﻿ Lecture 3. Chromatic number | Stochastic Analysis Seminar Lecture 3. Chromatic number – Stochastic Analysis Seminar

## Lecture 3. Chromatic number

Let ${C}$ be set of integers that represent colors. A vertex coloring of a graph ${G=([n], E)}$ is an assignment ${c\,:\,\{1, \ldots, n\} \rightarrow C}$ of a color ${c(i) \in C}$ to each vertex ${i \in [n]}$. Furthermore a vertex coloring ${c}$ is said to be proper if ${c(i)\neq c(j)}$ for all ${(i,j) \in E}$. The chromatic number ${\chi(G)}$ of a graph ${G}$ is the smallest number ${|C|}$ of colors needed for a proper vertex coloring to exist. In this lecture, any coloring is a proper vertex coloring. All logs are in base ${2}$.

Theorem [Bollobas, 1988]
The chromatic number ${\chi(G)}$ of a random graph ${G\sim G(n,\frac12)}$ satisfies

$\displaystyle \chi(G) =\frac{n}{2\log n}(1+o_P(1))$

1. Basic properties

In this section, we begin by reviewing some basic properties of the chromatic number, and in particular how it relates to independent sets. Recall that an independent set ${S \subset [n]}$ of a graph ${G=([n], E)}$ is a set of vertices such that ${i,j \in S \Rightarrow (i,j) \notin E}$. Independent sets are sometimes called stable sets. Denote by ${\bar G=([n], \bar E)}$ the complement graph of ${G}$ where ${(i,j) \in \bar E}$ iff ${(i,j) \notin E}$. If ${S}$ is an independent set of ${G}$, then it is a clique in ${\bar G}$. The largest independent set in ${G}$ is called independence number of ${G}$ and denoted by ${\alpha(G)}$.

Proposition For any graph ${G=([n],E)}$ with independence number ${\alpha(G)}$, clique number ${\omega(G)}$ and chromatic number ${\chi(G)}$ the following holds

1. ${1\le \chi(G) \le n}$
2. ${\chi(G) \ge \omega(G)}$
3. ${n \le \chi(G)\alpha(G)}$

Proof: Trivial. $\Box$

2. Chromatic, clique and independence numbers

Our goal is to study the chromatic number of a random graph ${G \sim G(n,\frac12)}$. It actually builds upon the clique number that we studied during last lecture. Indeed, we will use the following observation: if ${G \sim G(n,\frac12)}$ then so does the complement graph ${\bar G}$ of ${G}$. It implies that the independence number ${\alpha(G)}$ has the same distribution as the clique number ${\omega(G)}$. We know from last lecture that ${\omega(G)=(2\log n)(1+o(1))}$. Therefore, it follows from the above proposition that

$\displaystyle \chi(G) \ge \frac{n}{\alpha(G)}= \frac{n}{2\log n}(1+o(1))\,.$

Our main theorem suggests that this simple bound is, in fact, tight. Proving this requires a stronger lower bound on clique of a random graph.

Define ${m=\lfloor n/\ln^2 n\rfloor}$ and let ${V \subset [n]}$ be a subset of ${|V|=m}$ vertices and observe that induced subgraph ${G_V=(V, E \cap [m]^2)}$ restricted to vertices in ${V}$ has distribution ${G(m,\frac12)}$.

Define

$\displaystyle \bar \alpha_m=\min_{\substack{V\subset [n]\\ |V|=m}} \alpha(G_{V})$

That is: every ${m}$ vertex subgraph contains an independent set of size ${\bar \alpha_m}$.

Consider now the following coloring procedure to produce a proper coloring. Pull out independent sets of size ${\bar \alpha_m}$ one by one and give each a distinct color. According to the definition of ${\bar \alpha_m}$, as long as there are at least ${m}$ vertices, this is possible. If there are less than ${m}$ vertices left, give each a distinct color.

