons. Let M be a perfect matching of F . A cycle C of F is M -alternating if and only if the edges of C appear alternately in and off M . A set H of disjoint hexagons of F is called a resonant pattern if F has a perfect matching M such that all hexagons in H are M -alternating. A fullerene graph F is k-resonant if any i (0 ≤ i ≤ k) mutually disjoint hexagons of F form a resonant pattern. In this paper, we prove that every hexagon of a fullerene graph is resonant and all leapfrog fullerene graphs are 2-resonant. Finally, we show that a k-resonant (k ≥ 3) fullerene graph has at most 60 vertices and construct all nine 3-resonant fullerene graphs, which are also k-resonant (k > 3). Keywords: Fullerene graph; Resonant pattern; k-resonant; Perfect matching AMS 2000 subject classification: 05C70, 05C90
1
Introduction
A fullerene graph is a cubic 3-connected plane graph with exact 12 pentagonal faces and other hexagonal faces. Fullerene graphs has been studied in mathematics as trivalent polyhedra for a long time [9, 11], for example, the dodecahedron is the fullerene graph with 20 vertices. Fullerene graphs have been studied in chemistry as fullerene molecules which have extensive applications in physics, chemistry and material science [6]. Let G be a plane 2-connected graph. A perfect matching or 1-factor M of G is a set of independent edges such that every vertex of G is incident with exact one edge in M. A cycle of G is M-alternating if edges of C is alternately in and off M. For a fullerene graph F , every edge of F belongs to a perfect matching [4]. A hexagon h of a fullerene graph F is ∗ †
This work is supported by NSFC. Corresponding author.
1
resonant if F has a perfect matching M such that h is M-alternating. It was proved that every hexagon of a normal benzenoid system is resonant [22]. This result was generalized to normal coronoid systems [2] and plane elementary bipartite graphs [24]. However a fullerene graph is a non-bipartite graph. It is natural to ask if every hexagon of a fullerene graph is resonant. The present paper first uses Tutte’s Theorem to give a positive answer to this question. A set H of disjoint hexagons of a fullerene graph F is a resonant pattern (or sextet pattern) if F has a perfect matching M such that every hexagon in H is M-alternating; equivalently, if F − H has a perfect matching, where F − H denotes the subgraph obtained from F by deleting all vertices of H together with their incident edges. The maximum cardinality of resonant patterns of F is called Clar number of F [3, 17], and the maximum number of M-alternating hexagons over all perfect matchings M of F is called the Fries number of F [8]. For fullerene graphs, Graver [10] explored some connections among Clar number, face independence number and Fries number of a fullerene graph, and obtained a lower bound for the Clar number of leapfrog fullerene graphs with icosahedral group. Zhang and Ye [25] ⌋ showed that Clar number of a fullerene graph Fn with n vertices satisfies c(Fn ) ≤ ⌊ n−12 6 which is sharp for infinite many fullerene graphs, including C60 whose Clar number is 8 [1]. Shiu, Lam and Zhang [18] computed the Clar polynomial and sextet polynomial of C60 by showing that every face independent set of C60 is also a resonant pattern. A fullerene graph is k-resonant if any i (0 ≤ i ≤ k) disjoint hexagons form a resonant pattern. Hence a fullerene graph with each hexagon being resonant is 1-resonant. Zheng [28, 29] characterized general k-resonant benzenoid systems. In particular, he showed that every 3-resonant benzenoid system is also k-resonant (k ≥ 3). This result also holds for coronoid systems [14], open-ended nanotubes [23], toroidal polyhexes [19, 26] and Kleinbottle polyhexes [20]. For a recent survey on k-resonant benzenoid systems, refer to [12]. Next the paper mainly consider general k-resonant fullerene graphs. We show that all leapfrog fullerene graph are 2-resonant and a 3-resonant fullerene graph has at most 60 vertices. Finally, we construct all 3-resonant fullerene graphs, which are all k-resonant (k ≥ 3).
2
1-resonance of fullerene graphs
Let G be a plane graph admitting a perfect matching. For a face f of G, let V (f ) and E(f ) be the sets of vertices and edges of f , respectively. Use ∂G to denote the boundary of G. Let G1 and G2 be two subgraphs of G. We say G1 and G2 are adjacent if ∂G1 ∩ ∂G2 6= ∅ and (G1 − ∂G1 ) ∩ (G2 − ∂G2 ) = ∅. A subgraph H of G is nice if G − H has a perfect matching. So a resonant pattern of G is also a nice subgraph of G. A graph G is cyclically k-edge connected if deleting fewer than k edges of G can not separate G into two components such 2
that each of them contains a cycle. By Tutte’s Theorem on perfect matching of graphs ([15], Theorem 3.1.1), we have following result. Theorem 2.1. A subgraph H of a graph G is nice if and only if for any S ⊆ V (G − H), Co (G − H − S) ≤ |S|, where Co (G − H − S) is the number of odd components of G − H − S. Theorem 2.2. Let G be a cyclically 4-edge connected cubic graph with a 6-length cycle. Then for every 6-length cycle H of G, either H is nice or G − H is bipartite. Proof: Let H be a 6-length cycle in G. If G − H has a perfect matching, then the theorem holds. If not, then by Theorem 2.1 there exists an S ⊂ V (G −H) such that Co (G −H −S) ≥ |S| + 2 by parity, i.e. |S| ≤ Co (G − H − S) − 2. Since G is cubic, S sends out at most 3|S| ≤ 3Co (G − H − S) − 6 edges. Let G1 , G2 , ..., Gk be all odd components of G−H −S, where k = Co (G−H −S). Because G is cyclically 4-edge connected and cubic, it has no bridge. Every Gi (i = 1, 2, ..., k) sends odd number edges, hence at least three edges, to H ∪ S. So ∪ki=1 Gi sends out at least 3Co (G − H − S) edges to either S or H. Since H is a 6-length cycle, there are at most 6 edges between H and ∪ki=1 Gi . So ∪ki=1 Gi sends at least 3Co (G−H −S)−6 edges to S. Hence there are precisely 3Co (G − H − S) − 6 edges between S and ∪ki=1 Gi . So S is an independent set, and every Gi sends out exact 3 edges, and G − H − S has no even component. In addition, since G is cyclically 4-edge connected, every Gi is a tree. We claim that each Gi is a singular vertex. If not, then an odd component Gi has at least 2 vertices. So Gi has at least two leaves. Every leaf of Gi is adjacent to at least two vertices in S ∪ H. So Gi sends at least four edges out, contradicting the fact that every Gi sends precise three edges out. Therefore G − H is a bipartite graph with bipartition (S, G − H − S). This completes the proof of the theorem. Lemma 2.3. [5, 16] Every fullerene graph is cyclically 5-edge connected. By Lemma 2.3 and Theorem 2.2, we immediately have following result. Theorem 2.4. Every hexagon of a fullerene graph is resonant. Proof: Let F be a fullerene graph and H be a hexagon. It is obvious that F − H is not bipartite. By Theorem 2.2 and Lemma 2.3, H is nice. That means H is resonant.
