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Clustering
Consider the Erdos-Renyi random graph G1(n, p) with mean degree a.
(a) Show that in the limit of large n, the expected number of triangles in the network
is a constant. Call Δi,j,k as the indicator random variable that denotes a triangle
between i, j, k. Then
b) Calculate the clustering coefficient C in the limit of large n. Note: The clustering
coefficient is defined as three times the number of triangles divided by the number of
connected triplets. A “connected triplet” means three vertices uvw with edges (u, v)
and (v,w). The edge (u,w) can be present or not.
Call as the indicator random variable that denotes a triple between i, j, k.
Clustering Coefficient = p
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limn→∞ p = 0.
(c) Calculate the clustering coefficient C for the Erdos-Renyi random graph G2(n, p)
with p(n) = a log(n)/n. Compare your answer with part (b) in the limit of large n.
As shown before, it is just p, and still 0. But since p is larger in the latter case,
there is a higher chance that a triple is actually a triangle.
d) (2pt) Compare the ratio of the diameters of G1 and G2 in the limit of large n.
For the value of p in G1 the graph is almost surely disconnected. As a result it
has infinite diameter. On the other hand, G2 has a diameter that is Ω(log(n))
(and is almost surely connected). Therefore, the ratio G2/G1 = 0.
(e) Now, consider a different construction for a random graph model. We take n
vertices and go through each distinct trio of three vertices and with
independent probability p = connect the trio using three edges
to form a triangle. Compute the mean vertex degree and clustering coefficient for
this network model.
Pick an arbitrary vertex i, let Δij be the indicator random variable denoting a
link between i, j. Then

Then the average degree is E[Δij ](n-1) = a. Observe that in this case the number of
triangles grow linearly as the number of nodes
Number of Triangle =
Also, the number of triples of uvw occurs when either uvw is a triangle or there are
triangles uvj and vwk and no triangle uvw where j, k =6 u, v,w. Call the event of no
triangle in uvw but there exists a triple between them as
The ratio is roughly 3/(3a + 1) and points have been awarded to show that the
clustering coefficient is a constant (does not decay to 0 as n → ∞).
2. Centrality in Infinite Graphs.

In this problem, you will demonstrate an example that shows that eigenvector centrality
can be very sensitive to minimal changes in a network. The problem is broken into
different components that finally lead to the conclusion.
Part I
First, consider the infinite ring network as in Figure 1.
Assume that xi is the eigenvector centrality measure of node i.
(a) Show that the xi’s are computed by finding the largest λ for which there exists a set
of xi for i = 0, 1, 2, . . . such that:

λxi = xi-1 + xi+1, ∀i = 1, 2, . . .
Note that we can always normalize the eigenvector centrality by dividing xi by x0 for all
i, so that x0 = 1.
This follows by representing the ring network as Ax = λx where λ is the largest
eigenvalue and A is the adjacency matrix.
(b) Show that all the nodes have equal ranking. (Hint: Show that the is xi = 1 for all i.
This should be a straightforward conclusion and can be proved by inspection.).
One can argue by symmetry or show that λ = 2. If λ = 2 all xi = 1 is a solution
and hence the eigenvector. λ cannot be larger than 2. If so, and xt was the
largest then
λxt = xt-1 + xt+1 ≤ 2xt
λ ≤ 2 indeed.
Part II
Next, as shown in Figure 2, we add an edge between two nodes so that the infinite ring is
divided into two symmetric halves. We will examine the eigenvector centrality of this new
network. By symmetry, we only need to find the eigenvector centrality

Figure 2:
measures indexed by x0, x1, x2, . . . . As before, we always normalize it so that x0 = 1.
(c) Write down the system of equations which characterize the eigenvalue centrality.
λx0 = 2x1 + x0
λxt = xt-1 + xt+1 t > 0
(d) Show that the eigenvector centrality must satisfy:
x0 ≥ x1 ≥ x2 ≥ · · · ≥ 0
(Hint: start with an initial guess xi = 1 for all i, and try to iteratively compute the
eigenvector. You can prove the inequalities by induction.) Take initial guess as xi t = 1,
where

Clearly for t = 0 this is true (after many iterations this will converge to the true
centrality). Assume this for t = k. Now observe for
The last inequality follows from inductive hypothesis.
(e) Show that the largest eigenvalue λ must satisfy:
2 ≤ λ ≤ 3
(Hint: observe the system of equations you wrote down for (c).) λ ≥ 2 because if we
sum up the equation in (c) we get
Since x0 is the largest, we also have that
λx0 = 2x1 + x0 ≤ 3x0

(f) We have shown in Part (e) that xn is positive and decreasing in n. Please prove that
limn→∞ xn = 0. (Hint: consider writing the system of equations in Q2 into the form of
a linear dynamical system with state y[n] given by:
write down the recursive equation y[n + 1] = Ay[n] that describes the evolution of the
linear dynamics, and think about the equilibrium.) Now you have demonstrated that by
adding a single edge one can change the relative centrality measure drastically
Observe that
for all n ≥ 1. Then since xn ≤ xn-1 we only need to show that there is an eigenvalue of A
that is < 1. As it turns out that eigenvalue is

It can be shown that λA < 1 whenever λ > 2. This implies that xn → 0.
3. Synchronization.
An oscillator is a simple dynamical system that can be modeled by a first order
differential equation. A network of n oscillators can be modeled by a system of
differential equations of the form:
where θi represents the phase angle and is the state of the oscillator on vertex i, ω is a
constant, and the function g(x) has g(0) = 0 and respects the rotational symmetry of the
phases, meaning that g(x + 2π) = g(x) for all x.
(a) Characterize all solutions of the form θi(t) = ait+bi to the set of dynamical equations,
i.e., find ai, bi, i = 1, . . . , n.
ai = ω, bi = b + 2kiπ
(b) Consider a small perturbation away from the state θi = ωt + i and show that the
vector = (1, 2, . . . ,) satisfies

Your solution should specify L in terms of [Aij ], the adjacency matrix of an undirected
graph. (Hint: Using the Taylor series approximation of g(·) around x0, i.e., g(x) = g(x0)
+ (x - x0)g (x0) + maybe helpful) Substitute θi = ωt + i. Then the Taylor
approximation around 0 can be written as
Now stacking the i we get
(D - A) = L.
L is also called the graph Laplacian. We will prove that L has only non–negative
eigenvalues.

(c) Show that L = MTM where M is the incidence matrix, i.e., the rows correspond to
the edges and columns correspond to the vertices. Therefore, for every edge e = (i, j)
between i, j where i < j we have that
Mev = -1 if v = i
Mev = 1 if v = j
Mev = 0 otherwise
The proof is here
(d) Argue that L is a symmetric matrix and that for any vector x we have that xTLx ≥ 0.
Conclude from this that all the eigenvalues of L are non–negative. (Hint: Use the fact that
all eigenvalues, λ, of a symmetric matrix, P, are of the form
where v is the corresponding eigenvector.)
(e) For what values of g 0 (0) is the system stable to small perturbations around the
origin?

ensures system is stable.
(Hint: You can use a Lyapunov argument with quadratic Lyapunov function V (x) = xT x to
examine stability of the linearized system in part (b)).

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