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 GH, GG(n,p)?

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

m(H)=max{e(F)|F|:FH,|F|1}.

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

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

  • H a graph

  • GG(n,p)

Then
limn(GH)={1when pn1m(H)0when pn1m(H)0

Proof. We use a second moment argument. Let

X# of copies of H in G,

where GG(n,p). Then

𝔼X=Θ(n|H|pe(H)).

Since VarX=𝔼X2(𝔼X)2, we will compute 𝔼X2.

𝔼X2=(H𝟙HG)2=H1,H2(H1G,H2G)(𝔼X)2+(𝔼X)+FHH1,H2H1H2=Fp2e(H)e(F)(𝔼X)2+(𝔼X)+CFHn2|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 GG(n,p), then

limn(G is connected)={1p(1+𝜀) log nn0p(1𝜀) log nn

and

limn(GH)={1pm(H)n10pm(H)n1

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 GG(n,p) then

c1(G)={O𝜀(logn)if p1𝜀nΩ𝜀(n)if p1+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)=ABP 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 uvG 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+knXik1.

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

k2+k(nk)kn

exposures. And during this time we must have seen k1 1’s since V(C) is a component.

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

c1(G)8logn𝜀2.

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 XiBer(p).

So let

Yt=i=tt+knXi

for tn2.

We will use Chernoof’s inequality:

Let XBin(n,p) defined by

(X=k)=nkpk(1p)nk.

Then for 0tpn,

(|Xpn|t)2exp(t23pn).

Note YtBin(kn,p) so

(Ytk1)(|Ytpkn|𝜀k1)2exp((𝜀k1)23k) exp (𝜀2k3.5).

Note that if k8𝜀2logn, then the above is n2c. Then union bound over all n2 choices for t to see no such Yt satisfies Ytk1.

Theorem 10.9 (Ajtai, Komlós, Szemeredi). For 0<𝜀<13, if GG(n,p) where p=1+𝜀n, then GPl where l𝜀5n.

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

i=1TXipT𝜀5n,

where T=𝜀kn2. Then GPl where l𝜀5n.

Proof. If at time T we have |B|n3, then since |V(P)|𝜀5n, assuming for contradiction PlG, 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

()pT𝜀5n(1+𝜀)𝜀3n2n𝜀5n(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)Tn𝜀5n)(1+𝜀2)Tn(12𝜀3)(1+𝜀2)T>T

contradiction.

Reminder of Chernoff bound:

Lemma 10.11. Let XBin(n,p). Then, for 0tpn, we have

(|X𝔼X|t)2exp(t23pn).

Proof of Theorem 10.9. Let GG(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

(i=1TXi<pT𝜀5n)(|i=1TXipT|>𝜀5n)2exp(𝜀10n23(1+𝜀)𝜀3n)0

since 𝜀5n<pT=(1+𝜀)𝜀3n. So apply Lemma 10.10 to see that with probability 1o(1), GPl, where l𝜀5n.

Theorem 10.12. Let 𝜀0. Let w be some function with w. Let GG(n,p). Then

c1(G)={log(𝜀3n)𝜀2𝜀<n13wΘ(n23𝜀=Θ(n13)(2+o(1))𝜀n𝜀=wn13