Characterization of Forbidden Subgraphs for the Existence of Even Factors in a Graph

Home , Bipartite graph, Complete bipartite graph

Discrete Applied Mathematics 223 (2017) 135–139

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Discrete Applied Mathematics

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Characterization of forbidden subgraphs for the existence of even factors in a graph Liming Xiong School of Mathematics and Statistics, Beijing Key Laboratory on MCAACI, Beijing Institute of Technology, Beijing 100081, PR China article info a b s t r a c t ∼ Article history: In 2008, it is shown that if H = K1,3 (the claw), then every H-free graph G has an even Received 25 October 2015 factor if and only if δ(G) ≥ 2 and every odd branch-bond of G having an edge-branch. In Received in revised form 15 February 2017 this paper, we characterize all connected graphs H with order at least three such that every Accepted 16 February 2017 H-free graph has an even factor if and only if δ(G) ≥ 2 and every odd branch-bond of G has Available online 9 March 2017 an edge-branch. © 2017 Elsevier B.V. All rights reserved. Keywords: Even factor Forbidden graph Chair-free Claw-free Branch-bond

1. Introduction

All graphs considered in this paper are simple graphs. For the notation or terminology not defined here, see [2]. We say that a graph G is H-free if G contains no induced subgraph isomorphic to H; H is also called a forbidden subgraph. We often require H to be connected; if H has order two, then H-free graph is a graph without any edge; these graphs are of little interest. Thus, we need H to have order at least three. Using a forbidden subgraph condition to characterize the graphs with some properties is a common one; while it often imposes some necessary conditions, for example, hamiltonian graphs with 2-connectivity, traceable graphs with connectivity, 2-factor graph with minimum degree at least two; see the following three theorems. Here a path Pk is a path of order k.

Theorem 1 (Faudree and Gould, [3]). Suppose A is a connected graph of order at least 3. Then every 2-connected A-free graph is hamiltonian if and only if A = P3.

Theorem 2 (Faudree and Gould, [3]). Suppose A is a connected graph of order at least 3. Then every connected A-free is traceable if and only if A = P3. Although a graph to be hamiltonian (must have a 2-factor, with exactly one component) seems to be stronger than it has a 2-factor (may not be hamiltonian), the following result shows the forbidden subgraph condition is the same (comparing it with Theorem 1).

Theorem 3 (Aldred, Fujisawa and Saito, [1]). Suppose A is a connected graph of order at least 3. Then every connected graph with δ(G) ≥ 2 contains 2-factor if and only if A = P3.

In [5], Lai considered the same problem in a slight different way. A graph is called supereulerian if it has a spanning eulerian subgraph. A graph H is called an induced minor if it is isomorphic to a contraction image of an induced subgraph of G.A wheel Wn is the graph obtained from an n-cycle Cn = u1u2 ... unu1 and an additional vertex u by joining u with ′ ui for all i. Define the subdivided wheel Wn to be the graph obtained from Wn by subdividing each edge in E(Cn) once. Let = { ′ : ≥ } Ω Wn n 2 . Note that 2-edge connectivity is a necessary condition for a graph to be supereulerian.

Theorem 4 (Lai, [5]). Let G be a 2-edge connected graph. Then every 2-edge connected induced subgraph of G is supereulerian if, and only if, G has no induced minor isomorphic to a member in Ω.

A graph is called trivial if it has only one vertex, nontrivial otherwise. A subgraph of a graph G is called even if its degrees are all even. An even subgraph H is called nontrivial if each of its components is nontrivial. A subgraph of G is called an even factor if it is an even nontrivial spanning subgraph of G. A supereulerian graph has an even factor with exactly one component. A nontrivial path is called a branch if it has only internal vertices of degree two and end vertices of degree not two. The length of a branch is the number of its edges. Note that a branch of length one has no internal vertex, is called an edge branch. We denote by B(G) the set of branches of G. For any subset S of B(G), we denote by G−S the subgraph obtained from G by deleting all edges and internal vertices of branches of S. A subset S of B(G) is called a branch cut if G − S has more components than G. A minimal branch cut is called a branch-bond. A branch-bond is called odd if it has an odd number of branches. The idea of branch-bond comes originally from [9] which has also been applied in [4,8,7]. A star is the complete bipartite graph K1,m.A K1,3 is also called a claw. If G has an even factor, then δ(G) ≥ 2 and G has no odd branch-bonds with a shortest branch of length at least 2. The following result shows that the two obvious necessary conditions for the existence of even factors are also sufficient for claw-free graphs. Note that there is a polynomial-time algorithm to determine whether a graph satisfies the hypothesis that every odd branch-bond has an edge branch, see, e.g., [6].

