MATH6142 Complex Networks Exam 2016

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1 MATH642 Complex Networks Exam 26 Solution Problem a) The network is directed because the adjacency matrix is not symmetric. b) The network is shown in Figure. (4 marks)[unseen] 4 3 Figure : The directed network with adjacency matrix A. c) The network contains two weakly connected components formed respectivelly by the nodes {, 2} and {3, 4}. (2 marks)[unseen] d) The network contains one strongly connected component formed by the nodes {3, 4}. e) The network does not have any out-component. f) The in-degree sequence is fiven by {,,, } the out-degree sequence is given by {,,, }. (4 marks) [Unseen] g) The in-degree distribution is given by P in () = /4, P in () = 3/4P in (k) = for k = 2, 3. (2 marks) [Unseen] The out-degree distribution is given by P out () = /4, P out () = 3/4P out (k) = for k = 2, 3. (2 mark2) [Unseen] h) The eigenvector centrality x can be found as following.

2 Given the initial guess x () = N given by x () = 4 () let us is calculate the result of the iteration We have x () = x (2) = x (3) = x n = Ax n. (2) = 4 = 4 = 4... Therefore for every n 3. It follows that x (n) = 4 x = 2. (3). (4) 2

3 i) The result of point h) is expected because nodes and 2 are in an in-component of the network and therefore have zero eigenvector centrality. Nodes 3 and node 4 instead have the same eigevcetor centrality because they form a dimer. Solution Problem 2 a) The undirected network of N nodes with smallest diameter is the complete graph, formed by N nodes each one connected to all the others. The diameter of this network is D =. This network has the small-world distance property. In fact lim D ln N = lim ln N =. (5) b) The undirected network of N nodes which is connected and has the largest diameter is the (open) linear chain of N nodes that has diameter D = N. This network does not have the small-world distance property. In fact lim D ln N = lim N ln N =. (6) c) i) The degree distribution P (k) of the Poisson network with average degree k = c = 4 is given by P (k) = k! ck e c. (7) ( marks)[bookwork] The generating function G(x) is given by G(x) = P (k)x k = e c+cx. (8) k= The first and second moment of the Poisson degree distribution can be obtained by differentiating the generating function G(x). In particular we have k = dg B (x) dx k(k ) = d2 G B (x) dx 2 3 = c x= = c 2. (9) x=

4 It follows that b = k(k ) k = c = 4. () ( marks)[bookwork] ii) Using the result of point i) we get ( ) d k(k ) N d k = k d () k Therefore N d 4 d. (2) ( marks)[unseen] iii) Since at distance d = from the random node there is only one node (the node itself) we have N d 4 d (3) for d. ( marks)[unseen] The total number of nodes N d l at distance d l from any random node of the Poisson network is given by Therefore we have N d l = l d= N = 3 4 d = 4l+ 4. (4) [ 4 l+ ] (5) ( marks)[unseen] iv) By imposing that all the nodes of the network are found at distance d l from a random node we have N = N d l = [ 4 l+ ] (marks) 3 3N = 4 l+ (mark) 3N + = 4 l+ (mark) l = ln [3N + ].(mark) (6) 2 ln 2 4

5 v) Using Eq. (6) and considering the leading term for N we have l ln[3n] 2 ln 2 ln N 2 ln 2. (7) (4 marks) [Unseen] vi) The local clustering coefficient C i of a generic node i of a Poisson network can be estimated to be Number of triangles passing through node i C i = k i (k i )/2 pk i(k i )/2 k i (k i )/2 (8) where k i is the degree of node i and p = k /N is the probability of a link between any two nodes of the network. (2 marks) [Bookwork] In fact each node i belongs in average to pk i (k i )/2 triangles since any pair of its neighbor is connected with probability p. (2 marks) [Bookwork] Solution Problem 3 a) The average number of links Π i added to node i in timestep t is given by Π i = π (i,r) = r= r= A ir L. (9) Using we get L = 2 j= k j (2 marks) [Unseen] k i = A ir (2) r= Π i = 2k i N j= k j (2) (4 marks) [Unseen] b)since initially the number of links is m = and at each time we add 5

6 two links we have L = + 2t. ( mark) [Unseen] Since initially we have n = 2 nodes and at each time we add a node we have N = 2 + t. ( mark) [Unseen] c)the average degree k is given by k = 2L N 2( + 2t) =. (22) 2 + t (4 marks) [Bookwork] d)in the mean-field approximation, the degree k i (t) of node i at time t satisfies the following differential equation dk i (t) dt = Π i = 2k i N j= k j (23) In the limit t, we have j k j = 2L 4t. Therefore we can write the dynamical mean-field equation for the degree k i (t) of node i, getting with initial condition k i (t i ) = 2. This equation has solution dk i dt = 2k i 4t = k i 2t, (24) ( ) /2 t k i (t) = 2. (25) t i e) The probability P (k i (t) > k) that a random node has degree k i (t) > k, in the mean-field approximation can be calculated as follows ( ( ) ( (/2) ( ) ) 2 ( ) 2 t 2 2 P (k i (t) > k) = P 2 > k) = P t i < t =. (26) k k t i Therefore, the degree distribution P (k) is given by P (k) = dp (k i(t) < k) dk = d dk ( ) 2 2 = k ( ) 3 2. (27) k f) The master equation for N k (t) reads N k (t + ) = N k (t) + k N k (t)( δ k,2 ) k 2t 2t N k(t) + δ k,2 (28) 6

7 g) Assuming N k (t) (t + n )P (k) for large t we get P (k) = k P (k )( δ k,2 ) k 2 2 P (k) + δ k,2. (29) giving P (k) = k P (k ) (3) k + 2 for k > 2 and P (2) = 2. (3) Solving these equations we get P (k) = 2 k(k + )(k + 2). (32) 7

ECS 289 / MAE 298, Network Theory Spring 2014 Problem Set # 1, Solutions. Problem 1: Power Law Degree Distributions (Continuum approach)

ECS 289 / MAE 298, Network Theory Spring 204 Problem Set #, Solutions Problem : Power Law Degree Distributions (Continuum approach) Consider the power law distribution p(k) = Ak γ, with support (i.e.,