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Implicit degree condition for hamiltonicity of 2-heavy graphs
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Implicit degree condition for hamiltonicity of 2-heavy graphs Xing Huang Huaerzhi Education and Technology Company Limited, Beijing, 100036, China
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Article history: Received 11 April 2016 Received in revised form 18 September 2016 Accepted 17 October 2016 Available online xxxx Keywords: Implicit degree Hamilton cycle f-implicit-heavy 2-heavy
abstract Let id(v) denote the implicit degree of a vertex v in a graph G. An induced subgraph S of G is called f-implicit-heavy if max{id(x), id(y)} ≥ |V (G)|/2 for every pair of vertices x, y ∈ V (S ) at distance 2 in S. For a given graph R, G is called R-f-implicit-heavy if every induced subgraph of G isomorphic to R is f-implicit-heavy. For a family R of graphs, G is called R-f-implicit-heavy if G is R-f-implicit-heavy for every R ∈ R. G is called 2-heavy if there are at least two end-vertices of every induced claw (K1,3 ) in G have degree at least |V (G)|/2. In this paper, we prove that: Let G be a 2-connected 2-heavy graph. If G is {P7 , D}-f-implicit-heavy or {P7 , H }-f-implicit-heavy, then G is hamiltonian. © 2016 Elsevier B.V. All rights reserved.
1. Introduction In this paper, we consider only undirected, finite and simple graphs. Notation and terminology not defined here can be found in [2]. Let G be a graph and H be a subgraph of G. For a vertex u ∈ V (G), the neighborhood of u in H is denoted by NH (u) = {v ∈ V (H ) : uv ∈ E (G)} and the degree of u in H is denoted by dH (u) = |NH (u)|. For two vertices u, v ∈ V (H ), the distance between u and v in H, denoted by dH (u, v), is the length of a shortest (u, v)-path in H. When there is no danger of ambiguity, we can use N (u), d(u) and d(u, v) in place of NG (u), dG (u) and dG (u, v), respectively. We use N2 (v) to denote the set of vertices which are at distance 2 with v , i.e. N2 (v) = {u ∈ V (G) : d(u, v) = 2}. Set m2 (v) = min{d(u) : u ∈ N2 (v)} and M2 (v) = max{d(u) : u ∈ N2 (v)}. An induced subgraph K1,3 of G with vertex set {u, v, w, x} and edge set {uv, uw, ux} is called a claw, with center u and end-vertices v, w, x. For a given graph R, G is called R-free if G contains no induced subgraph isomorphic to R. For a family R of graphs, G is called R-free if G is R-free for every R ∈ R. A graph G is called hamiltonian if it contains a Hamilton cycle, i.e. a cycle that contains all vertices of G. Degree condition is an important type of sufficient conditions for a graph to be hamiltonian. The following two results due to Ore and Fan, respectively, are well-known. Theorem 1 ([10]). Let G be a graph on n ≥ 3 vertices. If d(x) + d(y) ≥ n for each pair of nonadjacent vertices x and y in G, then G is hamiltonian. Theorem 2 ([5]). Let G be a 2-connected graph on n ≥ 3 vertices. If max{d(x), d(y)} ≥ n/2 for every pair of vertices x and y at distance 2 in G, then G is hamiltonian. In 1989, Zhu, Li and Deng [11] found that though some vertices may have small degrees, we can use some large degree vertices to replace small degree vertices in the right position considered in the proofs, so that we may construct a longer cycle. This idea leads to the definition of implicit degree, denoted by id(v) of a vertex v .
E-mail address: [email protected]. http://dx.doi.org/10.1016/j.dam.2016.10.031 0166-218X/© 2016 Elsevier B.V. All rights reserved.
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Fig. 1. Forbidden subgraph.
Definition 1 ([11]). Let v be a vertex of a graph G and d(v) = l + 1. If N2 (v) ̸= ∅ and d(v) ≥ 2, then let d1 (v) ≤ d2 (v) ≤ d3 (v) ≤ · · · ≤ dl (v) ≤ dl+1 (v) ≤ · · · be the degree sequence of vertices in N (v) ∪ N2 (v). Define m2 (v), dl+1 (v), dl (v),
if dl (v) ≤ m2 (v); if dl+1 (v) > M2 (v); otherwise.
d (v) = ∗
Then the implicit degree of v is defined as id(v) = max{d(v), d∗ (v)}. If N2 (v) = ∅ or d(v) ≤ 1, then define id(v) = d(v). Clearly, id(v) ≥ d(v) for every vertex v . The authors in [11] gave a sufficient condition for a 2-connected graph to be hamiltonian under implicit degree condition. Theorem 3 ([11]). Let G be a 2-connected graph on n ≥ 3 vertices. If id(x) + id(y) ≥ n for every pair of nonadjacent vertices x and y in G, then G is hamiltonian. Remark 1. The introduction of implicit degree is very useful, since many classic results by considering degree conditions for the hamiltonicity of graphs can be generalized. We just give one example to show this. Fan’s theorem (Theorem 2) can be easily obtained from Theorem 3. And the authors in [11] gave a simple proof to show this. Forbidden subgraph condition is another type of sufficient conditions for hamiltonicity of graphs. The following two results are of this type, where P7 , D and H are graphs in Fig. 1. Theorem 4 ([4]). Let G be a 2-connected graph. If G is {K1,3 , P7 , D}-free, then G is hamiltonian. Theorem 5 ([6]). Let G be a 2-connected graph. If G is {K1,3 , P7 , H }-free, then G is hamiltonian. A vertex v of a graph G on n vertices is called heavy (or implicit-heavy) if d(v) ≥ n/2 (or id(v) ≥ n/2). If v is not heavy (or not implicit-heavy), we call it light (or implicit-light). A claw of G is called 2-heavy if at least two of its end-vertices are heavy. A subgraph S of G is called f -heavy (or f -implicit-heavy) if max{d(x), d(y)} ≥ n/2 (or max{id(x), id(y)} ≥ n/2) for every pair of vertices x, y ∈ V (S ) at distance 2 in S. For a given graph R, G is called R-f -heavy (or R-f -implicit-heavy) if every induced subgraph of G isomorphic to R is f -heavy (or f -implicit-heavy). For a family R of graphs, G is called R-f -heavy (or R-f -implicit-heavy) if G is R-f -heavy (or R-f -implicit-heavy) for every R ∈ R. Clearly, every R-free graph is also R-f -heavy, and every R-f -heavy graph is also R-f -implicit-heavy. In 1997, Broersma et al. [3] generalized existing results by combining degree conditions and forbidden subgraph conditions. More precisely, they restricted Fan’s condition (Theorem 2) to certain subgraphs and got the following result. Theorem 6 ([3]). Let G be a 2-connected graph on n ≥ 3 vertices. If G is 2-heavy and |N (u) ∩ N (v)| ≥ 2 for every pair of vertices u, v with d(u, v) = 2 and max{d(u), d(v)} ≤ n/2, then G is hamiltonian. Ning extended Theorem 4 and Theorem 5 by relaxing forbidden subgraph conditions to conditions in which the subgraphs are allowed. Theorem 7 ([9]). Let G be a 2-connected graph on n ≥ 3 vertices. If G is {K1,3 , P7 , D}-f -heavy or {K1,3 , P7 , H }-f -heavy, then G is hamiltonian. We use implicit degree instead of degree in Theorem 7 and get the following result. Theorem 8. Let G be a 2-connected 2-heavy graph. If G is {P7 , D}-f -implicit-heavy or {P7 , H }-f -implicit-heavy, then G is hamiltonian. Remark 2. Since id(v) ≥ d(v) for every vertex v , it is easy to see that every graph satisfying the condition of Theorem 7 also satisfies the condition of Theorem 8, but the converse is not true. Furthermore, the following graph is a hamiltonian graph satisfying the condition of Theorem 8, but not the condition of Theorem 7. Let n ≥ 18 be an even integer and K n ∪ K n −7 denotes the union of two complete graphs K n and K n −7 . Set V (K n ) = 2 2 2 2 2 {u1 , u2 , . . . , u n } and V (K n −7 ) = {v1 , v2 , . . . , v n −7 }. We construct a graph G with V (G) = V (K n ∪ K n −7 ) ∪ {x1 , x2 , . . . , x7 } 2
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Fig. 2. Graph in Remark 2.