Theorem 5 (Xiong, [7]). Let G be a claw-free graph. Then G has an even factor if and only if δ(G) ≥ 2 and every odd branch-bond of G has an edge branch.

The results above motivate the research of extending Theorem 5. In this paper, we shall characterize all connected graphs H of order at least three guaranteeing that those obvious necessary conditions for the existence of even factors in G are also sufficient.

2. Main results and their proofs

We start with the following auxiliary result.

Theorem 6 (Xiong, [8]). Let ℓ be a positive integer. Then a graph G has a set of vertex-disjoint circuits containing all branches of length not ℓ if every odd branch-bond of G has a shortest branch of length ℓ.

The unique tree with a degree sequence 1,1,1,2,3 is called a chair.A circuit is a connected even graph with at least three vertices. In the proof of main theorem, we frequently take the symmetric difference of a graph and a cycle. Let H be a subgraph of a graph G, and let C be a cycle in G. Then we define H1C by H1C = V (H) ∪ V (C), E(H)1E(C), where A1B denotes the symmetric difference of the sets A and B. Note that if H is an even graph, then H1C is also an even graph, but H1C may have more isolated vertices than H. We first prove the following result.

′ ≥ Theorem 7. Let G be either a P5-free graph or a chair-free graph other than W3. Then G has an even factor if and only if δ(G) 2 and every odd branch-bond of G has an edge branch.

Proof of Theorem 7. Necessity. An even factor should have minimum degree at least two and a graph has no even factor if it has an odd branch-bond which contains a shortest branch of length at least two. ′ ≥ Sufficiency. Let G be a P5-free or chair-free graph other than W3 with δ(G) 2 in which every odd branch-bond has an edge branch. By Theorem 6, there exists an even subgraph C of G which contains all branches of length at least 2. Then C contains all the vertices of degree 2. Now choose an even subgraph C of G containing all the vertices of degree 2 so that C contains as many vertices of possible. We claim that C is a spanning subgraph of G. Assume V (C) ̸= V (G). By the maximality of C, G \ V (C) does not contain an even subgraph. Let x ∈ V (G) \ V (C). By the choice of C, dG(x) ≥ 3.

Claim 1. NG(x) is independent in G.

Proof. Assume y1y2 ∈ E(G) for some {y1, y2} ⊂ NG(x). Let C1 = C∆(xy1y2x). Then C1 is an even subgraph of G. Moreover, since x ̸∈ V (C), {xy1, xy2} ∩ E(C) = ∅. Then {x, y1, y2} ⊂ V (C1) and hence V (C1) = V (C) ∪ {x}. This contradicts the maximality of C. L. Xiong / Discrete Applied Mathematics 223 (2017) 135–139 137

Claim 2. Let y1 and y2 be distinct neighbors of x and let z ∈ (NG(y1) ∩ NG(y2)) \{x}. Then

(a) dC (z) = 2 and {y1z, y2z} ⊂ E(C), and ′ ′ ′ ′ ′ (b) for each y ∈ NG(x) (possibly y ∈ {y1, y2}) and z ∈ NG(y ) \{x}, zz ̸∈ E(G).

Proof. (a) Let C2 = C∆(xy1zy2x). Then C2 is an even subgraph of G. Since {xy1, xy2}∩ E(C) = ∅, we have {x, y1, y2} ⊂ V (C2). If z ∈ V (C2), then V (C2) = V (C) ∪ {x}, which contradicts the maximality of C. Hence z ̸∈ V (C2). This implies that y1z and y2z are the only edges in C that are incident with z. This proves (a). ′ ′ ′ ′ (b) Assume zz ∈ E(G). We may assume y ̸= y1. Let C3 = C∆(xy1zz y x). Then C3 is an even subgraph of G with ′ ′ ′ {xy1, xy } ⊂ E(C3). Moreover, by (a), we also have zz ̸∈ E(C). Therefore, zz ∈ E(C3) and V (C3) = V (C) ∪ {x}. This contradicts the maximality of C. Hence (b) follows.

