二維曲面上的花樣

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(1)國立交通大學應用數學系碩士論文. 二維曲面上的花樣 Patterns Generation in Two-Dimensional Surfaces. 研究生：胡文貴指導老師：林松山. 教授. 中華民國九十四年六月.

(2) 二維曲面上的花樣 Patterns Generation in Two-Dimensional Surfaces. 研究生：胡文貴. Student：Wen-Kuei Hu. 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 2005 Hsinchu, Taiwan, Republic of China. 中華民國九十四年六月.

(3) 二維曲面上的花樣學生：胡文貴. 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. 在這一篇碩士論文中，主要研究的是在二維曲面的花樣，這裡討論的二維曲面包含有圓柱(cylinder)、環面(torus)和球面 (sphere)。研究的主要目的是去找這些二維曲面上花樣生成的遞迴公式。. i.

(4) Patterns Generation in Two-Dimensional Surfaces student：Wen-Kuei Hu. Advisors：Dr. Song-Sun Lin. Department﹙Institute﹚of Applied Mathematics National Chiao Tung University. ABSTRACT. In this thesis, we study the patterns generation of two-dimensional surfaces. The surfaces include cylinder, torus and sphere. Our purpose is to find the recursive formulas for patterns generation in two-dimensional surfaces.. ii.

(5) 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、石至文老師及楊定揮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；接著感謝同窗好友佩君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、忠澤學長、耀漢學長以及學妹們，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，將是我永遠的回億。最後，要感謝我最敬愛的父母，感謝你們在我學習的路上給予我最大的支持以及自由的學習環境，讓我可以無後顧之憂的學習。再次感謝所有幫助過我、關心我的人，謝謝你們！. iii.

(6) 目中文提要英文提要誌謝目錄 1. 2. 2.1 2.2 2.3 2.4 3. 4. 4.1 4.2 4.3 5. References. 錄. ……………………………………………………………… ……………………………………………………………… ……………………………………………………………… ……………………………………………………………… Preliminaries ………………………………………… The cylindrical patterns ………………………… Ordering matrices …………………………………… The relation between C_{n}^{∗} and X_{n} ……… The transition matrices ………………………… Spatial entropy …………………………………… The toric patterns ………………………………… The spherical patterns …………………………… The upper and lower parts of the boundary condition The boundary with corners ………………………… The boundary without corners …………………… Conclusion …………………………………………… ………………………………………………………………. iv. i ii iii iv 1 5 6 8 10 11 14 16 17 20 23 17 28.

(7) 1. Preliminaries. Since this study bases on some part of [4], we use some de…nitions and some results in [4]. Therefore we list them in following. Let S be a …nite set of p elements (symbols, colors or letters of an alphabet). Where Zd denotes the integer lattice on Rd , and d 1 is a positive integer representing the lattice dimension. Then, function U : Zd ! S is called a global pattern. For each 2 Zd , we write U ( ) as u . The set of all patterns U : Zd ! S is denoted by d. d p. SZ ;. i.e., dp is the set of all patterns with p di¤erent colors in d-dimensional lattice. As for local patterns, i.e., functions de…ned on (…nite) sublattices, for a given d-tuple N = (N1 ; N2 ; ; Nd ) of positive integers, let ZN = f( be an N1 N2 fk N Nk for all 1 denoted by. 1;. 2;. ;. d). :1. Nk ; 1. k. k. dg. e Nd …nite rectangular lattice. Denoted by N N if k d. The set of all local patterns de…ned on ZN is N. fU jZN : U 2. N;p. d p g:. Under many circumstances, only a(proper) subset B of N is admissible (allowable or feasible). In this case, local patterns in B are called basic patterns and B is called the basic set. In a one dimensional case, S consists of letters of an alphabet, and B is also called a set of allowable words of length N. Consider a …xed …nite lattice ZN and a given basic set B N . For larger …nite lattice ZNe ZN , the set of all local patterns on ZNe which can be generated by B is denoted as Ne (B). Indeed, Ne (B) can be characterized by e (B) N. =f. 2 Zd with Z. U 2 Ne : U +N = VN f or any and some VN 2 Bg;. +N. where + N = f(. 1. +. 1;. ;. d. 1. +. d). :(. 1;. ;. d). 2 N g;. ZNe.

(8) and U. +N. = VN means u. = v f or each. +. 2 ZN :. Similarly, the set of all global patterns which can be generated by B is denoted by (B) = fU 2. d p. :U. +N. 2 Zd with some VN 2 Bg:. = VN f or any. For clarity, we begin by the studying two symbols, i.e., S = f0; 1g. On a …xed …nite lattice Zm1 m2 , we …rst give a ordering = m1 m2 on Zm1 m2 by (( 1 ; 2 )) = m2 ( 1 1) + 2 ; i.e., m2 .. .. 2m2 .. .. 1. m2 + 1. m1 m2 .. .. .. . (m1. 1)m2 + 1. The ordering of (2.1) on Zm1 m2 can now be passed to for each U = (u 1 ; 2 ) 2 m1 m2 , de…ne (U ). m1 m2. . Indeed,. m1 m2 (U ). = 1+. m1 P m2 P 1 =1. All patterns in. :. 2 n. u. 1 2. 2m2 (m1. 1 )+(m2. 2). :. 2 =1. can be arranged by the ordering matrix Xn =. xn;i1 i2. a 2n 2n matrix with entry xn;i1 i2 = xn;i1 (U2 ) = i2 , 1 i1 ; i2 2n .. ; xn;i2 , where. (U1 ) = i1 and. Theorem For any n 2, 2 n = fyj1 jn g, where yj1 jn is given in (2.26). Furthermore, the ordering matrix Xn can be decomposed by n Zmaps successively as 2 3 Yn;1 Yn;2 Xn = 4 Yn;3 Yn;4 5 ; 2.

