三維面著色的熵

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三維面著色的熵

Spatial Entropy of 3- dimensional Face Coloring

研 究 生：張育慈

三維面著色的熵

Spatial Entropy of 3- dimensional Face Coloring

研 究 生：張育慈 Student：Yu-Tzu Chang

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

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 2011

三維面著色的熵

三維面著色的熵

三維面著色的熵

三維面著色的熵

學生：張育慈

指導教授

林松山教授

國立交通大學應用數學學系﹙研究所﹚碩士班

摘 要

這個研究主要是要去計算三維度兩個顏色的熵，但首先必須利用有序矩

陣以及矩陣自乘的性質所發展出來的遞迴公式去解決三維度兩個顏色下面

著色的花樣生成問題。

接下來，給一個限制集則就可以定義出轉移矩陣而且它的遞迴公式也會

被表現出來。最後，只需去計算矩陣的最大特徵值即可計算出熵的問題。

Spatial Entropy of 3- dimensional Face

Coloring

student：Yu-Tzu Chang

Advisors：Dr.

Department﹙Institute﹚of

National Chiao Tung University

ABSTRACT

The work investigates spatial Entropy of 3-dimensional face coloring, but we

need to solve three-dimensional pattern generation problem with edge- coloring

by using the properties of ordering and self-multiply matrices to establish some

recursive formulas, first.

Now, given admissible set of local patterns then the transition matrix is defined

and the recursive formulas are presented. Finally the spatial entropy is obtained

by computing the maximum eigenvalues of a sequence of transition matrices.

致

謝

首先，很感謝指導教授─ 林松山教授 一路上無論是研究心態、待人之

道、人生道理總是耳提面命，好讓我這一路上走得很順遂，這是有錢也買

不到的真心對待，所以身為學生的我心裡很感激。

其次，要感謝的是胡文貴學長的幫忙，每次學長都會很有耐心的與我討

論問題，有時也會以過來人的角色與我經驗分享，這些都讓我受益良多。

接著，很感激我的家人一路上的支持與鼓勵，因為有了他(她)們的陪伴

與鼓勵，這兩年我不孤單。

最後，要謝謝交大一起打排球的朋友們，因為排球所以才有機會在這麼

短的時間內認識這們多不同科系的人，也因為打排球這兩年才過的這麼精

彩，真的很謝謝他(她)們。

目

錄

中文提要 ………

i

英文提要 ………

ii

致謝

………

iii

目錄

………

iv

一、

Introduction ………

1

二、

Three-Dimensional Pattern Generation Problem ……

2

2.1

Ordering Matrix

………

4

三、

Transition Matrices and Spatial Entropy …………

16

3.1

Transition Matrices ………

16

3.2

_{Computation of}

λ

and entropy ………

21

Reference ………

23

1

Introduction

Here, we consider the problem of 2 symbols, then will get the set of all patterns Σ3

2, first, give any admissible set B ⊆ Σ32 and denote Σ32(B) be Σ32

which is restricted in B, secondly, denote Γ_m×n×k(B) is the quality of Σ3 2(B),

and finally, need to calculate spatial entropy of 3-dimensional face coloring,

h(B) = lim

m,n,k→∞

log Γ_m×n×k(B)

mnk (1.1)

clearly, how to calculate Γ_m×n×k(B) is the first problem we encountered from the equation (1.1). In order to solve the problem we face, we must study the problem of 3-dimensional pattern generation of 2 symbols in section 2 and find a way to control the colors of different directions by the matrix,

Y2.

Now, split section 2 into 3 steps as following:

Step 1 : find the recursive formula, Y2×n×2, of y-direction by Y2×2×2, for n ≥ 3

Step 2 : denote X2×n×2 ≡ Y2×n×2 and find the recursive formula , Xm×n×2by

X2×n×2, for n ≥ 3

Step 3 : denote Z_m×n×2 ≡ Ym×n×2and we will get Zm×n×k by Zm×n×2

which is self-multiply.

