Forbidden subgraphs generating almost all claw-free graphs with high connectivity Michitaka Furuya1∗ Maho Yokota2† 1

College of Liberal Arts and Science, Kitasato University,

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

Department of Applied Mathematic, Tokyo University of Science,

1-3 Kagurazaka, Shinjuku-ku, Tokyo 162-8601, Japan

Abstract For a family H of connected graphs and an integer k ≥ 1, let Gk (H) denote the family of k-connected graphs which contain no element of H as an induced subgraph. Let H+ be the family of those connected graphs of order 5 which contain K1,3 as an induced subgraph. In this paper, for each integer k ≥ 1, we characterize the families H ⊆ H+ such that the symmetric difference of Gk ({K1,3 }) and Gk (H) is finite.

Key words and phrases. claw-free graph, forbidden subgraph, Matthews-Sumner conjecture. AMS 2010 Mathematics Subject Classification. 05C75, 05C38.

1

Introduction

For two graphs G and H, we write H ≺ G if G contains H as an induced subgraph. For a family H of connected graphs, a graph G is said to be H-free if G contains no element of H as an induced subgraph; that is to say, H ̸≺ G for all H ∈ H. For an integer k ≥ 1, let Gk (H) be the family of k-connected H-free graphs. If H = {H}, a graph G is simply said to be H-free instead of saying that G is H-free, and we let Gk (H) := Gk (H). We let Km1 ,m2 denote the complete bipartite graph with partite ∗ †

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

1

Figure 1: Graph A

sets having cardinalities m1 and m2 . For an integer m ≥ 2, the vertex of K1,m having degree m is called the center of K1,m . A graph isomorphic to K1,3 is called a claw. In this paper, we take up a topic discussed by Fujisawa, Ota, Ozeki and Sueiro in [9], and study families H of connected graphs such that the families Gk ({K1,3 }) and Gk (H) are almost equal in the sense that their symmetric difference Gk ({K1,3 }) △ Gk (H) is a finite family. Our motivation comes from the following well-known conjecture posed by Matthews and Sumner. Conjecture 1 (Matthews and Sumner [16]) Every 4-connected claw-free graph is Hamiltonian. Conjecture 1 is still open, and Ryj´aˇcek [18] proved that Conjecture 1 is equivalent to Thomassen’s conjecture [20]: Every 4-connected line graph is Hamiltonian. This fact has motivated many researchers to give other conjectures equivalent to Conjecture 1 (for example, see [1, 4, 7, 11, 12, 19] or a survey [3]). On the other hand, there is another approach to Conjecture 1 focusing on forbidden subgraph conditions. Broersma, Kriesell and Ryj´aˇcek [2] proved that every 4-connected {claw, hourglass}-free graph is Hamiltonian, and Pfender [17] refined the result and proved that every 4-connected {claw, A}-free graph is Hamiltonian, where A is the graph depicted in Figure 1. Furthermore, various families of forbidden subgraphs are known to ensure the existence of a Hamiltonian cycle in 3-connected graphs (see [8, 10, 13, 14, 15]). However, when we carelessly choose a forbidden subgraph condition, it sometimes happens that the forbidden subgraph condition is almost equivalent to claw-freeness. Recently, based on the fact mentioned at the end of the preceding paragraph, Fujisawa, Ota, Ozeki and Sueiro [9] completely characterized the families H of connected graphs such that every H-free graph is also claw-free. However, some of the families H described in [9] have redundancy when we confine ourselves to graphs with high connectivity; that is to say, for some k ≥ 2, there exists a member H of H such that Gk (K1,3 ) △ Gk (H − {H}) is finite. In this paper, we give families with no redundancy. To simplify the problem, we restrict the order of forbidden subgraphs. + For a connected graph K, let HK be the family of connected graphs H with K ≺ H

and |V (H)| = |V (K)| + 1. We start with an easy lemma. 2

+ )⊆ Lemma 1.1 For an integer k ≥ 1 and a connected graph K, Gk (K) △ Gk (HK

{K}. + Proof. Since all graphs in HK contain K as an induced subgraph, every K-free + + graph is also HK -free, and hence Gk (K) − Gk (HK ) = ∅. Thus it suffices to show + that Gk (HK ) − Gk (K) ⊆ {K}. By way of contradiction, suppose that there exists + a graph G ∈ Gk (HK ) − Gk (K) with G ̸= K. Let X be a subset of V (G) inducing

K in G. Since G is connected and G ̸= K, there exists a vertex u ∈ V (G) − X + with NG (u) ∩ X ̸= ∅. Then by the definition of HK , the subgraph of G induced by + + X ∪ {u} belongs to HK , which contradicts the fact that G is HK -free.

