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# 15 networks lecture 6 preferential attachment degree

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Unformatted text preview: degrees below a certain level d at time t , we need to identify which node is exactly at level d at time t . Let i (d ) be the node that has degree d at time t , or di (d ) (t ) = d . 15 Networks: Lecture 6 Preferential Attachment Degree Distribution From the degree expression, this yields � m �2 i (d ) = , t d leading to the distribution function F (d ) = 1 − m2 d −2 , with a corresponding density function P (d ) = 2m2 d −3 . Thus, the (expected) degree distribution is a power law with exponent −3. This is the argument given by Barabasi and Albert (1999). Networks generated by preferential attachment look very diﬀerent from earlier models with similar average degree. 16 Networks: Lecture 6 Master Equation Method—1 In subsequent work, Dorogovstev, Mendes and Samukhin (2000), took a diﬀerent approach, using what they call the “master equation” to obtain rigorous asymptotics for the mean degree of the nodes. Let pk denote the fraction of nodes in the network with degree k . The probability that a new edge attaches to a node of degree k is kpk kp = k, 2m ∑d dpd since the mean degree of the network is 2m (there are m edges added for each node, and each edge contributes two ends to the degrees of nodes). Thus, the mean number of nodes of degree k that gain an edge when a single new node with m edges is added is m kpk = kpk . 2m 2 The number of nodes with degree k , given by npk , thus decreases by this amount (since the nodes that get new edges become nodes with degree k + 1). 17 Networks: Lecture 6 Master Equation Method—2 The number of nodes with degree k also increases because of inﬂux from nodes of degree k − 1 that have just acquired a new edge (except for nodes of degree m, which have an inﬂux of exactly equal to 1 due to the addition of the new node with m edges). Let pk ,n denote the value of pk when the graph has n nodes. Then we can write the dynamics as (n + 1)pk ,n+1 − npk ,n = 1 1 (k − 1)pk −1,n − kpk ,n , 2 2 1 (n + 1)pm,n+1 − npm,n = 1 − mpm,n , 2 for k > m, for k = m. Focusing on stationary solutions pk ,n+1 = pk ,n = pk , it follows that �1 1 2 (k − 1)pk −1 − 2 kpk for k > m, pk = 1 mp 1− 2 m for k = m. 18 Networks: Lecture 6 Master Equation Method—3 Rearranging for pk , we ﬁnd pm = 2/(m + 2) and pk = pk −1 (k − 1)/(k + 2), or pk = (k − 1)(k − 2) · · · m 2m (m + 1) pm = . (k + 2)(k + 1) · · · (m + 3) (k + 2)(k + 1)k In the limit of large k , this gives a power law degree distribution pk ∼ k −3 . 19...
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