Wide enough Latin rectangles are perfects

TL;DR

The paper proves that for any fixed number of rows and sufficiently large odd number of columns, there exists a perfect Latin rectangle, advancing understanding of a longstanding conjecture about their existence.

Contribution

It establishes the existence of perfect Latin rectangles for any fixed number of rows and large enough odd columns, partially confirming a well-known conjecture.

Findings

Existence of perfect Latin rectangles for fixed m and large odd n

Progress towards the conjecture that all odd m ≤ n admit perfect Latin rectangles

Asymptotic partial answer to the longstanding conjecture

Abstract

Given two integers $m$ and $n$ with $m\leq n$, a Latin rectangle of size $m\times n$ is a bi-dimensional array with $m$ rows and $n$ columns filled with symbols from an alphabet with $n$ symbols, such that each row contains a permutation of the alphabet and each column contains no repeated symbols. Two rows $a$ and $b$ of a Latin rectangle $R$ define a permutation $R_{a,b}$ assigning the symbol $y$ to the symbol $x$ if they are in the same column, $x$ is in row $a$ and $y$ is in row $b$. A Latin rectangle $R$ is perfect is the permutation $R_{a,b}$ is cyclic, for each pair of rows $a$ and $b$. We prove that for each integer $m$ and each large enough odd integer $n$ there is a perfect Latin rectangle $R$ of size $m\times n$. It is a partial (asymptotic) answer to a well-known conjecture which says that the same property holds for each odd integer $m\leq n$.