3
2-resonant fullerene graphs
Let F be a fullerene graph. The leapfrog operation on F is defined [7] as follows: for any face f ∈ F (F ), add a new vertex vf in f and join vf to all vertices in V (∂f ) to obtain a 3
new triangular graph F ′ ; Take the geometry dual of the graph F ′ and denote it by F ∗ (see Figure 1). Clearly, F ∗ is a fullerene graph since every vertex of F ′ is 6-degree excluding exact 12 5-degree vertices and every face of F ′ is a triangle. The edges of F ∗ cross the edges of F ⊂ F ′ in the geometry dual operation form a perfect matching M 0 of F ∗ . A fullerene graph is called leapfrog fullerene graph if it arises from a fullerene graph by the leapfrog operation. Fowler and Pisanski [7] explored the connection between leapfrog fullerenes and the Clar theory of aromatic sextet had been explored and showed that a fullerene graph is a leapfrog fullerene if and only if it has a Fries structure, i.e. a perfect matching which avoids edges in pentagons and is alternating on the maximal number n/3 hexagons.
Figure 1: The leapfrog operation on the dodecahedron F20 and the perfect matching M 0 of C60 (double edges).
Let F ∗ be a leapfrog fullerene graph arising from F . A face f of F ∗ is called a heritable face if it lies completely in some face of F , and a fresh face, otherwise. For example, C60 is the leapfrog fullerene graph of the dodecahedron and every pentagon is a heritable face and all hexagons are fresh faces. The perfect matching M 0 is corresponding to the Fries structure of C60 (see Figure 1). For a leapfrog fullerene graph, we have following result. Lemma 3.1. Let F be a leapfrog fullerene graph. Then every fresh face is M 0 -alternating and all heritable faces are independent. Let F be a leapfrog fullerene and f be a heritable face. A subgraph of F consisting of f together with all adjacent fresh faces is called the territory of f and denoted by T [f ]. For two heritable faces f1 and f2 , it is easily seen that there are at most 2 common fresh faces in their territories, which are adjacent. Theorem 3.2. Every leapfrog fullerene graph is 2-resonant. Proof: Let F be a leapfrog fullerene graph and f1 , f2 be any two disjoint hexagons. If both f1 and f2 are fresh faces, then clearly M 0 is alternating on both of them by Lemma 3.1. So 4
suppose that at least one of them is a heritable face, say f1 . If f2 is fresh, then f2 * T [f1 ] and it is adjacent to at most one face in T [f1 ]. Let us denote clockwise the 6 fresh hexagons in T [f1 ] by h0 , h1 , ..., h5 . Let M1 := M 0 ⊕ h1 ⊕ h3 ⊕ h5 if f2 adjoins none of h1 , h3 and h5 ; otherwise we choose h0 , h2 and h4 instead. Then M1 is a perfect matching and alternating on both f1 and f2 . So, in the following, we suppose both f1 and f2 are heritable. Let h′0 , h′1 , ..., h′5 be the six fresh hexagons of T [f2 ] in clockwise. First suppose T [f1 ] and T [f2 ] has a common hexagon; that is, there are i0 , j0 ∈ Z6 such that hi0 and h′j0 are the same hexagon. Let M2 := M 0 ⊕ hi0 ⊕ hi0 +2 ⊕ hi0 +4 ⊕ h′j0 +2 ⊕ h′j0 +4 . It is clear that M2 is a perfect matching alternating on both f1 and f2 . Now suppose T [f1 ] and T [f2 ] have no common hexagons. If no face in T [f2 ] adjoins one of h1 , h3 and h5 , let M3 := M 0 ⊕ h1 ⊕ h3 ⊕ h5 ⊕ h′1 ⊕ h′3 ⊕ h′5 ; otherwise, let M3 := M 0 ⊕ h0 ⊕ h2 ⊕ h4 ⊕ h′1 ⊕ h′3 ⊕ h′5 . Then M3 is also a perfect matching alternating on both f1 and f2 . So the theorem holds.
Figure 2: The dodecahedron F20 (left) and the Fullerene graph F24 with a perfect matching M (right).
There exist non-leapfrog 2-resonant fullerene graphs. The dodecahedron F20 is a trivial example. F24 (see Figure 2) is also 2-resonant since the two hexagons in it are simultaneously M-alternating with respect to the perfect matching M illustrated by double edges in Figure 2. Another non-trivial example is C70 . It has two perfect matchings M1 and M2 as shown in Figure 3. Let M3 := M1 ⊕ h2 ⊕ h10 and M4 := M1 ⊕ h2 ⊕ h4 ⊕ h6 ⊕ h8 ⊕ h10 . Let H := {h, h′ } be a set of any two disjoint hexagons in C70 . If h, h′ ∈ / {h1 , h3 , h5 , h7 , h9 }, then h and h′ are simultaneously M1 -alternating. So suppose h ∈ {h1 , h3 , h5 , h7 , h9 }. By symmetry, we may assume h = h1 . If h′ ∈ {h1 , h3 , h5 , h7 , h9 }, then both h and h′ are M4 -alternating. If h′ ∈ {h11 , h12 }, let h′ = h11 by the symmetry of h11 and h12 . Then both h and h′ are M2 -alternating. Finally, if h′ ∈ / {h3 , h5 , h7 , h9 , h11 , h12 }, then h and h′ are simultaneously M3 -alternating. By symmetry, we have C70 is 2-resonant. On the other hand, we can construct infinitely many fullerene graphs which are not 2resonant. Let H be a set of at most k disjoint facial cycles of a plane graph G. By Theorem 2.1, if there exists a vertex set S ⊆ V (G − H) such that Co (G − H − S) > |S|, then H is not a resonant pattern and G is not k-resonant. For convenience, the vertices in such S and the faces in such H are always illustrated in grey color and the vertices of odd components of G − H − S are illustrated in black color from now on. Let R5 and R6 be the graphs arising from deleting the outer pentagonal facial cycle from 5
Figure 3: C70 with two perfect matchings M1 (left) and M2 (right).
F20 and deleting the outer hexagonal cycle from F24 (see Figure 2), respectively.
Figure 4: R5 and R6 and the illustration for the proof of Theorem 3.3.
Theorem 3.3. Let F be a fullerene graph different from F20 and F24 . If F contains R5 or R6 as subgraphs, then F is not 2-resonant. Proof: First suppose R5 ⊂ F . Since F is different from F20 , there are at least two faces adjacent to R5 are hexagons and disjoint. Let H be the set of these two hexagons (grey hexagons in Figure 4). Then the four grey vertices of R5 in Figure 4 form a vertex set S such that such that F − H − S contains five isolated vertices (black vertices of R5 in Figure 4). So H is not a resonant pattern. Now suppose R6 ⊂ F . Since F is different from F24 , at least one of the faces of F adjacent to R6 is hexagonal. Let H be the set consisting of this hexagon together with the center hexagon of R6 . It is easy to see that H is not a resonant pattern (see Figure 4). Using R5 and R6 as caps, we can construct infinitely many fullerene graphs which are 1-resonant but not 2-resonant. It is interesting to characterize 2-resonant fullerene graphs. Here, we present a conjecture about 2-resonant fullerene graphs which is supported by Theorem 3.2. Conjecture 3.4. Every fullerene graph without adjacent pentagons is 2-resonant.
