Hamiltonian walks on the Sierpinski gasket

Transcription

1 Hamiltonian walks on the Sierpinski gasket Shu-Chiuan Chang Department of Physics National Cheng Kung University Tainan 70101, Taiwan Lung-Chi Chen Department of Mathematics Fu Jen Catholic University Taipei 24205, Taiwan April 1, 2013

2 Outline 1. Introduction 2. Definitions 3. Number of Hamiltonian cycles on SG 2 (n) 4. Number of Hamiltonian walks with one end at origin 5. Number of Hamiltonian walks 6. Notation for the vertices 7. Distribution F(n,x m ) with x {a,b,c}, 0 m n 8. Distribution F(n,x m,2 ) with x {a,b,c} and 1 m < n 9. Distribution F(n,x γ ) with γ 3 Acknowledgments Thanks the support from Taiwan NSC grant.

3 1 Introduction 1.1 Definitions A graph G(V,E) is defined by its vertex (node) set V and edge (bond) set E [Harary (1969); Biggs (1993)]. Denote n(g) = v(g) = V as the number of vertices and e(g) = E as the number of edges of G, respectively. Denote the number of edges attached to the vertex v i as degree k i. A k-regular graph is a graph that each of its vertices has the same degree k. A planar graph G has a dual graph G.

4 1.2 Hamiltonian cycle and walk The origin of the name, Hamiltonian cycle or Hamiltonian circuit, was coined in 1857 by Sir William Rowan Hamilton when he asked for the construction of a cycle containing all the vertices of a dodecahedron. A Hamiltonian cycle is a cycle in a graph which visits each vertex exactly once and returns to the starting vertex. A Hamiltonian walk is a path in a graph which visits each vertex of the graph exactly once [Harary (1969); Bollobás (1998)]. A Hamiltonian walk corresponds to a complete self-avoiding walk. The enumeration of Hamiltonian cycles or walks is a fundamental problem in physics [Kasteleyn (1963); Duplantier (1987)] and computer science [Rubin (1974); Rahman and Kaykobad (2005)].

5 1.3 Sierpinski gasket SG d (n) Consider self-similar fractal lattices which have scaling invariance. A well-known example of fractal is the Sierpinski gasket. The first four stages n = 0,1,2,3 of the two-dimensional Sierpinski gasket SG 2 (n): SG 2 (0) SG 2 (1) SG 2 (2) SG 2 (n) at stage n = 0 is an equilateral triangle. SG 2 (3) Stage n+1 is obtained by the juxtaposition of three n-stage structures.

6 For SG 2 (n), denote o as the leftmost vertex, a n as the rightmost vertex, and b n as the topmost vertex. SG d (n) can be built in any Euclidean dimension d. b n o a n SG d (0)atstagen = 0 isacompletegraphwith (d+1)vertices, eachofwhichisadjacent to all of the other vertices. Fractal dimensionality for SG d [Gefen et al. (1981)]: D(SG d ) = ln(d+1) ln2 The numbers of edges and vertices for SG d (n): ( ) d+1 e(sg d (n)) = (d+1) n = d 2 2 (d+1)n+1, v(sg d (n)) = d+1 2 [(d+1)n +1]. Except the (d + 1) outmost vertices which have degree d, all other vertices of SG d (n) have degree 2d. Therefore, SG d is 2d-regular in the large n limit..

7 2 Definitions Define f 1 (n) as the number of Hamiltonian walks whose end vertices are located at two specific outmost vertices, e.g., o and a n. Define f 2 (n) as the number of Hamiltonian walks whose one end vertex is located at a specific outmost vertex, e.g., o, while the other end vertex is not at the other two outmost vertices. Define f 3 (n) as the number of Hamiltonian walks whose both end vertices are not located at any of the outmost vertices. The total number of Hamiltonian walks on SG(n) is H(n) = 3f 1 (n)+3f 2 (n)+f 3 (n). The number of Hamiltonian walks with one end at o is H 0 (n) = 2f 1 (n)+f 2 (n). The number of Hamiltonian cycles on SG(n) is HC(n) = f 1 (n 1) 3. f 1 (n) = f 2 (n) f 3 (n) Consider the two-dimensional Sierpinski gasket without one specific outmost vertex, e.g., SG(n)\{b n }.

