aa r X i v : . [ m a t h . C O ] M a r k -tree connectivity of line graphs Shasha LiSchool of Mathematics and Statistics, Ningbo University, Ningbo 315211, Zhejiang, ChinaEmail: [email protected]
Abstract
For a graph G = ( V, E ) and a set S ⊆ V ( G ) of size at least 2, an S -Steiner tree T is a subgraph of G that is a tree with S ⊆ V ( T ). Two S -Steiner trees T and T ′ are internally disjoint (resp. edge-disjoint ) if E ( T ) ∩ E ( T ′ ) = ∅ and V ( T ) ∩ V ( T ′ ) = S (resp. if E ( T ) ∩ E ( T ′ ) = ∅ ). Let κ G ( S ) (resp. λ G ( S )) denote the maximumnumber of internally disjoint (resp. edge-disjoint) S -Steiner trees in G . The k -treeconnectivity κ k ( G ) (resp. k -tree edge-connectivity λ k ( G )) of G is then defined asthe minimum κ G ( S ) (resp. λ G ( S )), where S ranges over all k -subsets of V ( G ). In[H. Li, B. Wu, J. Meng, Y. Ma, Steiner tree packing number and tree connectivity,Discrete Math. 341(2018), 1945–1951], the authors conjectured that if a connectedgraph G has at least k vertices and at least k edges, then κ k ( L ( G )) ≥ λ k ( G ) for any k ≥
2, where L ( G ) is the line graph of G . In this paper, we confirm this conjectureand prove that the bound is sharp. Keywords:
Steiner tree; tree connectivity; tree edge-connectivity; line graph;
MSC Subject Classification 2020:
The connectivity κ ( G ) of a graph G is the minimum cardinality of a subset V ′ ofvertices such that G − V ′ is disconnected or trivial. The edge-connectivity λ ( G ) of a graph G is the minimum cardinality of a subset E ′ of edges such that G − E ′ is disconnected.An equivalent definition of connectivity was given in [13]. For each 2-subset S = { u, v } of vertices of G , let κ ( S ) denote the maximum number of internally vertex-disjoint ( u, v )-paths in G . Then κ ( G ) =min { κ ( S ) | S ⊆ V and | S | = 2 } . Similarly, the edge-connectivityalso has two equivalent definitions. Let λ ( S ) denote the maximum number of edge-disjoint( u, v )-paths in G . Then λ ( G ) =min { λ ( S ) | S ⊆ V and | S | = 2 } .1s a means of strengthening the connectivity, the tree connectivity was introduced byHager [5, 6] (or generalized connectivity by Chartrand et al. [2]) to meet wider applica-tions. Given a graph G = ( V, E ) and a set S ⊆ V ( G ) of size at least 2, an S -Steiner tree or a Steiner tree connecting S is such a subgraph T = ( V ′ , E ′ ) of G that is a tree with S ⊆ V ′ . Two S -Steiner trees T and T ′ are said to be internally disjoint if E ( T ) ∩ E ( T ′ ) = ∅ and V ( T ) ∩ V ( T ′ ) = S . Let κ G ( S ) denote the maximum number of internally disjoint S -Steiner trees in G . The k -tree connectivity (or generalized k -connectivity ) of G , denoted by κ k ( G ), is then defined as κ k ( G ) =min { κ G ( S ) | S ⊆ V ( G ) and | S | = k } , where 2 ≤ k ≤ n .Clearly, when k = 2, κ ( G ) is exactly the classical connectivity κ ( G ).As a natural counterpart of the tree-connectivity, the tree edge-connectivity (or gen-eralized edge-connectivity) was introduced by Li et al. [10]. For S ⊆ V ( G ) and | S | ≥ λ G ( S ) denote the maximum number of edge-disjoint S -Steiner trees in G . The k -treeedge-connectivity (or generalized k -edge-connectivity ) of G , denoted by λ k ( G ), is then de-fined as λ k ( G ) =min { λ G ( S ) | S ⊆ V ( G ) and | S | = k } , where 2 ≤ k ≤ n . It is also clearthat when k = 2, λ ( G ) = λ ( G ).There have been many results