[Illustration: DIAGRAM 1.]
[Illustration: DIAGRAM 2.]
The smallest possible number of knights with which this puzzle can be
solved is fourteen.
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.
[Illustration: DIAGRAM 3.]
[Illustration: DIAGRAM 4.]
[Illustration: DIAGRAM 5.]
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.