and E (G) = E (K n ∪ K n −7 ) ∪{x4 x7 , x4 x3 , x1 x2 , x1 x3 , x2 x3 , x2 x5 , x5 x6 } ∪{xi uj : i = 1, 2 and j = 1, 2, . . . , 2n } ∪{x3 uj : j = 2 2 1, 2, . . . , 2n − 3} ∪{xi vj : i = 6, 7 and j = 1, 2, . . . , 2n − 7} (see Fig. 2). It is easy to see {x1 , x2 , . . . , x7 } induces a P7 and n max{d(x3 ), d(x7 )} = 2 − 3 < 2n with d(x3 , x7 ) = 2. So G does not satisfy the condition of Theorem 7, but it is hamiltonian. However, since d(x3 ) = 2n −3 and the degree sequence of N (x3 )∪N2 (x3 ) is (2, 2n −7, 2n , 2n , 2n +1, 2n +1, 2n +1, 2n +2, . . . , 2n +2), by the definition of implicit degree, id(x3 ) ≥ 2n . Therefore, G satisfies the condition of Theorem 8. 2. Lemmas For a cycle C in G with a given orientation and a vertex x in C , x+ and x− denote the successor and the predecessor of x in C , respectively. Let x+2 = (x+ )+ and x−2 = (x− )− . For any I ⊆ V (C ), let I − = {x : x+ ∈ I } and I + = {x : x− ∈ I }. For two vertices x, y ∈ V (C ), xCy denotes the subpath of C from x to y. We use yC¯ x for the path from y to x in the reversed direction of C . A similar notation is used for paths. A cycle C is called implicit-heavy if it contains all implicit-heavy vertices of G; it is called extendable if there exists a longer cycle in G containing all vertices of C . For a path P = x1 x2 . . . xp with x1 = x and xp = y of a graph G, we call x and y two end-vertices of P. A y-path is a path which has y as one of its end-vertices. We use E ∗ (G) to denote the set {xy : xy ∈ E (G) or d(x) + d(y) ≥ n, x, y ∈ V (G)}. Let k ≥ 3 be an integer. Based on Ore’s condition (Theorem 1), we call a sequence of vertices C = x1 x2 . . . xk x1 an Ore-cycle or briefly, o-cycle of G, if for every i ∈ {1, 2, . . . , k}, there holds that xi xi+1 ∈ E ∗ (G), where the indices are taken modulo k. Our proof of Theorem 8 is based on the following lemmas. Lemma 1 ([1]). Let G be a non-hamiltonian 2-connected graph on n vertices and C be a nonextendable cycle of G. If P is a path with two end-vertices x, y such that V (C ) ⊂ V (P ), then xy ̸∈ E (G) and d(x) + d(y) < n, i.e. xy ̸∈ E ∗ (G). Lemma 2 ([8]). Let G be a graph and C be an o-cycle of G. Then there exists a cycle C ′ of G such that V (C ) ⊆ V (C ′ ). Lemma 3 ([7]). Let G be a 2-connected graph, P = x1 x2 . . . xp with x1 = x and xp = y be a path of G. If d(u) < id(x) for every vertex u ∈ NG−V (P ) (x) ∪ {x} and xy ̸∈ E (G), then either (a) there is some xj ∈ NP (x)− such that d(xj ) ≥ id(x); or (b) NP (x) = {x2 , x3 , . . . , xl }, d(xj ) < id(x) for any vertex xj ∈ NP (x)− and id(x) = m2 (x) = min{d(v) : v ∈ N2 (x)}. Lemma 4. Let G be a 2-connected 2-heavy graph on n vertices, P = x1 x2 . . . xp with x1 = x and xp = y be a path of G such that id(x) ≥ n/2. If d(x) < id(x) and G has no cycle containing all vertices of V (P ), then there exists a y-path P ′ such that V (P ) ⊆ V (P ′ ) and the other end-vertex of P ′ has degree at least n/2. Proof. Let R = G − V (P ). If there exists some vertex u ∈ NR (x) such that d(u) ≥ id(x), then P ′ = ux1 x2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(u) ≥ n/2. Next, we suppose d(u) < id(x) for every vertex u ∈ NR (x). If there exists some xj ∈ NP (x)− such that d(xj ) ≥ id(x), then ′ P = xj xj−1 . . . x1 xj+1 xj+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xj ) ≥ n/2. Thus, we may restrict our attention to the case d(xj ) < id(x) for every vertex xj ∈ NP (x)− and d(u) < id(x) for every vertex u ∈ NR (x). By Lemma 3, we have NP (x) = {x2 , x3 , . . . , xl } and d(u) ≥ id(x) for every vertex u ∈ N2 (x). If there exists some vertex xi such that d(xi ) ≥ n/2 with 1 ≤ i ≤ l − 1, then P ′ = xi xi−1 . . . x1 xi+1 xi+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xi ) ≥ n/2. Suppose d(xi ) < n/2 for every vertex xi with 1 ≤ i ≤ l − 1. If there exists some i with 2 ≤ i ≤ l − 1 such that NR (xi )\ NR (x) ̸= ∅, then u ∈ N2 (x) and d(u) ≥ id(x) for u ∈ NR (xi )\ NR (x). Thus, P ′ = uxi xi−1 . . . x1 xi+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(u) ≥ n/2. Next we assume NR (xi ) ⊆ NR (x) for
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every i with 2 ≤ i ≤ l − 1. If there exists some vertex v ∈ N2 (x) ∩ R, then P ′ = vw x1 x2 . . . xp with w ∈ NR (v) ∩ NR (x) is a y-path such that V (P ) ⊆ V (P ′ ) and d(v) ≥ id(x) ≥ n/2. Next we suppose N2 (x) ∩ R = ∅. Since G − xl is connected, there exists an edge x0 xs ∈ E (G) with x0 ∈ NR (x) and l < s or an edge xr xs ∈ E (G) with r < l < s. Case 1. There exists an edge x0 xs ∈ E (G) with x0 ∈ NR (x) and l < s. We choose such an edge x0 xs such that s is as large as possible. Since G has no cycle containing all vertices of V (P ), s ≤ p − 1. Then xs ∈ N2 (x) and d(xs ) ≥ id(x) ≥ n/2. Case 1.1. s ≥ l + 2. If x0 xs−1 ∈ E (G), then xs−1 ∈ N2 (x) and d(xs−1 ) ≥ id(x) ≥ n/2. Thus, P ′ = xs−1 xs−2 . . . x1 x0 xs xs+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs−1 ) ≥ n/2. Suppose x0 xs−1 ̸∈ E (G). By the choice of s, x0 xs+1 ̸∈ E (G). If xs−1 xs+1 ∈ E (G), then P ′ = xs x0 x1 . . . xs−1 xs+1 xs+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs ) ≥ n/2. If xs−1 xs+1 ̸∈ E (G), then {xs , x0 , xs−1 , xs+1 } induces a claw. Since G is 2-heavy and d(x0 ) < n/2, d(xs−1 ) ≥ n/2. Therefore, P ′ = xs−1 xs−2 . . . x0 xs xs+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs−1 ) ≥ n/2. Case 1.2. s = l + 1. Then xl+1 ∈ N2 (x) and d(xl+1 ) ≥ id(x) ≥ n/2. Since G − xl+1 is connected, there exists an edge xi xt ∈ E (G) with 2 ≤ i ≤ l and l + 2 ≤ t ≤ p − 1. We choose such an edge xi xt ∈ E (G) such that t is as small as possible. Now xt ∈ N2 (x) and d(xt ) ≥ id(x) ≥ n/2. If t = l + 2, then P ′ = xl+1 xl . . . xi+1 x1 x2 . . . xi xl+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xl+1 ) ≥ n/2. Suppose t ≥ l + 3. If xi xt +1 ∈ E (G), then P ′ = xt xt −1 . . . xi+1 x1 x2 . . . xi xt +1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt ) ≥ n/2. Suppose xi xt +1 ̸∈ E (G). By the choice of t , xi xt −1 ̸∈ E (G). If xt −1 xt +1 ∈ E (G), then P ′ = xt xi xi−1 . . . x1 xi+1 . . . xt −1 xt +1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt ) ≥ n/2. If xt −1 xt +1 ̸∈ E (G), then {xt , xi , xt −1 , xt +1 } induces a claw. Since G is 2-heavy and d(xi ) < n/2, d(xt −1 ) ≥ n/2. Then P ′ = xt −1 xt −2 . . . xi+1 x1 x2 . . . xi xt . