Collapsible graphs and Hamilton cycles of line graphs

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Collapsible graphs and Hamilton cycles of line graphs Xiangwen Li ∗ , Yan Xiong Department of Mathematics, Huazhong Normal University, Wuhan 430079, China

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Article history: Received 10 February 2014 Received in revised form 4 May 2015 Accepted 24 May 2015 Available online xxxx Keywords: Collapsible graph Hamilton cycle Line graph Eulerian graph

abstract For a graph G, let O(G) denote the set of odd degree vertices of G. A graph G is collapsible if for any subset R ⊆ V (G) with |R| ≡ 0 (mod 2), G has a spanning connected subgraph HR such that O(HR ) = R. The reduction of G is the graph obtained from G by contracting all maximally collapsible subgraphs until no collapsible subgraph is left. Let G be a graph on n ≥ 8 vertices. In this paper, we prove that if d(x) + d(y) ≥ n − 2 − p(n) for each xy ∈ E (G), then G is collapsible or G is one of 43 special graphs or the reduction of G is K1,t where t ≥ 2 or G is a class of well-characterized graphs, where p(n) = 0 for n even and p(n) = 1 for n odd, which generalizes the earlier results by Catlin (1987), and by Li and Yang (2012). © 2015 Elsevier B.V. All rights reserved.

1. Introduction The graphs we consider here are finite, connected simple graphs without loops. We follow [1] for terminology and notations unless stated otherwise. For a vertex v of a graph G, let NG (v) denote the set of the neighbors of v in G and for A ⊆ V (G), let NG (A) denote the set ∪v∈A NG (v)\A. For a subgraph H of a graph G and v ∈ V (G), denote by dH (v) = |N (v) ∩ V (H )| the number of the neighbors of v in H. If H = G, then dG (v) is the degree of v and we write d(v) for it. For two vertex-disjoint subgraphs A and B of G, let eG (A, B) (or briefly e(A, B)) denote the number of edges with one end in A and the other end in B. The concept of collapsible graphs was introduced by Catlin in [4]. For a graph G, let O(G) denote the set of odd degree vertices of G. A graph G is collapsible if for any subset R ⊆ V (G) with |R| ≡ 0 (mod 2), G has a spanning connected subgraph HR such that O(HR ) = R. A graph is eulerian if it has a closed trail through all the edges of G, that is, G is eulerian if it is connected with O(G) = ∅. A graph G is supereulerian if G has a spanning eulerian subgraph. Note that every collapsible graph is supereulerian. For a subgraph H of a graph G, the graph G/H is obtained from G by identifying the two ends of each edge in H and then deleting the resulting loops. The contraction G/H is called the reduction of G if H is the maximally collapsible subgraph of G, that is, there is no non-trivial collapsible subgraph in G/H. A vertex u in G/H is called non-trivial if the vertex is obtained by contracting a non-trivial collapsible subgraph Hu . We call Hu the preimage of u. A graph is reduced if it is the reduction of itself. For a graph G and an edge e = uv , d(u) + d(v) is called the edge degree of e. Define ξ (G) = min{d(x) + d(y) : xy ∈ E (G)}. The function p(n) is defined as follows: p(n) = 0 if n even and p(n) = 1 otherwise. Thomassen [14] conjectured that every 4-connected line graph is Hamiltonian. Harary and Nash-Williams [11] proved the classic result that if a graph G is not a star, the line graph L(G) is Hamiltonian if and only if G has a dominating closed trail. Clearly, both collapsible graphs and supereulerian graphs have dominating closed trails. It is well-known that the line graph of 4-edge-connected graph is Hamiltonian. Thus, the collapsibility and supereulerianicity of k-edge-connected graphs where k ∈ {1, 2, 3} are investigated (see [5,7–9,12,15,16] and others). In this paper, we still ficus on the collapsibility of connected graphs.

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Corresponding author. E-mail addresses: [email protected], [email protected] (X. Li).

http://dx.doi.org/10.1016/j.dam.2015.05.030 0166-218X/© 2015 Elsevier B.V. All rights reserved.

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Fig. 1. G′7 and G′′7 .

Theorem 1.1 (Catlin [3]). Let G be a connected graph on n vertices and let u, v ∈ V (G). If ξ (G) ≥ n, then exactly one of the following holds: (1) G has a spanning (u, v)-trail. (2) d(z ) = 1 for some vertex z ̸∈ {u, v}. (3) G = K2,n−2 , u = v and n is odd. (4) G = K2,n−2 , u ̸= v, uv ̸∈ E (G), where n is even and d(u) = d(v) = n − 2. (5) u = v , and u is the only vertex with degree 1 in G. Theorem 1.1 was improved as follows. Theorem 1.2. Let G be a connected graph on n ≥ 4 vertices. If ξ (G) ≥ n, then exactly one of the following holds: (1) G is collapsible. (2) The reduction of G is K1,t −1 for t ≥ 3 such that all of the vertices of degree 1 are trivial and they have the same neighbor in G, t ≤ 2n . Moreover, if t = 2, then G − v is collapsible for a vertex v in the K2 . (3) G is K2,n−2 . Theorem 1.2 was strengthened by Li and Yang [13] as follows. Theorem 1.3. Let G be a connected graph on n ≥ 4 vertices. If ξ (G) ≥ n − 1 − p(n), then exactly one of the following holds: (1) G is collapsible. (2) The reduction of G is K1,t −1 for t ≥ 3 such that all of the vertices of degree 1 are trivial and they have the same neighbor in G, t ≤ 2n . Moreover, if t = 2, then G − v is collapsible for a vertex v in the K2 . (3) G is one of {C5 , G′7 , G′′7 , K1,n−1 , K2,n−2 , K2′ ,n−3 }, where K2′ ,n−3 is obtained from K2,n−3 by adding a pendant edge on one of the vertices of degree n − 3, where G′7 and G′′7 are shown in Fig. 1. An edge is pendant if it is incident with a vertex of degree 1. Denote by K2′′,n−4 the graph obtained from K2,n−4 by adding two pendant edges on one of the vertices of degree n − 4. Denote by K2′′′,n−5 the graph obtained from K2,n−5 by adding three pendant edges on one of the vertices of degree n − 5. Denote by K2∗,n−4 the graph obtained from K2,n−4 by adding one pendant edge on each vertex of degree n − 4. Denote by K2∗,n−5 the graph obtained from K2,n−5 by adding two pendant edges on one vertex of degree n − 5 and adding one pendant edge on the other vertex of degree n − 5. Denote by K2∗,n−6 the graph obtained from K2,n−6 by adding two pendant edges on each vertex of degree n − 6. Define a family H of graphs such that if G ∈ H , then G is the graph obtained from two vertex disjoint graphs H1 and H2 by adding an edge with one endpoint in H1 and the other in H2 . Moreover, let |V (G)| = n. If n = 2l, then H1 = H2 = Kl ; if n = 2l + 1, then H1 = Kl and H2 is the graph obtained from Kl+1 by deleting a set of independent edges. Motivated by Theorems 1.2 and 1.3, we present the following result in this paper. Theorem 1.4. Let G be a connected graph on n ≥ 8 vertices. If ξ (G) ≥ n − 2 − p(n), then exactly one of the following holds: (1) G is collapsible. (2) The reduction of G is K1,t −1 for t ≥ 3 such that all of the vertices of degree 1 are trivial and they have the same neighbor in G, t ≤ 2n . Moreover, if t = 2, then G − v is collapsible for a vertex v in the K2 . (3) G = Gi , where 1 ≤ i ≤ 35, or G is one of {K1,n−1 , K2,n−2 , K2′ ,n−3 , K2′′,n−4 , K2∗,n−4 , K2′′′,n−5 , K2∗,n−5 , K2∗,n−6 }. (4) G is a class of well-characterized graphs H . The paper is organized as follows: In Section 2, the former related results are presented and some lemmas are established. In Section 3, Theorem 1.4 is proved. Applications of Theorem 1.4 will be presented in Section 4. 2. Lemmas Reduction method will play a key role in our proof of Theorem 1.4. It is known that any cycle of length less than 4 is collapsible. Some of the previous results concerning collapsible graphs in [4] are summarized as follows: Theorem 2.1. Let G be a connected graph. Each of the following holds: (1) If H is a collapsible subgraph of G, then G is collapsible if and only if G/H is collapsible; G is supereulerian if and only if G/H is supereulerian. (2) A reduced graph does not have a cycle of length less than 4. (3) If G is reduced and |E (G)| ≥ 3, then δ(G) ≤ 3 and 2|V (G)| − |E (G)| ≥ 4. (4) If G is reduced and violates (3), then G is either K1 or K2 .

