Ryser’s Latin square conjecture

Can you pick one cell from each row of a square while also using every column and every symbol exactly once? Consider the three rows ABC, BCA and CAB. Each symbol appears once in every row and column. The diagonal picks A, C and B, so all three demands are met.

Choosing instead the A in each row uses three different rows and columns but repeats the symbol. The unknown is whether every odd-sized square with this row-and-column rule has a successful full selection, however its symbols are arranged.[1]

Can all rows, columns and symbols be represented at once by a single selection from every odd-sized Latin square? In the square with rows ABC, BCA and CAB, choosing cells (1,1), (2,2), (3,3) uses all three of each. Choosing (1,1), (2,3), (3,2) uses all rows and columns but only symbol A.

An order-n Latin square L has each of its n symbols exactly once in each row and column. A full transversal can be described by a permutation $\pi$ of the columns with the values $L(i,\pi(i))$, for $1\leq i\leq n$, all distinct. The conjecture requires such a permutation for every odd n and every order-n Latin square; it does not impose this universal claim for even orders.[1]

The successful three-by-three example shows what a full selection looks like, not what every odd square permits. The September 2026 preprint counts many selections of size one less than the full order for sufficiently large squares. Leaving one row, column and symbol unused does not establish that a full selection exists; the full odd-order guarantee remains the target.[1]

The cited September 2026 preprint studies near-spanning matchings and counts size $n-1$ partial transversals for sufficiently large n. Its odd-order full-transversal conjecture remains separate. Removing the last deficit can require a different global selection, so the existence of almost-complete choices is not itself a full-choice theorem.[1]

A cell simultaneously consumes three resources: its row, its column and its symbol. Satisfying two uniqueness rules can still violate the third, as the all-A selection shows. The problem isolates how several individually well-balanced arrangements can constrain one another when a single compatible selection is required.[1]

The row-column bipartite graph turns a transversal into a matching whose edges also have distinct symbol labels. The Latin rule balances each label across the rows and columns, yet this local balance must be reconciled with global disjointness. The conjecture studies precisely that interaction between a perfect matching and an additional uniqueness constraint.[1]