Vol.
Xxxx
(xxx)
ACTA PHYSICA POLONICA A
No. Xxx
Proceedings of the 3nd International Congress APMAS2013, April 24_28, 2013, Antalya, Turkey
One Dimensional Cellular Automata
w
ith Reflective Boundary Conditions
a
nd Radius Three
H. AKIN
a
, I. SIAP
b
, AND S. UGUZ
c
a
Department of Mathematics, Education Faculty, Zirve University, 27260, Gaziantep

Turkey,
b
Department of Mathematics, Faculty of Art and Sciences, Yildiz Technical University,
34210,
Istanbul

Turkey
,
c
Department of Mathematics, Faculty of Art and
Sciences, Harran University, 63000,
Istanbul

Turkey,
Abstract:
A family of one dimensional finite linear cellular automata with reflective boundary condition over the field
p
Z
is defined. The generalizations are the radius and the field that states take values. Here, we establish a connection
between reversibility of cellular automata and the rule matrix
of the cellular automata with
radius
three
.
Also, we prove
that the revers
e CA of this fami
ly again falls into this family.
DOI: xxxxxxxxxx/APhysPolA.xxx.xxx
PACS: 02.10.Yn, 07.05.Kf, 02.10.Ox
Key words: Cellular automata, diagonal matrix, reversibility, matrix algebra
1.
Preliminaries
In this section, we define 1D finite CAs with reflective boundary conditions (shortly RBC) over primitive fields with
p
elements
={0,1,,1}
p
p
Z
, where
2
p
is prime. This definition is a natural generalization of particular 1D null boundary CAs. As a
special case with
= 2
p
the structure and reversibility problem over binary fields is studied by del Rel et al. in [3] and primitive
fields
is studied by Cinkir et al. in [4] and Siap et al. in [5] respectively. The approach of studying the algebraic structure and
their
reversibility property for this general case is generalized from [5]. Depending on the neighborhood of the extreme cells, th
ere
exist three approaches:
1)
The p
eriodic
b
oundary
CA
[4, 5].
2)
The n
ull
b
oundary CA
[2, 3].
3)
The r
eflective
b
oundary CA
[1].
In this paper, we will only deal with reflective boundary conditions. In [3], the reversibility problem for LCA with reflecti
ve
boundary defined by a rule matrix in the form of a diagonal matrix has been studied over the binary field recently, this work
ha
s
been extended mainly to ternary fields and partial answers regarding the reversibility are addressed also therein. A doubly

infinite
sequence
=
= ( )
n n
x x
, where the entries are from
p
Z
where
p
is a prime number is an element of the space
.
p
Z
Z
A CA is a
map defined by
:
f p p
T
Z Z
Z Z
where
( ) = (,,),
f i i r i r
T x f x x
where
2 1
:
r
p p
f
Z Z
is called local rule. Martin et al. [7]
showed that if a local rule
f
is linear, i.e.,
=
(,,) = (mod ),
r
r r i i
i r
f x x a x p
(1.1)
where at least one of
,,
r r
a a
is nonzero mod
p
, then then
f
T
is 1D linear CA.
A 1D linear cellular automaton (LCA)
]
,
[
r
r
f
T
is defined by the local rule
f
given in Eq. (1.1):
[,] =
=
( ) = ( ),= (,,) = (mod ),
r
f r r n n n n r n r i n i
i r
T x y y f x x a x p
(1.2)
where
,,
r r p
a a
Z
(see [7] for details). In this paper, we will only consider the 1D finite LCA defined by local rule (1.2)
under modulo