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Lemma 2.2 ([6]). The graphs K3,3 and K3,3 − e are collapsible, where K3,3 − e is the graph obtained by removing an edge from K3,3 . We follow the idea of Li and Yang [13] and prove the following lemma. Lemma 2.3. Let G be a graph on n ≥ 8 vertices and G′ be the reduction of G. If ξ (G) ≥ n − 2 − p(n), then G′ contains at most two non-trivial vertices. Proof. Suppose otherwise that G′ contains at least three nontrivial vertices u1 , u2 , u3 . Let Hu1 , Hu2 , Hu3 be the preimages of u1 , u2 and u3 , respectively. Assume that V (G′ )\{u1 , u2 , u3 } ̸= ∅. We assume, without loss of generality, that xy is the edge of G such that x ∈ V (Hu3 ) and y ̸∈ V (Hu1 ) ∪ V (Hu2 ) ∪ V (Hu3 ). Since G′ is reduced, x and y have at most two neighbors in Hu1 ∪ Hu2 . By Theorem 2.1(2), x and y have no common neighbor. Thus, d(x) + d(y) ≤ |V (G)\{V (Hu1 ) ∪ V (Hu2 ) ∪ V (Hu3 ) ∪ {y}}| + |V (Hu3 )\{x}| + 2 + 2 ≤ n − |V (Hu1 )| − |V (Hu2 )| − |V (Hu3 )| − 1 + |V (Hu3 )| − 1 + 2 + 2 = n − |V (Hu1 )| − |V (Hu2 )| + 2. Since |VHui | ≥ 3 for i = 1, 2, 3, d(x) + d(y) ≤ n − 4, contrary to that ξ (G) ≥ n − 3. Thus, assume that V (G′ )\{u1 , u2 , u3 } = ∅. We assume, without loss of generality, that |V (Hu1 )| ≤ |V (Hu2 )| ≤ |V (Hu3 )|. By Theorem 2.1(2), there are at most two edges among Hu1 , Hu2 , Hu3 . Since G is simple, each Hui contains at least three edges for i ∈ {1, 2, 3}. Pick an edge xy in Hu1 such that x and y have at most one neighbor outside of Hu1 . In this case, d(x) + d(y) ≤ 2(|V (Hu1 )| − 1) + 1 = n − |V (Hu2 )| − |V (Hu3 )| + |V (Hu1 )| − 1 ≤ n − 4, a contradiction.  Lemma 2.4. Let G be a connected graph on n ≥ 8 vertices and G′ be the reduction of G. Suppose that ξ (G) ≥ n − 2 − p(n), vH1 and vH2 are non-trivial vertices obtained by contracting H1 and H2 , respectively. If e = vH1 vH2 is a bridge in G′ , then G ∈ H or G = G1 in Fig. 2. Proof. Since e is a bridge in G′ , there exist x ∈ H1 , y ∈ H2 , such that xy is a bridge of G. Let H1∗ and H2∗ be two components of G − xy such that Hi∗ contains Hi for i = 1, 2. Since vH1 and vH2 are non-trivial, pick an edge ai bi of Hi such that x, y ̸∈ {a1 , a2 , b1 , b2 }. Let m1 = |V (H1∗ )| and m2 = |V (H2∗ )|. We assume, without loss of generality, that m1 ≤ m2 . If n is even, then by the given edge degree condition, n − 2 ≤ d(a1 ) + d(b1 ) ≤ 2(m1 − 1), which implies that m1 ≥ 2n . Since m1 ≤ m2 and m1 + m2 = n, m1 = m2 = 2n . By the given edge degree condition again, n − 2 ≤ d(a1 ) + d(b1 ) ≤ 2(m1 − 1) = n − 2. It follows that H1∗ is a complete graph and H1∗ = H1 . Similarly, we can prove that H2 = H2∗ and H2 is a complete graph. This means that G ∈ H . 1 . If n is odd, then by the given edge degree condition, n − 3 ≤ d(a1 ) + d(b1 ) ≤ 2(m1 − 1), which implies that m1 ≥ n− 2

1 1 Since m1 ≤ m2 and m1 + m2 = n, m1 = n− and m2 = n+ . 2 2 ∗ Observe the subgraph H1 . By the given edge degree condition, n − 3 ≤ d(a1 ) + d(b1 ) ≤ 2(m1 − 1) = n − 3, which implies that H1∗ is a complete graph and H1∗ = H1 . Observe the subgraph H2∗ . By the given edge degree condition, n − 3 ≤ d(a2 ) + d(b2 ) ≤ 2(m2 − 1) = n − 1. Since |V (H2∗ )| = m2 = n+2 1 and n ≥ 8, for each edge uv of H2∗ , by the given edge degree condition, d(u) + d(v) ≥ n − 3 ≥ |V (H2∗ )|. By Theorem 1.2, H2∗ is collapsible or the reduction of H2∗ is K1,t −1 or H2∗ is K2,m2 −2 . If H2∗ is K2,m2 −2 , let xy be an edge of H2∗ 1 ≥ n − 3, which implies that such that d(x) = 2 and d(y) = m2 − 2. It follows that d(x) + d(y) = m2 ≥ n − 3, that is, n+ 2 ∗ n ≤ 7, a contradiction. Suppose that the reduction of H2 is K1,t −1 . If t ≥ 3, then by Theorem 1.2, vH2 is the vertex of degree t − 1 in K1,t −1 . In this case, let u ∈ V (H2 ) and d(v) = 1 such that uv ∈ E (G). If u ̸= y, then n − 3 ≤ d(u) + d(v) ≤ t − 1 + |V (H2 )| − 1 + 1 = |V (H2∗ )|, which implies that n ≤ 7, a contradiction. If u = y, then n − 3 ≤ d(u) + d(v) ≤ t − 1 + |V (H2 )| − 1 + 1 + 1 = |V (H2∗ )| + 1 which implies n ≤ 9. Thus, n = 9 and G = G1 . If t = 2, then K1,t −1 = K1,1 . In this case, let the reduction of H2∗ is induced by {vH2 , v}. It follows that d(v) = 1. Let u ∈ H2 such that uv ∈ E (G). Note that |V (H2∗ )| = |V (H2 )| + 1. If u = y, then n − 3 ≤ d(u) + d(v) ≤ 2 + |V (H2 )| − 1 + 1 which implies n ≤ 9. Thus, n = 9 and G = G1 . If u ̸= y, then n − 3 ≤ d(u) + d(v) ≤ 1 + |V (H2 )| − 1 + 1 which implies n ≤ 7, a contradiction. Thus, assume that H2∗ is collapsible. This means that H2∗ = H2 is a complete graph short of at most a perfect matching. This implies that G ∈ H . 

Lemma 2.5. Assume that G is a connected graph on n ≥ 8 vertices and G′ is the reduction of G. If ξ (G) ≥ n − 2 − p(n), then G′ contains precisely two non-trivial vertices if and only if G ∈ {G1 , G2 } or G ∈ H . Proof. Assume that G′ contains two non-trivial vertices u1 , u2 and Hu1 and Hu2 are the preimages of u1 and u2 , respectively. By Lemma 2.4, we assume that u1 u2 is not a bridge of G′ . We assume, without loss of generality, that |V (Hu1 )| = m1 , |V (Hu2 )| = m2 . Note that G is simple, mi ≥ 3 for i = 1, 2. Claim 1. The graph G′ contains no trivial edge and each vertex of V (G′ )\{u1 , u2 } has degree at most 2.

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Fig. 2. 35 graphs.

Proof of Claim 1. Suppose otherwise that G′ contains a trivial edge xy, that is, x and y are both trivial. Since Hu1 and Hu2 are both maximum collapsible subgraphs, e({x, y}, Hu1 ∪ Hu2 ) ≤ 2 and for each vertex u ∈ V (G′ )\{x, y, u1 , u2 }, e({x, y}, u) ≤ 1. Thus, d(x) + d(y) ≤ 2 + 2 + n − m1 − m2 − 2 < n − 3, a contradiction.

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Since G is connected, and so is G′ . Since G′ contains no trivial edge, all the neighbors of each vertex of V (G′ )\{u1 , u2 } are in {u1 , u2 }. Thus, it has degree at most 2. This proves Claim 1. Claim 2. If G′ contains a vertex of degree 1, then G = G1 . Proof of Claim 2. Assume that G′ contains a vertex w of degree 1. Assume first that w is a non-trivial vertex of degree 1, say w = u1 . Note that G′ contains precisely two non-trivial vertices. By assumption, G′ has no bridge between two non-trivial vertices. Since G is connected, by Claim 1, there is a vertex u of degree 2 in V (G)\(V (Hu1 )∪ V (Hu2 )) such that uv1 , uv2 ∈ E (G), where v1 ∈ V (Hu1 ), v2 ∈ V (Hu2 ). Note that |V (Hu1 )| ≥ |NG (v1 ) ∪ {v1 }\{u}| ≥ n − 2 − p(n) − d(u) = n − 4 − p(n) and |V (G)\(V (Hu1 ) ∪ {u})| ≥ |NG (v2 ) ∪ {v2 }\{u}| ≥ n − 2 − p(n) − d(u) = n − 4 − p(n). This implies that n ≥ |V (Hu1 )| + |V (G)\(V (Hu1 ) ∪ {u})| + 1 ≥ 2(n − 4) + 1 for n even and n ≥ 2(n − 5) + 1 for n odd. This leads to that n ≤ 7 for n even and n ≤ 9 for n odd. Since n ≥ 8, n is odd and n = 9. Thus, G is G1 . Next, assume that w is a trivial vertex of degree 1. Since G′ has no trivial edge and G′ is connected, we assume, without loss of generality, that w u1 ∈ E (G′ ). Let z1 ∈ V (Hu1 ) such that w z1 ∈ E (G). We claim that u1 and u2 are not adjacent in G′ . Suppose otherwise that u1 is adjacent to u2 in G′ . We first assume that G′ contains a cycle in which u1 and u2 are adjacent. Since G′ has no trivial edge, such a cycle is of length 3, which is collapsible by Theorem 2.1, contrary to that G′ is reduced. We then assume that G′ has no cycle containing both u1 and u2 . By Claim 1, G′ has no cycle. In this case, we further assume that there are s vertices of degree 1 adjacent to u1 and t vertices of degree 1 adjacent to u2 and s ≥ t. It follows that n − 2 − p(n) ≤ d(w) + d(z1 ) ≤ 1 + |V (Hu1 )| + s. If t > 0, let x be a vertex of degree 1 adjacent to a vertex z2 in Hu2 . Similarly, n − 2 − p(n) ≤ d(x) + d(z2 ) ≤ 1 + |V (Hu2 )| + t. Thus, 2(n − 2 − p(n)) ≤ 2 + |V (Hu1 )| + |V (Hu2 )| + s + t = 2 + n, which implies that n ≤ 6 + 2p(n), a contradiction. If t = 0, pick an edge xy in Hu2 such that none of x and y has neighbor outside of Hu2 . Thus, n − 2 − p(n) ≤ d(x)+ d(y) ≤ 2(|V (Hu2 )|− 1), which implies that n−2−p(n) 2

