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On semicomplete multipartite digraphs whose king sets are semicomplete digraphs

机译:在半填充数字上的多亚特数字,其王套是半完整的数字

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Reid [Every vertex a king, Discrete Math. 38 (1982) 93–98] showed that a non-trivial tournament H is contained in a tournament whose 2-kings are exactly the vertices of H if and only if H contains no transmitter. Let T be a semicomplete multipartite digraph with no transmitters and let Kr(T) denote the set of r-kings of T. Let Q be the subdigraph of T induced by K4(T). Very recently, Tan [On the kings and kings-of-kings in semicomplete multipartite digraphs, Discrete Math. 290 (2005) 249–258] proved that Q contains no transmitters and gave an example to show that the direct extension of Reid's result to semicomplete multipartite digraphs with 2-kings replaced by 4-kings is not true. In this paper, we (1) characterize all semicomplete digraphs D which are contained in a semicomplete multipartite digraph whose 4-kings are exactly the vertices of D. While it is trivial that K4(Q)K4(T), Tan [On the kings and kings-of-kings in semicomplete multipartite digraphs, Discrete Math. 290 (2005) 249–258] showed that K3(Q)K3(T) and K2(Q)=K2(T). Tan [On the kings and kings-of-kings in semicomplete multipartite digraphs, Discrete Math. 290 (2005) 249–258] also provided an example to show that K3(Q) need not be the same as K3(T) in general and posed the problem: characterize all those semicomplete multipartite digraphs T such that K3(Q)=K3(T). In the course of proving our result (1), we (2) show that K3(Q)=K3(T) for all semicomplete multipartite digraphs T with no transmitters such that Q is a semicomplete digraph.
机译:里德[每个顶点是一个国王,离散数学。 38(1982)93-98]显示非琐事锦标赛H包含在锦标赛中,其2国钟正恰好是H的顶点,如果H不包含发射机。假设没有发射器的半完整性数字,并且让KR(T)表示T的R-kings一组。Let Q是K4(T)诱导的T的底层。最近,Tan [在半填充数字上的国王和国王国王,离散数学。事实证明,Q Q不包含发射器并给出一个例子,以表明Reid的直接延伸到半夸克替换为4国王的半夸粒式数字。在本文中,我们(1)表征了包含在半完整的多档数字中的全半完整数字D,其4-kings正好是D的顶点。虽然它是k4(q)k4(t),tan [典型的多鹦鹉数字中的国王和国王,离散数学。 290(2005)249-258]显示K3(Q)K3(T)和K2(Q)= K2(T)。棕褐色的半填充数字上的王子和国王王。 290(2005)249-258]还提供了一个示例,以证明K3(Q)通常与K3(T)相同,并提出了问题:表征了所有那些半夸大的多分层D,使得K3(Q)= K3(T)。在证明我们的结果(1)的过程中,我们(2)显示了所有半完整多方数字的K3(Q)= K3(T),没有发射器,使得Q是半完整的数字。

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