Closure and Spanning k-Trees

In this paper, we propose a new closure concept for spanning k-trees. A k-tree is a tree with maximum degree at most k. We prove that: Let G be a connected graph and let u and v be nonadjacent vertices of G. Suppose that $${\sum_{w \in S}d_G(w) \geq |V(G)| -1}$$ for every independent set S in G of order k with $${u,v \in S}$$. Then G has a spanning k-tree if and only if G + uv has a spanning k-tree. This result implies Win’s result (Abh Math Sem Univ Hamburg, 43:263–267, 1975) and Kano and Kishimoto’s result (Graph Comb, 2013) as corollaries.

paper, we consider a closure concept for spanning k-trees, and refer the reader to the survey [3] on closure concept. A k-tree is a tree with maximum degree at most k. Win [6] obtained a degree sum condition for the existence of spanning k-trees.
Theorem 1 (Win [6]) Let k ≥ 2 be an integer, and let G be a connected graph. If v∈S d G (v) ≥ |V (G)| − 1 for every independent set S in G of order k, then G has a spanning k-tree.
Recently, Kano and Kishimoto [4] considered a closure concept for spanning ktrees, and proved the following theorem.
Theorem 2 (Kano and Kishimoto [4]) Let k ≥ 2 be an integer, and let G be an m-connected graph. Let u and v be two nonadjacent vertices of G. Suppose that Then G has a spanning k-tree if and only if G + uv has a spanning k-tree.
In this paper, we give a closure result which implies the above theorems as corollaries.
Theorem 3 Let k ≥ 2 be an integer, and let G be a connected graph. Let u and v be two nonadjacent vertices of G. Suppose that w∈S d G (w) ≥ |V (G)| − 1 for every independent set S in G of order k such that u, v ∈ S. Then G has a spanning k-tree if and only if G + uv has a spanning k-tree.
We now show that a graph satisfying the condition of Theorem 2 also satisfies that of Theorem 3.
Proof of Theorem 2 Assume that G is an m-connected graph and satisfies Hence G satisfies the condition of Theorem 3.

Proof of Theorem 3
We prove a slightly stronger theorem than Theorem 3. For a graph G and If there is no confusion, then we abbreviate k (S; G) to k (S). Theorem 4 Let k ≥ 2 be an integer, and let G be a connected graph. Let u and v be two nonadjacent vertices of G. Suppose that there exists no independent set of order k + 1 containing both u and v, or k (S) ≥ |V (G)| − 1 for every independent set S in G of order k + 1 such that u, v ∈ S. Then G has a spanning k-tree if and only if G + uv has a spanning k-tree.
1. The degree condition of Theorem 4 is best possible in the following sense.
Let G be a complete bipartite graph K n,n(k−1)+2 with partite sets X and Y such that |X | = n and |Y | = n(k − 1) + 2, where n ≥ 1 and k ≥ 2. Let u and v be two vertices of Y . Then k (S) = nk = |V (G)| − 2 for every independent set S of order k + 1 such that u, v ∈ S, and G + uv has a spanningk-tree. But G has no spanning k-tree, because if G has a spanning k-tree T , then Let G 1 and G 2 be graphs obtained from G by recursively joining pairs of nonadjacent vertices which satisfy the condition of Theorem 4 until there exists no such a pair. Let e 1 , e 2 , . . . , e m and f 1 , f 2 , . . . , f n be the sequences of edges added to G in obtaining G 1 and G 2 , respectively. Suppose that e 1 , e 2 , . . . , e l ∈ E(G 2 ) and e l+1 ∈ E(G 2 ). Let e l+1 := uv and H := G + e 1 + · · · + e l . Then, by the definition of G 2 , there exists an independent set S in G 2 of order k + 1 such that u, v ∈ S and k (S; . This implies that G 1 = G 2 , and so cl (G) is well-defined. 3. Theorem 4 implies a result due to Neumann-Lara and Rivera-Campo.
Neumann-Lara and Rivera-Campo [5] obtained an independence number condition for the existence of spanning k-trees. (In fact, they proved a stronger result as we mention in Sect. 3.) Theorem 5 (Neumann-Lara and Rivera-Campo [5]) Let k ≥ 2 be an integer, and let G be a connected graph. If there exists no independent set of order k + 1, then G has a spanning k-tree.
If a graph G satisfies the hypothesis of Theorem 5, then cl (G) is complete, and hence Theorem 4 implies Theorem 5. If G has a spanning k-tree, then trivially also G + uv has a spanning k-tree. Hence we prove the converse.
Suppose that G + uv has a spanning k-tree T and G does not have a spanning k-tree. Then T − uv consists of two trees T 1 and T 2 such that u ∈ V (T 1 ) and v ∈ V (T 2 ). Note that for i = 1, 2, T i is a k-tree in G, and is as small as possible. Since G does not have a spanning k-tree, it follows that for some i = 1, 2, there exists no k-tree S i such that Without loss of generality, we may assume that there exists no k-tree S 1 Let T 3 := T 1 ∪T 2 +w 1 w 2 and let F 0 , . . . , F k be k+1 components of T 3 −w 1 . Since F i is a tree, there exists a vertex We regard F 0 as a rooted tree with root z 0 and F i as a rooted tree with root x i for 1 ≤ i ≤ k.

Claim 1 Let i, j be integers with
By Claim 1 and the choice of x 0 , we obtain the following.

Claim 2 X is an independent set in G, and k
We define For 0 ≤ i ≤ k and z ∈ V (F i ), we denote the parent and the children of z in F i by z − and ch(z), respectively and we let Y + i := y∈Y i ch(y). This contradicts (1).
Proof Suppose that z p ∈ N G (x q ) for some 0 ≤ p = q ≤ k. Assume that x p = v. Then z p = w 2 and the minimality of Assume that x q = v. Then note that d G (w 2 ) ≤ k − 1 by the choice of w 1 and w 2 . Thus, T 3 − z p w 1 + x q z p is a spanning k-tree in G, a contradiction. If v ∈ {x p , x q }, then S 1 := T 1 − z p w 1 + x q z p is a k-tree with V (S 1 ) = V (T 1 ) and d S 1 (w 1 ) = k − 1, which contradicts (1).
Since F i is a tree, ch(y 1 ) ∩ ch(y 2 ) = ∅ for every y 1 , y 2 ∈ Y i with y 1 = y 2 . Therefore we obtain Since

Problem
In this section, we propose a problem concerning a closure involving the independence number and the connectivity. Let α(G) and κ(G) be the independence number and the connectivity of G, respectively. Neumann-Lara and Rivera-Campo [5] obtained the following result.
We can consider the following problem as a closure result for Theorem 6. For a graph G and u, v ∈ V (G) with uv ∈ E(G), let α(u, v; G) be the cardinality of a maximum independent set containing u and v. For a graph G and u, v ∈ V (G), the local connectivity κ(u, v; G) is defined to be the maximum number of internallydisjoint paths connecting u and v in G.
Problem 7 Let k ≥ 2 be an integer, and let G be a graph. Let u and v be two nonadjacent vertices of G. Assume that α(u, v; G) ≤ (k − 1)κ(u, v; G) + 1. Then G has a spanning k-tree if and only if G + uv has a spanning k-tree.
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