+ 1. In this case, n − 2 − p(n) ≤ d(w)+ d(z1 ) ≤ 1 +|V (Hu1 )|+ s ≤ 1 + n −|V (Hu2 )| ≤ 1 + n − n−2−2 p(n) − 1, which implies that n ≤ 6 + 3p(n). This leads to that G is G1 . Since G′ is connected, G′ has a path from u1 to u2 . Since u1 and u2 are not adjacent, there is a vertex x1 ∈ V (G′ )\{u1 , u2 } such that x1 u1 , x1 u2 ∈ E (G′ ). Let y1 ∈ V (Hu2 ) such that x1 y1 ∈ E (G). Note that G is simple and m1 ≥ 3. Since G′ is reduced, e(y1 , Hu1 ) = 0. This implies that N (y1 ) ⊆ V (G)\(V (Hu1 )∪{w}). By Claim 1, d(x1 )+ d(y1 ) ≤ 2 +|V (G)\(V (Hu1 )∪{w})|− 1 ≤ 2 +(n − 4)− 1 = n − 3. On the other hand, by the given condition, d(x1 )+ d(y1 ) ≥ n − 2 − p(n). This leads to that n is odd and m1 = 3. Since Hu1 is collapsible, Hu1 is a 3-cycle denoted by {z1 , z2 , z3 }. By the given degree condition d(w) + d(z1 ) ≥ n − 3 and d(z2 ) + d(z3 ) ≥ n − 3. This implies that e({z1 , z2 , z3 }, V (G)\V (Hu1 ) ∪ {w, x1 }) ≥ (n − 4) + (n − 3) − 8 = 2n − 15. On the other hand, since Hu2 is collapsible and G is simple, m2 ≥ 3. Note that u1 and u2 are not adjacent in G′ . Thus, e(Hu1 , Hu2 ) = 0. Thus, e({z1 , z2 , z3 }, V (G)\(V (Hu1 ) ∪ {w, x1 })) ≤ n − 3 − 2 − 3 = n − 8. This implies that n − 8 ≥ 2n − 15, which implies that n ≤ 7. This contradicts that n ≥ 8.  |V (Hu2 )| ≥

By Claims 1 and 2, assume that each vertex of V (G′ )\{u1 , u2 } has degree 2 and hence G′ is K2,t −2 , where t ≥ 4. Let v1 , v2 , . . . , vt −2 be the vertices of degree 2 and u1 , u2 be the vertices of degree t − 2 in G′ . Note that mi ≥ 3 for i = 1, 2. For i = 1, 2, since G′ is reduced, each vertex of v1 , v2 , . . . , vt −2 is adjacent to at most one vertex of Hui , and Hui contains 2(t −2)

at least three edges. Thus, choose xi yi ∈ E (Hui ) such that 3 + mi − 1 + mi − 1 ≥ d(xi ) + d(yi ) ≥ n − 3. This implies that t ≤ 5. Assume that t = 4. If m1 = 3, then Hu1 is a K3 and n = 8. Since e(Hu1 , {v1 , v2 }) ≤ 2, let z1 z2 ∈ E (G) such that e({z1 , z2 }, {v1 , v2 }) is minimized. This implies that d(z1 ) + d(z2 ) ≤ 2 + 3 = 5, contrary to that ξ (G) ≥ n − 2 = 6. If m1 ≥ 4, then pick x3 y3 ∈ E (Hu1 ) such that e({x3 , y3 }, {v1 , v2 }) = 0 since e(Hu1 , {v1 , v2 }) ≤ 2. By the hypothesis, 1 1 . This implies that n = m1 + m2 + 2 ≥ 2 n− + 2 = n + 1, n − 3 ≤ d(x3 )+ d(y3 ) ≤ m1 − 1 + m1 − 1, which implies that m1 ≥ n− 2 2 a contradiction. Thus, t = 5. In this case, n ≥ 9. If Hu1 contains an edge z1 z2 such that e({z1 , z2 }, {v1 , v2 , v3 }) ≤ 1, then d(z1 ) + d(z2 ) ≤ m1 − 1 + m1 − 1 + 1 ≤ m1 + m2 − 1 = n − 4, contrary to that ξ (G) ≥ n − 3. Thus, for each edge z1 z2 in Hu1 , e({z1 , z2 }, {v1 , v2 , v3 }) = 2 since G′ is reduced. This implies that m1 = 3 and each vertex of Hu1 has precisely one neighbor in {v1 , v2 , v3 }. We assume, without loss of generality, that z1 v1 ∈ E (G). In this case, d(z1 ) + d(v1 ) = 2 + 1 + 2 = 5, contrary to that ξ (G) ≥ 9 − 3 = 6.  Lemma 2.6. Let G be a graph on n ≥ 8 vertices and G′ be the reduction of G with t vertices, only one of which is a non-trivial vertex. Let ξ (G) ≥ n − 2 − p(n). If G′ is not in {K2,t −2 , K2′ ,t −3 } for t ≥ 4 if and only if G is not one of {G3 , G4 , . . . , G20 }.

Proof. If G is one of {G3 , . . . , G20 }, then the reduction of G is in {K2,t −2 , K2′ ,t −3 } for some integer t ≥ 4. Thus, assume that u1 is the non-trivial vertex and Hu1 is the preimage of u1 . Suppose otherwise that G′ is K2′ ,t −3 or K2,t −2 . In the former case, let V (K2′ ,t −3 ) = {u, v, w, v1 , . . . , vt −3 } such that d(u) = t − 3, d(v) = t − 2, d(w) = 1 and d(vi ) = 2 for i ∈ {1, 2, . . . , t − 3}. Assume first that u1 = v . Then G′ contains a trivial edge e = uvi such that d(u) + d(vi ) = t − 1 for i ∈ {1, 2, . . . , t − 2}. By the hypothesis, t − 1 ≥ n − 2 − p(n), that is, t ≥ n − 1 − p(n). On the other hand, since G is simple, |V (Hu1 )| ≥ 3. Note that G′ contains only one non-trivial vertex. This implies that n = t − 1 + |V (Hu1 )| ≥ n − 1 − p(n) − 1 + 3. This leads to that p(n) = 1 and |V (Hu1 )| = 3. Thus, n is odd, t = n − 2 and Hu1 is a 3-cycle z1 z2 z3 z1 such that w z1 ∈ E (G). By the hypothesis, d(w) + d(z1 ) ≥ n − 3. This implies that e(z1 , {v1 , v2 , . . . , vt −3 }) ≥ n − 7 = t − 5. Thus, by hypothesis, n − 3 ≤ d(z2 ) + d(z3 ) ≤ 4 + 2, which implies that n ≤ 9. Since n is odd and n ≥ 8, n = 9. In this case, G is G3 . Assume that u1 = u. In this case, G′ contains a trivial edge e = vw such that d(v) + d(w) = t − 1. By the hypothesis, t − 1 ≥ n − 2 − p(n). As the proof above, n is odd, t = n − 2 and Hu1 is a 3-cycle z1 z2 z3 z1 . We assume, without loss of

