Complex Graphs and Networks

Lecture 4: The rise of the giant component

Linyuan Lu [email protected] University of South Carolina

BASICS2008 SUMMER SCHOOL July 27 – August 2, 2008 Overview of talks

■ Lecture 1: Overview and outlines ■ Lecture 2: Generative models - preferential attachment schemes ■ Lecture 3: Duplication models for biological networks ■ Lecture 4: The rise of the giant component ■ Lecture 5: The small world phenomenon: average distance and diameter ■ Lecture 6: Spectrum of random graphs with given degrees

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 2 / 47 Random graphs

A random graph is a set of graphs together with a probability distribution on that set.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 3 / 47 Random graphs

A random graph is a set of graphs together with a probability distribution on that set. Example: A random graph on 3 vertices and 2 edges with the uniform distribution on it.

j j j

j j j j j j 1 1 1 Probability 3 Probability 3 Probability 3

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 3 / 47 Random graphs

A random graph is a set of graphs together with a probability distribution on that set. Example: A random graph on 3 vertices and 2 edges with the uniform distribution on it.

j j j

j j j j j j 1 1 1 Probability 3 Probability 3 Probability 3 A random graph G almost surely satisfies a property P , if

Pr(G satisfies P ) = 1 o (1). − n Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 3 / 47 Erd˝os-R´enyi model G(n, p)

- n nodes

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 4 / 47 Erd˝os-R´enyi model G(n, p)

- n nodes - For each pair of vertices, create an edge independently with probability p.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 4 / 47 Erd˝os-R´enyi model G(n, p)

- n nodes - For each pair of vertices, create an edge independently with probability p. n e (2) e - The graph with e edges has the probability p (1 p) − . −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 4 / 47 Erd˝os-R´enyi model G(n, p)

- n nodes - For each pair of vertices, create an edge independently with probability p. n e (2) e - The graph with e edges has the probability p (1 p) − . −

j@ j @ @ @ @ j j

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 4 / 47 Erd˝os-R´enyi model G(n, p)

- n nodes - For each pair of vertices, create an edge independently with probability p. n e (2) e - The graph with e edges has the probability p (1 p) − . −

j@ j @ p @ @p p @ jp j

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 4 / 47 Erd˝os-R´enyi model G(n, p)

- n nodes - For each pair of vertices, create an edge independently with probability p. n e (2) e - The graph with e edges has the probability p (1 p) − . − 1 p − j@ppppppp j p @ p @ p p p1 p @p p p − p @ p jp j

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 4 / 47 Erd˝os-R´enyi model G(n, p)

- n nodes - For each pair of vertices, create an edge independently with probability p. n e (2) e - The graph with e edges has the probability p (1 p) − . − 1 p − The probability of this j@ppppppp j p @ p graph is @ p p p1 p @p p p p − 4 2 @ p p (1 p) . jp j −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 4 / 47 1 A example: G(3, 2)

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 5 / 47 The birth of random graph theory

Paul Erd˝os and A. R´enyi, On the evolution of random graphs Magyar Tud. Akad. Mat. Kut. Int. Kozl. 5 (1960) 17-61.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 6 / 47 The birth of random graph theory

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 7 / 47 Evolution of G(n, p)

0 the empty graph. disjoint union of trees. c n cycles of any size. 1 n the double jumps. c′ n one giant component, others are trees. p log n n G(n,p) is connected. log n Ω( n ) connected and almost regular. ǫ 1 Ω(n − ) finite diameter. Θ(1) dense graphs, diameter is 2. 1 the complete graph.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 8 / 47 Evolution of G(n, p)

Range I p = o(1/n) The random graph Gn,p is the disjoint union of trees. In fact, trees on k vertices, for k = 3, 4,... only appear k/(k 1) when p is of the order n− − .

