Distance-restricted matching extendability of fullerene graphs Michitaka Furuya1∗

Masanori Takatou2†

Shoichi Tsuchiya3‡ 1

College of Liberal Arts and Science, Kitasato University,

1-15-1 Kitasato, Minami-ku, Sagamihara, Kanagawa 252-0373, Japan 2

Department of Civil and Environmental Engineering, Hiroshima Institute of Technology 2-1-1 Miyake, Saeki-ku, Hiroshima 731-5193, Japan 3

School of Network and Information, Senshu University,

2-1-1 Higashimita, Tama-ku, Kawasaki-shi, Kanagawa 214-8580, Japan

Abstract A fullerene graph is a 3-connected cubic plane graph whose all faces are bounded by 5- or 6-cycles. In this paper, we show that a matching M of a fullerene graph can be extended to a perfect matching if the following hold: (i) Three faces around each vertex in {x, y : xy ∈ M } are bounded by 6-cycles and (ii) the edges in M lie at distance at least 13 pairwise.

Key words and phrases. fullerene graph, perfect matching, matching extendability. AMS 2010 Mathematics Subject Classification. 05C10, 05C70, 92E10.

1

Introduction

1.1

Definitions and notations

Let G be a graph. We let V (G) and E(G) denote the vertex set and the edge set of G, respectively. We let V (M ) = {x, y : xy ∈ M } for M ⊆ E(G), and let ∗

e-mail:[email protected] e-mail:[email protected] ‡ e-mail:[email protected] †

1

V (e) = V ({e}) for e ∈ E(G). For a vertex x ∈ V (G), we let NG (x) and dG (x) denote the neighborhood and the degree of x, respectively; thus NG (x) = {y ∈ V (G) : xy ∈ ∪ E(G)} and dG (x) = |NG (x)|. For X ⊆ V (G), we let NG (X) = ( x∈X NG (x)) − X. For vertices x1 , x2 ∈ V (G), we let dG (x1 , x2 ) denote the distance between x1 and x2 in G. For X1 , X2 ⊆ V (G), we let dG (X1 , X2 ) = min{dG (x1 , x2 ) : x1 ∈ X1 , x2 ∈ X2 }. For a subgraph H of G and a subset X of V (G), let H[X] be the subgraph of H induced by X ∩ V (H). We let co (G) denote the number of components of G having odd order. A subset M of E(G) is a matching if no two distinct edges in M have a common end-vertex. A matching M of G is perfect if V (M ) = V (G). For terms and symbols not defined in this paper, we refer the reader to [6].

1.2

Matching problem for fullerene graphs

A fullerene graph is a 3-connected cubic plane graph whose all faces are bounded by 5- or 6-cycles. By Petersen’s theorem, fullerene graph has a perfect matching. It has been known that fullerene graphs can be regarded as chemical graphs of fullerenes, where carbon atoms are represented by vertices of the graph, whereas the edges represent bonds between adjacent atoms. Furthermore, a perfect matching of a fullerene graph is corresponding to the Kekul´e structure of a fullerene. By such a reason, some properties of perfect matchings in fullerene graphs have been widely studied (see, for example, [5, 10, 17]). In this paper, we focus on matching extendability concept for fullerene graphs. For k ≥ 0, a graph G of order n ≥ 2k + 2 is k-extendable if every matching of G having size k can be extended to a perfect matching of G. Note that a graph is 0-extendable if and only if the graph has a perfect matching. Doˇsli´c [7] proved the following theorem concerning matching extendability for fullerene graphs. Theorem A (Doˇ sli´ c [7]) Every fullerene graph is 2-extendable. For a graph G, a set F ⊆ E(G) is a cycle-separating edge-cut if G − F has at least two components which contain cycles. For m ≥ 1, a graph is cyclically medge-connected if every cycle-separating edge-cut of the graph has size at least m. Theorem A is a corollary of the following two theorems. Theorem B (Steinitz [13]) Let G be a 3-connected and cyclically 4-edge-connected cubic plane graph. If G has no face bounded by 4-cycle, then G is 2-extendable. Theorem C (Doˇ sli´ c [7]) Every fullerene graph is cyclically 4-edge-connected. Doˇsli´c [8] improved Theorem C as follows. 2

