AUTOMATA 2015
Turku - Finland
Merging Cellular Automata Rules to Optimise a
Solution to the Modulo-n Problem
Claudio Martins
Pedro de Oliveira
Universidade Presbiteriana Mackenzie
São Paulo, SP
Brazil
1. The Modulo-n Problem in Cellular Automata
Definitions
• Definition 1: The MODn Problem
It consists of determining whether the number of 1-bits in a
cyclic binary string is multiple of n or not.
• Example 1: Solution for the MOD2 Problem
Initial Configuration
MOD2
Final Configuration
>>>
>>>
• Example 2: Solution for the MOD3 Problem
Initial Configuration
MOD3
Final Configuration
>>>
>>>
>>>
3
1. The Modulo-n Problem in Cellular Automata
Examples of ill-defined problem
• MOD2 (ill-defined Problem)
Initial Configuration
MOD2
Final Configuration
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
>>>
• MOD3 (ill-defined Problem)
Initial Configuration
MOD3
Final Configuration
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
>>>
It is always ill-defined for lattice sizes multiple of n !
4
1. The Modulo-n Problem in Cellular Automata
Definitions
• Definition 2: Rule Composition (temporal sequences of Rules)
Rjaj … R3a3 R2a2 R1a1
• Examples
Solution for the MOD2 Problem [Lee, Xu and Chau, 2001]
’ = S() =
N

 R132  2 

R
N
2
222  



N
2
 

Solution for the MOD3 Problem [Lee, Xu and Chau, 2003]
’ = S() =
N
N

N  
N  
N
N
3
3

N 
   R

R
R
R
E 254   4 2   3 1    
 
 

N

5
2. The Partition of a Binary String
Definitions
• Definition 3: The Partition number
The number of groups of 0s that is equal to the number of
groups of 1s (because of the periodic boundary condition).
Configuration
Partition
3
0
0
1
7
3. Active State Transition
Definitions
• Definition 4: Active State Transition
By active transition, we mean those that change the state value
of the cell at the centre of the neighbourhood.
Rule 238
Transition
7
6
5
4
3
2
1
0
Neighbourhood
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
Output
bit
1
1
1
0
1
1
1
0
23810 = 111011102
8
4. The Replacement Rules (R)
Definition
• A rule that can replace n end bits, of a sequence of n or more
consecutive identical bits, with n opposite bits.
Left end bits
Right end bits
1
2
3
... n -2 n -1 n
↓
↓
↓
↓
↓
↓
....
n n -1 n -2 ...
3
2
1
↓
....
↓
↓
↓
↓
↓
↓
↓
....
↓
↓
↓
↓
↓
↓
↓
↓
....
....
....
↓
↓
↓
↓
↓
↓
↓
↓
↓
↓
↓
↓
↓
↓
↓
↓
↓
↓
....
↓
....
↓
....
9
4. The Replacement Rules (R)
Representation
• Rn0 to replace n 0s with n 1s, and R1n to the opposite.
R30,0
1 2 3
.... ....
3 2 1
↓ ↓ ↓
.... ....
R00,3
1 2 3
R10,2
1 2 3
.... ....
3 2 1
↓ ↓ ↓
.... ....
R30
R20,1
1 2 3
.... ....
3 2 1
↓ ↓
↓
.... ....
.... ....
3 2 1
↓
↓ ↓
.... ....
• Rn0,0 or R00,n require radius n.
• Rn01,1 or Rn02,2 … or … R20,n2 or R10,n1 require radius n-1.
• MODn-conserving rules.
10
4. The Replacement Rules (R)
Observations
R20,1
Transition
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
↓
1 0 0 0 .... 0 0 0 1
Rule 4.059.296.252
Neighbourhood
1 1 1 1 1
1 1 1 1 0
1 1 1 0 1
1 1 1 0 0
1 1 0 1 1
1 1 0 1 0
1 1 0 0 1
1 1 0 0 0
1 0 1 1 1
1 0 1 1 0
1 0 1 0 1
1 0 1 0 0
1 0 0 1 1
1 0 0 1 0
1 0 0 0 1
1 0 0 0 0
0 1 1 1 1
0 1 1 1 0
0 1 1 0 1
0 1 1 0 0
0 1 0 1 1
0 1 0 1 0
0 1 0 0 1
0 1 0 0 0
0 0 1 1 1
0 0 1 1 0
0 0 1 0 1
0 0 1 0 0
0 0 0 1 1
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
Output bit / Note
1
1
1
1
0
0
0
1
1
1
1
1
0
0
1
1
1
1
1
1
0
0
0
1
1
1
1
1
1
1
0
0
Radius =1
=2
=2
↓
1 0 0 0 .... 0 0 0 1
1 st 0 from the left
=2
Radius
=3
Radius
=1
↓
1 0 0 0 .... 0 0 0 1
=1
=1
↓
1 0 0 0 .... 0 0 0 1
2 nd 0 from the left
2 nd 0 from the left
↓
1 0 0 0 .... 0 0 0 1
=2
↓
1 0 0 0 .... 0 0 0 1
=0
=0
Simplified Representation
R20,1
=3
Rule 4.059.296.252
1 st 0 from the left
1 st 0 from the right
1 st 0 from the right
Active
Transition
24
17
16
8
3
2
Neighbourhood
1
1
1
0
0
0
1
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
1
0
1
0
0
1
0
Output bit / Note
1
1
1
1
1
1
1 st 0 from the left
2 nd 0 from the left
2 nd 0 from the left
1 st 0 from the left
1 st 0 from the right
1 st 0 from the right
Active
Transitions
24 & 8
17 & 16
3&2
Neighbourhood
*
1
0
1
0
0
0
0
0
0
0
1
0
*
*
Output bit / Note
1
1
1
11
1 st 0 from the left
2 nd 0 from the left
1 st 0 from the right
4. The Replacement Rules (R)
Observations
• The Partition can be reduced.
Appying R20,1 on the Configuration
1
2
3
4
5
6
7
8
Partition
9 10 11 12
t0
t1
t2
t3
2
2
1
1
• There is no effect without n or more consecutive identical
bits, or on a string with only 0s or only 1s.
Appying R20,1 on the Configuration
1
t0
t1
t2
t3
2
3
4
5
6
7
8
Partition
9 10 11 12
3
3
3
3
12
5. The Grouping Rules (G)
Definition and Representation
• A rule that can shift, to the left or to the right, strings of m
bits (10m1 or 01m0), in order to group them with larger strings
of the bit at concern.

