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Forbidden subgraphs on Hamiltonian index
Discrete Mathematics 343 (2020) 111841
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Forbidden subgraphs on Hamiltonian index Xia Liu a , Liming Xiong b a b
School of Mathematics and Statistics, Beijing Institute of Technology, Beijing 100081, PR China School of Mathematics and Statistics, Beijing Key Laboratory on MCAACI, Beijing Institute of Technology, Beijing 100081, PR China
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Article history: Received 9 February 2019 Accepted 20 January 2020 Available online xxxx Keywords: Forbidden subgraph Hamiltonian index Collapsible DCT
a b s t r a c t Let G be a graph other than a path. The m-iterated line graph of a graph G is Lm (G) = L(Lm−1 (G)). where L1 (G) denotes the line graph L(G) of G. Define the Hamiltonian index h(G) of G to be the smallest integer m such that Lm (G) contains a Hamiltonian cycle. For a connected graph set H, G is said to be H-free if G does not contain H as an induced subgraph for all H ∈ H. In this paper, we characterize all forbidden graphs H for any integer k ≥ 2 and partial forbidden graphs H for k = 1 such that a connected (or 2-edge-connected or 2-connected) H-free graph G satisfies that h(G) ≤ k for the case when |H| ≤ 2. What is more, we settle four conjectures proposed in Holub (2014). © 2020 Elsevier B.V. All rights reserved.
1. Introduction All graphs considered in this paper are simple graphs. For the notation or terminology not defined here, see [1]. A graph is called trivial if it has only one vertex, nontrivial otherwise. We use κ (G), κ ′ (G) to denote the connectivity, edge connectivity of G, respectively. The line graph L(G) of G = (V (G), E(G)) has E(G) as its vertex set, and two vertices are adjacent in L(G) if and only if the corresponding edges share a common end vertex in G. The m-iterated line graph Lm (G) is defined recursively by L0 (G) = G, L1 (G) = L(G) and Lm (G) = L(Lm−1 (G)). The Hamiltonian index of a graph G, denoted by h(G), is the smallest integer m such that Lm (G) contains a Hamiltonian cycle. For a connected graph G that is not a path, Ryjáček, Woeginger and Xiong [10] showed that the problem to decide whether the Hamiltonian index of a given graph is less than or equal to a given constant is NP-complete, Chartrand [4] showed that the Hamiltonian index of G always exists and Saražin [11] showed h(G) ≤ n − ∆(G) if G is connected. For its upper bounds and stability, see [3], [13] and [15] and its survey paper, see [16]. An edge of G is said to be subdivided when it is deleted and replaced by a path of length 2 connecting its ends, the internal vertex of this path being a new vertex. For r ≥ 3 and i1 ≥ · · · ≥ ir ≥ 0, define Ti1 ,...,ir , θ (i1 , . . . , ir ) be the graphs obtained from K1,r or K2,r by subdividing i1 − 1, . . . , ir − 1 times of the r incident edges of K1,r or K2,r , respectively. Especially, if i1 = · · · = ir = j, then we denote Ti1 ,...,ir = Tj,r and θ (i1 , . . . , ir ) = θ (j, r) for convenience. Let Pn be the path of order n. We shall prove the following result, which is stronger than four conjectures proposed in [9] and considered only their sufficiency. Theorem 1. Let k ≥ 0, l, m, n ≥ 2 be integers and H be a connected graph. Then 2-edge-connected (or 2-connected) graph G does not contain H as its subgraph (not necessarily induced) implying h(G) ≤ k if and only if H is isomorphic to one of {P3k+5 , T1,1,2k+2 } ∪ {T1,m,n : m + n = 3k + 3} ∪ {Tl,m,n : l + m + n = 3k + 4}. E-mail addresses: [email protected] (X. Liu), [email protected] (L. Xiong). https://doi.org/10.1016/j.disc.2020.111841 0012-365X/© 2020 Elsevier B.V. All rights reserved.
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X. Liu and L. Xiong / Discrete Mathematics 343 (2020) 111841
Fig. 1. The common induced subgraphs.
Note that the graph H in above theorem is generally not necessarily induced, it is natural to consider the case that H is an induced subgraph of G. They maybe more popular in the literature. A graph G is said to be H-free if G does not contain H as an induced subgraph for all H ∈ H, and we call each graph H of H a forbidden subgraph. Especially, if {H } = H, then we simply say that G is H-free and we call H a forbidden pair if |H| = 2. For two sets H1 and H2 of connected graphs, we write H1 ⪯ H2 if for every graph H2 in H2 , there is a graph H1 in H1 such that H1 is an induced subgraph of H2 . By the ′ definition of the relation ‘‘⪯’’, if H1 ⪯ H2 , then every H1 -free graph is also H2 -free (see [7]). The graphs Zi , Bi,j , Ni,j,k , C2i +4 are depicted in Fig. 1. First of all, we characterize all forbidden subgraphs H with |H| = 1, 2 such that a connected graph G is H-free implying h(G) ≤ k for any integer k ≥ 0. Chartrand and Wall [5] determined the Hamiltonian index of a tree. So we may assume that G is not a tree. Theorem 2.
Let G, H be connected graphs and H be a connected forbidden pair. Then for any integer k ≥ 0,
(1) H-free graph G implies h(G) ≤ k if and only if H is an induced subgraph of Pk+3 , (2) H-free graph G (other than a tree) implies h(G) ≤ k for k ≥ 1 if and only if H ⪯ {Zk+1 , Pk+4 } or {Zk+1 , Tk+1,1,1 }. Note that Theorem 2 (1) for k = 0 and (2) are clear and also true for the case that k = 0 and κ ′ (G) ≥ 2. The following result solves the case that κ (G) ≥ 2 and k = 0, while Theorem 4 reveals the case when κ ′ (G) ≥ 2 and κ (G) ≥ 2 for k ≥ 2. Theorem 3 (Faudree and Gould, [6]). Let H be a connected graph and H be a connected forbidden pair. Then (1) 2-connected H-free graph G implies h(G) = 0 if and only if H = P3 , (2) 2-connected H-free graph G of sufficiently large order implies h(G) = 0 if and only if H ⪯ {K1,3 , P6 }, {K1,3 , Z3 }, {K1,3 , B1,2 } or {K1,3 , N1,1,1 }. Theorem 4. Let k ≥ 2, i1 , i2 , j1 , j2 ≥ 1 be integers with i1 + j1 = i2 + j2 − 1 = 2k + 2, H (H, respectively) be a connected graph (connected forbidden pair, respectively). Then (1) every 2-edge-connected (or 2-connected) H-free graph G implies h(G) ≤ k if and only if H is an induced subgraph of P3k+3 , (2) every 2-connected H-free graph G of order at least r(k + 1) + 2 ≥ 4k + 6 implies h(G) ≤ k if and only ′ if H ⪯ {C2k +4 , P3k+5 }, {Tk+1,k+1,k+1 , Nk+1,k+1,k+3 }, {Tk+1,k+1,k+1 , Bk+1,2k+4 }, {Ti1 ,j1 ,k , Nk+1,k+1,k+3 }, {Ti1 ,j1 ,k , Bk+1,2k+4 }, {Tk+1,k+1,k+1 , Ni2 ,j2 ,k+1 }, {Ti1 ,j1 ,k , Ni2 ,j2 ,k+1 } or {P3k+4 , Tk+1,r }, ′ (3) every 2-edge-connected H-free graph G of order at least 3k + 10 implies h(G) ≤ k if and only if H ⪯ {C2k +4 , P3k+5 }, {Tk+1,k+1,k+1 , Nk+1,k+1,k+3 }, {Tk+1,k+1,k+1 , Bk+1,2k+4 }, {Ti1 ,j1 ,k , Nk+1,k+1,k+3 }, {Ti1 ,j1 ,k , Bk+1,2k+4 }, {Tk+1,k+1,k+1 , Ni2 ,j2 ,k+1 }, {Ti1 ,j1 ,k , Ni2 ,j2 ,k+1 } or {P3k+4 , Tk+1,k+1,k+1,1 }. The proofs of Theorem 2 for k ≥ 2 and Theorem 4 will be exhibited in Section 3 and Theorem 2 for k = 1 is proved by Theorems 12 (1) and 13 in Section 4. What is more, we also characterize forbidden subgraphs and partial pairs forcing a graph G to satisfy that h(G) ≤ 1 in Section 3. Comparing Theorem 4 and the results in Section 3, we know as follows.
• Theorem 4 (1) for 2-connected graph is true for k = 1, • Theorem 4 (1) for 2-edge-connected graph and Theorem 4 (3) are not true for k = 1. We conjecture that Theorem 4 (2) would be true for k = 1 and leave one conjecture in the last section. Although we cannot completely prove it now, we may show the following results. Theorem 5.
Let G be a 2-edge-connected graph satisfying one of the following:
(1) G is T2,1,1 -free and H-free for any H ∈ {N2,2,4 , B2,6 } other than P(2, 2, 2), (2) G is 2-connected {T3,1,1 , H }-free for any H ∈ {N2,2,4 , B2,6 } other than P(2, 2, 2), (3) G is 2-connected P7 -free other than θ (2, r) for any odd integer r. Then h(G) ≤ 1.
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2. A useful result Use K (i, j, k) or P(i, j, k) to denote the graphs obtained from a K4 or the triangular prism by subdividing i, j, k times of the three edges which are incident in K4 or are not in a K3 of the triangular prism, respectively. Here is our main result in this section, which would be applied to the proofs of other main results. Theorem 6. Let k ≥ 2 be an integer and G be a 2-edge-connected graph with h(G) ≥ k + 1. Then G has an induced subgraph G0 ∈ {θ (l, m, n), K (l, m, n), P(l, m, n)} for some l ≥ m ≥ n ≥ k + 1 such that dG (u, v ) ≥ k + 1 for any pairs of disjoint vertices u, v ∈ V3 (G0 ) in G0 . Furthermore, V2 (G0 ) ⊆ V2 (G) if l = m = n = k + 1 ≥ 4. Before presenting its proof, we shall introduce some notations. Let u be a vertex of G. Use NG (u) to denote the neighborhood and dG (u) to denote the degree of u. Call u a k-vertex if dG (u) = k. Define Vi (G) = {u ∈ V (G) : dG (u) = i} and V≥i (G) = {u⋃∈ V (G) : dG (u) ≥ i}. Let S be a subset of V (G)(or E(G)). The induced subgraph of G is denoted by G[S ]. Define NG [S ] = x∈S NG (x), NG (S) = NG [S ]\S, E(u, S) = {uv ∈ E(G) : v ∈ S } and E(u, H) = E(u, V (H)) if H is a subgraph of G. For two graphs F , G, we write F ⊆ G if F is a subgraph of G, F ∼ = G if F and G are isomorphic and NG (F ) = NG (V (F )). Let P(u, v ) to denote a path between u and v . A branch in G is a nontrivial path with ends not in V2 (G) and with internal vertices, if any, that have degree 2. We denote by B(G) the set of branches of G. Define B1 (G) = {B ∈ B(G) : V (B)∩V1 (G) ̸ = ∅. The distance dH (H1 , H2 ) between two subgraphs H1 and H2 of H is defined to be min{dH (v1 , v2 ) : v1 ∈ V (H1 ), v2 ∈ V (H2 )}, where dH (v1 , v2 ) denotes the number of edges of a shortest path between v1 and v2 in H. The graph H1 ∪ S is a graph with V (H1 ) ∪ S and E(H1 ) as its vertex set and edge set. We then will define some operations which may be different from generic notations such that the resulting graphs have no isolated vertex. Define H1 ∪ H2 = H [E(H1 ) ∪ E(H2 )], H1 ∩ H2 = H [E(H1 ) ∩ E(H2 )], H1 − H2 = H [E(H1 )\E(H2 )], H1 △H2 = H [E(H1 )△E(H2 )] = H [(E(H1 ) ∪ E(H2 ))\(E(H1 ) ∩ E(H2 ))], respectively. Xiong and Liu [14] characterized the graphs for which the k-iterated line graph is Hamiltonian for any integer k ≥ 2. Theorem 7 (Xiong and Liu, [14]). Let G be a connected simple graph and let k ≥ 2 be an integer. Then h(G) ≤ k if and only if EUk (G) ̸ = ∅, where EUk (G) denotes the set of those subgraphs H of a graph G that satisfy the following conditions: (I) (II) (III) (IV) (V)
dH (x) ≡ 0(mod 2) for every x ∈ V (H), V0 (H) ⊆ V≥3 (G) ⊆ V (H), dG (H1 , H − H1 ) ≤ k − 1 for every subgraph H1 of H, |E(B)| ≤ k + 1 for every branch B ∈ B(G) with E(B) ∩ E(H) = ∅, |E(B)| ≤ k for every branch B ∈ B1 (G).
Theorem 8 (Wang and Xiong, [12]). Let G be a connected graph with three pendent vertices v1 , v2 , v3 . Then G either has an induced Ti,j,k or an induced Ni′ ,j′ ,k′ with leaves v1 , v2 , v3 for some i ≥ j ≥ k ≥ 1 and i′ ≥ j′ ≥ k′ ≥ 1. For two connected graphs L1 , L2 , we say a graph G belongs to the graph set θL1 ,L2 (i, j, k) if and only if G is obtained by connecting graphs L1 , L2 by adding three internal-disjoint paths of order i + 2, j + 2, k + 2, respectively. The proof of Theorem 6. Choose a subgraph H of G satisfying that: (1) dH (x) ≡ 0( mod 2) for every x ∈ V (H), (2) V0 (H) ⊆ V≥3 (G) ⊆ V (H), (3) subject to (1), (2), |B(G; H)| is minimized, where B(G; H) = {B ∈ B(G) : E(B) ∩ E(H) = ∅, |E(B)| ≥ k + 2},
(2.1)
(4) subject to (1), (2) and (3), |(G; H)| is minimized, where (G; H) = {H1 ⊆ H : dG (H1 , H − H1 ) ≥ k}. If B(G; H) = ∅ and (G; H) = ∅, then H ∈ EUk (G) since B1 (G) = ∅ and then h(G) ≤ k by Theorem 7, a contradiction. Then we assume that B(G; H) ̸ = ∅, i.e., there is a branch B(x, y) ∈ B(G; H) whose end vertices x, y are in at most ′ ′ two components H1 (and H2 ) of H. Let G1 = G[V (H ( 1 )]. Choose paths P1 (x, x ) ⊆ G1)and P2 (x , y) ⊆ G − B(x, y) − G1 (if H1 = H2 , then x′ = y and P2 (x′ , y) = y). Let H 1 = H △(B(x, y) ∪ P1 (x, x′ ) ∪ P2 (x′ , y)) ∪ V≥3 (G). Then H 1 satisfies (1), (2) of 2.1. Besides, there is a branch B1 (x1 , y1 ) ⊆ P1 (x, x′ ) ∩ H1 or B1 (x1 , y1 ) ⊆ P2 (x′ , y) ∩ H2 with length at least k + 1; for otherwise, B(G; H 1 ) ⊆ B(G; H)\{B(x, y)}, contradicting the choice of H. By symmetry, assume that B1 (x1 , y1 ) ⊆ P1 (x, x′ )∩H1 . Since κ ′ (G1 ) ≥ 2, there are at least two edge-disjoint paths between x and x′ in G1 , which means that G1 − B1 (x1 , y1 ) is connected. Repeat above operation, we can obtain branches B2 (x2 , y2 ), . . . , Bs (xs , ys ) with length at least k + 1 for some s ≥ 2 until G1 − B1 (x1 , y1 ) − · · · Bs (xs , ys ) has at least two connected subgraph L1 , L2 containing x, x′ , respectively. Then dG (L1 , L2 ) ≥ k + 1 and G0 = G[V (B(x, x′ )) ∪ V (B(x′ , y)) ∪ V (B1 (x1 , y1 )) ∪ V (B2 (x2 , y2 )) ∪ V (L1 ) ∪ V (L2 )] ∈ θL1 ,L2 (a1 , a2 , a3 ) for some a1 , a2 , a3 ≥ k + 1. By Theorem 8, G0 has an induced subgraph G0 ∈ {θ (l, m, n), K (l, m, n), P(l, m, n)} for some l ≥ m ≥ n ≥ k + 1 and V2 (G0 ) ⊆ V2 (G).
