# Hamilton cycles in almost distance-hereditary graphs

• Bing Chen and Bo Ning
From the journal Open Mathematics

## Abstract

Let G be a graph on n ≥ 3 vertices. A graph G is almost distance-hereditary if each connected induced subgraph H of G has the property dH(x, y) ≤ dG(x, y) + 1 for any pair of vertices x, yV(H). Adopting the terminology introduced by Broersma et al. and Čada, a graph G is called 1-heavy if at least one of the end vertices of each induced subgraph of G isomorphic to K1,3 (a claw) has degree at least n/2, and is called claw-heavy if each claw of G has a pair of end vertices with degree sum at least n. In this paper we prove the following two theorems: (1) Every 2-connected, claw-heavy and almost distance-hereditary graph is Hamiltonian. (2) Every 3-connected, 1-heavy and almost distance-hereditary graph is Hamiltonian. The first result improves a previous theorem of Feng and Guo [J.-F. Feng and Y.-B. Guo, Hamiltonian cycle in almost distance-hereditary graphs with degree condition restricted to claws, Optimazation57 (2008), no. 1, 135–141]. For the second result, its connectedness condition is sharp since Feng and Guo constructed a 2-connected 1-heavy graph which is almost distance-hereditary but not Hamiltonian.

MSC 2010: 05C38; 05C45

## 1 Introduction

In this paper, we only consider the graphs which are finite, undirected and without multi-edges and loops. For terminology and notation not defined here, we refer to Bondy and Murty [1].

Let G be a graph with vertex set V(G) and H be a subgraph of G. For a vertex υV(G), we denote by NH(υ) the set of vertices which are adjacent to υ in H, and by dH(υ) = |NH(υ)| the degree of υ in H. For two vertices x, yV(G), an (x, y)-path in H is a path starting from x to y with all vertices in H. The distance of x and y in H, denoted by dH(x, y), is defined as the length of a shortest (x, y)-path in H. When there is no danger of ambiguity, we use N(υ), d(υ) and d(x, y) instead of NG(υ), dG(υ) and dG(x, y), respectively.

A graph is called Hamiltonian if it contains a Hamilton cycle, i.e., a cycle passing through all the vertices of the graph. The study of cycles, especially Hamilton cycles, may be one of the most important and most studied areas of graph theory. It is well-known that to determine whether a given graph contains a Hamilton cycle is NP-complete, which is shown by Karp [2]. However, if we only consider some restricted graph classes, then the situation is completely changed. A graph G is called distance-hereditary if each connected induced subgraph H has the property that dH(x, y) = dG(x, y) for any pair of vertices x, y in H. This concept was introduced by Howorka [3] and a complete characterization of distance-hereditary graphs could be found in [3]. In 2002, Hsieh, Ho, Hsu and Ko [4] obtained an O(|V| + |E|)-time algorithm to solve the Hamiltonian problem on distance-hereditary graphs. Some other optimization problems can also be solved in linear time for distance-hereditary graphs although they are proved to be NP-hard for general graphs. For references in this direction, we refer to [5, 6].

A graph G is called almost distance-hereditary if each connected induced subgraph H of G has the property dH(x, y) ≤ dG(x, y) + 1 for any pair of vertices x, yV(H). For some properties and a characterization of almost-distance hereditary graphs, we refer to [7].

Let G be a graph. An induced subgraph of G isomorphic to K1,3 is called a claw. The vertex of degree 3 in the claw is called its center and the other vertices are its end vertices. G is called claw-free if G contains no claw. Throughout this paper, whenever the vertices of a claw are listed, its center is always the first one.

Many results about the existence of Hamilton cycles in claw-free graphs have been obtained. In particular, Feng and Guo [8] gave the following result on Hamiltonicity of almost distance-hereditary claw-free graphs.

Theorem 1.1

(Feng and Guo [8]). Let G be a 2-connected claw-free graph. If G is almost distance-hereditary, then G is Hamiltonian.

Let G be a graph on n vertices. A vertex υ of G is called heavy if d(υ) ≥ n/2. Broersma et al. [9] introduced the concepts of 1-heavy graph and 2-heavy graph. Later, Fujisawa and Yamashita [10] and Čada [11] introduced the concept of claw-heavy graphs, independently. Following [911], we say that a claw in G is 1-heavy (2-heavy) if at least one (two) of its end vertices is (are) heavy. G is called 1-heavy (2-heavy) if every claw of it is 1-heavy (2-heavy), and called claw-heavy if every claw of it has two end vertices with degree sum at least n. It is easily seen that every claw-free graph is 1-heavy (2-heavy, claw-heavy), every 2-heavy graph is claw-heavy and every claw-heavy graph is 1-heavy. But not every claw-heavy graph is 2-heavy, and not every 1-heavy graph is claw-heavy.

