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On king–serf pair in tournaments
Discrete Mathematics ( IF 0.8 ) Pub Date : 2020-07-01 , DOI: 10.1016/j.disc.2020.111903
Xiaoyun Lu

Abstract For a directed graph G , a vertex x is a king if every other vertex can be reached from x by a directed path of length at most 2 and is a serf if x can be reached from every other vertex by a directed path of length at most 2. A tournament T with vertex set V is called X -hamiltonian crossing if every hamiltonian path must have one end in X and the other in X = V − X . Our main results are the following: • A tournament T ≠ T 4 is X -hamiltonian crossing if and only if T has a strong component decomposition V = V 1 ∪ ⋯ ∪ V k such that V 1 ⊆ X and V k ⊆ X , or V 1 ⊆ X and V k ⊆ X . • Let T be a tournament and ( x , y ) a king–serf pair in T − x y . Then there exists no hamiltonian ( x , y ) -path in T if and only if T is an x → y special tournament with U ( S 1 ; D 1 , D 2 ) ≠ 0 and U ( S 2 ; D 1 , D 2 ) ≠ 0 . • Let T be a tournament of order n ≥ 3 . If x is a vertex of maximum out-degree and y a vertex of maximum in-degree then there exists a hamiltonian ( x , y ) -path if and only if x y is not exceptional.

中文翻译:

关于锦标赛中的国王-农奴对

摘要 对于有向图 G ,如果每个其他顶点都可以通过长度为 2 的有向路径从 x 到达,则顶点 x 是国王,如果 x 可以从每个其他顶点通过长度为有向路径到达,则它是一个奴隶。至多 2. 如果每个哈密顿路径的一端必须在 X 中,另一个在 X = V − X 中,则具有顶点集 V 的锦标赛 T 称为 X -哈密顿交叉。我们的主要结果如下: • 当且仅当 T 具有强分量分解 V = V 1 ∪ ⋯ ∪ V k 使得 V 1 ⊆ X 和 V k ⊆ X ,锦标赛 T ≠ T 4 是 X 哈密顿交叉,或 V 1 ⊆ X 和 V k ⊆ X 。• 设T 是一个锦标赛,并且(x, y) 是T − xy 中的国王-农奴对。那么在 T 中不存在哈密顿 ( x , y ) 路径当且仅当 T 是一个 x → y 特殊锦标赛,其中 U ( S 1 ; D 1 , D 2 ) ≠ 0 且 U ( S 2 ; D 1 , D 2)≠0。• 令T 为n ≥ 3 阶的锦标赛。如果 x 是最大出度的顶点和最大入度的 ya 顶点,则存在哈密顿 ( x , y ) -路径当且仅当 xy 不例外。
更新日期:2020-07-01
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