Home Mathematics Connected even factors in k-tree
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Connected even factors in k-tree

  • Yinkui Li EMAIL logo , Xiaoxiao Qin , Wen Li , Xiaoling Wang and Haicheng Ma
Published/Copyright: December 29, 2020

Abstract

A connected even [ 2 , 2 s ] -factor of a graph G is a connected factor with all vertices of degree i ( i = 2 , 4 , , 2 s ) , where s 1 is an integer. In this paper, we show that a k + 1 s + 2 -tough k-tree has a connected even [ 2 , 2 s ] -factor and thereby generalize the result that a k + 1 3 -tough k-tree is Hamiltonian in [Hajo Broersma, Liming Xiong, and Kiyoshi Yoshimoto, Toughness and hamiltonicity in k-trees, Discrete Math. 307 (2007), 832–838].

MSC 2010: 05C45

1 Introduction

All graphs considered in this paper are finite and simple. We use Bondy and Murty [1] for notation and terminology not defined here.

Let G = ( V , E ) be a graph with vertex set V and edge set E. We denote by G H and G [ S ] the induced subgraphs of G by V ( G ) V ( H ) and S V ( G ) , respectively. For a vertex v of G, the neighborhood of v is the set of all vertices that are adjacent to v and is denoted by N G ( v ) , simplified by N ( v ) . We use d G ( v ) to denote the degree of vertex v in graph G, simplified by d ( v ) . Clearly, d ( v ) = | N ( v ) | .

A connected factor H of a graph G is a spanning subgraph of G and H is connected. A connected factor H is called an even [ 2 , 2 s ] -factor if the degree of vertex of H is i ( i = 2 , 4 , , 2 s ) , where s 1 is an integer. Clearly, if G has a connected even [ 2 , 2 ] -factor H, then G is Hamiltonian. Some results of the existence of connected even factor on forbidden subgraphs have been reported in many literature studies, such as [2,3,4,5,6].

A k-tree is a graph formed by starting with a ( k + 1 ) -vertex complete graph and then repeatedly adding vertices in such a way that each added vertex has exactly k neighbors that, together, the k + 1 vertices form a clique. It is clear that all of the maximal cliques of k-tree have the same size k + 1 and all of the minimal clique separators have the same size k. A vertex v of k-tree is called a simplicial vertex if all neighbors of v are mutually adjacent. A graph G is t-tough if | S | t ω ( G S ) for every subset S of the vertex set V ( G ) with ω ( G S ) > 1 . The toughness of G, denoted by τ ( G ) , is the maximum value of t for which G is t-tough (taking τ ( K n ) = for all n 1 ). Hence, if G is not complete, τ ( G ) = min { | S | / ω ( G S ) } , where the minimum is taken over all cutsets S of vertices in G.

In [7], Broersma et al. discussed the relationship between the toughness and hamiltonicity in k-trees and obtained the following results.

Theorem 1

(Broersma [7]) Let G K 2 be a k-tree. Then G is Hamiltonian if and only if G contains a 1-tough spanning 2-tree.

Theorem 2

(Broersma [7]) Let G K 2 be a k + 1 3 -tough k-tree ( k 2 ). Then G is hamiltonian.

In this paper, we partially generalize Theorem 1 and get the following Theorem 3.

Theorem 3

Let G K 2 be a 2 s + 1 -tough 2-tree. Then G contains a connected even [ 2 , 2 s ] -factor.

Since 3 s + 2 2 s + 1 for s 1 , by Theorem 3, we get the following Corollary 4.

Corollary 4

Let G be a 3 s + 2 -tough 2-tree. Then G contains a connected even [ 2 , 2 s ] -factor.

We further generalize Theorem 2 and Corollary 4 to get the following Theorem 5.

Theorem 5

Let G K k be a k + 1 s + 2 -tough k-tree. Then G contains a connected even [ 2 , 2 s ] -factor.

2 Preliminaries

We start with some additional notations. Let G = G 1 and S 1 ( K k ) = ϕ , and for a k-tree G K k , S 1 ( G ) denotes the set of simplicial vertices of G if G K k + 1 and a set of one arbitrary vertex of G if G = K k + 1 . S i ( G ) and G i are defined as follows: G i = G i 1 S 1 ( G i 1 ) and S i ( G ) = S 1 ( G i ) for i = 2 , 3 , as long as S i ( G ) (i.e., G i 1 K k ). N i ( v ) denotes the set of neighbors of v in G i .