The above algorithm gives one possible way of coloring graph ${G}$ and thus the output number is an upper bound on the chromatic number ${\chi(G)}$. Let us now compute this upper bound. By definition of ${\bar \alpha_m}$, we have

$\displaystyle \chi(G)\le \Big\lceil \frac{n-m}{\bar \alpha_m}\Big\rceil + m \le \frac{n}{\bar \alpha_m} +m\,. \ \ \ \ \ (1)$

Therefore, to find an upper bound on ${\chi(G)}$, we need to find a lower bound on ${\bar \alpha_m}$ for an appropriately chosen ${m}$. To that end, observe that

$\displaystyle \begin{array}{rcl} \mathbb{P}(\bar \alpha_m

Therefore, we need to prove a bound on ${\mathbb{P}(\omega(G_{V}), uniformly in ${V}$. This can be done using concentration properties of a suitable martingale.

3. Azuma-Hoeffding inequality

Martingales have useful concentration properties. Specifically the following lemma holds.

Lemma [Azuma-Hoeffding inequality.]
Let ${(\Omega, \mathcal{F}, \{\mathcal{F}_i\}_{i \ge 0}, \mathbb{P})}$ be a filtered space and let ${X}$ be a random variable on ${(\Omega, \mathcal{F}, \mathbb{P})}$. Assume that the martingale ${X_i=\mathbb{E}[X|\mathcal{F}_i], i\ge 0}$ is such that for all ${i \ge 1}$,

$\displaystyle |X_{i}-X_{i-1}|\le 1\,, \qquad \text{a.s.}$

Then for any positive integer ${N}$ and any ${t>0}$, it holds,

$\displaystyle \mathbb{P}(X_N-X_0\ge t) \le \exp\Big(-\frac{t^2}{2N}\Big)$

and

$\displaystyle \mathbb{P}(X_N-X_0\le -t) \le \exp\Big(-\frac{t^2}{2N}\Big)$

Proof: We assume the following inequality due to Hoeffding. For any random variable ${Z}$ such that ${|Z| \le 1}$ a.s. and any ${s>0}$, it holds

$\displaystyle \mathbb{E}[e^{sZ}]\le e^{\frac{s^2}{2}}\,.$

Now, using a Chernoff bound, we get for any ${s>0}$,

$\displaystyle \begin{array}{rcl} \mathbb{P}(X_N-X_0>t)&=&\mathbb{P}\Big(\sum_{i=1}^N\{X_i-X_{i-1}\}>t\Big)\\ &\le& e^{-st}\mathbb{E}\exp\Big(s\sum_{i=1}^N\{X_i-X_{i-1}\}\Big)\\ &=&e^{-st}\mathbb{E}\prod_{i=1}^N\exp\Big(s\{X_i-X_{i-1}\}\Big)\\ &=&e^{-st}\mathbb{E}\prod_{i=1}^N\mathbb{E}\Big[\exp\Big(s\{X_i-X_{i-1}\}\Big)\Big|\mathcal{F}_{i-1}\Big]\\ \end{array}$

where ${\mathcal{F}_{i}=\sigma(X_1, \ldots, X_i)}$. The above two displays yield that

$\displaystyle \mathbb{P}(X_N-X_0>t)\le \exp\Big(\inf_{s>0}\Big\{\frac{Ns^2}{2} -st\Big\}\Big)=\exp\Big(-\frac{t^2}{2N}\Big)\,.$

The second part of the proof follows by applying the same argument to the martingale ${(-X_i)_{i\ge 0}}$. $\Box$

Consider the lexicographic ordering of ${[n]^2}$ and let ${E_1, \ldots, E_N \in \{0,1\}}$ be ${N={n \choose 2}}$ iid Bernoulli random variables with parameter ${1/2}$ indicating the presence of an edge in a graph ${G\sim G(n,\frac12)}$. Define the ${\sigma}$-algebra ${\mathcal{F}_i=\sigma(E_1, \ldots E_i), i=1, \ldots, N}$. We are going to apply the Azuma-Hoeffding inequality to the (Doob) martingale ${Y_i=\mathbb{E}[Y|\mathcal{F}_i]\,, \quad i=1, \ldots, N}$ where ${Y}$ is maximal number of edge-disjoint ${k}$-cliques in ${G}$.

Note first that ${|Y_i-Y_{i-1}|\le 1}$ a.s., since adding/removing one edge can add/remove at most one edge disjoint ${k}$-clique. Moreover, this inequality would not hold if we had chosen ${Y}$ to be the number cliques of size ${k}$ (not necessarily disjoint). Indeed, adding or removing an edge can create or destroy many overlapping cliques.