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt −1 ) ≥ n/2. Case 2. There exists an edge xr xs ∈ E (G) with 1 ≤ r < l < s. We choose such an edge xr xs ∈ E (G) such that (i) s is as large as possible and (ii) r is as large as possible, subject to (i). Clearly, we have s ≤ p − 1, xs ∈ N2 (x) and d(xs ) ≥ id(x) ≥ n/2. Case 2.1. s ≥ l + 2. If xr xs−1 ∈ E (G), then xs−1 ∈ N2 (x) and d(xs−1 ) ≥ id(x) ≥ n/2. Thus, P ′ = xs−1 xs−2 . . . xr +1 x1 x2 . . . xr xs xs+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs−1 ) ≥ n/2. Suppose xr xs−1 ̸∈ E (G). By the choice of s, xr xs+1 ̸∈ E (G). If xs−1 xs+1 ∈ E (G), then P ′ = xs xr xr −1 . . . x1 xr +1 xr +2 . . . xs−1 xs+1 xs+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs ) ≥ n/2. If xs−1 xs+1 ̸∈ E (G), then {xs , xr , xs−1 , xs+1 } induces a claw. Since G is 2-heavy and d(xr ) < n/2, d(xs−1 ) ≥ n/2. Therefore, P ′ = xs−1 xs−2 . . . xr +1 x1 x2 . . . xr xs xs+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs−1 ) ≥ n/2. Case 2.2. s = l + 1. Then xl+1 ∈ N2 (x) and d(xl+1 ) ≥ id(x) ≥ n/2. Since G − xl+1 is connected, there exists an edge xl xt ∈ E (G) with l + 2 ≤ t ≤ p − 1. We choose such an edge xl xt ∈ E (G) such that t is as small as possible. Now xt ∈ N2 (x) and d(xt ) ≥ id(x) ≥ n/2. If t = l + 2, then P ′ = xl+1 xr xr −1 . . . x1 xr +1 xr +1 . . . xl xl+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xl+1 ) ≥ n/2. Suppose t ≥ l + 3. If xl xt +1 ∈ E (G), then P ′ = xt xt −1 . . . xl+1 xr xr −1 . . . x1 xr +1 xr +2 . . . xl xt +1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt ) ≥ n/2. Suppose xl xt +1 ̸∈ E (G). By the choice of t , xl xt −1 ̸∈ E (G). If xt −1 xt +1 ∈ E (G), then P ′ = xt xl xl−1 . . . xr +1 x1 x2 . . . xr xl+1 xl+2 . . . xt −1 xt +1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt ) ≥ n/2. Suppose xt −1 xt +1 ̸∈ E (G), then {xt , xl , xt −1 , xt +1 } induces a claw. If d(xl ) ≥ n/2, then P ′ = xl xl−1 . . . xr +1 x1 x2 . . . xr xl+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xl ) ≥ n/2. Suppose d(xl ) < n/2. Since G is 2-heavy, d(xt −1 ) ≥ n/2. Then P ′ = xt −1 xt −2 . . . xl+1 xr xr −1 . . . x1 xr +1 xr +2 . . . xl xt . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt −1 ) ≥ n/2. Now we complete the proof of Lemma 4. Lemma 5. Let G be a 2-connected 2-heavy graph on n vertices and x, y be two nonadjacent implicit-heavy vertices. If G + xy has a cycle C containing all implicit-heavy vertices of G, then G has a cycle containing all vertices of C . Proof. Assume G does not have a cycle containing all vertices of C . Consider a path P from x to y in G containing all vertices of C . Clearly, x and y have no common neighbor in V (G) \ V (P ). By using Lemma 4 twice, we can get a path P ′ from u to v such that V (P ) ⊆ V (P ′ ) and d(u) + d(v) ≥ id(x) + id(y) ≥ n/2 + n/2 = n. Then C ′ = P ′ + uv is an o-cycle of G. By Lemma 2, there is a cycle C ′′ such that V (C ′ ) ⊆ V (C ′′ ). But since V (C ) ⊆ V (P ) ⊆ V (C ′ ), V (C ) ⊆ V (C ′′ ), a contradiction.
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3. Proof of Theorem 8 Suppose to the contrary that G is not hamiltonian. Let A = {x ∈ V (G) : id(x) ≥ n/2}. Claim 1. There exists a cycle C in G that contains all vertices of A. Proof. Suppose to the contrary that there exists no cycle in G that contains all vertices of A. Choose a cycle C in G such that C contains as many vertices of A as possible and give C an orientation. Then there is a vertex x ∈ (V (G) \ V (C )) ∩ A. Since G is 2-connected, there exist two internally disjoint paths Pi (x, xi ) from x to C with V (Pi ) ∩ V (C ) = {xi } for i = 1, 2. Suppose without loss of generality that the occurrence of x1 and x2 on C agrees with the given orientation of C . By the choice of C , − + − there exists at least one vertex of A on x+ 1 Cx2 . Let y be the first vertex of A occurring on x1 Cx2 . Set Q = xP1 x1 C¯ y. Then |V (Q ) ∩ A| > |V (C ) ∩ A|. Clearly, x and y have no common neighbors in V (G) \ V (Q ) and xy ̸∈ E (G). By using Lemma 4 twice, we can obtain a path P with both end-vertices u and v having degree at least n/2 and V (Q ) ⊆ V (P ). Thus, d(u) + d(v) ≥ n. By the choice of C , we can easily get that u and v have no common neighbors in V (G) \ V (P ). Then there is a vertex z ∈ V (P ) such that uz + ∈ E (G) and v z ∈ E (G), where z + is the successor of z on P. ¯ + P v z is a cycle in G such that |V (C ′ ) ∩ A| > |V (C ) ∩ A|, a contradiction. Therefore, C ′ = z Puz Then by Claim 1, G contains an implicit-heavy cycle. Let C be a longest implicit-heavy cycle of G and give C a clockwise orientation. Then V (G) \ V (C ) ̸= ∅. Since G is 2-connected, there exists a path P connecting two vertices x1 ∈ V (C ) and x2 ∈ V (C ) internally disjoint with C and such that |V (P )| ≥ 3. Without loss of generality, we assume that x1 , x2 appear in the order along C . Let P = x1 u1 u2 . . . ur x2 be such a path of minimum length. − ∗ ∗ Claim 2. uk x+ i ̸∈ E (G) and uk xi ̸∈ E (G) for every k = 1, 2, . . . , r and i = 1, 2. ′ ¯ 1 Cx− Proof. Since for every k = 1, 2, . . . , r , P ′ = uk Px 1 is a path containing all vertices of C and |V (P )| > |V (C )|, we have − ∗ uk x1 ̸∈ E (G) by Lemma 1. The other assertions can be proved similarly. + − + ∗ ∗ Claim 3. x− 1 x1 ∈ E (G) and x2 x2 ∈ E (G). + − + Proof. If x− 1 x1 ̸∈ E (G), then {x1 , x1 , x1 , u1 } induces a claw by Claim 2. By the choice of C , d(u1 ) ≤ id(u1 ) < n/2. Since G is − + + − + ∗ ∗ 2-heavy, d(x1 ) + d(x1 ) ≥ n. This implies that x− 1 x1 ∈ E (G). Similarly, we can prove x2 x2 ∈ E (G). − + + − + ∗ ∗ ∗ ∗ Claim 4. x− 1 x2 ̸∈ E (G), x1 x2 ̸∈ E (G), xi x3−i ̸∈ E (G) and xi x3−i ̸∈ E (G) for i = 1, 2. − − − ′ ∗ ¯ ¯ Proof. Since P ′ = x− 1 C x2 Px1 Cx2 is a path containing all vertices of C and |V (P )| > |V (C )|, we have x1 x2 ̸∈ E (G) by + + ∗ Lemma 1. Similarly, x1 x2 ̸∈ E (G) can be proved. −¯ + −¯ ′ ∗ ′ ¯ If x1 x− 2 ∈ E (G), then C = x1 x2 C x1 x1 C x2 Px1 is an o-cycle containing all the vertices of C and |V (C )| > |V (C )|. By − ∗ Lemma 2, there exists a longer implicit-heavy cycle in G, a contradiction. So x1 x2 ̸∈ E (G). The other assertions can be proved similarly. − + − By Claim 4, there exists some vertex in x+ i Cx3−i not adjacent to xi in G for i = 1, 2. Let yi be the first vertex in xi Cx3−i not adjacent to xi in G for i = 1, 2. Let u be a vertex in {u1 , u2 , . . . , ur } and let zi be an arbitrary vertex in x+ i Cyi , i = 1, 2.