(9) Yn;j1 for 1. k. n. 2. Yn;j1 4 = Yn;j1. jk. 2, and Yn;j1. jn. 1. 2. yj1 4 = yj1. Yn;j1 Yn;j1. jk 1 jk 3. jn. 11. jn. 13. yj1 yj1. 3. jk 2. 5;. jk 4. jn. 12. jn. 14. 3. 5:. From the proof of the above Theorem, we have in+1;1 = 2in;1. 1+[. jn+1 2. and 1 + fjn+1. in+1;2 = 2in;2. 1. 2[. 1. ];. jn+1 2. 1. ]g:. De…ne vj1 j2. jn. = vj1 j2 vj2 j3. vjn. 1 jn. ;. and Hn = [vj1 j2. jn ];. then the transition matrix Hn for B de…ned on Z2 n is a 2n 2n matrix with entries vj1 jn , which are either 1 or 0, by substituting yj1 jn by vj1 jn in Xn .. For any two matrices A = (aij ) and B = (bkl ), the Kronecker product (tensor product) of A B is de…ned by A On the other hand, for any two n. B = (aij B): n matrices. C = (cij ) and D = (dij );. 3.

(10) where cij and dij are numbers or matrices. Then, Hadamard product of C D is de…ned by C D = (cij dij ); where the product cij dij of cij and dij may be multiplication of numbers, numbers and matrices or matrices whenever it is well-de…ned. For instance, cij is number and dij is matrix.. Let H2 be a transition matrix. Then, for higher order transition matrices Hn , n 3, we have the following three equivalent expressions (I) Hn can be decomposed into n successive 2 2matrices (or n-successive Z-maps) as follows: 2 3 Hn;1 Hn;2 Hn = 4 Hn;3 Hn;4 5 ; Hn;j1 for 1. k. n. 2 and Hn;j1. Furthermore, Hn;k (II) Starting from. with. jk. jn. 2. Hn;j1 = 4 Hn;j1. 1. 2. vj1 4 = vj1. 2. vk1 Hn 4 = vk3 Hn. Hn;j1 Hn;j1. jk 1 jk 3. jn. 11. jn. 13. 1;1 1;3. vj1 vj1. vk2 Hn vk4 Hn. 0. 1 H1 H2 H2 = @ H3 H4 A ;. 1 vk1 vk2 Hk = @ vk3 vk4 A ; 0. 4. jk 2 jk 4. jn. 12. jn. 14. 1;2 1;4. 3. 5;. 3. 3. 5:. 5:.

(11) Hn can be obtained from Hn (3.14).. 1. (III) Hn = (Hn 1 )2n where E2k is the 2k. 2. 1. 2n. 1. by replacing Hk by Hk. 0. B E2n B @. H2 according to. 0. 1 1 H1 H2 @ H3 H4 A C C; A. 2. 2k matrix with 1 as its entries.. The cylindrical patterns. In geometry, we can construct a cylinder by pasting one pair of subtenses of a rectangle. So, consider the cylindrical patterns as the …nite two-dimensional patterns which have a periodic boundary condition in one direction. For clarity, we study two symbols, i.e., S = f0; 1g. According to above, de…ne Xc. m1 m2. =. U c = (u. 1. 2. )2. X. m1 (m2 +1). j ui;1 = ui;(m2 +1) ; 1. i. m1. to represent the …nite two-dimensional patterns which have periodic boundary condition in vertical direction. (We only study the two-dimensional patterns which have periodic boundary condition in vertical direction, since the two-dimensional patterns which have periodic boundary condition in horizontalPdirection is similar) We …rst de…ne an ordering of patterns for cm1 m2 as lexicographical ordering in one-dimensional case. On a …xed …nite lattice Zm1 (m2 +1) , we …rst give a ordering c = cm1 m2 on Zm1 (m2 +1) by c (( 1 ; 2 )) = (m2 + 1) ( 1 1) + 2 , i.e., m2 + 1 .. .. 2(m2 + 1) .. .. 1. (m2 + 1) + 1. m1 (m2 + 1) .. .. .. . (m1 5. 1)(m2 + 1) + 1. :.

(12) The ordering c onPZm1 each U c = (u 1 ; 2 ) 2 cm1 c. (U c ) =. c m1. (m2 +1) can. now be passed to. m2 ,. de…ne X c m2 (U ) = 1 + 1. 1. X. m1 1. u. 1; 2. Pc. m1 m2 .. 2m2 (m1. Indeed, for. 1 )+(m2. 2). :. m2. 2. Obviously, there is an one-to-one correspondence between local patterns P in cm1 m2 and positive integers in the set N2m1 m2 = fk 2 N j 1 k 2m1 m2 g.. 2.1. Ordering matrices. For 1 (n + 1) pattern U c = (uk ) ; 1 is assigned the number i=. c. (U c ) = 1 +. n + 1 in. k X. uk 2(n. k). Xc. 1 m2. ,as above, U c. :. 1 k n. As denoted by the 1 (n + 1)column pattern xcn;i , 2 3 un+1 un+1 6 .. 7 .. c xn;i = 4 . ,where u1 = un+1 . 5 or . u1 u1. xci = xc2;i , i = 1 + 2u1 + u2 and. In particular, when n = 2, as denoted by 2 3 u3 xci = 4 u2 5 or u1. u3 u2 u1. .. A 2 3 pattern U c = (u 1;P2 ) can now be obtained by a horizontal direct sum of two 1 3 patterns in c2 2 , i.e., xci1 ;i2. xci1. xci2. 2 where. 3 u1;3 u2;3 4 u1;2 u2;2 5 or u1;1 u2;1 ik = 1 + 2uk;1 + uk;2 ; 6. 1. u1;3 u2;3 u1;2 u2;2 u1;1 u2;1 k. 2:. ,.

(13) P Therefore, the complete set of all 16(= 22 (3 1) ) 2 3 patterns in c2 2 can be listed by a 4 4 matrix C2 = [xci1 ;i2 ] with 2 3 pattern xci1 ;i2 as its entries, i.e.,. 7.