□

By Lemma 1.1, we can restate Conjecture 1. Lemma 1.2 Conjecture 1 is equivalent to the following: + (C0) Every 4-connected HK -free graph is Hamiltonian. 1,3

The assumption of (C0) has redundancy. Indeed, by using an implicit result in [9] (or Theorem 1.3(i) in this paper), we see that the statement that every 4-connected + (HK − {K2,3 })-free graph is Hamiltonian is equivalent to Conjecture 1. Further, 1,3 + in view of the 4-connectedness of the graph under consideration, (HK − {K2,3 })1,3

freeness still has redundancy (see Corollary 1.4 below). + We now proceed to state the main theorem. In this paper, the family HK 1,3 + frequently appears. Thus, for convenience, we use the notation H+ instead of HK . 1,3

Let H1 , . . . , H7 be the graphs depicted in Figure 2. We can easily show that H+ = {Hi : 1 ≤ i ≤ 7} as follows. We clearly have H+ ⊇ {Hi : 1 ≤ i ≤ 7}. Let H ∈ H+ . Then H is obtained from K1,3 by adding one vertex u and joining u and K1,3 with l edges (1 ≤ l ≤ 4). If l = 1, then H is isomorphic to either H3 or H5 ; if l = 2, then H is isomorphic to either H4 or H6 ; if l = 3, then H is isomorphic to either H2 or H7 ; if l = 4, then H is isomorphic to H1 . Thus H+ = {Hi : 1 ≤ i ≤ 7}. Our main result is the following. Theorem 1.3 For H ⊆ H+ , the following hold. (i) The family G1 (K1,3 ) △ G1 (H) is finite if and only if {H1 , H2 , H3 , H4 , H5 , H6 } ⊆ H. (ii) The family G2 (K1,3 ) △ G2 (H) is finite if and only if either {H1 , H2 , H3 , H4 , H7 } or {H1 , H2 , H3 , H4 , H5 , H6 } is a subset of H. (iii) The family G3 (K1,3 ) △ G3 (H) is finite if and only if either {H1 , H2 , H3 , H4 , H5 } or {H1 , H2 , H3 , H4 , H7 } is a subset of H. 3

H1

H2

H5 (= K1,4 )

H4

H3

H6

H7 (= K2,3 )

Figure 2: The 7 elements of H+

(iv) For an integer k ≥ 4, the family Gk (K1,3 ) △ Gk (H) is finite if and only if one of {H1 , H2 , H5 , H6 }, {H1 , H2 , H3 , H6 , H7 }, {H1 , H2 , H3 , H4 , H5 } and {H1 , H2 , H3 , H4 , H7 } is a subset of H. Remark 1 As we mentioned above, Theorem 1.3(i) is implicitly stated in [9]. However, to keep the paper self-contained, we include the proof of Theorem 1.3(i). In the course of the proof of Theorem 1.3, we show that for k ≥ 4, if Gk (K1,3 ) △ Gk (H) is finite, then Gk (K1,3 ) △ Gk (H) = ∅. Thus we obtain the following as a corollary of Theorem 1.3. Corollary 1.4 Conjecture 1 is equivalent to all of the following: (C1) Every 4-connected {H1 , H2 , H5 , H6 }-free graph is Hamiltonian. (C2) Every 4-connected {H1 , H2 , H3 , H6 , H7 }-free graph is Hamiltonian. (C3) Every 4-connected {H1 , H2 , H3 , H4 , H5 }-free graph is Hamiltonian. (C4) Every 4-connected {H1 , H2 , H3 , H4 , H7 }-free graph is Hamiltonian. The assumptions of (C1)–(C4) are more complicated than claw-freeness. Thus we cannot expect that conjectures (C1)–(C4) will directly help us to solve Conjecture 1. However, those conjectures might be useful in judging whether or not a given forbidden subgraph condition is a good one as an approach to Conjecture 1. We will use the following notation and terminology. Let G be a graph. For x ∈ V (G), we let NG (x) denote the neighborhood of x in G. We let G denote the compliment of G. For X ⊆ V (G), we let G[X] denote the subgraph of G induced by X. For two graphs H1 and H2 , we let H1 + H2 denote the join of H1 and H2 . For

4

x

x

y

y +

z

w

Qz

+ +

Qw

Gex (H, {z, w}; p)

H

Figure 3: Construction of Gex (H, X; p)

an integer n ≥ 1, Kn denotes the complete graph of order n. For terms and symbols not defined here, we refer the reader to [6].

