Chapter 14, Complex networks Video Solutions, A Kinetic View of Statistical Physics

Problem 1

Show that regular random graphs with $z=1$ and $z=2$ can be defined dynamically and that the planarity condition is automatically obeyed. For $z=1$, the graph is a collection of dimers, while for $z=2$, the graph is a collection of loops. Show that for $z=1$ the final number of components is $N / 2$, while for $z=2$ the final number of components scales logarithmically with system size.

Hoan Nguyen

Numerade Educator

03:48

Problem 2

Solve the evolution equations (14.1) for the degree distribution of the ER graph, subject to the initial condition $n_{k}(t=0)=\delta_{k, 0}$, and show that the solution is given by (14.2). Try different methods (directly solving the equations one by one, the Laplace transform method, and the generating function technique).

Ryan Williams

Numerade Educator

01:31

Problem 3

Derive the governing equations for the moments $U_{1}=\sum_{k \geq 1} k u_{k}$ and $U_{2}=$ $\sum_{k \geq 1} k^{2} u_{k}$ and solve these equations. Show that in the sol phase
$$
U_{1}=\frac{t}{2(1-t)^{2}}, \quad U_{2}=\frac{t(1+t)}{2(1-t)^{4}}
$$

Xiaomeng Zhang

Numerade Educator

01:48

Problem 4

Find a solution to the recurrence (14.18). One way is via generating functions:
(a) Using the exponential generating functions, $\mathcal{A}(z)=\sum_{k \geq 1} A_{k} e^{k z}$ and $\mathcal{H}(z)=$ $\sum_{k \geq 1}\left(k^{k-1} / k !\right) e^{k z}$, recast the recurrence (14.18) into the differential equation
$$
\mathcal{A}=(1-\mathcal{H})^{-1} \frac{d \mathcal{H}}{d z}
$$
(b) The generating function $\mathcal{H}(z)$ encapsulates the cluster size distribution at the percolation point. Using the results from Chapter 5 show that $\mathcal{H} e^{-\mathcal{H}}=e^{z}$ Combine this with the above differential equation to yield $\mathcal{A}=(1-\mathcal{H})^{-2} \mathcal{H}$.
(c) Employ the Lagrange inversion method to compute $A_{k}$.

Jack Chen

Numerade Educator

01:40

Problem 5

Determine the large- $k$ asymptotic behavior of the sum $\sum_{0 \leq n \leq k-1} k^{n-1} / n !$. Here is one possible line of attack:
(a) Notice that the last term in the sum is the largest. This suggests reversing the summation order and writing the sum in the form
$$
\frac{k^{k-2}}{(k-1) !} S, \quad S=1+\frac{k-1}{k}+\frac{(k-1)(k-2)}{k^{2}}+\frac{(k-1)(k-2)(k-3)}{k^{3}}+\cdots
$$
(b) To compute $S$ show that in the large- $k$ limit it is possible to use the continuum approximation to give
$$
\begin{aligned}
S=\sum_{j=0}^{k-1} \prod_{i=0}^{j}\left(1-\frac{i}{k}\right) & \rightarrow \int_{0}^{k} d j \exp \left[-\frac{j(j+1)}{2 k}\right] \\
& \rightarrow \int_{0}^{\infty} d j \exp \left[-\frac{j^{2}}{2 k}\right]=\sqrt{\pi k / 2}
\end{aligned}
$$

Linh Vu

Numerade Educator

02:30

Problem 6

Show that, for a large but finite evolving random graph, the probability $s_{n}(N)$ to have $n$ unicyclic clusters at the percolation point is given by
$$
s_{n}(N) \sim \frac{1}{n !} N^{-1 / 6}\left[\frac{1}{6} \ln N\right]^{n}
$$

Mengchun Cai

Numerade Educator

01:43

Problem 7

Compute the exact average degree distribution $\left\langle N_{k}(N)\right\rangle$ for the random recursive tree for the first few $N=1,2,3, \ldots$, by solving the recursion
$$
\left\langle N_{k}(N+1)\right\rangle-\left\langle N_{k}(N)\right\rangle=\frac{\left\langle N_{k-1}(N)\right\rangle-\left\langle N_{k}(N)\right\rangle}{N}+\delta_{k, 1}
$$
Here you should apply the generating function $G_{k}(w)=\sum_{N \geq 1}\left\langle N_{k}(N)\right\rangle w^{N-1}$ to convert this recursion into soluble equations. Next, expand the solutions for $G_{k}(w)$ in a power series in $N$ to obtain the degree distributions. In particular, show that the average number of nodes of degree one is given by (14.27). Finally, compare your results with the asymptotic average degree distribution $N_{k}(N) \simeq N / 2^{k}$.

