Minimum Number of Components of 2-Factors in Iterated Line Graphs

Abstract

It is well known that it is NP-hard to determine the minimum number of components of a 2-factor in a graph, even for iterated line graphs. In this paper, we determine the minimum number of components of 2-factors in iterated line graphs of some special tree-like graphs. It extends some known results.

1 Introduction

Throughout this paper, all graphs considered are simple and finite graphs. We follow the most common graph-theoretical terminology and for concepts and notations not defined here, see [1].

A 2-factor of a graph G is a spanning subgraph whose components are cycles. In particular, a hamiltonian graph has a 2-factor with exactly one component. There are many results on the existence of 2-factors with a given number of components, mainly on the existence of hamiltonian graphs, see the survey paper [5]. The line graph $L\!(G)$ of a graph G is the graph with vertex set $E\!(G)$, in which two vertices are adjacent if, and only if, the corresponding edges have a common end vertex in G. The n-time iterated line graph $L^n\!(G)$ is defined to be $L\!(L^{n-1}(G))$, and we assume that $E\!(L^{n-1}(G))$ is not empty. The hamiltonian index of a graph G is the minimum nonnegative integer n such that $L^n\!(G)$ is hamiltonian, denoted by $h\!(G)$, the interested readers can consult [4]. The Hamilton-connected index of a graph G is the minimum nonnegative integer n such that $L^n\!(G)$ is Hamilton-connected, i.e., any two vertices in $L^n\!(G)$ are connected by a Hamilton path. We know that the Hamilton problem, i.e., the problem to decide whether a given graph is hamiltonian, is one of the classical NP-complete problems. In [8], the authors have proved that it is NP-hard to determine whether $L^k\!(G)$ is hamiltonian even for any large integer k. Thus, it is also NP-hard to determine the minimum number of components of a 2-factor in $L^k\!(G)$ for any large integer k. Wang and Xiong [10] provided an upper bound of minimum number of components of 2-factors in iterated line graph. In present paper, we consider the similar problem and determine the minimum number of components of 2-factors in iterated line graph of some special graphs. Before presenting our main results, we first introduce some additional terminology and notation.

A branch is a nontrivial path whose internal vertices have degree two and end vertices have degree other than two. The number of edges in a branch B is said to be its length, denoted by $l\!(B)$. We denote by ${\mathcal {B}}\!(G)$ the set of branches of G. Note that a branch of length one has no internal vertex. A branch B of a graph G is called a cut-branch if the subgraph obtained from G by deleting all edges and internal vertices of B has more components than G, and we denote by ${\mathcal {CB}}\!(G)$ the set of all cut-branches of G. Let ${\mathcal {B}}_2\!(G)=\{ B\in {\mathcal {CB}}\!(G): \mathrm{\ both\ end\ vertices\ of\ } B\mathrm{\ have\ degree\ at\ least\ 3 \ in\ } G \}$ and ${\mathcal {B}}_1\!(G)=\{ B\in {\mathcal {CB}}\!(G): \mathrm{\ at\ least\ one\ end\ vertex\ of\ } B\mathrm{\ has\ degree\ 1\ in\ } G \}$, then it is obvious that ${\mathcal {CB}}\!(G)={\mathcal {B}}_1\!(G)\cup {\mathcal {B}}_2\!(G).$

For $i\in \{1,2\},$ define

$$\begin{aligned} h_i(G)={\left\{ \begin{array}{ll} max\{l(B){:}B\in {\mathcal {B}}_i(G)\},&{}\quad \text {if }{\mathcal {B}}_i(G) \hbox { is not empty,}\\ 0, &{}\quad \text {otherwise}. \end{array}\right. } \end{aligned}$$

Now we state the main results as follows.

Theorem 1

Let G be a connected graph with $h_1\!(G)\le h_2\!(G)-1$ and $h_2\!(G)\ge 2$, such that every nontrivial component of the graph obtained from G by deleting all cut edges and by attaching at least three pendent edges at all vertices of degree at least 3 in G has hamiltonian index at most $h_2\!(G)-1$. Then

• (1) $L^{h_2(G)-2}(G)$ has no 2-factor;

• (2) if $h_2\!(G)\ge 3,$ then the minimum number of components of 2-factors in $L^{h_2(G)-1}(G)$ is

  $$\begin{aligned} \Big |\big \{B\in {\mathcal {B}}_2(G){:}l(B)\in \{h_2(G)-1, h_2(G)\}\big \}\Big |+1; \end{aligned}$$

• (3) the minimum number of components of 2-factors in $L^{h_2(G)}(G)$ is

  $$\begin{aligned} \Big |\big \{B\in {\mathcal {B}}_2(G){:}l(B)= h_2(G)\big \}\Big |+1; \end{aligned}$$

• (4) $L^{h_2(G)+1}(G)$ is hamiltonian.

Theorem 2

Let G be a connected graph with $h_1\!(G)\ge h_2\!(G)\ge 1$, such that every nontrivial component of the graph obtained from G by deleting all cut edges and by attaching at least three pendent edges at all vertices of degree at least three in G has hamiltonian index at most $h_1\!(G)$. Then

• (1) $L^{h_1(G)-1}(G)$ and $L^{h_2(G)-1}(G)$ have no 2-factor;

• (2) if $h_1(G)=h_2(G)\ge 2,$ then the minimum number of components of 2-factors in $L^{h_1(G)}(G)$ is $|\{B\in {\mathcal {B}}_2(G): l(B)= h_2(G)\}|+1;$

• (3) if $h_1(G)=h_2(G)\ge 1,$ then $L^{h_1(G)+1}(G)$ is hamiltonian;

• (4) if $h_1(G)>h_2(G)\ge 1,$ then $L^{h_1(G)}(G)$ is hamiltonian.

The proofs of Theorems 1 and 2 will be given in Sect. 3.

The authors in [4] gave a formula of the hamiltonian index $h\!(T)$ for a tree T. We shall extend the result. Eminjan and Elkin [3] gave a relation between hamiltonian index and Hamilton-connected index of trees.