Lemma 1

(Broersma [7]) Let G K k be a k-tree ( k 2 ). Then

  1. G is k-connected;

  2. S 1 ( G ) ;

  3. S 1 ( G ) is an independent set;

  4. Every simplicial vertex (if any) of G S 1 ( G ) is adjacent in G to at least one vertex of S 1 ( G ) ;

  5. τ ( G v ) τ ( G ) for a simplicial vertex v S 1 ( G ) ;

  6. τ ( G S 1 ( G ) ) τ ( G ) .

Lemma 2

(Broersma [7]) Let G K k be a k-tree ( k 2 ). For any vertex u S 2 ( G ) , there exists a vertex v S 1 ( G ) such that u v E ( G ) , N 1 ( u ) \ N 2 ( u ) S 1 ( G ) , and | N 2 ( u ) N 1 ( v ) | = k 1 .

Lemma 3

Let G K k be a k-tree ( k 2 ) and u S 2 ( G ) . If w N 1 ( u ) \ N 2 ( u ) , then N 1 ( w ) N 2 ( u ) { u } .

Proof

If not, suppose that there exists a vertex x N 1 ( w ) but x N 2 ( u ) { u } , then u x E ( G ) since N 1 ( w ) is a clique. Thus, x N 1 ( u ) , i.e., x N 1 ( u ) \ N 2 ( u ) . By Lemma 2, x , w S 1 ( G ) . Note x N 1 ( w ) , this contradicts to (iii) of Lemma 1.□

Lemma 4

Let G K k be a k-tree ( k 2 ) and K k be a k clique of G. Then ω ( G V ( K k ) ) s + 1 .

Proof

In fact, suppose that there exists a k clique K k of G such that ω ( G V ( K k ) ) s + 2 , then by deleting all k vertices of K k we will get at least s + 2 connected components in G, this contradicts to the fact G is k + 1 s + 2 -tough.□

3 The proof of Theorem 3

Assume G be a 2 s + 1 -tough 2-tree with order n. By Theorem 1, the conclusion holds for the case s = 1 . Here we suppose s 2 and distinguish two cases to complete the proof.

Case 1. S 2 ( G ) = .

We proceed by induction on n = | V ( G ) | . Clearly, the conclusion holds for n = 4 since G contains a Hamilton cycle this time. First suppose n 5 and the conclusion holds for all 2 s + 1 -tough 2-tree G with fewer than n vertices. Let G be a 2 s + 1 -tough 2-tree with S 2 ( G ) = , by the definition of 2-trees, G S 1 ( G ) is K 2 and denote V ( G ) \ S 1 ( G ) = { p , q } . By Lemma 4, we get | S 1 ( G ) | s + 1 . Let v S 1 ( G ) and G = G v , by Lemma 1 (v), G is also 2 s + 1 -tough 2-tree with | V ( G ) | < n . By the induction hypothesis, G has a connected even [ 2 , 2 s ] -factor, named F . Now based on F we constructed a connected even [ 2 , 2 s ] -factor of G.

Subcase 1. p q E ( F ) .

Since p q E ( F ) , replace edge pq by path pvq of G and thus get a connected even factor of G, named F, i.e., F = F { p q } + { p v , q v } . Clearly, d F ( v ) = 2 and d F ( x ) = d F ( x ) for any vertex x V ( G ) . This means that F is a connected even [ 2 , 2 s ] -factor of G.

Subcase 2. p q E ( F ) .

Consider G is 2 s + 1 -tough 2-tree and | S 1 ( G ) | s + 1 , for x { p , q } , we get d G ( x ) = d G ( x ) 1 = | S 1 ( G ) | s + 1 . Note that p q E ( F ) but p q E ( G ) , combining this with s 2 , we have d F ( x ) d G ( x ) 1 s 2 s 2 . Thus, by adding triangle pqv to F we get a connected even factor F of G, i.e., F = F + { p v , p q , q v } . Clearly, d F ( x ) = d F ( x ) + 2 2 s for x { p , q } , d F ( v ) = 2 , and d F ( y ) = d F ( y ) for y V ( G ) \ { p , q , v } . This means F is a connected even [2 , 2 s ] -factor of G.

Case 2. S 2 ( G ) .

Recalling that G is a 2 s + 1 -tough 2-tree with S 2 ( G ) , we prove that G has a connected even [2 , 2 s ] -factor. In fact, here we prove that G has a connected even [2 , 2 s ] -factor containing all edges ab of G such that ω ( G { a , b } ) s . We proceed by induction on n = | V ( G ) | . Clearly, the conclusion holds for n = 4 since G contains a Hamilton cycle which contains all edges ab of G such that ω ( G { a , b } ) 1 . Now suppose n 5 and the conclusion holds for all 2 s + 1 -tough 2-tree with fewer than n vertices.