Claim 5. uz1 , uz2 , z1 x2 , z2 x1 , z1 z2 ̸∈ E ∗ (G). − Proof. Suppose uz1 ∈ E ∗ (G). By Claim 2, we have z1 ̸= x+ 1 . Then x1 z1 ∈ E (G) by the choice of yi . Therefore, by Claim 3, − + − ′ C = x1 Puz1 Cx1 x1 Cz1 x1 is an o-cycle containing all vertices of C and |V (C ′ )| > |V (C )|. By Lemma 2, there exists a longer implicit-heavy cycle in G, a contradiction. Hence uz1 ̸∈ E ∗ (G). Similarly, uz2 ̸∈ E ∗ (G). − + − + − ′ Suppose z1 x2 ∈ E ∗ (G). By Claim 4, z1 ̸= x+ 1 . Then by Claim 3, C = x1 Px2 z1 Cx2 x2 Cx1 x1 Cz1 x1 is an o-cycle containing ′ all vertices of C and |V (C )| > |V (C )|. By Lemma 2, there exists a longer implicit-heavy cycle in G, a contradiction. Hence z1 x2 ̸∈ E ∗ (G). Similarly, z2 x1 ̸∈ E ∗ (G). + −¯ + −¯ − + − + ′ Suppose z1 z2 ∈ E ∗ (G). By Claim 4, z1 ̸= x+ 1 or z2 ̸= x2 . Then C = x1 Px2 z2 C x2 x2 C z1 z2 Cx1 x1 Cz1 x1 (if z1 ̸= x1 and + −¯ + −¯ + + − + − + + z2 ̸= x2 ) or x1 Px2 z2 C x2 x2 C z1 z2 Cx1 (if z1 = x1 and z2 ̸= x2 ) or x1 Px2 C¯ z1 z2 Cx1 x1 Cz1 x1 (if z1 ̸= x1 and z2 = x2 ) is an o-cycle containing all vertices of C and |V (C ′ )| > |V (C )|. By Lemma 2, there exists a longer implicit-heavy cycle in G, a contradiction. Hence z1 z2 ̸∈ E ∗ (G). + − + Claim 6. x− 1 x1 ∈ E (G) or x2 x2 ∈ E (G). + − + − + Proof. Suppose to the contrary that x− 1 x1 ̸∈ E (G) and x2 x2 ̸∈ E (G). By Claim 3, we have d(x1 ) + d(x1 ) ≥ n and − + − − + + d(x2 ) + d(x2 ) ≥ n. Thus, we can obtain that d(x1 ) + d(x2 ) ≥ n or d(x1 ) + d(x2 ) ≥ n, contradicting to Claim 4. + By Claim 6, without loss of generality, we may assume that x− 1 x1 ∈ E (G).
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Claim 7. x1 x2 ∈ E (G). − Proof. Suppose x1 x2 ̸∈ E (G). Now by the choice of P and Claim 5, we have {y1 , y− 1 , x1 , u1 , u2 , . . . , ur , x2 , y2 , y2 } induces a Pr +6 , where r ≥ 1. Since G is P7 -f -implicit-heavy, G is also Pr +6 -f -implicit-heavy. By the choice of C , u1 − − − and ur are implicit-light. It follows that id(y− ̸∈ E (G). Then C ′ = 1 ) ≥ n/2 and id(y2 ) ≥ n/2. By Claim 5, y1 y2 − − + −2 ¯ −2 ¯ + − ¯ − − − − y1 Cx2 x2 Cy2 x2 Px1 y1 C x1 x1 C y2 y1 is an o-cycle containing all vertices of C of G + y1 y2 and |V (C )| < |V (C ′ )|, by Lemma 2, − ′′ ′′ G + y− 1 y2 has a cycle C containing all implicit-heavy vertices of G, and such that C is longer than C . By Lemma 5, G has a ′′ cycle containing all vertices of C , a contradiction to the choice of C .
Claim 8. r = 1. + ∗ ∗ Proof. Suppose r ≥ 2. By the choice of P , u1 x2 ̸∈ E (G) and ur x1 ̸∈ E (G). By Claim 4, x1 x+ 2 ̸∈ E (G) and x1 x2 ̸∈ E (G). + + + ∗ ∗ By Claim 2, u1 x1 ̸∈ E (G) and ur x2 ̸∈ E (G). Note that x1 x2 ∈ E (G) by Claim 7. Thus {x1 , x1 , u1 , x2 } induces a claw and + + {x2 , ur , x+ 2 , x1 } induces a claw. Since G is 2-heavy and u1 , ur are implicit-light, d(x1 ) ≥ n/2 and d(x2 ) ≥ n/2. This implies + + ∗ that x1 x2 ∈ E (G), contradicting to Claim 4.
Note that G is D-f -implicit-heavy or H-f -implicit-heavy. Suppose G is D-f -implicit-heavy. By Claims 5 and 7,
− − − {y1 , y− 1 , x1 , u1 , x2 , y2 , y2 } induces a D. Since u1 is implicit-light, y1 and y2 are implicit-heavy. By similar arguments as in the
proof of Claim 7, we can get a contradiction. + Next, we assume G is H-f -implicit-heavy. By Claims 2 and 4, {x− 1 , x1 , x1 , u1 , x2 } induces an H. Since u1 is implicit− + − + + light, x1 and x1 are implicit-heavy. If x2 x2 ∈ E (G), then by Claims 2 and 4, {x− 2 , x2 , x2 , u1 , x1 } induces an H. Since u1 is − + − + + implicit-light, x2 and x2 are implicit-heavy. If x2 x2 ̸∈ E (G), since G is 2-heavy and u1 is light, both x+ 1 and x2 are heavy. + + Therefore, id(x1 ) ≥ n/2 and id(x2 ) ≥ n/2. The arguments we used in the proof of Claim 7 can now be applied to the graph + ′ ′ ′ G′ = G + x+ 1 x2 to conclude that G has a cycle C containing all implicit-heavy vertices of G and such that C is longer than ′ C . By Lemma 5 G has a cycle containing all vertices of C , contradicting to the choice of C . This completes the proof of Theorem 8. Acknowledgment The author is very grateful to the anonymous referee for carefully reading the manuscript and providing comments and suggestions which led to a substantial improvement of the paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
J.A. Bondy, Large cycles in graphs, Discrete Math. 1 (1971) 121–132. J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, Macmillan, London, 1976. H.J. Broersma, Z. Ryjá˘cek, I. Schiermeyer, Dirac’s minimum degree condition restricted to claws, Discrete Math. 167/168 (1997) 155–166. H.J. Broersma, H.J. Veldman, Restrictions on induced subgaphs ensuring hamiltonicity or pancyclicity of K1,3 -free graphs, in: R. Bodendiek (Ed.), Contemporary Methods in Graph Theory, BI-Wiss.-Verl, Mannhein-Wien-Zürich, 1990, pp. 181–194. G. Fan, New sufficient conditions for cycles in graphs, J. Combin. Theory Ser. B 37 (1984) 221–227. R.J. Faudree, Z. Ryjáček, I. Schiermeyer, Forbidden subgraphs and cycle extendability, J. Combin. Math. Combin. Comput. 19 (1995) 109–128. H. Li, W. Ning, J. Cai, An implicit degree condition for hamiltonian graphs, Discrete Math. 312 (2011) 2190–2196. B. Li, Z. Ryjáček, Y. Wang, S. Zhang, Pairs of heavy subgraphs for Hamiltonicity of 2-connected graphs, SIAM J. Discrete Math. 26 (2012) 1088–1103. B. Ning, Fan-type degree condition restricted to triples of induced subgraphs ensuring Hamiltonicity, Inform. Process. Lett. 113 (2013) 823–826. O. Ore, Note on hamilton circuits, Amer. Math. Monthly 67 (1960) 55. Y. Zhu, H. Li, X. Deng, Implicit-degrees and circumferences, Graphs Combin. 5 (1989) 283–290.
