For any integer k≥2,a spanning k-ended tree is a spanning tree with at most k leaves.In this paper,we provide tight spectral radius conditions for the existence of a spanning k-ended tree in t-connected graphs,which ...For any integer k≥2,a spanning k-ended tree is a spanning tree with at most k leaves.In this paper,we provide tight spectral radius conditions for the existence of a spanning k-ended tree in t-connected graphs,which generalizes a result of Ao,Liu and Yuan(2023).展开更多
Let T be a tree. The set of leaves of Τ is denoted by Leaf(Τ). The subtree Τ—Leaf(Τ) of T is called the stem of Τ. A stem is called a k-ended stem if it has at most k-leaves in it. In this paper, we prove the fo...Let T be a tree. The set of leaves of Τ is denoted by Leaf(Τ). The subtree Τ—Leaf(Τ) of T is called the stem of Τ. A stem is called a k-ended stem if it has at most k-leaves in it. In this paper, we prove the following theorem. Let G be a connected graph and k≥2 be an integer. Let u and ν be a pair of nonadjacent vertices in G. Suppose that |NG(u)∪NG(v)|≥|G|-k-1. Then G has a spanning tree with k-ended stem if and only if G+uv has a spanning tree with k-ended stem. Moreover, the condition on |NG(u)∪NG(v)| is sharp.展开更多
基金supported by the National Natural Science Foundation of China(Grant No.11901540).
文摘For any integer k≥2,a spanning k-ended tree is a spanning tree with at most k leaves.In this paper,we provide tight spectral radius conditions for the existence of a spanning k-ended tree in t-connected graphs,which generalizes a result of Ao,Liu and Yuan(2023).
文摘Let T be a tree. The set of leaves of Τ is denoted by Leaf(Τ). The subtree Τ—Leaf(Τ) of T is called the stem of Τ. A stem is called a k-ended stem if it has at most k-leaves in it. In this paper, we prove the following theorem. Let G be a connected graph and k≥2 be an integer. Let u and ν be a pair of nonadjacent vertices in G. Suppose that |NG(u)∪NG(v)|≥|G|-k-1. Then G has a spanning tree with k-ended stem if and only if G+uv has a spanning tree with k-ended stem. Moreover, the condition on |NG(u)∪NG(v)| is sharp.
