On an ordinary chessboard, 8 by 8, every square can be guarded—that is,
either occupied or attacked—by 5 queens, the fewest possible. There are exactly
91 fundamentally different arrangements in which no queen attacks another queen.
If every queen must attack (or be protected by) another queen, there are at
fewest 41 arrangements, and I have recorded some 150 ways in which some of the
queens are attacked and some not, but this last case is very difficult to
On an ordinary chessboard every square can be guarded by 8 rooks (the fewest
possible) in 40,320 ways, if no rook may attack another rook, but it is not
known how many of these are fundamentally different. (See solution to No.
Eight Rooks.") I have not enumerated the ways in which every rook shall be
protected by another rook.
On an ordinary chessboard every square can be guarded by 8 bishops (the
fewest possible), if no bishop may attack another bishop. Ten bishops are
necessary if every bishop is to be protected. (See Nos.
unguarded" and "Bishops
On an ordinary chessboard every square can be guarded by 12 knights if all
but 4 are unprotected. But if every knight must be protected, 14 are necessary.
(See No. 319,
Dealing with the queen on n2 boards generally, where
n is less than 8, the following results will be of interest:—
1 queen guards 22 board in 1 fundamental way.
1 queen guards 32 board in 1 fundamental way.
2 queens guard 42 board in 3 fundamental ways (protected).
3 queens guard 42 board in 2 fundamental ways (not protected).
3 queens guard 52 board in 37 fundamental ways (protected).
3 queens guard 52 board in 2 fundamental ways (not protected).
3 queens guard 62 board in 1 fundamental way (protected).
4 queens guard 62 board in 17 fundamental ways (not
4 queens guard 72 board in 5 fundamental ways (protected).
4 queens guard 72 board in 1 fundamental way (not protected).
NON-ATTACKING CHESSBOARD ARRANGEMENTS.
We know that n queens may always be placed on a square board of n2
squares (if n be greater than 3) without any queen attacking another queen. But
no general formula for enumerating the number of different ways in which it may
be done has yet been discovered; probably it is undiscoverable. The known
results are as follows:—
Where n = 4 there is 1 fundamental solution and 2 in all.
Where n = 5 there are 2 fundamental solutions and 10 in all.
Where n = 6 there is 1 fundamental solution and 4 in all.
Where n = 7 there are 6 fundamental solutions and 40 in all.
Where n = 8 there are 12 fundamental solutions and 92 in all.
Where n = 9 there are 46 fundamental solutions.
Where n = 10 there are 92 fundamental solutions.
Where n = 11 there are 341 fundamental solutions.
Obviously n rooks may be placed without attack on an n2 board in
n! ways, but how many of these are fundamentally different I have only worked
out in the four cases where n equals 2, 3, 4, and 5. The answers here are
respectively 1, 2, 7, and 23. (See No. 296,
We can place 2n-2 bishops on an n2 board in 2n ways.
(See No. 299,
in Convocation.") For boards containing 2, 3, 4, 5, 6, 7, 8 squares, on a
side there are respectively 1, 2, 3, 6, 10, 20, 36 fundamentally different
arrangements. Where n is odd there are 2½(n-1) such arrangements,
each giving 4 by reversals and reflections, and
2n-3 - 2½(n-3) giving 8. Where n is even there
are 2½(n-2), each giving 4 by reversals and reflections, and
2n-3 - 2½(n-4), each giving 8.
We can place ½(n2+1) knights on an n2 board without
attack, when n is odd, in 1 fundamental way; and ½n2 knights on an
n2 board, when n is even, in 1 fundamental way. In the first case we
place all the knights on the same colour as the central square; in the second
case we place them all on black, or all on white, squares.
THE TWO PIECES PROBLEM.
On a board of n2 squares, two queens, two rooks, two bishops, or
two knights can always be placed, irrespective of attack or not, in
½(n4 - n2) ways. The following formulæ will show
in how many of these ways the two pieces may be placed with attack and
||5n3 - 6n2 + n
||3n4 - 10n3 + 9n2 - 2n|
||n3 - n2
||n4 - 2n3 + n2|
||4n3 - 6n2 + 2n
||3n4 - 4n3 + 3n2 - 2n|
||4n2 - 12n + 8
||n4 - 9n2 + 24n|
(See No. 318,