In section 3, we defined T_y;2×2×2;iy as the transition matrix of Y2×2×2;iy, for

1 ≤ iy ≤ 4 and find that the main problem will be converted into finding

Γ

z;m×n×2(B) by Peronn Fubini’s theorem. Finally, using the result to calculus

2

Three-Dimensional Pattern Generation

Problems

This section describes three-dimensional pattern generation problem. Here

m, n, k ≥ 2 are fixed and indices for brevity. Let S be a set of p colors, and

Z_m×n×k be a fixed finite rectangular sublattice of Z3_{, where Z}3 _{denotes the}

integer lattice on R3_{and (m, n, k) be a three-tuple of positive integer.}

Func-tion U : Z3 −→ S and Um×n×k : Zm×n×k −→ S are called global patterns

and locally patterns respectively. The set of all patterns U is denoted by Σ3

p ≡ SZ

3

, such thatΣ3p is the set of all patterns with p different colors in

a three-dimensional lattice. For clarity, two symbols, S = {0, 1} are con-sidered. Let x, y and z coordinate represent 1st-, 2st- and 3st-coordinates respectively as in Fig.1. Six orderings [w] ordering are represented as the following: [x] : [1]  [2]  [3] [y] : [2]  [1]  [3] [z] : [3]  [1]  [2] [bx] : [1]  [3]  [2] [by] : [2]  [3]  [1] [bz] : [3]  [2]  [1] (2.1) X Y z

On a fixed lattice Z_m×n×k, an ordering [w]  [j]  [k] is obtained on

Z_m×n×k, which is any one of the above ordering on Z_m×n×k. Therefore, the six ordering of Z_2×2×2 are presented as Fig.2, where αi = {0, 1}.

[ ] orderingx 2 1 6 3 4 5 [ ] orderingx 4 [ ] orderingy 2 1 3 5 6 [ ] orderingz 2 1 3 4 5 6 [ ] orderingy 6 1 3 4 5 2 [ ] orderingz 2 1 3 5 6 4 2 1 3 4 5 6 x Y z x x x x x Y _Y Y Y Y z z z z z , Figure 2.

2.1

Ordering Matrices

By using the six forms can get six different ordering matrices of Z2×2×2and

denote the order of matrices as equation (2.2)

iα = 1 + 6 Σ i=1αi2 6−i_{, where α} i ∈ {0, 1}. (2.2)

Here, we choose [z] as the order for convenience, and we can denote the order of x, y and z-directions by ix, iyand iz (2.3) respectively, where

1 6 ix, iyand iz ≤ 4.    ix = 1 + α2+ α1× 2 iy = 1 + α4+ α3× 2 iz = 1 + α6+ α5× 2 (2.3)

For convenience again, we have to define the matrix Y2×2×2 (2.4) below

which present the relation between colors and each directions.

2 2 2 Y = 3 3 2 2 2;1 2 2 2;2 2 2 2;i;j 2 2 2 2 2;3 2 2 2;4 Y Y = [y ] = Y Y ! " # $ % & (2.4)

It’s not different to discover that the colors of each direct of Z2×2×2

be controlled in each layer of Y2×2×2 respectively. It means that Y2×2×2 is

divided into three layers by matrix partitioning as figure 3 and the colors of y-, x-and z-direction are controlled in first, second and the third layer respectively as figure 3 below.

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The process of investigating the pattern generation problem should be broken down in the following steps:

Step 1. find the recursive formula Y2×2×2→ Y2×3×2 → ... → Y2×n×2,

that is extend on the y-direction.

Step 2. here replaces X2×n×2 with Y2×n×2, it means that X2×n×2 is really

to extend the x-direction and we get the recursive formula like X2×n×2 → X3×n×2 → ... → Xm×n×2that is extend the x-direction by

Y2×n×2.

Step 3. here replaces Z_m×n×2 with X_m×n×2, it means that X_m×n×2 is really to extend the z − direction. By using the matrix to self-multiply, we can generate Z_m×n×2→ Zm×n×3 → ... → Zm×n×k,

that is extend the z-direction by X_m×n×2. STEP 1.

2 2 2

STEP 2.

3 2

X _n X_{4 n2} Xm n 2

Proposition 2.1.

(1) If we only consider the colers of the z-direction of those matrices, Z2×2×2;1,

Z2×2×2;2, Z2×2×2;3and Z2×2×2;4with ignoring x- and y- directions, we can represent

this scenario as Z2. Moreover, we denote Zn =Z2⊗ Z2⊗ ... ⊗ Z2and discover that

Znis self-multiply, where Z_2×2×2 = Z2×2×2;1 Z2×2×2;2 Z2×2×2;3 Z2×2×2;4 ! , Z2×2×2;l =     Y(rez) 2×2×2;1;l  4×4  Y(rez) 2×2×2;2;l  4×4  Y(rez) 2×2×2;3;l  4×4  Y(rez) 2×2×2;4;l  4×4    2×2 , 1≤ l ≤ 4, ¯ Z2 = , and ¯Zn=