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 9 / 47 Evolution of G(n, p)

Range I p = o(1/n) The random graph Gn,p is the disjoint union of trees. In fact, trees on k vertices, for k = 3, 4,... only appear k/(k 1) when p is of the order n− − . k/(k 1) Furthermore, for p = cn− − and c > 0, let τk(G) denote the number of connected components of G k 1 k 2 formed by trees on k vertices and λ = (2c) − k − /k!. Then, λje λ Pr(τ (G ) = j) − k n,p → j! for j = 0, 1,... as n . → ∞

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 9 / 47 Evolution of G(n, p)

Range II p c/n for 0 < c < 1 ∼ ■ In this range of p, Gn,p contains cycles of any given size with probability tending to a positive limit.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 10 / 47 Evolution of G(n, p)

Range II p c/n for 0 < c < 1 ∼ ■ In this range of p, Gn,p contains cycles of any given size with probability tending to a positive limit. ■ All connected components of Gn,p are either trees or unicyclic components. Almost all (i.e., n o(n)) vertices are in components which are trees.−

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 10 / 47 Evolution of G(n, p)

Range II p c/n for 0 < c < 1 ∼ ■ In this range of p, Gn,p contains cycles of any given size with probability tending to a positive limit. ■ All connected components of Gn,p are either trees or unicyclic components. Almost all (i.e., n o(n)) vertices are in components which are trees.− ■ The largest connected component of Gn,p is a tree 1 5 and has about α(log n 2 log log n) vertices, where α = c 1 log c. − − −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 10 / 47 Evolution of G(n, p)

Range III p 1/n + µ/n, the double jump ∼ ■ If µ < 0, the largest component has size 1 (µ log(1 + µ))− log n + O(log log n). −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 11 / 47 Evolution of G(n, p)

Range III p 1/n + µ/n, the double jump ∼ ■ If µ < 0, the largest component has size 1 (µ log(1 + µ))− log n + O(log log n). − ■ If µ = 0, the largest component has size of order n2/3.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 11 / 47 Evolution of G(n, p)

Range III p 1/n + µ/n, the double jump ∼ ■ If µ < 0, the largest component has size 1 (µ log(1 + µ))− log n + O(log log n). − ■ If µ = 0, the largest component has size of order n2/3. ■ If µ > 0, there is a unique giant component of size 1 αn where µ = α− log(1 α) 1. − − −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 11 / 47 Evolution of G(n, p)

Range III p 1/n + µ/n, the double jump ∼ ■ If µ < 0, the largest component has size 1 (µ log(1 + µ))− log n + O(log log n). − ■ If µ = 0, the largest component has size of order n2/3. ■ If µ > 0, there is a unique giant component of size 1 αn where µ = α− log(1 α) 1. − − − ■ Bollob´as showed that a component of size at least 2/3 n in Gn,p is almost always unique if p exceeds 1/2 4/3 1/n + 4(log n) n− .

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 11 / 47 Evolution of G(n, p)

Range IV p c/n for c > 1 ∼ ■ Except for one “giant” component, all the other components are relatively small, and most of them are trees.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 12 / 47 Evolution of G(n, p)

Range IV p c/n for c > 1 ∼ ■ Except for one “giant” component, all the other components are relatively small, and most of them are trees. ■ The total number of vertices in components which are trees is approximately n f(c)n + o(n). −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 12 / 47 Evolution of G(n, p)

Range IV p c/n for c > 1 ∼ ■ Except for one “giant” component, all the other components are relatively small, and most of them are trees. ■ The total number of vertices in components which are trees is approximately n f(c)n + o(n). − ■ The largest connected component of Gn,p has approximately f(c)n vertices, where

k 1 1 ∞ k − c k f(c) = 1 (ce− ) . − c k! Xk=1

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 12 / 47 Evolution of G(n, p)

Range V p = c log n/n with c 1 ≥ ■ The graph Gn,p almost surely becomes connected.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 13 / 47 Evolution of G(n, p)

Range V p = c log n/n with c 1 ≥ ■ The graph Gn,p almost surely becomes connected. ■ If log n (k 1) log log n y 1 p = + − + + o( ), kn kn n n then there are only trees of size at most k except for the giant component. The distribution of the number of trees of k vertices again has a Poisson e−ky distribution with mean value k k! . ·

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 13 / 47 Evolution of G(n, p)

Range VI p ω(n) log n/n where ω(n) . ∼ → ∞ In this range, Gn,p is not only almost surely connected, but the degrees of almost all vertices are asymptotically equal.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 14 / 47 Model G(w1, w2,...,wn)

Random graph model with given expected degree sequence

- n nodes with weights w1, w2,...,wn.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 15 / 47 Model G(w1, w2,...,wn)

Random graph model with given expected degree sequence

- n nodes with weights w1, w2,...,wn. - For each pair (i,j), create an edge independently with 1 probability pij = wiwjρ, where ρ = n . i=1 wi P

