## 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