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X. Liu and L. Xiong / Discrete Mathematics 343 (2020) 111841
We then assume that (G; H) ̸ = ∅, i.e., there is a connected subgraph H1 of H such that dG (H1 , H − H1 ) ≥ k. Then there are at least two branches P1 (x1 , y1 ), P2 (x2 , y2 ) with length at least k between H1 and H − H1 such that x1 , x2 ∈ V (H1 ), y1 , y2 ∈ V (H − H1 ) since κ ′ (G) ≥ 2. A path P satisfies the property Γ (H) if H has a connected subgraph H0 such that G either has a branch B ⊆ P ∩ H0 with |E(B)| ≥ k + 2 or a subgraph ˜ H and two branches B1 , B2 ⊆ P ∩ H0 satisfying that NG (˜ H) ⊆ V (B1 ∪ B2 ) and dG (˜ H , H − B1 − B2 − ˜ H) ≥ k. Let G1 = G[V (H1 )] and G2 = G[(V (G)\V (H1 ∪ P1 (x1 , y1 ))) ∪ V (P2 (x2 , y2 )) ∪ {y1 , y2 }]. Then G2 is connected and either each path of G1 with end vertices x1 , x2 or each path of G2 with end vertices y1 , y2 satisfies the property Γ (H). Suppose otherwise. Let C2 = P1 (x1 , y1 ) ∪ P4 (y1 , y2 ) ∪ P2 (x2 , y2 ) ∪ P3 (x1 , x2 ) and H 2 = (H △C2 ) ∪ V≥3 (G) if there are two paths P3 (x1 , x2 ) ⊆ G1 , P4 (y1 , y2 ) ⊆ G2 do not satisfy the property Γ (H). Then H 2 is a subgraph satisfying (1), (2) of 2.1 and |(G; H 2 )| < |(G; H)|, contradicting the choice of H. By symmetry, assume that each path with end vertices y1 , y2 in G2 satisfies the property Γ (H). Then there is a non-trivial connected subgraph H0 ⊆ H − H1 such that each path of H0 with end vertices z1 , z2 satisfies the property Γ (H) for any P(y1 , y2 ) ⊆ G2 and z1 , z2 ∈ V (H0 ) ∩ V (P(y1 , y2 )). Choose all paths P1 (z1 , z2 ), . . . , Pt (z1 , z2 ) ⊆ H0 between z1 and z2 for some t ≥ 2. Then there are paths P1 ⊆ ˜ 1 with |E(Pi )| ≥ 2k, P1 (z1 , z2 ), . . . , Pt ⊆ Pt (z1 , z2 ) satisfy that either |E(Pi )| ≥ k and Pi is a branch of G or Pi ⊆ B11 ∪ B21 ∪ H ˜ i is a connected subgraph such that NG (H˜ i ) ⊆ V (B1 ∪ B2 ) for i ∈ {1, . . . , t }. Hence for where B11 , B21 are two branches and H i i any i, j ∈ {1, . . . , t }, Pi , Pj are either identical or internal vertex-disjoint. Without loss of generality, we assume that P1 and P2 are internal vertex-disjoint. Then G[V (H0 )] − P1 − · · · − Ps has two connected subgraphs L0 , L′0 such that G0 = G[V (P1 ∪ P2 ∪ L0 ∪ L′0 ∪ P1 (x1 , y1 ) ∪ P2 (x2 , y2 ) ∪ P1 (x1 , x2 ) ∪ P(y1 , z1 ) ∪ P(y2 , z2 ))] ∈ θL0 ,L′ (a1 , a2 , a3 ) for some 0
a1 , a2 , a3 ≥ min{k + 1, 2k − 1} = k + 1 for any induced paths P(y1 , z1 ), P(y2 , z2 ) ⊆ G2 . By Theorem 8, G0 has an induced subgraph G0 ∈ {θ (l, m, n), K (l, m, n), P(l, m, n)} for some l ≥ m ≥ n ≥ k + 1. Especially, if l = m = n = k + 1 ≥ 4, then P1 , . . . , Ps are branches of G since 2k − 1 > k + 1 for k ≥ 3 and then V2 (G0 ) ⊆ V2 (G). □ 3. The proofs of Theorems 2 and 4 We prove the following result by Theorem 6.
Theorem 9. Let k ≥ 2, i1 , i2 , j1 , j2 ≥ 1 be integers with i1 + j1 = i2 + j2 − 1 = 2k + 2 and G be a 2-edge-connected graph satisfying exactly one of the following: (1) (2) (3) (4) (5)
G G G G G
is is is is is
′ {P3k+5 , C2k +4 }-free, {H1 , Ni2 ,j2 ,k+1 }-free for H1 ∈ {Tk+1,k+1,k+1 , Ti1 ,j1 ,k }, {H1 , H2 }-free for H1 ∈ {Tk+1,k+1,k+1 , Ti1 ,j1 ,k } and H2 ∈ {Nk+1,k+1,k+3 , Bk+1,2k+4 } other than P(k + 1, k + 1, k + 1), {P3k+4 , Tk+1,k+1,k+1,1 }-free other than θ (k + 1, k + 1, k + 1), 2-connected P3k+4 -free graph other than θ (k + 1, r) for any r ≥ 3 and r is odd.
Then h(G) ≤ k. Proof. We argue by contradiction. Let G be a 2-edge-connected graph with h(G) ≥ k + 1. Then G has an induced subgraph G0 ∈ {θ (l, m, n), K (l, m, n), P(l, m, n)} for some l ≥ m ≥ n ≥ k + 1 by Theorem 6. Besides, G0 is H-free if G is H-free. ′ (1) Since G0 is P3k+5 -free, G0 ∼ = θ (k + 1, k + 1, k + 1). However, G0 has an induced subgraph C2k +4 , a contradiction. (2) Since G0 is Ni2 ,j2 ,k+1 -free, G0 ∈ {θ (k + 1, k + 1, k + 1), K (k + 1, k + 1, k + 1)}. However, G0 has induced subgraphs Tk+1,k+1,k+1 , Ti1 ,j1 ,k , a contradiction. (3) Assume that G0 is either Tk+1,k+1,k+1 -free or Ti,j,k -free. Then G0 ∼ = P(k + 1, k + 1, k + 1) and V2 (G0 ) ⊆ V2 (G); for otherwise, G0 has induced subgraphs Nk+1,k+1,k+3 and Bk+1,2k+4 , a contradiction. Assume that xyzx, x′ y′ z ′ x′ ⊆ G0 . Then dG ({x, y, z }, {x′ , y′ , z ′ }) = k + 1. Besides, V (G − G0 ) ̸ = ∅ since G ≇ P(k + 1, k + 1, k + 1). By symmetry, assume that there is a vertex x0 ∈ NG (x) with E(x0 , {x′ , y′ , z ′ }) = ∅ for x′ y′ z ′ ̸ = xyz and x′ y′ z ′ ⊆ G0 and hence G[V (G0 ) ∪ {x0 }] has an induced Nk+1,k+1,k+3 . Then E(x0 , {y, z }) = ∅ since G[V (G0 ) ∪ {x0 }] does not have an induced Bk+1,2k+4 . Note that if there is an induced path P ′ between x0 and x′ y′ z ′ x′ in G − G0 , then |E(P ′ )| ≥ k and there are induced Tk+1,k+1,k+1 and Ti,j,k with x as their 3-vertex, a contradiction. Hence, there is an induced path P(x0 , y0 ) of order at least 2 between x0 and {y, z } in G − G0 such that NG ({y, z }) ∩ V (P(x0 , y0 )) = {y0 } since G is 2-edge-connected. Then there is an induced Bk+1,2k+4 , a contradiction. (4) Since G0 is P3k+4 -free, G0 ∼ = θ (k + 1, k + 1, k + 1). If there is a vertex x0 ∈ NG (x) for some x ⊆ xyz ⊆ G0 with E(x0 , {x′ , y′ , z ′ }) = ∅ for x′ y′ z ′ ̸ = xyz and x′ y′ z ′ ⊆ G0 , then G[V (G0 ) ∪ {x0 }] has an induced Tk+1,k+1,k+1,1 , a contradiction. Then V (G) = V (G0 ) and G ∼ = θ (k + 1, k + 1, k + 1) if k ≥ 3, a contradiction. Then k = 2 and G ≇ θ (3, 3, 3). So there is a vertex x′2 ∈ NG (x2 )\{x1 , x3 } for some x1 , x2 , x3 ∈ V2 (G0 ) and x1 x2 x3 ⊆ G0 such that |E(x′2 , G0 )| = 1. Then G[V (G0 ) ∪ {x′2 }] has an induced P10 , a contradiction. (5) By the same argument, we firstly prove that G ∼ = θ (3, 3, 3) for k = 2 since G is P10 -free. For any k ≥ 3, G0 ∼ = θ (k + 1, k + 1, k + 1) since G0 is P3k+4 -free. Assume that V3 (G0 ) = {u, v}. Since G ≇ θ (k + 1, k + 1, k + 1), NG ({u, v}) ̸ ⊆ V (G0 ). By symmetry, assume NG−G0 (u) = {x1 , . . . , xt } ̸ = ∅. Since κ (G) ≥ 2, there is an induced path Pi between xi and v in G − u with length at least k + 1 for any i ∈ {1, . . . , t }. Then Pi is a branch of G with length k + 1 since G is P3k+4 -free. Let G1 = G[V (G0 ) ∪ V (P1 ) · · · ∪ V (Pt )]. Then NG−G1 (u) = ∅, NG−G1 (v ) = ∅ and G = G1 ∼ = θ (k + 1, r) for some r ≥ 3. If r is even, then G is supereulerian, a contradiction. Then r is odd. □
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Fig. 2. The auxiliary graphs with Hamiltonian index at least k + 1 for any integers l ≥ m ≥ n ≥ k + 1.
Note that the graphs G7 and G11 in Fig. 2 are connected graphs with l ≥ m ≥ n ≥ k + 1. Then all graphs in Fig. 2 are connected with h(G) > k. We now present the proofs of Theorems 2 and 4. The proof of Theorem 2 when k ≥ 2. The sufficiency We may choose a subgraph H of G satisfying the conditions in 2.1. By Theorem 7, it suffices to prove that H ∈ EUk (G). (1) We claim that B(G; H) = ∅. Suppose otherwise. Then there is a branch B(x, y) whose end vertices are x, y connecting at most two components of H with |E(B(x, y))| ≥ k, dG (x) ≥ 3 and dG (y) ≥ 3. Let x′ ∈ NG (x)\V (B(x, y)) and y′ ∈ NG (y)\V (B(x, y)). Then x′ B(x, y)y′ would be an induced path of order at least k + 3, a contradiction. Similarly, we can prove that (G; H) = ∅. If there is a branch B = ux1 · · · xl ∈ B1 (G) such that l ≥ k + 1, then there is a vertex u′ ∈ NG (u)\{x1 }, and then u′ ux1 · · · xl is an induced path of G with length at least k + 3, a contradiction. This implies that B1 (G) = ∅ and then H ∈ EUk (G). (2) We claim that B(G; H) = ∅. Suppose otherwise. Then there is a branch B(x, y) whose end vertices are x, y connecting at most two components of H with |E(B(x, y))| ≥ k, x1 , x2 ∈ NG (x)\V (B(x, y)) and y1 , y2 ∈ NG (y)\V (B(x, y)). Since G is Zk+1 free, x1 x2 , y1 y2 ∈ / E(G). Then G[{x1 , x2 , y1 } ∪ V (B(x, y))] ∼ = T1,1,l for some l ≥ k, a contradiction. Similarly, we can prove that (G; H) = ∅. If there is a branch B = ux1 · · · xl ∈ B1 (G) such that l ≥ k + 1, then there are three vertices u1 , u2 ∈ NG (u)\{x1 } and u′1 ∈ NG (u1 )\{u}. Then G either has an induced Zl (if u1 u2 ∈ E(G) or u′1 = u2 or uu′1 ∈ E(G)) or has induced Tl,1,1 and Pl+4 (if u1 u2 ∈ / E(G), u′1 ̸= u2 and uu′1 ∈ / E(G) ), a contradiction. This implies that B1 (G) = ∅ and then H ∈ EUk (G). The necessity (1) Since G1 is K1,3 -free and G3 is K3 -free, ∆(H) ≤ 2. Then H should be a path since they have no common cycle, and hence H should be the subgraph of Pk+3 since G1 is Pk+4 -free. (2) Let H = {R, S }. Then each graph of {G1 , G2 , G3 , G4 , G5 } contains at least one of R, S as an induced subgraph. Without loss of generality, we may assume that R is an induced subgraph of G1 . If R is a tree, then R is an induced subgraph of Pk+3 , a contradiction. Thus we distinguish the following two cases: Assume that R contains K4 as a subgraph. Since G2 , G3 , G4 are K4 -free, they should contain S as an induced subgraph. Note that G4 is K1,3 -free, G3 is K3 -free and G2 is Pk+4 -free. Then H should be the subgraph of Pk+3 , a contradiction. Next assume that R is K4 -free. Then R contains K3 as a subgraph. Since G3 , G5 are K3 -free and they have no common induced cycle, S should be a tree. Since |V3 (G5 )| = 1, |V3 (S)| ≤ 1. Then the maximal common subgraph of G3 and G5 are Tk+1,1,1 and Pk+4 . Hence {R, S } ⪯ {Zk+1 , Tk+1,1,1 }, {Zk+1 , Pk+4 } since G1 is Zk+2 -free. □ The proof of Theorem 4. By Theorem 9, the sufficiency clearly holds. It remains to show the necessity. Let H = {R, S }. Then (1) Since G6 is K3 -free and G10 is K1,3 -free, ∆(H) ≤ 2. Then H should be a path since they have no common cycle, and hence H should be the subgraph of P3k+3 since G6 is P3k+4 -free. (2) Note that each graph of {G6 , G7 , G8 , G9 , G10 , G11 } contains at least one of R, S as an induced subgraph. Without loss of generality, we may assume that R is an induced subgraph of G6 . If ∆(R) = 2, then R is an induced subgraph of P3k+3 , a contradiction. Thus R either is a tree with |V≥3 (R)| = 1 or contains C2k+4 as an induced subgraph. Now we distinguish the following two cases:
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Assume that R contains C2k+4 as an induced subgraph. Note that G7 , G8 , G10 are C2k+4 -free, they should contain S as an induced subgraph. Since G7 is K3 -free, G10 is K1,3 -free and G8 is P3k+6 -free, S is an induced subgraph of P3k+5 . On the ′ ′ other hand, G9 is P3k+5 -free, then R is a common subgraph of G6 and G9 , which is C2k +4 . Therefore, {R, S } ⪯ {C2k+4 , P3k+5 }. Next assume that R is a tree. If ∆(R) = 3, then R contains K1,3 as an induced subgraph. Note that G10 , G11 are K1,3 -free, they should contain S as an induced subgraph. Since G12 is K4 -free and it has no induced subgraph contains two triangles as subgraph, S is an induced subgraph of G12 with at most one triangle as a subgraph, that is, Nk+1,k+1,k+3 , Ni2 ,j2 ,k+1 or Bk+1,2k+4 . Since G6 is S-free, R is an induced subgraph of G6 , that is Tk+1,k+1,k+1 or {Ti1 ,j1 ,k . Therefore, {R, S } ⪯ {H1 , H2 } for any H1 ∈ {Nk+1,k+1,k+3 , Ni2 ,j2 ,k+1 , Bk+1,2k+4 } and H2 ∈ {Tk+1,k+1,k+1 , Ti1 ,j1 ,k }. If ∆(R) ≥ 4, then R contains K1,4 as an induced subgraph. Note that G7 , G9 , G11 are K1,4 -free, they should contain S as an induced subgraph. Since G7 is K3 -free, G11 is K1,3 -free and they have no common cycle and G9 is P3k+5 -free, S is an induced subgraph of P3k+4 . Therefore, {R, S } ⪯ {P3k+4 , Tk+1,r } for some r ≥ 4. (3) Note that graphs G6 , G7 , G8 , G9 , G10 , G11 and the graph G12 are 2-edge-connected graphs with Hamiltonian ′ index at least k + 1. Besides, since G12 is {P3k+4 , Tk+1,k+1,k+1,2 }-free, {R, S } ⪯ {C2k +4 , P3k+5 }, {Tk+1,k+1,k+1 , Nk+1,k+1,k+3 }, {Tk+1,k+1,k+1 , Bk+1,2k+4 }, {Ti1 ,j1 ,k , Nk+1,k+1,k+3 }, {Ti1 ,j1 ,k , Bk+1,2k+4 }, {Tk+1,k+1,k+1 , Ni2 ,j2 ,k+1 }, {Ti1 ,j1 ,k , Ni2 ,j2 ,k+1 }, {P3k+4 , Tk+1,k+1,k+1,1 }. □ 4. Forbidden graphs force a graph G to satisfy h(G) ≤ 1 A dominating closed trail (abbreviated DCT) in a graph G is a closed trail (or, equivalently, an eulerian subgraph) T in G such that every edge of G has at least one vertex on T . In this section, we use the following theorem by Harary and Nash-Williams to transform the problem that guarantees h(G) ≤ 1 for a given graph G to one equivalent proposition, that is, guarantee whether G has a DCT. Theorem 10 (Harary and Nash-Williams, [8]). Let G be a graph with at least three edges. Then h(G) ≤ 1 if and only if G has a DCT. We call H a minor of G if H is isomorphic to the contraction image of a subgraph of G. Proof of Theorem 1. The necessity Note the graphs θ (l, m, n) and θ (k + 1, r) for any integer l ≥ m ≥ n ≥ k + 1, r ≥ 3 have Hamiltonian index at least k + 1. Then H is a tree with |V3 (H)| ≤ 1 and the necessity holds. The sufficiency We argue by contradiction. Assume that h(G) ≥ k + 1. By Theorem 6, G has a θ (k + 1, k + 1, k + 1)minor Hk for k ≥ 2. For k = 0, G is non-Hamiltonian and then G has a θ (1, 1, 1)-minor H0 . Hence we can find subgraphs P3k+5 , T1,1,2k+2 , T1,k+1,2k+1 and Tk+1,k+1,k+2 in Hk for k ̸ = 1, a contradiction. We then only consider the case that k = 1. By Theorem 10, G has no DCT. Choose a closed trail T of G such that |V (T )| is maximized. Then there is a non-trivial component H1 of G − V (T ) and two edges uu′ , vv ′ ∈ E(H1 , T ) such that u, v ∈ V (H1 ), u′ ̸ = v ′ ∈ V (T ) and either u ̸ = v or u = v and |V (H1 )| ≥ 3. Choose a path P(u′ , v ′ ) ⊆ T . Then by the choice of T , either |V (P(u′ , v ′ ))| ≥ 3 or P(u′ , v ′ ) = u′ x0 v ′ , x0 is a cut vertex of T and hence there is a cycle C1 ⊆ T with x0 ∈ V (C1 ) and T − E(P(u′ , v ′ )) − C1 is connected. Choose a path P1 (u′ , v ′ ) ⊆ T − E(P(u′ , v ′ )) − C1 and either |V (P1 (u′ , v ′ ))| ≥ 3 or P(u′ , v ′ ) = u′ x1 v ′ and there is a cycle C2 ⊆ T with x1 ∈ V (C2 ). We can easily find subgraphs P8 , T1,2,4 and T2,2,3 , a contradiction. □ For a subset X ⊆ E(G), the contraction G/X is the graph obtained from G by identifying the ends of each edge in X and then deleting the resulting loops. Note that the edges in E(G/X ) can be regarded as edges in E(G). If H is a subgraph, then we use G/H for G/E(H). Note that by this definition, if H is a connected subgraph of G, then G/H = G/G[V (H)]. For a graph G, let O(G) denote the set of odd degree vertices in G. In [2], Catlin defined collapsible graphs. A graph G is collapsible if for any even subset R of V (G), G has a connected spanning subgraph ΓR with O(Γ ) = R. The reduction of G, denoted by G′ , is obtained from G by contracting all maximal collapsible subgraphs of G. The vertex vH in G′ is called non-trivial if it is a contraction image of a non-trivial collapsible subgraph H of G. The following result would play an important role in our proofs. Theorem 11 (Catlin, [2]). Let G be a connected graph and G′ the reduction of G. Then G has a DCT if and only if G′ has a DCT containing all non-trivial vertices of G′ . For any two subsets S1 , S2 of V (G) and two subgraph T1 , T2 , define E(S1 , S2 ) = ∪x∈S1 E(x, S2 ), E(T1 , T2 ) = E(V (T1 ), V (T2 )) and D[T1 ] = {uv : {u, v} ∩ V (T1 ) ̸ = ∅}. Proof of Theorem 5. We argue by contradiction. By Theorem 10, assume that G has no DCT. Choose a closed trail T of G such that: (1.1) |V (T )| is maximized; (1.2) subject to (1.1), |D[T ]| is maximized.
