The Discrete Baker Transformation

The Discrete Baker Transformation

Burton Voorhees Center for Science Athabasca University

ReferencesBulitko, V., Voorhees, B., & Bulitko, V. (2006) Discrete baker transformation for linear cellular automata analysis. Journal of Cellular Automata 1(1) 40 - 70.

Voorhees, B. (2007) Discrete baker transformation for binary valued cylindrical cellular automata. Lecture Notes in Computer Science (4173) 182 - 191.

Bulitko, V. & Voorhees, B. (2008) Index permutations and classes of additive cellular automata rules with isomorphic STD. Journal of Cellular Automata (to appear)

Bulitko, V. & Voorhees, B. (2008) Cycle structure of baker transformation in one dimension. Journal of Cellular Automata (submitted, in revision).

The Baker Transformation

The baker transformation (or Bernoulli shift) in dynamical systems theory is the transformation of the interval [0,1] by

x --> 2x mod(1)

Where mod(1) means that only digits to the right of the decimal are retained. E.g., .75 --> .5 --> 0; 1/3 <--> 2/3. All rationals with denominator a power of 2 iterate to 0, other rationals to cycles. 1/7 --> 2/7 --> 4/7 --> 1/7 is a period 3 cycle, and period 3 implies chaos.

Additive Cellular Automata in One

DimensionAn additive cellular automaton rule in one dimension, with periodic boundary conditions and alphabet {0,…,p-1}, is defined in terms of the left shift operator by:

Equivalently, by the string (a0,…,an-1). The set of pn such strings defines the rule space of additive CA, and also the state space on which these CA operate.

X ass

s0

n 1

0 as p 1

Discrete Baker Transformation

For a rule T = (a0,…,an-1) with alphabet {0,…,p-1} (p prime), the discrete baker transformation is defined by:

Bp T i a j

j:pji mod(n) if {k | pk i mod(n)}

0 otherwise

Discrete Baker TransformationExamples for p = 2

n=6: B2T = (a0+a3,0,a1+a4,0,a2+a5,0)

n=7: B2T = (a0,a4,a1,a5,a2,a6,a3)

n=9: B2T = (a0,a5,a1,a6,a2,a7,a3,a8,a4)

Examples for p = 3

n=6: B2T = (a0+a2+a4,0,0,a1+a3+a5,0,0)

n=7: B2T = (a0,a5,a3,a1,a6,a4,a2)

n=9: B2T = (a0+a3+a6,0,0,a1+a4+a7,0,0a2+a5+a8,0,0)

Discrete Baker Transformation

If T is an additive rule acting on a d-dimensional torus with

Then

T ai1id 0ai1id p 1, 0is ns 1

B p T i1id

a j1 jdjs :pjsis mod(ns ) if {ks | pks is mod(n)}

0 otherwise

Discrete Baker Transformation

T = (a0,…an-1) defines a right circulant matrix C(T) that gives an expression of the update rule on a state µ by µ --> C(T)µ. The baker transformation exponentially speeds up rule evolution:

For p = 2 this is just B2T = T2.

C(T ) pr

C Bpr T

Properties of Baker Transformation

1. If p and n are relatively prime then Bp is a permutation (analogous results hold for all dimensions).

2. Let k be the largest integer such that pk|n. For all r > k, Bp is a permutation on the indices of

Bpr 1 T

Properties of Baker Transformation

Example: p = 2, n = 6 = 2x3, 12 = 22x3

(a0,a1,a2,a3,a4,a5) -->

(a0+a3,0,a1+a4,0,a2+a5,0) <--> (a0+a3,0,a2+a5,0,a1+a4,0)

(a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11) -->

(a0+a6,0,a1+a7,0,a2+a8,0,a3+a9,0,a4+a10,0,a5+a11,0) -->

(a0+a3+a6+a9,0,0,0,a1+a4+a7+a10,0,0,0,a2+a5+a8+a1

1,0,0,0 ) <-->

(a0+a3+a6+a9,0,0,0,a2+a5+a8+a11,0,0,0,a1+a4+a7+a1

0,0,0,0 )

Properties of Baker Transformation

Let n = pkm with p and m relatively prime and c = ordmp. Then

In addition, all cycle periods in STD(T) divide pk(pc - 1) and the height of trees in STD(T) does not exceed pk.