6
4
k-resonant fullerene graphs (k ≥ 3)
Let f be a face of a fullerene graph F . A vertex v ∈ V (F − V (f )) is incident to f if N(v) ∩ V (f ) 6= ∅, where N(v) is the neighbor set of v. Let G∗ be a graph as shown in Figure 5. Then G∗ is a forbidden subgraph of k-resonant (k ≥ 3) fullerene graphs since deleting the three hexagons of G∗ isolates the vertex v and hence the three hexagons do not form a resonant pattern by Theorem 2.1.
Figure 5: A forbidden subgraph G∗ for 3-resonant fullerene graphs.
Theorem 4.1. Let F be a k-resonant (k ≥ 3) fullerene graph. Then |V (F )| ≤ 60. Proof: Since F is k-resonant (k ≥ 3), G∗ * F . So any v ∈ V (F ) is incident to at least one pentagon of F . On the other hand, for any pentagon f of F , there are at most 5 vertices in V (F − f ) incident to it. Hence |V (F )| ≤ 12 × 5 = 60 since F has exact 12 pentagons. So the theorem holds. Since C60 has the Fries Kekul´e structure M 0 which is alternating on every hexagon of C60 (see Figure 1), it is k-resonant (k ≥ 1). By Theorem 4.1, C60 is the unique k-resonant leapfrog fullerene graph for k ≥ 3 since it is the unique leapfrog fullerene graph with no more than 60 vertices. We restate this fact as follows: Corollary 4.2. C60 is the unique k-resonant (k ≥ 3) leapfrog fullerene graph. A fragment B of a fullerene graph F is a subgraph of F consisting of a cycle together with its interior. A face of F adjacent to B but not in B is called a neighboring face of B. The territory T [B] of B is the subgraph of F consisting of B together with its all neighboring faces. The following useful lemma is obtained by Ye and Zhang [21].
K
2
K 1, 3
T0
Figure 6: Trees: K2 , K1,3 and T0 .
Lemma 4.3. ([21], Lemma 2.1) Let B be a fragment of F and W be the vertex set consisting of all 2-degree vertices on ∂B. If 0 < |W | ≤ 4, then T = F − (V (B) \ W ) is a forest and 7
(1) T is K2 if |W | = 2; (2) T is K1,3 if |W | = 3; (3) T is the union of two K2 s, or a 3-length path, or T0 , if |W | = 4. For a fragment B of a fullerene graph F , let FB := F − (B − ∂B). Let B1 , B2 , B3 and B4 be the four fragments shown in Figure 7. Lemma 4.4. Let B be a fragment of a fullerene graph F such that there are six 2-degree vertices and six 3-degree vertices appearing alternately on ∂B. Then either all neighboring faces of B are hexagons or FB is isomorphic to R6 or one of Bi for i = 1, 2, 3, 4.
Figure 7: Five possible isomorphic graphs of FB in Lemma 4.4.
Proof: Let x be the number of the hexagonal neighboring faces of B and T [B] be the territory of B. Then T [B] is a fragment of F with x 2-degree vertices where 0 ≤ x ≤ 6 and x 6= 1 since there are six 2-degree vertices and six 3-degree vertices appearing alternately on ∂B. If x = 6, then the lemma is true. So suppose that x < 6. If x = 0, all neighboring faces of B are pentagons and therefore ∂T [B] is a 6-cycle without 2-degree vertex. So FB is isomorphic to R6 . If x = 2, up to isomorphism, T [B] has three possible cases which are showed in Figure 8. By Lemma 4.3, the two 2-degree vertices u1 and u2 on ∂T [B] are adjacent in F . So only the graph (c) is possible since F does not has a 3-length cycle or 4-length cycle. Hence FB is isomorphic to B1 . A similar analysis for x = 3 and x = 4 will bring us to the conclusion that FB is isomorphic to B2 and FB is isomorphic to B3 , respectively. Here we omit the detailed analysis. u2
u1
B
u1
u1
B
u2
B u2
Figure 8: Three possible cases of T [B] for x = 2.
8
Now suppose x = 5. Let u1 , u2 , u3, u4 and u5 be the five 2-degree vertices on ∂T [B] (see Figure 9 (a)). It is clear that u1 and u5 are not adjacent in F . If they have a common neighbor vertex w, then we obtain a fragment with 4 2-degree vertices (see Figure 9 (b)). So by Lemma 4.3, one pair of vertices u2 ,u3 and vertices u3 ,u4 will be adjacent or have a common neighbor vertex in F − T [B], contradicting that F has no 3-length cycle and 4length cycle. So u1 and u5 have no common neighbor vertex. That is the neighboring face f of T [B] with u1 and u5 on its boundary must be a hexagon (see Figure 9 (c)). Let B ′ be the territory of T [B] and y be the number of 2-degree vertices on ∂B ′ . Then 0 ≤ y ≤ 4 and y 6= 1. If y > 0, by Lemma 4.3, there exist at least a pair of 2-degree vertices on ∂B ′ such that either they are adjacent in F or they have a common neighbor vertex in F − V (B ′ ), also contracting that F has no 3-length cycle and 4-length cycle. So y = 0 and FB is isomorphic to B4 . w
u5
u1
u1
u4 u3
u5
B u4
u2 u3
u5
u1
B
B u2
f
u4
u2 u3
Figure 9: The illustration for the proof of Lemma 4.4 for x = 5. A fragment B of a fullerene graph F is maximal if all its neighboring faces are hexagonal. Lemma 4.5. Let F be a k-resonant (k ≥ 3) fullerene graph. Then F contains none of R6 and B1 as maximal fragments and F does not contain B2 and B4 . Proof: Theorem 3.3 implies that F can not contain R6 as its maximal fragment. If F contains B1 as a maximal fragment or B2 , then F contains the subgraph G∗ as subgraph which is forbidden in a k-resonant k ≥ 3 fullerene graphs, a contradiction (see Figure 10 (a) and (b)). If F contains B4 , then the three independent grey hexagonal faces of B4 in Figure 10 (c) is not a resonant pattern since deleting them will result in a component isomorphic to K1,3 which has no perfect matching. Hence F is not 3-resonant, also a contradiction. Hence the lemma is true. A fragment B of a fullerene graph F is said to be a pentagon fragment if its every inner face is a pentagon. Let γ(B) be the minimum number of pentagons adjoining the same pentagon in B. Let B ∗ be the inner dual of B. Then γ(B) is the the minimum degree of B ∗ . For example, γ(R5 ) = 3. Lemma 4.6. ([21], Lemma 3.6) Let B be a pentagon fragment of a fullerene graph F . Then: (1) R5 ⊆ B if γ(B) ≥ 3; (2) B has a pentagon adjoining exactly two adjacent pentagons of B if γ(B) = 2. 9
Figure 10: Illustration for the proof of Lemma 4.5.
A turtle is a pentagon fragment consisting of six pentagons as illustrated in Figure 11. Clearly, γ(B) = 1 if B is a turtle. The fragment B3 in Figure 7 contains a turtle. Following is a theorem characterizing maximal pentagon fragments of k-resonant (k ≥ 3) fullerene graphs.
Figure 11: Pentagon fragment turtle.