Gm0

1
Gm

G30

G40
1 2 3 .... m
↓
↓
1 2 3 .... m
1 2 3 .... m
↓
↓
1 2 3 .... m
1 0 0 0 1
↓
↓
0 0 0 1 1
1 0 0 0 0 1
↓
↓
0 0 0 0 1 1

Gm0

1
Gm

G30

G40
1 2 3 .... m
↓
↓
1 2 3 .... m
1 2 3 .... m
↓
↓
1 2 3 .... m
1 0 0 0 1
↓
↓
1 1 0 0 0
1 0 0 0 0 1
↓
↓
1 1 0 0 0 0
13
5. The Grouping Rules (G)
Observations
0 or G1 require radius m+1.
• Gm
m

G30
↓
1 0 0 0 1
=4
Radius = 0

G40
↓
1 0 0 0 1
=3
↓
↓
1 0 0 0 0 1
Radius = 0
=1
=5
1 0 0 0 0 1
=4
=1
• There is no effect without m consecutive identical bits,
or on a string with only 0s or only 1s.
• MODn-conserving rules.
14
5. The Grouping Rules (G)
Observations
• The Partition can be reduced.

Appying G10 on the Configuration
1
2
3
4
5
6
7
8
Partition
9 10 11 12
t0
t1
t2
t3
3
3
2
2
• They can cause only Shifts on the lattice (only identical blocks).

