A Sufficient Condition for 2-Distance-Dominating Cycles

A cycle C of a graph G is a m-distance-dominating cycle if for all vertices of . Defining denotes the minimum value of the degree sum of any k independent vertices of G. In this paper, we prove that if G is a 3-connected graph on n vertices, and if , then every longest cycle is m-distance-dominating cycles.

• Degree Sums
• Distance Dominating Cycles
• Insertible Vertex

1. Introduction

Let G = ( V , E ) be a graph and H be a subgraph of G, for a S ⊆ V ( G ) , let N H ( S ) = N ( S ) ∩ V ( H ) . For any x , y ∈ V ( G ) , Y ⊆ V ( G ) , x y denotes the edge with ends x and y, an ( x , Y ) -path denotes a path starting at x and ending at Y. We denote by α ( G ) and κ ( G ) the independence number and the connectivity of G, respectively.

Let C be a cycle of G, and denote by C → the cycle C with a given orientation. For v ∈ V ( C ) , define v + and v − to be the successor and predecessor of v on C, define v + i and v − i to be the i-th successor and predecessor of v on C, respectively. In particular, we write A + i = { a + i | a ∈ A } and A − i = { a − i | a ∈ A } . If u , v ∈ V ( G ) , we denote by u C → v the consecutive vertices of C from u to v in the direction specified by C → . The same path, in reverse order, is denoted by v C ← u . We will consider u C → v and v C ← u both as paths and as vertex sets.

We use min { d G ( v 1 , v 2 ) : v 1 ∈ V ( H 1 ) , v 2 ∈ V ( H 2 ) } to denote the distance d G ( H 1 , H 2 ) between H 1 and H 2 , H 1 and H 2 are all the subgraphs of G, where d G ( v 1 , v 2 ) denotes the length of a shortest path between v 1 and v 2 in G. A subgraph H of G is m-dominating if for all x ∈ V ( G ) , d G ( x , H ) ≤ m . For an integer k ≥ 2 , define

σ k = min { ∑ i = 1 k     d ( x i ) | { x 1 , ⋯ , x k } is an independent set of G }

In 1987, Bondy [1] considered the existence of k-connected graphs of order n.

Theorem 1 [1] Let G be a k-connected graph on n vertices, where k ≥ 2 . If any k + 1 independent vertices x i ( 0 ≤ i ≤ k ) with N ( x i ) ∩ N ( x j ) = ∅ ( 0 ≤ i ≠ j ≤ k ) have degree-sum ∑ i = 0 k     d ( x i ) ≥ n − 2 k , then G has a 1-distance-dominating cycle.

In 1988, Broersma [2] and Fraisse [3] proved some results about m-distance-dominating cycles.

Theorem 2 [2] [3] Let G be a k-connected graph with no set of cardinality k + 1 , whose vertices are pairwise at distance at least 2 m + 2 . Then G has an m-distance-dominating cycle.

The circumference c ( G ) of a graph G is the length of the longest cycle on the graph. In 2021, Xiong [4] considered the relation between the graph circumference and m-distance-dominating cycle, and proved a sufficient condition that every longest cycle in k-connected graph is m-distance-dominating cycle.

Theorem 3 [4] Let G be a graph with κ ( G ) = k ≥ 2 . If c ( G ) ≥ ( 2 m + 2 ) k − 1 , then every longest cycle of G is a m-distance-dominating cycle.

A cycle C is m-edge-dominating if for all e ∈ E ( G ) , d G ( e , C ) ≤ m . Clearly, a cycle is 0-edge-dominating (or simply dominating) if every edge of G is incident with a vertex of C, G − V ( C ) is edgeless. It is very popular to decide whether a longest cycle is (0-edge) dominating. Bondy [5] gave a sufficient condition such that every longest cycle of 2-connected graph is (0-edge) dominating.

Theorem 4 [5] Let G be a 2-connected graph on n vertices. If σ 3 ( G ) ≥ n + 2 , then every longest cycle is dominating.