2

Necessary conditions for |Gk (H) − Gk (K1,3 )| < ∞

In this section, we give some necessary conditions for |Gk (H) − Gk (K1,3 )| < ∞. We use the following lemma. Lemma 2.1 (Wormald [21] and Egawa [5]) For integers k ≥ 2 and g ≥ 3, there exist infinitely many k-connected k-regular graphs with girth at least g. Let H be a graph, and let X ⊆ V (H). For an integer p ≥ 1, let Gex (H, X; p), called the p-expansion of H with respect to X, be the graph obtained from H − X by adding |X| complete graphs Qx (x ∈ X) of order p and the edge set {uv : u ∈ V (Qx ), v ∈ NH (x) ∩ (V (H) − X), x ∈ X} ∪ {uu′ : u ∈ V (Qx ), u′ ∈ V (Qx′ ), x, x′ ∈ X with xx′ ∈ E(H)} (see Figure 3). We show the following lemma. Lemma 2.2 For an integer k ≥ 1 and a subfamily H of H+ , suppose that Gk (H) − Gk (K1,3 ) is finite. Then the following hold. (a) We have {H1 , H2 } ⊆ H. (b) At least one of {H5 , H6 }, {H3 , H6 , H7 }, {H3 , H4 , H5 } and {H3 , H4 , H7 } is a subset of H. (c) If k ≤ 3, then {H3 , H4 } ⊆ H. (d) If k ≤ 2, then H6 ∈ H or H7 ∈ H. (e) If k = 1, then {H5 , H6 } ⊆ H. 5

u0 v2

v3 v1 x1

x2

Figure 4: Graph G∗1,2

Proof. Since Gk (H) − Gk (K1,3 ) is finite, there exists a positive integer n = n(k, H) such that every graph in Gk (H)−Gk (K1,3 ) has at most n vertices. Set p = max{k, n, 3}. (a) Let G1,1 = Kp + K3 . Then G1,1 is a k-connected graph of order p + 3 (> n) and K1,3 ≺ G1,1 . Furthermore, each induced subgraph of G1,1 isomorphic to K1,3 contains all vertices of K3 and exactly one vertex of Kp as the center. This implies that all induced subgraphs of G1,1 belonging to H+ are isomorphic to H1 . Consequently G1,1 ∈ Gk (H − {H1 }) − Gk (K1,3 ). By the definition of n, it follows that H1 ∈ H. Let G∗1,2 be the graph depicted in Figure 4, and let G1,2 = Gex (G∗1,2 , {x1 , x2 }; p). Then G1,2 is a k-connected graph of order 2p + 4 (> n). Furthermore, the subgraph of G1,2 induced by {u0 , v1 , v2 , v3 } is the unique induced subgraph of G1,2 isomorphic to K1,3 . This implies that all induced subgraphs of G1,2 belonging to H+ are isomorphic to H2 . Consequently G1,2 ∈ Gk (H − {H2 }) − Gk (K1,3 ). By the definition of n, it follows that H2 ∈ H. (b) It follows from Lemma 2.1 that there exists a p-connected p-regular graph G2,1 with girth at least 5 having at least n + 1 vertices. Note that K1,3 ≺ G2,1 . Furthermore, since G2,1 has no cycle of order 3 or 4, no induced subgraph of G2,1 belonging to H+ has a cycle. This implies that each induced subgraph of G2,1 belonging to H+ is isomorphic to either H3 or H5 . Consequently G2,1 ∈ Gk (H − {H3 , H5 }) − Gk (K1,3 ). By the definition of n, it follows that H3 ∈ H or H5 ∈ H.

