Colouring and list-colouring

Staying close in theme to last month’s post, this month I’m writing about decomposing matroids into independent sets.

The colouring number (or covering number), $\text{col}(M)$ of a matroid $M$ is defined as the smallest $t$ for which the ground set of $M$ can be partitioned into $t$ independent sets. I’ve written before about how I like the term colouring number since it highlights the analogy with the chromatic number of graphs (which is the smallest number of vertex-independent sets in which the graph can be partitioned). For this blog post, though, it is more illuminating to compare the colouring number to the edge-chromatic number of a graph.

The chromatic number for graphs has many variations that are studied for their own sake, such as the list-chromatic number. Similarly, there is a list-colouring version of the colouring number, which is defined as follows.

Definition. Let $M$ be a loopless matroid, and suppose that for each element $e$ of $M$ we are given a list $L(e)$ of colours available to element $e$. An $L$-colouring of $M$ is a function $c$ that maps each element $e$ of $M$ to an element of $L(e)$, with the additional property that $c^{-1}(i)$ is an independent set of $M$ for all $i \in \bigcup_{e} L(e)$. The matroid $M$ is $k$-list-colourable if it has an $L$-colouring whenever $|L(e)| \ge k$ for each $e$. We write $\text{col}_\ell(M)$ for the smallest $k$ such that $M$ is $k$-list-colourable.

When $L(e) = \{1,2,\ldots,k\}$ for each $e$, $L$-colouring corresponds to colouring the matroid with $k$ colours. This implies that $\text{col}_\ell(M) \ge \text{col}(M)$. For graphs the gap between chromatic number and list-chromatic number can be arbitrarily large. Surprisingly, the colouring and list-colouring numbers for matroids are equal, as shown by Seymour (using an application of matroid union!).

Theorem (Seymour). $\text{col}_\ell(M) = \text{col}(M)$ for all loopless matroids $M$.

It is clear from this theorem that nothing is to be gained by considering the list-colouring problem for matroids instead of the colouring problem.

However, it is not known if an analogue of Seymour’s result holds for matroid intersections. Let $M_1$ and $M_2$ be two matroids on a common ground set $E$. The colouring number $\text{col}(M_1 \wedge M_2)$ is the smallest $t$ for which $E$ can be partitioned into $k$ sets that are independent in both $M_1$ and $M_2$ (here, $M_1 \wedge m_2$ refers to the matroid intersection of $M_1$ and $M_2$). The list-colouring number for $\chi_\ell(M_1 \wedge M_2)$ is similarly obtained as a generalisation of the list-colouring number for matroids.

As in the case of (list-)colouring matroids, a straightforward argument shows that $\text{col}_\ell(M_1 \wedge M_2) \ge \text{col}(M_1 \wedge M_2)$, but it is not known if equality holds for all matroid intersections.

1-partition matroids and Galvin’s theorem

A 1-partition matroid is a matroid whose ground set can be partitioned into subsets such that the independent sets of the matroid contain at most one element from each of the subsets. In other words, a loopless matroid $M$ is a 1-partition matroid if and only if it is the direct sum of $r(M)$ parallel classes.

Such matroids arise naturally in matching theory. A bipartite graph $G$ with bipartition $L \cup R$ comes with two 1-partition matroids on $E$: one in which the parallel classes are formed by the edges incident with each of the vertices in $L$, and one whose parallel classes similarly correspond to the vertices in $R$. In fact, the matchings of $G$ are precisely the common independent sets of these two matroids.

Galvin showed that the list-edge-chromatic number and the edge-chromatic number of bipartite graphs coincide, which is equivalent to the following result.

Theorem (Galvin). $\text{col}_\ell(M_1 \wedge M_2) = \text{col}(M_1 \wedge M_2)$ for all loopless 1-partition matroids.

It is natural to ask if the conclusion of the theorem still holds when the 1-partition matroids are replaced by matroids that are the direct sum of uniform matroids whose rank may be larger than 1 (such matroids are called partition matroids). It turns out that the answer is ‘yes’, a result that was proved only a few months ago by Guo.

Theorem (Guo). $\text{col}_\ell(M_1 \wedge M_2) = \text{col}(M_1 \wedge M_2)$ for all loopless partition matroids.

When is $\text{col}_\ell(M_1 \wedge M_2) = \text{col}(M_1 \wedge M_2)$?

In light of Seymour’s theorem for the list-colouring number of matroids and the above results by Galvin and Guo, it is tempting to ask whether the list-colouring number is always equal to the colouring number for matroid intersections – as Király did.

Question (Király). Is it always true that $\text{col}_\ell(M_1 \wedge M_2) = \text{col}(M_1 \wedge M_2)$?

Very little appears to be known about this problem, with the exception of the results by Galvin and Guo and the following result by Király and Pap.

Theorem (Király and Pap). $\text{col}_\ell(M_1 \wedge M_2) = \text{col}(M_1 \wedge M_2)$ in each of the following cases:

• $M_1$ and $M_2$ are both transversal matroids; or
• $M_1$ is the graphic matroid and $M_2$ is the 1-partition matroid whose parts are formed by the in-stars of the graph that is the union of two arborescences with the same root; or
• $M_1$ and $M_2$ both have rank 2.

(The third case of of their result actually extends to the situation where $M_1$ is a rank-2 matroid and $M_2$ is an arbitrary matroid.)

As a step towards resolution of Király’s question, the following problem may be more accessible:

Question. Is it true that $\text{col}_\ell(M_1 \wedge M_2) = \text{col}(M_1 \wedge M_2)$ for all loopless rank-3 matroids $M_1$ and $M_2$?