Appying G10 on the Configuration
1
t0
t1
t2
t3
2
3
4
5
6
7
8
Partition
9 10 11 12
3
3
3
3
15
5. The Grouping Rules (G)
Observations

G10
Transition
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Rule 3.704.675.536
Neighbourhood
1 1 1 1 1
1 1 1 1 0
1 1 1 0 1
1 1 1 0 0
1 1 0 1 1
1 1 0 1 0
1 1 0 0 1
1 1 0 0 0
1 0 1 1 1
1 0 1 1 0
1 0 1 0 1
1 0 1 0 0
1 0 0 1 1
1 0 0 1 0
1 0 0 0 1
1 0 0 0 0
0 1 1 1 1
0 1 1 1 0
0 1 1 0 1
0 1 1 0 0
0 1 0 1 1
0 1 0 1 0
0 1 0 0 1
0 1 0 0 0
0 0 1 1 1
0 0 1 1 0
0 0 1 0 1
0 0 1 0 0
0 0 0 1 1
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
Output bit / Note
1
1
0 1 on the left
1
1 Isolated 0
1 Isolated 0
0
0
1
1
0 1 on the left
1
0
0
0
0
1
1
0 1 on the left
1
1 Isolated 0
1 Isolated 0
0
0
1
1
0 1 on the left
1
0
0
0
0

G10
↓
↓
1 0 1
Radius = 0
1 0 1
=2
=1 =1

G10
Simplified Representation
Active
Transition
29
27
26
21
13
11
10
5
Neighbourhood
1
1
1
1
0
0
0
0
1
1
1
0
1
1
1
0
1
0
0
1
1
0
0
1
0
1
1
0
0
1
1
0
1
1
0
1
1
1
0
1
Output bit / Note
0
1
1
0
0
1
1
0
1 on the left
Isolated 0
Isolated 0
1 on the left
1 on the left
Isolated 0
Isolated 0
1 on the left
Active
Transitions
27, 26, 11 & 10
29, 21, 13 & 5
Rule 3.704.675.536
Neighbourhood
*
*
1
*
0
1
1
0
Output bit / Note
*
1
1
0
16
Isolated 0
1 on the left
6. Composing Replacement and Grouping Rules
The Correct Sequence of the Rules
• The correct temporal composition of the Replacement and
the Grouping rules is:
’ = S() =

 G1
 n2

N
2
 
N
2
 
...G11
R1n
N
n
 
Gn02
N
2
 
...G10
N
2
 
Rn0
N
n
 




N
n
 

or
N
N
N
N
N
N 











 
’ = S() =  Gn02  2  ...G10  2  Rn0  n  Gn12  2  ...G11 2  R1n  n  


N
n
 

0 and n-2 rules G1 are necessary.
• n-2 rules Gm
m
18
6. Composing Replacement and Grouping Rules
Necklace Configurations
• Necklaces are configurations of the form (0A1B)C :
 for positive integers A, B and C,
 where A < n and B < n,
 or A < n and B > n, but B is not multiple of n,
 or B < n and A > n, but A is not multiple of n,
 or B > n and A > n, but A and B are not multiple of n.
• For Necklaces
 Grouping rules just cause Shifts on the lattice (only identical
blocks).
 Replacement rules cause no effect when A < n and B < n,
or may lead to periodic regimes (resulting Shifts) only when
A > n or B > n, by continuously transforming (0A1B )C into
(0A’1B’ )C, back and forth.
21
6. Composing Replacement and Grouping Rules
Necklace Configurations
• Example 2: A Replacement
rule for the MOD4 Problem
• Example 1: A Replacement
rule for the MOD3 Problem
(0712)2  (0316)2
(0711)2(0414)2(0117)2
t0
t1
t2
t3
t4
t5
t6
t7
t8
t9
t 10
t 11
t 12
Necklace
Initial
Configuration (0711)2
R20,1
R12,1
R20,1
R12,1
(0414)2
(0117)2
(0117)2
(0414)2
(0711)2
(0711)2
(0414)2
(0117)2
(0117)2
(0414)2
(0711)2
(0711)2
t0
Necklace
Initial
Configuration (0712)2
t1
R30,1
t2
t3
t4
R31,1
t5
t6
t7
R30,1
t8
t9
t 10
R31,1
t 11
t 12
22
(0316)2
(0316)2
(0316)2
(0712)2
(0712)2
(0712)2
(0316)2
(0316)2
(0316)2
(0712)2
(0712)2
(0712)2
6. Composing Replacement and Grouping Rules
What these compositions may cause in the lattice?
• Let ’ =