(2.1)

Let G2,2 = Kp,p . Then G2,2 is a k-connected graph of order 2p (> n) and K1,3 ≺ G2,2 . Furthermore, each induced subgraph of G2,2 isomorphic to K1,3 contains three vertices in a partite set and exactly one vertex in the other partite set. This implies that each induced subgraph of G2,2 belonging to H+ is isomorphic to either H5 or H7 . Consequently G2,2 ∈ Gk (H − {H5 , H7 }) −

6

Q1

Q2

Q3

yp

y1 y2

Figure 5: Graph G∗2,3

Gk (K1,3 ). By the definition of n, it follows that H5 ∈ H or H7 ∈ H.

(2.2)

For i (1 ≤ i ≤ 3), let Qi be a complete graph of order p, and write V (Qi ) = {xij : 1 ≤ j ≤ p}. Let G∗2,3 be the graph obtained from Q1 ∪ Q2 ∪ Q3 by adding p vertices y1 , . . . , yp and the edge set {xij yj : 1 ≤ i ≤ 3, 1 ≤ j ≤ p} (see Figure 5). Let G2,3 = Gex (G∗2,3 , V (Q1 ) ∪ V (Q2 ) ∪ V (Q3 ); p). Then G2,3 is a k-connected graph of order 3p2 + p (> n) and K1,3 ≺ G2,3 . Furthermore, each induced subgraph of G2,3 isomorphic to K1,3 contains exactly one vertex in {yj : 1 ≤ j ≤ p} as the center. This implies that each induced subgraph of G2,3 belonging to H+ is isomorphic to either H3 or H6 . Consequently G2,3 ∈ Gk (H − {H3 , H6 }) − Gk (K1,3 ). By the definition of n, it follows that H3 ∈ H or H6 ∈ H.

(2.3)

Let G∗2,4 be the graph depicted in Figure 6, and let G2,4 = Gex (G∗2,4 , {x1 , . . . , x6 }; p). Then G2,4 is a k-connected graph of order 6p + 1 (> n) and K1,3 ≺ G2,4 . Furthermore, each induced subgraph of G2,4 isomorphic to K1,3 contains x0 as the center. This implies that each induced subgraph of G2,4 belonging to H+ is isomorphic to either H4 or H6 . Consequently G2,4 ∈ Gk (H−{H4 , H6 })−Gk (K1,3 ). By the definition of n, it follows that H4 ∈ H or H6 ∈ H.

(2.4)

Now if H5 , H6 ∈ H, then clearly {H5 , H6 } ⊆ H; if H5 ∈ / H and H6 ∈ H, then it follows from (2.1) and (2.2) that {H3 , H6 , H7 } ⊆ H; if H5 ∈ H and H6 ∈ / H, then it follows from (2.3) and (2.4) that {H3 , H4 , H5 } ⊆ H; if H5 , H6 ∈ / H, then it follows from (2.1), (2.2) and (2.4) that {H3 , H4 , H7 } ⊆ H. In any case, we obtain the desired conclusion.

7

x1 x2

x6 x0

x3

x5 x4

Figure 6: Graph G∗2,4

(c) It follows from Lemma 2.1 that there exists a 3-connected 3-regular graph G3,1 with girth at least 5 having at least n + 1 vertices. Note that K1,3 ≺ G3,1 and H5 ̸≺ G3,1 . Furthermore, since G3,1 has no cycle of order 3 or 4, no induced subgraph of G3,1 belonging to H+ has a cycle. This implies that all induced subgraphs of G3,1 belonging to H+ are isomorphic to H3 . Consequently G3,1 ∈ Gk (H − {H3 }) − Gk (K1,3 ). By the definition of n, it follows that H3 ∈ H. Let again G∗2,4 be the graph depicted in Figure 6, and let G3,2 = Gex (G∗2,4 , {x2 }; n). Then G3,2 is a 3-connected graph of order n + 6 and K1,3 ≺ G3,2 . Furthermore, each induced subgraph of G3,2 isomorphic to K1,3 contains x0 as the center. This implies that all induced subgraphs of G3,2 belonging to H+ are isomorphic to H4 . Consequently G3,2 ∈ Gk (H − {H4 }) − Gk (K1,3 ). By the definition of n, it follows that H4 ∈ H. (d) Let ({x1 , x2 }, {y1 , y2 , y3 }) be the bipartition of K2,3 , and let G4,1 = Gex (K2,3 , {y1 }; n). Then G4,1 is a 2-connected graph of order n + 4 and K1,3 ≺ G4,1 . Furthermore, each induced subgraph of G4,1 isomorphic to K1,3 contains exactly one of x1 and x2 as the center. This implies that each induced subgraph of G4,1 belonging to H+ is isomorphic to either H6 or H7 . Consequently G4,1 ∈ Gk (H − {H6 , H7 }) − Gk (K1,3 ). By the definition of n, it follows that H6 ∈ H or H7 ∈ H. (e) Let G5,1 = K1,n . Then G5,1 is a connected graph of order n + 1, and clearly all induced subgraphs of G5,1 belonging to H+ are isomorphic to H5 . Consequently G5,1 ∈ G1 (H − {H5 }) − G1 (K1,3 ). By the definition of n, it follows that H5 ∈ H. Let ({x}, {y1 , y2 , y3 }) be the bipartition of K1,3 , and let G5,2 = Gex (K1,3 , {y1 }; n). Then G5,2 is a connected graph of order n + 3 and K1,3 ≺ G5,2 . Furthermore, each induced subgraph of G5,2 isomorphic to K1,3 contains x as the center. This implies that all induced subgraphs of G5,2 belonging to H+ are isomorphic to H6 . Consequently G5,2 ∈ G1 (H − {H6 }) − G1 (K1,3 ). By the definition 8

