Ryser–Brualdi–Stein conjecture on Latin square transversals

Prove that every n × n Latin square has a transversal of size n − 1, and furthermore a full transversal when n is odd.

Background

Montgomery proved the existence of a transversal of size n−1 for all sufficiently large n, settling the first part asymptotically. The full-transversal existence when n is odd remains open in general and is the central remaining part of the combined Ryser–Brualdi–Stein conjecture.

References

Conjecture [Ryser--Brualdi--Stein conjecture ] Every n × n Latin square has a transversal of size n-1, and a full transversal if n is odd.

— Sublinear expanders and their applications  (2401.10865 - Letzter, 2024) in Transversals in Latin squares (Section 12)

In striking contrast, the celebrated Ryser conjecture asserts that every Latin square of odd order has at least one transversal.

— Latin Squares with Few Transversals  (2609.08624 - Luria, 8 Sep 2026) in Section 1, Introduction

The transversal assertion for odd order remains open.

— Counting Near-Spanning Matchings in Latin Squares and Steiner Triple Systems  (2609.11006 - Tang et al., 10 Sep 2026) in Section 1, subsection “Latin squares”