二維網格模型中多符號的花樣生成問題

國 立 交 通 大 學

應 用 數 學 系

碩 士 論 文

二維網格模型中多符號的

花樣生成問題

Two-Dimensional Patterns Generation

Problem with Many Symbols

研 究 生：王 怡 菁

指導老師：林 松 山 教授

二維網格模型中多符號的

花樣生成問題

Two-Dimensional Patterns Generation

Problem with Many Symbols

研 究 生: 王怡菁 Student： Yi-Ching Wang

指導教授: 林松山 Advisor： Song-Sun Lin

國 立 交 通 大 學

應 用 數 學 系

碩 士 論 文

A Thesis

Submitted to Department of Applied Mathematics College of Science

National Chiao Tung University in Partial Fulfillment of the Requirements

for the Degree of Master

in

Applied Mathematics June 2006

Hsinchu, Taiwan, Republic of China

二維網格模型中多符號的花樣生成問題

學生：王怡菁 指導老師：林松山 教授

國立交通大學應用數學系(研究所)碩士班

摘 要

此篇論文主要是研究在二維網格模型下三個符號的花

樣生成問題。研究的主要目的是想找一些特別的置換矩陣 A

而能將熵明確地算出來。

Two-Dimensional Patterns Generation

Problem with Many Symbols

student：Yi-Ching Wang

Advisors：Dr. Song-Sun Lin

Department﹙Institute﹚of Applied Mathematics

National Chiao Tung University

ABSTRACT

In this paper we discuss patterns generation problems with three symbols mainly.

The main result is to find some special transition matrix A

such that their spatial

誌

謝

這篇論文的完成必須感謝許多協助與支持我的人。首先，感謝我

的指導教授 林松山老師這兩年來的指導與勉勵，在學問與待人處世

方面，都讓我受惠良多，謹此致上我最誠摯的敬意與謝意。口試期間，

承蒙林文偉老師、李明佳老師及許正雄老師費心審閱並提供許多寶貴

之意見，使本論文之完稿得以更加齊備，永誌於心。

研究所求學過程中，感謝榮超學長和吟衡學姊總在我遇到問題的

時候，給予我意見以及耐心的指導；接著感謝同窗好友園芳和倖綺與

研究所同學們的關心與協助，這些日子裡的互相幫忙、互相砥礪，深

厚的情誼永難忘懷。此外，謝謝志鴻學長、其儒學長、耀漢學長以及

文貴學長，因為有你們碩士班兩年生涯能夠如此充實愉快的度過，與

你們一起的點點滴滴，將是我永遠的回憶。

最後，要感謝的是陪伴了我二十多年的家人們，你們不但讓我有

良好的生活環境，讓我可以在求學的路上走得更專心，也總是不辭辛

苦的照料我幫忙我，讓我能夠克服種種的困難繼續向前邁進，因為有

你們，也讓我多了一份必須更上進的責任，如果沒有家人的支持，這

一切皆為空談，自然也沒有現在的成就。再次地感謝所有幫助過我及

關心過我的人，謝謝你們！

目

錄

中文提要 ………．

i

英文提要 ………．

ii

誌謝

………．

iii

目錄

………．

iv

1.

Introduction ………．

1

1.1

Transition matrices and spatial entropy ………

1

1.2

Computation of maximum eigenvalue and spatial entropy

3

2.

Three-symbols problems ………．

4

2.1

Transition matrices and spatial entropy ………

4

2.2

Computation of maximum eigenvalue and spatial entropy

6

3.

Spatial entropy of cyclic cases ………

14

1.

Introduction

Lattices are important in scientifically modeling underlying spatial struc-tures. Investigations in this field have covered phase transition [8], [11], [33], [34], [31], [32], [35], [42], [43], [44], [45], chemical reaction [6], [7], [23], biol-ogy [9], [10], [20], [21], [22], [28], [29], [30] and image processing and pattern recognition [15], [16], [17], [18], [19], [24]. In the field of lattice dynamical systems (LDS) and cellular neural networks (CNN), the complexity of the set of all global patterns recently attracted substantial interest. In particular, its spatial entropy has received considerable attention [1], [2], [5], [3], [4], [12], [14], [13], [25], [26], [27], [36], [37], [38], [39], [40], [41].