In [12], Feng and Guo extended Theorem 1.1 to a larger graph class of 2-heavy graphs.

Theorem 1.2

(Feng and Guo [12]). Let G be a 2-connected 2-heavy graph. If G is almost distance-hereditary, then G is Hamiltonian.

Feng and Guo [12] also constructed a 2-connected 1-heavy graph which is almost distance-hereditary but not Hamiltonian. Thus it is natural to ask which is the minimum connectivity for a 1-heavy almost distance-hereditary graph under this connectivity condition to be Hamiltonian.

Motivated by [9, 1315], in this paper we obtain the following two theorems which extend Theorem 1.1 and Theorem 1.2. In particular, Theorem 1.3 improves Theorem 1.2, and Theorem 1.4 answers the problem proposed above.

Theorem 1.3.

Let G be a 2-connected claw-heavy graph. If G is almost distance-hereditary, then G is Hamiltonian.

Theorem 1.4.

Let G be a 3-connected 1-heavy graph. If G is almost distance-hereditary, then G is Hamiltonian.

We emphasize that our technique of proofs is different from Feng and Guo [12]. One of our main tools is the so called "Ore-cycle" (motivated by Lemma 3 in [13]) introduced by Li et al. [16].

Remark 1.5.

The graph in Fig. 1 shows that the result in Theorem 1.3 indeed strengthen that in Theorem 1.2. As shown in [13, Fig. 2], let n ≥ 10 be an even integer and Kn/2 + Kn/2−3denote the join of two complete graphs Kn/2and Kn/2−3. Choose a vertex yV(Kn/2) and construct a graph G with V(G) = V(Kn/2 + Kn/2−3) ∪ {υ, u, x} and E(G) = E(Kn/2 + Kn/2−3) ∪ {, uy, ux} ∪ {υw, xw : wV(Kn/2−3)}. It is easy to see that G is a Hamiltonian graph satisfying the condition of Theorem 1.3, but not the condition of Theorem 1.2.

We postpone the proofs of Theorem 1.3 and 1.4 to Section 3.

Fig. 1.

A graph which shows the result in Theorem 1.3 strengthens that in Theorem 1.2.

## 2 Preliminaries

Let G be a graph on n vertices and k ≥ 3 be an integer. Recall that a vertex of degree at least n/2 in G is a heavy vertex; otherwise it is light. A claw in G is called a light claw if all its end vertices are light, and is called an o-light claw if any pair of end vertices has degree sum less than n. A cycle in G is called a heavy cycle if it contains all heavy vertices of G. Following [16], we use (G) to denote the set { : E(G) or d(u) + d(υ) ≥ n, u, υV(G)}. A sequence of vertices C = υ1υ2υk υ1 is called an Ore-cycle or briefly, o-cycle of G, if we have υiυi+1(G) for every i ∈ {1, 2, · , k}, where υ1 = υk+1.

Let G be a graph and k be a nonnegative integer. For a cycle C of G and a vertex uV(G)\V(C), a subgraph F of G is said to be a (u, C ; k)-fanif F is a union of paths P1, P2, … , Pk, where Pi is a (u, wi)-path (1 ≤ ik), PiC = {wi} (1 ≤ ik) and PiPj = {u} for 1 ≤ i < jk. In the following, we use F = (u: P1, P2, … , Pk) to denote the fan. The vertices in V(F)\{w1, w2, … , wk} are called internal vertices of F.

We need some notations from [13]. Let H be a path or a cycle with a given orientation. We denote by H the same graph as H but with the reverse orientation. For a vertex υV(H), we use υH+ to denote the successor of υ on H, and υH to denote its predecessor. If SV(H), then define S+H = {υ+H : υS} and SH = {υH : υS}. If there is no danger of ambiguity, we denote υ+H, υH, S+H and SH by υ+, υ, S+ and S, respectively. For two vertices u, υV(H), we denote by H[u, υ] the segment of H from u to υ, and denote H(u, υ), H [u, υ) and H(u, υ] by the paths H[u, υ] − {u, υ}, H[u, υ] − {υ} and H[u, υ] − {u}, respectively.

To prove Theorems 1.3 and 1.4, the following six lemmas are needed. In particular, similar proofs of the facts in Lemma 2.3 can be found in [13, 16] (for example, see Claims 1-4 of Theorem 8 in [16]). For the convenience of the readers, we write the detailed proofs here.

Lemma 2.1

(Bollobás and Brightwell [17], Shi [18]). Every 2-connected graph contains a heavy cycle.