Wu [6] considered the same problem for k-connected graphs and established the following.

Theorem 5 [6] Let G be k-connected graph on n vertices with k ≥ 2 . If σ k + 1 ( G ) > ( n + 1 ) ( k + 1 ) / 3 , then every longest cycle is dominating.

In this paper, we consider the general version for degree sums condition that guarantees that every longest cycle is a 2-distance-dominating cycle in 3-connected graphs. Our main result is the following.

Theorem 6 Let G be a 3-connected graph on n vertices. If σ 4 ( G ) > 4 n / 3 − 4 / 3 , then every longest cycle is a 2-distance-dominating cycle.

1. Key Lemmas

Lemma 1 [6] Let P = u 1 u 2 ⋯ u l and Q 1 , ⋯ , Q m be m + 1 pairwise vertex disjoint paths of a graph G. If for any v ∈ V ( Q i ) , there are u k , u k + 1 ∈ N ( v ) such that { u k , u k + 1 } ⊈ N ( v ′ ) for any v ′ ∈ V ( Q j ) with j ≠ i , then G has a ( u 1 , u l ) -path with V ( P ) ∪ ( ∪ i = 1 m V ( Q i ) ) as its vertex set.

A k-fan from x to Y is a family of k internal disjoints ( x , Y ) -paths whose terminal vertices are distinct. The following lemma known as Fan Lemma establishes an useful property of k-connected graphs.

Lemma 2 [7] Let G be a k-connected graph, let x be a vertex of G, and let Y ⊆ V ( G ) \ x be a set of at least k vertices of G. Then there exists a k-fan in G from x to Y.

Next, we assume G be a k-connected non-hamiltonian graph of order n, k ≥ 2 . Let C be a longest cycle of G with a given orientation. Let R = V ( G ) − V ( C ) , assume H is a component of G − V ( C ) and N C ( H ) = { h 1 , h 2 , ⋯ , h t } , where the subscripts increase with the orientation of C.

A vertex u ∈ h i + C → h i + 1 − is insertible if there exist vertices v , v + ∈ h i + 1 C → h i such that u v , u v + ∈ E ( G ) and the edge v v + ∈ E ( G ) is called an insertion edge of u, and noninsertible otherwise.

For any i with 1 ≤ i ≤ t , if each vertex of h i + C → h i + 1 − is insertible, then by Lemma 1, G has an ( h i , h i + 1 ) -path P such that V ( P ) = V ( C ) . Thus, there is a ( h i , h i + 1 ) -path L with internal vertices in H and | L | ≥ 3 . We find ℂ = h i P h i + 1 L h i is a cycle longer than C, contradiction. Thus, h i + C → h i + 1 − contains at least one noninsertible vertex. Write a i as the first noninsertible vertex occurring on h i + C → h i + 1 − , A i = h i + C → a i . For any v ∈ V ( H ) , we let A v = { a i | h i ∈ N ( v ) } .

Lemma 3 [6] 1) There is no ( x , y ) -path without internal vertices in V ( C ) ∪ V ( H ) for any x ∈ A i and y ∈ A j with i ≠ j ;

2) N P − ( A i ) ∩ N P ( A j ) = N Q ( A i ) ∩ N Q − ( A j ) = ∅ , where P = a i + C → h j and Q = a j + C → h i .

Lemma 4 [6] Suppose v 1 , v 2 ∈ V ( H ) with v 1 ≠ v 2 , a i ∈ A v 1 and a j ∈ A v 2 with i ≠ j . Then 1) a i + ∉ N ( A j ) and a j + ∉ N ( A i ) ;

2) N P − 2 ( A i ) ∩ N P ( A j ) = N Q ( A i ) ∩ N Q − 2 ( A j ) = ∅ , where P = a i + C → h j and Q = a j + C → h i .