8 (i) Define g 1 (n) as the number of Hamiltonian walks whose end vertices are located at the remaining two outmost vertices, e.g., o and a n. (ii) Define g 2 (n) as the number of Hamiltonian walks whose one end vertex is located at one specific remaining outmost vertex, e.g., o, while the other end vertex is not at the other outmost vertex. g 1 (n) g 2 (n) Divide the vertices of the two-dimensional Sierpinski gasket into two subsets V 1 (n) and V 2 (n). One ofthem containstwooutmostverticeswhile the othercontainsonlyone, e.g., o,a n V 1 (n), b n V 2 (n). (i) Define g 3 (n) as the number of two Hamiltonian walks separately visiting all the vertices of V 1 (n) and V 2 (n), such that one of them has end vertices located at the two outmost vertices in V 1 (n), e.g., o and a n, while the g 3 (n) other one has one end vertex located at the other outmost vertex in V 2 (n), e.g., b n. (ii) Define g 4 (n) as the number of two Hamiltonian walks separately visiting all the vertices V 1 (n) and V 2 (n), such that one of them has only g one end vertex located at a specific outmost vertex in V 1 (n), e.g., o, while 4 (n) the other one has one end vertex located at the outmost vertex in V 2 (n), e.g., b n.

9 Consider the two-dimensional Sierpinski gasket without two specific outmost vertices, e.g., SG(n)\{o,a n }. Define t 1 (n) as the number of Hamiltonian walks whose one end vertex is located at the remaining outmost vertices, e.g., b n. t 1 (n) Consider the two-dimensional Sierpinski gasket without one specific outmost vertex, e.g., SG(n)\{b n }, and divide its vertices into two subsets V 1(n) and V 2(n). Both of these subsets contain one outmost vertex, e.g., o V 1, a n V 2(n). Define t 2 (n) as the number of two Hamiltonian walks separately visiting all the vertices of V 1(n) and V 2(n), t such that both of them have one end 2 (n) vertex located at an outmost vertex, e.g., o and a n. At stage n = 0, only f 1 (0) and g 1 (0) are equal to one, and f 2 (0) = f 3 (0) = g 2 (0) = g 3 (0) = g 4 (0) = t 1 (0) = t 2 (0) = 0. We shall only consider n 1 for the following discussion.

10 3 Number of Hamiltonian cycles on SG 2 (n) Initial values are f 1 (1) = 2, g 1 (1) = 3. = For n 1, such that + = + + δ n0 f 1 (n+1) = 2f 1 (n) 2 g 1 (n), g 1 (n+1) = 2g 1 (n) 2 f 1 (n), f 1 (n+1) g 1 (n+1) = f 1(n) g 1 (n) = f 1(1) g 1 (1) = 2 3, f 1 (n) = 3f 1 (n 1) 3 = = 3 3n f 1 (1) 3n 1 = 3 3n (2 3n 1 ) = g 1 (n) = 3f 1(n) 3(2 3) 3 n 1 = 2 2 HC(n) is one for n = 1 and HC(n) = f 1 (n 1) 3 = f 1(n) 3(2 3) 3 n 1 = 3 9 for n 2 [Bradley (1986)]. 3(2 3) 3 n 1 3

11 4 Number of Hamiltonian walks with one end at origin Initial values are f 2 (1) = 2, g 2 (1) = 3, g 3 (1) = 4, t 1 (1) = 4. Recursion relation for n 1 and illustration for f 2 (n+1): f 2 (n+1) = 2f 1 (n) 2 [g 1 (n)+g 2 (n)+g 3 (n)]+2f 1 (n)f 2 (n)g 1 (n). = 2 ( ) Recursion relation for n 1 and illustration for g 2 (n+1): g 2 (n+1) = 2f 1 (n)g 1 (n)[g 1 (n)+g 2 (n)+g 3 (n)]+f 2 (n)g 1 (n) 2 +f 1 (n) 2 t 1 (n). = δ n0

12 Recursion relation for n 1 and illustration for g 3 (n+1): g 3 (n+1) = 2f 1 (n)g 1 (n)[2g 1 (n)+g 2 (n)+3g 3 (n)]+f 2 (n)g 1 (n) 2 +f 1 (n) 2 t 1 (n). = 2 ( ) + + Recursion relation for n 1 and illustration for t 1 (n+1): t 1 (n+1) = 2g 1 (n) 2 [g 1 (n)+g 2 (n)+g 3 (n)]+2f 1 (n)g 1 (n)t 1 (n). = 2 ( δ n0 ) Define the ratios α 2 (n) = f 2(n) f 1 (n), β 2(n) = g 2(n) f 1 (n), β 3(n) = g 3(n) f 1 (n), γ 1(n) = t 1(n) f 1 (n). Using g 1 (n) = 3f 1 (n)/2 for n 1 and f 1 (n+1) = 3f 1 (n) 3, they satisfy