Theorem D (Doˇ sli´ c [8]) Every fullerene graph is cyclically 5-edge-connected. Since every k-extendable graph is (k + 1)-connected (see [11]), every fullerene graph is not 3-extendable. In this sense, Theorem A is best possible. Here, one problem naturally arises: For a (non-k-extendable) graph G, find a condition such that a matching M of G with |M | = k satisfying the condition can be extended to a perfect matching. One of answers of the problem, the concept of distance matching extendability was introduced by Aldred and Plummer [2]. For k ≥ 0 and d ≥ 1, a graph G of order n ≥ 2k + 2 is distance d k-extendable if every matching M of G such that |M | = k and the edges of M lie pairwise distance at least d in G can be extended to a perfect matching of G. The distance extendability has been widely studied for graphs on closed surfaces (see [1, 3, 4, 9]). In this paper, we prove the following theorem which will serve as an initial attempt to develop the distance extendability for fullerene graphs. Theorem 1.1 Let G be a fullerene graph, and let M be a matching of G satisfying the following: (M1) The three faces around each vertex in V (M ) are bounded by 6-cycles and (M2) dG (V (e), V (e′ )) ≥ 13 for all e, e′ ∈ M with e ̸= e′ . Then G has a perfect matching M0 with M ⊆ M0 . Remark 1 Since the prescribed edges in Theorem 1.1 are forced the condition (M1), Theorem 1.1 cannot strictly assure that fullerene graphs have the distance extendability. (We expect that the condition (M1) can be dropped because it derives from our proof technique). However, it is known that every fullerene graph has exactly twelve 5-cycles as faces, and so Theorem 1.1 assure that fullerene graphs are “almost” distance extendable. Furthermore, we also expect that the distance condition in (M2) can be improved. Thus we leave the problem to give a sharp distance condition instead of (M2) for the readers.

1.3

Application of main result

Let G be a graph, and let M be a perfect matching of G. A subset M ′ of M is a forcing set of M if M is the unique perfect matching of G with M ′ ⊆ M . The forcing number of M is the minimum cardinality of a forcing set of M . As mentioned in [12, 14, 15], it is important to research a fullerene graph with a perfect matching having a large forcing number (for example, see [16]).

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To analyze the forcing number, it seems to be useful to judge whether a given matching is contained in at least two perfect matching or not. Our result contribute to such a problem as follows: Theorem 1.2 Let G be a fullerene graph. Let M be a matching of G, and let e1 , e2 ∈ E(G). If the following hold, then M is a forcing set of no perfect matching of G. (M’0) The edges e1 and e2 are adjacent (i.e., {e1 , e2 } is not a matching), (M’1) the three faces around each vertex in V (M ) ∪ V (e1 ) ∪ V (e2 ) are bounded by 6-cycles, (M’2) dG (V (e), V (e′ )) ≥ 13 for all e, e′ ∈ M with e ̸= e′ , and (M’3) dG (V (ei ), V (e)) ≥ 13 for i ∈ {1, 2} and e ∈ M . Proof. For i ∈ {1, 2}, it follows from Theorem 1.1 that there exists a perfect matching Mi such that M ∪ {ei } ⊆ Mi . Since e1 and e2 are adjacent, M1 ̸= M2 . Thus M is a forcing set of no perfect matching of G.