2. Proof of Theorem 6

Let G be a graph of order n with connectivity κ ( G ) ≥ 3 , satisfying σ 4 ( G ) > ( 4 n − 4 ) / 3 . Let C be a longest cycle with a given orientation and | C | ≥ 3 since G is 3-connected. Assuming there is a vertex u ∈ V ( G ) \ C such that d ( u , C ) ≥ 3 . By Lemma 2, there is a 3-fan B = { P 1 , P 2 , P 3 } in G from u to V ( C ) and the length of P i is at least 3, i = 1 , 2 , 3 . Let R = V ( G ) − V ( C ) , and H is a component of G − V ( C ) containing u. By the definition of k-fan, we get | H | ≥ 7 . Let x i = V ( P i ) ∩ V ( C ) , i = 1 , 2 , 3 , a i is the noninsertible vertex as the same definition on the previous section, A i = x i + C → a i , i = 1 , 2 , 3 . And v i is x i ’s first neighbor on the path P i , i = 1 , 2 , 3 .

Claim 1 A i ∩ N C ( H ) = ∅ , i = 1 , 2 , 3 .

proof Without loss of generality, suppose there is a vertex x ∈ A 1 such that x ∈ N C ( H ) . Let Q = x 1 + C → x , P = x + C → x 1 , then all the vertex on Q are insertible. Thus we can get a ( x 1 , x ) -path P ′ such that V ( P ′ ) = V ( C ) by Lemma 1. Let L denote a ( x 1 , x ) -path with internal vertices in H, and | L | ≥ 3 . We can get a new cycle ℂ = x 1 P ′ x L x 1 longer than C. (Contradiction) □

Claim 2 d ( v 1 ) + d ( a 1 ) + d ( a 2 ) ≤ n .

proof We first find that { a 1 , a 2 } ∩ N C ( H ) = ∅ by Claim 1, a 1 and a 2 have no common neighbors on R − H since Lemma 3(1). Thus, N R ( v 1 ) , N R ( a 1 ) , N R ( a 2 ) are pairwise disjoint. And since d ( u , C ) ≥ 3 , u v 1 ∉ E ( G ) . Therefore, we have the inequality as follows.

d R ( v 1 ) + d R ( a 1 ) + d R ( a 2 ) ≤ | R | − 2.

Similarly, by Claim 1 and Lemma 3(1), we have

d A 1 ∪ A 2 ( v 1 ) + d A 1 ∪ A 2 ( a 1 ) + d A 1 ∪ A 2 ( a 2 ) ≤ | A 1 | − 1 + | A 2 | − 1.

Next let P = a 1 + C → x 2 , Q = a 2 + C → x 1 , U 1 = A v 1 ∩ V ( P ) , U 2 = A v 1 ∩ V ( Q ) . Since N C ( v 1 ) ∩ A i = ∅ , i = 1 , 2 , 3 , then A v 1 \ { a 1 , a 2 } ⊆ U 1 ∪ U 2 . Note that d C ( v 1 ) = | A v 1 | , thus | U 1 | + | U 2 | ≥ d C ( v 1 ) − 2 . Let U 1 = { a v 11 , a v 12 , ⋯ , a v 1 t } , t ≤ n . We will analyse d P ( a 1 ) + d P ( a 2 ) by considering the following cases.

Case 1. For any a v 1 j + ∈ U 1 + , a v 1 j + ∉ N P ( a 1 ) , which implies a v 1 j ∉ N P − ( a 1 ) , j = 1 , 2 , ⋯ , t .

By Lemma 3(1), we have a v 1 j ∉ N P ( a 2 ) , and thus for any a v 1 j ∈ U 1 , j = 1 , 2 , ⋯ , t , we have a v 1 j ∉ N P − ( a 1 ) ∪ N P ( a 2 ) . And by Lemma 3(2), N P − ( a 1 ) ∩ N P ( a 2 ) = ∅ . Hence N P − ( a 1 ) ∪ N P ( a 2 ) ⊆ V ( P ) ∪ { a 1 } \ U 1 . Therefore, d P ( a 1 ) + d P ( a 2 ) = | N P − ( a 1 ) | + | N P ( a 2 ) | ≤ | P | + 1 − | U 1 | .