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generality, that z1 v1 ∈ E (G). By the given edge degree condition, d(z1 ) + d(v1 ) ≥ n − 3. Thus, e(z1 , {v1 , . . . , vt −3 }) ≥ n − 7. Since G′ is reduced, each vertex of v1 , v2 , . . . , vt −3 is adjacent to at most one vertex of {z1 , z2 , z3 }. Thus, n − 5 − (n − 7) + 4 ≥ d(z2 ) + d(z3 ) ≥ n − 3, which implies that n ≤ 9. Since n is odd, n = 9. In this case, G is G4 . Next, assume that u1 = w . In this case, G′ contains a trivial edge e = uv1 . As the proof above, n is odd, t = n − 2 and Hu1 is a 3-cycle z1 z2 z3 z1 such that z3 v ∈ E (G). By the given edge degree condition, 4 = d(z2 ) + d(z1 ) ≥ n − 3, which implies that n ≤ 7. This contradicts that n ≥ 8. Finally, assume that u1 = vi . In this case, G′ contains a trivial edge e = vw such that d(v)+d(w) = t −1. By the hypothesis, t − 1 ≥ n − 2 − p(n). As the proof above, n is odd, t = n − 2 and Hu1 is a 3-cycle z1 z2 z3 z1 . We assume, without loss of generality, that e({z1 , z2 }{u, v}) is minimized. In this case by the given edge degree condition, 5 ≥ d(z1 ) + d(z2 ) ≥ n − 3, which implies that n ≤ 8. This contradicts that n ≥ 8 and n is odd. In the latter case, let V (K2,t −2 ) = {u, v, v1 , v2 , . . . , vt −2 } such that d(vi ) = 2 for 1 ≤ i ≤ t − 2, and d(u) = d(v) = t − 2. Since G′ has precisely one non-trivial vertex, G′ contains a trivial edge wvi for i ∈ {1, 2, . . . , t − 2}, where w ∈ {u, v}. By the given edge degree condition, t = d(w) + d(vi ) ≥ n − 2 − p(n). Assume that n is even. Then t = n − 2, which implies that |V (Hu1 )| = 3 since G is simple. Let Hu1 be a 3-cycle z1 z2 z3 z1 . It this case, if u1 = vi for some i ∈ {1, 2, . . . , t − 2}, as the proof above, we may assume that 5 ≥ d(zs ) + d(zt ) ≥ n − 2 for some {s, t } ⊂ {1, 2, 3}, which implies that n ≤ 7, contrary to that n ≥ 8. Thus, by symmetry, assume that u1 = v . We assume, without loss of generality, that z1 v1 ∈ E (G). By the given edge degree condition, d(z1 ) + d(v1 ) ≥ n − 2. This implies that d(z1 ) ≥ n − 4 and e(z1 , {v1 , v2 , . . . , vt −2 }) ≥ n − 6. Since G′ is reduced, n − 4 − (n − 6) + 4 ≥ d(z2 ) + d(z3 ) ≥ n − 2, which implies that n ≤ 8. In this case, G is G5 . Assume that n is odd. Then t = n − 2 or n − 3, which implies that |V (Hu1 )| = 3 or 4. If |V (Hu1 )| = 3, let Hu1 be a 3-cycle z1 z2 z3 z1 . As the proof above, it is easy to show that u1 ̸= vi for each i ∈ {1, 2, . . . , t − 2}. By symmetry, assume that u1 = v . We assume, without loss of generality, that z1 v1 ∈ E (G). By the given edge degree condition, d(z1 ) + d(v1 ) ≥ n − 3. This implies that d(z1 ) ≥ n − 5 and e(z1 , {v1 , v2 , . . . , vt −2 }) ≥ n − 7. Since G′ is reduced, n − 4 −(n − 7)+ 4 ≥ d(z2 )+ d(z3 ) ≥ n − 3, which implies that n ≤ 10. Since n is odd, n ≥ 9 and so G is G6 . If |V (Hu1 )| = 4, let V (Hu1 ) = {z1 , z2 , z3 , z4 }. In this case, we need to deal with two cases: Hu1 is a K4 or K4− , which is the graph obtained from K4 by deleting one edge. Case 1. Hu1 is a K4 Assume that u1 = u or v . If Hu1 contains only one vertex adjacent to {v1 , . . . , vt −2 }, we assume, without loss of generality, that z1 v1 ∈ E (G). By the given edge degree condition, d(z1 ) + d(v1 ) ≥ n − 3. Thus, e(z1 , {v1 , v2 , . . . , vt −2 }) ≥ n − 8. In this case, by the given edge degree condition, 6 = d(z2 ) + d(z3 ) ≥ n − 3, which implies that n ≤ 9. Since n is odd and n ≥ 8, n = 9 and G = G7 . If Hu1 contains precisely two vertices adjacent to {v1 , . . . , vt −2 }, we assume, without loss of generality, that e({z3 , z4 }, {v1 , v2 , . . . , vt −2 }) = 0. In this case, by the given edge degree condition, 6 = d(z2 ) + d(z3 ) ≥ n − 3, which implies that n ≤ 9. Similarly, n = 9 and G is G8 or G9 . If Hu1 contains precisely three vertices adjacent to {v1 , . . . , vt −2 }, we may assume that e(zi , {v1 , . . . , vt −2 }) ≥ 1 for i = 1, 2, 3. In this case, as the proof above, we can show that for = 1, 2, e(zi , {v1 , v2 , . . . , vt −2 }) ≥ n − 8. By the given edge degree condition, n − 4 − 2(n − 8) + 6 ≥ d(z3 ) + d(z4 ) ≥ n − 3, which implies that n ≤ 10. Thus, n = 9 and G is G10 . If each vertex of Hu1 is adjacent to {v1 , . . . , vt −2 }, similarly we obtain that n = 9 and G is G11 . Now assume that u1 = vi for some i ∈ {1, 2, . . . , vt −1 }. In this case, since |V (Hu1 )| = 4, we may assume that e({z3 , z4 }, {u, v}) = 0. By the given edge degree condition, 6 = d(z3 ) + d(z4 ) ≥ n − 3, which implies that n ≤ 9. Since n is odd, n = 9 and G is G12 or G13 . Case 2. Hu1 is a K4− and z1 z2 ̸∈ E (G) Assume first that u1 = u or v . If Hu1 contains only one vertex zi adjacent to {v1 , v2 , . . . , vt −2 } for i ∈ {1, 2, 3, 4}, by symmetry, we consider that zi = z1 or zi = z3 . No matter zi = z1 or zi = z3 , by the given edge degree condition, we have 2 + 3 = d(z2 ) + d(z4 ) ≥ n − 3, which implies that n ≤ 8, contrary to that n ≥ 8 and n is odd. If Hu1 contains precisely two vertices {zi , zj } adjacent to {v1 , . . . , vt −2 } for i, j ∈ {1, 2, 3, 4}, by symmetry, we consider {zi , zj } = {z1 , z2 }, {zi , zj } = {z1 , z3 } or {zi , zj } = {z3 , z4 }. If zi = z1 , zj = z2 , without loss of generality, assume that zi vi ∈ E (G) for i ∈ {1, 2}. By the given degree condition, 2 + 2 + e(z1 , {v1 , v2 , . . . , vt −2 }) = d(z1 ) + d(v1 ) ≥ n − 3, which implies that e(z1 , {v1 , v2 , . . . , vt −2 }) ≥ n − 7. Since G′ is reduced, by the given edge degree condition, 2 + 2 + (n − 5) − (n − 7) ≥ d(z2 ) + d(v2 ) ≥ n − 3, which implies that n ≤ 9. Since n is odd, n = 9 and G is G14 . If zi = z1 , zj = z3 , by the given edge degree condition, we have 2 + 3 = d(z2 )+ d(z4 ) ≥ n − 3, which implies that n ≤ 8, a contradiction. If zi = z3 , zj = z4 , without loss of generality, assume that zi vi ∈ E (G) for i ∈ {3, 4}. By the given edge degree condition, 3 + 2 + e(z3 , {v1 , v2 , . . . , vt −2 }) = d(z3 ) + d(v3 ) ≥ n − 3, which implies that e(z3 , {v1 , v2 , . . . , vt −2 }) ≥ n − 8. Since G′ is reduced, n − 5 − (n − 8) + 5 ≥ d(z3 ) + d(z2 ) ≥ n − 3, which implies that n ≤ 11. Since n ≥ 8 is odd, n = 9 or 11. If n = 9, then G is G15 or G16 . If n = 11, then G is G17 . Let Hu1 contain precisely three vertices, by symmetry, say {z1 , z2 , z3 } or {z2 , z3 , z4 }, adjacent to {v1 , v2 , . . . , vt −2 }. In the former case, without loss of generality, assume that z1 v1 , z2 v2 , z3 v3 ∈ E (G). By the given edge degree condition, 2 + e(z1 , {v1 , v2 , . . . , vt −2 }) + 2 = d(z1 ) + d(v1 ) ≥ n − 3. This implies that e(z1 , {v1 , v2 , . . . , vt −2 }) ≥ n − 7. Similarly, e(z2 , {v1 , v2 , . . . , vt −2 }) ≥ n − 7. Note that e(z3 , {v1 , v2 , . . . , vt −2 }) ≥ 1. Thus, n − 5 − 2(n − 7) ≥ 1, which implies that n ≤ 8, a contradiction. In the latter case, by the given edge degree condition, e(zi , {v1 , v2 , . . . , vt −2 }) ≥ n − 8 for i = 3, 4. Since z2 is adjacent to {v1 , . . . , vt −2 } and G′ is reduced, (n − 5) − 2(n − 8) ≥ e(z2 , {v1 , v2 , . . . , vt −2 }) ≥ 1, which implies that n ≤ 10. Since n is odd, G is G18 . If each vertex of Hu1 is adjacent to {v1 , . . . , vt −2 }, without loss of generality, assume that zi vi ∈ E (G) for 1 ≤ i ≤ 4. By the given edge degree condition, 2 + e(zj , {v1 , v2 , . . . , vt −2 }) + 2 = d(vj ) + d(zj ) ≥ n − 3

