Lecture 3. Chromatic number
Let be set of integers that represent colors. A vertex coloring of a graph is an assignment of a color to each vertex . Furthermore a vertex coloring is said to be proper if for all . The chromatic number of a graph is the smallest number 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 .
Theorem [Bollobas, 1988]
The chromatic number of a random graph satisfies
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 of a graph is a set of vertices such that . Independent sets are sometimes called stable sets. Denote by the complement graph of where iff . If is an independent set of , then it is a clique in . The largest independent set in is called independence number of and denoted by .
Proposition For any graph with independence number , clique number and chromatic number the following holds
2. Chromatic, clique and independence numbers
Our goal is to study the chromatic number of a random graph . It actually builds upon the clique number that we studied during last lecture. Indeed, we will use the following observation: if then so does the complement graph of . It implies that the independence number has the same distribution as the clique number . We know from last lecture that . Therefore, it follows from the above proposition that
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 and let be a subset of vertices and observe that induced subgraph restricted to vertices in has distribution .
That is: every vertex subgraph contains an independent set of size .
Consider now the following coloring procedure to produce a proper coloring. Pull out independent sets of size one by one and give each a distinct color. According to the definition of , as long as there are at least vertices, this is possible. If there are less than vertices left, give each a distinct color.
The above algorithm gives one possible way of coloring graph and thus the output number is an upper bound on the chromatic number . Let us now compute this upper bound. By definition of , we have
Therefore, we need to prove a bound on , uniformly in . 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 be a filtered space and let be a random variable on . Assume that the martingale is such that for all ,
Then for any positive integer and any , it holds,
Proof: We assume the following inequality due to Hoeffding. For any random variable such that a.s. and any , it holds
Now, using a Chernoff bound, we get for any ,
where . The above two displays yield that
The second part of the proof follows by applying the same argument to the martingale .
Consider the lexicographic ordering of and let be iid Bernoulli random variables with parameter indicating the presence of an edge in a graph . Define the –algebra . We are going to apply the Azuma-Hoeffding inequality to the (Doob) martingale where is maximal number of edge-disjoint –cliques in .
Note first that a.s., since adding/removing one edge can add/remove at most one edge disjoint –clique. Moreover, this inequality would not hold if we had chosen to be the number cliques of size (not necessarily disjoint). Indeed, adding or removing an edge can create or destroy many overlapping cliques.
It follows that,
It remains to show that is sufficiently large. Note first that , where is is the number of –cliques that do not share an edge with any other –clique of the graph.
4. A strong lower bound on the clique number
Fix positive integers and
and let denote the number of cliques in that do not share an edge with any other –clique of the graph. Then
for all . Here the asymptotic notation notations are as .
Proof: Define (for large enough) for some to be chosen later.
Moreover, note that for large enough,
so that where is the number of cliques in that do not share an edge with any other –clique of the graph . Let .
For any define the indicator:
Moreover for any set , let denote the number of –cliques in , other than itself, that share at least two vertices with . Observe that
We bound as follows
The following bound holds for all ,
Moreover, for , , it holds
To conclude the proof, observe that by Stirling, since ,
Let and observe that
We can therefore apply the previous proposition together with~(2), to get
5. Proof of the main theorem
we get for ,
Thus . Together with~(1), it yields
Which completes the proof of our theorem.
Lecture and scribing by Philippe Rigollet