Download Feb 20 Part 1

2/20/2008 CS683
Part I
Haibo Lu
Grown graph
Nk t = number of components of size k
Let’s write Nk t as Nk t = ak t , t is time/steps elapsed
Generating function for Nk t
∞
kak x k
g x =
k=0
Use g(x) to solve for ak as recurrence equation:
g = −2δxg ′ + 2δxgg ′ + x
g
1−x 1
g′ =
∗
1 − g 2δ
g ′ 1 = expected size of finite components
(1) δ > δcritical : Giant component apppears
1
g 1 =1 ,
g′ 1 =
2δ
(2) δ ≤ δcritical : Only finite components
g′ x − g
g(x)
1− x
1
1
2
g ′ 1 = 𝑙𝑖𝑚
=
𝑙𝑖𝑚 − x ′ (use L^Hospital Law)
x→1 2δ 1 − g(x)
2δ x→1
−g
1
= 2δ 𝑙𝑖𝑚x→1
1
= 2δ ∗
[g ′ 1 ]2 −
g′ 1 =
g 1 −g ′ (1)
−g ′ (1)
g ′ 1 −g(1)
g ′ (1)
1 ′
1
g 1 + g 1 =0
2δ
2δ
1∓ 1−8δ
4δ
(only retain ‘-‘ term)
1
1
We can see g ′ 1 only has real solution if δ ≤ 8. δcritical = 8.
Derivation of Molley-Reed condition
Consider four generating functions:
 g 0 = degree of vertex chosen uniformly at random
 g1 = out degree of vertex at end of random edge
 h0 = size of component containing vertex chosen uniformly at random
 h1 = size of component at end of edge
∞
pk x k ,
g0 =
k=0
where pk is probability that a vertex chosen a random is of degree k
g1 =
1
x
∞
k
k=1 kpk x
∞
k=1 kpk
=
∞
k−1
k=0 kpk x
∞
k=0 kpk
=
g ′0 (x)
g ′0 (1)
(A)
Probability distribution for number of vertices at distance 2 from randomly
chosen vertices:
∞
pk (g1 (x))k = g 0 (g1 (x))
k=0
Equation for h1 (x)
Let q k be probability of k outgoing edges
x[q 0 + q1 h1 x + q 2 h1 x
h1 x = xg1 (h1 x )
2
+ ⋯]
Equation for h0 (x)
∞
pk (h1 x )k = xg 0 (h1 x )
h0 x = x
k=0
(Continue on Part II)