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Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
Implicit degree condition for hamiltonicity of 2-heavy graphs Xing Huang Huaerzhi Education and Technology Company Limited, Beijing, 100036, China
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Article history: Received 11 April 2016 Received in revised form 18 September 2016 Accepted 17 October 2016 Available online xxxx Keywords: Implicit degree Hamilton cycle f-implicit-heavy 2-heavy
abstract Let id(v) denote the implicit degree of a vertex v in a graph G. An induced subgraph S of G is called f-implicit-heavy if max{id(x), id(y)} ≥ |V (G)|/2 for every pair of vertices x, y ∈ V (S ) at distance 2 in S. For a given graph R, G is called R-f-implicit-heavy if every induced subgraph of G isomorphic to R is f-implicit-heavy. For a family R of graphs, G is called R-f-implicit-heavy if G is R-f-implicit-heavy for every R ∈ R. G is called 2-heavy if there are at least two end-vertices of every induced claw (K1,3 ) in G have degree at least |V (G)|/2. In this paper, we prove that: Let G be a 2-connected 2-heavy graph. If G is {P7 , D}-f-implicit-heavy or {P7 , H }-f-implicit-heavy, then G is hamiltonian. © 2016 Elsevier B.V. All rights reserved.
1. Introduction In this paper, we consider only undirected, finite and simple graphs. Notation and terminology not defined here can be found in [2]. Let G be a graph and H be a subgraph of G. For a vertex u ∈ V (G), the neighborhood of u in H is denoted by NH (u) = {v ∈ V (H ) : uv ∈ E (G)} and the degree of u in H is denoted by dH (u) = |NH (u)|. For two vertices u, v ∈ V (H ), the distance between u and v in H, denoted by dH (u, v), is the length of a shortest (u, v)-path in H. When there is no danger of ambiguity, we can use N (u), d(u) and d(u, v) in place of NG (u), dG (u) and dG (u, v), respectively. We use N2 (v) to denote the set of vertices which are at distance 2 with v , i.e. N2 (v) = {u ∈ V (G) : d(u, v) = 2}. Set m2 (v) = min{d(u) : u ∈ N2 (v)} and M2 (v) = max{d(u) : u ∈ N2 (v)}. An induced subgraph K1,3 of G with vertex set {u, v, w, x} and edge set {uv, uw, ux} is called a claw, with center u and end-vertices v, w, x. For a given graph R, G is called R-free if G contains no induced subgraph isomorphic to R. For a family R of graphs, G is called R-free if G is R-free for every R ∈ R. A graph G is called hamiltonian if it contains a Hamilton cycle, i.e. a cycle that contains all vertices of G. Degree condition is an important type of sufficient conditions for a graph to be hamiltonian. The following two results due to Ore and Fan, respectively, are well-known. Theorem 1 ([10]). Let G be a graph on n ≥ 3 vertices. If d(x) + d(y) ≥ n for each pair of nonadjacent vertices x and y in G, then G is hamiltonian. Theorem 2 ([5]). Let G be a 2-connected graph on n ≥ 3 vertices. If max{d(x), d(y)} ≥ n/2 for every pair of vertices x and y at distance 2 in G, then G is hamiltonian. In 1989, Zhu, Li and Deng [11] found that though some vertices may have small degrees, we can use some large degree vertices to replace small degree vertices in the right position considered in the proofs, so that we may construct a longer cycle. This idea leads to the definition of implicit degree, denoted by id(v) of a vertex v .
E-mail address: [email protected]. http://dx.doi.org/10.1016/j.dam.2016.10.031 0166-218X/© 2016 Elsevier B.V. All rights reserved.
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Fig. 1. Forbidden subgraph.
Definition 1 ([11]). Let v be a vertex of a graph G and d(v) = l + 1. If N2 (v) ̸= ∅ and d(v) ≥ 2, then let d1 (v) ≤ d2 (v) ≤ d3 (v) ≤ · · · ≤ dl (v) ≤ dl+1 (v) ≤ · · · be the degree sequence of vertices in N (v) ∪ N2 (v). Define m2 (v), dl+1 (v), dl (v),
if dl (v) ≤ m2 (v); if dl+1 (v) > M2 (v); otherwise.
d (v) = ∗
Then the implicit degree of v is defined as id(v) = max{d(v), d∗ (v)}. If N2 (v) = ∅ or d(v) ≤ 1, then define id(v) = d(v). Clearly, id(v) ≥ d(v) for every vertex v . The authors in [11] gave a sufficient condition for a 2-connected graph to be hamiltonian under implicit degree condition. Theorem 3 ([11]). Let G be a 2-connected graph on n ≥ 3 vertices. If id(x) + id(y) ≥ n for every pair of nonadjacent vertices x and y in G, then G is hamiltonian. Remark 1. The introduction of implicit degree is very useful, since many classic results by considering degree conditions for the hamiltonicity of graphs can be generalized. We just give one example to show this. Fan’s theorem (Theorem 2) can be easily obtained from Theorem 3. And the authors in [11] gave a simple proof to show this. Forbidden subgraph condition is another type of sufficient conditions for hamiltonicity of graphs. The following two results are of this type, where P7 , D and H are graphs in Fig. 1. Theorem 4 ([4]). Let G be a 2-connected graph. If G is {K1,3 , P7 , D}-free, then G is hamiltonian. Theorem 5 ([6]). Let G be a 2-connected graph. If G is {K1,3 , P7 , H }-free, then G is hamiltonian. A vertex v of a graph G on n vertices is called heavy (or implicit-heavy) if d(v) ≥ n/2 (or id(v) ≥ n/2). If v is not heavy (or not implicit-heavy), we call it light (or implicit-light). A claw of G is called 2-heavy if at least two of its end-vertices are heavy. A subgraph S of G is called f -heavy (or f -implicit-heavy) if max{d(x), d(y)} ≥ n/2 (or max{id(x), id(y)} ≥ n/2) for every pair of vertices x, y ∈ V (S ) at distance 2 in S. For a given graph R, G is called R-f -heavy (or R-f -implicit-heavy) if every induced subgraph of G isomorphic to R is f -heavy (or f -implicit-heavy). For a family R of graphs, G is called R-f -heavy (or R-f -implicit-heavy) if G is R-f -heavy (or R-f -implicit-heavy) for every R ∈ R. Clearly, every R-free graph is also R-f -heavy, and every R-f -heavy graph is also R-f -implicit-heavy. In 1997, Broersma et al. [3] generalized existing results by combining degree conditions and forbidden subgraph conditions. More precisely, they restricted Fan’s condition (Theorem 2) to certain subgraphs and got the following result. Theorem 6 ([3]). Let G be a 2-connected graph on n ≥ 3 vertices. If G is 2-heavy and |N (u) ∩ N (v)| ≥ 2 for every pair of vertices u, v with d(u, v) = 2 and max{d(u), d(v)} ≤ n/2, then G is hamiltonian. Ning extended Theorem 4 and Theorem 5 by relaxing forbidden subgraph conditions to conditions in which the subgraphs are allowed. Theorem 7 ([9]). Let G be a 2-connected graph on n ≥ 3 vertices. If G is {K1,3 , P7 , D}-f -heavy or {K1,3 , P7 , H }-f -heavy, then G is hamiltonian. We use implicit degree instead of degree in Theorem 7 and get the following result. Theorem 8. Let G be a 2-connected 2-heavy graph. If G is {P7 , D}-f -implicit-heavy or {P7 , H }-f -implicit-heavy, then G is hamiltonian. Remark 2. Since id(v) ≥ d(v) for every vertex v , it is easy to see that every graph satisfying the condition of Theorem 7 also satisfies the condition of Theorem 8, but the converse is not true. Furthermore, the following graph is a hamiltonian graph satisfying the condition of Theorem 8, but not the condition of Theorem 7. Let n ≥ 18 be an even integer and K n ∪ K n −7 denotes the union of two complete graphs K n and K n −7 . Set V (K n ) = 2 2 2 2 2 {u1 , u2 , . . . , u n } and V (K n −7 ) = {v1 , v2 , . . . , v n −7 }. We construct a graph G with V (G) = V (K n ∪ K n −7 ) ∪ {x1 , x2 , . . . , x7 } 2
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Fig. 2. Graph in Remark 2.