Since every nontrivial component of the graph obtained from T by deleting all cut edges and by attaching at least three pendent edges at all vertices of degree at least 3 in G has hamiltonian index at most $h_2\!(T)-1$ under the condition $h_1\!(T)\le h_2\!(T)-1$ and $h_2\!(T)\ge 2$, which satisfies the condition of Theorem 1. Let $l=max\{l(B){:}B\in {\mathcal {B}}_2(T)\},$ and ${\mathcal {B}}_2(T)={\mathcal {B}}(T){\setminus }{\mathcal {B}}_1(T)$, then we consequently have the following conclusion.

Corollary 3

Let $l\ge 2$ and let T be a tree with $l(B)\le l-1$ for any $B\in {\mathcal {B}}_1(T)$. Then

• (1) $L^{l-2}(T)$ has no 2-factor;

• (2) if $l\ge 3,$ then the minimum number of components of 2-factors in $L^{l-1}(T)$ is

  $$\begin{aligned} \Big |\big \{B\in {\mathcal {B}}(T)\setminus {\mathcal {B}}_1(T):\big |E(B)\big |\in \{l-1, l\}\big \}\Big |+1; \end{aligned}$$

• (3) the minimum number of components of 2-factors in $L^{l}(T)$ is

  $$\begin{aligned} \Big |\big \{B\in {\mathcal {B}}(T)\setminus {\mathcal {B}}_1(T):\big |E(B)\big |= l\big \}\Big |+1; \end{aligned}$$

• (4) $L^{l+1}(T)$ is hamiltonian.

By Corollary 3 and Theorem 2, we can obtain the following result:

Corollary 4

(Chartrand and Wall [4]) If T is a tree which is not a path, then

$$\begin{aligned} h(T)= max\{h_2(T)+1, h_1(T)\}. \end{aligned}$$

Proof

By Corollary 3 (4), we have $h(T)=h_2(T)+1$ if $h_1(T)\le h_2(T)-1$ and $h_2(T)\ge 2$. Since any tree T with $h_1(T)\ge h_2(T)\ge 1$ satisfies the condition of Theorem 2, then we have $h(T)\le h_2(T)+1$ if $h_1(T)= h_2(T)\ge 1$, and $h(T)\le h_1(T)$ if $h_1(T)> h_2(T)\ge 1$ by (3) and (4) of Theorem 2. Again by Theorem 2 (1), $h(T)>h_1(T)-1$ and $h(T)>h_2(T)-1.$ Above all, we can obtain $h(T)=max\{h_2(T)+1, h_1(T)\}$. $\square$

2 Preliminaries and Notations

As noted in the first section, for graph-theoretic notation not explained in this paper, we refer readers to [1]. Let $G=(V(G),E(G))$ be a graph with vertex set V(G) and edge set E(G). For a nonnegative integer k, we define $V_k(G)$ by $V_k(G)=\{x\in V(G):d_G(x)=k\}$, where $d_G(x)$ is the degree of x in G. Given two subgraphs $G_1$ and $G_2$, we define the distance $d_G(G_1, G_2)$ between $G_1$ and $G_2$ by $d_G(G_1, G_2)=\min \{d_G(x_1, x_2):x_1\in V(G_1), x_2\in V(G_2)\}$. For subgraphs $G_1$, $G_2,\dots , G_k$, their union $G_1\cup G_2\cup \dots \cup G_k$ is the graph whose vertex set and edge set are $V(G_1)\cup V(G_2)\cup \dots \cup V(G_k)$ and $E(G_1)\cup E(G_2)\cup \dots \cup E(G_k)$, respectively. For $S\subseteq V(G)$, we denote by G[S] the subgraph of G induced by S.

A circuit is a connected graph with at least three vertices in which every vertex has even degree. A set of vertices S is said to dominate G if each edge of G has at least one end vertex in S. A circuit of G is called a dominating circuit of G if every edge of G either belongs to the circuit or is adjacent to an edge of the circuit. For a graph G of order at least three, its subgraph H is called a k-system that dominates if it comprises k edge-disjoint stars ($K(1,s),s\ge 3$) and circuits, such that each edge of G is either contained in one of the circuits or stars, or is adjacent to one of the circuits. Harary and Nash-Williams [7] showed that for a connected graph G with at least three edges, L(G) has a hamiltonian cycle if and only if G has a dominating circuit. This characterization has been widely employed to study the properties of cycles in line graphs and iterated line graphs, see [2]. In 1999, Gould and Hynds presented a necessary and sufficient condition for the line graph L(G) of a graph G that has a 2-factor with exactly k components.

Lemma 5

(Gould and Hynds [6]) Let G be a graph such that each component of G has at least three edges. Then L(G) has a 2-factor with exactly k components if and only if G has a k-system that dominates.

We let $EU_n^k(G)$ denote the set of subgraphs H of G satisfying the following five conditions:

• (I) H is an even graph;

• (II) $V_0(H)\subseteq \bigcup \limits _{i\ge 3} V_i(G) \subseteq V(H)$;

• (III) $|E(B)|\le n+1$ for any branch B with $E(B)\cap E(H) =\emptyset$;

• (IV) $|E(B)|\le n$ for any branch B in ${\mathcal {B}}_1(G)$;

• (V) H can be decomposed into at most k pairwise vertex-disjoint subgraphs $H_1,\dots , H_t$ ($t\le k$) such that for every j and for every induced subgraph F of $H_j$ with $\emptyset \ne V(F)\subsetneq V(H_j)$, it holds $d_G(F, H_j-V(F))\le n-1$.

In [11], Xiong and Liu considered iterated line graphs and gave a characterization of the graphs G for which $L^n(G)$ is hamiltonian for $n\ge 2$, which has been used to study the hamiltonian index. We state it as follows.

Lemma 6

(Xiong and Liu [11]) Let G be a connected graph with at least three edges. Then for $n\ge 2$, $L^n(G)$ is hamiltonian if and only if $EU_n^1(G)\not =\emptyset .$

Saito and Xiong in [9] showed that the following result, which extends Lemma 6.

