# Colouring matroid reductions

## Covering a graph by forests

Given a loopless graph $G = (V,E)$ with at least one edge, what is the smallest possible number of forests needed to cover all edges? This number is known as the arboricity $a(G)$ of $G$.

For example, consider the following 6-vertex graph, which is obtained from the complete graph $K_5$ by removing one edge, and adding a new pendant edge to one of the vertices of degree 3.

As the graph is connected and has six vertices, each forest in $G$ contains at most five edges. As the graph has ten edges in total, we need at least two forests to cover all edges.

But try as we might, we’ll never be able to cover all edges with only two forests. The reason is that similar bounds must hold in every subgraph of $G$. If we remove the unique degree-1 vertex and the edge incident with it, we are left with a connected graph on five vertices and nine edges. As in this subgraph every forest has at most four edges, we need at least three forests to cover the whole graph. I’ll leave it as an exercise to show that the original graph can actually be covered by three forests.

So, each subgraph $G’$ of $G$ (with at least one edge) provides a lower bound for $a(G)$, as $a(G) \ge \left\lceil \frac{|E(G’)|}{|V(G’)|-c(G’)}\right\rceil$ (here, $c(G’)$ denotes the number of components of $G’$).

Nash-Williams showed that we don’t need more forests than suggested by the worst possible bound obtained in this way: $$a(G) = \max_{G’\subseteq G : E(G’) \neq \emptyset} \left\lceil \frac{|E(G’)|}{|V(G’)|-c(G’)} \right\rceil.$$

## Colouring matroids by independent sets

Each of the definitions above can be generalised to matroids, in which case the corresponding question becomes: Given a matroid $M$, how many independent sets do we need to cover the entire ground set of $M$?

The minimum number of independent sets that we need is known as the colouring number (or covering number) of the matroid, and we write $\text{col}(M)$. As loops are not contained in any independent set, we will assume that $M$ has no loops. Edmonds showed (PDF) that the following generalisation of Nash-Williams’ formula holds: $$\text{col}(M) = \max_{E’ \subseteq E: r(E’) > 0} \left\lceil\frac{|E’|}{r_M(E’)}\right\rceil.$$

We say that $M$ is $k$-colourable if $\text{col}(M) \le k$. Note that $a(G) = \text{col}(M(G))$ for any graph $G$, so the colouring number is a proper generalisation of arboricity.

(I like to use “colouring number” rather than “covering number”, as the definition is analoguous to that of “chromatic number” in graphs, which, after all, is the smallest number of independent sets of vertices into which the graph can be partitioned.)

## Colouring partition matroids

The colouring number of the uniform matroid $U_{r,n}$ is $\lceil n/r\rceil$. It is clear that this is a lower bound, as $U_{r,n}$ has $n$ elements, and each independent set contains at most $r$ elements.

It is also an upper bound, for we can arbitrarily partition $U_{r,n}$ into $\lfloor n/r\rfloor$ independent sets of size $r$ and, if $r$ does not divide $n$, one more independent set of size less than $r$.

It is an exercise to show that $\text{col}(M_1 \oplus M_2) = \max\{\text{col}(M_1), \text{col}(M_2)\}$.

A matroid is a partition matroid if it is a direct sum of uniform matroids. The following result follows immediately from the above observations:

Lemma.If $M = \bigoplus\limits_{i=1}^k U_{r_i, n_i}$, then $\text{col}(M) = \max\limits_{i=1}^k \lceil n_i / r_i \rceil$.

A 1-partition matroid is a partition matroid in which each of the summands $U_{r_i, n_i}$ has rank 1. The colouring number of a 1-partition matroid is just the size of its largest component.

While colouring 1-partition matroids sounds trivial, the problem occurs naturally in discrete optimisation. For example, edge-colouring bipartite graphs can be described as finding an optimal colouring of the intersection of two 1-partition matroids corresponding to the two sides of the bipartition:

## Weak maps and bounds

Suppose that $M$ is some complicated matroid and we want to know or bound $\text{col}(M)$. If $N$ is another matroid on the same ground set such that every independent set in $N$ is independent in $M$ (that is, $N$ is a weak map image of $M$), then $\text{col}(M) \le \text{col}(N)$.

This is particularly useful when $N$ is a 1-partition matroid, as in that case $\text{col}(M)$ is bounded by the size of the largest uniform summand in $N$.

Thus, we have the problem of finding a 1-partition reduction: given a matroid $M$, can we find a 1-partition matroid $N$ that is a weak map image of $M$ whose colouring number is not much larger than that of $M$?

In particular, does there exist a constant $c$ such that we can always find a 1-partition reduction $N$ such that $\text{col}(M) \le \text{col}(N) \le c\cdot\text{col}(M)$? We call a matroid $(1,c)$-decomposable if admits such a 1-partition reduction.

## A conjecture

This question was asked by Bérczi, Schwartz and Yamaguchi, who conjectured that every matroid is (1,2)-decomposable; that is, for every $k$-colourable matroid $M$, there exists a partition $E(M) = X_1 \cup X_2 \cup \ldots \cup X_\ell$ such that each $X_i$ contains at most $2k$ elements, and each transversal of $\{X_1, X_2, \ldots, X_\ell\}$ is independent.

While this sounds like a very strong statement, it is true for a large number of natural classes of matroids, including graphic matroids and transversal matroids. (In those cases, an actual 1-partition reduction can be found in polynomial time, based on the natural representations of such matroids).

Unfortunately, the general conjecture was recently shown to fail for large binary projective geometries by Abdolazimi, Karlin, Klein and Oveis Gharan (in fact, their argument shows that there are matroids that are not $(1,c)$-decomposable for arbitrary $c$; a stronger result was later obtained by Leichter, Moseley and Pruhs). I was later able to show the slightly stronger result that almost every $\mathbb{F}_q$-representable matroid is not $(1,c)$-decomposable.

However, not all is lost. While there are counterexamples to the conjecture, the number of such counterexamples may be small; more precisely, the conjecture may still hold for almost all matroids.

Let $\mathcal{P}$ be a matroid property that is closed under isomorphism. We say that almost every matroid has property $\mathcal{P}$ if the fraction of matroids on ground set $[n]$ that fail $\mathcal{P}$ vanishes as $n$ tends to infinity, that is $$\lim_{n \to \infty} \frac{\#\{\text{matroids on [n] with \mathcal{P}}\}}{\#\{\text{matroids on [n]}\}} = 1.$$

Bérczi, Schwarcz, and Yamaguchi proved that paving matroids are (1,2)-decomposable. It is widely believed that almost every matroid is paving; proving this conjecture would imply that almost every matroid is (1,2)-decomposable.

In fact, since we know a little more about random matroids, the factor 2 can be improved slightly.

First, we know that almost every matroid has rank close to half its size. More precisely, for $\varepsilon > 0$, almost every matroid satisfies $$\left(\frac{1}{2}-\varepsilon\right)|E(M)| \le r(M) \le \left(\frac{1}{2}+\varepsilon\right)|E(M)|.$$

Second, Bérczi, Schwarcz and Yamaguchi proved that a paving matroid of rank $r \approx n/2$ is 2- or 3-colourable, and that if it is $k$-colourable, it has a $\lceil\frac{r}{r-1}k\rceil$-colourable partition reduction. This leads to the following updated version of the conjecture, which may still be true:

Conjecture. Almost every matroid is (1, 3/2)-decomposable.

This site uses Akismet to reduce spam. Learn how your comment data is processed.