(2) As the ideal of (1) above, we consider the colors of the x-direction of those matri-ces, X2×2×2;1, X2×2×2;2, X2×2×2;3and X2×2×2;4with ignoring y- and z- directions, then

we can represent this scenario as X2. Moreover, we define Xn= X2⊗ X2⊗ ... ⊗ X2

and discover that Xnis a matrix which control the colors of the front and rear faces

by the columns and rows respectively, where

X_2×2×2 = X2×2×2;1 X2×2×2;2 X2×2×2;3 X2×2×2;4 ! ; X2×2×2;l =     Y(rez) 2×2×2;1;l  4×4  Y(rez) 2×2×2;2;l  4×4  Y(rez) 2×2×2;3;l  4×4  Y(rez) 2×2×2;4;l  4×4    2×2 , 1≤ l ≤ 4 X2 = , and X2;n=

Example 2.2. As n = 3

(1) We focus on the colors of z-direction, and the order be presented as Z2:

Z2 =

Now, consider

Z3 =

(2) We focus on the colors of x-direction, and the order be presented as X2:

X2 =

Now, consider

X3=

Here X3is a matrix which controls the the colors of front and rear faces

Now, we talk about the details of those steps. First, the recursive formula can be developed by using the properties of Y2×2×2as the following:

Y_2×3×2 =       Y2×2×2;1⊗ Y2×2×2;1 +_{Y2×2×2;2⊗ Y2×2×2;3} Y2×2×2;1⊗ Y2×2×2;2 +_{Y2×2×2;2⊗ Y2×2×2;4} Y2×2×2;3⊗ Y2×2×2;1 +_{Y2×2×2;4⊗ Y2×2×2;3} Y2×2×2;3⊗ Y2×2×2;2 +_{Y2×2×2;4⊗ Y2×2×2;4}       2×2 = Y2×3×2;1 Y2×3×2;2 Y2×3×2;3 Y2×3×2;4 ! 2×2 , where Y2×3×2;l is 42× 42 matrix, 1 ≤ l ≤ 4. Y2×n×2 =       Y_{2×2×2;1⊗ Y2×n−1×2;1} +_{Y2×2×2;2⊗ Y2×n−1×2;3} Y_{2×2×2;1⊗ Y2×n−1×2;2} +_{Y2×2×2;2⊗ Y2×n−1×2;4} Y2×2×2;3⊗ Y2×n−1×2;1 +_{Y2×2×2;4⊗ Y2×n−1×2;3} Y2×2×2;3⊗ Y2×n−1×2;2 +_{Y2×2×2;4⊗ Y2×n−1×2;4}       2×2 = Y2×n×2;1 Y2×n×2;2 Y2×n×2;3 Y2×n×2;4 ! 2×2 , where Y2×n×2;l is a 4n−1× 4n−1 matrix , 1 ≤ l ≤ 4. (2.5) Secondly, we denote X2×n×2 = 4 Σ iy=1

Y2×n×2;iy and find that the order of the

colors of each layers which be controlled by X2×n×2 as iz1 → ix1 → iz2 →

ix2 → ... → izn−2 → ixn−2 → izn−1 → ixn−1, for n ≥ 3, so need to rearrange the

order as iz1 → iz2 → · · · → izn−2 → ix1 → ix2 → · · · → ixn−2 → izn−1 → ixn−1.

For fixing any z and it’s the fact that change the view of x- into y-direction by property 1 ,where the method of the detail will be presented by items below:

• (1.) Rearrange iz1 (X(riz2) 2×n×2)iz1 = (X2×n×2)ix1; iz1 • (2.) Rearrange iz2, for iz1 (X_2×n×2(riz2) )i_z1;iz2 = (X2×n×2)i_z1;i_x1;i_z2 • (3.) Rearrange iz3, for iz1, iz2 (X_2×n×2(riz2 ;iz3))i_z1;i_z2;i_z3 = (X (r_iz2) 2×n×2)iz1;iz2;ix1;ix2;iz3 .. . • (n-1.) Rearrange izn−1, for iz1, iz2,· · · , izn−2

(Xr_2×n×2iz1 ;iz2 ;··· ;izn−1)i_z1;i_z2;··· ;i_zn−1

= (X_2×n×2(reiz2 ;iz3 ;··· ;izn−2))i_z1;i_z2;··· ;i_zn−2;i_x1;i_x2;··· ;i_xn−1;i_zn−1

here (X_2×n×2(reiz2;iz3;...;iz n−2))i_z1;i_z2;...;i_zn−2 is abbreviated as (Xr_2×n×2)z, for n ≥ 4.