The one dimensional spatial entropy h can be found from an associated transition matrix T. The spatial entropy h equals logρ(T), where ρ(T) is the maximum eigenvalue of T.

In this paper, we study the two-dimensional patterns generation problems with many symbols. We first recall the results of two symbols in [3], [4].

1.1.

Transition Matrices and Spatial Entropy

In two-dimensional situation, higher transition matrices have been dis-covered in [27] and developed systematically in [3] by studying the pattern generation problem. For simplicity, two symbols on 2 × 2 lattice Z2×2 are

considered.

A transition matrix in the horizontal (or vertical) direction A2 =     a11 a12 a13 a14 a21 a22 a23 a24 a31 a32 a33 a34 a41 a42 a43 a44     , (1.1) which is linked to a set of admissible local patterns on Z2×2 is considered,

where aij ∈ {0, 1} for 1 ≤ i, j ≤ 4. The associated vertical (or horizontal)

transition matrix B2 is given by

B2 =     b11 b12 b13 b14 b21 b22 b23 b24 b31 b32 b33 b34 b41 b42 b43 b44     , (1.2) A2 and B2 are connected to each other as follows.

A2 =     b11 b12 b21 b22 b13 b14 b23 b24 b31 b32 b41 b42 b33 b34 b43 b44     = A2;1 A2;2 A2;3 A2;4  , (1.3) 1

and B2 =     a11 a12 a21 a22 a13 a14 a23 a24 a31 a32 a41 a42 a33 a34 a43 a44     = B2;1 B2;2 B2;3 B2;4  . (1.4) Notably if A2 represents the horizontal (or vertical) transition matrix

then B2 represents the vertical (or horizontal) transition matrix. Results

that hold for A2 are also valid for B2. Therefore, for simplicity, only A2 is

presented herein.

The recursive formulae for n-th order transition matrices An defined on

Z2×n are obtained in [4] as follows

An+1=     b11An;1 b12An;2 b21An;1 b22An;2 b13An;3 b14An;4 b23An;3 b24An;4 b31An;1 b32An;2 b41An;1 b42An;2 b33An;3 b34An;4 b43An;3 b44An;4     , (1.5) whenever An =  An;1 An;2 An;3 An;4  , (1.6) for n ≥ 2, or equivalently, An+1;α=  bα1An;1 bα2An;2 bα3An;3 bα4An;4  , (1.7) for α ∈ {1, 2, 3, 4}.

The number of all admissible patterns defined on Zm×n which can be

generated from A2 is now defined by

Γm,n(A2) = |Am−1n |

= the sum of all entries in 2n× 2n _{matrix A}m−1 n .

(1.8) The spatial entropy h(A2) is defined as

h(A2) = lim m,n→∞ 1 mnlogΓm,n(A2) = lim m,n→∞ 1 mnlog |A m−1 n |. (1.9) The existence of the limit (1.9) has been shown in [3], [12], [13]], [27]. When h(A2) > 0, the number of admissible patterns grows exponentially with the

lattice size m × n. In this situation, spatial chaos arises. When h(A2) = 0,

patterns formation occurs.

To compute the double limit in (1.9), n ≥ 2 can be fixed initially and m is allowed to tend to infinity as in [3] and [27]; then Perron-Frobenius theorem is applied; lim m→∞ 1 mlog |A m−1 n | = logρ(An), (1.10) which implies h(A2) = lim n→∞ 1 nlogρ(An), (1.11) where ρ(An) is the maximum eigenvalue of matrix An.

1.2.

Computation of Maximum Eigenvalue and Spatial

Entropy

Anis a 2n× 2nmatrix, so computing ρ(An) is usually quite difficult when

n is large. However, for a class of A2, the recursive formulae for ρ(An) can

be computed explicitly, along with a limiting equation to ρ∗ =exp(h(A2)),

as in [3]. This class of A2 has the form of

 A B B A  , i.e., A2 =  A B B A  , (1.12) where A = a a2 a3 a  , B = b b2 b3 b  and a, a2, a3, b, b2, b3 ∈ {0, 1}.

The results in [3] are recalled as follows.

