Catherine M. Walsh (Dr. Michael J Bardzell),

designpadAI and Robotics

Dec 1, 2013 (3 years and 6 months ago)

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Catherine M. Walsh (Dr. Michael J
Bardzell
),

Department of Mathematics and Computer
Science, Salisbury University, Salisbury, Maryland



Cellular Automata(CA)


an example of Discrete Dynamical System which
consists of:


A finite set called the
alphabet
; can be a group, ring or
field


A lattice of
cells
, finite or infinite of any dimension,
take on values from that alphabet


The system is updated in discrete time steps,
t=1,2,3… according to a
local rule
.


Let T
0

denote the set of all states


Let T
1

will be the states that remain after one time step


Let T
i

will be the states that remain after
i

time steps.



Update Rule


Determines how neighboring cells will determine
the value of a cell in the following time step. The
update rule defines a function ɸ: T
0

→ T
1
.










Null Boundary Conditions


At the end of lattice, used the additive identity,
usually zero, as the value for your update rule.

This update rule will be called Rule 1.


Each “node” in the Diagram represents one
state of the CA.


Each “arrow” represents the evolution of the
state in the next time step.

The State Transition Diagram for a 12 cell CA over the
alphabet {0,1} with periodic boundary conditions,
where you use a wrapping technique at the end of a
lattice from Rule 1.


The kernel of ɸ is the
preimage

of the identity,
which can be represented as (0,0,…,0), of the CA.


The state transition diagram’s geometry ties in
with the kernel of the evolution homomorphism
and determines whether the diagram is a
collection of rooted trees, cycles, and/or
products of cycles with rooted trees.


If the kernel of ɸ is trivial, then the evolution
map is one
-
to
-
one and the CA is reversible,
which means all nodes are on cycles.



The kernel can suggest how long or how many
steps until each state within the system will hit a
fixed point, or hit a cycle.




First looked at the update rule under an
Abelian

group alphabet:


Is the evolution corresponding to this this rule homomorphism?

Seen above will be called Rule 2.

Rule 3 will be the mirror image; so
step t+1 will be given by the cell
above added to the cell to the right.



We found that this update rule is actually a
homomorphism by showing,


If S
1
=(a
1
,a
2
,a
3
,…,a
n
) and S
2
=(b
1
,b
2
,b
3
,…,
b
n
)


Then, ɸ(S
1
+ S
2
)=ɸ(S
1
)+ɸ(S
2
).


If ɸ: T
0

→ T
1

is defined using the update Rule
2, with an alphabet that is any
Abelian

group,
then ɸ will be a homomorphism.


If the kernel is trivial, {(0,0,…,0)}, and it is a
homomorphism then ɸ: T
0

→ T
1

is one
-
to
-
one.


Theorem:


If ɸ: T
0

→ T
1
, where ɸ is defined using the update
Rule 2, with null boundary conditions, |Ker(ɸ)|=1,
and that ɸ is a homomorphism, then the CA is
reversible.


Additionally


Rule 3 is equivalent, since it is a mirror image of
update Rule 2 and thus, is also reversible.


Observe update Rule 2 with an alteration:









Such that
r,s

are elements of the chosen
alphabet, over the ring
Z
n
.


Looking at which
r,s

will produce
Ker(ɸ)≠{(0,0,…,0)}.


If working over
Z
p
, where p is prime, do we always
get a trivial kernel?


When
r,s

are zero divisors, what interesting
qualities occur?


Working under
Z
6

such that r=3 and s=3 over 4 cells
you can find {(0, 2, 2, 2)}
є

Ker(ɸ) and hence, the
system is not reversible.


“The Evolution Homomorphism and
Permutation Actions on Group Generated
Cellular Automata,” Nicole Miller & Michael
Bardzell
.


“A Classification of Sample 1
-
Dimensional
Automata Generated over Cyclic Groups,”
Stephan
Gymnich
.


“Linear Finite Dynamical Systems,” René
Hernández

Toledo.