Manik Pulyani

Numerade Educator

06:02

Problem 8

Extend the genealogical tree construction to the RRT with redirection.
(a) First show that the master equation for $L_{g}$, the number of nodes in generation $g$, is given by
$$
\frac{d L_{g}}{d N}=\frac{(1-r) L_{g-1}+r L_{g}}{N}
$$
(b) Show that the solution for $L_{g}$ is
$$
L_{g+1}(\tau)=\int_{0}^{\tau} d x \frac{[(1-r) x]^{g}}{g !} e^{x r}
$$
where $\tau=\ln N$.

Chris Trentman

Numerade Educator

03:23

Problem 9

Show that the limiting behavior of $\mu$ in $A(N)=\mu N$ is given by
$$
\begin{array}{ll}
\mu=1+B_{0} \gamma+\mathcal{O}\left(\gamma^{2}\right), & \gamma \downarrow 0 \\
\mu=2-B_{1}(1-\gamma)+\mathcal{O}\left((1-\gamma)^{2}\right), & \gamma \uparrow 1
\end{array}
$$vwith
$$
B_{0}=\sum_{j \geq 1} \frac{\ln j}{2^{j}} \doteq 0.5078, \quad B_{1}=4 \sum_{j \geq 1} \frac{\ln j}{(j+1)(j+2)} \div 2.407
$$
Here $\gamma$ is the exponent in the attachment rate $A_{k}$ defined by $A_{k}=k^{\gamma}$.

Surendra Kumar

Numerade Educator

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Problem 10

For linear preferential attachment, $A_{k}=k$, calculate $N_{k}$ explicitly from the master equations (14.34) for $k=1,2,3$, and 4 .

Shu Naito

Numerade Educator

07:16

Problem 11

Determine the degree distribution for the shifted linear attachment rate $A_{k}=k+\lambda$. First simplify the normalization factor $A=\sum_{j} A_{j} N_{j}$ to $A(N)=M_{1}+\lambda M_{0}=$ $(2+\lambda) N$, implying that $\mu=2+\lambda$. Using these results show that the degree distribution is
$$
n_{k}=(2+\lambda) \frac{\Gamma(3+2 \lambda)}{\Gamma(1+\lambda)} \frac{\Gamma(k+\lambda)}{\Gamma(k+3+2 \lambda)}
$$
Show that asymptotically this distribution decays as $k^{-3-\lambda}$. Notice also that Eq. (14.34) with the shifted linear attachment rate $A_{k}=k+\lambda$ is identical to that of the redirection model, Eq. (14.32), when the correspondence $\lambda=1 / r-2$ is made,

Amany Waheeb

Numerade Educator

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Problem 12

Determine the degree distribution for the growth process in which, with probability $p$, a new node attaches to an existing node of the network by linear preferential attachment and, with probability $1-p$, a new node attaches to an existing node by uniform random attachment.

Victor Salazar

Numerade Educator

01:16

Problem 13

Consider the attachment rate $A_{1}=1$ and $A_{k}=a k$ for $k \geq 2$. Show that the resulting degree distribution is asymptotically a power law, $n_{k} \sim k^{-v}$, with $v=$ $(3+\sqrt{1+8 / a}) / 2$, which can indeed be tuned to any value larger than 2 .

Manik Pulyani

Numerade Educator

04:49

Problem 14

Generalize linear preferential attachment networks to the case where each new node links to $m$ pre-existing nodes. Write the master equation for this process, and by applying the same approach as that used for Eq. (14.34), find the degree distribution.

