Every latin hypercube of order 5 has transversals

TL;DR

This work resolves Wanless's conjecture for order q=5 by proving that every latin n-cube with order 5 has a transversal for all n>1. It combines theoretical analysis of layer-latin cubes with heavy computer-aided enumeration to classify all 2-layer latin cuboids of order 5 lacking transversals, and uses these classifications to construct larger nontransversal examples in higher-layer cuboids. The authors also determine the full set of nonextendible and noncompletable latin cuboids of order 5, and provide exhaustive appendices listing representative cuboids. Overall, the paper advances understanding of transversals in multidimensional latin structures and clarifies the landscape of extendibility and completeness at order 5, with implications for permanents and polystochastic matrices.

Abstract

We prove that for all n>1 every latin n-dimensional cube of order 5 has transversals. We find all 123 paratopy classes of layer-latin cubes of order 5 with no transversals. For each $n\geq 3$ and $q\geq 3$ we construct a (2q-2)-layer latin n-dimensional cuboid with no transversals. Moreover, we find all paratopy classes of nonextendible and noncompletable latin cuboids of order 5.

Theorems & Definitions (29)

• Proposition 1
• Corollary 1
• Proposition 2: KP, Corollary 2
• Proposition 3: AAT16, Proposition 13
• Corollary 2
• Proposition 4
• Theorem 1
• Proposition 5
• Theorem 2
• Lemma 1