2

□

Proof of Theorem 1.1

Let G be a fullerene graph, and let M be a matching of G satisfying (M1) and (M2). By way of contradiction, suppose that G has no perfect matching M0 with M ⊆ M0 . This implies that G − V (M ) has no perfect matching. This together with Tutte’s 1-factor theorem implies that there exists S ⊆ V (G) − V (M ) such that co (G − (V (M ) ∪ S)) ≥ |S| + 1. We choose M and S so that (S1) |M | is as small as possible, and (S2) subject to (S1), |S| is as small as possible. By Theorem A, |M | ≥ 3. Write M = {e1 , . . . , em } and eh = x1h x2h (1 ≤ h ≤ m), and let G′ := G − V (M ). For each h (1 ≤ h ≤ m), by the condition (M1), G′ has the 14facial cycle Hh whose one region contains only x1h and x2h in G. We may assume that x1h and x2h belongs to the interior of Hh . Write Hh = yh1,1 yh1,2 · · · yh1,7 yh2,1 yh2,2 · · · yh2,7 yh1,1 where NG (xih ) ∩ V (Hh ) = {yhi,2 , yhi,6 } (see Figure 1). The following claim will be implicitly used in the proof. Claim 2.1 The graph G′ is a connected graph of even order and S ̸= ∅.

4

yh1,2 yh1,4

yh1,3

yh1,1 x1h

yh1,5

yh1,7

yh2,7 x2h

yh2,6 yh2,5 yh2,3

yh2,1

yh1,6

yh2,4

yh2,2

Figure 1: Cycle Hh

Proof. Since G is cubic and |V (M )| is even, |V (G′ )| (= |V (G)| − |V (M )|) is even. Suppose that G′ is disconnected. Then V (M ) is a cutset of G. Let X ⊆ V (M ) be a minimal cutset of G, and take xih ∈ X. For each h′ (h′ ̸= h), it follows from (M2) that (V (Hh ) ∪ V (eh )) ∩ (V (Hh′ ) ∪ V (eh′ )) = ∅. In particular, Hh is a cycle of G − X. Hence, no matter whether x3−i ∈ X or not, all vertices in NG−X (xih ) h belong to the same component of G − X. Then X − {xih } is also a cutset of G, which contradicts the minimality of G. Consequently, G′ is connected. Since G′ is a connected graph of even order and co (G′ − S) ≥ |S| + 1 ≥ 1, we have S ̸= ∅.

□

For S1 ⊆ S, let C(S1 ) = {C : C is a component of G′ − S with NG′ (V (C)) ∩ S1 ̸= ∅}. We give the following claim whose proof does not use the planarity of G. Claim 2.2 The following hold: (i) Let S1 ⊆ S with S1 ̸= ∅. Then there exist |S1 | + 2 odd components C of G′ − S such that NG′ (V (C)) ∩ S1 ̸= ∅. In particular, |C(S1 )| ≥ |S1 | + 2. (ii) Let u ∈ S. Then u is adjacent to exactly three different odd components of G′ − S. In particular, u is adjacent to no vertex in V (M ) ∪ S in G. (iii) Every component of G′ − S has odd order. In particular, C(S) is the set of odd components of G′ − S. Proof. Since |V (G′ )| is even, |S| ≡ co (G′ − S) (mod 2), and hence co (G′ − S) ≥ |S| + 2.

5

(2.1)

If there exists a non-empty subset S1 of S such that the number of odd components C of G′ − S with NG′ (V (C)) ∩ S1 ̸= ∅ is at most |S1 | + 1, then by (2.1), co (G′ − (S − S1 )) ≥ co (G′ − S) − (|S1 | + 1) ≥ (|S| + 2) − (|S1 | + 1) = |S − S1 | + 1, which contradicts (S2). Thus (i) holds. Let u ∈ S. Since dG′ (u) ≤ 3, u is adjacent to at most three odd components of G′ − S. This together with (i) implies that u is adjacent to exactly three different odd components of G′ − S. Thus (ii) holds. Since G′ is connected, if a component C of G′ − S has even order, then a vertex u ∈ S is adjacent to C in G′ , which contradicts (ii). Thus (iii) holds.