1. For any z, (Xr

2×n×2)z controls the orders i1, i2 by rows and columns

respectively, where i2and i1are the orders of the front and rear faces

respectively, 1 ≤ i1, i2≤ 2n−1. [(Xr 2×n×2)z]2n−1×2n−1 =        a1,1 a1,2 · · · a1,2n−1 a2,1 a2,2n−1 .. . . .. ... a2n−1_,1 · · · a₂n−1_,2n−1        2n−1_×2n−1

Now, the order becomes z1 → z2 → · · · zn−2 → zn−1 → i1 → i2 from the

previous discussion, but is not well enough to control the order, so we still need to complete the following items as below before entering second step.

2. For any z and i1, we denote [(Xr_2,n)z;i1]

(r_i2)_. [(Xr_2×n×2)z;i1] (r_i2) ₌       ai1,1 · · · ai1,2k2 .. . . .. ... a_i1,(2k1₋₁₎₍₂k2)+1 · · · ai1,2k1+k2       2k1_×2k2 , where    k1= n−1₂ , k2 = n−1₂ , n is odd. k1= n₂, k2 = n−2₂ , n is even.

3. To assign the seats of i1is the goal in this item. [(Xr_2×n×2)z](ri2) =       [(Xr 2×n×2)z;1] (r_i2)_[(Xr 2×n×2)z;2] (r_i2) · · · [(Xr 2×n×2)z;22 k2](ri2) .. . ... h (X_2×n×2r )z;(2k1₋₁₎₍₂k2₎₊₁ i(r_i2) · · · h(Xr_2×n×2)z;2k1+k2 i(r_i2)       2k1_×2k2 , ∀z ,where    k1 = n−1₂ , k2 = n−1₂ n is odd. k1 = n−2₂ , k2 = n₂ n is even.

Here, we denote the process from (1) to (3) as (Xr

2;n)

(r_i2) _{which controls the}

color of each direction in each layer respectively, iz1 → iz2 → · · · → izn−2 → izn−1 → i1→ i2. For convenience again, we need to rearrange i1and i2as the

define as following:

•

[(X_2×n×2r )(ri2)_](ri1)_{is called rearrangement matrix of i1 of}

(Xr 2×n×2) (r_i2)_{, if ([(X}r 2×n×2) (r_i2)_](r_i1)₎ i1;z = [(X r 2×n×2) (r_i2)_] z;i1, for 1 ≤ i1, izl ≤ 4, 1 ≤ l ≤ n − 2. • ([(Xr 2×n×2)

(r_i2)_](r_i1)₎(r_i2)_{is called rearrangement matrix of i} 2 of [(Xr_2×n×2)(ri2)_]r(i1) if ([(Xr 2×n×2) (r_i2)_](r_i1)₎(ri2) i1;i2;z = [(X r 2×n×2) (r_i2)_](rei1) i1;z;i2 for 1 ≤ i1, i2, izl ≤ 4, 1 ≤ l ≤ n − 2, n ≥ 3.

By doing step 2, we need to denote X(a)_2×n×2 = ([(Xr_2×n×2)(ri2)_](ri1)₎(ri2) _and X(b)_2×n×2 =2

n−1

Σ

i2=1

X(a)_2×n×2;i

1;i2 to develop the recursive formula of X

(b)

Finally, denote Z_m×n×2 = 2

n−1

Σ

i1=1

X(b)_m×n×2;i

1 which records the colors of the

z-direction as iz1 → iz2 → . . . → izn−1 and we know that Zm×n×2 is a

self-multiply matrix by property 1, therefore Z_m×n×k =_Zk−1

m×n×2. Here, we give a

method to solve the problem of 3-d pattern generation.

3

Transition Matrices and Spatial Entropy

3.1

Transition Matrices

Based on the process of the ordering matrix, we have to define transition matrix as the following :

1. Given an admissible set B ⊆ Σz3

2×2×2. 2.Define ₍ t_y;2×2×2;i = 1 , y2×2×2;i;j;k ∈ B t_y;2×2×2;i = 0 , y2×2×2;i;j;k <B 3. Define Tr y2;iy;iz;ix =_Ty 2;iy;ix;iz,where 1 ≤ iy; ix; iz ≤ 4.

