[[Latin square]]
# Completion of a Latin square from its diagonal

If $\Delta_{i} = M_{ii}$ is fixed for $i=1,\dots,n$, there exists an $n \times n$ [[Latin square]] $(M_{ij})$ except if the diagonal $\Delta_{i}$ consists of exactly $n-1$ occurances of one symbol.[^1979] #m/thm/comb 

  [^1979]: 1979\. [[Sources/@changCompleteDiagonalsLatin1979|Complete Diagonals of Latin Squares]]

> [!missing]- Proof
> #missing/proof 
> See op. cit.

#
---
#state/develop | #lang/en | #SemBr