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A Combinatorial Proof of Hilton's Conjecture and Beyond
Abstract
Using refined absorption, we prove that for every integer g≥ 1 and real γ> 0, and for sufficiently large n, there exists an n−1+γ-spread distribution on Latin squares of order n and girth at least g that have no proper subsquares. This implies a combinatorial proof of Hilton's conjecture from the 1970s (recently proved algebraically by Allsop and Wanless) that for all sufficiently large n, there exists a subsquare-free Latin square of order n; indeed, it implies there exist at least n(1−o(1))n2 subsquare-free squares. Simultaneously it also implies the existence of high girth Latin squares (recently proved by Kwan, Sah, Sawhney and Simkin) and even an n−1+γ-spread distribution on high girth Latin squares.
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BibTeX
@misc{lesgourgues2026combinatorial,
title = {{A Combinatorial Proof of Hilton's Conjecture and Beyond}},
author = {Thomas Lesgourgues and Luke Postle},
howpublished = {arXiv},
year = {2026},
month = sep,
doi = {10.48550/arxiv.2609.28966},
eprint = {2609.28966},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2609.28966},
}
RIS
TY - UNPB
TI - A Combinatorial Proof of Hilton's Conjecture and Beyond
AU - Thomas Lesgourgues
AU - Luke Postle
PY - 2026
DA - 2026/09/24
PB - arXiv
DO - 10.48550/arxiv.2609.28966
UR - https://arxiv.org/abs/2609.28966
AB - Using refined absorption, we prove that for every integer g≥ 1 and real γ> 0, and for sufficiently large n, there exists an n^−1+γ-spread distribution on Latin squares of order n and girth at least g that have no proper subsquares. This implies a combinatorial proof of Hilton's conjecture from the 1970s (recently proved algebraically by Allsop and Wanless) that for all sufficiently large n, there exists a subsquare-free Latin square of order n; indeed, it implies there exist at least n^(1−o(1))n^2 subsquare-free squares. Simultaneously it also implies the existence of high girth Latin squares (recently proved by Kwan, Sah, Sawhney and Simkin) and even an n^−1+γ-spread distribution on high girth Latin squares.
ER -