Abstract
Let \(\alpha \ge 0\) and \(k \ge 2\) be integers. For a graph G, the total kexcess 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 kexcess. We prove that: Let G be a connected graph, and let u and v be two nonadjacent 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)
\(\max \{ \sum _{x \in X} d_G(x): X \text{ is } \text{ a } \text{ subset } \text{ of } S \text{ with } X=k \} \ge G1 \text{ for } \text{ every } \text{ independent } \text{ set } S \text{ in } G \text{ of } \text{ order } k+1 \text{ such } \text{ that } \{ u,v \} \subseteq S\); 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.
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Notes
 1.
We can generalize \(G_1\) by replacing each vertex in \(V_1\) with a complete graph.
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Acknowledgements
The authors would like to thank Professor Hikoe Enomoto, Professor Haruhide Matsuda and the anonymous referees for their valuable comments and suggestions. This work was supported by JSPS KAKENHI Grant number JP20J15332 (to S.M.), JP20K14353 (to T.Y).
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Maezawa, Si., Tsugaki, M. & Yashima, T. Closure and Spanning Trees with Bounded Total Excess. Graphs and Combinatorics (2021). https://doi.org/10.1007/s0037302102283z
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Keywords
 Spanning tree
 ktree
 kended tree
 Total excess
 Degree sum
 Closure
Mathematics Subject Classification
 05C05
 05C50