586 Proceedings of the Royal Society 
and (§ 8) we have already virtually got it from the first of the three, 
viz., 
^3 = 1, and fe , . . . , ^,-3 , - 1) = , . . . , q ,_^) , 
because in the change which is the test of the symmetry of the 
equation and are interchanged letters. In short, if we con- 
sider the case where ^3 = 1, we do not require to consider the case 
where 
12. Taking the case of a cycle of 16 elements, 
a, h, c, d, e,f g, h, g,f c, d, c, b, a , 
the equation of condition is 
(b, c, d, e, /, g) = 2{a, b, c, d, ej ) . 
The solution leads to the following “ tree ” : — 
<5r 
CM \ 
II 
<aT 
-o' 
'o' 
(M 
'7 = 2 «, 
^ r 
b = 2f ^ I 
II -I e = 2 c+ 1 , = 1 
+ 
^ i 
Si 
e = l 
+ 
c/ = 2a-l- 1^5 
b = \ 
<;r 
'j' 
c = 2 
/=2 
CM 
1 
-I f=2d + 2,e=\ 
7 f 
^ I 
i d = e--.‘l 
II I 
