THE KNIGHT-GUARDS.
(
The Guarded Chessboard)
The knight is the irresponsible low comedian of the chessboard. "He is a
very uncertain, sneaking, and demoralizing rascal," says an American
writer. "He can only move two squares, but makes up in the quality of
his locomotion for its quantity, for he can spring one square sideways
and one forward simultaneously, like a cat; can stand on one leg in the
middle of the board and jump to any one of eight squares he chooses; can
get on one side of a fence and blackguard three or four men on the
other; has an objectionable way of inserting himself in safe places
where he can scare the king and compel him to move, and then gobble a
queen. For pure cussedness the knight has no equal, and when you chase
him out of one hole he skips into another." Attempts have been made over
and over again to obtain a short, simple, and exact definition of the
move of the knight--without success. It really consists in moving one
square like a rook, and then another square like a bishop--the two
operations being done in one leap, so that it does not matter whether
the first square passed over is occupied by another piece or not. It is,
in fact, the only leaping move in chess. But difficult as it is to
define, a child can learn it by inspection in a few minutes.
I have shown in the diagram how twelve knights (the fewest possible that
will perform the feat) may be placed on the chessboard so that every
square is either occupied or attacked by a knight. Examine every square
in turn, and you will find that this is so. Now, the puzzle in this case
is to discover what is the smallest possible number of knights that is
required in order that every square shall be either occupied or
attacked, and every knight protected by another knight. And how would
you arrange them? It will be found that of the twelve shown in the
diagram only four are thus protected by being a knight's move from
another knight.
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 n squared boards generally, where n is less
than 8, the following results will be of interest:--
1 queen guards 2 squared board in 1 fundamental way.
1 queen guards 3 squared board in 1 fundamental way.
2 queens guard 4 squared board in 3 fundamental ways (protected).
3 queens guard 4 squared board in 2 fundamental ways (not protected).
3 queens guard 5 squared board in 37 fundamental ways (protected).
3 queens guard 5 squared board in 2 fundamental ways (not protected).
3 queens guard 6 squared board in 1 fundamental way (protected).
4 queens guard 6 squared board in 17 fundamental ways (not protected).
4 queens guard 7 squared board in 5 fundamental ways (protected).
4 queens guard 7 squared 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 squared
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 squared 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 squared board in 2^{n} 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^{1/2(n-1)} such arrangements, each giving 4 by reversals and
reflections, and 2^{n-3} - 2^{1/2(n-3)} giving 8. Where n is even there
are 2^{1/2(n-2)}, each giving 4 by reversals and reflections, and 2^{n-3}
- 2^{1/2(n-4)}, each giving 8.
We can place 1/2(n squared+1) knights on an n squared board without attack, when n
is odd, in 1 fundamental way; and 1/2n squared knights on an n squared 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 n squared squares, two queens, two rooks, two bishops, or two
knights can always be placed, irrespective of attack or not, in 1/2(n^{4}
- n squared) 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.
2 Queens 5n cubed - 6n squared + n 3n^{4} - 10n cubed + 9n squared - 2n
------------------- ------------------------------
3 6
2 Rooks n cubed - n squared n^{4} - 2n cubed + n squared
----------------------
2
2 Bishops 4n cubed - 6n squared + 2n 3n^{4} - 4n cubed + 3n squared - 2n
-------------------- -----------------------------
6 6
2 Knights 4n squared - 12n + 8 n^{4} - 9n squared + 24n
--------------------
2
(See No. 318, " Lion Hunting.")
DYNAMICAL CHESS PUZZLES.
"Push on--keep moving."
THOS. MORTON: _Cure for the Heartache_.
Answer:
[Illustration: DIAGRAM 1.]
[Illustration: DIAGRAM 2.]
The smallest possible number of knights with which this puzzle can be
solved is fourteen.
It has sometimes been assumed that there are a great many different
solutions. As a matter of fact, there are only three arrangements--not
counting mere reversals and reflections as different. Curiously enough,
nobody seems ever to have hit on the following simple proof, or to have
thought of dealing with the black and the white squares separately.
[Illustration: DIAGRAM 3.]
[Illustration: DIAGRAM 4.]
[Illustration: DIAGRAM 5.]
Seven knights can be placed on the board on white squares so as to
attack every black square in two ways only. These are shown in Diagrams
1 and 2. Note that three knights occupy the same position in both
arrangements. It is therefore clear that if we turn the board so that a
black square shall be in the top left-hand corner instead of a white,
and place the knights in exactly the same positions, we shall have two
similar ways of attacking all the white squares. I will assume the
reader has made the two last described diagrams on transparent paper,
and marked them _1a_ and _2a_. Now, by placing the transparent Diagram
_1a_ over 1 you will be able to obtain the solution in Diagram 3, by
placing _2a_ over 2 you will get Diagram 4, and by placing _2a_ over 1
you will get Diagram 5. You may now try all possible combinations of
those two pairs of diagrams, but you will only get the three
arrangements I have given, or their reversals and reflections. Therefore
these three solutions are all that exist.
Informational