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for j = 1, 2. Thus, e(zj , {v1 , v2 , . . . , vt −2 }) ≥ n − 7 for j = 1, 2. On the other hand, e(zk , {v1 , v2 , . . . , vt −2 }) ≥ 1 for k = 3, 4. Since G′ is reduced, n − 5 = e({z1 , z2 , z3 , z4 }, {v1 , v2 , . . . , vt −2 }) ≥ 2(n − 7) + 2, which implies that n ≤ 7, a contradiction. Next, we assume that u1 = vi for some i ∈ {1, 2, . . . , t − 2}. If Hu1 contains only one vertex adjacent to both u and v , then Hu1 contains an edge zi zj such that d(zi ) + d(zj ) = 5, where i, j ∈ {1, 2, 3, 4}. By the given edge degree condition, d(zi ) + d(zj ) ≥ n − 3. Thus, n ≤ 8, a contradiction. It remains for us to assume that Hu1 contains two vertices zi , zj such that zi u, zj v ∈ E (G), by symmetry, we consider {zi , zj } = {z1 , z2 }, {zi , zj } = {z1 , z3 } or {zi , zj } = {z3 , z4 }. If zi = z1 , zj = z2 , by the given edge degree condition, 3 + 3 = d(z3 )+ d(4) ≥ n − 3, which implies that n ≤ 9. In this case, G is G19 . If zi = z1 , zj = z3 , by the given edge degree condition, 2 + 3 = d(z2 ) + d(z4 ) ≥ n − 3, which implies that n ≤ 8, a contradiction. If zi = z3 , zj = z4 , by the given edge degree condition, 2 + 4 = d(z2 ) + d(z4 ) ≥ n − 3, which implies that n ≤ 9, In this case, n = 9 and G is G20 .  Lemma 2.7. Let G be a graph on n ≥ 8 vertices and G′ be the reduction of G. If ξ (G) ≥ n − 2 − p(n) and G′ has only one non-trivial vertex, then G′ is K1,t −1 for some integer t or G is one of {G3 , . . . , G20 }. Proof. Let u be the only one non-trivial vertex of G′ and Hu be the preimage of u. We only show that if G is not in {G3 , . . . , G20 }, then G′ is K1,t −1 for some integer t. It suffices to show that G′ contains no trivial edge. Suppose otherwise that xy ∈ E (G′ ) is trivial edge and xu ∈ E (G′ ). We first establish a claim. Claim. Let |V (Hu )| = k for k = 3, 4. If for each trivial edge xy, d(x) + d(y) = n − k + 1, then G′ is isomorphic to Kd(x),d(y) . Proof of Claim. Since |V (Hu )| = k, by Theorem 2.1 (2), dG′ (x) + dG′ (y) ≤ n − k + 1. By assumption, d(x) + d(y) = dG′ (x) + dG′ (y) = n − k + 1. This means that each vertex of V (G)\(V (Hu ) ∪ {x, y}) must be adjacent to one of x and y, and only one vertex of Hu is adjacent to x or y. This implies that V (G)\(V (Hu ) ∪ {x, y}) ⊆ N ({x, y}). Define S = V (G′ )\{u, x, y} = {y1 , y2 , . . . , yn−k−2 }. Since G′ is reduced, by Theorem 2.1, we may assume that NG′ (y) = {x, y1 , y2 , . . . , yl } and NG′ (x) = {y, u, yl+1 , yl+2 , . . . , yn−k−2 }. For i ∈ {1, 2, . . . , l}, yyi is a trivial edge. By assumption, dG′ (y) + dG′ (yi ) = n − k + 1. This ′ implies that NG′ (yi ) = NG′ (x) for i = 1, 2, . . . , l, and G = Kd(x),d(y) . This proves the Claim. Assume that n is even. By the given edge degree condition, d(x) + d(y) ≥ n − 2. On the other hand, since |V (Hu )| = k, ′ by Theorem 2.1 (2), dG′ (x) + dG′ (y) ≤ n − 2. Thus, d(x) + d(y) = dG′ (x) + dG′ (y) = n − 2. By the Claim, G = Kd(x),d(y) . By ′ Lemma 2.2, either d(x) ≤ 2 or d(y) ≤ 2. Noting that G = K2,n−4 = K2,t −2 contradicts Lemma 2.6. This implies that G′ is K1,t −1 for some t. Assume that n is odd. In this case, similarly, n − 3 ≤ d(x) + d(y) = dG′ (x) + dG′ (y) ≤ n − 2. This implies that |V (Hu )| = 3 or 4. If |V (Hu )| = 4, then d(x) + d(y) = dG′ (x) + dG′ (y) = n − 3. By the Claim, G′ is K1,t −1 for some t. Thus, assume that |V (Hu )| = 3 and V (Hu ) = {z1 , z2 , z3 } with d(z1 ) ≥ d(z2 ) ≥ d(z3 ). By the given edge degree condition, d(z2 ) + d(z3 ) ≥ n − 3. 3(n−7) This shows that n − 3 ≥ e(Hu , G − V (Hu )) ≥ 2 , which implies n ≤ 15. We claim that ξ (G′ ) ≥ |V (G′ )| − 1. Let xy be an ′ edge of G . If none of x and y is u, then d(x) + d(y) ≥ n − 3 = |V (G′ )| − 1. Thus, by symmetry, let x = u. Then Hu contains a vertex zi such that zi y ∈ E (G). By the given edge degree condition, d(zi ) + d(y) ≥ n − 3. Let {i, j, k} = {1, 2, 3}. Note that zj zk ∈ E (G) and by the hypothesis d(zj )+ d(zk ) ≥ n − 3. Thus, d(x)+ d(y) = d(zi )− 2 + d(y)+ d(zj )+ d(zk )− 4 ≥ 2(n − 3)− 6. Since n ≥ 9 and |V (G′ )| = n − 2, it follows that d(x) + d(y) ≥ |V (G′ )| − 1. Since G′ is reduced, by Theorem 1.3, G′ is K1,n−3 or K2,n−4 or K2′ ,n−5 . By Lemma 2.6, G′ is K1,t −1 for some t.  Lemma 2.8. Assume that G is a reduced graph on n ≥ 8 vertices and that ξ (G) ≥ n − 2 − p(n) and δ(G) = 2. Let u ∈ V (G) and N (u) = {u1 , u2 }. If either max{d(u1 ), d(u2 )} ≥ n − 2 or min{d(u1 ), d(u2 )} ≥ n − 4, then G is K2,n−2 . Proof. Since G is reduced, {u1 , u2 } is an independent set. Let V (G) = {u, u1 , u2 , v1 , v2 , . . . , vn−3 }. If one of u1 and u2 has degree n − 2, then G is K2,n−2 since δ(G) = 2. Thus, assume that d(u1 ) ≤ n − 3 and d(u2 ) ≤ n − 3. Assume that d(u1 ) = d(u2 ) = n − 3. If N (u1 ) = N (u2 ) = {u, v1 , v2 , . . . , vn−4 }, then N (vn−3 ) ⊆ {v1 , . . . , vn−4 } since G has no 3-cycle. Since δ(G) = 2, G contains a subgraph isomorphic to K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Thus, N (u1 ) ̸= N (u2 ). We assume, without loss of generality, that N (u1 ) = {u, v1 , v2 , . . . , vn−4 } and N (u2 ) = {u, v2 , v2 , . . . , vn−3 }. Since δ(G) = 2, v1 vn−3 ∈ E (G). By the given edge degree condition, 4 = d(v1 ) + d(vn−3 ) ≥ n − 3, which implies that n ≤ 7, a contradiction. Next, assume that d(u1 ) = n − 3 and d(u2 ) = n − 4. If N (u2 ) ⊂ N (u1 ), then we may assume that N (u1 ) = {u, v1 , v2 , . . . , vn−4 } and N (u2 ) = {u, v1 , v2 , . . . , vn−5 }. If vn−3 is adjacent to at least two vertices of {v1 , v2 , . . . , vn−5 }, then G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Thus, vn−3 is adjacent to at most one vertex of {v1 , v2 , . . . , vn−5 }. Since δ(G) = 2, vn−3 is adjacent to exactly one vertex of {v1 , v2 , . . . , vn−5 } and vn−4 vn−3 ∈ E (G). By the given edge degree condition, 4 = d(vn−3 ) + d(vn−4 ) ≤ n − 3, which implies that n ≤ 7, a contradiction. Assume that N (u2 ) ̸⊂ N (u1 ). We assume, without loss of generality, that N (u1 ) = {u, v1 , v2 , . . . , vn−4 }, N (u2 ) = {u, v3 , v4 , . . . , vn−3 }. Since δ(G) = 2, v1 vn−3 , v2 vn−3 ∈ E (G). By the given edge degree condition, 5 = d(v1 ) + d(vn−3 ) ≥ n − 2 if n is even and 5 = d(v1 ) + d(vn−3 ) ≥ n − 3 if n is odd, which n ≤ 7 if n is even and n ≤ 8 if n is odd, a contradiction. It remains for us to consider the case that d(u1 ) = d(u2 ) = n − 4. We assume, without loss of generality, that N (u1 ) = {u, v1 , v2 , . . . , vn−5 }. Let N (u2 ) = {u, v1 , v2 , . . . , vn−5 }. If vn−4 vn−3 ̸∈ E (G), then vn−4 is adjacent to at least two vertices of {v1 , v2 , . . . , vn−5 }. In this case, G contains K3,3 −e which is collapsible by Lemma 2.2, a contradiction. If vn−4 vn−3 ∈ E (G), then