□

For a component C of G′ − S, let I(C) = {h : 1 ≤ h ≤ m, V (C) ∩ V (Hh ) ̸= ∅}. Claim 2.3 For a component C of G′ − S and an index h ∈ I(C), Hh [V (C)] is a path. Proof. We first suppose that a component of Hh [V (C)] is not a path. Since Hh is a cycle, this implies that V (Hh ) ⊆ V (C). Let M ′ = M − {eh } and G′′ = G − V (M ′ ). Since NG (V (eh )) ⊆ V (Hh ), C ′ is a component of G′′ − S if and only if either C ′ = G[V (C) ∪ V (eh )] or C ′ is a component of G′ − S except for C. By Claim 2.2(iii), |V (C)| is odd, and hence G[V (C) ∪ V (eh )] is an odd component of G′′ − S. Hence co (G′′ − S) = co (G′ − S) ≥ |S| + 1, which contradicts (S1). Thus every component of Hh [V (C)] is a path. To prove the claim, it suffices to show that Hh [V (C)] is connected. By way of contradiction, suppose that Hh [V (C)] has two components C1 and C2 . Let P be a shortest path of C joining V (C1 ) and V (C2 ), and let pi be the vertex in V (P )∩V (Ci ) for i ∈ {1, 2}. Let u1 and u2 be the vertices in V (Hh ) ∩ S with NHh (ui ) ∩ V (C1 ) ̸= ∅. Since a closed curve obtained from P by adding a curve joining p1 and p2 through the interior of Hh separates u1 and u2 , G′ − V (C) has two distinct components D1 and D2 with ui ∈ V (Di ). For each i ∈ {1, 2}, set Si = V (Di ) ∩ S. Since D1 is a component of G′ − V (C), NG′ (V (D1 )) ⊆ V (C). For C ′ ∈ C(S1 ) − {C}, this together with the fact that V (C ′ ) ⊆ V (D1 ) implies that C ′ is a component of G′ − S1 . Hence by Claim 2.2(i), co (G′ − S1 ) ≥ |C(S1 )| − 1 ≥ |S1 | + 1. Furthermore, since u2 ∈ S2 and S1 ∩ S2 = ∅, we have |S1 | ≤ |S| − |S2 | < |S|, which contradicts (S2).

□

For a component C of G′ − S, let F (C) be the set of edges of G′ joining V (C) and S. Claim 2.4 For a component C of G′ − S with |I(C)| ≥ 2, |F (C)| ≥ 5|I(C)|. 6

Proof. Since C is a plane graph, there exists a boundary oriented closed walk W of C such that each arc of W appears in W exactly once, where two arcs (v, v ′ ) and (v ′ , v) are regarded as different. Let h ∈ I(C). Then there exists an oriented subwalk Qh of W with the initial vertex vh and the terminal vertex vh+ such that V (Qh ) ∩ V (Hh ) = {vh } and V (Qh ) ∩ ∪ ( h′ ̸=h V (Hh′ )) = {vh+ } (note that W might have plural oriented subwalks with the ∪ initial vertex in V (Hh ) and the terminal in h′ ̸=h V (Hh′ )). Let th be the index such that vh+ ∈ V (Hth ). Let Ph = vh1 vh2 · · · vhlh be a path on Qh joining vh and vh+ where ∪ vh1 = vh and vhlh = vh+ , and let P˜h = Ph −{vh1 , vhlh }. Since V (P˜h )∩( 1≤h≤m V (Hh )) = ∅, NG′ (v) − V (Ph ) consists of one vertex wv for v ∈ V (P˜h ). Recall that each arc of W appears in W exactly once. Thus we can construct a closed curve Clh obtained from Ph by adding a curve Ah joining vh and vh+ so that (C1) for h ∈ I(C), every vertex in V (Qh ) − V (Ph ) belongs to the interior of Clh , (C2) for h ∈ I(C), Ah and each edge in {vwv : v ∈ P˜h } has no edge-crossing, and (C3) for h, h′ ∈ I(C) with h ̸= h′ , the interior of Clh and the interior of Clh′ have no intersection. A vertex v ∈ V (P˜h ) is called an in-vertex with respect to Ph if the edge vwv intersects with the interior of Clh . Now we show that for i (2 ≤ i ≤ lh − 2), at least one of vhi and vhi+1 is an in-vertex with respect to Ph . (2.2) Suppose that for some i (2 ≤ i ≤ lh − 2), neither vhi nor vhi+1 is an in-vertex with respect to Ph . Then there exists a face f of G′ with the facial cycle D such that vhi−1 , vhi , vhi+1 , vhi+2 are consecutive vertices in D. Since W is a boundary closed walk of C, we see that V (D) ∩ S ̸= ∅. Note that D consists of 5 or 6 vertices because f is a face of G. If |V (D) ∩ S| = 1, then the unique vertex in V (D) ∩ S is adjacent to two vertices of C, which contradicts Claim 2.2(ii). Thus |V (D) ∩ S| ≥ 2. Since D contains four vertices vhi−1 , vhi , vhi+1 , vhi+2 of C, |V (D) ∩ S| = 2 and the vertices in V (D) ∩ S are adjacent, which again contradicts Claim 2.2(ii). Thus (2.2) holds. Since dG (V (eh ), V (et )) ≥ 13, P˜h has at least 6 vertices. This together with (2.2) h

