论文标题
在代表部分拉丁广场的国际象棋板上对子系统的分析（第2部分）。
Analysis of subsystems with rooks on a chess-board representing a partial Latin square (Part 2.)
论文作者
论文摘要
部分拉丁订单$ n $的部分拉丁广场可以由$ 3 $维的棋盘大小$ n \ times n \ times n $，最多最多$ n^2 $ n^2 $ nontakearing Rooks。在拉丁方块中，子系统及其最遥远的伴侣在一起具有与其能力一样多的菜。这意味着要完成部分拉丁正方形的简单容量条件，这实际上是Cruse对于特征矩阵的必要条件。安德森·希尔顿（Andersen-Hilton）证明，除了某些列出的案例外，如果仅包含$ n $符号，则可以完成订单$ n $的PL。安德森（Andersen）以$ n+1 $的符号证明了这一点，列出了要排除的案例。确定可以将$ n $ n $或$ n+1 $ roks重载的国际象棋棋盘结构，因此，如果满足容量条件，则可以从最多$ n+1 $ nontake的Rook衍生出的PLS，最多可确切地完成。在拉丁广场的一层中，一对远程夫妇的两个子系统处于平衡状态。因此，可以制定完成层的必要条件，即平衡条件。对于LSC，棋盘的每个1维子空间都完全包含一个Rook。因此，对于源自部分拉丁正方形的PLSC，我们检查了某些1维子空间集，因为它们表明了缺失的Rooks数量。
A partial Latin square of order $n$ can be represented by a $3$-dimensional chess-board of size $n\times n\times n$ with at most $n^2$ non-attacking rooks. In Latin squares, a subsystem and its most distant mate together have as many rooks as their capacity. That implies a simple capacity condition for the completion of partial Latin squares which is in fact the Cruse's necessary condition for characteristic matrices. Andersen-Hilton proved that, except for certain listed cases, a PLS of order $n$ can be completed if it contains only $n$ symbols. Andersen proved it for $n+1$ symbols, listing the cases to be excluded. Identifying the structures of the chess-board that can be overloaded with $n$ or $n+1$ rooks, it follows that a PLS derived from a chess-board with at most $n+1$ non-attacking rooks can be completed exactly if it satisfies the capacity condition. In a layer of a Latin square, two subsystems of a remote couple are in balance. Thus, a necessary condition for completion of a layer can be formulated, the balance condition. For an LSC, each 1-dimensional subspace of the chess-board contains exactly one rook. Consequently, for the PLSCs derived from partial Latin squares, we examine certain sets of 1-dimensional subspaces because they indicate the number of missing rooks.
