For five letters (adding e) I subdivide the 16 Cells of the previous Diagram by oblique partitions, assigning all the upper portions to e, and all the lower portions to e′. Here, I admit, we lose the advantage of having the e-Class all together, “in a ring-fence”, like the other 4 Classes. Still, it is very easy to find; and the operation, of erasing it, is nearly as easy as that of erasing any other Class. We have now got 32 Cells.
For six letters (adding h, as I avoid tailed letters) I substitute upright crosses for the oblique partitions, assigning the 4 portions, into which each of the 16 Cells is thus divided, to the four Classes eh, eh′, e′h, e′h′. We have now got 64 Cells.
[pg178]For seven letters (adding k) I add, to each upright cross, a little inner square. All these 16 little squares are assigned to the k-Class, and all outside them to the k′-Class; so that 8 little Cells (into which each of the 16 Cells is divided) are respectively assigned to the 8 Classes ehk, ehk′, &c. We have now got 128 Cells.
For eight letters (adding l) I place, in each of the 16 Cells, a lattice, which is a reduced copy of the whole Diagram; and, just as the 16 large Cells of the whole Diagram are assigned to the 16 Classes abcd, abcd′, &c., so the 16 little Cells of each lattice are assigned to the 16 Classes ehkl, ehkl′, &c. Thus, the lattice in the N.W. corner serves to accommodate the 16 Classes abc′d′ehkl, abc′d′eh′kl′, &c. This Octoliteral Diagram (see [next page]) contains 256 Cells.
For nine letters, I place 2 Octoliteral Diagrams side by side, assigning one of them to m, and the other to m′. We have now got 512 Cells.