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each vertex of {v1 , . . . , vn−5 } is adjacent at most one vertex of vn−3 and vn−4 since G has no 3-cycle. Thus, d(vn−3 )+ d(vn−4 ) ≤ 2 + n − 5 = n − 3. In this case, by the given edge degree condition, n is odd and d(vn−3 ) + d(vn−4 ) = n − 3. Since n ≥ 9, one of {vn−3 , vn−4 } is adjacent to at least two vertices of {v1 , v2 , . . . , vn−5 }. This implies that G contains a K3,3 − e which is collapsible by Lemma 2.2, a contradiction. Let N (u2 ) = {u, v2 , v3 , . . . , vn−4 }. Assume that v1 vn−3 ∈ E (G). If n ≥ 9, by the given degree condition, d(v1 ) + d(vn−3 ) ≥ n − 3 which implies that d(vn−3 ) ≥ n − 5 if v1 vn−4 ̸∈ E (G), and d(vn−3 ) ≥ n − 6 if v1 vn−4 ∈ E (G). Since G has no 3-cycle, vn−3 is adjacent to at least two vertices of {v2 , v3 , . . . , vn−5 }. If n = 8, by the given degree condition, d(v1 ) + d(vn−3 ) ≥ n − 2 which implies that d(vn−3 ) ≥ n − 4 if v1 vn−4 ̸∈ E (G), and d(vn−3 ) ≥ n − 5 if v1 vn−4 ∈ E (G). Since G has no 3-cycle, vn−3 is adjacent to at least two vertices of {v2 , v3 , . . . , vn−5 }. The proof is similar if vn−3 vn−4 ∈ E (G). If v1 vn−3 , vn−4 vn−3 ̸∈ E (G), vn−3 is adjacent to at least two vertices of {v2 , v3 , . . . , vn−5 } since δ(G) ≥ 2. Thus, in each case, G contains K3,3 − e which is collapsible by Lemma 2.2, a contradiction. Let N (u2 ) = {u, v3 , v4 , . . . , vn−3 }. By the given degree condition and δ(G) ≥ 2, v1 vn−4 , v1 vn−3 , v2 vn−4 , v2 vn−3 ∈ E (G) and n = 9. It is easy to verify that G contains K3,3 − e which is collapsible by Lemma 2.2, a contradiction.  3. Proof of Theorem 1.4 Assume that G is a graph on n ≥ 8 vertices and that G satisfies ξ (G) ≥ n − 2 − p(n). Let G′ be the reduction of G. By Lemma 2.3, G′ contains at most two non-trivial vertices. By Lemmas 2.5 and 2.7, we assume that G′ has no non-trivial vertices. This means that G′ = G. By Theorem 2.1, |E (G)| ≤ 2n − 4 and δ(G) ≤ 3. We now distinguish the following two cases for n even or odd. Case 1. n is even. Assume first that δ(G) = 3. Let d(u) = 3 and N (u) = {u1 , u2 , u3 }. Since G contains no triangle, {u1 , u2 , u3 } is an independent set. Since n is even, d(u)+ d(ui ) ≥ n − 2, which implies that d(ui ) ≥ n − 5 and |E (G)| ≥ d(u1 )+ d(u2 )+ d(u3 ) ≥ 3(n − 5). = 12 since This means that 2n − 4 ≥ |E (G)| ≥ 3(n − 5) which implies n ≤ 11. If n = 8, then 12 = 2n − 4 ≥ |E (G)| ≥ 3n 2 δ(G) = 3. This means that G is cubic and |E (G)| = 12. In this case, G is G21 or G22 . If n = 10, then 16 = 2n − 4 ≥ |E (G)| ≥ 3n = 15. It follows that G contains at least 8 vertices of degree 3 and denote by the set S all vertices of degree 3. We claim that 2 G[S ] contains edges for otherwise, each edge of G contains at most one vertex in S and hence |E (G)| ≥ 3|S | ≥ 3 × 8 = 24, a contradiction. Pick an edge e = xy of G[S ] and d(x) = d(y) = 6 < 10 − 2 = 8, a contradiction. Next, assume that δ(G) = 2. Let d(u) = 2, N (u) = {u1 , u2 } and denote by V (G)\{u, u1 , u2 } = {v1 , v2 , . . . , vn−3 }. By the given edge degree condition, d(ui ) ≥ n − 4. By Lemma 2.8, G is a K2,n−2 . Finally, we assume that δ(G) = 1. Let d(u) = 1, N (u) = {u1 } and V (G)\{u, u1 } = {v1 , v2 , . . . , vn−2 }. Since ξ (G) ≥ n − 2, d(u1 ) ≥ n − 3. If d(u1 ) = n − 1, then G is a K1,n−1 . If d(u1 ) = n − 2, let N (u1 ) = {v1 , v2 , . . . , vn−3 }. Without loss of generality, we assume that vn−2 v1 ∈ E (G). By the given edge degree condition, d(vn−2 ) ≥ n − 4. Thus, G is a K2′ ,n−3 or a K2′′,n−4 . Thus, assume that d(u1 ) = n − 3 and N (u1 ) = {u, v1 , v2 , . . . , vn−4 }. In the case that vn−3 vn−2 ∈ E (G), N (vn−3 ) ∪ N (vn−2 ) = {v1 , v2 , . . . , vn−4 } since ξ (G) ≥ n − 2. If N (vn−3 ) = {v1 , v2 , . . . , vn−4 }, then G is K2∗,n−4 . If not, let vn−3 v1 , vn−2 vn−4 ∈ E (G). Since ξ (G) ≥ n − 2, d(vn−2 ) ≥ n − 4 and d(vn−3 ) ≥ n − 4. By Theorem 2.1(2), N (vn−2 ) ∩ N (vn−3 ) = ∅. This means that 2(n − 4) ≤ n − 2 which implies that n ≤ 6, a contradiction. Thus, assume that vn−3 vn−2 ̸∈ E (G). Since G is connected, without loss of generality, we assume that vn−2 v1 ∈ E (G). By the given edge degree condition, d(vn−2 ) + d(v1 ) ≥ n − 2, which implies that e(vn−2 , {v1 , v2 , . . . , vn−4 }) ≥ n − 5 ≥ 3. Similarly, e(vn−3 , {v1 , v2 , . . . , vn−4 }) ≥ n − 5 ≥ 3. This implies that G contains K3,3 − e which is collapsible by Lemma 2.2, a contradiction. Case 2. n is odd. Assume first that δ(G) = 3. Let d(u) = 3 and N (u) = {u1 , u2 , u3 }. By Theorem 2.1 (3), |E (G)| ≤ 2n − 4. Since ξ (G) ≥ n − 3, d(ui ) ≥ n − 6 for each i ∈ {1, 2, 3}. By Theorem 2.1 (2), G has no 3-cycle. This means that each edge of G is incident with at most one vertex of {u1 , u2 , u3 }. Thus, 3(n − 6) ≤ d(u1 ) + d(u2 ) + d(u3 ) ≤ |E (G)| ≤ 2n − 4 which implies n ≤ 14. On the 3n 3n+1 other hand, since n ≥ 8 and n is odd, n ≥ 9. Since δ(G) = 3, |E (G)| = 12 v∈V (G) d(v) ≥ 2 . Since n is odd, |E (G)| ≥ 2 . If n = 9, then

3n+1 2

= 14 = 2n − 4 = |E (G)|. This implies that G has one vertex of degree 4 and the other 8 vertices of degree

3. Thus, G is in {G23 , G24 , G25 }. If n = 11, then 3n2+1 = 17 ≤ |E (G)| ≤ 2n − 4 = 18. This means that G has either one vertex of degree 5 and one vertex of degree 4, and the other 9 vertices each of which is degree 3 or one vertex of degree 4 and the other 10 vertices each of which is degree 3. In each case, G contains one edge xy with two end-vertices each of which is of degree 3. By the given edge degree condition, 6 = d(x) + d(y) ≥ n − 3 = 11 − 3 = 8, a contradiction. If n = 13, then 3n+1 = 20 ≤ |E (G)| ≤ 2n − 4 = 22. As the proof above, G contains one edge xy with two end-vertices each of which is of 2 degree 3. By the given edge degree condition, 6 = d(x) + d(y) ≥ n − 3 = 13 − 3 = 10, a contradiction. Next, assume that δ(G) = 2. Let d(u) = 2 and N (u) = {u1 , u2 } and denote by V (G)\{u, u1 , u2 } = {v1 , v2 , . . . , vn−3 }. Since ξ (G) ≥ n − 3, d(ui ) ≥ n − 5. If either max{d(u1 ), d(u2 )} ≥ n − 2 or min{d(u1 ), d(u2 )} ≥ n − 4, by Lemma 2.8, G is a K2,n−2 . Thus, d(u1 ) ≤ n − 5 or d(u2 ) ≤ n − 5, together with max{d(u1 ), d(u2 )} ≤ n − 3. We assume, without loss of generality, that d(u1 ) ≥ d(u2 ). Consider that d(u1 ) = n − 3 and d(u2 ) = n − 5. We assume, without loss of generality, that N (u1 ) = {u, v1 , v2 , . . . , vn−4 }. We only need to consider the following two cases. If N (u2 ) = {u, v1 , v2 , . . . , vn−6 }, observe vertex vn−3 . Since δ(G) ≥ 2, vn−3 is adjacent to vn−4 . By the given edge degree condition, vn−3 is adjacent to at least n − 5 vertices of {v1 , . . . , vn−4 }. Thus, G contains a K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. If N (u2 ) = {u, v4 , v5 , . . . , vn−3 }, then