implies that there exist at least 3 in-vertices with respect to Ph .

(2.3)

Let X = {v : v is an in-vertex with respect to Ph for some h ∈ I(C)}. By (C2) and (C3), if a vertex v is an in-vertex with respect to Ph , then v is not an in-vertex 7

with respect to Ph′ for h′ ∈ I(C) − {h}. Hence by (2.3), we have |X| =

∑

|{v : v is an in-vertex with respect to Ph }| ≥ 3|I(C)|.

(2.4)

h∈I(C)

For v ∈ X, let Cv be the component of Qh − (V (Ph ) − {v}) containing v where h is the index such that v is an in-vertex with respect to Ph . For v, v ′ ∈ X with v ̸= v ′ , suppose that V (Cv ) ∩ V (Cv′ ) ̸= ∅. Then wv ∈ V (C). Let h be the index such that v is an in-vertex with respect to Ph . It follows from (C1) and (C3) that v ′ is also an in-vertex with respect to Ph and there exists a path L of C with vwv ∈ E(L) joining v and v ′ , which contradicts the fact that W is a boundary closed walk of C. Thus V (Cv ) ∩ V (Cv′ ) = ∅ for v, v ′ ∈ X with v ̸= v ′ .

(2.5)

Now we define an edge e∗v ∈ F (C) such that V (e∗v ) ∩ V (Cv ) ̸= ∅ as follows: If wv ∈ S, let e∗v = vwv . We assume that wv ∈ / S. Then by (2.5), vwv is a bridge of ∪ C. Since G is 3-connected and Cv contains no vertices in 1≤h≤m V (Hh ), we have NG′ (V (Cv ) − {v}) ∩ S ̸= ∅. Let e∗v be an edge joining V (Cv ) − {v} and S. By (2.5), e∗v ̸= e∗v′ for v, v ′ ∈ X with v ̸= v ′ . This together with (2.4) implies that |{e∗v : v ∈ X}| = |X| ≥ 3|I(C)|.

(2.6)

For each h ∈ I(C), it follows from Claim 2.3 that Hh [V (C)] is a path, and hence |F (C) ∩ E(Hh )| ≥ 2. For v ∈ X, since the unique vertex in V (e∗v ) − S does not ∪ belong to h∈I(C) V (Hh ), (F (C) ∩ E(Hh )) ∩ {e∗v : v ∈ X} = ∅. This together with (2.6) implies that ∪ ∗ |F (C)| ≥ |{ev : v ∈ X}| + (F (C) ∩ E(Hh )) ≥ 3|I(C)| + 2|I(C)| = 5|I(C)|, h∈I(C) as desired.

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Claim 2.5 If a component C of G′ − S contains a vertex v such that dC (v) = 1 and ∪ v∈ / 1≤h≤m V (Hh ), then there exist two edges in F (C) incident with v. Proof. Since v ∈ /

∪

1≤h≤m V

(Hh ), dG′ (v) = 3. Hence the number of edges in F (C)

incident with v is equal to dG′ (v) − dC (v) = 3 − 1 = 2.