13 α 2 (n+1) = 1+α 2 (n)+ 2 3 β 2(n)+ 2 3 β 3(n), β 2 (n+1) = α 2(n)+β 2 (n)+β 3 (n)+ 1 3 γ 1(n), β 3 (n+1) = α 2(n)+β 2 (n)+3β 3 (n)+ 1 3 γ 1(n), γ 1 (n+1) = β 2(n)+ 3 2 β 3(n)+γ 1 (n), Define the vectors b = (1, 3 2,3, 9 4 )t, X(n) = (α 2 (n),β 2 (n),β 3 (n),γ 1 (n)) t, and construct A = such that X(1) = (1, 3 2,2,2)t and for general n 1,, X(n) = AX(n 1)+b = A 2 X(n 2)+Ab+b = ( 717 )4 n 2 1 ( )δ n n1 = A n 1 X(1)+ A j ( 717 )4 n 7 +( 1 b = )δ n1 ( 717 )4 n 47 j= , ( )4n δ 2 4 n1 3(2 3) 3 n 1 {( ) ( } 717 H 0 (n) = 4 n )δ n1.

14 5 Number of Hamiltonian walks Initial values are f 3 (1) = 0, g 4 (1) = 1 and t 2 (1) = 6. Recursion relation for n 1 and illustration for f 3 (n+1): f 3 (n+1) = 6f 1 (n)[f 2 (n)g 2 (n)+f 1 (n)g 2 (n)+f 1 (n)g 4 (n)]. = 3 ( )

15 Recursion relation for n 1 and illustration for g 4 (n+1): g 4 (n+1) = f 1 (n)[g 1 (n) 2 +g 2 (n) 2 ]+4f 1 (n)g 1 (n)[g 2 (n)+g 3 (n)+g 4 (n)] +f 1 (n)g 3 (n)[2g 2 (n)+3g 3 (n)]+f 1 (n) 2 [t 1 (n)+t 2 (n)] +f 2 (n)g 1 (n)[g 1 (n)+2g 2 (n)+2g 3 (n)]+f 1 (n)f 2 (n)t 1 (n), =

16 Recursion relation for n 1 and illustration for t 2 (n+1): t 2 (n+1) = 2g 2 1(n)[g 1 (n)+2g 2 (n)+4g 3 (n)+2g 4 (n)] +2g 1 (n)g 3 (n)[2g 2 (n)+3g 3 (n)]+2g 1 (n)[g 2 (n) 2 +f 2 (n)t 1 (n)] +4f 1 (n)t 1 (n)[2g 1 (n)+g 2 (n)+g 3 (n)]+4f 1 (n)g 1 (n)t 2 (n). = 2 ( δ n0 + δ n0 )

17 Define the ratios α 3 (n) = f 3(n) f 1 (n), β 4(n) = g 4(n) f 1 (n), γ 2(n) = t 2(n) f 1 (n), Using g 1 (n) = 3f 1 (n)/2 for n 1, they satisfy α 3 (n+1) = 2[α 2 (n)β 2 (n)+β 2 (n)+β 4 (n)], β 4 (n+1) = 3 4 +β 2(n)[2+ β 2(n) ]+β 3 3 (n)[ β 2(n)+β 3 (n)] +α 2 (n)[ 3 4 +β 2(n)+β 3 (n)+ γ 1(n) ]+2β 3 4 (n)+ 1 3 [γ 1(n)+γ 2 (n)], γ 2 (n+1) = 9 4 +β 2(n)[3+β 2 (n)]+β 3 (n)[6+2β 2 (n)+3β 3 (n)] +γ 1 (n)[4+α 2 (n)+ 4 3 β 2(n)+ 4 3 β 3(n)]+3β 4 (n)+2γ 2 (n). Define vectors Y(n) = (α 3 (n),β 4 (n),γ 2 (n)) t and K(n) = (K 1 (n),k 2 (n),k 3 (n)) t where K 1 (n) = α 3 (n+1) 2β 4 (n), K 2 (n) = β 4 (n+1) 2β 4 (n) γ 2 (n)/3, K 3 (n) = γ 2 (n+1) 3β 4 (n) 2γ 2 (n), and the matrix B = PDP 1, such that Y(n+1) = K(n)+BY(n) for n 1, where D = , P = , P 1 =

18 Initial value Y(1) = (0,1/2,3) t, and for n 2, Y(n) = = P = n 1 j=1 n 2 B n 1 j K(j)+B n 1 Y(1) D n 1 j P 1 K(j)+K(n 1)+PD n 1 P 1 Y(1) j= n 2 j= (3K 2(j) K 3 (j)) 3 n 3 j (3K 2 2 (j)+k 3 (j)) K 1 (n), K 2 (n), K 3 (n) for n 1 can be solved: K 1 (n) = K 2 (n) = K 3 (n) = ( ( )(16) n + )(16) n ( )(16) 2 n ( ) ( 5 ) n n 283 ( ) n ( n )δ n1, ( 19 )δ n1, ( 19 )δ n1. K 1 (n 1) K 2 (n 1) K 3 (n 1)