Lemma 1.1. Let A and B be non-negative and non-zero m × m matrices, respectively, and α and β are positive numbers. The maximum eigenvalue of 

A αB βB A



is then the maximum eigenvalue of

A +pαβB. (1.13) Theorem 1.2. Assume that A2 =

 A B B A  and A =  a a2 a3 a  and B =  b b2 b3 b 

where a, b, a2, a3, b2, b3 ∈ {0, 1}. For n ≥ 2, let λn be the

largest eigenvalue of

|An− λ| = 0.

Then

λn= αn−1+ βn−1,

where αk and βk satisfy the following recursive relations:

αk+1 = aαk+ bβk,

βk+1 =

p

(a2αk+ b2βk)(a3αk+ b3βk),

for k ≥ 0, and α0 = β0 = 1.

Furthermore, the spatial entropy h(A2) is equal to

logξ∗, where ξ∗ is the maximum root of the following polynomials Q(ξ):

(I) if a2 = a3 = 1, Q(ξ) ≡ 4ξ2(ξ − a)2+ (γ2− 4δ)(ξ − a)2− γ2ξ2− 2γ(2b − aγ)ξ − (2b − aγ)2, where γ = b2+ b3 and δ = b2b3. (II) if a2a3 = 0 and a2b3+ a3b2 = 1, Q(ξ) ≡ ξ3− aξ2_{− δξ + aδ − b.}

Moreover, if a2a3 = 0 and a2b3+ a3b2 = 0, then h(A2) = 0.

The proofs of above two theorems are shown in [3].

2.

Three-Symbols Problems

In this section,we focus our study on three-symbols problems. We try to generalize the result of Theorem 1.2 to the three-symbols cases.

2.1.

Transition Matrices and Spatial Entropy

By the same reason as two symbols on lattice Z2×2, we take a transition

matrix A2 of three symbols on lattice Z2×2 as A2 =               a11 a12 a13 a14 a15 a16 a17 a18 a19 a21 a22 a23 a24 a25 a26 a27 a28 a29 a31 a32 a33 a34 a35 a36 a37 a38 a39 a41 a42 a43 a44 a45 a46 a47 a48 a49 a51 a52 a53 a54 a55 a56 a57 a58 a59 a61 a62 a63 a64 a65 a66 a67 a68 a69 a71 a72 a73 a74 a75 a76 a77 a78 a79 a81 a82 a83 a84 a85 a86 a87 a88 a89 a91 a92 a93 a94 a95 a96 a97 a98 a99               =               b11 b12 b13 b21 b22 b23 b31 b32 b33 b14 b15 b16 b24 b25 b26 b34 b35 b36 b17 b18 b19 b27 b28 b29 b37 b38 b39 b41 b42 b43 b51 b52 b53 b61 b62 b63 b44 b45 b46 b54 b55 b56 b64 b65 b66 b47 b48 b49 b57 b58 b59 b67 b68 b69 b71 b72 b73 b81 b82 b83 b91 b92 b93 b74 b75 b76 b84 b85 b86 b94 b95 b96 b77 b78 b79 b87 b88 b89 b97 b98 b99               =   A2;1 A2;2 A2;3 A2;4 A2;5 A2;6 A2;7 A2;8 A2;9  , (2.1)

which is linked to a set of admissible patterns on Z2×2, where ai,j, bi,j ∈ {0, 1}.

The recursive formulae for n+1-th order transition matrices An+1 defined

on Z2×(n+1) are An+1=               b11An;1 b12An;2 b13An;3 b21An;1 b22An;2 b23An;3 b31An;1 b32An;2 b33An;3 b14An;4 b15An;5 b16An;6 b24An;4 b25An;5 b26An;6 b34An;4 b35An;5 b36An;6 b17An;7 b18An;8 b19An;9 b27An;7 b28An;8 b29An;9 b37An;7 b38An;8 b39An;9 b41An;1 b42An;2 b43An;3 b51An;1 b52An;2 b53An;3 b61An;1 b62An;2 b63An;3 b44An;4 b45An;5 b46An;6 b54An;4 b55An;5 b56An;6 b64An;4 b65An;5 b66An;6 b47An;7 b48An;8 b49An;9 b57An;7 b58An;8 b59An;9 b67An;7 b68An;8 b69An;9 b71An;1 b72An;2 b73An;3 b81An;1 b82An;2 b83An;3 b91An;1 b92An;2 b93An;3 b74An;4 b75An;5 b76An;6 b84An;4 b85An;5 b86An;6 b94An;4 b95An;5 b96An;6 b77An;7 b78An;8 b79An;9 b87An;7 b88An;8 b89An;9 b97An;7 b98An;8 b99An;9               , (2.2) where An=   An;1 An;2 An;3 An;4 An;5 An;6 An;7 An;8 An;9  , (2.3) 5