 G1 N ...G1N R
n
1
 n2

N 
1  n 
N
N
    n 
0  n  
Gn0 2 N ...G10 N Rn


,
• Then ’ will be one of the following:
 0N (Partition = 0);
 1N (Partition = 0);
 0N-MODn(||1)1MODn(||1), (Partition = 1);
||1 is the quantity of 1-bits inside string 
N-MODn(||1) ≠ multiple of n
MODn(||1) ≠ 0
 Necklaces (0A1B )C, (Partition > 1).
C > 1, B < n, A ≠ multiple of n, and (A+B)C = N.
23
6. Composing Replacement and Grouping Rules
Final Configurations for Lattice Size Not Multiple of N or Not Multiple of Any Factor of n
 If MODn(N) = MODn(||1):
MODn(||1) ≠ 0:
 ’ = 1N
 If MODn(N) ≠ MODn(||1):
MODn(||1) = 0:
 ’ = 0N
MODn(||1) ≠ 0:
 ’ = 0N-MODn(||1)1MODn(||1) or
 ’ =(0A1B)C (necklaces)
• For all initial configurations:
 When MODn(||1) = 0, the problem is solved, as defined.
 However, in order for all the other initial configurations to
lead to 1N, elementary rule 254 can be used.
26
7. The Modulo-n Problem Solution
The Generic Solution
• The generalisation of the solution for MODn finally leads to:
N
 2 
 
Sn = E254
1
G
n
2


N
2
  ...
N 
1 2 
N
1  n 
G1 Rn Gn0 2
N 
2
  ...
N
0  2 
G1
N
 N    n 
0  n  
Rn



• 2 Replacement rules, Rn0 and R1n (radius n-1)
0 and G1
• n-2 Grouping rules, Gm
m
0
1
• G1 and G1 shift isolated bits (radius 2)
• G20 and G21 shift isolated pair of bits (radius 3)
• and so on, up to
• Gn02 and Gn12 shift isolated strings of n-2 bits (radius n-1)
• All these rules may have radius n-1.
27
8. The Merging Operation
Joining and Separating the Active State Transitions
• What happens when we separate the active transitions of a
rule or simply join active transitions of some rules in a single
one?
Rule 238
Transition
Neighbourhood
7
6
5
4
3
2
1
0
N/2
R252
1
1
1
1
0
0
0
0
N/2
R238
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
N/2
R238
Rule 252
Output
bit
1
1
1
0
1
1
1
0
N/2
R252
Neighbourhood
+
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
N/2
(R252 R238)
Rule 254
Output
bit
1
1
1
1
1
1
0
0
Neighbourhood
=
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
N/2
(R238 R252)
1
0
1
0
1
0
1
0
Output
bit
1
1
1
1
1
1
1
0
?
N
R254
t0
t1
t2
t3
t4
t5
t6
t7
t8
t9
t 10
t 11
t 12
28
8. The Merging Operation
Decomposing the Active State Transitions of the Elementary Inverting Rule
• Of course it will not always work!
Rule 51
Transition
7
6
5
4
3
2
1
0
Neighbourhood
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
Rule 76
Output
bit
0
0
1
1
0
0
1
1
Neighbourhood
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
Rule 140
Output
bit
0
1
0
0
1
1
0
0
Neighbourhood
1
1
1
1
0
0
0
0
Rule 196
Transition
7
6
5
4
3
2
1
0
Neighbourhood
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
Rule 236
Output
bit
1
0
0
0
1
1
0
0
Neighbourhood
1
1
1
1
0
0
0
0
Rule 200
Output
bit
1
1
0
0
0
1
0
0
Neighbourhood
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
Rule 220
Output
bit
1
1
1
0
1
1
0
0
Neighbourhood
1
1
1
1
0
0
0
0
Rule 206
Output
bit
1
1
0
0
1
0
0
0
Neighbourhood
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
Output
bit
1
1
0
1
1
1
0
0
Rule 205
Output
bit
1
1
0
0
1
1
1
0
Neighbourhood
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
29
Output
bit
1
1
0
0
1
1
0
1
8. The Merging Operation
Decomposing the Active State Transitions of the Elementary Inverting Rule
• Even the ordering of these rules can change the final result!
E 5116
(E205 E206 E200 E196 E220 E236 E140 E76) 16
(E76 E206 E200 E196 E220 E236 E140 E205)16
t0
t1
t2
t3
t4
t5
t6
t7
t8
t9
t 10
t 11
t 12
t 13
t 14
t 15
t 16
30
9. Merging Replacement with Grouping Rules
The MOD3 Case
• How about merging Replacement and Grouping rules?
R20,1
Transition
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0

