Homeomorphically Irreducible Spanning Trees, Halin Graphs, and Long Cycles in 3-Connected Graphs with Bounded Maximum Degrees

Georgia State University ScholarWorks @ Georgia State University

Mathematics Dissertations Department of Mathematics and Statistics

5-11-2015

Homeomorphically Irreducible Spanning Trees, Halin Graphs, and Long Cycles in 3-connected Graphs with Bounded Maximum Degrees

Songling Shan

Follow this and additional works at: https://scholarworks.gsu.edu/math_diss

Recommended Citation Shan, Songling, "Homeomorphically Irreducible Spanning Trees, Halin Graphs, and Long Cycles in 3-connected Graphs with Bounded Maximum Degrees." Dissertation, Georgia State University, 2015. https://scholarworks.gsu.edu/math_diss/23

This Dissertation is brought to you for free and open access by the Department of Mathematics and Statistics at ScholarWorks @ Georgia State University. It has been accepted for inclusion in Mathematics Dissertations by an authorized administrator of ScholarWorks @ Georgia State University. For more information, please contact [email protected]. Homeomorphically Irreducible Spanning Trees, Halin Graphs, and Long Cycles in 3-connected Graphs with Bounded Maximum Degrees

Songling Shan

Under the Direction of Guantao Chen, PhD

ABSTRACT

A tree T with no vertex of degree 2 is called a homeomorphically irreducible tree (HIT) and if T is spanning in a graph, then T is called a homeomorphically irreducible spanning

tree (HIST). Albertson, Berman, Hutchinson and Thomassen asked if every triangulation

of at least 4 vertices has a HIST and if every connected graph with each edge in at least

two triangles contains a HIST. These two questions were restated as two conjectures by

Archdeacon in 2009. The ﬁrst part of this dissertation gives a proof for each of the two conjectures. The second part focuses on some problems about Halin graphs, which is a class of graphs closely related to HITs and HISTs. A Halin graph is obtained from a plane embedding of a HIT of at least 4 vertices by connecting its leaves into a cycle following the cyclic order determined by the embedding. And a generalized Halin graph is obtained from a HIT of at least 4 vertices by connecting the leaves into a cycle. Let G be a suﬃciently large n-vertex graph. Applying the Regularity Lemma and the Blow-up Lemma, it is shown that G contains a spanning Halin subgraph if it has minimum degree at least (n +1)/2 and

G contains a spanning generalized Halin subgraph if it is 3-connected and has minimum degree at least (2n + 3)/5. The minimum degree conditions are best possible. The last part estimates the length of longest cycles in 3-connected graphs with bounded maximum degrees. In 1993 Jackson and Wormald conjectured that for any positive integer d 4, there ≥ exists a positive real number α depending only on d such that if G is a 3-connected n-vertex graph with maximum degree d, then G has a cycle of length at least αnlogd−1 2. They showed that the exponent in the bound is best possible if the conjecture is true. The conjecture is conﬁrmed for d 425. ≥

INDEX WORDS: Homeomorphically irreducible spanning tree, Halin graph, Genaralized Halin graph, 3-connected graphs, Tutte decomposition. Homeomorphically Irreducible Spanning Trees, Halin Graphs, and Long Cycles in 3-connected Graphs with Bounded Maximum Degrees

Songling Shan

A Dissertation Submitted in Partial Fulﬁllment of the Requirements for the Degree of

Doctor of Philosophy in the College of Arts and Sciences Georgia State University 2015 Copyright by Songling Shan 2015 Homeomorphically Irreducible Spanning Trees, Halin Graphs, and Long Cycles in 3-connected Graphs with Bounded Maximum Degrees

Songling Shan

Committee Chair: Guantao Chen

Committee: Ron Gould Frank Hall Zhongshan Li Hendricus Van der Holst Yi Zhao

Electronic Version Approved:

Oﬃce of Graduate Studies College of Arts and Sciences Georgia State University May 2015 iv

DEDICATION

This dissertation is dedicated to my family and my advisor Professor Guantao Chen. v

ACKNOWLEDGEMENTS

This dissertation work would not have been possible without the support of many people. First and foremost, I would like to express my sincere gratitude to my advisor Professor Guantao Chen. He serves not just as an academic advisor, but also like a friend and a father. What I appreciate most is his always-open office door, whenever I have questions, I can just walk in, talk to him and discuss with him. His guidance and suggestions keep me going forward for tracking research problems and living positively. On Chinese new year and other Chinese holidays, he invites us to his home for dinner. For students like me who are thousands of miles away from home, he offers us the home-like comfort. There are so many words of gratitude that I want to express, but I just don’t know where and how to start. Above all, it is the biggest treasure in my life to have Professor Chen as my advisor. Thanks also to Professor Frank Hall and Professor Zhongshan Li. Their help and encouragement lightened my life at GSU. I wish to thank the other members of my dissertation committee: Professors Ron Gould, Hendricus Van der Holst and Yi Zhao. In addition, I give my gratitude to all the other faculty and staff members in the department. Also, I would like to thank my group members Nana Li, Ping Yang, and Amy Yates. Beyond metropolitan Atlanta, I sincerely thank Professor Mark Ellingham of Vanderbilt University, whom I got to know in my first year at GSU, and who has helped me as a mentor; I also sincerely thank Professor Akira Saito, he is like my secondary advisor. Further back, I would like to thank my first graph theory enlightenment teacher Professor Bing Yao. It was he who helped me in writing the first research paper; and that was the very primary motivation of my stepping on the research journey in graph theory. Then it is the time to say thanks to my Master of Science advisor Professor Han Ren, and one of my favorite professors at East China Normal University, Professor Xingzhi Zhan. Finally and most importantly, I thank my family members both in China and in the US for their love and support. Thinking of them, deep in my heart, can always calm me down, and is always a wonderful thing to do. vi

TABLE OF CONTENTS

ACKNOWLEDGEMENTS ...... v

LIST OF FIGURES ...... viii

LIST OF ABBREVIATIONS ...... ix

PART 1 INTRODUCTION ...... 1

PART 2 THE EXISTENCE OF HISTS IN SURFACE TRIANGU- LATIONS AND CONNECTED GRAPHS WITH EACH EDGE IN AT LEAST TWO TRIANGLES ...... 3 2.1 Proof of Conjecture 2.1 ...... 4 2.1.1 ProofofTheorem2.1.1...... 4 2.2 Proof of Conjecture 2.2 ...... 7 2.2.1 ProofofTheorem2.2.1...... 8

PART 3 MINIMUM DEGREECONDITION FOR SPANNINGHALIN GRAPHS AND SPANNINGGENERALIZED HALIN GRAPHS 16 3.1 Notations and deﬁnitions ...... 16 3.2 The Regularity Lemma and the Blow-up Lemma ...... 17 3.3 Dirac’s sondition for spanning Halin graphs ...... 18 3.3.1 Introduction...... 18 3.3.2 Ladders and “ladder-like” Halin graphs ...... 21 3.3.3 ProofofTheorem3.3.1...... 23 3.3.3.1 Proof of Theorem 3.3.2 ...... 26 3.3.3.2 Proof of Theorem 3.3.3 ...... 30 3.3.3.3 Proof of Theorem 3.4.3 ...... 38 vii

3.4 Minimum degree condition for spanning generalized Halin graphs 48 3.4.1 Introduction...... 48 3.4.2 Proof of Theorem 3.4.1 and the sharpness of Theorem 3.4.2 . . . . 50 3.4.3 ProofofTheorem3.4.2...... 52 3.4.3.1 Proof of Theorem 3.4.3 ...... 53 3.4.3.2 Proof of Theorem 3.4.4 ...... 75 3.4.3.3 Proof of Theorem 3.4.5 ...... 84

PART 4 A LOWER BOUND ON CIRCUMFERENCES OF 3- CONNECTED GRAPHS WITH BOUNDED MAXIMUM DEGREES ...... 102 4.1 Introduction ...... 102 4.2 Paths in block-chains ...... 104

4.3 Lower bounds of mlogb 2 + nlogb 2 ...... 108 4.4 Long paths in block-chains ...... 111 4.5 Proofs of Theorem (4.1.1) (a) and (b) ...... 126 4.5.1 Case 1: t 2 or t 2...... 129 1 ≥ 2 ≥ 4.5.2 Case 2. t1 = t2 =1...... 131 4.6 Reduction of Theorem (4.1.1)(c) ...... 133 4.6.1 Case 1 x / p, q ...... 140 ∈{ } 4.6.1.1 Subcase1.1. p, q = a , b = V (H H )...... 141 { } 6 { t t} t ∩ t+1 4.6.1.2 Case p, q = a , b = V (H H )...... 143 { } { t t} t ∩ t+1 4.6.2 Case 2 x p, q ...... 149 ∈{ }

REFERENCES ...... 167 viii

LIST OF FIGURES

Figure2.1 G4, a locally connected graph without a spanning 2-tree . . . . . 5 Figure2.2 Θ-graph,∆-operation,Θ-operation ...... 9 Figure2.3 uw / E(T ′) or uw E(T ′) E(G) ...... 13 1 ∈ 1 ∈ ∩ Figure2.4 uw E(T ′) E(G)...... 15 1 ∈ − Figure3.1 L4, Hi constructed from L4, and Ti associated with Hi for each i = 1, 2, , 5 ...... 24 · · · Figure3.2 Ladder L oforder14...... 26 Figure3.3 A Halin graph H ...... 29

Figure3.4 The HIT TB ...... 64

Figure3.5 The tree TA ...... 66

Figure3.6 The tree T11 ...... 77 Figure3.7 T with I =1...... 92 W | | Figure3.8 F with M =3...... 93 M | | ix

LIST OF ABBREVIATIONS

GSU - Georgia State University •

HIT - Homeomorphically irreducible tree •

HIST - Homeomorphically irreducible spanning tree •

SGHG - Spanning generalized Halin subgraph • 1

PART 1

INTRODUCTION

Finding longest cycles, in particular a hamiltonian cycle in a graph, is one of a few fundamental yet very difficult problems in graph theory. In fact, to determine whether a graph is hamiltonian is a classic NP-complete problem. Moreover, Karger, Motwani, and Ramkumar [35] showed that, unless = , it is impossible to find, in polynomial time, a P N P path of length n nǫ in an n-vertex hamiltonian graph for any ǫ < 1. On the other hand, − inspired by classic results obtained by Dirac [19] in 1954 and Tutte [51] in 1956, respectively, many sufficient conditions for hamiltonian graphs have been obtained. For examples, see [25]. Accompanying with each of these sufficient conditions, various stronger results such as being hamiltonian connected and pancyclic have also been established. As an antithetical class to hamiltonian paths/cycles, homeomorphically irreducible graphs, graphs which have no vertex of degree 2, were introduced by graph theorists in 1970s. A homeorphically irreducible tree is called a HIT, and a homeomorphically irreducible spanning tree of a graph is called a HIST of the graph. As graphs of at most three vertices contain no HIST, we assume the graphs in consideration are of at least four vertices when we considering HISTs. In the first part of this dissertation, we show the existence of a HIST in surface triangulations and connected graphs with each edge contained in at least two triangles. This confirms the two conjectures raised by Albertson, Berman, Hutchinson, and Thomassen [1]. Another class of graphs which is closely related to HITs and HISTs is the class of Halin graphs. Let T be a HIT of at least 4 vertices. Then a Halin graph H is obtained from a plane embedding of T by connecting the leaves into a cycle C following the cyclic order determined by the plane embedding. In this notation, we may write the Halin graph as H = T C. A wheel is an example of a Halin graph. Since a HIT of at least 4 vertices ∪ 2 contains two leaves sharing the same parent, a Halin graph contains a triangle, and thus is not bipartite. Moreover, cubic Halin graphs are in one-to-one correspondence (via weak duality) with the plane triangulations of the disc. Halin constructed Halin graphs in [27] for the study of minimally 3-connected graphs. Lovász and Plummer named such graphs as Halin graphs in their study of planar bicritical graphs [40], which are planar graphs having a 1-factor after deleting any two vertices. It was conjectured by Lovász and Plummer [40] that every 4-connected plane triangulation contains a spanning Halin subgraph (disproved in [10]). Although the conjecture is not true, it inspires new questions and problems. We may ask, can we find any other class of graphs which contain a spanning Halin subgraph or a spanning generalized Halin subgraph? The second part of this dissertation considers the existence of spanning Halin subgraphs and spanning generalized Halin subgraphs in graphs with large minimum degree. Halin graphs possess very nice hamiltonicity properties. Hence finding the existence of a spanning Halin subgraph can be viewed as a generalization of finding hamiltonian paths/cycles in graphs. Finally, in the last part, the problem of finding longest cycles in 3-connected graphs with bounded maximum degrees is investigated. In 1993 Jackson and Wormald conjectured that for any positive integer d 4, there exists a positive real number α depending only on ≥ d such that if G is a 3-connected n-vertex graph with maximum degree d, then G has a cycle of length at least αnlogd−1 2. They showed that the exponent in the bound is best possible if the conjecture is true. The conjecture is confirmed for d 425. ≥ Throughout this report, we limit our attention to simple and connected graphs, and further assume graphs to be finite unless we specify otherwise; and refer to Bondy and Murty [7] for notations and terminologies used but not defined. The vertex set and edge set of a graph G are denoted by V (G) and E(G), respectively. For S V (G), let G[S] ⊆ denote the subgraph of G induced by S. Similarly, G[F ] is the subgraph induced on F if F E(G). The minimum degree and maximum degree of G are denoted by δ(G) and ∆(G), ⊆ respectively. Other specified notations are introduced in each chapters. 3

PART 2

THE EXISTENCE OF HISTS IN SURFACE TRIANGULATIONS AND CONNECTED GRAPHS WITH EACH EDGE IN AT LEAST TWO TRIANGLES

Recall that a tree is called homeomorphically irreducible if it does not contain vertices of degree 2 and a homeomorphically irreducible spanning tree of a graph G is called a HIST of G. Albertson, Berman, Hutchinson, and Thomassen [1] obtained various sufficient conditions for a graph to contain a HIST. They also showed that it is NP -complete to decide whether a graph G contains a HIST. Hill [28] conjectured that every triangulation of the plane contains a HIST. Malkevitch [41] conjectured that the same result hold for near-triangulations of the plane (2-connected plane graphs such that all, but at most one, faces are triangles). Albert- son, Berman, Hutchinson, and Thomassen [1] confirmed the conjecture. Furthermore, they asked whether every graph that triangulates some surface has a HIST, and more generally if every connected graph with each edge contained in two triangles contains a HIST. To estab- lish a strategy to tackle the problem, Ellingham [20] asked whether every triangulation of a given surface with sufficiently large representativity contains a HIST. Huneke observed that every triangulation of the projective plane contains a spanning plane subgraph such that every face is a triangle with one possible exception, so every triangulation of the projective plane contains a HIST. Davidow, Hutchinson, and Huneke [18] showed that every triangulation of the torus contains a HIST. In 2009, Achdeacon [4] (Chapter 15) restated the above two questions as two conjectures.

Conjecture 2.1. Every surface triangulation contains a HIST.

Conjecture 2.2. Every connected graph with each edge in at least two triangles contains a HIST.

We conﬁrm the two conjectures in this Chapter. The proofs can also be found in [11] 4 and [13].

2.1 Proof of Conjecture 2.1

A graph G is locally connected if for every vertex v V (G), the subgraph induced by ∈ the neighborhood N(v) is connected. Ringel [46] showed that every triangulation (includes orientable and nonorientable ) is a connected and locally connected graph. In this section, we prove the following much more general result, which conﬁrms the conjecture by Archdeacon and answers the ﬁrst question asked by Albertson, Berman, Hutchinson, and Thomassen positively.

Theorem 2.1.1. Every connected and locally connected graph with order at least four contains a HIST.

Corollary 2.1.1. Let Π be a surface (orientable or nonorientable). Then every triangulation of Π with at least four vertices contains a HIST.

Let G be a graph. Write v G if v V (G) and similarly e G if e E(G). ∈ ∈ ∈ ∈

2.1.1 Proof of Theorem 2.1.1

Let k be a positive integer. A graph G is called a k-tree if there is an ordering v v 1 ≺ 2 ≺ v of V (G) such that (i) G[ v , v ,...,v ] is a complete graph and (ii), for each i>k, ···≺ n { 1 2 k} N(v ) v , v ,...,v induces a clique of order k. Clearly, 1-trees are the same as trees. i ∩{ 1 2 i−1} Hwang, Richards, and Winter [31] proved that 2-trees are maximal series-parallel graphs. As shown in Lemma 2.2 and 2.3, we observe that every 2-tree with more than three vertices contains a HIST. However, not every connected and locally connected graph contains a 2- tree as a spanning subgraph. Let Wn := K1 + Cn be a wheel of order n + 1 and let Gn be obtained from Wn by adding n new vertices such that each is adjacent to a distinct pair of two consecutive vertices on the cycle Cn. It is not diﬃcult to verify that Gn does not contain a spanning 2-tree. G4 is depicted below. 5

Figure (2.1) G4, a locally connected graph without a spanning 2-tree

Let G be a graph and v be a vertex not in V (G). We write H = G v if there ⊕ exist two distinct vertices u ,u G such that V (H) = V (G) v and E(H) = E(G) 1 2 ∈ ∪{ } ∪ u v,u v,u u . Note that the edge u u may already exist in G. We let P (v) := u ,u { 1 2 1 2} 1 2 { 1 2} and call u1 and u2 the parents of v.

Deﬁnition 2.1.1. A graph T of order n 3 is called a weak 2-tree (W2-tree) if there is an ≥ ordering : v v v of vertices of T and a sequence of graphs G G ≺ 1 ≺ 2 ≺···≺ n 3 ⊂ 4 ⊂···⊂ Gn = T such that the following properties hold.

(1) G = T [ v , v , v ] = K , and 3 { 1 2 3} ∼ 3

(2) for each i =3, 4, ... , n 1, G = G v . − i+1 ∼ i ⊕ i+1 In addition, we call the ordering a W2-tree ordering of T . ≺ Clearly, every 2-tree is a W2-tree. However, the converse is not true, for example, the

above graphs Gn are W2-trees but not 2-trees. Given a W2-tree with a W2-tree ordering , if we shift a degree 2 vertex to the end ≺ and keep the remaining ordering unchanged, we obtain another W2-tree ordering. So, the following result holds.

Lemma 2.1.1. Let G be a W2-tree with n 4 vertices. Let w G be a degree 2 vertex and ≥ ∈ N(w)= u, v . Then either G w or G w uv is a W2-tree. { } − − − 6

Lemma 2.1.2. Let T be W2-tree with n 4 vertices. Then, there exist two vertices u and ≥ v such that T = (T ′ u) v and N[u] N(v) = , where T ′ is a W2-tree, K , or K . In ⊕ ⊕ ∩ 6 ∅ 3 2 this case, u, v is called a removable pair of T . { }

Proof. We prove Lemma 2.1.2 by applying induction on n = V (G) . Since K− (K | | 4 4 minus an edge) is the unique W2-tree with 4 vertices, Lemma 2.1.2 holds for n = 4. Suppose n 5 and that Lemma 2.1.2 holds for all W2-trees with less than n vertices. ≥ Let T be a W2-tree with n vertices and w be the last vertex in a W2-ordering of T . Moreover, we assume that T = T ′ w, where T ′ is a W2-tree with n 1 vertices. Suppose that u, v ⊕ − { } is a removable pair of T ′ and T ′ = (T ∗ u) v, where T ∗ is a W2-tree, K , or K . We ⊕ ⊕ 3 2 complete the proof by considering the following ﬁve cases regarding N(w) u, v . ∩{ }

if N(w) u, v = , then T = [(T ∗ w) u] v, so u, v is a removable pair of T ; • ∩{ } ∅ ⊕ ⊕ ⊕ { }

if N(w) u, v = v , then T = [(T ∗ u) v] w, so v,w is a removable pair of • ∩{ } { } ⊕ ⊕ ⊕ { } T ;

if N(w) u, v = u and uv / E(T ′), then T = [(T ∗ v) u] w, so u,w is a • ∩{ } { } ∈ ⊕ ⊕ ⊕ { } removable pair of T ;

if N(w) u, v = u and uv E(T ′), then T = [(T ∗ u) v] w, so v,w is a • ∩{ } { } ∈ ⊕ ⊕ ⊕ { } removable pair of T ;

if N(w)= u, v , then T = [(T ∗ u) v] w, so v,w is a removable pair of T . • { } ⊕ ⊕ ⊕ { }

Lemma 2.1.3. A W2-tree with at least 4 vertices contains a HIST.

Proof. Let G be a W2-tree with n 4 vertices. We proceed by induction on n. If ≥ − n = 4, then G = K4 , which contains a spanning star. If n = 5, by case analysis, we can show that G contains a spanning star, so a HIST . 7

Assume n > 6 and let G be a W2-tree with n vertices. By Lemma 2.1.2, let u, v be a { } removable pair of G and assume G =(G′ u) v, where G′ is a W2-tree with n 2 vertices. ⊕ ⊕ − By the induction hypothesis, G′ contains a HIST, say, T ′. Since u, v is a removable pair { } of G, N[u] N(v) = . If uv / E(G), then N(u) N(v) = ; if uv E(G), by the ∩ 6 ∅ ∈ ∩ 6 ∅ ∈ deﬁnition of , the other neighbor of v is adjacent to u. In either case, there exists a vertex ⊕ w N(u) N(v). Then, T := T ′ wu,wv is a HIST of G. ∈ ∩ ∪{ }

Lemma 2.1.4. Every connected and locally connected graph with at least three vertices contains a spanning W2-tree.

Proof. Let G be a connected and locally connected graph of order n 3. Since every ≥ triangle is a W2-tree, G contains W2-trees as subgraphs. Let T G be a W2-tree such that ⊆ V (T ) is maximum. We claim that V (T ) = V (G). Otherwise, W := V (G) V (T ) = . | | − 6 ∅ Since G is connected, there is a vertex v V (T ) such that N (v) = , where N (v) is the ∈ W 6 ∅ W set of neighbors of v in W . Since T is a W2-tree, N(v) V (T ) N (v) = . Since G[N(v)] ∩ ⊇ T 6 ∅ is connected, there is an edge uw E(G) with u N (v) and w N (v). Then, T w is ∈ ∈ T ∈ W ⊕ a W2-tree containing more vertices than T , where P (w)= u, v . Since uv,wv,uw E(G) { } ∈ and T G, we have T w G, which contradicts the maximality of V (T ) . ⊆ ⊕ ⊆ | | So, the proof of Theorem 2.1.1 is completed.

2.2 Proof of Conjecture 2.2

We now answer the second question raised by Albertson et al. positively as follows, whose proof will be given in the next section. We would like to mention that the main proof technique used in the proof is similar to that for Conjecture 2.1 in the first section. However, the induction proceeded on the spanning Θ patch graph H (we will give the definition very − shortly) of G is not straightforward. In fact, when H has property Q2(defined in subsection 2), we can not directly proceed the induction. The new approach in dealing with this case, looks easy and natural, yet really took efforts to come out. 8

Theorem 2.2.1. Let G be a graph with every edge in at least two triangles. Then G contains a HIST.

2.2.1 Proof of Theorem 2.2.1

The proof consists of three main components: (1) deﬁne a class of graphs called Θ- patch graphs(we will deﬁne this class of graphs very shortly), and show that every graph with each edge in at least two triangles contains a spanning Θ-patch graph; (2) prove a rearrangeability of Θ-patch graphs; and (3) show every Θ-patch graph contains a HIST.

− Throughout this section, a graph isomorphic to K4 (K4 with exactly one edge removed) is called a Θ-graph.

Deﬁnition 2.2.1. Given a graph H and a vertex v / V (H), let H∆v be a graph with ∈ V (H∆v)= V (H) v and E(H∆v)= E(H) u v,u v,u u , where u , u V (H) are ∪{ } ∪{ 1 2 1 2} 1 2 ∈ two distinct vertices. That is, H∆v is obtained from H by adding a new vertex v and edges u v, u v, and u u if u u / E(H). We name such an operation ∆-operation and denote by 1 2 1 2 1 2 ∈ A(v) := u ,u , the set of attachments of v on H. Moreover, we let A[v] := A(v) v . { 1 2} ∪{ } Note that u1u2 may or may not be an edge of H.

Definition 2.2.2. Given a graph H and a Θ-graph F with a specified degree 3 vertex, let HΘF be the graph obtained by identifying the specified vertex of F with a vertex u in H. Let A(F ) = u be the set of the attachment of F on H. Such an operation is called a { } Θ-operation.

We use to denote either a ∆-operation or a Θ operation. ⊕ − Deﬁnition 2.2.3. A graph G is called a Θ-patch graph if there exists a subgraph sequence G G G = G with s > 2 such that 1 ⊂ 2 ⊂···⊂ s (1) G1 ∼= K3, and

(2) G is obtained from G by a -operation for each i (1 6 i 6 s 1 ). i+1 i ⊕ − By the above deﬁnition, a Θ-patch graph has at least 4 vertices, and a Θ-patch graph with exactly 4 vertices is a Θ-graph. 9

H v H u u

Figure (2.2) Θ-graph, ∆-operation, Θ-operation

Lemma 2.2.1. A connected graph with every edge in at least two triangles contains a Θ-patch graph as a spanning subgraph. Proof. Let G be a graph such that every edge is in at least two triangles. Since two triangles sharing a common edge induce a Θ-graph, G contains a Θ-graph, which is also a Θ-patch graph by Deﬁnition 2.2.3. Let H G be a Θ-patch graph such that ⊂ V (H) is maximum. If V (H) = V (G), the proof is completed. So assume the contrary: | | W = V (G) V (H) = . Since G is connected, there is an edge uw E(G) such that − 6 ∅ ∈ u V (H) and w W . Let v uwv and v uwv be two distinct triangles containing uw. ∈ ∈ 1 1 2 2 If v V (H) for some i = 1, 2, then H∆w with A(w) = u, v is a Θ-patch graph larger i ∈ { i} than H, contradicting the maximality of H. Hence, we have both v , v W . Clearly, 1 2 ∈ G[ u, v , v ,w ], the subgraph induced on u, v , v ,w , contains a Θ-graph F . So HΘF { 1 2 } { 1 2 } with A(F )= u is a Θ-patch graph larger than H, contradicting the maximality of H. { } It will be shown in the following lemma that the ordering of subgraph sequence in the deﬁnition of Θ-patch graphs can be rearranged to preserve a nice recursive property.

Lemma 2.2.2. Let G be a Θ-patch graph of order n > 5. Then there exist a subgraph H

which is either a Θ-patch graph or isomorphic to K3 such that one of the following properties 10

holds:

P : G =(H∆x )∆x and A(x ) A[x ] = ; 1 2 2 ∩ 1 6 ∅ Q (0 k 3) : There exist vertices x , x , , x such that k ≤ ≤ 1 2 ··· k G =(HΘF )∆x ∆x ∆x (G = HΘF when k =0) with A[x ] A[x ]= for all 1 2 ··· k i ∩ j ∅ i = j and A(x ) (V (F ) V (H)) = for i =1, 2, ,k. 6 i ∩ − 6 ∅ ··· Proof. If n = 5, from the deﬁnition of Θ-patch graphs, there exist two vertices x1 and x such that G = K ∆x ∆x and A(x ) A[x ] = , so P holds. We assume that n > 6 and 2 3 1 2 2 ∩ 1 6 ∅ Lemma 2.2.2 holds for graphs with order < n. By the deﬁnition of Θ-patch graphs, G = H∗ F ∗, where H∗ is a Θ-patch graph, and ⊕ ∗ ∗ F is either a single vertex or a Θ-graph. If F is a Θ-graph, then Q0 holds. So, we assume F ∗ is a single vertex graph, and say V (F ∗) = w . By applying Lemma 2.2.2 to H∗, we { } divide the remaining proof into two cases below.

Case P . H∗ =(H∆x )∆x and A(x ) A[x ] = . 1 2 2 ∩ 1 6 ∅ If A(w) x , x = , let H′ := H∆w, which is a Θ-patch graph and a subgraph of G. ∩{ 1 2} ∅ ′ Then G =(H ∆x1)∆x2, so P holds. Suppose A(w) x , x = . If x A(x ) or x A(w), H′ := H∆x G is a Θ- ∩{ 1 2} 6 ∅ 1 ∈ 2 2 ∈ 1 ⊂ patch graph. Then, we have G =(H′∆x )∆w and either x A(w) A[x ] or x A(w), so 2 1 ∈ ∩ 2 2 ∈ P holds. We may assume that x / A(x ) and x / A(w). In this case, we have x A(w). 1 ∈ 2 2 ∈ 1 ∈ ′ ′ Let H = H∆x2, which is a Θ-patch graph and a subgraph of G. Then G = H ∆x1∆w, so P holds.

Case Q . H∗ =(HΘF )∆x ∆x ∆x , where F is a Θ-graph and x is a vertex in H∗. k 1 2 ··· k i If A(w) ((V (F ) V (H)) x , x , , x )= , then ∩ − ∪{ 1 2 ··· k} ∅

G = ((H∆w)ΘF ) ∆x ∆x ∆x , 1 2 ··· k

so Q holds. If A(w) A[x ] = , w.l.o.g., say A(w) A[x ] = , then k ∩ i 6 ∅ ∩ k 6 ∅

G =(HΘF ∆x ∆x ∆x ) ∆x ∆w, 1 2 ··· k−1 k 11

so P holds. Hence, we assume A(w) (V (F ) V (H)) = and A(w) A[x ] = for ∩ − 6 ∅ ∩ i ∅ i =1, 2, ,k. Under this assumption together with the assumption that A[x ] A[x ]= ··· i ∩ j ∅ for i = j and A(x ) (V (F ) V (H)) = for i =1, 2, ,k, we have k 2. Then, we have 6 i ∩ − 6 ∅ ··· ≤

G =(HΘF )∆x ∆x ∆x ∆w, 1 2 ··· k

so Qk+1 holds.

Lemma 2.2.3. Every Θ-patch graph contains a HIST. Proof. We use induction on n = V (G) . When n = 4, G = K− is a Θ-graph. Clearly, | | ∼ 4 G contains a HIST. Suppose n > 5, and assume that Lemma 2.2.3 holds for graphs of order < n. We divide the remaining proof into ﬁve cases according to the ﬁve properties given in Lemma 2.2.2.

If G has property Qi for some i = 0, 1, 2 or 3, we follow the notations given in Lemma 2.2.2, and assume that A(F ) = u and V (F ) V (H) = v , v , v . If G has { } − { 1 2 3} property P then u is a specially selected vertex in H. We let T be a HIST of H if H is a

Θ-patch graph, and let T ∼= P3 with dT (u) = 2 if H ∼= K3. The case that G satisﬁes property

Q2 is the most complicated one, and we can not straightforwardly play induction on it, so we defer this case to the end.

Property P holds. Suppose that G = H∆x ∆x and A(x ) A[x ] = . 1 2 2 ∩ 1 6 ∅ In this case, we ﬁrst show that N(x ) N(x ) V (H) = . This is clearly true if 1 ∩ 2 ∩ 6 ∅ A(x ) A(x ) = , so we may assume x A(x ). Let u be the other vertex in A(x ). 1 ∩ 2 6 ∅ 1 ∈ 2 2 Since E(G)= E((H∆x )∆x )= E(H∆x ) x u, x x , ux , we have ux E(G), that is, 1 2 1 ∪{ 2 2 1 1} 1 ∈ u N(x ) N(x ). ∈ 1 ∩ 2 Let u N(x ) N(x ). Then, it is readily seen that T ux , ux is a HIST of G. ∈ 1 ∩ 2 ∪{ 1 2}

Property Q0 holds. Let G = HΘF . In this case, T uv , uv , uv is a HIST of G. ∪{ 1 2 3} Property Q holds. Let G = (HΘF )∆x , and assume, without loss of generality, v 1 1 1 ∈ A(x ) (V (F ) V (H)), and let w be another vertex of A(x ). 1 ∩ − 1 1 12

In this case, T w v ,w x , uv , uv is a HIST of G regardless of whether w V (F ) ∪{ 1 1 1 1 2 3} 1 ∈ or not.

Property Q holds. Let G = (HΘF )∆x ∆x ∆x and assume that A(x ) = v ,w for 3 1 2 3 i { i i} each i =1, 2, 3 with w ,w ,w V (H). 1 2 3 ∈ By the deﬁnition of ∆-operation, all three edges w1v1,w2v2,w3v3 are in E(G). Then, T w x ,w v ,w x ,w v ,w x ,w v is a HIST in G. ∪{ 1 1 1 1 2 2 2 2 3 3 3 3}

Property Q holds. Let G = (HΘF )∆x ∆x such that A(x ) (V (F ) V (H)) = for 2 1 2 i ∩ − 6 ∅ each i =1, 2, and A[x ] A[x ]= . Assume that A(x )= v ,w for i =1, 2. 2 ∩ 1 ∅ i { i i} We may assume w = u for each i = 1, 2; otherwise, say w = u, then T i 6 1 ∪ w v ,w x , uv , ux , uv is a HIST of G. Since A[x ] A[x ] = , we may assume that { 2 2 2 2 1 1 3} 2 ∩ 1 ∅ w V (H) u . Moreover, under the assumption that w V (H) u , let notation 1 ∈ −{ } 1 ∈ −{ } be chosen so that v is the degree 2 vertex in F u whenever it is possible, that is, if 1 − w V (H) u and v is the degree two vertex in F u, we rename x , v and w as x , 2 ∈ −{ } 2 − 2 2 2 1 v1 and w1, and vice versa. Let z / V (G) be a vertex and G′ := H∆z with A(z) = u,w . Clearly, uw E(G′) ∈ { 1} 1 ∈ although uw may not be in E(G). Clearly, G′ is a Θ-patch graph and V (G′ < n, so it 1 | | ′ ′ contains a HIST T . Since dG′ (z) = 2, z is a degree 1 vertex of T . So, we have either w z E(T ′) or uz E(T ′) but not both. Let T := T ′ z. 1 ∈ ∈ H −

Subcase 1. uw / E(T ′) or uw E(T ′) E(G). 1 ∈ 1 ∈ ∩ ′ ∗ Note that d ′ (z)=1. If uz E(T ), let T := T uv ,w v ,w x ,w v ,w x , as T ∈ H ∪{ 3 1 1 1 1 2 2 2 2} depicted in Figure 2.3. It is routine to check that T ∗ is a spanning tree of G and the following 13

w1 w1 x1 v1 x1 v1 H v3 H v u u 3 v2 x2 v2 w2 x2 w2 uz E(T ′) w z E(T ′) ∈ 1 ∈ Figure (2.3) uw / E(T ′) or uw E(T ′) E(G) 1 ∈ 1 ∈ ∩

equalities/inequalities hold.

d ∗ (u) = d ′ (u) uz + uv = d ′ (u) =2 T T − |{ }| |{ 3}| T 6

d ∗ (w ) = d ′ (w )+ w v ,w x = d ′ (w )+2 =2 T 1 T 1 |{ 1 1 1 2}| T 1 6 dT ′ (w2)+ w2v2,w2x2 = dT ′ (w2)+2 =2, if w2 V (H); dT ∗ (w2) = |{ }| 6 ∈  w v ,w x , uv =3, if w = v .  |{ 2 2 2 2 3}| 2 3 d ∗ (x) = d ′ (x) = 2 for all other vertices x V (H), and T T 6 ∈

d ∗ (x) = 2 for each vertex x v , v , v , x , x . T 6 ∈{ 1 2 3 1 2}

Consequently, T ∗ is a HIST of G. If w z E(T ′), let T ∗ := T w x , uv , uv ,w v ,w x , as depicted in Figure 2.3. 1 ∈ H ∪{ 1 1 1 3 2 2 2 2} ∗ (w2 = v3 may occur.) As in the previous case, we can show that T is a HIST of G. 14

Subcase 2. uw E(T ′) E(G). 1 ∈ − In this case, T := T uw has exactly two components. We construct a HIST of G 1 H − 1 from T according to whether uz E(T ′) or w z E(T ′). 1 ∈ 1 ∈ If uz E(T ′), let T ∗ = T uv , uv , v w , v x ,w v ,w x , as depicted in Figure 2.4. ∈ 1 ∪{ 3 1 1 1 1 1 2 2 2 2} It is routine to check that T ∗ is a spanning tree of G and the following equalities/inequalities hold.

d ∗ (u) = d ′ (u) uw , uz + uv , uv = d ′ (u) =2 T T − |{ 1 }| |{ 1 3}| T 6

d ∗ (w ) = d ′ (w ) uw + v w = d ′ (w ) =2 T 1 T 1 − |{ 1}| |{ 1 1}| T 1 6 dT ′ (w2)+ w2v2,w2x2 = dT ′ (w2)+2 =2, if w2 V (H); dT ∗ (w2) = |{ }| 6 ∈  w v ,w x , uv =3, if w = v .  |{ 2 2 2 2 3}| 2 3 d ∗ (x) = d ′ (x) = 2 for all other vertices x V (H), and T T 6 ∈

d ∗ (x) = 2 for each vertex x v , v , v , x , x . T 6 ∈{ 1 2 3 1 2}

So, T ∗ is a HIST of G. In the case w z E(T ′), if v v E(G), let 1 ∈ 1 3 ∈

T ∗ = T w x ,w v , v u, v v ,w v ,w x , 1 ∪{ 1 1 1 1 1 1 3 2 2 2 2} as depicted in Figure 2.4. As in the previous case, we can show that T ∗ is a HIST of G. To complete the proof, we show that the vertex v can be chosen such that v v E(G). If 1 1 3 ∈ v v / E(G), then both v and v are degree 1 vertices in F u = P . So, v is the degree 2 1 3 ∈ 1 3 − ∼ 3 2 vertex in F u. If w V (H), we would pick x as x and v as our v in the very beginning. − 2 ∈ 2 1 2 1 So, w = v . In this case, we can simply swap v and v (also w ) to ensure that v v E(G). 2 3 2 3 2 1 3 ∈

Clearly, the combination of the above three Lemmas gives Theorem 2.2.1. 15

w1 w1 x1 v1 x1 v1 H v3 H v u u 3 v2 x2 v2 w2 x2 w2 uz E(T ′) w z E(T ′) ∈ 1 ∈ Figure (2.4) uw E(T ′) E(G) 1 ∈ − 16

PART 3

MINIMUM DEGREE CONDITION FOR SPANNING HALIN GRAPHS AND SPANNING GENERALIZED HALIN GRAPHS

3.1 Notations and deﬁnitions

We consider simple and ﬁnite graphs only. Let G be a graph. Denote by e(G) the cardinality of E(G). Let v V (G) be a vertex and S V (G) a subset. The notation Γ (v,S) ∈ ⊆ G denotes the set of neighbors of v in S, and deg (v,S) = Γ (v,S) . We let Γ (v,S) = G | G | G S Γ (v,S) and deg (v,S)= Γ (v,S) . Given another set U V (G), deﬁne Γ (U, S)= − G G | G | ⊆ G Γ (u,S), deg (U, S) = Γ (U, S) , and N (U, S) = Γ (u,S). When U = ∩u∈U G G | G | G ∪u∈U G u ,u , ,u , we may write Γ (U, S), deg (U, S), and N (U, S) as Γ (u ,u , ,u ,S), { 1 2 ··· k} G G G G 1 2 ··· k deg (u ,u , ,u ,S), and N (u ,u , ,u ,S), respectively, in specifying the vertices in G 1 2 ··· k G 1 2 ··· k U. When S = V (G), we only write Γ (U), deg (U), and N (U). Let U , U V (G) G G G 1 2 ⊆ be two disjoint subsets. Then δ (U , U ) = min deg (u , U ) u U and ∆ (U , U ) = G 1 2 { G 1 2 | 1 ∈ 1} G 1 2 max deg (u , U ) u U . Notice that the notations δ (U , U ) and ∆ (U , U ) are not { G 1 2 | 1 ∈ 1} G 1 2 G 1 2 symmetric with respect to U1 and U2. We denote by EG(U1, U2) the set of edges with one

end in U1 and the other in U2, the cardinality of EG(U1, U2) is denoted as eG(U1, U2). We may omit the index G if there is no risk of confusion. Let u, v V (G) be two vertices. We ∈ write u v if u and v are adjacent. A path connecting u and v is called a (u, v)-path. If G ∼ is a bipartite graph with partite sets A and B, we denote G by G(A, B) in emphasizing the two partite sets. A matching in G is a set of independent edges; a -matching is a set of ∧ vertex-disjoint copies of K1,2; and a claw-matching is a set of vertex-disjoint copies of K1,3. The set of degree 2 vertices in a -matching is called the center of the -matching ; and the ∧ ∧ set of degree 3 vertices in a claw-matching is called the center of the claw-matching. A cycle C in a graph G is dominating if G V (C) is an edgeless graph. − 17

3.2 The Regularity Lemma and the Blow-up Lemma

The Regularity Lemma of Szemerédi [50] and Blow-up lemma of Komlós et al.[36] are main tools used in finding a spanning Halin subgraph or spanning generalized Halin subgraph. For any two disjoint non-empty vertex-sets A and B of a graph G, the density of A and

e(A,B) B is the ratio d(A, B) := |A||B| . Let ε and δ be two positive real numbers. The pair (A, B) is called ε-regular if for every X A and Y B with X > ε A and Y > ε B , ⊆ ⊆ | | | | | | | | d(X,Y ) d(A, B) <ε holds. In addition, if deg(a, B) > δ B for each a A and deg(b, A) > | − | | | ∈ δ A for each b B, we say (A, B) an (ε, δ)-super regular pair. | | ∈

Lemma 3.2.1 (Regularity lemma-Degree form [50]). For every ε > 0 there is an M = M(ε) such that if G is any graph with n vertices and d [0, 1] is any real number, then ∈ there is a partition of the vertex set V (G) into l +1 clusters V ,V , ,V , and there is a 0 1 ··· l spanning subgraph G′ G with the following properties. ⊆ l M; • ≤

V εn, all clusters V = V εn for all 1 i = j l; • | 0| ≤ | i| | j| ≤ ⌈ ⌉ ≤ 6 ≤

deg ′ (v) > deg (v) (d + ε)n for all v V (G); • G G − ∈

e(G′[V ]) = 0 for all i 1; • i ≥

all pairs (V ,V ) (1 i < j l) are ε-regular, each with a density either 0 or greater • i j ≤ ≤ than d.

Lemma 3.2.2 (Blow-up lemma-weak version [36]). Given a graph R of order r and positive parameters δ, ∆, there exists a positive ε = ε(δ, ∆,r) such that the following holds. Let n , n , , n be arbitrary positive integers and let us replace the vertices v , v , , v 1 2 ··· r 1 2 ··· r with pairwise disjoint sets V ,V , ,V of sizes n , n , , n (blowing up). We construct 1 2 ··· r 1 2 ··· r two graphs on the same vertex set V = Vi. The ﬁrst graph K is obtained by replacing

each edge vivj of R with the complete bipartiteS graph between the corresponding vertex sets

Vi and Vj. A sparser graph G is constructed by replacing each edge vivj arbitrarily with an 18

(ε, δ)-super regular pair between V and V . If a graph H with ∆(H) ∆ is embeddable into i i ≤ K then it is already embeddable into G.

Lemma 3.2.3 (Blow-up lemma-strengthened version [36]). Given c > 0, there are positive numbers ε = ε(δ, ∆,r,c) and γ = γ(δ, ∆,r,c) such that the Blow-up lemma in the equal size case(all V are the same) remains true if for every i there are certain vertices x | i| to be embedded into V whose images are a priori restricted to certain sets C V provided i x ⊆ i that (i) each C within a V is of size at least c V ; x i | i| (ii) the number of such restrictions within a V is not more than γ V . i | i| Besides the above two lemmas, we also need the two lemmas below regarding regular pairs.

Lemma 3.2.4. If (A, B) is an ε-regular pair with density d, then for any A′ A with ⊆ A′ >ε A , there are at most ε B vertices b B such that deg(b, A′) (d ε) A′ . | | | | | | ∈ ≤ − | | Lemma 3.2.5 (Slicing lemma). Let (A, B) be an ε-regular pair with density d, and for some ν >ε, let A′ A and B′ B with A′ ν A , B′ ν B . Then (A′, B′) is an ⊆ ⊆ | | ≥ | | | | ≥ | | ε′-regular pair of density d′, where ε′ = max ε/ν, 2ε and d′ >d ε. { } − The following two results on hamiltonicity are used in ﬁnding hamiltonian cycles in the proofs.

Lemma 3.2.6 ([45]). If G is a graph of order n satisfying d(x)+ d(y) n +1 for every pair ≥ of nonadjacent vertices x, y V (G), then G is hamiltonian-connected. ∈ Lemma 3.2.7 ([42]). Let G be a balanced bipartite graph with 2n vertices. If d(x)+ d(v) ≥ n +1 for any two non-adjacent vertices x, y V (G), then G is hamiltonian. ∈

3.3 Dirac’s sondition for spanning Halin graphs

3.3.1 Introduction

A classic theorem of Dirac [19] from 1952 asserts that every graph on n vertices with minimum degree at least n/2 is hamiltonian if n 3. Following Dirac’s result, numerous ≥ 19 results on hamiltonicity properties on graphs with restricted degree conditions have been obtained (see, for instance, [26] and [25]). Traditionally, under similar conditions, results for a graph being hamiltonian, hamiltonian-connected, and pancyclic are obtained separately. We may ask, under certain conditions, if it is possible to uniformly show a graph possessing several hamiltonicity properties. The work on finding the square of a hamiltonian cycle in a graph can be seen as an attempt in this direction. However, it requires quite strong degree conditions for a graph to contain the square of a hamiltonian cycle, for examples, see [21], [22], [37], [9], and [49]. For bipartite graphs, finding the existence of a spanning ladder is a way of simultaneously showing the graph having many hamiltonicity properties (see [16] and [17]). In this paper, we introduce another approach of uniformly showing the possession of several hamiltonicity properties in a graph: we show the existence of a spanning Halin graph in a graph under given minimum degree condition. A tree with no vertex of degree 2 is called a homeomorphically irreducible tree (HIT). A Halin graph H is obtained from a HIT T of at least 4 vertices embedded in the plane by connecting its leaves into a cycle C following the cyclic order determined by the embedding. According to the construction, the Halin graph H is denoted as H = T C, and the ∪ HIT T is called the underlying tree of H. A wheel graph is an example of a Halin graph, where the underlying tree is a star. Halin constructed Halin graphs in [27] for the study of minimally 3-connected graphs. Lovász and Plummer named such graphs as Halin graphs in their study of planar bicritical graphs [40], which are planar graphs having a 1-factor after deleting any two vertices. Intensive researches have been done on Halin graphs. Bondy [5] in 1975 showed that a Halin graph is hamiltonian. In the same year, Lovász and Plummer [40] showed that not only a Halin graph itself is hamiltonian, but each of the subgraph obtained by deleting a vertex is hamiltonian. In 1987, Barefoot [2] proved that Halin graphs are hamiltonian-connected, i.e., there is a hamiltonian path connecting any two vertices of the graph. Furthermore, it was proved that each edge of a Halin graph is contained in a hamiltonian cycle and is avoided by another [48]. Bondy and Lovász [6], and Skowrońska [47], independently, in 1985, showed that a Halin graph is almost pancyclic and is pancyclic if the 20 underlying tree has no vertex of degree 3, where an n-vertex graph is almost pancyclic if it contains cycles of length from 3 to n with the possible exception of a single even length, and is pancyclic if it contains cycles of length from 3 to n. Some problems that are NP-complete for general graphs have been shown to be polynomial time solvable for Halin graphs. For example, Cornuéjols, Naddef, and Pulleyblank [15] showed that in a Halin graph, a hamiltonian cycle can be found in polynomial time. It seems so promising to show the existence of a spanning Halin subgraph in a given graph in order to show the graph having many hamiltonicity properties. But, nothing comes for free, it is NP-complete to determine whether a graph contains a (spanning) Halin graph [30]. Despite all these nice properties of Halin graphs mentioned above, the problem of deter- mining whether a graph contains a spanning Halin subgraph has not yet well studied except a conjecture proposed by Lovász and Plummer [40] in 1975. The conjecture states that every 4-connected plane triangulation contains a spanning Halin subgraph (disproved recently [10]). In this paper, we investigate the minimum degree condition for implying the existence of a spanning Halin subgraph in a graph, and thereby giving another approach for uniformly showing the possession of several hamiltonicity properties in a graph under a given minimum degree condition. We obtain the following result.

Theorem 3.3.1. There exists n > 0 such that for any graph G with n n vertices, if 0 ≥ 0 δ(G) (n + 1)/2, then G contains a spanning Halin subgraph. ≥

Note that an n-vertex graph with minimum degree at least (n + 1)/2 is 3-connected if n 4. Hence, the minimum degree condition in Theorem 3.3.1 implies the 3-connectedness, ≥ which is a necessary condition for a graph to contain a spanning Halin subgraph, since every Halin graph is 3-connected. A Halin graph contains a triangle, and bipartite graphs are triangle-free. Hence, K n n contains no spanning Halin subgraph. Immediately, we see ⌊ 2 ⌋,⌈ 2 ⌉ that the minimum degree condition in Theorem 3.3.1 is best possible. 21

3.3.2 Ladders and “ladder-like” Halin graphs

In constructing Halin graphs, we use ladder graphs and a class of “ladder-like” graphs as substructures. We give the description of these graphs below.

Deﬁnition 3.3.1. An n-ladder Ln = Ln(A, B) is a balanced bipartite graph with A = a , a , , a and B = b , b , , b such that a b iﬀ i j 1. We call a b { 1 2 ··· n} { 1 2 ··· n} i ∼ j | − | ≤ i i the i-th rung of L . If 2n(mod 4) 0, we call each of the shortest (a , b )-path and (b , a )- n ≡ 1 n 1 n path a side of Ln; otherwise we call each of the shortest (a1, an)-path and (b1, bn)-path a side of Ln.

Let L be a ladder with xy as one of its rungs. For an edge gh, we say xy and gh are adjacent if x g,y h or x h, y g. Suppose L has its ﬁrst rung as ab and its last ∼ ∼ ∼ ∼ rung as cd. We denote L by ab L cd in specifying the two rungs, and we always assume − − that the distance between a and c is V (L) /2 (we make this assumption for being convenient | | in constructing other graphs based on ladders). Under this assumption, we denote L as −→ab L −→cd. Let A and B be two disjoint vertex sets. We say the rung xy of L is contained − − in A B if either x A, y B or x B,y A. Let L′ be another ladder vertex-disjoint × ∈ ∈ ∈ ∈ with L. If the last rung of L is adjacent to the ﬁrst rung of L′, we write LL′ for the new ladder obtained by concatenating L and L′. In particular, if L′ = gh is an edge, we write LL′ as Lgh.

We now define five types of “ladder-like” graphs, call them H1,H2,H3,H4 and H5, respectively. Let Ln be a ladder with a1b1 and anbn as the first and last rung, respectively, and x,y,z,w,u five new vertices. Then each of Hi is obtained from Ln by adding some specified vertices and edges as follows. Additionally, for each i with 1 i 5, we define a ≤ ≤ graph Ti associated with Hi.

H1: Adding two new vertices x, y and the edges xa1, xb1, yan, ybn and xy.

Let T = H [ x, y, a , b , a , b ]. 1 1 { 1 1 n n}

H2: Adding three new vertices x, y, z and the edges za1, zb1, xz, xb1, yan, ybn and xy. 22

Let T = H [ x,y,z,a , b , a , b ]. 2 2 { 1 1 n n}

H3: Adding three new vertices x, y, z and the edges xa1, xb1, yan, ybn, either zai or zbi for some 2 i n 1 and xz, yz. ≤ ≤ − Let T = H [ x,y,z,a , b , a , b ]. 3 3 { 1 1 n n}

H4: Adding four new vertices x,y,z,w and the edges wa1, wb1, xw, xb1, yan, ybn, either zai or zb for some 2 i n 1 and xz, yz. i ≤ ≤ − Let T = H [ x,y,z,w,a , b , a , b ]. 4 4 { 1 1 n n}

H5: Adding ﬁve new vertices x,y,z,w,u.

