(related to Problem: The Knight-guards)

The smallest possible number of knights with which this puzzle can be solved is fourteen.

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It has sometimes been assumed that there are a great many different solutions. As a matter of fact, there are only three arrangements — not counting mere reversals and reflections as different. Curiously enough, nobody seems ever to have hit on the following simple proof or to have thought of dealing with the black and the white squares separately.

a319c a319d a319e

Seven knights can be placed on the board on white squares so as to attack every black square in two ways only. These are shown in Diagrams $1$ and $2.$ Note that three knights occupy the same position in both arrangements. It is therefore clear that if we turn the board so that a black square shall be in the top left-hand corner instead of a white, and place the knights in exactly the same positions, we shall have two similar ways of attacking all the white squares. I will assume the reader has made the two last described diagrams on transparent paper and marked them $1a$ and $2a.$ Now, by placing the transparent Diagram $1a$ over $1$ you will be able to obtain the solution in Diagram $3,$ by placing $2a$ over $2$ you will get Diagram $4,$ and by placing $2a$ over $1$ you will get Diagram $5.$ You may now try all possible combinations of those two pairs of diagrams, but you will only get the three arrangements I have given, or their reversals and reflections. Therefore these three solutions are all that exist.

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Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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