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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.

Lieutenant Zipp
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  • In other words, "knight-closedness" is the diameter of the graph? – saulspatz Oct 02 '19 at 20:25
  • I suppose if the correct conditions were supplied, then knight-closedness is a diameter- unfortunately, I do not know how to apply that concept to this particular problem. – Lieutenant Zipp Oct 02 '19 at 20:37
  • I'm just trying to make sure I understand the problem you describe. There's no magic way to compute the diameter of a graph. The diameter of a graph is the longest length of a shortest path between two vertices. Is that what you are saying? – saulspatz Oct 02 '19 at 20:40
  • Yes, for a knight. Learning that there is no closed form is disappointing- oh well, what can I do? – Lieutenant Zipp Oct 02 '19 at 20:50
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    I think the word "optimal" in the question should be replaced with "minimal". As in "maximum possible length of a minimal knight route from one square to another on that board." The word optimal could mean a lot of things.

    Cool question. +1. I wonder if this has a nice generalization to $m\times n$ boards.

     – Mason Oct 02 '19 at 22:12

1 Answers1

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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.

Lieutenant Zipp
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    What does $int$ mean here? Are you rounding up or down or nearest? – Mason Oct 02 '19 at 22:09
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    integer part i.e. rounding down. –  Oct 02 '19 at 22:54
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    Thanks. I hope you spotted my follow up about a $m\times n$ board above. Could be interesting. – Mason Oct 02 '19 at 22:56
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    To the extent that I have carried out the labelling (and assuming I've not made any errors) the pattern looks easier for rectangles when $|m-n|$ is greater than 3 i.e. the slight variations in patterns seem to happen near the diagonal. –  Oct 02 '19 at 23:07
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    It would be good if someone was interested who could program a computer to carry out the labelling and check that no complications occur for large $m,n$. –  Oct 02 '19 at 23:15