PreprintCC BY 4.0arXiv
Latin Squares with Few Transversals
Abstract
Let t(n) denote the minimum number of transversals in a Latin square of odd order n. Improving upon a recent bound of Dai, Divoux and Kelly, we prove that for every n such that n ≡ 3 pmod 6, t(n) ≤ ( (1+o(1)) 2n/3e2)n . Our proof is based on a family of 3 × 3 block Latin squares whose transversals are constrained to either lie entirely in the diagonal blocks or avoid them altogether.
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BibTeX
@misc{luria2026latin,
title = {{Latin Squares with Few Transversals}},
author = {Zur Luria},
howpublished = {arXiv},
year = {2026},
month = sep,
doi = {10.48550/arxiv.2609.08624},
eprint = {2609.08624},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2609.08624},
note = {CC BY 4.0},
}
RIS
TY - UNPB
TI - Latin Squares with Few Transversals
AU - Zur Luria
PY - 2026
DA - 2026/09/08
PB - arXiv
DO - 10.48550/arxiv.2609.08624
UR - https://arxiv.org/abs/2609.08624
AB - Let t(n) denote the minimum number of transversals in a Latin square of odd order n. Improving upon a recent bound of Dai, Divoux and Kelly, we prove that for every n such that n ≡ 3 pmod 6, t(n) ≤ ( (1+o(1)) 2n/3e^2)^n . Our proof is based on a family of 3 × 3 block Latin squares whose transversals are constrained to either lie entirely in the diagonal blocks or avoid them altogether.
ER -