and E (G) = E (K n ∪ K n −7 ) ∪{x4 x7 , x4 x3 , x1 x2 , x1 x3 , x2 x3 , x2 x5 , x5 x6 } ∪{xi uj : i = 1, 2 and j = 1, 2, . . . , 2n } ∪{x3 uj : j = 2 2 1, 2, . . . , 2n − 3} ∪{xi vj : i = 6, 7 and j = 1, 2, . . . , 2n − 7} (see Fig. 2). It is easy to see {x1 , x2 , . . . , x7 } induces a P7 and n max{d(x3 ), d(x7 )} = 2 − 3 < 2n with d(x3 , x7 ) = 2. So G does not satisfy the condition of Theorem 7, but it is hamiltonian. However, since d(x3 ) = 2n −3 and the degree sequence of N (x3 )∪N2 (x3 ) is (2, 2n −7, 2n , 2n , 2n +1, 2n +1, 2n +1, 2n +2, . . . , 2n +2), by the definition of implicit degree, id(x3 ) ≥ 2n . Therefore, G satisfies the condition of Theorem 8. 2. Lemmas For a cycle C in G with a given orientation and a vertex x in C , x+ and x− denote the successor and the predecessor of x in C , respectively. Let x+2 = (x+ )+ and x−2 = (x− )− . For any I ⊆ V (C ), let I − = {x : x+ ∈ I } and I + = {x : x− ∈ I }. For two vertices x, y ∈ V (C ), xCy denotes the subpath of C from x to y. We use yC¯ x for the path from y to x in the reversed direction of C . A similar notation is used for paths. A cycle C is called implicit-heavy if it contains all implicit-heavy vertices of G; it is called extendable if there exists a longer cycle in G containing all vertices of C . For a path P = x1 x2 . . . xp with x1 = x and xp = y of a graph G, we call x and y two end-vertices of P. A y-path is a path which has y as one of its end-vertices. We use E ∗ (G) to denote the set {xy : xy ∈ E (G) or d(x) + d(y) ≥ n, x, y ∈ V (G)}. Let k ≥ 3 be an integer. Based on Ore’s condition (Theorem 1), we call a sequence of vertices C = x1 x2 . . . xk x1 an Ore-cycle or briefly, o-cycle of G, if for every i ∈ {1, 2, . . . , k}, there holds that xi xi+1 ∈ E ∗ (G), where the indices are taken modulo k. Our proof of Theorem 8 is based on the following lemmas. Lemma 1 ([1]). Let G be a non-hamiltonian 2-connected graph on n vertices and C be a nonextendable cycle of G. If P is a path with two end-vertices x, y such that V (C ) ⊂ V (P ), then xy ̸∈ E (G) and d(x) + d(y) < n, i.e. xy ̸∈ E ∗ (G). Lemma 2 ([8]). Let G be a graph and C be an o-cycle of G. Then there exists a cycle C ′ of G such that V (C ) ⊆ V (C ′ ). Lemma 3 ([7]). Let G be a 2-connected graph, P = x1 x2 . . . xp with x1 = x and xp = y be a path of G. If d(u) < id(x) for every vertex u ∈ NG−V (P ) (x) ∪ {x} and xy ̸∈ E (G), then either (a) there is some xj ∈ NP (x)− such that d(xj ) ≥ id(x); or (b) NP (x) = {x2 , x3 , . . . , xl }, d(xj ) < id(x) for any vertex xj ∈ NP (x)− and id(x) = m2 (x) = min{d(v) : v ∈ N2 (x)}. Lemma 4. Let G be a 2-connected 2-heavy graph on n vertices, P = x1 x2 . . . xp with x1 = x and xp = y be a path of G such that id(x) ≥ n/2. If d(x) < id(x) and G has no cycle containing all vertices of V (P ), then there exists a y-path P ′ such that V (P ) ⊆ V (P ′ ) and the other end-vertex of P ′ has degree at least n/2. Proof. Let R = G − V (P ). If there exists some vertex u ∈ NR (x) such that d(u) ≥ id(x), then P ′ = ux1 x2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(u) ≥ n/2. Next, we suppose d(u) < id(x) for every vertex u ∈ NR (x). If there exists some xj ∈ NP (x)− such that d(xj ) ≥ id(x), then ′ P = xj xj−1 . . . x1 xj+1 xj+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xj ) ≥ n/2. Thus, we may restrict our attention to the case d(xj ) < id(x) for every vertex xj ∈ NP (x)− and d(u) < id(x) for every vertex u ∈ NR (x). By Lemma 3, we have NP (x) = {x2 , x3 , . . . , xl } and d(u) ≥ id(x) for every vertex u ∈ N2 (x). If there exists some vertex xi such that d(xi ) ≥ n/2 with 1 ≤ i ≤ l − 1, then P ′ = xi xi−1 . . . x1 xi+1 xi+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xi ) ≥ n/2. Suppose d(xi ) < n/2 for every vertex xi with 1 ≤ i ≤ l − 1. If there exists some i with 2 ≤ i ≤ l − 1 such that NR (xi )\ NR (x) ̸= ∅, then u ∈ N2 (x) and d(u) ≥ id(x) for u ∈ NR (xi )\ NR (x). Thus, P ′ = uxi xi−1 . . . x1 xi+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(u) ≥ n/2. Next we assume NR (xi ) ⊆ NR (x) for
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every i with 2 ≤ i ≤ l − 1. If there exists some vertex v ∈ N2 (x) ∩ R, then P ′ = vw x1 x2 . . . xp with w ∈ NR (v) ∩ NR (x) is a y-path such that V (P ) ⊆ V (P ′ ) and d(v) ≥ id(x) ≥ n/2. Next we suppose N2 (x) ∩ R = ∅. Since G − xl is connected, there exists an edge x0 xs ∈ E (G) with x0 ∈ NR (x) and l < s or an edge xr xs ∈ E (G) with r < l < s. Case 1. There exists an edge x0 xs ∈ E (G) with x0 ∈ NR (x) and l < s. We choose such an edge x0 xs such that s is as large as possible. Since G has no cycle containing all vertices of V (P ), s ≤ p − 1. Then xs ∈ N2 (x) and d(xs ) ≥ id(x) ≥ n/2. Case 1.1. s ≥ l + 2. If x0 xs−1 ∈ E (G), then xs−1 ∈ N2 (x) and d(xs−1 ) ≥ id(x) ≥ n/2. Thus, P ′ = xs−1 xs−2 . . . x1 x0 xs xs+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs−1 ) ≥ n/2. Suppose x0 xs−1 ̸∈ E (G). By the choice of s, x0 xs+1 ̸∈ E (G). If xs−1 xs+1 ∈ E (G), then P ′ = xs x0 x1 . . . xs−1 xs+1 xs+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs ) ≥ n/2. If xs−1 xs+1 ̸∈ E (G), then {xs , x0 , xs−1 , xs+1 } induces a claw. Since G is 2-heavy and d(x0 ) < n/2, d(xs−1 ) ≥ n/2. Therefore, P ′ = xs−1 xs−2 . . . x0 xs xs+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs−1 ) ≥ n/2. Case 1.2. s = l + 1. Then xl+1 ∈ N2 (x) and d(xl+1 ) ≥ id(x) ≥ n/2. Since G − xl+1 is connected, there exists an edge xi xt ∈ E (G) with 2 ≤ i ≤ l and l + 2 ≤ t ≤ p − 1. We choose such an edge xi xt ∈ E (G) such that t is as small as possible. Now xt ∈ N2 (x) and d(xt ) ≥ id(x) ≥ n/2. If t = l + 2, then P ′ = xl+1 xl . . . xi+1 x1 x2 . . . xi xl+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xl+1 ) ≥ n/2. Suppose t ≥ l + 3. If xi xt +1 ∈ E (G), then P ′ = xt xt −1 . . . xi+1 x1 x2 . . . xi xt +1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt ) ≥ n/2. Suppose xi xt +1 ̸∈ E (G). By the choice of t , xi xt −1 ̸∈ E (G). If xt −1 xt +1 ∈ E (G), then P ′ = xt xi xi−1 . . . x1 xi+1 . . . xt −1 xt +1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt ) ≥ n/2. If xt −1 xt +1 ̸∈ E (G), then {xt , xi , xt −1 , xt +1 } induces a claw. Since G is 2-heavy and d(xi ) < n/2, d(xt −1 ) ≥ n/2. Then P ′ = xt −1 xt −2 . . . xi+1 x1 x2 . . . xi xt . