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首页> 外文期刊>Quaestiones mathematicae >TRANSVERSALS IN 5-UNIFORM HYPERGRAPHS AND TOTAL DOMINATION IN GRAPHS WITH MINIMUM DEGREE FIVE
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TRANSVERSALS IN 5-UNIFORM HYPERGRAPHS AND TOTAL DOMINATION IN GRAPHS WITH MINIMUM DEGREE FIVE

机译:最小程度为5的图形的5一致超图的横向和总控制

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摘要

In [9] Thomasse and Yea pose the following problem: Find the minimum c(k) for which every k-uniform hypergraph with n vertices and n edges has a transversal of size at most c(k)n. A direct consequence of this result is that every graph of order n with minimum degree at least k has a total dominating set of cardinality at most ckn. It is known that c(2) = 2/3, c(3) = 1/2, and c(4) = 3/7. Thomasse and Yeo show that 4/11 <= c(5) and conjecture that c(5) = 4/11. In this paper we show that c(5) <= 17/44. Thus, 16/44 <= c(5) <= 17/44. More generally we prove that every 5-uniform hypergraph on n vertices and m edges has a transversal with no more than (10n+7m)/44 vertices. Consequently, every graph on n vertices with minimum degree at least five has total domination number at most 17n/44.
机译:在[9]中,Thomasse和Yea提出了以下问题:找到最小c(k),对于该最小c(k),具有n个顶点和n个边的每个k均匀超图的横向尺寸最大为c(k)n。该结果的直接结果是,阶数为n且最小度至少为k的每个图的最大基数总集合为ckn。已知c(2)= 2/3,c(3)= 1/2,c(4)= 3/7。 Thomasse和Yeo证明4/11 <= c(5)并猜想c(5)= 4/11。在本文中,我们证明c(5)<= 17/44。因此,16/44 <= c(5)<= 17/44。更笼统地说,我们证明n个顶点和m个边上的每5个均匀超图都有一个不超过(10n + 7m)/ 44个顶点的横向。因此,最小度至少为5的n个顶点上的每个图形的总控制数最多为17n / 44。

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