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v1 vn−3 , v2 vn−3 , v3 vn−3 ∈ E (G) since δ(G) = 2. By the given edge degree condition, 6 = d(v1 ) + d(vn−3 ) ≥ n − 3 since G has no 3-cycle, which implies that n ≤ 9. Since n ≥ 8 and n is odd, n = 9. In this case, G is G26 . Consider that d(u1 ) = n − 4 and d(u2 ) = n − 5. We assume, without loss of generality, that N (u1 ) = {u, v1 , v2 , . . . , vn−5 }. Let N (u2 ) = {u, v1 , v2 , . . . , vn−6 }. If vn−3 vn−4 ∈ E (G), then by the given degree condition, d(vn−3 ) + d(vn−4 ) ≥ n − 3. By Theorem 2.1, G has no 3-cycle. Thus, N (vn−3 ) ∩ N (vn−4 ) = ∅ and N (vn−3 ) ∪ N (vn−4 ) = {v1 , v2 , . . . , vn−5 }. We assume, without loss of generality, that vn−5 vn−4 ∈ E (G). In this case, d(vn−5 ) = 2. By the given degree condition, d(vn−5 ) + d(vn−4 ) ≥ n − 3. This implies that vn−4 is adjacent to at least n − 6 vertices of {v1 , . . . , vn−5 }. Thus, G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. If vn−3 vn−4 ̸∈ E (G), then vn−5 is adjacent to at least one of {vn−4 , vn−3 } since δ(G) = 2. We assume, without loss of generality, that vn−5 vn−4 ∈ E (G). This implies that vn−4 is adjacent to at least n − 6 vertices of {v1 , . . . , vn−5 }. Thus, G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Let N (u2 ) = {u, v3 , v4 , . . . , vn−4 }. If for each i ∈ {1, 2}, vi vn−4 ∈ E (G), then vn−3 is adjacent to one of {v3 , . . . , vn−5 } since δ(G) ≥ 2. By the given edge degree condition, vn−3 is adjacent to at least n − 7 vertices of {v3 , v4 , . . . , vn−5 } since d(vj ) = 3 for j ∈ {v3 , v4 , . . . , vn−5 }. Thus, G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Thus, assume that vn−4 is adjacent to at most one vertex of {v1 , v2 }. Thus, since δ(G) ≥ 2 and G has no 3-cycle, vn−3 must be adjacent to one of {v1 , v2 }. Let v1 vn−3 ∈ E (G). If for each i ∈ {1, 2}, vi vn−4 ̸∈ E (G), then v1 vn−3 , v2 vn−3 , vn−4 vn−3 ∈ E (G). By the given edge degree condition, d(v1 ) + d(vn−3 ) ≥ n − 3, which implies that vn−3 is adjacent to at least n − 8 vertices of {v3 , v4 , . . . , vn−5 }. If n ≥ 11, then G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. If n = 9, then G is G27 . Let N (u2 ) = {u, v4 , v5 , . . . , vn−3 }. Since δ(G) = 2 and G has no 3-cycle, there is vi vj ∈ E (G), where i ∈ {1, 2, 3} and j ∈ {n − 4, n − 3}. By the given edge degree condition, 7 ≥ d(vi ) + d(vj ) ≥ n − 3, which implies that n ≤ 10. Since n is odd, n = 9. In this case, G contains K3,3 which is collapsible by Lemma 2.2, a contradiction. Consider that d(u1 ) = d(u2 ) = n − 5. We assume, without loss of generality, that N (u1 ) = {u, v1 , v2 , . . . , vn−6 }. If N (u1 ) = N (u2 ), we claim that F = G[{vn−5 , vn−4 , vn−3 }] has no edge. Since G has no 3-cycle, F contains at most two edges. If F has two edges, say vn−5 vn−4 , vn−4 vn−3 ∈ E (G). By the given degree condition, d(vn−5 ) + d(vn−4 ) ≥ n − 3, which implies that e({vn−5 , vn−4 }, {v1 , . . . , vn−6 }) ≥ n − 6. Since G has no 3-cycle, N (vn−4 ) ∩ N (vn−5 ) = ∅. Thus, vn−3 cannot be adjacent to any vertex of {v1 , . . . , vn−6 }, contrary to that δ(G) = 2. If F has only one edge, say vn−5 vn−4 ∈ E (G). By the given degree condition, d(vn−5 ) + d(vn−4 ) ≥ n − 3, which implies that n − 6 ≥ e({vn−5 , vn−4 }, {v1 , . . . , vn−6 }) ≥ n − 5, a contradiction. Thus, F has no edge. Since δ(G) = 2, d(vj ) ≥ 2 for j ∈ {n − 3, n − 4, n − 5}. This implies that G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Let N (u2 ) = {u, v2 , v3 , . . . , vn−5 }. Assume that G contains the edge vn−3 vn−4 . If v1 vn−4 , vn−5 vn−3 ∈ E (G), then by the given edge degree condition, d(v1 ) + d(vn−4 ) ≥ n − 3 which implies that e(vn−4 , {v1 , . . . , vn−5 }) ≥ n − 7. Similarly, e(vn−3 , {v1 , . . . , vn−5 }) ≥ n − 7. Since G has no 3-cycle, n − 5 ≥ 2(n − 7). This shows that n ≤ 9. Since n ≥ 8 and n is odd, n = 9. In this case, G is G28 . If v1 vn−3 , vn−3 vn−5 ∈ E (G), then vn−4 vj ∈ E (G) since δ(G) ≥ 2, where j ∈ {2, . . . , n − 6}. Since G has no 3-cycle, d(vj ) = 3. By the given edge degree condition, d(vj ) + d(vn−4 ) ≥ n − 3, which implies that vn−4 is adjacent to at least n − 5 vertices of {v2 , . . . , vn−6 }. This is a contradiction. The proof is similar for the case that v1 vn−4 , vn−4 vn−5 ∈ E (G). Thus, assume that vn−3 vn−4 ̸∈ E (G). If v1 vn−5 ∈ E (G), then by given edge degree condition, 6 ≥ d(v1 ) + d(vn−5 ) ≥ n − 3, which implies that n ≤ 9. Since n ≥ 8 and n is odd, n = 9. In this case, G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. If v1 vn−5 ̸∈ E (G), let v1 vn−4 , vn−5 vn−3 ∈ E (G) since δ(G) = 2. By the given edge degree condition, d(v1 ) + d(vn−4 ) ≥ n − 3 and hence e(vn−4 , {v1 , v2 , . . . , vn−5 }) ≥ n − 6. Similarly, e(vn−3 , {v1 , v2 , . . . , vn−5 }) ≥ n − 6. If n ≥ 11, then G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. If n = 9, then G is G29 . Let N (u2 ) = {u, v3 , v4 , . . . , vn−4 }. Denote by W the subgraph induced by {v1 , v2 , vn−5 , vn−4 }. If vn−3 is adjacent to all four vertices of {v1 , v2 , vn−5 , vn−4 }, then W has no edge. By the given edge degree condition, d(v1 ) + d(vn−3 ) ≥ n − 3, which implies that e(vn−3 , {v1 , v2 , . . . , vn−4 }) ≥ n − 5. If n ≥ 11, then G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. If n = 9, then G is G30 or G31 . Assume that vn−3 is adjacent to three vertices of {v1 , v2 , vn−5 , vn−4 }. Without loss of generality, we assume that v1 vn−3 ̸∈ E (G). Since δ(G) = 2, by symmetry, let v1 vn−4 ∈ E (G). By the given degree condition, 3 + 3 ≥ d(v1 ) + d(vn−4 ) ≥ n − 3, which implies that n ≤ 9. Since n ≥ 8 and n is odd, n = 9. By the given degree condition, d(v2 ) + d(vn−3 ) ≥ n − 3, which implies that vn−3 is adjacent to each of {v2 , . . . , vn−4 }. It is easy to verify that G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Assume that vn−3 is adjacent to two vertices of {v1 , v2 , vn−5 , vn−4 }. By symmetry, there are two cases: vn−3 v1 , vn−3 v2 ∈ E (G) or vn−3 v1 , vn−3 vn−4 ∈ E (G). In the former case, similarly, let v1 vn−4 ∈ E (G). By the given degree condition, 3 + 4 = d(v1 ) + d(vn−4 ) ≥ n − 3, which implies that n ≤ 10. Since n ≥ 8 and n is odd, n = 9. Similarly, we can prove that G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. In the latter case, similarly we can prove that n = 9 and G is G32 . Assume that vn−3 is adjacent to at most one vertex of {v1 , v2 , vn−5 , vn−4 }. In this case, since δ(G) = 2, let v1 vn−3 , vn−4 vn−3 ̸∈ E (G) and v1 vn−4 ∈ E (G). By the given edge degree condition, 3 + 3 ≥ d(v1 ) + d(vn−4 ) ≥ n − 3, which implies that n ≤ 9. Since n ≥ 8 and n is odd, n = 9. Since δ(G) = 2, vn−3 vn−6 ∈ E (G) but d(vn−3 ) + d(vn−6 ) = 5 < n − 3 = 6, a contradiction. Let N (u2 ) = {u, v4 , v5 , . . . , vn−3 }. Since v1 is adjacent to at most all the three vertices of {vn−3 , vn−4 , vn−5 }, d(v1 ) ≤ 4. Similarly, d(vn−3 ) ≤ 4. By the given degree condition, 4 + 4 ≥ d(v1 ) + d(vn−4 ) ≥ n − 3, which implies that n ≤ 11. Since n ≥ 8 and n is odd, n = 9 or n = 11. If n = 11, then the subgraph induced by {v1 , v2 , v3 , vn−5 , vn−4 , vn−3 } is a K3,3 , which is collapsible by Lemma 2.2, a contradiction. If n = 9, denote by F the subgraph induced by {v1 , v2 , v3 , vn−5 , vn−4 , vn−3 }. Since G has no 3-cycle, F is a bipartite graph with one part V1 = {v1 , v2 , v3 } and the other part V2 = {vn−5 , vn−4 , vn−3 }. Since δ(G) = 2, the minimum degree of F is 1. If F contains three vertices of degree 3 or two vertices of degree 3 in one part, then