Bpkc T Bp

k T

Baker Diagram

1. STD(X) and STD(Y) are isomorphic as graphs

2. Tree heights in STD(X) do not exceed 1

3. For all s, cs|(pL - 1), and L|c

L lcm ordc1 p,,ordcr p

The baker transformation is a mapping on the space of additive CA rules. The state transition diagram for this mapping is the Baker Diagram. Suppose that X and Y are two additive rules belonging to the same cycle in the Baker Diagram, with cycle length L and let (c1,…,cr) be the set of cycle lengths in STD(X). Then:

Baker DiagramLet T be an additive rule, h(T) the height of T in the baker diagram (i.e., its distance from an attractor) and c(T) the period of the attractor (cycle or fixed point). Then the length of all cycles in STD(T) divides ph(T)(pc(T) - 1).

Example: p = 2, n = 6, T = (1,1,0,0,0,0) = I + (rule 102)

Under the baker transformation

(1,1,0,0,0,0) --> (1,0,1,0,0,0) <--> (1,0,0,0,1,0)

h(T) = 1, c(T) = 2 so ph(T)(pc(T) - 1) = 6 and the cycle periods in STD(I + ) are 1, 2, 3, and 6.

Baker DiagramFor n = 5 the baker diagram consists of four fixed points, two period 2 cycles, and six period 4 cycles. with X = (a0a1a2a3a4) Fixed Points: (00000), (11111), (01111), and (10000) Period 2: {(01001), (00110)}, {(11001), (10110)} Period4: {(00001), (00010), (01000), (00100)},

{(01100), (00101), (00011), (01010)},

{(01110), (01101), (00111), (01011)},

{(10001), (10010), (11000), (10100)},

{(11100), (10101), (10011), (11010)},

{(11110), (11101), (10111), (11011)}

Baker DiagramFixed Points: 0, 1, I, and I + 1 mod(2)Period 2: }, {I+ I+} Period4: {},

{}, {}, {I+}, {I+},

{I+}

Baker DiagramFor n = 6 the baker diagram consists of four fixed points and two period 2 cycles with all states on cycles the root of height 1 trees containing 7 states. with X = (a0a1a2a3a4,a5) Fixed Points: (000000), (100000), (001010), (101010)

0, I, Period 2: {(001000), (000010)}, {(101000), (100010)}

{

Rules having isomorphic STDs are at the same level in the baker diagram (but rules at the same level but not on cycles do not necessarily have isomorphic STDs).

Index Baker Transformation

For any rule T with h(T) its height in the baker diagram, the rule is on a cycle so Bp acts as a permutation on T*. Thus, Bp can be represented by a permutation on the index space: bp:{0,…,n-1} --> {0,…,n-1}. This is called the index baker transformation. It has the same cycle structure as the baker transformation.

T * Bph(T ) T

Index Baker Transformation

In a recent paper (Bulitko, V. & Voorhees, B., Cycle structure of baker transformation in one dimension) we show the complete cycle structure for Bp in terms of the bp cycles and their multiplicities on the prime factors of n. Thus the index baker transformation only needs to be considered on values of n that are relatively prime to p.

In one dimension, all non-singular transformations that map every additive CA into an additive CA with an isomorphic STD are compositions of shifts and powers of baker (e.g., if n = 7 the powers of baker showing up are 1/2, 1, 3/2 and their inverses).

Brief Number Theoretic Excursions

If n is prime, all non-trivial cycles of bp on {0,…,n-1} have the same cycle length.

Let s be a positive integer, p and q prime. Then s is the length of a non-trivial cycle of bp on {0,…,q-1} if and only if s|(q-1) and there exist positive integers m, t such that (a) s = mt and (b) q|(pt - 1)

For any p, s there are only a finite number of primes q with non-trivial bp cycles of length s.

