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<k)&\le& \sum_{\substack{V\subset [n]\\ |V|=m}} \mathbb{P}(\alpha(G_{V})<k)\\ &\le& {n \choose m} \max_{\substack{V\subset [n]\\ |V|=m}}\mathbb{P}(\omega(G_{V})<k)\,. \end{array}

Therefore, we need to prove a bound on {\mathbb{P}(\omega(G_{V})<k)}, 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)<k]&=&\mathbb{P}[Y=0]\\ &=&\mathbb{P}[Y-\mathbb{E}[Y]=-\mathbb{E}[Y]]\\ &=&\mathbb{P}[Y_N-Y_0=-\mathbb{E}[Y]]\\ \end{array}

Therefore, applying the Azuma-Hoeffding inequality, we get

\displaystyle \mathbb{P}[\omega(G)<k]\le e^{-\frac{(\mathbb{E}[Y])^2}{2N}}\,. \ \ \ \ \ (2)

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<k)&\le &{n \choose m} \exp\Big(-\frac{m^{2(2-\varepsilon)a+o(1)}}{n(n-1)}\Big)\\ &\le& \exp\Big(m^{1+o(1)}(\ln 2)-\frac{m^{2(2-\varepsilon)a+o(1)}}{n(n-1)}\Big)=o(1)\,. \end{array}

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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