The K-partitions of the R-gon. in 



face having two, and only two, non-marginal diagonals, is 

 an irreducible partition. 



H has 23 summits and 14 diagonals. Its 15 faces are 

 9 marginal triangles, 4 sub-marginal faces, and 2 faces, a 

 triangle and a pentagon, each of which has for edges 3 

 non-marginal diagonals that are the 5 dark edges of the 

 figure. 



An irreducible H may have any number of faces that 

 have each for edges 3 + z non-marginal diagonals, i being 

 different or not in any two such faces ; and such edges 

 will be called dark edges in H. 



It is equally true that triangles and squares having two, and 

 two only, non-marginal diagonals for edges, and carrying no 

 marginal triangle, may fall out and leave only an irreducible 

 H. Thus, if in Fig. H the dark line 65 be completed into 

 either the triangle 655' or the 4-gon 6'6$$'. carrying no 

 marginal triangle, either 655' or '665 5' would have for edges 

 two non-marginal diagonals d d\ and both could fall out, 

 leaving, after union of each d and d\ exactly the irreducible 

 H. We shall have to consider both the insertion and the 

 dropping out of such simple faces. They form, when out 

 of the R-gon, not a belt, but a tape, like Fig. 8. 



The dark and dotted lines in Figs. (1, 2, 3, 4, 5), viz., the 

 pairs 65, 6 / 5 / , 31, 3V, &c, are the diagonals d d' that could 

 unite if the primes between them were to fall out. The 

 dark line on the right in A, Fig. 1 is an error — it should be 

 not dark. 



This brings out a novel and useful notion — that H can 

 be completed into a k-partitioned R-gon by simply splitting 

 its dark diagonals to receive one or more primes out of a 

 given belt. And it is evident to the reader that there lies 

 the secret of the construction upon H of Figs. (1, 2, 3, 4, 5). 



4. If^rbe the number of the primes in the belt, and k 

 the number of the dark diagonals of H, we have only to 

 form all the k-partitions of x, as a lf o 2 , . . . a k , whose sum of 