(14) 2 2. 3 0 0 4 0 5 B B B 0 B B 2 3 B B 0 B 4 1 5 B B B 0 B B 2 3 B B 1 B 4 0 5 B B B 1 B B 2 3 B B 1 B 4 1 5 @ 1. 2. 3 0 4 0 5 0. 2. 3 0 4 1 5 0. 2. 3 1 4 0 5 1. 2. 3 1 4 1 5 1. 3 2 3 2 3 2 3 1 0 0 0 0 0 1 0 1 4 0 0 5 4 0 1 5 4 0 0 5 4 0 1 5 C C 0 0 0 0 0 1 0 1 C C C 2 3 2 3 2 3 2 3 C 0 0 0 0 0 1 0 1 C C 4 1 0 5 4 1 1 5 4 1 0 5 4 1 1 5 C C 0 0 0 0 0 1 0 1 C C C 2 3 2 3 2 3 2 3 C 1 0 1 0 1 1 1 1 C C 4 0 0 5 4 0 1 5 4 0 0 5 4 0 1 5 C C 1 0 1 0 1 1 1 1 C C C 2 3 2 3 2 3 2 3 C 1 1 1 1 C 1 0 1 0 C 4 1 0 5 4 1 1 5 4 1 0 5 4 1 1 5 A 1 0 1 0 1 1 1 1. c It is easy to verify xci1 ;i2 = 4 (i1 1) + i2 , i.e., we are counting Pc that table. local patterns in 2 2 by going through each row successively in aboveP c Correspondingly, C2 canPbe referred to as an ordering matrix for 2 2. c Similarly, all patterns in 2 n can be arranged by the ordering matrix Cn = xcn;i1 ;i2 a 2n 2n matrix with entry xcn;i1 ;i2 = xcn;i1 xcn;i2 , where c (U1c ) = i1 and c (U2c ) = i2 , 1 i1 ; i2 2n .. 2.2. The relation between Cn and Xn. The de…nitions of Xn and are P according to the Preliminaries. We consider U c = (u 1 ; 2 ) 2 cm1 m2 =) ui;1 = ui;m2 +1 ; 1 i m1 ;then ^ P 1 we can decompose U c as U 1 U 2 whereU 1 = (u 1 ; 2 )11 12 m m1 m2 and m2 2 P 1 m 1 U 2 = (u 1 ; 2 )m2 1 2 m 2 m1 2 ; 2 +1. 8.

(15) i.e., 2 6 6 6 6 6 4. u1;1 u2;1 u1;m2 u2;m2 .. .. . . u1;2 u2;2 u1;1 u2;1. um1 ;1 um1 ;m2 .. . um1 ;2 um1 ;1. 3. 2 u1;m2 u2;m2 7 7 6 .. .. 6 . 7 . = 6 7 7 4 u1;2 u2;2 5 u1;1 u2;1. 3 um1 ;m2 7 .. 7 . 7 um1 ;2 5 um1 ;1. u1;1 u2;1 u1;m2 u2;m2. ^. um1 ;1 um1 ;m2. Therefore, we get c (U c ) = (U 1 ) and is an one-to-one correspondence c c 1 between U c and U 1 , i.e., we can determine the 2 (U ) by 3 (U ). u1;1 u2;1 6 u1;n u2;n 7 6 7 Xc 6 .. 7 . c . In particular, consider the pattern U2 n = 6 . 2 , then . 7 v;2 n 6 7 4 u1;2 u2;2 5 u1;1 u2;1 3 2 u1;n u2;n 6 .. .. 7 P ^ 6 . . 7 1 2 1 c decompoes U2 n into U2 n U2 n where U2 n = 6 7 2 2 n and 4 u1;2 u2;2 5 u1;1 u2;1 P u1;1 u2;1 U22 n = 2 2 2: u1;n u2;n Suppose U2c. U2c. 2. n;1. u1;1 u1;n .. .. 6 6 6 =6 6 4 u1;2 u1;1. n. = U2c. n;1. 3. 7 7 7 c 7 ; U2 7 5. U2c. 2. n;2. n;2. u2;1 u2;n .. .. 6 6 6 =6 6 4 u2;2 u2;1. and U21. n. 3. 7 7 7 1 7 ; U2 7 5. And suppose c (U2c n;1 ) = i1 and and (U21 n;2 ) = i2 (By the de…nition of and U21 n = xn;i1 ;i2 :. 9. n;1. c. = U21. U21. n;2. where. 3 u1;n 6 .. 7 6 7 = 6 . 7 and U21 4 u1;2 5 u1;1. (U2c and. 2. n;1. n;2. 2. 3 u2;n 6 .. 7 7 6 = 6 . 7: 4 u2;2 5 u2;1. = i2 , then (U21 n;1 ) = i1 ).Therefore, U2c n = xcn;i1 ;i2 ;. n;2 ) c.

(16) According to Preliminaries, we represent xn;i1 ;i2 as yj1 ;j2 :::jn where xn;i1 ;i2 and yj1 ;j2 :::jn de…ned in [4]. Since the top and bottom layers are the same, then ^ ^ xcn;i1 ;i2 = xn;i1 ;i2 yjn ;j1 = yj1 ;j2 :::jn yjn ;j1 = yj1 ;j2 :::jn ;j1 : By the Theorem in the Preliminaries and the argument as above, we can get the following Theorem easily. P Theorem 1 For any n 2, c2 n = fyj1 jn ;j1 g. Furthermore, the ordering matrix Cn can be decomposed by n Z-maps successively as 2 c 3 c Yn;1 Yn;2 c c 5 Yn;4 Cn = 4 Yn;3 ; c Yn;j 1. for 1. k. n. 2, and c Yn;j 1. 2.3. jk. jn. 1. 2. c Yn;j 1 c = 4 Yn;j 1. 2. yj1 = 4 yj1. jk ;1 jk ;3. jn. 1 ;1;j1. jn. 1 ;3;j1. c Yn;j 1 c Yn;j 1. yj1 yj1. jk ;2 jk ;4. 3. 5;. jn. 1 ;2;j1. jn. 1 ;4;j1. 3. 5:. The transition matrices. P This part derives the transition matrix C for a given basic set B n 2 2: P Given a basic set B , the transition matrix C can be de…ned by n 2 2 n n Cn = [cn;i1 ;i2 ]1 i1 ;i2 2n , the 2 2 matrix with entries either 0 or 1, according to the following rules: Since Cn = [xcn;i1 ;i2 ] and xcn;i1 ;i2 = yj1 ;j2 :::jn ;j1 = yj1 ;j2. ^. yj2 ;j3. ^. yjn. 1 ;jn. ^. yjn ;j1 ;. where j1 jn are determined uniquely by i1 and i2 .Then 8 < cn;i1 ;i2 = 1 if all yj1 ;j2 , yjn 1 ;jn and yjn ;j1 belong to B = 0 otherwise. : 10.

