Closure and Spanning Trees with Bounded Total Excess

Abstract

Let $\alpha \ge 0$ and $k \ge 2$ be integers. For a graph G, the total k-excess of G is defined as $\text{ te }(G;k)=\sum _{v \in V(G)}\max \{d_G(v)-k,0\}$. In this paper, we propose a new closure concept for a spanning tree with bounded total k-excess. We prove that: Let G be a connected graph, and let u and v be two non-adjacent vertices of G. If G satisfies one of the following conditions, then G has a spanning tree T such that $\text{ te }(T;k) \le \alpha$ if and only if $G+uv$ has a spanning tree $T'$ such that $\text{ te }(T';k) \le \alpha$:

• (i) $\max \{ \sum _{x \in X} d_G(x): X \text{ is } \text{ a } \text{ subset } \text{ of } S \text{ with } |X|=k \} \ge |G|-1 \text{ for } \text{ every } \text{ independent } \text{ set } S \text{ in } G \text{ of } \text{ order } k+1 \text{ such } \text{ that } \{ u,v \} \subseteq S$; or

• (ii) $\max \{ \sum _{x \in X} d_G(x): X \text{ is } \text{ a } \text{ subset } \text{ of } S \text{ with } |X|=k \} \ge |G|-\alpha -1 \text{ for } \text{ every } \text{ independent } \text{ set } S \text{ in } G \text{ of } \text{ order } k+\alpha +1 \text{ such } \text{ that } S \cap \{ u,v \} \ne \emptyset .$

We also show examples to show that these conditions are sharp.