Fibonacci äÿàÿ š f 1,2 ÜŠö: Ü 2, Öwý 3, o~ 3 1 Institute for Cognitive Neurodynamics, East China University of Science and Technology 2 School of Science, Hangzhou Dianzi University 3 Department of Mathematics, Shanghai University Email: wgsun@hdu.edu.cn 2010. 10. 17
Outline of Topics 1 Introduction 2 Fibonacci Network 3 Topological properties 4 Conclusion
Fibonacci Number The original problem that Fibonacci investigated (in the year 1202) was about how fast rabbits could breed in ideal circumstances. Suppose a newly-born pair of rabbits, one male, one female, are put in a field. Rabbits are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits. Suppose that our rabbits never die and that the female always produces one new pair (one male, one female) every month from the second month on. How many pairs will there be in one year?
Illustration of Fibonacci Number
Model Presentation Degree distribution APL Model Fibonacci network is extracted from the reproduction principle of Fibonacci rabbits. We regard each pair of Fibonacci rabbits as one node and one parent-child relation as one edge. Let F t be Fibonacci network of t generation, and F t is obtained from F t 1 by adding new nodes and new edges as follows: Generation t = 1 : We start with an adult node, called the main root of our network. Generation t = 2 : The main root gives birth to one baby node, and this baby node and its mother node are linked. Generation t 3 : Suppose that we have F t 1. F t is obtained from F t 1 when each of its existing adult nodes is replaced with F 2, moreover each of its existing baby nodes grows into an adult node. The illustration of F t with t = 1, 2, 3, 4 is shown in Figure.
Model Presentation Degree distribution APL Figure: Illustration of Fibonacci network F t with t = 1, 2, 3, 4.
Model Presentation Degree distribution APL Node number and average degree Let N t and E t be the order and size of F t, we obtain N t = N t 1 + N t 2, for t 2. (1) Suppose that the particular solution of Eq. (1) is N t = r t. We get the characteristic equation with regard to r: r 2 r 1 = 0. (2) The solutions of Eq. (2) are r 1 = 1+ 2 and r 2 = 1 2. Thus the general solution of Eq. (1) is N t = G 1 r1 t + G 2r2 t, where G 1 and G 2 are arbitrary constants. Using initial conditions N 1 = 1 and N 2 = 2, we obtain N t = 1 (r1 t+1 r2 t+1 ). (3) Thus the average degree is k = 2E t N t 2. (4)
Model Presentation Degree distribution APL Degree distribution It is obvious that F t has N t 2 baby nodes. These baby nodes will grow into adult nodes in the next generation, and produce new nodes from the second generation. Let node v be such a baby node of F t. The degree of node v is 1 in t + 1 generation, and its degree becomes k in t + k generation with k 2. Consequently the number of nodes of degree k in F t+k is N t 2. For degree k with k 2 and t k 2 1, there are N t k 2 nodes of degree k in F t. Thus degree distribution P (k, t) in t generation is P (k, t) = N t k 2 N t. According to Eq. (3), when t tends to +, we have P (k) = r (k+2) 1 = e (k+2)α e kα, () and by the similar calculation, we easily get P (1) = 1 r 1 = e α, (6) where α = ln r 1 and k 2.
Model Presentation Degree distribution APL Average Path Length Let d t denote the average path length of Fibonacci network F t. We represent all the shortest path length of F t as a matrix in which the entries d ij is the shortest distance from node i to node j, then d t = D t N t (N t 1)/2 (7) is defined as the mean of d ij over all couples of nodes, where D t = i F t, j F t i j d ij (8) denotes the sum of the shortest path length between two nodes over all couples.
Then we write the sum D t+1 as D t+1 = D t + D t 1 + t, t 2, (9) with the initial conditions D 1 = 0 and D 2 = 1, where t is the sum over all shortest paths between one end point in F t branch and the other end point in F t 1 branch. In order to find t, we define s t = i F t d ix, (10) where X is the main root of F t and d XX is 0 for the sake of convenience. Considering the self-similar network structure, we easily find that s t has the following recursive relation: s t+1 = s t + s t 1 + N t 1 (11) for t 2 and initial conditions are s 1 = 0 and s 2 = 1.
Figure: The structure of F t+1 with two parts F α t and F α t 1.