(17) Therefore, Cn also canP be represented by Cn = [vj1 j2 jn ;j1 ], and the trann sition matrix Cn for B 2n matrix with 2 2 de…ned on Z2 (n+1) is a 2 entries vj1 jn ;j1 , which are either 1 or 0, by substituting yj1 jn ;j1 by vj1 jn ;j1 in Cn : " # e2;1 Ve2;2 V e2;i = v1i v2i 1 De…nition 2 Ve2 = e where V i 4, and v3i v4i V2;3 Ve2;4 e1 =. 1 0 0 0. ; e2 =. 0 1 0 0. P. Theorem 3 Cn = Hn. 0 0 1 0. ; e3 =. ei. E2n. and e4 =. 2. 1 i 4. Proof. Since Cn = [cn;i1 ;i2 ]1 i1 ;i2 2n = [vj1 j2 Observe the Theorem1.1, we can get j1 j1 j1 j1. = 1; = 2; = 3; = 4;. if if if if. Ve2;i. jn ;j1 ]. = [vj1 j2. :. jn vjn j1 ] :. 1 i1 ; i2 2n 1 1 i1 2n 1 ; 2n 1 i2 2n 2n 1 i1 2n ; 1 i2 2n 1 2n 1 i1 ; i2 2n :. and the jn in the Cn are 2 1 2 6 3 4 6 6 .. 6 . 6 4 1 2 3 4. 3. 1 2 3 4 .. .. .. .. 7 7 7 7 7 5. 1 2 3 4. 2n 2 n. Then the result is easily to prove.. 2.4. 0 0 0 1. Spatial entropy. First, we save the all components of cm n;i;j 1 i;j 2n ; and the entries cm. n;i;j. = xcn;i;k1 ;k2. km. 2 ;j. j1 11. Pc. m n. in a 2n. k1 ; k2. ; km. 2n matrix Cm. 2. 2n. n. =.

(18) are subsets of. Pc. m n. :We know. [. 1 i;j. disjoint to each other.. 2n. cm. n;i;j. =. Pc. m n. and cm. n;i;j. are. P n Given a basic set B 2n matrix Cm n can be de…ned 2 2 ; the 2 P Cm n = [cm n;i;j ]1 i;j 2n where cm n;i;j = card cm n;i;j \ cm n (B) :Then P Pc cm n;i;j = cm n (B) where cm n (B) = card m n (B) :By the con1 i;j 2n P struction of Cn = [cn;i1 ;i2 ]1 i1 ;i2 2n for the same B 2 2 , we can get cm. n;i;j. =. 1 k1 ;k2. X. ;km. cn;i;k1 cn;k1 ;k2 : : : cn;km. 2. In particular, when m = 2, cn;i;j = c2 Cn = C2. Theorem 4 Cm. n. = (Cn )m. 1. for Pm. 1 i;j 2n. 2 ;j. 2n. n;i;j. for 1. i; j. 2n , i.e.,. n:. 2. Furthermore, (Cn )m 1 i;j = cm n (B):. Proof. We prove this by induction. (1). When m = 2, Cn = C2 n , clearly. (2). Suppose, when m = l , Cl n = (Cn )l 1 holds, i.e., h i cl n;i;j = (Cn )l 1 ; 1 i; j 2n i;j. (3). When m = l + 1,. 12.

(19) c(l+1). n;i;j. = 1 k1 ;k2. P. =. 1 kl. 1 kl. 1 kl. P. h. 1. P. =. Since. 1. P. =. =. P. 1. (Cn )l. cm. ;kl. 2n. ". cn;i;k1 cn;k1 ;k2 : : : cn;kl. 1 ;j. 1 k1 ;k2. h. 2n. n;i;j. cn;kl. cn;kl. 2n. i. 2n. 1. 1 ;j. cl. (Cn )l. ; f or1. n;i;kl. 1. i. i;kl. P. ;kl. 1 ;j. 2n. 2. cn;i;k1 cn;k1 ;k2 : : : cn;kl. 2 ;kl 1. !#. 1. cn;kl. 1 ;j. 1. i; j. 2n :. i;j. =. 1 i;j 2n. c m n. (B) and Cn = C2 P (Cn )m 1. 1 i;j. The proof is complet.. 2n. n,. i;j. then =. c m n. (B) :. P The spatial entropy hc (B) of c (B) is de…ned as follows : P P Let cm n (B) = card( cm n (B));the number of distinct patterns in cm The spatial entropy hc (B) is de…ned as 1 log m;n!1 mn. hc (B) = lim sup. n (B).. c m n (B):. As for spatial entropy hc (B), we have the following theorem.. P c Theorem 5 Given a basic set B 2 2 , let n be the largest eigenvalue of the associated transition matrix Cn which is de…ned above. Then, hc (B) = lim sup n!1. 13. log n. c n. :.

(20) Proof.. From the construction of Cn , we know that for m P c (Cn )m 1 i;j m n (B) =. 2,. 1 i;j 2n. #(Cnm 1 ): As in a one dimensional case, we have log #(Cnm 1 ) = log m!1 m lim. c n:. Therefore, 1 log m;n!1 mn. hc (B) = lim sup. c m n (B). log cm n (B) 1 = lim sup ( lim ) m n!1 n m!1 = lim sup n!1. log n. c n. :. The proof is complete. Remark 6 Similarly,the two-dimensional patterns which have periodic boundary condition in horizontal direction have the same arguments.. 3. The toric patterns. In geometry, we can construct a torus by pasting the two pair of subtenses of a rectangle. So, consider the toric patterns as the …nite two-dimensional patterns which have double-periodic boundary condition in vertical and horizontal directions. For clarity, we study two symbols, i.e., S = f0; 1g. According to above, de…ne n Pt P U t = (u 1 ; 2 ) 2 (m1 +1) (m2 +1) j ui;1 = ui;(m2 +1) ; u1;j = u(m1 +1);j ; m1 m2 = f or 1 i m1 + 1; 1 j m2 + 1g 14.