p
algebra
and we
further
assume that radius
r
and
2
p
is a prime number. CAs can also be defined by special
local rules; next state cell is determined by all neighboring cells within radius
r
such that the first and last cell are assumed to be
next
to
each other
in a cyclic manner.
Let us consider
the
vector
1 2 3 1
=[,,,,,]
t t t t t t n
n n p
C x x x x x
Z
The vector
t
C
is called a configuration of the 1D finite LCA at time
t
,
therefore
0
C
is the initial configuration. The configuration length of a vector will be assumed to be
n
in this paper. Hence,
it is
clear that
7.
n
Reflective boundary conditions are obtained by reflecting the lattice at the boundary. In this case, in order to
obtain neighboring cell values for them the cells states at the extremities are repeated. In one dimension, reflective bounda
ries and
their ite
rations under the 1D finite CA
n
A
can be arranged by
3 2 1 1 2 1 1 2
][ ][
t t t t t t t t t t
n n n n n
x x x x x x x x x x
( )
A
1 1 1 1 1 1 1 1 1 1
3 2 1 1 2 1 1 2
][ ][.
r
n
t t t t t t t t t t
n n n n n
x x x x x x x x x x
(1.3)
Let us define the 1D finite CA
A
n
( 2 1)
n r
with RBC. Therefore, we have to define the local rule with RBC.
Proposition 1.1
The local rule under reflective boundary condition (1.3) is given by the following expression:
0 1 1 1 2 2
2 3 3 3 4
1 2 1 0 3 2
1 3 2 4 3 5
6
3 2 1 3
=2
6
1
3 1
=0
4
3 4 2 3
=0
( ) ( )
=1;
( ) mod,
( ) ( )
= 2;
mod,
( ) mod,= 3;
mod,4 3;
( ) mod
t t
t t
t t
t t t
t t
k k
k
t
t
i
k k i
k
t t
k n k n
k
a a x a a x
i
a a x a x p
a a x a a x
i
a x a x a x p
a a x a x p i
x
a x p i n
a x a a x
3 4 2 3 1 2
0 3 1 1 2
3 3 2 3 2
1 2 1 0 1
,= 2;
= 1;
( ) ( ) mod,
( )
=
( ) ( ) mod,
t t t
n n n
t t
n n
t t
n n
t t
n n
p i n
a x a x a x
i n
a a x a a x p
a x a a x
i n
a a x a a x p
(1.4)
where
\{0}
i p
a
Z
and
t
i
x
is a symbol of the state of the
th
i
cell at time
t
. A common approach for defining a CA which
merely depends on its local rule is to introduce a numbering label. By examining the relation among the states we see that th
e next
cell
1
t
i
x
state is obtained by the relation of the previous an
d the
neigboring
state of the cells within the radius
r
. So accordingly
we introduce the following ordering and hence the numbering of the rules:
p
r
r
k
r
k
r
r
k
t
j
i
j
r
r
j
t
i
a
a
a
a
a
p
a
RN
x
a
x
1
0
1
=
=
1
=
=
=
(1.5)
For instance, if
=2,
r
= 3,
p
and
2 1 0 1 2
(,,,,) = (2,2,1,1,1)
a a a a a
then this rule will be named as rule number (RN)
4 3 2 1 0
3
3 3 3 2 3 2 3 = 11122 =125
or if
=1,
r
= 5,
p
and
1 0 1
(,,) = (2,1,1)
a a a
then this rule will be named as rule
number (RN)
2 1 0
5
5 5 2 5 = 112 = 32.
This presentation not only identifies the rules uniquely but it is also practical.
Further, since each cell is assumed to effect the transitions we always assume that
\0
i
a
p
Z
.
Theorem 1.2
The rule matrix
(3)
n
M
corresponding to an 1D finite CA
(3)
n
A
generated by the local rule (1.4) with RBC is given
by
0 1 1 2 2 3 3
1 2 0 3 1 2 3
2 3 1 0 1 2 3
3 2 1 0 1 2 3
3 2 1 0 1 2 3
3 2 1 0 1 2 3
3 2 1 0 1 2 3
3 2 1 0 1 2 3
3 2 1 0
0 0 0 0 0
0 0 0 0
0 0 0
0 0
0 0
0 0
0 0
0 0 0
0 0 0 0
a a a a a a a
a a a a a a a
a a a a a a a
a a a a a a a
a a a a a a a
a a a a a a a
a a a a a a a
a a a a a a a
a a a a
3 1 2
3 2 3 1 2 0 1
0 0 0 0 0
a a a
a a a a a a a
F
E
D
C
B
A
=
(1.6)
where
=0
E
(i.e. zero matrix) and
C
is a non

singular (full rank) upper triangular matrix (i.e.
6
=
)
(
n
C
rank
). This is a
heptagonal band matrix.
2.
Reversibility of the 1D finite
CA with RBC and radius
three
In this section, we study the inverse of the rule matrix
(3)
n
M
corresponding to the 1D finite LCA
n
A
. To obtain the inverse of
the matrix
(1.6)
, we will co
mpute the rank of the matrix (1.6
)
.
Our strategy to find the rank of
n
M
is as follows
: First we
c
onsider
n
M
as
(3)
0
= 0
n
A B A B B
M C D E F F
E F C D C D
(2.1)
where
B
and
F
are matrices obtained after from a suitable row