Lemma 2.2

(Li, Wang, Ryjáček, Zhang [16]). Let G be graph and let C′ be an o-cycle of G. Then there exists a cycle C of G such that V(C′) ⊆ V(C).

Lemma 2.3.

Let G be a non-Hamiltonian graph on n vertices, C be a longest cycle (a longest heavy cycle) of G, R a component of GV(C), and A = {υ1, υ2, … , υ3} the set of neighbors of R on C. Let uV(R) andυi, υjA. Then there hold

(a) i(G), +i(G); (b) υiυj(G), υ+iυ+j(G); and (c)ifυiυ+i(G), thenυiυj(G), υiυ+j(G).

Furthermore, if G is a 2-connected claw-heavy graph, then

(d) υiυ+i(G) andυjυ+j(G); (e) υiυ+iE(G) orυjυ+jE(G).

Proof. (a) Suppose that i(G). Then C′ = iC[υi, υi]υiu is an o-cycle of length longer than C, a contradiction. The other assertion can be proved similarly.

(b) Suppose that υiυj(G). Then C=uυjC[υj,υi]υiυjC[υj,υi]υiu is an o-cycle of length longer than C, a contradiction. The other assertion can be proved similarly.

(c) Suppose that υiυj(G). Then C′ = υijC[υj, υi]υiυ+iC[υ+i, υi]υiυi is an o-cycle of length longer than C, a contradiction. The other assertion can be proved similarly.

(d) If υiυ+iE(G), then by (a), we know iE(G), +iE(G). Thus {υi, u, υi, υi+}, induces a claw. Since G is claw-heavy, by (a), we have d(υi) + d(υ+i) ≥ n. This implies that υiυ+i(G). If υiυ+iE(G), then obviously υiυ+i(G). The other assertion can be proved similarly.

(e) Suppose that υiυ+iE(G) and υjv+jE(G). By (d), we have d(υi) + d(υ+i) ≥ n and d(υj) + d(υ+j) ≥ n. This implies that d(υi) + d(υj) ≥ n or d(υ+i) + d(υ+j) ≥ n. Thus υiυj(G) or υ+iυ+j(G), a contradiction to (b).

Lemma 2.4.

Let G be a non-Hamiltonian graph, C be a longest cycle (a longest heavy cycle) of G, R a component of G − V(C), and A = {υ1, υ2, . . . , υk} the set of neighbors of R on C. Let υi, υjA. Then there hold

(a) forlV(C(υi, υj]), ifυil(G), thenlυ+j(G) andl+υ+j(G);

(b) forlV(C(υi, υj]) ∩ N(υi), ifυiυ+i(G), thenlυj(G) andl+υ+j(G); and

(c) forlV(C(υi, υj]) ∩ N(υi) ∩ N(υj), ifυiυ+i(G), thenll+(G).

Proof. Let P be a (υi, υj)-path with all internal vertices in R.

(a) Suppose lυ+j(G). Then C'=PC[υi,l]lυj+C[υj+,υi]υilC[l,υj] is an o-cycle such that V(C) ⊂ V(C′). By Lemma 2.2, there is a longer cycle C″ containing all vertices in C, that is, a longer cycle (a longer heavy cycle) in G, contradicting the choice of C. Suppose l+υ+j(G). Then C'=PC[υj,l+]l+υj+C[υj+,υi]υilC[l,υi] is an o-cycle such that V(C) ⊂ V(C′), a contradiction.

(b) Suppose lυj(G). Then C'=PC[υj,υi]υiυi+C[υi+,l]lυjC[υj,l]lυi is an o-cycle such that V(C) ⊂ V(C′), a contradiction. Suppose l+υ+j(G). Then C'=PC[υj,l+]l+υj+C[υj+,υi]υiυi+C[υi+,l]lυi is an o-cycle such that V(C) ⊂ V(C′), a contradiction.

(c) Suppose ll+(G). Then C′ = PC[υj, υi]υiυ+iC[υ+i, l]ll+C[l+, υj]υji is an o-cycle such that V(C) ⊂ V(C′), a contradiction. □

Lemma 2.5.

Let G be a 3-connected 1-heavy non-Hamiltonian graph and C be a longest heavy cycle of G, R a component of GV(C). Let uV(R), and υ0, υ1, υ2be three neighbors of u which are in the order around C andυ1, υ+1are light. LetliC[υ+i, υi+1) such thatυi+1liE(G) and li υiE(G), siC(υ+i−1, υi] such thatυ+i−1siE(G) and si υiE(G) (where the indices are taken modulo 3).

(a) Ifυ2υ+2E(G) andυ0υ+0E(G), then (i) υ1υ2E(G), (ii) υ1l1E(G) andυ1s+1E(G), (iii) υ1, l+1, l1, s1, s+1, l1, s1are light;

(b) Ifυ2υ+2E(G) andυ0υ+0E(G), then {υ+1, l+0, s+2} induces an independent set.