(21) to represent the …nite two-dimensional patterns which have double-periodic boundary condition . h We can i save the all components of d^m n;i;j and the entries 1 i;j 2n ;. d^m. = xcn;i;k1 ;k2 km Pt are subsets of m n : We know n;i;j. 2 ;j;i. Pt. j1. [. 1 i;j 2n. disjoint to each other.. P. m n. k1 ; k2 d^m. ^m 2n matrix D. in a 2n. n;i;j. =. ; km Pt. n. =. 2n. 2. and d^m. m n. n;i;j. are. 2n matrix Dm n can be deP …ned Dm n = [dm n;i;j ]1 i;j 2n where dm n;i;j = card d^m n;i;j \ tm n (B) ; P Pt dm n;i;j = tm n (B) where tm n (B) = card then m n (B) : Given a basic set B. the 2n. 2 2;. 1 i;j 2n. By the construction of Cn = [cn;i1 ;i2 ]1 i1 ;i2 2n for the same B above, we can get X cn;i;k1 cn;k1 ;k2 : : : cn;km 2 ;j cn;j;i : dm n;i;j = 1 k1 ;k2. ;km. 2. P. 2 2. as. 2n. T Theorem 7 Dm n = (Cn )hm 1 (Cn )T ; where i (Cn ) means the transpose of P Cn , for m 2,and (Cn )m 1 (Cn )T = tm n (B) : i;j. 1 i;j 2n. Proof.. dm. n;i;j. = 1 k1 ;k2. P. ;km. = 1 k1 ;k2. = cm =. P. 2. ;km. 2n. 2. cn;i;k1 cn;k1 ;k2 : : : cn;km. 2n. 2 ;j. cn;i;k1 cn;k1 ;k2 : : : cn;km. 2 ;j. n;i;j cn;j;i. (C2. m 1 n) i;j. cn;j;i ; f or 1 15. i; j. cn;j;i. 2n :. !. cn;j;i.

(22) Then, Since. P. 1 i;j. Dm dm 2n. n;i;j. P. 1 i;j. = 2n. n. = (Cn )m. 1. (Cn )T :. t m n. (B) ;we can get h i (Cn )m 1 (Cn )T =. t m n. i;j. (B) :. P The spatial entropy ht (B) of t (B) is de…ned as follows : P P Let tm n (B) = card( tm n (B));the number of distinct patterns in tm The spatial entropy ht (B) is de…ned as 1 log m;n!1 mn. ht (B) = lim sup. n (B).. t m n (B):. Then, X h 1 log( (Cn )m m;n!1 mn 1 i;j 2n. ht (B) = lim sup. 4. 1. (Cn )T. i. ). i;j. The spherical patterns. In geometry, we can suppose that a ball constructed by pasting the four edges of a rectangle into a point. So , consider the spherical patterns as the …nite two-dimensional patterns whose boundaries’ symbols are all the same. In here, we consider two kinds of the spherical patterns: one kind of the spherical patterns is the …nite two-dimensional patterns whose boundaries with four corners(e.x. Fig.1),and the other kind of the spherical patterns is the …nite two-dimensional patterns whose boundaries without four corners(e.x. Fig.2).For charity, we study two symbols , i.e, S = f0; 1g :. 16.

(23) Fig.1 Boundary with four corners (the …rst kind of the spherical patterns). 4.1. Fig.2 Boundary without four corners (the second kind of the spherical patterns). The upper and lower parts of the boundary condition. Before we study the spherical patterns, we consider the top and bottom layers of boundaries …rstly. De…nition 8 n Pc0 = u m1 m2 n Pc1 m1 m2 = u. 1; 2. ju. 1; 2. ju. 1; 2. 2. 1; 2. 2. P. m1. (m2 +2) ; ui;1 = ui;m2 +2 = 0; 1. i. m1. m1. (m2 +2) ; ui;1 = ui;m2 +2 = 1; 1. i. m1. P. o. o. Observe the matrix Cn+1 = xcn+1;i1 ;i2 1 i ;i 2n+1 : By the Preliminaries 1 2 and the relation between Cn+1 and Xn+1 , we can get 1. i1 ; i2. 2n + 1. i1 ; i2. 2n. , j1 = 1 () the top and bottom layers are 0 0. 2n+1 , j4 = 4 () the top and bottom layers are 1 1. Therefore,. X c0. 2 n. and. X c1. 2 n. Furthermore,. X c0. m n. = xcn+1;i1 ;i2 j 1. = xcn+1;i1 ;i2 j 2n + 1 = xcn+1;i1 ;i2. im. j1. 17. i1 ; i2 i1 ; i2. i1 ; i2. 2n 2n+1 :. im. 2n.

(24) and. X c1. m n. We can save entries. Pc0. m n. = xcn+1;i1 ;i2. in a 2n. j1. i1 ; i2. 0 2n matrix C^m. = xcn+1;i;k1 ;k2 km Pc0 are subsets of m n : We know c^0m. im. n;i;j. 2 ;j. j1. [. 1 i;j 2n. disjoint to each other.. n. = c^0m. k1 ; k2 c^0m. 2n. im. n;i;j. =. n;i;j 1 i;j 2n ,. ; km Pc0. 2. m n. and the. 2n and c^0m. n;i;j. are. P n 0 Given a basic set B 2n matrix Cm can be de…ned n 2 2 ; the 2 Pc0 0 0 Cm n = [c0;m n;i;j ]1 i;j 2n where c0;m n;i;j = card c^m n;i;j \ m n (B) :Then P Pc0 c0;m n;i;j = cm0 n (B) where cm0 n (B) = card m n (B) :. 1 i;j 2n. P By the construction of Cn+1 = [cn+1;i1 ;i2 ]1 2 2 and the above arguments, we can get c0;m. n;i;j. =. 1 k1 ;k2. X. ;km. i1 ;i2 2n+1. for the same B. cn+1;i;k1 cn+1;k1 ;k2 : : : cn+1;km. 2. 2 ;j. 2n. P 1 Similarly,we can save cm1 n in a 2n 2n matrix C^m ^1m n;i;j 1 i;j 2n , n = c and the entries n o 1 c n n+1 c^m n;i;j = xn+1;i+2n ;k1 ;k2 km 2 ;j+2n j 2 + 1 k1 ; k2 ; km 2 2 are Pc1 Pc1 1 1 subsets of [ n c^m n;i;j = m n and c^m n;i;j are disjoint m n : We know 1 i;j 2. to each other.. P n 1 Given a basic set B 2n matrix Cm n can be de…ned 2 2 ; the 2 P 1 1 Cm n = [c1;m n;i;j ]1 i;j 2n wherec1;m n;i;j = card c^m n;i;j \ cm1 n (B) :Then Pc1 P 1 c1;m n;i;j = cm1 n (B) where c;m n (B) = card m n (B) : 1 i;j 2n. 18.