Theorem 4.7. Let F be a k-resonant (k ≥ 3) fullerene graph different from F20 and B be a maximal pentagon fragment of F . Then B is isomorphic to either a pentagon or a turtle. Proof: Let B be a maximal pentagon fragment of F . By Theorem 3.3, we may assume that B dose not contain R5 . Lemma 4.6 implies γ(B) ≤ 2. If γ(B) = 0, then B is a pentagon. So suppose that γ(B) > 0. Let H be a set of disjoint of hexagons. Since F is k-resonant (k ≥ 3), F − H has a perfect matching; that is, F contains no H such that F − H has no perfect matchings.
(∗)
Case 1. γ(B) = 1. That means B has a pentagonal face f0 which has only one neighboring pentagonal face f1 ⊂ B. Let h1 , h2 , h3 and h4 be the four neighboring hexagonal faces of B such that hi is adjacent to hi+1 (1 ≤ i ≤ 3) and both h1 and h4 are also adjacent to f1 . Further, let f2 and f3 be the other two neighboring faces of f1 as illustrated in Figure 12 (a). If one of f2 and f3 is hexagonal, say f2 , then H := {h2 , h4 , f2 } contradicts (∗) since F −H has an isolate vertex. So suppose both f2 and f3 are pentagons of B. Let f4 6= f1 be the face adjacent to both f2 and f3 . If f4 is a hexagon, then H := {h1 , h4 , f4 } contradicts (∗) since F − H has an isolate vertex. So f4 must be a pentagonal face of B. Let h5 and h6 be the two faces adjacent to a pair pentagons f2 , f4 and a pair pentagons f3 , f4 , respectively (see Figure 12 (b)). Let f5 be the face adjacent to h5 , f4 and h6 . If f5 is a hexagon, then H := {h1 , h4 , f5 } 10
also contradicts (∗) since one component of F − H is K1,3 . So f5 is a pentagonal face of B. Thus G := ∪5i=0 fi ⊆ B is isomorphic to a turtle. It sufficient to show that B = G. That is all neighboring faces of G are hexagons. Beside h1 , . . . , h6 , let h7 and h8 be other are two neighboring faces of G as illustrated in Figure 12 (b). Let G′ := G ∪ (∪8i=1 hi ). Because h1 , h2 , h3 and h4 are already hexagonal, we need only to show that h5 , h6 , h7 and h8 are hexagonal. Since R5 * B, one of h5 and h6 must be hexagonal, say h5 by symmetry. Then h6 is either a pentagon or a hexagon.
Figure 12: Illustration for the proof of Subcase 1.1 of Lemma 4.7.
Subcase 1.1. h6 is a pentagon. If h7 is a hexagon, then H := {h1 , h4 , h7 } contradicts (∗) since F − H has an odd component with five vertices (see Figure 12 (b)). So h7 must be a pentagon. If h8 is a pentagon, then G′ is a fragment with only two 2-degree vertices w1 and w2 on h2 (see Figure 12 (c)). By Lemma 4.3, w1 and w2 are adjacent in F − (G′ − {w1 , w2 }), resulting in a 2-length cycle of F , a contradiction. So h8 must be a hexagon. Then the fragment G′ contains four 2-degree vertices (see Figure 12 (d)). If G′ ⊂ F , it is easy to check that there will be at least one face of F with size of at most 4 by Lemma 4.3, also a contradiction. The contradiction implies that h6 is not a pentagon.
Figure 13: Illustration for the proof of Subcase 1.2 of Lemma 4.7.
Subcase 1.2. h6 is a hexagon. Let h9 , h10 be other two neighboring faces of G′ as shown 11
in Figure 13 (a) and let G′′ := G′ ∪ h9 ∪ h10 . We claim that both h7 and h8 are hexagons. Suppose to the contrary that at least one of them is a pentagon, say h7 . If at least one of h8 , h9 and h10 is a pentagon, then G′′ is a fragment with at most four 2-degree vertices. By Lemma 4.3, it is readily checked that G′′ can not be a subgraph of F . So all of h8 , h9 and h10 are hexagons. Then H := {h3 , h8 , h9 } contradicts (∗) since F − H has an odd component with 15 vertices (see Figure 13 (b)). Hence both h7 and h8 are hexagons. So all neighboring faces of G are hexagonal and hence B = G. Case 2. γ(B) = 2. Lemma 4.6 implies that B contains a pentagonal face f0 which has two adjacent pentagonal neighboring faces f1 and f2 . Let h1 , h2 and h3 be the other three hexagonal neighboring faces of f0 as shown in Figure 14 (a).
Figure 14: Illustration for the proof of Case 2 of Lemma 4.7.
Let f3 6= f0 be the face adjacent to both f1 and f2 . If f3 is a hexagon, then H := {h1 , h3 , f3 } contradicts (∗) since F − H has an isolate vertex. Therefore f3 must be a pentagon in B. Let h4 , f4 , h5 be the other three neighboring faces of f3 as shown in Figure 14 (a). If f4 is hexagonal, then H := {h1 , h3 , f4 } also contradicts (∗) since one component of F − H is K1,3 . Since R5 * B, at least one of h4 and h5 is a hexagon, say h4 by symmetry. If h5 is also a hexagon, then H := {h2 , h4 , h5 } contradicts (∗) since one component of F − H is K1,3 . So h5 is a pentagon. Let h6 and h7 be other two neighboring faces of f4 as shown in Figrue 14. If h6 is a hexagon, then H := {h1 , h3 , h6 } contradicts (∗) since F − H has an odd component with size seven (see Figure 14 (b)). Hence h6 is a pentagon. Similarly, h7 is also a pentagon (see Figure 14 (c)). Now, we have a fragment G := (∪4i=0 fi ) ∪ (∪7j=1 hj ) with four 2-degree vertices w1 , w2 , w3 , w4 (see Figure 14 (d)). By Lemma 4.3, G * F . This completes the proof of the theorem.
5
Construction of k-resonant (k ≥ 3) fullerene graphs
Let {fi |i ∈ Zl , l ≥ 5} be a set of l faces of a fullerene graph F such that fi ∩ fi+1 (i ∈ Zl ) is an edge, denoted by ei . The subgraph R := ∪i∈Zl fi is called a ring if {ei |i ∈ Zl } is an 12
independent edge set and l is called the size of the ring R, denoted by l(R). A ring is called a pentagon ring if every fi (i ∈ Zl ) is a pentagon. The R5 and R6 in Figure 4 are two pentagon rings with size five and six, respectively. Note that R5 is the unique pentagon ring with size five. C60 is the unique fullerene graph with no more than 60 vertices and without adjacent pentagons. By Theorem 4.1 and Theorem 4.7, a k-resonant (k ≥ 3) fullerene graph different from C60 contains either maximal pentagon fragments each of which is isomorphic to a turtle or pentagon rings. Lemma 5.1. Let F be a k-resonant (k ≥ 3) fullerene graph containing no pentagon rings. 1 Then F is isomorphic to C60 or the graph F36 shown in Figure 15.