If 2(n 1)(mod 4) 2, adding the edges wa , wb , xw, xb , ua , ub ,yu,yb , either za − ≡ 1 1 1 n n n i or zb for some 2 i n 1 and xz, yz ; i ≤ ≤ − and if 2(n 1)(mod 4) 0, adding the edges wa , wb , xw, xb , ua , ub ,yu,ya , either − ≡ 1 1 1 n n n za or zb for some 2 i n 1 and xz, yz. i i ≤ ≤ − Let T = H [ x,y,z,w,u,a , b , a , b ]. 5 5 { 1 1 n n}

Let i =1, 2, , 5. Notice that each of H is a Halin graph and except H , each H has ··· i 1 i a unique underlying tree. Notice also that xy is an edge on the cycle along the leaves of any underlying tree of Hi. For each Hi, call x the left end and y the right end, and call a vertex of degree at least 3 in the underlying tree of Hi a Halin constructible vertex. By analyzing the structure of H , we see that each of the vertices on one side of the ladder H x,y,z,w,u i i −{ } is a Halin constructible vertex. Noting that any vertex in V (H ) x, y can be a Halin 1 −{ } constructible vertex. In Figure 3.1, we depict a ladder L4, H1,H2,H3,H4,H5 constructed from L4, and the graph Ti associated with Hi. We call a1b1 the head link of Ti and anbn the tail link of Ti, and for each of T3, T4, T5, we call the vertex z not contained in any triangles the pendent vertex . The notations of Hi and Ti are ﬁxed hereafter. Let T T , , T be a subgraph of a graph G. Suppose that T has head link ab, ∈ { 1 ··· 5} tail link cd, and possibly the pendent vertex z. It is clear that if G V (T ) contains a − 23

spanning ladder L with ﬁrst rung c1d1 and last rung cndn such that c1d1 is adjacent to ab,

′ cndn is adjacent to cd, and z is adjacent some vertex z on some internal rung of L if z exists, then abLcd T or abLcd T zz′ when z exists is a spanning Halin subgraph of ∪ ∪ ∪{ } G. This technique is frequently used later on in constructing a Halin graph. The following proposition gives another way of constructing a Halin graph based on H1 and H2.

Proposition 3.3.1. For i = 1, 2, let G H ,H with left end x and right end y i ∈ { 1 2} i i be deﬁned as above, and let u V (G ) be a Halin constructible vertex, then G G i ∈ i 1 ∪ 2 − x y , x y x x ,y y ,u u is a Halin graph spanning on V (G ) V (G ). { 1 1 2 2}∪{ 1 2 1 2 1 2} 1 ∪ 2

Proof. For i =1, 2, let Gi be embedded in the plane, and let TGi be a underlying plane tree of G . Then T ′ := T T u u is a homeomorphically irreducible tree spanning i G1 ∪ G2 ∪{ 1 2} on V (G ) V (G ). Moreover, we can draw the edge u u such that T T u u is 1 ∪ 2 1 2 G1 ∪ G2 ∪{ 1 2} a plane graph. Since G [E(G T ) x y ] is an (x ,y )-path spanning on the leaves of i i − Gi −{ i i} i i

TGi obtained by connecting the leaves following the order determined by the embedding, we see G [E(G T ) x y ] G [E(G T ) x y ] x x ,y y is a cycle spanning 1 1 − G1 −{ 1 1} ∪ 2 2 − G2 −{ 2 2} ∪{ 1 2 1 2} on the leaves of T ′ obtained by connecting the leaves following the order determined by the embedding of T ′. Thus G G x y , x y x x ,y y ,u u is a Halin graph. 1 ∪ 2 −{ 1 1 2 2}∪{ 1 2 1 2 1 2}

3.3.3 Proof of Theorem 3.3.1

In this section, we prove Theorem 3.3.1. Following the standard setup of proofs applying the Regularity Lemma, we divide the proof into non-extremal case and extremal cases. For this purpose, we deﬁne the two extremal cases in the following. Let G be an n-vertex graph and V its vertex set. Given 0 β 1, the two extremal ≤ ≤ cases are deﬁned as below. Extremal Case 1. G has a vertex-cut of size at most 5βn. Extremal Case 2. There exists a partition V V of V such that V (1/2 7β)n and 1 ∪ 2 | 1| ≥ − ∆(G[V ]) βn. 1 ≤ Non-extremal case. We say that an n-vertex graph with minimum degree at least (n+1)/2 is in non-extremal case if it is in neither of Extremal Case 1 and Extremal Case 2. 24

a1 b2 a3 b4 a1 b4 a1 a4 z x y x y b a b1 a2 b3 a4 1 4 b1 b4 L 4 H1 H2 a1 b4 a1 b4 a1 b4 w w u x y x y x y b1 a4 b1 a4 b1 a4 z z z H3 H4 H5

z z z x y x y x y x y x y

T1 T2 T3 T4 T5

Figure (3.1) L , H constructed from L , and T associated with H for each i = 1, 2, , 5 4 i 4 i i · · ·

The following three theorems deal with the non-extremal case and the two extremal cases, respectively, and thus give a proof of Theorem 3.3.1.

Theorem 3.3.2. Suppose that 0 < β 1/(20 173) and n is a suﬃciently large integer. ≪ · Let G be a graph on n vertices with δ(G) (n + 1)/2. If G is in Extremal Case 1, then G ≥ contains a spanning Halin subgraph.

Theorem 3.3.3. Suppose that 0 < β 1/(20 173) and n is a suﬃciently large integer. ≪ · Let G be a graph on n vertices with δ(G) (n + 1)/2. If G is in Extremal Case 2, then G ≥ contains a spanning Halin subgraph.

Theorem 3.3.4. Let n be a suﬃciently large integer and G an n-vertex graph with δ(G) ≥ (n + 1)/2. If G is in the Non-extremal case, then G has a spanning Halin subgraph.

We need the following “Absorbing Lemma” in each of the proofs of Theorems 3.3.2 - 3.4.3 in dealing with “garbage” vertices.

Lemma 3.3.1 (Absorbing Lemma). Let F be a graph such that V (F ) is partitioned as S R. ∪ Suppose that (i) δ(R,S) 3 R , (ii) for any two vertices u, v N(R,S), deg(u,v,S) 6 R , ≥ | | ∈ ≥ | | 25

and (iii) for any three vertices u,v,w N(N(R,S),S), deg(u,v,w,S) 7 R . Then there ∈ ≥ | | is a ladder spanning on R and some other 7 R 2 vertices from S. | | − Proof. Let R = w ,w , ,w . Consider ﬁrst that r = 1. Choose x , x , x { 1 2 ··· r} | | 11 12 13 ∈ Γ(w ,S). By (ii), there are distinct vertices y1 Γ(x , x ,S) and y1 Γ(x , x ,S). 1 12 ∈ 11 12 23 ∈ 12 13 Then the graph L on w , x , x , x ,y1 ,y1 with edges in { 1 11 12 13 12 23}

w x ,w x ,w x ,y1 x ,y1 x ,y1 x ,y1 x { 1 11 1 12 1 13 12 11 12 12 23 12 23 13} is a ladder covering R with V (L) = 6. Suppose now r 2. For each i with 1 i r, choose | | ≥ ≤ ≤ distinct (and unchosen) vertices x , x , x Γ(w ,S). This is possible since deg(x, S) i1 i2 i3 ∈ i ≥ 3 R for each x R. By (ii), we choose distinct vertices y1 ,y1 , ,yr ,yr diﬀerent from the | | ∈ 12 23 ··· 12 23 existing vertices already chosen such that yi Γ(x , x ,S) and yi Γ(x , x ,S) for each 12 ∈ i1 i2 23 ∈ i2 i3 i, and at the same time, we chose distinct vertices z , z , , z from the unchosen vertices 1 2 ··· r−1 in S such that z Γ(x , x ,S) for each 1 i r 1. Finally, by (iii), choose distinct i ∈ i3 (i+1),1 ≤ ≤ − vertices u ,u , ,u from the unchosen vertices in S such that u Γ(x , x , z ,S). 1 2 ··· r−1 i ∈ i3 i+1,1 i Let L be the graph with

V (L)= R x , x , x ,yi ,yi , z ,u , x , x , x ,yr ,yr 1 i r 1 and ∪{ i1 i2 i3 12 23 i i r1 r2 r3 12 23 | ≤ ≤ − }

r r r r E(L) consisting of the edges wrxr1,wrxr2,wrxr3,y12xr1,y12xr2,y23xr2,y23xr3 and the edges indicated below for each 1 i r 1: ≤ ≤ −

w x , x , x ; yi x , x ; yi x , x ; z x , x ; u x , x , z . i ∼ i1 i2 i3 12 ∼ i1 i2 23 ∼ i2 i3 i ∼ i3 i+1,1 i ∼ i3 i+1,1 i

It is easy to check that L is a ladder covering R with V (L) = 8r 2. Figure 3.2 gives a | | − depiction of L for R = 2. | | The following simple observation is heavily used in the proofs explicitly or implicitly.

Lemma 3.3.2. Let U = u ,u ,u ,S V (G) be subsets. Then deg(u ,u , ,u ,S) { 1 2 ··· k} ⊆ 1 2 ··· k ≥ S (deg (u ,S)+ + deg (u ,S)) S k( S δ(U, S)). | | − G 1 ··· G k ≥| | − | | − 26

w1 w2

x11 x13 x21 x23

x12 x22

z1 y1 y1 2 2 12 23 y12 y23

Figure (3.2) Ladder L of order 14

Extremal Case 1 is relatively easy among the three cases, therefore we prove Theo- rem 3.3.2 ﬁrst below.

3.3.3.1 Proof of Theorem 3.3.2 We assume that G has a vertex-cut W such that W 5βn. As δ(G) (n + 1)/2, by simply counting degrees we see G W has exactly | | ≤ ≥ − two components. Let V1 and V2 be the vertex set of the two components, respectively. Then (1/2 5β)n V (1/2+5β)n. We partition W into two subsets as follows: − ≤| i| ≤

W = w W deg(w,V ) (n + 1)/4 2.5βn and W = W W . 1 { ∈ | 1 ≥ − } 2 − 1

As δ(G) (n + 1)/2, we have deg(w,V ) (n + 1)/4 2.5βn for any w W . Since G ≥ 2 ≥ − ∈ 2 is 3-connected and (1/2 5β)n > 3, there are three independent edges p p , q q , and r r − 1 2 1 2 1 2 between G[V W ] and G[V W ] with p , q ,r V W and p , q ,r V W . 1 ∪ 1 2 ∪ 2 1 1 1 ∈ 1 ∪ 1 2 2 2 ∈ 2 ∪ 2 For i = 1, 2, by the partition of W , we see that δ(W ,V ) 3 W + 3. As δ(G) i i i ≥ | i| ≥ (n + 1)/2, we have δ(G[V ]) (1/2 5β)n. Then, as V (1/2+5β)n, for any u, v V , i ≥ − | i| ≤ ∈ i deg(u,v,V ) (1/2 25β)n 6 W +2, and for any u,v,w V , deg(u,v,w,V ) (1/2 i ≥ − ≥ | i| ∈ i i ≥ − 35β)n 7 W + 2. By Lemma 3.4.2, we can ﬁnd a ladder L spans W p , q and another ≥ | i| i i −{ i i} 7 W p , q 2 vertices from V p , q if W p , q = . Denote a b and c d the | i −{ i i}| − i −{ i i} i −{ i i} 6 ∅ i i i i 27

ﬁrst and last rung of Li (if Li exists), respectively. Let

G = G[V V (L )] and n = V (G ) . i i − i i | i |

Then for i =1, 2, n (n+1)/2 5βn 7 W (n+1)/2 40βn and δ(G ) δ(G[V ]) 7 W (n+1)/2 40βn. i ≥ − − | i| ≥ − i ≥ i − | i| ≥ −

Let i =1, 2. We now show that Gi contains a spanning subgraph isomorphic to either H or H as deﬁned in the beginning of this section. Since n (1/2+5β)n and δ(G ) 1 2 | i| ≤ i ≥ (n+1)/2 40βn, any subgraph of G induced on at least (1/4 40β)n vertices has minimum − i − degree at least (1/4 85β)n, and thus has a matching of size at least 2. Hence, when n is − i even, we can choose independent edges ei = xiyi and fi = ziwi with

x ,y Γ (p ) q and z ,w Γ (q ) p . i i ∈ Gi i −{ i} i i ∈ Gi i −{ i}

(Notice that p or q may be contained in W , and in this case we have deg (p ),deg (q ) i i i Gi i Gi i ≥ (1/4 40β)n.) And if n is odd, we can choose independent edges g y and f = z w with − i i i i i i

g , x ,y Γ (p ) q , x Γ (g ,y ) p , q and z ,w Γ (q ) x ,p , i i i ∈ Gi i −{ i} i ∈ Gi i i −{ i i} i i ∈ Gi i −{ i i}

where the existence of the vertex xi is possible since the subgraph of Gi induced on ΓGi (pi) has minimum degree at least (1/2 40β)n ((1/2+5β)n Γ (p ) ) Γ (p ) 45βn, − − −| Gi i | ≥ | Gi i | − and hence contains a triangle. In this case, again, denote ei = xiyi. Let

′ Gi = Gi pi, qi , if ni is even;  −{ }  ′ Gi = Gi pi, qi,gi , if ni is odd. −{ }   By the deﬁnition above, V (G′ ) is even. | i | The following claim is a modiﬁcation of (1) of Lemma 2.2 in [17]. 28

Claim 3.3.1. For i =1, 2, let a′ b′ ,c′ d′ E(G′ ) be two independent edges. Then G′ contains i i i i ∈ i i ′ two vertex disjoint ladders Qi1 and Qi2 spanning on V (Gi) such that Qi1 has ei = xiyi as ′ ′ ′ ′ its first rung, aibi as its last rung, and Qi2 has cidi as its first rung and fi = ziwi as its last rung, where ei and fi are defined prior to this claim.

Proof. We only show the claim for i = 1 as the case for i = 2 is similar. Notice that by the definition of G′ , V (G′ ) is even. Since V (G′ ) (1/2+5β)n and δ(G′ ) (n +1)/2 1 | 1 | | 1 | ≤ 1 ≥ − 40βn 2 V (G′ ) /2 + 8, G′ has a perfect matching M containing e , f , a′ b′ ,c′ d′ . We − ≥ | 1 | 1 1 1 1 1 1 1 ′ ′ ′ ′ ′ ′ identify a1 and c1 into a vertex called s , and identify b1 and d1 into a vertex called t . Denote G′′ as the resulting graph and let s′t′ E(G′′) if the two vertices are not adjacent. Partition 1 ∈ 1 V (G′′) arbitrarily into U and V with U = V such that x , z ,s′ U, y ,w , t′ V , and let 1 | | | | 1 1 ∈ 1 1 ∈ ′ ′ ′ ′ ′ ′ ′ ′ M := M a b ,c d s t EG′ (U, V ). Define an auxiliary graph H with vertex set −{ 1 1 1 1}∪{ } ⊆ 1 ′ ′ M and edge set defined as follows. If xy, uv M with x, u U then xy ′ uv if and only ∈ ∈ ∼H if x G′ v and y G′ u (we do not include the case that x G′ u and y G′ v as we defined ∼ 1 ∼ 1 ∼ 1 ∼ 1 ′ ′ ′ ′ ′ a bipartition here). Particularly, for any pq M s t with p U, pq ′ s t if and ∈ −{ } ∈ ∼H ′ ′ ′ ′ ′ only if p G′ b ,d and q G′ a ,c . Notice that a ladder with rungs in M is corresponding ∼ 1 1 1 ∼ 1 1 1 to a path in H′ and vice versa. Since (1/2 40β)n 2 V (G′ ) (1/2+5β)n 2 and − − ≤ | 1 | ≤ − δ(G′ ) (n + 1)/2 40βn 2, any two vertices in G′ has at least (1/2 130β)n common 1 ≥ − − 1 − neighbors. This together with the fact that U = V V (G′′) /2 (1/4+2.5β)n gives | | | |≤| 1 | ≤ that δ(U, V ), δ(V, U) (1/4 132.5β)n. Hence ≥ −

δ(H′) (1/4 132.5β)n ((1/4+2.5β)n (1/4 132.5β)n)=(1/4 267.5β)n V (H′) /2+1, ≥ − − − − − ≥| |

′ since β < 1/2200 and n is very large. Hence H has a hamiltonian path starting with e1,

′ ′ ′ ′ ′ ′ ending with f1, and having s t as an internal vertex. The path with s t replaced by a1b1 ′ ′ ′ and c1d1 is corresponding to the required ladders in G1.

We may assume n1 is even and n2 is odd and construct a spanning Halin subgraph of G (the construction for the other three cases follow a similar argument). Recall that p p , q q ,r r are the three prescribed independent edges between G[V W ] and G[V W ], 1 2 1 2 1 2 1 ∪ 1 2 ∪ 2 29

where p , q ,r V W and p , q ,g ,r V W . For a uniform discussion, we may 1 1 1 ∈ 1 ∪ 1 2 2 2 2 ∈ 2 ∪ 2 assume that both of the ladders L1 and L2 exist. Let i = 1, 2. Recall that Li has aibi

′ ′ as its ﬁrst rung and cidi as its last rung. Choose a ΓG′ (ai), b ΓG′ (bi) such that i ∈ i i ∈ i ′ ′ ′ ′ ′ ′ a b E(G) and c ΓG′ (ci), d ΓG′ (di) such that c d E(G). This is possible as i i ∈ i ∈ i i ∈ i i i ∈ δ(G′ ) (n+1)/2 40βn 2. Let Q and Q be the ladders of G′ given by Claim 3.3.1. Set i ≥ − − 1i 2i i H = Q L Q p x ,p y , q z , q w . Assume Q L Q is a ladder can be denoted as a 11 1 12 ∪{ 1 1 1 1 1 1 1 1} 21 2 22 x y Q L Q z w . To make r a Halin constructible vertex, we let H = Q L Q −−→2 2 − 21 2 22 − −−→2 2 2 b 21 2 22 ∪ g x ,g y ,p g ,p y , q z , q w if r is on the shortest (y ,w )-path in Q L Q , and let { 2 2 2 2 2 2 2 2 2 2 2 2} 2 2 2 21 2 22 H = Q L Q g x ,g y ,p g ,p x , q z , q w if r is on the shortest (x , z )-path (recall b 21 2 22 ∪{ 2 2 2 2 2 2 2 2 2 2 2 2} 2 2 2 that g , x ,y Γ (p )). Let H = H H p p ,r r , q q . Then H is a spanning Halin 1 1 1 ∈ G1 1 a ∪ b ∪{ 1 2 1 2 1 2} subgraph of G by Proposition 3.3.1 as H p q = H and H p q = H . Figure 3.3 a ∪{ 1 1} ∼ 1 b ∪{ 2 2} ∼ 2 gives a construction of H for the above case when r2 is on the shortest (y2,w2)-path in

Q21L2Q22.

a2 d2

b ′ c2 x2 2 ′ d2 z2 a2 g2 Q22 Q21 p2 ′ q2 y b2 ′ w 2 r2 ′ c2 2 y1 a1 ′ r1 w1 d1 p1 q1

Q11 Q12 x1 ′ ′ z b1 c1 1 a1 d1

b1 c1

Figure (3.3) A Halin graph H 30

3.3.3.2 Proof of Theorem 3.3.3 Recall Extremal Case 2: There exists a partition V V of V such that V (1/2 7β)n and ∆(G[V ]) βn. Since δ(G) (n + 1)/2, the 1 ∪ 2 | 1| ≥ − 1 ≤ ≥ assumptions imply that

(1/2 7β)n V (1/2+ β)n and (1/2 β)n V (1/2+7β)n. − ≤| 1| ≤ − ≤| 2| ≤

Let β and α be real numbers satisfying β α/20 and α (1/17)3. Set α = α1/3 and ≤ ≤ 1 2/3 α2 = α We ﬁrst repartition V (G) as follows.

V ′ = v V deg(v,V ) (1 α ) V ,V = v V V ′ deg(v,V ′) (1 α ) V ′ , 2 { ∈ 2 | 1 ≥ − 1 | 1|} 01 { ∈ 2 − 2 | 2 ≥ − 1 | 2 |} V ′ = V V , and V = V V ′ V . 1 1 ∪ 01 0 2 − 2 − 01

Claim 3.3.2. V , V V V ′ α V . | 01| | 0|≤| 2 − 2 | ≤ 2| 2|

Proof. Notice that e(V ,V ) (1/2 7β)n V 1/2−7β V V (1 α) V V as 1 2 ≥ − | 2| ≥ 1/2+β | 1|| 2| ≥ − | 1|| 2| β α/20. Hence, ≤

(1 α) V V e(V ,V ) e(V ,V ′)+ e(V ,V V ′) V V ′ + (1 α ) V V V ′ . − | 1|| 2| ≤ 1 2 ≤ 1 2 1 2 − 2 ≤| 1|| 2 | − 1 | 1|| 2 − 2 |

This gives that V V ′ α V , and thus V , V V V ′ α V . | 2 − 2 | ≤ 2| 2| | 01| | 0|≤| 2 − 2 | ≤ 2| 2| As a result of moving vertices from V2 to V1 and by Claim 3.3.2, we have the following.

∆(G[V ′]) βn + V βn + α V , 1 ≤ | 01| ≤ 2| 2| δ(V ′,V ′) (1/2 β)n V V ′ (1/2 β)n α V , 1 2 ≥ − −| 2 − 2 | ≥ − − 2| 2| δ(V ′,V ′) (1 α ) V (1 α )(1/2 7β)n, (3.1) 2 1 ≥ − 1 | 1| ≥ − 1 − δ(V ,V ′) (n + 1)/2 (1 α ) V ′ V 3α n +8 3 V + 10, 0 1 ≥ − − 1 | 2 |−| 0| ≥ 2 ≥ | 0| δ(V ,V ′) (n + 1)/2 (1 α ) V V 3α n +8 3 V + 10, 0 2 ≥ − − 1 | 1|−| 0| ≥ 2 ≥ | 0|

where the last two inequalities hold because we have 7β + 10/n α, and α (1/8)3. ≤ ≤ 31

Claim 3.3.3. We may assume that ∆(G) < n 1. −

Proof. Suppose on the contrary and let w V (G) such that deg(w) = n 1. Then ∈ − by δ(G) (n + 1)/2 we have δ(G w) (n 1)/2, and thus G w has a hamiltonian ≥ − ≥ − − cycle. This implies that G has a spanning wheel subgraph, in particular, a spanning Halin subgraph of G.

Claim 3.3.4. There exists a subgraph T G such that V (T ) n (mod 2), where T is ⊆ | | ≡ isomorphic to some graph in T , T , , T . Assume that T has head link x x and tail { 1 2 ··· 5} 1 2 link y y . Let m = n V (T ) . Then G V (T ) contains a balanced spanning bipartite graph 1 2 −| | − G′ with partite sets U and U and a subset W of U U with at most α n vertices such 1 2 1 ∪ 2 2 that the following holds:

′ (i) deg ′ (x, V (G ) W ) (1 α 2α )m for all x W ; G − ≥ − 1 − 2 6∈

(ii) There exists x′ x′ ,y′ y′ E(G′) such that x′ ,y′ U W , x′ x , and y′ y , for 1 2 1 2 ∈ i i ∈ i − 3−i ∼ i 3−i ∼ i i =1, 2; and if T has a pendent vertex, then the vertex is contained in V ′ V ′ W . 1 ∪ 2 −

(iii) There are W vertex-disjoint 3-stars (K s) in G′′ x′ , x′ ,y′ ,y′ with the vertices | | 1,3 −{ 1 2 1 2} in W as their centers.

Proof. By (3.1), for i =1, 2, we notice that for any u,v,w V ′, ∈ i

deg(u,v,w,V ′ ) V ′ 3( V ′ δ(V ′,V ′ )) (1/2 28β 3α )n>n/4.(3.2) 3−i ≥ | 3−i| − | 3−i| − i 3−i ≥ − − 1

We now separate the proof into two cases according to the parity of n. Case 1. n is even. Suppose ﬁrst that max V ′ , V ′ n/2. We arbitrarily partition V into V and {| 1 | | 2 |} ≤ 0 10 V such that V ′ V = V ′ V = n/2. Suppose G[V ′] contains an edge x u and 20 | 1 ∪ 10| | 2 ∪ 20| 1 1 1 there is a vertex u Γ(u ,V ′) such that u is adjacent to a vertex y V ′. By (3.2), 2 ∈ 1 2 2 2 ∈ 2 there exist distinct vertices x Γ(x ,u ,V ′) y ,u ,y Γ(y ,u ,V ′) x ,u . Then 2 ∈ 1 1 2 −{ 2 1} 1 ∈ 2 2 1 −{ 1 1} G[ x ,u , x ,y ,u ,y ] contains a subgraph T isomorphic to T . So we assume G[V ′] { 1 1 2 1 1 2} 1 1 32

′ ′ contains an edge x1u1 and no vertex in Γ(u1,V2 ) is adjacent to any vertex in V2 . As δ(G) (n + 1)/2, δ(G[V ′ V ]) 1. Let u Γ(u ,V ′) and u y E(G[V ′ V ]). ≥ 2 ∪ 20 ≥ 2 ∈ 1 2 2 2 ∈ 2 ∪ 20 Since deg(u ,V ′) (n + 1)/2 V > V ′ V V and deg(y ,V ′) 3 V + 10, 2 1 ≥ − | 0| | 1 ∪ 10|−| 0| 2 1 ≥ | 0| deg(u ,y ,V ′ V ) 2 V +10. Let x Γ(x ,u ,V ′) y ,u ,y Γ(y ,u ,V ′) x ,u . 2 2 1 ∪ 10 ≥ | 0| 2 ∈ 1 1 2 −{ 2 2} 1 ∈ 2 2 1 −{ 1 1} Then G[ x ,u , x ,y ,u ,y ] contains a subgraph T isomorphic to T . By symmetry, we { 1 1 2 1 2 2} 1 ′ ′ ′ can ﬁnd T ∼= T1 if G[V2 ] contains an edge. Hence we assume that both V1 and V2 are independent sets. Again, as δ(G) (n + 1)/2, δ(G[V ′ V ]), δ(G[V ′ V ]) 1. Let ≥ 1 ∪ 10 2 ∪ 20 ≥ x u E(G[V ′ V ]) and y u E(G[V ′ V ]) such that x V ′ and u Γ(u ,V ′). 1 1 ∈ 1 ∪ 10 2 2 ∈ 2 ∪ 20 1 ∈ 1 2 ∈ 1 2 Since deg(x ,V ′) (n + 1)/2 V > V ′ V V and deg(u ,V ′) 3 V + 10, we 1 2 ≥ −| 0| | 2 ∪ 20|−| 0| 1 2 ≥ | 0| have deg(x ,u ,V ′) 2 V + 10. Hence, there exists x Γ(x ,u ,V ′) y ,u . Similarly, 1 1 2 ≥ | 0| 2 ∈ 1 1 2 −{ 2 2} there exists y Γ(y ,u ,V ′) x ,u . Then G[ x ,u , x ,y ,u ,y ] contains a subgraph 1 ∈ 2 2 1 −{ 1 1} { 1 1 2 1 2 2} T isomorphic to T . Let m =(n 6)/2, U =(V ′ V (T )) V and U =(V ′ V (T )) V , 1 − 1 1 − ∪ 10 2 2 − ∪ 20 and W = V V (T ). We then have U = U = m. 0 − | 1| | 2| Let G′ = (V (G) V (T ), E (U , U )) be the bipartite graph with partite sets U and − G 1 2 1 ′ U . Notice that W V α V < α n. By (3.1), we have deg ′ (x, V (G ) W ) 2 | |≤| 0| ≤ 2| 2| 2 G − ≥ (1 α 2α )m for all x / W . This shows (i). By the construction of T above, we have − 1 − 2 ∈ x ,y V ′. Let i =1, 2. By (3.1), we have δ(V , U W ) 3 V + 6. Applying statement 1 1 ∈ 1 0 i − ≥ | 0| (i), we have e ′ (Γ ′ (x , U W ), Γ ′ (x , U W )), e ′ (Γ ′ (y , U W ), Γ ′ (y , U W )) G G 1 2 − G 2 1 − G G 1 2 − G 2 1 − ≥ (3 V + 4)(1 2α 4α )m > 2m. Hence, we can ﬁnd independent edges x′ x′ and y′ y′ | 0| − 1 − 2 1 2 1 2 such that x′ ,y′ U W , x′ x , and y′ y . This gives statement (ii). Finally, as i i ∈ i − 3−i ∼ i 3−i ∼ i δ(V , U W ) 3 V + 6, we have δ(V , U W x′ , x′ ,y′ ,y′ ) 3 V + 2. Hence, there 0 i − ≥ | 0| 0 i − −{ 1 2 1 2} ≥ | 0| are W vertex-disjoint 3-stars with their centers in W . | | Otherwise we have max V ′ , V ′ > n/2. Assume, w.l.o.g., that V ′ n/2 + 1. {| 1 | | 2|} | 1 | ≥ Then δ(G[V ′]) 2 and thus G[V ′] contains two vertex-disjoint paths isomorphic to P 1 ≥ 1 3 and P , respectively. Let m = (n 8)/2. We consider three cases here. Case (a): 2 − V ′ 5 m. Then let x u w ,y v G[V ′] be two vertex-disjoint paths, and let | 1 | − ≤ 1 1 1 1 1 ⊆ 1 x Γ(x ,u ,w ,V ′),y Γ(y , v ,V ′) and z Γ(w , v ,V ′) be three distinct vertices. 2 ∈ 1 1 1 2 2 ∈ 1 1 2 ∈ 1 1 2 Then G[ x ,u ,w , x ,z,y , v ,y ] contains a subgraph T isomorphic to T . Notice that { 1 1 1 2 1 1 2} 4 33

V ′ V (T ) m. We arbitrarily partition V into V and V such that V ′ V = | 2 − | ≤ 0 10 20 | 1 ∪ 10| V ′ V = m. Let U = (V ′ V (T )) V , U = (V ′ V (T )) V , and W = V . Hence | 2 ∪ 20| 1 1 − ∪ 10 2 2 − ∪ 20 0 we assume V ′ 5= m + t for some t 1. This implies that V ′ n/2+ t + 1 and thus | 1 | − 1 1 ≥ | 1 | ≥ 1 δ(G[V ′]) t +2. Let V 0 be the set of vertices u V ′ such that deg(u,V ′) α m. Case 1 ≥ 1 1 ∈ 1 1 ≥ 1 (b): V 0 V ′ 5 m. Then we form a set W with V ′ 5 m vertices from V 0 and all | 1 |≥| 1 | − − | 1 | − − 1 the vertices of V . Then V ′ W = m +5+ t ( V ′ 5 m) = m +5= n/2+1, and 0 | 1 − | 1 − | 1 | − − hence δ(G[V ′ W ]) 2. Similarly as in Case (a), we can ﬁnd a subgraph T of G contained 1 − ≥ in G[V ′ W ] isomorphic to T . Let U = V ′ V (T ) W , U = (V ′ V (T )) W . 1 − 4 1 1 − − 2 2 − ∪ Then U = U = m. Thus we have Case (c): V 0 < V ′ 5 m. Suppose | 1| | 2| | 1 | | 1 | − − that V ′ V 0 = m +5+ t′ = n/2 + t′ + 1 for some t′ 1. This implies that | 1 − 1 | 1 1 1 ≥ δ(G[V ′ V 0]) t′ + 2. We show that G[V ′ V 0] contains t′ + 2 vertex-disjoint 3-stars. 1 − 1 ≥ 1 1 − 1 1 To see this, suppose G[V ′ V 0] contains a subgraph M of at most s < t′ + 2 3-stars. By 1 − 1 1 counting the number of edges between V (M) and V ′ V 0 V (M) in two ways, we get that 1 − 1 − ′ ′ 0 ′ 0 ′ 0 t V V V (M) eG−V 0 (V (M),V V V (M)) 4s∆(G[V V ]) 4sα1m. Since 1| 1 − 1 − | ≤ 1 1 − 1 − ≤ 1 − 1 ≤ V ′ V 0 = m+5+t′ = n/2+t′ +1, V ′ V 0 V (M) m 3t′ m 6α m, where the last | 1 − 1 | 1 1 | 1 − 1 − | ≥ − 1 ≥ − 2 inequality holds as V ′ (1/2+ β)n + α V ′ implying that t′ V ′ m 5 2α m. This, | 1 | ≤ 2| 2 | 1 ≤| 1 | − − ≤ 2 together with the assumption that α (1/8)3 gives that s t′ + 2, showing a contradiction. ≤ ≥ 1 Hence we have s t′ +2. Let x u w and y v be two paths taken from two 3-stars in M. ≥ 1 1 1 1 1 1 Then we can ﬁnd a subgraph T of G isomorphic to T4 the same way as in Case (a). We

′ take exactly t1 3-stars from the remaining ones in M and denote the centers of these stars by W ′. Let U = V ′ V 0 V (T ) W ′, W = W ′ V 0 V , and U = (V ′ V (T )) W . 1 1 − 1 − − ∪ 1 ∪ 0 2 2 − ∪ Then U = U = m. | 1| | 2| For the partition of U1 and U2 in all the cases discussed in the paragraph above, we let G′ =(V (G) V (T ), E (U , U )) be the bipartite graph with partite sets U and U . Notice − G 1 2 1 2 that W V α n if Case (a) occurs, W V + V ′ m 5 (1/2+β)n+ V n/2 | |≤| 0| ≤ 2 | |≤| 0| | 1 |− − ≤ | 0|− ≤ α n if Case (b) occurs, and W = W ′ V 0 V = V ′ U V (T ) + V (1/2+β)n (1/2 2 | | | ∪ 1 ∪ 0| | 1 − 1− | | 0| ≤ − − 4)n + V α n if Case (c) occurs. Since δ(V ′,V ′) (1 α ) V from (3.1) and V ′ U | 0| ≤ 2 2 1 ≥ − 1 | 1| | 1 − 1| ≤ 2α m, we have δ(U W, U W ) (1 α 2α )m. On the other hand, from (3.1), 2 2 − 1 − ≥ − 1 − 2 34

δ(V ′,V ′) (1/2 β)n α V . This gives that δ(U W, U W ) (1 α 2α )m. 1 2 ≥ − − 2| 2| 1 − 2 − ≥ − 1 − 2 ′ Hence, we have deg ′ (x, V (G ) W ) (1 α 2α )m for all x / W . Applying statement G − ≥ − 1 − 2 ∈ (i), we have e ′ (Γ ′ (x , U W ), Γ ′ (x , U W )), e ′ (Γ ′ (y , U W ), Γ ′ (y , U W )) G G 1 2 − G 2 1 − G G 1 2 − G 2 1 − ≥ (3 V +4)(1 2α 4α )m> 2m. Hence, we can find independent edges x′ x′ and y′ y′ such | 0| − 1 − 2 1 2 1 2 that x′ ,y′ U W , x′ x , and y′ y . By the construction of T , T is isomorphic i i ∈ i − 3−i ∼ i 3−i ∼ i to T , and the pendent vertex z V ′ V V ′ W . This gives statement (ii). Finally, as 4 ∈ 2 ⊆ 1 ∪ 2 − δ(V , U W ) 3α n +5 3 W + 5, we have δ(V , U W x′ , x′ ,y′ ,y′ ) 3 W + 1. 0 1 − ≥ 2 ≥ | | 0 1 − −{ 1 2 1 2} ≥ | | By the definition of V 0, we have δ(V 0,V ′ W x′ , x′ ,y′ ,y′ ) α m α n 4 3 W . 1 1 1 − −{ 1 2 1 2} ≥ 1 − 2 − ≥ | | For the vertices in W ′ in Case (c), we already know that there are vertex-disjoint 3-stars in G′ with centers in W ′. Hence, regardless of the construction of W , we can always find W | | vertex-disjoint 3-stars with their centers in W . Case 2. n is odd. Suppose first that max V ′ , V ′ (n + 1)/2 and let m = (n 7)/2. We arbitrarily {| 1 | | 2 |} ≤ − partition V into V and V such that, w.l.o.g., say V ′ V = (n + 1)/2 and V ′ 0 10 20 | 1 ∪ 10| | 2 ∪ V = (n 1)/2. We show that G[V ′ V ] either contains two independent edges or is 20| − 1 ∪ 10 isomorphic to K . As δ(G) (n + 1)/2, we have δ(G[V ′ V ]) 1. Since n is 1,(n−1)/2 ≥ 1 ∪ 10 ≥ sufficiently large, (n +1)/2 > 3. Then it is easy to see that if G[V ′ V ] = K , then 1 ∪ 10 6∼ 1,(n−1)/2 G[V ′ V ] contains two independent edges. Furthermore, we can choose two independent 1 ∪ 10 edges x u and y v such that u , v V ′. This is obvious if V 1. So we assume 1 1 1 1 1 1 ∈ 1 | 10| ≤ V 2. As δ(V ,V ′) 3 V + 10, by choosing x ,y V , we can choose distinct | 10| ≥ 0 1 ≥ | 0| 1 1 ∈ 10 vertices u Γ(x ,V ′) and v Γ(y ,V ′). Let x Γ(x ,u ,V ′),y Γ(y , v ,V ′) and 1 ∈ 1 1 1 ∈ 1 1 2 ∈ 1 1 2 2 ∈ 1 1 2 z Γ(u , v ,V ′). Then G[ x ,u , x ,y , v ,y , z ] contains a subgraph T isomorphic to T . ∈ 1 1 2 { 1 1 2 1 1 2 } 3 We assume now that G[V ′ V ] is isomorphic to K . Let u be the center of the 1 ∪ 10 1,(n−1)/2 1 star K . Then each leave of the star has at least (n 1)/2 neighbors in V ′ V . 1,(n−1)/2 − 2 ∪ 20 Since V ′ V = (n 1)/2, we have Γ(v,V ′ V ) = V ′ V if v V ′ V u . | 2 ∪ 20| − 2 ∪ 20 2 ∪ 20 ∈ 1 ∪ 10 −{ 1} By the definition of V , ∆(V ,V ′) < (1 α ) V and ∆(V ,V ′) < (1 α ) V ′ , and so 0 0 1 − 1 | 1| 0 2 − 1 | 2 | u V ′, V = and V = . We claim that V ′ is not an independent set. Otherwise, by 1 ∈ 1 10 ∅ 20 ∅ 2 δ(G) (n + 1)/2, for each v V ′, Γ(v,V ′) = V ′. This in turn shows that u has degree ≥ ∈ 2 1 1 1 35 n 1, showing a contradiction to Claim 3.3.3. So let y v E(G[V ′]) be an edge. Let w − 2 2 ∈ 2 1 ∈ Γ(v ,V ′) u and w u x the path containing w . Choose y Γ(y , v ,V ′) w ,u , x 2 1 −{ 1} 1 1 1 1 1 ∈ 2 2 1 −{ 1 1 1} and x Γ(x ,u ,w ,V ′) y , v . Then G[ x ,u , x ,w , v ,y ,y ] contains a subgraph T 2 ∈ 1 1 1 2 −{ 1 1} { 1 1 2 1 2 2 1} isomorphic to T . Let U =(V ′ V (T )) V and U =(V ′ V (T )) V and W = V V (T ). 2 1 1 − ∪ 10 2 2 − ∪ 20 0 − We have U = U = m and W V α n. | 1| | 2| | |≤| 0| ≤ 2 Otherwise we have max V ′ , V ′ (n + 1)/2 + 1. Assume, w.l.o.g., that V ′ {| 1 | | 2 |} ≥ | 1 | ≥ (n + 1)/2 + 1. Then δ(G[V ′]) 2 and thus G[V ′] contains two independent edges. Let 1 ≥ 1 m =(n 7)/2 and V 0 be the set of vertices u V ′ such that deg(u,V ′) α m. We consider − 1 ∈ 1 1 ≥ 1 three cases here. Since V ′ (n +1)/2+1 > m + 4, we assume V ′ = m +4+ t for some | 1 | ≥ | 1 | 1 t 1. Case (a): V 0 V ′ m 4. Then we form a set W with V ′ 4 m vertices from V 0 1 ≥ | 1 |≥| 1 |− − | 1 |− − 1 and all the vertices of V . Then V ′ W = m+4+t ( V ′ 4 m)= m+4=(n+1)/2+1. 0 | 1 − | 1 − | 1 |− − Then we have δ(G[V ′ W ]) 2. Hence G[V ′ W ] contains two independent edges. Let 1 − ≥ 1 − x u ,y v E(G[V ′ W ]) be two independent edges, and let x Γ(x ,u ,V ′),y 1 1 1 1 ⊆ 1 − 2 ∈ 1 1 2 2 ∈ Γ(y , v ,V ′) and z Γ(w , v ,V ′) be three distinct vertices. Then G[ x ,u , x ,z,y , v ,y ] 1 1 2 ∈ 1 1 2 { 1 1 2 1 1 2} contains a subgraph T isomorphic to T . Let U = V ′ V (T ) W , U =(V ′ V (T )) W . 3 1 1 − − 2 2 − ∪ Then U = U = m and W V + V ′ U V V ′ + βn +4α n. Thus we | 1| | 2| | |≤| 0| | 1 − 1|≤| 2 − 2 | 2 have V 0 < V ′ 4 m. Suppose that V ′ V 0 = m +4+ t′ = (n + 1)/2+ t′ for | 1 | | 1 | − − | 1 − 1 | 1 1 some t′ 1. This implies that δ(G[V ′ V 0]) t′ +1. Case (b): t′ 2. We show that 1 ≥ 1 − 1 ≥ 1 1 ≥ G[V ′ V 0] contains t′ + 2 vertex-disjoint 3-stars. To see this, suppose G[V ′ V 0] contains a 1 − 1 1 1 − 1 ′ subgraph M of at most s vertex disjoint 3-stars. We may assume that s < t1 + 2. Then we ′ 0 ′ 0 ′ 0 have (t1 1) V V V (M) eG−V 0 (V (M),V V V (M)) 4s∆(G[V V ]). Since − | 1 − 1 − | ≤ 1 1 − 1 − ≤ 1 − 1 V ′ V 0 = m+5+t′ =(n+1)/2+t′ , V ′ V 0 V (M) m 3t′ m 6α m, where the last | 1 − 1 | 1 1 | 1 − 1 − | ≥ − 1 ≥ − 2 inequality holds as V ′ (1/2+ β)n + α V ′ implying that t′ V ′ m 5 2α m. This, | 1 | ≤ 2| 2 | 1 ≤| 1 | − − ≤ 2 together with the assumption that α (1/8)3 gives that s t′ + 2, showing a contradiction. ≤ ≥ 1 Hence we have s t′ +2. Let x u and y v be two paths taken from two 3-stars in M, ≥ 1 1 1 1 1 and we can find a subgraph T of G isomorphic to T3 the same way as in Case (a). We take

′ ′ exactly t1 3-stars from the remaining ones in M and denote the centers of these stars by W . Let U = V ′ V 0 V (T ) W ′, W = W ′ V 0 V , and U = (V ′ V (T )) W . Then 1 1 − 1 − − ∪ 1 ∪ 0 2 2 − ∪ 36

U = U = m. Case (c): t′ = 1. In this case, we let m =(n 9)/2. If G[V ′ V 0] contains | 1| | 2| 1 − 1 − 1 a vertex adjacent to all other vertices in V ′ V 0, we take this vertex to V ′. This gets back 1 − 1 2 to Case (a). Hence, we assume that G[V ′ V 0] has no vertex adjacent to all other vertices 1 − 1 in V ′ V 0. Then by the assumptions that δ(G) (n +1)/2 and V ′ V 0 =(n +1)/2 + 1, 1 − 1 ≥ | 1 − 1 | we can ﬁnd two copies of vertex disjoint P s in G[V ′ V 0]. Let x u w and y v z be two 3 1 − 1 1 1 1 1 1 1 P s in G[V ′]. There exist distinct vertices x Γ(x ,u ,w ,V ′),y Γ(y , v , z ,V ′) and 3 1 2 ∈ 1 1 1 2 2 ∈ 1 1 1 2 z Γ(w , z ,V ′). Then G[ x ,u ,w , x ,y , v , z ,y , z ] contains a subgraph T isomorphic ∈ 1 1 2 { 1 1 1 2 1 1 1 2 } to T . Let U = V ′ V 0 V (T ), W = V 0 V , and U = V ′ V (T ). Then U = U = m. 5 1 1 − 1 − 1 ∪ 0 2 2 − | 1| | 2| For the partition of U and U in all the cases discussed in Case 2, we let G′ =(V (G) 1 2 − V (T ), EG(U1, U2)) be the bipartite graph with partite sets U1 and U2. Similarly as in Case 1, we can show that all the statements (i)-(iii) hold.

Let W = U W and W = U W . For i =1, 2, by the definition of W , we see that 1 1 ∩ 2 2 ∩ δ(W , U x′ ,y′ , x′ ,y′ ) 3 W . And for any u, v U , Γ(u,v,U ) 6 W , and for any i i −{ 1 1 2 2} ≥ | i| ∈ i 3−i ≥ | i| u,v,w U , Γ(u,v,w,U ) 7 W . By Lemma 3.4.2, we can find ladder L spanning on ∈ i 3−i ≥ | i| i W and another 7 W 2 vertices from U x′ , x′ ,y′ ,y′ if W = . Denote a a and i | i| − i −{ 1 2 1 2} i 6 ∅ 1i 2i b b the first and last rungs of L (if L exists), respectively, where a , b U . Let 1i 2i i i 1i 1i ∈ 1

U ′ = U V (L ), m′ = U ′ = U ′ , and G′′ = G′′(U ′ U ′ , E (U ′ , U ′ )). i i − i | 1| | 2| 1 ∪ 2 G 1 2

Since W α n, m (n 9)/2, and n is suﬃciently large, we have 1/n +7 W 15α m. | | ≤ 2 ≥ − | | ≤ 2 As δ(G′ W ) (1 α 2α )m and α (1/17)3, we obtain the following: − ≥ − 1 − 2 ≤

δ(G′′) 7m′/8+1. ≥

Let a′ Γ(a , U ′ ), a′ Γ(a , U ′ ) such that a′ a′ E(G); and b′ Γ(b , U ′ ), 2i ∈ 1i 2 1i ∈ 2i 1 1i 2i ∈ 2i ∈ 1i 2 b′ Γ(b , U ′ ) such that b′ b′ E(G). We have the claim below. 1i ∈ 2i 1 1i 2i ∈ ′′ Claim 3.3.5. The balanced bipartite graph G contains three vertex-disjoint ladders Q1, Q2,

′′ ′ ′ and Q3 spanning on V (G ) such that the ﬁrst rung of Q1 is x1x2 and the last rung of Q1 is 37

′ ′ ′ ′ ′ ′ a11a21, the ﬁrst rung of Q2 is b11b21 and the last rung of Q2 is a12a22, the ﬁrst rung of Q3 is ′ ′ ′ ′ b12b22 and the last rung of Q3 is y1y2.

Proof. Since δ(G′′) 7m′/8+5, G′′ has a perfect matching M containing the following ≥ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ edges: x1x2, a11a21, b11b21, a12a22, b12b22,y1y2. We identify a11 and b11, a21 and b21, a12 and b12, ′ ′ ′ ′ ′ ′ ∗ ∗ ∗ ∗ and a22 and b22 as vertices called c11, c21, c12, and c22, respectively. Denote G = G (U1 , U2 ) as the resulting graph and let c′ c′ ,c′ c′ E(G∗) if they do not exist in E(G∗). Denote 11 21 12 22 ∈ M ′ := M a′ a′ , b′ b′ , a′ a′ , b′ b′ c′ c′ ,c′ c′ . Deﬁne an auxiliary graph H′ on −{ 11 21 11 21 12 22 12 22}∪{ 11 21 12 22} ′ ′ ′ M as follows. If xy, uv M with x, u U then xy ′ uv if and only if x ′ v and ∈ ∈ 1 ∼H ∼G ′ ′ ′ ′ ′ ′ ′ ′ y ′ u. Particularly, for any pq M c c ,c c with p U , pq ′ c c (resp. ∼G ∈ −{ 11 21 12 22} ∈ 2 ∼H 11 21 ′ ′ ′ ′ ′ ′ ′ ′ pq ′ c c ) if and only if p ′ a , b and q ′ a , b (resp. p ′ a , b and ∼H 12 22 ∼G 11 11 ∼G 21 21 ∼G 12 12 ′ ′ q ′ a , b ). Notice that there is a natural one-to-one correspondence between ladders ∼G 22 22 ′ ′ ∗ ∗ ∗ ∗ ′ with rungs in M and paths in H . Since δ ∗ (U , U ), δ ∗ (U , U ) 3m /4+1, we get G 1 2 G 2 1 ≥ δ(H′) m′/2 + 1. Hence H′ has a hamiltonian path starting with x′ x′ , ending with y′ y′ , ≥ 1 2 1 2 ′ ′ ′ ′ ′ ′ and having c11c21 and c12c22 as two internal vertices. The path with the vertex c11c21 replaced ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ by a11a21 and b11b21, and with the vertex c12c22 replaced by a12a22 and b12b22 is corresponding to the required ladders in G′′. If T T , T , then ∈{ 1 2} H = x x Q L Q L Q y y T. 1 2 1 1 2 2 3 1 2 ∪

is a spanning Halin subgraph of G. Suppose now that T T , T , T and z is the pendent ∈{ 3 4 5} vertex. Then z V ′ V ′ W by Claim 3.3.4. By (3.1) and the deﬁnition of U ′ and U ′ , ∈ 1 ∪ 2 − 1 2 we get deg (z, U ′ ),deg (z, U ′ ) (1 α 9α )m > m′/2. So z has a neighbor on each G 1 G 2 ≥ − 1 − 2 side of the ladder Q L Q L Q . Let H′ be obtained from x x Q L Q L Q y y T by 1 1 2 2 3 1 2 1 1 2 2 3 1 2 ∪ suppressing the degree 2 vertex z. Then H′ is a Halin graph such that each vertex on one

′ side of Q1L1Q2L2Q3 is a degree 3 vertex on its underlying tree. Let z be a neighbor of z such that z′ has degree 3 in the underlying tree of H′. Then

H = x x Q L Q L Q y y T zz′ , 1 2 1 1 2 2 3 1 2 ∪ ∪{ } 38

is a spanning Halin subgraph of G.

3.3.3.3 Proof of Theorem 3.4.3 We ﬁrst show that G contains a subgraph T isomorphic to T if n is even and to T if n is odd. Then by showing that G V (T ) contains 1 2 − a spanning ladder L with its ﬁrst rung adjacent to the head link of T and its last rung adjacent to the tail link of T , we get a spanning Halin subgraph H of G formed by L T . ∪

Finding a subgraph T

Claim 3.3.6. Let n be a suﬃcient large integer and G an n-vertex graph with δ(G) ≥ (n + 1)/2. If G is not in Extremal Case 2, then G contains a subgraph T isomorphic to T1

if n is even and to T2 if n is odd.

Proof. Suppose first that n is even. Let xy E(G) be an edge. We show that ∈ G[N(x) y ] contains an edge x x and G[N(y) x ] contains an edge y y such that the −{ } 1 2 −{ } 1 2 two edges are independent. Since G is not in Extremal Case 2, it has no independent set of size at least (1/2 7β)n. Hence, we can find the two desired edges, and G[ x, y, x , x ,y ,y ] − { 1 2 1 2} contains a subgraph T isomorphic to T1. Then assume that n is odd. We show in the first step that G contains a subgraph isomorphic to K− (K with one edge removed). Let yz E(G). 4 4 ∈ As δ(G) (n+1)/2, there exists y Γ(y, z). If there exists y Γ(y, z) y , we are done. ≥ 1 ∈ 2 ∈ −{ 1} Otherwise, (Γ(y) y , z ) (Γ(z) y ,y )= . As δ(G) (n + 1)/2, y is adjacent to a −{ 1 } ∩ −{ 1 } ∅ ≥ 1 vertex y Γ(y) Γ(z) y ,y,z . Assume y Γ(z) y ,y . Then G[ y,y ,z,y ] contains 2 ∈ ∪ −{ 1 } 2 ∈ −{ 1 } { 1 2} a copy of K−. Choose x Γ(y) z,y ,y and choose an edge x x G[Γ(x) y,y ,y , z ]. 4 ∈ −{ 1 2} 1 2 ∈ −{ 1 2 } Then G[ y,y ,z,y , x, x , x ] contains a subgraph T isomorphic to T . { 1 2 1 2} 2 Let T be a subgraph of G as given by Claim 3.3.6. Suppose the head link of T is x1x2 and the tail link of T is y y . Let G′ = G V (T ). We show in next section that G′ contains 1 2 − a spanning ladder with first rung adjacent to x1x2 and its last rung adjacent to y1y2. Let n′ = V (G′) . Then we have δ(G′) (n + 1)/2 7 n′/2 4 (1/2 β)n′. | | ≥ − ≥ − ≥ −

Finding a spanning ladder of G′ with prescribed end rungs 39

Theorem 3.3.5. Let n′ be a suﬃciently large even integer and G′ the subgraph of G obtained by removing vertices in T . Suppose that δ(G′) (1/2 β)n′ and G = G[V (G′) V (T )] is in ≥ − ∪ ′ Non-extremal case, then G contains a spanning ladder with ﬁrst rung adjacent to x1x2 and

its last rung adjacent to y1y2.

