Given a bipartite graph with bipartition $(A,B)$ where B is equipartitioned into k 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 $v_1,\dots ,v_n$ of the vertices in A such that for every $j\in \{1,\dots ,n\}$, the numbers of vertices with a neighbor in $\{v_1,\dots ,v_j\}$ 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 This is related to a well-known lemma of Steinitz, and partially answers an unpublished question of Scott and Seymour.

给定一个二分图 $(\mathrm{一种},乙)$其中B被均分为k个块，A 中的顶点是否可以一个接一个地选取，以便在每一步中，选取的顶点在这些块中的每一个中都覆盖大致相同数量的顶点？我们证明，如果每个块的基数为m，则B 中的顶点具有相同的度数，并且A 中的每个顶点在每个块中最多有cm 个邻居，其中$C>0$ 是一个小的常数，那么有一个排序 $v_1,\dots ,v_n$A中的顶点使得对于每个$j\in \{1,\dots ,n\}$, 具有邻居的顶点数 $\{v_1,\dots ,v_j\}$ 在每两个块中最多相差 $\sqrt{2(克-1)C}\cdot 米$. 这与著名的 Steinitz 引理有关，并部分回答了 Scott 和 Seymour 未发表的问题。