A polynomial embedding of pairs of orthogonal partial Latin squares

Abstract

We show that a pair of orthogonal partial latin squares of order n can be embedded in a pair of orthogonal latin squares of order at most 16n 4 and all orders greater than or equal to 48n 4 . This paper provides the first direct polynomial embedding construction.

Constructing and embedding mutually orthogonal Latin squares: reviewing both new and existing results

“…In 2014, Donovan and Yazıcı [15] revisited Jenkins' work, extending it to obtain a polynomial order embedding of a pair of MOPLS. Their approach was to begin with pair of MOPLS, P and Q, such that all the elements in P are distinct.…”

“…Their approach was to begin with pair of MOPLS, P and Q, such that all the elements in P are distinct. From there they used techniques similar to Jenkins to prove: The use of the elementary Abelian 2-group also allows Donovan and Yazıcı [15] to remove the restriction that all the elements in P are distinct to obtain a more general embedding than that given in Theorem 4.9, but at the price of increasing the order of the embedding.…”

“…Theorem 4.10. [15,16] Let P and Q be a pair of MOPLS(n). Then P and Q can be embedded in a pair of MOLS(k 4 ) and any order greater than or equal to 3k 4 where 2 a = k 2n > 2 a−1 for some integer a.…”

Embedding partial Latin squares in Latin squares with many mutually orthogonal mates

“…In this section we make use of the embedding result of Donovan and Yazıcı, [3], to show that a pair of orthogonal partial Latin squares can be embedded in pair of orthogonal Latin square which have many orthogonal mates. 1 ([3]).…”

“…Since C α is a Latin square substituting Equation (18) Proof. Let A 1 and A 2 be two orthogonal partial Latin squares of order n. By [3] we can embed them into two orthogonal Latin squares A 1 and A 2 of order k 4 where 2 a = k 2n > 2 a−1 . As k is a power of a prime, there are at least k 4 − 1 MOLS(k 4 ).…”