Proof. We ﬁx the following sequence of parameters

0 <ε d β 1 ≪ ≪ ≪

and specify their dependence as the proof proceeds. Let β be the parameter deﬁned in the two extremal cases. Then we choose d β and ≪ choose 1 ε = ǫ(d/2, 3, 2,d/4) 4 following the deﬁnition of ǫ in the Blow-up Lemma. Applying the Regularity Lemma to G′ with parameters ε and d, we obtain a partition of V (G′) into ℓ+1 clusters V ,V , ,V for some ℓ M M(ε), and a spanning subgraph G′′ 0 1 ··· ℓ ≤ ≤ of G′ with all described properties in the Regularity Lemma. In particular, for all v V (G′), ∈

′ ′ ′ deg ′′ (v) > deg ′ (v) (d + ε)n (1/2 β ε d)n (1/2 2β)n (3.3) G G − ≥ − − − ≥ − provided that ε + d β. On the other hand, ≤

(d + ε) e(G′′) e(G′) (n′)2 > e(G′) d(n′)2 ≥ − 2 −

by ε

We further assume that ℓ = 2k is even; otherwise, we eliminate the last cluster Vℓ by removing all the vertices in this cluster to V . As a result, V 2εn′, and 0 | 0| ≤

(1 2ε)n′ ℓN =2kN n′, (3.4) − ≤ ≤ 40

where N = V for 1 i ℓ. | i| ≤ ≤ For each pair i and j with 1 i = j ℓ, we write V V if d(V ,V ) d. As ≤ 6 ≤ i ∼ j i j ≥ in other applications of the Regularity Lemma, we consider the reduced graph Gr, whose vertex set is 1, 2, ,r and two vertices i and j are adjacent if and only if V V . From { ··· } i ∼ j δ(G′′) > (1/2 2β)n′, we claim that δ(G ) (1/2 2β)ℓ. Suppose not, and let i V (G ) − r ≥ − 0 ∈ r be a vertex with deg (i ) < (1/2 2β)ℓ. Let V be the cluster in G corresponding to i . Gr 0 − i0 0 Then we have

′ ′ ′ (1/2 β)n V E ′ (V ,V V ) < (1/2 2β)ℓN V +2εn V < (1/2 β)n V . − | i0 |≤| G i0 − i0 | − | i0 | | i0 | − | i0 |

This gives a contradiction by ℓN n′ from inequality (3.4). ≤ Let x V (G′) be a vertex and A a cluster. We say x is typical to A if deg(x, A) ∈ ≥ (d ε) A , and in this case, we write x A. − | | ∼

Claim 3.3.7. Each vertex in x , x ,y ,y is typical to at least (1/2 2β)l clusters in { 1 2 1 2} − V , ,V . { 1 ··· l}

Proof. Suppose on the contrary that there exists x x , x ,y ,y such that x is ∈ { 1 2 2 2} typical to less than (1/2 2β)l clusters in V , ,V . Then we have deg ′ (x) < (1/2 − { 1 ··· l} G − 2β)lN +(d + ε)n′ (1/2 β)n′ by lN n′ and d + ε β. ≤ − ≤ ≤ Let x V (G′) be a vertex. Denote by the set of clusters to which x typical. ∈ Vx

Claim 3.3.8. There exist V and V such that d(V ,V ) d. x1 ∈Vx1 x2 ∈Vx2 x1 x2 ≥

We may assume that is an independent set in G . For otherwise, we are done Vx1 ∩Vx2 r by ﬁnding an edge within . Also we may assume that E ( , )= Vx1 ∩Vx2 Gr Vx1 ∩Vx2 Vx1 −Vx2 ∅ and E ( , )= . Since δ(G ) (1/2 2β)l and δ ( , ) = 0, Gr Vx1 ∩Vx2 Vx2 −Vx1 ∅ r ≥ − Gr Vx1 ∩Vx2 Vx1 ∪Vx2 we know that l (1/2 2β)l. Hence, = + −|Vx1 ∪Vx2 | ≥ − |Vx1 ∪Vx2 | |Vx1 | |Vx2 |−|Vx1 ∩Vx2 | ≤ (1/2+2β)l. This gives that + (1/2+2β)l (1/2 2β)l + |Vx1 ∩ Vx2 | ≥ |Vx1 | |Vx2 | − ≥ − (1/2 2β)l (1/2+2β)l (1/2 6β)l. Let be the union of clusters in . Then − − ≥ − U Vx1 ∩Vx2 (1/2 7β)n and ∆(G[ ]) (d + ε)n′ βn. This shows that G is in Extremal Case |U| ≥ − U ≤ ≤ 2. Similarly, we have the following claim.

Claim 3.3.9. There exist V V ,V and V V ,V such that y1 ∈ Vy1 − { x1 x2 } y2 ∈ Vy2 − { x1 x2 } d(V ,V ) d. y1 y2 ≥ Claim 3.3.10. The reduced graph G has a hamiltonian path X Y X Y such that r 1 1 ··· k k X ,Y = V ,V and X ,Y = V ,V . { 1 1} { x1 x2 } { k k} { y1 y2 }

Proof. We contract the edges Vx1 Vx2 and Vy1 Vy2 in Gr. Denote the two new vertices

′ ′ ′ ′ as Vx and Vy respectively, and denote the resulting graph as Gr. Then we show that Gr ′ ′ contains a hamiltonian (Vx,Vy )-path. This path is corresponding to a required hamiltonian path in Gr.

′ ′ ′ To show Gr has a hamiltonian (Vx,Vy )-path, we need the following generalized version of a result due to Nash-Williams [44] : Let Q be a 2-connected graph of order m. If δ(Q) max (m +2)/3+1,α(Q)+1 , then Q is hamiltonian connected, where α(Q) is the ≥ { } size of a largest independent set of Q. We claim that G′ is 2βl-connected. For otherwise, let S be a vertex-cut of G′ with S < r r | | 2βl and the vertex set corresponding to S in G . Then V V (T ) 2βn′ +2εn′ < 5βn, S |S ∪ 0 ∪ | ≤ showing that G is in Extremal Case 1. Since n′ = Nl+ V (l+2)εn′, we have l 1/ε 2 | 0| ≤ ≥ − ≥ 1/β. Hence, G′ is 2-connected. As G is not in Extremal Case 2, α(G′ ) (1/2 7β)l. By r r ≤ − δ(G ) (1/2 2β)l, we have δ(G′ ) (1/2 2β)l 2 max (l +2)/3+1, (1/2 7β)l +1 . r ≥ − r ≥ − − ≥ { − } ′ Thus, by the result on hamiltonian connectedness given above, we know that Gr contains a ′ ′ hamiltonian (Vx,Vy )-path. 42

Following the order of the clusters on the hamiltonian path given in Claim 3.3.10, for i =1, 2, ,k, we call X ,Y partners of each other and write P (X )= Y and P (Y )= X . ··· i i i i i i Claim 3.3.11. For each 1 i k, there exist X′ X and Y ′ Y such that (X′,Y ′) is ≤ ≤ i ⊆ i i ⊆ i i i (2ε,d 3ε)-super-regular, Y ′ = X′ +1, Y ′ = X′ +1, and X′ = Y ′ for 2 i k 1. − | 1 | | 1| | k| | k| | i| | i | ≤ ≤ − Additionally, each pair (Y ′,X′ ) is 2ε-regular with density at least d ε for i =1, 2, ,k, i i+1 − ··· ′ ′ where Xk+1 = X1.

Proof. For each 1 i k, let ≤ ≤

X′′ = x X deg(x, Y ) (d ε)N , and i { ∈ i | i ≥ − } Y ′′ = y Y deg(y,X ) (d ε)N . i { ∈ i | i ≥ − }

If necessary, we either take a subset X′ of X′′ or take a subset Y ′ of Y ′′ such that Y ′ = i i i i | 1 | X′ +1, Y ′ = X′ +1, and X′ = Y ′ for 2 i k 1. Since (X ,Y ) is ε-regular, we have | 1| | k| | k| | i| | i | ≤ ≤ − i i X′′ , Y ′′ (1 ε)N. This gives that X ′, X′ (1 ε)N 1 and X′ = Y ′ (1 ε)N | i | | i | ≥ − | 1| | k| ≥ − − | i| | i | ≥ − for 2 i k 1. As a result, we have deg(x, Y ′) (d 2ε)N for each x X′ and ≤ ≤ − i ≥ − ∈ i deg(y,X′) (d 2ε)N 1 (d 3ε)N for each y Y ′. By Slicing lemma (Lemma 3.2.5), i ≥ − − ≥ − ∈ i (X′,Y ′) is 2ε-regular. Hence (X′,Y ′) is (2ε,d 3ε)-super-regular for each 1 i k. By i i i i − ≤ ≤ Slicing lemma again, we know that (X′,Y ′ ) is 2ε-regular with density at least d ε. i i+1 − ′ ′ For 1 i k, we call (Xi,Yi ) a super-regularized cluster (sr-cluster). Denote R = k ≤ ≤ ′ ′ ′ ′ V0 ( ((Xi Yi) (Xi Yi ))). Since (Xi Yi) (Xi Yi ) 2εN for 2 i k 1 and ∪ i=1 ∪ − ∪ | ∪ − ∪ | ≤ ≤ ≤ − (X SY ) (X′ Y ′) , (X Y ) (X′ Y ′) 2εN +1, we have R 2εn+2kεN +2 3εn′. | 1 ∪ 1 − 1 ∪ 1 | | k ∪ k − k ∪ k | ≤ | | ≤ ≤ As n′ is even and X′ + Y ′ + + X′ + Y ′ is even, we know R is even. We arbitrarily group | 1| | 1 | ··· | k| | k| | | vertices in R into R /2 pairs. Given two vertices u, v R, we deﬁne a (u, v)-chain of length | | ∈ 2t as distinct clusters A , B , , A , B such that u A B A B v and 1 1 ··· t t ∼ 1 ∼ 1 ∼···∼ t ∼ t ∼ each A and B are partners, in other words, A , B = X ,Y for some i 1, ,k . j j { j j} { ij ij } j ∈{ ··· } We call such a chain of length 2t a 2t-chain.

Claim 3.3.12. For each pair (u, v) in R, we can ﬁnd a (u, v)-chain of length at most 4 such that every sr-cluster is used in at most d2N/5 chains. 43

Proof. Suppose we have found chains for the ﬁrst m< 2εn′ pairs of vertices in R such that no sr-cluster is contained in more than d2N/5 chains. Let Ω be the set of all sr-clusters that are used exactly by d2N/5 chains. Then

d2N 2kN Ω 4m< 8εn′ 8ε , 5 | | ≤ ≤ 1 2ε − where the last inequality follows from (3.4). Therefore,

80kε 80lε Ω βl/2, | | ≤ d2(1 2ε) ≤ d2 ≤ − provided that 1 2ε 1/2 and 80ε d2β/2. − ≥ ≤ Consider now a pair (w, z) of vertices in R which does not have a chain found so far, we want to ﬁnd a (w, z)-chain using sr-clusters not in Ω. Let be the set of all sr-clusters U adjacent to w but not in Ω, and let be the set of all sr-clusters adjacent to z but not in V Ω. We claim that , (1/2 2β)l. To see this, we ﬁrst observe that any vertex x R |U| |V| ≥ − ∈ is adjacent to at least (1/2 3β/2)l sr-clusters. For instead, −

′ ′ (1/2 β)n deg ′ (x) < (1/2 3β/2)lN +(d 2ε)lN +3εn , − ≤ G − − (1/2 3β/2+ d +2ε)n′ ≤ − < (1/2 β)n′ (provided that d +2ε<β/2 ), −

showing a contradiction. Since Ω βl/2, we have , (1/2 2β)l. Let P ( ) and | | ≤ |U| |V| ≥ − U P ( ) be the set of the partners of clusters in and , respectively. By the deﬁnition of the V U V chains, a cluster A Ω if and only its partner P (A) Ω. Hence, (P ( ) P ( )) Ω = . ∈ ∈ U ∪ V ∩ ∅ Notice also that each cluster has a unique partner, and so we have P ( ) = (1/2 2β)l | U | |U| ≥ − and P ( ) = (1/2 2β)l. | V | |V| ≥ − If E (P ( ),P ( )) = , then there exist two adjacent clusters B P ( ), A P ( ). Gr U V 6 ∅ 1 ∈ U 2 ∈ V If B and A are partners of each other, then w A B z gives a (w, z)-chain of length 1 2 ∼ 2 ∼ 1 ∼ 2. Otherwise, assume A = P (B ) and B = P (A ), then w A B A B z 1 1 2 2 ∼ 1 ∼ 1 ∼ 2 ∼ 2 ∼ 44

gives a (w, z)-chain of length 4. Hence we assume that E (P ( ),P ( )) = . We may Gr U V ∅ assume that P ( ) P ( ) = . Otherwise, let be the union of clusters contained in U ∩ V 6 ∅ S V (G ) (P ( ) P ( )). Then R V (T ) with R V (T ) 4βn′ +3εn′ +7 5βn r − U ∪ V S ∪ ∪ |S ∪ ∪ | ≤ ≤ (provided that 3ε+7/n′ < β) is a vertex-cut of G, implying that G is in Extremal Case 1. As E (P ( ),P ( )) = , any cluster in P ( ) P ( ) is adjacent to at least (1/2 2β)l clusters in Gr U V ∅ U ∩ V − V (G ) (P ( ) P ( )) by δ(G ) (1/2 2β)l. This implies that P ( ) P ( ) (1/2+2β)l, r − U ∪ V r ≥ − | U ∪ V | ≤ and thus P ( ) P ( ) P ( ) + P ( ) P ( ) P ( ) (1/2 6β)l. Then P ( ) P ( ) | U ∩ V |≥| U | | V |−| U ∪ V | ≥ − U ∩ V is corresponding to a subset V of V (G) such that V (1/2 6β)lN (1/2 7β)n 1 | 1| ≥ − ≥ − and ∆(G[V ]) (d + ε)n′ βn. This implies that G is in Extremal Case 2, showing a 1 ≤ ≤ contradiction. For each cluster Z X′ ,Y ′, ,X′ ,Y ′ , let R (Z) denote the set of vertices in R ∈ { 1 1 ··· k k} 2 using Z in the 2-chains and R4(Z) denote the set of vertices in R using Z in the 4-chains given by Claim 3.3.12. By the deﬁnition of 2-chains and 4-chains, we have the following holds.

Claim 3.3.13. For each i = 1, 2, ,k, if R (X′) = , then R (X′) = R (Y ′) ; and if ··· 2 i 6 ∅ | 2 i | | 2 i | R (X′) = , then R (X′) = R (Y ′ ) . 4 i 6 ∅ | 4 i | | 4 i+1 |

Claim 3.3.14. For each i =1, 2, ,k, if R (X′) = , then there exist vertex-disjoint lad- ··· 2 i 6 ∅ ders Li and Li covering all vertices in R (X′) R (Y ′) such that X′ V (Li Li ) = 2x 2y 2 i ∪ 2 i | i ∩ 2x ∪ 2y | Y ′ V (Li Li ) ; and if R (X′) = , then there exist three vertex disjoint ladders | i ∩ 2x ∪ 2y | 4 i 6 ∅ Li , Li , Li+1 covering all vertices in R (X′) R (Y ′ ) such that V (Li ) X′ Y ′, 4x 4xy 4y 4 i ∪ 4 i+1 4x ⊆ i ∪ i V (Li ) Y ′ X′ , and V (Li+1) X′ Y ′ , and that X′ V (Li Li ) = 4xy ⊆ i ∪ i+1 4y ⊆ i+1 ∪ i+1 | i ∩ 4x ∪ 4xy | Y ′ V (Li Li Li+1) = X′ V (Li Li Li+1) = Y ′ V (Li Li+1) . | i ∩ 4x ∪ 4xy ∪ 4y | | i+1 ∩ 4x ∪ 4xy ∪ 4y | | i+1 ∩ 4xy ∪ 4y |

Proof. Notice that by Claim 3.3.11, (X′,Y ′) is (2ε,d 3ε)-super-regular and (Y ,X ) i i − i i+1 is 2ε-regular. Assume R (X′) = . By Claim 3.3.12 and Claim 3.3.13, we have 2 i 6 ∅ R (X′) = R (Y ′) d2N/5. Let R (X′) = x , , x . For each j = 1, ,r, | 2 i | | 2 i | ≤ 2 i { 1 ··· r} ··· since Γ(x ,X′) (d 2ε) X′ > 2ε X′ , by Lemma 3.2.4, there exists a vertex set | j i | ≥ − | i| | i| B Y ′ with B (1 2ε) Y ′ such that B is typical to Γ(x ,X′). If r 2, for j ⊆ i | j| ≥ − | i | j j i ≥ 45

j = 1, ,r 1, there also exists a vertex set B Y ′ with B (1 4ε) Y ′ ··· − j,j+1 ⊆ i | j,j+1| ≥ − | i | ′ ′ such that Bj,j+1 is typical to both Γ(xj,Xi) and Γ(xj+1,Xi). That is, for each vertex b B , we have deg(b , Γ(x ,X′)) (d 5ε) Γ(x ,X′) > 4 R , and for each vertex 1 ∈ j 1 j i ≥ − | j i | | | b B , we have deg(b , Γ(x ,X′)),deg(b , Γ(x ,X′)) (d 5ε) Γ(x ,X′) > 4 R . 2 ∈ j,j+1 2 j i 2 j+1 i ≥ − | j i | | | When r 2, since B , B , B (d 4ε) Y ′ > 2ε Y ′ , there is a set A X′ with ≥ | j| | j,j+1| | j+1| ≥ − | i | | i | ⊆ i A (1 6ε) X′ R such that A is typical to each of B , B and B . Notice that | | ≥ − | i|≥| | j j+1 j+1 (d 5ε) B , (d 5ε) B , (d 5ε) B (d 5ε)(1 4ε) Y ′ > 3 R . Hence we can choose − | j| − | j,j+1| − | j+1| ≥ − − | i | | | distinct vertices u ,u , ,u A such that deg(u , B ),deg(u , B ),deg(u , B ) 1 2 ··· r−1 ∈ j j j j,j+1 j j+1 ≥ 3 R . Then we can choose distinct vertices yj Γ(u , B ), z Γ(u , B ) and | | 23 ∈ j j j ∈ j j,j+1 yj+1 Γ(u , B ) for each j, and choose distinct and unchosen vertices y1 B and 12 ∈ j j+1 12 ∈ 1 yr B . Finally, as for each vertex b B , we have deg(b , Γ(x ,X′)) > 4 R and for each 23 ∈ r 1 ∈ j 1 j i | | vertex b B , we have deg(b , Γ(x ,X′)),deg(b , Γ(x ,X′)) > 4 R , we can choose 2 ∈ j,j+1 2 j i 2 j+1 i | | x , x , x Γ(x ,X′) u , ,u such that yj Γ(x , x ,Y ′), yj Γ(x , x ,Y ′), j1 j2 j3 ∈ j i −{ 1 ··· r−1} 12 ∈ j1 j2 i 23 ∈ j2 j3 i and z Γ(x , x ,Y ′). Let Li be the graph with j ∈ i3 i+1,1 i 2x

V (Li )= R (X′) x , x , x ,yi ,yi , z ,u , x , x , x ,yr ,yr 1 i r 1 and 2x 2 i ∪{ i1 i2 i3 12 23 i i r1 r2 r3 12 23 | ≤ ≤ − }

i r r r r E(L2x) consisting of the edges xrxr1, xrxr2, xrxr3,y12xr1,y12xr2,y23xr2,y23xr3 and the edges indicated below for each 1 i r 1: ≤ ≤ −

x x , x , x ; yi x , x ; yi x , x ; z x , x ; u x , x , z . i ∼ i1 i2 i3 12 ∼ i1 i2 23 ∼ i2 i3 i ∼ i3 i+1,1 i ∼ i3 i+1,1 i

It is easy to check that Li is a ladder spanning on R (X′), 4 R (X′) 1 vertices from 2x 2 i | 2 i | − X′ and 3 R (X′) 1 vertices from Y ′. Similarly, we can ﬁnd a ladder Li spanning on i | 2 i | − i 2y R (Y ′), 4 R (Y ′) 1 vertices from X′ and 3 R (X′) 1 vertices from X′. Clearly, we have 2 i | 2 i | − i | 2 i | − i X′ V (Li Li ) = Y ′ V (Li Li ) . | i ∩ 2x ∪ 2y | | i ∩ 2x ∪ 2y | Assume now that R (X′) = . Then by Claim 3.3.12, we have R (X′) = R (Y ′ ) . 4 i 6 ∅ | 4 i | | 4 i+1 | By the similar argument as above, we can ﬁnd ladder Li , Li+1 such that R (X′) 4x 4y 4 i ⊆ 46

V (Li ), R (Y ′ ) V (Li+1). Furthermore, we have 4x 4 i+1 ⊆ 4y

X′ V (Li ) = 4 R (X′) 1, Y ′ V (Li ) = 3 R (X′) 1; | i ∩ 4x | | 4 i | − | i ∩ 4x | | 4 i | − Y ′ V (Li+1) = 4 R (Y ′ ) 1, X′ V (Li+1) = 3 R (Y ′ ) 1. | i+1 ∩ 4y | | 4 i+1 | − | i+1 ∩ 4y | | 4 i+1 | −

Finally, we claim that we can find a ladder Li between (Y ′,X′ ) such that Y ′ V (Li ) = 4xy i i+1 | i ∩ 4xy | X′ V (Li ) = R (Y ′ ) and is vertex-disjoint from Li Li+1. Since 3 R (Y ′ ) | i+1 ∩ 4xy | | 4 i+1 | 4x ∪ 4y | 4 i+1 | ≤ 3d2N/5 and (Y ′,X′ ) is 2ε-regular with density at least d ε by Claim 3.3.11, a similar i i+1 − argument as in the proof of Lemma 3.3.11, we can find Y ′′ Y ′ V (Li ) and X′′ i ⊆ i − 4x i+1 ⊆ X′ V (Li+1) such that (Y ′′,X′′ ) is (4ε,d 5ε)-super-regular and Y ′′ = X′′ , and i+1 − 4y i i+1 − | i | | i+1| thus is (4ε,d/2)-super-regular (provided that ε d/10). Notice that there are at least ≤ (d 9ε) Y ′′ d Y ′′ /4 vertices typical to X′′ , and there are at least (d 9ε) X′′ − | i | ≥ | i | i+1 − | i+1| ≥ d2 X′′ /4 vertices typical to Y ′′. Applying the Below-up Lemma (Lemma 3.2.2), we can find | i+1| i a ladder Li within (Y ′′,X′′ ) such that Y ′ V (Li ) = X′ V (Li ) = R (Y ′ ) . It 4xy i i+1 | i ∩ 4xy | | i+1 ∩ 4xy | | 4 i+1 | i i+1 i is routine to check that L4x, L4y , L4xy are the desired ladders. For each i = 1, 2, ,k, let X∗∗ = X′ V (Li Li Li Li Li ) and Y ∗∗ = ··· i i − 2x ∪ 2y ∪ 4x ∪ 4xy ∪ 4y i Y ′ V (Li Li Li Li Li ). Using Lemma 3.2.4, for i 1, ,k 1 , choose y∗ Y ∗∗ i − 2x∪ 2y ∪ 4x∪ 4xy ∪ 4y ∈{ ··· − } i ∈ i such that A dN/4, where A := X∗∗ Γ(y∗). This is possible, as (Y ∗∗,X∗∗ ) is | i+1| ≥ i+1 i+1 ∩ i i i+1 4ε-regular (applying Slicing lemma based on (Y ′,X′ )). Similarly, choose x∗ A such i i+1 i+1 ∈ i+1 that B dN/4, where B := Y ∗∗ Γ(x∗ ). Let S = y∗, x∗ 1 i k 1 , and let | i| ≥ i i ∩ i+1 { i i+1 | ≤ ≤ − } X∗ = X∗∗ S and Y ∗ = Y ∗∗ S. We have the following holds. i i − i i − Claim 3.3.15. For each i =1, 2, ,k, (X∗,Y ∗) is (4ε,d/2)-super-regular such that Y ∗ = ··· i i | 1 | X∗ +1, Y ∗ = X∗ +1, and X∗ = Y ∗ for 2 i k 1. | 1 | | k | | k | | i | | i | ≤ ≤ − Proof. Since R (X′) , R (Y ′ ) d2N/5 for each i, we have X∗ , Y ∗ (1 ε | 2 i | | 4 i+1 | ≤ | i | | i | ≥ − − d2)N 1. As ε,d 1, we can assume that 1 ε d2 1/N < 1/2. Thus, by Slicing lemma − ≪ − − − ′ ′ ∗ ∗ based on the 2ε-regular pair (Xi,Yi ), we know that (Xi ,Yi ) is 4ε-regular. Recall from Claim 3.3.11 that (X′,Y ′) is (2ε,d 3ε)-super-regular, as 4 R (X′) , 4 R (Y ′ ) < d2 Y ∗ , i i − | 2 i | | 4 i+1 | | i | we know that for each x X∗, deg(x, Y ∗) (d 3ε d2) Y ∗ > d Y ∗ /2. Similarly, we ∈ i i ≥ − − | i | | i | 47

have for each y Y ∗, deg(y,X∗) d X∗ /2. Thus (X∗,Y ∗) is (4ε,d/2)-super-regular. ∈ i i ≥ | i | i i Finally, Combining Claims 3.3.11 and 3.3.14, we have Y ∗ = X∗ + 1, Y ∗ = X∗ + 1, and | 1 | | 1 | | k | | k | X∗ = Y ∗ for 2 i k 1. | i | | i | ≤ ≤ − For each i =1, 2, ,k 1, now set B := Y ∗ Γ(x∗ ) and C := X∗ Γ(y∗). Since ··· − i+1 i ∩ i+1 i i ∩ i (X∗,Y ∗) is (4ε,d/2)-super-regular, we have B , C d X∗ /2 > d X∗ /4. Recall from i i | i+1| | i| ≥ | i | | i | Claim 3.3.10 that X ,Y = V ,V and X ,Y = V ,V . We assume, w.l.o.g., { 1 1} { x1 x2 } { k k} { y1 y2 } that X = V and X = V . Let A = X∗ Γ(x ), B = Y ∗ Γ(x ), C = X∗ Γ(y ), 1 x1 k y1 1 1 ∩ 1 1 1 ∩ 2 k k ∩ 1 and D = Y ∗ Γ(y ). Since deg(x ,X ) (d ε)N, we have deg(x ,X∗) (d ε 2ε k k ∩ 2 1 1 ≥ − 1 1 ≥ − − − d2)N d X∗ /4, and thus A d X∗ /4. Similarly, we have B , C , D d X∗ /4. ≥ | 1 | | 1| ≥ | 1 | | 1| | k| | k| ≥ | 1 | For each 1 i k, we assume that Li = ai bi Li ci di , Li = ai bi Li ci di , ≤ ≤ 2x 1 1 − 2x − 1 1 2y 2 2 − 2y − 2 2 Li = ai bi Li ci di , Li = ai bi Li ci di , and Li = ai bi Li ci di , where 4x 3 3 − 4x − 3 3 4xy 4 4 − 4xy − 4 4 4y 5 5 − 4y − 5 5 ai ,ci Y ′ Y for j =1, 2, , 5. For j =1, 2, , 5, let Ai = X∗ Γ(ai ), Ci = X∗ Γ(ci ), j j ∈ i ⊆ i ··· ··· j i ∩ j j i ∩ j Bi = Y ∗ Γ(bi ), and Di = Y ∗ Γ(di ). Since (X′,Y ′) is (2ε,d 3ε)-super-regular, for j i ∩ j j i ∩ j i i − j = 1, 2, 3, 5, we have Γ(ai ,X′) , Γ(ci ,X′) (d 3ε) X′ and Γ(bi ,Y ′) , Γ(di ,Y ′) | j i | | j i | ≥ − | i| | j i | | j i | ≥ (d 3ε) Y ′ . From the proof of Claim 3.3.14, the pair (Y ′′ ,X ′′ ) is (4ε,d 5ε)-super-regular. − | i | i i+1 − Hence, Γ(ai ,X′ ) , Γ(ci ,X′ ) (d 4ε) X′ and Γ(bi ,Y ′) , Γ(di ,Y ′) (d 4ε) Y ′ . | 4 i+1 | | 4 i+1 | ≥ − | i+1| | 4 i | | 4 i | ≥ − | i | Thus, we have Ai , Bi , Ci , Di (d 4ε) X′ d2N d X∗ /4= d Y ∗ /4. | j| | j| | j| | j| ≥ − | i| − ≥ | i | | i | ∗ ∗ i We now apply the Blow-up lemma on (Xi ,Yi ) to find a spanning ladder L with its first and last rungs being contained in A B and C D , respectively, and for j =1, 2, , 5, i × i i × i ··· its (2j)-th and (2j +1)-th rungs being contained in Ai Bi and Ci Di , respectively. We can j × j j × j i i i then insert L2x between the 2nd and 3rd rungs of L , L2y between the 4th and 5th rungs of i i i i i L , L4x between the 6th and 7th rungs of L , L4xy between the 8th and 9th rungs of L , and Li between the 10th and 11th rungs of Li to obtained a ladder i spanning on X Y S. 4y L i ∪ i − Finally, 1y∗x∗ 2 y∗ x∗ k is a spanning ladder of G′ with its first rung adjacent to x x L 1 2L ··· k−1 kL 1 2 and its last rung adjacent to y1y2. The proof is then complete. 48

3.4 Minimum degree condition for spanning generalized Halin graphs

3.4.1 Introduction

A tree with no vertex of degree 2 is called a homeomorphically irreducible tree (HIT), and a spanning tree with no vertex of degree 2 is a homeomorphically irreducible spanning tree (HIST). A Halin graph, constructed by Halin in 1971 [27], is a graph formed from a plane embedding of a HIT T of at least 4 vertices by connecting its leaves into a cycle following the cyclic order determined by the embedding. Relaxing the planarity requirement, a generalized Halin graph is obtained from a HIT T of at least 4 vertices by connecting its leaves into a cycle. We call the HIT T the underlying tree or underlying HIST of the resulting (generalized) Halin graph. Halin graphs possess many hamiltonicity properties. For examples, Halin graphs are hamiltonian [5], hamiltonian-connected [2] (there is a hamiltonian path between any two distinct vertices), and almost pancyclic [6] (contains all possible cycle lengths with one possible exception of a single even length). Compared to Halin graphs, generalized Halin graphs are less studied. Kaiser et al. in [34] showed that a generalized Halin graph is prism hamiltonian; that is, the Cartesian product of a generalized Halin graph and K2 is hamiltonian. Since a tree with no degree 2 vertices has more leaves than the non-leaves, a generalized Halin graph contains a cycle of length at least half of its order. Also, one can notice that by contracting the non-leaves of the underlying tree of a generalized Halin graph into a singe vertex, a wheel graph is resulted with the contracted vertex as the hub, where a minor of a graph is obtained from the graph by deleting edges/contracting edges, or deleting vertices. Therefore, a generalized Halin graph contains a wheel-minor of order at least half of its order. Although a generalized Halin graph may not be hamiltonian, we conjecture that the lengths of a longest cycle in a generalized Halin graph is large.

Conjecture 3.1. Let G be an n-vertex generalized Halin graph. Then the length of a longest cycle of G is at least 4n/5.

It was shown by Horton, Parker, and Borie [30] that it is NP-complete to determine 49

whether a graph contains a (spanning) Halin graph. For generalized Halin graphs we obtain the following.

Theorem 3.4.1. It is NP-hard to determine whether a graph contains a spanning generalized Halin graph.

A classic theorem of Dirac [19] from 1952 asserts that every graph on n vertices with minimum degree at least n/2 is hamiltonian if n 3. As a continuous “generalization” of ≥ Dirac’s Theorem as well as an approach of showing many hamiltonicity properties simultaneously in a graph, the existence of a spanning Halin graph in graphs with large minimum degree was investigated in the previous section, and it was shown that any suﬃciently large n-vertex graph with minimum degree at least (n+1)/2 contains a spanning Halin graph. We here determine the minimum degree threshold for a graph to contain a spanning generalized Halin graph.

Theorem 3.4.2. There exists a positive integer n0 such that every 3-connected graph with n n vertices and minimum degree at least (2n + 3)/5 contains a spanning generalized ≥ 0 Halin graph. The result is best possible in the sense of the connectivity and minimum degree constraints.

Since a generalized Halin graph of order n contains a wheel-minor of order at least n/2, we get the following corollary.

Corollary 3.4.1. There exists a positive integer n0 such that every 3-connected graph with n n vertices and minimum degree at least (2n + 3)/5 contains a wheel-minor of order at ≥ 0 least n/2.

For notational convenience, for a graph T , we denote by L(T ) the set of degree 1 vertices of T and S(T )= V (T ) L(T ). Also we abbreviate spanning generalized Halin graph − as SGHG in what follows, and denote a generalized Halin graph as H = T C, where T is ∪ the underlying HIST of H and C is the cycle spanning on L(T ). 50

3.4.2 Proof of Theorem 3.4.1 and the sharpness of Theorem 3.4.2

Proof of Theorem 3.4.1. To show the problem is NP-hard we assume the existence of a polynomial algorithm to test for an SGHG and use it to create a polynomial algorithm to test for a hamiltonian path between two vertices in an arbitrary graph. The decision problem for such hamiltonian paths is a classic NP-complete problem [24]. Let G be a graph and x, y V (G). We want to determine whether there exists a ∈ hamiltonian path connecting x and y. We ﬁrst construct a new graph G′ and show that G contains a hamiltonian path between x and y if and only if G′ contains a HIST (the proof of this part is the same as the proof of Albertson et al. in [1]). Then based on G′, we construct a graph G′′ and show that G′ contains a HIST if and only if G′′ contains an SGHG. Let z , z , , z = V (G) x, y . Then G′ is formed by adding new vertices { 1 2 ··· t} − { } z′ , z′ , , z′ and new edges z z′ : 1 i t . It is clear that if P is a hamiltonian { 1 2 ··· t} { i i ≤ ≤ } path between x and y, then P z z′ : 1 i t is a HIST of G′. Conversely, let T ∪{ i i ≤ ≤ } ′ ′ ′ ′ be a HIST of G . Since 1 d (z ) d ′ (z ) = 1, we get d (z ) = 1 for each i. Since ≤ T i ≤ G i T i ′ ′ ′ ′ N ′ (z ) = z and T is a HIST, we have d (z ) 3. Hence T z , z , , z is a tree G i { i} T i ≥ −{ 1 2 ··· t} with leaves possibly in x, y . Since each tree has at least 2 leaves and a tree with exactly { } two leaves is a path, we conclude that T z′ , z′ , , z′ is a path between x and y. −{ 1 2 ··· t} Then based on G′, we construct a graph G′′. First, for each i with 1 i t, we ≤ ≤ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ add new vertices zi1, zi2, zi3 and new edges zizi1, zizi2, zizi3, zi1zi2, zi2zi3. Then we connect all vertices in x, y z′ , z′ , z′ : 1 i t into a cycle C′′ such that z′ z′ , z′ z′ : 1 { }∪{ i1 i2 i3 ≤ ≤ } { i1 i2 i2 i3 ≤ i t E(C′′). If T ′ is a HIST of G′, then T ′′ := T ′ z′z′ , z′z′ , z′z′ : 1 i t ≤ } ⊆ ∪{ i i1 i i2 i i3 ≤ ≤ } is a HIST of G′′ and T ′′ C′′ is an SGHG of G′′. Conversely, suppose H = T C is an ∪ ∪ SGHG of G′′. We claim that C = C′′. This in turn gives that T = T ′′ and therefore T ′′ z′ , z′ , z′ : 1 i t is a HIST of G′. To show that C = C′′, we ﬁrst show that −{ i1 i2 i3 ≤ ≤ } z′ L(T ) for each i. Suppose on the contrary and assume, without loss of generality, that i2 ∈ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ z S(T ). Then as N ′′ (z ) = z , z , z , we get z z , z z , z z E(T ). Since 12 ∈ G 12 { 1 11 13} { 12 1 12 11 12 13} ⊆ T is acyclic, z′ z′ , z′ z′ E(T ). This in turn shows that z′ , z′ , z′ L(T ). However, 11 1 13 1 6∈ { 1 11 13} ⊆ z′ z′ , z′ z′ , z′ z′ forms a component of T , showing a contradiction. Then we show that { 12 1 12 11 12 13} 51

z′ , z′ L(T ) for each i. Suppose on the contrary and assume, without loss of generality, i1 i3 ∈ that z′ S(T ). By the previous argument, we have z′ L(T ). Then z′ , z′ L(T ) 11 ∈ 12 ∈ 1 13 ∈ ′ ′ ′ ′ as z12 is on C and z1 and z13 are the only two neighbors of z12 which can be on the cycle ′ ′ ′ ′ ′ ′ ′ ′ ′ C. As d ′′ (z ) = 3 and z , z N ′′ (z ), z z , z z E(T ). Since z L(T ) and G 11 { 12 1} ⊆ G 11 11 12 11 1 ∈ 12 ∈ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ z , z L(T ), we get z z , z z , z z E(T ). Since d ′′ (z ) = d ′′ (z ) = 3, we have 1 13 ∈ 12 13 12 1 1 13 6∈ G 12 G 13 z′ z′ , z′ z′ , z′ z′ E(C). However, z′ z′ , z′ z′ , z′ z′ forms a triangle but V (C) 4, 12 13 12 1 1 13 ∈ 12 13 12 1 1 13 | | ≥ showing a contradiction. So we have shown that z′ , z′ , z′ : 1 i t L(T ). This { i1 i2 i3 ≤ ≤ } ⊆ indicates that in the tree T z′ , z′ , z′ : 1 i t , each vertex z′ has degree 1 and no −{ i1 i2 i3 ≤ ≤ } i vertices of degree 2. Hence T z′ , z′ , z′ : 1 i t is a HIST of G′. −{ i1 i2 i3 ≤ ≤ } Combining the arguments in the two paragraphs above, we see that G has a hamiltonian path between x and y if and only if G′′ has an SGHG. Hence a polynomial SGHG-tester becomes a polynomial path-tester.

Since a generalized Halin graph is 3-connected, the connectivity requirement in Theo- rem 3.4.2 is necessary. To show that the minimum degree requirement is best possible, we show the following proposition.

Proposition 3.4.1. Let G(A, B) = K be a complete bipartite graph with A = a and a,b | | B = b. Then G(A, B) has no HIST T with L(T ) A = L(T ) B if b> 3(a−1) . | | | ∩ | | ∩ | 2 If a bipartite graph G(A, B) contains an SGHG H = T C, then L(T ) A = L(T ) B . ∪ | ∩ | | ∩ | Thus, by Proposition 3.4.1, it is easy to see that the complete bipartite graphs Ka,b with

3a−1 3a−2 b = 2 when a is odd and b = 2 when a is even does not have an SGHG. Let n = a + b. 2n+1 3a−1 2n+2 By direct computation, we get δ(Ka,b) = 5 when b = 2 and δ(Ka,b) = 5 when 3a−2 b = 2 . We now prove Proposition 3.4.1.

Proof of Proposition 3.4.1. Suppose on the contrary that G(A, B) contains a HIST T such that L(T ) A = L(T ) B . Then | ∩ | | ∩ |

S(T ) B S(T ) A = B L(T ) B ( A L(T ) A) | ∩ |−| ∩ | | |−| ∩ | − | |−| ∩ | 3(a 1) a 3 = B A > − a = − . | |−| | 2 − 2 52

Since G(A, B) is bipartite and T is a HIST of G(A, B), we have S(T ) A 1. Thus, from the | ∩ | ≥ inequalities above, we obtain S(T ) B > (a 1)/2. Since T is a HIST, we have d (y) 3 | ∩ | − T ≥ for each y S(T ) B. Let E = e E(T ): e is incident to a vertex in S(T ) B . ∈ ∩ B { ∈ ∩ } ′ ′ Denote by T the subgraph of T induced on EB. Notice that T is a forest of at least 3 S(T ) B edges. Hence T ′ has at least 3 S(T ) B + 1 vertices. As T ′ is a bipartite | ∩ | | ∩ | graph with one partite set as S(T ) B, and another as a subset of A, we conclude that ∩ V (T ) A = V (T ) S(T ) B 2 S(T ) B + 1. Since S(T ) B > (a 1)/2, we then | ∩ | | |−| ∩ | ≥ | ∩ | | ∩ | − have V (T ) A > a. This gives a contradiction to the assumption A = a. | ∩ | | |

3.4.3 Proof of Theorem 3.4.2

Given 0 β α 1, we deﬁne the two extremal cases with parameters α and β as ≤ ≪ ≪ follows. Extremal Case 1. There exists a partition of V (G) into V and V such that V 1 2 | i| ≥ (2/5 4β)n and d(V ,V ) <α. Furthermore, deg(v ,V ) 2βn for each v V . − 1 2 1 2 ≤ 1 ∈ 1 Extremal Case 2. There exists a partition of V (G) into V and V such that V > 1 2 | 1| (3/5 α)n and d(V ,V ) 1 3α. Furthermore, deg(v ,V ) (2n + 3)/5 2βn for each − 1 2 ≥ − 1 2 ≥ − v V . 1 ∈ 1 Then Theorem 3.4.2 is shown through the following three theorems.

Theorem 3.4.3 (Non-extremal Case). For every α> 0, there exists β > 0 and a positive integer n such that if G is a 3-connected graph with n n vertices and δ(G) (2n + 0 ≥ 0 ≥ 3)/5 βn, then G contains an SGHG or G is in one of the two extremal cases. − Theorem 3.4.4 (Extremal Case 1). Suppose that 0 < β α 1 and n is a suﬃciently ≪ ≪ large integer. Let G be a 3-connected graph on n vertices with δ(G) (2n +3)/5. If G is in ≥ Extremal Case 1, then G contains an SGHG.

Theorem 3.4.5 (Extremal Case 2). Suppose that 0 < β α 1 and n is a suﬃciently ≪ ≪ large integer. Let G be a 3-connected graph on n vertices with δ(G) (2n +3)/5. If G is in ≥ Extremal Case 2, then G contains an SGHG. 53

We show Theorems 3.4.3-3.4.5 separately in the following three subsections.

3.4.3.1 Proof of Theorem 3.4.3 We ﬁx the following sequence of parameters,

0 <ε d β α< 1, (3.5) ≪ ≪ ≪ and specify their dependence as the proof proceeds. We let β α be the same α and β as ≪ deﬁned in the two extremal cases. Then we choose d β. Finally we choose ≪

1 d 2 d 1 d 3 1 d d ε = min ε , , 2, , ε , , 3 , ε , 2, 2, , (3.6) 4 2 d3 2 9 2 d3 4 2 2

d 3 where ε 2 , d3 , 3 follows from the definition of the ε in the weak version of the Blow- d 2 d d d up lemma and ε 2, d3 , 2, 2 and ε 2 , 2, 2, 2 follow from the definition of the ε in the strengthened version of the Blow-up lemma. Choose n to be sufficiently large. In the proof, we omit non-necessary ceiling and floor functions. Let G be a graph of order n such that δ(G) (2n+3)/5 βn and suppose that G is not ≥ − in any of the two extremal cases. Applying the regularity lemma to G with parameters ε and d, we obtain a partition of V (G) into l+1 clusters V ,V , ,V for some l M = M(ε), and 0 1 ··· l ≤ a spanning subgraph G′ of G with all described properties in Lemma 3.2.1 (the Regularity lemma). In particular, for all v V , ∈

deg ′ (v) > deg (v) (d + ε)n (2/5 β d ε)n G G − ≥ − − − (2/5 2β)n (provided that ε + d β), (3.7) ≥ − ≤

and (d + ε) e(G′) e(G) n2 e(G) dn2, ≥ − 2 ≥ −

by using ε

We further assume that l = 2k is even; otherwise, we eliminate the last cluster Vl by 54

removing all the vertices in this cluster to V . As a result, V 2εn and 0 | 0| ≤

(1 2ε)n lN =2kN n, (3.8) − ≤ ≤

here we assume that V = N for i 1. | i| ≥ For each pair i and j with 1 i < j l, we write V V if d(V ,V ) d. We now ≤ ≤ i ∼ j i j ≥ consider the reduced graph G , whose vertex set is 1, 2, ,l , and two vertices i and j are r { ··· } adjacent if and only if V V . We claim that δ(G ) (2/5 2β)l. Suppose not, and let i ∼ j r ≥ − i V (G ) such that deg(i ,V (G )) < (2/5 2β)l. Then, for the corresponding cluster V 0 ∈ r 0 r − i0 ′ we have e ′ (V ,V (G ) V ) < V (2/5 2β)lN. On the other hand, by using (3.7), we have G i0 − i0 | i0 | − ′ e ′ (V ,V (G ) V ) V (2/5 2β)n. As lN n from (3.8), we obtain a contradiction. G i0 − i0 ≥ | i0 | − ≤ The rest of the proof consists of the following steps. Step 1. Show that G contains a dominating cycle C and there is a -matching in G with r ∧ r all vertices in V (G ) V (C) as its center. We distinguish two cases in Step 1, and each of r − the other steps will be separated into two cases correspondingly. Case A. C = X Y X Y X Y is an even cycle for some t k. 1 1 2 2 ··· t t ≤ Case B. C = X X Y X Y X Y is an odd cycle for some t

Since C is not necessarily hamiltonian in Gr, we need to take care of the clusters of G which are not represented on C. For each vertex F V (G ) V (C), we partition the ∈ r − corresponding cluster F into two small clusters F and F such that 1 F F 1. We 1 2 − ≤| 1|−| 2| ≤ call each F1 and F2 a half-cluster. Then we group all the original clusters and the partitioned clusters into pairs (A, B) and triples (C,D,F ) with F as a half-cluster such that each pair (A, B) and (C,D) is still ε-regular with density d and the pair (D, F ) is 2.1ε-regular with 55

density d ε. Having the cluster groups like this, in the end, we will ﬁnd “small” HITs − within each pair (A, B) or among each triple (C,D,F ). Step 2. For each 1 i t 1, initiate two independent edges connecting Y and X . In ≤ ≤ − i i+1 Case A, also initiate two independent edges connecting X1 and Yt; and in Case B, initiate two independent edges connecting the clusters in each pair of X0 and X1, and X0 and Yt. Step 3. Make each regular pair in the new grouped pairs and triples given in Step 1 super- regular.

Step 4. Construct HITs covering all vertices in V0 using vertices from the super-regular pairs obtained from Step 3, and obtain new super-regular pairs. Step 5. Apply the Blow-up lemma to ﬁnd a HIT between a super-regular pair resulted from Step 4 or among a triple (A, B, F ), where both (A, F ) and (A, B) are super-regular pairs resulted from Step 4, and F is a half cluster. In addition, in the construction, for each triple (A, B, F ), we require the HIT to use as many vertices as possible from F as non-leaves. Step 6. Apply the Blow-up Lemma again on the regular-pairs induced on the leaves of each HIT obtained in Step 5 to ﬁnd two disjoint paths covering all the leaves. Then connect all the HITs into a HIST of G using edges guaranteed by the regularity and connect the disjoint paths into a cycle using the edges initiated in Step 2. The union of the HIST and the cycle gives an SGHG of G. We now give details of each step. The assumption that G is not in any of the two extremal cases leads to the following claim, which will be used in Step 1.

Claim 3.4.1. Each of the following holds for Gr.

(a) Gr contains no cut-vertex set of size at most βl; (b) G contains no independent set of size more than (3/5 α/2)l . r −

Proof. (a) Suppose instead that Gr contains a vertex-cut W of size at most βl. As δ(G ) (2/5 2β)l, then each component of G W has at least (2/5 3β)l vertices. Let r ≥ − r − − U be the vertex set of one of the components of G W , A = V , and B = V (G) A. r − i∈U i − S 56

We see that A , B (2/5 3β)lN (2/5 4β)n, and since e(G) e(G′)+ dn2, we have | | | | ≥ − ≥ − ≤

2 2 e (A, B) e ′ (A, B)+ dn W A + dn G ≤ G ≤| || | βlN(3/5+3β)lN + dn2 (3β/5+3β2 + d)n2 (as A (3/5+3β)lN and ln n) ≤ ≤ | | ≤ ≤ 25 (3β/5+3β2 + d) A B (since A B 3n2/25) ≤ 3 | || | | || | ≥ 25 < α A B (provided that (3β/5+3β2 + d) <α). | || | 3

This shows that d(A, B) <α. Since deg (u,V (G ) U)= deg (u, W ) βl for each u U, Gr r − Gr ≤ ∈ we see that deg (a, B) βlN +(d + ε)n 2βn for each a A provided that d + ε β. G ≤ ≤ ∈ ≤ However, the above argument shows that G is in Extremal Case 1, showing a contradiction. (b) Suppose instead that G contains an independent set U of size larger than (3/5 r − α/2)l. Let U ′ = V (G ) U, A = V , and B = V (G) A. Then A (3/5 α/2)lN r − i∈U i − | | ≥ − ≥ (3/5 α)n. For each vertex v AS, since deg (v, A) deg ′ (v, A)+(d + ε)n βn, we have − ∈ G ≤ G ≤ deg (v, B) (2n + 3)/5 βn βn (2n + 3)/5 2βn. This gives that G ≥ − − ≥ −

(2/5 2β)n (2/5 2β)n d(A, B) − − 1 3α, ≥ B ≥ (2/5+ α)n ≥ − | |

provided that β α/10+3α2/2. We see that G is in Extremal Case 2. ≤ Step 1. Show that G contains a dominating cycle C, and there is a -matching in G with r ∧ r all vertices in V (G ) V (C) as its center. r − We need some results on longest cycles and paths as follows.

Lemma 3.4.1 ([44]). Let G be a 2-connected graph on n vertices with δ(G) (n + 2)/3. ≥ Then every longest cycle in G is a dominating cycle.

Lemma 3.4.2 ([3]). Let G be a 2-connected graph on n vertices with δ(G) (n + 2)/3. ≥ Then G contains a cycle of length at least min n, n + δ(G) α(G) , where α(G) is the size { − } of a largest independent set in G.

Lemma 3.4.3 ([38]). If G is a 3-connected graph of order n such that the degree sum of any 57 four independent vertices is at least 3n/2+1, then the number of vertices on a longest path and that on a longest cycle diﬀers at most by 1.

By (a) of Claim 3.4.1, G is βl-connected. Since n = Nl + V (l + 2)εn, we get r | 0| ≤ l 1/ε 2. Since 1/ε 2 3/β (provided that β 3ε/(1 2ε)), we then have βl 3. ≥ − − ≥ ≥ − ≥ So Gr is 3-connected. By Claim 3.4.1 (b), Gr has no independent set of size more than (3/5 α/2)l. Notice that δ(G ) (2/5 2β)l > (l + 2)/3. Applying Lemma 3.4.1 and − r ≥ − Lemma 3.4.2 on Gr, we see that there is a cycle C in Gr which is longest, dominating, and has length at least (4/5+ α/2 2β)l. Let = V (G ) V (C). In Case B, we order and − W r − label the vertices of C such that X is adjacent to a vertex, say Y (recall that = 0 0 ∈ W W 6 ∅ in this case). We ﬁx (X ,Y ) as a pair at the ﬁrst place(X Y E(G ), as cluster in G, 0 0 0 0 ∈ r (X0,Y0) is an ε-regular pair with density d). Let

, if in Case A; ′ = W W  Y , if in Case B.  W−{ 0}  We have ′ (1/5 α/2+2β)l if in Case A and ′ (1/5 α/2+2β)l 1 if in |W | ≤ − |W | ≤ − − Case B. So 2 ′ (2/5 α +4β)l< (2/5 2β)l (provided that β <α/6) if in Case A and |W | ≤ − − 2 ′ (2/5 α +4β)l 2 < (2/5 2β)l 1 (provided that β <α/6) if in Case B. Thus |W | ≤ − − − − there is a -matching centered in all vertices in ′; furthermore, if in Case B, we can choose ∧ W the matching such that X0 is not covered by it. Let M∧ be such a matching. For a vertex X ′, denote by M (X) the two vertices from V (C) to which X is adjacent in M . Then ∈ W ∧ ∧ we have two facts about vertices in M∧(X).