Case 2. There exist some a v 1 j + ∈ U 1 + , a v 1 j + ∈ N P ( a 1 ) , say { a v i 1 + , a v i 2 + , ⋯ , a v i r + } , r ≤ t .

Then we can note that a v i j + + ∉ N ( a 1 ) . Since a 1 is noninsertible, which implies a v i j + ∉ N P − ( a 1 ) , j = 1 , 2 , ⋯ , r , r ≤ t . And by Lemma 4(1), we know a v i j + ∉ N P ( a 2 ) . Thus a v i j + ∉ N P − ( a 1 ) ∪ N P ( a 2 ) . On the other hand, for the remaining vertices, a v 1 j ∈ U 1 \ { a v i 1 , a v i 2 , ⋯ , a v i r } , similar to case 1, we have a v 1 j ∉ N P − ( a 1 ) ∪ N P ( a 2 ) . In addition, a v 1 j + ≠ a v 1 j + 1 , since there are some x ∈ N C ( H ) on a v 1 j C → a v 1 j + 1 . And by Lemma 3(2), N P − ( a 1 ) ∩ N P ( a 2 ) = ∅ . Therefore, we have the inequality as follows.

d P ( a 1 ) + d P ( a 2 ) = | N P − ( a 1 ) | + | N P ( a 2 ) | ≤ | P | + 1 − | U 1 | .

By Lemma 3(1) and Lemma 4(1), for any a 1 j ∈ U 2 , we have a 1 j ∉ N Q ( a 1 ) ∪ N Q − ( a 2 ) . So we have the inequality as follows.

d Q ( a 1 ) + d Q ( a 2 ) = | N Q ( a 1 ) | + | N Q − ( a 2 ) | ≤ | Q | + 1 − | U 2 | .

Therefore,

d ( v 1 ) + d ( a 1 ) + d ( a 2 ) ≤ ( d P ( v 1 ) + | P | + 1 − | U 1 | ) + ( d Q ( v 1 ) + | Q | + 1 − | U 2 | ) + ( | A 1 | + | A 2 | − 2 ) + ( | R | − 2 ) = d C ( v 1 ) + ( | P | + | Q | + | A 1 | + | A 2 | + | R | ) − ( | U 1 | + | U 2 | ) − 2 ≤ n + d C ( v 1 ) − ( d C ( v 1 ) − 2 ) − 2 = n .

□

By a similar argument as Claim 2, Claim 3 holds.

Claim 3 d ( v 1 ) + d ( a 1 ) + d ( a 3 ) ≤ n .

Claim 4 d ( v 1 ) + d ( a 2 ) + d ( a 3 ) ≤ n .

proof Similarly, by Claim 1 and Lemma 3(1), we have

d R ( v 1 ) + d R ( a 2 ) + d R ( a 3 ) ≤ | R | − 2.

d A 2 ∪ A 3 ( v 1 ) + d A 2 ∪ A 3 ( a 2 ) + d A 2 ∪ A 3 ( a 3 ) ≤ | A 2 | − 1 + | A 3 | − 1.

Let P = a 2 + C → x 3 , Q = a 3 + C → x 2 . U 1 = A v 1 ∩ V ( P ) , U 2 = A v 1 ∩ V ( Q ) . Then we have | U 1 | + | U 2 | ≥ d C ( v 1 ) − 2 . And by Lemma 3(2), N P − ( a 2 ) ∩ N P ( a 2 ) = ∅ .

