Given an otherwise empty $n\times n$ chessboard with a knight on one of the squares, define the “knight-closedness” of this board as the maximum possible length of a minimal knight route from one square to another on that board. Determine a closed form for the knight-closedness of such a board in terms of $n$.
I came up with this problem on the train, and I think I’ve determined a few isolated values: for example, I have that the knight-closedness of an $8\times 8$ board is 6. However, I am not learned enough in graph theory to thoroughly resolve this question.
I don't think there's any need to be disappointed!
I have not written out a formal proof but, for an $n$ x $n$ board, with $n\ge 6$,the knight-closeness appears to be $n-\lfloor\frac {n}{3}\rfloor$.
The method I applied was to work outwards from a square labelled $0$. All squares which can be reached directly are labelled $1$. Then all previously unlabelled squares which can be reached from squares labelled $1$ are labelled $2$ and so on.
If you have a go at doing this you will see that what makes the problem solvable is that a clear pattern emerges across the board once one is a few squares away from the start square. For example, as you circle the start square, the squares which have a maximum $x$ or $y$ distance of $5$ from the start square are all labelled $3$ or $4$ alternately.