Claim 3. For every z ∈ V (G) \{x}, |NG(x) \ NG(z)| ≤ 2.

Proof. Assume |NG(x)∩ NG(z)| ≥ 3 and {y1, y2, y3} ⊆ NG(x)∩ NG(z). By applying Claim 2(a) to y1 and y2, we have dC (z) = 2 and {y1z, y2z} ⊂ E(C). However, by applying Claim 2(a) to y1 and y3, we also have y3z ∈ E(C). This is a contradiction.

Claim 4. For each distinct neighbors y1 and y2 of x, |NG(y1) ∩ NG(y2)| ≤ 2.

Proof. Assume |NG(y1)∩NG(y2)| ≥ 3 and let {x, z1, z2} ⊂ NG(y1)∩NG(y2). By Claim 1, {z1, z2}∩NG(x) = ∅. Since dG(x) ≥ 3, we can take y3 ∈ NG(x) \{y1, y2}. By Claim 2(b), we have z1z2 ̸∈ E(G), and by Claim 3, we have {y3z1, y3z2} ∩ E(G) = ∅. Then {y1, z1, z2, x, y3} induces a chair. This implies that G is not chair-free and hence it is P5-free. Since dG(y3) ≥ 2, NG(y3) \{x} ̸= ∅. Let z3 ∈ NG(y3) \{x}. By Claims 1 and3, z3 ∈ {y1, y2, z1, z2}. Moreover, by Claim 2(b) and Claim 3, {z1z3, z1y3} ∩ E(G) = ∅. On the other hand, {z1, y1, x, y3, z3} does not induce P5. This is possible only if y1z3 ̸∈ E(G). Then by Claim 3, y2z3 ̸∈ E(G). However, again by Claim 2(b) and Claim 3, {z2z3, z2y3}∩ E(C) = ∅. Therefore, {z2, y2, x, y3, z3} induces P5. This is a contradiction, and the claim follows.

Claim 5. There exist three independent edges y1z1, y2z2 and y3z3 in G − x with {y1, y2, y3} ⊂ NG(x).

Proof. Let y1, y2 and y3 be three distinct neighbors of x. Let Y = {y1, y2, y3} and Z = (NG(y1) ∪ NG(y2) ∪ NG(y3)) \{x}. Let H be the graph defined by V (H) = Y ∪ Z and E(H) = {yz : y2 ∈ Y , z ∈ Z, yz ∈ E(G)}. Then H is a bipartite graph with partite sets Y and Z. If H contains a matching which saturates Y , it constitutes required independent edges. Thus, we may ′ ′ ′ assume that H does not contain a matching saturating Y . Then by Hall’s Theorem, |NH (Y )| < |Y | for some Y ⊂ Y . Choose ′ ′ ′ ′ Y so that |Y | is as small as possible. Since dG(yi) ≥ 2 for i ∈ {1, 2, 3}, NH (Y ) ̸= ∅. Therefore, we have 2 ≤ |Y | ≤ 3. If ′ ′ ′ |Y | = 2, then |NH (Y )| = 1. We may assume Y = {y1, y2}. Let NH (y1) = NH (y2) = {z}. Then NG(y1) = NG(y2) = {x, z}. By Claim 2(a), {y1z, y2z} ⊂ E(C). Then since C is an even factor, {xy1, xy2} ⊂ E(C). This contradicts x ̸∈ V (C). ′ ′ If |Y | = 3, then Y = Y = {y1, y2, y3} and NH (Y ) = Z. By Claim 3, |Z| = 2. Let Z = {z1, z2}. If dG(yi) = 2 for every i ′′ ′′ ′′ ′ with 1 ≤ i ≤ 3, then for some Y ⊂ Y with |Y | = 2, |NH (Y )| = 1. This contradicts the minimality of Y . Hence we may assume dG(y3) = 3 and hence NG(y3) = {x, z1, z2}. Then by Claim 4, we may assume NG(y1) = {x, z1} and NG(y2) = {x, z2}. However, by Claim 2(a), we have {y1z, y3z} ⊂ E(C). Since C is an even factor, {xy1, xy3} ⊂ E(C). This again contradicts x ̸∈ V (C). Therefore, the claim follows.