X. Liu and L. Xiong / Discrete Mathematics 343 (2020) 111841
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Let H = G[V (T )]. Then there is a non-trivial component H1 of G[V (G)\V (T )] and a longest induced path P(u, v ) of H1 such that {uu′ , vv ′ } ⊆ E(P(u, v ), H) = E({u, v}, H) for some u′ , v ′ ∈ V (H). Then uv ′ , v u′ , u′ v ′ ∈ / E(G); for otherwise, T ∪ uv u′ u, T ∪ uvv ′ u or T △ uvv ′ u′ u is a closed trail with larger number of order than T , respectively, contradicting to (1.1). Choose two induced collapsible subgraphs F1 and F2 of G such that: (1.3) u′ ∈ V (F1 ), v ′ ∈ V (F2 ); (1.4) subject to (1.3), |V (F1 )| and |V (F2 )| are maximized. Let T0 = T [E(T )\E(F1 ∪F2 )], T1 = T [E(T )\E(F1 )], T2 = T [E(T )\E(F2 )] and Xi = O(T0 )∩V (Fi ) for i ∈ {1, 2}. Then |Xi | is even since |NT (Fi )| is even. By the definition of collapsible, for any {u1 , u2 } ⊆ V (Fi ), we have Fi has a spanning trail Ti (u1 , u2 ) such that O(Ti (u1 , u2 )) = Xi △{u1 , u2 }.
(4.1)
Then V (H1 ) ∩ V (F1 ∪ F2 ) = ∅; for otherwise, F1 , F2 have spanning trails T1 , T2 , respectively, such that O(T1 ) ∪ O(T2 ) = O(T0 ) and then T0 ∪ T1′ ∪ T2′ is a closed trail with larger number of order than T , a contradiction. Besides, V (F1 ) ∩ V (F2 ) = ∅; for otherwise, F1 ∪ F2 has a spanning trail T (u′ , v ′ ) such that O(T (u′ , v ′ )) = O(T0 )△{u′ , v ′ } since F1 ∪ F2 is collapsible, and then T0 ∪ T (u′ , v ′ ) ∪ u′ uP(u, v )vv ′ is a closed trail with larger number of order than T , a contradiction. Furthermore, E(F1 , F2 ) = ∅. Otherwise, assume that u1 u2 ∈ E(F1 , F2 ) for u1 ∈ V (F1 ), u2 ∈ V (F2 ). Then exactly one of T △uP(u, v )vv ′ u′ u (if u′ = u1 , v ′ = u2 ) or (T1 ∪ T1 (u′ , u1 )) △ u′ uP(u, v )vv ′ u1 (if u′ ̸ = u1 , v ′ = u2 ) or (T0 ∪ T1 (u′ , u1 ) ∪ T2 (v ′ , u2 )) △ (u′ uP(u, v )vv ′ ∪ u1 u2 ) (if u′ ̸ = u1 , y′ ̸ = u2 ) is a closed trail with larger number of order than T , a contradiction. Note that T0 is non-trivial. We can choose two edge-disjoint paths P1 = xx1 · · · xm x′ ⊆ T0 and P2 = yy1 · · · yn y′ ⊆ T0 such that |V (P1 ) ∩ V (P2 )| is minimized for any x, y ∈ V (F1 ), x′ , y′ ∈ V (F2 ) and m, n ≥ 1 since T is 2-edge-connected. Then x1 ̸ = y1 , xm ̸ = yn and x1 y1 , xm yn ∈ / E(G) by the choice of F1 , F2 . Then we have ′
E({x1 , y1 }, H1 ) = E({xm , yn }, H1 ) = ∅.
′
′
′
(4.2)
(Suppose otherwise. By symmetry, assume that x1 v1 ∈ E(G) for some v1 ∈ V (H1 ). Then there is a path P(u, v1 ) ⊆ H1 such that either T △x1 v1 P(u, v1 )uu′ x1 (if u′ = x) or (T1 ∪ T1 (u′ , x)) △ u′ uP(u, v1 )v1 x1 x (if u′ ̸ = x) is a closed trail with larger number of order than T , respectively, a contradiction.) Besides, E(v2 , H1 ) = ∅ for any v2 ∈ NG−P1 −P2 ({x1 , y1 }) ∪ NG−P1 −P2 ({xm , yn }).
(4.3)
(Suppose otherwise. By symmetry, assume that v1 v2 ∈ E(G) for some v1 ∈ V (H1 ) and v2 ∈ NG (x1 ) \ {x, x2 }. Then there is a path P(u, v1 ) ⊆ H1 such that either T △x1 v2 v1 P(u, v1 )uu′ x1 (if u′ = x) or (T1 ∪ T1 (u′ , x)) △ u′ uP(u, v1 )v1 v2 x1 x (if u′ ̸ = x) is a closed trail with larger number of order than T , respectively, a contradiction.) (1) We assume that G is 2-edge-connected T2,1,1 -free and then either obtain induced N2,2,4 , B2,6 or prove that G∼ = P(2, 2, 2) to reach a contradiction. By Eq. (4.2), E({x1 , y1 }, {u, u1 }) = ∅ for any u1 ∈ NH1 (u). Then G[V (F1 ) ∪ {u, u1 , x1 , y1 }] either has an induced T2,1,1 or is isomorphic to Ni,j,k for some i ≥ j ≥ k ≥ 1 by Theorem 8. Since G is T2,1,1 -free, |V (F1 )| ≥ 3 and u′ , x, y are three distinct vertices. By symmetry, |V (F2 )| ≥ 3 and v ′ , x′ , y′ are three distinct vertices. Then |E(P(u, v ))| ≥ 1 and dG (u) = dG (v ) = 2; for otherwise, G[{u, u1 , u′ , x, v ′ }] ∼ = T2,1,1 for u1 ∈ NH1 (u1 )\{v1 }, a contradiction. We claim that dG (F1 , F2 ) ≥ 3 and then m ≥ 2, n ≥ 2. Suppose otherwise. Without loss of generality, assume that m = 1 and there is a vertex x′1 ∈ NG (x1 )\{x, x′ }. By the choice of F1 , F2 , E(x′1 , F1 ∪ F2 ) = ∅ and then G[{x′1 , x1 , x, x′ , v1 }] ∼ = T2,1,1 for any v1 ∈ NF1 (x), a contradiction. Besides, x1 and y1 have no common neighbor with degree at least 3; for otherwise, E(x′12 , {x1 , y1 , x}) = ∅ by the choice of F1 for any x12 ∈ NG (x1 ) ∩ NG (y1 ) and x′12 ∈ NG (x12 )\{x1 , y1 }, and hence G[{u1 , x1 , x, y1 , x3 }] ∼ = T2,1,1 , a contradiction. By symmetry, xm and yn have no common neighbor with degree at least 3. Then x1 , x2 , y1 , y2 are four distinct vertices and x1 y2 , y1 x2 ∈ / E(G). Claim 1.
dG (F1 , F2 ) ≥ 4.
Proof. Without loss of generality, assume that m = 2. Then E(y2 , F2 ) ̸ = ∅; for otherwise, either G[{x2 , x1 , x, x′ , y2 }] ∼ = T2,1,1 (if x2 y2 ∈ E(G)) or G1 = G[V (F1 ∪F2 )∪V (P(u, v ))∪{x1 , x2 , y1 , y2 }] has induced N2,2,4 and B2,6 (if x2 y2 ∈ / E(G)), a contradiction. Therefore either G1 ∼ = P(2, 2, 2) or G1 has induced N2,2,4 and B2,6 . Besides, NG (G1 ) = ∅ and then G = G1 = P(2, 2, 2), a contradiction. □ Claim 2.
G has an induced B2,6 .
Proof. We claim that either x1 , x2 , yn , yn−1 , yn−2 or y1 , y2 , xm , xm−1 , xm−2 are five distinct vertices. If P1 and P2 are vertex disjoint, then it holds; otherwise, let i1 , j1 be the smallest integers such that xi1 = yj1 and i2 , j2 be the largest integers such that xi2 = yj2 . Then max{i1 , j1 } ≥ 3 and max{m − i2 , n − j2 } ≥ 3 and hence it holds. Without loss of generality, assume that x1 , x2 , yn , yn−1 , yn−2 are five distinct vertices. Then x1 yn , x1 xm , y1 yn ∈ / E(G) and P(u, v ) = uv . If x1 yn−1 , x2 yn ∈ / E(G), then either G[V (F1 ) ∪ {u, v, x1 , x2 , yn−1 , yn , y′ , y′′ }] for any y′′ ∈ NF2 (y′ ) (if x2 yn−1 ∈ E(G)) or G[V (F1 ) ∪ V (P(v ′ , y′ )) ∪ {u, v, x1 , x2 , yn , yn−1 }] for any induced path P(v ′ , y′ ) of F2 (if x2 yn−1 ∈ / E(G)) has an induced B2,6 . Otherwise either G[V (F1 ) ∪ V (P(x′ , y′ )) ∪ {u, v, x1 , yn−1 , yn , xm }] for any induced ′ ′ path P(x , y ) of F2 (if x1 yn−1 ∈ E(G)) or G[V (F2 ) ∪ V (P(x, y)) ∪ {u, v, yn , x2 , x1 , y1 }] for any induced path P(x, y) of F1 (if x2 yn ∈ E(G)) has an induced B2,6 . □
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Fig. 3. The graph G1 and its reduction G′ in Theorem 5 (1).
We then finish the proof of (1) by the following claim. Claim 3.
G has an induced N2,2,4 .
Proof. We argue by contradiction. Recall x1 , x2 , y1 , y2 are four distinct vertices, then E({x1 , x2 , y1 , y2 }, F2 ) = ∅ and y1 x2 , x1 y2 ∈ / E(G). Then x2 y2 ∈ E(G); for otherwise, G[V (F1 ) ∪ V (P(u, v )) ∪ {x1 , x2 , y1 , y2 , v ′ , v2 }] has induced N2,2,4 for any v2 ∈ NF2 (v ′ ), a contradiction. Besides, {x1 , y1 } ⊆ V2 (G); for otherwise, assume that dG (x1 ) ≥ 3, then either G[{x1 , x′1 , x, x2 , y2 }] ∼ = T2,1,1 or G[V (F1 ) ∪ {x1 , x2 , y1 , y2 , x′1 }] is collapsible for any x′1 ∈ NG (x1 )\{x, x2 }, contradicting the choice of F1 . To sum up, NG (F1 ) ⊆ V≤2 (T ) is an independent set of G and V1 = NG (NG (F1 ))\F1 is a clique. Besides, there is a vertex v2 ∈ NG (V1 )\NG (F1 ) with |E(v2 , V1 )| ≥ 2 since G is T2,1,1 -free. Then G[V1 ∪ {v2 }] is collapsible. Choose a collapsible subgraph L1 containing V1 such that |V (L1 )| is maximized. Repeat the above operation by replacing F1 by F1 ∪ L1 · · · ∪ Li for i ≥ 1, we can obtain a series of induced collapsible subgraphs L1 , . . . , Lt of G for some t ≥ 1. Let F1 = L0 and F2 = Lt +1 . Then G ∼ = G1 , which depicted in Fig. 3. Then the reduction G′ of G has a DCT T ′ containing all non-trivial vertices. Hence G has a DCT by Theorem 11, a contradiction. □ (2) Note that G is 2-connected. Then |E(P(u, v ))| ≥ 1 and we can replace all the graphs T2,1,1 by T3,1,1 in the proof of (1) and the remainder discussion is similar but easier than the proof of (1). So we omit it here. (3) We assume that G is 2-connected P7 -free and will prove that G ∼ = θ (2, r) to reach a contradiction. Claim 4.
V (P1 ) ∩ V (P2 ) = ∅.
Proof. We argue by contradiction. Let i, j be the smallest integers such that ui,j = xi = yj . Then either ui,j ∈ / {x1 , y1 } or ui,j ∈ / {xm , yn }. Suppose otherwise. By symmetry, assume that x1 = ui,j = xm . Then exactly one of T △uvv ′ x1 u′ u (if u′ = x, v ′ = x′ ) or (T1 ∪T1 (u′ , x))△u′ uvv ′ x1 x (if u′ ̸ = x, v ′ = x′ ) or (T0 ∪T1 (u′ , x)∪T2 (v ′ , x′ ))△(u′ uvv ′ ∪xx1 x′ ) (if u′ ̸ = x, v ′ ̸ = x′ ) is a closed trail with larger number of order than T , a contradiction. By symmetry, assume that ui,j ∈ / {xm , yn }. Besides i + j ≥ 4 by the choice of F1 . Then either x1 , x2 , xm , yn (if i ≥ 2) or y1 , y2 , xm , yn (if j ≥ 2) are four distinct vertices. Without loss of generality, assume that x1 , x2 , xm , yn are four distinct vertices. By the choice of F1 , E(x2 , F1 ) = ∅. Besides E(x1 , F2 ) = ∅; for otherwise, exactly one of T △x1 u′ uvv ′ x1 x1 (if u′ = x, v ′ = v1 ) or (T1 ∪ T1 (u′ , x)) △ xx1 v ′ v uu′ (if u′ ̸ = x, v ′ = v1 ) or (T0 ∪ T1 (u′ , x) ∪ T2 (v ′ , v1 )) △ (u′ uvv ′ ∪ xx1 v1 ) (if u′ ̸ = x, v ′ ̸ = v1 ) for any x1 v1 ∈ E(x1 , F2 ) is a closed trail with larger number of order than T , a contradiction. We claim that dG (x1 ) = 2. Suppose otherwise. Then there is a vertex x′1 ∈ NG (x1 )\{x, x2 } with E(x′1 , F1 ) = ∅ by the choice of F1 and E(x′1 , P(u, v )) = ∅ by Eq. (4.3). By the choice of F2 , |E(x′1 , F2 )| ≤ 1. Then u′ = x; for otherwise, there is a non-trivial induced path P(x, u′ ) in F1 and then x′1 x1 xP(x, u′ )u′ uvv ′ is an induced path of order at least 7. So |V (F2 )| = 1; for otherwise, there is a vertex v1 ∈ NF2 (v ′ ) such that x′1 v1 ∈ / E(G) and G[{x′1 , x1 , u′ , u, v, v ′ , v1 }] ∼ = P7 , a contradiction. By symmetry, ′ ′ E({yn , xm }, V (P(u, v )) ∪ {u }) = ∅. Then {x1 yn , x′1 xm } ̸ ⊆ E(G); for otherwise, at least one of T △uP(u, v )vv ′ yn x′1 x1 u′ u or T △uP(u, v )vv ′ xm x′1 x1 u′ u is a closed trail with larger number of order than T , a contradiction. By symmetry, assume that x′1 yn ∈ / E(G). Besides, x1 yn ∈ / E(G); for otherwise, T △uvv ′ yn x1 u′ u is a closed trail with larger number of order than T , a contradiction. However, G[{x′1 , x1 , u′ , u, v, v ′ , yn }] ∼ = P7 , a contradiction. Then E(x2 , P(u, v ) ∪ F2 ) = ∅; for otherwise, either T △uP(u, v1 )v1 x2 x1 u′ (if u′ = x) or (T1 ∪ T1 (u′ , x))△u′ uP(u, v1 )v1 x2 x1 x (if u′ ̸ = x) for any P(u, v1 ) ⊆ P(u, v ) and x1 v1 ∈ E(G), exactly one of T △uvv ′ x2 x1 u′ u (if u′ = x, v ′ = v1 ) or (T1 ∪ T1 (u′ , x))△u′ uvv ′ x2 x1 x (if u′ ̸ = x, v ′ = v1 ) or (T0 ∪ T1 (u′ , x) ∪ T2 (v ′ , x2 ))△(u′ uvv ′ ∪ xx1 x2 v1 ) (if u′ ̸ = x, v ′ ̸ = v1 ) for any x2 v1 ∈ E(x2 , F2 ) is a closed trail with larger number of order than T , respectively, a contradiction. So x = u′ and |V (F2 )| = 1 since G is P7 -free. By symmetry, |V (F1 )| = 1 and x2 u′ ∈ / E(G). Besides, x2 yn ∈ / E(G); for otherwise, T △uvv ′ yn x2 x1 u′ u is a closed trail with larger number of order than T , a contradiction. However G[{x2 , x1 , u′ , u, v, v ′ , yn }] ∼ = P7 , a contradiction. □ We then consider the following two cases to finish our proof. Case 1. dG (F1 , F2 ) = 2.
