of Edinburgh, Session 1879-80. 
665 
path, for it is a rook’s upon a new chess-board formed by the alter- 
nate diagonals of the squares on the original board.) That there 
may be circulation in the channel, one of its squares must be the 
blank one. 
Hence an odd number of pieces lies along the channel, and, 
therefore, when they are anyhow displaced along it, so that the 
blank square finally remains unchanged, the number of inter- 
changes is essentially even. 
Thus to test whether any given arrangement can be solved, all 
we need know is how many interchanges of two pieces will reduce 
it to the normal one. If this number be even, the solution is 
possible. To find the number of interchanges, we have only 
to write in pairs the numbers occupying the same square in 
each arrangement, and divide them into groups, such as 
? ^ ® ^ , which form closed cycles. Here there are four pairs 
o c a a 
in the group, which correspond to three interchanges, because 
^ ^ is one interchange. 
Dr Crum Brown suggests the term Aryan for the normal arrange- 
ment, with the corresponding term Semitic for its perversion. 
Similarly Chinese would signify the Aryan rotated right-handedly 
through a quadrant, and Mongol Semitic rotated left-handedly 
through a quadrant. 
How it is easily seen that Aryan is changed into Semitic, and 
Chinese into Mongol, or vice versa , by an odd number of inter- 
changes. Similarly Aryan and Mongol, and Semitic and Chinese, 
differ by an even number of interchanges. 
Hence any given arrangement must be either Aryan or Semitic. 
The former can be changed into Mongol, the latter into Chinese. 
Unless the 6 and 9 be carefully distinguished from one another 
every case is solvable, for if it be Semitic the mere turning these 
figures upside down effects one interchange and makes it Aryan. 
The principle above stated is, of course, easily applicable to the 
conceivable, but scarcely realisable, case of a rectangular arrangement 
of equal cubes with one vacant space. 