Proof. By Lemma 2.3 (a), 1E(G) and +1E(G). If υ1υ+1E(G), then since u, υ1, υ+1 are light, {υ1, u, υ1, υ+1} induces a light claw, a contradiction. Thus υ1υ+1E(G).

(a) Since υ2υ+2E(G) and υ0υ+0E(G), we have υ2, υ+0 are heavy or υ+2, υ0 are heavy by Lemma 2.3 (b).

(i) Suppose υ1υ2E(G) and υ2, υ+0 are heavy. Let C′ = υ1υ2C[υ2, υ0]υ21C[υ1, υ2]υ2υ+0C[υ+0, υ1]. Then C′ is an o-cycle such that V(C) ⊂ V(C′), a contradiction.

Suppose υ1υ2E(G) and υ+2, υ0 are heavy. Now {υ2, υ2, u, υ1} induces a light claw, a contradiction.

(ii) Suppose υ1l1E(G). Note that υ1l+1(G) and l1l+1(G) by Lemma 2.4 (b) and (c). Since υ2, υ+0 are heavy or υ+2, υ0 are heavy, by Lemma 2.3 (c) and Lemma 2.4 (b), υ1, l+1, l1 are light. Now {l1, l+1, υ1, l1} induces a light claw, a contradiction. Similarly, we can prove that υ1s+1E(G).

(iii) By Lemma 2.3 (c) and Lemma 2.4 (b), υ1, l+1, l1, s1, s+1 are light. Since υ1l1E(G), we obtain υ+2l1(G) and υ+0l1(G) by Lemma 2.4 (b). Note that either υ+0 or υ+2 is a heavy vertex. This implies l1 is a light vertex. The other assertion that s1 is light can be proved similarly.

(b) Since υ0l0E(G) and υ2s2E(G), υ+1l+0(G) and υ+1s+2(G) by Lemma 2.4 (b). Furthermore, we can prove that l+0s+2E(G). (Otherwise, C'=υ0uυ2s2C[s2,l0+]l0+s2+C[s2+,υ2]υ2υ2+C[υ2+,υ0]υ0υ0+C[υ0+,l0]l0υ0 is an o-cycle such that V(C) ⊂ V(C′), a contradiction.)

Lemma 2.6.

Let G be a non-Hamiltonian almost distance-hereditary graph, C be a longest cycle (a longest heavy cycle) of G, R a component of GV(C). If there exists a vertex uV(R) such that NC(u) = {υ1; υ2, … , υr}, then there hold

(a) for any induced (u, υ)-path P, whereυNC(u) orυN+C(u), the length of P is at most 3;

(b) ifυiυ+i(G), then there exists a vertexliC[υ+i, υi+1) such thatυi+1liE(G) and liυiE(G), and there exists a vertexsiC(υ+i−1, υ1] such thatυ+i−1siE(G) and si υiE(G);

(c) ifυiυ+iE(G) andυi+1υ+i+1E(G), thenυ+i+1liE(G); and

(d) ififυiυ+iE(G) andυi+1υ+i+1E(G), then both {li, li, υi, υi+1} and {li, l+i, υi, υ+i+1} induce claws.

Proof. (a) Suppose there exists an induced (u, υ)-path P such that the length of P is at least 4. It follows that dP (u, υ) ≥ 4, contradicting the fact that dG(u, υ) = 2 and G is almost distance-hereditary.

(b) Let H = G[{u} ∪ V(C[υi, υi+1])] − {υi+1}. Since dG(υi+1, u) = 2 and G is almost distance-hereditary, we have dH(υi+1, u) ≤ 3. Since υiυ+i(G), by Lemma 2.3 (c), we have υiυi+1E(G) and dH(υi+1, u) = 3. It follows that dH(υi+1, υi) = 2. So there exists a vertex liC[υ+i, υi+1) such that υi+1liE(G) and li υiE(G). The other assertion can be proved similarly.

(c) Suppose υ+i+1liE(G). Let H = G[{υ+i+1, υi+1, li, υi, u}]. By Lemma 2.3 (c), υiυi+1E(G) and υiυ+i+1E(G). We can see H is an induced (u, υ+i+1)-path of length 4 in G, contradicting Lemma 2.6 (a).