Fact 3.4.1. Let X ′. Then the two vertices in M (X) are non-consecutive on C. (By ∈ W ∧ the assumption that C is longest.)

Fact 3.4.2. Let X = Y ′. Then no two vertices from M (X) M (Y ) are adjacent on 6 ∈ W ∧ ∪ ∧ C. (By applying Lemma 3.4.3.)

For a complete bipartite graph, if it contains an SGHG, then the ratio of the cardinalities of the two partite sets should be greater than 2/3 as shown in Proposition 3.4.1. Since a 58

longest dominating cycle in Gr is not necessarily hamiltonian, we need to take care of the clusters of G which are not represented by the vertices on C. One possible consideration is that for each F V (G ) V (C), suppose F is adjacent to A V (C), recall P (A) is the ∈ r − ∈ partner of A. Then as clusters, we consider the bipartite graph of G with partite sets A and P (A) F . However, A / P (A) F is about 1/2, which is less than 2/3. For this reason, ∪ | | | ∪ | we partition F V (G ) V (C) into two parts to attain the right ratio in the corresponding ∈ r − bipartite graphs. Suppose M (F ) = D ,D V (C). As a cluster of G, we partition F ∧ { 1 2} ⊆ into F1 and F2 arbitrarily such that

F N F N F = | | = and F = | | = . | 1| 2 2 | 2| 2 2

We call each Fi a half-cluster of G. Then we create two pairs (Di, Fi), and call Di the dominator of Fi, and Fi the follower of Di, and (Di, Fi) a DF-pair, for i =1, 2. We have the following fact about a DF-pair.

Fact 3.4.3. Each DF-pair (D, F ) is 2.1ε-regular with density at least d ε. (By Slicing − lemma.)

Also, by Fact 3.4.1 and Fact 3.4.2, if D V (C) is a dominator, then P (D), the partner ∈ of D, is not a dominator for any followers. As X V ( ′), we know that X is not a 0 6∈ W 0 dominator for any half-clusters. We group the clusters and half-clusters of G into H-pairs and H-triples in a way below. For each pair (X ,Y ) on C, if X ,Y V (M )= , we take i i { i i} ∩ ∧ ∅ (X ,Y ) as an H-pair. Otherwise, X ,Y V (M ) = 1 by Fact 3.4.1 and Fact 3.4.2. Since i i |{ i i} ∩ ∧ | there is no diﬀerence for the proof for the case that X V (M ) or the case that Y V (M ), i ∈ ∧ i ∈ ∧ throughout the remaining proof, we always assume that Y V (M ) if X ,Y V (M ) = . i ∈ ∧ { i i}∩ ∧ 6 ∅ In this case, there is a unique half-cluster F with Yi as its dominator. Then we take (Xi,Yi, F )

as an H-triple. We assign (X0,Y0) as an H-pair. Step 2. Initiating connecting edges. Given an ε-regular pair (A, B) of density d and a subset B′ B, we say a vertex a A ⊆ ∈ typical to B′ if deg(a, B′) (d ε) B′ . Then by the regularity of (A, B), the fact below ≥ − | | 59 holds.

Fact 3.4.4. If (A, B) is an ε-regular pair, then at most ε A vertices of A are not typical to | | B′ B whenever B′ >ε B . ⊆ | | | | For each 1 i t 1, choose y∗ Y typical to both X and X , and y∗∗ Y typical ≤ ≤ − i ∈ i i i+1 i ∈ i to each of X , X , and Γ(y∗,X ). Correspondingly, choose x∗ Γ(y∗,X ) typical to i i+1 i i i+1 ∈ i i+1 Y , and x∗∗ Γ(y∗∗,X ) typical to both Y and Γ(x∗ ,Y ). For i = t, we choose y∗ i+1 i+1 ∈ i i+1 i+1 i+1 i+1 t and y∗∗ the same way as for i < t, but if in Case A, choose x∗ Γ(y∗∗,X ) typical to Y , and t 1 ∈ t 1 1 x∗∗ Γ(y∗,X ) typical to both Y and Γ(x∗,Y ); and if in Case B, choose x∗ Γ(y∗∗,X ) 1 ∈ t 1 1 1 1 0 ∈ t 0 typical to X , and x∗∗ Γ(y∗,X ) typical to both X and Γ(x∗,X ). Furthermore, in Case 1 0 ∈ t 0 1 0 1 B, we choose y∗ X typical to both Y and X , and y∗∗ X typical to each of Y , X , t+1 ∈ 0 0 1 t+1 ∈ 0 0 1 and Γ(y∗ ,Y ). Correspondingly, choose x∗ Γ(y∗ ,X ) typical to Y and x∗∗ Γ(y∗∗ ,X ) t+1 0 1 ∈ t+1 1 1 1 ∈ t+1 1 typical to both Y and Γ(x∗,Y ). Additionally, we choose y∗ Γ(y∗ ,Y ) such that y∗ is 1 1 1 0 ∈ t+1 0 0 typical to X , and choose y∗∗ Γ(y∗∗ ,Y ) such that y∗∗ is typical to X . Notice that by the 0 0 ∈ t+1 0 0 0 choice of these vertices above, we have the following.

y∗x∗ ,y∗∗x∗∗ E(G), for 1 i t 1; i i+1 i i+1 ∈ ≤ ≤ −  x∗y∗∗, x∗∗y∗ E(G), in Case A;  1 t 1 t ∈  x∗y∗∗, x∗∗y∗, x∗y∗ , x∗∗y∗∗ ,y∗y∗ ,y∗∗y∗∗ E(G), in Case B. 0 t 0 t 1 t+1 1 t+1 0 t+1 0 t+1 ∈   By Fact 3.4.4, for each 0 i t, we have Γ(x∗,Y ) Γ(x∗∗,Y ) , Γ(y∗,X ) Γ(y∗∗,X ) ≤ ≤ | i i ∩ i i | | i i ∩ i i | ≥ (d ε)2N, and Γ(y∗ ,Y ) Γ(y∗∗ ,Y ) (d ε)2N. − | t+1 0 ∩ t+1 0 | ≥ − Step 3. Super-regularizing the regular pairs in each H-pair and H-triple given in Step 1. For each 0 i t, if (X ,Y ) is an H-pair, let ≤ ≤ i i

X′ = x X : deg(x, Y ) (d ε)N and Y ′ = y Y : deg(y,X ) (d ε)N . i { ∈ i i ≥ − } i { ∈ i i ≥ − }

By Fact 3.4.4, we have X′ , Y ′ (1 ε)N. Recall that x∗, x∗∗ X and y∗,y∗∗ Y are | i| | i | ≥ − i i ∈ i i i ∈ i the initiated vertices in Step 2. For 1 i t, if X′ x∗, x∗∗ = Y ′ y∗,y∗∗ , say ≤ ≤ | i −{ i i }| 6 | i −{ i i }| X′ x∗, x∗∗ > Y ′ y∗,y∗∗ , we then remove X′ x∗, x∗∗ Y ′ y∗,y∗∗ vertices | i −{ i i }| | i −{ i i }| | i −{ i i }|−| i −{ i i }| 60

out from X′ x∗, x∗∗ , and denote the remaining set still as X′. Denote Y ′ y∗,y∗∗ still i −{ i i } i i −{ i i } as Y ′. We see that X′ = Y ′ . As Y ′ (1 ε)N (to be precise, the lower bound should i | i| | i | | i | ≥ − be (1 ε)N 2, however, the constant 2 can be made vanished by adjusting the ε factor, − − we ignore the slight diﬀerent of the ε-factor here), we have that X Y (X′ Y ′) 2εN. | i ∪ i − i ∪ i | ≤ For i = 0, if X′ x∗, x∗∗,y∗ ,y∗∗ = Y ′ y∗,y∗∗ , say X′ x∗, x∗∗,y∗ ,y∗∗ > | i −{ i i t+1 t+1}| 6 | i −{ i i }| | i −{ i i t+1 t+1}| Y ′ y∗,y∗∗ , then we remove X′ x∗, x∗∗,y∗ ,y∗∗ Y ′ y∗,y∗∗ vertices out from | i −{ i i }| | i −{ i i t+1 t+1}|−| i −{ i i }| X′ x∗, x∗∗,y∗ ,y∗∗ and denote the remaining set still as X′. Denote Y ′ y∗,y∗∗ still i −{ i i t+1 t+1} i i −{ i i } as Y ′. We see that X′ = Y ′ . We call the resulting H-pairs supper-regularized H-pairs. By i | i| | i | ′ ′ Slicing lemma (Lemma 3.2.5) and the deﬁnitions of Xi,Yi , we see that

Fact 3.4.5. Each supper-regularized H-pair (X′,Y ′) is a (2ε,d 2ε)-super-regular pair. i i −

For each H-triple (Xi,Yi, F ), by Fact 3.4.3, (Yi, F ) is 2.1ε-regular with density at least d ε. Let −

X′ = x X : deg(x, Y ) (d ε)N , i { ∈ i i ≥ − } Y ′ = y Y : deg(y,X ) (d ε)N,deg(y, F ) (d 3.1ε) F , and i { ∈ i i ≥ − ≥ − | |} F ′ = f F : deg(f,Y ) (d 3.1ε)N . { ∈ i ≥ − }

Recall that x∗, x∗∗ X and y∗,y∗∗ Y are the initiated vertices in Step 2. We remove i i ∈ i i i ∈ i ∗ ∗∗ ′ ∗ ∗∗ ′ ′ xi , xi out from Xi, and remove yi ,yi out from Yi . Still denote the resulted clusters as Xi and Y ′, respectively. Remove d3N vertices out from F , which consists of all vertices in i ⌈ ⌉ F F ′ and any d3N F F ′ vertices from F ′ (we need to increase the ratio Y ′ / X′ F ′ − ⌈ ⌉−| − | | i | | i ∪ | ′ a little as later on we may use vertices in Yi in constructing HITs covering vertices in V0). Denote the resulting set still by F ′. Then we see that X′ (1 ε)N, Y ′ (1 3.1ε)N, | i| ≥ − | i | ≥ − and F ′ (1 2.1ε) F d3N (1 2.1ε 2d3) F . We call the resulted H-triples supper- | | ≥ − | | − ≥ − − | | regularized H-triples. By the Slicing Lemma and the deﬁnitions above, the following is true.

Fact 3.4.6. For each super-regularized H-triple (X′,Y ′, F ′), (X′,Y ′) is (2ε,d 3.1ε)- super- i i i i − regular, and (Y ′, F ′) is (4.2ε,d 3.1ε 2d3)-super-regular. i − − 61

Let V 1 be the union of the set of vertices from each (X Y (X′ Y ′)) 0 i ∪ i − i ∪ i − x∗, x∗∗,y∗,y∗∗ y∗ ,y∗∗ ( y∗ ,y∗∗ exists only if in Case B), where (X ,Y ) is an { i i i i }−{ t+1 t+1} { t+1 t+1} i i H-pair, and let V 2 be the union of the set of vertices from each (X Y F (X′ 0 i ∪ i ∪ − i ∪ Y ′ F ′)) x∗, x∗∗,y∗,y∗∗ , where (X ,Y , F ) is an H-triple. Notice that for each H-pair i ∪ −{ i i i i } i i (X ,Y ), we have X Y (X′ Y ′) 2εN; and for each H-triple (X ,Y , F ), we have i i | i ∪ i − i ∪ i | ≤ i i X X′ εN, Y Y ′ (ε +2.1ε)N, and F F ′ d3N. Hence by using the facts that | i − i| ≤ | i − i | ≤ | − | ≤ ′ (1/5 α/2+2β)l, t = l/2, and Nl n from inequality (3.8), we get |W | ≤ − ≤

V 1 + V 2 2εNl/2+2(1/5 α +2β)l(d3N +2.1εN) 2d3Nl/5+2εNl 2d3n/5+2εn. | 0 | | 0 | ≤ − ≤ ≤

Let V ′ = V V 1 V 2. Then 0 0 ∪ 0 ∪ 0

V ′ 2εn +2d3n/5+2εn d3n/2 (provided that ε d3/40). (3.9) | 0 | ≤ ≤ ≤

′ Step 4. Construct small HITs covering all vertices in V0 . Consider a vertex x V ′ and a cluster or a half-cluster A, we say that x is adjacent to ∈ 0 A, denoted by x A, if deg(x, A) (d ε) A . We call A the partner of x. ∼ ≥ − | | Claim 3.4.2. For each vertex x V ′, there is a cluster or a half-cluster A such that x A, ∈ 0 ∼ ′ where A is not a dominator, and we can assign all vertices in V0 to their partners which are d2N not dominators such that each of the cluster or half-cluster is used by at most 20 vertices ′ from V0 .

Proof. Suppose we have found partners for the ﬁrst m

d2N Ω m

by inequality (3.8). Therefore,

20d3k 20d3l Ω + | | ≤ d2 d2(1 2ε) − 10dl + 40dl (provided that 1 2ε 1/2 ) ≤ − ≥ βl (provided that 50d β ). ≤ ≤

Consider now a vertex v V ′ not having a partner found so far. Let be the set of all ∈ 0 U non-dominator clusters and half-clusters adjacent to v not contained in Ω. We claim that (α 7β)l. To see this, we ﬁrst observe that any vertex v V ′ is adjacent to at least |U| ≥ − ∈ 0 (α 6β)l non-dominator clusters and half-clusters. For instead, as v may adjacent to 2 ′ − |W | dominators, vertices in V ′, or clusters A with deg(v, A) < (d ε) A , we have 0 − | |

(2/5 β)n deg (v) < (α 6β)lN + (2/5+4β α)lN + d3n/2+(d ε)lN − ≤ G − − − (2/5 2β + d3/2+ d ε)n ≤ − − < (2/5 3β/2)n (provided that d ε + d3/2 < β/2 ), − −

showing a contradiction. Since Ω βl, we have (2α 7β)l. | | ≤ |U| ≥ − ′ ′ Now for each non-dominator cluster A (A is either a cluster Xi, Yi , or a half cluster ′ ′ F ), let I(A) be the set of vertices from V0 such that each of them has A as its partner. By

2 Claim 3.4.2, we have I(A) d N . | | ≤ 20 ′ We need three operations below for constructing small HITs covering vertices in V0 . Operation I Let (A, B) be an (ε′, δ)-super-regular pair, and I a set of vertices disjoint from A B. Suppose that (i) deg(x, B) d′ B > ε′ B and deg(x, B) d′ B 3 I for ∪ ≥ | | | | ≥ | | ≥ | | any x I; (ii) (δ ε′)d′ B 3 I ; (iii) (δ ε′) A > I ; and (iv) δ A > 4 I . Then we can ∈ − | | ≥ | | − | | | | | | | | do the following operations on (A, B) and I. Let I = x , x , , x . We first assume that I 2. { 1 2 ··· |I|} | | ≥ Since (A, B) is (ε′, δ)-super-regular, for each v Γ(x , B), Γ(v, A) δ A . By condition ∈ i | | ≥ | | (i), we have Γ(x , B) >ε′ B for each i. Applying Fact 3.4.4, we then know that there are | i | | | 63 at least (δ ε′) A > I vertices from Γ(v, A) typical to Γ(x , B) for each 1 i I 1. − | | | | i+1 ≤ ≤| | − That is, there exists A Γ(v, A) with A (δ ε′) A > I such that for each a A , 1 ⊆ | 1| ≥ − | | | | 1 ∈ 1 Γ(a , Γ(x , B)) (δ ε′)d′ B 3 I . As deg(x, B) d′ B 3 I for any x I and | 1 i+1 | ≥ − | | ≥ | | ≥ | | ≥ | | ∈ (δ ε′)d′ B 3 I , combining the above argument, we know there is a claw-matching M − | | ≥ | | I from I to B centered in I such that one vertex from Γ(xi,V (MI )) and one vertex from Γ(x ,V (M )) have at least (δ ε′) A > I common neighbors in A. Let x , x , x be the i+1 I − | | | | i1 i2 i3 three neighbors of x in M (in fact in B) and suppose that Γ(x , A) Γ(x , A) I . i I | i3 ∩ i+1,1 |≥| | For 1 i I 1, we then choose distinct vertices y Γ(x , A) Γ(x , A). By ≤ ≤ | | − i ∈ i3 ∩ i+1,1 condition (iv), there is a -matching M between the vertex set x : 1 i I 1 and ∧ 2 { i3 ≤ ≤| | − } the vertex set A y : 1 i I 1 centered in the first set, a matching M between −{ i ≤ ≤ | | − } 3 x : 1 i I 1 and A y : 1 i I 1 V (M ) covering the first set, { i+1,1 ≤ ≤ | | − } −{ i ≤ ≤ | | − } − 2 and a matching M between the vertex set y : 1 i I 1 and B V (M ) covering 4 { i ≤ ≤| | − } − I the first set. Finally, by using (iv) again, we can find three distinct vertices y ,y ,y 31 32 33 ∈ Γ(x , A) y : 1 i I 1 V (M ) V (M ). Let T be the graph with 13 −{ i ≤ ≤| | − } − 2 − 3 B

V (T )= V (M ) y : 1 i I 1 V (M ) V (M ) V (M ) y ,y ,y B I ∪{ i ≤ ≤| | − } ∪ 2 ∩ 3 ∪ 4 ∪{ 31 32 33} and

E(T )= M y x ,y x : 1 i I 1 M M M x y , x y , x y . B I ∪{ i i3 i i+1,1 ≤ ≤| | − } ∪ 2 ∪ 3 ∪ 4 ∪{ 13 31 13 32 13 33}

If I = 1, we choose x , x , x Γ(x , B) and y ,y ,y Γ(x , A). Then let T be the | | 11 12 13 ∈ 1 31 32 33 ∈ 13 B graph with V (T )= x , x , x , x ,y ,y ,y B { 1 11 12 13 31 32 33} and E(T )= x x , x x , x x , x y , x y , x y . B { 1 11 1 12 1 13 13 31 13 32 13 33} 64

In any case, we see that TB is a HIT satisfying

V (T ) B = V (T ) A =4 I 1, | B ∩ | | B ∩ | | | − L(T ) B = min 2 I +1, 3 I 1 , L(T ) A =3 I . (3.10) | B ∩ | { | | | | − } | B ∩ | | |

We call TB the insertion HIT associated with B. Figure 3.4 gives a depiction of TB for I =1, 3, respectively. | |

B B A A

I =1 I =3 | | | | Figure (3.4) The HIT TB

Operation II Let (A, B) be an (ε′, δ)-super-regular pair, and I a set of vertices disjoint from A B. Suppose that (i) deg(x, A) d′ A >ε′ A and deg(x, A) d′ A 3 I for any ∪ ≥ | | | | ≥ | | ≥ | | x I; (ii) (δ ε′)d′ A 3 I ; (iii) (δ 2ε′) B > I ; and (iv) δ B > 3 I . Then we can do ∈ − | | ≥ | | − | | | | | | | | the following operations on (A, B) and I. Let I = x , x , , x . We ﬁrst assume that I 3. { 1 2 ··· |I|} | | ≥ Since (A, B) is (ε′, δ)-super-regular, for each v Γ(x , A), Γ(v, B) δ B . By condition ∈ i | | ≥ | | (i), we have Γ(x , A) >ε′ A for each i. Applying Fact 3.4.4, we then know that there are at | i | | | least (δ 2ε′) B > I vertices from Γ(v, B) typical to both Γ(x , A) and Γ(x , A) for each − | | | | i+1 i+2 1 i I 2. That is, there exists B Γ(v, B) with B (δ 2ε′) B > I such that ≤ ≤| | − 1 ⊆ | 1| ≥ − | | | | for each b B , Γ(b , Γ(x , A)) , Γ(b , Γ(x , A)) (δ ε′)d′ A 3 I . As deg(x, A) 1 ∈ 1 | 1 i+1 | | 1 i+2 | ≥ − | | ≥ | | ≥ d′ A 3 I for any x I and (δ ε′)d′ A 3 I , combining the above argument, we | | ≥ | | ∈ − | | ≥ | | know there is a claw-matching MI from I to A centered in I such that any one vertex from

Γ(xi,V (MI )), any one vertex from Γ(xi+1,V (MI )), and any one vertex from Γ(xi+2,V (MI )) have at least I common neighbors in B. Let x , x , x be the three neighbors of x in M (in | | i1 i2 i3 i I 65

fact in A). For i = 1, choose y Γ(x , A) Γ(x , A) Γ(x , A). Let h = ( I 3)/2 . For 0 ∈ 13 ∩ 23 ∩ 33 ⌈ | |− ⌉ 1 k h, we then choose distinct vertices y Γ(x , A) Γ(x , A) Γ(x , A) ≤ ≤ k ∈ 1+2k,2 ∩ 2+2k,3 ∩ 3+2k,3 (if I = 2+2k, let Γ(x , A)= A). By condition (iv), there is a matching M between the | | 3+2k,3 vertex set x , x : 1 i I , 1 k h and the vertex set B y ,y : 1 k h { i3 1+2k,2 ≤ ≤| | ≤ ≤ } −{ 0 k ≤ ≤ } covering the ﬁrst set. If I is even, choose y ,y Γ(x , B) such that they have not been | | 31 32 ∈ 13 chosen before; if I is odd, choose y ,y ,y Γ(x , B) such that they have not been | | 31 32 33 ∈ 13 chosen before. Let TA be the graph with

V (MI ) V (M) y0,yk : 1 k h y31,y32 , if I is even; V (TA)= ∪ ∪{ ≤ ≤ }∪{ } | |  V (M ) V (M) y ,y : 1 k h y ,y ,y , if I is odd;  I ∪ ∪{ 0 k ≤ ≤ }∪{ 31 32 33} | |  and E(T ) containing all edges in M M y x ,y x ,y x and all edges in A I ∪ ∪{ 0 13 0 23 0 33}

x y , x y , x y , x y , x y : 1 k h 1 y ,y , if I is even; { 1+2k,2 k 2+2k,2 k 3+2k,2 k 1+2h,2 h 2+2h,2 h ≤ ≤ − }∪{ 31 32} | |  x y , x y , x y : 1 k h y ,y ,y , if I is odd.  { 1+2k,2 k 2+2k,2 k 3+2k,2 k ≤ ≤ }∪{ 31 32 33} | |  If I = 1, we choose x , x , x Γ(x , A) and y ,y Γ(x , B), and then let T be the | | 11 12 13 ∈ 1 31 32 ∈ 13 A graph with

V (T )= x , x , x , x ,y ,y and E(T )= x x , x x , x x , x y , x y . B { 1 11 12 13 31 32} B { 1 11 1 12 1 13 13 31 13 32}

If I = 2, we choose x , x , x Γ(x , A), x , x , x Γ(x , A), y Γ(x , B) Γ(x , B), | | 11 12 13 ∈ 1 11 12 13 ∈ 2 ∈ 13 ∩ 21 y ,y Γ(x , B), and y ,y Γ(x , B) such that they are all distinct, then let T be 11 12 ∈ 13 21 22 ∈ 21 A the graph with

V (T )= x , x , x , x ,y,y ,y : i =1, 2 and B { i i1 i2 i3 i1 i2 }

E(T )= x x , x x , x x , x y, x y, x y , x y , x y , x y . B { i i1 i i2 i i3 13 21 13 11 13 12 21 21 21 22}

We see that T is a tree which has a degree 2 vertex y only if I = 2 and a degree 2 vertex A | | 66 y only if I > 2 and I is even. In addition, T satisﬁes the following. h | | | | A

2 I , if I =1, 2; V (TA) A = 3 I and L(TA) A = | | | | | ∩ | | | | ∩ |  2 I |I|−3 , if I 3; and  | | − 2 | | ≥ l m 2, if I = 1;  V (TA) B = | | | ∩ |  2 I +1, if I 2; and  | | | | ≥  2 I , if I =1, 2; L(TA) B = | | | | (3.11) | ∩ |  2 I |I|−3 , if I 3.  | | − 2 | | ≥ l m  In this case, we call T the insertion tree associated with A. Notice that L(T ) A = A | A ∩ | L(T ) B always holds. Figure 3.4 gives a depiction of T for I =1, 2, 5, 6, respectively. | A ∩ | A | |

A A A B B B

I =1 I =2 I =5 | | | | | |

I =6 | | Figure (3.5) The tree TA

Operation III Let (B, F ) be an (ε′, δ)-super-regular pair, and I a set of vertices disjoint from B F . Suppose that deg(x, F ) d′ F 3 I for any x I and δ B 6 I . Then we ∪ ≥ | | ≥ | | ∈ | | ≥ | | can do the following operations on (A, B) and I. Let I = x , x , , x . Since deg(x, B) d′ B 3 I for any x I, there is a { 1 2 ··· |I|} ≥ | | ≥ | | ∈ claw-matching M from I to F centered in I. Then as δ B 6 I , there is a -matching I | | ≥ | | ∧ 67

M from V (M ) F to B centered in V (M ) F . Let T be the graph with ∧ I ∩ I ∩ F

V (T )= V (M ) V (M ) and E(T )= M M . B I ∪ ∧ B I ∪ ∧

We see that TF is a forest with no vertex of degree 2 satisfying

V (T ) F = S(T ) F = 3 I and V (T ) B = L(T ) B =6 I . (3.12) | F ∩ | | F ∩ | | | | F ∩ | | F ∩ | | |

We call TF the insertion forest associated with F . Now for each H-pair (X′,Y ′), we may assume that I(X′) = and I(Y ′) = for a i i i 6 ∅ i 6 ∅ uniform discussion, as the consequent argument is independent of the assumptions. Recall that (X′,Y ′) is (2ε,d 2ε)-super-regular by Fact 3.4.5. Notice that deg(x, X′) (d ε) X′ i i − i ≥ − | i| 2 for each x I(X′), I(X′) d N , and X′ , Y ′ (1 ε)N. By simple calculations, ∈ i | i | ≤ 20 | i| | i | ≥ − we see that (i) deg(x, X′) (d ε) X′ > 2ε X′ and (d ε) X′ 3d2N/20 for each i ≥ − | i| | i| − | i| ≥ x I(X′); (ii) (d 2ε 2ε)(d ε) X′ > 3d2N/20; (iii) (d 4ε) Y ′ > d2N/20; and (iv) ∈ i − − − | i| − | i | (d 2ε) Y ′ >d2N/5 4I(X′). Thus all the conditions in Operation I are satisfied. So we − | i | ≥ i ′ ′ ′ ′ d2N can find a HIT TX′ associated with X . As V (TX′ ) X = V (TX′ ) Y 4 I(X ) , we i i | i ∩ i| | i ∩ i | ≤ | i | ≤ 5 ′ ′ 2 ′ know that (X V (TX′ ),Y V (TX′ )) is (4ε,d 2ε d N/5)-super regular. Since deg(y,Y ) i − i i − i − − i ≥ ′ ′ ′ 2 ′ (d ε) Y for each y I(Y ), we get deg(y,Y V (TX′ )) (d ε d /5) Y for each − | i | ∈ i i − i ≥ − − | i | y I(Y ′). By direct checking, conditions (i) (iv) of Operation I are satisfied by the pair ∈ i ∼ ′ ′ ′ ′ ′ (X V (TX′ ),Y V (TX′ )) and I(Y ). Then we use Operation I on (X V (TX′ ),Y V (TX′ )) i − i i − i i i − i i − i ′ ′ and I(Y ) to get a HIT TY ′ associated with Y V (TX′ ). Denote i i i − i

∗ ′ ∗ ′ X = X V (TX′ ) V (TY ′ ) and Y = Y V (TX′ ) V (TY ′ ). i i − i − i i i − i − i

By using (3.10) in Operation I, we have X∗ = Y ∗ (1 2d2/5 ε)N N/2. By Slicing | i | | i | ≥ − − ≥ lemma (Lemma 3.2.5) and Fact 3.4.5, we have the following.

Fact 3.4.7. For each H-pair (X ,Y ), (X∗,Y ∗) is (4ε,d 2ε 2d2/5)-super-regular with i i i i − − X∗ = Y ∗ . We call (X∗,Y ∗) a ready H-pair. | i | | i | i i 68

Then for each H-triple (X′,Y ′, F ′), we may assume that I(X′) = and I(F ′) = (recall i i i 6 ∅ 6 ∅ that Y is assumed to be the dominator of F, so I(Y ′) = by the distribution principle of i i ∅ vertices in V ′ from Claim 3.4.2). By Fact 3.4.6, we know that (X′,Y ′) is (2ε,d 3.1ε)- 0 i i − super-regular and (Y ′, F ′) is (4.2ε,d 3.1ε 2d3)-super-regular. Notice also that X′ i − − | i| ≥ (1 ε)N, Y ′ (1 3.1ε)N, F ′ (1 2.1ε 2d3)N/2, and deg(x, X′) (d ε) X′ and − | i | ≥ − | | ≥ − − i ≥ − | i| 2 deg(y, F ′) (d ε) F ′ for each x I(X′) and each y I(F ′). Since I(X′) , I(F ′) d N ≥ − | | ∈ i ∈ | i | | | ≤ 20 and ε d 1, the conditions of Operation III are satisﬁed by (Y ′, F ′) and I(F ′) by direct ≪ ≪ i ′ calculations. Let TF ′ be the insertion forest associated with F . Then we use Operation II

′ ′ ′ ′ on (X ,Y V (TF ′ )) and I(X ) to get a tree TX′ associated with X . Denote i i − i i i

∗ ′ ∗ ′ ∗ ′ X = X V (TX′ ),Y = Y V (TF ′ ) V (TX′ ), and F = F V (TF ′ ). i i − i i i − − i −

By using (3.11) and (3.12) in Operation II and Operation III, respectively, we have X∗ , Y ∗ (1 3.1ε 9d2/20)N N/2 and F ∗ (1 2.1ε 2d3)N/2 3d2N/20 | i | | i | ≥ − − ≥ | | ≥ − − − ≥ (1 2.1ε 2d3 3d2/10)N/2. By Slicing lemma and Fact 3.4.6, we have the following. − − − Fact 3.4.8. For each H-triple (X ,Y , F ), (X∗,Y ∗) is (4ε,d 3.1ε 9d2/20)-super-regular i i i i − − and (Y ∗, F ∗) is (8.4ε,d 2.1ε 3d2/10 2d3)-super-regular. We call (X∗,Y ∗, F ∗) a ready i − − − i i H-triple.

Step 5. Apply the Blow-up lemma to ﬁnd a HIT within each ready H-pair and among each ready H-triple. In order to apply the Blow-up Lemma, we ﬁrst give two lemmas which assure the existence of a given subgraph in a complete bipartite graph.

Lemma 3.4.4. Suppose 0 <ε d 1 and N is a large integer. If G(A, B) is a balanced ≪ ≪ complete bipartite graph with (1 ε d2/2)N A = B N, then G(A, B) contains a − − ≤ | | | | ≤ HIST T with ∆(T ) 2/d3 and L(T ) A L(T ) B = ℓ for any given pair pair ≤ ⌈ ⌉ || pair ∩ |−| pair ∩ || non-negative integer ℓ with ℓ d2N. ≤ Proof. By the symmetry, we only show that we can construct a HIST T such that 69

′ 3 L(T ) A L(T ) B = ℓ. Let ∆ = d N . We choose distinct a , a , , a ′ A and | ∩ |−| ∩ | ⌈ ⌉ 1 2 ··· ∆ ∈ distinct b , b , , b ′ B. Then we decompose all vertices in B into B , B , , B ′ such 1 2 ··· ∆ −1 ∈ 1 2 ··· ∆ that 3 B 1/d3, B B = b for 1 i ∆′ 1, and B B = for i j > 1. ≤| i| ≤ i ∩ i+1 { i} ≤ ≤ − i ∩ j ∅ | − | ′ Now we choose ℓ + 1 distinct vertices b ′ , b ′ , , b ′ from B b : 1 i ∆ 1 . ∆ ∆ +1 ··· ∆ +ℓ −{ i ≤ ≤ − } As ∆′ = d3N , ℓ + ∆′ (d2 + d3)N + 1, and thus ⌈ ⌉ ≤

2(ℓ + ∆′) (2d2 +2d3)N +2 (1 d2/2 ε)N d3N A d3N . ≤ ≤ − − − ⌈ ⌉≤| | − ⌈ ⌉

Thus we can use all of the vertices in b : 1 i ∆′ + ℓ to cover all vertices in { i ≤ ≤ } A a 1 i ∆′ 1 such that each b can be adjacent to at least two distinct vertices. We −{ i | ≤ ≤ − } i ′ 3 partition A a 1 i ∆ 1 arbitrarily into A , A , , A ′ such that 2 A 1/d . −{ i | ≤ ≤ − } 1 2 ··· ℓ+∆ ≤| i| ≤ Now let T be a spanning subgraph of G(A, B) such that

E(T )= a b b B , 1 i ∆′ b a a A , 1 j ∆′ + ℓ . { i | ∈ i ≤ ≤ }∪{ j | ∈ j ≤ ≤ }

Clearly, ∆(T ) 2/d3 . As A = B , S(T ) A = ∆′, and S(T ) B = ∆′ + ℓ, we then ≤ ⌈ ⌉ | | | | | ∩ | | ∩ | have that L(T ) A L(T ) B = ℓ. We denote T as T . | ∩ |−| ∩ | pair Lemma 3.4.5. Suppose 0 < ε d 1 and N is a large integer. Let G = G(A, B, F ) ≪ ≪ be a tripartite graph with V (G) partitioned into A B F such that both G[A B] and ∪ ∪ ∪ G[B F ] are complete bipartite graphs. If (i) (1 4ε d2/2)N A , B N, (ii) ∪ − − ≤ | | | | ≤ (1/2 2.1ε 3d2/20 d3)N F (1/2 d3)N, and (iii) for any given non-negative − − − ≤ | | ≤ − integer l 3d2N/10, we have B 2( A F B l) 3d3N/2 holds, then G contains a ≤ | | − | ∪ |−| | − ≥ HIST Ttriple and a path Ptriple spanning on a subset of L(Ttriple) such that (a) T is a HIST of G with ∆(T ) 3/d3 ; triple triple ≤ ⌈ ⌉

(b) L(T ) B = L(T ) (A F ) l. | triple ∩ | | triple ∩ ∪ | −

(c) P is a (b, f)-path on L(T ) F and any L(T ) F vertices from L(T ) B, triple triple ∩ | triple ∩ | triple ∩ and V (P ) F 5d2N/6. | triple ∩ | ≤ 70

′ 3 Proof. Let ∆ = d N/2 . We choose distinct b , b , , b ′ B and partition all ⌈ ⌉ 1 2 ··· ∆ ∈ 3 vertices in F into F , F , , F ′ such that 3 F 1/d . Then we choose distinct 1 2 ··· ∆ ≤ | i| ≤ a , a , , a ′ A and decompose all vertices in A into A , A , , A ′ such that 3 1 2 ··· ∆ −1 ∈ 1 2 ··· ∆ ≤ A 2/d3, A A = a for 1 i ∆′ 1, and A A = for i j > 1. Choose one | i| ≤ i ∩ i+1 { i} ≤ ≤ − i ∩ j ∅ | − | ′ ′ ′ more vertex, say a ′ A a 1 i ∆ 1 . Let l = A F B l. Notice that l > 0. ∆ ∈ −{ i | ≤ ≤ − } | ∪ |−| |− ′ ′ Now we choose l distinct vertices f , f , , f ′ from A a : 1 i ∆ F (choose as 1 2 ··· l −{ i ≤ ≤ } ∪ many as possible from F ﬁrst) and partition any 2l′ vertices of B b : 1 i ∆′ into −{ i ≤ ≤ } 3 B , B , , B ′ such that B = 2. By (iii), we see that there are at least d N vertices left 1 2 ··· l | i| ⌊ ⌋ ′ ′ l′ ′ ′ ′ ′ in B = B b : 1 i ∆ B . Hence we can partition B = B B B ′ −{ i ≤ ≤ } − i=1{ i} 1 ∪ 2 ∪···∪ ∆ ′ ′ ′ such that B ′ 2 and B S1 for j = ∆ . We let T be a subgraph of G on A B F | ∆ | ≥ | j| ≥ 6 ∪ ∪ with

E(T )= b f, b a, a b′ : f F , a A , b′ B′, 1 i ∆′ f b : b B , 1 i l′ . { i i i ∈ i ∈ i ∈ i ≤ ≤ }∪{ i ∈ i ≤ ≤ }

By the construction, T is a HIST of G, which clearly satisﬁes (a). Since S(T ) B = ∆′ | ∩ | and S(T ) (A F ) = ∆′ + l′ = ∆′ + A F B l, we then see that T satisﬁes | ∩ ∪ | | ∪ |−| | − (b). If L(T ) F = , let f L(T ) F and b L(T ) B, we can then take a (b, f)-path ∩ 6 ∅ ∈ ∩ ∈ ∩ P with V (P ) F = L(T ) F and V (P ) = 2 L(T ) F . By (i) and (ii), we see that ∩ ∪ | | | ∩ | l′ = A F B l (1/2 6.1ε 4d2/5 d3)N. Hence V (P ) F = F l′ 5d2N/6. | ∪ |−| | − ≥ − − − | ∩ | | | − ≤ Denote T as Ttriple and P as Ptriple.

Now for 1 i t and for each ready H-pair (X∗,Y ∗), suppose, without of loss generality, ≤ ≤ i i ′ ′ ′ ′ ′ that (L(TX′ ) Y ) ((L(TY ′ ) Y ) ((L(TX′ ) X ) ((L(TY ′ ) X ) = l , where TX′ | i ∩ i ∪ i ∩ i |−| i ∩ i ∪ i ∩ i | i ′ ′ is the insertion HIT associated with X and T ′ is the insertion HIT associated with Y . i Yi i ′ 2 ′ Notice that l d N from (3.10) and (3.11). Let xa S(TX′ ) X be a non-leaf of TX′ and ≤ ∈ i ∩ i i ′ ′ ′ yb S(TY ′ ) Y a non-leaf of TY ′ . Since (X ,Y ) is (2ε,d 2ε)-super-regular by Fact 3.4.5 ∈ i ∩ i i i i − and Y ′ Y ∗ 2d2N/5, we have deg(x ,Y ∗) (d 2ε d2/2)N dN/2. Similarly, | i − i | ≤ a i ≥ − − ≥ deg(y ,X∗) (d 2ε d2/2)N dN/2. Also, from Step 2, we have Γ(x∗,Y ), Γ(x∗∗,Y ) b i ≥ − − ≥ i i i i ≥ (d 3ε)N. So, Γ(x∗,Y ∗), Γ(x∗∗,Y ∗) (d 3ε d2/2)N dN/2. Similarly, we have − i i i i ≥ − − ≥ 71

Γ(y∗,X∗), Γ(y∗∗,X∗) (d 3ε d2/2)N dN/2. Recall that (X∗,Y ∗) is (4ε,d 2ε 8d2/20)- i i i i ≥ − − ≥ i i − − ∗ ∗ super-regular by Fact 3.4.7, and therefore (Xi ,Yi ) is (4ε,d/2)-super-regular. By the the strengthened version of the Blow-up lemma and Lemma 3.4.4 (the conditions are certainly satisﬁed by X∗ and Y ∗), we can ﬁnd a HIST T i = T on X∗ Y ∗ such that there exist i i 1 ∼ pair i ∪ i y S(T i) Γ(x ,Y ∗), x S(T i) Γ(y ,X∗), y′ S(T i) Γ(x∗,Y ), y′′ S(T i) Γ(x∗∗,Y ), a ∈ 1 ∩ a i b ∈ 1 ∩ b i i ∈ 1 ∩ i i i ∈ 1 ∩ i i and x′ S(T i) Γ(y∗,X ), x′′ S(T i) Γ(y∗∗,X ) such that L(T i) X∗ L(T i) Y ∗ = l′. i ∈ 1 ∩ i i i ∈ 1 ∩ i i | 1 ∩ i |−| 1 ∩ i | i ∗ ′ ′ i ∗ ′ ′ Hence L(T ) X + L(TX′ ) X + L(TY ′ ) X = L(T ) Y + L(TX′ ) Y + L(TY ′ ) Y . | 1 ∩ i | | i ∩ i| | i ∩ i| | 1 ∩ i | | i ∩ i | | i ∩ i | i i ∗ ′ ∗∗ ′′ ∗ ′ ∗∗ ′′ i Let T = T TX′ TY ′ xaya,ybxb x y , x y ,y x ,y x . It is clear that T is a HIST 1 ∪ i ∪ i ∪{ }∪{ i i i i i i i i } on X′ Y ′ I(X′) I(Y ′) such that i ∪ i ∪ i ∪ i

x∗, x∗∗,y∗,y∗∗ L(T i) and L(T i) X′ = L(T i) Y ′ . { i i i i } ⊆ | ∩ i| | ∩ i |

∗ ∗ ′ ′ For the ready H-pair (X ,Y ), let xa S(TX′ ) X be a non-leaf of TX′ and yb S(TY ′ ) Y a 0 0 ∈ 0 ∩ 0 0 ∈ 0 ∩ 0

′ non-leaf of TY0 . By the the strengthened version of the Blow-up lemma and Lemma 3.4.4 (the conditions are certainly satisﬁed by X∗ and Y ∗), we can ﬁnd a HIST T 0 = T on X∗ Y ∗ 0 0 1 ∼ pair 0 ∪ 0 such that there exist y′ S(T 0) Γ(x∗,Y ), y′′ S(T 0) Γ(x∗∗,Y ), x′ S(T 0) Γ(y∗ ,Y ), 0 ∈ 1 ∩ 0 0 0 ∈ 1 ∩ 0 0 t+1 ∈ 1 ∩ t+1 0 x′′ S(T 0) Γ(y∗∗ ,Y ), and x′ S(T 0) Γ(y∗,X ), x′′ S(T i) Γ(y∗∗,X ) such that t+1 ∈ 1 ∩ t+1 0 0 ∈ 1 ∩ 0 0 0 ∈ 1 ∩ 0 0 0 ∗ ′ ′ 0 ∗ ′ ′ L(T ) X + L(TX′ ) X + L(TY ′ ) X = L(T ) Y + L(TX′ ) Y + L(TY ′ ) Y +2. | 1 ∩ 0 | | 0 ∩ 0| | 0 ∩ 0| | 1 ∩ 0 | | 0 ∩ 0 | | 0 ∩ 0 | 0 0 ∗ ′ ∗∗ ′′ ∗ ′ ∗∗ ′′ ∗ ′ ∗∗ ′′ Let T = T TX′ TY ′ xaya,ybxb x y , x y ,y x ,y x ,y x ,y x . It is 1 ∪ 0 ∪ 0 ∪{ }∪{ 0 0 0 0 0 0 0 0 t+1 t+1 t+1 t+1} clear that T 0 is a HIST on X′ Y ′ I(X′ ) I(Y ′) such that 0 ∪ 0 ∪ 0 ∪ 0

x∗, x∗∗,y∗,y∗∗,y∗ ,y∗∗ L(T 0) and L(T 0) X′ = L(T i) Y ′ +2. { 0 0 0 0 t+1 t+1} ⊆ | ∩ 0| | ∩ 0 |

For each ready triple (X∗,Y ∗, F ∗), we know that (X∗,Y ∗) is (4ε,d 3.1ε 9d2/20)- i i i i − − super-regular and (Y ∗, F ∗) is (8.4ε,d 2.1ε 3d2/10 2d3)-super-regular by Fact 3.4.8. i − − − Notice that (1 4ε 9d2/20)N X∗ , Y ∗ N and (1/2 2.1ε 3d2/30 d3)N F ∗ − − ≤| i | | i | ≤ − − − ≤| | ≤ (1/2 d3)N. Let I(X′) = l′ and I(F ′) = l/6 for some integer l. By Operation II we have − | i | | | ′ ′ ′ ′ ′ V (TX′ ) X 3l and V (TX′ ) Y 2l +1. By Operation III we have V (TF ′ ) F = l/2 | i ∩ i| ≤ | i ∩ i | ≤ | ∩ i | 72

′ ′ ′ and V (TF ′ ) Y = l. Notice that L(TX′ ) X = l(TX′ ) Y . Hence, | ∩ i | | i ∩ i| | i ∩ i |

Y ∗ 2( X∗ F ∗ Y ∗ l) 3( Y ′ 2l′ l 1) 2( X′ 3l′) 2( F ′ l/2)+2l | i | − | i ∪ |−| i | − ≥ | i | − − − − | i| − − | | − = 3 Y ′ 2 X′ 2 F ′ 3 | i | − | i| − | | − 3(1 3.1ε)N 2N N +2d3N 3 > 3d3N/2. ≥ − − − −

By the weak version of the Blow-up lemma (Lemma 3.2.2) and Lemma 3.4.5, we then can ﬁnd a HIT T i = T on X∗ Y ∗ F ∗ and a path P = P spanning on L(T i) F ∗ and 1 ∼ triple i ∪ i ∪ i ∼ triple 1 ∩ i ∗ ∗ ′ other L(T ) F vertices from Y . Let ya S(TX′ ) Y be a non-leaf of TX′ (take ya as | 1 ∩ | i ∈ i ∩ i i ′ ′ ′ ′ the degree 2 vertex if TX′ has one) and y S(TF ′ ) Y a non-leaf of TF ′ . Then as (Y , F ) i a ∈ ∩ i i is (4.1ε,d 2.1ε 2d3)-super-regular, we have Γ(y , F ′) , Γ(y′ , F ′) (d 2.1ε 2d3)N/2. − − | a | | a | ≥ − − Since F ′ F ∗ 3d2N/20, we then know that Γ(y , F ∗) , Γ(y′ , F ∗) (d 2.1ε 3d2/10 | − | ≤ | a | | a | ≥ − − − 2d3)N/2. Since F ∗ L(T i) = V (P ) F ∗ 5d2N/6 < (d 2.1ε 3d2/10 2d3)N/2, | ∩ 1 | | i ∩ | ≤ − − − there exist f (S(T i) F ∗) Γ(y , F ∗) and f ′ (S(T i) F ∗) Γ(y′ , F ∗). For each a ∈ 1 ∩ ∩ a a ∈ 1 ∩ ∩ a x I(F ′), since deg(x, F ′) (d ε) F ′ (d ε)(1 2.1ε d3)N/2, we know there exists ∈ ≥ − | | ≥ − − − f ′ (S(T i) F ∗) Γ(x, F ∗). From Step 2, we have Γ(x∗,Y ) Γ(x∗∗,Y ) (d ε)2N and ∈ 1 ∩ ∩ | i i ∩ i i | ≥ − Γ(y∗,X ) Γ(y∗∗,X ) (d ε)2N. Hence Γ(x∗,Y ′) Γ(x∗∗,Y ′) ((d ε)2 3.1ε)N. | i i ∩ i i | ≥ − | i i ∩ i i | ≥ − − i ′ 2 ′ ∗ ∗∗ Since S(T TX′ TF ′ ) X

Let Hi = T i P . We call P the accompany path of T i. ∪ i i Step 6. Apply the Blow-up Lemma again on the regular-pairs induced on the leaves of each HIT obtained in Step 5 to ﬁnd two vertex-disjoint paths covering all the leaves. Then connect all the HITs into a HIST of G and connect the disjoint paths into a cycle using the edges initiated in Step 2. 73

Suppose 1 i t. For each H-pair (X ,Y ), let XL = X′ L(T i) x∗, x∗∗ and Y L = ≤ ≤ i i i i ∩ −{ i i } i Y ′ L(T i) y∗,y∗∗ , and for each H-triple (X ,Y , F ), let XL = X′ L(T i P ) x∗, x∗∗ and i ∩ −{ i i } i i i i ∩ ∪ i −{ i i } Y L = Y ′ L(T i P ) y∗,y∗∗ , where T i is the HIST found in Step 5, and P is the accompany i i ∩ ∪ i −{ i i } i path of T i. By Operations I, II and III, and the proofs of the Lemmas 3.4.4 and 3.4.5, we have I(X′) I(Y ′) S(T ) and F ′ I(F ′) S(T P ). Thus, XL Y L = L(T i) x∗, x∗∗,y∗,y∗∗ i ∪ i ⊆ i ∪ ⊆ i ∪ i i ∪ i −{ i i i i } for each H-pair and XL Y L = L(T i P ) x∗, x∗∗,y∗,y∗∗ for each H-triple. Furthermore, i ∪ i ∪ i −{ i i i i } we have XL = Y L . For the H-pair (X ,Y ), let XL = X′ L(T 0) x∗, x∗∗,y∗ ,y∗∗ | i | | i | 0 0 0 0 ∩ −{ 0 0 t+1 t+1} and Y L = Y ′ L(T 0) y∗,y∗∗ . We have XL Y L = L(T 0) x∗, x∗∗,y∗ ,y∗∗ and 0 0 ∩ −{ 0 0 } 0 ∪ 0 −{ 0 0 t+1 t+1} XL = Y L since from Step 5 we have L(T 0) X′ = L(T 0) Y ′ + 2. By the construction | 0 | | 0 | | ∩ 0| | ∩ 0 | of T and H , we see that S(T ) X′ , S(T ) Y ′ d2N. Since each H-pair (X′,Y ′) is pair triple | i ∩ i| | i ∩ i | ≤ i i (2ε,d 2ε)-super-regular, and each pair (X′,Y ′) from an H-triple (X′,Y ′, F ′) is (2ε,d 3.1ε)- − i i i i − super-regular, by Slicing Lemma, we then know that (XL,Y L) is (4ε,d 3.1ε d2)-super- i i − − regular and hence is (4ε,d/2)-super-regular. For each 1 i t, by the choice of x∗, x∗∗,y∗,y∗∗, we have Γ(x∗,Y ) , Γ(x∗∗,Y ) ≤ ≤ i i i i | i i | | i i | ≥ (d ε)N and Γ(y∗,X ) , Γ(y∗∗,X ) (d ε)N. Hence, Γ(x∗,Y L) , Γ(x∗∗,Y L) (d ε − | i i | | i i | ≥ − | i i | | i i | ≥ − − 3.1ε d2)N > dN/2 and Γ(y∗,XL) , Γ(y∗∗,XL) (d ε 3.1ε d2)N > dN/2. Similar − | i i | | i i | ≥ − − − results hold for the vertices x∗, x∗∗,y∗ ,y∗∗ . For each 0 i t, we choose distinct vertices 0 0 t+1 t+1 ≤ ≤ y′ Γ(x∗,Y L), y′′ Γ(x∗∗,Y L) and x′ Γ(y∗,XL), x′′ Γ(y∗∗,XL). If T i does not have i ∈ i i i ∈ i i i ∈ i i i ∈ i i the accompany path, then by the strengthened version of the Blow-up lemma, we can find an (x′ ,y′)-path P i and an (x′′,y′′)-path P i such that P i P i is spanning on XL Y L. If T i i i 1 i i 2 1 ∪ 2 i ∪ i has the accompany (b, f)-path P , we see that deg(b, XL),deg(f,Y L) dN/2 as (X′,Y ′) is i i i ≥ i i (2ε,d 3.1ε)- super-regular, and (Y ′, F ′) is (4.2ε,d 3.1ε 2d3)-super-regular. Applying the − i − − ′ ′ i ′′ ′′ strengthened version of the Blow-up lemma, we can find an (xi,yi)-path P11 and an (xi ,yi )- path P i such that P i P i is spanning on XL Y L, and two consecutive internal vertices 2 11 ∪ 2 i ∪ i a′, b′ V (P i ) with b′ Γ(f,Y L), and a′ Γ(b, XL). Let P i = P i P fb′, ba′ a′b′ . ∈ 11 ∈ i ∈ i 1 11 ∪ i ∪{ }−{ } ∗ ∗∗ Notice that for the H-pair (X0,Y0), the two vertices yt+1,yt+1 are not used in this step, but ∗ ∗∗ we will connect them to y0 and y0 , respectively, in next step. We now connect the small HITs and paths together to find an SGHG of G. In Case A, 74

for 1 i t 1, we have S(T i) Y d3N/2 > εN and S(T i+1) X d3N/2 > εN. ≤ ≤ − | ∩ i| ≥ | ∩ i+1| ≥ Since (Yi,Xi+1) is an ε-regular pair with density d, we see that there is an edge ei connecting S(T i+1) X and S(T i+1) X . Let ∩ i+1 ∩ i+1

t T = T i e 1 i t 1 . ∪{ i | ≤ ≤ − } i=1 [

Then T is a HIST of G. Let C be the cycle formed by all the paths in t (P i P i) and all i=1 1 ∪ 2 edges in the following set S