(25) P By the construction of Cn+1 = [cn+1;i1 ;i2 ]1 2 2 and the above arguments, we can get c1;m. n;i;j. X. =. 2n +1. k1 ;k2. ;km. 2. n;i;j. = cn+1;i;j and c1;2. Cn;1 Cn;2 Cn;3 Cn;4. Therefore, C20. = Cn+1;1 and C21. n. 1 i;j 2n. 1 i;j. n;i;j. n. = cn+1;i+2n ;j+2n ,. = Cn+1;4 :. n. m 1 = (Cn+1;1 )m 1 ;and Cm n = (Cn+1;4 ) (Cn+1;1 )m 1 i;j = cm0 n (B) ; and. X. 2 ;j+2. 2n+1. De…nition 9 Cn =. 0 Theorem 10 Cm P Furthermore,. for the same B. cn+1;i+2n ;k1 cn+1;k1 ;k2 : : : cn+1;km. In particular m = 2, c0;2 1 i; j 2n :. n. i1 ;i2 2n+1. (Cn+1;4 )m. 1 i;j. =. c1 m n (B). f orm. 1. for m. 2:. 2:. 2n. Proof. We know c0;m. n;i;j. = 1 k1 ;k2. = 1 k1 ;k2. and c1;m. n;i;j. P. ;km. 2. 2n. ;km. 2. 2n. P. P. =. 2n +1 k1 ;k2. = 1 k1 ;k2. ;km. 0 We prove the part Cm. n. 2 ;j. cn+1;1;i;k1 cn+1;1;k1 ;k2 : : : cn+1;1;km. cn+1;i+2n ;k1 cn+1;k1 ;k2 : : : cn+1;km. ;km. P. cn+1;i;k1 cn+1;k1 ;k2 : : : cn+1;km. 2. 2. 2n. 2n+1. cn+1;4;i;k1 cn+1;4;k1 ;k2 : : : cn+1;4;km. = (Cn+1;1 )m. 1. by induction. m 1 1 is similar.) (The other part Cm n = (Cn+1;4 ) 0 (1). When m = 2, C2 n = Cn+1;1 , clearly.. 19. 2 ;j. 2 ;j. 2 ;j+2. n.

(26) (2). Suppose, when m = l , Cl0 n = (Cn+1;1 )l 1 holds, i.e., h i c0;l n;i;j = (Cn+1;1 )l 1 ; 1 i; j 2n i;j. (3). When m = l + 1,. c0;(l+1). n;i;j. = 1 k1 ;k2. =. P. 1 kl. 1 kl. P. =. 1 kl. = Since. P. h. 1. 1. l. n;i;j. 2n , then. 2n. cn+1;1;i;k1 cn+1;1;k1 ;k2 : : : cn+1;1;kl. cn+1;1;kl. =. i. h. 4.2. 1 ;j. c0;l. (Cn+1;1 )l. ; for 1. 1. i. n;i;kl. i;kl. i; j. P. ;kl. 2. 2n. 1 ;j. cn+1;1;i;k1 cn+1;1;k1 ;k2 : : : cn+1;1;kl. 2 ;kl 1. 1. cn+1;1;kl. 1 ;j. 1. 2n :. i;j c0 m n. (B) and c0;m. P. 1 i;j. The proof is complet.. 1 ;j. 1 k1 ;k2. 2n. (Cn+1;1 ). c0;m. 1. cn+1;1;kl. 2n. 1 i;j 2n. i; j. ;kl. 2n. 1. P. =. P. n;i;j. (Cn+1;1 )m 2n. = (Cn+1;1 )m. 1 i;j. =. c0 m n. 1 i;j. ;1. (B). The boundary with corners. Now, we add the left and right layers of boundaries for the …rst kind of the spherical patterns(i.e, whose boundaries contain four corners), we do some de…nition as following: n PS0 P = u 1 ; 2 j u 1 ; 2 2 (m1 +2) (m2 +2) ; ui;1 = ui;m2 +2 = u1;j = m1 m2 um1 +2;j = 0; 1 i m1 + 2; 1 j m2 + 2g and 20.

(27) PS1. m1 m2. n P = u 1 ; 2 j u 1 ; 2 2 (m1 +2) (m2 +2) ; ui;1 = ui;m2 +2 = u1;j = um1 +2;j = 1; 1 i m1 + 2; 1 j m2 + 2g. And then, we can save and the entries s^0m. n;i;j. are subsets of. PS0. m n. = xcn+1;1;i;k1 ;k2 PS0. m n. 0 in a 2n 2n matrix S^m. km. : We know. disjoint to each other... 2 ;j;1. [. 1 i;j. j1 2n. s^0m. k1 ; k2 =. n;i;j. n. = s^0m. ; km PS0. 2. m n. n;i;j 1 i;j 2n ,. 2n. and s^0m. n;i;j. P. n 0 2n matrix Sm n can be 2 2 ; the 2 PS0 0 0 0 0 Sm n = sm n;i;j 1 i;j 2n wheresm n;i;j = card s^m n;i;j \ m n (B) P 0 Ps0 sm n;i;j = sm0 n (B) where sm0 n (B) = card m n (B) : 1 i;j 2n. Given a basic set B. P By the construction of Cn+1;1 = [cn+1;1;i1 ;i2 ]1 2 2 and the above arguments, we can get s0m. n;i;j. =. 1 k1 ;k2. X. = ( 1 k1 ;k2. ;km. X. ;km. i1 ;i2 2n+1. are. de…ned :Then. for the same B. cn+1;1;1;i cn+1;1;i;k1 cn+1;1;k1 ;k2 : : : cn+1;1;km. 2 ;j. cn+1;1;j;1. 2n. 2. cn+1;1;i;k1 cn+1;1;k1 ;k2 : : : cn+1;1;km 2. 2 ;j. 2n. = c0;m n;i;j cn+1;1;1;i cn+1;1;j;1 = (Cn+1;1 )m 1 i;j cn+1;1;1;i cn+1;1;j;1 for 1. 21. i; j. 2n. )cn+1;1;1;i cn+1;1;j;1.