19 For n 2, α 3 (n) = = n 2 [ (1+3 n 2 j )K 2 (j)+ 1 ] 3 (3n 2 j 1)K 3 (j) j=1 ( ) ( ) 717 (16) n 4 n + 2( ) δ n2 + 3n 1 1 +K 1 (n 1) 2 ) 3 n ( H(n) is twelve for n = 1 and 3(2 3) 3 n 1 ( ) ( ) H(n) = { (16) n + ( 3 2) n ( ) δ n2 } for n 2. When n is large, H(n) H 0 (n) ( 5 2 ) n n

20 6 Notation for the vertices Fix the origin o as the leftmost vertex, and all the Hamiltonian walks considered in the following discussions have one end at o. ConsiderSG(m) with 0 m < n always has o as its leftmost vertex. Denote a m and b m as its rightmost and topmost vertices, respectively. c m is defined such that the vertices a m, b m and c m demarcate the largest lacunary triangle of SG(m+1). Define the vertex in the middle of the line connecting a m and a m+1 with m 1 as a m,1. a m,1 and the associated b m,1 and c m,1 demarcate a lacunary triangle with b m,1 on the left and c m,1 on the right. For m 2, append the subscript (m,1,0) for the vertices of the largest lacunary inside the triangle with outmost vertices a m, a m,1, b m,1 ; (m,1,1) for the vertices of the largest lacunary inside the triangle with outmost vertices a m,1, a m+1, c m,1 ; (m,1,2) for the vertices of the largest lacunary inside the triangle with outmost vertices b m,1, c m,1, c m. In general, use the notation x γ where x {a,b,c} and the subscript γ = (γ 1,γ 2,...,γ s ) has γ = s components with 1 s n, 1 γ 1 < n and γ k {0,1,2} for k {2,3,...,s}.

21 For the vertices above the line connecting o and c n 1 on SG(n), we will also use the notation x γ such that it is the reflection of the vertex x γ with respect to this line. For the vertices above the line connecting a n and b n 1 on SG(n), we use the notation ˆx γ such that it is the reflection of the vertex x γ with respect to this line. For examples, a 22 = b 21, b 221 = ã 212, c 222 = c 211, and b 21 = ˆb 21, c 210 = â 212, a 211 = ĉ 211 on SG(3). Denote the number of Hamiltonian walks with one end at o and the other end at specific vertex x on SG(n) as f(n,x), and the distribution ratio F(n,x) = f(n,x)/f 1 (n). Denote the factor R(n) = ( ) n ( ) 1 δ n1 such that H 0 (n) = f 1 (n)r(n), then define the probability measure p(n,x) = f(n,x) H 0 (n) = F(n,x) R(n) The summation of p(n,x) over all the vertices of SG(n) should give one..

22 b 4 b 3222 c 3222 b 322 c 322 a 3222 b 3220 c b c 3221 b 32 c 32 a 3220 a 322 a 3221 b 3202 c 3202 b3212 c 3212 b 320 c 320 b321 c 321 a 3202 a 3212 b 3200 c b c b c b c 3211 b 3 c 3 a 3200 a 320 a 3201 a 32 a 3210 a 321 a 3211 b 222 c 222 b3122 c 3122 b 22 c 22 b312 c 312 a 222 a 3122 b 220 c 220 b 221 c 221 b3120 c b c 3121 b 2 c 2 b31 c 31 a 220 a 22 a 221 a 3120 a 312 a 3121 b 12 c 12 b212 c 212 b3102 c 3102 b3112 c 3112 b 1 c 1 b21 c 21 b310 c 310 b311 c 311 a 12 a 212 a 3102 a 3112 b 0 c 0 b11 c 11 b 210 c210 b 211 c211 b 3100 c b c b c 3110 b 3111 c 3111 o a 0 a 1 a 11 a 2 a 210 a 21 a 211 a 3 a 3100 a 310 a 3101 a 31 a 3110 a 311 a 3111 a 4

23 g 2 (n,x) denotes the number of g 2 (n) such that the end points of the Hamiltonian walk are located at o and x, and G 2 (n,x) = g 2 (n,x)/g 1 (n). g 3 (n,x) denotes the number of g 3 (n) such that one Hamiltonian walk has end points located at o and a n, while the other Hamiltonian walk has end points located at b n and x. Define u(n,x) = g 3 (n,x)+g 3 (n, x) and U(n,x) = u(n,x)/g 1 (n) t 1 (n,x) denotes the number of t 1 (n) such that the end points of the Hamiltonian walk are located at b n and x, and T 1 (n,x) = t 1 (n,x)/g 1 (n).

