Finding intercalates within a (reduced) latin square

Question

I need confirmation on if my intuition on finding intercalates is correct, suppose we have the following reduced latin square

\begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 1 & 3 \\ 3 & 1 & 4 & 2\\ 4 & 3 & 2 & 1 \end{bmatrix}

we know from a theorem that for a latin square of order n(even), there is at most $\frac{n^2(n-1)}{4}$ intercalates; in this case 12.

I am not sure if two 2 x 2 sub matrixes can be considered the same intercalate if they are flipped and rotated versions of each other. From the square above we get: (e.g R1 = row 1)

R1,R2 \begin{bmatrix} 1 & 2 & | 3&4\\ 2 & 4 & |1&3 \end{bmatrix} R1,R3 \begin{bmatrix} 1 & 2 & | 3&4\\ 3 & 1 & |4&2 \end{bmatrix} R1,R4\begin{bmatrix} 1 & 2 & | 3&4\\ 4 & 3 & |2&1 \end{bmatrix} R2,R3 \begin{bmatrix} 2 & 4 & | 1&3\\ 3 & 1 & |4&2 \end{bmatrix} R2,R4 \begin{bmatrix} 2 & 4 & | 1&3\\ 4 & 3 & |2&1 \end{bmatrix}

R3,R4 \begin{bmatrix} 3 & 1 & | 4&2\\ 4 & 3 & |2&1 \end{bmatrix}

And only keeping ones that are not flips and rotations of each other (I keep 1 copy). These should be the only intercalates right? So 6 of them.

\begin{bmatrix} 1&2&|3&4&|&3&4&|&1&2&|1&2&|&2&4&\\ 2&4&|1&3&|&4&2&|&3&1&|4&3&|&3&1& \end{bmatrix}

Answer 1

None of these are intercalates. As defined in this paper (for example), we're looking for two rows $i,j$ and two columns $x,y$ such that $L_{i,x} = L_{j,y}$ and $L_{j,x} = L_{i,y}$. There are four of these in your example:

• $\{i,j\} =\{1,4\}$ and $\{x,y\} = \{1,4\}$: $$\begin{bmatrix}1 & 4 \\ 4 & 1\end{bmatrix}$$
• $\{i,j\} =\{2,3\}$ and $\{x,y\} = \{2,3\}$: $$\begin{bmatrix}4 & 1 \\ 1 & 4\end{bmatrix}$$
• $\{i,j\} =\{1,4\}$ and $\{x,y\} = \{2,3\}$: $$\begin{bmatrix}2 & 3 \\ 3 & 2\end{bmatrix}$$
• $\{i,j\} =\{2,3\}$ and $\{x,y\} = \{1,4\}$: $$\begin{bmatrix}2 & 3 \\ 3 & 2\end{bmatrix}$$