By Claim 5, we can take three independent edges y1z1, y2z2 and y3z3 in G − x with {y1, y2, y3} ⊂ NG(x).

Claim 6. There exists an induced P5 in G.

Proof. Assume G is P5-free. Suppose |NG(zi) ∩ {y1, y2, y3}| ≥ 2 for some i ∈ {1, 2, 3}. We may assume y2z1 ∈ E(G). Then by Claim 1, Claim 2(b) and Claim 3, we have {y1y3, z1z3, y3z1}∩E(G) = ∅. Since {z1, y1, x, y3, z3} does not induce P5, y1z3 ∈ E(G). Then again by Claim 1, Claim 2(b) and Claim 3, {y2y3, z2z3, y2z3} ∩ E(G) = ∅, and since {z2, y2, x, y3, z3} does not induce P5, y3z2 ∈ E(G). However, this implies y2z3 ̸∈ E(G) and hence {z1, y2, x, y3, z3} induces P5. This is a contradiction. Therefore, we have NG(zi) ∩ {y1, y2, y3} = {yi} for each i ∈ {1, 2, 3}. Since {z1, y1, x, y2, z2} does not induce P5, we have z1z2 ∈ E(G). Similarly, we have {z1z3, z2z3} ⊂ E(G). However, now {y1, x, y2, z2, z3} induces P5, a contradiction. Therefore, the claim follows.

By Claim 6, G is not P5-free and hence it is chair-free. Since {x, y1, y2, y3, z1} does not induce a chair, NG(z1)∩{y2, y3} ̸= ∅. We may assume y2z1 ̸∈ E(G). By Claim 4, y1z2 ̸∈ E(G). Then since {x, y1, y2, y3, z2} does not induce a chair, y3z2 ∈ E(G). Again by Claim 4, y2z3 ̸= E(G) and since {x, y1, y2, y3, z3} does not induce a chair, we have y1z3 ∈ E(G). By Claim 2(b), {z1, z2, z3} is independent.

Claim 7. For each i, 1 ≤ i ≤ 3, dG(yi) = 3. ′ ′ Proof. Assume the contrary. Then we may assume dG(y1) ≥ 4. Let z ∈ NG(y1)\{x, z1, z3}. By Claim 3, z ̸= z2. By Claim 2(b), ′ ̸∈ { ′ } ∩ = ∅ { ′ } z1z3 E(G), and by Claim 3, y2z1, y2z3 E(G) . Then y1, x, z1, z3, y2 induces a chair. This is a contradiction, and the claim follows. 138 L. Xiong / Discrete Applied Mathematics 223 (2017) 135–139

Fig. 1. Two graphs having no even factor.

Claim 8. dG(x) = 3.

Proof. Assume dG(x) ≥ 4 and take y4 ∈ NG(x) \{y1, y2, y3}. By Claim 3, NG(y4) ∩ {z1, z2, z3} = ∅. Let z4 ∈ NG(y4) \{x}. Then by Claim 4, NG(z4) ∩ {y1, y2, y3} = ∅, and {x, y1, y2, y4, z4} induces a chair. This is a contradiction.

Claim 9. For each i, 1 ≤ i ≤ 3, dG(zi) = 2.

Proof. Assume the contrary. Then we may assume d(z1) ≥ 3. Take u1 ∈ NG(z1) \{y1, y2}. By Claim 7, {u1y1, u1y2, y1z2} ∩ E(G) = ∅, while {z1, u1, y1, y2, z2} does not induce a chair. This implies u1z2 ∈ E(G). Then by applying the same argument to {z2, u1, y2, y3, z3}, we also have u1z3 ∈ E(G). On the other hand, by Claim 2(a), {u1z2, u1z3} ∩ E(C) = ∅. Let C4 = C∆(xy1z1u1z2y2x). Then C4 is an even subgraph of G with V (C4) = V (C) ∪ {x, u1} (Note that possibly u1 is already in C). This contradicts the maximality of C. ′ By Claims 7–9, G is isomorphic to W3. This is a final contradiction, and the theorem follows. The following is our main result.

