More than six months late, but here are notes from the combinatorial nullsetllensatz talk I gave at the student colloquium at MIT. This was also my term paper for 18.434, “Seminar in Theoretical Computer Science”.
1. Introducing the choice number
One of the most fundamental problems in graph theory is that of a graph coloring, in which one assigns a color to every vertex of a graph so that no two adjacent vertices have the same color. The most basic invariant related to the graph coloring is the chromatic number:
In this exposition we study a more general notion in which the set of permitted colors is different for each vertex, as long as at least colors are listed at each vertex. This leads to the notion of a so-called choice number, which was introduced by Erdös, Rubin, and Taylor.
We are thus naturally interested in how the choice number and the chromatic number are related. Of course we always have
Näively one might expect that we in fact have an equality, since allowing the colors at vertices to be different seems like it should make the graph easier to color. However, the following example shows that this is not the case.
This surprising behavior is the subject of much research: how can we bound the choice number of a graph as a function of its chromatic number and other properties of the graph? We see that the above example requires exponentially many vertices in .
One of the most major open problems in this direction is the following.
If is a claw-free graph, then . In particular, this conjecture implies that for edge coloring, the notions of “chromatic number” and “choice number” coincide.
In this exposition, we prove the following result of Alon.
In particular, recall that a planar bipartite graph with vertices contains at most edges. Thus for such graphs we have and deduce:
This corollary is sharp, as it applies to which we have seen in Example 4 has .
The rest of the paper is divided as follows. First, we begin in §2 by stating Theorem 9, the famous combinatorial nullstellensatz of Alon. Then in §3 and §4, we provide descriptions of the so-called graph polynomial, to which we then apply combinatorial nullstellensatz to deduce Theorem 18. Finally in §5, we show how to use Theorem 18 to prove Theorem 7.
2. Combinatorial Nullstellensatz
The main tool we use is the Combinatorial Nullestellensatz of Alon.
3. The Graph Polynomial, and Directed Orientations
Motivated by Example 10, we wish to apply a similar technique to general graphs . So in what follows, let be a (simple) graph with vertex set .
We observe that coefficients of correspond to differences in directed orientations. To be precise, we introduce the notation:
Proof: Consider expanding . Then each expanded term corresponds to a choice of or from each , as in Example 13. The term has coefficient is the orientation is even, and if the orientation is odd, as desired.
Thus we have an explicit combinatorial description of the coefficients in the graph polynomial .
4. Coefficients via Eulerian Suborientations
We now give a second description of the coefficients of .
Eulerian suborientations are brought into the picture by the following lemma.
Proof: Consider any orientation , Then we define a suborietation of , denoted , by including exactly the edges of whose orientation in is in the opposite direction. It’s easy to see that this induces a bijection
Moreover, remark that
- is even if and are either both even or both odd, and
- is odd otherwise.
The lemma follows from this.
We now arrive at the main result:
5. Finding an orientation
Armed with Theorem 18, we are almost ready to prove Theorem 7. The last ingredient is that we need to find an orientation on in which the maximal degree is not too large. This is accomplished by the following.
Proof: This is an application of Hall’s marriage theorem.
Let . Construct a bipartite graph
Connect and if is an endpoint of . Since we satisfy Hall’s condition (as is a condition for all subgraphs ) and can match each edge in to a (copy of some) vertex in . Since there are exactly copies of each vertex in , the conclusion follows.