column operations. It can be seen from 2.1 that the rank of
(3)
n
M
is
( ) = ( ) = ( 6).
n
B B
rank M rank C rank n rank
F F
So the problem of finding the rank reduces dramatically.
No
w let us compute this relation explicitly.
A Computational Method For The Rank Of
(3)
n
M
.
Let
i
T
denote the
i
th row and
[ ]
i
T j
denote the
j
the entry of the
i
th row of matrix
T
respectively. By definition, we have
1 0 1 1 2 2 3 3 1
=[,,,,0,0,0,0,0] ( )
n p
T a a a a a a a M
Z
2 1 2 0 3 1 2 3 1
=[,,,,,0,...,0] ( )
n p
T a a a a a a a M
Z
3 2 3 1 0 1 2 3 1
=[,,,,,,0,0,...,0] ( )
n p
T a a a a a a a M
Z
4 3 2 1 0 1 2 3 1
=[,,,,,,,0,0,0,...,0] ( )
n p
T a a a a a a a M
Z
……
2 3 2 1 0 1 2 3 1
=[0,0,0,...,0,0,0,,,,,,] ( )
n n p
T a a a a a a a M
Z
1 3 2 1 0 3 1 2 1
=[0,0,...,0,0,,,,,] ( )
n n p
T a a a a a a a M
Z
3 2 3 1 2 0 1 1
=[0,0,...,0,0,,,,] ( )
n n p
T a a a a a a a M
Z
Define
the map σ a
s follows:
1 1
:( ) ( ),
n p n p
M M
Z Z
1 2 3 1 1 2 3 1
([,,,...,,]) =[0,,,,...,]
n n n
x x x x x x x x x
Also, if
1 2 3 1
=[,,,...,,],
n n
X x x x x x
then
[ ] =
i
X i x
represents the
i
th entry of
.
X
Further
1 2 1 2
[,,...,] =[,,...,].
n n
y x x x yx yx yx
The
following theorem presents an algorithm for computing the rank of
(3)
n
M
.
Theorem 2.1
Let the matrix
(3)
n
M
with
7
n
represent the one dimensional cellular automata over
p
Z
fields. Let
(1) ( 1) ( ) ( 1) ( )
1 1 1 1 4 1
3
1
=,= [ ] ( ) for 1 6,
k k k k
T T T T k T T k n
a
(1) ( 1) ( ) ( 1) ( )
2 2 2 2 4 2
3
1
=,= [ ] ( ) for 1 6,
k k k k
T T T T k T T k n
a
(1) ( 1) ( ) ( 1) ( )
3 3 3 3 4 3
3
1
=,= [ ] ( ) for 1 6,
k k k k
T T T T k T T k n
a
(1) ( 1) ( ) ( 1) ( )
2 2 2 2 4 2
3
1
=,= [ ] ( ) for 1 6,
k k k k
n n n n n
T T T T k T T k n
a
(1) ( 1) ( ) ( 1) ( )
1 1 1 1 4 1
3
1
=,= [ ] ( ) for 1 6,
k k k k
n n n n n
T T T T k T T k n
a
(1) ( 1) ( ) ( 1) ( )
4
3
1
=,= [ ] ( ) for 1 6.
k k k k
n n n n n
T T T T k T T k n
a
Define the following
6 ( 6)
n
block matrix consisting of blocks of square matrices of order
.
n
( 6) ( 6) ( 6) ( 1)
1 1 1 1
( 6) ( 6) ( 6) ( 1)
2 2 2 2
( 6) ( 6) ( 6) ( 1)
3 3 3 3
( 6) ( 6) ( 6) ( 6)
2 2 2 2
[ 6] [ 5] [ 1] [ ]
[ 6] [ 5] [ 1] [ ]
[ 6] [ 5] [ 1] [ ]
=
[ 6] [ 5] [ 1] [ ]
n n n n
n n n n
n n n n
n n n n
n n n n
n
T n T n T n T n
T n T n T n T n
T n T n T n T n
B
T n T n T n T n
T
( 6) ( 6) ( 6) ( 6)
1 1 1 1
( 6) ( 6) ( 6) ( 6)
1
[ 6] [ 5] [ 1] [ ]
[ 6] [ 5] [ 1] [ ]
n n n n
n n n
n n n n
n n n
n T n T n T n
T n T n T n T n
Then,
(3)
( ) = ( 6) ( ).
n
rank M n rank B
A straightforward corollary which gives a lower bound for the rank of a cellular automaton
is presented below.
Corollary
Let
(3)
n
M
be
the rule matrix corresponding to
a cellular automaton defined above. Then,
(3)
( 6) ( ).
n
n rank M n
An Algorithm for Computing the Rank of
(3)
n
M
:
Now we can summarize the method introduced above for computing the rank of
(3)
n
M
as follows:
Step 1:
Determine the first three rows
1 2 3
,,
T T T
and the last three rows
2 1
,,
n n n
T T T
of
n
M
which consists of block of matrices.
Set
(1) (1) (1)
1 1 2 2 3 3
=,=,=
T T T T T T
and
(1) (1) (1)
2 2 1 1
=,=,=.
n n n n n n
T T T T T T
Step 2:
For
1,
1
n
k
compute
( 1) ( ) ( 1) ( )
1 1 4 1
3
1
= [ ] ( ),
k k k k
T T k T T
a
( 1) ( ) ( 1) ( )
2 2 4 2
3
1
= [ ] ( )
k k k k
T T k T T
a
and the other
( )
k
i
T
for
.
Hence, determin
ing
the matrix
B
. For every iterations, compute
( )
k
i
T
(
=1,2,3,2,1,),
i n n n
since
B
is a matrix, arithmetics
can be carried out modulo
p
, (prime number p) the characteristic polynomial of
B
which saves reasonable time.
Step 3:
Compute the rank of
B
.
Therefore,
).
(
6)
(
=
)
(
(3)
B
rank
n
M
rank
n
3.
Conclusion
In this paper we studied
the reversibility problem of the family of 1D finite CAs
with ref
lectiv
e boundary
. The reversibility
problem becomes easy to solve computing the rank of the matrix.
If the rule matrix corresponding to the 1D finite linear CA has
fu
ll rank, then the CA is reversible, otherwise the CA is irreversible.
Some other interesting results and
further connections on this
direction wait to be explored.
Acknowledgements
This work is supported by "The Scientific and Technological Research
Council of Turkey" (TÜB
İ
TAK) (Project Number:
110T713).
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