of n, it follows that H6 ∈ H.

3

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Proof of Theorem 1.3

We first prove a technical lemma. Lemma 3.1 Let k ≥ 1 be an integer, and let H ⊆ H+ with {H1 , H2 , H3 , H4 , H5 } ⊆ H. Let G ∈ Gk (H), and let m1 ≥ 2 and m2 ≥ 3 be integers, and suppose that there exist disjoint subsets X 1 and X 2 of V (G) with |X i | = mi (i ∈ {1, 2}) such that G[X 1 ∪ X 2 ] is isomorphic to Km1 ,m2 with bipartition (X 1 , X 2 ). Then one of the following holds: (a) G ∈ {K2,3 , K3,3 }; or (b) m1 = 2, m2 = 3 and every vertex u ∈ V (G) − (X 1 ∪ X 2 ) with NG (u) ∩ (X 1 ∪ X 2 ) ̸= ∅ satisfies X 1 ⊆ NG (u) and |NG (u) ∩ X 2 | = 1. Proof. We choose m1 and m2 so that m1 + m2 is as large as possible. It suffices to show that (a) or (b) holds under this condition. Write X i = {xij : 1 ≤ j ≤ mi } for i ∈ {1, 2}. Since G is H5 -free, we have m1 ∈ {2, 3} and m2 = 3. Since G is connected, if there exists no vertex in V (G) − (X 1 ∪ X 2 ) adjacent to a vertex in X 1 ∪X 2 , then (a) holds. Thus we may assume that there exists u ∈ V (G)−(X 1 ∪X 2 ) with NG (u) ∩ (X 1 ∪ X 2 ) ̸= ∅. If NG (u) ∩ X 2 = ∅, then NG (u) ∩ X 1 ̸= ∅, and hence {x1j , x21 , x22 , x23 , u} induces H5 in G where x1j ∈ NG (u) ∩ X 1 , which is a contradiction. Thus NG (u) ∩ X 2 ̸= ∅. We may assume that ux21 ∈ E(G). Suppose that NG (u)∩X 1 = ∅. If X 2 ̸⊆ NG (u), say ux22 ∈ / E(G), then {x11 , x21 , x22 , x23 , u} induces H4 or H3 in G according as ux23 ∈ E(G) or not, which is a contradiction. Thus X 2 ⊆ NG (u). This implies that X 1 ∪ X 2 ∪ {u} induces Km1 +1,m2 in G, which contradicts the choice of m1 and m2 . Thus NG (u) ∩ X 1 ̸= ∅. We may assume that ux11 ∈ E(G). If |NG (u) ∩ X 2 | ≥ 2, say ux22 ∈ E(G), then {x11 , x21 , x22 , x23 , u} induces H1 or H2 in G according as ux23 ∈ E(G) or not, which is a contradiction. Thus |NG (u) ∩ X 2 | = 1 (i.e., NG (u) ∩ X 2 = {x21 }). For i (2 ≤ i ≤ m1 ), since {x1i , x21 , x22 , x23 , u} does not induce H3 in G, it follows that ux1i ∈ E(G). Hence X 1 ⊆ NG (u). Recall that m1 ∈ {2, 3} and m2 = 3. If m1 = 3, then {x21 , x11 , x12 , x13 , u} induces H1 in G, which is a contradiction. Therefore m1 = 2, and (b) holds.