′′ ′′ x∗y′, x∗∗y ,y∗x′ ,y∗∗x :1 i t y∗x∗ ,y∗∗x∗∗ :1 i t 1 y∗x∗∗,y∗∗x∗ , { i i i i i i i i ≤ ≤ }∪{ i i+1 i i+1 ≤ ≤ − }∪{ t 1 t 1} notices that the edges in y∗x∗ ,y∗∗x∗∗ :1 i t 1 y∗x∗∗,y∗∗x∗ above are guaranteed { i i+1 i i+1 ≤ ≤ − }∪{ t 1 t 1} in Step 2. It is easy to see that C is a cycle on L(T ). Hence H = T C is an SGHG of G. ∪ In Case B, for 1 i t 1, we have S(T i) Y d3N/2 > εN and S(T i+1) X ≤ ≤ − | ∩ i| ≥ | ∩ i+1| ≥ 3 d N/2 > εN. Since (Yi,Xi+1) is an ε-regular pair with density d, we see that there is an edge e connecting S(T i+1) X and S(T i+1) X . Similarly, there is an edge e connecting i ∩ i+1 ∩ i+1 0 S(T ) X and S(T 1) X . Let 0 ∩ 0 ∩ 1

t T = T i e 0 i t 1 . ∪{ i | ≤ ≤ − } i=1 [

Then T is a HIST of G. Let C be the cycle formed by all paths in t (P i P i) and all i=1 1 ∪ 2 edges in the set y∗y∗ ,y∗∗y∗∗ ,y∗ x∗,y∗∗ x∗∗, x∗y∗∗, x∗∗y∗ and in theS following set { 0 t+1 0 t+1 t+1 1 t+1 1 0 t 0 t }

′′ ′′ x∗y′, x∗∗y ,y∗x′ ,y∗∗x :0 i t y∗x∗ ,y∗∗x∗∗ :1 i t 1 . { i i i i i i i i ≤ ≤ }∪{ i i+1 i i+1 ≤ ≤ − }

It is easy to see that C is a cycle on L(T ). Hence H = T C is an SGHG of G. ∪ The proof of Theorem 3.4.3 is now ﬁnished. 75

3.4.3.2 Proof of Theorem 3.4.4 By the assumption that deg(v ,V ) 2βn for 1 2 ≤ each v V and the assumption that δ(G) (2n + 3)/5 in Extremal Case 1, we see that 1 ∈ 1 ≥

δ(G[V ]) (2n + 3)/5 2βn. (3.13) 1 ≥ −

Then (3.13) implies that

V (2n + 3)/5 2βn and V 3n/5+2βn. (3.14) | 1| ≥ − | 2| ≤

Also, by V (2/5 4β)n in the assumption, | 2| ≥ −

V (3/5+4β)n. (3.15) | 1| ≤

We will construct an SGHG of G following several steps below. Step 1. Repartitioning

1/3 2/3 Set α1 = α and α2 = α . Let

V ′ = V and V ′ = v V deg(v,V ) α V . 1 1 2 { ∈ 2 | 1 ≤ 1| 1|}

Then by d(V ,V ) α, we have 1 2 ≤

α V V V ′ e(V ,V ′)+ e(V ,V V ′)= e(V ,V ) α V V . 1| 1|| 2 − 2 | ≤ 1 2 1 2 − 2 1 2 ≤ | 1|| 2|

This gives that V V ′ α V . (3.16) | 2 − 2 | ≤ 2| 2|

Denote V 0 = V V ′. Then by the deﬁnition of V ′, we have 12 2 − 2 2

δ(V 0 ,V ′) >α V ′ and δ(G[V ′]) (2n +3)/5 α V ′ (2/5 α (3/5+4β))n, (3.17) 12 1 1| 1 | 2 ≥ − 1| 1 | ≥ − 1 76

where the last inequality follows from (3.15). Let n = V ′ for i =1, 2. Then by (3.13) and (3.15), i | i |

2/5 2β δ(G[V ′]) (2n + 3)/5 2βn − n (2/3 8β)n , (3.18) 1 ≥ − ≥ 3/5+4β 1 ≥ − 1 and by (3.14) and the second inequality in (3.17),

(2/5 α (3/5+4β)) δ(G[V ′]) (2/5 α (3/5+4β))n − 1 n (2/3 1.1α )n , 2 ≥ − 1 ≥ 3/5+2β 2 ≥ − 1 2

provided that β 0.3α1 . ≤ 9α1+20/3 Step 2. Finding three connecting edges

1 1 2 2 AS G is 3-connected, there are 3 independent edges xLyL, xLyL and xN yN connecting V ′ V 0 and V ′ such that x1 , x2 , x V ′ V 0 and y1 ,y2 ,y V ′. In the remaining steps, 1 ∪ 12 2 L L N ∈ 1 ∪ 12 L L N ∈ 2 we will ﬁnd a HIST T in G[V ′ V 0 ] with x as a non-leaf and x1 , x2 as leaves, and a HIST 1 1 ∪ 12 N L L T of G[V ′] with y as a non-leaf and y1 ,y2 as leaves. Then T = T T x y is a HIST 2 2 N L L 1 ∪ 2 ∪{ N N } 1 2 1 2 of G. By ﬁnding a hamiltonian (xL, xL)-path P1 on L(T1), and a hamiltonian (yL,yL)-path

on L(T2), we see that C := P P x1 y1 , x2 y2 1 ∪ 2 ∪{ L L L L}

forms a cycle on L(T ). Hence H := T C is an SGHG of G. ∪ Step 3. Initiating two HITs In this step, we ﬁrst initiate a HIT in G[V ′ V 0 ] containing X as a non-leaf and x1 1 ∪ 12 N L 2 ′ 1 and xL as leaves. Then, we initiate a HIT in G[V2 ] containing yN as a non-leaf and yL and 2 yL as leaves. For x1 , x2 , x V ′ V 0 , by (3.13) and (3.17), each of them has at least α V ′ 9 L L N ∈ 1 ∪ 12 1| 1 | ≥ neighbors in V ′. Thus, we choose distinct z1 , z1, z2 , z2, z1 , z2 , z3 V ′ such that 1 L L N N N ∈ 1

x1 z1 , z1, x2 z2 , z2, x z1 , z2 , z3 . L ∼ L L ∼ L N ∼ N N N

1 2 0 (Note that xL and xL may be from V12, and therefore they may not have too many neighbors 77

′ 1 2 ′ in V1 , we then choose zL and zL from V1 as their neighbors, respectively.) By (3.18), we see that any two vertices in G[V ′] have at least (1/3 16β)n 14 1 − 1 ≥ neighbors in common. Thus, we can choose distinct vertices z11, z22, z12, vR V ′ 1 ∈ 1 − x1 , x2 , x , z1 , z1, z2 , z1 , z2 , z3 such that { L L N L L N N N }

z11 z1 , z1, z22 z2 , z2, z12 z11, z22, vR z12, z1 . ∼ L ∼ L ∼ 1 ∼ N

Furthermore, by (3.18) again, we have δ(G[V ′]) (2/3 8β)n 17. Choose z1, z2, z11 V ′ 1 ≥ − 1 ≥ 1 2 N ∈ 1 not chosen above such that z1 z1, z2 z2, z11 z1 . 1 ∼ 2 ∼ N ∼ N

Let T11 be the graph with

V (T )= x1 , x2 , x , z1 , z1 , z1, z2 , z11, z12, z22, z2, z2 , z3 , vR, z1, z2, z11 11 { L L N N L L N N 1 1 2 N }

1 1 2 2 and with edges indicated above except the edges xLzL and xLzL. We see that T11 is a tree with vR as the only degree 2 vertex, and V (T ) = 17 and L(T ) = 9. Notice that in T , 1 | 11 | | 11 | 11 1 1 2 2 zL, xL and zL, xL are leaves, and xN is a non-leaf. Figure 3.6 gives a depiction of T11.

1 1 2 2 xL zL xL zL 11 1 zN z1 2 z2 2 3 1 2 zN zN z z 1 zN 22 z11 z xN z12 R v1 Figure (3.6) The tree T11

1 1 2 2 Notice that the edges xLzL and xLzL are not used in T11. We will first construct a HIST T in G[V 1 V 0 ] containing T as a subgraph, then find a hamiltonian (z1 , z2 )-path on 1 1 ∪ 12 11 L L L(T ) x1 , x2 by Lemma 3.2.6, finally by adding x1 z1 and x2 z2 to the path, we get a 1 −{ L L} L L L L 78

1 2 1 2 hamiltonian (xL, xL)-path on L(T1). The reason that we avoid using xL and xL is that when x1 , x2 V 0 , we may not be able to have the condition of Lemma 3.2.6 on G[L(T )] in our L L ∈ 12 1 ﬁnal construction.

′ 1 2 Then we initiate a HIT in G[V2 ] containing yL,yL as leaves, and yN as a non-leaf. As y1 ,y2 ,y V ′, by (3.19) and the fact that each two vertices from V ′ have at least L L N ∈ 2 2 (1/3 2.2α )n 7 common neighbors implied from (3.19), we can choose distinct vertices − 1 2 ≥

y12,y1 ,y2 ,y3 , vR V ′ y1 ,y2 ,y N N N 2 ∈ 2 −{ L L N } such that y12 y1 ,y2 , y y1 ,y2 ,y3 , vR y12,y . (3.19) ∼ L L N ∼ N N N 2 ∼ N

Let T be the graph with V (T ) = y1 ,y2 ,y ,y12,y1 ,y2 ,y3 , vR and with E(T ) de- 21 21 { L L N N N N 2 } 21 scribed as in (3.19). We see that T is a tree with vR the only degree 2 vertex and y1 ,y2 L(T ), y 21 2 L L ∈ 21 N ∈ S(T21) and V (T ) V ′ =8, L(T ) V ′ =5. (3.20) | 21 ∩ 2 | | 21 ∩ 2 |

Denote U = V ′ V (T ), U = V ′ V (T ), and V = V 0 V (T ). 1 1 − 11 2 2 − 21 12 12 − 11

0 Step 4. Absorbing vertices in V12 We may assume that V 0 = . For otherwise, we skip this step. Let V = n and 12 6 ∅ | 12| 12 V 0 = x , x , , x . 12 { 1 2 ··· n12 } Since V (T ) = 17, by (3.17), we get | 11 |

δ(V 0 , U ) >α V ′ 17 3α V 3 V V ′ 3 V 0 . 12 1 1| 1 | − ≥ 2| 2| ≥ | 2 − 2 | ≥ | 12|

Thus, there is a claw-matching M from V 0 to U centered in V 0 . For i =1, 2, , n , let c 12 1 12 ··· 12 xi1, xi2 and xi3 be the three neighbors of xi in Mc. If n12 = 1, let Ta = Mc, and we ﬁnish 79

this step. Thus we assume n 2. 12 ≥ ′ By (3.18), each two vertices in in V1 have at least

(1/3 16β)n 6α V 0 + 17 (3.21) − 1 ≥ 2| 12| neighbors in common. The above inequality holds as n 2n/5 2βn, V 3n/5+2βn 1 ≥ − | 2| ≤ by (3.14), and we can assume that 18α /5+106β/15+12α β + 18/n 32β2 2/15. 2 2 − ≤ Thus, for each i = 1, 2, , n 1, we can ﬁnd distinct vertices xi , xi , x3 , x3 in ··· 12 − 13 23 i3 i+1,1 U V (M ) such that 1 − c

xi x , x , xi xi , x3 x , x3 x . (3.22) 13 ∼ i3 i+1,1 23 ∼ 13 i3 ∼ i3 i+1,1 ∼ i+1,1

Let T be the graph with V (T ) = V (M ) xi , xi , x3 , x3 : 1 i n 1 , and a a c ∪{ 13 23 i3 i+1,1 ≤ ≤ 12 − } E(Ta) including all edges indicated in (3.22) for all i and all edges in Mc. It is easy to see,

by the construction, that Ta is a HIT with

V (T ) U =7n 4 and L(T ) U =4n 1. | a ∩ 1| 12 − | a ∩ 1| 12 −

Using (3.21) again, we can ﬁnd x11 U V (T ) such that x11 vR, x , where N ∈ 1 − a N ∼ 1 11 vR V (T ) and x V (T ). By (3.18), 1 ∈ 11 11 ∈ a

δ[G[V ′]] (2n + 3)/5 2βn 6α V 0 + 20, 1 ≥ − ≥ 2| 12|

since V 2n/5+2βn, and we can assume that 2β 12α β 18α /5 21/n 2/5. So | 2| ≤ − 2 − 2 − ≤ we can ﬁnd distinct vertices x12, x1 U V (T ) x11 such that x12 x11, x1 x . N 11 ∈ 1 − a −{ N } N ∼ N 11 ∼ 11 1 Let T1 be the graph with

V (T 1)= V (T ) V (T ) x11, x12, x1 and E(T 1)= E(T ) E(T ) x11v1 , x11x , x12x11, x1 x . 1 11 ∪ a ∪{ N N 11} 1 11 ∪ a ∪{ N R N 11 N N 11 11} 80

1 Then T1 is a HIT such that

V (T 1) U =7n + 16 and S(T 1) U =3n +7. (3.23) | 1 ∩ 1| 12 | 1 ∩ 1| 12

Denote U ′ = U V (T 2) and U ′ = U V (T 2). 1 1 − 1 2 2 − 1 Step 5. Completion of HITs T1 and T2

′ i In this step, we construct a HIST Ti in G[Vi ](i = 1, 2) containing T1 as an induced subgraph. The following lemma guarantees the existence of a speciﬁed HIST in a graph with n vertices and minimum degree at least (2/3 α′)n for some 0 <α′ 1. − ≪ Lemma 3.4.6. Let H be an n-vertex graph with δ(H) (2/3 α′)n for some constant ≥ − 0 <α′ 1. Then H has a HIST T satisfying ≪ H (i) T has a vertex v of degree at least (2/3 α′)n 1, and v can be chosen arbitrarily H R − − R from V (H);

(ii) S(T ) (1/6+ α′/2)n +2. | H | ≤

Proof. Let v V (H) be an arbitrary vertex. If n(mod 2) deg(v )+1(mod 2), R ∈ ≡ R then we let N = N (v ). For otherwise, let N be a subset of N(v ) with N (v ) 1 R H R R R | H R | − elements. Let T be the star with V (T ) = v N and E(T ) = E( v , N ). Let vR vR { R} ∪ R R { R} R V = V (H) V (T ). By δ(H) (2/3 α′)n, we have V (1/3+ α′)n + 1. By the choice 0 − vR ≥ − | 0| ≤ of N , we have V 0(mod 2). If V = , then let T = T . For otherwise, we claim as R | 0| ≡ 0 ∅ H vR follows.

Claim 3.4.3. Let V V (H) be a subset with V (2/3 α′)n 1 and V (mod 2) 1 ⊆ | 1| ≥ − − | 1| ≡ n(mod 2). Then there exist two vertices from V = V (H) V such that they have a common 0 − 1 neighbor in V1.

Proof of Claim 3.4.3. We assume that V (2/3+2α′)n. For otherwise, V < (1/3 2α′)n. | 1| ≤ | 0| − Since δ(H) (2/3 α′)n, any two vertices of H have at least (1/3 2α′)n neighbors in ≥ − − common. By V < (1/3 2α′)n, any two vertices from V have a common neighbor from | 0| − 0 81

V . We are done. Thus V (2/3+2α′)n, and hence V (1/3 2α′)n 3. By the 1 | 1| ≤ | 0| ≥ − ≥ assumption that V (2/3 α′)n 1, we have V (1/3+ α′)n + 1. This implies that | 1| ≥ − − | 0| ≤ deg(v ,V ) (1/3 2α′)n 2 for each v V . As V 3 and 3((1/3 2α′)n 2) > 0 1 ≥ − − 0 ∈ 0 | 0| ≥ − − (2/3+2α′)n> V (provided that 8α′ +6/n < 1/3), we see that there must be two vertices ≥| 1| from V0 such that they have a neighbor in common in V1. By Claim 3.4.3, there exist two vertices v11, v12 V such that they have a common 0 0 ∈ 0 11 12 neighbor in TvR . Adding v0 and v0 to TvR and two edges connecting them to one of their 1 1 common neighbor in V (TvR ). Let TvR be the resulting graph. Then we see that TvR is 1 a HIT with V (T ) = V (Tv ) + 2, and hence ( V (Tv ) + 2)(mod 2) n(mod 2). Also | vR | | R | | R | ≡ 1 ′ V (T ) V (Tv ) (2/3 α )n 1. So we can use Claim 3.4.3 again to ﬁnd another | vR |≥| R | ≥ − − pair of vertices from V v11, v12 such that they have a common neighbor in V (T 1 ). 0 −{ 0 0 } vR Adding the new pair of vertices and two edges connecting them to one of their common

1 1 2 neighbor in V (TvR ) into TvR , we get a new HIT TvR . By repeating the above process another

l0 l0 l0 =( V0 4)/2 times, we get a HIT T . Let TH = T . We claim that TH has the required | | − vR vR properties in Lemma 3.4.6. Notice ﬁrst that d (v ) (2/3 α′)n 1. Then since T has TH R ≥ − − H v and at most V /2 distinct vertices as non-leaves and V (1/3+ α′)n + 1, we see that R | 0| | 0| ≤ S(T ) (1/6+ α′/2)n +2. | H | ≤ Let H = G[U ′ v1 ]. Recall that v1 is a non-leaf in T 1. By (3.18) and (3.23), and 1 1 ∪{ R} R 1 by noticing that n V V ′ α V 3α n /2 (by (3.14)), we see that 12 ≤| 2 − 2 | ≤ 2| 2| ≤ 2 1

δ(H ) (2/3 8β)n (7n + 19) 1 ≥ − 1 − 12 (2/3 8β)n 21α n /2 19 ≥ − 1 − 2 1 − (2/3 11α ) V (H ) . (3.24) ≥ − 2 | 1 |

Let α′ = 11α 1 (by assuming α 1). By Lemma 3.4.6, we can ﬁnd a HIT T ′ in 2 ≪ ≪ 1 H with v1 as the prescribed vertex in condition(i). It is easy to see that T := T 1 T ′ is a 1 R 1 1 ∪ 1 82

HIST of G[V ′ V 0 ] and 1 ∪ 12

s = S(T ) V ′ = S(T 1) V ′ + S(T ′) V ′ 1 | 1 ∩ 1 | | 1 ∩ 1 | | 1 ∩ 1 | 3n +7+(1/6+5.5α ) V (H ) + 2(by(3.23) and Lemma 3.4.6) ≤ 12 2 | 1 | 3n +9+(1/6+5.5α )n ≤ 12 2 1 (1/6+10.5α )n (by n 3α n /2). (3.25) ≤ 2 1 12 ≤ 2 1

Let H = G[U ′ v2 ]. By(3.19) and (3.20), we see that 2 2 ∪{ R}

δ(H ) (2/3 1.1α )n 8 (2/3 1.2α ) V (H ) . 2 ≥ − 1 2 − ≥ − 1 | 2 |

′ ′ 2 By letting α = 1.2α1, we can ﬁnd a HIT T2 in H2 with vR as the prescribed vertex in condition (i) of Lemma 3.4.6. Then T := T 2 T ′ is a HIST of G[V ′]. Also, notice that 2 1 ∪ 2 2

s = S(T ) V ′ = S(T 2) V ′ + S(T ′) V ′ 2 | 2 ∩ 2 | | 1 ∩ 2 | | 2 ∩ 2 | 3+(1/6+0.6α ) V (H ) +2 ≤ 1 | 2 | (1/6+0.7α )n , (3.26) ≤ 2 2 where the last inequality holds by assuming 5/n 0.1α . 2 ≤ 2 Step 6. Finding two long paths In this step, we first find a hamiltonian (z1 , z2)-path P 1 in G[L(T ) x1 , x2 ]; then L 2 1 1 −{ L L} find a hamiltonian (y1 ,y2 )-path P in G[L(T )]. Let G = G[L(T ) x1 , x2 ] and n = L L 2 2 11 1 −{ L L} 11 V (G ) . We will show that δ(G ) > 1 n . We may assume s (1/6 8β)n 2. For | 11 | 11 2 11 1 ≥ − 1 − otherwise, if s < (1/6 8β)n 2, then by (3.18), we get 1 − 1 −

δ(G ) δ(G[V ′]) s 2 11 ≥ 1 − 1 − (2/3 8β)n ((1/6 8β)n 1 2) 2 ≥ − 1 − − 1 − − − 1 1 n +1 n +1. ≥ 2 1 ≥ 2 11 83

Hence, s (1/6 8β)n 2, implying that 1 ≥ − 1 −

n 2 n (5/6+8β)n + 2 and thus n 11 − . (3.27) 11 ≤ 1 1 ≥ 5/6+8β

Hence, by (3.25)

δ(G ) δ(G[V ′]) s 2 (2/3 8β)n (1/6+10.5α )n 2 11 ≥ 1 − 1 − ≥ − 1 − 2 1 − 1/2 2β 11α (1/2 8β 11α )n − − 2 (n 2) > n /2, ≥ − − 2 1 ≥ 5/6+2β 11 − 11

the last inequality holds by assuming 3β + 11α2 +2/n11 < 1/12. By applying Lemma 3.4.6 on G , we ﬁnd a hamiltonian (z1 , z2 )-path P 1 in G . Let P = P 1 z1 x1 , z2 x2 . We see 11 L L 1 11 1 1 ∪{ L L L L} 1 2 that P1 is a a hamiltonian (xL, xL)-path on L(T1). Let G = G[L(T )] and n = V (G ) . We will show that δ(G ) > n /2. We may 22 2 22 | 22 | 22 22 assume that s (1/6 1.1α )n 2. For otherwise, if s < (1/6 1.1α )n 2, then by 2 ≥ − 1 2 − 2 − 1 2 − (3.19), we see that

δ(G ) δ(G[V ′]) s 2 22 ≥ 2 − 2 − > (2/3 1.1α )n ((1/6 1.1α )n 2) 2 − 1 2 − − 1 2 − − > n /2 n /2. 2 ≥ 22

Thus, s (1/6 1.1α )n 2, implying that 2 ≥ − 1 2 −

n22 2 n22 n1 s2 (5/6+1.1α1)n2 +2 givesthat n2 − . ≤ − ≤ ≥ 5/6+1.1α1 84

By (3.19) and (3.26),

δ(G ) δ(G[V ′]) s 2 22 ≥ 2 − 2 − (2/3 1.1α )n (1/6+0.7α )n 2 ≥ − 1 2 − 1 2 − 1/2 1.9α2 (1/2 1.9α1)n2 − (n22 2) ≥ − ≥ 5/6+1.1α2 −

> n22/2.

The last inequality follows by assuming that 2.45α1 +2/n11 < 1/12. Hence, by Lemma 3.4.6,

1 2 there is a hamiltonian (yL,yL)-path P2 in G22. Step 7. Forming an SGHG Let T = T T x y and C = P P x1 y1 , x2 y2 . We see that T is a HIST 1 ∪ 2 ∪{ N N } 1 ∪ 2 ∪{ L L L L} of G with L(T ) = V (P ) V (P ) and C is a cycle spanning on L(T ). Hence H = T C is 1 ∪ 2 ∪ an SGHG of G.

3.4.3.3 Proof of Theorem 3.4.5 Notice that the assumption of Extremal Case 2 implies that V > (3/5 α)n and V (2/5 2β)n. | 1| − | 2| ≥ −

We may assume that the graph G is minimal with respective to the number of edges. This implies that no two adjacent vertices both have degree larger than (2n + 3)/5. (For otherwise, we could delete any edges incident to two vertices both with degree larger than (2n + 3)/5.) We construct an SGHG in G step by step. Step 1. Repartitioning

1/3 2/3 Set α1 = α and α2 = α . Let

V ′ = v V deg(v,V ) (1 3α ) V , 2 { ∈ 2 | 1 ≥ − 1 | 1|} V ′ = v V V ′ deg(v,V ) α V /6 , 0 { ∈ 2 − 2 | 1 ≤ 1| 2| } V ′ = V V ′, V 0 = V V ′ V ′. 1 1 ∪ 0 12 2 − 2 − 0 85

As d(V ,V ) 1 3α, the following holds, 1 2 ≥ −

(1 3α) V V e (V ,V )= e (V ,V ′)+ e (V ,V V ′) − | 1|| 2| ≤ G 1 2 G 1 2 G 1 2 − 2 V V ′ + (1 3α ) V V V ′ . ≤ | 1|| 2 | − 1 | 1|| 2 − 2 |

The inequality implies that

V V ′ α V . (3.28) | 2 − 2 | ≤ 2| 2|

As a consequence of moving vertices in V V ′ out from V , by (3.28) we get 2 − 2 2

δ(V ,V ′) (2n + 3)/5 2βn α V 1 2 ≥ − − 2| 2| (2n + 3)/5 6β V α V ≥ − | 2| − 2| 2| (2n + 3)/5 2α V , (3.29) ≥ − 2| 2| provided that 6β α . And as a consequence of moving vertices in V ′ to V , ≤ 2 0 1

δ(V ′,V ′) δ(G) ∆(V ′,V ) ∆(V ′,V V ′) 0 2 ≥ − 0 1 − 0 2 − 2 (2n + 3)/5 α V /6 α V ≥ − 1| 2| − 2| 2| (2n + 3)/5 α V /3 (provided that α α /6), (3.30) ≥ − 1| 2| 2 ≤ 1 and α V /6 < δ(V 0 ,V ′) < (1 3α ) V . (3.31) 1| 2| 12 1 − 1 | 1|

From (3.29) and (3.30), we have

δ(V ′,V ′) (2n + 3)/5 α V /3. (3.32) 1 2 ≥ − 1| 2|

As δ(V ′,V ′) (1 3α ) V (1 3α )(3/5 α)n> (2n + 3)/5 , (3.33) 2 1 ≥ − 1 | 1| ≥ − 1 − ⌈ ⌉ 86 we get that deg(v′ )= (2n + 3)/5 (3.34) 1 ⌈ ⌉ for each v′ V ′, by the minimality assumption of e(G). Hence (3.32) and (3.34) give that 1 ∈ 1

∆(G[V ′]) α V /3. (3.35) 1 ≤ 1| 2|

∗ ′ ′ Step 2. Finding a vertex v2 from V2 with large degree in V1 Let

′ ein = e(G[V1 ]) (3.36)

′ be the number of edges within V1 , notice that ein maybe 0. Then

e (V ′,V ′ V 0 )= V ′ (2n + 3)/5 2e . (3.37) G 1 2 ∪ 12 | 1 |⌈ ⌉ − in

Let d = e / V ′ and n = V ′ V 0 (2n + 3)/5 . (3.38) in in | 1 | | 0| || 2 ∪ 12| − ⌈ ⌉|

By (3.35) and the deﬁnition of din in (3.38), we have

d α V /6. ⌊ in⌋ ≤ 1| 2|

In fact, since ∆(V ,V ′) ∆(V ,V )+∆(V ,V ′) 2βn+ V ′ 2βn+α V , and ∆(V ′,V ′) 1 1 ≤ 1 1 1 0 ≤ | 0 | ≤ 2| 2| 0 1 ≤ α V /6+ α V , more precisely, we have 1| 2| 2| 2|

2d = 2e / V ′ (2βn + α V ) V / V ′ +(α V /6+ α V ) V ′ / V ′ in in | 1 | ≤ 2| 2| | 1| | 1 | 1| 2| 2| 2| | 0 | | 1 | (2βn + α V )+ α (α V /6+ α V ) (as V ′ α V and V , V V ′ ) ≤ 2| 2| 2 1| 2| 2| 2| | 0 | ≤ 2| 2| | 1| | 2|≤| 1 | (6β + α + α/6+ α2) V (as βn 3β V ) ≤ 2 2 | 2| ≤ | 2| 2α V (provided that 6β + α/6+ α2 α ). (3.39) ≤ 2| 2| 2 ≤ 2

Set 87

Case A. (2n + 3)/5 V ′ V 0 = n 0; ⌈ ⌉−| 2 ∪ 12| 0 ≥ Case B. V ′ V 0 (2n + 3)/5 = n 1. | 2 ∪ 12| − ⌈ ⌉ 0 ≥ We have

(2n + 3)/5 V ′ V 0 2βn + α V (6β + α ) V 2α V , Case A, ⌈ ⌉−| 2 ∪ 12| ≤ 2| 2| ≤ 2 | 2| ≤ 2| 2|

n0 =  (3.40)   V ′ V 0 (2n + 3)/5 (2/5+ α)n (2n + 3)/5 αn, Case B. | 2 ∪ 12| − ⌈ ⌉ ≤ − ⌈ ⌉ ≤   Then in case A,

e (V ′,V ′ V 0 ) = V ′ (2n + 3)/5 2e (by (3.34)) G 1 2 ∪ 12 | 1 |⌈ ⌉ − in = V ′ ( V ′ V 0 + n 2d ) | 1 | | 2 ∪ 12| 0 − in V ′ V 0 ( V ′ +1.4n 3.2d ), ≥ | 2 ∪ 12| | 1 | 0 − in as 1.4 V ′ V 0 1.4((2n+3)/5+αn) (3/5 α)n< V ′ and 1.6 V ′ V 0 1.6((2n+3)/5 | 2 ∪ 12| ≤ ≤ − | 1 | | 2 ∪ 12| ≥ − 2β α )n (3/5+2β + α )n) > V ′ provided that 2.4α < 1/25 and 5.2β +2.6α 1/25 − 2 ≥ 2 | 1 | 2 ≤ respectively. Since e (V ′,V ′ V 0 ) V ′ V 0 V ′ , we have V ′ +1.4n 3.2d V ′ , and G 1 2 ∪ 12 ≤| 2 ∪ 12|| 1 | | 1 | 0 − in ≤| 1 | thus 1.4n 3.2d . 0 ≤ in In Case B,

e (V ′,V ′ V 0 ) = V ′ (2n + 3)/5 2e (by (3.34)) G 1 2 ∪ 12 | 1 |⌈ ⌉ − in = V ′ ( V ′ V 0 n 2d ) | 1 | | 2 ∪ 12| − 0 − in V ′ V 0 ( V ′ 1.6n 3.2d ), ≥ | 2 ∪ 12| | 1 | − 0 − in as 1.6 V ′ V 0 1.6((2n + 3)/5 2β α n) (3/5+2β + α )n > V ′ provided that | 2 ∪ 12| ≥ − − 2 ≥ 2 | 1 | 5.2β +2.6α 1/25. 2 ≤ Let

3.2d 1.4n , if Case A, ⌊ in − 0⌋ dl = ( 1.6n +3.2d , if Case B. (3.41) ⌊ 0 in⌋ 88

By (3.39) and (3.40), we see that

3.2α V , if Case A, 2| 2| dl ≤ ( 6.4α V , if Case B. (3.42) 2| 2|

Then there is a vertex v∗ in V ′ V 0 of degree at least V ′ d . We will ﬁx this vertex 2 2 ∪ 12 | 1 | − l ∗ ′ in what follows. In fact, such a vertex v2 is in V2 by the facts that

δ(V 0 ,V ′) < (1 3α ) V and V ′ d (1 3α ) V , (3.43) 12 1 − 1 | 1| | 1 | − l ≥ − 1 | 1|

where V ′ d (1 3α ) V holds because of (3.42). | 1 | − l ≥ − 1 | 1| ∗ ′ Step 3. Finding a matching M within G[Γ(v2 ,V1 )] In this step, if e 1, we ﬁrst ﬁnd a matching within G[V ′] of size at least e /(2 in ≥ 1 in △ ′ (G[V1 ])). We assume this by giving the following lemma.

Lemma 3.4.7. If G is a graph with maximum degree ∆, then G contains a matching of size

|E(G)| at least 2∆ .

Proof. We use induction on V (G) . We may assume that the graph is connected. | | For otherwise, we are done by the induction hypothesis. Let e = xy E(G) be an edge and ∈ G′ = G x, y . Since N (x) N (y) x, y 2(∆ 1), we have −{ } | G ∪ G |−|{ }| ≤ −

e(G′) e(G) 2(∆ 1) 1 e(G) 2∆. ≥ − − − ≥ −

Hence, by the induction hypothesis, G′ has a matching of size at least e(G)−2∆ = e(G) 1. 2∆ 2∆ − ′ e(G) Adding e to the matching obtained in G gives a matching of size at least 2∆ in G. In case A, we take a matching in G[V ′] of size at least max 11d , 11n . This is 1 {⌊ in⌋ 0} possible because e e 3d in in = in 11d 2 (G[V ′]) ≥ 2α V ′ /3 2α ≥ in △ 1 1| 1 | 1 89

provided that α ( 3 )3, and ≤ 22

2e V ′ (2n + 3)/5 V ′ V ′ (1 3α ) V V 0 in ≥ | 1 |⌈ ⌉−| 1 || 2 | − − 1 | 1|| 12| V ′ (2n + 3)/5 V ′ ( (2n + 3)/5 n V 0 ) V V 0 +3α V V 0 ≥ | 1 |⌈ ⌉−| 1 | ⌈ ⌉ − 0 −| 12| −| 1|| 12| 1| 1|| 12| V ′ n +3α V V 0 (3.44) ≥ | 1 | 0 1| 1|| 12|

implying that e e V ′ n /2 3n in in | 1 | 0 0 11n 2 (G[V ′]) ≥ 2α V ′ /3 ≥ 2α V ′ /3 ≥ 4α ≥ 0 △ 1 1| 1 | 1| 1 | 1 provided that α ( 3 )3. ≤ 44 By (3.41), V ′ Γ(v∗,V ′) d 3.2d , we can then choose a matching M from | 1 | − 2 1 ≤ l ≤ ⌊ in⌋ ∗ ′ Γ(v2,V1 ) such that M = max 7d , 7n . (3.45) | | {⌊ in⌋ 0}

In case B, we take a matching in G[V ′] of size at least 8d . This is possible as 1 ⌊ in⌋

e e 3d in in = in 8d (G[V ′]) ≥ 2α V ′ /3 2α ≥ ⌊ in⌋ △ 1 1| 1 | 1

provided that α ( 3 )3. ≤ 16 By the second equality of (3.41), V ′ Γ(v∗,V ′) 3.2d +1.6n . If n < 2d , then | 1 | − 2 1 ≤ ⌊ in 0⌋ 0 in 3.2d +1.6n 7d . Thus, there is a matching M within Γ(v∗,V ′) such that ⌊ in 0⌋ ≤ ⌊ in⌋ 2 1

din, if n0 < 2din, M = ⌊ ⌋ | | ( 0, if n 2d . (3.46) 0 ≥ in We ﬁx M for denoting the matching we deﬁned in this step hereafter. Step 4. Insertion In this step, we insert vertices in V 0 into V ′ V (M). Let I = V 0 = x , x , , x , 12 1 − 12 { 1 2 ··· I } 90

U = Γ(v∗,V ′) V (M), and U = V ′. Then (i) 1 2 1 − 2 2

δ(I, U ) δ(I,V ′) V (M) V ′ Γ(v∗,V ′) 1 ≥ 1 −| |−| 1 − 2 1 | α V /6 max 7d , 7n 1.6n +3.2d , ≥ 1| 2| − {⌊ in⌋ 0} − ⌊ 0 in⌋ α V /6 20.4α V (by (3.39)and(3.40)) ≥ 1| 2| − 2| 2| 3α V 3 I (provided that 23.4α α /6), ≥ 2| 2| ≥ | | 2 ≤ 1 and from (3.32), we have (ii)

δ(U , U v∗ ) (2n + 3)/5 α V /3 1 >α V I . 1 2 −{ 2} ≥ ⌈ ⌉ − 1| 2| − 2| 2|≥| |

By condition (i), there is a claw-matching M1 between I and U1 centered in I. Suppose that Γ(x , M ) = x , x , x . We denote by P the path x x x . By (ii), there is a i 1 { i0 i1 i2} xi i1 i i2 matching M between x 1 i I and U v∗ covering x 1 i I . So far, 2 { i0 | ≤ ≤ | |} 2 −{ 2} { i0 | ≤ ≤ | |} we get two matchings M1 and M2. Next we delete three types of edges not contained in

|I| ( E(P )) x x : 1 i I . xi ∪{ i i0 ≤ ≤| |} i=1 [ Those edges include edges incident to a vertex in I, edges incident to a vertex in

|I| ((Γ(x ) Γ(x )) (Γ(x ) Γ(x ))) , i1 − i2 ∪ i2 − i1 i=1 [ and one edge from the two edges connecting a vertex in Γ(x ) Γ(x ) to both x and x , i1 ∩ i2 i1 i2 for each i =1, 2, , I . ··· | | For the resulting graph after the deletion of edges above, we contract each path P (1 xi ≤ i I ) into a single vertex v . We call each v a wrapped vertex and call P the preimage ≤| | xi xi xi ∗ of vxi . Denote by G the graph obtained by deleting and contracting the same edges as above, and let U ∗ = V ′ and U ∗ = V (G∗) U ∗. (We will need the following degree condition 2 2 1 − 2 91 in the end of this proof.) Since U ∗ = V ′ (2/5+ α)n, combining with (3.32), we have | 2 | | 2 | ≤

deg(v , U ∗) Γ(x , U ∗) Γ(x , U ∗) 1 2n/5 α V . xi 2 ≥| i1 2 ∩ i2 2 | − ≥ − 1| 2|

By the above inequality and (3.32), we get the ﬁrst inequality below in (3.47). Since one edge from the two edges connecting a vertex in Γ(x ) Γ(x ) to both x and x is deleted i1 ∩ i2 i1 i2 in G∗ for each i = 1, 2, , I , combining with (3.33), we have the second inequality as ··· | | follows.

δ(U ∗, U ∗) 2n/5 α V , 1 2 ≥ − 1| 2| δ(U ∗, U ∗) δ(V ′,V ′) 1 (1 3α ) V 1. (3.47) 2 1 ≥ 2 1 − ≥ − 1 | 1| −

′ ′ Let U1 and U2 be the corresponding sets of U1 and U2, respectively, after the contraction.

Let TW be the graph with

V (T )= x , v : 1 i I (V (M ) U ) and E(T )= x v : 1 i I E(M ). W { i0 xi ≤ ≤| |}∪ 2 ∩ 2 W { i0 xi ≤ ≤| |}∪ 2

By the construction,

V (T ) U ′ = x , v : 1 i I =2 I , L(T ) U ′ = v : 1 i I = I , and | W ∩ 1| |{ i0 xi ≤ ≤| |}| | | | W ∩ 1| |{ xi ≤ ≤| |}| | |

V (T ) U ′ = L(T ) U ′ = V (M ) U ′ = I . | W ∩ 2| | I ∩ 2| | 2 ∩ 2| | |

Notice that T is a forest with I components and each vertex x (1 i I ) has W | | i0 ≤ ≤ | | ∗ degree 2 in TW . (We will make TW connected in the end by connecting each xi0 to v2.) See a depiction of this operation with I = 1 in Figure 3.7 below. | | Let U 1 =(V ′ U ) U ′ V (T ), U 2 = U ′ V (T ), and G the resulting graph with I 1 − 1 ∪ 1 − W I 2 − W I 92

x11

x10 x1 Wrap x10 vx1

x12

Figure (3.7) T with I =1 W | |

V (G )= U 1 U 2. We have that I I ∪ I

U 1 = V ′ 3 I = V ′ 3n0 , U 2 = V ′ I = V ′ n0 , | I | | 1 | − | | | 1 | − 12 | I | | 2 |−| | | 2 | − 12 δ(U 1, U 2) δ(V ′,V ′) n0 (2n + 3)/5 α V /3 n0 , I I ≥ 1 2 − 12 ≥ ⌈ ⌉ − 1| 2| − 12 δ(U 2, U 1) δ(V ′,V ′) 3n0 (1 3α ) V 3n0 . (3.48) I I ≥ 2 1 − 12 ≥ − 1 | 1| − 12

Step 5. Matching Extension

In this step, in the graph GI , we do some operation on the matching M found in Step 3. Notice that the vertices in M are unused in Step 4. Recall that M max 7n , 7d . | | ≤ { 0 ⌊ in⌋} By d α V from (3.39) and n 2α V from (3.40), we get ⌊ in⌋ ≤ 2| 2| 0 ≤ 2| 2|

M 14α V . (3.49) | | ≤ 2| 2|

Hence, δ(U 1, U 2 v∗ ) (2n + 3)/5 α V /3 n0 1 M . Let V be the I I −{ 2} ≥ ⌈ ⌉ − 1| 2| − 12 − ≥ | | M set of vertices containing exactly one end of each edge in M. Then there is a matching M ′ between V and U v∗ covering V . Let F be a forest with M 2 −{ 2} M M

V (F )= V (M) (V (M ′) U ) and E(F )= E(M) E(M ′). M ∪ ∩ 2 M ∪

Notice that

V (F ) U =2 M , L(F ) U = V (M) V = M , | M ∩ 1| | | | M ∩ 1| | − M | | |

V (F ) U = L(F ) U = M . | M ∩ 2| | M ∩ 2| | | 93

Notice that F has M components, and all vertices in V has degree 2. (We will make M | | M F a HIT later on by connecting each vertex in V to the vertex v∗ U ) See Figure 3.8 M M 2 ∈ 2 for a depiction of F with M = 3. M | |

U2 Figure (3.8) F with M =3 M | |

Let U 1 = U 1 V (F ) and U 2 = U 2 V (F ). M I − M M I − M

Notice that

U 1 = U 1 2 M = V ′ 3n0 2 M , | M | | I | − | | | 1 | − 12 − | | U 2 = U 2 M = V ′ n0 M , (3.50) | M | | I |−| | | 2 | − 12 −| | and

δ(U 1 , U 2 ) = (2n + 3)/5 α V /3 n0 M , M M ⌈ ⌉ − 1| 2| − 12 −| | δ(U 2 , U 1 ) (1 3α ) V 3n0 2 M . (3.51) M M ≥ − 1 | 1| − 12 − | |

Step 6. Distribute Remaining vertices in U 1 Γ(v∗,V ′) M − 2 1 Let

We may assume n 1. For otherwise, we skip this step. By (3.42), we have l ≥ 3.2α V , Case A, 2| 2| nl ≤ ( 6.4α V , CaseB. (3.52) 2| 2| 94

By n0 α V from (3.28) and M 14α V from (3.49), we have (i) 12 ≤ 2| 2| | | ≤ 2| 2|

δ(U 1 , U 2 ) (2n + 3)/5 α V /3 n0 M (2n + 3)/5 α V /3 15α V M M ≥ ⌈ ⌉ − 1| 2| − 12 −| | ≥ ⌈ ⌉ − 1| 2| − 2| 2| (1 3α) V α V /3 15α V (as (2n + 3)/5 (1 3α)(2/5+ α)n) ≥ − | 2| − 1| 2| − 2| 2| ⌈ ⌉ ≥ − (1 3α α /3 15α ) V ) (1 α ) V (provided 3α + 15α 2α /3) ≥ − − 1 − 2 | 2| ≥ − 1 | 2| 2 ≤ 1 (1 α ) U 2 . (3.53) ≥ − 1 | M |

By (3.50) and (3.52), we have (ii)

U 2 10α V n /10 1 V ′ n0 M 16α V 0.64α V 2 | M | − 1| 2| − ⌈ l ⌉ − ≥ | 2 | − 12 −| | − 1| 2| − 2| 2| − (1 α 14α 10α 0.64α V /2) V ≥ − 2 − 2 − 1 − 2 −| 2| | 2| (1 11α ) V (provided 15.64α + V /2 α ) ≥ − 1 | 2| 2 | 2| ≤ 1

> 0 (provided 11α1 < 1).

Let U = U 1 Γ(v∗,V ′) and denote |UR| = l. Suppose ﬁrst that U 2. We partition R M − 2 1 10 | R| ≥ U = U U U arbitraryl suchm that each set contains at least 2 and at most R R1 ∪ R2 ∪···∪ Rl U /10 vertices. Then by the conditions (i) and (ii), for each 1 i l, there is a vertex | R| ≤ ≤ y U v∗ which is common to all vertices in U , and is not used by any other U with i ∈ 2 −{ 2 } Ri Rj j = i. Let T be the graph with 6 R

V (T )= U y : 1 i l and E(T )= xy : x U , 1 i l . R R ∪{ i ≤ ≤ } R { i ∈ Ri ≤ ≤ }

Suppose now U = 1, let U = x . Choose x′ U 1 U and y U 2 v∗ be a | R| R { R} R ∈ M − R R ∈ M −{ 2} ′ vertex common to xR and xR. Let TR be a tree with

V (T )= x , x′ ,y and E(T )= x y , x′ y . R { R R R} R { R R R R} 95

By the construction,

V (T ) U 1 = L(T U 1 = max U , 2 , V (T ) U 2 = l, and L(T U 2 =0. | R ∩ M | | R ∩ M | {| R| } | R ∩ M | | R ∩ M |

Notice that T is not connected when U 17 and that T may have degree 2 vertices in R | R| ≥ R V (T ) U 2 . Later on, by joining each vertex in T U 2 to a vertex of a tree, we will make R ∩ M R ∩ M the resulting graph connected, and thereby eliminating the possible degree 2 vertices in TR. Let U 1 = U 1 V (T ) and U 2 = U 2 V (T ). R M − R R M − R

Then we have

U 1 = U 1 n = V ′ 3n0 2 M max 2, n , | R| | M | − l | 1 | − 12 − | | − { l} U 2 = U 2 n /10 = V ′ n0 M n /10 , (3.54) | R| | M | − ⌈ l ⌉ | 2 | − 12 −| | − ⌈ l ⌉ and

δ(U 1 , U 2 ) (2n + 3)/5 α V /3 n0 M n /10 , R R ≥ ⌈ ⌉ − 1| 2| − 12 −| | − ⌈ l ⌉ δ(U 2 , U 1 ) = (1 3α ) V 3n0 2 M max 2, n . (3.55) R R − 1 | 1| − 12 − | | − { l}

Let G be the subgraph of G induced on U 1 U 2 . R R ∪ R Step 7. Completion of a HIST in GR

In this step, we ﬁnd a HIST Tmain in GR such that

L(T ) U 1 = L(T ) /2+ L(F ) U 1 + L(T ) U 1 = | main ∩ R| | W | | M ∩ I | | R ∩ M | L(T U 2 ) = L(T ) /2+ L(F ) U 2 + L(T ) U 1 . | main ∩ R | | W | | M ∩ I | | R ∩ M |

By the construction of F and T , we have L(F ) U 1 = L(F ) U 2 and L(T ) M R | M ∩ I | | M ∩ I | | R ∩ 96

U 1 L(T ) U 2 = max 2, n , respectively. So M |−| R ∩ M | { l}

L(T ) U 2 L(T ) U 1 = max 2, n . (3.56) | main ∩ R|−| main ∩ R| { l}

Notice that v∗ U 2 , v∗ is adjacent to each vertex in U 1 , and V ′ Γ(v∗,V ′) V (U 1 ). 2 ∈ R 2 R 1 − 2 1 ⊆ R We now construct Tmain step by step. Step 7.1

1 Let Tmain be the graph with

V (T 1 )= v∗ U 1 and E(T 1 )= v∗x x U 1 . main { 2} ∪ R main { 2 | ∈ R}

To make the requirement of (3.56) possible, we need to make at least

d = U 1 U 2 + max 2, n , 3 | R|−| R| { l} = V ′ V ′ 2n0 M + n /10 (3.57) | 1 |−| 2 | − 12 −| | ⌈ l ⌉

1 vertices in UR with degree at least 3 in Tmain, where the last inequality above follows from (3.54). Hereinafter, we assume that max 2, n = n . Since the proof for max 2, n = 2 { l} l { l} follows the same idea, we skip the details.

1 1 1 Since all vertices in UR are included in Tmain and Tmain is connected, each vertex in T 1 needs to join to at least two distinct vertices from U 2 v∗ to have degree no less main R −{ 2} than 3. Hence, to make a desired HIST Tmain, it is necessary that

d = U 2 1 2d f∗ | R| − − 3 = V ′ n0 e n /10 1 2d | 2 | − 12 − M − ⌈ l ⌉ − − 3 = 3 V ′ 2 V ′ +3n0 + M 3 n /10 1 | 2 | − | 1 | 12 | | − ⌈ l ⌉ − 0. (3.58) ≥ 97

We show (3.58) is true, separately, for each of Case A and Case B. For Case A, notice that

V ′ = n (2n + 3)/5 + n and V ′ = (2n + 3)/5 n n0 . | 1 | − ⌈ ⌉ 0 | 2 | ⌈ ⌉ − 0 − 12

Hence,

3 V ′ =3 (2n + 3)/5 3n 3n0 and 2 V ′ =2n 2 (2n + 3)/5 +2n . | 2 | ⌈ ⌉ − 0 − 12 | 1 | − ⌈ ⌉ 0

Thus,

d = 5 (2n + 3)/5 2n 5n 3n0 +3n0 + M 3 n /10 1 f∗ ⌈ ⌉ − − 0 − 12 12 | | − ⌈ l ⌉ − 2 5n + M 3 3.2d /10 (by n d 3.2d 1.4n from (3.41)) ≥ − 0 | | − ⌈⌊ in⌋ ⌉ l ≤ l⌊ in − 0⌋ = 2 5n + max 7n , 7d 3 3.2d /10 − 0 { 0 ⌊ in⌋} − ⌈⌊ in⌋ ⌉ 0. ≥

Now we show (3.58) is true for case B. Notice that

V ′ = n (2n + 3)/5 + n and V ′ = (2n + 3)/5 + n n0 . | 1 | − ⌈ ⌉ 0 | 2 | ⌈ ⌉ 0 − 12

So 3 V ′ =3 (2n + 3)/5 +3n 3n0 and 2 V ′ =2n 2 (2n + 3)/5 +2n . | 2 | ⌈ ⌉ 0 − 12 | 1 | − ⌈ ⌉ 0

Recall that n 1 in this case. We have 0 ≥

d = 5 (2n + 3)/5 2n + n 3n0 +3n0 + M 3 n /10 1 f∗ ⌈ ⌉ − 0 − 12 12 | | − ⌈ l ⌉ − 2+ n + M 3 n /10 ≥ 0 | | − ⌈ l ⌉ = 2+ n + M 3 3.2d +1.6n /10 (by n 3.2d +1.6n from (3.41)) 0 | | − ⌈⌊ in 0⌋ ⌉ l ≤ ⌊ in 0⌋ 2+ n + d 9.2d /10 4.8n /10 1 0, if n < 2d ; 0 ⌊ in⌋ − ⌊ in ⌋ − ⌊ 0 ⌋ − ≥ 0 in ≥  2+ n 9.2d /10 4.8n /10 1 0, if n 2d .  0 − ⌊ in ⌋ − ⌊ 0 ⌋ − ≥ 0 ≥ in  98

1 We now in Step 2 below show that there is a way to make exactly df∗ vertices in Tmain with degree 3 by joining each to two distinct vertices from U 2 v∗ . R −{ 2} Step 7.2 We ﬁrst take 2d vertices from U 2 v∗ . For those 2d vertices, pair them up into d 3 R −{ 2} 3 3 1 pairs. We show that for each pair of vertices, they have at least d3 common neighbors in UR. Using (3.55), M 14α V from (3.49), n d 6.4α V from (3.42), we have | | ≤ 2| 2| l ≤ l ≤ 2| 2|

δ(U 2 , U 1 ) (1 3α ) V 3n0 2 M max 2, n R R ≥ − 1 | 1| − 12 − | | − { l} V 3α V 3α V 28α V 6.4α V ≥ | 1| − 1| 1| − 2| 2| − 2| 2| − 2| 2| V ′ V V 37.4α V 3α V . (3.59) ≥ | 1 |−| 1 − 1| − 2| 2| − 1| 1|

Since U 1 V ′ , we know that any two vertices in U 2 have at least | R|≤| 1 | R

n = V 2 V ′ V 74.8α V 6α V c | 1| − | 1 − 1| − 2| 2| − 1| 1| (3/5 α)n 76.8α V 6α V (by V ′ V = V ′ V V ′ α V ) ≥ − − 2| 2| − 1| 1| | 1 − 1| | 0 |≤| 2 − 2 | ≤ 2| 2| 3n/5 10α V (provided that 76.8α +3α 4α ) ≥ − 1| 1| 2 ≤ 1

1 common neighbors in UR. On the other hand,

d = V ′ V ′ 2n0 M n /10 3 | 1 |−| 2 | − 12 −| | − ⌈ l ⌉ (3/5 α)n (2n/5 2βn V V ′ )+(1.6n +3.2 d )/10+1 ≤ − − − −| 2 − 2 | 0 ⌊ in⌋ = n/5 αn +2βn + V V ′ + (3.2α V +3.2α V )/10 (by(3.39)and(3.40)) − | 2 − 2 | 2| 2| 2| 2| n/5 αn +2βn + α V +0.64α V ≤ − 2| 2| 2| 2| n/5+2α V < n (provided 12α < 2/5). ≤ 1| 2| c 1

Denote by u1,u2 , u1,u2 , , u1 ,u2 the d pairs of vertices from U 2 v∗ . Then by { 1 1} { 2 2} ··· { d3 d3 } 3 R −{ 2} the above argument, we can choose d distinct vertices say v , v , , v from L(T 1 ) such 3 1 2 ··· d3 main that v u1,u2 for all 1 i d . i ∼ i i ≤ ≤ 3 99

2 Let Tmain be the graph with

V (T 2 )= V (T 1 ) u1,u2 : 1 i d and E(T 2 )= E(T 1 ) v u1, v u2 : 1 i d . main main ∪{ i i ≤ ≤ 3} main main ∪{ i i i i ≤ ≤ 3}

If V (G ) V (T 2 ) = , we let T = T 2 . For otherwise, we need one more step to R − main ∅ main main ﬁnish constructing Tmain. Step 7.3 For the remaining vertices in U 2 V (T 2 ), we show that each of them has a neighbor R − main in S(T 2 ) U 1 ; that is, a neighbor in U 1 of degree 3 in V (T 2 ). This is clear, as by main ∩ R R main (3.59), we have

δ(U 2 , U 1 ) V ′ V ′ V 37.4α V 3α V R R ≥ | 1 |−| 1 − 1| − 2| 2| − 1| 1| U 1 38.4α V 3α V (by V ′ V V V ′ α V ). ≥ | R| − 2| 2| − 1| 1| | 1 − 1|≤| 2 − 2 | ≤ 2| 2|

Since S(T 2 ) U 1 = d , and | main ∩ R| 3

d = V V ′ 2n0 2 M + n /10 3 | 1|−| 2 | − 12 − | | ⌈ l ⌉ (3/5 α)n (2/5+ α)n 2α V 28α V +0.64α V ≥ − − − 2| 2| − 2| 2| 2| 2| n/5 2αn 29.36α V ≥ − − 2| 2| > 38.4α V +3α V (provided 2α + 67.76α +3α < 1/5). 2| 2| 1| 1| 2 1

Now, we join an edge between each vertex in U 2 V (T 2 ) and a neighbor of the vertex in R − main S(T 2 ) U 1 . Let T be the resulting tree. By the construction procedure, it is easy to main ∩ R main verify that Tmain is a HIST of GR.

