SEMI-REGULAR NETWORKS OF THE PLANE IN ABSOLUTE GEOMETRY. 733 
conditions, then, we cannot find any relation between the k’s, which are therefore inde- 
terminate. ‘This shows that in general a composite type admits of infinite variation, the 
ratios of the number of polygons being indefinite. 
Composite Types. 
§ 13. We have defined two networks to be of the same type when they involve 
only nodes of the same species, whatever may be their form and structure. The 
determination of the composite types in any group thus reduces to finding all the 
possible combinations of nodes admitted by the group. In order to eliminate the 
impossible combinations, we can find what combinations must occur. We shall take 
each group in turn. 
Class B. I. Triangles and Squares (1, 2, 4). 
(a) 2 must be accompanied by 4 (fig. 11). 
Hence the only combinations are 1, 4; 2,4; 1, 2, 4, each of which gives 
a composite type. Examples of each are given by joining together infinite 
parallel strips of the triangular and square networks. 
II. Triangles and Hexagons (1, 3, 5, 6). 
(b) 3 must be accompanied by 5 (fig. 12). 
Fence the only combinations are 1, 5; 1,65 3,5; 5,6; 1), 3,°5; 1,5,6; 
3,5,6; 1,38,5,6. There are thus eight composite types. Hvery variety of 
them can be obtained from the triangular network by replacing groups of six 
triangles by hexagons. 
III. Triangles and Dodecagons (1, 7). 
IV. Squares and Octagons (2, 8). 
Each of these is unique and admits of no composite types. 
§ 14. Class C. I. Triangles, Squares, and Hexagons (1, 2, 3, 4, 5, 6, 10). 
The number of combinations of these numbers, two or more together, so 
as to include the three polygons, is 105. We proceed to establish rules for the 
rejection of impossible combinations. 
(a) 2 must be accompanied by 4 or 10 (fig. 11). 
(b) 8 - ‘ 5 5; and either 6, or 10 and 4 (fig. 12). 
Exclude 6; then we must have 10, since squares and hexagons come 
together ; and since there are always two triangles together, we must also 
have 4. 
Exclude 10; then squares cannot be introduced till the hexagons have 
been surrounded by triangles, and we have 6. 
(c) 5 must be accompanied by either 10, or 4 and 6 (fig. 12). 
Exclude 10; before squares can be added we get 6, and the further 
addition of squares gives 4. 
TRANS. ROY. SOC, EDIN., VOL, XLI. PART. IIT. (NO. 29). 109 