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 15 / 47 Model G(w1, w2,...,wn)

Random graph model with given expected degree sequence

- n nodes with weights w1, w2,...,wn. - For each pair (i,j), create an edge independently with 1 probability pij = wiwjρ, where ρ = n . i=1 wi - The graph H has probability P

p (1 p ). ij − ij ij YE(H) ij YE(H) ∈ 6∈

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 15 / 47 Model G(w1, w2,...,wn)

Random graph model with given expected degree sequence

- n nodes with weights w1, w2,...,wn. - For each pair (i,j), create an edge independently with 1 probability pij = wiwjρ, where ρ = n . i=1 wi - The graph H has probability P

p (1 p ). ij − ij ij YE(H) ij YE(H) ∈ 6∈

- The expected degree of vertex i is wi.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 15 / 47 Model G(w1, w2,...,wn)

Random graph model with given expected degree sequence

- n nodes with weights w1, w2,...,wn. - For each pair (i,j), create an edge independently with 1 probability pij = wiwjρ, where ρ = n . i=1 wi - The graph H has probability P

p (1 p ). ij − ij ij YE(H) ij YE(H) ∈ 6∈

- The expected degree of vertex i is wi.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 15 / 47 An example: G(w1, w2, w3, w4)

  @@  @ @ @ @ @ @ @    

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 16 / 47 An example: G(w1, w2, w3, w4)

w3 w4 @@  @ @ @ @ @ @ @ w1 w2  

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 16 / 47 An example: G(w1, w2, w3, w4)

w3 w4 @@  @ @ w1w3ρ @ @ w2w3ρ @ w1w4ρ @ @ w1 w2 w1w2ρ  

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 16 / 47 An example: G(w1, w2, w3, w4)

1 w3w4ρ  w3 − w4 qqqqqqq @@  @ q @ q @ q w1w3ρ 1 w2w4ρ @ w2w3ρ q − @ q w1w4ρ @ q @ q w1 w2 w1w2ρ  

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 16 / 47 An example: G(w1, w2, w3, w4)

2 1 w3ρ q q − q q q 1 w3w4ρ q 2 q w3 − w4 1q w4ρ q qqqqqqq q − @@  @ q @ q @ q w1w3ρ 1 w2w4ρ @ w2w3ρ q − @ q w1w4ρ @ q q @ 2 w1 w2 1 w2ρ q w1w2ρ q − q  q q q 2 q q 1q w1ρ q −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 16 / 47 An example: G(w1, w2, w3, w4)

2 1 w3ρ q q − q q q 1 w3w4ρ q 2 q w3 − w4 1q w4ρ q qqqqqqq q − @@  @ q @ q @ q w1w3ρ 1 w2w4ρ @ w2w3ρ q − @ q w1w4ρ @ q q @ 2 w1 w2 1 w2ρ q w1w2ρ q − q  q q q 2 q q 1q w1ρ q − The probability of the graph is

4 w3w2w2w ρ4(1 w w ρ) (1 w w ρ) (1 w2ρ). 1 2 3 4 − 2 4 × − 3 4 − i Yi=1 Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 16 / 47 A example: G(1, 2, 1)

Loops are omitted here.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 17 / 47 Notations

For G = G(w1,...,wn), let 1 n - d = i=1 wi n n 2 i=1 wi - d˜= Pn . Pi=1 wi P - The volume of S: Vol(S) = i S wi. ∈ P

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 18 / 47 Notations

For G = G(w1,...,wn), let 1 n - d = i=1 wi n n 2 i=1 wi - d˜= Pn . Pi=1 wi P - The volume of S: Vol(S) = i S wi. ∈ We have P d˜ d ≥ “=” holds if and only if w = = w . 1 · · · n

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 18 / 47 Notations

For G = G(w1,...,wn), let 1 n - d = i=1 wi n n 2 i=1 wi - d˜= Pn . Pi=1 wi P - The volume of S: Vol(S) = i S wi. ∈ We have P d˜ d ≥ “=” holds if and only if w = = w . 1 · · · n A connected component S is called a giant component if

vol(S) = Θ(vol(G)).

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 18 / 47 Classical result on G(n, p)

■ If np < 1, almost surely there is no giant component.