for any n ≥ 2.

The definition of spatial entropy h(A2) of three symbols on lattice Z2×2

is the same as two symbols which can also be proved as h(A2) = lim

n→∞

1

n log ρ(An), (2.4) see [3].

2.2.

Computation of Maximum Eigenvalues and

En-tropy

For three symbols, An is a 3n× 3n matrix, so computing the maximum

eigenvalue of An (ρ(An)) is harder than it is for two symbols. We begin with

the study of A2 of the form

A2 =   A B C B C A C A B  , (2.5) where A =   a a12 a13 a21 a a23 a31 a32 a  , B =   b b12 b13 b21 b b23 b31 b32 b  , C =   c c12 c13 c21 c c23 c31 c32 c  , (2.6) and a, b, c, aij, bij, cij ∈ {0, 1}, i, j ∈ {1, 2, 3}, i 6= j. Consider An =   An Bn Cn Bn Cn An Cn An Bn   (2.7) Let λn be the eigenvalue of An, and Un be the corresponding eigenvector of

λn, i.e., AnUn= λnUn, (2.8) where Un=   un vn wn  . Therefore,   An Bn Cn Bn Cn An Cn An Bn     un vn wn  = λn   un vn wn  . (2.9) 6

Assume

un = vn= wn, (2.10)

then (2.9) implies

(An+ Bn+ Cn)un = λnun. (2.11)

Conversely, under the assumption (2.10), (2.11) implies (2.9). Therefore, (2.8) and (2.11) are equivalent.

We first prove the following lemma which is a generalization of Lemma 1.1.

Lemma 2.1. Let A2 be given as in (2.5) and (2.6). If

a + b + c = a12+ b12+ c12= a13+ b13+ c13 = a21+ b21+ c21= a23+ b23+ c23 = a31+ b31+ c31= a32+ b32+ c32 (2.12) holds. Then An+ Bn+ Cn = (a + b + c)   An−1 Bn−1 Cn−1 Bn−1 Cn−1 An−1 Cn−1 An−1 Bn−1  . (2.13) Proof . Let A2 =   A2 B2 C2 B2 C2 A2 C2 A2 B2 

, where A2, B2, and C2 are given in (2.6).

By (2.2), An=   An Bn Cn Bn Cn An Cn An Bn   =            aAn−1 a12Bn−1 a13Cn−1 bAn−1 b12Bn−1 b13Cn−1 cAn−1 c12Bn−1 c13Cn−1 a21Bn−1 aCn−1 a23An−1 b21Bn−1 bCn−1 b23An−1 c21Bn−1 cCn−1 c23An−1 a31Cn−1 a32An−1 aBn−1 b31Cn−1 b32An−1 bBn−1 c31Cn−1 c32An−1 cBn−1 bAn−1 b12Bn−1 b13Cn−1 cAn−1 c12Bn−1 c13Cn−1 aAn−1 a12Bn−1 a13Cn−1 b21Bn−1 bCn−1 b23An−1 c21Bn−1 cCn−1 c23An−1 a21Bn−1 aCn−1 a23An−1 b31Cn−1 b32An−1 bBn−1 c31Cn−1 c32An−1 cBn−1 a31Cn−1 a32An−1 aBn−1 cAn−1 c12Bn−1 c13Cn−1 aAn−1 a12Bn−1 a13Cn−1 bAn−1 b12Bn−1 b13Cn−1 c21Bn−1 cCn−1 c23An−1 a21Bn−1 aCn−1 a23An−1 b21Bn−1 bCn−1 b23An−1 c31Cn−1 c32An−1 cBn−1 a31Cn−1 a32An−1 aBn−1 b31Cn−1 b32An−1 bBn−1            . Now An+Bn+Cn=   (a + b + c)An−1 (a12+ b12+ c12)Bn−1 (a13+ b13+ c13)Cn−1 (a21+ b21+ c21)Bn−1 (a + b + c)Cn−1 (a23+ b23+ c23)An−1 (a31+ b31+ c31)Cn−1 (a32+ b32+ c32)An−1 (a + b + c)Bn−1  . By the assumption (2.12), (2.13) follows. The proof is complete.