1 Figure 15: The k-resonant (k ≥ 3) fullerene graph F36 with a perfect matching M
Proof: C60 is clearly k-resonant (k ≥ 3) since the Fries Kekul´e structure M 0 is alternating 1 on every hexagon (see Figure 1). For F36 showed in Figure 15, the perfect matching M 1 illustrated in Figure 15 is also alternating on its every hexagon. So F36 is also k-resonant (k ≥ 3). Let F be a k-resonant (k ≥ 3) fullerene graph without pentagon rings as subgraphs. By Theorem 4.7, every maximal pentagon fragment of F is isomorphic to a pentagon or a turtle. If F does not contain a maximal pentagon fragment isomorphic to a turtle, then F is isomorphic to C60 by Theorem 4.1. So suppose that F contains a maximal pentagon fragment B isomorphic to a turtle. Denote clockwise the hexagonal neighboring faces of B by h1 , h2 , ..., h8 as shown in Figure 16. Let G0 := B ∪ h3 ∪ h4 ∪ h7 ∪ h8 . Then G0 is is a fragment isomorphic to B3 in Figure 7. Clearly, G0 satisfies the conditions of Lemma 4.4. So either FG0 := F − (G0 − ∂G0 ) is isomorphic to one of R6 and Bi (i = 1, 2, 3, 4) in Figure 7 or all neighboring faces of G0 is hexagonal. By Theorem 3.3 and Lemma 4.5, either FG0 is isomorphic to B3 or all neighboring faces of G0 are hexagonal. 1 If FG0 is isomorphic to B3 , then F is isomorphic to F36 since B is maximal. Now suppose that all neighboring faces of G0 are hexagonal. Let G1 be the fragment consisting of G0 together with its all hexagonal neighboring faces. A similar analysis for G1 yields that either 13
Figure 16: The fragments G0 , G1 and G2 in the proof of Lemma 5.1.
FG1 := F − G0 is isomorphic to B3 or all neighboring faces of G1 are hexagonal. If the former holds, then there are two fullerene graphs with 48 vertices containing two subgraphs isomorphic to B3 as shown in Figure 17. Both of the two fullerene graphs are not 3-resonant since each of them has three disjoint hexagons whose deleting results in a odd component (grey hexagons in Figure 17).
Figure 17: Two fullerene graphs with 48 vertices and two subgraphs isomorphic to B3 . Now suppose that all neighboring faces of G1 are hexagonal. Let G2 be the fragment consisting of G1 together with its all hexagonal neighboring faces. Let FG2 := F − G1 . Since |V (G1 )| = 36 and |V (F )| ≤ 60, |V (FG2 )| ≤ 24. By Theorem 3.3, Lemma 4.4 and Lemma 4.5, FG2 is isomorphic to B3 since |V (B3 )| = 24. Then we have two fullerene graphs with 60 vertices as shown in Figure 18. Both of the two fullerene graphs are not 3-resonant since each of them has three disjoint hexagons whose deletion results in a single vertex (the grey hexagons in Figure 18).
Figure 18: Two fullerene graphs with 60 vertices and two subgraphs isomorphic to B3 . 1 Summarizing above discussions, F is isomorphic to C60 or F36 .
14
1 Let F be a k-resonant (k ≥ 3) fullerene graph different from C60 and F36 . By Lemma 5.1, F should contain a pentagon ring. Let R ⊂ F be a pentagon ring with size l(R). Clearly, R has two faces different from the pentagons of it. Let C and C ′ be the boundaries of the two faces, respectively. Suppose that C and C ′ have s(R) and s′ (R) 2-degree vertices, respectively. Assume s(R) ≤ s′ (R). We always embed R in the plane such that C ′ = ∂R. The cycles C and C ′ are called inner cycle and outer cycle of R, respectively. Since s′ (R) + s(R) = l(R) and s(R) ≤ s′ (R), s(R) ≤ ⌊ l(R) ⌋. Clearly, s(R) 6= 1 and s′ (R) 6= 1 2 since F is 3-connected. Let ψl (F ) := min{s(R)| all pentagon rings of F with size l}. For example, ψ5 (F20 ) = 0 and ψ6 (F24 ) = 0. Let r(R), n6 (R) and n5 (R) be the numbers of vertices, hexagons and pentagons within C, respectively. By Euler formula,
n5 (R) = 6 + s(R) − l(R)
(1)
and
1 (2) n6 (R) = l(R) + (r(R) − s(R)) − 5. 2 Equation (2) implies that r(R) ≡ s(R) (mod 2). For a fullerene graph F , let τ (F ) := min{l(R)|R is a pentagon ring of F }. For example, τ (F20 ) = 5 and τ (F24 ) = 6. Lemma 5.2. Let F be a k-resonant (k ≥ 3) fullerene with τ (F ) = 5 or 6. Then F is isomorphic to F20 or F24 . Proof: Since both F20 and F24 are 2-resonant and contains no more than three pentagons, they are also k-resonant (k ≥ 3). Now let F be a k-resonant (k ≥ 3) fullerene graph. If τ (F ) = 5, Theorem 3.3 implies that F is isomorphic to F20 since R5 is the unique pentagon ring with size five. Now suppose τ (F ) = 6. Let R be a pentagon ring with size l(R) = τ (R) = 6 and let C and C ′ be the inner cycle and outer cycle of R, respectively. Let f0 , f1 , ..., f5 be the 6 pentagons of R in clockwise. Then s(R) ≤ ⌊ 62 ⌋ = 3 and s(R) 6= 1.
Figure 19: Illustration for the proof of Lemma 5.2
If s(R) = 3, there are three 2-degree vertices on C and also three 2-degree vertices on C ′ . By Lemma 4.3, the three 2-degree vertices on C have a common neighbor vertex within C 15
(see Figure 19 (a)). Hence F contains a R5 . So τ (F ) = 5, a contradiction. If s(R) = 2, there are two 2-degree vertices v1 , v2 on C and four 2-degree vertices on C ′ . By Lemma 4.3, v1 and v2 are adjacent in F . Since every face of F is a pentagon or a hexagon, v1 and v2 lie on fi and fi+3 (i ∈ Z6 ), respectively (see Figure 19 (b)). Thus F contains a R5 , contradicting τ (F ) = 6. So s(R) = 0. Then R is isomorphic to R6 . By Theorem 3.3, F is isomorphic to F24 . Lemma 5.3. There is no fullerene graphs with τ (F ) = 7. Proof: Suppose to the contrary that pentagon ring of F with size l(R) = Lemma 4.3, whether s(R) = 2 or 3, yellow). So τ (F ) = 6, a contradiction
F is a fullerene graph with τ (F ) = 7. Let R be the 7. Then s(R) ≤ ⌊ l(R) ⌋ = 3. So s(R) = 2 or 3. By 2 F contains a R6 (see Figure 20, R6 is illustrated in to τ (F ) = 7.
Figure 20: Illustration for the proof of Lemma 5.3
Lemma 5.4. Let F be a k-resonant (k ≥ 3) fullerene graph with τ (F ) = 8. Then F is isomorphic to F28 shown in Figure 21. Proof: Let R be a pentagon ring of F with size l(R) = τ (F ) = 8 and s(R) = ψ8 (F ) and let C and C ′ be the inner cycle and outer cycle of R, respectively. Let f0 , f1 , · · · , f7 be the 8 pentagons of R in clockwise ordering. Obviously, 2 ≤ ψ8 (F ) = s(R) ≤ ⌊ l(R) ⌋ = 4. 2
Figure 21: The illustration for Case 1 of the proof of Lemma 5.4.