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Without loss of generality, assume that m = 1. Then dG (x1 ) ≥ dT (x1 ) ≥ 4 by the maximality of |V (T )|. Let X0 = V (F1 ∪ F2 ∪ P(u, v )) ∪ {y1 , yn }. By the choice of F1 , F2 and Eqs. (4.2) and (4.3), |E(x1 , X0 )| = 2, E({y1 , yn }, P(u, v )) = ∅ and
|E(v1 , X0 )| = 1 for any v1 ∈ / X0 and E(v1 , X0 \V (P(u, v ))) ̸= ∅.
(4.4)
Let V1 = NG (x1 )\{x, x′ }, V2 = NG (V1 )\{x1 } and V3 = NG (V2 )\V1 . Then E(V1 , X0 ) = ∅ by Eq. (4.3) and V2 ̸ = ∅ since x1 is not a cut vertex. If E(V2 , X0 ) = ∅, then x = y = u′ and x′ = y′ = v ′ since G is P7 -free and V3 ̸ = ∅ since x1 is not a cut vertex. Then there is a vertex v3 ∈ V3 ∩ NG ({u′ , v ′ , u, v}) such that either |E(v3 , {u′ , v ′ , u, v, z1 , z2 })| ≤ 1 or E(v3 , X0 ) = {v3 u, v3 v} and then there is an induced P7 , a contradiction. So E(V2 , X0 ) ̸ = ∅ and there are two vertices v1 ∈ V1 , v2 ∈ V2 such that v1 v2 ∈ E(G). Then |E(v2 , X0 )| ≥ 2, E(v2 , X0 ) = E(v2 , P(u, v )) = {v2 w : w ∈ V (P(u, v ))}, x = y = u′ and x′ = y′ = v ′ since G is P7 -free. Hence G[V (P(u, v )) ∪ {v2 }] is collapsible. Then dT (v1 ) ≥ 4; for otherwise, T △u′ uvv ′ v2 v1 x1 u′ is a closed trail with larger number of order than T , a contradiction. By symmetry, there is a vertex v2′ ∈ V2 such that G[V (P(u, v )) ∪ {v2′ }] is collapsible and then G[V (P(u, v )) ∪ {u′ , v ′ , x1 , v1 , v2 , v2′ }] is collapsible, contradicting the choice of F1 , F2 . Case 2. dG (F1 , F2 ) ≥ 3. Then min{m, n} ≥ 2. We claim that x1 yn , y1 xm ∈ / E(G). Suppose otherwise. By symmetry, assume that x1 yn ∈ E(G). Then exactly one of T △uP(u, v )vv ′ yn x1 u′ u (if u′ = x and v ′ = y′ ), (T1 ∪ T1 (u′ , x)) △ (u′ uP(u, v )vv ′ ∪ xx1 yn y′ ) (if u′ ̸ = x and v ′ = y′ ) or (T0 ∪ T1 (u′ , x) ∪ T2 (v ′ , y′ )) △ (u′ uP(u, v )vv ′ ∪ xx1 yn y′ ) (if u′ ̸= x and v ′ ̸= y′ ) is a closed trail with larger number of order than T , a contradiction. Then u′ = x = y, v ′ = x′ = y′ and P(u, v ) = uv since G is P7 -free. We claim that {x1 , y1 , xm , yn } ⊆ V2 (G). / E(G) by the Suppose otherwise. By symmetry, assume that there is a vertex x′1 ∈ NG (x1 )\{u′ , x2 }. Then x′1 u′ ∈ choice of F1 and E(x′1 , {u′ , u, v}) = ∅ by the choice of F1 and Eq. (4.3). Besides, E(x′1 , {v ′ , yn }) = ∅; for otherwise, T △uvv ′ x′1 x1 u′ u, T △uvv ′ yn x′1 x1 u′ u is a closed trail with larger number of order than T , respectively, a contradiction. However G[{x′1 , x1 , u′ , u, v, v ′ , yn }] ∼ = P7 , a contradiction. Thus m = n = 2; for otherwise, assume that n ≥ 3, then E(x2 , {u′ , u, v, v ′ , yn }) = ∅ and G[{x2 , x1 , u′ , u, v, v ′ , yn }] ∼ = P7 , a contradiction. This implies that there is no induced path of order at least 5 between u′ and v ′ . If there is a vertex u1 ∈ NG (u)\{u′ , v}, then E(u1 , {u′ , x1 , x2 , v ′ , y2 }) = ∅ and G[{u1 , u, u′ , x1 , x2 , v ′ , y2 }] ∼ = P7 , a contradiction. Thus dG (u′ ) = dG (v ′ ) = 2 and V≥3 (G) = {u′ , v ′ } and then G ∼ = θ (2, r) for r = dG (u′ ) = dG (v ′ ) and r is odd. □ The following two theorems completely characterize forbidden subgraphs that force a connected (2-edge-connected or 2-connected) graph to have a DCT and forbidden pairs that force a connected graph to have a DCT. Theorem 12. Let H and G be connected graphs of order at least 3. Then H-free graph G implies G has a DCT if and only if H satisfies one of the following: (1) H is an induced subgraph of P4 , (2) H is an induced subgraph of P5 when κ ′ (G) ≥ 2, (3) H is an induced subgraph of P6 when κ (G) ≥ 2. Proof. The sufficiency By contradiction, assume that G has no DCT and choose a closed trail T such that |V (T )| is maximized. Then there is an induced path P(u′ , v ′ ) ⊆ G − T such that {u′ } ⊆ V (P(u′ , v ′ )) ∩ V (T ) and {u′ , v ′ } ⊆ V (P(u′ , v ′ )) ∩ V (T ) if κ ′ (G) ≥ 2. Furthermore, |E(P(u′ , v ′ ))| ≥ 2 or |E(P(u′ , v ′ ))| ≥ 3 if κ ′ (G) ≥ 2 or |E(P(u′ , v ′ ))| ≥ 4 if κ (G) ≥ 2. So there is an induced P4 or an induced P5 if κ ′ (G) ≥ 2 or an induced P6 if κ (G) ≥ 2. The necessity The graph G′1 is the graph obtaining from K2,3 by replacing each vertex degree 2 in K2,3 by a clique Km for any m ≥ 3 such that |NG′ ({u, v}) ∩ V (Km )| = 1, where dK2,3 (u) = dK2,3 (v ) = 3. Note that G1 , G6 , G10 depicted in Fig. 2 1 with k = 2 and the graph G′1 have no DCT. They contain H as an induced subgraph. Since G6 is K3 -free and G10 is K1,3 -free and they have no common cycle, H is a subgraph of path. Then H ⊆ P4 since G1 is P5 -free or H ⊆ P5 since G′1 is P6 -free and κ ′ (G′1 ) ≥ 2 or H ⊆ P6 since G6 is P7 -free and κ (G6 ) ≥ 2. □ Theorem 13.
Every connected {R, S }-free graph G other than a tree has a DCT if and only if {R, S } ⪯ {Z2 , P5 }, {Z2 , T2,1,1 }.
Proof. The necessity is clearly true. The sufficiency Assume that G has no DCT. Choose a closed trail T of G such that |V (T )| is maximized. Let H = G[V (T )]. Then there is an edge uv ⊆ G − H such that ux ∈ E(G) and x1 xx2 x3 ⊆ T . Then E(v, {x, x1 , x2 , x3 }) = ∅ by the maximality of T . Since G is Z2 -free, x1 ̸ = x3 and x1 x2 , xx3 ∈ / E(G). Hence G[{x, x1 , x2 , u, v}] ∼ = T2,1,1 and G[{x3 , x2 , x, u, v}] ∼ P . □ = 5 Here the graph Z ′ is the graph obtained by identifying one vertex of K3 and the leaf of T2,2,1 whose neighbor has degree 3. Theorem 14.
Every 2-edge-connected {P6 , Z ′ }-free graph G has a DCT.
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Proof. By contradiction, assume that G has no DCT. Choose a closed trail T of G such that |V (T )| is maximized. Let H = G[V (T )]. Then G[V (G)\V (H)] has at least one non-trivial component H1 . Choose two induced paths P(u, v ) ⊆ H1 and P(u′ , v ′ ) = u′ x1 · · · xt v ′ ⊆ H of G such that (1) E(P(u, v ), P(u′ , v ′ )) = {uu′ , vv ′ }; (2) subject to (1), |E(P(u, v ))| is maximized; (3) subject to (1) and (2), |E(P(u′ , v ′ ))| is maximized. Then u′ v ′ ∈ / E(G); for otherwise, T △uP(u, v )vv ′ u′ u is a closed trail with larger number of order than T , a contradiction. Note that P(u, v ) ∪ P(u′ , v ′ ) ∪ {uu′ , vv ′ } is an induced cycle of G, 1 ≤ t ≤ 3 since G is P6 -free. Besides, there is a vertex z1 ∈ NH (u′ ) since H is 2-edge-connected. Claim 5.
E(z1 , {u, v, v ′ , xt }) = ∅.
Proof. Suppose otherwise. Then T △uu′ z1 u (if z1 u ∈ E(G)), T △v uu′ z1 v (if z1 v ∈ E(G)), T △uu′ z1 v ′ v u (if z1 v ′ ∈ E(G)) or T △uu′ z1 xt v ′ v u (if z1 xt ∈ E(G)) is a closed trail of G with larger number of order than T , respectively, a contradiction. □ If t = 3, then P(u, v ) = u = v since G is P6 -free. Since H1 is non-trivial and by the choice of P(u, v ), there is a vertex u1 ∈ NH1 (u) with E(u1 , H) = ∅ . However G[{u1 , u, u′ , x1 , x2 , x3 }] ∼ = P6 , a contradiction. If t = 2, then P(u, v ) = u = v , since otherwise P(u, v ) = uv and then G[{z1 , u′ , u, v, v ′ , x2 }] ∼ = P6 by Claim 5, a contradiction. Besides, V (H1 ) = NH1 [u]. Then there are at least two vertices u1 , u2 ∈ V (H1 ) with u1 u2 ∈ E(G) and hence H1 is collapsible and H1 has a spanning closed trail T0 with |V (T0 )| ≥ 3. Then at least one of x1 , x2 is a cut vertex of T ; for otherwise, T1 = (T ∪ T0 )△uu′ x1 x2 v ′ u is a closed trail with |V (T1 )| > |V (T )|, a contradiction. Without loss of generality, assume that x2 is a cut vertex of T , x1 x2 v ′ ⊆ T and there is a vertex x′2 ∈ NT (x2 )\{v ′ , x1 }. Then E(x′2 , {v ′ , u, u′ , z1 }) = ∅; for otherwise, (T ∪ T0 )△x2 x′2 v ′ x2 , (T ∪ T0 )△x2 x′2 uv ′ x2 , (T ∪ T0 )△x2 x′2 u′ uv ′ x2 , (T ∪ T0 )△x2 x′2 z1 u′ uv ′ x2 is a closed trail with larger number of order than T , respectively, a contradiction. However G[{x′2 , x2 , v ′ , u, u′ , z1 }] ∼ = P6 , a contradiction. Thus t = 1, u′ x1 v ′ ⊆ T and x1 is a cut vertex of T by the choice of T . Then there is a cycle C1 ⊆ T such that V (C1 ) ∩ {u′ , v ′ , x1 } = {x1 }. We claim that V (C1 ) ⊆ NG [x1 ]. Suppose otherwise. There is an induced path x′′1 x′1 x1 ⊆ G[V (C1 )]. Note that E(x′1 , {u′ , v ′ } ∪ V (H1 )) = ∅; for otherwise, T △x′1 u′ uP(u, v )vv ′ x1 x′1 (if x′1 u′ ∈ E(G)), T △x′1 v ′ v P(u, v )uu′ x1 x′1 (if x′1 v ′ ∈ E(G)) or T △x′1 x1 u′ uP(u, u1 )u1 x′1 (if x′1 u1 ∈ E(G) for some u1 ∈ V (H1 )) is a closed trail of G with larger number of order than T , respectively, a contradiction. Since G is P6 -free, E(x′′1 , {u′ , v ′ } ∪ V (H1 )) ̸ = ∅. By the same argument, we ′′′ ′′ ′ ′ ′ can prove that x′1 is a cut vertex of T and there is a vertex x′′′ 1 ∈ NT (x1 )\{x1 , x1 } with E(x1 , {u , v , x1 } ∪ V (H1 )) = ∅ and ′ ′ ∼ (u), a contradiction. G[{x′′′ , x , x , u , u , u }] P for any u ∈ N = 1 1 6 1 H1 1 1 Note that there is a path P(u′ , v ′ ) = u′ y1 · · · yl v ′ ⊆ T − C1 − u′ x1 v ′ for some l ≥ 1. By symmetry, l = 1, y1 is a cut vertex of T and there is a cycle C2 ⊆ T such that V (C2 ) ⊆ NG [y1 ]. Hence G[{x1 , x′1 , x′′1 , y1 , y′1 , u′ , u, u1 }] ∼ = Z ′ for some u1 ∈ NH1 (u), y′1 ∈ NC2 (y1 ) and x′1 , x′′1 ∈ NC1 (x1 ) with x′1 x′′1 ∈ E(G), a contradiction. □ Theorem 15.
Let R, S be connected graphs. Then
(1) if every 2-connected {R, S }-free graph G of order at least 2r + 3 ≥ 13 implies G has a DCT, then {R, S } ⪯ {T2,2,2 , N2,2,4 }, {T2,2,2 , B2,6 }, {T3,1,1 , B2,6 }, {T3,1,1 , N2,2,4 }, {P8 , C6′ }, {P7 , T2,r }, (2) if every 2-edge-connected {R, S }-free graph G other than P(2, 2, 2) implies G has a DCT, then {R, S } ⪯ {T2,2,2 , N2,2,4 }, {T2,2,2 , B2,6 }, {P6 , Z ′ }. Proof. (1) can be immediately obtained by Theorem 4 (2) when k = 1. (2) The graph G′2 is the graph obtaining from θ (2, 2, 1) by replacing the common neighbor of two vertices of degree 3 with a clique Km for any m ≥ 3 such that |NG′ (Km )| = 2. The graph G′3 is the graph obtained from K2,t by replacing each 2 vertex degree 2 in K2,t by K3 such that |NG′ (K3 )| = 2 for any odd integer t ≥ 3. Then G′2 and G′3 have no DCT. Note that 3 G′3 is K4 -free, P6 -free, C6 -free and T1,1,3 -free. Then if R has K1,3 as a subgraph, then S is an induced subgraph of N2,2,4 or B2,6 . Otherwise, {R, S } ⪯ {P6 , Z ′ } since the common induced cycle of G′2 and G′3 is C3 . □ By comparing Theorems 5, 13, 14 (2) and 15, we know if the following conjecture is true, we may accomplish the complete characterization of forbidden pairs forcing a 2-edge-connected or 2-connected graph to have a DCT, i.e., h(G) ≤ 1. Conjecture 16.