Let U 1 = { a v 11 , a v 12 , ⋯ , a v 1 t } , for any a v 1 j ∈ U 1 , we have a v 1 j + ∉ N P ( a 2 ) by Lemma 4(1), that is a v 1 j ∉ N P − ( a 2 ) , j = 1 , 2 , ⋯ , t . And for any a v 1 j ∈ U 1 , we have a v 1 j ∉ N P ( a 3 ) by Lemma 3(1). Therefore, a v 1 j ∉ N P − ( a 2 ) ∪ N P ( a 3 ) , j = 1 , 2 , ⋯ , t .

By Lemma 3(2), N P − ( a 2 ) ∩ N P ( a 3 ) = ∅ . Hence,

N P − ( a 2 ) ∪ N P ( a 3 ) ⊆ V ( P ) ∪ { a 2 } \ U 1 .

Thus we have the inequality as follows,

d P ( a 2 ) + d P ( a 3 ) = | N P − ( a 2 ) | + | N P ( a 3 ) | ≤ | P | + 1 − | U 1 | .

Furthermore, according to the symmetry of P and Q,

d Q ( a 2 ) + d Q ( a 3 ) ≤ | Q | + 1 − | U 2 | .

Therefore,

d ( v 1 ) + d ( a 2 ) + d ( a 3 ) ≤ ( d P ( v 2 ) + | P | + 1 − | U 1 | ) + ( d Q ( v 1 ) + | Q | + 1 − | U 2 | ) + ( | A 2 | + | A 3 | − 2 ) + ( | R | − 2 ) = d C ( v 1 ) + ( | P | + | Q | + | A 2 | + | A 3 | + | R | ) − ( | U 1 | + | U 2 | ) − 2 ≤ n + d C ( v 1 ) − ( d C ( v 1 ) − 2 ) − 2 = n .

□

Claim 5 d ( a 1 ) + d ( a 2 ) + d ( a 3 ) ≤ n − 4 .

proof Let P = a 1 + C → x 2 , Q = a 2 + C → x 3 , M = a 3 + C → x 1 . By Lemma 3(2) and Lemma 4(2), note that N P − 2 ( a 1 ) , N P ( a 2 ) , N P − ( a 3 ) are pairwise disjoint. So N P − 2 ( a 1 ) ∪ N P ( a 2 ) ∪ N P − ( a 3 ) ⊆ a 1 − C → x 2 , which implies

d P ( a 1 ) + d P ( a 2 ) + d P ( a 3 ) ≤ | P | + 2.

According to the symmetry of P, Q and R, we have

d Q ( a 1 ) + d Q ( a 2 ) + d Q ( a 3 ) ≤ | Q | + 2.

d M ( a 1 ) + d M ( a 2 ) + d M ( a 3 ) ≤ | M | + 2.

By Lemma 3(1), we have

d A 1 ∪ A 2 ∪ A 3 ( a 1 ) + d A 1 ∪ A 2 ∪ A 3 ( a 2 ) + d A 1 ∪ A 2 ∪ A 3 ( a 3 ) ≤ | A 1 | − 1 + | A 2 | − 1 + | A 3 | − 1.

At last, by Claim 1 and Lemma 3(1), we have

d R ( a 1 ) + d R ( a 2 ) + d R ( a 3 ) ≤ | R | − | H | .

Note that | H | ≥ 7 , thus

d ( a 1 ) + d ( a 2 ) + d ( a 3 ) ≤ ( | P | + 2 ) + ( | Q | + 2 ) + ( | M | + 2 ) + ( | A 1 | + | A 2 | + | A 3 | − 3 ) + ( | R | − | H | ) = ( | P | + | Q | + | M | + | A 1 | + | A 2 | + | A 3 | + | R | ) + 3 − | H | = n + 3 − | H | ≤ n − 4.

□

By Lemma 3(1) and Claim 1, { v 1 , a 1 , a 2 , a 3 } is an independent set in G. By Claim 2-5, we have

d ( v 1 ) + d ( a 1 ) + d ( a 2 ) + d ( a 3 ) ≤ 4 3 n − 4 3 ,

a contradiction.

This completes the proof of Theorem 6.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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