Shape

298 II Seeing How It Works
are all made up of triangles. Their numbers provide a nice summary of what happens
as the rule is applied from step to step. Consider the following numerical series
The first of the trio shows the maximum number of triangles—these can be picked out
using an identity—in each of the shapes. The next gives the number of triangles in a
shape after the rule has been applied, and the last gives the number of triangles in a
shape as the rule is being applied. In the second and third series, the number of triangles
depends on what the rule does, either by rotating an existing triangle or by seeing
a new one to rotate, and then counting in terms of the triangle and its complement.
The resulting inconsistencies in the fourth and sixth shapes tell the story. Counting
goes awry. Three triangles can’t be two, and neither number is five. (It’s blind luck
that two and three make five. Usually, there’s no easy relation to find.) But this way
of calculating is a continuous process when topologies are given for the nine shapes.
The topologies for these shapes are defined as Boolean algebras with the atoms
shown in table 11. This is practicable using the mapping h 1 ðxÞ ¼x tðAÞ and my formulas
for closed parts. First a topology is given for the final shape. Any finite Boolean
algebra works. I’ve used the trivial topology—the final shape is an atom—because the
rule doesn’t divide it. There’s nothing to say—the final shape is without finer parts or
definite purpose, ready for calculating to go on in any way at all. Once the topology
of the final shape is decided, the topologies of the preceding shapes are defined in reverse
order to keep the application of the rule continuous. Each topology contains the
triangle tðAÞ resolved by the rule—the topmost atom in table 11—and its complement
C tðAÞ to form a Boolean algebra.
Keeping track of how rules work is a good way to see what’s happening. I can
record the divisions in the triangle in the right side of the rule
formed with respect to my topologies. This shows—again in the terminology of my
formulas—what tðBÞ does in C þ . The triangle is cut in alternative ways