It would be true that
(1) Every 2-connected {P8 , C6′ }-free graph G has a DCT. (2) Every 2-edge-connected {T2,2,2 , H }-free graph G for H ∈ {N2,2,4 , B2,6 } other than P(2, 2, 2) has a DCT. Declaration of competing interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
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Acknowledgments This work is supported by Nature Science Funds of China (Nos. 11871099 and 11671037) and by Natural Science Foundation of Qinghai Province, China (No. 2018-ZJ-717). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
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Forbidden subgraphs on Hamiltonian index Xia Liu a , Liming Xiong b a b
School of Mathematics and Statistics, Beijing Institute of Technology, Beijing 100081, PR China School of Mathematics and Statistics, Beijing Key Laboratory on MCAACI, Beijing Institute of Technology, Beijing 100081, PR China
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Article history: Received 9 February 2019 Accepted 20 January 2020 Available online xxxx Keywords: Forbidden subgraph Hamiltonian index Collapsible DCT
a b s t r a c t Let G be a graph other than a path. The m-iterated line graph of a graph G is Lm (G) = L(Lm−1 (G)). where L1 (G) denotes the line graph L(G) of G. Define the Hamiltonian index h(G) of G to be the smallest integer m such that Lm (G) contains a Hamiltonian cycle. For a connected graph set H, G is said to be H-free if G does not contain H as an induced subgraph for all H ∈ H. In this paper, we characterize all forbidden graphs H for any integer k ≥ 2 and partial forbidden graphs H for k = 1 such that a connected (or 2-edge-connected or 2-connected) H-free graph G satisfies that h(G) ≤ k for the case when |H| ≤ 2. What is more, we settle four conjectures proposed in Holub (2014). © 2020 Elsevier B.V. All rights reserved.
1. Introduction All graphs considered in this paper are simple graphs. For the notation or terminology not defined here, see [1]. A graph is called trivial if it has only one vertex, nontrivial otherwise. We use κ (G), κ ′ (G) to denote the connectivity, edge connectivity of G, respectively. The line graph L(G) of G = (V (G), E(G)) has E(G) as its vertex set, and two vertices are adjacent in L(G) if and only if the corresponding edges share a common end vertex in G. The m-iterated line graph Lm (G) is defined recursively by L0 (G) = G, L1 (G) = L(G) and Lm (G) = L(Lm−1 (G)). The Hamiltonian index of a graph G, denoted by h(G), is the smallest integer m such that Lm (G) contains a Hamiltonian cycle. For a connected graph G that is not a path, Ryjáček, Woeginger and Xiong [10] showed that the problem to decide whether the Hamiltonian index of a given graph is less than or equal to a given constant is NP-complete, Chartrand [4] showed that the Hamiltonian index of G always exists and Saražin [11] showed h(G) ≤ n − ∆(G) if G is connected. For its upper bounds and stability, see [3], [13] and [15] and its survey paper, see [16]. An edge of G is said to be subdivided when it is deleted and replaced by a path of length 2 connecting its ends, the internal vertex of this path being a new vertex. For r ≥ 3 and i1 ≥ · · · ≥ ir ≥ 0, define Ti1 ,...,ir , θ (i1 , . . . , ir ) be the graphs obtained from K1,r or K2,r by subdividing i1 − 1, . . . , ir − 1 times of the r incident edges of K1,r or K2,r , respectively. Especially, if i1 = · · · = ir = j, then we denote Ti1 ,...,ir = Tj,r and θ (i1 , . . . , ir ) = θ (j, r) for convenience. Let Pn be the path of order n. We shall prove the following result, which is stronger than four conjectures proposed in [9] and considered only their sufficiency. Theorem 1. Let k ≥ 0, l, m, n ≥ 2 be integers and H be a connected graph. Then 2-edge-connected (or 2-connected) graph G does not contain H as its subgraph (not necessarily induced) implying h(G) ≤ k if and only if H is isomorphic to one of {P3k+5 , T1,1,2k+2 } ∪ {T1,m,n : m + n = 3k + 3} ∪ {Tl,m,n : l + m + n = 3k + 4}. E-mail addresses: [email protected] (X. Liu), [email protected] (L. Xiong). https://doi.org/10.1016/j.disc.2020.111841 0012-365X/© 2020 Elsevier B.V. All rights reserved.
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X. Liu and L. Xiong / Discrete Mathematics 343 (2020) 111841
Fig. 1. The common induced subgraphs.
Note that the graph H in above theorem is generally not necessarily induced, it is natural to consider the case that H is an induced subgraph of G. They maybe more popular in the literature. A graph G is said to be H-free if G does not contain H as an induced subgraph for all H ∈ H, and we call each graph H of H a forbidden subgraph. Especially, if {H } = H, then we simply say that G is H-free and we call H a forbidden pair if |H| = 2. For two sets H1 and H2 of connected graphs, we write H1 ⪯ H2 if for every graph H2 in H2 , there is a graph H1 in H1 such that H1 is an induced subgraph of H2 . By the ′ definition of the relation ‘‘⪯’’, if H1 ⪯ H2 , then every H1 -free graph is also H2 -free (see [7]). The graphs Zi , Bi,j , Ni,j,k , C2i +4 are depicted in Fig. 1. First of all, we characterize all forbidden subgraphs H with |H| = 1, 2 such that a connected graph G is H-free implying h(G) ≤ k for any integer k ≥ 0. Chartrand and Wall [5] determined the Hamiltonian index of a tree. So we may assume that G is not a tree. Theorem 2.
Let G, H be connected graphs and H be a connected forbidden pair. Then for any integer k ≥ 0,
(1) H-free graph G implies h(G) ≤ k if and only if H is an induced subgraph of Pk+3 , (2) H-free graph G (other than a tree) implies h(G) ≤ k for k ≥ 1 if and only if H ⪯ {Zk+1 , Pk+4 } or {Zk+1 , Tk+1,1,1 }. Note that Theorem 2 (1) for k = 0 and (2) are clear and also true for the case that k = 0 and κ ′ (G) ≥ 2. The following result solves the case that κ (G) ≥ 2 and k = 0, while Theorem 4 reveals the case when κ ′ (G) ≥ 2 and κ (G) ≥ 2 for k ≥ 2. Theorem 3 (Faudree and Gould, [6]). Let H be a connected graph and H be a connected forbidden pair. Then (1) 2-connected H-free graph G implies h(G) = 0 if and only if H = P3 , (2) 2-connected H-free graph G of sufficiently large order implies h(G) = 0 if and only if H ⪯ {K1,3 , P6 }, {K1,3 , Z3 }, {K1,3 , B1,2 } or {K1,3 , N1,1,1 }. Theorem 4. Let k ≥ 2, i1 , i2 , j1 , j2 ≥ 1 be integers with i1 + j1 = i2 + j2 − 1 = 2k + 2, H (H, respectively) be a connected graph (connected forbidden pair, respectively). Then (1) every 2-edge-connected (or 2-connected) H-free graph G implies h(G) ≤ k if and only if H is an induced subgraph of P3k+3 , (2) every 2-connected H-free graph G of order at least r(k + 1) + 2 ≥ 4k + 6 implies h(G) ≤ k if and only ′ if H ⪯ {C2k +4 , P3k+5 }, {Tk+1,k+1,k+1 , Nk+1,k+1,k+3 }, {Tk+1,k+1,k+1 , Bk+1,2k+4 }, {Ti1 ,j1 ,k , Nk+1,k+1,k+3 }, {Ti1 ,j1 ,k , Bk+1,2k+4 }, {Tk+1,k+1,k+1 , Ni2 ,j2 ,k+1 }, {Ti1 ,j1 ,k , Ni2 ,j2 ,k+1 } or {P3k+4 , Tk+1,r }, ′ (3) every 2-edge-connected H-free graph G of order at least 3k + 10 implies h(G) ≤ k if and only if H ⪯ {C2k +4 , P3k+5 }, {Tk+1,k+1,k+1 , Nk+1,k+1,k+3 }, {Tk+1,k+1,k+1 , Bk+1,2k+4 }, {Ti1 ,j1 ,k , Nk+1,k+1,k+3 }, {Ti1 ,j1 ,k , Bk+1,2k+4 }, {Tk+1,k+1,k+1 , Ni2 ,j2 ,k+1 }, {Ti1 ,j1 ,k , Ni2 ,j2 ,k+1 } or {P3k+4 , Tk+1,k+1,k+1,1 }. The proofs of Theorem 2 for k ≥ 2 and Theorem 4 will be exhibited in Section 3 and Theorem 2 for k = 1 is proved by Theorems 12 (1) and 13 in Section 4. What is more, we also characterize forbidden subgraphs and partial pairs forcing a graph G to satisfy that h(G) ≤ 1 in Section 3. Comparing Theorem 4 and the results in Section 3, we know as follows.
• Theorem 4 (1) for 2-connected graph is true for k = 1, • Theorem 4 (1) for 2-edge-connected graph and Theorem 4 (3) are not true for k = 1. We conjecture that Theorem 4 (2) would be true for k = 1 and leave one conjecture in the last section. Although we cannot completely prove it now, we may show the following results. Theorem 5.
Let G be a 2-edge-connected graph satisfying one of the following:
(1) G is T2,1,1 -free and H-free for any H ∈ {N2,2,4 , B2,6 } other than P(2, 2, 2), (2) G is 2-connected {T3,1,1 , H }-free for any H ∈ {N2,2,4 , B2,6 } other than P(2, 2, 2), (3) G is 2-connected P7 -free other than θ (2, r) for any odd integer r. Then h(G) ≤ 1.
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2. A useful result Use K (i, j, k) or P(i, j, k) to denote the graphs obtained from a K4 or the triangular prism by subdividing i, j, k times of the three edges which are incident in K4 or are not in a K3 of the triangular prism, respectively. Here is our main result in this section, which would be applied to the proofs of other main results. Theorem 6. Let k ≥ 2 be an integer and G be a 2-edge-connected graph with h(G) ≥ k + 1. Then G has an induced subgraph G0 ∈ {θ (l, m, n), K (l, m, n), P(l, m, n)} for some l ≥ m ≥ n ≥ k + 1 such that dG (u, v ) ≥ k + 1 for any pairs of disjoint vertices u, v ∈ V3 (G0 ) in G0 . Furthermore, V2 (G0 ) ⊆ V2 (G) if l = m = n = k + 1 ≥ 4. Before presenting its proof, we shall introduce some notations. Let u be a vertex of G. Use NG (u) to denote the neighborhood and dG (u) to denote the degree of u. Call u a k-vertex if dG (u) = k. Define Vi (G) = {u ∈ V (G) : dG (u) = i} and V≥i (G) = {u⋃∈ V (G) : dG (u) ≥ i}. Let S be a subset of V (G)(or E(G)). The induced subgraph of G is denoted by G[S ]. Define NG [S ] = x∈S NG (x), NG (S) = NG [S ]\S, E(u, S) = {uv ∈ E(G) : v ∈ S } and E(u, H) = E(u, V (H)) if H is a subgraph of G. For two graphs F , G, we write F ⊆ G if F is a subgraph of G, F ∼ = G if F and G are isomorphic and NG (F ) = NG (V (F )). Let P(u, v ) to denote a path between u and v . A branch in G is a nontrivial path with ends not in V2 (G) and with internal vertices, if any, that have degree 2. We denote by B(G) the set of branches of G. Define B1 (G) = {B ∈ B(G) : V (B)∩V1 (G) ̸ = ∅. The distance dH (H1 , H2 ) between two subgraphs H1 and H2 of H is defined to be min{dH (v1 , v2 ) : v1 ∈ V (H1 ), v2 ∈ V (H2 )}, where dH (v1 , v2 ) denotes the number of edges of a shortest path between v1 and v2 in H. The graph H1 ∪ S is a graph with V (H1 ) ∪ S and E(H1 ) as its vertex set and edge set. We then will define some operations which may be different from generic notations such that the resulting graphs have no isolated vertex. Define H1 ∪ H2 = H [E(H1 ) ∪ E(H2 )], H1 ∩ H2 = H [E(H1 ) ∩ E(H2 )], H1 − H2 = H [E(H1 )\E(H2 )], H1 △H2 = H [E(H1 )△E(H2 )] = H [(E(H1 ) ∪ E(H2 ))\(E(H1 ) ∩ E(H2 ))], respectively. Xiong and Liu [14] characterized the graphs for which the k-iterated line graph is Hamiltonian for any integer k ≥ 2. Theorem 7 (Xiong and Liu, [14]). Let G be a connected simple graph and let k ≥ 2 be an integer. Then h(G) ≤ k if and only if EUk (G) ̸ = ∅, where EUk (G) denotes the set of those subgraphs H of a graph G that satisfy the following conditions: (I) (II) (III) (IV) (V)
dH (x) ≡ 0(mod 2) for every x ∈ V (H), V0 (H) ⊆ V≥3 (G) ⊆ V (H), dG (H1 , H − H1 ) ≤ k − 1 for every subgraph H1 of H, |E(B)| ≤ k + 1 for every branch B ∈ B(G) with E(B) ∩ E(H) = ∅, |E(B)| ≤ k for every branch B ∈ B1 (G).
Theorem 8 (Wang and Xiong, [12]). Let G be a connected graph with three pendent vertices v1 , v2 , v3 . Then G either has an induced Ti,j,k or an induced Ni′ ,j′ ,k′ with leaves v1 , v2 , v3 for some i ≥ j ≥ k ≥ 1 and i′ ≥ j′ ≥ k′ ≥ 1. For two connected graphs L1 , L2 , we say a graph G belongs to the graph set θL1 ,L2 (i, j, k) if and only if G is obtained by connecting graphs L1 , L2 by adding three internal-disjoint paths of order i + 2, j + 2, k + 2, respectively. The proof of Theorem 6. Choose a subgraph H of G satisfying that: (1) dH (x) ≡ 0( mod 2) for every x ∈ V (H), (2) V0 (H) ⊆ V≥3 (G) ⊆ V (H), (3) subject to (1), (2), |B(G; H)| is minimized, where B(G; H) = {B ∈ B(G) : E(B) ∩ E(H) = ∅, |E(B)| ≥ k + 2},
(2.1)
(4) subject to (1), (2) and (3), |(G; H)| is minimized, where (G; H) = {H1 ⊆ H : dG (H1 , H − H1 ) ≥ k}. If B(G; H) = ∅ and (G; H) = ∅, then H ∈ EUk (G) since B1 (G) = ∅ and then h(G) ≤ k by Theorem 7, a contradiction. Then we assume that B(G; H) ̸ = ∅, i.e., there is a branch B(x, y) ∈ B(G; H) whose end vertices x, y are in at most ′ ′ two components H1 (and H2 ) of H. Let G1 = G[V (H ( 1 )]. Choose paths P1 (x, x ) ⊆ G1)and P2 (x , y) ⊆ G − B(x, y) − G1 (if H1 = H2 , then x′ = y and P2 (x′ , y) = y). Let H 1 = H △(B(x, y) ∪ P1 (x, x′ ) ∪ P2 (x′ , y)) ∪ V≥3 (G). Then H 1 satisfies (1), (2) of 2.1. Besides, there is a branch B1 (x1 , y1 ) ⊆ P1 (x, x′ ) ∩ H1 or B1 (x1 , y1 ) ⊆ P2 (x′ , y) ∩ H2 with length at least k + 1; for otherwise, B(G; H 1 ) ⊆ B(G; H)\{B(x, y)}, contradicting the choice of H. By symmetry, assume that B1 (x1 , y1 ) ⊆ P1 (x, x′ )∩H1 . Since κ ′ (G1 ) ≥ 2, there are at least two edge-disjoint paths between x and x′ in G1 , which means that G1 − B1 (x1 , y1 ) is connected. Repeat above operation, we can obtain branches B2 (x2 , y2 ), . . . , Bs (xs , ys ) with length at least k + 1 for some s ≥ 2 until G1 − B1 (x1 , y1 ) − · · · Bs (xs , ys ) has at least two connected subgraph L1 , L2 containing x, x′ , respectively. Then dG (L1 , L2 ) ≥ k + 1 and G0 = G[V (B(x, x′ )) ∪ V (B(x′ , y)) ∪ V (B1 (x1 , y1 )) ∪ V (B2 (x2 , y2 )) ∪ V (L1 ) ∪ V (L2 )] ∈ θL1 ,L2 (a1 , a2 , a3 ) for some a1 , a2 , a3 ≥ k + 1. By Theorem 8, G0 has an induced subgraph G0 ∈ {θ (l, m, n), K (l, m, n), P(l, m, n)} for some l ≥ m ≥ n ≥ k + 1 and V2 (G0 ) ⊆ V2 (G).