24 7 Distribution F(n,x m ) with x {a,b,c}, 0 m n f(n,a m ) = f(n,b m ) for all m n by symmetry, and f(n,a n ) = f(n,b n ) = f 1 (n) by definition. For m = n 1 0, f(n,a n 1 ) = f(n,b n 1 ) = f 1 (n 1) 2 g 1 (n 1) = 1 2 f 1(n), f(n,c n 1 ) = 0, { 2f1 (n 1)g g 3 (n,b n 1 ) = g 3 (n,c n 1 ) = 1 (n 1) 2 = g 1 (n) for n > 1, 2 = 2 3 g 1(1) for n = 1, g 3 (n,a n 1 ) = 0, where the expression for g 3 (n,b n 1 ): g 2 (n,a n 1 ) = g 2 (n,b n 1 ) = + { f1 (n 1)g 1 (n 1) 2 = 1 2 g 1(n) for n > 1, 1 = 1 3 g 1(1) for n = 1, g 2 (n,c n 1 ) = { 0 for n > 1, 1 = 1 3 g 1(1) for n = 1.

25 For 0 m n 2 and x {a,b,c}, { f(n,xm ) = f 1 (n 1) 2 u(n 1,x m ), g 3 (n,x m ) = 2f 1 (n 1)g 1 (n 1)g 3 (n 1,x m ) + + g 3 (n,x m ) for 0 m n 2 can be written in general as g 3 (n,x m ) = r(n)g 3 (n 1,x m ), where r(n) = g 1(n) g 1 (n 1), such that g 3 (n,a m ) = 0 and g 3 (n,b m ) = g 3 (n,c m ) = g 3 (m+1,b m ) = n j=m+2 r(j) { g1 (m+1) n j=m+2 r(j) = g 1(n) (0 < m < n 1), 2 3 g 1(1) n j=2 r(j) = 2 3 g 1(n) (0 = m < n 1),

26 We obtain U(n,a m ) = U(n,b m ) = G 2 (n,a m ) = G 2 (n,b m ) = { 1 (0 < m < n), 2 3 (0 = m < n), { 1 (0 < m < n), (0 = m < n), U(n,c m ) = Since g 2 (n,x m ) = f 1 (n 1)g 1 (n 1)u(n 1,x m ) for 0 m n 2, G 2 (n,c m ) = { 2 (0 < m < n), 4 3 (0 = m < n). 1 (0 < m < n 1), 0 (0 < m = n 1), 2 3 (0 = m < n 1), 1 3 (0 = m = n 1), f(n,x m ) for m n 2: { 1 f(n,a m ) = f(n,b m ) = f 1 (n 1) 2 g 3 (n 1,b m ) = 2 f 1(n) (0 < m < n 1), 1 { 3 f 1(n) (0 = m < n 1), f(n,c m ) = 2f 1 (n 1) 2 f1 (n) (0 < m < n 1), g 3 (n 1,c m ) = 2 3 f 1(n) (0 = m < n 1), Distribution ratio F(n,x m ) with x {a,b,c} for integer 0 m n: 1 (0 m = n), 1 F(n,a m ) = F(n,b m ) = 2 (0 < m < n), 1 2 (m = 0, n = 1), 1 3 (m = 0, n > 1), F(n,c m ) = 0 (0 m = n 1), 1 (0 < m < n 1), 2 3 (m = 0, n > 1).

27 For any vertex x v(sg(n)), denote the distance between o and x as x, and set the distance between the vertices o and a 0 equal to one. For any positive l > 0, define the mean l displacement for the vertices x m with x {a,b,c} and 0 m n as ξ x (n,l) j ;j=0,1,2,...,n;x=a,b,c x j l p(n,x j ) x x j ;j=0,1,2,...,n;x=a,b,c p(n,x = j ;j=0,1,2,...,n;x=a,b,c x j l F(n,x j ) j) x j ;j=0,1,2,...,n;x=a,b,c F(n,x, j) where p(n,c n ) = 0 is assumed. For any integer n > 1, ξ (n,l) = 2{ n 1 j=1 2jl +12 nl } n 2 j=1 (2j 3) l 2{ (n 1)+1} (n 2) = 2nl (1+ 3)+ 2nl (1+( 2n ) l ) 2 l (2 3) l 2 l 1 When n is large, we have the asymptotic expression ( ( )) ξ (n,l) ( 3 2 )l 2 2 l 1. 2nl n.