■ If np > 1, almost surely there is a unique giant component.

d˜= d = np.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 19 / 47 Four questions

■ Is it true that G(w1,...,wn) almost surely has no giant component if d < 1?

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 20 / 47 Four questions

■ Is it true that G(w1,...,wn) almost surely has no giant component if d < 1?

■ Is it true that G(w1,...,wn) almost surely has a giant component if d˜ > 1?

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 20 / 47 Four questions

■ Is it true that G(w1,...,wn) almost surely has no giant component if d < 1?

■ Is it true that G(w1,...,wn) almost surely has a giant component if d˜ > 1?

■ Is it true that G(w1,...,wn) almost surely has no giant component if d˜ < 1?

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 20 / 47 Four questions

■ Is it true that G(w1,...,wn) almost surely has no giant component if d < 1?

■ Is it true that G(w1,...,wn) almost surely has a giant component if d˜ > 1?

■ Is it true that G(w1,...,wn) almost surely has no giant component if d˜ < 1?

■ Is it true that G(w1,...,wn) almost surely has a giant component if d > 1?

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 20 / 47 Case d < 1

Is it true that G(w1,...,wn) almost surely has no giant component if d < 1?

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 21 / 47 Case d < 1

Is it true that G(w1,...,wn) almost surely has no giant component if d < 1? No.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 21 / 47 Case d < 1

Is it true that G(w1,...,wn) almost surely has no giant component if d < 1? n n 3 No. A counter-example: G( 2 , 0) + G( 2 , n). n 3 Since G( 2 , n) has n 3 3 n′p′ = = > 1. 2 n 2 It has a giant component. But as the whole graph, the 3 average degree is d = 4 < 1.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 21 / 47 Case d˜ > 1

Is it true that G(w1,...,wn) almost surely has a giant component if d˜ > 1?

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 22 / 47 Case d˜ > 1

Is it true that G(w1,...,wn) almost surely has a giant component if d˜ > 1? No.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 22 / 47 Case d˜ > 1

Is it true that G(w1,...,wn) almost surely has a giant component if d˜ > 1? No. A counter-example: V = S T (with S = log n), weights are defined as follows. ∪ | |

√n if vi S wi = ∈  1 ǫ otherwise. −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 22 / 47 Case d˜ > 1

Is it true that G(w1,...,wn) almost surely has a giant component if d˜ > 1? No. A counter-example: V = S T (with S = log n), weights are defined as follows. ∪ | |

√n if vi S wi = ∈  1 ǫ otherwise. − Every component in G T has size at most O(log n). Adding S can join at most O(√| n log2 n) vertices in T . The volume of maximum component is at most O(√n log2 n).

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 22 / 47 Case d˜ > 1

Is it true that G(w1,...,wn) almost surely has a giant component if d˜ > 1? No. A counter-example: V = S T (with S = log n), weights are defined as follows. ∪ | |

√n if vi S wi = ∈  1 ǫ otherwise. − Every component in G T has size at most O(log n). Adding S can join at most O(√| n log2 n) vertices in T . The volume of maximum component is at most O(√n log2 n).

n log n + (1 ǫ)(n log n) d˜= − − > log n. √n log n + (1 ǫ)(n log n) − − p Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 22 / 47 Case d˜ < 1

Is it true that G(w1,...,wn) almost surely has no giant component if d˜ < 1?

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 23 / 47 Case d˜ < 1

Is it true that G(w1,...,wn) almost surely has no giant component if d˜ < 1? Yes. Chung and Lu (2001) Suppose that d˜ < 1 δ. For ˜2 − any α > 0, with probability at least 1 dd , a random α2(1 d˜) − − graph G in G(w1,...,wn) has all connected components with volume at most α√n.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 23 / 47 Proof

Let x = Pr( a component S, vol(S) α√n). ∃ ≥

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 24 / 47 Proof

Let x = Pr( a component S, vol(S) α√n). ∃ ≥ Choose two vertices u and v randomly with probability proportional to their weights.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 24 / 47 Proof

Let x = Pr( a component S, vol(S) α√n). ∃ ≥ Choose two vertices u and v randomly with probability proportional to their weights. Two ways to estimate z = Pr(u v) the probability that u and v are connected by a path. ∼