Cory Glover

Numerade Educator

02:11

Problem 15

Suppose that each node is assigned an initial "attractiveness" $\eta>0$ from a distribution $p_{0}(\eta)$. The attachment rate of a node with degree $k$ and attractiveness $\eta$ is defined as $A_{k}(\eta)$. Let $N_{k}(\eta)$ be the number of nodes of degree $k$ and attractiveness $\eta$. Show that this joint degree-attractiveness distribution evolves according to
$$
\frac{d N_{k}(\eta)}{d N}=\frac{A_{k-1}(\eta) N_{k-1}(\eta)-A_{k}(\eta) N_{k}(\eta)}{A}+p_{0}(\eta) \delta_{k 1}
$$
where $A=\int d \eta \sum_{k} A_{k}(\eta) N_{k}(\eta)$ is the total attachment rate. By using the substitutions $A=\mu N$ and $N_{k}(\eta)=N n_{k}(\eta)$ reduce the above master equation to a recursion whose solution is
$$
n_{k}(\eta)=p_{0}(\eta) \frac{\mu}{A_{k}(\eta)} \prod_{1 \leq j \leq k}\left(1+\frac{\mu}{A_{j}(\eta)}\right)^{-1}
$$For the specific example where the attachment rate is linear in the degree and attractiveness, $A_{k}(\eta)=\eta k$, show that the degree distribution is
$$
n_{k}(\eta)=\frac{\mu p_{0}(\eta)}{\eta} \frac{\Gamma(k) \Gamma(1+\mu / \eta)}{\Gamma(k+1+\mu / \eta)}
$$

Manik Pulyani

Numerade Educator

09:12

Problem 16

Consider a growing network in which nodes are introduced one by one and each can link to one pre-existing node of degree $k$, with rate $A_{k}=k^{\gamma}$, and with $\gamma>1$.
(a) Start with the master equation for $N_{1}$ and show generally that $N_{1}$ is proportional to $N$ for any value of $\gamma$
(b) Next study the master equation for $N_{2}$. Show that asymptotically, $d N_{2} / d N \sim$ $N^{1-\gamma}$, and thereby conclude that for $1<\gamma<2, N_{2}$ grows as $N^{2-\gamma}$.
(c) Continue this line of reasoning to show that for $\frac{j}{j-1}<\gamma<\frac{j+1}{j}, N_{k} \sim$ $N^{k-(k-1) \gamma}$ for $k \leq j$, while $N_{k} \sim \mathcal{O}(1)$ for $k>j$

Chris Trentman

Numerade Educator

02:42

Problem 17

Fill the gaps in the derivation of the Gaussian distribution that was outlined in (14.46)-(14.49). In particular, derive the differential equation for the scaling function $\Phi(\eta)$ and solve it.

Amany Waheeb

Numerade Educator

33:09

Problem 18

The goal of this problem is to solve Eq. (14.50).
(a) Show that the homogeneous version of (14.50) admits the following solution:
$$
\prod_{1 \leq j \leq N}\left(1+\frac{2 p}{j}\right)=\frac{\Gamma(2 p+N)}{\Gamma(2 p+1) \Gamma(N)}
$$
b) Argue that it is a good idea to use the solution of the homogeneous version of (14.50) as an integrating factor, that is, to seek the solution in the form
$$
L(N)=U(N) \frac{\Gamma(2 p+N)}{\Gamma(2 p+1) \Gamma(N)}
$$
Show that the above ansatz recasts (14.50) into
$$
U(N+1)=U(N)+\frac{\Gamma(2 p+1) \Gamma(N+1)}{\Gamma(2 p+N+1)}
$$
(c) Solve the above recurrence to yield
$$
L(N)=\frac{\Gamma(2 p+N)}{\Gamma(N)} \sum_{2 \leq j \leq N} \frac{\Gamma(j)}{\Gamma(2 p+j)}
$$
d) Using the well-known asymptotic relation
$$
\frac{\Gamma(2 p+x)}{\Gamma(x)} \simeq x^{2 p} \quad \text { when } \quad x \gg 1
$$
confirm the asymptotic behavior (14.52) when $p<1 / 2$.
e) Show that, when $p=1 / 2$, the exact result is $L(N)=N\left(H_{N}-1\right)$ where $H_{N}=\sum_{1 \leq j \leq N} j^{-1}$ is the harmonic number. Using the asymptotics of the harmonic numbers confirm the prediction of Eq. (14.52) for $p=1 / 2$.(f) Show that the sum on the right-hand side of the equation in (c) converges. Compute this sum using the identity
$$
\sum_{k \geq 0} \frac{\Gamma(b+k)}{\Gamma(c+k)}=\frac{\Gamma(b)}{(c-b-1) \Gamma(c-1)}
$$
Confirm Eq. (14.52) for $p>1 / 2$ with $A(p)=1 /[(2 p-1) \Gamma(1+2 p)]$.

Lucas Finney

Numerade Educator