G10
Rule 4.059.296.252
Neighbourhood
1 1 1 1 1
1 1 1 1 0
1 1 1 0 1
1 1 1 0 0
1 1 0 1 1
1 1 0 1 0
1 1 0 0 1
1 1 0 0 0
1 0 1 1 1
1 0 1 1 0
1 0 1 0 1
1 0 1 0 0
1 0 0 1 1
1 0 0 1 0
1 0 0 0 1
1 0 0 0 0
0 1 1 1 1
0 1 1 1 0
0 1 1 0 1
0 1 1 0 0
0 1 0 1 1
0 1 0 1 0
0 1 0 0 1
0 1 0 0 0
0 0 1 1 1
0 0 1 1 0
0 0 1 0 1
0 0 1 0 0
0 0 0 1 1
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
Output bit / Note
1
1
1
1
0
0
0
1
1
1
1
1
0
0
1
1
1
1
1
1
0
0
0
1
1
1
1
1
1
1
0
0
1 st 0 from the left
2 nd 0 from the left
2 nd 0 from the left
1 st 0 from the left
1 st 0 from the right
1 st 0 from the right
Transition
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
R20,1
Rule 3.704.675.536
Neighbourhood
1 1 1 1 1
1 1 1 1 0
1 1 1 0 1
1 1 1 0 0
1 1 0 1 1
1 1 0 1 0
1 1 0 0 1
1 1 0 0 0
1 0 1 1 1
1 0 1 1 0
1 0 1 0 1
1 0 1 0 0
1 0 0 1 1
1 0 0 1 0
1 0 0 0 1
1 0 0 0 0
0 1 1 1 1
0 1 1 1 0
0 1 1 0 1
0 1 1 0 0
0 1 0 1 1
0 1 0 1 0
0 1 0 0 1
0 1 0 0 0
0 0 1 1 1
0 0 1 1 0
0 0 1 0 1
0 0 1 0 0
0 0 0 1 1
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
Output bit / Note
1
1
0 1 on the left
1
1 Isolated 0
1 Isolated 0
0
0
1
1
0 1 on the left
1
0
0
0
0
1
1
0 1 on the left
1
1 Isolated 0
1 Isolated 0
0
0
1
1
0 1 on the left
1
0
0
0
0
Active
Transition
24
17
16
8
3
2
1
1
1
0
0
0
Active
Transitions
24 & 8
17 & 16
3&2
*
1
0
Rule 4.059.296.252
Neighbourhood
1
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
1
Active
Transitions
27, 26, 11 & 10
29, 21, 13 & 5
1
1
1
1
1
1
Neighbourhood
1
0
0
0
0
0
0
0
1

G10
Active
Transition
29
27
26
21
13
11
10
5
Output bit / Note
0
1
0
0
1
0
Output bit / Note
0
*
*
1
1
1
1
1
1
0
1
1
1
0
1
0
0
1
1
0
0
1
0
1
1
0
0
1
1
0
1
1
0
1
1
1
0
1
Neighbourhood
*
*
1
*
1 st 0 from the left
2 nd 0 from the left
1 st 0 from the right
Rule 3.704.675.536
Neighbourhood
1
1
1
1
0
0
0
0
1 st 0 from the left
2 nd 0 from the left
2 nd 0 from the left
1 st 0 from the left
1 st 0 from the right
1 st 0 from the right
0
1
1
0
*
1
Output bit / Note
0
1
1
0
0
1
1
0
1 on the left
Isolated 0
Isolated 0
1 on the left
1 on the left
Isolated 0
Isolated 0
1 on the left
Output bit / Note
1
0
Isolated 0
1 on the left
31
9. Merging Replacement with Grouping Rules
The MOD3 Case

and the Grouping rule G 0 have
1
R20,1
• Replacement rule
different active state transitions.