4. The recursive formula for y-direction is as following:

Tr y;2×3×2=       T_y;2×2×2;1⊗ Ty;2×2×2;1 +_T y;2×2×2;2⊗ Ty;2×2×2;3 T_y;2×2×2;1⊗ Ty;2×2×2;2 +_T y;2×2×2;2⊗ Ty;2×2×2;4 T_y;2×2×2;3⊗ Ty;2×2×2;1 +_T y;2×2×2;4⊗ Ty;2×2×2;3 T_y;2×2×2;3⊗ Ty;2×2×2;2 +_T y;2×2×2;4⊗ Ty;2×2×2;4       2×2 = Ty;2×3×2;1 Ty;2×3×2;2 T_y;2×3×2;3 T_y;2×3×2;4 ! 2×2 ,

where T_y;2×3×2;l is 42× 42matrix, 1 ≤ l ≤ 4. T_y;2×n×2 =      T_{y;2×2×2;1⊗ Ty;2×n−1×2;1} +_T y;2×2×2;2⊗ Ty;2×n−1×2;3 T_{y;2×2×2;1⊗ Ty;2×n−1×2;2} +_T y;2×2×2;2⊗ Ty;2×n−1×2;4 T_{y;2×2×2;3⊗ Ty;2×n−1×2;1} +_T y;2×2×2;4⊗ Ty;2×n−1×2;3 T_{y;2×2×2;3⊗ Ty;2×n−1×2;2} +_T y;2×2×2;4⊗ Ty;2×n−1×2;4      2×2 = Ty;2×n×2;1 Ty;2×n×2;2 T_y;2×n×2;3 T_y;2×n×2;4 ! 2×2 ,

where Y_y;2×n×2;lis 4n−1_{× 4}n−1_{matrix, 1 ≤ l ≤ 4.}

5. Rearrangement. • (1.) Rearrange iz1 (T_x;2×n×2(riz2) )i_z1 = (Tx;2×n×2)ix1; iz1 • (2.) Rearrange iz2, for iz1 (T_x;2×n×2(riz2) )i_z1;iz2 = (Tx;2×n×2)i_z1;i_x1;i_z2 • (3.) Rearrange iz3, for iz1, iz2 (T_x;2×n×2(riz2 ;iz3))i_z1;i_z2;i_z3 = (T (r_iz2) x;2×n×2)iz1;iz2;ix1;ix2;iz3 .. . • (n-1.) Rearrange iz_n−1, for iz1, iz2,· · · , izn−2

6. For convenience, we need to define some forms of matrices as the fol-lowing: (1) [(Tr x;2×n×2)z]2n−1×2n−1 =       t1,1 t1,2 · · · t1,2n−1 t2,1 t2,2n−1 .. . . .. ... t2n−1_,1 · · · t₂n−1_,2n−1       2n−1_×2n−1

∀z, here the colors of the front and rear faces are represent as i2and i1respectively,

where 1 ≤ i1, i2≤ 2n−1. (2) [(Tr x;2×n×2)z;i1] (r_i2)₌       ti1,1 · · · ti1,2k2 .. . . .. ... ti 1,(2k1 −1)(2k2 )+1 · · · ti1,2k1+k2       2k1×2k2 , ∀z, i1,where    k1 = n−1₂ , k2= n−1₂ and n is odd. k1 = n₂, k2= n−2₂ and n is even.

(3) [(Tr x;2×n×2)z] (r_i2) ₌       [(Tr x;2×n×2)z;1] (r_i2) _[(Tr x;2×n×2)x;2×n×2] (r_i2) · · · [(Tr x;2×n×2)z;22 k2](r) .. . . .. ... h (Tr x;2×n×2)z;(2k1₋₁₎₍₂k2)+1 i(r_i2) · · · h(Tr x;2×n×2)z;2k1+k2 i(r_i2)       2k1×2k2 , ∀z ,where    k1= n−1₂ , k2 = n−1₂ and n is odd. k1= n−2₂ , k2 = n₂ and n is even.

7.Rearrange i1, and i2 as the define as following:

[(Tr x;2×n×2)

(r_i2)_](r_i1)_{is called rearrangement matrix of i}

1of (Tr_2;n)(ri2), if ([(Tr x;2×n×2) (r_i2)_](re_i1)₎ i1;z = [(T r x;2×n×2) (r_i2)_] z;i1, for 1 ≤ i1, izl ≤ 4,1 ≤ l ≤ n − 2.

