Now that we have all the derivatives for the second generation of Rule 18, it is possible to calculate its factors using the representation that they are intersections of the derivatives. The easiest way to calculate an intersection is to represent each intersectand via its diagram, collect corresponding nodes into pairs, and work out the regular expression representing the terminal nodes in this new diagram. A terminal pair is one which is terminal for both of its components, signified by the presence of 's in their own diagrams.

Accordingly the exit diagram for the intersections of the left
derivatives can be gotten from a table of links, which can be written
in a compact form if certain preparations are made. Define then deduce the links for **e** and its derivatives by writing
them in canonical form, and finally display all the pairwise
intersections in a matrix.

To read this array, suppose that the intersection is to be
calculated. Then from the row **ae,** column **be** we see from that the node is not terminal, that 0 leads from
to and that 1 leads from to . If further
links are traced onward from these, a diagram with eight nodes is
obtained, but none of them is a terminal state. Thus this intersection
is empty, a conclusion which is confirmed by
observing that **ae** always contains an even number of 1's, while the
number of 1's in **be** is always odd. Since the node occurs in
the same diagram with we can also conclude that that
intersection is empty.

It is evident from the definition of **xe** that The diagram
for reads

These equations can be solved by inspection to obtain

The only triple intersection not deducible from the foregoing results is

finally giving us all the possible right factors. By examining
canonical forms for the derivatives these results can be somewhat
simplified to the forms shown in parentheses. As was to be expected,
they all involve incomplete segments of the basic unit in
All told, including the null intersection they are seven
in number--- **ae,** **be,**
**0ae,** and

By repeating essentially the same calculation with
the left factors are found to be **ef,** **eg,**
**ef0,** and

In general, the **n**-fold intersections can be deduced from a table of
**n**-tuples, and in each case a diagram can be made up and solved to
determine the regular expression corresponding to each intersection.
The only slight doubt that might remain is whether the diagram could be
reduced, due to the unsuspected equality of some of its nodes. Sometimes
this is evident from inspection; as when two different nodes solve the
same equation, or when the expression is fairly short with a clear
interpretation. Otherwise it might be necessary to resort to testing the
diagram for equivalence classes.

Having developed this theory to a fair degree of elaboration, the question naturally arises as to its applicability. For studying infinite or cyclic systems formed from symmetric rules, it does not seem likely that there will be a useful distinction between left factors and right factors; they seem more related to starting up or ending a finite expression, as the analysis which we have made for Rule 18 shows. They are also likely to be of importance for rules which admit fuses or gliders.

On the other hand, it cannot be denied that the combination of Arden's lemma and the calculus of regular expressions yields a very concise and elegant way to describe diagrams symbolically.

E-mail:mcintosh@servidor.unam.mx