Lemma 7

(Saito and Xiong [9]) Let G be a connected graph with at least three edges and let k be a positive integer. Then for $n\ge 2$, $L^n(G)$ has a 2-factor with at most k components if and only if $EU_n^k(G)\not =\emptyset .$

Next, we provide the proofs of Theorem 1 and Theorem 2.

3 Proofs of Main Results

Proof of Theorem 1

(1) Let $B\in {\mathcal {B}}_2(G)$ and $l(B) = h_2(G).$ Then B becomes a branch $B^{\prime }$ of length 2 in $L^{h_2(G)-2}(G),$ whose end vertices belong to two distinct components. Let u be the vertex of degree 2 in $B^{\prime }$, then u does not belong to any 2-factor component of $L^{h_2(G)-2}(G).$ Thus, $L^{h_2(G)-2}(G)$ has no 2-factor. $\square$

(2) Assume that the minimum number of components of 2-factors in $L^{h_2(G)-1}(G)$ is k and let $\big |\big \{B\in {\mathcal {B}}_2(G): l(B)\in \{h_2(G)-1, h_2(G)\}\big \}\big |=s.$

Let $G_1, G_2,\ldots , G_t$ be the components of the graph obtained from G by deleting all edges and inner vertices of any branch in $\big \{B\in {\mathcal {B}}_2(G): l(B)\in \{h_2(G)-1, h_2(G)\}\big \}.$ It is obvious that $G_1, G_2,\ldots , G_t$ are pairwise vertex-disjoint and $l(B)\le h_2(G)-2$ for any branch $B\in {\mathcal {B}}_2(G_j)\cap {\mathcal {B}}_2(G), 1\le j\le t.$ Contracting $G_j$ to a vertex, the resulting graph becomes a tree, then $t=s+1$ since $|V(T)|=|E(T)|+1$ in a tree T. Since any possible 2-factor has at least one component in every $L^{h_2(G)-1}(G_j),$ then $k\ge s+1.$ Now we verify that $k\le s+1.$ By Lemma 7, we need to prove that $EU_{h_2(G)-1}^{s+1}(G)\not =\emptyset .$

Assume that every $G_j~(1\le j\le s+1)$ is composed of $i_j$ nontrivial components $G_{j,1},\;G_{j,2},\ldots ,G_{j,i_{j}}$ of the graph obtained from G by deleting all cut edges and by attaching at least three pendent edges at all vertices of degree at least 3 in G, i.e., $G_j=\bigcup _{k=1}^{i_j}G_{j,~k}.$ Then $G_{j,1},G_{j,2},\ldots ,G_{j,i_{j}}$ are pairwise vertex-disjoint and $h(G_{j,k})\le h_2(G)-1$ for $1\le k\le i_j, 1\le j\le s+1$ by the hypothesis of Theorem 1. By Lemma 6, $EU_{h_2(G)-1}^{1}(G_{j,k})\not =\emptyset .$ Suppose $H_{j,k}\in EU_{h_2(G)-1}^{1}(G_{j,k}),~ 1\le j\le s+1,~1\le k\le i_j.$ Then $H_{j,k}$ satisfies (I)-(V) in the definition of $EU_{h_2(G)-1}^{1}(G_{j,k})$, and $H_{j,1},\; H_{j,2},\; \ldots ,\;H_{j,i_j}$ are pairwise vertex-disjoint. Let $H_j=\bigcup _{k=1}^{i_j}H_{j,k}, 1\le j\le s+1.$ Next, we verify that $H_{j}\in EU_{h_2(G)-1}^{1}(G_{j}),~ 1\le j\le s+1.$

Since $H_{j,k}$ is even graph and ${H_{j,k}}^{,}s$ are mutually vertex-disjoint subgraphs, then $H_j$ is even graph, and (I) follows. Since $V_0(H_{j,k})\subseteq \bigcup _{i\ge 3}V_i(G_{j,k})\subseteq V(H_{j,k})$, $1\le k\le i_j,\;\bigcup _{k=1}^{i_j}V_0(H_{j,k})\subseteq \bigcup _{k=1}^{i_j}\bigcup _{i\ge 3}V_i(G_{j,k})\subseteq \bigcup _{k=1}^{i_j}V(H_{j,k}),$ then $V_0(H_j)\subseteq \bigcup _{i\ge 3}V_i(G_j)\subseteq V(H_j),$ and (II) follows. Let $B^{\prime }$ be any branch with $E(B^{\prime })\cap E(H_j)=\emptyset .$ Assume that $|E(B^{\prime })|\ge h_2(G)+1,$ which contradicts that $H_{j,k}$ satisfies (III) of the definition of $EU_{h_2(G)-1}^{1}(G_{j,k})$, then $|E(B^{\prime })|\le h_2(G)$, and (III) follows. By the hypothesis of Theorem 1, we have $|E(B^{\prime \prime })|\le h_2(G)-1$ for any branch $B^{\prime \prime }\in {\mathcal {B}}_1(G_j)$, and (IV) follows. For any induced subgraph F of $H_j$ with $\emptyset \not = V(F)\subsetneq V(H_{j}).$ If $F\cap H_{j,k}\not = \emptyset ,$ then $d_{G_{j}}(F, H_{j}-V(F))= d_{G_{j,k}}(F, H_{j,k}-V(F))\le h_2(G)-2$ by the fact that $H_{j,k}$ satisfies (V) in the definition of $EU_{h_2(G)-1}^{1}(G_{j,k}).$ If $F\cap H_{j,k}= \emptyset ,$ then $d_{G_{j}}(F, H_{j}-V(F))\le h_2(G)-2$. Thus, $d_{G_{j}}(F, H_{j}-V(F))\le h_2(G)-2$ for any induced subgraph F of $H_j$ with $\emptyset \not = V(F)\subsetneq V(H_{j}),$ and (V) follows. Then $H_{j}\in EU_{h_2(G)-1}^{1}(G_{j}),\;1\le j \le s+1.$ Then $H_j$ satisfies (I)-(V) in the definition of $EU_{h_2(G)-1}^{1}(G_j)$, and $H_1, H_2, \ldots , H_{s+1}$ are pairwise vertex-disjoint. Let $H=\bigcup _{j=1}^{s+1}H_j.$ Then we shall prove $H\in EU_{h_2(G)-1}^{s+1}(G)$, i.e., we shall show that H satisfies the conditions (I)-(V) for $n=h_2(G)-1$ and $k=s+1$ in Lemma 7.