Now, we can prove our first theorem. 7

Theorem 2.2. Assume (2.12) holds, and a + b + c ≥ 1, then

h(A2) = log(a + b + c). (2.14)

Proof . Under the assumption (2.12) and by Lemma 2.1, An+ Bn+ Cn= (a + b + c)   An−1 Bn−1 Cn−1 Bn−1 Cn−1 An−1 Cn−1 An−1 Bn−1   which implies λn = (a + b + c)λn−1, (2.15)

for any n ≥ 3. Now

A2+ B2+ C2 = (a + b + c)   1 1 1 1 1 1 1 1 1  . which implies λ2 = 3(a + b + c). (2.16) Combining (2.15) with (2.16), λn = 3(a + b + c)n−1. (2.17) Hence h(A2) = lim n→∞ 1 nlog λn = lim n→∞ n − 1 n log(a + b + c) = log(a + b + c),

(2.14) follows. The proof is complete.

Remark 2.3. (i) It is of interest to study the case when A2 is of the form

(2.5) but (2.12) fails. A lemma like Lemma 1.1 need to be established, some progress has been made.

(ii) Result of Theorem 2.2 also holds for any number of symbols provides that the assumptions like (2.12) hold.

Lemma 2.4. Let A and B be non-negative and non-zero n × n matrices, re-spectively, and a1, a2, a3, and a4 are positive numbers. The maximum

eigen-value of   a1A a2B a2B a3B a4B a1A a3B a1A a4B 

 is then the maximum eigenvalue of a1A +

a4+pa24+ 8a2a3

2 B. (2.18) 8

Proof . Consider

a1A − λ a2B a2B

a3B a4B − λ a1A

a3B a1A a4B − λ

= 0. There are two cases:

Case I. If |a1A − λ| = 0, it is clear that λ = a1A.

Case II. For |a1A − λ| 6= 0, the last equation is equivalent to

a1A − λ a2B a2B

0 (a4B − λ) − a2a3B(a1A − λ)−1B a1A − a2a3B(a1A − λ)−1B

0 0 P

= 0, where

P =[(a4B − λ) − a2a3B(a1A − λ)−1B] − [a1A − a2a3B(a1A − λ)−1B]

[(a4B − λ) − a2a3B(a1A − λ)−1B]−1[a1A − a2a3B(a1A − λ)−1B],

and we could simplify it to

|I − {[a1A − a2a3B(a1A − λ)−1B][(a4B − λ) − a2a3B(a1A − λ)−1B]−1}2| = 0.

Then, we have

|I + {[a1A − a2a3B(a1A − λ)−1B][(a4B − λ) − a2a3B(a1A − λ)−1B]−1}| = 0

or |I − {[a1A − a2a3B(a1A − λ)−1B][(a4B − λ) − a2a3B(a1A − λ)−1B]−1}| = 0.

Since A and B are non-negative and a1, a2, a3, and a4 are positive,

veri-fying that the maximum eigenvalue λ of   a1A a2B a2B a3B a4B a1A a3B a1A a4B   and a1A + a4+ √ a2 4+8a2a3

2 B are equal is relatively easy. The proof is complete.

Theorem 2.5. Assume that A2 =

  A B B B B A B A B   and A =   a1 a2 a2 a3 a4 a1 a3 a1 a4   and B =   b1 b2 b2 b3 b4 b1 b3 b1 b4 

 where ai, bi ∈ {0, 1}, and i ∈ {1, 2, 3, 4}. For n ≥ 2,

let λn be the largest eigenvalue of

|An− λ| = 0.

