Page:Amusements in mathematics.djvu/108

96 3 queens guard 6$2$ board in 1 fundamental way (protected).

4 queens guard 6$2$ board in 17 fundamental ways (not protected).

4 queens guard 7$2$ board in 5 fundamental ways (protected).

4 queens guard 7$2$ 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 n$2$ 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 n$2$ 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 $$n^2$$ 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 $$2^{\frac{1}{2}(n-1)}$$ such arrangements, each giving 4 by reversals and reflections, and 2^n-3- 2{\frac{1}{2}(n-3)} giving 8. Where $$n$$ is even there are $$2^{\frac{1}{2}(n-3)}$$ each giving 4 by reversals and reflections, and $$2^{\frac{1}{2}(n-4)}$$, each giving 8.

We can place $$\frac{1}{2}(n^2+1)$$ knights on an $$n^2$$ board without attack, when n is odd, in i fundamental way ; and ^n^ knights on an n^ board, when n is even, in i 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 n^ squares, two queens, two rooks, two bishops, or two knights can always be placed, irrespective of attack or not, in ways. The following formulae will show in how many of these ways the two pieces may be placed with attack and without : — With Attack. Without Attack. 3 6 2 Kooks w — n^ ! — . 2 Queens 2 Ri<;h <; 4^3 — Gw^ + 2W 3^* — 4»3 -)- 3n2 — 2n 6 6 2 Knights 4m2_i2m-|-8 (See No. 318, " Lion Hunting.") n'^ — gn^+24n DYNAMICAL CHESS PUZZLES. " Push on — ^keep moving." Thos. Morton : Cure for the Heartache 320.— THE ROOK'S TOUR. The puzzle is to move the single rook over the whole board, so that it shall visit every square of the board once, and only once, and end its tour on the square from which it starts. You have to do this in as few moves as possible, and unless you are very careful you wiU take just one move too many. Of course, a square is regarded equally as " visited " whether you merely pass over it or make it a stopping-place, and we wiU not quibble over the point whether the original square is actually visited twice. We will assume that it is not. 321.— THE ROOK'S JOURNEY. This puzzle I call " The Rook's Journey," be- cause the word " tour " (derived from a turner's wheel) implies that we return to the point from which we set out, and we do not do this in the present case. We should not be satisfied with