Latin squares with non-partitioning disjoint subsquares
Fuente:
arXiv
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| Format: | Preprint |
| Published: |
2026
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| _version_ | 1866917360999333888 |
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| author | Kemp, Tara |
| author_facet | Kemp, Tara |
| contents | A latin square of order $n$ with pairwise disjoint subsquares of orders $h_1,\dots,h_k$ such that $h_1+\dots+h_k = n$ is known as a realization. The existence of realizations is a partially solved problem with a few general results for an arbitrary number of subsquares, $k$. Requiring only that $h_1+\dots+h_k\leq n$ gives a variation of the problem that has few known results. In this paper we prove a general necessary condition for existence and completely determine existence when there are at most three subsquares or the subsquares are all of the same order. Importantly, we prove that if $h_1\geq h_2\geq\dots\geq h_k$ and $n\geq h_1+\sum_{i=1}^kh_i$ then such a latin square always exists. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2603_23894 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | Latin squares with non-partitioning disjoint subsquares Kemp, Tara Combinatorics A latin square of order $n$ with pairwise disjoint subsquares of orders $h_1,\dots,h_k$ such that $h_1+\dots+h_k = n$ is known as a realization. The existence of realizations is a partially solved problem with a few general results for an arbitrary number of subsquares, $k$. Requiring only that $h_1+\dots+h_k\leq n$ gives a variation of the problem that has few known results. In this paper we prove a general necessary condition for existence and completely determine existence when there are at most three subsquares or the subsquares are all of the same order. Importantly, we prove that if $h_1\geq h_2\geq\dots\geq h_k$ and $n\geq h_1+\sum_{i=1}^kh_i$ then such a latin square always exists. |
| title | Latin squares with non-partitioning disjoint subsquares |
| topic | Combinatorics |
| url | https://arxiv.org/abs/2603.23894 |