8. Rearrange i2as the define as following:

([(Tr x;2×n×2)

(r_i2)_](re_i1)₎(re_i2)_{is called rearrangement matrix of i}

2 of [(Tr_x;2×n×2)(ri2)]re(i1) if ([(Tr_x;2×n×2)r](rei1₎₎(rei2) i1;i2;z = [(T r x;2×n×2) (r_i2)_]rei1 i1;z;i2 for 1 ≤ i1, i2, izl ≤ 4, 1 ≤ l ≤ n − 2, n ≥ 3. 9.

T(a)_x;2×n×2 = ([(Tr_2×n×2)(ri2)_]re(i1)₎re(i2) _{and T}(b)

x;2×n×2 = 2n−1 Σ i2=1 T(a)_x;2,n;i 1;i2 (b)

10. T_z;m×n×2 =2 n−1 Σ i1=1 T_x;m×n×2;i(b)

1 which record the colors of the z-direction as

iz1 → iz2 → . . . → izn−1 , by property 1. Tz;m×n×2is self-multiply, therefore Tz;m×n×k =T_z;m×n×2k−1 .

Theorem 3.1. Let λTz;m×n×2be the maximum eigenvalue of Tz;m×n×2, then

h(B) = lim

m,n→∞

log λTz;m×n×2 mn

Proof. By the same arguments as in [Chow et al., 1996a], the limit Eq. (1) is well-defined and exists.

From

T_z;m×n×k(B) =X(T_z;m×n×k)i, j

1≤i,j≤2(m−1)(n−1)

= |(T_z;m×n×2k−1 B)|

As in the one-dimensional case, lim

m→∞

log |(Tk−1 z;m×n×2)|

m = log λTz;m,n,k,

as for example[Ban Lin,2005]. Hence,

h(B) = lim m,n,k→∞ log |Tk−1 z;m×n×2| mnk = _lim m,n→∞ 1 mnlim k→∞ log |Tk−1 z;m×n×2| k = _lim m,n→∞ log λ_Tz;m×n×2 mn

3.2

Computation of λ

and entropy

Here, ρ(E) = 2, ρ(G) = g, where g = 1+

√ 5 2 . Let Tiy;i =E ⊗G and T r iy;i

=_{G ⊗E, then we will get the following formula:} 1. Tr_y;2×n×2;iy = 2(n−2)_{(G ⊗ E)}(n−2)_{⊗ (E ⊗ G).} 2. Tr_x;2×n×2 = Σ4 iy=1 Ty;n;iy = 2 n (G ⊗ E)(n−2) ⊗ (E ⊗ G). 3. Tr x;2×n×2 = 2 n_(G)n−2_{⊗ (E)}n−1_{⊗ G.} 4. (Tr x;2×n×2) (r_i2) _{= 2}n_[(G)n−2_{⊗ (E)}n−1_{⊗ G].} 5. [(T_x;2×n×2r )(ri2)_](rei1) =   

2n_[(E)n−1_{2 ⊗ (G)}n−2_{⊗ (E)}n−12 ⊗ (G)n−2] , n is odd.

2n_[E 2n−22 _×2n2 ⊗ (G) n−2_{⊗ E} 2n2_×2n−22 ⊗ G] , n is even. 6. ([(Tr x;2×n×2)

(r_i2)_](re_i1)₎(re_i2) _{= 2}n_[(E)n−1_{⊗ (G)}n−1_{], n ≥ 3.}

7. (T(a)_x;2×n×2)i1;i2 = 2 n_(G)n−1_. 8. (T(b)_x;m×n×2)i1 = 2n−1 Σ i2=1 (T_x;2×n×2(a) )i1;i2 = 2(m−1)(2n−1)(G)(m−1)(n−1).

10. T_z;m×n×k =_Tk−1

z;m×n×2 = 2(k−1)(m−1)n+m(n−1)G(m−1)(n−1)(k−1).

Now, we calculate the entropy ,

h(B) = lim m,n→∞ 1 mnlog(ρ(Tz;m×n×2)) = _lim m,n→∞ 1 mnlog(ρ(2(2mn−n−m)G(m−1)(n−1))) = _lim m,n→∞ 1 mnlog(2(2mn−n−m)( 1+√5 2 )(m−1)(n−1))) = _{2 log 2 + log g ≥ 0}

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