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We now prove lemmas concerning sufficient conditions. Lemma 3.2 If {H1 , H2 , H3 , H4 , H5 , H6 } ⊆ H, then G1 (H)−G1 (K1,3 ) ⊆ {K1,3 , K2,3 , K3,3 }. 9

y1

y2

z1

x

z2

Figure 7: Labels for each vertex of G[X] (≃ H6 )

Proof. Suppose that there exists G ∈ G1 (H)−G1 (K1,3 ) with G ∈ / {K1,3 , K2,3 , K3,3 }. If H7 ̸≺ G (i.e., G ∈ G1 (H+ ) − G1 (K1,3 )), then by Lemma 1.1, G = K1,3 , which is a contradiction. Thus H7 ≺ G. In particular, Km1 ,m2 ≺ G for some m1 ≥ 2 and m2 ≥ 3. Let X 1 and X 2 be subsets of V (G) with |X i | = mi (i ∈ {1, 2}) such that G[X 1 ∪ X 2 ] is isomorphic to Km1 ,m2 with bipartition (X 1 , X 2 ). Since G∈ / {K2,3 , K3,3 }, it follows from Lemma 3.1 that m1 = 2 and m3 = 3. Again since G ̸= K2,3 , there exists a vertex u ∈ V (G) − (X 1 ∪ X 2 ) with NG (u) ∩ (X 1 ∪ X 2 ) ̸= ∅. Then it follows from Lemma 3.1 that X 1 ⊆ NG (u) and |NG (u) ∩ X 2 | = 1, and hence {x1 , u} ∪ X 2 induces H6 in G where x1 ∈ X 1 , which is a contradiction.

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Lemma 3.3 If {H1 , H2 , H3 , H4 , H7 } ⊆ H, then G2 (H) − G2 (K1,3 ) = ∅. Proof. Suppose that there exists G ∈ G2 (H) − G2 (K1,3 ). If H5 ̸≺ G and H6 ̸≺ G (i.e., G ∈ G2 (H+ ) − G2 (K1,3 )), then by Lemma 1.1, G = K1,3 , which contradicts the fact that G is 2-connected. Thus either H5 ≺ G or H6 ≺ G. Suppose that H6 ≺ G. Let X be a subset of V (G) such that G[X] ≃ H6 . We label each vertex of G[X] as in Figure 7. Since G is 2-connected, there exists a path P of G − x joining {y1 , y2 } and {z1 , z2 }. We choose X and P so that the length of P is as small as possible. Write P = u0 u1 · · · um , where u0 ∈ {y1 , y2 } and um ∈ {z1 , z2 }. We may assume that u0 = y1 . Note that u1 ∈ / X. If u1 x ∈ / E(G) and u1 y2 ∈ E(G), then {x, y1 , y2 , z1 , u1 } induces H7 or H4 in G according as u1 z1 ∈ E(G) or not; if u1 x, u1 y2 ∈ / E(G), then {x, y1 , y2 , z1 , u1 } induces H4 or H3 in G according as u1 z1 ∈ E(G) or not. In either case, we obtain a contradiction. Thus u1 x ∈ E(G). If u1 y2 ∈ E(G), then {x, y1 , y2 , z1 , u1 } induces H1 or H2 in G according as u1 z1 ∈ E(G) or not, which is a contradiction. Thus u1 y2 ∈ / E(G). If u1 zi ∈ E(G) for some i ∈ {1, 2}, then {x, y1 , y2 , zi , u1 } induces H2 in G, which is a contradiction. Thus u1 z1 , u1 z2 ∈ / E(G). Consequently {x, u1 , y2 , z1 , z2 } induces H6 in G and P − y1 (= u1 · · · um ) is a path of G − x with length m − 1 joining u1 and z1 , which contradicts the choice of X and P . Therefore H6 ̸≺ G, and hence H5 ≺ G. Let X ′ be a subset of V (G) such that G[X ′ ] ≃ H5 (= K1,4 ). Let x′0 be the center of G[X ′ ] and write X ′ = {x′0 , x′1 , . . . , x′4 }. Since G is 2-connected, there exists a 10

