I have been having trouble figuring out exactly visually when a certain reduced Latin square does not fit the quadrangle criterion; especially when you can flip things to match. So this is my attempt at finding bits in a square that doesn't fit it. Can I get a confirmation that I am doing it right or not
For an order 5 \begin{bmatrix} 0 & 1 & 2 & 3 & 4 \\ 1 & 0 & 3 & 4 & 2 \\ 2 & 3 & 4 & 0& 1 \\ 3 & 4 & 1 & 2 & 0 \\ 4 & 2 & 0 & 1 & 3 \\ \end{bmatrix}
Counter examples of the criterion
\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 2 & 0 \end{bmatrix}
and \begin{bmatrix} 1 &0 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 4 & 2 \\ 0 & 1 \end{bmatrix}
Examples that do fit it
\begin{bmatrix} 3 &4 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ 4 & 2 \end{bmatrix}
and not in the square but something like this would satisfy the criterion right, by a flip and reflect. \begin{bmatrix} 1 &2 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}
I think I have also seen examples where people take 1 by 2 sub matrixes and combine it with another 1 by 2 in the same square and do that again to either find a counter or example like:
\begin{bmatrix} 0 &2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}
in this case with the sub matrixes taken from the order 5 square it does not fit the criterion