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 24 / 47 Proof

Let x = Pr( a component S, vol(S) α√n). ∃ ≥ Choose two vertices u and v randomly with probability proportional to their weights. Two ways to estimate z = Pr(u v) the probability that u and v are connected by a path. ∼ One the one hand,

z Pr(u v, a component S vol(S) α√n) ≥ ∼ ∃ ≥ = Pr(u v a component S vol(S) α√n)x ∼ |∃ ≥ Pr(u, v S a component S vol(S) α√n)x ≥ ∈ |∃ ≥ α2nρ2x. ≥

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 24 / 47 Proof

On the other hand, the probability Pk(u, v) of u and v being connected by a path of length k + 1 is at most

P (u, v) (w w ρ) (w w ρ) (w w ρ) k ≤ u i1 i1 i2 · · · ik v i1,iX2,...,ik k = wuwvρd˜ .

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 25 / 47 Proof

On the other hand, the probability Pk(u, v) of u and v being connected by a path of length k + 1 is at most

P (u, v) (w w ρ) (w w ρ) (w w ρ) k ≤ u i1 i1 i2 · · · ik v i1,iX2,...,ik k = wuwvρd˜ .

The probability that u and v are connected is at most

n 1 P (u, v) w w ρd˜k = w w ρ. k ≤ u v ˜ u v Xk=0 Xk 0 1 d ≥ −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 25 / 47 Proof

1 d˜2 z wuρ wvρ wuwvρ = ρ. ≤ 1 w˜ 1 d˜ Xu,v − − Combining this with z xα2nρ2 we have ≥ d˜2 α2xnρ2 ρ ≤ 1 d˜ − which implies that

dd˜2 x . ≤ α2(1 d˜) − The proof is finished. 

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 26 / 47 Case d > 1

Gap theorem: ■ Almost surely G has a unique giant component (GCC).

(1 2 + o(1))Vol(G) if d 4. vol(GCC) − √de ≥ e ≥  (1 1+log d + o(1))Vol(G) if d < 2. − d ■ The second largest component almost surely has size at most (1 + o(1))µ(d) log n, where

1 1+log d log 4 if d > 4/e; µ(d) = 1 −  d 1 log d if 1 < d < 2. − −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 27 / 47 Matrix-tree theorem

Kirchhofff (1847) The number of spanning trees in a graph G is equal to any cofactor of L = D A, where − D = diag(d1,...,dn) is the diagonal degree matrix and A is the adjacency matrix.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 28 / 47 Matrix-tree theorem

Kirchhofff (1847) The number of spanning trees in a graph G is equal to any cofactor of L = D A, where − D = diag(d1,...,dn) is the diagonal degree matrix and A is the adjacency matrix.

The matrix-tree theorem holds for weighted graphs.

w = det M . e | | XT f YE(T ) ∈ Here M is obtained by deleting one row and one column from D A. −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 28 / 47 A set S as component

Let S = vi1 , vi2 ...,vik with weights wi1 , wi2 ,...,wik . The probability{ that there is} no edge leaving S is

vi S,vj S (1 wiwjρ) ∈ 6∈ − ρ wiwj Q − vi∈S,vj 6∈S e P ≈ ρvol(S)(vol(G) vol(S)) = e− − .

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 29 / 47 A set S as component

Let S = vi1 , vi2 ...,vik with weights wi1 , wi2 ,...,wik . The probability{ that there is} no edge leaving S is

vi S,vj S (1 wiwjρ) ∈ 6∈ − ρ wiwj Q − vi∈S,vj 6∈S e P ≈ ρvol(S)(vol(G) vol(S)) = e− − .

The probability G is connected is at most |S k 2 k 1 w w ρ = w w w vol(S) − ρ − . ij il i1 i2 · · · ik XT (vi viY) E(T ) j l ∈ Computation is done by matrix-tree theorem.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 29 / 47 Detail computation

Let 0 w w ρ w w ρ i1 i2 · · · i1 ik  wi2 wi1 ρ 0 wi2 wik ρ  A = . . · · · ......    w w ρ w w ρ 0   ik i1 ik i2 · · ·  and D is the diagonal matrix

diag(wi1 (vol(S) wi1 )ρ,...,wik (vol(S)wik wik )ρ). Then compute the− determinant of any k 1− k 1 sub-matrix. − × −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 30 / 47 A set S as component

