Latin squares with restricted transversals

Abstract

Abstract:We prove that for all odd m ≥ 3 there exists a latin square of order 3m that contains an (m−1)×m latin subrectangle consisting of entries not in any transversal. We prove that for all even n ≥ 10 there exists a latin square of order n in which there is at least one transversal, but all transversals coincide on a single entry. A corollary is a new proof of the existence of a latin square without an orthogonal mate, for all odd orders n ≥ 11. Finally, we report on an extensive computational study of transversal-free entries and sets of disjoint transversals in the latin squares of order n ≤ 9. In particular, we count the number of species of each order that possess an orthogonal mate. q

Monogamous latin squares

“…The following simple lemma is crucial to our work, just as related results have been in [1,[6][7][8][9][10]13]. …”

“…Without condition (C1) the definition would be uninteresting. It is known [6,9,13] that for every order n 4 there exists a latin square containing an entry γ 1 for which there is no T ∈ S i with γ 1 ∈ T .…”

“…However, the name [12] and the four distinct proofs [6,9,10,13] of their existence for all orders 4, are comparatively recent.…”

Bachelor latin squares with large indivisible plexes

“…Other new general constructions of confirmed bachelor latin squares are described in [10]. We refer the reader to that article for a comparison of the known constructions for bachelor latin squares [10][11][12]18].…”

Covers and partial transversals of Latin squares

“…y + xz n and (7) z + xy n, where the first inequality holds because x 2 + y 2 + z 2 xy + xz + yz and the second follows from (6)- (8). Since α 0 and α + α 2 n, it follows that 3n − (x + y + z) = 3n − 3α 3 n + 1/2 − n + 1/4 .…”