SEMI-REGULAR NETWORKS OF THE PLANE IN ABSOLUTE GEOMETRY. 735 
The number of combinations of two or more together of these numbers which 
involve all four polygons is 856, but we have the following rules for the rejection of 
impossible combinations : 
(a) 2 must be accompanied by 4 or 10. 
(b) 3 i a A 5; and either 6, or 10 and 4. 
(Os ay - s either 10, or 4 and 6. 
(d) 6 x i ‘3 4 or 5. 
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The previous proofs of these still hold. 
(e) 1 must be accompanied by either 4, 5 and 6, or 11. 
(9) 
(h 
ee 
— 
If we exclude 11, triangles or hexagons must be in combination with 
squares, and we have seen that squares can never be introduced if we exclude 
4 and either 5 or 6. But 1 may be continued by 11 (fig. 16). 
We must have either 11, or 10 and 12. 
For, excluding 11, we must have 12 at least, for dodecagons are only given 
by 7, 11, and 12, and 7 is excluded by (/). 
Starting therefore with 12, we must have fig. 13. The concavities can now 
be filled either with dodecagons, or with squares and triangles. The latter 
gives 10, the former never introduces triangles. 
If 4 and 12 be excluded, the only combination is 7, 10, 11. 
Excluding 4 and 12, we must have 11. Let us start therefore with fig. 14 
(the heavy lines). a cannot be a square, for that gives a hexagon at b; nor 
a hexagon; nor a dodecagon, since 12 is excluded. Hence a must be a 
triangle, and we get the figure with dotted lines. 
Again, if we start with fig. 15 (the heavy lines), it must be continued as in 
the dotted lines. 
Lastly, let us start with fig. 16. A dodecagon at @ gives us a variation of 
fig. 14 with the hexagon turned through 60°, a square gives fig. 15, and a 
triangle fig. 16 with the dotted lines. 
The continuation of any of these figures (under the given conditions) will 
introduce no angles other than 1, 7, 10, 11; and fig. 16 must be excluded, 
since it does not contain hexagons. Hence the only possible combination is 
nO, 11. 
If 4 and 10 be excluded, the only combination is 1, 11, 12. 
We must have 12, by (h); and 11, by (9g). 
Starting with 12 we get fig. 17. At awe must put either a hexagon or 
six triangles, hence the figure can only be continued as in the dotted lines, 
where some of the hexagons, but not all, must be filled up with triangles. 
This is the combination 1, 11, 12. 
Rejecting according to these rules, there are left 222 combinations. I have tested 