Since $H_j$ is even graph and disjoint union of even graphs is still even graph, H is even graph, and (I) follows. Further, $V_0(H_j)\subseteq \bigcup _{i\ge 3}V_i(G_j)\subseteq V(H_j)$, $1\le j\le s+1,\;\bigcup _{j=1}^{s+1}V_0(H_j)\subseteq \bigcup _{j=1}^{s+1}\bigcup _{i\ge 3}V_i(G_j)\subseteq \bigcup _{j=1}^{s+1}V(H_j),$ then $V_0(H)\subseteq \bigcup _{i\ge 3}V_i(G)\subseteq V(H),$ and (II) follows. Let B be a branch of G with $E(B)\cap E(H)=\emptyset .$ Assume that $|E(B)|\ge h_2(G)+1,$ which contradicts $h_2(G)= max\{l(B): B\in {\mathcal {B}}_2(G)\},$ hence (III) follows. (IV) is obvious by the hypothesis of Theorem 1.

Next, we verify that H satisfies (V). By the construction of H, H can be decomposed into $s+1$ pairwise vertex-disjoint subgraphs $H_1, H_2, \ldots , H_{s+1},$ and for every j and for every induced subgraph F of $H_j$ with $\emptyset \not = V(F)\subsetneq V(H_j),$ we have $d_{G_j}(F, H_j-V(F))\le h_2(G)-2$ by the condition (V) of $EU_{h_2(G)-1}^{1}(G_{j}).$ Furthermore, $d_G(F, H_j-V(F))=d_{G_j}(F, H_j-V(F))\le h_2(G)-2$ for every j and the same F, then (V) follows. Hence, $H\in EU_{h_2(G)-1}^{s+1}(G).$ Therefore, $k=s+1.$ $\square$

(3) The proof is similar to the proof of (2). Assume that the minimum number of components of 2-factors in $L^{h_2(G)}(G)$ is $k^{\prime }$ and let $\big |\big \{B\in {\mathcal {B}}_2(G): l(B)= h_2(G)\big \}\big |=s^{\prime }.$

Let $G_1^{\prime }, G_2^{\prime }, \ldots , G_t^{\prime }$ be the components of the graph obtained from G by deleting all edges and inner vertices of any branch in $\big \{B\in {\mathcal {B}}_2(G):l(B)=h_2(G)\big \},$ then $G_1^{\prime }, G_2^{\prime }, \ldots , G_t^{\prime }$ are pairwise vertex-disjoint and $l(B)\le h_2(G)-1$ for any branch $B\in {\mathcal {B}}_2(G_j^{\prime })\cap {\mathcal {B}}_2(G), 1\le j\le t.$ Contracting $G_j^{\prime }$ to a vertex, the resulting graph becomes a tree, then $t=s^{\prime }+1.$ Since any possible 2-factor has at least one component in every $L^{h_2(G)}(G_j^{\prime }),$ then $k^{\prime }\ge s^{\prime }+1.$ It remains to verify that $k^{\prime }\le s^{\prime }+1.$ By Lemma 7, we need to prove that $EU_{h_2(G)}^{s^{\prime }+1}(G)\not =\emptyset .$

Assume that every $G_j^{\prime }~(1\le j\le s^{\prime }+1)$ is composed of $i_j$ nontrivial components $G_{j,1}^{\prime },\;G_{j,2}^{\prime },\;\ldots ,\;G_{j,i_{j}}^{\prime }$ of the graph obtained from G by deleting all cut edges and by attaching at least three pendent edges at all vertices of degree at least 3 in G, i.e., $G_j^{\prime }=\bigcup _{q=1}^{i_j}G_{j,~q}^{\prime }.$ Then $G_{j,1}^{\prime },G_{j,2}^{\prime },\ldots ,G_{j,i_{j}}^{\prime }$ are pairwise vertex-disjoint, and $h(G_{j,q}^{\prime })\le h_2(G)-1$ for $1\le q\le i_j, 1\le j\le s^{\prime }+1$ by the hypothesis of Theorem 1. This means that $L^{h_2(G)-1}(G_{j,q}^{\prime })$ has a hamiltonian cycle. Then $L^{h_2(G)}(G_{j,q}^{\prime })$ is hamiltonian. By Lemma 6, there exists a subgraph $H_{j,q}^{\prime }$ of $G_{j,q}^{\prime }$ such that $H_{j,q}^{\prime }\in EU_{h_2(G)}^{1}(G_{j,q}^{\prime }),~1\le q\le i_j,~ 1\le j\le s^{\prime }+1.$ Let $H_j^{\prime }=\bigcup _{q=1}^{i_j}H_{j,q}^{\prime }, 1\le j\le s^{\prime }+1.$ Now, we shall show that $H_{j}^{\prime }\in EU_{h_2(G)}^{1}(G_{j}^{\prime }),~ 1\le j\le s^{\prime }+1.$

