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## THE GUARDED CHESSBOARD.

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 enumerate exactly.

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. 295, "The 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. 297 and 298, "Bishops unguarded" and "Bishops guarded.")

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, "The Knight-Guards.")

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 protected).

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, "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 without:—

 With Attack. Without Attack. 2 Queens 5n3 - 6n2 + n 3n4 - 10n3 + 9n2 - 2n 3 6 2 Rooks n3 - n2 n4 - 2n3 + n2 2 2 Bishops 4n3 - 6n2 + 2n 3n4 - 4n3 + 3n2 - 2n 6 6 2 Knights 4n2 - 12n + 8 n4 - 9n2 + 24n 2

(See No. 318, "Lion Hunting.")