生长平衡覆盖集
Growing balanced covering sets
摘要：
给定一个具有二分图$(A,B)$的二分图，其中$B$被等分为$k\ge2$块，能否逐个拾取$A$中的顶点，使得在每一步中，拾取的顶点覆盖这些块中每个块的大致相同数量的顶点？我们证明了，如果每个块有基数$m$，$B$中的顶点有相同的度，并且$A$中的每个顶点在每个块中至多有$cm$邻居，其中$c>0$是一个小常数，那么$A$中的顶点有一个排序$v_1,\ldots,v_n$，使得对于每个$j\in\{1,\ldots,n\}$，每两个块中具有$\{v_1,\ldots,v_j\}$邻居的顶点的数量相差至多$\sqrt{2(k-1)c}\cdot m$。这是一个著名的引理斯坦尼茨，并部分回答了一个未发表的问题斯科特和西摩。
英文摘要：
Given a bipartite graph with bipartition $(A,B)$ where $B$ is equipartitioned into $k\ge2$ blocks, can the vertices in $A$ be picked one by one so that at every step, the picked vertices cover roughly the same number of vertices in each of these blocks? We show that, if each block has cardinality $m$, the vertices in $B$ have the same degree, and each vertex in $A$ has at most $cm$ neighbors in every block where $c>0$ is a small constant, then there is an ordering $v_1,\ldots,v_n$ of the vertices in $A$ such that for every $j\in\{1,\ldots,n\}$, the numbers of vertices with a neighbor in $\{v_1,\ldots,v_j\}$ in every two blocks differ by at most $\sqrt{2(k-1)c}\cdot m$. This is related to a well-known lemma of Steinitz, and partially answers an unpublished question of Scott and Seymour.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