0
0
• Merging these two rules, R and G1 , results the rule M 0 .
2,1
2,1

M 20,1
Rule 3.721.649.628
Active
Transitions
27, 26, 11 & 10
29, 21, 13 & 5
24 & 8
*
*
*
1
*
0
1
1
0
*
1
1
0
Is ol a ted 0
1 on the l eft
1
0
0
1
1st 0 from the l eft
17 & 16
1
0
0
0
0
*
1
2nd 0 from the l eft
3&2
0
0
0
1
*
1
1 0 from the ri ght
Neighbourhood
Output bit / Note
st
• Can these 3 alternatives perform the same tasks?
N
  N
 0
0
 G1 R2,1







N
,
N  N

 0
0
 R2,1 G1







N
, or
N
 
 0
 M 2,1







N

32
9. Merging Replacement with Grouping Rules
The MOD3 Case
13
  13
 0
0
G
R
2,1
 1

t0
t1
t2
t3
t4
t5
t6
t7
t8
t9
t 10
t 11
t 12
t 13
t 14
t 15
t 16
t 17
t 18
t 19
t 20
t 21
t 22
t 23
t 24
t 25
t 26
t 27
t 28
t 29
t 30
t 31
t 32
t 33
t 34
t 35
t 36
t 37
t 38
t 39
t 40
t 41
t 42
t 43
t 44
t 45
t 46
t 47
t 48
t 49
t 50
t 51
t 52




2
13  13

 0
0
R
 2,1 G1





2
  13

0
 M 2,1





2
33
9. Merging Replacement with Grouping Rules
The MOD3 Case
• From a superficial analysis, we would say no!
• But, going back to the initial goals, these three alternatives
transform the lattice into a final configuration as desired.
• All final configurations have
 at most two consecutive 0s, and
 there are no isolated 0s and isolated pairs of 0s occurring
simultaneously.
34
9. Merging Replacement with Grouping Rules
The MOD3 Case
• Applying :
 rule M 30 instead of R30 and G10 ,
 and rule M 31 instead of G11 and R31 ,
 the solution S3 is simplified to Ss3 :
S3 = E254
N
2
 

 G1
 1

Ss3 = E254
N
2
 
N
2
 
R31
N
3
 
G10
N
2
 
R30
N
 1N
0
M
 3 M3





N
3
 
N
3
 




N
3
 


35
9. Merging Replacement with Grouping Rules
The MOD3 Case
• Comparing S3’ and Ss3’
S3’ =

 G1
 1

N
2
 
R31
N
3
 
G10
N
2
 
R30
N
3
 




N
3
 

Ss3’ =
N
 1N
0
M
 3 M3





N
3
 

• Summary of all final configurations left, after applying these
two solutions to all possible initial configurations with N=16.
Original Sequence S3'
Amount
of 1s
Final Configuration
with some Necklaces
0
0000000000000000
0000000000000011
0000000100000001
2
Simplified Sequence Ss3'
No. Of
Configs.
21845
19232
2608
Amount
of 1s
Final Configuration
with some Necklaces
0
0000000000000000
0000000000000011
0000000100000001
2
No. Of
Configs.
21845
14576
656
Amount
of 1s
2
5
8
16
0011001100110011
0101010101010101
1111111111111111
TOTAL
4
2
21845
65536
8
16
0011001100110011
0101010101010101
1111111111111111
TOTAL
Final Configuration with
some Partial Necklaces
0000000000000101
0000000000100001
0000000101010101
0000100001010101
0000101000010101
4
2
21845
65536
36
No. Of
Configs.
4768
1568
144
64
64
9. Merging Replacement with Grouping Rules
The MOD3 Case – Partial Necklaces
• Partial Necklaces in the MOD3 Problem, are configurations of
the form ((03)+(01)+ )+ or ((13)+(10)+ )+.
• The lattice “gets locked” on these configurations when the
partial simplified solution is applied.
0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 1
= (03)2
= (01)5
Necklace
0 0 0 0 1 0 0 0 0 1 0 1 0 1 0 1
= 03
= 01
= 03
Necklace
= 01
= (01)3
Necklace
37
9. Merging Replacement with Grouping Rules
The MOD3 Case – Partial Necklaces
S'3 in 0000000101010101
S'3 in 0000100001010101
Ss'3 in 0000000101010101
06(01)5 <<>> 16(10)5
Ss'3 in 0000100001010101
(0301)2(01)3 <<>> (1310)2(10)3
38
9. Merging Replacement with Grouping Rules
The MOD4 Case
Merging Rules
Results
?
R40 + G10 + G20
=
M 40
R14 + G11 + G21
=
M 14
• The first problem is related to the Grouping rules (G 0 and G 0 )
1
2
because of possible shifts of isolated 0s and isolated pairs of
0s simultaneously.
• Next figure shows the joining of these rules and also the
temporal evolution of an initial configuration where the
problem occurs.
39
9. Merging Replacement with Grouping Rules
The MOD4 Case and the Problem of Shifts
R20,2 +