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X. Liu and L. Xiong / Discrete Mathematics 343 (2020) 111841
We then assume that (G; H) ̸ = ∅, i.e., there is a connected subgraph H1 of H such that dG (H1 , H − H1 ) ≥ k. Then there are at least two branches P1 (x1 , y1 ), P2 (x2 , y2 ) with length at least k between H1 and H − H1 such that x1 , x2 ∈ V (H1 ), y1 , y2 ∈ V (H − H1 ) since κ ′ (G) ≥ 2. A path P satisfies the property Γ (H) if H has a connected subgraph H0 such that G either has a branch B ⊆ P ∩ H0 with |E(B)| ≥ k + 2 or a subgraph ˜ H and two branches B1 , B2 ⊆ P ∩ H0 satisfying that NG (˜ H) ⊆ V (B1 ∪ B2 ) and dG (˜ H , H − B1 − B2 − ˜ H) ≥ k. Let G1 = G[V (H1 )] and G2 = G[(V (G)\V (H1 ∪ P1 (x1 , y1 ))) ∪ V (P2 (x2 , y2 )) ∪ {y1 , y2 }]. Then G2 is connected and either each path of G1 with end vertices x1 , x2 or each path of G2 with end vertices y1 , y2 satisfies the property Γ (H). Suppose otherwise. Let C2 = P1 (x1 , y1 ) ∪ P4 (y1 , y2 ) ∪ P2 (x2 , y2 ) ∪ P3 (x1 , x2 ) and H 2 = (H △C2 ) ∪ V≥3 (G) if there are two paths P3 (x1 , x2 ) ⊆ G1 , P4 (y1 , y2 ) ⊆ G2 do not satisfy the property Γ (H). Then H 2 is a subgraph satisfying (1), (2) of 2.1 and |(G; H 2 )| < |(G; H)|, contradicting the choice of H. By symmetry, assume that each path with end vertices y1 , y2 in G2 satisfies the property Γ (H). Then there is a non-trivial connected subgraph H0 ⊆ H − H1 such that each path of H0 with end vertices z1 , z2 satisfies the property Γ (H) for any P(y1 , y2 ) ⊆ G2 and z1 , z2 ∈ V (H0 ) ∩ V (P(y1 , y2 )). Choose all paths P1 (z1 , z2 ), . . . , Pt (z1 , z2 ) ⊆ H0 between z1 and z2 for some t ≥ 2. Then there are paths P1 ⊆ ˜ 1 with |E(Pi )| ≥ 2k, P1 (z1 , z2 ), . . . , Pt ⊆ Pt (z1 , z2 ) satisfy that either |E(Pi )| ≥ k and Pi is a branch of G or Pi ⊆ B11 ∪ B21 ∪ H ˜ i is a connected subgraph such that NG (H˜ i ) ⊆ V (B1 ∪ B2 ) for i ∈ {1, . . . , t }. Hence for where B11 , B21 are two branches and H i i any i, j ∈ {1, . . . , t }, Pi , Pj are either identical or internal vertex-disjoint. Without loss of generality, we assume that P1 and P2 are internal vertex-disjoint. Then G[V (H0 )] − P1 − · · · − Ps has two connected subgraphs L0 , L′0 such that G0 = G[V (P1 ∪ P2 ∪ L0 ∪ L′0 ∪ P1 (x1 , y1 ) ∪ P2 (x2 , y2 ) ∪ P1 (x1 , x2 ) ∪ P(y1 , z1 ) ∪ P(y2 , z2 ))] ∈ θL0 ,L′ (a1 , a2 , a3 ) for some 0
a1 , a2 , a3 ≥ min{k + 1, 2k − 1} = k + 1 for any induced paths P(y1 , z1 ), P(y2 , z2 ) ⊆ G2 . By Theorem 8, G0 has an induced subgraph G0 ∈ {θ (l, m, n), K (l, m, n), P(l, m, n)} for some l ≥ m ≥ n ≥ k + 1. Especially, if l = m = n = k + 1 ≥ 4, then P1 , . . . , Ps are branches of G since 2k − 1 > k + 1 for k ≥ 3 and then V2 (G0 ) ⊆ V2 (G). □ 3. The proofs of Theorems 2 and 4 We prove the following result by Theorem 6.
Theorem 9. Let k ≥ 2, i1 , i2 , j1 , j2 ≥ 1 be integers with i1 + j1 = i2 + j2 − 1 = 2k + 2 and G be a 2-edge-connected graph satisfying exactly one of the following: (1) (2) (3) (4) (5)
G G G G G
is is is is is
′ {P3k+5 , C2k +4 }-free, {H1 , Ni2 ,j2 ,k+1 }-free for H1 ∈ {Tk+1,k+1,k+1 , Ti1 ,j1 ,k }, {H1 , H2 }-free for H1 ∈ {Tk+1,k+1,k+1 , Ti1 ,j1 ,k } and H2 ∈ {Nk+1,k+1,k+3 , Bk+1,2k+4 } other than P(k + 1, k + 1, k + 1), {P3k+4 , Tk+1,k+1,k+1,1 }-free other than θ (k + 1, k + 1, k + 1), 2-connected P3k+4 -free graph other than θ (k + 1, r) for any r ≥ 3 and r is odd.
Then h(G) ≤ k. Proof. We argue by contradiction. Let G be a 2-edge-connected graph with h(G) ≥ k + 1. Then G has an induced subgraph G0 ∈ {θ (l, m, n), K (l, m, n), P(l, m, n)} for some l ≥ m ≥ n ≥ k + 1 by Theorem 6. Besides, G0 is H-free if G is H-free. ′ (1) Since G0 is P3k+5 -free, G0 ∼ = θ (k + 1, k + 1, k + 1). However, G0 has an induced subgraph C2k +4 , a contradiction. (2) Since G0 is Ni2 ,j2 ,k+1 -free, G0 ∈ {θ (k + 1, k + 1, k + 1), K (k + 1, k + 1, k + 1)}. However, G0 has induced subgraphs Tk+1,k+1,k+1 , Ti1 ,j1 ,k , a contradiction. (3) Assume that G0 is either Tk+1,k+1,k+1 -free or Ti,j,k -free. Then G0 ∼ = P(k + 1, k + 1, k + 1) and V2 (G0 ) ⊆ V2 (G); for otherwise, G0 has induced subgraphs Nk+1,k+1,k+3 and Bk+1,2k+4 , a contradiction. Assume that xyzx, x′ y′ z ′ x′ ⊆ G0 . Then dG ({x, y, z }, {x′ , y′ , z ′ }) = k + 1. Besides, V (G − G0 ) ̸ = ∅ since G ≇ P(k + 1, k + 1, k + 1). By symmetry, assume that there is a vertex x0 ∈ NG (x) with E(x0 , {x′ , y′ , z ′ }) = ∅ for x′ y′ z ′ ̸ = xyz and x′ y′ z ′ ⊆ G0 and hence G[V (G0 ) ∪ {x0 }] has an induced Nk+1,k+1,k+3 . Then E(x0 , {y, z }) = ∅ since G[V (G0 ) ∪ {x0 }] does not have an induced Bk+1,2k+4 . Note that if there is an induced path P ′ between x0 and x′ y′ z ′ x′ in G − G0 , then |E(P ′ )| ≥ k and there are induced Tk+1,k+1,k+1 and Ti,j,k with x as their 3-vertex, a contradiction. Hence, there is an induced path P(x0 , y0 ) of order at least 2 between x0 and {y, z } in G − G0 such that NG ({y, z }) ∩ V (P(x0 , y0 )) = {y0 } since G is 2-edge-connected. Then there is an induced Bk+1,2k+4 , a contradiction. (4) Since G0 is P3k+4 -free, G0 ∼ = θ (k + 1, k + 1, k + 1). If there is a vertex x0 ∈ NG (x) for some x ⊆ xyz ⊆ G0 with E(x0 , {x′ , y′ , z ′ }) = ∅ for x′ y′ z ′ ̸ = xyz and x′ y′ z ′ ⊆ G0 , then G[V (G0 ) ∪ {x0 }] has an induced Tk+1,k+1,k+1,1 , a contradiction. Then V (G) = V (G0 ) and G ∼ = θ (k + 1, k + 1, k + 1) if k ≥ 3, a contradiction. Then k = 2 and G ≇ θ (3, 3, 3). So there is a vertex x′2 ∈ NG (x2 )\{x1 , x3 } for some x1 , x2 , x3 ∈ V2 (G0 ) and x1 x2 x3 ⊆ G0 such that |E(x′2 , G0 )| = 1. Then G[V (G0 ) ∪ {x′2 }] has an induced P10 , a contradiction. (5) By the same argument, we firstly prove that G ∼ = θ (3, 3, 3) for k = 2 since G is P10 -free. For any k ≥ 3, G0 ∼ = θ (k + 1, k + 1, k + 1) since G0 is P3k+4 -free. Assume that V3 (G0 ) = {u, v}. Since G ≇ θ (k + 1, k + 1, k + 1), NG ({u, v}) ̸ ⊆ V (G0 ). By symmetry, assume NG−G0 (u) = {x1 , . . . , xt } ̸ = ∅. Since κ (G) ≥ 2, there is an induced path Pi between xi and v in G − u with length at least k + 1 for any i ∈ {1, . . . , t }. Then Pi is a branch of G with length k + 1 since G is P3k+4 -free. Let G1 = G[V (G0 ) ∪ V (P1 ) · · · ∪ V (Pt )]. Then NG−G1 (u) = ∅, NG−G1 (v ) = ∅ and G = G1 ∼ = θ (k + 1, r) for some r ≥ 3. If r is even, then G is supereulerian, a contradiction. Then r is odd. □
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Fig. 2. The auxiliary graphs with Hamiltonian index at least k + 1 for any integers l ≥ m ≥ n ≥ k + 1.
Note that the graphs G7 and G11 in Fig. 2 are connected graphs with l ≥ m ≥ n ≥ k + 1. Then all graphs in Fig. 2 are connected with h(G) > k. We now present the proofs of Theorems 2 and 4. The proof of Theorem 2 when k ≥ 2. The sufficiency We may choose a subgraph H of G satisfying the conditions in 2.1. By Theorem 7, it suffices to prove that H ∈ EUk (G). (1) We claim that B(G; H) = ∅. Suppose otherwise. Then there is a branch B(x, y) whose end vertices are x, y connecting at most two components of H with |E(B(x, y))| ≥ k, dG (x) ≥ 3 and dG (y) ≥ 3. Let x′ ∈ NG (x)\V (B(x, y)) and y′ ∈ NG (y)\V (B(x, y)). Then x′ B(x, y)y′ would be an induced path of order at least k + 3, a contradiction. Similarly, we can prove that (G; H) = ∅. If there is a branch B = ux1 · · · xl ∈ B1 (G) such that l ≥ k + 1, then there is a vertex u′ ∈ NG (u)\{x1 }, and then u′ ux1 · · · xl is an induced path of G with length at least k + 3, a contradiction. This implies that B1 (G) = ∅ and then H ∈ EUk (G). (2) We claim that B(G; H) = ∅. Suppose otherwise. Then there is a branch B(x, y) whose end vertices are x, y connecting at most two components of H with |E(B(x, y))| ≥ k, x1 , x2 ∈ NG (x)\V (B(x, y)) and y1 , y2 ∈ NG (y)\V (B(x, y)). Since G is Zk+1 free, x1 x2 , y1 y2 ∈ / E(G). Then G[{x1 , x2 , y1 } ∪ V (B(x, y))] ∼ = T1,1,l for some l ≥ k, a contradiction. Similarly, we can prove that (G; H) = ∅. If there is a branch B = ux1 · · · xl ∈ B1 (G) such that l ≥ k + 1, then there are three vertices u1 , u2 ∈ NG (u)\{x1 } and u′1 ∈ NG (u1 )\{u}. Then G either has an induced Zl (if u1 u2 ∈ E(G) or u′1 = u2 or uu′1 ∈ E(G)) or has induced Tl,1,1 and Pl+4 (if u1 u2 ∈ / E(G), u′1 ̸= u2 and uu′1 ∈ / E(G) ), a contradiction. This implies that B1 (G) = ∅ and then H ∈ EUk (G). The necessity (1) Since G1 is K1,3 -free and G3 is K3 -free, ∆(H) ≤ 2. Then H should be a path since they have no common cycle, and hence H should be the subgraph of Pk+3 since G1 is Pk+4 -free. (2) Let H = {R, S }. Then each graph of {G1 , G2 , G3 , G4 , G5 } contains at least one of R, S as an induced subgraph. Without loss of generality, we may assume that R is an induced subgraph of G1 . If R is a tree, then R is an induced subgraph of Pk+3 , a contradiction. Thus we distinguish the following two cases: Assume that R contains K4 as a subgraph. Since G2 , G3 , G4 are K4 -free, they should contain S as an induced subgraph. Note that G4 is K1,3 -free, G3 is K3 -free and G2 is Pk+4 -free. Then H should be the subgraph of Pk+3 , a contradiction. Next assume that R is K4 -free. Then R contains K3 as a subgraph. Since G3 , G5 are K3 -free and they have no common induced cycle, S should be a tree. Since |V3 (G5 )| = 1, |V3 (S)| ≤ 1. Then the maximal common subgraph of G3 and G5 are Tk+1,1,1 and Pk+4 . Hence {R, S } ⪯ {Zk+1 , Tk+1,1,1 }, {Zk+1 , Pk+4 } since G1 is Zk+2 -free. □ The proof of Theorem 4. By Theorem 9, the sufficiency clearly holds. It remains to show the necessity. Let H = {R, S }. Then (1) Since G6 is K3 -free and G10 is K1,3 -free, ∆(H) ≤ 2. Then H should be a path since they have no common cycle, and hence H should be the subgraph of P3k+3 since G6 is P3k+4 -free. (2) Note that each graph of {G6 , G7 , G8 , G9 , G10 , G11 } contains at least one of R, S as an induced subgraph. Without loss of generality, we may assume that R is an induced subgraph of G6 . If ∆(R) = 2, then R is an induced subgraph of P3k+3 , a contradiction. Thus R either is a tree with |V≥3 (R)| = 1 or contains C2k+4 as an induced subgraph. Now we distinguish the following two cases:
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Assume that R contains C2k+4 as an induced subgraph. Note that G7 , G8 , G10 are C2k+4 -free, they should contain S as an induced subgraph. Since G7 is K3 -free, G10 is K1,3 -free and G8 is P3k+6 -free, S is an induced subgraph of P3k+5 . On the ′ ′ other hand, G9 is P3k+5 -free, then R is a common subgraph of G6 and G9 , which is C2k +4 . Therefore, {R, S } ⪯ {C2k+4 , P3k+5 }. Next assume that R is a tree. If ∆(R) = 3, then R contains K1,3 as an induced subgraph. Note that G10 , G11 are K1,3 -free, they should contain S as an induced subgraph. Since G12 is K4 -free and it has no induced subgraph contains two triangles as subgraph, S is an induced subgraph of G12 with at most one triangle as a subgraph, that is, Nk+1,k+1,k+3 , Ni2 ,j2 ,k+1 or Bk+1,2k+4 . Since G6 is S-free, R is an induced subgraph of G6 , that is Tk+1,k+1,k+1 or {Ti1 ,j1 ,k . Therefore, {R, S } ⪯ {H1 , H2 } for any H1 ∈ {Nk+1,k+1,k+3 , Ni2 ,j2 ,k+1 , Bk+1,2k+4 } and H2 ∈ {Tk+1,k+1,k+1 , Ti1 ,j1 ,k }. If ∆(R) ≥ 4, then R contains K1,4 as an induced subgraph. Note that G7 , G9 , G11 are K1,4 -free, they should contain S as an induced subgraph. Since G7 is K3 -free, G11 is K1,3 -free and they have no common cycle and G9 is P3k+5 -free, S is an induced subgraph of P3k+4 . Therefore, {R, S } ⪯ {P3k+4 , Tk+1,r } for some r ≥ 4. (3) Note that graphs G6 , G7 , G8 , G9 , G10 , G11 and the graph G12 are 2-edge-connected graphs with Hamiltonian ′ index at least k + 1. Besides, since G12 is {P3k+4 , Tk+1,k+1,k+1,2 }-free, {R, S } ⪯ {C2k +4 , P3k+5 }, {Tk+1,k+1,k+1 , Nk+1,k+1,k+3 }, {Tk+1,k+1,k+1 , Bk+1,2k+4 }, {Ti1 ,j1 ,k , Nk+1,k+1,k+3 }, {Ti1 ,j1 ,k , Bk+1,2k+4 }, {Tk+1,k+1,k+1 , Ni2 ,j2 ,k+1 }, {Ti1 ,j1 ,k , Ni2 ,j2 ,k+1 }, {P3k+4 , Tk+1,k+1,k+1,1 }. □ 4. Forbidden graphs force a graph G to satisfy h(G) ≤ 1 A dominating closed trail (abbreviated DCT) in a graph G is a closed trail (or, equivalently, an eulerian subgraph) T in G such that every edge of G has at least one vertex on T . In this section, we use the following theorem by Harary and Nash-Williams to transform the problem that guarantees h(G) ≤ 1 for a given graph G to one equivalent proposition, that is, guarantee whether G has a DCT. Theorem 10 (Harary and Nash-Williams, [8]). Let G be a graph with at least three edges. Then h(G) ≤ 1 if and only if G has a DCT. We call H a minor of G if H is isomorphic to the contraction image of a subgraph of G. Proof of Theorem 1. The necessity Note the graphs θ (l, m, n) and θ (k + 1, r) for any integer l ≥ m ≥ n ≥ k + 1, r ≥ 3 have Hamiltonian index at least k + 1. Then H is a tree with |V3 (H)| ≤ 1 and the necessity holds. The sufficiency We argue by contradiction. Assume that h(G) ≥ k + 1. By Theorem 6, G has a θ (k + 1, k + 1, k + 1)minor Hk for k ≥ 2. For k = 0, G is non-Hamiltonian and then G has a θ (1, 1, 1)-minor H0 . Hence we can find subgraphs P3k+5 , T1,1,2k+2 , T1,k+1,2k+1 and Tk+1,k+1,k+2 in Hk for k ̸ = 1, a contradiction. We then only consider the case that k = 1. By Theorem 10, G has no DCT. Choose a closed trail T of G such that |V (T )| is maximized. Then there is a non-trivial component H1 of G − V (T ) and two edges uu′ , vv ′ ∈ E(H1 , T ) such that u, v ∈ V (H1 ), u′ ̸ = v ′ ∈ V (T ) and either u ̸ = v or u = v and |V (H1 )| ≥ 3. Choose a path P(u′ , v ′ ) ⊆ T . Then by the choice of T , either |V (P(u′ , v ′ ))| ≥ 3 or P(u′ , v ′ ) = u′ x0 v ′ , x0 is a cut vertex of T and hence there is a cycle C1 ⊆ T with x0 ∈ V (C1 ) and T − E(P(u′ , v ′ )) − C1 is connected. Choose a path P1 (u′ , v ′ ) ⊆ T − E(P(u′ , v ′ )) − C1 and either |V (P1 (u′ , v ′ ))| ≥ 3 or P(u′ , v ′ ) = u′ x1 v ′ and there is a cycle C2 ⊆ T with x1 ∈ V (C2 ). We can easily find subgraphs P8 , T1,2,4 and T2,2,3 , a contradiction. □ For a subset X ⊆ E(G), the contraction G/X is the graph obtained from G by identifying the ends of each edge in X and then deleting the resulting loops. Note that the edges in E(G/X ) can be regarded as edges in E(G). If H is a subgraph, then we use G/H for G/E(H). Note that by this definition, if H is a connected subgraph of G, then G/H = G/G[V (H)]. For a graph G, let O(G) denote the set of odd degree vertices in G. In [2], Catlin defined collapsible graphs. A graph G is collapsible if for any even subset R of V (G), G has a connected spanning subgraph ΓR with O(Γ ) = R. The reduction of G, denoted by G′ , is obtained from G by contracting all maximal collapsible subgraphs of G. The vertex vH in G′ is called non-trivial if it is a contraction image of a non-trivial collapsible subgraph H of G. The following result would play an important role in our proofs. Theorem 11 (Catlin, [2]). Let G be a connected graph and G′ the reduction of G. Then G has a DCT if and only if G′ has a DCT containing all non-trivial vertices of G′ . For any two subsets S1 , S2 of V (G) and two subgraph T1 , T2 , define E(S1 , S2 ) = ∪x∈S1 E(x, S2 ), E(T1 , T2 ) = E(V (T1 ), V (T2 )) and D[T1 ] = {uv : {u, v} ∩ V (T1 ) ̸ = ∅}. Proof of Theorem 5. We argue by contradiction. By Theorem 10, assume that G has no DCT. Choose a closed trail T of G such that: (1.1) |V (T )| is maximized; (1.2) subject to (1.1), |D[T ]| is maximized.