Since ${H_{j,q}^{\prime }}^{,}s$ ($1\le j\le s^{\prime }+1,\;1\le q\le i_j$) are mutually vertex-disjoint even subgraphs, and hence $H_j^{\prime }$ is also even subgraph, and (I) follows. Since $V_0(H_{j,q}^{\prime })\subseteq \bigcup _{i\ge 3}V_i(G_{j,q}^{\prime })\subseteq V(H_{j,q}^{\prime })$, $1\le q\le i_j,\;\bigcup _{q=1}^{i_j}V_0(H_{j,q}^{\prime })\subseteq \bigcup _{q=1}^{i_j}\bigcup _{i\ge 3}V_i(G_{j,q}^{\prime })\subseteq \bigcup _{q=1}^{i_j}V(H_{j,q}^{\prime }),$ then $V_0(H_j^{\prime })\subseteq \bigcup _{i\ge 3}V_i(G_j^{\prime })\subseteq V(H_j^{\prime }),$ and (II) follows. Let $B^{\prime }$ be any branch with $E(B^{\prime })\cap E(H_j^{\prime })=\emptyset .$ Assume that $|E(B^{\prime })|\ge h_2(G)+2,$ which contradicts the fact that $H_{j,q}^{\prime }$ satisfies (III) of the definition of $EU_{h_2(G)}^{1}(G_{j,q}^{\prime })$, then $|E(B^{\prime })|\le h_2(G)+1$, and hence (III) follows. (IV) is obviously true by the hypothesis of Theorem 1. By the fact that $H_{j,q}^{\prime }$ satisfies (V) of the definition of $EU_{h_2(G)}^{1}(G_{j,q}^{\prime })$, we have $d_{G_{j}^{\prime }}(F, H_{j}^{\prime }-V(F))= d_{G_{j,q}^{\prime }}(F, H_{j,q}^{\prime }-V(F))\le h_2(G)-1$ for any induced subgraph F of $H_j^{\prime }$ with $\emptyset \not = V(F)\subsetneq V(H_{j}^{\prime })$ if $F\cap H_{j,q}^{\prime }\not = \emptyset$, and $d_{G_{j}^{\prime }}(F, H_{j}^{\prime }-V(F))\le h_2(G)-1$ if $F\cap H_{j,q}^{\prime }= \emptyset$. Thus, $d_{G_{j}^{\prime }}(F, H_{j}^{\prime }-V(F))\le h_2(G)-1$ for any induced subgraph F of $H_j^{\prime }$ with $\emptyset \not = V(F)\subsetneq V(H_{j}^{\prime }),$ and (V) follows. Then $H_{j}^{\prime }\in EU_{h_2(G)}^{1}(G_{j}^{\prime }),\;1\le j \le s^{\prime }+1.$ Let $H^{\prime }=\bigcup _{j=1}^{s^{\prime }+1}H_j^{\prime }.$ Then we shall verify that $H^{\prime }$ satisfies the conditions (I)-(V) in the definition of $EU_{h_2(G)}^{s^{\prime }+1}(G).$

$H^{\prime }$ is even graph since it is composed of those vertex-disjoint even subgraphs ${H_j^{\prime }}$, and (I) follows. Since $H_j^{\prime }$ satisfies (II) in the definition of $EU_{h_2(G)}^{1}(G_j^{\prime }),$ we have $V_0(H_j^{\prime })\subseteq \bigcup _{i\ge 3}V_i(G_j^{\prime })\subseteq V(H_j^{\prime })$, $1\le j\le s^{\prime }+1.$ Then $\bigcup _{j=1}^{s^{\prime }+1}V_0(H_j^{\prime })\subseteq \bigcup _{j=1}^{s^{\prime }+1}\bigcup _{i\ge 3}V_i(G_j^{\prime })\subseteq \bigcup _{j=1}^{s^{\prime }+1}V(H_j^{\prime }),$ that is, $V_0(H^{\prime })\subseteq \bigcup _{i\ge 3}V_i(G)\subseteq V(H^{\prime }),$ and (II) follows. Let B be any branch of G with $E(B)\cap E(H^{\prime })=\emptyset .$ Then $|E(B)|\le h_2(G)\le h_2(G)+1,$ (III) follows. (IV) is obvious by the hypothesis of Theorem 1.

By the construction of $H^{\prime }$ and $H_j^{\prime }$ satisfies (V) in the definition of $EU_{h_2(G)}^{1}(G_j^{\prime }),\;H^{\prime }$ can be decomposed into $s^{\prime }+1$ pairwise vertex-disjoint subgraphs $H_1^{\prime }, H_2^{\prime }, \cdots , H_{s^{\prime }+1}^{\prime },$ and for every j and for every induced subgraph F of $H_j^{\prime }$ with $\emptyset \not = V(F)\subsetneq V(H_j^{\prime }),$ it holds $d_{G_j^{\prime }}(F, H_j^{\prime }-V(F))\le h_2(G)-1.$ Since $d_G(F, H_j^{\prime }-V(F))=d_{G_j^{\prime }}(F, H_j^{\prime }-V(F))\le h_2(G)-1$ for every j and the same F, $H^{\prime }$ satisfies (V). Hence, $H^{\prime }\in EU_{h_2(G)}^{s^{\prime }+1}(G).$ $\square$

(4) By Theorem 1 (2), we have that $L^{h_2(G)-1}(G)$ has a 2-factor which has at least $s+1$ components. Let $L^{h_2(G)-1}(G)=G^*$, we need to prove that $EU_{2}^{1}(G^*)\not =\emptyset .$

Assume that $H_j$ is a 2-factor component of $L^{h_2(G)-1}(G),~1\le j\le s+1.$ Let $H=\bigcup _{j=1}^{s+1}H_j$. Then H is a 2-factor of $G^*$ since $\{H_j,~1\le j\le s+1\}$ is pairwise vertex-disjoint. It is obvious that H is an even graph of $G^*$, and (I) follows. Since $V_0(H)=\emptyset$ and H is a 2-factor of $G^*$, $\emptyset =V_0(H)\subseteq \bigcup _{i\ge 3}V_i(G^*)\subseteq V(H)$, and (II) follows. Let B be a branch with $E(B)\cap E(H)=\emptyset .$ Then $|E(B)|=1\le 2+1$ by H is a 2-factor of $G^*$, hence (III) follows. (IV) holds by the hypothesis of Theorem 1. Now, we prove (V). Assume, by contradiction, that $d_{G^*}(F, H-V\!(F))\ge 2$ for some induced subgraph F of H with $\emptyset \not = V\!(F)\subsetneq V\!(H).$ This implies that there exists a branch B of length at least 2 between F and $H-F$. Let x be the vertex of degree 2 in B, then $x\in V(B)\setminus V(H)$. This contradicts the fact that H is a 2-factor of $G^*$, hence H satisfies (V). Therefore, $H\in EU_{2}^{1}(G^*).$ By Lemma 6, $L^{2}(G^*)=L^{h_2(G)+1}(G)$ is hamiltonian. $\square$

