There are8tuples of cellular automata inEthat can not be distinguished by the invariants inAppendix A.
The following pairs of cellular automata are conjugate by the mapϑ :{0, 1}Z→ {0, 1}Z
that flips every second bit, i.e.,
ϑ : { 0, 1 }Z→ { 0, 1 }Z, ϑ(x)k:=¨1 − xx k ifkis odd
k ifkis even
(a) (15, 170), w15 = σ ◦ υ,w170 = σ. Notice that w15 and w170 can not be strongly
conjugate because every cellular automaton commutes withσ.
(b) (23, 178),
(c) (77, 232).
Next we have the three rules w90, w105, w150:FZ2 → FZ2 with
w90,loc(x−1, x0, x1) = x−1+ x1,
w105,loc(x−1, x0, x1) = 1 + x−1+ x0+ x1,
w150,loc(x−1, x0, x1) = x−1+ x0+ x1.
These, together with their conjugates with respect toυ, are exactly the left- and right-
permutive elementary cellular automata. Therefore, by a result of Kurka [Kur03], they are conjugate to the one-sided full shift with alphabet{ 1, . . . , 4 }and in particular they
6.2. The Special Cases 121 11 10 00 01 0 0 0 1 0 0 0 1 (a) w36 11 10 00 01 0 1 0 0 0 0 1 0 (b) w72
Figure 6.1.: De Bruijn graphs forw36andw72.
are conjugate to each other. A conjugacy between w105 andw150 is given by
ϕ :{0, 1}Z→ {0, 1}Z, ϕ(x)k:=¨1 − xk ifk ∈ { 4ℓ | ℓ ∈ Z } ∪ { 4ℓ + 1 | ℓ ∈ Z }
xk otherwise
,
while the conjugacy betweenw90and w150 is significantly more complicated. We will discuss these cellular automata in more detail inSection 6.3andSection 6.4.
We will show on a case by case basis that all cellular automata in the remaining classes are pairwise non-conjugate. To do so, we define a further invariant of topological conjugacy.
Definition 6.1(Fixed points withkpreimages). For a map f : X → X let Fixk( f )to be the set of all fixed points of f withkpreimages, i.e.,
Fixk( f ) :=x ∈Fix( f ) |f−1[{x}]| = k
.
For every cellular automaton f with local rule floc: { 0, 1 }3 → { 0, 1 }we draw the De Bruijn graph over the alphabet{0, 1}with block length3and annotate its edges by floc. An edge is drawn with a bold line if f (x−1, x0, x1) = x0. The edge shift of the subgraph
defined by the bold edges thus equalsψ[Fix( f )] ∼=Fix( f )whereψ :{0, 1}Z→ ({0, 1}3)Z
is the higher-block map with block length3.
Rules36 and 72
Because of the horizontal symmetry of the annotated De Bruijn graph inFigure 6.1(a), we see that Fix1(w36) = ;. On the other hand∞(011)∞∈Fix1(w72).
Rules78 and 140
From Figure 6.2we derive that ∞1∞ ∈Fix
1(w140). On the other hand, Fix1(w78) = ;
since w−1
78[{∞0∞}] = {∞0∞,∞1∞)and each occurrence of01010or10110might be
replaced by, respectively,01110or10010in fixed points ofw78without changing the image.
122 Chapter 6. Elementary Cellular Automata 11 10 00 01 0 1 0 0 0 1 1 1 (a) w78 11 10 00 01 1 0 0 0 0 0 1 1 (b) w140
Figure 6.2.: De Bruijn graphs forw78andw140.
11 10 00 01 0 0 1 0 0 0 1 0 (a) w24 11 10 00 01 0 0 0 1 0 1 1 1 (b) w46
Figure 6.3.: De Bruijn graphs forw24andw46.
Rules24 and 46
Both cellular automata agree with the shift, eitherσorσ−1, on their eventual image.
Consider the sets M24 := w−1
24[Fix(W24)] = w−124[{∞0∞}] and M46 := w−146[Fix(w46)] =
w−1
46[{∞0∞}]. Both are countable subshifts of finite type.M24is generated by∞0.(10)∞
and∞1.(01)∞, whileM
46is generated by∞1.0∞. ThereforeM24has the four accumula-
tion points∞0∞,∞1∞,∞(01)∞ and∞(10)∞, while M
46has only two, namely∞0∞
and∞1∞.