Brief Number Theoretic Excursions

For p and q prime, define the height of p modulo q as

with ordqp the smallest integer s such that ps = 1 mod(q). Define:

This represents all primes having non-trivial bp cycles of length s.

hq (p)max k qk | pordq p 1

p (s) qhq ( p)q primesordq p

Brief Number Theoretic Excursions

Some properties of this “universal” number:

1.

2. If s ≠ r

3.

4. non-trivial bp cycles on {0,…,m-1} have length s and is the largest number for which this is true.

p (s) ps 1 gcd p (s), p (r) 1

gcd p (s), p (r) 1 sr or p (s)1

p (s)m p (s)

Brief Number Theoretic Excursions

For s a positive integer and p prime set i(q) to the largest integer such that qi(q)|(ps - 1) and define

M(2,s) = ps - 1 is the s-th Mersenne number.

If s is prime, Thus, for prime s the s-th Mersenne number is the largest number such that all non-trivial cycles of b2 have length s.

N (p, s) qi(q)q|( p 1) M (p, s)

ps 1N (p, s)

p (s)M (p, s)

Brief Number Theoretic Excursions

There are similar results for the Fermat numbers: Define

F(2,s) is the s-th Fermat number and hence F(2,s) is the largest number for which b2 has non-trivial cycles of length 2s+1.

F(p, s) prps

r0

p 1

, so F(2, s)22s

1

p ps1 F(p, s)

Extension to General Rules

In one dimension (and perhaps higher dimensions) the baker transformation can be extended to all rules acting on cylinders. Doing this requires a messy, but theoretically useful extension of the CA rule table to maximal neighborhoods; i.e., on a cylinder of size n, the neighborhoods are also of size n and in line with the expression of a rule in terms of the left shift, are left justified. The rule table has size pn.

From here on, p = 2.

Extension to General Rules

For k-site additive rules X = (a0,…,an-1) there is a relation between the rule coefficients as and the rule table entries xi (0 ≤ i ≤ 2k-1). For k = n, taking the binary form i = i0…in-1

In terms of the components xi the condition that X be additive is

as 1 x

2n s 11 0sn 1

0 otherwise

xi in s 1x2s mod(2)s0

n 1

Extension to General Rules

The baker transformation can be given in matrix form, both in terms of the rule coefficients as and the rule table components xi:

The second form of this doesn’t depend on additivity, hence gives a generalization of the baker transformation to all CA. The matrix b is size n, the matrix B is size 2n. If n is odd, both b and B are non-singular permutation matrices.

a bax Bx xi xi b

Extension to General Rules

Example: n = 5

b

1 0 0 0 00 0 0 1 00 1 0 0 00 0 0 0 10 0 1 0 0

B

A 0 0 00 A 0 0D 0 0 00 D 0 00 0 A 00 0 0 A0 0 D 00 0 0 D

A

1 0 0 0 0 0 0 00 0 0 0 1 0 0 00 1 0 0 0 0 0 00 0 0 0 0 1 0 0

D

0 0 1 0 0 0 0 00 0 0 0 0 0 1 00 0 0 1 0 0 0 00 0 0 0 0 0 0 1

Extension to General Rules

Let n be odd and let A(X) be the adjacency matrix for STD(X): Aij = 1 iff X(i0…in-1) = j0…jn-1.

Then A(B2X) = B2A(X)B2-1.

Since for odd n B2 is a permutation, B2-1

exists and this defines an isomorphism between STD(X) and STD(B2X).

This is the equivalent non-linear expression for the linear equation [b2C(T)b2

-1](b2µ) = b2ß where C(T) is the circulant matrix representing T with C(T)µ = ß.

Extension to General Rules

It is no longer true that (B2T) = T2.

For rule 18 with n = 5, for example, B2T equals T2 on states 00000, 00001, 00010, 00100, 01000, 10000, 01111, 10111, 11011, 11101, 11110, 11111. It equals T on states 00011, 00110, 01100, 11000, 10001, 00101, 01010, 10100, 01001, 10010.

With substitutions f:00111 --> 11010, f-

1:11010 --> 00111 and all shifts of this (B2T)(fµ) = T2µ.