Case 1. ψ8 (F ) = 2. Then C contains two 2-degree vertices v1 and v2 . By Lemma 4.3, v1 and v2 are adjacent. Since each face of F is either a pentagon or a hexagon, v1 and v2 lie on two pentagons fi and fi+4 (i ∈ Z8 ) (f2 , f6 in Figure 21 (a)), respectively. So there are 16
exact two adjacent hexagons h′ and h′′ within C. Let h1 and h2 be the two faces out of C ′ such that h1 is adjacent to faces fi−1 , fi and fi+1 , while h2 is adjacent to faces fi+3 , fi+4 and fi+5 (see Figure 21 (a)). If both h1 and h2 are hexagons, let S := {v} and H := {h1 , h2 , h′ }. Then F − H − S has two isolated vertices v1 and v2 (see Figure 21 (a)). By Theorem 2.1, H is not a resonant pattern, contradicting that F is k-resonant (k ≥ 3). So at least one of h1 and h2 is a hexagon, say h2 . If h1 is a pentagon, let h3 and h4 be other two neighboring faces of h1 as shown in Figure 21 (b). By Lemma 4.3, it is easy to check that both h3 and h4 are hexagons. Hence F is isomorphic to the fullerene graph F30 shown in Figure 21 (b). Clearly, H := {h1 , h4 , h′′ } is not a resonant pattern since F − V (H) has a singular vertex (the vertex v in Figure 21 (b)), also contradicting that F is not k-resonant (k ≥ 3). So both h1 and h2 are pentagons. By Lemma 4.3, F is isomorphic to the fullerene graph F28 shown in Figure 21. The perfect matching of F28 consisting all double edges illustrated in Figure 21 alternates on and off its all hexagons. So F is k-resonant.
Figure 22: The illustration for Case 2 of the proof of Lemma 5.4.
Case 2. ψ8 (F ) = 3. By Lemma 4.3 and the equations (1) and (2), we have r(R) = 1, n5 (R) = 1 and n6 (R) = 2. Immediately, we have a subgraph of F as shown in Figure 22. Hence F contains a pentagon ring R′ with size 8 and s(R′ ) = 2 (see Figure 22, R′ is illustrated in yellow). So 3 = ψ8 (F ) ≤ s(R′ )=2, a contradiction. The contradiction implies that there is no k-resonant (k ≥ 3) fullerene graph F with ψ8 (F ) = 3.
Figure 23: The illustration for Case 3 of the proof of Lemma 5.4.
Case 3. ψ8 (F ) = 4. By Lemma 4.3, the subgraph G of F induced by R together with all vertices within C is isomorphic to one of the four graphs illustrated in Figure 23. If G is isomorphic to the graph (a) or the graph (b), then F contains a pentagon ring with size six, contradicting τ (F ) = 8. If G is isomorphic to the graph (c) or the graph (d), then F contains a pentagon ring R′ with size eight and s(R′ ) = 2, contradicting
17
ψ8 (F ) = 4. The contradictions imply that there is no k-resonant (k ≥ 3) fullerene graph F with ψ8 (F ) = 4. Lemma 5.5. Let F be a k-resonant (k ≥ 3) fullerene graph with τ (F ) = 9. Then F is isomorphic to the fullerene graph F32 as shown in Figure 24. Proof: Let R be a pentagon ring of F with size l(R) = 9 and s(R) = ψ9 (F ) and let C and C ′ be the inner cycle and outer cycle of R, respectively. If s(R) ≤ 2, then F has a face with size more than six within C by Lemma 4.3, contradicting that F is a fullerene graph. So ⌋ = 4. Let f0 , f1 , · · · , f8 be the nine pentagons of R in clockwise. 3 ≤ ψ9 (F ) = s(R) ≤ ⌊ l(R) 2
Figure 24: Illustration for Case 1 of the proof of Lemma 5.5.
Case 1. ψ9 (F ) = 3. By Lemma 4.3 and the equations (1) and (2), we have r(R) = 1, n5 (R) = 0 and n6 (R) = 3. Denote the three hexagons within C by h1 , h2 and h3 . The three 2-degree vertices on C may lie on the pentagons fj , fj+3 and fj+6 (j, j + 3, j + 6 ∈ Z9 ) (see Figure 24 (a) where j = 1). Let h′1 , h′2 , · · · , h′6 be the six faces adjacent to R clockwise along C ′ such that h′1 is adjacent to fj−1, fj and fj+1 (see Figure 24 (a)). If at least two of h′1 , h′3 and h′5 are hexagons, say h′1 and h′5 , then H := {h′1 , h′5 , h1 } is not a resonant pattern since F − V (H) has an isolated vertex v (see Figure 24 (a)), contradicting the assumption that F is k-resonant (k ≥ 3). So suppose that only one of h′1 , h′3 and h′5 is a hexagon. Then by Lemma 4.3, F should has a face with size either four or seven, a contradiction. So all of h′1 , h′3 and h′5 are pentagons. By Lemma 4.3 again, F is isomorphic to the graph (b) in Figure 24, also the fullerene graph F32 in Figure 24. Conversely, we need to show that F32 is k-resonant k ≥ 3. Since there are no more than two disjoint hexagons in F32 , it suffices to show that any two disjoint hexagons of F32 are simultaneously resonant. By symmetry, we need only to show that {h1 , h′4 } and {h1 , h′6 } are resonant patterns. Let M be the perfect matching of F32 consisting of the double edges illustrated in Figure 24. Clearly, all of h1 , h′4 and h′6 are M-alternating. Hence both {h1 , h′4 } and {h1 , h′6 } are resonant patterns of F32 . So F is isomorphic to F32 . Case 2. ψ9 (F ) = 4. Let G be the subgraph of R induced by the vertices of R together with the vertices with in C. By Lemma 4.3 and the equations (1) and (2), either r(R) = 0, 18
n5 (R) = 1 and n6 (R) = 2, or r(R) = 2, n5 (R) = 1 and n6 (R) = 3. By Lemma 4.3, G is isomorphic to the graph (a) in Figure 25 if the former holds, and G is isomorphic to the graph (b) or the graph (c) in Figure 25 if the latter holds.
Figure 25: The illustration for Case 2 of the proof Lemma 5.5.