It follows that,

$\displaystyle \begin{array}{rcl} \mathbb{P}[\omega(G)

Therefore, applying the Azuma-Hoeffding inequality, we get

$\displaystyle \mathbb{P}[\omega(G)

It remains to show that ${\mathbb{E}[Y]}$ is sufficiently large. Note first that ${Y \ge X}$, where is ${Y}$ is the number of ${k}$-cliques that do not share an edge with any other ${k}$-clique of the graph.

4. A strong lower bound on the clique number

Fix positive integers ${m}$ and

$\displaystyle k=(2-\varepsilon+o(1))\log m$

and let ${X}$ denote the number of ${k}$ cliques in ${G \sim G(m,\frac12)}$ that do not share an edge with any other ${k}$-clique of the graph. Then

$\displaystyle \mathbb{E}[X]\ge m^{(2-\varepsilon)a+o(1)}$

for all ${a\in (0,1)}$. Here the asymptotic notation notations ${o(1)}$ are as ${m \rightarrow \infty}$.

Proof: Define ${M=\left\lceil 2^ak2^{\frac{k-1}{2}}\right\rceil\ge 4k}$ (for ${m}$ large enough) for some ${a \in (1,2)}$ to be chosen later.

Moreover, note that for ${m}$ large enough,

$\displaystyle M\le 8m^{1-\frac{\varepsilon}2 +o(1)}\le m$

so that ${X \ge X'}$ where ${X'}$ is the number of ${k}$ cliques in ${G \sim G(M,\frac12)}$ that do not share an edge with any other ${k}$-clique of the graph ${G}$. Let ${G \sim G(M, \frac12)}$.

For any ${S \in [M]}$ define the indicator:

$\displaystyle I_S=1\{S \text{ is a clique of size} \ k \ \text{in} \ G\}.$

Moreover for any set ${S}$, let ${Z(S)}$ denote the number of ${k}$-cliques in ${G}$, other than ${S}$ itself, that share at least two vertices with ${S}$. Observe that

$\displaystyle \begin{array}{rcl} \mathbb{E}[X']&=&\mathbb{E}\sum_{\substack{S \subset [M] \\ |S| =k}}I_S1\{Z(S)=0\}\\ &=&{M\choose k}\mathbb{P}(I=1, Z=0)\qquad \text{where } I=I_{[k]}\,, \ Z=Z([k])\\ &=&{M \choose k}\Big(\frac12\Big)^{k \choose 2}\mathbb{P}(Z=0|I=1)\,, \end{array}$

We bound ${\mathbb{P}(Z=0|I=1)}$ as follows

$\displaystyle \begin{array}{rcl} \mathbb{P}(Z=0|I=1)&=&1-\mathbb{P}(Z\ge 1|I=1)\\ &\ge& 1-\mathbb{E}(Z|I=1)\\ &=&1-\sum_{\substack{S \subset [M], |S|=k\\ |S\cap [k]| \ge 2}}\mathbb{P}(I_S=1)\\ &=&1-\sum_{s=2}^{k-1}\underbrace{{k\choose s}{M-k \choose k-s}\Big(\frac12\Big)^{{k\choose 2}-{s\choose 2}}}_{F_s}\\ \end{array}$

The following bound holds for all ${2\le s \le k/2}$,

$\displaystyle \begin{array}{rcl} \frac{F_s}{F_2}&=&\frac{(M-2k+2)!}{(M-2k+s)!}\Big[ \frac{(k-2)!}{(k-s)!} \Big]^2\frac2{s!}2^{\frac{(s+1)(s-2)}{2}}\\ &\le& \frac2{s!}\Big(\frac{(k-2)^22^{\frac{s+1}2}}{M-2k+2} \Big)^{s-2}\\ &\le&\frac1{(s-2)!}\Big(\frac{2k^22^{\frac{k/2+1}{2}}}{M} \Big)^{s-2}\qquad \text{for} \ M\ge 4k\\ &\le&\frac1{(s-2)!}(2^{2+a}k2^{-\frac{k}4})^{s-2} \end{array}$