X. Liu and L. Xiong / Discrete Mathematics 343 (2020) 111841
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Let H = G[V (T )]. Then there is a non-trivial component H1 of G[V (G)\V (T )] and a longest induced path P(u, v ) of H1 such that {uu′ , vv ′ } ⊆ E(P(u, v ), H) = E({u, v}, H) for some u′ , v ′ ∈ V (H). Then uv ′ , v u′ , u′ v ′ ∈ / E(G); for otherwise, T ∪ uv u′ u, T ∪ uvv ′ u or T △ uvv ′ u′ u is a closed trail with larger number of order than T , respectively, contradicting to (1.1). Choose two induced collapsible subgraphs F1 and F2 of G such that: (1.3) u′ ∈ V (F1 ), v ′ ∈ V (F2 ); (1.4) subject to (1.3), |V (F1 )| and |V (F2 )| are maximized. Let T0 = T [E(T )\E(F1 ∪F2 )], T1 = T [E(T )\E(F1 )], T2 = T [E(T )\E(F2 )] and Xi = O(T0 )∩V (Fi ) for i ∈ {1, 2}. Then |Xi | is even since |NT (Fi )| is even. By the definition of collapsible, for any {u1 , u2 } ⊆ V (Fi ), we have Fi has a spanning trail Ti (u1 , u2 ) such that O(Ti (u1 , u2 )) = Xi △{u1 , u2 }.
(4.1)
Then V (H1 ) ∩ V (F1 ∪ F2 ) = ∅; for otherwise, F1 , F2 have spanning trails T1 , T2 , respectively, such that O(T1 ) ∪ O(T2 ) = O(T0 ) and then T0 ∪ T1′ ∪ T2′ is a closed trail with larger number of order than T , a contradiction. Besides, V (F1 ) ∩ V (F2 ) = ∅; for otherwise, F1 ∪ F2 has a spanning trail T (u′ , v ′ ) such that O(T (u′ , v ′ )) = O(T0 )△{u′ , v ′ } since F1 ∪ F2 is collapsible, and then T0 ∪ T (u′ , v ′ ) ∪ u′ uP(u, v )vv ′ is a closed trail with larger number of order than T , a contradiction. Furthermore, E(F1 , F2 ) = ∅. Otherwise, assume that u1 u2 ∈ E(F1 , F2 ) for u1 ∈ V (F1 ), u2 ∈ V (F2 ). Then exactly one of T △uP(u, v )vv ′ u′ u (if u′ = u1 , v ′ = u2 ) or (T1 ∪ T1 (u′ , u1 )) △ u′ uP(u, v )vv ′ u1 (if u′ ̸ = u1 , v ′ = u2 ) or (T0 ∪ T1 (u′ , u1 ) ∪ T2 (v ′ , u2 )) △ (u′ uP(u, v )vv ′ ∪ u1 u2 ) (if u′ ̸ = u1 , y′ ̸ = u2 ) is a closed trail with larger number of order than T , a contradiction. Note that T0 is non-trivial. We can choose two edge-disjoint paths P1 = xx1 · · · xm x′ ⊆ T0 and P2 = yy1 · · · yn y′ ⊆ T0 such that |V (P1 ) ∩ V (P2 )| is minimized for any x, y ∈ V (F1 ), x′ , y′ ∈ V (F2 ) and m, n ≥ 1 since T is 2-edge-connected. Then x1 ̸ = y1 , xm ̸ = yn and x1 y1 , xm yn ∈ / E(G) by the choice of F1 , F2 . Then we have ′
E({x1 , y1 }, H1 ) = E({xm , yn }, H1 ) = ∅.
′
′
′
(4.2)
(Suppose otherwise. By symmetry, assume that x1 v1 ∈ E(G) for some v1 ∈ V (H1 ). Then there is a path P(u, v1 ) ⊆ H1 such that either T △x1 v1 P(u, v1 )uu′ x1 (if u′ = x) or (T1 ∪ T1 (u′ , x)) △ u′ uP(u, v1 )v1 x1 x (if u′ ̸ = x) is a closed trail with larger number of order than T , respectively, a contradiction.) Besides, E(v2 , H1 ) = ∅ for any v2 ∈ NG−P1 −P2 ({x1 , y1 }) ∪ NG−P1 −P2 ({xm , yn }).
(4.3)
(Suppose otherwise. By symmetry, assume that v1 v2 ∈ E(G) for some v1 ∈ V (H1 ) and v2 ∈ NG (x1 ) \ {x, x2 }. Then there is a path P(u, v1 ) ⊆ H1 such that either T △x1 v2 v1 P(u, v1 )uu′ x1 (if u′ = x) or (T1 ∪ T1 (u′ , x)) △ u′ uP(u, v1 )v1 v2 x1 x (if u′ ̸ = x) is a closed trail with larger number of order than T , respectively, a contradiction.) (1) We assume that G is 2-edge-connected T2,1,1 -free and then either obtain induced N2,2,4 , B2,6 or prove that G∼ = P(2, 2, 2) to reach a contradiction. By Eq. (4.2), E({x1 , y1 }, {u, u1 }) = ∅ for any u1 ∈ NH1 (u). Then G[V (F1 ) ∪ {u, u1 , x1 , y1 }] either has an induced T2,1,1 or is isomorphic to Ni,j,k for some i ≥ j ≥ k ≥ 1 by Theorem 8. Since G is T2,1,1 -free, |V (F1 )| ≥ 3 and u′ , x, y are three distinct vertices. By symmetry, |V (F2 )| ≥ 3 and v ′ , x′ , y′ are three distinct vertices. Then |E(P(u, v ))| ≥ 1 and dG (u) = dG (v ) = 2; for otherwise, G[{u, u1 , u′ , x, v ′ }] ∼ = T2,1,1 for u1 ∈ NH1 (u1 )\{v1 }, a contradiction. We claim that dG (F1 , F2 ) ≥ 3 and then m ≥ 2, n ≥ 2. Suppose otherwise. Without loss of generality, assume that m = 1 and there is a vertex x′1 ∈ NG (x1 )\{x, x′ }. By the choice of F1 , F2 , E(x′1 , F1 ∪ F2 ) = ∅ and then G[{x′1 , x1 , x, x′ , v1 }] ∼ = T2,1,1 for any v1 ∈ NF1 (x), a contradiction. Besides, x1 and y1 have no common neighbor with degree at least 3; for otherwise, E(x′12 , {x1 , y1 , x}) = ∅ by the choice of F1 for any x12 ∈ NG (x1 ) ∩ NG (y1 ) and x′12 ∈ NG (x12 )\{x1 , y1 }, and hence G[{u1 , x1 , x, y1 , x3 }] ∼ = T2,1,1 , a contradiction. By symmetry, xm and yn have no common neighbor with degree at least 3. Then x1 , x2 , y1 , y2 are four distinct vertices and x1 y2 , y1 x2 ∈ / E(G). Claim 1.
dG (F1 , F2 ) ≥ 4.
Proof. Without loss of generality, assume that m = 2. Then E(y2 , F2 ) ̸ = ∅; for otherwise, either G[{x2 , x1 , x, x′ , y2 }] ∼ = T2,1,1 (if x2 y2 ∈ E(G)) or G1 = G[V (F1 ∪F2 )∪V (P(u, v ))∪{x1 , x2 , y1 , y2 }] has induced N2,2,4 and B2,6 (if x2 y2 ∈ / E(G)), a contradiction. Therefore either G1 ∼ = P(2, 2, 2) or G1 has induced N2,2,4 and B2,6 . Besides, NG (G1 ) = ∅ and then G = G1 = P(2, 2, 2), a contradiction. □ Claim 2.
G has an induced B2,6 .
Proof. We claim that either x1 , x2 , yn , yn−1 , yn−2 or y1 , y2 , xm , xm−1 , xm−2 are five distinct vertices. If P1 and P2 are vertex disjoint, then it holds; otherwise, let i1 , j1 be the smallest integers such that xi1 = yj1 and i2 , j2 be the largest integers such that xi2 = yj2 . Then max{i1 , j1 } ≥ 3 and max{m − i2 , n − j2 } ≥ 3 and hence it holds. Without loss of generality, assume that x1 , x2 , yn , yn−1 , yn−2 are five distinct vertices. Then x1 yn , x1 xm , y1 yn ∈ / E(G) and P(u, v ) = uv . If x1 yn−1 , x2 yn ∈ / E(G), then either G[V (F1 ) ∪ {u, v, x1 , x2 , yn−1 , yn , y′ , y′′ }] for any y′′ ∈ NF2 (y′ ) (if x2 yn−1 ∈ E(G)) or G[V (F1 ) ∪ V (P(v ′ , y′ )) ∪ {u, v, x1 , x2 , yn , yn−1 }] for any induced path P(v ′ , y′ ) of F2 (if x2 yn−1 ∈ / E(G)) has an induced B2,6 . Otherwise either G[V (F1 ) ∪ V (P(x′ , y′ )) ∪ {u, v, x1 , yn−1 , yn , xm }] for any induced ′ ′ path P(x , y ) of F2 (if x1 yn−1 ∈ E(G)) or G[V (F2 ) ∪ V (P(x, y)) ∪ {u, v, yn , x2 , x1 , y1 }] for any induced path P(x, y) of F1 (if x2 yn ∈ E(G)) has an induced B2,6 . □
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Fig. 3. The graph G1 and its reduction G′ in Theorem 5 (1).
We then finish the proof of (1) by the following claim. Claim 3.
G has an induced N2,2,4 .
Proof. We argue by contradiction. Recall x1 , x2 , y1 , y2 are four distinct vertices, then E({x1 , x2 , y1 , y2 }, F2 ) = ∅ and y1 x2 , x1 y2 ∈ / E(G). Then x2 y2 ∈ E(G); for otherwise, G[V (F1 ) ∪ V (P(u, v )) ∪ {x1 , x2 , y1 , y2 , v ′ , v2 }] has induced N2,2,4 for any v2 ∈ NF2 (v ′ ), a contradiction. Besides, {x1 , y1 } ⊆ V2 (G); for otherwise, assume that dG (x1 ) ≥ 3, then either G[{x1 , x′1 , x, x2 , y2 }] ∼ = T2,1,1 or G[V (F1 ) ∪ {x1 , x2 , y1 , y2 , x′1 }] is collapsible for any x′1 ∈ NG (x1 )\{x, x2 }, contradicting the choice of F1 . To sum up, NG (F1 ) ⊆ V≤2 (T ) is an independent set of G and V1 = NG (NG (F1 ))\F1 is a clique. Besides, there is a vertex v2 ∈ NG (V1 )\NG (F1 ) with |E(v2 , V1 )| ≥ 2 since G is T2,1,1 -free. Then G[V1 ∪ {v2 }] is collapsible. Choose a collapsible subgraph L1 containing V1 such that |V (L1 )| is maximized. Repeat the above operation by replacing F1 by F1 ∪ L1 · · · ∪ Li for i ≥ 1, we can obtain a series of induced collapsible subgraphs L1 , . . . , Lt of G for some t ≥ 1. Let F1 = L0 and F2 = Lt +1 . Then G ∼ = G1 , which depicted in Fig. 3. Then the reduction G′ of G has a DCT T ′ containing all non-trivial vertices. Hence G has a DCT by Theorem 11, a contradiction. □ (2) Note that G is 2-connected. Then |E(P(u, v ))| ≥ 1 and we can replace all the graphs T2,1,1 by T3,1,1 in the proof of (1) and the remainder discussion is similar but easier than the proof of (1). So we omit it here. (3) We assume that G is 2-connected P7 -free and will prove that G ∼ = θ (2, r) to reach a contradiction. Claim 4.
V (P1 ) ∩ V (P2 ) = ∅.
Proof. We argue by contradiction. Let i, j be the smallest integers such that ui,j = xi = yj . Then either ui,j ∈ / {x1 , y1 } or ui,j ∈ / {xm , yn }. Suppose otherwise. By symmetry, assume that x1 = ui,j = xm . Then exactly one of T △uvv ′ x1 u′ u (if u′ = x, v ′ = x′ ) or (T1 ∪T1 (u′ , x))△u′ uvv ′ x1 x (if u′ ̸ = x, v ′ = x′ ) or (T0 ∪T1 (u′ , x)∪T2 (v ′ , x′ ))△(u′ uvv ′ ∪xx1 x′ ) (if u′ ̸ = x, v ′ ̸ = x′ ) is a closed trail with larger number of order than T , a contradiction. By symmetry, assume that ui,j ∈ / {xm , yn }. Besides i + j ≥ 4 by the choice of F1 . Then either x1 , x2 , xm , yn (if i ≥ 2) or y1 , y2 , xm , yn (if j ≥ 2) are four distinct vertices. Without loss of generality, assume that x1 , x2 , xm , yn are four distinct vertices. By the choice of F1 , E(x2 , F1 ) = ∅. Besides E(x1 , F2 ) = ∅; for otherwise, exactly one of T △x1 u′ uvv ′ x1 x1 (if u′ = x, v ′ = v1 ) or (T1 ∪ T1 (u′ , x)) △ xx1 v ′ v uu′ (if u′ ̸ = x, v ′ = v1 ) or (T0 ∪ T1 (u′ , x) ∪ T2 (v ′ , v1 )) △ (u′ uvv ′ ∪ xx1 v1 ) (if u′ ̸ = x, v ′ ̸ = v1 ) for any x1 v1 ∈ E(x1 , F2 ) is a closed trail with larger number of order than T , a contradiction. We claim that dG (x1 ) = 2. Suppose otherwise. Then there is a vertex x′1 ∈ NG (x1 )\{x, x2 } with E(x′1 , F1 ) = ∅ by the choice of F1 and E(x′1 , P(u, v )) = ∅ by Eq. (4.3). By the choice of F2 , |E(x′1 , F2 )| ≤ 1. Then u′ = x; for otherwise, there is a non-trivial induced path P(x, u′ ) in F1 and then x′1 x1 xP(x, u′ )u′ uvv ′ is an induced path of order at least 7. So |V (F2 )| = 1; for otherwise, there is a vertex v1 ∈ NF2 (v ′ ) such that x′1 v1 ∈ / E(G) and G[{x′1 , x1 , u′ , u, v, v ′ , v1 }] ∼ = P7 , a contradiction. By symmetry, ′ ′ E({yn , xm }, V (P(u, v )) ∪ {u }) = ∅. Then {x1 yn , x′1 xm } ̸ ⊆ E(G); for otherwise, at least one of T △uP(u, v )vv ′ yn x′1 x1 u′ u or T △uP(u, v )vv ′ xm x′1 x1 u′ u is a closed trail with larger number of order than T , a contradiction. By symmetry, assume that x′1 yn ∈ / E(G). Besides, x1 yn ∈ / E(G); for otherwise, T △uvv ′ yn x1 u′ u is a closed trail with larger number of order than T , a contradiction. However, G[{x′1 , x1 , u′ , u, v, v ′ , yn }] ∼ = P7 , a contradiction. Then E(x2 , P(u, v ) ∪ F2 ) = ∅; for otherwise, either T △uP(u, v1 )v1 x2 x1 u′ (if u′ = x) or (T1 ∪ T1 (u′ , x))△u′ uP(u, v1 )v1 x2 x1 x (if u′ ̸ = x) for any P(u, v1 ) ⊆ P(u, v ) and x1 v1 ∈ E(G), exactly one of T △uvv ′ x2 x1 u′ u (if u′ = x, v ′ = v1 ) or (T1 ∪ T1 (u′ , x))△u′ uvv ′ x2 x1 x (if u′ ̸ = x, v ′ = v1 ) or (T0 ∪ T1 (u′ , x) ∪ T2 (v ′ , x2 ))△(u′ uvv ′ ∪ xx1 x2 v1 ) (if u′ ̸ = x, v ′ ̸ = v1 ) for any x2 v1 ∈ E(x2 , F2 ) is a closed trail with larger number of order than T , respectively, a contradiction. So x = u′ and |V (F2 )| = 1 since G is P7 -free. By symmetry, |V (F1 )| = 1 and x2 u′ ∈ / E(G). Besides, x2 yn ∈ / E(G); for otherwise, T △uvv ′ yn x2 x1 u′ u is a closed trail with larger number of order than T , a contradiction. However G[{x2 , x1 , u′ , u, v, v ′ , yn }] ∼ = P7 , a contradiction. □ We then consider the following two cases to finish our proof. Case 1. dG (F1 , F2 ) = 2.