The probability that S is a component is at most

k 2 k 1 vol(S)(1 vol(S)/vol(G)) w w w vol(S) − ρ − e− − . i1 i2 · · · ik XS The probability that there exists a connected component on size k with volume less than ǫvol(G) is at most

k 2 k 1 vol(S)(1 ǫ) f(k,ǫ) = w w w vol(S) − ρ − e− − . i1 i2 · · · ik XS =k | |

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 31 / 47 4 Case d > e(1 ǫ)2 −

k 2 k 1 vol(S)(1 ǫ) f(k,ǫ) = w w w vol(S) − ρ − e− − i1 i2 · · · ik XS k 1 ρ − 2k 2 vol(S)(1 ǫ) vol(S) − e− − ≤ kk XS k 1 ρ − 2k 2 2k 2 (2k 2) ( − ) − e− − ≤ kk 1 ǫ XS − k k 1 n ρ − 2k 2 2k 2 (2k 2) ( − ) − e− − ≤ k! kk 1 ǫ 1 − 4 ( )k ≤ 4ρ(k 1)2 de(1 ǫ)2 − −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 32 / 47 1 2 Case 1 ǫ < d < 1 ǫ − −

First, we split f(k,ǫ) into two parts as follows:

f(k,ǫ) = f1(k,ǫ) + f2(k,ǫ)

where

k 2 k 1 vol(S)(1 ǫ) f (k,ǫ) = w w w vol(S) − ρ − e− − 1 i1 i2 · · · ik vol(XS)

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 33 / 47 Bounding f1(k, ǫ)

k 2 k 1 vol(S)(1 ǫ) f (k,ǫ) = w w vol(S) − ρ − e− − 1 i1 · · · ik vol(XS)

n d k 2 ( d(1 ǫ) 1 ) ≤ dk e − −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 34 / 47 Bounding f2(k, ǫ)

k 2 k 1 vol(S)(1 ǫ) f (k,ǫ) = w w w vol(S) − ρ − e− − 2 i1 i2 · · · ik vol(XS) dk ≥ k 1 k 2 dk(1 ǫ) w w ρ − (dk) − e− − ≤ i1 · · · ik vol(XS) dk ≥ k 1 k 2 dk(1 ǫ) w w w ρ − (dk) − e− − ≤ i1 i2 · · · ik XS k vol(G) k 1 k 2 dk(1 ǫ) < ρ − (dk) − e− − k! n d k 2 ( (d(1 ǫ) 1) ) ≤ dk e − −

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 35 / 47 Put together

4 If d > e(1 ǫ)2 , then − 1 4 f(k,ǫ) ( )k. ≤ 4ρ(k 1)2 de(1 ǫ)2 − − 1 2 If 1 ǫ < d < 1 ǫ, then − −

n d k f(k,ǫ) 2 2 ( (d(1 ǫ) 1) ) . ≤ dk e − − Choose k = µ(d) log n, then f(k,ǫ) = o(1). The gap theorem is proved.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 36 / 47 Volume of Giant Component

Chung and Lu (2004) If the average degree is strictly greater than 1, then almost surely the giant component in a graph G in G(w) has 3 5 log . n volume (λ0 + O √n Vol(G) ) Vol(G), where λ0 is the unique positive root of the following
equation:

n n wiλ w e− = (1 λ) w . i − i Xi=1 Xi=1

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 37 / 47 Sketch proof

k ■ With probability at least 1 2n− , a vertex with weight greater than max 8k, 2(k +− 1 + o(1))µ(d) log n is in the GCC. { }

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 38 / 47 Sketch proof

k ■ With probability at least 1 2n− , a vertex with weight greater than max 8k, 2(k +− 1 + o(1))µ(d) log n is in the GCC. { } k+2 ■ For any k > 2, with probability at least 1 6n− , we have Vol(GCC) E(Vol(GCC)) − | 2 − 2.5 |≤ 2C1(k + 1) √k 2√n log n, where − 2 C1 = 10µ(d) + 2µ(d) .

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 38 / 47 Sketch proof

k ■ With probability at least 1 2n− , a vertex with weight greater than max 8k, 2(k +− 1 + o(1))µ(d) log n is in the GCC. { } k+2 ■ For any k > 2, with probability at least 1 6n− , we have Vol(GCC) E(Vol(GCC)) − | 2 − 2.5 |≤ 2C1(k + 1) √k 2√n log n, where − 2 C1 = 10µ(d) + 2µ(d) . ■ Vol(G) E(vol(GCC)) = − 3.5 wvE(Vol(GCC))ρ 3√ wv

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 38 / 47 Lagrange inversion formula

Suppose that z is a function of x and y in terms of another analytic function φ as follows:

z = x + yφ(z).