′ ≥ Theorem 8. Let H be a connected graph of order at least three and G be a graph other than the W3 such that δ(G) 2 and every odd branch-bond of G has an edge branch. Then G being H-free implies that G has an even factor if, and only if, H is a subgraph of either a P5 or a chair, i.e., H = P3, P4, P5, K1,3, or the chair. Proof of Theorem 8. Theorem 7 shows that the sufficiency of this theorem holds. Now we prove the necessary of this theorem. Now we need define two graphs family. Let G1 be the graph obtained from three vertex-disjoint complete graphs and one additional vertex w by making w to be adjacent to exactly three vertices from three complete graphs respectively, see the first graph which is depicted in Fig. 1. Let G2 be the graph obtained from a complete bipartite graph K2,2t with one ′ ′ part V1 = {x, y} of degree 2t and three additional vertices w, x , z1, z2 by jointing w to the triple of vertices x, y, x and by ′ jointing y to both z1 and z2 and by jointing x to both z1 and z2, see the right graph that is depicted in Fig. 1 (where Kqi is a clique of order at least three; the right graph has exactly four odd degree vertices and the other vertices have degree exactly two) satisfies the conditions that δ(G) ≥ 2 and every odd branch-bond of G has an edge branch; but they all have no even factors. Since they have no even factor, they are not H-free, or equivalently, H should be an induced subgraph of both G1 and G2. Since G1 contains a unique induced cycle that is a triangle but G2 contains no induced triangle, H must be a tree. Every induced tree in G1 is a subtree of T2,2,2 (where a Ts1,s2,...,st is the tree with an exactly vertex of maximum degree at least 3 and the other t paths of length s1, s2,..., st attaching the vertex of degree at least three; thus a chair is in fact a T2,1,1). On the other hand, every induced tree in G2 is a subgraph of T3,1,1,...,1. Therefore, the common induced subgraph of T3,1,1,...,1 and T2,2,2 must be a subgraph of T2,1,1 or P5. Thus, H can only be a P3, a P4, a P5, a claw K1,3 or a chair T2,1,1.

3. Concluding remarks

It is known that the property that every odd branch-bond has an edge branch are also necessary for a graph to be hamiltonian and to have a 2-factor. The following result shows that the conclusions of Theorems 1 and2 do not change even if we impose the necessary condition. In the proof, we shall use the line graph concept. The line graph of a graph G = (V (G), E(G)) is defined to be the graph with vertex set V (L(G)) = E(G) and two vertices e, f ∈ E(G) are adjacent if and only if they have a vertex in common, i.e., V (e) ∩ V (f ) ̸= ∅.

Theorem 9. Let G be the set of all the finite simple connected graphs of minimum degree at least 2 in which every odd branch- bond contains an edge branch. Let A be a connected graph of order at least 3. Then

(1) there exists a positive integer n1 such that every A-graph G in G of order at least n1 contains a 2-factor if and only if A = P3, and (2) there exists a positive integer n2 such that every 2-connected A-free graph G in G of order at least n2 is hamiltonian if and only if A = P3. L. Xiong / Discrete Applied Mathematics 223 (2017) 135–139 139

