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.
Example 4 (Erdös)
Let be an integer and define
We claim that for any integer we have
The latter equality follows from being partite.
Now to see the first inequality, let have vertex set , where is the set of functions and . Then consider colors for . On a vertex , we list colors , , \dots, . On a vertex , we list colors , , \dots, . By construction it is impossible to properly color with these colors.
The case is illustrated in the figure below (image in public domain).
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 .
Theorem 5 (Noel, West, Wu, Zhu)
If is a graph with vertices then
In particular, if then .
One of the most major open problems in this direction is the following.
A claw-free graph is a graph with no induced . For example, the line graph (also called edge graph) of any simple graph is claw-free.
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.
Theorem 7 (Alon)
A bipartite graph is choosable, where
is half the maximum of the average degree of subgraphs .
In particular, recall that a planar bipartite graph with vertices contains at most edges. Thus for such graphs we have and deduce:
A planar bipartite graph is -choosable.
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.
Let us give a second proof that
for every positive integer . Our proof will be an application of the Nullstellensatz.
Regard the colors as real numbers, and let be the set of colors at vertex (hence , and ). Consider the polynomial
The coefficient of is . Therefore, one can select a color from each so that does not vanish.
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 .
The graph polynomial of is defined by
We observe that coefficients of correspond to differences in directed orientations. To be precise, we introduce the notation:
In the graph polynomial of , the coefficient of is
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 .
Let , viewed as a directed graph. An Eulerian suborientation of is a subgraph of (not necessarily induced) in which every vertex has equal indegree and outdegree. We say that such a suborientation is
- even if it has an even number of edges, and
- odd if it has an odd number of edges.
Note that the empty suborientation is allowed. We denote the even and odd Eulerian suborientations of by and , respectively.
Eulerian suborientations are brought into the picture by the following lemma.
Assume . Then there are natural bijections
Similarly, if then there are bijections
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.
In the graph polynomial of , the coefficient of is
where is arbitrary.
Proof: Combine Lemma 14 and Lemma 16.
We now arrive at the main result:
Proof: Combine Corollary 17 with Theorem 9.
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.
Let as in Theorem 7. Then has an orientation in which every indegree is at most .
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.
Now we can prove Theorem 7. Proof: According to Lemma 19, pick where . Since is bipartite, we obviously have , since cannot have any odd cycles. So Theorem 18 applies and we are done.