(d) By Lemma 2.3 (c) and Lemma 2.4 (b), we have υiυi+1(G) and liυi+1(G). By Lemma 2.6 (c) and Lemma 2.4 (b), we have υiliE(G). So {li, li, υi, υi+1} induces a claw. The other assertion can be proved similarly. □

## 3 Proofs of Theorems 1.3 and 1.4

Proof of Theorem 1.3

Let G be a graph satisfying the condition of Theorem 1.3. Let C be a longest cycle of G and assign an orientation to it. Suppose G is not Hamiltonian. Then V(G)\V(C)0. Let R be a component of GC, and A = {υ1, υ2, … , υk} be the set of neighbors of R on C. Since G is 2-connected, there exists a (υi, υj)-path P = υiu1ur υj with all internal vertices in R, and υi, υjA. Choose P such that:

1. |V(C(υi, υj))| is as small as possible;

2. |V(P)| is as small as possible subject to (1).

Claim 1

There is no o-cycle Cin G such that V(C) ⊂ V(C′).

Proof. Otherwise, C′ is an o-cycle such that V(C) ⊂ V(C′). By Lemma 2.2, there exists a cycle containing all vertices in C′ and longer than C, contradicting the choice of C. □

By Lemma 2.3 (e), without loss of generality, assume that υiυ+iE(G).

Claim 2

r = 1, that is, V(P) = {υj, u1, υj}.

Proof. Suppose r ≥ 2. Consider H = G[V(P) ∪ V(C[υi, υj])] − {υj}. Since υiυ+iE(G), υiυjE(G) by Lemma 2.3 (c). Thus dH(υj, υi) ≥ 2. By the choice condition of P and Lemma 2.3 (a), we have dP(υi, ur) ≥ 2 and dH(υj, ur) = dH(υj, υi) + dH(υj, ur) ≥ 4, which yields a contradiction to the fact G is almost distance-hereditary and dG(υj, ur) = 2. Hence V(P) = {υi, u1, υj}. □

Claim 3

|V(C[υi, υj])| ≥ 5.

Proof. Suppose |V(C[υi, vj])| = 4 or |V(C[υi, υj])| = 3. This means C[υi, υj] = υiυ+iυiυj or C[υi, υj] = υiυiυj. Let C′ = υiu1υjυjυ+jC[υ+j, υi]υiυ+iυi or C′ = υiu1υjυjυ+jC[υ+j, υi]. Then C′ is an o-cycle such that V(C) ⊂ V(C′) by Lemma 2.3 (d), contradicting Claim 1. □

Recall that υiυ+iE(G). Let H = G[{u1, υi} ∪ V(C[υi, υj])] – {υi}. Since dG(υi, u1) = 2 and G is almost distance-hereditary, dH(υi, u1) ≤ 3. By Lemma 2.3 (c) and (d), we have υiυjE(G). By the choice of P, u1υE(G), where υC[υ+i, υj]. It follows that dH(υi, u1) = 3 and dH(υi, υj) = 2. By Lemma 2.3 (b), (c) and (d), υiυjE(G) and υ+iυjE(G). Thus there exists a vertex wC(υ+i, υj) such that υiwE(G) and jE(G). Note that w is well-defined.

Claim 4

+j(G).

Proof. Suppose +j(G). By Lemma 2.4 (a), we obtain υiw+(G). Since υiwE(G), we have υjw+(G) by Lemma 2.4 (b) and by symmetry. Note that υjυi(G) by Lemma 2.3 (c). Thus {w, w+, υj, υi} induces an o-light claw in G, a contradiction.

Next we will show that {υj, u1, w, υ+j} induces an o-light claw and get a contradiction. Before proving this fact, the following claim is needed.

Claim 5

u1w1(G).

Proof. First we will show that wυi(G). Since υjυ+j(G) and υj wE(G), we have wυi(G) by Lemma 2.4 (b) and symmetry.

Next we will show that wυj(G). Suppose not. Consider the subgraph induced by {w, w, υj, υi}. Note that υjυi(G) by Lemma 2.3 (c) and wυi(G) by the analysis above. Then {w, w, υj, υi} induces an o-light claw, a contradiction.

Now we will show that u1w(G), since otherwise, C′ = u1wC[w, υj]υjυ+jC[υ+j, w]wυju1 is an o-cycle such that V(C) ⊂ V(C′), contradicting Claim 1. □

By Claims 4, 5 and Lemma 2.3 (a), {υj, u1, w, υ+j} induces an o-light claw, contradicting the fact G is claw-heavy. The proof of Theorem 1.3 is complete. □

Proof of Theorem 1.4.

Let G be a graph satisfying the condition of Theorem 1.4. By Lemma 2.1, there exists a heavy cycle in G. Now choose a longest heavy cycle C of G and assign an orientation to it. Suppose G is not Hamiltonian. Then V(G)\V(C)0. Let R be a component of GC and A = {w1, w2, … , wk} be the set of neighbors of R on C. Since G is 3-connected, for any vertex u of R, there exists a (u, C; 3)-fan F such that F = (u; Q1, Q2, Q3), where Q1 = ux1xr1wi, Q2 = uy1yr2wj and Q3 = uz1zr3wk, and wi, wj, wk are in the order of the orientation of C.