Step 8. Connecting TW , FM , TR, and V (Tmain) into a connected graph

In this step, we connect TW , FM , TR, and V (Tmain) into a connected graph. Recall that

∗ ∗ each degree 2 vertex in TW and FM is a neighbor of v2. We join an edge connecting v2 in

V (Tmain) and each degree 2 vertex in TW and FM . By the argument in step 7.3 above, we know each vertex in V (T ) U 2 has a neighbor in S(T ) U 1 . Thus, we join an edge R ∩ M main ∩ R 100

between each vertex in V (T ) U 2 to exactly one of its neighbor in S(T ) U 1 . Let T ∗ R ∩ M main ∩ R be the ﬁnal resulting graph. Notice that I = V 0 = x , x , , x L(T ∗) is the set of 12 { 1 2 ··· I } ⊆ the wrapped vertices from Step 4. Recall that G∗ is the graph obtained from G be deleting and contracting edges from Step 4. Then by the constructions of TW , FM , TR, and Tmain, we see that T ∗ is a HIST of G∗ with L(T ∗) U ∗ = L(T ∗) U ∗ . | ∩ 1 | | ∩ 2 | Step 9. Finding a cycle on L(T ∗) Denote

U 1 = L(T ∗) U ∗, U 2 = L(T ∗) U ∗ and G = G[E (U 1, U 2)]. L ∩ 1 L ∩ 2 L G L L

Notice that GL is a balanced bipartite graph. And

S(T ∗) U ∗ = d n/5+2α V (by (3.60)) | ∩ 1 | 3 ≤ 1| 2| S(T ∗) U ∗ = 1+ n /10 2+0.64α V (by n d 6.4α V from (3.42)). | ∩ 2 | ⌈ l ⌉ ≤ 2| 2| l ≤ l ≤ 2| 2|

Thus by (3.47),

1 2 2 δ ∗ (U , U ) 2n/5 α V (2+0.64α V ) > 3n/10 > U /2+1, G L L ≥ − 1| 2| − 2| 2| | L| 2 1 1 δ ∗ (U , U ) (1 3α ) V 1 (n/5+2α V ) > n/3 > U /2+1. G L L ≥ − 1 | 1| − − 1| 2| | L|

′ By Lemma 3.2.7, GL contains a hamiltonian cycle C . Step 10. Unwrap vertices in V (C′) v , v , , v ∩{ x1 x2 ··· x|I| } On C′, replace each vertex v with its preimage P = x x x for each i =1, 2, , I . xi xi i1 i i2 ··· | | Denote the resulting cycle by C. Recall that x , x Γ(v∗) by the choice of x and x . i1 i2 ∈ 2 i1 i2 Let T be the graph on V (G) with

E(T )= E(T ∗) v∗x , v∗x : i =1, 2, , I . ∪{ 2 i1 2 i2 ··· | |}

Then T is a HIST of G. Let H = T C. Then H is an SGHG of G. ∪ 101

The proof of Extremal Case 2 is ﬁnished. 102

PART 4

A LOWER BOUND ON CIRCUMFERENCES OF 3-CONNECTED GRAPHS WITH BOUNDED MAXIMUM DEGREES

4.1 Introduction

In 1980, Bondy and Simonovits [8] showed that the best general lower bound on the

length of a longest cycle in an n-vertex 3-connected cubic graph is between exp(c1√log n)

c2 and n for some positive constants c1 and c2, and they also obtained similar bounds for

3-connected graphs with bounded degrees. The lower bound exp(c1√log n) for cubic graphs was improved to n0.69 by Jackson [32] and was further improved to n0.8 by Liu, Yu and Zhang [39]. In 1993, Jackson and Wormald [33] proved that every 3-connected n-vertex

1 logb 2 2 graph with maximum degree at most d has a cycle of length at least 2 n +1 with b =6d . They also conjectured that for d 4 the correct value for b should be d 1, and they gave ≥ − an inﬁnite class of graphs showing that b = d 1 is the best possible value that one can − hope for.

Recently there has been considerable interest in approximating longest cycles in 3- connected graphs with bounded degrees. Karger, Motwani, and Ramkumar [35] showed that unless = , it is impossible to ﬁnd, in polynomial time, a path of length n nǫ (for P N P − any ǫ < 1) in an n-vertex Hamiltonian graph. They conjectured that it is hard even for graphs with bounded degrees. On the positive side, Feder, Motwani, and Subi [23] showed

that there is a polynomial time algorithm for ﬁnding a cycle of length at least n(log3 2)/2 in any 3-connected cubic n-vertex graph, and they asked the same question for 3-connected graphs with bounded degrees. Chen, Xu, and Yu [14] provided a cubic-time algorithm that, given a 3-connected n-vertex graph with maximum degree at most d, ﬁnds a cycle of length 103

at least nlogb 2 + 3 with b = 2(d 1)2 + 1. This result was improved to b =4d + 1 by Chen, − Gao, Yu, and Zang [12].

Before stating the main result, we introduce some notation. For any graph G, we denote by G the number of vertices of G, G z the graph obtained from G by deleting the vertex | | − z V (G), and N (z) the set of neighbors of z in G. If G is a path or cycle, then ℓ(G) ∈ G denotes the length of G. Let S ,S V (G) be two disjoint sets. An (S ,S )-path is a path 1 2 ⊆ 1 2 P connecting one vertex in S and one vertex in S such that V (P ) S = V (P ) S = 1. 1 2 | ∩ 1| | ∩ 2| When S = x is a singleton, we simply write as (x, S )-path. The main result of this paper 1 { } 2 is the following:

Theorem (4.1.1). Let d 425 be an integer, and r = log 2. Let G be either a cycle or ≥ d−1 a 3-connected graph and e = xy E(G) be an edge. ∈

(a) If G is 3-connected, then for any z V (G) x, y such that ∆(G z) d and z has ∈ −{ } − ≤ at most t neighbors distinct from x and y, there is a cycle C in G z through xy such r − 1 d 2.1 G that ℓ(C) − | | +2. ≥ 4 d 1 t − (b) If ∆(G) d, then for any f E(G) e , there is a cycle C in G through e and f ≤ ∈ r −{ } 1 d 2.1 such that ℓ(C) − G +2. ≥ 4 (d 1)2 | | − (c) If ∆(G) d, then there is a cycle C through e in G such that ℓ(C) 1 G r +2. ≤ ≥ 4 | |

We ﬁrst note that Theorem (4.1.1) holds trivially when G d; hence, throughout the | | ≤ rest of this part, we assume G d+1 426. Also note that Theorem (4.1.1) holds trivially | | ≥ ≥ when G is a cycle. However, we include cycles in the statement of Theorem (4.1.1) for the following reason: cycles occur in our inductive arguments, and their inclusion makes many arguments less cumbersome.

The rest of this part is organized as follows. In Section 2, we recall Tutte decomposition [52] for decomposing a 2-connected graph into 3-connected components and some results 104 from [12] concerning paths in 2-connected graphs. In Section 3, we prove a useful inequality about the function f(x)= xlogb 2. In Section 4, we state lemmas concerning paths in a chain of 3-connected components, and in Section 5, we inductively prove Theorem (1.1) (a) and (b). Section 6 is the most signiﬁcant part of the paper, where we prove Theorem (4.1.1)(c) inductively.

4.2 Paths in block-chains

We recall Tutte decomposition for decomposing a 2-connected graph into 3-connected components. A detailed description can be found in [14] and [29]. Let denote the set D of all 3-connected (simple) graphs, denote the set of cycles (with at least three vertices), C and denote the set of bonds (a bond is a multigraph with two vertices and at least three B edges between them). Tutte [52] proved that every 2-connected graph G can be uniquely decomposed into 3-connected components, which belong to . We call such a de- B∪C∪D composition as the Tutte decomposition. Those 3-connected components are linked together by virtual edges to form a tree-like structure. More precisely, if we deﬁne a graph whose vertices are the 3-connected components of G obtained from the Tutte decomposition and two vertices are adjacent if the corresponding two 3-connected components share a common virtual edge, then such a graph is a tree, which we call the block-bond tree of G. Hopcroft and Tarjan [29] gave a linear time algorithm for decomposing any 2-connected graph into 3-connected components.

Recall that in the block-cut tree of a connected graph there is a cut-vertex between two consecutive blocks. However, in a block-bond tree, it is not necessarily true that there is a bond between any two 3-connected components. For example, let G1 and G2 be two

3-connected graphs such that each Gi contains two nonadjacent vertices ui and vi for each i = 1, 2. Let G be obtained from G1 and G2 by identifying u1 with v1 and u2 with v2, respectively. According to Tutte’s decomposing algorithm, G1 + u1v1 and G2 + u2v2 are the 105

only two 3-connected components of G. Clearly, in the block-bond tree, they are adjacent but there is no bond between them.

For convenience, 3-connected components that are not bonds are called 3-blocks, consisting of cycles and simple 3-connected graphs. An extreme 3-block is a 3-block that contains at most one virtual edge. That is, either it is the only 3-connected component (in which case G is either a cycle or a 3-connected simple graph), or it corresponds to a degree one vertex in the block-bond tree.

A block-chain in G is a sequence H1H2 ...Hh of 3-blocks of G for which there exist B , B ,...,B such that for each 1 j h 1, B = or B is a bond, and 1 2 h−1 ≤ ≤ − j ∅ j H1B1H2B2 ...Bh−1Hh is a path in the block-bond tree of G. A detailed description with examples can be found in [14]. For convenience, we sometimes write := H H ...H H 1 2 h for this situation. In the rest of the paper, unless stated otherwise, we will always assume that each virtual edge in E(H H ) is deleted from if at least one of H and H is i ∩ i+1 H i i+1 3-connected and there are exactly two components in G V (H H ). Because in this − i ∩ i+1 case it is not possible to replace the virtual edge by a path in G outside of . However, if H both Hi and Hi+1 are cycles, then the virtual edge shared by Hi and Hi+1 can always be replaced by a path outside of . Throughout this section, we adopt the convention that an H object is empty if it is not deﬁned. For example, if = H H ...H is a block-chain under H 1 2 h consideration, then H0 and Hh+1 are both empty graphs.

The following result is proved in the proof of Lemma (3.6) in [14], which will be used to link together long paths from diﬀerent block-chains. We note that the path stated in the lemma can be found in linear time by using a result from [43].

Lemma (4.2.1). Let = H H ...H be a block-chain in a 2-connected graph G, x H 1 2 h ∈ V (H ) V (H ), and f E(H ) E(H ) such that f is not incident with x, and let pq,vw 1 − 2 ∈ h − h−1 be two distinct edges in E( ) f . Then there is a path P in through f from x to some H −{ } H z p, q v,w such that if z p, q then pq / E(P ) and vw E(P ) and if z v,w ∈{ }∪{ } ∈{ } ∈ ∈ ∈{ } 106

then vw / E(P ) and pq E(P ). ∈ ∈ Lemma (4.2.2). Let = H H ...H be a block-chain in a 2-connected graph G, x H 1 2 h ∈ V (H ) V (H ), and pq,f E(H ) E(H ) be distinct such that neither pq nor f is 1 − 2 ∈ h − h−1 incident with x. If H is 3-connected and q is not incident with f, then in q, there is h H − an (x, p)-path through f.

Proof. We use induction on h. If h = 1, then H is 3-connected and H q is 2-connected, 1 1 − so H q contains an (x, p)-path through f. Suppose h 2 and let a, b = V (H H ). 1 − ≥ { } h−1 ∩ h Since pq / E(H ), we may assume a / p, q . Let P be an (a, p)-path through f in ∈ h−1 ∈ { } h H q. h −

If ab / P , let P bean(x, a)-path in H H ...H b; then P := P P is the desired ∈ h 1 1 2 h−1 − h 1 ∪ h path. If ab P , let P be an (x, b)-path in H H ...H a; then P := P (P ab) is ∈ h 1 1 2 h−1 − 1 ∪ h − the desired path.

Lemma (4.2.3). Let = H H ...H be a block-chain. Let xy,pq,uv be three edges such H 1 2 h that xy E(H ) E(H ) and pq,uv E(H ) E(H ), where pq = xy = uv but it is ∈ 1 − 2 ∈ h − h−1 6 6 possible that pq = uv. Then there is a path P in from some z x, y to w p, q u, v H ∈{ } ∈{ }∪{ } such that x, y V (P ), uv E(P ) if w p, q , and pq E(P ) if w u, v . { } 6⊆ ∈ ∈{ } ∈ ∈{ }

Proof. We ﬁrst consider h = 1, that is = H . The result is trivial if H is a cycle. Suppose H 1 1 that H is 3-connected. Then H y is 2-connected, and thus contains an (x, p, q )- path 1 1 − { } P through uv.

We now assume h 2. Let a, b = V (H ) V (H ). By the same argument as for ≥ { } h ∩ h−1 the case where h = 1, there is a path P from z∗ a, b to w p, q u, v such that H ∈ { } ∈ { }∪{ } a, b V (P ), uv E(P ) if w p, q and pq E(P ) if w u, v . Clearly, there is { } 6⊆ H ∈ H ∈ { } ∈ H ∈ { } a path Q in H ...H ab from some z x, y to z∗ such that x, y V (Q). Then 1 h−1 − ∈ { } { } 6⊆ Q P is the desired path. ∪ H 107

Lemma (4.2.4). Let = H H ...H be a block-chain, u, v V (H ) be distinct, and H 1 2 h ∈ h x V ( v). Then there is a path from x to u avoiding v. ∈ H −

Proof. We use induction on h. The result is clearly true when h = 1. Assume the claim holds for block-chains with fewer than h blocks. Let a, b = V (H H ). If x V (H ), { } h−1 ∩ h ∈ h let P be a path in H from x to u avoiding v. If ab / E(P ), let P := P ; if ab E(P ), let h h ∈ h h ∈ h P be obtained from Ph by replacing ab by a path in H1 ...Hh−1 from a to b.

Suppose x / V (H ). Assume, without loss of generality, that a = v, and let P be a path ∈ h 6 h in Hh from a to u avoiding v. By induction, let P1 be an (x, a)-path avoiding b in H1 ...Hh−1 if ab E(P ), and let P be an (x, b)-path avoiding a in H ...H if ab E(P ). Then 6∈ h 1 1 h−1 ∈ h P := P (P ab) is the desired path. 1 ∪ h − Lemma (4.2.5). Let = H H ...H be a block-chain, and let xx′ E(H ) E(H ) and H 1 2 h ∈ 1 − 2 uv E(H ) be two edges in . Then there is an (x′, u, v )-path in x. ∈ h H { } H −

Proof. We use induction on h. The statement is clearly true when h = 1 as H x is 1 − connected. So we assume h 2, and let V (H ) V (H ) = a, b . Suppose ﬁrst that ≥ 1 ∩ 2 { } x / a, b . By induction, we let P be an (x′, a, b )-path, say (x′, a)-path, in H x, and ∈ { } 1 { } 1 − let P be an (a, u, v )-path in H H H b. Then P P is the desired path. Then, 2 { } 2 2 ··· h − 1 ∪ 2 suppose, without loss of generality, that x = a. Let P be an (x′, b)-path in H x, and 1 1 − by induction, let P be a (b, u, v )-path in H H H a. Then P P is the desired 2 { } 2 2 ··· h − 1 ∪ 2 path.

We conclude this section by recalling two graph operations from [14]. Let G be a graph and let e, f be distinct edges of G. An H-transform of G at e, f is an operation { } that subdivides e and f by vertices x and y, respectively, and then adds the edge xy. Let x V (G) such that x is not incident with e. A T -transform of G at x, e is an operation ∈ { } that subdivides e with a vertex y and then adds the edge xy. Let G′ be a graph obtained 108

from a 3-connected graph G by an H-transform or a T -transform. It is easy to see that G′ is also 3-connected (see, e.g., Lemma (3.3) in [14] for a proof).

4.3 Lower bounds of mlogb 2 + nlogb 2

Fix b = d 1 hereinafter, where d 4 is an integer and let r = log 2. Clearly, 0

Lemma (4.3.1). Let m and n be two positive real numbers such that m bβn> 0 for some ≥ real number β. Then,

−β r −β r mr + nr m + bβ blog2(1+2 ) 1 n bβblog2(1+2 ) n . (4.1) ≥ − ≥

Proof. Deﬁne f(t)= 1 (1 + tr)1/r 1 . It is easy to verify that t − 1 mr + nr =(m + f(n/m)n)r and f ′(t)= 1 (1 + tr)(1−r)/r . t2 − Since b 3, we have 0 0. Therefore f(t) is a ≥ decreasing function on the interval (0, ). For m bβn > 0, we have n/m b−β, and so ∞ ≥ ≤ (since br = 2)

1/r −β f(n/m) f(b−β)= bβ 1+2−β 1 = bβ blog2(1+2 ) 1 . ≥ − − Here, the ﬁrst inequality in (4.1) follows from mr + nr (m + f(n/m)n)r, and the 2nd ≥ inequality in (4.1) follows from m bβn. ≥

Taking β = 0, log 1.1, 1, 2, 1, 2, we get the following inequalities from (4.1). It is b − − 109

straightforward to verify (4.1a) - (4.1f)

(m +(b 1)n)r, if m n; (4.1a) − ≥ − log 1.1 (m +1.1(blog2(1+2 b ) 1)n)r, if m 1.1n; (4.1b)  − ≥  (m + b(blog2 3/2 1)n)r, if m bn; (4.1c)  mr + nr  − ≥ ≥  2 log 5/4 r 2 (m + b (b 2 1)n) , if m b n; (4.1d)  − ≥ log 3/2 r (b 2 n) , if m n/b; (4.1e)  ≥   log2 5/4 r 2 (b n) , if m n/b . (4.1f)  ≥   In the proofs, the following elementary inequality will be used frequently for any two positive real numbers x and y,

xr + yr 2√xryr = (d 1)2xy r/2 . (4.2) ≥ − Lemma (4.3.2). The following inequalities hold:

xr +1 (x + d 1)r provided x 1. ≥ − ≥

Lemma (4.3.3). Let b 23 be an integer. If m and n are two positive real numbers such ≥ that m 1.1n, then mr + nr (m + bn)r. ≥ ≥

Proof. Applying Lemma (4.3.1) for β = log 1.1 log 1.1, we have mr + nr (m + b ≤ 23 ≥ − logb 1.1 − logb 1.1 1.1(blog2(1+2 ) 1)n)r. So, we only need to show that 1.1(blog2(1+2 ) 1) b − − ≥ provided b 23. For any x 1.1, let τ := τ(x) = log 1.1, φ := φ(τ) = log (1+2−τ ) and ≥ ≥ x 2 f(x) = xφ(τ(x)). It is clearly that lim (1.1f(x) x) = . It is suﬃcient to show that x→∞ − ∞ 1.1f(x) x is an increasing function for x 23, which is equivalent to d f(x) 10/11 for − ≥ dx ≥ x 23. ≥ 110

Simple calculations show that d τ(x)= ln 1.1 = τ(x) and d φ(τ)= 1 . So, dx − x ln2 x − x ln x dτ − 1+2τ

d dφ dτ d ln x f(x) τ f(x)= f(x) (ln x) + φ = + φ(τ) . dx dτ dx dx x 1+2τ

d Since limx→∞ τ(x) = 0 and limτ→0 φ(τ) = 1, limx→∞ dx f(x) = 1. It is suﬃcient to show d d that dx f(x) is decreasing as x increasing. Writing dx f(x) in terms of τ, we have

d φ(τ)−1 τ (φ(τ)−1) ln 1.1 τ f(x)= x ( + φ(τ)) = e τ ( + φ(τ)). dx 1+2τ 1+2τ

We only need to show d f(x) is increasing as τ increasing when τ τ(23), which is equivalent dx ≤ d df to dτ dx > 0. Taking derivative, we obtain

τ τ τ d df (φ(τ)−1) ln 1.1 ln 1.1(1 φ 1+2τ )(φ + 1+2τ ) τ2 ln 2 = e τ − − . dτ dx τ 2 − (1+2τ )2

So, we only need to show that

τ τ τ ln 1.1(1 φ τ )(φ + τ ) τ2 ln 2 g(τ)= − − 1+2 1+2 > 0 if τ τ(23). τ 2 − (1+2τ )2 ≤

We deﬁne the value of g(τ) at τ =0 as

ln 1.1 ln 2 g(0) = lim g(τ)= . τ→0 8

Then g(τ) is a continuous function on the closed interval [0, τ(23)]. To show g(τ) > 0 within [0, τ(23)], as g(0) > 0, by the intermediate zero theorem, instead, we show that g(τ) has no zero in [0, τ(23)]. To do so, using the bisection method, with tolerance as 1 10−10, a numerical search within [0, 1] interval gives 0.04765221 as the root of g(τ). Since × τ(23) < 0.0304 < 0.04765221, we conclude that g(τ) > 0 when 0 τ τ(23). The proof is ≤ ≤ completed. 111

4.4 Long paths in block-chains

In this section, we will give a few lower bounds of long paths connecting special vertices in a block-chain. Throughout this section, we assume that n 4, Theorem (4.1.1) holds ≥ for graphs with at most n 1 vertices, and := H H ...H is a block-chain such that − H 1 2 h n 1 such that |H| ≤ −

∆(H ) d for each 1 i h. • i ≤ ≤ ≤ As of a subgraph of G, contains at most 2d 1 vertices of degree 2. • H −

Recall, for convention, we also denote by the graph with vertex set V (H ) and edge set H i E(Hi) with the deletion of virtual edges. S

SLemma (4.4.1). For any edge uv E(H ) E(H ), there is a (u, v)-path P in such that ∈ 1 − 2 H

1 d 2.1 r ℓ(P ) − ( + 1) +1, ≥ 4 d 1 |H| − provided that d 23. ≥

Proof. Since n 1, it follows from our assumption that Theorem (4.1.1) holds for |H| ≤ − each H . We proceed with induction on h. Suppose h = 1. Then is either a cycle or a i H 3-connected graph. Since the case d is trivial, we may assume d + 1, and hence |H| ≤ |H| ≥

d 2.1 > − ( + 1). |H| d 1 |H| −

By Theorem (4.1.1)(c), = H contains a cycle C through uv such that H 1

1 1 d 2.1 r ℓ(C) r +2 − ( + 1) +2. ≥ 4|H| ≥ 4 d 1 |H| −

Hence P := C uv gives the desired path. −{ } 112

Therefore, assume h 2. Let ′ := H H and a , b := V (H H ). We consider ≥ H 2 ··· h { 1 1} 1 ∩ 2 two cases.

First, assume H d−2.1 ( + 1). By Theorem (4.1.1)(c), we can ﬁnd a cycle C in | 1| ≥ d−1 |H| 1 H1 through uv such that

1 1 d 2.1 r ℓ(C ) H r +2 − ( + 1) +2. 1 ≥ 4| 1| ≥ 4 d 1 |H| −

If C does not contain a b , then P := C uv is the desired path. If C contains a b , 1 1 1 1 −{ } 1 1 1 then let C be a cycle in ′ through a b . It is clear that P := (C C ) a b , uv is the 2 H 1 1 1 ∪ 2 −{ 1 1 } desired path.

Now assume H < d−2.1 ( + 1). Then | 1| d−1 |H|

d 1 1.1 H 1.1 H ′ +1= H +3 > − H H +2 > | 1| > | 1|. |H | |H|−| 1| d 2.1| 1|−| 1| d 2.1 d 1 − − −

Applying Theorem (4.1.1)(b), we ﬁnd a cycle C1 in H1 through uv and a1b1 such that

1 (d 2.1) H r C − | 1| +2. | 1| ≥ 4 (d 1)2 −

By induction, we ﬁnd a path P ′ in ′ between a and b such that H 1 1

1 d 2.1 r ℓ(P ′) − ( ′ + 1) +1. ≥ 4 d 1 |H | −

Hence P := (C P ′) uv,a b is a path between u and v in such that 1 ∪ −{ 1 1} H

1 (d 2.1) H r 1 d 2.1 r ℓ(P ) − | 1| + − ( ′ + 1) +1 ≥ 4 (d 1)2 4 d 1 |H | − − 1 d 2.1 H r > − ( ′ +1)+(d 1) | 1| + 1 (by Lemma (4.3.3)) 4 d 1 |H | − d 1 − − 1 d 2.1 r > − ( + 1) +1, 4 d 1 |H| − 113

where in the 2nd inequality above, the inequality ′ +1 1.1|H1| is used. |H | ≥ d−1 Lemma (4.4.2). Let x V (H ) V (H ) such that d (x) = d (x) d 1 and uv ∈ h − h−1 H Hh ≤ − ∈ E(H ) E(H ) such that x / u, v when h =1. Then there exists a path P in v from 1 − 2 ∈{ } H − u to x such that

1 h d 2.1 r 1 1 d 2.1 r 1 ℓ(P ) − H + − + . ≥ 4 (d 1)2 | i| 2 ≥ 4 (d 1)2 |H| 2 i=1 X − −

Moreover, if H1 is 3-connected, we can improve the constant 1/2 to 1:

1 h d 2.1 r 1 d 2.1 r ℓ(P ) − H +1 − +1. ≥ 4 (d 1)2 | i| ≥ 4 (d 1)2 |H| i=1 X − −

Proof. Note that the second inequality in each of the lower bounds above for ℓ(P ) is a simple application of Lemma (4.3.1). So we only show the ﬁrst part of the lower bounds.

We apply induction on h. Suppose h =1. If H is a cycle, then H 2d 1, which 1 | 1| ≤ − in turn gives 1 ( d−2.1 H )r < 1/2 and 1 ( d−2.1 H )r + 1 < 1. On the other hand, since 4 (d−1)2 | 1| 4 (d−1)2 | 1| 2 x / u, v , there is an (x, u)-path P in H v with ℓ(P ) 1. Hence, the assertion holds. ∈ { } 1 − ≥ Now assume H = is 3-connected. Note that ∆(H + xu v) d and v has at most d 1 1 H 1 − ≤ − neighbors in H v. By applying Theorem (4.1.1)(a) to H + xu we ﬁnd a cycle C through 1 − 1 xu in (H + xu) v such that ℓ(C) 1 ((d 2.1) H /(d 1)2)r + 2. Hence P := C xu 1 − ≥ 4 − | 1| − −{ } gives the desired path.

Now we assume h 2. Let a , b := V (H H ) and := H H H . By induction, ≥ { 1 1} 1 ∩ 2 I 2 3 ··· h in H v, there exists a path P from u to some vertices in a , b , say to a (notice 1 − 1 { 1 1} 1 that a b may be on P ) such that ℓ(P ) 1 ((d 2.1) H /(d 1)2)r + 1 unless H is a 1 1 1 1 ≥ 4 − | 1| − 2 1 cycle and u / a , b (in this case, P may only contain one vertex). Moreover, ℓ(P ) ∈ { 1 1} 1 1 ≥ 1 ((d 2.1) H /(d 1)2)r + 1 when H is 3-connected. We will consider the case that H is 4 − | 1| − 1 1 a cycle and u a , b at the end. ∈{ 1 1} 114

Applying induction again we ﬁnd a path P in b from x to a such that 2 I − 1 1

1 h (d 2.1) H r 1 ℓ(P ) − | i| + . 2 ≥ 4 (d 1)2 2 i=2 X −

Moreover, when H2 is 3-connected,

1 h (d 2.1) H r ℓ(P ) − | i| +1. 2 ≥ 4 (d 1)2 i=2 X −

If a b E( ) or a b / E(P ), P := P P is the desired path. Thus, we may assume 1 1 ∈ H 1 1 ∈ 1 1 ∪ 2 that a b is a virtual edge in H , a b E(P ) and a b / E( ). 1 1 1 1 1 ∈ 1 1 1 ∈ H

If H is a cycle, then H must be 3-connected since a b / E( ). Let P in a from 1 2 1 1 ∈ H 2 I − 1 x to b1 such that 1 h (d 2.1) H r ℓ(P ) − | i| +1. 2 ≥ 4 (d 1)2 i=2 X − Then, P := (P a ) P satisfying 1 − 1 ∪ 2

1 h (d 2.1) H r 1 h (d 2.1) H r 1 ℓ(P ) ℓ(P ) − | i| +1 − | i| + , ≥ 2 ≥ 4 (d 1)2 ≥ 4 (d 1)2 2 i=2 i=1 X − X − so P is the desired path.

1 (d−2.1)|H1| r We may assume that H1 is 3-connected. In this case, we have ℓ(P1) 4 ( (d−1)2 ) + 1. ≥r Let P be an (x, b )-path in a such that ℓ(P ) 1 h (d−2.1)|Hi| + 1 . Moreover, in 2 1 I − 1 2 ≥ 4 i=2 (d−1)2 2 this case, we can ﬁnd the desired path if H2 is 3-connected.P So, we may additionally assume that H is a cycle. Since P is a (u, a )-path, a b E(P ), and ℓ(P ) > 1, u / a , b . 2 1 1 1 1 ∈ 1 1 ∈{ 1 1}

We now, under the assumption that H is 3-connected, H is a cycle, u / a , b , and 1 2 ∈{ 1 1} a b E(P ), construct a path P according to the following two cases. 1 1 ∈ 1

Suppose ﬁrst that h = 2. If H is a triangle, that is, V (H ) = a , b , x . Applying 2 2 { 1 1 } 115

Theorem (4.1.1)(a) to ( +ux) v, we obtain the desired (x, u)-path P in v. So we may H − H− assume that V (H ) 4, which in turn shows that H contains a path P with ℓ(P ) 2, | 2 | ≥ 2 2 2 ≥ which is either an (x, a1)-path or an (x, b1)-path. Assume, without not loss of generality, P2 is an (x, a )-path and Q is the (x, b )-path in H a b . Let Q bea(u, b )-path in H v 1 2 1 2 − 1 1 1 1 1 − such that ℓ(Q ) 1 ( (d−2.1)|H1| )r +1. If a b / E(Q ), then P := Q Q is the desired path. 1 ≥ 4 (d−1)2 1 1 ∈ 1 2 ∪ 1 If a b E(Q ), then P (Q a ) is the desired path since H 2d 1. 1 1 ∈ 1 2 ∪ 1 − 1 | 2| ≤ −

We now assume that h 3 and let a , b := V (H H ) and ′′ := H H H . ≥ { 2 2} 2 ∩ 3 H 3 4 ··· h Applying induction, there is a (u, a2)-path P1 in H1H2 avoiding v such that

1 2 (d 2.1) H r ℓ(P ) − | i| +1. 1 ≥ 4 (d 1)2 i=1 X −

If a b / E(P ), by the induction hypothesis, we ﬁnd an (a , x)-path P ′ in ′′ b such 2 2 ∈ 1 2 H − 2 that 1 h d 2.1 r 1 ℓ(P ′) − H + . ≥ 4 (d 1)2 | i| 2 i=3 X − Then P := P P ′ gives the desired path. Thus, we assume a b E(P ). If H is 3- 1 ∪ 2 2 ∈ 1 3 ′ ′′ ′ connected, then by induction there is a (b2, x)-path P avoiding a2 in such that ℓ(P ) r H ≥ 1 h (d−2.1)|Hi| + 1 in ′′. Hence, P := (P a ) P ′ gives the desired path. Thus, 4 i=3 (d−1)2 H 1 − 2 ∪ weP have a2b2 E(P1) and H3 is a cycle. Since both H2 and H3 are cycles, a2b2 E( ). ∈ r ∈ H We ﬁnd an (a , x)-path P ′ in ′′ b such that ℓ(P ′) 1 h (d−2.1)|Hi| + 1 . Then, 2 H − 2 ≥ 4 i=3 (d−1)2 2 P := P P ′ is the desired path. P 1 ∪

Let U and W be two vertex sets. By deﬁnition, an (U, W )-path P is a (u,w)-path for some u U and w W , and V (P ) U = 1 and V (P ) W = 1. We call P a path from U ∈ ∈ | ∩ | | ∩ | to W if P is a (u, v)-path from some u U and w W while V (P ) U or V (P ) W may ∈ ∈ ∩ ∩ not be singleton.

Lemma (4.4.3). Suppose that n 2 and Theorem (4.1.1) holds for graphs with less |H| ≤ − than n vertices. Let x V (H ) V (H ) such that d (x) d 1, f E(H ) E(H ) ∈ h − h−1 H ≤ − ∈ 1 − 2 116

h and pq E i=1Hi f . Then there exists a path P in from x to z p, q through ∈ ∪ −{ } r H ∈ { } f such that pq / E(P ) and ℓ(P ) 1 d−2.1 . ∈ ≥ 4 (d−1)2 |H|

Proof. We use induction on h and consider the base case h = 1 ﬁrst. In this case, if H1 is a cycle, then there exists a path P from x to p, q through f. Since H1 < 2d 1, r { } | | − ℓ(P ) 1 1 d−2.1 . We may assume that H is 3-connected and consider two cases ≥ ≥ 4 (d−1)2 |H| 1 according to whether x p, q . ∈{ }

If x / p, q , let H′ be the graph obtained from H by a T -transform at x, pq , and ∈ { } 1 1 { } let x′ be the new vertex. Since V (H′ ) V ( ) +1 n 1, we use Theorem (1.1)(b) to | 1 |≤| H | ≤ − ﬁnd a cycle C in H′ through xx′ and f such that ℓ(C) 1 ((d 2.1) H′ /(d 1)2)r +2. It 1 ≥ 4 − | 1| − is clear that C x′ gives a desired path. If x p, q , we use Theorem (4.1.1)(b) to ﬁnd a − ∈{ } cycle in H through pq and f. Then C pq is the desired path. 1 1 −{ }

Assume h 2. Let ′ = H H H and a , b := V (H ) V (H ). We consider the ≥ H 2 3 ··· h { 1 1} 1 ∩ 2 following three cases.

Case 1. pq / E(H ). We use Theorem (4.1.1)(b) to ﬁnd a cycle C in H through f and ∈ 1 1 a1b1 such that 1 (d 2.1) H r ℓ(C) − | 1| +2. ≥ 4 (d 1)2 − We apply induction to ﬁnd a path P ′ in ′ from x to p, q such that a b E(P ′), H { } 1 1 ∈ pq / E(P ′), and ∈ r 1 (d 2.1) ′ ℓ(P ′) − |H | . ≥ 4 (d 1)2 − Then (C a b ) P ′ is also the desired path. −{ 1 1} ∪

′ ′ Case 2. p, q = a1, b1 . We use Lemma (4.4.2) to ﬁnd a path P in q from x to p { } { } r H − avoiding q such that ℓ(P ′) 1 d−2.1 ′ + 1 and let C be the cycle in H as in Case 1. ≥ 4 (d−1)2 |H | 2 1 Then P := P ′ (C pq ) is the desired path. ∪ −{ } 117

Case 3. pq E(H ) a b . We assume, without loss of generality, that a / V ( H ). ∈ 1 −{ 1 1} 1 ∈ ∪i≥3 i By induction, there is a path P1 in H1 from a1 to p, q such that f E(P1), pq / E(P1), r { } ∈ ∈ and ℓ(P ) d−2.1 H . Since x V (H ) V (H ) and a , b = V (H ) V (H ), we 1 ≥ (d−1)2 | 1| ∈ h − h−1 { 1 1} 1 ∩ 2 have x / a , b . By Lemma (4.4.2), there is a path P ′ in ′ b from x to a such that ∈{ 1 1} H − 1 1

1 d 2.1 r 1 ℓ(P ′) − ′ + . ≥ 4 (d 1)2 |H | 2 −

Moreover, if H2 is 3-connected then

1 d 2.1 r ℓ(P ′) − ′ +1. ≥ 4 (d 1)2 |H | −

Hence P := P P ′ is the desired path in if a b / E(P ), a b E( ), or H is 3- 1 ∪ H 1 1 ∈ 1 1 1 ∈ H 2 connected. So, we assume a b E(P ), a b is a virtual edge in H , a b / E( ), and H 1 1 ∈ 1 1 1 1 1 1 ∈ H 2 is a cycle.

If h = 2 and V (H ) 4, then in H , we can ﬁnd an (x, a , b )-path P ′ such that | 2 | ≥ 2 { 1 1} ℓ(P ′) 2. If P ′ is an (x, b )-path, then P := (P a ) P ′ is the desired path by noting that ≥ 1 1 − 1 ∪ H 2d 1. So assume that P ′ is an (x, a )-path. In H , let P be a path from b to p, q | 2| ≤ − 1 1 1 1 { } such that f E(P ), pq / E(P ) and ℓ(P ) 1 ((d 2.1) H /(d 1)2)r. We may assume ∈ 1 ∈ 1 1 ≥ 4 − | 1| − that a b E(P ) (otherwise, let P ′ be an (x, b )-path in H a , then P := P P ′ is the 1 1 ∈ 1 1 2 − 1 1 ∪ desired path, as x V (H ) V (H ), ℓ(P ′) 1). Then P := (P b ) P ′ is the desired ∈ 2 − 1 ≥ 1 − 1 ∪ path. So we assume V (H ) =3 or V (H ) V (H ) = x . Let H∗ be the graph obtained | 2 | 2 − 1 { } from H H by a T -transform at x, pq , and let x′ be the new vertex. Since V (H∗) n 1, 1 2 { } | | ≤ − we can then apply Theorem (1.1)(b) to ﬁnd a cycle C in H∗ through xx′ and f such that ℓ(C) 1 ((d 2.1) H′ /(d 1)2)r + 2. It is clear that C x′ gives the desired path. ≥ 4 − | 1| − −

If h 3, let a , b := V (H ) V (H ) and ′′ := H H H . Assume, without loss ≥ { 2 2} 2 ∩ 3 H 3 4 ··· h of generality, that a V (H ) V (H ). Applying induction, there is a path P in H H 2 ∈ 2 − 1 1 1 2 118

from a to p, q such that f E(P ) and pq / E(P ), and 2 { } ∈ 1 ∈ 1

1 2 (d 2.1) H r ℓ(P ) − | i| . 1 ≥ 4 (d 1)2 i=1 X −

′ ′′ If a2b2 / E(P1), by Lemma (4.4.2), we ﬁnd an (a2, x)-path P in b2 such that ∈ r H − ℓ(P ′) 1 h (d−2.1)|Hi| + 1 . Then P := P P ′ gives the desired path. Thus, we ≥ 4 i=3 (d−1)2 2 1 ∪ ′ ′′ assume a2bP2 E(P1). If H3 is 3-connected, then there is a (b2, x)-path P in such that ∈ r H a / V (P ′) and ℓ(P ′) 1 h (d−2.1)|Hi| +1 by Lemma (4.4.2). Hence, P := (P a ) P ′ 2 ∈ ≥ 4 i=3 (d−1)2 1− 2 ∪ gives the desired path. Thus,P we have a b E(P ) and H is a cycle. Recall that H is a 2 2 ∈ 1 3 2 ′ cycle by our earlier assumption. We use Lemma (4.4.2) to ﬁnd an (a2, x)-path P of desired length in ′′ b . Let P := P P ′ (since a b E( ) in this case). Then P is the desired H − 2 1 ∪ 2 2 ∈ H path.

Lemma (4.4.4). Assume that Theorem (4.1.1) holds for graphs with less than n vertices. Let = H H ...H be a block-chain in G y such that < n, x V (H ) V (H ) H 1 2 h − |H| ∈ 1 − 2 with d (x) d 1, w V (H ) V (H ) x for some k with d (w) d 1, and let H ≤ − ∈ k − k−1 −{ } H ≤ − 1 m h be ﬁxed. Then there is a (w, x)-path P in such that ≤ ≤ H

max{k,m} 1 1 d 2.1 r ℓ(P ) H r + − H ; (4.3) ≥ 4| m| 4 (d 1)2 | i| i=1X,=6 m − particularly, when k = h,

1 1 d 2.1 r ℓ(P ) H r + − H . (4.4) ≥ 4| m| 4 (d 1)2 | i| Xi6=m −

Proof. Let V (H H )= a , b for i =1, 2,...,h 1 such that each H a b a b i ∩ i+1 { i i} − i − i i − i−1 i−1 contains two vertex-disjoint paths connecting ai−1 to ai and bi−1 to bi, respectively. We consider the following cases.

Case 1. m = k =1 119

If Hm is 3-connected, then by Theorem (4.1.1) (c), Hm + wx contains a cycle Cm through wx such that C 1 H r + 2. So, C wx is the desired path. If H is a cycle, the | m| ≥ 4 | m| m −{ } m (x, w)-path in H a b with length at least 1 is the desired path (as x = w). m − m m 6

Case 2. m =1 and k > 1

If Hm is 3-connected, we perform a T -transform on (x, ambm) and let z be the resulting new

vertex. Then, by Theorem (4.1.1) (c), there is a cycle Cm in the T -transformation through xz such that C 1 H r +2. Let P = C z. Clearly, a b E(P ) and ℓ(P ) 1 H r. | m| ≥ 4 | m| m m − m m 6∈ m m ≥ 4 | m| Assume P is from x to a . If H is a cycle, let P be a path from x to a , b , say to m m m m { m m} a , which has length at least 1. Let Q bea(w, a )-path in H H ...H b given by m m m+1 m+2 k − m Lemma (4.4.2). Then P := P Q is the desired path. m ∪

Case 3. m> 1 and k = m Then w H and w / a , b . If H is 3-connected, we do a T -transformation ∈ m ∈ { m−1 m−1} m on (w, a b ) and, use Thorem (4.1.1) (c) to obtain a path P from a , b , say m−1 m−1 m { m−1 m−1} a , to w with ℓ(P ) 1 H r; if H is a cycle, let P be a path from w to a , b , m−1 m ≥ 4 | m| m m { m−1 m−1} say am−1 of length at least 1. Let P1 be an (x, am−1)-path in H1 ...Hm−1 bm−1 with r − ℓ(P ) 1 d−2.1 H given by Lemma (4.4.2). Then, P := P P is the desired 1 ≥ 4 i

Case 4. m> 1 and k < m Applying Theorem (4.1.1)(c), we ﬁnd an (a , b )-path P in H with ℓ(P ) 1 H r + m−1 m−1 m m m ≥ 4 | m| 1. In case that a b E(P ) and a b / E(G), the edge a b on P is replaced by an m m ∈ m m m ∈ m m m (a , b )-path in H H H . m m m+1 m+2 ··· h

For each i with k

If w ak, bk , say w = ak, applying Lemma (4.4.2), we ﬁnd an (x, bk)-path P1 in ∈ { } r H ...H a with ℓ(P ) 1 d−2.1 H . Clearly, P ( m−1(P Q )) P gives 1 k − k 1 ≥ 4 i≤k (d−1)2 | i| 1 ∪ ∪i=k i ∪ i ∪ m the desired path. So assume wP / a , b . If H is 3-connected, we do a T -transformation ∈ { k k} k ′ on (w, ak−1bk−1) and let w denote the new vertex. Applying Theorem (4.1.1) (b), we ﬁnd a cycle C in H through w′w and a b such that ℓ(C ) 1 ( (d−2.1)|Hk| )r +2. Let P and k k k k k ≥ 4 (d−1)2 k Q be the two components of C w′, a b . Without loss of generality, we may assume k k −{ k k} that P is a (w, a )-path. Note that, ℓ(P )+ ℓ(Q ) 1 ( d−2.1 H )r 1. If H is a cycle, k k−1 k k ≥ 4 (d−1)2 | k| − k let P ′ be the path from w to a , b , say a , in H a b through a b . Let { k−1 k−1} k−1 k − k−1 k−1 k k P and Q be the two components of P ′ a b . Then ℓ(P )+ ℓ(Q ) 1 ( d−2.1 H )r as k k −{ k k} k k ≥ 4 (d−1)2 | k| w / a , b , a , b and H 2d 1. ∈{ k−1 k−1 k k} | k| ≤ −

Applying Lemma (4.4.2), we ﬁnd an (x, ak−1)-path P1 in H1 ...Hk−1 bk−1 with ℓ(P1) r − ≥ 1 d−2.1 H . Clearly, P ( m−1(P Q )) P gives the desired path. 4 im> 1

We start by ﬁnding a desired path in Hm and ﬁrst consider the case that Hm is 3-connected.

′ Let Hm be obtained by an H-transform of Hm over (am−1bm−1, ambm) and let cm−1 and cm be ′ new vertices. By Theorem (4.1.1) (c), we ﬁnd a cycle Cm in Hm through cm−1cm such that C 1 H r + 2. Then C c ,c gives a path P from a , b to a , b , | m| ≥ 4 | m| m −{ m m−1} m { m−1 m−1} { m m} say from a to b , such that ℓ(P ) 1 H r 1. If H is a cycle, let P be a nontrivial m−1 m m ≥ 4 | m| − m m path from a , b to a , b , say from a to b , not containing the edges a b { m−1 m−1} { m m} m−1 m m−1 m−1 and ambm.

Applying Lemma (4.4.2), we ﬁnd an (x, a )-path P in H H ...H b with m−1 1 1 2 m−1 − m−1 ℓ(P ) 1 ( d−2.1 H )r + 1 ; and ﬁnd a (b ,w)-path in H H ...H a with 1 ≥ 4 i

lemma, we will say H ab is a block-chain even in the case that H is a cycle. We include − this trivial case just for notational convenience.

Lemma (4.4.5). Let = H H ...H be a block-chain in G y such that n 1, H 1 2 h − |H| ≤ − and xx′ E(H ) E(H ) and y′ V (H ) V (H ) x, x′ . Suppose H = max H : ∈ 1 − 2 ∈ h − h−1 −{ } k {| i| H . Let a, b = V (H ) V (H ), where a = x′ and b = x if k = 1. Let H := i ∈ H} { } k ∩ k−1 0 H H H H be the block-chain H ab (when H is a cycle, H is a copy of K k1 k2 ··· kk0 ··· kk1 k − k ki 2 for each 1 i k ) such that a H , b H , and H = max H : 1 i k . ≤ ≤ 1 ∈ k1 ∈ kk1 | kk0 | {| ki| ≤ ≤ 1} Then there is a path P in x from x′ to y′ with H −

1 1 k0−1 d 2.1 r 1 h d 2.1 r 1 ℓ(P ) H r + − H + − H , (4.5) ≥ 4| kk0 | 4 (d 1)2 | ki| 4 (d 1)2 | i| − 2 i=1 X − i=Xk+1 −

and a path Q in from x to x′ with H

1 1 d 2.1 r 1 d 2.1 r ℓ(Q) H r + − H + − H ; (4.6) ≥ 4| kk0 | 4 (d 1)2 | ki| 4 (d 1)2 | i| iX6=k0 − Xi

moreover, if H1 is a cycle,

1 1 d 2.1 r 1 d 2.1 r 1 ℓ(Q) H r + − H + − H + . (4.7) ≥ 4| kk0 | 4 (d 1)2 | ki| 4 (d 1)2 | i| 2 iX6=k0 − Xi

Proof. We prove the ﬁrst statement ﬁrst.