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt −1 ) ≥ n/2. Case 2. There exists an edge xr xs ∈ E (G) with 1 ≤ r < l < s. We choose such an edge xr xs ∈ E (G) such that (i) s is as large as possible and (ii) r is as large as possible, subject to (i). Clearly, we have s ≤ p − 1, xs ∈ N2 (x) and d(xs ) ≥ id(x) ≥ n/2. Case 2.1. s ≥ l + 2. If xr xs−1 ∈ E (G), then xs−1 ∈ N2 (x) and d(xs−1 ) ≥ id(x) ≥ n/2. Thus, P ′ = xs−1 xs−2 . . . xr +1 x1 x2 . . . xr xs xs+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs−1 ) ≥ n/2. Suppose xr xs−1 ̸∈ E (G). By the choice of s, xr xs+1 ̸∈ E (G). If xs−1 xs+1 ∈ E (G), then P ′ = xs xr xr −1 . . . x1 xr +1 xr +2 . . . xs−1 xs+1 xs+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs ) ≥ n/2. If xs−1 xs+1 ̸∈ E (G), then {xs , xr , xs−1 , xs+1 } induces a claw. Since G is 2-heavy and d(xr ) < n/2, d(xs−1 ) ≥ n/2. Therefore, P ′ = xs−1 xs−2 . . . xr +1 x1 x2 . . . xr xs xs+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xs−1 ) ≥ n/2. Case 2.2. s = l + 1. Then xl+1 ∈ N2 (x) and d(xl+1 ) ≥ id(x) ≥ n/2. Since G − xl+1 is connected, there exists an edge xl xt ∈ E (G) with l + 2 ≤ t ≤ p − 1. We choose such an edge xl xt ∈ E (G) such that t is as small as possible. Now xt ∈ N2 (x) and d(xt ) ≥ id(x) ≥ n/2. If t = l + 2, then P ′ = xl+1 xr xr −1 . . . x1 xr +1 xr +1 . . . xl xl+2 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xl+1 ) ≥ n/2. Suppose t ≥ l + 3. If xl xt +1 ∈ E (G), then P ′ = xt xt −1 . . . xl+1 xr xr −1 . . . x1 xr +1 xr +2 . . . xl xt +1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt ) ≥ n/2. Suppose xl xt +1 ̸∈ E (G). By the choice of t , xl xt −1 ̸∈ E (G). If xt −1 xt +1 ∈ E (G), then P ′ = xt xl xl−1 . . . xr +1 x1 x2 . . . xr xl+1 xl+2 . . . xt −1 xt +1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt ) ≥ n/2. Suppose xt −1 xt +1 ̸∈ E (G), then {xt , xl , xt −1 , xt +1 } induces a claw. If d(xl ) ≥ n/2, then P ′ = xl xl−1 . . . xr +1 x1 x2 . . . xr xl+1 . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xl ) ≥ n/2. Suppose d(xl ) < n/2. Since G is 2-heavy, d(xt −1 ) ≥ n/2. Then P ′ = xt −1 xt −2 . . . xl+1 xr xr −1 . . . x1 xr +1 xr +2 . . . xl xt . . . xp is a y-path such that V (P ) ⊆ V (P ′ ) and d(xt −1 ) ≥ n/2. Now we complete the proof of Lemma 4. Lemma 5. Let G be a 2-connected 2-heavy graph on n vertices and x, y be two nonadjacent implicit-heavy vertices. If G + xy has a cycle C containing all implicit-heavy vertices of G, then G has a cycle containing all vertices of C . Proof. Assume G does not have a cycle containing all vertices of C . Consider a path P from x to y in G containing all vertices of C . Clearly, x and y have no common neighbor in V (G) \ V (P ). By using Lemma 4 twice, we can get a path P ′ from u to v such that V (P ) ⊆ V (P ′ ) and d(u) + d(v) ≥ id(x) + id(y) ≥ n/2 + n/2 = n. Then C ′ = P ′ + uv is an o-cycle of G. By Lemma 2, there is a cycle C ′′ such that V (C ′ ) ⊆ V (C ′′ ). But since V (C ) ⊆ V (P ) ⊆ V (C ′ ), V (C ) ⊆ V (C ′′ ), a contradiction.
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3. Proof of Theorem 8 Suppose to the contrary that G is not hamiltonian. Let A = {x ∈ V (G) : id(x) ≥ n/2}. Claim 1. There exists a cycle C in G that contains all vertices of A. Proof. Suppose to the contrary that there exists no cycle in G that contains all vertices of A. Choose a cycle C in G such that C contains as many vertices of A as possible and give C an orientation. Then there is a vertex x ∈ (V (G) \ V (C )) ∩ A. Since G is 2-connected, there exist two internally disjoint paths Pi (x, xi ) from x to C with V (Pi ) ∩ V (C ) = {xi } for i = 1, 2. Suppose without loss of generality that the occurrence of x1 and x2 on C agrees with the given orientation of C . By the choice of C , − + − there exists at least one vertex of A on x+ 1 Cx2 . Let y be the first vertex of A occurring on x1 Cx2 . Set Q = xP1 x1 C¯ y. Then |V (Q ) ∩ A| > |V (C ) ∩ A|. Clearly, x and y have no common neighbors in V (G) \ V (Q ) and xy ̸∈ E (G). By using Lemma 4 twice, we can obtain a path P with both end-vertices u and v having degree at least n/2 and V (Q ) ⊆ V (P ). Thus, d(u) + d(v) ≥ n. By the choice of C , we can easily get that u and v have no common neighbors in V (G) \ V (P ). Then there is a vertex z ∈ V (P ) such that uz + ∈ E (G) and v z ∈ E (G), where z + is the successor of z on P. ¯ + P v z is a cycle in G such that |V (C ′ ) ∩ A| > |V (C ) ∩ A|, a contradiction. Therefore, C ′ = z Puz Then by Claim 1, G contains an implicit-heavy cycle. Let C be a longest implicit-heavy cycle of G and give C a clockwise orientation. Then V (G) \ V (C ) ̸= ∅. Since G is 2-connected, there exists a path P connecting two vertices x1 ∈ V (C ) and x2 ∈ V (C ) internally disjoint with C and such that |V (P )| ≥ 3. Without loss of generality, we assume that x1 , x2 appear in the order along C . Let P = x1 u1 u2 . . . ur x2 be such a path of minimum length. − ∗ ∗ Claim 2. uk x+ i ̸∈ E (G) and uk xi ̸∈ E (G) for every k = 1, 2, . . . , r and i = 1, 2. ′ ¯ 1 Cx− Proof. Since for every k = 1, 2, . . . , r , P ′ = uk Px 1 is a path containing all vertices of C and |V (P )| > |V (C )|, we have − ∗ uk x1 ̸∈ E (G) by Lemma 1. The other assertions can be proved similarly. + − + ∗ ∗ Claim 3. x− 1 x1 ∈ E (G) and x2 x2 ∈ E (G). + − + Proof. If x− 1 x1 ̸∈ E (G), then {x1 , x1 , x1 , u1 } induces a claw by Claim 2. By the choice of C , d(u1 ) ≤ id(u1 ) < n/2. Since G is − + + − + ∗ ∗ 2-heavy, d(x1 ) + d(x1 ) ≥ n. This implies that x− 1 x1 ∈ E (G). Similarly, we can prove x2 x2 ∈ E (G). − + + − + ∗ ∗ ∗ ∗ Claim 4. x− 1 x2 ̸∈ E (G), x1 x2 ̸∈ E (G), xi x3−i ̸∈ E (G) and xi x3−i ̸∈ E (G) for i = 1, 2. − − − ′ ∗ ¯ ¯ Proof. Since P ′ = x− 1 C x2 Px1 Cx2 is a path containing all vertices of C and |V (P )| > |V (C )|, we have x1 x2 ̸∈ E (G) by + + ∗ Lemma 1. Similarly, x1 x2 ̸∈ E (G) can be proved. −¯ + −¯ ′ ∗ ′ ¯ If x1 x− 2 ∈ E (G), then C = x1 x2 C x1 x1 C x2 Px1 is an o-cycle containing all the vertices of C and |V (C )| > |V (C )|. By − ∗ Lemma 2, there exists a longer implicit-heavy cycle in G, a contradiction. So x1 x2 ̸∈ E (G). The other assertions can be proved similarly. − + − By Claim 4, there exists some vertex in x+ i Cx3−i not adjacent to xi in G for i = 1, 2. Let yi be the first vertex in xi Cx3−i not adjacent to xi in G for i = 1, 2. Let u be a vertex in {u1 , u2 , . . . , ur } and let zi be an arbitrary vertex in x+ i Cyi , i = 1, 2.
