ECS 289 / MAE 298, Network Theory
Spring 2014
Problem Set # 1 b, Solutions
Problem 1
Consider BA but with uniform rather than preferential attachment, Pij = 1/
P
dj = 1/t.
Let nk,t denote number of nodes of degree k at time t. We need to consider nodes of degree
k > 1 and k = 1 separately:
1
1
k > 1; nk,t+1 = nk,t + nk−1,t − nk,t .
t
t
1
k = 1; n1,t+1 = n1,t + 1 − n1,t .
t
Assumption 1: pk,t =
nk,t
nt
=
nk,t
,
t
(1)
thus nk,t = pk,t · t.
Assumption 2: steady state pk,t → pk .
Writing Eqns (1) in terms using nk,t = pk · t:
1
1
k > 1 : pk · (t + 1) = pk · t + pk−1 · t − pk · t.
t
t
1
k = 1; p1 · (t + 1) = p1 · t + 1 − p1 · t.
t
(2)
(3)
Solving Eqn (3) gives the recurrence:
1
pk = pk−1
2
Solving Eqn (4) we find p1 = 1/2. Thus p2 = 12 p1 =
j
1
pj =
2
(4)
1 2
,
2
p3 = 12 p2 =
1 3
,
2
etc. So in general
(5)
This is not a power law. It is a geometric distribution (the discrete analog of an exponential
distribution).
1
Problem 2
(a) Let probability of picking a particular node of degree ki be pki = Aki . Since this is a probability,
the sum over all nodes present should sum to one. Using this we can determine the normalization
constant A.
X
pki = 2Am = 1 ⇒ A =
i
1
2m
(6)
ki
Therefore, probability of picking a node i with degree ki is pki = 2m
. Each time we add an edge
we have two times we pick a node, therefore, the probability of picking a node i when an edge is
added is given by kmi .
(b) We would like to track the probability of finding a node of degree k as we add edges. The
expected number of nodes with a given degree, k, after addition of an edge would be equal to the
expected number of nodes with degree k before the edge was added, plus the expected number
of nodes of degree k − 1 which would acquire an edge, minus the expected number of nodes with
degree k that would acquire an edge (and would have degree k + 1 after the addition of the edge).
Writing the above argument in terms of an equation,
pk (m + 1) = pk (m) +
k
(k − 1)
pk−1 (m) − pk (m)
m
m
(7)
(c) Rearranging the terms in the above equation and substituting m = nc/2,
k
(k − 1)
pk−1 (m) − pk (m)
pk (m + 1) − pk (m) =
m
m
nc
nc 2(k − 1)
nc 2k nc pk
+ 1 − pk
=
pk−1
− pk
.
2
2
nc
2
nc
2
Let P̃k (c) = pk (nc/2). Using the new variable, the above equation can be written as
h
i
2
c P̃k c + n − P̃k (c)
= (k − 1)P̃k−1 (c) − k P̃k (c) .
2
(8)
(9)
n
Taking the limit n → ∞ we get
c
dP̃k
= (k − 1)P̃k−1 (c) − k P̃k (c) .
dc
(10)
Since the symbol used to denote the variable is not relevant, we have
c
dpk
= (k − 1)pk−1 − kpk .
dc
2
(11)