(28) i.e.,. 0 Sm n. 2 6 6 6 4. m 1. = (Cn+1;1 ). 2 6 6 6 4. = (Cn+1;1 ). 1 Sm. n. = (Cn+1;4 )m. 1. n;i;j. 2 6 6 6 4. 2. m 1. = (Cn+1;4 ) P 0 Since sm n;i;j 1 i;j 2n. get. P. 6 6 6 4. 1 i;j. cn+1;1;2n ;1 cn+1;1;2n ;1 .. .. .. .. cn+1;1;2n ;1 2n. 3. 7 7 7 5. 7 7 7 5. cn+1;4;2n ;i cn+1;4;j;2n for 1. cn+1;4;2n ;1 cn+1;4;2n ;2 .. .. cn+1;4;1;2n cn+1;4;2;2n cn+1;4;1;2n cn+1;4;2;2n .. .. . .. cn+1;4;2n ;1 cn+1;4;2n ;2 .. .. .. .. cn+1;4;2n ;2n. .. .. cn+1;4;2n ;2n cn+1;4;2n ;2n .. .. 3. 3 7 7 7 5. 7 7 7 5. cn+1;4;1;2n cn+1;4;2;2n cn+1;4;2n ;2n [cn+1;4;2n ;i cn+1;4;j;2n ]1 i;j 2n P 1 = sm0 n (B) and sm n;i;j = sm1 n (B) ;we can 1 i;j 2n. (Cn+1;1 )m. 1. (Cn+1;4 )m. 1. Remark 11 If we de…ne Hn = i. = (Cn+1;4 )m. cn+1;1;1;2n. cn+1;4;2n ;2n cn+1;4;2n ;2n. 1 i;j 2n. for 1. .. .. cn+1;1;1;1 cn+1;1;2;1 cn+1;1;1;1 cn+1;1;2;1 .. .. . .. 3. cn+1;1;1;1 cn+1;1;1;2 .. .. cn+1;1;1;2n cn+1;1;1;2n. cn+1;4;2n ;1 cn+1;4;2n ;2 .. .. 1 i;j 2n. P. cn+1;1;1;1 cn+1;1;1;2 .. .. cn+1;1;1;1 cn+1;1;2;1 [cn+1;1;1;i cn+1;1;j;1 ]1 i;j. m 1. Similarly, we can get s1m i; j 2n ;i.e.,. cn+1;1;1;1 cn+1;1;1;2 .. .. i;j. cn+1;1;1;i cn+1;1;j;1 =. i;j. cn+1;4;2n ;i cn+1;4;j;2n = Hn;1 Hn;2 Hn;3 Hn;4. 4: 22. s0 m n. (B). s1 m n. (B) :. ; then Cn;i = Hn;i. E2n. 2. Ve2;i ;.

(29) Ps0 s0 The spatial entropy h (B) of (B) is de…ned as follows : P Ps0 s0 Let m n (B) = card( m n (B));the number of distinct patterns in sm0 The spatial entropy hs0 (B) is de…ned as 1 log m;n!1 mn. hs0 (B) = lim sup. n (B).. s0 m n (B):. Then, X 1 log( (Cn+1;1 )m m;n!1 mn 1 i;j 2n. hs0 (B) = lim sup. 1 i;j. cn+1;1;1;i cn+1;1;j;1 ).. P Similarly, the spatial entropy hs1 (B) of s1 (B) is de…ned as follows P Ps1 : s1 s1 Let m n (B) = card( m n (B));the number of distinct patterns in m n (B). The spatial entropy hs1 (B) is de…ned as 1 log m;n!1 mn. hs1 (B) = lim sup. s1 m n (B):. Then, 1 log( (Cn+1;4 )m mn m;n!1. hs1 (B) = lim sup. 4.3. 1 i;j. cn+1;1;2n ;i cn+1;1;j;2n ).. The boundary without corners. Now, we add the left and right layers of boundaries for the second kind of the spherical patterns (i.e, whose boundaries do NOT contain four corners), we do some de…nition as following:. 23.

(30) Let U 1 = u11; and U 3 = u31;. 2. 1. 1. P 2 1 n ; U 2 = u2 1 ; 1 n P 2 1 n ,de…ne. 1. 2. 2. 1. m;1. 1. 1. n+2. 2. P. m (n+2) ;. n. u11;n .. .. U1 ~ U2 ~ U3 =. u21;n+2 u21;n+1 .. .. u22;n+2 u22;n+1 .. .. u21;2 u21;1. u22;2 u22;1. u11;1. .. .. u2m;n+2 u2m;n+1 .. .. :::. u2m;2 u2m;1. u31;n .. : . u31;1. and v s0 P. m1 m2. n u11;. (m2 +2) ;. n = u11;. (m2 +2) ;. P P 2 3 1 2 ~ u ~ u j u 2 ; u 2 ; ; 1; 1; 1 m m1 1 2 1 2 2 1 2 1 P u31; 2 2 1 m2 ; u11;i = u2j;1 = u2j;m2 +2 = u31;i = 0; for 1. =. and v s1 P. m1 m2. P P 2 3 1 2 ~ u ~ u j u 2 ; u 2 ; ; 1; 1; 1 m2 m1 1 2 1 2 1 2 1 P 3 1 2 2 3 u1; 2 2 1 m2 ; u1;i = uj;1 = uj;m2 +2 = u1;i = 1; for 1. And then, we can save and the entries s~0;m. n;i;j. ^. v s0 P. m n ;where. u21;n .. .. (. u21;1. We know. [. m n. in a 2n 2n matrix S~0;m. xn;1;i ~ xcn+1;i;k1 ;k2. =. are subsets of we know. v s0 P. 1 i;j 2. km. 2 ;j. ^. ~ xn;j;1 j 1. n;i;j. = s~0;m. k1 ; k2. i. m2; 1. m2; 1. n;i;j 1 i;j 2n ,. ; km. 2. 2n. ^. =. v s0 P. m n and. 24. s~0;m. n;i;j. 0 u2m;n c. (. .. .. ):. u2m;1 0. are disjoint to each other.. j. m1. o. j. m1. o. and. ~ means "the ~ need to overlap one layer". Since. 0 u21;n u2m;n ) = c ( ... ) and ( ... ) = u21;1 u2m;1 0. s~ n 0;m. n. i. and.