Claim 5. uz1 , uz2 , z1 x2 , z2 x1 , z1 z2 ̸∈ E ∗ (G). − Proof. Suppose uz1 ∈ E ∗ (G). By Claim 2, we have z1 ̸= x+ 1 . Then x1 z1 ∈ E (G) by the choice of yi . Therefore, by Claim 3, − + − ′ C = x1 Puz1 Cx1 x1 Cz1 x1 is an o-cycle containing all vertices of C and |V (C ′ )| > |V (C )|. By Lemma 2, there exists a longer implicit-heavy cycle in G, a contradiction. Hence uz1 ̸∈ E ∗ (G). Similarly, uz2 ̸∈ E ∗ (G). − + − + − ′ Suppose z1 x2 ∈ E ∗ (G). By Claim 4, z1 ̸= x+ 1 . Then by Claim 3, C = x1 Px2 z1 Cx2 x2 Cx1 x1 Cz1 x1 is an o-cycle containing ′ all vertices of C and |V (C )| > |V (C )|. By Lemma 2, there exists a longer implicit-heavy cycle in G, a contradiction. Hence z1 x2 ̸∈ E ∗ (G). Similarly, z2 x1 ̸∈ E ∗ (G). + −¯ + −¯ − + − + ′ Suppose z1 z2 ∈ E ∗ (G). By Claim 4, z1 ̸= x+ 1 or z2 ̸= x2 . Then C = x1 Px2 z2 C x2 x2 C z1 z2 Cx1 x1 Cz1 x1 (if z1 ̸= x1 and + −¯ + −¯ + + − + − + + z2 ̸= x2 ) or x1 Px2 z2 C x2 x2 C z1 z2 Cx1 (if z1 = x1 and z2 ̸= x2 ) or x1 Px2 C¯ z1 z2 Cx1 x1 Cz1 x1 (if z1 ̸= x1 and z2 = x2 ) is an o-cycle containing all vertices of C and |V (C ′ )| > |V (C )|. By Lemma 2, there exists a longer implicit-heavy cycle in G, a contradiction. Hence z1 z2 ̸∈ E ∗ (G). + − + Claim 6. x− 1 x1 ∈ E (G) or x2 x2 ∈ E (G). + − + − + Proof. Suppose to the contrary that x− 1 x1 ̸∈ E (G) and x2 x2 ̸∈ E (G). By Claim 3, we have d(x1 ) + d(x1 ) ≥ n and − + − − + + d(x2 ) + d(x2 ) ≥ n. Thus, we can obtain that d(x1 ) + d(x2 ) ≥ n or d(x1 ) + d(x2 ) ≥ n, contradicting to Claim 4. + By Claim 6, without loss of generality, we may assume that x− 1 x1 ∈ E (G).
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Claim 7. x1 x2 ∈ E (G). − Proof. Suppose x1 x2 ̸∈ E (G). Now by the choice of P and Claim 5, we have {y1 , y− 1 , x1 , u1 , u2 , . . . , ur , x2 , y2 , y2 } induces a Pr +6 , where r ≥ 1. Since G is P7 -f -implicit-heavy, G is also Pr +6 -f -implicit-heavy. By the choice of C , u1 − − − and ur are implicit-light. It follows that id(y− ̸∈ E (G). Then C ′ = 1 ) ≥ n/2 and id(y2 ) ≥ n/2. By Claim 5, y1 y2 − − + −2 ¯ −2 ¯ + − ¯ − − − − y1 Cx2 x2 Cy2 x2 Px1 y1 C x1 x1 C y2 y1 is an o-cycle containing all vertices of C of G + y1 y2 and |V (C )| < |V (C ′ )|, by Lemma 2, − ′′ ′′ G + y− 1 y2 has a cycle C containing all implicit-heavy vertices of G, and such that C is longer than C . By Lemma 5, G has a ′′ cycle containing all vertices of C , a contradiction to the choice of C .
Claim 8. r = 1. + ∗ ∗ Proof. Suppose r ≥ 2. By the choice of P , u1 x2 ̸∈ E (G) and ur x1 ̸∈ E (G). By Claim 4, x1 x+ 2 ̸∈ E (G) and x1 x2 ̸∈ E (G). + + + ∗ ∗ By Claim 2, u1 x1 ̸∈ E (G) and ur x2 ̸∈ E (G). Note that x1 x2 ∈ E (G) by Claim 7. Thus {x1 , x1 , u1 , x2 } induces a claw and + + {x2 , ur , x+ 2 , x1 } induces a claw. Since G is 2-heavy and u1 , ur are implicit-light, d(x1 ) ≥ n/2 and d(x2 ) ≥ n/2. This implies + + ∗ that x1 x2 ∈ E (G), contradicting to Claim 4.
Note that G is D-f -implicit-heavy or H-f -implicit-heavy. Suppose G is D-f -implicit-heavy. By Claims 5 and 7,
− − − {y1 , y− 1 , x1 , u1 , x2 , y2 , y2 } induces a D. Since u1 is implicit-light, y1 and y2 are implicit-heavy. By similar arguments as in the
proof of Claim 7, we can get a contradiction. + Next, we assume G is H-f -implicit-heavy. By Claims 2 and 4, {x− 1 , x1 , x1 , u1 , x2 } induces an H. Since u1 is implicit− + − + + light, x1 and x1 are implicit-heavy. If x2 x2 ∈ E (G), then by Claims 2 and 4, {x− 2 , x2 , x2 , u1 , x1 } induces an H. Since u1 is − + − + + implicit-light, x2 and x2 are implicit-heavy. If x2 x2 ̸∈ E (G), since G is 2-heavy and u1 is light, both x+ 1 and x2 are heavy. + + Therefore, id(x1 ) ≥ n/2 and id(x2 ) ≥ n/2. The arguments we used in the proof of Claim 7 can now be applied to the graph + ′ ′ ′ G′ = G + x+ 1 x2 to conclude that G has a cycle C containing all implicit-heavy vertices of G and such that C is longer than ′ C . By Lemma 5 G has a cycle containing all vertices of C , contradicting to the choice of C . This completes the proof of Theorem 8. Acknowledgment The author is very grateful to the anonymous referee for carefully reading the manuscript and providing comments and suggestions which led to a substantial improvement of the paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
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