Case 1. h = k =1

′ ′ ′ ′ If H1 is a cycle, then since y / x, x , we can ﬁnd an (x ,y )-path P in H1 x such that ∈ { } r − ℓ(P ) 1 1 H r + 1 k0−1 d−2.1 H (using H 2d 1). So assume H is ≥ ≥ 4 | kk0 | 4 i=1 (d−1)2 | ki| | 1| ≤ − 1 ′ ′ 3-connected, then apply LemmaP (4.4.4) on H0 = Hk ab, there is a path P from x to y r − such that ℓ(P ) 1 H r + 1 k0−1 d−2.1 H . ≥ 4 | kk0 | 4 i=1 (d−1)2 | ki| P Case 2. h> 1 and k < h 122

Let a , b = V (H H ). Suppose k = 1. If H is a cycle, then let P be a path in { k k} k ∩ k+1 1 1 H x from x′ to a , b , say to a , such that ℓ(P ) 1(notice that a b may be on P ). 1 − { k k} k 1 ≥ k k 1 If a b / E(P ), then let P be an (a ,y′)-path in H H H b given by Lemma (4.4.2) k k ∈ 1 2 k 2 3 ··· h − k such that ℓ(P ) 1 ( d−2.1 H )r + 1 . Then P := P P gives the desired path. 2 ≥ 4 i≥2 (d−1)2 | i| 2 1 ∪ 2 Hence we assume thatPa b E(P ). If H is also a cycle, then a b E(G). We let k k ∈ 1 2 k k ∈ P := P P as in the previous case. So assume H is 3-connected. Let P bea(b ,y′)-path 1 ∪ 2 2 2 k in H H H a given by Lemma (4.4.2) such that ℓ(P ) 1 ( d−2.1 H )r + 1. 2 3 ··· h − k 2 ≥ 4 i≥2 (d−1)2 | i| Then (P a b ) P gives the desired path. Suppose H is 3-connected.P Since a, b = 1 −{ k k} ∪ 2 1 { } V (H ) V (H ) and a , b = V (H ) V (H ), we have a, b = a , b . Assume that k ∩ k−1 { k k} k ∩ k+1 { } 6 { k k} a = ak. By applying Lemma (4.4.4) on H0 = H1 ab, there is a path P1 from a to ak such 6 r − that ℓ(P ) 1 H r + 1 k0−1 d−2.1 H . If a b / E(P ), let P be an (a ,y′)-path in 2 ≥ 4 | kk0 | 4 i=1 (d−1)2 | ki| k k ∈ 1 2 k H H H b given byP Lemma (4.4.2) such that ℓ(P ) 1 ( d−2.1 H )r + 1 . Then 2 3 ··· h − k 2 ≥ 4 i≥2 (d−1)2 | i| 2 P := P P gives the desired path. Hence we assume that a b PE(P ). Let P bea(b ,y′)- 1 ∪ 2 k k ∈ 1 2 k path in H H H a given by Lemma (4.4.2) such that ℓ(P ) 1 ( d−2.1 H )r + 1 . 2 3 ··· h − k 2 ≥ 4 i≥2 (d−1)2 | i| 2 Then (P a b ) P gives the desired path. P 1 −{ k k} ∪ 2

If k 2, let P be an (x′, a, b )-path, say (x′, a)-path, in H H H x given by ≥ 1 { } 1 2 ··· k−1 − Lemma (4.2.5). Again assume that a = ak. By applying Lemma (4.4.4) on H0 = Hk ab, 6 −r there is a path P from a to a such that ℓ(P ) 1 H r + 1 k0−1 d−2.1 H . If 2 k 2 ≥ 4 | kk0 | 4 i=1 (d−1)2 | ki| akbk / E(P2), then by Lemma (4.4.2) we ﬁnd a path P3 in Hk+1HPk+2 Hh bk from ak ∈ r ··· − to y′ such that ℓ(P ) 1 d−2.1 H + 1 . Then P := P P P is the desired 3 ≥ 4 i≥k+1 (d−1)2 | i| 2 1 ∪ 2 ∪ 3 path. Hence assume akbk PE(P2). By Lemma (4.4.2), let P3 in Hk+1Hk+2 Hh ak from ∈ r ··· − b to y′ such that ℓ(P ) 1 d−2.1 H + 1 . Then P := P (P a b ) P is k 3 ≥ 4 i≥k+1 (d−1)2 | i| 2 1 ∪ 2 − k k ∪ 3 the desired path. P

Case 3. h = k > 1 Since y′ V (H ) V (H ) and a, b = V (H ) V (H ), we have y′ a, b . Applying ∈ h − h−1 { } k ∩ k−1 6∈ { } Lemma (4.4.4) on H = H ab, we obtain an (a, y′)-path P such that ℓ(P ) 1 H r + 0 k − 2 2 ≥ 4 | kk0 | 123

r 1 k0−1 d−2.1 H . Then P := P P is the desired path. 4 i=1 (d−1)2 | ki| 1 ∪ 2 P For the second statement, we first apply Lemma (4.4.4) to H0 = Hk ab to find an(a, b)- r − 1 r 1 d−2.1 path Pk of length at least Hkk + 2 Hki (the virtual edge may contained 4 | 0 | 4 i6=k0 (d−1) | | in E(H H ) will be replaced by aP path in H H H ). Denote a , b := x, x′ . k ∩ k+1 k+1 k+2 ··· h { 0 0} { } Then for each 1 i k 1 we use Theorem (4.1.1) (b) to find a cycle C in H through ≤ ≤ − i i a b , and a b such that ℓ(C ) 1 ( (d−2.1)|Hi| )r +2. Let P and Q be the two components i−1 i−1 i i i ≥ 4 (d−1)2 i i of C a b , a b . Then Q := k−1(P Q ) P gives the desired path. i −{ i−1 i−1 i i} i=1 i ∪ i ∪ k S ′ Particularly, suppose H1 is a cycle. If k = 1, we can find an (x, x )-path Q in H1 such that ℓ(Q) 2 (the virtual edge may contained in E(H1 H2) will be replaced by a ≥ ∩ r 1 r 1 d−2.1 path in H2H2 Hh). Then ℓ(Q) 2 Hkk + 2 Hki +1/2. If k > 1, ··· ≥ ≥ 4 | 0 | 4 i6=k0 (d−1) | | then we apply Lemma (4.4.4) to H0 = Hk ab toP find an (a, b)-path Pk of length at r − 1 r 1 d−2.1 least Hkk + 2 Hki (the virtual edge may contained in E(Hk Hk+1) 4 | 0 | 4 i6=k0 (d−1) | | ∩ will be replaced byP a path in H H H ). Denote a , b := x, x′ . Then for each k+1 k+2 ··· h { 0 0} { } 1 i k 1 we use Theorem (4.1.1) (b) to find a cycle C in H through a b , ≤ ≤ − i i i−1 i−1 1 (d−2.1)|Hi| r and aibi such that ℓ(Ci) 4 ( (d−1)2 ) +2. Let Pi and Qi be the two components of ≥ r C a b , a b . As H is a cycle, in particular, ℓ(P )+ℓ(Q ) 1 1 d−2.1 H +1/2. i−{ i−1 i−1 i i} 1 1 1 ≥ ≥ 4 (d−1)2 | 1| Then Q := k−1(P Q ) P gives the desired path. i=1 i ∪ i ∪ k Lemma (4.4.6).S Assume that Theorem (4.1.1) holds for graphs with less than n vertices. Let G be a 3-connected graph with ∆(G) d , G < n and xy E(G). Suppose = ≤ | | ∈ H H H ...H and = L L ...L are two block-chains in G y such that (a) x (V (H ) 1 2 h L 1 2 ℓ − ∈ 1 − V (H )) (V (L ) V (L )); (b) xw E(H ) E(H ) and xw′ E(L ) E(L ); and 2 ∩ 1 − 2 ∈ 1 − 2 ∈ 1 − 2 (c) x = V ( ) V ( ) when w = w′, and x, w = V ( ) V ( ) otherwise. Let y′ { } H ∩ L 6 { } H ∩ L ∈ V (H ) V (H ) x, w and y′′ V (L ) V (L ) x, w′ . Then, provided that d 25, h − h−1 −{ } ∈ l − l−1 −{ } ≥ either there is a path P from w to y′ in x, and a path P from w′ to x in or H H − L L there is a path P from w to x in , and a path P from w′ to y′ in x such that H H L L − ℓ(P )+ ℓ(P ) 1 r + 1 r 1/2; moreover, if H is a cycle and L is a cycle, we can H L ≥ 4 |H| 4 |L| − 1 1 124

have ℓ(P )+ ℓ(P ) 1 r + 1 r. H L ≥ 4 |H| 4 |L|

Proof. Let 1 k h and 1 p ℓ such that H = max H : 1 i h and ≤ ≤ ≤ ≤ | k| {| i| ≤ ≤ } L = max L : 1 i ℓ . Let a, b = V (H ) V (H ), where a, b = x, w | p| {| i| ≤ ≤ } { } k ∩ k−1 { } { } when k = 1; and c,d = V (L ) V (L ), where c,d = x, w′ when p = 1. Let { } p ∩ p−1 { } { } H := H H H H be the block-chain H ab and L := L L L P 0 k1 k2 ··· kk0 ··· kk1 k − 0 p1 p2 ··· pp0 ··· pp1 be the block-chain L cd, such that (i) H = max H : H H and L = p − | kk0 | {| ki| ki ∈ 0} | pp0 | max L : L L , and (ii) a H , b H and c L ,d L be distinct. Denote {| pi| pi ∈ 0} ∈ k1 ∈ kk1 ∈ p1 ∈ pp1

r r h+ = d−2.1 H , h− = d−2.1 H ; • i>k (d−1)2 | i| ik0 (d−1) | | 0 ip (d−1)2 | i| i

p0 (d−1) | | 0 i

By symmetry between and (both H and L are cycles for the statement in the H L 1 1 “moreover” part), we may assume

h+ + l− + l+ l+ + h− + h+. 0 ≥ 0

Let P be a path in x from w to y′ given by Lemma (4.4.5) (see (4.5)) such that H H −

1 1 1 ℓ(P ) H r + h+ + h− 1/2, H ≥ 4| kk0 | 4 4 0 −

and P be a path in from w′ to x given by Lemma (4.4.5) (see (4.6)) such that L L

1 1 1 1 ℓ(P ) L r + l+ + l− + l−. L ≥ 4| pp0 | 4 0 4 0 4 125

In particular, if L1 is a cycle, then

1 1 1 1 ℓ(P ) L r + l+ + l− + l− +1/2. L ≥ 4| pp0 | 4 0 4 0 4

Since h+ + ℓ− + ℓ+ ℓ+ + h− + h+, 0 ≥ 0

h+ + l− + l+ 1/2(h+ + l− + l+)+1/2(l+ + h− + h+). 0 ≥ 0 0

Using (d 1)log2 5/4 1 d−1 when d 25 and xr + yr (x +(d 1)2((d 1)log2 5/4 1)y)r − − ≥ d−2.1 ≥ ≥ − − − if x (d 1)2y (equality (1d)), we have ≥ −

1 1 1 1 1 1 1 1 1 H r + ( )rh+ + h− + ( )rh− + ( )rh+ r, 4| kk0 | 4 d 1 4 0 4 d 1 4 d 1 0 ≥ 4|H| − − − and, 1 1 1 1 1 1 1 1 1 L r + ( )rl+ + l− + ( )rl− + ( )rl+ r. 4| pp0 | 4 d 1 4 0 4 d 1 4 d 1 0 ≥ 4|L| − − −

Hence,

ℓ(PH )+ ℓ(PL) 1 1 1 1 1 1 1 1 H r + h+ + h− + L r + l+ + l− + l− ≥ 4| kk0 | 4 4 0 4| pp0 | 4 0 4 0 4 − 2 1 1 1 1 1 1 1 1 = H r + h− + L r + l− +( h+ + l− + l+) 4| kk0 | 4 0 4| pp0 | 4 0 4 4 4 0 − 2 1 1 1 1 1 1 H r + h− + L r + l− + (h+ + ℓ− + ℓ+ + ℓ+ + h− + h+) ≥ 4| kk0 | 4 0 4| pp0 | 4 0 4 0 0 − 2 1 1 1 1 1 1 1 1 H r + h− + L r + ℓ− + ( )rh+ + ( )rh− + ≥ 4| kk0 | 4 0 4| pp0 | 4 0 4 d 1 4 d 1 0 1 1 1 1 1 1 − 1 1 − 1 + ( )rl+ + ( )rh+ + ( )rl+ + ( )rl−) since ( 1 )r = 1 ) 4 d 1 0 4 d 1 0 4 d 1 4 d 1 − 2 d−1 2 1 − 1 1 − 1 1 −1 1 − 1 = H r + ( )rh+ + h− + ( )rh− + ( )rh+ + 4| kk0 | 4 d 1 4 0 4 d 1 4 d 1 0 1 1 1− 1 1 1− 1 1 − 1 L r + ( )rl+ + l− + ( )rl− + ( )rl+ 4| pp0 | 4 d 1 4 0 4 d 1 4 d 1 0 − 2 1 1 −1 − − r + r ; ≥ 4|H| 4|L| − 2 126

and when both H and L are cycles, ℓ(P )+ ℓ(P ) 1 r + 1 r. 1 1 H L ≥ 4 |H| 4 |L|

4.5 Proofs of Theorem (4.1.1) (a) and (b)

The following two lemmas state that parts (a) and (b) of Theorem (4.1.1) can be reduced to Theorem (4.1.1) for smaller graphs. The proof of (a) is essentially the same as that in [12], but the proof of (b) needs more work.

Lemma (4.5.1). Let n 4 be an integer. If Theorem (4.1.1) holds for graphs with at most ≥ n 1 vertices, then Theorem (4.1.1)(a) holds for graphs with n vertices. −

Proof. Let G be an arbitrary 3-connected graph with n vertices, let xy E(G) and z ∈ ∈ V (G) x, y , and assume that ∆(G z) d. Let t denote the number of neighbors of z −{ } − ≤ in G x, y . Since G is 3-connected, t 1. −{ } ≥

Let = H ...H be a block-chain in G z such that xy E(H ) E(H ) and, subject H 1 h − ∈ 1 − 2 to this, is maximum. Therefore, H is an extreme 3-block of G z. Since each extreme |H| h − 3-block of G z must contain a neighbor of z, there are at most t 1 extreme 3-blocks of − − G z diﬀerent from H , and hence V (G z) is covered by at most t block-chains starting − h − from H and ending with an extreme 3-block of G z. It then follows that (n 1)/t. 1 − |H| ≥ −

Note that ∆(G z) d implies that ∆(H ) d for 1 i h. By Lemma (4.4.1), − ≤ i ≤ ≤ ≤ there is a path P in from x to y such that H

1 d 2.1 r 1 d 2.1 n r ℓ(P ) − ( + 1) +1 − +1. ≥ 4 d 1 |H| ≥ 4 d 1 · t − −

Then C := P + xy is a cycle through xy in G z with ℓ(C)= ℓ(P ) + 1, giving the desired − cycle.

Lemma (4.5.2). Let n 4 be an integer. Suppose Theorem (4.1.1) holds for graphs with ≥ 127 at most n 1 vertices. Then Theorem (4.1.1)(b) holds for graphs with n vertices. −

Proof. We note that by Lemma (4.5.1), Theorem (4.1.1)(a) holds for graphs with n vertices. Also note that Theorem (4.1.1)(b) holds trivially for cycles. So it suﬃces to show that Theorem (4.1.1)(b) holds for 3-connected graphs. Let G be a 3-connected graph with n vertices, let e = xy, f be two distinct edges of G, and assume ∆(G) d. ≤

Suppose e and f share a common vertex. Let f = yz. We note that G′ := G + xz is 3-connected, ∆(G′ y) d, and that y has at most d 2 neighbors distinct from x and z. − ≤ − By employing Theorem (4.1.1)(a) to G′, which has n vertices, there is a cycle C′ through xz in G′ y such that −

1 (d 2.1)n r 1 (d 2.1)n r ℓ(C′) − +2 − +2. ≥ 4 (d 1)(d 2) ≥ 4 (d 1)2 − − −

Then C := (C′ xz ) e, f gives a cycle through e and f such that −{ } ∪{ }

1 (d 2.1)n r ℓ(C) − +3. ≥ 4 (d 1)2 −

Therefore, we may assume that e and f are not adjacent. Let := H ...H be a H 1 h block-chain in G y such that x V (H ) V (H ) and f E(H ) E(H ). Note − ∈ 1 − 2 ∈ h − h−1 that the degree of x is at most d 1 in and ∆(H ) d for all 1 i h. Suppose − H i ≤ ≤ ≤ V ( ) = V (G y). If is a cycle, then every vertex of is adjacent to y. Let x′ be a H − H H neighbor of x in and P be the path in from x to x′ through f. Then P yx, yx′ is H H ∪{ } the desired cycle for Theorem (4.1.1)(b).

Now assume that is not a cycle. If H is a cycle, we choose x′ to be an endvertex H h of f which has degree 2 in ; otherwise, let x′ (V (H ) V (H )) N (y) such that x′ H ∈ h − h−1 ∩ G is incident with f whenever possible (for the choice of f ′ in the following). Let H′ be the graph obtained from by joining x to x′, and then suppressing all the remaining degree H 128

2 vertices. It is clear that H′ is 3-connected, H′ n 1 (d 1), and ∆(H′) d. | | ≥ − − − ≤ Let f ′ = f if f E(H′), otherwise let f ′ denote the new edge incident with x′ in H′. ∈ We use Theorem (4.1.1)(b) to ﬁnd a cycle C′ in H′ through xx′ and f ′ such that ℓ(C′) ≥ 1 ( d−2.1 H′ )r + 2. Then (C′ xx′ ) yx, yx′ (adding back the suppressed vertices if 4 (d−1)2 | | −{ } ∪{ } necessary) gives a cycle C in G through xy and f such that

1 d 2.1 r 1 d 2.1 r ℓ(C) − H′ +3 − n + 2 (by Lemma (4.3.2)). ≥ 4 (d 1)2 | | ≥ 4 (d 1)2 − −

So we may assume that V ( ) = V (G y). Then there is a 3-block B of G y such H 6 − − that V (B) V ( ) = 2. Let p, q := V (B) V ( ) and G be the graph obtained from G | ∩ H | { } ∩ H 1 by deleting those components of G y,p,q containing a vertex of . We choose p, q so −{ } H { } that G is maximum. Then, | 1| (d 1) G + n. (4.8) − | 1| |H| ≥

If V (G)= V (G ), we let G = . Otherwise, there is a 3-block B′ of G y such that 1 ∪H 2 ∅ − V (B′) V ( G )= v,w for some v,w = p, q . (Note that v,w V ( ).) Deﬁne ∩ H ∪ 1 { } { } 6 { } { } ⊆ H G as the graph obtained from G by deleting those components of G y,v,w containing 2 −{ } a vertex of G . We choose G such that G is maximum. Then 1 ∪ H 2 | 2|

(d 2) G + G + n. (4.9) − | 2| | 1| |H| ≥

Clearly, G G . Let G′ be the graph obtained from G by adding the edges yp,yq, | 1|≥| 2| 1 1 ′ ′ ′ and pq if they are not already in G1. Deﬁne G2 similarly from G2. We note that G1 and G2 (if nonempty) are both 3-connected. We shall ﬁnd the desired cycle for Theorem (4.1.1)(b) by combining long paths in the two largest graphs among , G , and G . Let t := N(y) H 1 2 i | ∩ V (G ) ( p, q v,w ) for i =1, 2, respectively. We divide the remaining proof into two i − { }∪{ } | cases. 129

4.5.1 Case 1: t 2 or t 2. 1 ≥ 2 ≥

In this case, inequalities (4.8) and (4.9) can be improved (by exactly the same reasons) to

(d 2) G + n and (4.10) − | 1| |H| ≥

(d 3) G + G + n. (4.11) − | 2| | 1| |H| ≥

Suppose G . Then from (4.11), we have |H| ≥ | 2|

G +(d 2) n. (4.12) | 1| − |H| ≥

If pq = f, we use Lemma (4.4.3) to ﬁnd a path P in from x to z p, q , say z = p, 6 H ∈ { } through f such that pq / E(P ), and ∈

1 (d 2.1) r ℓ(P ) − |H| . ≥ 4 (d 1)2 −

Since e = xy is not adjacent to f, x p, q . Hence if pq = f, we can apply Lemma (4.4.2) 6∈ { } to ﬁnd a path P ′ in p from x to q such that H −

1 (d 2.1) r 1 ℓ(P ′) − |H| + , ≥ 4 (d 1)2 2 − and set P := P ′ pq in this case. Since ∆(G′ q) d, we may use Theorem (4.1.1)(a) ∪{ } 1 − ≤ 130

to ﬁnd a cycle C through py in G′ q such that 1 1 −

1 d 2.1 r ℓ(C ) − G +2. 1 ≥ 4 (d 1)2 | 1| −

Then P (C py ) xy gives a cycle C in G through xy and f such that ∪ 1 −{ } ∪{ }

1 (d 2.1) G r 1 (d 2.1) r ℓ(C) − | 1| + − |H| +2 ≥ 4 (d 1)2 4 (d 1)2 − − r 1 d−2.1 4 (d−1)2 ((d 2) G1 + ) +2, if G1 ; − | | |H| r |H| ≥ | | (by (1a)) ≥  1 d−2.1  2 ( G1 +(d 2) ) +2, if < G1 ;  4 (d−1) | | − |H| |H| | | 1 (d 2.1)n r  − + 2 (by (4.10) and (4.12)). ≥ 4 (d 1)2 −

Now assume < G , and hence G = . Let P be a path in from x to z |H| | 2| 2 6 ∅ 1 H ∈ p, q v,w through f as given by Lemma (4.2.1) such that (i) exactly one of pq and vw { }∪{ } is in E(P ); (ii) if pq E(P ) then z v,w ; and (iii) if vw E(P ) then z p, q . 1 ∈ 1 ∈ { } ∈ 1 ∈ { } Assume, without loss of generality, that pq E(P ) and z = v. Let P be a (p, q)-path ∈ 1 2 ′ ′ in G1 y given by Theorem (4.1.1)(a), and let P3 be a (v,y)-path in G2 w given by − r − r Theorem (4.1.1)(a). Then ℓ(P ) 1 (d−2.1) G′ + 1 and ℓ(P ) 1 (d−2.1) G + 1. 2 ≥ 4 (d−1)2 | 1| 3 ≥ 4 (d−1)2 | 2| Now C := (P pq ) P P xy is a cycle through xy and f in Gsuch that 1 −{ } ∪ 2 ∪ 3 ∪{ }

1 (d 2.1) G r 1 (d 2.1) G r ℓ(C) − | 1| + − | 2| +2 ≥ 4 (d 1)2 4 (d 1)2 − − 1 d 2.1 r − ( G +(d 2) G ) +2 (by (1a) and G G ) ≥ 4 (d 1)2 | 1| − | 2| | 1|≥| 2| − 1 d 2.1 r − ( G +(d 3) G + ) + 2 (since G > ) ≥ 4 (d 1)2 | 1| − | 2| |H| | 2| |H| − 1 (d 2.1)n r − +2 (by (10)) . ≥ 4 (d 1)2 − 131

4.5.2 Case 2. t1 = t2 =1.

Let P be a path in from x to z p, q v,w through f as given by Lemma (4.2.1) H ∈{ }∪{ } such that if pq E(P ) then z v,w and if vw E(P ) then z p, q . ∈ ∈{ } ∈ ∈{ }

Suppose (d 3) G . If vw E(P ) and z p, q , say z = p, then let P be a |H| ≤ − | 2| ∈ ∈ { } 1 longest path in G′ q from p to y and P a longest (v,w)-path in G′ y. By Theorem (4.1.1) 1 − 2 2 − (a) for P and Lemma (4.4.1) for P (using t = 1, in this case G′ y is a block-chain), we 1 2 2 2 − have the following lower bounds for ℓ(P1) and ℓ(P2).

1 (d 2.1) G r ℓ(P ) − | 1| +2, (4.13) 1 ≥ 4 (d 1)2 − 1 (d 2.1) G r ℓ(P ) − | 2| +1. (4.14) 2 ≥ 4 d 1 −

Let C := (P vw ) P P xy . Then C is a cycle in G though xy and f such −{ } ∪ 1 ∪ 2 ∪{ } that

1 (d 2.1) r 1 (d 2.1) r − ((d 2) G +(d 1) G ) +3 − n +3, if (d 1) G G 4 (d 1)2 − | 1| − | 2| ≥ 4 (d 1)2 − | 2|≥| 1| ℓ(C)  − r − r ≥  1 (d 2.1) 1 (d 2.1)  − ( G1 +(d 2)(d 1) G2 ) +3 − n +3, if (d 1) G2 < G1 ; 4 (d 1)2 | | − − | | ≥ 4 (d 1)2 − | | | | − −   If pq E(P ), z v,w , say z = w, let P be a longest path in G′ v from w to y, and ∈ ∈ { } 2 2 − P be a longest (p, q)-path in G′ y. Using Theorem (4.1.1) (a) for P and Lemmas (4.4.1) 1 1 − 2 for P (using t = 1, then in this case G′ y is a block-chain), we have the following lower 1 1 1 − bounds for ℓ(P1) and ℓ(P2).

1 (d 2.1) G r ℓ(P ) − | 1| + 1 (4.16) 1 ≥ 4 d 1 − 1 (d 2.1) G r ℓ(P ) − | 2| +2. (4.17) 2 ≥ 4 (d 1)2 − 132

Let C := (P pq ) P P xy . Since (d 2) G , by (7), (d 1) G + −{ } ∪ 1 ∪ 2 ∪{ } − | 2| ≥ |H| − | 1| (d 2) G n. Then as (d 1) G G always holds, by (1a) we have − | 2| ≥ − | 1|≥| 2|

1 (d 2.1) r 1 (d 2.1) r ℓ(C) − ((d 1) G +(d 2) G ) +3 − n +3. ≥ 4 (d 1)2 − | 1| − | 2| ≥ 4 (d 1)2 − −

So we may assume > (d 3) G . If pq = f, we use Lemma (4.4.3) to ﬁnd a path |H| − | 2| 6 P in from x to z p, q , say z = p, through f such that pq / E(P ) and H ∈{ } ∈

1 (d 2.1) r ℓ(P ) − |H| . ≥ 4 (d 1)2 −

Again x p, q . Hence if pq = f, we can apply Lemma (4.4.2) to ﬁnd a path P ′ in p 6∈ { } H − from x to q such that 1 (d 2.1) r ℓ(P ′) − |H| , ≥ 4 (d 1)2 − and set P := P ′ pq . Since ∆(G′ q) d, we may use Theorem (4.1.1)(a) to ﬁnd a cycle ∪{ } 1 − ≤ C through py in G′ q such that 1 1 −

1 d 2.1 r ℓ(C ) − G +2. 1 ≥ 4 (d 1)2 | 1| −

Then P (C py ) xy gives a cycle C in G through xy and f. If (d 4) G , ∪ 1 −{ } ∪{ } |H| ≥ − | 1| then by inequality (1c),

1 d 2.1 r C − ( +(d 4)((d 1)log2 3/2 1) G ) +2 | | ≥ 4 (d 1)2 |H| − − − | 1| − 1 (d 2.1) r − ( +(d 1) G ) +2 when d 8, (d 1)log2(3/2) 3 and 2(d 4) (d 1) ≥ 4 (d 1)2 |H| − | 1| ≥ − ≥ − ≥ − − 1 (d 2.1) r − n +2 (by (4.8)) . ≥ 4 (d 1)2 −

So (d 3) G < < (d 4) G . Then from (4.8) and (4.9), we have − | 2| |H| − | 1|

G +(d 2) n and (d 2) G + G +(d 2) G + n, | 1| − |H| ≥ − | 1| |H| ≥ | 1| − | 2| |H| ≥ 133

where the second inequality follows from (d 3) G > (d 2) G as (d 4) G > (d 3) G . − | 1| − | 2| − | 1| − | 2| By using the inequalities,

G r + r min ( G +(d 2) )r, ((d 2) G + )r , | 1| |H| ≥ { | 1| − |H| − | 1| |H| } we obtain that ℓ(C) 1 ((d 2.1)n/(d 1)2)r + 2. ≥ 4 − −

4.6 Reduction of Theorem (4.1.1)(c)

In this section, we prove the following result which reduces Theorem (4.1.1)(c) to The- orem (4.1.1) for smaller graphs. The part of proof is long and tedious, but contains a few crucial new ideas in estimating the lower bound of special paths.

Lemma (4.6.1). Let n 4 be an integer. If Theorem (4.1.1) holds for graphs with at most ≥ n 1 vertices, then Theorem (4.1.1)(c) holds for graphs with n vertices. −

To prove Lemma (4.6.1), let G be a 3-connected graph with n vertices and ∆(G) d, ≤ and let xy E(G). It is easy to see that when n 5, G contains a cycle through xy of length ∈ ≥ r at least 5 = 1 (d 1)log2 6 + 2. Hence, Theorem (4.1.1)(c) holds when n (d 1)log2 6. So 2 − ≤ − we assume n> (d 1)log2 6 hereafter. −

Let := H H H be a block-chain in G y such that x V (H ) V (H ) and H 1 2 ··· h − ∈ 1 − 2 subject to this, is maximum. We note that may contain only one block H . In |H| H 1 this case, all 3-blocks attached to H1 contain x and H1 may not be an extreme block. However, when h 2, H must be an extreme 3-block in G y and there is a vertex ≥ h − x′ (V (H ) V (H )) N (y). ∈ h − h−1 ∩ G Claim (4.6.0.1). We may assume V (G y) = V ( ). − 6 H

Proof. Suppose V (G y) = V ( ). Since G is 3-connected, there exists a vertex x′ − H ∈ 134

(V (H ) V (H )) (N (y) x). Let H′ be obtained from H by joining x′ and x and then h − h−1 ∩ G − suppressing all remaining degree 2 vertices. Clearly, H′ is 3-connected with ∆(H′) d and ≤ n> H′ (n 1) (d 2) = n d +1. Let C′ be a longest cycle in H′ through xx′. By | | ≥ − − − − Theorem (4.1.1) (c), we have C′ 1 H′ r +2. Let C =(C′ xx′ ) xy, x′y . Then by | | ≥ 4 | | −{ } ∪{ } Lemma (4.3.2), we have C C′ +1 1 (n d + 1)r +3 1 nr +2, so C is the desired | |≥| | ≥ 4 − ≥ 4 cycle.

A block-chain := L L ...L diﬀerent from is called an -leg if L L L 1 2 ℓ H H H∩L⊆ 1 − 2 and L is an extreme block. Note that, for each extreme block L not in , there is a unique ℓ H -leg containing L. H

Since = G y and G is 3-connected, there are -legs. Let := L L ...L be an -leg H 6 − H L 1 2 ℓ H with maximum. Suppose further that V ( ) V ( )= V (H ) V (H ) V (L )= p, q |L| H ∩ L t − t−1 ∩ 1 { } for some 1 t h. Since each -leg contains an extreme block and each extreme block ≤ ≤ H contains a neighbor of y, there are at most d 1 -legs. Hence, (d 1)( 2)+ n 1, − H − |L|− |H| ≥ − that is, (d 1) + n +2d 3. (4.18) − |L| |H| ≥ −

We will use the following parameters (which approach 0 as d ): →∞

d 1 1 ǫ1 := − , ǫ2 := . (d 2.1)((d 1)log2(3/2) 1) (d 1)log2(5/4) 1 − − − − −

Claim (4.6.0.2). We may assume (ǫ + ǫ )n. |H| ≤ 1 2

Proof. Suppose > (ǫ + ǫ )n. Let H such that H is maximum, ′ := |H| 1 2 m ∈ H | m| H H H ...H and ′′ := H H ...H . For each 2 i h, let a , b = 1 2 m−1 H m+1 m+2 h ≤ ≤ { i i} V (H ) V (H ). i ∩ i−1

If the vertex x′ (V (H ) V (H )) (N (y) x) is well deﬁned, we let P be a ∈ h − h−1 ∩ G − longest (x, x′)-path in and C := P xy, x′y . Clearly, C is a cycle containing edge xy. H ∪{ } 135

We will show that C is the desired cycle by estimating lower bounds of C in diﬀerent cases | | accordingly.

Suppose ′ + ′′ ǫ n > 0. Then, h 2 and H is an extreme block of G y, so |H | |H | ≥ 1 ≥ h − the vertex x′ (V (H ) V (H )) (N (y) x) is well deﬁned. Applying Lemma (4.4.4) ∈ h − h−1 ∩ G − (see (4)), we obtain a lower bound of C below. | |

1 1 d 2.1 r C H r + − H +2 | | ≥ 4| m| 4 (d 1)2 | i| Xi6=m − 1 d 2.1 r H + (d 1)log2(3/2) 1 − ( ′ + ′′ ) + 2 (by H H and (4.1c)) ≥ 4 | m| − − d 1 |H | |H | | m|≥| i| − 1 d 2.1 r − (d 1)log2(3/2) 1 ( ′ + ′′ ) +2 ≥ 4 d 1 − − |H | |H | − 1 nr +2. ≥ 4

Thus, we assume ′ + ′′ < ǫ n. Then, H > ǫ n as > (ǫ + ǫ )n. |H | |H | 1 | m| 2 |H| 1 2

We distinguish two cases by considering which one is bigger between and ′′ . |L| |H | 2 Suppose ﬁrst that ′′ . Then, using H H = (d−1) d−2.1 H for each 1 i |L| ≤ |H | | m|≥| i| d−2.1 (d−1)2 | i| ≤ ≤ m 1 and (4.1d), we have a lower bound of C below. − | |

′′ r 1 r 1 (d 2.1) C H + (d 1)log2(5/4) 1 ′ + − |H | +2. | | ≥ 4 | m| − − |H | 4 (d 1)2 − ′′ log (5/4) ′ If (d−2.1)|H | |Hm|+((d−1) 2 −1)|H | , then using inequality (4.1f) of Lemma (4.3.1) and the (d−1)2 ≥ (d−1)2 inequality (d 1)log2(5/4) 1 , − ≥ ǫ2

1 r 1 C (d 1)log2(5/4)( H + ((d 1)log2(5/4) 1) ′ ) +2 nr +2. | | ≥ 4 − | m| − − |H | ≥ 4 log (5/4) ′ (d−2.1)|H′′| |Hm|+((d−1) 2 −1)|H | Thus we may assume (d−1)2 < (d−1)2 . Using inequality (4.1d) in 136

Lemma (4.3.1), by noting that (d 1)log2(5/4) > 2 if d> 10, we have −

1 (d 1)2(d 2.1) ′′ r C H + (d 1)log2(5/4) 1 ′ + (d 1)log2(5/4) 1 − − |H | +2, | | ≥ 4 | m| − − |H | − − (d 1)2 − 1 r H + ′ + ′′ +(d 4)((d 1)log2(5/4) 1) ′′ +2 ≥ 4 | m| |H | |H | − − − |H | 1 r ( +(d 1) ) +2 since ′′ and (d 4)((d 1)log2(5/4) 1) d 1 ≥ 4 |H| − |L| |H | ≥ |L| − − − ≥ − 1 nr +2 (by (4.18)). ≥ 4

We now consider the case > ′′ . Since is maximum subject to x V (H ) |L| |H | |H| ∈ 1 − V (H ), we have either m 2 and 1 t m 1 or m = t = 1 and x p, q . 2 ≥ ≤ ≤ − ∈{ }

If m 2, let P be a path in H between a and b as given by Theorem (4.1.1)(c). ≥ m m m−1 m−1 If a , b = p, q , let P ′ be a path in ′ through a b from x to p, q as given { m−1 m−1} 6 { } H m−1 m−1 { } by Lemma (4.4.3), and let the notation be chosen so that P ′ is from x to p; otherwise, a , b = p, q , let P ′′ be a path in ′ p from x to q given by Lemma (4.4.2), and { m−1 m−1} { } H − let P ′ := P ′′ pq . Let P be a path in q from p to y′ (V (L ) V (L )) N (y) ∪{ } L L − ∈ l − l−1 ∩ G as given by Lemma (4.4.2). Then C := P P ′ P yx,yy′ a b is a cycle L ∪ ∪ m ∪{ }−{ m−1 m−1} through xy in G such that

1 1 (d 2.1) ′ r 1 (d 2.1) r C H r + − |H | + − |L| +2. | | ≥ 4| m| 4 (d 1)2 4 (d 1)2 − −

′ By the same argument as above and using (4.1d) and (4.1f) depending on whether (d−2.1)|H | (d−1)2 ≥ |Hm| (d−1)2 , we have

1 r 1 (d 2.1) r 1 C (d 1)log2(5/4) H + − |L| +2 nr +2 or | | ≥ 4 − | m| 4 (d 1)2 ≥ 4 − 137

1 (d 2.1) ′ r 1 (d 2.1) r C H +(d 1)2 (d 1)log2(5/4) 1 − |H | + − |L| +2 | | ≥ 4 | m| − − − (d 1)2 4 (d 1)2 − − 1 r 1 (d 2.1) r H + ′ + − |L| +2. ≥ 4 | m| |H | 4 (d 1)2 −

If H + ′ (d 1)2 (d−2.1)|L| , then | m| |H | ≥ − (d−1)2

1 r 1 (d 2.1) r H + ′ + − |L| +2 4 | m| |H | 4 (d 1)2 − 1 (d 2.1) r H + ′ +(d 1)2((d 1)log2(5/4) 1) − |L| +2 ≥ 4 | m| |H | − − − (d 1)2 − 1 H + ′ + ′′ +(d 1) )r + 2 (since > ′′ ) ≥ 4 | m| |H | |H | − |L| |L| |H | 1 nr +2. ≥ 4

Otherwise, H + ′ < (d 1)2 (d−2.1)|L| , then we get | m| |H | − (d−1)2

1 r 1 (d 2.1) r H + ′ + − |L| +2 4 | m| |H | 4 (d 1)2 − 1 r (d 1)log2(5/4) 1)( H + ′ ) +2 ≥ 4 − − | m| |H | 1 nr + 2 (since H > ǫ n). ≥ 4 | m| 2

We now assume m = t = 1 and, without loss of generality, x = p. Let Pm be a longest path from x to q in given by Lemma (4.4.4) and P a longest path in p from q to H L L − ′ ′ y (V (Ll) V (Ll−1)) NG(y) given by Lemma (4.4.2). Then C := Pm PL xy,y y ∈ − ∩ r ∪ ∪{ } is a cycle of length ℓ(C) 1 H r + 1 (d−2.1)|L| +2 1 nr + 2, where the last inequality ≥ 4 | m| 2 (d−1)2 ≥ 4 follows from a similar argument as above for m 2 and > ′′ . ≥ |L| |H |

= p, q if x / p, q An -leg is called a minor-leg of if V ( ) 6 { } ∈{ } ; or there H M H M ∩ H   x if x p, q 6∋ ∈{ } ∗ ∗ ∗ is another -leg such that both and intersect on p, q , V ( ) ( ) = p, q , H L M L H { } L ∩ M 6 { } and ∗ ǫ n/(d 2.1). We call the minor-leg deﬁned in the ﬁrst case an A-type |L −M| ≤ 2 − M 138

minor-leg; and in the later case a B-type minor-leg.

Note that if has an A-type minor-leg, then h 2. For an A-type minor-leg of , H ≥ M H we have the following claim. ǫ n Claim (4.6.0.3). If := M M is an A-type minor-leg of , then 2 . M 1 ··· m H |M| ≤ d 2.1 −

Proof. Suppose > ǫ2n . By our choice of H , t is the smallest positive integer such that |M| d−2.1 t p, q H . Similarly, let s be the smallest integer such that V ( ) V ( )= V ( ) V (H ). { } ⊆ t M ∩ H M ∩ s Let u, v = V (H ) V ( ) and g := max s, t . Moreover, let ′ = H H H and { } s ∩ M { } H 1 2 ··· g ′′ = H H H . By the maximality of and and the existence of , we H g+1 g+2 ··· h |H| |L| M have g < h, ′′ , and . Recall V (H ) V (H ) = a , b . Since |H | ≥ |M| |L| ≥ |M| g ∩ g+1 { g g} x V (H ) V (H ) and a , b V (H ) V (H ), we know x is not incident to a b . Hence, ∈ 1 − 2 g g ∈ g ∩ g+1 g g we can apply Lemma (4.2.1) to ﬁnd a path P ′ in ′ through a b from x to z p, q u, v . H g g ∈{ }∪{ } Let P ′′ be a path in ′′ between a and b as given by Lemma (4.4.1) such that ℓ(P ′′) H g g ≥ 1 ( d−2.1 ( ′′ + 1))r +1. If z u, v (say, z = u), let P be a path in v from u to 4 d−1 |H | ∈ { } M M − y′ N (y) (V (M ) V (M ) u, v ) (the vertex y′ exists by the 3-connectivity of G) ∈ G ∩ m − m−1 −{ } as given by Lemma (4.4.2) with ℓ(P ) 1 ( d−2.1 )r + 1 , and P be a path in between M ≥ 4 (d−1)2 |M| 2 L L p and q as given by Lemma (4.4.1) with ℓ(P ) 1 ( d−2.1 ( + 1))r + 1. The case z p, q L ≥ 4 d−1 |L| ∈{ } is treated similarly. Since and ′′ , the paths P ′,P ′′,P ,P , and edges |M| ≤ |L| |M| ≤ |H | L M yx,yy′ give rise to a cycle C in G through xy such that

1 d 2.1 r 1 d 2.1 r 1 d 2.1 r C − + − + − + 2 (4.19) | | ≥ 4 d 1 |M| 4 d 1 |M| 4 (d 1)2 |M| − − − 1 1 d 2.1 r = ((d 2.1) )r + − + 2 (4.20) 4 − |M| 4 (d 1)2 |M| − 1 r (d 1)log2(5/4)(d 2.1) ((1f) in Lemma (4.3.1)) ≥ 4 − − |M| 1 nr +2. ≥ 4 139

Recall that t is the minimum positive integer such that p, q V (H ). Let := { } ⊆ t I H H H and := H H H . We have a similar claim for a B-type -minor-leg. 1 2 ··· t J t+1 t+2 ··· h H Claim (4.6.0.4). Let ∗ and ∗∗ be two -legs with attachments p, q . If ∗ ∗∗ = p, q , L L H { } L ∩L 6 { } then we may assume that one of ∗ and ∗∗ is a B-type -minor-leg. L L H

Proof. Assume, without loss of generality, that ∗ ∗∗ . Let = ∗ ∗∗. Let |L | ≤ |L | L0 L ∩ L u∗, v∗ = V ( ) V (L∗) and u∗∗, v∗∗ = V ( ) V (L∗∗). { } L0 ∩ 1 { } L0 ∩ 1

∗ ∗∗ ∗∗ By Lemma (4.2.3), we may assume that there is a (p,u )-path P0 through edge u v in , but not the edge u∗v∗. Let ω∗ = ∗ +2. Let P be a path in p from x to L0 |L −L0| I I − q such that ℓ(P ) 1 ( (d−2.1)|I| )r given by Lemma (4.4.2), P a (q,p)-path in such that I ≥ 4 (d−1)2 J J ∗ ℓ(P ) 1 ( (d−2.1)|J| )r 1 ( (d−2.1)ω )r given by Lemma (4.4.1) (note that ω∗), J ≥ 4 d−1 ≥ 4 d−1 |J| ≥ |L| ≥ P ∗∗ be a path in ∗∗ ( u∗∗, v∗∗ ) from u∗∗ to v∗∗ given by Lemma (4.4.1) with L − L0 − { } ∗∗ ℓ(P ∗∗) 1 ( (d−2.1)|L | )r + 1, and P ∗ a path in G[ ∗ ( u∗, v∗ )] from x′ to u∗ avoiding ≥ 4 (d−1) L − L0 −{ } v∗ given by Lemma (4.4.2), where x′ is a neighbor of y in the last block of ∗. Then we L obtain a cycle C := P ∗ (P u∗∗v∗∗ ) P ∗∗ P P yx′, xy through xy such that ∪ 0 −{ } ∪ ∪ J ∪ I ∪{ }

1 (d 2.1)ω∗ (d 2.1)ω∗ 1 ℓ(C) 2( − )r +( − )r +2 ((d 2.1)(d 1)log2 5/4ω∗)r +2, ≥ 4 d 1 (d 1)2 ≥ 4 − − − −

where the last inequality follows from (4.1f). Noticing ǫ = 1 , we have (d 2 (d−1)log2(5/4)−1 − 2.1)(d 1)log2 5/4ω∗ n if ω∗ ǫ n/(d 2.1). So, we may assume ω∗ < ǫ n/(d 2.1). − ≥ ≥ 2 − 2 −

Let G be the subgraph of G obtained by deleting the components of G y,p,q that 0 −{ } contain a vertex in . By adding a few special edges to G , we deﬁne G′ as follows: H 0 0

G0 py,qy if Ht is a cycle and p, q = at, bt ; ′ ∪{ } { } 6 { } G0 :=   G0 y ] py,qy,pq if pq / E(G) and the above case false. ∪{ } ∪{ } ∈   Note that the diﬀerence is whether the edge pq is forced to be added. 140

Suppose that there are exactly ς -minor-legs. Then 0 ς d 3(as y is adjacent to H ≤ ≤ − at least two vertices in and at least one vertex in , there are at most d 3 neighbors of H L − y contained in -minor-legs ). Let be one of the largest minor-legs if there is one. Then, H M the following inequalities hold.

ςǫ n G n ς n 2 (1 ǫ 2ǫ )n, | 0| ≥ −|H|− |M| ≥ −|H|− d 2.1 ≥ − 1 − 2 n ς − − |M|, and |H| ≥ d 1 ς − − . |J| ≥ |L|≥|M|

Since < (ǫ + ǫ )n, we have n−|H| > (1−ǫ1−ǫ2)n . To complete our proof of |H| 1 2 |L| ≥ d−1 d−1 Lemma (4.6.1), we consider two cases according to whether x p, q or not. ∈{ }

4.6.1 Case 1 x / p, q . ∈{ }

In this case, by the maximality of , we have and 1 t h 1. |H| |H| ≥ |J| ≥ |L| ≤ ≤ − Consequently, we have h 2 and the vertex x′ (V (H ) V (H )) (N (y) x) is well ≥ ∈ h − h−1 ∩ G − deﬁned.

Claim (4.6.1.1). ∆(G′ ) d and G′ is 3-connected. 0 ≤ 0

Proof. Since dG′ (v) = dG(v) for every v V (G0) y,p,q , we only need to verify that 0 ∈ −{ } degrees dG′ (p), dG′ (q), and dG′ (y) are not bigger than d. Since > 0, Ht+1 exists. 0 0 0 |J| ≥ |L| Then both p and q have at least two neighbors in G V (G0), and thus dG′ (p) d and − 0 ≤ ′ dG′ (q) d. Furthermore, dG′ (y) dG(y)+ p, q x, x d. 0 ≤ 0 ≤ |{ }|−|{ }| ≤

For the connectivity, it is clear that if there exist at least three internally vertex-disjoint

′ (p, q)-path, then G0 is 3-connected. As G0 is connected, there is a (p, q)-path using only vertices of G ; pyq is another (p, q)-path which intersects V (G ) only on p, q . If pq E(G′ ), 0 0 { } ∈ 0 the edge pq gives the third (p, q)-path. Hence G′ is 3-connected if pq E(G′ ). So, we only 0 ∈ 0 141

need to show that G′ is 3-connected when G′ = G yp,yq and pq / E(G′ ). We suppose 0 0 0 ∪{ } ∈ 0 ′ ′ on the contrary that G0 has exactly two internally vertex-disjoint (p, q)-paths (as G0 + pq

is 3-connected). As y connecting p and q, G0 contains exactly one (p, q)-path. Denote by P [p, q] a shortest (p,q)-path in G . Then in G y, (H pq) P [p, q] is an induced cycle. 0 − t − ∪ According to Tutte’s decomposition algorithm, (H pq) P [p, q] forms a 3-block. This gives t − ∪ a contradiction to that Ht is a 3-block.

Claim (4.6.1.2). There is a path P in G′ with two endvertices in y,p,q such that ℓ(P ) 0 0 { } 0 ≥ 1 ( G +1)r and ( py,qy,pq E(G)) E(P )= (when pq E(G), we allow pq E(P )). 4 | 0| { } − ∩ 0 ∅ ∈ ∈ 0 Moreover, if H is a cycle, given z p, q , we can choose P such that one of the endvertices t ∈{ } 0 of P0 is z.

Proof. Since ∆(G′ ) d and G′ is 3-connected, G′ contains a (p,y)-path P ′ such that 0 ≤ 0 0 0 ℓ(P ′) 1 ( G +1)r + 1 by Theorem (4.1.1)(c). If qy P ′, then P := P ′ y is the desired 0 ≥ 4 | 0| ∈ 0 0 0 − (p, q)-path. Since G (1 ǫ 2ǫ )n (1 ǫ 2ǫ )(d 1)log2 6 > (d 1)2, we have | 0| ≥ − 1 − 2 ≥ − 1 − 2 − − ℓ(P ′) 3. Hence if pq P ′, then qy / E(P ′). So P := P ′ p is the desired path. 0 ≥ ∈ 0 ∈ 0 0 0 −

′ When Ht is a cycle, we use Theorem (4.1.1) (c) to ﬁnd a (z,y)-path P0 in G0 such that ℓ(P ) 1 ( G + 1)r +1. If V (P ) ( p, q z) = , then P itself is the desired path. 0 ≥ 4 | 0| 0 ∩ { } − ∅ 0 So assume p, q z V (P ). If pq / E(P ) , then P y is the desired path. Hence, { } − ⊆ 0 ∈ 0 0 − assume that pq E(P ), and so pq E(G′ ). We may assume pq / E(G); otherwise P is ∈ 0 ∈ 0 ∈ 0 the desired path. By the deﬁnition of G′ , we have p, q = a , b in this case. Let P be 0 { } { t t} J an (a , b )-path in given by Lemma (4.4.1). Then P := (P pq ) P is the desired t t J 0 0 −{ } ∪ J path with ℓ(P ) 1 ( G + 1)r + 1 ( (d−2.1)|J| )r + 1. 0 ≥ 4 | 0| 4 d−1

4.6.1.1 Subcase1.1. p, q = a , b = V (H H ). Using the inequalities { } 6 { t t} t ∩ t+1 max , n−|H| (1−ǫ1−ǫ2)n , we will consider a few cases to show that {|I| |J|} ≥ |J| ≥ |L| d−1 ≥ d−1 142

there exists a cycle C through xy such that

1 (d 2.1) max , C ( G + 1)r +( − · {|I| |J|})r +2 | | ≥ 4 | 0| (d 1)2 − 1 (1 ǫ ǫ )(d 2.1) r (1 ǫ 2ǫ )r + − 1 − 2 − nr +2 ≥ 4 − 1 − 2 (d 1)3 − 1 −β r (1 ǫ 2ǫ )+(1 ǫ 2ǫ )((d 1)log2(1+2 ) 1) nr +2 ≥ 4 − 1 − 2 − 1 − 2 − − 1 nr +2, ≥ 4

3 where we let β = log (d−1) (1−ǫ1−2ǫ2) for Lemma (4.3.1), which is greater than 1 but d−1 (d−2.1)(1−ǫ1 −ǫ2) less than 2 when d 42. We also use the inequalities 1 ǫ 2ǫ > 0 when d 43, and ≥ − 1 − 2 ≥ −β (1 ǫ 2ǫ )+(1 ǫ 2ǫ )((d 1)log2(1+2 ) 1) > 1 when d 68. − 1 − 2 − 1 − 2 − − ≥

We ﬁrst consider the case that there is a (p, q)-path P in G y such that ℓ(P ) 0 0 − 0 ≥ 1 r 4 ( G0 + 1) . Let PI be an ( at, bt , x), say (at, x)- path in through pq given by | | { } r I 1 (d−2.1)|I| Lemma (4.4.3) such that ℓ(PI ) 4 (d−1)2 (as p, q = at, bt ), and PJ be an ≥ { } 6 { } r (a , x′) path in b given by Lemma (4.4.2) such that ℓ(P ) 1 (d−2.1)|J| , where t J − t J ≥ 4 (d−1)2 x′ (V (H ) V (H )) N (y)(as H exists and G is 3-connected, x′ a , b ). Then, ∈ h − h−1 ∩ G t+1 6∈ { t t} C := (P pq ) P P xy is the desired path. I −{ } ∪ J ∪ 0 ∪{ }

Suppose that Ht is 3-connected. By Claim (4.6.1.2) and the discussion above, we may assume that there is a path P in G from p to y avoiding q such that ℓ(P ) 1 ( G +1)r. 0 0 0 ≥ 4 | 0| If , by Lemma (4.4.2), let P be a path in q from x to p such that ℓ(P ) |I| ≥ |J| H I − H ≥ 1 ( (d−2.1)|I| )r +1. If a b E(P ), then we replace a b by a path in from a to b . Then, 4 (d−1)2 t t ∈ H t t J t t C := P P xy is the desired cycle. If , let P be a path in q from x H ∪ 0 ∪{ } |I| ≤ |J| I I − to p through a b given by Lemma (4.4.3) and P be a path in from a to b such that t t J J t t ℓ(P ) 1 ( (d−2.1)|J| )r + 1 given by Lemma (4.4.1). Then C := (P a b ) P P xy J ≥ 4 (d−1) I −{ t t} ∪ J ∪ 0 ∪{ } is the desired cycle.

