Cycles in $5×5$ Latin squares (solution found) One of my favorite puzzle formats is KenKen, in which you are to find an $n×n$ Latin square to be filled with whole numbers from 1 to $n$, given sums, differences, products or quotients of various groups of entries.  The puzzles come in various sizes, including $5×5$.
When I solve a $5×5$ puzzle I always seem to find at the end that at least two rows are cyclic permutations of each other.  Is this always true in $5×5$ Latin squares, or is there some weird coincidence going on?
Turns out I found the solution.  Over 90% of 5×5 Latin squares actually have at least one pair of cyclically permuted rows.  So, a 5×5 KenKen without such a cyclic permutations turns out to be relatively rare!

A counter example where none of the rows are a cyclic permutation of another:
$$\begin{bmatrix}
1&2&3&4&5\\
2&3&5&1&4\\
3&5&4&2&1\\
4&1&2&5&3\\
5&4&1&3&2
\end{bmatrix}$$
But, it turns out that there are only $5$ reduced Latin squares out of $56$ total for $5×5$ squares.  So, less than 10% of all $5×5$ solutions should be expected to lack such a cyclic row permutation.  Hence such solutions are hard to find.
Here are the other four reduced squares with no cyclic row permutations:
$$\begin{bmatrix}
1&2&3&4&5\\
2&5&1&3&4\\
3&1&4&5&2\\
4&3&5&2&1\\
5&4&2&1&3
\end{bmatrix}$$
$$\begin{bmatrix}
1&2&3&4&5\\
2&5&4&1&3\\
3&4&2&5&1\\
4&1&5&3&2\\
5&3&1&2&4
\end{bmatrix}$$
$$\begin{bmatrix}
1&2&3&4&5\\
2&4&5&3&1\\
3&5&2&1&4\\
4&3&1&5&2\\
5&1&4&2&3
\end{bmatrix}$$
$$\begin{bmatrix}
1&2&3&4&5\\
2&4&1&5&3\\
3&1&5&2&4\\
4&5&2&3&1\\
5&3&4&1&2
\end{bmatrix}$$