on the k -tree (edge-)connectivity, see [3, 7, 9, 10, 12]and a book [8].The line graph L ( G ) of G is the graph whose vertex set can be put in one-to-onecorrespondence with the edge set of G in such a way that two vertices of L ( G ) areadjacent if and only if the corresponding edges of G are adjacent. The connectivity of theline graph of a graph G is closely related to the edge-connectivity of G . Lemma 1.1 (Chartrand and Stewart [4]) . If G is a connected graph, then κ ( L ( G )) ≥ λ ( G ) . Naturally, one would like to study the relationship between κ k ( L ( G )) and λ k ( G ), for k ≥
3. In [10], Li et al. showed that if G is a connected graph, then κ ( L ( G )) ≥ λ ( G ). In[11], Li et al. showed that if a graph G is connected, then κ ( L ( G )) ≥ λ ( G ). Furthermore,they proved that if a connected graph G has at least k vertices and at least k edges, then κ k ( L ( G )) ≥ ⌊ λ k ( G )4 ⌋ − k ≥
2. However, they suspect that their result is notsharp and proposed the following conjecture:
Conjecture 1.1 (Li, Wu, Meng and Ma [11]) . Let k ≥ be an integer. If a connectedgraph G has at least k vertices and at least k edges, then κ k ( L ( G )) ≥ λ k ( G ) . In this paper, we will confirm this conjecture and prove that the bound is sharp.2
Main result
Before proving our main result, we first introduce some concepts. A maximal con-nected subgraph of G is called a component of G . A connected acyclic graph is calleda tree . The vertices of degree 1 in a tree are called leaves . A connected graph G with | V ( G ) | = | E ( G ) | is called a unicyclic graph . A spanning subgraph of a graph G is asubgraph whose vertex set is the entire vertex set of G . We refer the reader to [1] for theterminology and notations not defined in this paper.By the definition of the tree edge-connectivity, the following result is obvious. Observation 2.1 ([11]) . For any integer ≤ s ≤ t , λ s ( G ) ≥ λ t ( G ) . Now, we give a confirmative solution to Conjecture 1.1.
Theorem 2.2.
Let k ≥ be an integer. If a connected graph G has at least k verticesand at least k edges, then κ k ( L ( G )) ≥ λ k ( G ) . Moreover, the bound is sharp.Proof. Let v e be the vertex of the line graph L ( G ) corresponding to the edge e of G .Assume that λ k ( G ) = m . For any k -subset S L = { v e , v e , . . . , v e k } of V ( L ( G )), by thedefinition of κ k , it suffices to show that κ L ( G ) ( S L ) ≥ m .Now, let S G = { e , e , . . . , e k } and then S G ⊆ E ( G ). Denote by G [ S G ] the edge-induced subgraph of G whose edge set is S G and whose vertex set consists of all ends ofedges of S G .We distinguish two cases: Case 1:
None of the components of G [ S G ] is a tree or unicyclic.Therefore, for each component C l of G [ S G ], | E ( C l ) | − | V ( C l ) | ≥ | E ( G [ S G ]) | = | S G | = k ≥ | V ( G [ S G ]) | + 1. Let V ( G [ S G ]) = Q ∗ . It follows that | Q ∗ | ≤ k −
1. Since G has at least k vertices, we can take a vertex v ∗ in V ( G ) \ Q ∗ and then let Q = Q ∗ ∪ { v ∗ } .Now, because | Q | ≤ k , by Observation 2.1, there are m edge-disjoint Q -Steiner trees T , T , . . . , T m in G .Next, in each tree T r (1 ≤ r ≤ m ), we will assign a specific edge to each vertex of Q ∗ .To see this, we let v ∗ be the root and define the level l ( v ) of a vertex v in T r to be thedistance from the root v ∗ to v . It is easy to see that, for each vertex v i ∈ Q ∗ = Q \ { v ∗ } ,there is a unique edge e connecting the vertex v i and a vertex of level l ( v i ) −