vertex u′ ∈ NG (x′1 ) − {x′0 }. Recall that G ∈ Gk (H ∪ {H6 }). If u′ x′0 ∈ / E(G), then {x′0 , x′1 , x′2 , x′3 , u′ } induces H3 , H4 or H7 in G according as |NG (u′ ) ∩ {x′2 , x′3 }| = 0, 1 or 2; if u′ x′0 ∈ E(G), then {x′0 , x′1 , x′2 , x′3 , u′ } induces H6 , H2 or H1 in G according as |NG (u′ ) ∩ {x′2 , x′3 }| = 0, 1 or 2. In either case, we obtain a contradiction.

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Lemma 3.4 If {H1 , H2 , H3 , H4 , H5 } ⊆ H, then G3 (H) − G3 (K1,3 ) ⊆ {K3,3 }. Proof. Suppose that G3 (H) − G3 (K1,3 ) ̸⊆ {K3,3 }, and let G ∈ G3 (H) − G3 (K1,3 ) with G ̸= K3,3 . If H7 ̸≺ G, then it follows from Lemma 3.3 that G ∈ G3 (H ∪ {H7 }) − G3 (K1,3 ) ⊆ G2 ({H1 , H2 , H3 , H4 , H7 }) − G2 (K1,3 ) = ∅, which is a contradiction. Thus H7 ≺ G. In particular, Km1 ,m2 ≺ G for some m1 ≥ 2 and m2 ≥ 3. Let X 1 and X 2 be subsets of V (G) with |X i | = mi (i ∈ {1, 2}) such that G[X 1 ∪ X 2 ] is isomorphic to Km1 ,m2 with bipartition (X 1 , X 2 ). Since G ̸= K3,3 and G is 3-connected, it follows from Lemma 3.1 that m1 = 2 and m2 = 3. Write X 1 = {x11 , x12 } and X 2 = {x21 , x22 , x23 }. Since G is 3-connected, there exists a path P of G − X 1 joining x21 and {x22 , x23 }. We choose X 1 , X 2 and P so that the length of P is as small as possible. Write P = u0 u1 · · · um , where u0 = x21 and um ∈ {x22 , x23 }. Note that u1 ∈ / X 1 ∪ X 2 . By Lemma 3.1, X 1 ⊆ NG (u1 ) and |NG (u1 ) ∩ X 2 | = 1. In particular, u1 x22 , u1 x23 ∈ / E(G). Consequently X 1 ∪ {u1 , x22 , x23 } induces H7 in G and P − x21 (= u1 · · · um ) is a path of G − X 1 with length m − 1 joining u1 and {x22 , x23 }, which contradicts the choice of X 1 , X 2 and P .

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Lemma 3.5 For k ≥ 4, if {H1 , H2 , H5 , H6 } ⊆ H, then Gk (H) − Gk (K1,3 ) = ∅. Proof. Suppose that there exists G ∈ Gk (H) − Gk (K1,3 ). Then there exists X ⊆ V (G) with G[X] ≃ K1,3 . Let x0 be the center of G[X]. Since G is 4-connected, there exists a vertex u ∈ NG (x0 ) − X. Then X ∪ {u} induces H5 , H6 , H2 or H1 in G according as |NG (u) ∩ (X − {x0 })| = 0, 1, 2 or 3, which is a contradiction.

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Lemma 3.6 For k ≥ 4, if {H1 , H2 , H3 , H6 , H7 } ⊆ H, then Gk (H) − Gk (K1,3 ) = ∅. Proof. Suppose that there exists G ∈ Gk (H) − Gk (K1,3 ). If H5 ̸≺ G (i.e., G ∈ Gk ({H1 , H2 , H5 , H6 }) − Gk (K1,3 )), then Lemma 3.5 leads to a contradiction. Thus H5 ≺ G. Hence there exists X ⊆ V (G) with G[X] ≃ H5 (= K1,4 ). Let x0 be the center of G[X] and write X = {x0 , x1 , x2 , x3 , x4 }. Since G is 4-connected, there exists a vertex u ∈ NG (x1 )−X. If ux0 ∈ E(G), then {x0 , x1 , x2 , x3 , u} induces H6 , H2 or H1 in G according as |NG (u)∩{x2 , x3 }| = 0, 1 or 2; if ux0 ∈ / E(G), then {x0 , x1 , x2 , x3 , u} induces H3 , H4 or H7 in G according as |NG (u) ∩ {x2 , x3 }| = 0, 1 or 2. Since 11

