有五个不相交的子正方形的拉丁正方形

IF 0.5 4区数学 Q3 MATHEMATICS

Journal of Combinatorial Designs Pub Date : 2024-10-16 DOI:10.1002/jcd.21960

Tara Kemp

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引用次数: 0

摘要

给定一个整数分区(h 1 h 2…h k)$ ({h}_{1}{h}_{2}{\rm{\ldots}}{h}_{k})$ ofN $ N $，有没有可能找到一个n阶拉丁方阵它有k阶k阶不相交的h阶子方阵1，…，h k ${h}_{1},{\rm{\ldots}},{h}_{k}$ ？这个问题是由Fuchs提出的，并且对于k≤4$ k\le 4$的所有分区都有答案。在本文中，我们回答了k=5$ k=5$的情况下的问题，并扩展了这一特殊情况的结果，例如当最大部分最多是最小部分的三倍时。

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Latin squares with five disjoint subsquares

Given an integer partition $(h_{1} h_{2} \dots h_{k})$ of $n$ , is it possible to find an order $n$ latin square with $k$ pairwise disjoint subsquares of orders $h_{1}, \dots, h_{k}$ ? This question was posed by Fuchs and has been answered for all partitions with $k \leq 4$ . In this paper, we answer the question in the case $k = 5$ and expand on results for special cases of this, such as when the largest part is at most three times the smallest part.

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来源期刊

Journal of Combinatorial Designs 数学-数学

CiteScore

1.60

自引率

14.30%

发文量

审稿时长

>12 weeks

期刊介绍： The Journal of Combinatorial Designs is an international journal devoted to the timely publication of the most influential papers in the area of combinatorial design theory. All topics in design theory, and in which design theory has important applications, are covered, including: block designs, t-designs, pairwise balanced designs and group divisible designs Latin squares, quasigroups, and related algebras computational methods in design theory construction methods applications in computer science, experimental design theory, and coding theory graph decompositions, factorizations, and design-theoretic techniques in graph theory and extremal combinatorics finite geometry and its relation with design theory. algebraic aspects of design theory. Researchers and scientists can depend on the Journal of Combinatorial Designs for the most recent developments in this rapidly growing field, and to provide a forum for both theoretical research and applications. All papers appearing in the Journal of Combinatorial Designs are carefully peer refereed.