1. Assignthe edge e to the vertex v i . We say that the edge e is the corresponding edge of v i in T r and denoted by ˆ e ri . Note that any two vertices of Q ∗ in T r have different correspondingedges. More precisely, ˆ e ri = ˆ e rj for any 1 ≤ i = j ≤ | Q ∗ | .3ow, for each tree T r (1 ≤ r ≤ m ) and each edge e = v i v j ∈ S G , do the followingoperation. Note that, by Lemma 1.1, L ( T r ) (1 ≤ r ≤ m ) is a connected subgraph of L ( G ). Moreover, since Q ∗ = V ( G [ S G ]), both ends of each edge in S G belong to Q ∗ andso v i , v j ∈ V ( T r ). Operation A: If e ∈ E ( T r ), it is done; otherwise e / ∈ E ( T r ), that is v e / ∈ V ( L ( T r )). Notethat, T , T , . . . , T m are edge-disjoint and so at most one of them contains the edge e .If e ∈ E ( T s ), where 1 ≤ s = r ≤ m , then e is the corresponding edge of one of itsends in T s . Without loss of generality, assume that e is the corresponding edge of v i in T s , that is, ˆ e si = e . Now, for T r , there is an edge ˆ e rj corresponding to the vertex v j , whichis the other end of e . Since e and ˆ e rj have the same end v j , they are adjacent and so v e v ˆ e rj ∈ E ( L ( G )). Add the vertex v e and the edge v e v ˆ e rj to L ( T r ).Otherwise, none of the trees T , T , . . . , T m contains the edge e = v i v j . In this case,we can add the vertex v e and either the edge v e v ˆ e ri or the edge v e v ˆ e rj to L ( T r ), where ˆ e ri and ˆ e rj are the corresponding edges of v i and v j in T r , respectively. (cid:3) Now, L ( T ) , L ( T ) , . . . , L ( T m ) are transformed into m connected subgraphs of L ( G ),each of which contains the vertex set S L . Next, for each of the obtained subgraphs of L ( G ), take a spanning tree T ∗ r (1 ≤ r ≤ m ). Because V ( T ∗ r ) ⊇ S L (1 ≤ r ≤ m ), it remainsto show that T ∗ , T ∗ , . . . , T ∗ m are internally disjoint. Note that, if v e / ∈ V ( L ( T r )), for some e ∈ S G and r ∈ { , , . . . , m } , v e must be a leaf of T ∗ r .Since T , T , . . . , T m are edge-disjoint in G , L ( T ) , L ( T ) , . . . , L ( T m ) are vertex-disjointin L ( G ). Moreover, the vertices added to L ( T r ) by Operation A are all from S L . Therefore, T ∗ , T ∗ , . . . , T ∗ m are vertex-disjoint except S L , that is, V ( T ∗ r ) ∩ V ( T ∗ s ) = S L , for any 1 ≤ r < s ≤ m .Now, assume that there are two trees T ∗ r and T ∗ s such that E ( T ∗ r ) ∩ E ( T ∗ s ) = ∅ (1 ≤ r < s ≤ m ). Let e ∗ ∈ E ( T ∗ r ) ∩ E ( T ∗ s ). Since V ( T ∗ r ) ∩ V ( T ∗ s ) = S L , both ends of e ∗ belong to S L . Thus, without loss of generality, let e ∗ = v e v e . If L ( T r ) contains neither v e nor v e , by Operation A, both v e and v e are leaves of T ∗ r and hence it is impossiblethat v e v e ∈ E ( T ∗ r ). So is L ( T s ). And L ( T r ) and L ( T s ) are vertex-disjoint ( T r and T s are edge-disjoint). Thus, without loss of generality, suppose that v e ∈ L ( T r ) and v e ∈ L ( T s ), and so v e / ∈ L ( T r ) and v e / ∈ L ( T s ). Since v e and v e are adjacent in L ( G ), e and e are adjacent in G . Assume that v i is the common end of e and e in G andlet e = v i v j . Since v e v e ∈ E ( T ∗ s ), we added the vertex v e and the edge v e v e to L ( T s ).So by Operation A, we know that e is exactly the corresponding edge of v i in T s , thatis, e = ˆ e si . Again by Operation A, since e / ∈ E ( T r ), we added the vertex v e and the4dge v e v ˆ e rj to L ( T r ), where e and ˆ e rj have the same end v j . Since e = e and e and e have the same end v i , it is impossible that ˆ e rj = e . Therefore, by Operation A, it isimpossible that v e v e = e ∗ ∈ E ( T ∗ r ), a contradiction. It follows that T ∗ , T ∗ , . . . , T ∗ m areedge-disjoint.Thus, in this case, T ∗ , T ∗ , . . . , T ∗ m are m internally disjoint trees connecting S L in L ( G ). Case 2:
There is a component of G [ S G ] which is either a tree or unicyclic.For each component C l of G [ S G ] which is neither a tree nor unicyclic, add all verticesof C l to the vertex set Q and add all edges of C l to the edge set S G . Clearly, if Q = ∅ , | S G | > | Q | .Next, for each component C t of G [ S G ] which is either a tree or unicyclic, if C t isunicyclic, choose an edge e t from C t such that C t − e t is a tree and let one end of e t asthe root r t ; otherwise, select an arbitrary vertex as the root r t . For C t (if C t is a tree)or C t − e t (if C t is unicyclic), define the level l ( v ) of a vertex v to be the distance fromthe root r t to v . Notice that each edge in the tree C t (or C t − e t if C t is unicyclic) joinsvertices on consecutive levels. Then, for each edge e = uv , where l ( u ) + 1 = l ( v ), weassign the vertex v which has higher level to the edge e and say that the vertex v is thecorresponding vertex of the edge e . If C t is unicyclic, let the root r t be the correspondingvertex of the remaining edge e t . Now, each edge of C t has a corresponding vertex. Bythe definition, it is obvious that any two edges of C t have different corresponding vertices.Add the corresponding vertices of all edges of C t to the vertex set Q and add all edgesof C t to the edge set S G . Clearly, | S G | = | Q | .Moreover, it is clear that Q ∩ Q = ∅ , S G ∩ S G = ∅ and S G = S G ∪ S G . Let S L = { v e | e ∈ S G } and S L = { v e | e ∈ S G } . Then S L = S L ∪ S L . Let Q = Q ∪ Q . We have | Q | = | Q | + | Q | ≤ | S G | + | S G | = | S G | = k . Since Q ⊆ V ( G ), by Observation 2.1, thereare m edge-disjoint Q -Steiner trees T , T , . . . , T m in G . Note that both ends of each edgein S G belong to Q , but there may be an edge in S G , only one end of which belongs to Q . Thus, we use different methods to deal with the edges in S G and S G .For every edge of S G , we take the same approach as Case 1. In each tree T r (1 ≤ r ≤ m ), since Q = ∅ (it is possible that Q = ∅ ), take an arbitrary vertex v ∗ in Q as theroot and define the level l ( v ) of a vertex v in T r to be the distance from the root v ∗ to v . For each vertex v i ∈ Q (if Q = ∅ ), there is a unique edge e connecting the vertex v i and a vertex of level l ( v i ) −
1. Let the edge e be the corresponding edge of v i in T r ,denoted by ˆ e ri . Any two vertices of Q in T r have different corresponding edges. Now,apply Operation A to each tree T r (1 ≤ r ≤ m ) and each edge e = v i v j ∈ S G . Then, each5ertex of S L is added to L ( T r ) (1 ≤ r ≤ m ).Next, for each edge e i of S G and each tree T r (1 ≤ r ≤ m ), do the following operation. Operation B: If e i ∈ E ( T r ), it is done; otherwise e i / ∈ E ( T r ), that is v e i / ∈ V ( L ( T r )). Bythe definitions of S G and Q , there is a corresponding vertex v i of e i , and v i ∈ Q ⊆ Q and so v i ∈ V ( T r ). Thus, there exists an edge ˜ e ri = e i incident with