28 8 Distribution F(n,x m,2 ) with x {a,b,c} and 1 m < n The expressions for x m,1 can be obtained from those for x m,2 by symmetry. For m = n 1, f(n,x n 1,2 ) = f 1 (n 1)g 1 (n 1)f(n 1,ˆ ˆxn 2 )+f 1 (n 1) 2 g 2 (n 1,ˆ x n 2 ), + Next consider m n 2, f(n,x m,2 ) = f 1 (n 1) 2 u(n 1,x m,2 ) and u(n 1,x m,2 ) = u(m+1,x m,2 ) n 1 j=m+2 r(j), where the product is set to one when n = m+2 and u(m+1,x m,2 ) = f 1 (m) 2 t 1 (m,x m 1 )+g 1 (m) 2 f(m, ˆx m 1 ) +f 1 (m)g 1 (m) ( g 2 (m, ˆx m 1 )+g 2 (m,ˆx m 1 )+2u(m,x m 1 ) )

29 The term t 1 (m,x m 1 ) is given by { g1 (m 1) t 1 (m,b m 1 ) = t 1 (m,c m 1 ) = 3 = 3 2 f 1(m 1)g 1 (m 1) 2 = 3 4 g 1(m) (m > 1) { 1 = 1 3 g 1(1) (m = 1) 0 (m > 1), t 1 (m,a m 1 ) = 2 = 2 3 g such that 1(1) (m = 1), u(m+1,a m,2 ) = f{ 1 (m) 2 t 1 (m,a m 1 )+g 1 (m) 2 f(m,c m 1 )+2f 1 (m)g 1 (m)(g 2 (m,c m 1 )+u(m,a m 1 )) g1 (m+1) (m > 1), = 44 = 11 9 g 1(2) (m = 1), where g 1 (2) = 36, and u(m+1,b m,2 ) = f 1 (m) 2 t 1 (m,b m 1 )+g 1 (m) 2 f(m,b m 1 ) +f { 1 (m)g 1 (m)(g 2 (m,a m 1 )+g 2 (m,b m 1 )+2u(m,b m 1 )) 2g1 (m+1) (m > 1), = 49 = g 1(2) (m = 1), u(m+1,c m,2 ) = f 1 (m) 2 t 1 (m,c m 1 )+g 1 (m) 2 f(m,a m 1 ) +f { 1 (m)g 1 (m)(g 2 (m,b m 1 )+g 2 (m,a m 1 )+2u(m,c m 1 )) 3g1 (m+1) (m > 1), = 73 = g 1(2) (m = 1).

30 For m n 2, u(n 1,a m,2 ) = u(n 1,c m,2 ) = { g1 (n 1) (m > 1), 11 { 9 g 1(n 1) (m = 1), 3g1 (n 1) (m > 1), g 1(n 1) (m = 1), u(n 1,b m,2 ) = { 2g1 (n 1) (m > 1), g 1(n 1) (m = 1), For n 2, the distribution ratio F(n,x n 1,γ2 ) with x {a,b,c} and γ 2 {1,2} is { 1 (n > 2), F(n,a n 1,2 ) = F(n,c n 1,2 ) = F(n,b n 1,1 ) = F(n,c n 1,1 ) = 2 5 { 12 (n = 2), 0 (n > 2), F(n,b n 1,2 ) = F(n,a n 1,1 ) = 1 (n = 2), 6 and the distribution ratio F(n,x m,γ2 ) with integer m n 2, x {a,b,c} and γ 2 {1,2} is { 1 { F(n,a m,2 ) = F(n,b m,1 ) = 2 (m > 1), 1 (m > 1), 11 F(n,b (m = 1), m,2 ) = F(n,a m,1 ) = 49 { (m = 1), (m > 1), F(n,c m,2 ) = F(n,c m,1 ) = (m = 1).

31 For m = n 1, g 2 (n,x n 1,2 ) = f 1 (n 1)g 1 (n 1)g 2 (n 1,ˆ ˆxn 2 )+f 1 (n 1) 2 t 1 (n 1, ˆx n 2 ), + The distribution ratio G 2 (n,x n 1,2 ) with x {a,b,c} is { 1 { (n > 2), G 2 (n,a n 1,2 ) = G 2 (n,c n 1,2 ) = 2 0 (n > 2), 5 G 18 (n = 2), 2 (n,b n 1,2 ) = 7 18 (n = 2). The values for G 2 (n,x n 1,1 ) and G 2 (n,x m,γ2 ) with m n 2, x {a,b,c} and γ 2 {1,2} and equal to F(n,x n 1,1 ) and F(n,x m,γ2 ), respectively. Distribution ratios F(1,x), G 2 (1,x) and U(1,x) for SG(1): 1 1/2 0 1/2 1 1/3 1/3 1/3 2/3 4/3 2/3 F(1,x) G 2 (1,x) U(1, x)

32 Distribution ratios F(2,x), G 2 (2,x) and U(2,x) for SG(2): 1 1/6 5/12 1/2 0 5/12 1/2 2/3 5/12 5/12 1/3 1/2 1/6 1 7/18 5/18 1/2 0 5/18 1/3 2/3 5/12 5/12 1/3 1/2 1/6 49/36 73/ /9 2/3 4/3 11/9 73/36 2/3 1 49/36 F(2,x) G 2 (2,x) U(2, x)