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G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Suppose that F contains two vertices of degree 3 such that each part has precisely one. If the other four vertices have at least one degree 2, then G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. If the other four vertices are of degree 1, then G is G33 . If F contains only one vertex of degree 3, then G contains an edge xy such that 5 = d(x) + d(y). By the given edge degree condition, d(x) + d(y) ≥ n − 3, which implies that n ≤ 8, a contradiction. Thus, assume that F has no vertex of degree 3. Since ξ (G) ≥ n − 3, each vertex of F is of degree 2 and F is a 6-cycle. Thus, G is G34 . Finally, assume that δ(G) = 1. Let d(u) = 1 and N (u) = {u1 }. Denote by V (G)\{u, u1 } = {v1 , v2 , . . . , vn−2 }. Since ξ (G) ≥ n − 3, d(u1 ) ≥ n − 4. If d(u1 ) = n − 1, then G is K1,n−1 . If d(u1 ) = n − 2, then G is K2′ ,n−3 or K2′′,n−4 or K2′′′,n−5 . Assume that d(u1 ) = n − 3 and N (u1 ) = {u, v1 , v2 , . . . , vn−4 }. In this case, vn−3 vn−2 ∈ E (G) or vn−3 vn−2 ̸∈ E (G). In the former case, by symmetry, assume that d(vn−2 ) = 1. In this case. G is K2∗,n−4 . Thus, assume that d(vn−2 ) ≥ 2 and d(vn−3 ) ≥ 2. Without loss of generality, assume that v1 vn−3 , vn−2 vn−4 ∈ E (G). By the given edge degree condition, d(v1 ) + d(vn−3 ) ≥ n − 3, which implies that e(vn−3 , {v1 , . . . , vn−4 }) ≥ n − 6. Similarly, e(vn−2 , {v1 , . . . , vn−4 }) ≥ n − 6. Since G has no 3-cycle, 2(n − 6) ≤ n − 4 which implies that n ≤ 8, a contradiction. In the latter case, since G is connected, let v1 vn−3 , vn−2 vn−4 ∈ E (G). Similarly, we obtain that d(vn−3 ) ≥ n − 6 and d(vn−2 ) ≥ n − 6. Thus, G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Assume that d(u1 ) = n − 4 and N (u1 ) = {u, v1 , v2 , . . . , vn−5 }. Denote by K the subgraph induced by {vn−4 , vn−3 , vn−2 }. Let K have no edges. Since G is connected, without loss of generality, we assume that vn−4 v1 , vn−3 v2 , vn−2 v3 ∈ E (G). By the given edge degree condition, d(v1 ) + d(vn−4 ) ≥ n − 3. Since d(v1 ) ≤ 4, d(vn−4 ) ≥ n − 7. Similarly, d(vj ) ≥ n − 7 for j = n − 3, n − 2. Recall that n is odd. If n ≥ 11 then G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Let n = 9. If d(vj ) = 2 for j = n − 4, n − 3, n − 2, then G is G35 . If d(vj ) ≥ 3 for some j ∈ {n − 4, n − 3, n − 2}, then G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Let K contain one edge, say vn−4 vn−3 ∈ E (G). By the given edge degree condition, d(vn−4 ) + d(vn−3 ) ≥ n − 3. Since G has no 3-cycle, N (vn−3 ) ∪ N (vn−4 ) = {v1 , . . . , vn−5 } and N (vn−4 )∩ N (vn−3 ) = ∅. Since ξ (G) ≥ n − 3, this forces that vn−2 is adjacent to at least n − 6 vertices of {v1 , . . . , vn−5 }. Thus, G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Let K contain two edges, say vn−4 vn−3 , vn−3 vn−2 ∈ E (G). If d(vn−4 ) = d(vn−2 ) = 1, d(vn−3 ) > 2, then G is {K2∗,n−6 , K2∗,n−5 }. If d(vn−4 ) = 1, d(vn−3 ) > 2, d(vn−2 ) > 1, then by the given degree condition, d(vn−4 )+ d(vn−3 ) ≥ n − 3, which implies that e(vn−3 , {v1 , . . . , vn−5 }) ≥ n − 6. Since G has no 3-cycle, vn−2 is adjacent to one vertex say v1 , of {v1 , . . . , vn−5 }. In this case, d(v1 ) + d(vn−2 ) = 4 ≥ n − 3, which implies that n ≤ 7, a contradiction. By the given edge degree condition, we can get a contradiction if d(vn−4 ) = 1, d(vn−3 ) = 2 or d(vn−3 ) = 2, d(vn−2 ) = 1. If d(vn−3 ) = 2, d(vn−2 ) > 1 and d(vn−4 ) > 1, then by the given degree condition, d(vn−3 ) + d(vn−4 ) ≥ n − 3, which implies that e(vn−4 , {v1 , . . . , vn−5 }) ≥ n − 6. Similarly, e(vn−2 , {v1 , . . . , vn−5 }) ≥ n − 6. In this case, G contains K3,3 − e, which is collapsible by Lemma 2.2, a contradiction. Finally, assume that d(vn−4 ) > 1, d(vn−3 ) > 2, d(vn−2 ) > 1. By the given degree condition, e({vn−4 , vn−3 }, {v1 , . . . , vn−5 }) ≥ n − 6, e({vn−2 , vn−3 }, {v1 , . . . , vn−5 }) ≥ n − 6. Since G has no 3-cycle, N (vn−2 ) ∩ N (vn−3 ) = ∅ and N (vn−4 ) ∩ N (vn−3 ) = ∅. Let n1 = e(vn−4 , {v1 , . . . , vn−5 }), n2 = e(vn−3 , {v1 , . . . , vn−5 }) and n3 = e(vn−2 , {v1 , . . . , vn−5 }). By the given edge degree condition, n1 ≥ n − 7 and n3 ≥ n − 7 and n2 ≤ 2. Note that vi can be adjacent to both vn−2 and vn−4 for i ∈ {1, . . . , n − 5}. Thus, G contains a K3,3 − e, then by Lemma 2.2, G is collapsible, a contradiction.  4. Applications A subgraph H of a graph G is dominating if G − V (H ) is edgeless. If a graph is collapsible, then it has a dominating eulerian subgraph. Harary and Nash-Williams [11] established a closed relationship between dominating eulerian subgraphs of a graph and Hamilton cycles in its line graph, as follows. Theorem 4.1. Let G be a graph on n ≥ 4 vertices. The line graph L(G) is Hamiltonian if and only if G has a dominating eulerian subgraph. Let G be a connected graph on n ≥ 8 vertices. Using Theorem 1.2, Brualdi and Shanny [2] proved that if ξ (G) ≥ n, then L(G) is Hamiltonian. This result was improved by Clark in [10] who proved if ξ (G) ≥ n − 1 − p(n), then L(G) is Hamiltonian. Using Theorems 1.4 and 4.1, we can get the following result. Corollary 4.2. Let G be a connected graph on n ≥ 8 vertices. If ξ (G) ≥ n−2−p(n), then L(G) is Hamiltonian except G = {G1 , G2 } or G ∈ H . By the definition of collapsible graph, Theorem 1.3 can be written as follows. Corollary 4.3 ([13]). Let G be a connected graph on n vertices and let u, v ∈ V (G). If ξ (G) ≥ n − 1 − p(n), then exactly one of the following holds: (1) G has a spanning (u, v)-trail. (2) G is one of {C5 , G′7 , G′′7 , K1,n−1 , K2′ ,n−3 } or the reduction of G is K1,t −1 for some integer t ≥ 2. (3) G = K2,n−2 , u = v and n is odd. (4) G = K2,n−2 , u ̸= v, uv ̸∈ E (G), n is even and d(u) = d(v) = n − 2.

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As a corollary of Theorem 1.4, we strengthen Theorem 1.3 as follows. Corollary 4.4. Let G be a connected graph on n ≥ 8 vertices. If ξ (G) ≥ n − 2 − p(n), then exactly one of the following holds: (1) G has a spanning (u, v)-trail. (2) The reduction of G is K1,t −1 for some t ≥ 2. (3) G is one in {G1 , G2 , . . . , G35 } or G is in {K1,n−1 , K2′ ,n−3 , K2′′,n−4 , K2∗,n−4 , K2′′′,n−5 , K2∗,n−5 , K2∗,n−6 }. (4) G = K2,n−2 , u = v and n is odd. (5) G = K2,n−2 , u ̸= v, uv ̸∈ E (G), n is even and d(u) = d(v) = n − 2. Acknowledgments This work is supported by the Natural Science Foundation of China (10571071) and by Doctoral Fund of Ministry of Education of China (20130144110001). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]

J.A. Bondy, U.S. R Murty, Graph Theory with Application, North-Holland, New York, 1976. R.A. Brualdi, R.F. Shanny, Hamilton line graphs, J. Graph Theory 5 (1981) 307–314. P.A. Catlin, Spanning trail, J. Graph Theory 11 (1987) 161–167. P.A. Catlin, A reduction method to find spanning Eulerian subgraphs, J. Graph Theory 12 (1988) 29–45. P.A. Catlin, Contractions of graphs with no spanning Eulerian subgraphs, Combinatorica 8 (1988) 313–327. P.A. Catlin, Super-Eulerian graphs, collapsible graphs and four-cycles, Congr. Number. 58 (1988) 233–246. P.A. Catlin, Spanning Eulerian subgraphs and matchings, Discrete Math. 76 (1989) 95–116. P.A. Catlin, X. Li, Supereulerian graphs of minimum degree at least 4, Adv. Math. (China) 29 (1999) 65–68. Z.-H. Chen, H.-J. Lai, Collapsible graphs and matchings, J. Graph Theory 17 (1993) 597–605. L. Clark, On Hamilton line graphs, J. Graph Theory 8 (1984) 303–307. F. Harary, C.St.J.A. Nash-Williams, On Eulerian and Hamilton graphs and line graphs, Canad. Math. Bull. 8 (1965) 701–709. H.-J. Lai, Y. Shao, H. Wu, J. Zhou, Every 3-connected, essentially 11-connected line graph is Hamiltonian, J. Combin. Thoery Ser. B 96 (2006) 571–576. H. Li, W. Yang, A note on collapsible graphs and super-Eulerian graphs, Discrete Math. 312 (2012) 2223–2227. C. Thomassen, Reflection on graph theory, J. Graph Theory 10 (1987) 309–324. W. Yang, H.-J. Lai, H. Li, X. Guo, Collapsible graphs and Hamiltonian connectedness of line graphs, Discrete Appl. Math. 160 (2012) 1837–1844. W. Yang, H.-J. Lai, H. Li, X. Guo, Collapsible graphs and Hamiltonicity of line graphs, Graphs Combin. 30 (2014) 501–510.