v i in the tree T r .Since e i and ˜ e ri have the same end v i , they are adjacent and so v e i v ˜ e ri ∈ E ( L ( G )). Addthe vertex v e i and the edge v e i v ˜ e ri to L ( T r ).Now, after applying Operations A and B, L ( T ) , L ( T ) , . . . , L ( T m ) are transformedinto m connected subgraphs of L ( G ), each of which contains the vertex set S L = S L ∪ S L .For each of the obtained subgraphs of L ( G ), take a spanning tree T ∗ r (1 ≤ r ≤ m ). Notethat, if v e / ∈ V ( L ( T r )), for some e ∈ S G and r ∈ { , , . . . , m } , whether e ∈ S G or S G ,that is, whether Operation A or Operation B is applied, v e must be a leaf of T ∗ r . Because V ( T ∗ r ) ⊇ S L for any 1 ≤ r ≤ m , it remains to show that T ∗ , T ∗ , . . . , T ∗ m are internallydisjoint.Since L ( T ) , L ( T ) , . . . , L ( T m ) are vertex-disjoint in L ( G ) and the vertices added to L ( T r ) by Operations A and B are all from S L , T ∗ , T ∗ , . . . , T ∗ m are vertex-disjoint except S L , that is, V ( T ∗ r ) ∩ V ( T ∗ s ) = S L , for any 1 ≤ r < s ≤ m .To complete the proof, it remains to show that T ∗ , T ∗ , . . . , T ∗ m are edge-disjoint. Bycontradiction, assume that there are two trees T ∗ r and T ∗ s such that E ( T ∗ r ) ∩ E ( T ∗ s ) = ∅ (1 ≤ r < s ≤ m ). Let e ∗ ∈ E ( T ∗ r ) ∩ E ( T ∗ s ). Since V ( T ∗ r ) ∩ V ( T ∗ s ) = S L , both endsof e ∗ belong to S L . Thus, without loss of generality, let e ∗ = v e v e . If L ( T r ) containsneither v e nor v e , then whether apply Operation A or Operation B, both v e and v e areleaves of T ∗ r , which is impossible. So is L ( T s ). And L ( T r ) and L ( T s ) are vertex-disjoint( T r and T s are edge-disjoint). Thus, without loss of generality, suppose that v e ∈ L ( T r )and v e ∈ L ( T s ), and so v e / ∈ L ( T r ) and v e / ∈ L ( T s ). Since v e and v e are adjacent in L ( G ), e and e are adjacent in G . Therefore, e and e belong to the same componentof G [ S G ]. Hence, by the definitions of S G and S G , both e and e belong to S G or S G .If both e and e belong to S G , by Operation A, it is impossible that v e v e = e ∗ ∈ E ( T ∗ r ) ∩ E ( T ∗ s ). The proof is the same as that of Case 1.If both e and e belong to S G , since e / ∈ E ( T r ), by Operation B, we added thevertex v e and the edge v e v ˜ e r to L ( T r ), where the common end v of e and ˜ e r in G isthe corresponding vertex of e . Similarly, since e / ∈ E ( T s ), by Operation B, we addedthe vertex v e and the edge v e v ˜ e s to L ( T s ), where the common end v of e and ˜ e s in G is the corresponding vertex of e . Since v = v by the definition of Q , at least one6f the equations ˜ e r = e and ˜ e s = e is not true. So v e v ˜ e r = v e v e or v e v ˜ e s = v e v e .It is impossible that e ∗ = v e v e ∈ E ( T ∗ r ) ∩ E ( T ∗ s ), a contradiction. It follows that T ∗ , T ∗ , . . . , T ∗ m are edge-disjoint.Thus, in both cases, there always exist m internally disjoint trees connecting S L in L ( G ) and so κ L ( G ) ( S L ) ≥ m . By the arbitrariness of S L , we conclude that κ k ( L ( G )) ≥ m .For a cycle C n with n ≥ k , since L ( C n ) = C n , κ k ( L ( C n )) = λ k ( C n ) = 1 for k ≥ κ ( L ( C n )) = λ ( C n ) = 2. Thus, the bound is sharp. The proof is complete. Acknowledgments.
The author’s work was supported by Zhejiang Provincial Natural Science Foundationof China (No. LY18A010002) and the Natural Science Foundation of Ningbo, China.