THE CANTERBURY PUZZLES
ing the example that I gave, not counting the reversals and reflec-
tions of these arrangements as different. Jaenisch, in his " Analyse
Mathematique au jeu des Echecs" (1862) quotes the statement
that there are just twenty-one solutions to the little problem on which
this puzzle is based. As I had myself only recorded seventeen, I
examined the matter again, and found that he was in error, and,
doubtless, had mistaken reversals for different arrangements.
Here are the seventeen answers. The figures indicate the rows,
and their positions show the columns. Thus, 104603 means that
we place a pig in the first row of the
first
column, in no row of the
second
column, in the fourth row of the
third
column, in the sixth
row of the
fourth
column, in no row of the
fifth
column, and in the
third row of the
sixth
column. The arrangement E is that which I
gave in diagram form :
A.
B.
C.
D.
E.
F.
G.
H.
I.
104603
136002
140502
140520
160025
160304
201405
201605
205104
J.
K.
L.
M.
N.
O.
P.
Q.
206104
241005
250014
250630
260015
261005
261040
306104
It will be found that forms N and Q are semi-symmetrical with
regard to the centre, and, therefore, give only two arrangements
each by reversal and reflection ; that form H is quarter-symmetrical,
and gives only four arrangements ; while all the fourteen others
yield by reversal and-reflection eight arrangements each. Therefore,
the pigs may be placed in (2 x 2) + (4 x 1) + (8 x 14) = 120 different
ways by reversing and reflecting all the seventeen forms.
Three pigs alone may be placed so that every sty is in line with a
pig, provided that the pigs are not forbidden to be in line with one
another, but there is only one way of doing it (if we do not count
reversals as different), and I will leave the reader to find it for him-
self,
184
