96 REV. T. P. KIRKMAN ON THE 
in the pentagon D, nor in the tetragon G. Hence it 
comes to pass that we read five units under the pentagon 
A, showing all its summits, and five units opposite the 
pentace a, showing all its faces, &c. 
The fifteen edges of the figure are represented thus: 
AjA,B,5;=% AjA.C,C,=1, A,A,D.D,=—1, A,A,G63=5 
A,A,H,H,=1, D,D.E,E.=1, B,B,C;=1, B,BzE,E,=1, 
C.C,D.Da=1, D,D,G,G,=1, D.D,F.F-=1, G,G,F,F;=1, 
G,G,0,0,=1, FP n,n = 1, 2b Bee 2) 
Each factor in every quadruplet is unity. The paradigm 
shows these edges each one as four points of a rectangle; 
thus A, A, D, D, are four such points, and F, F; H, H; 
are such also. The charming peculiarity of the paradigm 
is, that no displacement of any face and its column, or of 
any summit and its row, can break these edges, 7.e. spoil 
these rectangles ; so that the same polyedron is accurately 
exhibited, whatever be the order of the letters A B C---or 
abc--- And no rectangle occurs in the paradigm which 
is not an edge of the polyedron. The property of these 
quadruplets is, that no duad of two capitals or of two small 
letters appears a second time, while every duad of a capital 
and a small letter occurs exactly twice over, and always in 
two different quadruplets. 
If the reader will take any other 8 numbers, ABC.-. 
and some 9 adc.--., satisfying such an equation as (A), 
and endeavour by trial to form with them a paradigm 
showing fifteen edges, he will, whether he succeeds or not, 
obtain some idea of the tactical difficulty involved in this 
problem of the polyedron. Or he may try to satisfy him- 
self as to the possibility of forming other 9-acral 8-edra 
with the numbers above handled. 
The next step is to express all the conditions of the 
question in the shape of equations, which are to be true 
independently of all arrangement of their symbols. 