Then

λn = αn−1+ βn−1, (2.19)

where αk and βk satisfy the following recursive relations:

αk=a1αk−1+ b1βk−1, (2.20) βk= 1 2{(a4αk−1+ b4βk−1) + [(a4αk−1+ b4βk−1) 2₊ 8(a2αk−1+ b2βk−1)(a3αk−1+ b3βk−1)] 1 2}, (2.21) for k = 1, 2, . . . , n − 1, and α0 = 1, β0 = 2. (2.22)

Furthermore, the spatial entropy h(A2) is equal to logξ∗, where ξ∗ is the

maximum root of the following polynomials Q(ξ): (I) if a1 = b1 = 1,

Q1(ξ) ≡ξ4− (2 + b4)ξ3+ (1 − a4+ 2b4− 2b2b3)ξ2+

[(a4− b4) − 2b2(a3− b3) − 2b3(a2− b2)]ξ − 2(a2− b2)(a3− b3).

(2.23) (II) if a1 = 0, b1 = 1,

Q2(ξ) ≡ ξ4− b4ξ3− (a4+ 2b2b3)ξ2− 2(a2b3+ a3b2)ξ − 2a2a3. (2.24)

(III) if a1 = 1, b1 = 0,

Q3(ξ) ≡ ξ2− b4ξ − 2b2b3. (2.25)

Proof . Since the structure of A2 is special, it is easy to show that for any

k ≥ 2, we get Hk =   Ak Bk Bk Bk Bk Ak Bk Ak Bk  , and Hk+1 =   Ak+1 Bk+1 Bk+1 Bk+1 Bk+1 Ak+1 Bk+1 Ak+1 Bk+1  , here Ak+1 = Hk A =   a1Ak a2Bk a2Bk a3Bk a4Bk a1Ak a3Bk a1Ak a4Bk  , (2.26) 10

and Bk+1 = Hk B =   b1Ak b2Bk b2Bk b3Bk b4Bk b1Ak b3Bk b1Ak b4Bk  , (2.27) A2 = A and B2 = B. We know that |An+1− λn+1| = 0, so

|An+1+ 2Bn+1− λn+1| = 0. (2.28)

Let

α0 = 1 and β0 = 2. (2.29)

By induction on k, 1 ≤ k ≤ n, and using (2.26),(2.27),(2.28) and Lemma 2.4, it is straightforward to derive

|αkAn−k+1+ βkBn−k+1− λn+1| = 0, (2.30)

with αk and βk satisfy (2.20) and (2.21). In particular,

αn=a1αn−1+ b1βn−1, (2.31) βn= 1 2{(a4αn−1+ b4βn−1) + [(a4αn−1+ b4βn−1) 2 + 8(a2αn−1+ b2βn−1)(a3αn−1+ b3βn−1)] 1 2}, (2.32) and λn+1 = αn+ βn.

This proves the first part of the theorem.

The remainder of the proof, demonstrates that h(A2) = logλ∗ where λ∗ is

the maximum root of Q(λ). There are three cases: Case I. From (2.31), if a1 = b1 = 1, we have

βn−1 = αn− αn−1. (2.33)

Substituting (2.33) into (2.21), yields αn+1− αn= 1 2{[(a4− b4)αn−1+ b4αn] + [((a4 − b4)αn−1+ b4αn) 2 + 8((a2− b2)αn−1+ b2αn)((a3− b3)αn−1+ b3αn)] 1 2}. (2.34) Now, let ξn= αn αn−1 (2.35) 11

and after dividing (2.34) by αn, we have ξn+1− 1 = 1 2{[(a4 − b4) 1 ξn + b4] + [((a4− b4) 1 ξn + b4) 2 + 8((a2− b2) 1 ξn + b2)((a3− b3) 1 ξn + b3)] 1 2}. (2.36)

(2.36) can be written as the following iteration map:

ξn+1= G1(ξn), (2.37) where G1(ξ) =1 + 1 2{[(a4− b4) 1 ξ + b4] + [((a4− b4) 1 ξ + b4) 2 + 8((a2− b2) 1 ξ + b2)((a3− b3) 1 ξ + b3)] 1 2}. (2.38)

We first observe the fixed point ξ∗ of G1(ξ), i.e., ξ∗ = G1(ξ∗), is a root of

Q(ξ).