Proof of Theorem 2

(1) Let $B\in {\mathcal {B}}_1(G)$ and $l(B) = h_1(G).$ Then B becomes a branch $B^{\prime }$ with an end vertex of degree 1 in $L^{h_1(G)-1}(G)$ and $L^{h_2(G)-1}(G)$, respectively. Therefore, $L^{h_1(G)-1}(G)$ and $L^{h_2(G)-1}(G)$ have no 2-factor. $\square$

(2) Assume that the minimum number of components of 2-factors in $L^{h_1(G)}(G)$ is $k^{\prime \prime }$ and let $\big |\{B\in {\mathcal {B}}_2(G): l(B)= h_2(G)=h_1(G)\}\big |=s^{\prime \prime }.$

Let $G_1^{\prime \prime }, G_2^{\prime \prime }, \ldots , G_t^{\prime \prime }$ be the components of the graph obtained from G by deleting all edges and internal vertices of any branch in $\big \{B\in {\mathcal {B}}_2(G): l(B)= h_2(G)=h_1(G)\ge 2 \big \}.$ Then $G_1^{\prime \prime }, G_2^{\prime \prime }, \ldots , G_t^{\prime \prime }$ are pairwise vertex-disjoint, and $l(B)\le h_1(G)-1$ for any branch $B\in {\mathcal {B}}_2(G_j^{\prime \prime })\cap {\mathcal {B}}_2(G), 1\le j\le t.$ Contracting $G_j^{\prime \prime }$ to a vertex, the resulting graph becomes a tree, then $t=s^{\prime \prime }+1$ since $|V(T)|=|E(T)|+1$ in a tree T. Since any possible 2-factor has at least one component in every $L^{h_1(G)}(G_j^{\prime \prime }),$ then $k^{\prime \prime }\ge s^{\prime \prime }+1.$ Next, we shall prove that $k^{\prime \prime }\le s^{\prime \prime }+1.$ According to Lemma 7, we shall prove that $EU_{h_1(G)}^{s^{\prime \prime }+1}(G)\not =\emptyset .$

Assume that every $G_j^{\prime \prime }~(1\le j\le s^{\prime \prime }+1)$ comprises $i_j$ nontrivial components $G_{j,1}^{\prime \prime },\;G_{j,2}^{\prime \prime },\;\ldots ,\;G_{j,i_{j}}^{\prime \prime }$ of the graph obtained from G by deleting all cut edges and by attaching at least three pendent edges at all vertices of degree at least 3 in G, that is, $G_j^{\prime \prime }=\bigcup _{q=1}^{i_j}G_{j,q}^{\prime \prime }.$ It is obvious that $G_{j,1}^{\prime \prime },G_{j,2}^{\prime \prime },\ldots ,G_{j,i_{j}}^{\prime \prime }$ are pairwise vertex-disjoint. By the hypothesis of Theorem 2, $h(G_{j,q}^{\prime \prime })\le h_1(G)$ for $1\le q\le i_j, 1\le j\le s^{\prime \prime }+1.$ By Lemma 6, $EU_{h_1(G)}^{1}(G_{j,q}^{\prime \prime })\not =\emptyset .$ We may take $H_{j,q}^{\prime \prime }\in EU_{h_1(G)}^{1}(G_{j,q}^{\prime \prime }),\;1\le j\le s^{\prime \prime }+1,~1\le q\le i_j,$ then $H_{j,q}^{\prime \prime }$ satisfies (I)-(V) in the definition of $EU_{h_1(G)}^{1}(G_{j,q}^{\prime \prime })$, and ${H_{j,q}^{\prime \prime }}^{,}s$ are mutually vertex-disjoint subgraphs. Let $H_j^{\prime \prime }=\bigcup _{q=1}^{i_j}H_{j,q}^{\prime \prime }, 1\le j\le s^{\prime \prime }+1.$ Next, we shall show that $H_{j}^{\prime \prime }\in EU_{h_1(G)}^{1}(G_{j}^{\prime \prime }),\;1\le j\le s^{\prime \prime }+1.$