Proof of Theorem 9. Sufficiency of (1) and (2). It follows from the fact that every P3-free graph is complete. Necessary of (1). Let n = max{n1, |V (A)|} and let G0 be the multigraph consisting of two vertices x and y and four parallel edges joining x and y. Then for positive integers k, l and m, let G0(k, l, m) be the graph obtained from G0 by subdividing three edges with k, l and m vertices (or equivalently, replacing three edges with paths of order k + 1, l + 1 and m + 1). The resulting graph has order k + l + m + 2 with two vertices of degree 4 and k + l + m vertices of degree 2. Note that the vertices of degree 4 are adjacent with each other. Note also that G0(k, l, m) contains a unique branch-bond, is even and contains an edge branch. Thus G0(k, l, m) ∈ G for every positive triple of integers k, l and m. Moreover, G0(k, l, m) does not contain a 2-factor. Let G1 = G0(n, n, n). Then |V (G)| ≥ n1. Since every A-free graph in G of order at least n1 contains a 2-factor while G1 does not contain a 2-factor, G1 is not A-free and hence it contains an induced subgraph which is isomorphic to A. However, since the girth of G1 is at least |V (A)| + 2, no induced subgraph of G1 of order |V (A)| contains a cycle. Since A is connected, this implies that A is a tree. Note that ∆(A) ≤ 4 and A contains at most two vertices of degree at least 3. Assume A contains two vertices of degree at least 3. Let G2 = G0(1, 1, n). Then |V (G2)| ≥ n1 and hence G2 contains an { } ⊂ | ∩ | ≥ | ∩ | ≥ induced subgraph H2 which isomorphic to A. This yields x, y V (H2), NG2 (x) V (H2) 3 and NG2 (y) V (H2) 3. ∩ ∩ ̸= ∅ However, this implies NG2 (x) NG2 (y) V (H2) and hence H2 contains a triangle. This is a contradiction since A is a tree. Hence A contains at most one vertex of degree at least 3, which implies that A is either a path or obtained from K1,3 or K1,4 by subdividing its edges. Let H0 be the unique graph having the degree sequence 2, 2, 2, 2, 4. Let G3 (respectively, G4) be the graph obtained from H0 and Kn by identifying one vertex of Kn and the vertex of degree 4 (respectively, a vertex of degree 2) in H0. Every branch-bond of G3 and G4 consists only of edge branches and hence both G3 and G4 belong to G. Moreover, |V (G3)| ≥ n1 and |V (G4)| ≥ n1, but neither G3 nor G4 contains a 2-factor. Therefore, Gi contains an induced subgraph Hi which is isomorphic to A(i ∈ {3, 4}). Since H3 is an induced subgraph of G3, A is either K1,3 or K1,2. On the other hand, since G4 is K1,3-free, A K1,3. Thus, we have A = K1,2. Necessary of (2). Since both G1 and G2 defined above are all 2-connected non-hamiltonian graphs, from the discussion of (1), we know that A is either a path or obtained from K1,3 or K1,4 by subdividing its edges. ′ Let G5 be the graph obtained from H0 and Kn by identifying one vertex w of Kn and the vertex of degree 4 in H0, by jointing ′′ ′ one vertex w (̸=w ) of Kn and the four vertices of degree two of H0, respectively. Then every branch-bond of G5 has an edge branch and hence G5 ∈ G. Moreover, |V (G5)| ≥ n2, but G5 is not hamiltonian. Therefore, G5 has a subgraph H5 which is isomorphic to A. Since H5 is an induced subgraph of G5, A is K1,4-free and P4-free. Since H5 is an induced subgraph, A is either K1,3 or K1,2. 0 Now let G6 be the line graph of the graph G6 obtained from the bipartite graph K2,3 and three complete graphs Kq1 , Kq2 , Kq3 of order at least three by identifying the three different vertices of degree two in K2,3 and three vertices of the three complete graphs Kq1 , Kq2 , Kq3 , respectively. Then G6 is a 2-connected graph whose every branch-bond has an edge branch and hence G5 ∈ G. Moreover, |V (G6)| ≥ n2, but G6 is not hamiltonian. Therefore, G6 has a subgraph H6 which is isomorphic to A. Since H6 is an induced subgraph of G5, A is K1,3-free. Recall that A is either K1,3 or K1,2. Therefore, A = K1,2. As we remarked between Theorems 2 and3, there is no different thing of forbidden subgraph conditions guaranteeing a graph to be hamiltonian, traceable or to have a 2-factor. However, our result reports a different thing: there is a complete different characterizations of forbidden subgraphs guaranteeing a graph with δ(G) ≥ 2 and with the property that every odd branch-bond has an edge-branch to have an even factor (comparing Theorems 8 and9).

Acknowledgments

This research is supported by Nature Science Funds of China (Nos. 11471037 and 11671037). The author thanks one of the referees very much for his/her representation of the proofs of Theorems 7 and9 and clarified the original arguments.

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