The Evans conjecture ( which was proved later by Smetaniuk) states that for any $n$, if at most $n-1$ entries of a partial $n\times n$ latin square are filled, it can be completed to the full latin square.


My question pertains to whether this is applicable to symmetric (or commutative) latin square? That is, given $n\times n$ latin square, with at most $n-1$ prefilled entries ( which are symmetric with respect to diagonal), can it be completed to a symmetric latin square? Note that this corresponds to pre- total coloring of a complete graph of order $n$ with at most $n-1$ previous filled colors. Any hints? Thanks beforehand.



Evans conjecture for symmetric latin squares
No, when it comes to symmetric latin squares it is no longer true that as many as $n-1$ cells can be prescribed unconditionally. This is explained in the Ph.D. thesis of [Matthew Henderson.](https://mjh-phd.netlify.app/sec110.html)



> 
> The key point here is that in a symmetric latin square, precisely
> because of the symmetry, every symbol $\sigma$ occurs an even number
> of times in cells outside of the main diagonal. Therefore, as every
> symbol $\sigma$ occurs $n$ times in total it follows that the number
> of cells of the main diagonal containing symbol $\sigma$ is congruent
> to $n$ modulo 2. A partial latin square can be incompletable because
> there are more symbols which occur on the main diagonal a number of
> times (zero included) incongruent to $n$ modulo 2 than there are
> empty cells on the main diagonal.  
> 
> If this obstruction does not occur the diagonal is called "admissible". For $n$ odd a diagonal is admissible if and only if no symbol occurs more than once on it.
> 
> 
> 


The generalization of Evans theorem to symmetric latin squares by [Andersen and Hilton](https://www.academia.edu/20306267/Symmetric_latin_square_and_complete_graph_analogues_of_the_evans_conjecture) is that the symmetric latin square is completable if the diagonal is admissible and at most $n-1$ cells in total are prescribed.


In response to the comment by Richard Stanley: it is not sufficient to only count cells on or above the main diagonal, as this counter example shows. The diagonal is admissible ($n=5$ is odd and no symbol appears more than once on the diagonal), and the number of prescribed cells on or above the main diagonal is $n-1=4$, and yet the latin square is not completable.


![](https://ilorentz.org/beenakker/MO/latin_square_symmetric.png)