33 9 Distribution F(n,x γ ) with γ 3 The subscript of a vertex x γ on SG(n) is given as γ = (γ 1,γ 2,γ 3,,γ s ) where 1 γ 1 n 1 and γ k {0,1,2} for k = 2,3,...,s. We should derive F(n,x γ ) for the vertices with γ 1 = n 1, γ 2 = 2 on SG(n) and the vertices with γ 1 < n 1, i.e., the vertices within the upper SG(n 1) and the lower-left SG(n 1) that constitute SG(n). Fortheverticeswithγ 1 = n 1,γ 2 = 1,i.e.,theverticeswithinthelower-rightSG(n 1), we use F(n,x γ ) = F(n, x γ ) by symmetry. AquantitylikeF(n,x γ )onsg(n)willbeexpressedintermsofthequantitiesonsg(n 1) as the recursion relation. Denote the subscript γ for the corresponding vertex x γ on SG(n 1). When the first component of γ, i.e., γ 1, is equal to n 1 and γ 2 {1,2}, we have γ = (n 2,γ 3,,γ s ) with γ = γ 1; while the first component of γ is smaller than n 1, we have γ = γ.

34 Using f(n,x γ ) = f 1 (n 1)g 1 (n 1)f(n 1,ˆ ˆx γ )+f 1 (n 1) 2 g 2 (n 1,ˆ x γ ) if γ 1 = n 1, γ 2 = 2, f(n,x γ ) = f 1 (n 1) 2 u(n 1,x γ ) if γ 1 < n 1, the distribution ratio F(n,x γ ) with 3 γ n is { F(n,x γ ) = 1 2 F(n 1,ˆ ˆx γ )+ 1 2 G 2(n 1,ˆ x γ ) if γ 1 = n 1, γ 2 = 2, F(n,x γ ) = 1 2 U(n 1,x γ ) if γ 1 < n 1. The distribution ratio G 2 (n,x γ ): (i) When γ 1 = n 1 and γ 2 = 1 with γ 2, (ii) When γ 1 < n 1, G 2 (n,x γ ) = F(n,x γ ). G 2 (n,x γ ) = F(n,x γ ) = 1 2 U(n 1,x γ ), U(n,x γ) = U(n 1,x γ ). (iii) When γ 1 = n 1 and γ 2 = 2 with γ 3, G 2 (n,x γ ) = G 2 (n 1,ˆ ˆx γ ( ) ) if γ 3 = 1,2, G 2 (n 1,ˆ x γ )+G 2 (n 1,ˆ ˆx γ ) if γ 3 =

35 The distribution ratio U(n,x γ ) for γ 1 = n 1 and γ 2 = 2 with γ 3: { U(n 1,ˆx γ )+U(n 1,x U(n,x γ ) = γ ) if γ 3 = 2, G 2 (n 1,ˆx γ )+G 2 (n 1, ˆx γ )+U(n 1,x γ ) if γ 3 = 0,1. Distributions F(3,x), G 2 (3,x) and U(3,x) for SG(3): 1 5/12 1/6 0 1/2 5/12 25/72 25/72 2/3 1/3 1/2 0 5/18 1/2 1/3 49/72 73/72 1/3 1/3 1/2 1 1/2 1/2 11/18 2/3 1/3 2/3 11/18 73/72 5/18 25/72 5/12 1/6 1/3 1/2 49/72 1/2 25/72 0 5/12 1 5/18 7/18 0 1/2 5/18 25/72 25/72 2/3 1/3 1/2 0 5/18 1/2 1/3 49/72 73/72 1/3 1/3 1/2 1 1/2 1/2 11/18 2/3 1/3 2/3 11/18 73/72 5/18 25/72 5/12 1/6 1/3 1/2 49/72 1/2 25/72 0 5/12 F(3,x) G 2 (3,x)

36 For almost all the vertex x γ on the Sierpinski gasket SG(n) with n 2, we have F(n,x γ ) = G 2 (n,x γ ), except for the outmost vertices a n, b n, and the three vertices x γ 0 with γ 0 = (n 1,2,2,...,2) and x {a,b,c}. 73/36 97/ /9 11/9 73/36 23/12 31/ / /18 49/36 73/36 37/18 31/ /9 23/12 2/3 4/3 11/9 73/36 49/36 73/36 23/9 97/36 2/3 1 49/ /9 2 73/36 The values for the vertices x γ 0 are F(n,a γ 0) = 5 { 5 12, F(n,b γ 0) = 12 for odd n, 1 6 for even n, and G 2 (n,a γ 0) = 5 { 5 18, G 2(n,b γ 0) = 18 for odd n, 7 for even n, 18 F(n,c γ 0) = G 2 (n,c γ 0) = U(3, x) { 1 6 for odd n, 5 12 for even n, { 7 18 for odd n, 5 for even n. 18