Since ${H_{j,q}^{\prime \prime }}^{,}s$ are mutually vertex-disjoint even subgraphs, $H_j^{\prime \prime }$ is even graph, and (I) follows. Since $V_0(H_{j,q}^{\prime \prime })\subseteq \bigcup _{i\ge 3}V_i(G_{j,q}^{\prime \prime })\subseteq V(H_{j,q}^{\prime \prime })$, $1\le q\le i_j,$ and $\bigcup _{q=1}^{i_j}V_0(H_{j,q}^{\prime \prime })\;\subseteq \; \bigcup _{q=1}^{i_j}\bigcup _{i\ge 3}V_i(G_{j,q}^{\prime \prime })=\bigcup _{i\ge 3}V_i(G_j^{\prime \prime })\;\subseteq \; \bigcup _{q=1}^{i_j}V(H_{j,q}^{\prime \prime }),$ then $V_0(H_j^{\prime \prime })\subseteq \bigcup _{i\ge 3}V_i(G_j^{\prime \prime })\subseteq V(H_j^{\prime \prime }),$ and (II) follows. Let $B^{\prime }$ be any branch with $E(B^{\prime })\cap E(H_j^{\prime \prime })=\emptyset .$ Assume that $|E(B^{\prime })|\ge h_1(G)+2,$ which contradicts the fact that $H_{j,q}^{\prime \prime }$ satisfies (III) of $EU_{h_1(G)}^{1}(G_{j,q}^{\prime \prime })$, then $|E(B^{\prime })|\le h_1(G)+1$, and (III) follows. (IV) is obvious by the hypothesis of Theorem 2. By the fact that $H_{j,q}^{\prime \prime }$ satisfies (V) of the definition of $EU_{h_1(G)}^{1}(G_{j,q}^{\prime \prime })$, we have $d_{G_{j}^{\prime \prime }}(F, H_{j}^{\prime \prime }-V(F))= d_{G_{j,q}^{\prime \prime }}(F, H_{j,q}^{\prime \prime }-V(F))\le h_1(G)-1$ for any induced subgraph F of $H_j^{\prime \prime }$ with $\emptyset \not = V(F)\subsetneq V(H_{j}^{\prime \prime })$ when $F\cap H_{j,q}^{\prime \prime }\not = \emptyset$, and $d_{G_{j}^{\prime \prime }}(F, H_{j}^{\prime \prime }-V(F))\le h_1(G)-1$ when $F\cap H_{j,q}^{\prime \prime }= \emptyset$. Then $d_{G_{j}^{\prime \prime }}(F, H_{j}^{\prime \prime }-V(F))\le h_1(G)-1$ for any induced subgraph F of $H_j^{\prime \prime }$ with $\emptyset \not = V(F)\subsetneq V(H_{j}^{\prime \prime }),$ and hence (V) follows. Then $H_{j}^{\prime \prime }\in EU_{h_1(G)}^{1}(G_{j}^{\prime \prime }),\;1\le j \le s^{\prime \prime }+1.$ Let $H^{\prime \prime }=\bigcup _{j=1}^{s^{\prime \prime }+1}H_j^{\prime \prime }.$ Next, we show that $H^{\prime \prime }\in EU_{h_1(G)}^{s^{\prime \prime }+1}(G)$.

Since $H_j^{\prime \prime }$ satisfies (I) and (II) of the definition of $EU_{h_1(G)}^{1}(G_j^{\prime \prime })$, it is easy to deduce $H^{\prime \prime }$ is even graph and $V_0(H^{\prime \prime })\subseteq \bigcup _{i\ge 3}V_i(G)\subseteq V(H^{\prime \prime }),$ then (I) (II) follow. Let B be a branch of G with $E(B)\cap E(H^{\prime \prime })=\emptyset .$ Then $|E(B)|\le h_2(G)\le h_2(G)+1=h_1(G)+1,$ and (III) follows. It is obvious that $|E(B)|\le h_1(G)$ for any branch B in ${\mathcal {B}}_1(G)$, (IV) follows.

Now, we verify that (V) holds. By the construction of $H^{\prime \prime }$, $H^{\prime \prime }$ can be decomposed into $s^{\prime \prime }+1$ pairwise vertex-disjoint subgraphs $H_1^{\prime \prime }, H_2^{\prime \prime }, \cdots , H_{s^{\prime \prime }+1}^{\prime \prime },$ and for every j and for every induced subgraph F of $H_j^{\prime \prime }$ with $\emptyset \not = V(F)\subsetneq V(H_j^{\prime \prime }),$ we have $d_{G_j^{\prime \prime }}(F, H_j^{\prime \prime }-V(F))\le h_1(G)-1$ by the fact that $H_j^{\prime \prime }$ satisfies (V) of the definition of $EU_{h_1(G)}^{1}(G_{j}^{\prime \prime }).$ Furthermore, $d_G(F, H_j^{\prime \prime }-V(F))=d_{G_j^{\prime \prime }}(F, H_j^{\prime \prime }-V(F))\le h_1(G)-1$ for every j and the same F, then (V) follows. Hence, $H^{\prime \prime }\in EU_{h_1(G)}^{s^{\prime \prime }+1}(G).$ $\square$

(3) According to Lemma 6, we need to prove that $EU_{h_1(G)+1}^{1}(G)\not =\emptyset$ when $h_1(G)=h_2(G)\ge 1$. Let $G_1,G_2,\ldots ,G_q$ be those nontrivial components of the graph obtained from G by deleting all cut edges and by attaching at least three pendent edges at all vertices of degree at least 3 in G. Then we have $h(G_i)\le h_1(G)$ by the hypothesis of Theorem 2. It implies that $L^{h_1(G)+1}(G_i)$ is hamiltonian. By Lemma 6, there exists a subgraph $H_i\in EU_{h_1(G)+1}^{1}(G_i)$, $1\le i \le q$. Then $H_1,\;H_2$, $\ldots ,H_q$ (some of these subgraphs maybe isolated vertices because there maybe some vertices ${v_k}^{,}s$ in G such that $d_G(v_k)\ge 3$ and the edges incident to $v_k$ are all cut edges) are pairwise vertex-disjoint. Let $H=\bigcup _{i=1}^{q}H_i.$ Then H is even graph since ${H_{i}}^{,}s$ are mutually vertex-disjoint even subgraphs, and (I) follows. Since $V_0(H_i)\subseteq \bigcup _{i\ge 3}V_i(G_i)\subseteq V(H_i)$ ($1\le i \le q$), $\bigcup _{i=1}^{q}V_0(H_i)\subseteq \bigcup _{i=1}^{q}\bigcup _{i\ge 3}V_i(G_i)=\bigcup _{i\ge 3}V_i(G)\subseteq \bigcup _{i=1}^{q}V(H_i),$ then $V_0(H)\subseteq \bigcup _{i\ge 3}V_i(G)\subseteq V(H),$ (II) follows. Since $|E(B^{\prime })|\le h_1(G)+2$ for any branch $B^{\prime }$ with $E(B^{\prime })\cap E(H_i)=\emptyset$ by $H_i$ satisfies (III) in the definition of $EU_{h_1(G)+1}^{1}(G_i),$ and $|E(B^{\prime \prime })|\le h_2(G)=h_1(G)\le h_1(G)+2$ for any branch $B^{\prime \prime }\in {\mathcal {B}}_2(G)$, then $|E(B)|\le h_1(G)+2$ for any branch B of G with $E(B)\cap E(H)=\emptyset ,$ and (III) follows. (IV) follows from the fact that $|E(B)|\le h_1(G)\le h_1(G)+1$ for any branch B in ${\mathcal {B}}_1(G)$. For any induced subgraph F of H with $\emptyset \not = V(F)\subsetneq V(H),$ we have $d_{G}(F, H-V(F))= d_{G_i}(F, H_{i}-V(F))\le h_1(G)$ if $F\cap H_{i}\not = \emptyset$ by the fact that $H_{i}$ (Here $H_i$ is the subgraph which is not an isolated vertex) satisfies (V) of the definition of $EU_{h_1(G)+1}^{1}(G_{i}).$ If $F\cap H_{i}= \emptyset ,$ then $d_{G}(F, H-V(F))\le h_2(G)=h_1(G)$. Thus, $d_{G}(F, H-V(F))\le h_1(G)$ for any induced subgraph F of H with $\emptyset \not = V(F)\subsetneq V(H),$ and (V) follows. Then $H\in EU_{h_1(G)+1}^{1}(G).$ $\square$