If G is isomorphic to the graph (a), then F contains a pentagon ring with size eight, contradicting τ (F ) = 9. If G is isomorphic to the graph (b), then F contains a pentagon ring R′ with size 9 and s(R′ ) = 3. Further, s(R′ ) = 3 < 4 = ψ9 (F ) ≤ s(R′ ), a contradiction. So suppose G is isomorphic to the graph (c). Let f be the face adjacent to R along a 4-length path as shown in Figure 25 (d). If f is a pentagon, then F contains a pentagon ring with size 8 which consists of 7 pentagons of R and f , contradicting τ (F ) = 9. So f is a hexagon. Let H := {f, h1 , h2 } where h1 , h2 are two disjoint hexagons within C (see Figure 25 (d)). Then F − V (H) has an isolated vertex v (see Figure 25 (d)). So H is not a resonant pattern and F is k-resonant (k ≥ 3), a contradiction again. So there is no k-resonant (k ≥ 3) fullerene graph F with ψ9 (F ) = 4. The following lemma is due to Kutnar and Maruˇsiˇc [13]: Lemma 5.6. [13] Let F be a fullerene graph containing a ring R of five faces, and let C and C ′ be the inner cycle and the outer cycle of R, respectively. Then either (1) C or C ′ is the boundary of a face, or (2) both C and C ′ are of length 10, and the five faces of R are all hexagonal. Lemma 5.7. Let F be a k-resonant (k ≥ 3) fullerene graph with τ (F ) = 10. Then F is 2 isomorphic to F36 in Figure 26 or F40 in Figure 27. Proof: Let R be a pentagon ring of F with size l(R) = τ (F ) = 10 and s(R) = ψ10 (F ) and let C be the inner cycle of R. Let f0 , · · · , f9 be the 10 pentagons of R10 in clockwise. Clearly, ψ10 (F ) ≥ 2. If ψ10 (F ) = 2, then the two 2-degree vertices are adjacent by Lemma 4.3. Then there are two faces h0 , h1 of F within C and |h0 | + |h1 | = l(R) + s(R) + 2 = 14 since the edge within C is counted twice in |h0 | + |h1 |. So at least one of h0 and h1 has size more than seven, a contradiction. If ψ10 (F ) = 3, then the three 2-degree vertices of R on C together with vertices within C induce a K1,3 by Lemma 4.3. Hence there are three faces h0 , h1 , h2 19
P of F within C and i∈Z3 |hi | = l(R) + s(R) + 6 = 19. So one of h0 , h1 , h2 with size at least ⌋ = 5. seven, a contradiction. So 4 ≤ ψ10 (F ) = s(R) ≤ ⌊ l(R) 2
Figure 26: The illustration for Case 1 of the proof of Lemma 5.7.
Case 1. ψ10 (F ) = 4. By Lemma 4.3 and the equation (1) and (2), n5 (R) = 0 and r(R) = 0, n6 (R) = 3 or r(R) = 2, n6 (R) = 4. If r(R) = 0 and n6 (R) = 3, the four 2-degree vertices on C belong to the pentagons fj , fj+1, fj+5 and fj+6 for some j ∈ Z10 , say j = 3 (see Figure 26 (a)). Let h1 , h2 and h3 be the three hexagons within C such that h1 ∩ f2 6= ∅, h2 ∩ f3 6= ∅ and h3 ∩ f4 6= ∅. Let f be a common neighboring face of f2 , f3 , f4 and f5 (see Figure 26 (a)). If f is a pentagon, then F contains a pentagon ring (R − {f3 , f4 }) ∪ f with size 9, contradicting τ (F ) = 10. So suppose f is a hexagon. Then H := {f, h1 , h3 } is not a resonant pattern since F − V (H) has an isolated vertex (the black vertex in Figure 26 (a)), contradicting that F is k-resonant (k ≥ 3). So suppose r(R) = 2 and n6 (R) = 4. By Lemma 4.3, the four 2-degree vertices on C together with all vertices within C induce a T0 . Hence, the four 2-degree vertices belong to the pentagons fj , fj+3 , fj+5 and fj+8 for some j ∈ Z10 , say j = 1 (see Figure 26 (b)). Let h1 , h2 , h3 and h4 be the four hexagons within C in clockwise and h1 ∩f0 6= ∅. Let h′1 , h′2 , · · · h′6 be the six faces adjacent to R clockwise along its boundary such that h′1 ∩ f1 6= ∅ (see Figure 26 (b)). If h′4 and h′6 are both hexagons, then H := {h4 , h′4 , h′6 } is not a resonant pattern since F − V (H) has an isolated vertex v, contradicting that F is k-resonant (k ≥ 3). So one of h′4 and h′6 is a pentagon, say h′6 is a pentagon by symmetry. By the symmetry again, one of h′1 and h′3 is a pentagon. If h′1 is a pentagon, then F will has a face h′2 = h′5 with size 8 by Lemma 4.3, a contradiction. So h′3 is a pentagon, and both h′1 and h′4 are hexagons. 2 Immediately, F is isomorphic to the graph (c) in Figure 26, and also F36 in Figure 26. 2 In the following, it suffices to prove that F36 is k-resonant (k ≥ 3). Let M be a perfect 2 matching of F36 consisting of double edges illustrated in Figure 26. Let M1 := M ⊕ h2 ⊕ h4 , ′ M2 = M ⊕ h1 ⊕ h′4 and M3 := M ⊕ h2 ⊕ h4 ⊕ h′1 ⊕ h′4 . Let H a resonant pattern of F . If h1 , h′5 ∈ / H, then every hexagon in H is M-alternating. If h1 ∈ H but h′5 ∈ / H, then / H, then every hexagon in H is every hexagon of H is M1 -resonant. If h′5 ∈ H but h1 ∈ ′ ′ M2 -alternating. If h1 , h5 ∈ H, then H = {h1 , h5 } and hence they are H is M3 -alternating. 20
2 So F36 is k-resonant (k ≥ 3). Case 2. ψ10 (F ) = 5. By the equations (1) and (2), n5 (R) = 1 and n6 (R) = 5− 12 (5−r(R)). Subcase 2.1. There exist two vertices of R on C having a common neighbor vertex within C. Let G be the subgraph induced by R together with all vertices within C. By Lemma 4.3, G is isomorphic to the graph (a) or the graph (b) in Figure 27. If G is isomorphic to the graph (a), then F contains a pentagon ring R′ with size 10 and s(R) = 4. Hence s(R′ ) = 4 < ψ10 (F ) ≤ s(R′ ), a contradiction. So suppose G is isomorphic to the graph (b). Let f be a face adjacent to R along a 4-length path (see Figure 27, the common neighboring face f of f2 , ..., f5 ). If f is a pentagon, then F contains a pentagon ring with size 9 (the pentagon ring (R − {f3 , f4 }) ∪ f in Figure 27), contradicting τ (F ) = 10. So suppose f is a hexagon. Then F has a hexagon set H consisting of three mutually disjoint hexagons such that f ∈ H and F − H has an isolated vertex (the grey hexagons and the black vertex in Figure 27 (b)), contradicting that F is k-resonant (k ≥ 3). So there is no k-resonant fullerene graph satisfying the condition of Subcase 2.1.
Figure 27: The illustration for Subcase 2.1 of the proof of Lemma 5.7.