Then z can be written as a power series in y as follows:

k ∞ y (k 1) k z = x + D − φ (x) k! Xk=1

where D(t) denotes the t-th derivative.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 39 / 47 Apply it to G(n, p)

dλ For the G(n,p), the equation is simply e− = (1 λ). Let z d z − λ = 1 . We have z = de− e . − d

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 40 / 47 Apply it to G(n, p)

dλ For the G(n,p), the equation is simply e− = (1 λ). Let z d z − λ = 1 d. We have z = de− e . We apply Lagrange − d z inversion formula with x = 0, y = de− , and φ(z) = e . Then we have

k ∞ y (k 1) kx z = D − e k! |x=0 Xk=1 k 1 ∞ k − d k = (de− ) k! Xk=1 This is exactly Erd˝os and R´enyi’s result on G(n,p).

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 40 / 47 G(n, p) verse G(w1,...,wn)

Question: Does the random graph with equal expected degrees generates the smallest giant component among all possible degree distribution with the same volume?

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 41 / 47 G(n, p) verse G(w1,...,wn)

Question: Does the random graph with equal expected degrees generates the smallest giant component among all possible degree distribution with the same volume? Chung Lu (2004)

■ e Yes, for 1

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 41 / 47 G(n, p) verse G(w1,...,wn)

Question: Does the random graph with equal expected degrees generates the smallest giant component among all possible degree distribution with the same volume? Chung Lu (2004)

■ e Yes, for 1

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 41 / 47 G(n, p) verse G(w1,...,wn)

Question: Does the random graph with equal expected degrees generates the smallest giant component among all possible degree distribution with the same volume? Chung Lu (2004)

■ e Yes, for 1

1 4 1 + 1 + o(1) Vol(G). 2 r − de 
This is asymptotically best possible.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 41 / 47 Sizes and edges in GCC

Chung, Lu (2004) If the expected average degree is strictly greater than 1, then almost surely the giant component in a random graph of given expected degrees wi, i = 1,...,n, n wiλ0 3.5 has n e− + O(√n log n) vertices and − i=1 (λ 1λP2)Vol(G) + O( Vol(G) log3.5 n) edges. 0 − 2 0 p

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 42 / 47 In the collaboration graph

Vol(GCC) 248000 λ (2 λ ) . 0 − 0 ≈ Vol(G) ≈ 284000 We have λ 0.644. 0 ≈ Let nk denote the number of vertices of degree k. We have

k wi wi n E(n ) e− . k ≈ k ≈ k! Xi 0 ≥

n0 n1 n2 n3 n4 n5 n6 n7 n8 n9 166381 145872 34227 16426 9913 6670 4643 3529 2611 2032

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 43 / 47 Compute GCC | |

n λ0wi GCC n e− | | ≈ − Xi=1 n (1 λ0)wi wi = n e − e− − Xi=1 n ∞ k (1 λ0) k wi = n − w e− k − k! i Xk 0 Xi=1 Xk=0 ≥ n (1 (1 λ )k) ≈ k − − 0 Xk 0 ≥ = n (1 (1 λ )k). k − − 0 Xk 1 ≥ Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 44 / 47 Conclusion

The size of giant component is predicted to be about 177, 400 by our theory. This is rather close to the actual value 176, 000, within an error bound of less than 1%.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 45 / 47 References

■ Fan Chung and Linyuan Lu. Connected components in a random graph with given degree sequences, Annals of Combinatorics, 6 (2002), 125–145. ■ Fan Chung and Linyuan Lu, The volume of the giant component for a random graph with given expected degrees , SIAM J. Discrete Math., 20 (2006), No. 2, 395–411.

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 46 / 47 Overview of talks

■ Lecture 1: Overview and outlines ■ Lecture 2: Generative models - preferential attachment schemes ■ Lecture 3: Duplication models for biological networks ■ Lecture 4: The rise of the giant component ■ Lecture 5: The small world phenomenon: average distance and diameter ■ Lecture 6: Spectrum of random graphs with given degrees

Lecture 4: The rise of the giant component Linyuan Lu (University of South Carolina) – 47 / 47