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THE GREYHOUND PUZZLE.

(The Guarded Chessboard)
In this puzzle the twenty kennels do not communicate with one another by
doors, but are divided off by a low wall. The solitary occupant is the
greyhound which lives in the kennel in the top left-hand corner. When he
is allowed his liberty he has to obtain it by visiting every kennel once
and only once in a series of knight's moves, ending at the bottom
right-hand corner, which is open to the world. The lines in the above
diagram show one solution. The puzzle is to discover in how many
different ways the greyhound may thus make his exit from his corner
kennel.


Answer:

There are several interesting points involved in this question. In the
first place, if we had made no stipulation as to the positions of the
two ends of the string, it is quite impossible to form any such string
unless we begin and end in the top and bottom row of kennels. We may
begin in the top row and end in the bottom (or, of course, the reverse),
or we may begin in one of these rows and end in the same. But we can
never begin or end in one of the two central rows. Our places of
starting and ending, however, were fixed for us. Yet the first half of
our route must be confined entirely to those squares that are
distinguished in the following diagram by circles, and the second half
will therefore be confined to the squares that are not circled. The
squares reserved for the two half-strings will be seen to be symmetrical
and similar.
The next point is that the first half-string must end in one of the
central rows, and the second half-string must begin in one of these
rows. This is now obvious, because they have to link together to form
the complete string, and every square on an outside row is connected by
a knight's move with similar squares only--that is, circled or
non-circled as the case may be. The half-strings can, therefore, only be
linked in the two central rows.
Now, there are just eight different first half-strings, and consequently
also eight second half-strings. We shall see that these combine to form
twelve complete strings, which is the total number that exist and the
correct solution of our puzzle. I do not propose to give all the routes
at length, but I will so far indicate them that if the reader has
dropped any he will be able to discover which they are and work them out
for himself without any difficulty. The following numbers apply to those
in the above diagram.
The eight first half-strings are: 1 to 6 (2 routes); 1 to 8 (1 route);
1 to 10 (3 routes); 1 to 12 (1 route); and 1 to 14 (1 route). The eight
second half-strings are: 7 to 20 (1 route); 9 to 20 (1 route); 11 to 20
(3 routes); 13 to 20 (1 route); and 15 to 20 (2 routes). Every different
way in which you can link one half-string to another gives a different
solution. These linkings will be found to be as follows: 6 to 13 (2
cases); 10 to 13 (3 cases); 8 to 11 (3 cases); 8 to 15 (2 cases); 12 to
9 (1 case); and 14 to 7 (1 case). There are, therefore, twelve different
linkings and twelve different answers to the puzzle. The route given in
the illustration with the greyhound will be found to consist of one of
the three half-strings 1 to 10, linked to the half-string 13 to 20. It
should be noted that ten of the solutions are produced by five
distinctive routes and their reversals--that is, if you indicate these
five routes by lines and then turn the diagrams upside down you will get
the five other routes. The remaining two solutions are symmetrical
(these are the cases where 12 to 9 and 14 to 7 are the links), and
consequently they do not produce new solutions by reversal.










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