10 The binomial Random Graph

G(n,p)

We are interested in sliding p from 0 to 1. At what point do certain structures emerge?

Questions:

Fix some H. When do we expect G⊃H, G∼G(n,p)?

Definition 10.1 (m). For H a graph, define

m(H)=max⁡{e(F)|F|:F⊂H,|F|≥1}.

Example. H=K3, m(K3)=1.

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Theorem 10.2 (Bollobás, 80s). Assuming that:

  • H a graph

  • G∼G(n,p)

Then
lim⁡n→∞ℙ(G⊃H)={1when pn1m(H)→∞0when pn1m(H)→0

Proof. We use a second moment argument. Let

X# of copies of H in G,

where G∼G(n,p). Then

𝔼X=Θ(n|H|⋅pe(H)).

Since Var⁡X=𝔼X2−(𝔼X)2, we will compute 𝔼X2.

𝔼X2=(∑H𝟙H⊂G)2=∑H1,H2ℙ(H1⊂G,H2⊂G)≤(𝔼X)2+(𝔼X)+∑∅≠F⊊H∑H1,H2H1∩H2=Fp2e(H)−e(F)≤(𝔼X)2+(𝔼X)+C∑∅F⊊Hn2|H|−|F|p2e(H)−e(F)≤(𝔼X)2+(𝔼X)+Cn2|H|p2e(H)∑F1n|F|pe(F)

Now note

(1npe(F)|F|)|F|≤(1npm(H))|F|→0

when pn1m(H)→∞. So

𝔼X2=(1+o(1))(𝔼X)2+𝔼X.□

10.1 The Giant Component

Proposition 10.3. Let 𝜀>0. If G∼G(n,p), then

lim⁡n→∞ℙ(G is connected)={1p≥(1+𝜀) log ⁡⁡nn0p≤(1−𝜀) log ⁡⁡nn

and

lim⁡n→∞ℙ(G⊃H)={1pm(H)n≫10pm(H)n≪1

Remark. This “threshold” for p looks different from what we saw from the threshold of containing a given H – it has a “sharp” jump from 0 to 1.

Definition 10.4 (c1). For a graph G, let

c1(G)=# vertices in the largest connected component.

Theorem 10.5 (Erdös-Renyi). Let 𝜀>0. If G∼G(n,p) then

c1(G)={O𝜀(log⁡n)if p≤1−𝜀nΩ𝜀(n)if p≥1+𝜀n

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Definition 10.6 (DFS-sequence). Let G be a graph. I define the DFS-sequence from G

X1,X2,X3,X4,…∈{0,1}

as follows. We process the graph using a Depth first search. We also imagine V(G) have an order on them. As we saw before, we maintain a partition V(G)=A∪B∪P where P is a path.

We start with A=V(G), B=∅, P=∅. We then repeat the following:

  • (1) If P is empty, move a vertex from A to P.
  • (2) If P is not empty, let x be the last vertex of P, and now query the edges from x to A in order until I get a “yes”.

    Now move the neighbour of v to the end of P and append to our DFS sequence the outcomes of the queries.

    If v has no neighbour in A, then move v to B and append all the “no” outcomes to the DFS-sequence.

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Note.

  • We don’t query an edge more than once.

  • We don’t query all edges in G. But if uv∈G is not queried, then u,v are in the same component.

Lemma 10.7. Let G be a graph on n vertices, with a component of size k. Let X1,X2,… be the DFS-sequence. Then there exists t such that

∑i=tt+knXi≥k−1.

Proof. Let C={x1,…,xk} be a component of size k, C⊂V(G). Say we first encounter x1 at time t. From time t up until C⊂B, we only query edges incident to v∈C. So we make at most

k2+k(n−k)≤kn

exposures. And during this time we must have seen ≥k−1 1’s since V(C) is a component. □

Lemma 10.8. For 𝜀>0, let p=1−𝜀n. Then if G∼G(n,p),

c1(G)≤8log⁡n𝜀2.

Proof. Let X1,X2,… be the corresponding DFS-sequence. Since we never query an edge more than once in the definition of the DFS-sequence, we have that X1,X2,X3,… is a sequence of iid random variables, with each being Xi∼Ber⁡(p).

So let

Yt=∑i=tt+knXi

for t≤n2.

We will use Chernoof’s inequality:

Let X∼Bin⁡(n,p) defined by

ℙ(X=k)=nkpk(1−p)n−k.

Then for 0≤t≤pn,

ℙ(|X−pn|≥t)≤2exp⁡(−t23pn).

Note Yt∼Bin⁡(kn,p) so

ℙ(Yt≥k−1)≤ℙ(|Yt−pkn|≥𝜀k−1)≤2exp⁡(−(𝜀k−1)23k)≤ exp ⁡(−𝜀2k3.5).

Note that if k≥8𝜀2log⁡n, then the above is ≤n−2−c. Then union bound over all ≤n2 choices for t to see no such Yt satisfies Yt≥k−1. □

Theorem 10.9 (Ajtai, Komlós, Szemeredi). For 0<𝜀<13, if G∼G(n,p) where p=1+𝜀n, then G⊃Pl where l≥𝜀5n.

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Lemma 10.10. Let G be an n vertex graph with DFS-sequence X1,X2,… and

∑i=1TXi≥pT−𝜀5n,

where T=𝜀kn2. Then G⊃Pl where l≥𝜀5n.

Proof. If at time T we have |B|≥n3, then since |V(P)|≤𝜀5n, assuming for contradiction Pl⁄⊂G, for all steps. Then there must have been some step ≤T in the algorithm where |A|≥n3 and |B|≥n3. Then

𝜀2n2=T≥|A||B|≥n29,

contradiction.

So we may assume |B|≤n3. So we can assume at time T that the algorithm is still running. We have

𝜀5n+|B|≥|V(P)|+|B|≥∑i=1TXi=(∗)

since, each time we see Xi=1 we move a vertex from A into V(P)∪B, and it never returns. Now

(∗)≥pT−𝜀5n≥(1+𝜀)𝜀3n2n−𝜀5n≥(1+𝜀2)Tn+𝜀5n.

So combining we have

n3≥|B|≥(1+𝜀2)Tn.

Also note

|A|=(n−|B|−|V(P)|)

so

T≥|A||B|≥(n−|B|−𝜀5n)|B|≥(n−(1+𝜀2)Tn−𝜀5n)(1+𝜀2)Tn≥(1−2𝜀3)(1+𝜀2)T>T

contradiction. □

Reminder of Chernoff bound:

Lemma 10.11. Let X∼Bin⁡(n,p). Then, for 0≤t≤pn, we have

ℙ(|X−𝔼X|≥t)≤2exp⁡(−t23pn).

Proof of Theorem 10.9. Let G∼G(n,p), where p=(1+𝜀)n, for 0<𝜀<13. Let X1,X2,X3,… DFS-sequence. This is a sequence of iid Bernoulli random variables with probability p. So set T=𝜀2n2 and note

ℙ(∑i=1TXi<pT−𝜀5n)≤ℙ(|∑i=1TXi−pT|>𝜀5n)≤2exp⁡(−𝜀10n23(1+𝜀)𝜀3n)→0

since 𝜀5n<pT=(1+𝜀)𝜀3n. So apply Lemma 10.10 to see that with probability 1−o(1), G⊃Pl, where l≥𝜀5n. □

Theorem 10.12. Let 𝜀→0. Let w be some function with w→∞. Let G∼G(n,p). Then

c1(G)={log⁡(𝜀3n)𝜀2𝜀<n−13wΘ(n23𝜀=Θ(n−13)(2+o(1))𝜀n𝜀=w⋅n−13