Finally, we assume that Ht is a cycle and P0 given by Claim (4.6.1.2) is a (p,y)-path. Since in this case, the edge a b exists. As a , b = p, q , we can assume, |J| ≥ |L| t t { t t} 6 { } 143

without loss of generality, that a ...a b ...p...b lie in this order along H a b . t−1 t t t−1 t − t−1 t−1 Let ∗ := H H ...H . Applying Lemma (4.4.2), we ﬁnd a path P ∗ in ∗ b from x to I 1 2 t−1 H I − t−1 ∗ a such that ℓ(P ∗ ) 1 ( (d−2.1)|I | )r + 1 . Extending this path along H and , we obtain t−1 H ≥ 4 (d−1)2 2 t J a path P in from x to p avoiding q such that ℓ(P ) ℓ(P ∗ ) + 2. Since the number of H H H ≥ H degree 2 vertices in is no more than 2d 1, we have ℓ(P ) 1 ( (d−2.1)|I| )r +1. Let P be H − H ≥ 4 (d−1)2 J a path in from a to b such that ℓ(P ) 1 ( (d−2.1)|J| )r + 1 given by Lemma (4.4.1). Note J t t J ≥ 4 (d−1) that in this case, we can choose P tobea(p,y)-path such that ℓ(P ) 1 ( G +1)r. Then, 0 0 ≥ 4 | 0| C := (P a b ) P P xy is the desired cycle. H −{ t t} ∪ J ∪ 0 ∪{ }

4.6.1.2 Case p, q = a , b = V (H H ) In this case, G′ := G py,qy,pq . { } { t t} t ∩ t+1 0 0 ∪{ } Assume, without loss of generality, that dG (p) dG (q). Let tp = NG′ (p) q,y . Clearly, 0 ≤ 0 | 0 −{ }| t d (p) 2 d 2. Let P be an (x, p)-path in q given by Lemma (4.4.2), P a(p, q)- p ≤ G − ≤ − I I − J path in given by Lemma (4.4.1), and P a(q,y)-path in G′ p given by Theorem (4.1.1) J 0 0 − (a). Let C := P P P xy . Then we have I ∪ J ∪ 0 ∪{ }

1 (d 2.1) G (d 2.1) (d 2.1) ℓ(C) ( − | 0|)r +( − |J|)r +( − |I|)r +2. (4.21) ≥ 4 (d 1)t d 1 (d 1)2 − p − −

Claim (4.6.1.3). r +( |I| )r ( + )r r provided d 61. |J| d−1 ≥ |J| |I| ≥ |H| ≥

Proof. By Lemma (4.3.3), we only need to show that 1.1|I| . Otherwise, using |J| ≥ d−1 |L| ≥ (1−ǫ1−ǫ2)n d−1 , we have

(d 1) (d 1) (1 ǫ ǫ )n (ǫ + ǫ )n> > − |J| − |L| − 1 − 2 . 1 2 |H| ≥ |I| 1.1 ≥ 1.1 ≥ 1.1

However, when d 61, ǫ + ǫ 0.47 and (1−ǫ1−ǫ2) > 0.47, showing a contradiction. ≥ 1 2 ≤ 1.1 144

Consequently, by Lemma ?? we have

1 (d 2.1) G (d 2.1) ℓ(C) ( − | 0|)r +( − |H|)r + 2 (4.22) ≥ 4 (d 1)t d 1 − p − 1 (d 2.1)2 r/2 ( − (1 /n ςǫ /(d 2.1))( /n) nr +2. ≥ 4 t − |H| − 2 − |H| p

Clearly, C is the desired cycle if

(d 2.1)2(1 /n ςǫ /(d 2.1))( /n) − − |H| − 2 − |H| 1. (4.23) tp ≥

Assuming this is not the case, we will show that there are very few minor-legs of , which H reveals some properties of . H Claim (4.6.1.4). We may assume ς 3 (provided d 195). ≤ ≥

Proof. By plugging /n 1−ςǫ2/(d−2.1) and t d 2 in (4.23), we get |H| ≥ d−1−ς p ≤ −

(d 2.1)2(1 /n ςǫ /(d 2.1)) ( /n)/t − − |H| − 2 − · |H| 1 ςǫ /(d 2.1) ςǫ 1 ςǫ /(d 2.1) (d 2.1)2 1 − 2 − 2 − 2 − /(d 2) ≥ − − d 1 ς − d 2.1 d 1 ς − − − − − − (d 2.1)2(d 2.1 ςǫ )2(d 2 ς) = − − − 2 − − 1, (d 2.1)2(d 1 ς)2(d 2) ≥ − − − −

provided d 195 and ς 4. ≥ ≥

We now reﬁne the legs of contained in G′ . Let , , ... , attach at p, q H 0 L1 L2 Lℓ H { } such that V ( ) V ( ) = p, q for any i = j and, subject to this constraint, V ( ) Li ∩ Lj { } 6 i | Li | is maximum. We name them major-legs of . Clearly, all other -legs remainedP in G′ are H H 0 B-type minor-legs.

Claim (4.6.1.5). We may assume ℓ d 5 provided that d 195. ≥ − ≥ 145

Proof. Since , , ... , are all possible non-minor-legs, (1 ςǫ2 )n/(ℓ + 1). L1 L2 Lℓ |H| ≥ − d−2.1 Plugging this inequality in (4.23) and assuming ℓ d 6, we get the following ≤ −

1 (d 2.1)2 1 ςǫ2 1 ςǫ2 r/2 ℓ(C) ( − (1 − d−2.1 ςǫ /(d 2.1))( − d−2.1 ) nr +2 ≥ 4 t − ℓ +1 − 2 − ℓ +1 1 ℓ(d 2.1 ςǫ )2 r/2 − − 2 nr +2 ≥ 4 (ℓ + 1)2t 1 nr +2, ≥ 4

for each ς =0, 1, 2, 3 when d 195; where we used t d 2 and ℓ d 6 . ≥ ≤ − ≤ −

For each , let G be induced by the component of G p,q,y containing p, q , Li i −{ } Li −{ } and including two vertices p and q, that is, the union of and all B-type minor-legs sharing Li a vertex with p, q . Let G′ = G[V (G ) y ] pq,py,qy . Clearly, each G′ is 3- Li −{ } i i ∪{ } ∪{ } i ′ 1 connected with ∆(G ) d. Let ti(p)= dG′ (p) 2, ti(q)= dG′ (q) 2, and ti = (ti(p)+ti(q)). i ≤ i − i − 2 By counting the neighbors of p and q, respectively, we have

t (p)+ t (p)+ + t (p) d (p) N (p) q d 2, 1 2 ··· ℓ ≤ G −| H −{ }| ≤ − t (q)+ t (q)+ + t (q) d (q) N (q) p d 2, 1 2 ··· ℓ ≤ G −| H −{ }| ≤ − t + t + + t = (t (p)+ t (q)) d 2. 1 2 ··· ℓ i i ≤ − 1X≤i≤ℓ

We note that, for each i, t (p) 1, t (q) 1, and t 1. Assume, without loss of i ≥ i ≥ i ≥ generality, |G1| = max |Gi| . t1 1≤i≤ℓ ti

Claim (4.6.1.6). We may assume t =1 provided that d 194. 1 ≥

Proof. Otherwise, we have t 3/2 since both t (p) and t (q) are positive integers. Then 1 ≥ 1 1 either p or q has degree at least 2 in G ; and consequently, ℓ d 2 1 ς (each has a 1 ≤ − − − Li neighbor of p and a neighbor of q). By Claim (4.6.1.5) that ℓ d 5, we may assume that ≥ − ς 2 in this case. ≤ 146

Let T := i : t = t (p) = t (q)=1 and T := i : t > 1 and let ℓ := T and 1 { i i i } 2 { i } 1 | 1| ℓ := T . By Claim (4.6.1.5), we have ℓ + ℓ = ℓ d 5. On the other hand, we have 2 | 2| 1 2 ≥ − ℓ +3/2ℓ t d 2, which in turn gives ℓ 12. So, ℓ d 17. As d 180, 1 2 ≤ 1≤i≤ℓ i ≤ − 2 ≤ 1 ≥ − ≥ 2ℓ /3 ℓ . P 1 ≥ 2

ℓ For each i =1, 2,...,ℓ, let ωi = V (Gi) V ( i) . Clearly, i=1 ωi < | − L | M is an H-minor-leg |M| ≤ ςǫ n/(d 1). P P 2 −

For each i T , by the maximality of G /t , we have ∈ 1 | 1| 1

G = G /t G /t (2/3) G (2/3)( + ω ) (since when x / p, q ). | i| | i| i ≤| 1| 1 ≤ | 1| ≤ |H| 1 |L1| ≤ |H| ∈{ }

For each i T , we have ∈ 2 G = + ω + ω . | i| |Li| i ≤ |H| i

A simple calculation gives the following inequalities.

2ℓ 2ℓ G ( 1 + ℓ ) + 1 ω + ω | i| ≤ 3 2 |H| 3 1 i 1X≤i≤ℓ iX∈T2 2ℓ ℓ2 2ℓ1 ( + ) + (ω1 + max ωi ) (since ℓ2 2ℓ1/3) ≤ 3 3 |H| 3 i∈T2 { } ≤ 2ℓ ℓ2 2ℓ1 ςǫ2n ( + ) + (since ω ςǫ2n ) ≤ 3 3 |H| 3 d 2.1 i i ≤ d−2.1 − 2(d + 3) 2ς(d 3)ǫ2n P + − (since ℓ d 3,ℓ 12 ) ≤ 3 |H| 3(d 2.1) ≤ − 2 ≤ −

Since + G n ςǫ2n , we get the following inequality |H| 1≤i≤ℓ | i| ≥ − d−2.1 P 3 2(d−1.5)ςǫ2 − d−2.1 n. |H| ≥ 2(d +4.5)

3− 2(d−1.5)ςǫ2 n− ςǫ2n n− ςǫ2n When d 194, for each ς = 0, 1, 2, d−2.1 n > d−2.1 . Recall that d−2.1 is the lower ≥ 2(d+4.5) d−5 d−5 bound on used in the proofs of both Claim (4.6.1.4) and Claim (4.6.1.5), and so we are |H| done by the previous conclusions. 147

Let H be a block of with maximum number of vertices, that is, H = max H : k J | k| {| i| t +1 i h . Let := L L ...L and let L be a block of with maximum number ≤ ≤ } L1 1 2 s m L1 of vertices. Since t1 = 1, L1 is a cycle.

Claim (4.6.1.7). Let z′ (V (L ) V (L )) N (y). We may assume that there is a ∈ s − s−1 ∩ G (p, z′)-path P in q such that 1 L1 −

1 r d 2.1 r 1 r ℓ(P ) Lm + ( − 2 Li ) 1 1 1. ≥ 4 | | (d 1) | | ! − ≥ 4|L | − Xi6=m −

Proof. Assume V (L L )= a, b . In , we replace L by a triangle zabz and apply the 1 ∩ 2 { } L1 1 particular part of Lemma (4.4.4) to get a (z, z′)-path. Replacing either the edge za or zb by a path from p to a, b (we can ﬁx p as L is a cycle), and denote the resulted path by { } 1 P . We obtain the desired path; in case that p a, b , we may have the lower bound ℓ(P ) ∈{ } above 1 unit less than the bound given in Lemma (4.4.4).

Claim (4.6.1.8). d (p) 1 2, so both H and H are cycles. H − ≤ t t+1

Proof. Otherwise, we have ℓ d 3, which in turn shows that ≤ −

n ςǫ n/(d 2.1) − 2 − . |H| ≥ ℓ +1

Let P be an (x, q)-path in p given by Lemma (4.4.2), P bea(p, q)-path in given by I I − J J Lemma (4.4.1), and P be a (z′,p)-path given by Claim (4.6.1.7). Let C := P P P 1 I ∪ J ∪ 1 ∪ 148

yz′, xy . Then C is a cycle through xy and { }

1 1 (d 2.1) 1 (d 2.1) ℓ(C) r 1+ ( − |I|)r + ( − |J|)r +1+2 ≥ 4|L1| − 4 (d 1)2 4 d 1 − − 1 (d 2.1) r +( − |H|)r + 2 (by Claim (4.6.1.3)) ≥ 4 |L1| d 1 − 1 n ςǫ2n (d 2.1) ( −|H|− d−2.1 )r +( − |H|)r +2 ≥ 4 ℓ d 1 − ςǫ n r/2 n− 2 1 (d 1)2(n ℓ+1 ςǫ2n )(d 2.1)(n ςǫ2n ) − − d−2.1 − d−2.1 − − d−2.1 +2 ≥ 4  ℓ(d 1)(ℓ + 1)  − 1 (d 1)(d 2.1 ςǫ )2 r/2  = − − − 2 nr +2 4 (d 2.1)(ℓ + 1)2 − 1 (d 1)(d 2.1 ςǫ )2 r/2 − − − 2 nr + 2 (by ℓ + ς d 2) ≥ 4 (d 2.1)(d ς 1)2 ≤ − − − − 1 nr +2, ≥ 4

when d 41 and ς 1. Thus, we assume ς = 0. Then by ℓ d 3, we get ≥ ≥ ≤ −

1 (d 1)(d 2.1 ςǫ )2 r/2 ℓ(C) − − − 2 nr +2 ≥ 4 (d 2.1)(ℓ + 1)2 − 1 (d 1)(d 2.1)2 r/2 − − nr +2 ≥ 4 (d 2.1)(d 2)2 − − 1 nr +2. ≥ 4

Let P be an(x, p)-path in q given by Lemma (4.4.2) such that ℓ(P ) 1 (d−2.1)|I| )r + I I − I ≥ 4 (d−1)2 1 . Applying Lemma (4.4.6) on and , with taking the role of , p taking the role of 2 J L J H x and q taking the role of both w and w′ in the lemma, respectively. Let y′ (V (H ) ∈ h − V (H )) N (y). Then we can ﬁnd a (q,y′)-path P in p and a (p, q)-path P in h−1 ∩ G J J − L L such that ℓ(P )+ ℓ(P ) 1 r + 1 r 1/2. Then C := P P P xy,yy′ is a cycle J L ≥ 4 |J| 4 |L| − I ∪ J ∪ L ∪{ } 149

through xy such that

1 (d 2.1) 1 1 1 1 ℓ(C) − |I|)r + + r + r +2 ≥ 4 (d 1)2 2 4|J| 4|L| − 2 − 1 1 r + r + 2 (By Claim (4.6.1.3)) ≥ 4|H| 4|L| 1 n n n/(d 1) r/2 (d 1)2 − − +2 ≥ 4 − · d 1 · d 2 − − 1 = nr +2, 4

since in this case, n−|H| , and as > gives that n . |L| ≥ d−2 |H| |J| ≥ |L| |H| ≥ d−1

4.6.2 Case 2 x p, q . ∈{ }

Let the notation be chosen so that x = p. In this case, the notation = L L L is L 1 2 ··· m used to indicate an arbitrary -leg. We note that may no longer hold because it H |H| ≥ |L| is possible that x V (L ). An -leg is proper if x V (L ) V (L ). For a proper -leg ∈ 2 H L ∈ 1 − 2 H , still holds. L |L| ≤ |H|

If x, v is a 2-cut of G y for some v V (H ), let G be obtained from G y by { } − ∈ 1 v − deleting all components of G y, x, v containing a vertex of and adding the edge xv −{ } H |Gv0 | ′ when xv / E(G). Let v0 V (H1) such that d (x)−1 is maximum. Let Gv = Gv xy,vy . ∈ ∈ Gv0 ∪{ }

|Gv0 | ǫ2n Claim (4.6.2.1). d (x)−1 > d−2.1 provided d 93. Gv0 ≥

Proof. Notice that all -legs not containing x are A-type minor-legs, and there are at most H 3ǫ2n three -minor-legs by Claim (4.6.1.4). Hence Gv0 + Gv + + d−2.1 n. Thus we H | | v6=v0 | | |H| ≥ P 150

have

G G | v| n 3ǫ2n | v0 | v −|H|− d−2.1 d (x) 1 (dP (x) 1) d 2 Gv0 ≥ Gv ≥ − v − − 3ǫ2n Pn (ǫ1 + ǫ2)n ǫ n − − d−2.1 > 2 . ≥ d 2 d 2.1 − −

Let T denote the block-bond tree resulted in from G y+xv by the Tutte decomposition. − 0 We treat T as a rooted tree with the root at the bond B0 containing the edge xv0 (notice that as xv E(G y + xv ), and G y + xv x, v has at least two components, the 0 ∈ − 0 − 0 −{ 0} bond B0 exists). Except B0, we assume that all bonds are removed from T and two 3-blocks are adjacent if either one is the parent of the other one in the original tree or there is a bond B between them such that one is the parent of B and the other one is a child of B. We will follow the partial order of T generalized naturally by the parent-child relationship of the ≺ tree, that is B B if B is a descendent of B . 1 ≺ 2 1 2

A block of T is called an x-block if it contains x. Let be the union of all x-blocks. X A block-chain Y Y ...Y such that Y is a virtual edge not incident with x is called a 1 2 m 1 ∩ X y-chain. The following deﬁnition will play a key role in our proof.

Claim (4.6.2.2). If Y and Y are two distinct y-chains attached to the same x-block B / 1 2 ∈ H with Y Y , then Y < ǫ2n . | 1|≥| 2| | 2| d−2.1

Proof. Suppose to the contrary that Y ǫ2n . Suppose there are x-blocks B , B , , B | 2| ≥ d−2.1 1 2 ··· t such that BB B B is a block-chain and B H = x, v , x y = E(Y B) for each 1 2 ··· t t ∩ 1 { } i i i ∩ i =1, 2, and B B = x, u . ∩ 1 { }

By Lemma (4.2.1), we may assume that there is a path (u, x1)-path PB in B containing edge x y . Let P be a longest (x , z′)- path in Y y given by Lemma (4.4.2), where z′ is a 2 2 1 1 1 − 1 151

vertex in the extreme block in Y1 which is adjacent to y. Let P2 be a longest (x2,y2)-path in Y given by Lemma (4.4.1), and P be a longest (x, v)-path in given by Lemma (4.4.1). 2 H H Since BB B B is 2-connected, BB B B x is connected. Let P be a (u, v)-path 1 2 ··· t 1 2 ··· t − 3 in BB B B x. Let C = P P (P x y ) P P xy, yz′ . Since x / Y and 1 2 ··· t − H ∪ 3 ∪ B − 2 2 ∪ 2 ∪ 1 ∪{ } ∈ 1 x / Y , Y Y . Then ∈ 2 |H| ≥ | 1|≥| 2|

1 d 2.1 r 1 d 2.1 r 1 d 2.1 r ℓ(C) − + − Y + − Y +2 ≥ 4 d 1 |H| 4 d 1 | 2| 4 (d 1)2 | 1| − − − 1 d 2.1 r 1 2.5 − Y +2 nr +2, ≥ 4 · d 1 | 2| ≥ 4 −

since 2.5=((d 1)log2 5/2)r. −

We call a y-chain Y a small chain if Y < ǫ2n . | | d−2.1 Deﬁnition (4.6.2). A block B is called a giant block (GB) if ∈ X

|B| ǫ2n , or if there is a y-chain Y attached to B such that • dB (x)−1 ≥ d−2.1

Y ǫ2n , or • | | ≥ d−2.1

|BY | 2ǫ2n . • dBY (x)−1 ≥ d−2.1

If B is not a GB, we call B a small block (SB).

Let B be an x-block. If there exist y-chains attached to B, let Y be one of the y-chains with largest cardinality. Then BY is called a y-extension of B. Notice that BY is a proper -leg, and so BY . H | | ≤ |H|

Following the notation in the above deﬁnition, we have the following observation.

Claim (4.6.2.3). Let B be an x-block and BY a y-extension of B. Suppose that xb and xb′ are the virtual edges of B corresponding to its parent and one of its children, respectively. 152

Then, there is a (b, b′)-path P in BY x of length ℓ(P ) 1 ( d−2.1 Y )r +1 and there is a − ≥ 4 d−1 | | (b, y)-path Q in G[V (BY ) y ] x of length ℓ(Q) 1 ( d−2.1 Y )r +1. ∪{ } − ≥ 4 d−1 | |

Proof. Let u, v = V (B) V (Y ). If B is a cycle, let P be the unique path from b { } ∩ 1 to b′ through uv. If B is 3-connected, then B x is 2-connected. There is a path P − 1 ′ from b to b through uv. By Lemma (4.4.1),there is a (u, v)-path path P2 in Y such that ℓ(P ) 1 ( d−2.1 Y )r + 1. Then P := (P uv ) P gives the desired (b, b′)-path. 2 ≥ 4 d−1 | | 1 −{ } ∪ 2

To prove the second statement, if B is a cycle, let P be the unique path in B x from 1 − b to z u, v , say u, avoiding v; if B is 3-connected, B x v is connected, there is a ∈ { } − − path P from b to u. We may assume that d (v) 3. For otherwise, let Y ∗ be the graph 1 Y ≥ obtained from G[(V (Y ) y ] yu,yv,uv by suppressing all degree 2 vertices. Then, ∪{ } ∪{ } applying Theorem (4.1.1) (a) to Y ∗ v, we can ﬁnd a (u,y)-path P not containing uv − 2 such that ℓ(P ) 1 ( d−2.1 Y )r + 1. Thus, P P is the desired path. Hence, d (v) 3. 2 ≥ 4 d−1 | | 2 ∪ 1 Y ≥ This implies that the ﬁrst block of Y is 3-connected. Let Y ∗ be the graph obtained from

∗ G[(V (Y ) y ] yu,uv by suppressing all degree 2 vertices. If d ∗ (y) 3, then Y ∪{ } ∪{ } Y ≥ is 3-connected. Applying Theorem (4.1.1) (c) on Y ∗, we ﬁnd a (u,y)-path P of ℓ(P ) 2 2 ≥ 1 ( d−2.1 Y )r + 2. (We may assume that uv E(P ). As otherwise we can choose P to 4 d−1 | | 6∈ 2 1 be a (b, v)-path avoiding u and P := P (P uv ) gives the desired path.) Otherwise, 1 ∪ 2 −{ } ∗ dY ∗ (y) = 2; and thus Y is a block-chain with edge uy in one end-block and v at the other end. Applying Lemma (4.4.1) on Y ∗, we ﬁnd a (u,y)-path P of ℓ(P ) 1 ( d−2.1 Y )r + 1. 2 2 ≥ 4 d−1 | | In any case, P P is the desired path. 2 ∪ 1

We need to distinguish three diﬀerent types of degrees of x in B for each x-block: dB(x) is the degree of x in B, d(G,B)(x) is the number of edges of G incident with x in B, and

d(V,B)(x) is the number of virtual edges in B, that is, the degree of B, as a vertex in the subtree of T induced by all x-blocks. We have d (x) d (x)+ d (x) and the strick B ≤ (G,B) (V,B) inequality may hold (for example, an edge may be counted in both d(G,B)(x) and d(V,B)(x)). 153

For each x-block, we now associate it with a number t(B):

t(B)= d (x) 2. B −

This number is in correspondence of the parameter t in Theorem (4.1.1) (a), and it is used when applying Theorem (4.1.1) (a) on B. We will consider the ration B /t(B). For | | convention, we deﬁne B /t(B) = 0 when t(B) = 0. (In this case B is a cycle.) | | Claim (4.6.2.4). All GBs form an anti-chain(a set of vertices in the block-bond tree forms an anti-chain if, pairwise, they don’t have the parent-child relationship) in T.

Proof. Suppose there is an -leg B = B B B , where each B , 1 j k, is a 3-block H 1 ··· k−1 kL j ≤ ≤ containing x, and is the largest block-chain attached to B (so B contains an extreme L k kL block, and so has a neighbor of y). Let x, b := V (H ) V (B ), and V (B ) V (B ) = { 0} 1 ∩ 1 i ∩ i+1 x, b for 1 i k 1. Suppose there are indices i and m with i < m such that both B { i} ≤ ≤ − i and Bm are GBs, and m = k if Bm is an external GB.

For each 1 j = m k 1, if B is a cycle then let P be the path in B x from b ≤ 6 ≤ − j j j − j−1 to b ; otherwise let P be a path in B x from b to b as given by Theorem (4.1.1)(a). j j j − j−1 j Let Yi and Ym be the largest y-chains (if exist) attached to Bi and Bm, respectively.

|Bm| ǫ2n In the case k = m, let Pm be a longest (bm−1, bm)-path in Bm x if , and 6 − dBm (x)−2 ≥ d−2.1 let P be a longest (b , b )-path as guaranteed by Claim (4.6.2.3) if Y ǫ2n ; let P be m m−1 m | m| ≥ d−2.1 k a longest path in G[V (B ) y ]+b y x from b to y as given by Theorem (4.1.1)(a) (as kL ∪{ } k−1 − k−1 B has an extreme block which contains a neighbor of y, G[V (B ) y ]+ b y is 3- kL kL ∪{ } k−1 connected).

|BYk| 2ǫ2n |B| ǫ2n If k = m, we pick Pk as in the previous case if , or ; if t(Bk) ≥ d−2 dB (x)−2 ≥ d−2.1 Y ǫ2n , let P be the (b ,y)-path as guaranteed by Claim (4.6.2.3). Let P be | k| ≥ d−2.1 k k−1 0 a path in from x to b as given by Lemma (4.4.1). So, we have in either case that H 0 154

ℓ(P ) 1 ( d−2.1 ǫ2n )r. k ≥ 4 d−1 d−2.1

Then C := P ( k−1P ) P xy gives a cycle in G through xy. Noting B 0 ∪ ∪i=1 i ∪ k ∪{ } |H| ≥ | j| and B Y , thus |H| ≥ | k |

1 d 2.1 r 1 d 2.1 B r 1 d 2.1 ǫ n r ℓ(C) − + − | i| + − 2 +2 ≥ 4 d 1 |H| 4 d 1 d (x) 2 4 d 1 d 2 − − Bi − − − 1 d 2.1 r 1 d 2.1 B r 1 d 2.1 ǫ n r − B + − | i| + − 2 +2 ≥ 4 d 1 | i| 4 d 1 d (x) 2 4 d 1 d 2.1 − − Bi − − − 1 r |Bi| ǫ2n = n +2. (Provided d (x)−2 d−2.1 .) 4 Bi ≥

|Bi| ǫ2n So, we may assume < . Since Bi is a GB, by the deﬁnition, it has a y-chain dBi (x)−2 d−2 Y such that Y ǫ2n . Let P be a path in B x from b to b as guaranteed by i | i| ≥ d−2.1 i i − i−1 i Claim (4.6.2.3) such that ℓ(P ) 1 ( d−2.1 Y )r + 1. All other paths are as deﬁned in the i ≥ 4 d−1 | i| previous argument, we obtain a cycle C in G through xy such that

1 d 2.1 r 1 d 2.1 r 1 d 2.1 ǫ n r ℓ(C) − + − Y + − 2 +2 ≥ 4 d 1 |H| 4 d 1 | i| 4 d 1 d 2 − − − − 1 d 2.1 r 1 d 2.1 r 1 d 2.1 ǫ n r − Y + − Y + − 2 +2 ≥ 4 d 1 | i| 4 d 1 | i| 4 d 1 d 2 − − − − 1 = nr +2. (Provided Y ǫ2n .) 4 | i| ≥ d−2.1

Claim (4.6.2.5). We may assume 2ǫ2n provided that d 123. |H| ≥ d−2.1 ≥

Proof. Suppose that < 2ǫ2n . Then for each maximal proper -leg , we have |H| d−2.1 H L |L| ≤ < 2ǫ2 . As each maximal proper -leg either contains an extreme block (and thus has |H| d−2.1 H a neighbor of y), or it is an x-block (and thus has a neighbor of x), we then have at most 2(d 1) maximal proper -legs. All those -legs, together with , cover all the vertices of − H H H V (G) y. However, 4(d−1)ǫ2n < n 0.1 when d 123, showing a contradiction. − d−2.1 − ≥ 155

Claim (4.6.2.6). For any virtual edge xv with v = v , if is an -leg such that = xv, 6 0 L H L∩H then ǫ2n provided d 123. |L| ≤ d−2.1 ≥

Proof. We consider two cases according to whether H > H H ...H . If H | 1| | 2 2 h| | 1| ≥ H H ...H , then H ǫ2n by Claim (4.6.2.5). Let P be a longest (x, v)-path in , | 2 3 h| 1 ≥ d−2.1 L L P be a longest (v ,y)-path in G′ x given by Theorem (4.1.1)(a), C be a longest cycle in 0 0 V0 − H H through two edges xv and xv given by Theorem (4.1.1)(b), and let P = C xv . 1 0 H H −{ 0} From them, we can obtain a cycle C through xy such that

r r ′ r 1 d 2.1 1 d 2.1 Gv0 1 d 2.1 1 r ℓ(C) − + − | | + − 2 H1 +2 n +2, ≥ 4 d 1 |L| 4 d 1 dG′ (x) 2 4 (d 1) | | ≥ 4 − − v0 − ! −

|G′ | provided min , 0 , H ǫ2n . Since the other two already do by Claims (4.6.2.1) d ′ (x)−2 1 d−2.1 {|L| G0 | |} ≥ and (4.6.2.5), we may assume does not. |L|

If H < H H ...H , then H H ...H > ǫ2n . Deﬁne P and P the same way | 1| | 2 3 h| | 2 3 h| d−2.1 L 0 as above. Let ′ = H H ...H . Let C be a cycle in H through edges xv, xv , and ab, H 2 3 h 1 1 0 where ab = H ′, and P = C xv . Let P ′ be a longest (a, b)-path in ′ given 1 ∩ H 1 1 −{ 0} H H by Lemma (4.4.1). Then we have ℓ(P ′ ) 1 ( d−2.1 ′ )r + 1. Similarly, we can show that H ≥ 4 d−1 |H | the cycle C := (P ab, xv ) P ′ P xy passes through xy, and ℓ(C) 1 nr + 2 if 1 −{ } ∪ H ∪ L ∪{ } ≥ 4 ǫ2n . |L| ≥ d−2.1

Notice that there are at most d 2(as G has an extreme block, and thus has a neighbor − v0 of y) -legs with = xv . By Claim (4.6.0.3) and Claim (4.6.2.6), ǫ2n for H M M ∩ H 6 0 |M| ≤ d−2.1 each such -leg, which gives d−2 ǫ n. An immediate consequence is that H M |M| ≤ d−2.1 2 P ′ d−2 G n ǫ2n | v0 | −|H|− d−2.1 . (4.24) dG′ (x) ≥ d 2 v0 −

Claim (4.6.2.7). We may assume that < (1+1.5ǫ2)n provided d 125. |H| d−1 ≥ 156

Proof. Let P be a longest (v ,y)-path in G′ x given by Lemma (4.1.1) (a), and P be a 0 0 v0 − H longest (x, v )-path in given by Lemma (4.4.1). Then C := P P xy gives a cycle 0 H 0 ∪ H ∪{ } through xy such that

r 1 d 2.1 r 1 d 2.1 G′ ℓ(C) − + − | v0 | +2 ≥ 4 d 1 |H| 4 d 1 dG′ (x) 2 − − v0 − ! 1 (d 2.1)2 d 2 r/2 1 (d 1)2 − |H| (1 |H| − ǫ n) nr +2 nr +2, ≥ 4 − · (d 1)2(d 2) − n − d 2.1 2 ≥ 4 − − −

provided (1+1.5ǫ2)n and d 125. |H| ≥ d−1 ≥

Since all GBs form an anti-chain of T and the root B0 is not a GB, there is a subtree T of T containing the root such that it contains no GB and each branch of T T contains 0 − 0 at most one GB. (The subtree containing B0 obtained from T by deleting all of the GBs has the described property. ) We may assume T0 has this property with maximum cardinality. Let , , , be the block-trees corresponding to branches of T T . For each , we T1 T2 ··· Tm − 0 Ti call the block of which is an immediate child of the x-block to which attaching in T Ti Ti the ﬁrst block of . By the maximality of T and the fact that all GBs form an anti-chain Ti 0 in T, we have the following observations:

Excluding B , every x-block which is a leave of T is adjacent to a GB, which is the • 0 0 ﬁrst block of some (as if not, we can make T larger by adding the ﬁrst block of the Ti 0 to that leaf). Conversely, each GB is attached to an x-block which is a leaf of T ; Ti 0

Each virtual edge in B T which is adjacent to neither the parent of B nor the child • ∈ 0 of B is corresponding to at least one some branch . Ti

For each x-block B, let B := B be a maximal block-chain containing B as the ﬁrst L block such that B has the largest cardinality among all of such block-chains. Then by the maximality of B, it contains an extreme block. 157

For each i, let B be the GB contained in , we let := B . Then by Claim (4.6.0.3) i Ti Li i and Claim (4.6.0.4), we know each leg of is contained in an -minor-leg, and hence has Li H cardinality less than ǫ2n . By the construction of , it contains an extreme block. Note d−2.1 Li that each and the branch containing it, have exactly the same predecessors, that is Li Ti they are connected to B0 through exactly a same block-chain in T0. Also, notice that the ﬁrst block of each is an x-block. Let Li

L = , ,..., . {L1 L2 Lm}

Correspondingly, for each L, we let T be the chain of x-blocks which Li ∈ Mi ⊂ 0 connects to B . Thus, is a block-chain. Notice that may be empty in case that Li 0 MiLi Mi the ﬁrst block of is an immediate child of B . Li 0

We use the partial order generalized by T naturally, i.e., if B is a child of B , we ≺ 1 0 |B| have B1 B0. For each i, if i= , let η( i)= , and ≺ L M 6 ∅ L B∈Mi t(B) P ω( )= + η( ). Li |Li| Li

Note that by introducing η( ), (d (x) 2) |B| vertices in B are distributed into when Li (V,B) − · t(B) Li d (x) 3( B is not a cycle). As each virtual edge incident to x in B which is not incident (V,B) ≥ to the parent or the child of B is contained also in some , and there are (d (x) 2) Ti (V,B) − of such virtual edges. Let us see now which portion of vertices of G are not considered into ω( ). i Li P

(i) On a cycle-block B T , degree 2 vertices which are neighbors of y; ∈ 0

(ii) Small y-chains and legs of contained in the branch ; Li Ti

We estimate the number of vertices in the above three cases.

If (i), let δ be the number of such degree 2 vertices. • 2

If (ii), as each maximal block-chain has an extreme block, and each block-chain can • be extended to a maximal one, we suppose there are in total exactly s extreme blocks which are contained in some branch , but not in any one of . Then, Ti Lj

sǫ n + 2 . |Ti| ≤ |Li| d 2.1 i i X X −

For each B T , which is not in case (i), B (d (x) 2) |B| + d (x) |B| . • ∈ 0 | | ≤ (V,B) − · t(B) (G,B) · t(B) As each B T is a SB, we have B (d (x) 2) |B| + d (x) ǫ2n . ∈ 0 | | ≤ (V,B) − · t(B) (G,B) d−2.1

Let

′ s = δ2 + d(G,B)(x). T BX∈ 0

For each L, let τ ( ) = N (x) 1, τ ( ) = N (y) , and τ( ) = 1 (τ ( )+ L ∈ x L | L | − y L | G ∩ L| L 2 x L τ ( )). Note that the deﬁnition for τ ( ) is diﬀerent from that for τ ( ), as when we remove y L x L y L legs of in , it may be possible that in , x is only incident to virtual edges. However, Li Ti Li each virtual edge incident to x correspondences to at least one real edge incident to x in some legs of , so we let τ ( )= N (x) 1. The following inequalities hold. Li x L | L | −

τx( ) d(x) 1 d 1, L L ≤ − ≤ − L∈X τy( ) d(y) 1 d 1, and L L ≤ − ≤ − XL∈ 1 τ( ) (d(x)+ d(y) 2) d 1. L L ≤ 2 − ≤ − XL∈

We note that for each L, we have τ ( ) 1, τ ( ) 1, and τ( ) 1. By L ∈ x L ≥ y L ≥ L ≥ 159

relabeling the branches L , suppose we have L ∈ − H

ω( ) ω( ′) ω( ) L L Lm . τ( ) ≥ τ( ′) ≥···≥ τ( ) L L Lm

As xy E(G), and also by noticing that N (y) x 1 and d (x) 1 1, when ∈ | H −{ }| ≥ H − ≥ d 425, ≥

(s+s′)ǫ n ω( ) ω( ) n 2 n i i −|H|− d−2.1 , (4.25) L L 1 ′ − |H| τ( ) ≥ τ( i) ≥ d 1 (τ ( )+ τ ( )) (s + s )/2 ≥ d 2 L Pi L − − 2 x H y H − − P (1+1.5ǫ2)n ′ since under the assumption that d−1 and s + s 2(d 3)(xy E(G), and |H| ≤ ≤ − ′ ∈ n−|H|− (s+s )ǫ2n both x and y have at least one neighbor in each of and ), d−2.1 is an increasing L H d−2−(s+s′)/2 function of s + s′. The notations and ′ will be ﬁxed for the above deﬁnition hereafter. L L Claim (4.6.2.8). Let := M M ...M be the block-chain connecting some ′′ L to M 1 2 m L ∈ B0. Suppose V ( )= V (Mm )= x, vm and xv0 E(M1). Then in , there is a M∩L ∩L { }r ∈ M 1 d−2.1 |Mi| (v0, vm)-path PM with ℓ(PM ) . ≥ 4 i d−1 t(Mi) P Proof. For each i = 1, 2, , m 1, let M M = x, v . Let P be an (m , m )-path ··· − i ∩ i+1 { i} i i−1 i in Mi x given by Theorem (4.1.1) (a) (when Mi is a cycle, the assertion trivially holds) − r 1 d−2.1 |Mi| such that ℓ(Pi) . If Pi contains some virtual edges, which are supposed to ≥ 4 d−1 t(Mi) be replaced by a path connecting the two ends of the virtual edge in a y-chain of Mi with the ends as attachments (notice that this y-chain is a small-chain; and thus is not contained in any other block-chain in L ′′ by the construction of L′′). Let P := P , which is − {L } M ∪i i the desired path.

Claim (4.6.2.9). We have satisﬁes τ( )=1. In particular, if let = L L L , then L L L 1 2 ··· l x V (L ) V (L ) and L is a cycle provided that d 85. ∈ 1 − 2 1 ≥ 160

Proof. In the proof, we let α = τ( ). Suppose on the contrary that α > 1. Then α 1.5 L ≥ from the deﬁnition. So, by (4.25) we have

1.5(n ) ω( ) − |H| . L ≥ d 2 −

Let := M M ...M be the block chain connecting the root B of T and the block M 1 2 m 0 L in , and suppose L M = x, v . Let G′ be a graph obtained from G[V ( ) 1 L 1 ∩ m { m} L ∪ y ] yx, yvm by suppressing all degree 2 vertices. Then it is 3-connected, and then by { } ∪{ } r Theorem (4.1.1) (a), there is an (v ,y′)-path P in G′ x such that ℓ(P ) 1 d−2.1 +2, m L − L ≥ 4 (d−1)2 |L| ′ where y is a neighbor of y in the last block of . Let PH be an (x, v0)-path in given by L H r 1 d−2.1 |Mi| Lemma (4.4.1), and PM be a (v0, vm)-path in x such that ℓ(PM ) M − ≥ 4 i d−1 t(Mi) given by Claim (4.6.2.8). P

Set C := P P P yy′, xy , which is a cycle through xy such that H ∪ M ∪ L ∪{ }

1 d 2.1 r 1 d 2.1 M r 1 d 2.1 r ℓ(C) − + − | i| + − +2. ≥ 4 d 1 |H| 4 d 1 t(M ) 4 (d 1)2 |L| i i − X − −

(d−2)ǫ2n n− d−2.1 |Mi| As and for each Mi (as contains a GB and each Mi is |H| ≥ d−1 |L| ≥ t(Mi) L not a GB), by using (4.1c),

1 d 2.1 r 1 d 2.1 r ℓ(C) − ) + − ( +(d 1)η( )) +2 ≥ 4 d 1 |H| 4 (d 1)2 |L| − L − − 1 d 2.1 r 1 d 2.1 r − ) + − ω( ) +2 ≥ 4 d 1 |H| 4 d 1 L − − 1 (d 1)2(d 2.1)2 ω( ) r/2 − − |H| L nr +2 ≥ 4 (d 1)2 − 1 1.5(d 2.1 (d 2)ǫ )(d 2.1+ ǫ ) r/2 = − − − 2 − 2 nr +2 4 (d 1)2 − 1 nr +2, ≥ 4 161

when d 85. ≥ Claim (4.6.2.10). We may assume that ω( ) < n provided d 25. L d−2.1 ≥

Proof. Suppose not. Then we have also + η( ) n . Let := M M ...M be the |H| L ≥ d−2.1 M 1 2 m block chain connecting the root B of T and the block L in , and suppose L M = x, x′ . 0 1 L 1∩ m { } Let G′ be a graph obtained from G[V ( ) y ]+ yx, yx′ by suppressing all degree 2 vertices. L ∪{ } { } Then it is 3-connected, and by Theorem (4.1.1) (a), there is an (x′,y′)-path P in G′ x L − r such that ℓ(P ) 1 d−2.1 +2, as N (x) = 2, where y′ is a neighbor of y in L ≥ 4 d−1 |L| | G ∩ L| the last block of . Let PH be an (x, v0)-path in given by Lemma (4.4.1), and PM be a L H r ′ 1 d−2.1 |Mi| (v0, x )-path in x such that ℓ(PM ) . M − ≥ 4 i d−1 t(Mi) P Set C := P P P yy′, xy , which is a cycle through xy such that H ∪ M ∪ L ∪{ }

1 d 2.1 r 1 d 2.1 M r 1 d 2.1 r ℓ(C) − + − | i| + − +2. ≥ 4 d 1 |H| 4 d 1 t(M ) 4 d 1 |L| i i − X − −

r r r 1 d−2.1 |Mi| 1 d−2.1 |Mi| 1 d−2.1 |Mi| |Mi| As = 2 + 2 , and and 4 i d−1 t(Mi) 4 i (d−1) t(Mi) 4 i (d−1) t(Mi) |H| ≥ t(Mi) log (3/2) |Mi| (d−2.1)(d−1) 2 Pfor each Mi, by usingP (4.1c), and the fact thatP 1 when d 25, |L| ≥ t(Mi) (d−1) ≥ ≥ we have

1 d 2.1 r 1 d 2.1 r ℓ(C) − ( + η( )) + − ( + η( )) +2 ≥ 4 d 1 |H| L 4 d 1 |L| L − − 1 (d 1)2(d 2.1)2 r/2 − − nr +2 ≥ 4 (d 1)2(d 2.1)2 − − 1 = nr +2. 4

We now show that there is a cycle C through xy in G such that ℓ(C) 1 nr + 2. ≥ 4 162

Case 1. H is a cycle. 1 ∈ H

Recall that + η( ) n−|H| . This gives that + η( ) n by ( as is |L| L ≥ d−2 |H| L ≥ d−1 |H| ≥ |L| L a proper -leg). H

Let w := v , and let := M M ...M be the block-chain connecting the root B of 0 M 1 2 m 0 T and the block L in . Suppose that V ( ) = V (M L ) = x, w′ . Let y′ be a 1 L M∩L m ∩ 1 { } ′ neighbor of y in diﬀerent from x (y exists by the 3-connectivity of G). Let PM be a path H r ′ 1 d−2.1 |Mi| in x from w to w such that ℓ(PM ) given by Claim (4.6.2.8). As M − ≥ 4 i d−1 t(Mi) both H and L are cycles, apply the particularP part of Lemma (4.4.6) on and , without 1 1 H L loss of generality, assume that we ﬁnd a (w,y′)-path P in x, and an (x, w′)-path P H H − L in such that ℓ(P )+ ℓ(P ) 1 ( r + r). Let C := P P P yy′, xy . Then L H L ≥ 4 |H| |L| H ∪ M ∪ L ∪{ } C is a cycle through xy such that ℓ(C) 1 r + 1 r + ℓ(P ) + 2. By splitting the value ≥ 4 |H| 4 |L| M ℓ(PM ), we have

1 1 ℓ(C) r + r + ℓ(P )+2 ≥ 4|H| 4|L| M 1 1 ( + η( ))r + ( + η( ))r +2 ≥ 4 |H| L 4 |L| L 1 n n n/(d 1) r/2 (d 1)2 − − +2 ≥ 4 − · (d 1) · d 2 − − 1 = nr +2. 4

Case 2. H is 3-connected. 1 ∈ H

In this case, we have τ ( ) 2. Hence by (4.25), we have x H ≥

ω( ) n L − |H|. (4.26) τ( ) ≥ d 2.5 L − 163

As τ( ) 1, a similar argument as in (4.25) gives that L ≥

ω( ′) n ω( ) L −|H|− L . (4.27) τ( ′) ≥ d 3.5 L −

If η( ) η( ′), then we construct a cycle C through xy using and and a cycle H ≥ L 1 H L C through xy using and ′, and show that ℓ(C )+ ℓ(C ) 2( 1 nr +2). If η( ) < η( ′), 2 L L 1 2 ≥ 4 H L then we construct a cycle C through xy using and ′ and a cycle C through xy using 1 H L 2 and ′, and show that ℓ(C )+ ℓ(C ) 2( 1 nr + 2). Assume, without loss of generality, L L 1 2 ≥ 4 ′ ′ ′ ′ ′ that η( ) η( ). Suppose = L L L and = L L L ′ . Let := M M ...M H ≥ L L 1 2 ··· l L 1 2 ··· l M 1 2 m be the block-chain connecting the root B of T and the ﬁrst block L in , and let ′ := 0 1 L M ′ ′ ′ ′ ′ M M ...M ′ be the block-chain connecting the root B of T and the ﬁrst block L in . 1 2 m 0 1 L Furthermore, we suppose

′ ′ ′ M L = x, b and M ′ L = x, b ; • m ∩ 1 { } m ∩ 1 { } L = max L : L and L′ = max L′ : L′ ; • k { i i ∈L1} p { i i ∈L2} L L = a, b , L L = a , b , L′ L′ = c,d , and L′ L′ = c ,d ; • k ∩ k−1 { } k ∩ k+1 { k k} p ∩ p−1 { } p ∩ p+1 { k k} L := L L L L is the block-chain L ab, and • 0 k1 k2 ··· kk0 ··· kk1 k − L′ := L′ L′ L′ L′ is the block-chain L′ cd such that • 0 p1 p2 ··· pp0 ··· pp1 p − (i) L = max L : L L and L′ = max L′ : L′ L′ , kk0 { ki ki ∈ 0} pp0 { pi pi ∈ 0} (ii) a L , b L , c L′ , and d L′ , and ∈ k1 ∈ kk1 ∈ p1 ∈ pp1

(iii) given by Lemma (4.2.5), P is a path in L L L x from b to a, and P ′ L1 1 2 ··· k−1 − L 1 is a path in L′ L′ L′ x from b′ to c. 1 2 ··· k−1 −

′ We include the trivial case that Lk or Lp is a cycle in the above notations. Denote

r r l+ = d−2.1 L , l− = d−2.1 L ; • i>k (d−1)2 | i| i

r r + d−2.1 − d−2.1 l = 2 Lki , l = 2 Lki ; • 0 i>k0 (d−1) | | 0 ip (d−1)2 | i| i

p0 (d−1)2 | pi| 0 i

Let y′ be a neighbor of y in the last block of . We now construct a cycle C through L 1 xy by using paths in and as follows: H L

Let P be a path in from x to v given by Lemma (4.4.1) such that ℓ(P ) • H H 0 H ≥ 1 ( d−2.1 )r + 1, 4 d−1 |H| r 1 d−2.1 |Mi| PM be a path from v0 to b in x such that ℓ(PM ) given by • M − ≥ 4 i d−1 t(Mi) Claim (4.6.2.8), and P

P be a path in x from b to y′ given by Lemma (4.4.5) such that ℓ(P ) 1 L r + • L L− L ≥ 4 | kk0 | 1 ℓ− + 1 ℓ+ 1 . 4 0 4 − 2

Then C := P P P yy′, xy is a cycle through xy. Now we construct a cycle 1 H ∪ M ∪ L ∪{ } C in and ′. Assume, without loss of generality, that the following inequality holds. 2 L L

l+ + w− + w+ w+ + l− + l+. 0 ≥ 0

Let P be a path in x from b to y′ given by Lemma (4.4.5) such that L L −

1 1 1 1 ℓ(P ) L r + l+ + l− , L ≥ 4| kk0 | 4 4 0 − 2

′ ′ and P ′ be a path in x from b to x given by (4.6) of Lemma (4.4.5) such that L L −

1 ′ r 1 + 1 − 1 − ℓ(P ′ ) L + w + w + w . L ≥ 4| pp0 | 4 0 4 0 4 165

r 1 d−2.1 |Mi| Let PM be a path in x from b to v0 such that ℓ(PM ) given M − ≥ 4 i d−1 t(Mi) ′ ′ by Claim (4.6.2.8), and let P ′ be a path in x from b to vPsuch that ℓ(P ′ ) M M − 0 M ≥ ′ r 1 d−2.1 |Mi | ′ ′ ′ 4 i d−1 t(M ′) given by Claim (4.6.2.8). Then C2 := PL PM PM PL xy,yy i ∪ ∪ ∪ ∪{ } ′ 1 containsP a cycle through xy of length at least ℓ(P )+ ℓ(P )+2 (notice that P and P ′ L L − 2 M M may intersect). Then,

1 1 d 2.1 1 d 2.1 M r ℓ(P )+ ℓ(P ) = − ( ) + − | i| +1 H 4 M 4 d 1 |H| 4 (d 1)3 t(M ) − i − i X r 1 d 2.1 (d 2.1)((d 1)log2(5/4) 1) M − ( + − − − | i| ) +1 ≥ 4 d 1 |H| d 1 · t(M ) i i ! − X − 1 d 2.1 r − ( + η( )) +1, ≥ 4 d 1 |H| L1 −

log (5/4) as when d 12, (d−2.1)((d−1) 2 −1) > 1 and ≥ d−1

ℓ(P )+ 1 (l+ + l− + ℓ(P )) L 4 · 0 M r r r 1 r 1 d−2.1 1 d−2.1 1 d−2.1 |Mi| 1 Lpp + 4 Lpj + 4 Li + 3 ≥ 4 | 0 | 4 j (d−1) ·| | 4 i6=p (d−1) ·| | 4 i (d−1) · t(Mi) − 2 log (9/8) log (9/8) P(d−2.1)((d−1) 2 −1)|LpjP| (d−2.1)((d−1) 2 P−1)|L | 1 ( L + + i ≥ 4 | pp0 | j d−1 i6=p d−1 log (5/4) + (d−2.1)((Pd−1) 2 −1) |Mi| )r 1 P i d−1 · t(Mi) − 2 1 ωP( )r 1 , ≥ 4 L − 2 as (d 2.1)((d 1)log2(9/8) 1)/(d 1) > 1 when d 64. Therefore, − − − − ≥

1 1 1 d 2.1 r 1 1 ℓ(C )+ (l+ + l−) ℓ(P ) − ( + η( )) + ω( )r +2+ 1 4 · 0 − 2 M ≥ 4 d 1 |H| L 4 L 2 − 1 (d 1)(d 2 1.5ǫ ) r/2 1 − − − 2 nr +2+ ≥ 4 (d 1.5)(d 2.5) 2 − − 1 1 nr +2+ , ≥ 4 2

where the last inequality is obtained by using inequality (4.2), ω( ) n−|H| from (4.26), L ≥ d−2.5 + η( ) n following from + η( ) ω( ), and (1+1.5ǫ2)n from Claim |H| L ≥ d−1.5 |H| L ≥ L |H| ≤ d−1 166

(4.6.2.7). Similarly,

1 1 1 + − ℓ(P )+ ℓ(P )+ ℓ(P ′ )+ ℓ(P ) (l + l ) L 4 M L 4 M − 4 0 1 1 1 ( + η( ))r + ( ′ + η( ′))r . ≥ 4 |L| L 4 |L | L − 2

Thus

1 1 1 1 1 ℓ(C )+ (ℓ(P )+ ℓ(P )) (l+ + l−) ( + η( ))r + ( + η( ))r +2 2 4 · M M − 4 0 ≥ 4 |L1| L1 4 |L2| L2 − 2 1 r/2 1 (d 2 1.5ǫ2)(d 2 1.5ǫ2 ) 1 − − − − − d−2.1 +2 ≥ 4 (d 2.5)(d 3.5) − 2 − − ! 1 1 nr +2 , ≥ 4 − 2

provided that d 125, where the conditions that ω( ) n−|H| , ω( ′) n−|H|−ω(L) from ≥ L ≥ d−2.5 L ≥ d−3.5 (4.27), ω( ) n from Claim (4.6.2.10), and (1+1.5ǫ2)n from Claim (4.6.2.7) are L ≤ d−2.1 |H| ≤ d−1 used.

From above, we now can see ℓ(C )+ ℓ(C ) 2 ( 1 nr + 2), this implies that at least one 1 2 ≥ · 4 1 r of ℓ(C1) and ℓ(C2) is at least 4 n + 2. The proof is then completed. 167

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