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Claim 2.6 For an index h (1 ≤ h ≤ m), the following hold: (i) For i ∈ {1, 2} and j ∈ {2, 6}, if G′ − S has a component consisting of yhi,j , then the component C of G′ − S containing yh3−i,8−j satisfies |F (C)| ≥ 4. 8

(ii) For i ∈ {1, 2}, G′ − S has at most one component consisting of one vertex in {yhi,2 , yh3−i,6 }. In particular, the number of components of G′ − S consisting of one vertex in {yhi,j : i ∈ {1, 2}, j ∈ {2, 6}} is at most two. (iii) If G′ − S has two components consisting of one vertex in {yhi,j : i ∈ {1, 2}, j ∈ {2, 6}}, then one of the following holds: (a) there exist two components C1 , C2 of G′ − S such that h ∈ I(Ci ) and |F (Ci )| ≥ 4 for each i ∈ {1, 2}, or (b) there exists a component C of G′ − S such that h ∈ I(C) and |F (C)| ≥ 5. Proof. (i) By the symmetry of {yhi,j : i ∈ {1, 2}, j ∈ {2, 6}}, we may assume that G′ −S has a component consisting of yh1,2 . This implies that yh1,3 , yh1,1 ∈ S. It follows from Claim 2.2(ii) that yh2,6 , yh2,7 ∈ / S. Thus yh2,6 and yh2,7 belong to a same component C of G′ −S. We show that satisfies |F (C)| ≥ 4. By Claim 2.3, |F (C)∩E(Hh )| = 2. Thus it suffices to show that |F (C)| ≥ 4 or |F (C) − E(Hh )| ≥ 2. If |I(C)| ≥ 2, then by Claim 2.4, |F (C)| ≥ 10, as desired. Thus we may assume that I(C) = {h}. We first suppose that C is a tree. If C has a leaf v in V (C) − V (Hh ), then by Claim 2.5, v is incident with two edges in F (C) − E(Hh ), as desired. Thus we may assume that all leaves of C are in V (Hh ). Then C is equal to the path Hh [V (C)]. Since yh2,6 , yh2,7 ∈ V (C) and C is an odd component of G′ − S by Claim 2.2(iii), |V (C)| ≥ 3, and so yh2,5 ∈ V (C). Then yh2,j is incident with an edge in F (C) − E(Hh ) for each j ∈ {5, 7}, and hence |F (C) − E(Hh )| ≥ 2, as desired. Thus we may assume that C is not a tree, and hence C contains a cycle C ∗ . Recall that m ≥ 3. Let h′ be an index with 1 ≤ h′ ≤ m and h′ ̸= h, and let C1∗ be a cycle contained in G[V (Hh′ ) ∪ V (eh′ )]. Now we construct an edge-cut F of G having size |F (C)| + 1 and separating C ∗ and C1∗ . If yh2,2 ∈ / V (C), let F = F (C) ∪ {x2h yh2,6 }; if yh2,2 ∈ V (C) and yh1,6 ∈ / V (C), let F = F (C) ∪ {x1h x2h }; if yh2,2 , yh1,6 ∈ V (C), let F = F (C) ∪ {x1h yh1,2 }. Note that F (C) is an edge-cut of G′ and I(C) = {h}. This together with Claim 2.3 implies that F separates C ∗ and C1∗ in G. Since G is cyclically 5-edge-connected by Theorem D, this implies that |F (C)| + 1 = |F | ≥ 5, and so |F (C)| ≥ 4, as desired. (ii) It follows from (i) that (ii) holds.

