On completing three cyclic transversals to a latin square

TL;DR

This paper proves that any partial Latin square of prime order greater than 7, formed by three cyclic transversals, can be completed to a diagonally cyclic Latin square.

Contribution

It establishes a new result on completing specific partial Latin squares with three cyclic transversals to diagonally cyclic Latin squares.

Findings

Partial Latin squares with three cyclic transversals can be completed to diagonally cyclic Latin squares.

The result applies to prime order p > 7.

Provides a constructive proof for completion.

Abstract

Let $P$ be a partial latin square of prime order $p>7$ consisting of three cyclically generated transversals. Specifically, let $P$ be a partial latin square of the form: \[ P=\{(i,c+i,s+i),(i,c'+i,s'+i),(i,c''+i,s''+i)\mid 0 \leq i< p\} \] for some distinct $c,c',c''$ and some distinct $s,s',s''$. In this paper we show that any such $P$ completes to a latin square which is diagonally cyclic.