Rules4, 12, 76 and 200
These cellular automata are all equal to the identity on their eventual image, or, more specifically, Fix( f ) = f [{0, 1}Z]forf ∈ {w4, w12, w76, w200}. The set of fixed points of each
of these four cellular automata is a Cantor space. Notice that even Fix(w4) =Fix(w12).
As a last invariant we look at the possible cardinalities of the preimage of a point and
11 10 00 01 0 0 0 0 0 0 0 1 (a) w4 11 10 00 01 0 0 0 0 0 0 1 1 (b) w12 11 10 00 01 0 1 0 0 0 0 1 1 (c) w76 11 10 00 01 1 1 0 0 0 0 1 0 (d) w200
6.2. The Special Cases 123
(a) w76 (b) w200
Figure 6.5.: Space-Time-Diagrams ofw76, w200with random initial condition and periodic boundary, black represents0and gray represents1.
defineP( f ) := | f−1[{x}] | x ∈ AZ
. Let Fib be the set of Fibonacci numbers defined bya1= 1, a2= 2, ak+2= ak+1+ ak fork ∈ N. We will show that
P(w200) = P(w12)
= |R| ∪ { b1b2. . . bk | k ∈ N, bℓ∈Fib forℓ∈ {1, . . . , k }}.
In the case of w200, the ambiguity in forming the preimage comes from blocks of the form110k11, seeFigure 6.5(b). Since isolated ones are erased byw
200, the number of
preimages of∞1.0k1∞ equals the number of words of lengthk − 2containing no two
consecutive ones, which equals ak−1∈Fib. If more than one block of the form110k11
occurs, one can independently put isolated ones in these blocks without changing the image, hence the number of preimages is the product of those for the single blocks. The same is true for w76, but here we look at blocks which are terminated by11 on both sides and which contain only isolated ones, e.g.,11001001010001011. We can replace
010k10by01k+20without changing the image. But since we can not do this for adjacent
occurrences of010k10, again the number of preimages of∞10w01∞ withwcontaining
ℓisolated ones is aℓ.
On the other hand,
w−1
12[{∞(01).0∞}] = ∞
(01).1k0∞ k ∈ N0 ∪ {∞(01).1∞} ,
so |N| ∈ P(w12). But |N| ̸∈ P(w4). Any point having infinitely many preimages with
respect to w4 must contain infinitely many occurrences of blocks of the form10k1with
k ≥ 3or it must start with∞0or end with0∞. Thus it already has uncountably many
preimages. Consequently,w12is not conjugate to any ofw4, w76andw200. This leaves us with these three cellular automata. Next we look atw−1
4 [{x}]for
x =∞(01).000000(10)∞).
Each element of this set has to coincide with x everywhere except for the underlined block of four consecutive zeros. In this block we only have to ensure that no isolated ones occur. So we have to determine the number of{0, 1}blocks of length4where ones only occur in blocks of length at least two. There can be only either zero or one block of ones of length from2to4. This gives1 + 3 + 2 + 1 = 7possibilities, thusw−1
4 ({x}) = 7. But7is
not a product of Fibonacci numbers, hencew4 is not conjugate to eitherw76orw200. Finally we differentiate between these two cellular automata. Notice that Fix3(w200)
consists of all configurations in Fix(w200)containing the block11000011but no other
124 Chapter 6. Elementary Cellular Automata Fix3(w200) ∪Fix1(w200). On the other hand we have(∞0.10∞) ∈Fix3(w76), hence there
is (xn)n∈N in Fix3(w76) withxn→∞0∞ ∈Fix2(w76). With that we have finally shown
that w200 andw76are not topologically conjugate.
Notice, however, that|Fixk(w76)| = |Fixk(w200)| = |R|for allk ∈ P(w200) = P(w76). Since
w76and w200 are idempotent, they are thus conjugate when{0, 1}Zis endowed with the discrete topology instead of the product topology.