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(iii) By the symmetry of {yhi,j : i ∈ {1, 2}, j ∈ {2, 6}}, we may assume that G′ − S has a component consisting of yh1,2 . Then by (i), the component C of G′ − S containing yh2,6 satisfies |F (C)| ≥ 4 and G′ − S has a component consisting of one vertex in {yh1,6 , yh2,2 }. Suppose that G′ − S has a component consisting of yh2,2 . Then by (i), the component C0 of G′ − S containing yh1,6 satisfies |F (C0 )| ≥ 4. By Claim 2.3, we have C ̸= C0 , and hence (iii-a) holds. Thus we may assume that G′ − S has a component consisting of yh1,6 . By (i), the component C1 of G′ − S containing yh2,2 satisfies |F (C1 )| ≥ 4. If C ̸= C1 , then we obtain (iii-a). Thus we may assume that C = C1 . By Claim 2.3, V (C) ∩ V (Hh ) = {yh2,j : 1 ≤ j ≤ 7}. We show that C satisfies (iii-b). Since yh1,1 yh2,7 , yh2,1 yh1,7 ∈ F (C) ∩ E(Hh ), it suffices to show that |F (C)| ≥ 5 or |F (C) − E(Hh )| ≥ 3. If |I(C)| ≥ 2, then by Claim 2.4, |F (C)| ≥ 10, as desired. Thus we may assume that I(C) = {h}. Suppose that C is a tree. For each j ∈ {3, 4, 5}, let C j be the component of C − (V (Hh ) − {yh2,j }) containing yh2,j . If |V (C j )| = 1, then yh2,j is incident with an edge in F (C) − E(Hh ); if |V (C j )| ≥ 2, then C has a leaf v in V (C j ) − {yh2,j }, and hence it follows from Claim 2.5 that v is incident with an edge in F (C). In either case, there exists an edge in F (C) − E(Hh ) joining V (C j ) and S. Since ′

V (C j ) ∩ V (C j ) = ∅ for j ̸= j ′ , we have |F (C) − E(Hh )| ≥ 3, as desired. Thus we may assume that C is not a tree, and hence C contains a cycle C ∗ . Recall that m ≥ 3. Let h′ be an index with 1 ≤ h′ ≤ m and h′ ̸= h, and let C1∗ be a cycle contained in G[V (Hh′ ) ∪ V (eh′ )]. Let F = F (C) ∪ {x1h x2h }. Since F (C) is an edge-cut of G′ and I(C) = {h}, F separates C ∗ and C1∗ in G. Since G is cyclically 5-edge-connected by Theorem D, this implies that |F (C)| + 1 = |F | ≥ 5, and so |F (C)| ≥ 4. Suppose that |F (C)| = 4. Note that dG′ (yh2,2 ) = dG′ (yi2,6 ) = 2. Furthermore, since I(C) = {h}, dG′ (v) = 3 for all v ∈ V (C) − {yh2,2 , yh2,6 }. Hence ∑ ∑ dG′ (v) − |F (C)| = (3|V (C)| − 2) − 4 = 3|V (C)| − 6. dC (v) = v∈V (C)

v∈V (C)

It follows from Claim 2.2(iii) that |V (C)| is odd. Hence

∑

v∈V (C) dC (v)

is also

odd, which contradicts the shake hands lemma. Consequently |F (C)| ≥ 5.

□

Claim 2.7 Let C be a component of G′ − S with |F (C)| ≤ 2. Then C consists of one vertex in {yhi,j : 1 ≤ h ≤ m, i ∈ {1, 2}, j ∈ {2, 6}} and |F (C)| = 2. 10

Proof. Since |F (C)| ≤ 2, it follows from Claim 2.4 that |I(C)| ≤ 1. If I(C) = ∅, then F (C) is an edge-cut of G of size at most 2, which contradicts the fact that G is 3-connected. Thus |I(C)| = 1. Write I(C) = {h}. Since |F (C)| ≤ 2, it follows from Claim 2.3 that |F (C)| = 2 and F (C) ⊆ E(Hh ). Suppose that V (C) − {yhi,j : i ∈ {1, 2}, j ∈ {2, 6}} ̸= ∅. Then there exists a vertex y ∈ V (C)∩(V (Hh )−{yhi,j : i ∈ {1, 2}, j ∈ {2, 6}}). Choose u ∈ NHh (V (C))∩S and y so that dHh (y, u) is as small as possible. Then we see that 1 ≤ dHh (y, u) ≤ 2. Let z be the unique vertex in NG′ (y) − V (Hh ). Then there exists a face f of G′ incident with all of y, z and u. Let D be the facial cycle of f . Note that u ∈ V (D)∩S. In particular, V (D)∩S ̸= ∅. Then there exists an edge in E(D)−E(Hh ) joining V (C) and S, which contradicts the fact that F (C) ⊆ E(Hh ). Thus V (C) ⊆ {yhi,j : i ∈ {1, 2}, j ∈ {2, 6}}. This implies that C consists of one vertex in {yhi,j : i ∈ {1, 2}, j ∈ {2, 6}}.