To obtain an explicit expression for s t, we firstly try to seek the particular solution s 1 t of the difference equation s t+1 = s t + s t 1 + 1 r t 1. (12) We set s 1 t = A 1 t r1 t and insert it into Eq. (12), and then get A 1 = 1 r 1 r1 2 +1. Secondly, by the similar calculations, the particular solution s 2 t of the difference equation s t+1 = s t + s t 1 1 r2 t is s 2 t = A 2 t r t 2, where A 2 = 1 s t of Eq. (11) is r 2 r2 2 +1. Thus the particular solution s t = A 1 t r t 1 + A 2 t r t 2, (13) So the general solution s t of the difference equation s t+1 = s t + s t 1 is s t = A 3 r1 t + A 4r2 t, where A 3 and A 4 are arbitrary constants. So the general solution s t of Eq. (11) is s t = s t + s t, i.e., s t = A 1 t r t 1 + A 2 t r t 2 + A 3 r t 1 + A 4 r t 2. (14)
Using the initial conditions s 1 = 0 and s 2 = 1, we obtain A 3 = 1 r 2 r 1 (r 2 r 1 ) and A 4 = r 1 1 r 2 (r 2 r 1 ). On the other hand, let X be the main root of Ft α and Y be the main root of Ft 1 α respectively, by the above definition and d XY = 1, we have t = d ij = (d ix + d XY + d jy ) i F α t,j F α t 1 = N t 1 i F α t,j F α t 1 d ix + N t N t 1 d XY + N t d jy i F α t j F α t 1 = N t 1 s t + N t N t 1 + N t s t 1. (1)
Substituting Eqs. (3) and (14) into Eq. (1), we have t = C 1 r 2t 1 + C 2 r 2t 2 + C 3 (r 1 r 2 ) t +C 4 tr 2t 1 + C tr 2t 2 + C 6 t(r 1 r 2 ) t, (16) where C 1 = 1 r 1 + 1 A 3 + 1 (A 3 A 1 ) = 1 0 + 0 3, C 2 = 1 r 2 1 A 4 1 (A 4 A 2 ) = 1 0 0 3, C 3 = 1 (r 1 + r 2 ) + 1 (A 4 A 3 ) + 1 ( r 1(A 4 A 2 ) r 2 r 2(A 3 A 1 ) r 1 ) = 1 2, C 4 = 2 A 1 = 2, C = 2 A 2 = 2 C 6 = 1 (A 2 A 1 ) + 1 ( r 1A 2 r 2 r 2A 1 r 1 ) = 1.
Let D 1 t be the particular solution of the difference equation D t+1 = D t + D t 1 + C 1 r 2t 1. (17) Set Dt 1 = D 1 r1 2t, where D1 is a constant to be decided. Substituting this into Eq. (17), we get Dt 1 = D 1 r1 2t, where D 1 = C 1r1 2 r1 4 r2 1 1 = 4+ 0.
Proceeding similarly, we can obtain the particular solutions of the following difference equations: D t+1 = D t+1 = D t + D t 1 + C 2 r2 2t, D t + D t 1 + C 3 (r 1 r 2 ) t, D t+1 = D t + D t 1 + C 4 t(r 1 ) 2t, D t+1 = D t + D t 1 + C t(r 2 ) 2t, D t+1 = D t + D t 1 + C 6 t(r 1 r 2 ) t. (18) The corresponding particular solutions of Eq. (18) are: Dt 2 = D 2 r2 2t, D2 = 4 0, Dt 3 = D 3 (r 1 r 2 ) t, D 3 = 1 2, Dt 4 = (D 4 t + H 1 )r1 2t, D4 = + 0, H 1 = 1 9 Dt = (D t + H 2 )r2 2t, D = 0, H 2 = 1+9 Dt 6 = (D 6 t + H 3 )(r 1 r 2 ) t, D 6 = 1, H 3 = 2. 100, 100, (19)
With the results obtained for D i t, i = 1, 2,, 6, the particular solution D t of Eq. (9) is D t = 6 Dt. i (20) i=1 On the other hand, the general solution Dt equation D t+1 = D t + D t 1 is of the difference D t = B 1 r t 1 + B 2 r t 2, (21) where B 1 and B 2 are arbitrary constants. Hence the general solution of Eq. (9) is D t = Dt + Dt, i.e., D t = B 1 r t 1 + B 2 r t 2 + 6 Dt. i (22) i=1 From the initial conditions D 1 = 0 and D 2 = 1, we have B 1 = 20 and B 2 = + 20.
Then the analytical expression d t = B 1r1 t + B 2r2 t + 6 N t (N t 1)/2 i=1 Di t, t 2, (23) for the average path length is obtained, which approximates 10(D 4 t+d 1 +H 1 ) in the infinite t. The corresponding numerical r1 2 results are shown in Fig. 11.
Figure: Average path length d t versus network order N t.
Conclusions Inspired by the production principle of Fibonacci rabbits, this paper proposed Fibonacci network and studied the topological properties including the degree distribution and the average path length. We obtain the analytical results of degree distribution as well as the average path length. The distribution of this network follows exponential distribution, and it has the small-world characteristics. It is obvious that the clustering coefficient is zero. Our future work may include studying the generalized Fibonacci network and consider its dynamical processes such as trapping or random walks.
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