(4) Similar to the proof of Theorem 2 (3), we shall show that $EU_{h_1(G)}^{1}(G)\not =\emptyset$ when $h_1(G)>h_2(G)\ge 1$. We use $G_i^{\prime }$ ($1\le i \le q$) to denote those components of the graph obtained from G by deleting all cut edges and by attaching at least three pendent edges at all vertices of degree at least 3 in G. Then $h(G_i^{\prime })\le h_1(G)$ by the hypothesis of Theorem 2. This means that $L^{h_1(G)}(G_i^{\prime })$ is hamiltonian. By Lemma 6, $EU_{h_1(G)}^{1}(G_i^{\prime }) \not = \emptyset .$ Let $H_i^{\prime }\in EU_{h_1(G)}^{1}(G_i^{\prime })$, $1\le i \le q$, then $H_1^{\prime },\;H_2^{\prime }$, $\ldots ,H_q^{\prime }$ (some of these subgraphs maybe isolated vertices) are pairwise vertex-disjoint. Let $H^{\prime }=\bigcup _{i=1}^{q}H_i^{\prime }.$ Then $H^{\prime }$ is even graph and $V_0(H^{\prime })\subseteq \bigcup _{i\ge 3}V_i(G)\subseteq V(H^{\prime })$ by the fact that $H_i^{\prime }$ satisfies the conditions (I) and (II) in the definition of $EU_{h_1(G)}^{1}(G_i^{\prime })$, and hence (I), (II) follow. Since $|E(B^{\prime })|\le h_1(G)+1$ for any branch $B^{\prime }$ with $E(B^{\prime })\cap E(H_i^{\prime })=\emptyset$ by the fact that $H_i^{\prime }$ satisfies (III) in the definition of $EU_{h_1(G)}^{1}(G_i^{\prime }),$ and $|E(B^{\prime \prime })|\le h_2(G)< h_1(G)+1$ for any branch $B^{\prime \prime }\in {\mathcal {B}}_2(G)$, then $|E(B)|\le h_1(G)+1$ for any branch B of G with $E(B)\cap E(H^{\prime })=\emptyset ,$ and (III) follows. (IV) is obviously true. By the fact that $H_i^{\prime }$ (Here $H_i^{\prime }$ is the subgraph which is not an isolated vertex) satisfies the condition (V) in the definition of $EU_{h_1(G)}^{1}(G_i^{\prime }),$ we have $d_{G_i^{\prime }}(F, H_i^{\prime }-V(F))\le h_1(G)-1$ for every induced subgraph F of $H_i^{\prime }$ with $\emptyset \not = V(F)\subsetneq V(H_i^{\prime })$. Then $d_{G}(F, H^{\prime }-V(F))= d_{G_i^{\prime }}(F, H_{i}^{\prime }-V(F))\le h_1(G)-1$ for any induced subgraph F of $H^{\prime }$ if $F\cap H_{i}^{\prime }\not = \emptyset$. If $F\cap H_{i}^{\prime }= \emptyset ,$ then $d_{G}(F, H^{\prime }-V(F))\le h_2(G)\le h_1(G)-1$. Thus, $d_{G}(F, H^{\prime }-V(F))\le h_1(G)-1$ for any induced subgraph F of $H^{\prime }$ with $\emptyset \not = V(F)\subsetneq V(H^{\prime }),$ and hence (V) follows. Therefore, $H^{\prime }$ $\in EU_{h_1(G)}^{1}(G)$. $\square$

4 Remark

Our results in this paper provide two classes of graphs, such that as long as they satisfy the condition of Theorems 1 or 2, their iterated line graph has a 2-factor, and we determine the minimum number of components of 2-factors. Note that Theorem 1 is best possible in the sense: (1) The condition $h_2(G)\ge 2$ can not be replaced by $h_2(G)=1$. Otherwise, we have $h_1(G)=h_2(G)-1=0$. However, the graphs satisfying the condition “$h_1(G)=0,~ h_2(G)=1$” could not reach the condition “every nontrivial component of the graph obtained from G by deleting all cut edges and by attaching at least three pendent edges at all vertices of degree at least 3 in G has hamiltonian index at most $h_2(G)-1$”. (2) The condition $h_2(G)\ge 3$ in Theorem 1 (2) can not be replaced by $h_2(G)\le 2$. This can be seen from the graph $G^*$ in Fig. 1 with $h_2(G^*)=2$, but there is no 2-factor in $L^{h_2(G^*)-1}(G^*)=L(G^*)$. In addition, Theorem 2 is also best possible in the sense: The condition $h_1(G)=h_2(G)\ge 2$ in Theorem 2 (2) can not be replaced by $h_1(G)=h_2(G)=1$. This can be seen from the graph $G^{**}$ in Fig. 1 with $h_1(G^{**})=h_2(G^{**})=1$, but there is no 2-factor in $L(G^{**})$.

Fig. 1

Condition (2) in both Theorems 1 and 2 is sharp