It yields

$\displaystyle \begin{array}{rcl} \sum_{2\le s \le k/2}\frac{F_s}{F_2}&\le & \exp\Big(2^{2+a}k2^{-\frac{k}4}\Big)=1+o(1)\\ \end{array}$

Moreover, for ${s> k/2}$, ${M \ge k}$, it holds

$\displaystyle \begin{array}{rcl} {M-k \choose k-s}&\le& {M-k \choose s}\frac{s!(M-k-s)!}{(k-s)!(M-2k+s)!}\\ &\le& {M-k \choose s}\Big(\frac{s}{M-k+s}\Big)^k\\ &\le & {M-k \choose s}\Big(\frac{k}{M-k/2}\Big)^{k}\\ \end{array}$

Hence,

$\displaystyle \begin{array}{rcl} F_s&=& {k\choose k-s}{M-k \choose k-s}\Big(\frac12\Big)^{{k\choose 2}-{s\choose 2}}\\ &\le& F_{k-s}\Big(\frac{k}{M-k/2}\Big)^{k}\Big(\frac12\Big)^{{k-s\choose 2}-{s\choose 2}}\\ &\le & F_{k-s}\Big(\frac{k2^{\frac{k-1}2}}{M-k/2}\Big)^{k}\\ &\le & F_{k-s}\Big(\frac{2^{-a}M}{M-k/2}\Big)^{k}\\ &\le &F_{k-s}o(1)\,. \end{array}$

Therefore,

$\displaystyle \sum_{k/2\le s \le k-1}\frac{F_s}{F_2}\le \sum_{2\le s \le k/2}\frac{F_s}{F_2}o(1)=o(1)\\$

Thus,

$\displaystyle \mathbb{P}(Z=0|I=1)\ge 1-F_2(1+o(1))\,.$

Moreover,

$\displaystyle \begin{array}{rcl} F_2&=&2{k \choose 2}{M-k \choose k-2} \frac{1}{2^{k \choose 2}}\\ &=& k^2\Big(\frac{M-k}{k-2}\Big)^{k-2}2^{-\frac{k(k-1)}{2}}(1+o(1))\\ &=& \frac{k^2}{M^2}\Big(\frac{M}{k2^{\frac{k-1}{2}}}\Big)^{k}(1+o(1))\\ &=& \frac{k^22^{ak}}{M^2}(1+o(1))\\ &\le&\frac{2}{2^{2a}}\Big(\frac{2^a}{2}\Big)^k(1+o(1))=o(1)\,, \end{array}$

since ${2^a<2}$.

To conclude the proof, observe that by Stirling, since ${k=o(n)}$,

$\displaystyle \begin{array}{rcl} {M \choose k}\Big(\frac12\Big)^{k \choose 2}&\ge&\Big(\frac{M}{k2^{\frac{k-1}{2}}}\Big)^k(1+o(1))\\ &=&2^{ak}(1+o(1))=m^{(2-\varepsilon) a+o(1)}\,. \end{array}$

$\Box$

Let ${G \sim G(m, \frac12)}$ and observe that

$\displaystyle (2-\varepsilon+o(1))\log m=(2-\varepsilon+o(1))\log n$

We can therefore apply the previous proposition together with~(2), to get

$\displaystyle \mathbb{P}[\omega(G)<(2-\varepsilon+o(1))\log n]\le \exp\Big(-\frac{m^{2(2-\varepsilon)a+o(1)}}{n(n-1)}\Big)$

5. Proof of the main theorem

Since

$\displaystyle {n \choose m}\le 2^n=2^{m^{1+o(1)}}\,,$

we get for ${k=(2-\varepsilon+o(1))\log n}$,

$\displaystyle \begin{array}{rcl} \mathbb{P}(\bar \alpha_m

Thus ${\bar \alpha_m \ge (2\log n)(1+o_P(1))}$. Together with~(1), it yields

$\displaystyle \begin{array}{rcl} \chi(G)&\le&\frac{n}{k}(1+o_P(1))+m \\ &=&\frac{n}{2\log n}(1+o_P(1))+o\Big(\frac{n}{\log n}\Big)\\ &=&\frac{n}{2\log n}(1+o_P(1)) \end{array}$

Which completes the proof of our theorem.

Lecture and scribing by Philippe Rigollet

14. March 2013 by Philippe
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