Answer to Question #191456 in Complex Analysis for Milton

Question

We adopt the following notation: N and L indicate respectively the total number of

nodes and links of the network, Air indicates the generic element of the adjacency

matrix A of the network, ki

indicates the degree of node i and hki indicates the

average degree of the network.

At time t = 1 the network is formed by a n0 = 6 nodes m0 = 6 links.

At every time step t > 1 the network evolves according to the following rules:

- A link (r,s) between a node r and a node s is chosen randomly with uniform

probability

π(r,s) = Ar,s/L

and is removed from the network.

- A single new node joins the network and is connected to the rest of the network

by m links with m fixed to a time-independent integer constant satisfying

2 < m ≤ 6. Each of these new links connects the new node to a generic node j

chosen with probability

Πj =kj/(k)N

.a) Evaluate Π˜

i(t) indicating the expected increase in the number of links of node i

at any given time t and show that it follows the preferential attachment rule.

Answer 1

$\Pi_i=\sum_{r=1}^N \pi(i,r)=\sum _{r=1}^N\dfrac{A_{ir}}{L}$

As we know, $L=\dfrac{1}{2}\sum_{j=1}^NK_j$

         and $k_i=\sum_{r=1}^NA_{ir}$

So, $\Pi_i=\sum_{r=1}^N\dfrac{K_i}{\frac{1}{2}\sum_{j=1}^Nk_j}$

$\Pi_i=\dfrac{2k_i}{\sum_{j=1}^NK_j}$

So we have $\Pi_1,\Pi_2,\Pi_3..$ .

According to question-

$\dfrac{dk_i(t)}{dt}=\Pi_i=\dfrac{2k_i}{\sum_{j=1}^NK_j}$

We limit $t\ge 1$ , we have $\sum _jK_j =2L \sim ut$

Initilly $K_i(t_i)=6$

So , $k_i(t_i)=6(\dfrac{t}{t_1})^{0.5}$