By the choice of C, all internal vertices of F are not heavy. By Lemma 2.3 (b), there is at most one heavy vertex in N+C(R) and at most one heavy vertex in NC(R). Without loss of generality, assume that wi, w+i are light. Hence wiw+iE(G), otherwise {wi, wi, w+i, xr1} induces a light claw, contradicting G is 1-heavy.

Claim 1

There exists a (u, C; 3)-fan F such that V(F) = {u, wi, wj, wk}.

Proof. Now we choose the fan F in such a way that:

1. Q1 = uwi;

2. |V(C[wi, wj])| is as small as possible subject to (1);

3. |V(Q2)| is as small as possible subject to (1) and (2);

4. |V(C[wk, wi])| is as small as possible subject to (1), (2) and (3);

5. |V(Q3)| is as small as possible subject to (1), (2), (3) and (4).

Since G is 3-connected, for any neighbor of C in R, say u (with uwiE(G), where wiV(C)), there are three disjoint paths from u to C. Obviously, we can choose one such path as uwi. Thus (1) is well-defined, and furthermore, the choice condition of F is well-defined.

Claim 1.1.

V(Q2) = {U, wj}.

Proof. Suppose V(Q2)\{u,wj}0. Without loss of generality, set y = yr2. Let H = G[V(Q1) ∪ V(Q2) ∪ V(C[wi, wj])] − {wj}. Note that wiw+iE(G). By Lemma 2.3 (c), it is easy to see that wi wjE(G), so dH(wj, wi) ≥ 2. In the meantime, the choice condition (2) implies that N(V(Q2)\{wj})V(C(wj,wj))=0. This means that dH (wj, y) = dH (wj, wi) + dH (wi, y) ≥ 2 + dH (wi, y). Since G is almost distance-hereditary and dG (wj, y) = 2, we have dH (wj, y) = 3 and ywiE(G). Let F′ = (y; Q1, Q2, Q3) such that Q1 = ywi, Q2 = ywj and Q3 = Q2[y, u]Q3[u, wk]. Then F′ is a (y, C;3)-fan satisfying (1), (2) and |V(Q2)| = 2, contradicting the choice condition (3), a contradiction. □

Claim 1.2.

V (Q3) = {u, wk}.

Proof. Suppose V(Q3)\{u,wk}0. Without loss of generality, set z = zr3.

If zwiE(G), then set H = G[V (Q1) ∪ V (Q3) ∪ V(C[wk, wi])] − {wk}. Since wiw+iE(G), we obtain w+kwiE(G) by Lemma 2.3 (c). This means dH (w+k, wi) ≥ 2. By the choice condition (4), we have N(V(Q3)\{wk})V(C(wk,wi))=0, and hence dH (w+k, z) = dH (w+k, wi) + dH (wi, z). Since zwiE(G), dH (wi, z) ≥ 2 and we get dH (w+k, z) ≥ 4. It yields a contradiction to the fact G is almost distance-hereditary and dG (w+k, z) = 2.

If zwiE(G), then set H = G[V(C [wi, wj]) ∪ V(Q3 [u, z])] − {wi}. Note that NC(z)V(C(wi,wj])=0 (by the choice conditions (2), (5)) and NC(V(Q3)\{z,wk})V(C(wi,wj))=0 (by the choice condition (2)). Since dG(w+i, z) = 2 and G is almost distance-hereditary, dH(w+i, z) ≤ 3. But if w+i, wjE(G), then the distance from z to w+i in H is at least 4, where in such a shortest path, the path Q3 [z, u] contributes at least 1, the path Q2 [u, wj] contributes 1, a contradiction. Thus we have w+iwjE(G), and hence wjw+j(G) by Lemma 2.3 (c). Consider the subgraph induced by {wj, w+i, w+j, u}. Since G is 1-heavy and w+i, u are light, w+j is heavy. Now let H = G[{wi} ∪ V(C[wi, wj]) ∪ V(Q3[u, z])] – {wi}. Similarly, since dG(wi, z) = 2, we have dH(wi, z) = 3, and wi, wjE(G). Consider the subgraph induced by {wj, wi, wj, u}. Similarly, we can see wj is heavy, and hence wjw+j(G), a contradiction. Thus V(Q3) = {u, wk}. □

By Claims 1.1 and 1.2, the proof of Claim 1 is complete. □

By Claim 1, there exists a (u, C; 3)-fan F such that V(F)\V(C) = {u}. Suppose that NC (u) = {υ1, υ2, …, υr} (r ≥ 3) and υ1, υ2, …, υr are in the order of the orientation of C. In the following, all the subscripts of υ are taken modulo r, and υ0 = υr.