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Without loss of generality, assume that m = 1. Then dG (x1 ) ≥ dT (x1 ) ≥ 4 by the maximality of |V (T )|. Let X0 = V (F1 ∪ F2 ∪ P(u, v )) ∪ {y1 , yn }. By the choice of F1 , F2 and Eqs. (4.2) and (4.3), |E(x1 , X0 )| = 2, E({y1 , yn }, P(u, v )) = ∅ and
|E(v1 , X0 )| = 1 for any v1 ∈ / X0 and E(v1 , X0 \V (P(u, v ))) ̸= ∅.
(4.4)
Let V1 = NG (x1 )\{x, x′ }, V2 = NG (V1 )\{x1 } and V3 = NG (V2 )\V1 . Then E(V1 , X0 ) = ∅ by Eq. (4.3) and V2 ̸ = ∅ since x1 is not a cut vertex. If E(V2 , X0 ) = ∅, then x = y = u′ and x′ = y′ = v ′ since G is P7 -free and V3 ̸ = ∅ since x1 is not a cut vertex. Then there is a vertex v3 ∈ V3 ∩ NG ({u′ , v ′ , u, v}) such that either |E(v3 , {u′ , v ′ , u, v, z1 , z2 })| ≤ 1 or E(v3 , X0 ) = {v3 u, v3 v} and then there is an induced P7 , a contradiction. So E(V2 , X0 ) ̸ = ∅ and there are two vertices v1 ∈ V1 , v2 ∈ V2 such that v1 v2 ∈ E(G). Then |E(v2 , X0 )| ≥ 2, E(v2 , X0 ) = E(v2 , P(u, v )) = {v2 w : w ∈ V (P(u, v ))}, x = y = u′ and x′ = y′ = v ′ since G is P7 -free. Hence G[V (P(u, v )) ∪ {v2 }] is collapsible. Then dT (v1 ) ≥ 4; for otherwise, T △u′ uvv ′ v2 v1 x1 u′ is a closed trail with larger number of order than T , a contradiction. By symmetry, there is a vertex v2′ ∈ V2 such that G[V (P(u, v )) ∪ {v2′ }] is collapsible and then G[V (P(u, v )) ∪ {u′ , v ′ , x1 , v1 , v2 , v2′ }] is collapsible, contradicting the choice of F1 , F2 . Case 2. dG (F1 , F2 ) ≥ 3. Then min{m, n} ≥ 2. We claim that x1 yn , y1 xm ∈ / E(G). Suppose otherwise. By symmetry, assume that x1 yn ∈ E(G). Then exactly one of T △uP(u, v )vv ′ yn x1 u′ u (if u′ = x and v ′ = y′ ), (T1 ∪ T1 (u′ , x)) △ (u′ uP(u, v )vv ′ ∪ xx1 yn y′ ) (if u′ ̸ = x and v ′ = y′ ) or (T0 ∪ T1 (u′ , x) ∪ T2 (v ′ , y′ )) △ (u′ uP(u, v )vv ′ ∪ xx1 yn y′ ) (if u′ ̸= x and v ′ ̸= y′ ) is a closed trail with larger number of order than T , a contradiction. Then u′ = x = y, v ′ = x′ = y′ and P(u, v ) = uv since G is P7 -free. We claim that {x1 , y1 , xm , yn } ⊆ V2 (G). / E(G) by the Suppose otherwise. By symmetry, assume that there is a vertex x′1 ∈ NG (x1 )\{u′ , x2 }. Then x′1 u′ ∈ choice of F1 and E(x′1 , {u′ , u, v}) = ∅ by the choice of F1 and Eq. (4.3). Besides, E(x′1 , {v ′ , yn }) = ∅; for otherwise, T △uvv ′ x′1 x1 u′ u, T △uvv ′ yn x′1 x1 u′ u is a closed trail with larger number of order than T , respectively, a contradiction. However G[{x′1 , x1 , u′ , u, v, v ′ , yn }] ∼ = P7 , a contradiction. Thus m = n = 2; for otherwise, assume that n ≥ 3, then E(x2 , {u′ , u, v, v ′ , yn }) = ∅ and G[{x2 , x1 , u′ , u, v, v ′ , yn }] ∼ = P7 , a contradiction. This implies that there is no induced path of order at least 5 between u′ and v ′ . If there is a vertex u1 ∈ NG (u)\{u′ , v}, then E(u1 , {u′ , x1 , x2 , v ′ , y2 }) = ∅ and G[{u1 , u, u′ , x1 , x2 , v ′ , y2 }] ∼ = P7 , a contradiction. Thus dG (u′ ) = dG (v ′ ) = 2 and V≥3 (G) = {u′ , v ′ } and then G ∼ = θ (2, r) for r = dG (u′ ) = dG (v ′ ) and r is odd. □ The following two theorems completely characterize forbidden subgraphs that force a connected (2-edge-connected or 2-connected) graph to have a DCT and forbidden pairs that force a connected graph to have a DCT. Theorem 12. Let H and G be connected graphs of order at least 3. Then H-free graph G implies G has a DCT if and only if H satisfies one of the following: (1) H is an induced subgraph of P4 , (2) H is an induced subgraph of P5 when κ ′ (G) ≥ 2, (3) H is an induced subgraph of P6 when κ (G) ≥ 2. Proof. The sufficiency By contradiction, assume that G has no DCT and choose a closed trail T such that |V (T )| is maximized. Then there is an induced path P(u′ , v ′ ) ⊆ G − T such that {u′ } ⊆ V (P(u′ , v ′ )) ∩ V (T ) and {u′ , v ′ } ⊆ V (P(u′ , v ′ )) ∩ V (T ) if κ ′ (G) ≥ 2. Furthermore, |E(P(u′ , v ′ ))| ≥ 2 or |E(P(u′ , v ′ ))| ≥ 3 if κ ′ (G) ≥ 2 or |E(P(u′ , v ′ ))| ≥ 4 if κ (G) ≥ 2. So there is an induced P4 or an induced P5 if κ ′ (G) ≥ 2 or an induced P6 if κ (G) ≥ 2. The necessity The graph G′1 is the graph obtaining from K2,3 by replacing each vertex degree 2 in K2,3 by a clique Km for any m ≥ 3 such that |NG′ ({u, v}) ∩ V (Km )| = 1, where dK2,3 (u) = dK2,3 (v ) = 3. Note that G1 , G6 , G10 depicted in Fig. 2 1 with k = 2 and the graph G′1 have no DCT. They contain H as an induced subgraph. Since G6 is K3 -free and G10 is K1,3 -free and they have no common cycle, H is a subgraph of path. Then H ⊆ P4 since G1 is P5 -free or H ⊆ P5 since G′1 is P6 -free and κ ′ (G′1 ) ≥ 2 or H ⊆ P6 since G6 is P7 -free and κ (G6 ) ≥ 2. □ Theorem 13.
Every connected {R, S }-free graph G other than a tree has a DCT if and only if {R, S } ⪯ {Z2 , P5 }, {Z2 , T2,1,1 }.
Proof. The necessity is clearly true. The sufficiency Assume that G has no DCT. Choose a closed trail T of G such that |V (T )| is maximized. Let H = G[V (T )]. Then there is an edge uv ⊆ G − H such that ux ∈ E(G) and x1 xx2 x3 ⊆ T . Then E(v, {x, x1 , x2 , x3 }) = ∅ by the maximality of T . Since G is Z2 -free, x1 ̸ = x3 and x1 x2 , xx3 ∈ / E(G). Hence G[{x, x1 , x2 , u, v}] ∼ = T2,1,1 and G[{x3 , x2 , x, u, v}] ∼ P . □ = 5 Here the graph Z ′ is the graph obtained by identifying one vertex of K3 and the leaf of T2,2,1 whose neighbor has degree 3. Theorem 14.
Every 2-edge-connected {P6 , Z ′ }-free graph G has a DCT.
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Proof. By contradiction, assume that G has no DCT. Choose a closed trail T of G such that |V (T )| is maximized. Let H = G[V (T )]. Then G[V (G)\V (H)] has at least one non-trivial component H1 . Choose two induced paths P(u, v ) ⊆ H1 and P(u′ , v ′ ) = u′ x1 · · · xt v ′ ⊆ H of G such that (1) E(P(u, v ), P(u′ , v ′ )) = {uu′ , vv ′ }; (2) subject to (1), |E(P(u, v ))| is maximized; (3) subject to (1) and (2), |E(P(u′ , v ′ ))| is maximized. Then u′ v ′ ∈ / E(G); for otherwise, T △uP(u, v )vv ′ u′ u is a closed trail with larger number of order than T , a contradiction. Note that P(u, v ) ∪ P(u′ , v ′ ) ∪ {uu′ , vv ′ } is an induced cycle of G, 1 ≤ t ≤ 3 since G is P6 -free. Besides, there is a vertex z1 ∈ NH (u′ ) since H is 2-edge-connected. Claim 5.
E(z1 , {u, v, v ′ , xt }) = ∅.
Proof. Suppose otherwise. Then T △uu′ z1 u (if z1 u ∈ E(G)), T △v uu′ z1 v (if z1 v ∈ E(G)), T △uu′ z1 v ′ v u (if z1 v ′ ∈ E(G)) or T △uu′ z1 xt v ′ v u (if z1 xt ∈ E(G)) is a closed trail of G with larger number of order than T , respectively, a contradiction. □ If t = 3, then P(u, v ) = u = v since G is P6 -free. Since H1 is non-trivial and by the choice of P(u, v ), there is a vertex u1 ∈ NH1 (u) with E(u1 , H) = ∅ . However G[{u1 , u, u′ , x1 , x2 , x3 }] ∼ = P6 , a contradiction. If t = 2, then P(u, v ) = u = v , since otherwise P(u, v ) = uv and then G[{z1 , u′ , u, v, v ′ , x2 }] ∼ = P6 by Claim 5, a contradiction. Besides, V (H1 ) = NH1 [u]. Then there are at least two vertices u1 , u2 ∈ V (H1 ) with u1 u2 ∈ E(G) and hence H1 is collapsible and H1 has a spanning closed trail T0 with |V (T0 )| ≥ 3. Then at least one of x1 , x2 is a cut vertex of T ; for otherwise, T1 = (T ∪ T0 )△uu′ x1 x2 v ′ u is a closed trail with |V (T1 )| > |V (T )|, a contradiction. Without loss of generality, assume that x2 is a cut vertex of T , x1 x2 v ′ ⊆ T and there is a vertex x′2 ∈ NT (x2 )\{v ′ , x1 }. Then E(x′2 , {v ′ , u, u′ , z1 }) = ∅; for otherwise, (T ∪ T0 )△x2 x′2 v ′ x2 , (T ∪ T0 )△x2 x′2 uv ′ x2 , (T ∪ T0 )△x2 x′2 u′ uv ′ x2 , (T ∪ T0 )△x2 x′2 z1 u′ uv ′ x2 is a closed trail with larger number of order than T , respectively, a contradiction. However G[{x′2 , x2 , v ′ , u, u′ , z1 }] ∼ = P6 , a contradiction. Thus t = 1, u′ x1 v ′ ⊆ T and x1 is a cut vertex of T by the choice of T . Then there is a cycle C1 ⊆ T such that V (C1 ) ∩ {u′ , v ′ , x1 } = {x1 }. We claim that V (C1 ) ⊆ NG [x1 ]. Suppose otherwise. There is an induced path x′′1 x′1 x1 ⊆ G[V (C1 )]. Note that E(x′1 , {u′ , v ′ } ∪ V (H1 )) = ∅; for otherwise, T △x′1 u′ uP(u, v )vv ′ x1 x′1 (if x′1 u′ ∈ E(G)), T △x′1 v ′ v P(u, v )uu′ x1 x′1 (if x′1 v ′ ∈ E(G)) or T △x′1 x1 u′ uP(u, u1 )u1 x′1 (if x′1 u1 ∈ E(G) for some u1 ∈ V (H1 )) is a closed trail of G with larger number of order than T , respectively, a contradiction. Since G is P6 -free, E(x′′1 , {u′ , v ′ } ∪ V (H1 )) ̸ = ∅. By the same argument, we ′′′ ′′ ′ ′ ′ can prove that x′1 is a cut vertex of T and there is a vertex x′′′ 1 ∈ NT (x1 )\{x1 , x1 } with E(x1 , {u , v , x1 } ∪ V (H1 )) = ∅ and ′ ′ ∼ (u), a contradiction. G[{x′′′ , x , x , u , u , u }] P for any u ∈ N = 1 1 6 1 H1 1 1 Note that there is a path P(u′ , v ′ ) = u′ y1 · · · yl v ′ ⊆ T − C1 − u′ x1 v ′ for some l ≥ 1. By symmetry, l = 1, y1 is a cut vertex of T and there is a cycle C2 ⊆ T such that V (C2 ) ⊆ NG [y1 ]. Hence G[{x1 , x′1 , x′′1 , y1 , y′1 , u′ , u, u1 }] ∼ = Z ′ for some u1 ∈ NH1 (u), y′1 ∈ NC2 (y1 ) and x′1 , x′′1 ∈ NC1 (x1 ) with x′1 x′′1 ∈ E(G), a contradiction. □ Theorem 15.
Let R, S be connected graphs. Then
(1) if every 2-connected {R, S }-free graph G of order at least 2r + 3 ≥ 13 implies G has a DCT, then {R, S } ⪯ {T2,2,2 , N2,2,4 }, {T2,2,2 , B2,6 }, {T3,1,1 , B2,6 }, {T3,1,1 , N2,2,4 }, {P8 , C6′ }, {P7 , T2,r }, (2) if every 2-edge-connected {R, S }-free graph G other than P(2, 2, 2) implies G has a DCT, then {R, S } ⪯ {T2,2,2 , N2,2,4 }, {T2,2,2 , B2,6 }, {P6 , Z ′ }. Proof. (1) can be immediately obtained by Theorem 4 (2) when k = 1. (2) The graph G′2 is the graph obtaining from θ (2, 2, 1) by replacing the common neighbor of two vertices of degree 3 with a clique Km for any m ≥ 3 such that |NG′ (Km )| = 2. The graph G′3 is the graph obtained from K2,t by replacing each 2 vertex degree 2 in K2,t by K3 such that |NG′ (K3 )| = 2 for any odd integer t ≥ 3. Then G′2 and G′3 have no DCT. Note that 3 G′3 is K4 -free, P6 -free, C6 -free and T1,1,3 -free. Then if R has K1,3 as a subgraph, then S is an induced subgraph of N2,2,4 or B2,6 . Otherwise, {R, S } ⪯ {P6 , Z ′ } since the common induced cycle of G′2 and G′3 is C3 . □ By comparing Theorems 5, 13, 14 (2) and 15, we know if the following conjecture is true, we may accomplish the complete characterization of forbidden pairs forcing a 2-edge-connected or 2-connected graph to have a DCT, i.e., h(G) ≤ 1. Conjecture 16.
It would be true that
(1) Every 2-connected {P8 , C6′ }-free graph G has a DCT. (2) Every 2-edge-connected {T2,2,2 , H }-free graph G for H ∈ {N2,2,4 , B2,6 } other than P(2, 2, 2) has a DCT. Declaration of competing interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
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Acknowledgments This work is supported by Nature Science Funds of China (Nos. 11871099 and 11671037) and by Natural Science Foundation of Qinghai Province, China (No. 2018-ZJ-717). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
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