G10

G20
+
= Rule 321600197915107966665688126986998903292
Active State Transitions
R20,2

G10

G20
112
97
7
67
123
119
121
103
80
96
6
66
122
118
105
102
48
33
5
3
107
117
89
101
16
32
4
2
106
116
73
100
91
87
57
39
90
86
41
38
75
85
25
37
74
84
9
36




t0
t1
t2
t3
t4
t5
t6
t7
t8
t9
t 10
t 11
t 12
t 13
t 14
t 15
t 16
t 17
59
55
58
54
43
53
42
52
 7 7
3
G20 G10 R20,2 


27
23
26
22
11
21
10
20
*
*
0
*
*
*
*
*
*
1
0
0
*
*
*
1
Neighbourhood
1
0
0
0
0
0
0
0
1
0
0
0
*
1
0
1
0
1
*
1
0
0
0
1
0
0
*
1
1
*
0
*
0
*
*
*
*
*
1
*
  7
 R 0  G 0  G 0 
2, 2
1
2


40
Output bit
1
1
1
1
0
1
0
1
9. Merging Replacement with Grouping Rules
Removing and Inserting Active Transitions to Change the Displacement Step of the Rule

• A solution is to change the displacement step of rule G10 .
• Instead moving just 1 position, move 2 positions, whenever
possible.

• The new rule with this feature: G 0,2 .
1

G10
119 118 117 116 87
123 122 107 106 91
86
90
= Rule 324253482922812750238312970506082513664
85
75

G10,2
119 118 117 116 87
107 106
125 109
93
TA's
84 55
74 59
54
58
53
43
52
42
23
27
22
26
21
11
20
10
*
*
Neighbourhood
*
1
0
1
*
*
*
1
0
1
*
*
Output bit
1
0
*
*
1
Output bit
1
0
0
= Rule 297668273963817264613187722719825810176
Active State Transitions
86 85 84 55 54 53
75 74
43
77
61 45
52
42
23
22
29
13
21
11
20
10
*
*
*
Neighbourhood
*
1
0
1
*
*
0
1
0
1
*
*
1
1
0
41
9. Merging Replacement with Grouping Rules
Removing Active State Transitions to Eliminate Remaining Shifts
• Finally, the second problem: a simultaneous shift of a single
bit and its respective pair of bits will continue to occur when
there is a string as *100101*.
INSTEAD
BETTER
1 0 0 1 0 1
1 0 0 1 0 1
↓
↓
↓ ↓ ↓
0 0 1 0 1 1
↓
0 0 0 1 1 1
• Some active state transitions should be turned off.




• Rules G10,2 and G20 become G10*,2 and G20* .
42
9. Merging Replacement with Grouping Rules
Removing Active State Transitions to Eliminate Remaining Shifts
• For clarity purposes, notice that:
 String **0101* has been instantiated into strings
000101*, 010101*, 100101*, 110101*, and
 String *1001** into *100100, *100101, *100110, *100111.

G10,2
119 118 117 116 87
86
= Rule 297668273963817264613187722719825810176
Active State Transitions
85 84 55 54 53
52
43
42
75
23
22
29
13
74
107 106
125 109
93
121 105 89 73 57
103
39
102
101
100
77

G20
41
61
45
21
11
20
10
*
0
0
1
1
*
*
0
1
0
1
*
Neighbourhood
1
0
1
0
1
0
0
1
0
0
1
0
0
1
0
*
1
1
*
1
1
1
1
0
*
*
*
*
*
1
Output bit
1
0
0
1
0
0
0
1
1
0
0
1
1
0
1
0
Output bit
0
1
1
0
1
= Rule 336299833476273345402472786266733739264
Active State Transitions
25 9
38
37
36
*
*
*
*
*
*
1
1
1
1
Neighbourhood
*
1
0
0
0
1
0
0
1
0
0
1
0
0
1
43
9. Merging Replacement with Grouping Rules
Removing Active State Transitions to Eliminate Remaining Shifts



