The Evans conjecture ( which was proved later by Smetianuk) 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.

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.

The Evans conjecture 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.

The Evans conjecture ( which was proved later by Smetianuk) 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.