Uncovering the role of elementary processes in network evolution

The growth and evolution of networks has elicited considerable interest from the scientific community and a number of mechanistic models have been proposed to explain their observed degree distributions. Various microscopic processes have been incorporated in these models, among them, node and edge addition, vertex fitness and the deletion of nodes and edges. The existing models, however, focus on specific combinations of these processes and parameterize them in a way that makes it difficult to elucidate the role of the individual elementary mechanisms. We therefore formulated and solved a model that incorporates the minimal processes governing network evolution. Some contribute to growth such as the formation of connections between existing pair of vertices, while others capture deletion; the removal of a node with its corresponding edges, or the removal of an edge between a pair of vertices. We distinguish between these elementary mechanisms, identifying their specific role on network evolution.


Table of Contents
S1 Converting the rate equation to a differential equation The rate equation in the manuscript Eq.(3) is hard to treat analytically, primarily due to the presence of the joint probabilities π k,k and p k,k . This can be simplified, however, if we make the assumption that our network lacks degree correlations (or has vanishingly small values for the correlation). In this case, p k,k can be factorized as p k p k , while j e k,j = kp k / k , therefore Eq.
(3) can be written as, Defining the generating functions, multiplying by z k , summing over k and inserting the attachment kernels from Eqn. (1) and (2) we arrive at the differential equation, where, S2 Solving for the degree distribution p k The expression in Eq. (S2) can be solved via numerical methods. However, we are interested in the explicit form of p k . Unfortunately the equation in its complete form does not lend itself well to analytical techniques. We can make progress however by considering specific cases. We start by neglecting deletion processes, and consider the case of pure growth, which can be solved exactly. Following this, we will resort to approximation methods to solve the evolution process including deletion. S-2

S2.2 Growth with deletion
We have established that the pure growth process leads to a power law distribution. Previous work [27,45] indicates that the presence of deletion can potentially induce a transition from a power law to an exponential regime. To account for both regimes, we assume p k follows a power-law with an exponential correction, p k = Ck −γ Ω k . Next, we simplify the expression for the attachment kernels by setting b, t = 1, such that π k = A(a + k) and f k = B(t + k) and solve for γ and Ω in the limit k >> 1. To do so, we employ a high-degree expansion of the telescoping product p k /p k−1 to leading orders in 1/k thus, S-3 Multiplying Eq.(3) by 1/p k−1 , substituting this expansion and ignoring terms in 1/k, yields the equation, − Ω + (r + 2qm/ k )Ω 2 + γ[Ω(Ac + 2mB − 2Ω(r + 2qm/ k ) + (r + 2qm/ k )] = 0.
Since (S9) must be true for all k, we can set the coefficient of k to zero, which gives two solutions for Ω, namely Ω = 1 and Ω = Ac + 2mB r + 2qm/ k . (S10) If Ω < 1 then the solution (S10) is normalizable and p k decays exponentially (with a powerlaw correction). However if the ratio is greater than 1, it does not correspond to a normalizable probability distribution and therefore the correct solution must be Ω = 1, leading to a purely power law distribution p k ∼ k −γ .
To determine the expression for the power-law exponent γ, we set Ω = 1, the k-independent term in (S9) to zero, finding . (S11)

S2.3 Stretched exponential at the critical point a c
At the critical point a c (Eq. (9)), p k follows a stretched exponential. Making the ansatz, and once again employing the high-degree expansion, but this time expanding to order k −3/2 , we have As before we substitute this into Eq. (3) and find that for ζ = 0, the terms of order k and √ k force B = 1 and r + 2qm/ k = A(c + 2m). Thus ζ can be non-zero only at the critical point. The term of order 1 gives, ζ = 2 c + m(1 − q) cr + m(q + r) 1/2 (S13) and the term of order k −1/2 gives which reduces to the work of [27] in the special case a = 0, m = 0 and r = 1. S-4