Let u S 2 ( G ) , by Lemma 1(iv) and (v), there exists a simplicial vertex v S 1 ( G ) such that u v E ( G ) , let G = G v and G is also a 2 s + 1 -tough 2-tree. By the induction hypothesis, G has a connected even [2 , 2 s ] -factor F containing all edges ab of G such that ω ( G { a , b } ) s . Now select a vertex w N G ( v ) \ { u } , it is clear that u w E ( G ) . Note that G is 2 s + 1 -toughness and v is an isolated component of G { u , w } but not a component of G { u , w } , so we have ω ( G { u , w } ) ω ( G { u , w } ) 1 . Combining Lemma 4, we get ω ( G { u , w } ) s + 1 1 = s . By the induction hypothesis, edge u w F and thus, by replacing edge uw in F with path uvw, we get a connected even factor F of G, i.e., F = F { u w } + { u v , v w } . Clearly, d F ( v ) = 2 , d F ( x ) = d F ( x ) for x V ( G ) \ { v } . Combining this with the fact u v , v w F with ω ( G { v , u } ) = ω ( G { v , w } ) = 1 < s , it follows that F is a connected even [2 , 2 s ] -factor of G containing all edges ab such that ω ( G { a , b } ) s .□

The condition 2 s + 1 -tough in Theorem 3 is not necessary for a 2-tree with a connected even [2 , 2 s ] -factor. For example, G = K 2 + 4 K 1 is a 2-tree with a connected even [ 2 , 4 ] -factor, but G is 1 2 -toughness not 2 3 -toughness.

4 The proof of Theorem 5

Assume G K k be a k + 1 s + 2 -tough k-tree. By Theorem 2 and Corollary 4, the conclusion holds for the case when s = 1 or k = 2 . Here suppose s 2 and k 3 . Now we distinguish two cases to complete the proof.

Case 1. S 2 ( G ) = .

We proceed by induction on | V ( G ) | . Clearly, the conclusion holds for | V ( G ) | = k + 2 , in fact, G is k 2 -tough this time and thus k + 1 3 -tough, by Theorem 2, G contains a Hamilton cycle. So here we suppose | V ( G ) | k + 3 and the conclusion holds for | V ( G ) | is fewer than n. Note that G is a k + 1 s + 2 -tough k-tree with S 2 ( G ) = . It is clear that G S 1 ( G ) = K k . Note that | V ( G ) | k + 3 , by Lemma 4, we get | S 1 ( G ) | s + 1 . Take v S 1 ( G ) is a k-simplicial vertex of G, let G = G v , by Lemma 1 ( v ) , G is also a k + 1 s + 2 -tough k-tree. By the induction hypothesis, G contains a connected even [ 2 , 2 s ] -factor F . Now based on F we distinguish two subcases to construct a connected even [2 , 2 s ] -factor of G.

Subcase 1.1. E ( G S 1 ( G ) ) E ( F ) .

Suppose there exists an edge e = p q E ( G S 1 ( G ) ) E ( F ) , then replace edge pq with path pvq to get a connected even factor F of G, i.e., F = F { p q } + { p v , q v } . Clearly, d F ( v ) = 2 and d F ( x ) = d F ( x ) for all vertices x V ( G ) \ { v } . This means F is a connected even [2 , 2 s ] -factor of G.

Subcase 1.2. E ( G S 1 ( G ) ) E ( F ) = .

For any x V ( G S 1 ( G ) ) , since | S 1 ( G ) | s + 1 , then d G ( x ) = d G ( x ) 1 ( s + 1 + k 1 ) 1 = s + k 1 . Note that E ( G S 1 ( G ) ) E ( F ) = and s 2 , we get d F ( x ) d G ( x ) ( k 1 ) s 2 s 2 . Now choose two vertices x , y V ( G S 1 ( G ) ) . Considering x v , y v E ( G ) , then by adding triangle xvy to F we get a connected factor F of G, i.e., F = F + { x v , x y , y v } . Clearly, d F ( x ) = d F ( x ) + 2 2 s , d F ( y ) = d F ( y ) + 2 2 s , d F ( v ) = 2 , and d F ( z ) = d F ( z ) for z V ( G ) \ { x , v , y } . This means that F is a connected even [2 , 2 s ] -factor of G.

Case 2. S 2 ( G ) .

Let G be a k + 1 s + 2 -tough k-tree with order | V ( G ) | and S 2 ( G ) . It is suffice to prove that G contains a connected even [2 , 2 s ] -factor. We proceed by induction on | V ( G ) | .

Subcase 2.1. u has no neighbor in S 1 ( G ) \ { v } .

