Mathematical_Recreations-Kraitchik-2e

The Problem of the Queens 247
up into a set of smaller boards of higher order than 1, together
with empty square or rectangular boards, in such a
way that each board which is occupied contains a solution
for its order. A composite solution
may be called proper or
improper, according as the
pieces into which the board is
broken are equal squares or
not. Figure 118 shows an improper
composite solution of
order 8, and Figure 119 shows
a proper composite solution of
order 9. In a proper composite
solution the occupied and
FIGURE 120.
unoccupied squares themselves
form a solution, so that if the solutions in the smaller squares
are identical the composite solution may be regarded as the
product of the two solutions. Figure 120 is the square of the
solution in any occupied small square.
3. THE PROBLEM OF THE QUEENS
The solutions of the problem of the queens are now seen
to consist of those solutions of the problem of the rooks
which have no two queens on any diagonal (in the sense explained
on p. 238). If we classify these solutions as we did in
the previous section we find that there are no simply or doubly
diagonally symmetric solutions. Thus only the classes
0, C, and Q remain, and not all members of those classes.
(For example, the ordinary solutions of the problem of the
rooks for n = 4 are not solutions of the problem of the
queens.) But it is still true that all solutions of a set of equivalent
solutions belong to the same class.
There are no ordinary solutions for n < 5, and there is
just one complete set of ordinary solutions for n = 5, namely