Subcase 2.2. Any two 2-degree vertices on C have no common neighbor vertex within C. Then the five faces adjacent to R within C form a ring R′ of F with C as its outer cycle. Since |C| = 15, the inner cycle of R′ bounds a face f ′ of F by Lemma 5.6. Note that r(R) ≡ s(R) = ψ10 (F ) (mod 2). So f ′ is a pentagon. Therefore the subgraph G induced by R together with all inner vertices of C is isomorphic to the graph (c) in Figure 27. A similar discussion yields that the five faces adjacent G along its boundary are hexagons and F − G is a pentagon. So F is isomorphic to F40 as shown in Figure 27. Now, it suffices to prove that F40 is k-resonant (k ≥ 3). Let M be the perfect matching of F40 consisting of all double edges illustrated in Figure 27. Clearly, all hexagons of F40 are M-alternating. Hence F40 is k-resonant (k ≥ 3). Summarizing Case 1 and Case 2, a k-resonant (k ≥ 3) fullerene graph F with τ (F ) = 10 2 is isomorphic to F36 in Figure 26 or F40 in Figure 27. Lemma 5.8. A fullerene graph F with τ (F ) = 11 is not k-resonant (k ≥ 3). Proof: Let R be a pentagon ring of F with size l(R) = τ (F ) = 11 and s(R) = ψ11 (F ) and let C be the inner cycle of R. By equation (1), n5 (R) = s(R)−5 and further ψ11 (F ) = s(R) ≥ 5. 21
⌋ = 5. So ψ11 (F ) = s(R) = 5 and n5 = 0; that is, On the other hand, ψ11 (F ) = s(R) ≤ ⌊ l(R) 2 there is no pentagons within C. Let v1 , v2 , v3 , v4 and v5 be the five 2-degree vertices clockwise on C. If two of these five vertices are adjacent, let v1 v2 ∈ E(F ). By Lemma 4.3, v3 , v4 and v5 have a common neighbor vertex within C, denoted by w. Let h be the face containing v1 , v2 , v3 , w and v5 . Note that any two vertices of v1 , v2 , ..., v5 are not adjacent on C. So |h| ≥ 7, a contradiction. If any two vertices of v1 , · · · v5 have no a common vertex, then the five faces adjacent to R along C form a ring R′ with size five and C as its out cycle. Since |C| = 16, the inner cycle of R′ bounds a face f ′ of F by Lemma 5.6. Note s(R) ≡ r(R) (mod 2). So f ′ is a pentagon, contradicting n5 (R) = 0.
Figure 28: The illustration of Lemma 5.8
So there exist two vertices of v1 , ..., v5 with a common neighbor vertex within C, say v1 and v2 . By Lemma 4.3 and n5 (R) = 0, the subgraph of F induced by R together with all vertices within C is isomorphic to the graph (a) in Figure 28. Let f be a face adjacent to R along a 4-length path on the boundary of R (see Figure 28 (b)). If f is a pentagon, then F contains a pentagon ring R′ with size l(R′ ) = 10. Then 11 = τ (F ) ≤ l(R′ ) = 10, a contradiction. So suppose f is a hexagon. Then F contains a set H consisting of three mutually disjoint hexagons such that F − V (H) has an isolated vertex (the grey hexagons and the black vertex in Figure 28 (b)), a contradiction to the assumption that F is k-resonant (k ≥ 3). So no fullerene graph F with τ (F ) = 11 is k-resonant (k ≥ 3).
Figure 29: The fullerene graph F48 with a perfect matching M0 .
Lemma 5.9. Let F be a k-resonant (k ≥ 3) fullerene graph with τ (F ) = 12. Then F is isomorphic to F48 as shown in Figure 29. 22
Proof: Let R be the unique pentagon ring of F with size l(R) = 12. Let C be the inner cycle of R. Since F has 12 pentagons only, there is no pentagons within C and hence n5 (R) = 0. By the equation (1), s(R) = 6. Let v0 , v1 , · · · , v5 be the six 2-degree vertices clockwise on C. First suppose that r(R) = 0; that is, there is no vertex within C. Then the subgraph of F induced by R together with all vertices within C is isomorphic to the graph G1 (see Figure 30 (left)) and F is isomorphic to the fullerene graph shown in Figure 30 (right). Let H be the set consisting of the three mutually disjoint grey hexagons of F . Then F − V (H) has an isolated vertex and hence F is not k-resonant (k ≥ 3), a contradiction.
Figure 30: The subgraph G1 (left) and the fullerene graph containing G1 (right). So suppose r(R) ≥ 2 by r(R) ≡ s(R) (mod 2). Let G be the subgraph induced by the vertices within C and let n and m be the vertex number and edge number of G, respectively. Then n = r(R) ≥ 2. Since C has six 2-degree vertices, 3n − 2m = 6. If G is not connected, then G is a forest since F is 3-connected and cyclically 5-connected. Then m = n − w where w ≥ 2 is the number of components of G. So 3n − 2(n − w) = 6, and further n = 6 − 2w ≤ 2. So n = 2 and hence G is two isolated vertices, denoted by u, v. We may assume that N(u) = {v0 , v1 , v2 } and N(v) = {v3 , v4 , v5 }. Let f be the face within C containing vertices v0 , u, v2, v3 , v and v5 . Note that the six 2-degree vertices are not adjacent on C. So v2 v3 ∈ / E(f ) and v5 v0 ∈ / E(f ). Therefore |f | ≥ 8, a contradiction. So suppose that G is connected. Let ∂G be the boundary of G. Note that ∂G may be not a cycle and an edge which belongs only to the infinite face will contribute 2 to |∂G|. Let x be the number of inner faces of G. By Euler’s formula, m = n − 1 + x. So 3n − 2(n − 1 + x) = 6, and further n = 4 + 2x. On the other hand, since every inner face of G is also a face of F , every inner face of G is a hexagon. So |∂G| + 6x = 2m = 2n − 2 + 2x. Hence |∂G| = 6.
Figure 31: The subgraphs G2 and G3 in the proof of Lemma 5.9. If x = 0, then G is a tree. Since |∂G| = 6, G has three edges. So G is isomorphic to 23
a K1,3 or a 3-length path. Hence the subgraph of F induced by V (R ∪ G) is isomorphic to either G2 or G3 shown in Figure 31. Whether G2 ⊂ F or G3 ⊂ F , F has three mutually disjoint hexagons whose deletion results in an isolated vertex (the grey hexagons in Figure 31). So F is not k-resonant for k ≥ 3, a contradiction. So suppose x 6= 0. Since the length of every cycle of G is at least 6 and |∂G| = 6, ∂G is a 6-length cycle. Since F is cubic, there are six edges connecting 2-degree vertices v0 , v1 , · · · , v5 to vertices of G. So G is a hexagon. By symmetry of R, F is isomorphic to F48 shown in Figure 29. Now, it suffices to prove F48 is k-resonant (k ≥ 3). Let M0 be the perfect matching of F48 consisting of all double edges illustrated in Figure 29. Let f1 , f2 , f3 and f1′ , f2′ , f3′ be the faces marked in F48 and let M1 := M0 ⊕ f1 ⊕ f2 ⊕ f3 , M2 := M0 ⊕ f1′ ⊕ f2′ ⊕ f3′ and M3 := M0 ⊕ f1 ⊕ f2 ⊕ f3 ⊕ f1′ ⊕ f2′ ⊕ f3′ . For any set H of mutually disjoint hexagons, every hexagon of H is Mj -alternating for some j ∈ Z4 . So F48 is k-resonant (k ≥ 3). Summarizing above results, we have the following main theorem, Theorem 5.10. A fullerene graph F is k-resonant (k ≥ 3) if and only if F is isomorphic 1 2 to one of F20 , F24 , F28 , F32 , F36 , F36 , F40 , F48 and C60 . According to Theorem 5.10, a fullerene graph F is k-resonant (k ≥ 3) if and only if it is 3-resonant. This result is coincident with these for benzenoid systems [29], coronoid systems [2, 14], open-end nanotubes [23], toroidal polyhexes [26] and Klein-bottle polyhexes [20].
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