By Lemma 2.3 (b), there is at most one heavy vertex in N+C(u) and at most one heavy vertex in NC (u). Since r ≥ 3, we know that there exists υjNC (u), such that υj, υ+j are light, and hence υjυ+jE(G) by the fact G is 1-heavy. Without loss of generality, assume that υ1υ+1E(G) and υ1, υ+1 are light. By Lemma 2.6 (b), there exists a vertex l1C[υ+1, υ2) such that υ2l1E(G) and l1υ1E(G), and there exists a vertex s1C(υ+0, υ1] such that υ+0s1E(G) and s1v1E(G).

We divide the proof into two cases.

Case 1

υ2υ+2E(G) andυ0μ+0E(G).

Both {υ, υ2, υ+2, u} and {υ, υ0, υ+0, u} induce claws. By Lemma 2.3 (b) and the fact G is 1-heavy, υ2 and υ+0 are heavy or υ+2 and υ0 are heavy.

Claim 2

υ1l1E(G) andl1υ2E(G).

Proof. Suppose υ1l1E(G). Note that uv1(G) by Lemma 2.3 (a). By Lemma 2.5 (a), l1 is light. Now {υ1, l1, u, υ1} induces a light claw, a contradiction.

Suppose l1v2E(G). Let H = G[{υ1, l1, υ2, υ2, u}]. By Lemma 2.3, we get uv2E(G), uv1E(G) and υ1υ2E(G). Note that υ2υ1E(G) by Lemma 2.5 (a). Now G[{υ1, l1, υ2, υ2, u}] is an induced path of length 4 in G, contradicting Lemma 2.6 (a).

Now we consider the following two subcases.

Subcase 1.1.

υ2, υ+0are heavy vertices.

By Lemma 2.3 (a), +2(G). By Lemma 2.5 (a), we have υ1l1E(G). Note that l′ ≔ l1N(υ1) and υ1υ+1E(G). By Lemma 2.4 (b), l+υ+2 = l1υ+2(G). By Lemma 2.5 (a) and Lemma 2.3 (b), l1 and υ+2 are light. Now {υ2, l1, u, υ+2} induces a light claw, a contradiction.

Subcase 1.2.

υ+2, υ0are heavy vertices.

Consider the subgraph induced by {υ+0, s1, l1, υ2, υ}. It is easily to check that υ+0s1E(G), l1υ2E(G) (by Claim 2) and υ2uE(G). By Lemma 2.3 (a), υ+0uE(G). By Lemma 2.5 (a) and Lemma 2.4 (b), we know that υ1l1E(G) and υ+0l1E(G). Now we obtain s1l1E(G) or υ+0υ2E(G) or s1υ2E(G) (Otherwise, G[{υ+0, s1, l1, υ2, u}] is an induced path of length 4, contradicting Lemma 2.6 (a)).

Suppose s1l1E(G). By Lemma 2.5 (a), l1, s1 are light. Now {υ1, l1, s1, u} induces a light claw, contradicting G is 1-heavy.

Suppose υ+0υ2E(G). Consider the subgraph induced by {υ2, υ2, υ+0, u}. By Lemma 2.3 (a), we have 2E(G) and +0E(G). Since υ2, υ+0, u are light and G is 1-heavy, υ+0υ2E(G). By Lemma 2.5 (a) and Claim 2, υ1l1E(G) and l1υ2E(G). Now C=υ1l1C[l1,υ1+]υ1+υ1C[υ1,υ0+]υ0+υ2C[υ2,l1]l1υ2C[υ2,υ0]υ0uυ1 is an o-cycle such that V(C) ⊂ V(C′), a contradiction.

Suppose s1υ2E(G). Consider the subgraph induced by {υ2, υ2, s1, u}. By Lemma 2.5 (a) and Lemma 2.3 (b), s1 and υ2 are light. Since G is 1-heavy, s1υ2E(G). Now C=υ1uυ2C[υ2,s1]s1υ2C[υ2,υ1+]υ1+s1+C[s1+,υ1] is an o-cycle such that V(C) ⊂ V(C′), a contradiction. (First, we can prove s1υ+1E(G). Otherwise, {υ1, s1, υ+1, u} induces a light claw, a contradiction. Note that υ+0υ+1E(G) and υ+0s+1E(G) by Lemma 2.3 (b) and Lemma 2.4 (b). Then we obtain υ+1s+1E(G) since otherwise {s1, υ+0, υ+1, s+1} induces a light claw, a contradiction.)