0
,
2
0
0
M 2,2  R2,2  G1*  G20*
= Rule 295014986420811337252501870505639399932
Active State Transitions
112 80
97 96
7
6
67 66
48
33
5
3
Neighbourhood
16
32
4
2
119 118 117 116 87
86
85
84
55
54
53
52
43
42
23
22
29
13
107 106
125 109
121 105 89
103
102
73
93
77
57
39
41
61
25
9
38
100
36
45
21
11
20
10
Output bit
*
*
0
*
*
1
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
*
1
0
*
*
*
1
1
1
1
*
0
0
1
*
*
0
1
1
*
1
0
0
0
*
0
1
1
1
1
1
0
0
0
1
*
1
1
1
0
*
*
*
*
1
1
0
0
0
0
*
*
*
*
*
1
1
1
*
0
0
0
1
0
0
0
0
1
1
1
0
1
1
0
1
1
0
0
0
1
1
1

• The
leads us to rule M 12,2 , composed by R1 ,
 same rationale

2,2
1
,
2
1
G1* , and G .
2*
• Wolfram number:
255816358659918533639648036274123862084
44
10. The Modulo-n Problem Simplified Solution
The Generic Simplified Solution
• The solution S4 is simplified to Ss4 :
Ss4 = E254
N
2
 
 1N
0N
M
M
 4
4





N
4
 

• Testing the same procedure for MOD5 and obtaining similar
results, we conclude that the simplified solution that can
solve the MODn problem is :
Ssn = E254
N
2
 
N
 1N
 M n M n0






N
n
 

• It has only 3 rules, one elementary, and two with radius n-1
• The size N of the binary string  cannot be multiple of n
•
N and n have to be numbers relatively prime.
45
11. Concluding Remarks and Next Steps
Concluding Remarks
• To solve the MOD5 problem and to merge satisfactorily the
rules G10*,3, G20*,2 and G30* , or rules G11*,3 , G21,*2 and G31* we have
to disable active transitions in the rule pairs involved in
simultaneous shifts of strings, as shown below :
• G  stands for either G 0 or G1
46
11. Concluding Remarks and Next Steps
Concluding Remarks
• For the suitability of M 50 and M 51 , we then have to disable 5
pairs of groups of active transitions. In general, this number
depends on n, as below:
Pairs of groups of active state transitions to be disabled
n =5
1 Step
1 Step
2 Steps
G 3* G 2*
G 2*
G 2* 1
G 1* 1
1
1
2 Steps
G 1*
1
3 Steps
n =6
1 Step
2 Steps
G 4* G 3* G 2* G 3* G 2*
G 3* 1
G 2* 1
G 1* 1
1
1
G 2*
G 1*
1
1
1
1
1
G 1*
3 Steps
G 2*
1
G 4*
G 3*
G 2*
G 1*
1
1
1
G 3*
G 2*
G 1*
1
G 2*
G 1*
Total
1 Step
2 Steps
3 Steps
G 5* G 4* G 3* G 2* G 4* G 3* G 2* G 3* G 2*
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Cn -2,2
2(Cn -3,2)
Cn -2,2
3
2
6
5
2(C n -3,2) 3(Cn -4,2)
6
15
3
4 Steps
G 2*
1
1
1
1
1
1
1
1
1
1
1
1
1
1
G 1*
4 Steps
Subtotal
n =7
1
C n -2,2
2(C n -3,2)
10
12
3(C n -4,2) 4(C n -5,2)
9
4
35
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11. Concluding Remarks and Next Steps
Concluding Remarks
• The solution to the MODn problem was generalised for any
value of n.
• The generalised solution was simplified by merging cellular
automata rules through the joining of the active transitions.
• In order to merge rules for performing equivalent operation,
tasks were modified and conflicts solved, following a standard
that allowed us to generalise the MODn simplified solution.
• Much study is still necessary for the understanding and the
generalisation of these mergings.
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Acknowledgements
We are grateful to:
 IPM – Instituto Presbiteriano Mackenzie
 MackPesquisa – Fundo Mackenzie de Pesquisa
Thank You Very Much!
Muito Obrigado!
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