表示部分拉丁方的棋盘上的车的子系统分析（第2部分）
Analysis of subsystems with rooks on a chess-board representing a partial Latin square (Part 2.)
AI总结：
该论文分析部分拉丁方对应的棋盘车子系统，证明Cruse容量函数的闭式恒等式，得出非退化远砖块对的容量性质，推导部分拉丁方可完成的容量条件，并提出层可完成的平衡条件。
摘要：
n阶部分拉丁方可表示为n×n×n的三维棋盘，其中包含至多n²个互不攻击的车。在拉丁方中，一个子系统与其远配子系统的车的总数等于它们的容量，这为可完成性提供了一个简单的容量条件，即Cruse的特征矩阵必要条件。我们证明了Cruse容量函数的闭式恒等式，由此可直接得出其最小值。非退化远砖块对的容量至少为n-1，当且仅当其中一个边长为1、另一个边长为n-1时取等；第二小值为n，我们列出了达到该值的砖块对。几何上，容量对每个变量均为仿射函数，其在立方体[1,n-1]³上的最小水平集由既不与(1,1,1)也不与(n-1,n-1,n-1)相邻的六条边组成。Andersen和Hilton、Andersen分别列出了填充单元数为n和n+1时无法完成的部分拉丁方。通过识别可过载的结构，我们得出：来自含至多n+1个车的棋盘的部分拉丁方（PLS）可完成当且仅当它满足容量条件。最后，由于远对的两个子系统在一层内平衡，我们为一层的可完成性提出了另一个必要条件——平衡条件。
英文摘要：
A partial Latin square of order n can be represented by a 3-dimensional chess-board of size n x n x n with at most n^2 non-attacking rooks. In Latin squares, a subsystem and its remote mate together have as many rooks as their capacity, which yields a simple capacity condition for completion, in fact Cruse's necessary condition for characteristic matrices. We prove a closed-form identity for Cruse's capacity function, from which its smallest values follow directly. The capacity of a non-degenerated remote brick couple is at least n-1, with equality precisely when one edge length is 1 and another is n-1; the second smallest value is n, and we list the couples attaining it. Geometrically the capacity is affine in each variable, and its minimum level set on the cube [1,n-1]^3 consists of the six edges adjacent neither to (1,1,1) nor to (n-1,n-1,n-1). Andersen and Hilton, and then Andersen, listed the partial Latin squares with n, resp. n+1 filled cells that cannot be completed. Identifying the structures that can be overloaded, we obtain that a PLS coming from a chess-board with at most n+1 rooks is completable exactly if it satisfies the capacity condition. Finally, since the two subsystems of a remote couple are in balance within a layer, we formulate a further necessary condition for the completion of a layer, the balance condition.
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