Case 2

υ2υ+2E(G) orυ0υ+0E(G).

Without loss of generality (by symmetry), assume that υ2υ+2E(G).

Subcase 2.1.

υ0υ+0E(G).

By Lemma 2.3 (a), 0E(G) and +0E(G). Now {υ0, υ0, υ+0, u} induces a claw. Since G is 1-heavy and u is light, υ0 is heavy or υ+0 is heavy.

Suppose υ0 is heavy. By Lemma 2.3 (b), (c) and Lemma 2.4 (b), υ2, υ1 and l1 are light. By Lemma 2.6 (d), {l1, l1, υ1, υ2} induces a light claw, a contradiction.

Suppose υ+0 is heavy. By Lemma 2.3 (b), (c) and Lemma 2.4 (b), we can see υ+2, υ1 and l+1 are light. By Lemma 2.6 (d), {l1, l+1, υ1, υ+2} induces a light claw, a contradiction. (Note that υ1υ+1E(G) and υ2υ+2E(G). By Lemma 6 (c), l1υ+2E(G).

Subcase 2.2.

υ0υ+0E(G).

By Lemma 6 (b), there exists a vertex s2C(υ+1, υ2] such that υ+1s2E(G) and s2υ2E(G).

Claim 3

(i) υ1is heavy, (ii) l0, l+0, s2, s+2are light.

Proof. Recall that the definition of l0 occurred in the condition of Lemma 5 before. Let l0C[υ+0, υ1) such that υ0l0E(G) and l0υ0E(G).

(i) By Lemma 2.6 (d), each of {l0, l0, υ0, υ1} and {l1, l1, υ1, υ2} induces a claw. Since G is 1-heavy, at least one vertex of {l1, υ1, υ2} is heavy.

Suppose υ2 is heavy. By Lemma 2.3 (b), (c) and Lemma 2.4 (b), υ1, υ0 and l0 are light. Now {l0, l0, υ0, υ1} induces a light claw, contradicting G is 1-heavy.

Suppose l1 is heavy. By Lemma 2.6 (c), υ+2l1E(G). By Lemma 2.4 (a) and (b), υ1l1(G) and υ0l1(G). This implies that υ0, υ1 are light. At the same time, we can prove that l0 is light. (Otherwise, C=υ0uυ2C[υ2,l1]l1υ2C[υ2+,υ0]υ0υ0+C[υ0+,l0]l0l1C[l1,l0]lυ0 is an o-cycle such that V(C) ⊂ V(C′), a contradiction.) Now {l0, l0, υ0, υ1} induces a light claw, contradicting G is 1-heavy.

Note that {l1, l1, υ1, υ2} induces a claw and υ2, l1 are light. Since G is 1-heavy, υ1 is heavy.

(ii) Note that υ1 is heavy. If l+0 is heavy, then C′ = υ01l+0C[l+0, υ1]υ1υ+1C[υ+1, υ0]υ0υ+0C[υ+0, l0]l0υ0 is an o-cycle such that V(C) ⊂ V(C′), a contradiction. If l0 is heavy, then C′ = υ0l0C[l0, υ1]υ1υ+1C[υ+1, υ0]υ0υ+0C[υ+0, l0]l0υ10 is an o-cycle such that V(C) ⊂ V(C′), a contradiction. Similarly, by symmetry, we can prove that s2, s+2 are light. □

Claim 4

υ1l+0E(G) andυ1s+2E(G).

Proof. Suppose υ1l+0E(G). By Lemma 2.4 (b) and (c), υ1l0(G) and l0l+0(G). By Claim 3, l+0, l0 are light. Now {l0, l0, l+0, υ1} induces a light claw, a contradiction.

By Lemma 2.6 (c) and by symmetry, we obtain υ1s2E(G). Suppose υ1s+2E(G). By Lemma 2.4 (b) and (c), υ1s2(G) and s2s+2(G). By Claim 3, s+2, s2 are light. Now {s2, s2, s+2, υ1} induces a light claw, a contradiction. □

By Claim 3, Lemma 2.5 (b) and Claim 4, l+0, s+2 are light, {υ1+l0+,υ1+s2+,l0+s2+}E(G)=0 and {υ1l+0, υ1s+2} ⊂ E(G). It is proved that {υ1, υ+1, l+0, s+0} induces a light claw, a contradiction.

The proof of Theorem 1.4 is complete. □

## Acknowledgement

The authors are particularly grateful to the anonymous referee for his/her extensive comments that considerably improved the paper.

The first author is supported by NSFC (No. 11271300) and the Scientific Research Program of Shaanxi Provincial Education Department (No. 2013JK0580).

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Accepted: 2015-10-27
Published Online: 2016-2-9
Published in Print: 2016-1-1