Amusements in Mathematics HTML version
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,
"The Four Lions.")
We can place 2n-2 bishops on an n2 board in 2n ways. (See No. 299, "Bishops 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
With Attack. Without Attack.
5n3 - 6n2 + n 3n4 - 10n3 + 9n2 - 2n
n4 - 2n3 + n2
2 Rooks n3 - n2 2
4n3 - 6n2 + 2n 3n4 - 4n3 + 3n2 - 2n
2 Bishops 6
n4 - 9n2 + 24n
2 Knights 4n2 - 12n + 8