□

Let C∗ = {C ∈ C(S) : I(C) ̸= ∅}, and for h (1 ≤ h ≤ m), let C∗h = {C ∈ C∗ : h ∈ I(C)}. Claim 2.8 For h (1 ≤ h ≤ m),

∑

|F (C)| C∈C∗h |I(C)|

≥ 3|C∗h |.

Proof. We first suppose that there exists C0 ∈ C∗h such that |I(C0 )| ≥ 2. Then by |F (C0 )| |I(C0 )|

Claim 2.4,

≥

∑ |F (C)| = |I(C)| ∗

C∈Ch

5|I(C0 )| |I(C0 )|

= 5. This together with Claims 2.6 and 2.7 implies that

∑ C∈C∗h −{C0 }

|F (C)| |F (C0 )| + ≥ (3(|C∗h | − 1) − 2) + 5 = 3|C∗h |, |I(C)| |I(C0 )|

as desired. Thus we may assume that I(C) = {h} for all C ∈ C∗h . For a ≥ 1, let C∗h,a = {C ∈ C∗h : |F (C)| = a}. Then by Claims 2.6 and 2.7, we can easily verify that ∑

|F (C)| +

C∈C∗h,2

∑ a≥4

 

∑





|F (C)| ≥ 3 |C∗h,2 | +

∗ C∈Ch,a

∑

 |C∗h,a | = 3(|C∗h | − |C∗h,3 |).

a≥4

Consequently,     ∑ ∑ ∑ |F (C)| ∑  |C∗h,a | + 3|C∗h,3 | = 3|C∗h |, = |F (C)| ≥ 3 |C∗h,2 | + |I(C)| ∗ ∗ a≥2

C∈Ch

as desired.

a≥4

C∈Ch,a

□

Let A = {(C, h) : C ∈ C∗ , h ∈ I(C)}. Then by Claim 2.8,   ∑ |F (C)| ∑ ∑ ∑ |F (C)| ≥  = 3|C∗h | ≥ 3|C∗ |. |I(C)| |I(C)| ∗ (C,h)∈A

1≤h≤m

1≤h≤m

C∈Ch

11

(2.7)

On the other hand, ∑ (C,h)∈A

  ∑ ∑ |F (C)| ∑ |F (C)|  = = |F (C)|. |I(C)| |I(C)| ∗ ∗ C∈C

(2.8)

C∈C

h∈I(C)

By (2.7) and (2.8), we obtain ∑

|F (C)| ≥ 3|C∗ |.

(2.9)

C∈C∗

For C ∈ C(S)−C∗ , it follows from Claim 2.7 that |F (C)| ≥ 3. Thus 3(|C(S)| − ∑

|C∗ |).

|F (C)| =

C∈C(S)

∑

C∈C(S)−C∗

|F (C)| ≥

This together with (2.9) implies that ∑

|F (C)| +

C∈C∗

∑

|F (C)| ≥ 3|C∗ | + 3(|C(S)| − |C∗ |) = 3|C(S)|.

C∈C(S)−C∗

Furthermore, by Claim 2.2(ii), ∑ C∈C(S)

|F (C)| =

∑

|C({u})| =

u∈S

∑

3 = 3|S|,

u∈S

and hence 3|S| ≥ 3|C(S)|, which contradicts Claim 2.2(i). This completes the proof of Theorem 1.1.

Acknowledgment This work was supported by JSPS KAKENHI Grant number 26800086 (to M.F), JSPS KAKENHI Grant number JP16K17646 (to S.T) and research grant of Senshu University (to S.T).

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