G is {H1 , H2 , H3 , H6 , H7 }-free, this forces ux0 ∈ / E(G) and |NG (u) ∩ {x2 , x3 }| = 1. We may assume that ux2 ∈ E(G) and ux3 ̸∈ E(G). If ux4 ∈ E(G), then {x0 , x1 , x2 , x4 , u} induces H7 in G; if ux4 ∈ / E(G), then {x0 , x1 , x3 , x4 , u} induces H3 in G. In either case, we obtain a contradiction.

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Proof of Theorem 1.3. Note that for an integer k ≥ 1, Gk (K1,3 ) △ Gk (H) is finite if and only if Gk (H) − Gk (K1,3 ) is finite. (i) If G1 (K1,3 ) △ G1 (H) is finite, then it follows from Lemma 2.2(a)(c)(e) that {H1 , H2 , H3 , H4 , H5 , H6 } ⊆ H. Conversely, if {H1 , H2 , H3 , H4 , H5 , H6 } ⊆ H, then it follows from Lemma 3.2 that G1 (K1,3 ) △ G1 (H) is finite. (ii) Suppose that G2 (K1,3 ) △ G2 (H) is finite. If H7 ∈ H, then it follows from Lemma 2.2(a)(c) that {H1 , H2 , H3 , H4 , H7 } ⊆ H; if H7 ∈ / H, then it follows from Lemma 2.2(a)(b)(c)(d) that {H1 , H2 , H3 , H4 , H5 , H6 } ⊆ H. Conversely, if {H1 , H2 , H3 , H4 , H7 } ⊆ H, then it follows from Lemma 3.3 that G2 (K1,3 ) △ G2 (H) is finite; if {H1 , H2 , H3 , H4 , H5 , H6 } ⊆ H, then it follows from (i) that G2 (K1,3 ) △ G2 (H) (⊆ G1 (K1,3 ) △ G1 (H)) is finite. (iii) Suppose that G3 (K1,3 ) △ G3 (H) is finite. If H5 ∈ H, then it follows from Lemma 2.2(a)(c) that {H1 , H2 , H3 , H4 , H5 } ⊆ H. Thus we may assume that H5 ∈ / H. Then Lemma 2.2(b) leads to H7 ∈ H. This together with Lemma 2.2(a)(c) implies that {H1 , H2 , H3 , H4 , H7 } ⊆ H. Conversely, if {H1 , H2 , H3 , H4 , H5 } ⊆ H, then it follows from Lemma 3.4 that G3 (K1,3 )△G3 (H) is finite; if {H1 , H2 , H3 , H4 , H7 } ⊆ H, then it follows from (ii) that G3 (K1,3 ) △ G3 (H) (⊆ G2 (K1,3 ) △ G2 (H)) is finite. (iv) Let k ≥ 4 be an integer. If Gk (K1,3 ) △ Gk (H) is finite, then it follows from Lemma 2.2(a)(b) that {H1 , H2 , H5 , H6 } ⊆ H, {H1 , H2 , H3 , H6 , H7 } ⊆ H, {H1 , H2 , H3 , H4 , H5 } ⊆ H or {H1 , H2 , H3 , H4 , H7 } ⊆ H. Conversely, if {H1 , H2 , H5 , H6 } ⊆ H, then it follows from Lemma 3.5 that Gk (K1,3 )△Gk (H) is finite; if {H1 , H2 , H3 , H6 , H7 } ⊆ H, then it follows from Lemma 3.6 that Gk (K1,3 )△Gk (H) is finite; if {H1 , H2 , H3 , H4 , H5 } ⊆ H or {H1 , H2 , H3 , H4 , H7 } ⊆ H, then it follows from (iii) that Gk (K1,3 ) △ Gk (H) (⊆ G3 (K1,3 )△G3 (H)) is finite. This completes the proof of Theorem 1.3. □

Acknowledgment The authors would like to thank Professor Yoshimi Egawa for his assistance in correction of this paper. This work was supported by JSPS KAKENHI Grant number 12

26800086 (to M.F).

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