37 For any positive l > 0, define the mean l displacement ξ(n,l) as ξ(n,l) = x l p(n,x), and ψ(n,l) = x v(sg(n)) where ξ(n,l) = ψ(n,l)/r(n). Define φ(n,l) = x V(SG(n)) x l U(n,x), U(n) = x v(sg(n)) x γ v(sg(n)), γ >2 x l F(n,x), U(n,x). For n 3, the summation of F(n,x γ ) over most of the vertices in the upper SG(n 1): F(n,x γ ) = F(n 1,x γ )+O(1) = R(n 1)+O(1), x γ v(sg(n)), γ >2,γ 1 =n 1,γ 2 =2 x γ v(sg(n 1)) where R(n) = ( 717 )4 n + O(1) when n is large, and the number of neglected vertices with γ = 1,2 in the upper SG(n 1) has order one. We have F(n,x γ ) = R(n) = 1 2 U(n 1)+2R(n 1)+O(n), x γ v(sg(n)) because the number of neglected vertices with γ = 1,2 has order n, so when n is large U(n 1) = 2(R(n) 2R(n 1))+O(n) = R(n)+O(n).

38 Foralltheverticesx γ v(sg(n))with γ > 2,γ 1 = n 1,γ 2 = 2, thedistancebetween x γ and o, i.e., x γ, is larger than 2 n 1 and less than 2 n. Using the inequality: 2 (n 1)l x γ l 2 nl for l > 0, 1 2 φ(n 1,l)+2{2(n 1)l [R(n 1)+O(n)]} ψ(n,l) 1 2 φ(n 1,l)+2{2nl [R(n 1)+O(n)]}, i.e., Similarly, i.e., [( ) φ(n 1,l)+2nl l ) 1 2 φ(n 1,l)+2nl+1 [( ] 4 n 1 +O(n) ] 4 n 1 +O(n) ψ(n,l) φ(n 1,l)+2 (n 1)l 2{2R(n 1)+U(n 1)+O(n)} φ(n,l) φ(n 1,l)+2 nl 2{2R(n 1)+U(n 1)+O(n)}, ( ) ] 717 φ(n 1,l)+2 [6 (n 1)l+1 4 n 1 +O(n) φ(n,l) ( ) ] 717 φ(n 1,l)+2 [6 nl+1 4 n 1 +O(n)

39 By induction, ( ){ 717 (2 2+l ) n } ( ){ l (2 2+l ) n } l 1 +2(n 1)l O(n) φ(n,l) l 1 +2(n 1)l O(n) so that for large n. ( ) { (l+2)n l 1+l +1 2 ( ) } 2 2+l 1 +4 (n 1) O(n) } 2 (l+2)n { 2 1+l l 1 +4 (n 1) O(n) ψ(n,l) When n is large, the mean l displacement is bounded as { 2 2 (n 1)l 1+l } { l 1 +4 n O(n) ξ(n,l) 2 nl 1+l } l 1 +4 n O(n) Define the exponent ν such that ξ(n,l) is proportional to [lnl(w)] ν in the large n limit, where the length of w, denoted as L(w), is equal to (3 n+1 +1)/2. In the limit of infinite n, the exponent for the mean l displacement is ( ) lnξ(n,l) ln2 ν = lim n ln[(3 n+1 +1)/2] = l. ln3.

40 References [1] F. Harary, Graph Theory (Addison-Wesley, New York, 1969). [2] N. L. Biggs, Algebraic Graph Theory (2nd ed., Cambridge Univ. Press, Cambridge, 1993). [3] B. Bollobás, Modern Graph Theory (Springer, New York, 1998). [4] P. W. Kasteleyn, A soluble self-avoiding walk problem, Physica 29 (1963) [5] B. Duplantier, Critical exponents of Manhattan Hamiltonian walks in two dimensions, from Potts and O(n) models J. Stat. Phys. 49 (1987) 411. [6] F. Rubin, A search procedure for Hamilton paths an circuits, J. Assoc. Computing Machinery 21 (1974) 576. [7] M. Sohel Rahman and M. Kaykobad, On Hamiltonian cycle n Hamiltonian paths, Inform. Process. Lett. 94 (2005) 37. [8] Y. Gefen, A. Aharony, B. B. Mandelbrot and S. Kirkpartrick, Solvable fractal family, and its possible relation to the backbone at percolation, Phys. Rev. Lett. 47 (1981) [9] R. M. Bradley, Statistical mechanics of the travelling salesman on the Sierpinski gasket, J. Physique 47 (1986) 9.