(31) P n Given a basic set B 2 2 ; the 2 0 S~m ~0m n;i;j 1 i;j 2n ; where n = s s~0m P. Then. s~0;m. n;i;j. n;i;j. 1 i;j 2n. = card s~0;m. = ~ sm0. n;i;j. 0 2n matrix S~m. \. ~ s0 n (B) where m. v s0 X. m n. n. can be de…ned. !. (B) :. n (B) = card. v s0 P. m n. (B) :. By the construction P of Cn+1;1 = [cn+1;1;i1 ;i2 ]1 i1 ;i2 2n+1 and Hn = [hn;i1 ;i2 ]1 f or the same B 2 2 and the above arguments, we can get s~0m. n;i;j. =. 1 k1 ;k2. X. = ( 1 k1 ;k2. ;km. X. ;km. hn;1;i cn+1;1;i;k1 cn+1;1;k1 ;k2 : : : cn+1;1;km. 2 ;j. cn+1;1;i;k1 cn+1;1;k1 ;k2 : : : cn+1;1;km ;j )hn;1;i hn;j;1 2. 2n. = c0;m n;i;j hn;1;i hn;j;1 = (Cn+1;1 )m 1 i;j hn;1;i hn;j;1 for 1. 2n. i; j. i.e.,. S~0;m. n. hn;j;1. 2n. 2. = (Cn+1;1 )m. 1. 2 6 6 6 4. hn;1;1 hn;1;2 .. .. = (Cn+1;1 ). .. .. hn;1;2n hn;1;2n. 2. m 1. hn;1;1 hn;1;2 .. .. hn;1;1 hn;2;1 6 hn;1;1 hn;2;1 6 6 .. .. 4 . . hn;1;1 hn;2;1 [hn;1;i hn;j;1 ]1 i;j. 25. hn;1;1 hn;1;2 .. . hn;1;2n. .. .. hn;2n ;1 hn;2n ;1 .. . hn;2n ;1. 2n. 3 7 7 7 5. 3 7 7 7 5. i1 ;i2 2n+1.

(32) Similarly,. S~1;m. n. = (Cn+1;4 )m. 2 6 6 6 4. 1. hn;2n: ;1 hn;2n ;2 .. .. hn;2n: ;1 hn;2n ;2 .. .. .. .. hn;2n ;2n hn;2n ;2n. hn;2n ;2n. 2. hn;1;2n hn;2;2n 6 hn;1;2n hn;2;2n 6 6 .. .. 4 . . hn;1;2n hn;2;2n [hn;2n ;i hn;j;2n ]1 i;j. m 1. = (Cn+1;4 ) Since. P. 1 i;j. get. 2n. s~0m. P. = ~ sm0. n. (B) and. P. 1 i;j. 1 i;j. 2n. 1 i;j. 2n. P. n;i;j. (Cn+1;1 )m. 1. (Cn+1;4 )m. 1. hn;2n: ;1 hn;2n ;2 .. .. 2n. s~1m. hn;2n ;2n hn;2n ;2n .. .. .. .. hn;2n ;2n 2n. n;i;j. i;j. [hn;1;i hn;j;1 ] = ~ sm0. i;j. [hn;2n ;i hn;j;2n ] = ~ sm1. n. = ~ sm1. n. 3. 3 7 7 7 5. 7 7 7 5 (B) ;we can. (B) n. (B) :. v s0 ~ s0 (B) of P (B) is de…ned as follows : The spatial entropy h v s0 v s0 P P Let ~ sm0 n (B) = card( m n (B));the number of distinct patterns in m ~ s0 (B) is de…ned as The spatial entropy h. ~ s0 (B) = lim sup 1 log ~ s0 h m m;n!1 mn. n (B).. n (B):. Then, X ~ s0 (B) = lim sup 1 log( h (Cv;n+1;1 )m m;n!1 mn 1 i;j 2n. 1 i;j. [hn;1;i hn;j;1 ]).. v s1 ~ s1 (B) ofP (B) is de…ned as follows : Similarly, the spatial entropy h v s1 v s1 P P s1 ~ Let m n (B) = card( m n (B));the number of distinct patterns in m n (B).. 26.

(33) ~ s1 (B) is de…ned as The spatial entropy h ~ s1 (B) = lim sup 1 log ~ s1 h m m;n!1 mn. n (B):. Then, X ~ s1 (B) = lim sup 1 log( (Cv;n+1;4 )m h m;n!1 mn 1 i;j 2n. 5. 1 i;j. [hn;2n ;i hn;j;2n ]).. Conclusion. We get the recusive formulas for patterns generation in two-dimensional surfaces ( cylinder, torus and sphere). 1. Cylinder: Cm. n. = (Cn )m 1 for m 2. Furthermore, X (Cn )m 1 i;j = cm n (B):. 1 i;j 2n. 2. Torus: Dm. n. = (Cn )m 1 (Cn )T for m 2. Furthermore, i X h (Cn )m 1 (Cn )T = tm n (B) : i;j. 1 i;j 2n. 3. Sphere: (i) The boundary contains four corners. 0 Sm 1 Sm. n n. = (Cn+1;1 )m = (Cn+1;4 )m. Furthermore, P (Cn+1;1 )m n 1 i;j 2 P (Cn+1;4 )m 1 i;j. 2n. 1. 1 1. [cn+1;1;1;i cn+1;1;j;1 ] [cn+1;4;2n ;i cn+1;4;j;2n ]. i;j. cn+1;1;1;i cn+1;1;j;1 =. i;j. cn+1;1;2n ;i cn+1;1;j;2n =. 1. 27. s0 m n. (B). s1 m n. (B) :.

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