Indeed, by letting ξn= ξn+1= ξ∗ in (2.36), we have

ξ∗− 1 = 1 2{[(a4− b4) 1 ξ∗ + b4] + [((a4− b4) 1 ξ∗ + b4) 2 + 8((a2 − b2) 1 ξ∗ + b2)((a3− b3) 1 ξ∗ + b3)] 1 2},

which gives us Q(ξ∗) = 0. It can be proven that the maximum fixed point of

G1(ξ) or the maximum root ξ∗ of Q(ξ) = 0 satisfies 1 ≤ ξ∗ ≤ 2 and

ξn→ ξ∗ as n → ∞. (2.39)

Details are omitted here for brevity. By (2.19),(2.33) and (2.35), we can also prove

λn+1

λn

→ ξ∗ as n → ∞. (2.40)

Hence, h(T2) = log ξ∗.

Table 3.1 is to evaluate the value of λ∗, where 4a2+2a3+a4+1 = i, 1 ≤ i ≤ 8,

and 4b2+ 2b3+ b4+ 1 = j, 1 ≤ j ≤ 8.

ij 1 2 3 4 5 6 7 8 8 2 2.2938 2.2056 2.5214 2.2056 2.5214 2.6180 3 7 1.7900 2 2 2.2599 2 2.2599 2.4142 2.7693 6 1.6180 2 2 2.3593 1.6180 2 2.3028 2.7321 5 1 1 1.6956 2 1 1 2 2.4142 4 1.6180 2 1.6180 2 2 2.3593 2.3028 2.7321 3 1 1 1 1 1.6956 2 2 2.4142 2 1.6180 2 1.6180 2 1.6180 2 2 1 1 1 1 1 1 1 1 1.4142 2 Table 3.1

Case II. If a1 = 0 and b1 = 1, then, from (2.31), we have

αn= βn−1. (2.41)

Again, substituting (2.41) into (2.21) and letting ξn= _ββ_n−1n lead to

ξn= 1 2{(a4 1 ξn−1 + b4+ [(a4 1 ξn−1 + b4)2+ 8(a2 1 ξn−1 + b2)(a3 1 ξn−1 + b3)] 1 2}, (2.42) i.e., ξn = G2(ξn−1), where G2(ξ) = 1 2{(a4 1 ξ + b4+ [(a4 1 ξ + b4) 2 + 8(a2 1 ξ + b2)(a3 1 ξ + b3)] 1 2}. (2.43)

The maximum fixed point ξ∗ of (2.43) is the maximum root of Q(ξ) = 0 in

(2.24). It can be also be proven that (2.39) and (2.40) holds in this case. Table 3.2 is to show the value of λ∗, where 4a2+ 2a3+ a4+ 1 = i, 1 ≤ i ≤ 8,

and 4b2+ 2b3+ b4+ 1 = j, 1 ≤ j ≤ 8. ij 1 2 3 4 5 6 7 8 8 1.4142 1.8536 1.6956 2.1234 1.6956 2.1234 2.2695 2.7321 7 1.1892 1.5437 1.4945 1.8737 1.4945 1.8737 2.0907 2.5346 6 1 1.6180 1.5214 2 1 1.6180 2 2.5115 5 0 1 1.2599 1.6956 0 1 1.7693 2.2695 4 1 1.6180 1 1.6180 1.5214 2 2 2.5115 3 0 1 0 1 1.2599 1.6956 1.7693 2.2695 2 1 1.6180 1 1.6180 1 1.6180 1.7321 2.3028 1 0 1 0 1 0 1 1.4142 2 13

Table 3.2

Case III. If a1 = 1 and b1 = 0, then, from (2.31), we have

αn= αn−1. (2.44)

Repeating the above steps, hence we get ξn =

b4+pb24+ 8a2a3

2 . (2.45) The maximum fixed point ξ∗ of (2.45) is the maximum root of Q(ξ) = 0 in

(2.25). The proof is complete. Table 3.3 is to show the value of λ∗.

b2 b3 b4 λ∗ 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 √2 1 1 1 2 Table 3.3

3.