Clearly, the conclusion holds for | V ( G ) | = k + 2 , since G contains a Hamilton cycle. So here we suppose | V ( G ) | k + 3 and the conclusion holds for | V ( G ) | fewer than n. Now take u S 2 ( G ) , by Lemma 3, there exist v S 1 ( G ) such that u v E ( G ) . Let G = G v , By the induction hypothesis, G contains a connected even [2 , 2 s ] -factor F . Based on F , we prove that G also contains a connected even [2 , 2 s ] -factor.

Since u S 2 ( G ) and u N 1 ( v ) , by Lemma 3, we have | N 2 ( u ) \ N 1 ( v ) | = 1 and let v be vertex of N 2 ( u ) \ N 1 ( v ) . Consider u has no neighbor in S 1 ( G ) \ { v } and F is a connected even [2 , 2 s ] -factor of G , so there exists at least one edge u x E ( F ) E ( G [ N 1 ( v ) ] ) . Note that x v , u v E ( G ) , by replacing edge ux of F with path uvx, we get a connected even [2 , 2 s ] -factor F of G, as desired.

Subcase 2.2. u has at least one neighbor in S 1 ( G ) \ { v } .

In this case, we actually prove that G has a connected even [2 , 2 s ] -factor F, which satisfies E ( K k ) E ( F ) for any k clique K k of G if ω ( G V ( K k ) ) s . Here we still proceed by induction on | V ( G ) | .

Clearly, the conclusion holds when | V ( G ) | = k + 2 . In fact, G has Hamilton cycle C and satisfies E ( K k ) E ( C ) for any k clique K k of G with ω ( G V ( K k ) ) 1 . Now suppose that | V ( G ) | k + 3 and the conclusion holds for | V ( G ) | is fewer than n. Similarly, let G = G v , by the induction hypothesis, G contains a connected even [ 2 , 2 s ] -factor F , which satisfies that E ( K k ) E ( F ) for any k clique K k of G while ω ( G V ( K k ) ) s . Next, based on F , we prove that G contains a connected even [2 , 2 s ] -factor F, which satisfies that E ( K k ) E ( F ) for a k clique K k of G while ω ( G V ( K k ) ) s .

Note that u has a neighbor in S 1 ( G ) \ { v } , suppose that w ( S 1 ( G ) \ { v } ) N 1 ( u ) , by Lemma 4, we have N 1 ( w ) N 2 ( u ) { u } . At the same time, consider N 1 ( v ) is a k clique of G, by Lemma 4 again, we get ω ( G N 1 ( v ) ) s + 1 . This means ω ( G N 1 ( v ) ) s , by the induction hypothesis, G has a connected even [ 2 , 2 s ] -factor F , which satisfies E ( F ) E ( G [ N 1 ( v ) ] ) . Without loss of generality, suppose v 1 v 2 E ( F ) E ( G [ N 1 ( v ) ] ) , note v 1 v , v 2 v E ( G ) , then by replacing edge v 1 v 2 with path v 1 v v 2 we will get a connected even [ 2 , 2 s ] -factor F of G. Furthermore, by simple checking, F is a connected even [2 , 2 s ] -factor such that E ( K k ) E ( F ) for any k clique K k of G while ω ( G V ( K k ) ) s . Of course, here we only need to check in the induced subgraph G [ N 1 ( v ) { v } ] . In fact, suppose that K k be an arbitrary k clique in G [ N 1 ( v ) { v } ] , if v V ( K k ) , it is clear that ω ( G V ( K k ) ) s , note that v 1 v , v v 2 E ( F ) , so the conclusion holds. But if v V ( K k ) , note that K k = G [ N 1 ( v ) ] and thus either ω ( G V ( K k ) ) s or ω ( G V ( K k ) ) s + 1 , this cannot guarantee E ( F ) E ( K k ) holds, and thus so does the conclusion.

By the above argument, G contains a connected even [ 2 , 2 s ] -factor F such that E ( K k ) E ( F ) for any clique K k of G if ω ( G V ( K k ) ) s . This completes the proof of Theorem 5.

Acknowledgments

The authors would like to thank anonymous reviewers for their valuable comments and suggest to improve the quality of the article. Yinkui Li was partially supported by NSFC (No. 11661066), QHFRP (No. 2017-ZJ-701), and QHMUNSFC (No. 2019XJG10). Wen Li was partially supported by QHMUNSFC (No. 2020XJGH14).

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Received: 2020-02-25
Revised: 2020-09-24
Accepted: 2020-10-18
Published Online: 2020-12-29

© 2020 Yinkui Li et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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