Spatial Entropy of Cyclic Cases

In this section, we study the spatial entropy of A2 when A2 has certain cyclic

structure. Consider A2 =   A B C C A B B C A   (3.1) We first study A,B,and C with the following form, A, B, C ∈ {E, I, J, J0}, where E =   1 1 1 1 1 1 1 1 1  , I =   1 0 0 0 1 0 0 0 1  , J =   0 1 0 0 0 1 1 0 0  , J 0 =   0 0 1 1 0 0 0 1 0  . (3.2) Now, we have the following theorem for cyclic A2.

Theorem 3.1. Assume A2 is of the form (3.1). Then λ∗ ≡ lim n→∞λ 1 n n, (3.3)

satisfies the limiting equation Q(λ) as follows:

(I) A B C limiting equation Q(λ) λ∗ (i) E I J Q2(λ) λ∗2 (ii) E I J0 Q2(λ) λ∗2 (iii) E J I Q2(λ) λ∗2 (iv) E J J0 Q1(λ) λ∗1 (v) E J0 I Q3(λ) λ∗3 (vi) E J0 J Q1(λ) λ∗1 (II) A B C limiting equation Q(λ) λ∗ (i) I E J Q2(λ) λ∗2 (ii) I E J0 Q1(λ) λ∗1 (iii) I J E Q1(λ) λ∗1 (iv) I J0 E Q2(λ) λ∗2 (III) A B C limiting equation Q(λ) λ∗ (i) J E I Q3(λ) λ∗3 (ii) J E J0 Q2(λ) λ∗2 (iii) J I E Q1(λ) λ∗1 (iv) J J0 E Q3(λ) λ∗3 (IV) A B C limiting equation Q(λ) λ∗ (i) J0 E I Q1(λ) λ∗1 (ii) J0 E J Q3(λ) λ∗3 (iii) J0 I E Q3(λ) λ∗3 (iv) J0 J E Q2(λ) λ∗2 where, Q1(λ) = λ − 1, λ∗1 = 1 Q2(λ) = λ2− λ − 1, λ∗2 = 1+ √ 5 2 ∼= 1.618033988 Q3(λ) = λ3− λ2− λ − 1, λ∗3 ∼= 1.839286755.

Proof of Case(I)(i). Since the structure of A2 is similar to (2.5), it is easy

to verify that (2.11) is also right to (3.1) and for any k ≥ 2, we have Ak =   Ek Ik Jk Jk Ek Ik Ik Jk Ek  . 15

By (2.11), |An− λn| = 0, so

|En+ In+ Jn− λn| = 0. (3.4)

Let

α0 = 1 and β0 = 1. (3.5)

By induction on k, 1 ≤ k ≤ n, and using (2.11), it is straightforward to derive |αkEn−k+ βkIn−k+ Jn−k − λn| = 0, (3.6)

where αk and βk satisfy the following recursive relations:

αk =αk−1+ βk−1, (3.7)

βk =αk−1+ 1, (3.8)

and

λn= 3αn−2+ βn−2+ 1. (3.9)

By recursive formulae of (3.7) and (3.8), we get αn = 5 +√5 10 ( 1 +√5 2 ) n+1₊ 5 − √ 5 10 ( 1 −√5 2 ) n+1₊ −5 − 3 √ 5 10 (1 − 1 +√5 2 ) n−1₊ −5 + 3√5 10 (1 − 1 −√5 2 ) n−1 , (3.10) βn = 5 +√5 10 ( 1 +√5 2 ) n+1₊ 5 − √ 5 10 ( 1 −√5 2 ) n+1₊ −5 − √ 5 10 (1 − 1 +√5 2 ) n−1₊ −5 +√5 10 (1 − 1 −√5 2 ) n−1_{+ 1. (3.11)}

Substituting (3.10) and (3.11) into (3.9), we have λn = 11 + 5√5 2 ( 1 +√5 2 ) n−3₊11 − 5 √ 5 2 ( 1 −√5 2 ) n−3_{− 2. (3.12)}

By (1.11), it is obvious that h(A2) = lim n→∞ 1 n log λn = log 1 +√5 2 . Thus it is clear that Q(λ) = λ2− λ − 1.

The proofs of the following cases is similar to Case(I)(i). The proof is com-plete.

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