Happy New Year from all of us at the Matroid Union!

Decomposition of objects into smaller or simpler objects is a central theme in combinatorics. Think about decomposing a graph into its connected components or blocks, or into forests. In today’s post, I discuss decomposing finite projective geometries into projective geometries of smaller rank.

Definition: Let $q$ be prime power, let $n \ge t \ge 1$ be integers and let $M = \text{PG}(n-1,q)$ be the projective geometry over the field with $q$ elements. A $t$-spread of $G$ is a collection of rank-$t$ flats of $M$ that partition the points of $M$—that is, these flats are pairwise disjoint and their union is the set of all points of $M$.

(Geometers often use dimension instead of rank; the dimension of a projective geometry is one less than its rank. Their $t$-spreads are collections of $t$-dimensional subspaces of an $n$-dimensional space. We will however stick to the terminology that is more familiar in the matroid theory setting.)

Every projective geometry has a 1-spread: the flats making up the spread are simply the points of the geometry. It is much more interesting to look into the existence of $t$-spreads when $t \ge 2$.

The Fano plane $F_7 = \text{PG}(2,2)$ does not admit a decomposition into triangles. The quickest argument to see this is based on counting the number of points: any matroid that can be decomposed into triangles necessarily has a number of points that is a multiple of 3, but the Fano plane has seven points. By contrast, the projective geometry $\text{PG}(3,2)$, which has fifteen points, can be partitioned into five triangles (as encoded using different colours in the following representation:

The counting argument above can be generalised. In order for $M = \text{PG}(n-1,q)$ to have any hope of admitting a $t$-spread, the number of points in a projective geometry of rank $t$ should evenly divide the number of points in $M$, or $\frac{q^t-1}{q-1} \bigg| \frac{q^n-1}{q-1}$. It is a nice exercise to show that the latter happens precisely when $n$ is a multiple of $t$. So, if $\text{PG}(n-1,q)$ can be decomposed into rank-$t$ subgeometries, we must have $t|n$.

It turns out that this divisibility criterion is not only necessary—it is sufficient as well. We end this blog post with a slick proof of this fact. The proof can be found in many introductory texts on projective geometry.

Theorem: The projective geometry $M=\text{PG}(n-1,q)$ admits a $t$-spread if and only if $t|n$.

Proof: We already proved the implication from left to right. The implication in the other direction follows from a construction.

Suppose that $n = kt$ for some integer $k$. The points of $M$ can be identified with the 1-dimensional subspaces of the vector space $\mathbb{F}_q^n$. We will show that $\mathbb{F}_q^n$ can be decomposed into $t$-dimensional subpspaces that are pairwise disjoint (except that they all contain the zero-vector). As the 1-dimensional subspaces of $\mathbb{F}_q^t$ can be identified with the points of a $\text{PG}(t-1,q)$, this immediately implies the theorem.

Consider the $k$-dimensional $\mathbb{F}_{q^t}$-vector space $\mathbb{F}_{q^t}^k$. Its 1-dimensional subspaces intersect only in the zero-vector. These 1-dimensional subspaces give us our spread, as each such subspace can alternatively be viewed as a $t$-dimensional $\mathbb{F}_q$-vector space. $\Box$

Tuza’s Conjecture

One of my favourite open problems in Tuza’s Conjecture, which relates triangle packings and triangle hitting sets in graphs.

Let $G$ be a graph. A triangle packing is a set of triangles in $G$ no two of which have an edge in common. The maximum size of a triangle packing is called the triangle packing number of $G$, for which we write $\nu(G)$.

A triangle-hitting set is a subset $X \subseteq E(G)$ with the property that removing $X$ from $G$ results in a triangle-free graph. We write $\tau(G)$ for the size of the smallest triangle-hitting set in $G$.

A straightforward argument shows that $\nu(G) \le \tau(G) \le 3\nu(G)$. Tuza conjectured that the upper bound can be improved.

Conjecture (Tuza, 1981): $\tau(G) \le 2\nu(G)$ for all graphs $G$.

In other words, a graph without $t+1$ edge-disjoint triangles can be made triangle-free by removing at most $2t$ edges.

The factor 2 in the conjecture, if true, is best possible: $K_4$ and $K_5$ are examples for which $\tau(G) = 2\nu(G)$.

Geometric Tuza’s Conjecture

Tuza’s Conjecture, being formulated in terms in terms of edges and triangles only, can be naturally generalised to matroids. A triangle in a matroid $M$ is a restriction isomorphic to $U_{2,3}$. We define $\nu(M)$ to be the maximum number of disjoint triangles in $M$, and $\tau(M)$ to be the minimum size of a set $X$ such that $M\backslash X$ does not contain any triangles.

Unfortunately, the geometric version of Tuza’s Conjecture fails for matroids in general: For the Fano plane, $F_7 = \mathrm{PG}(2,2)$, we have $\tau(F_7) = 3$ and $\nu(F_7) = 1$.

Kazu Nomoto and I believe that, at least for binary matroids, the Fano plane is the only obstruction to a geometric version of Tuza’s Conjecture.

Conjecture (Nomoto–Van der Pol, 2021): Let $M$ be a simple binary matroid with no restriction isomorphic to $F_7$. Then $\tau(M) \le 2\nu(M)$.

As graphic matroids do not have restrictions isomorphic to $F_7$, the geometric version of Tuza’s conjecture implies the graphic version. We verified the conjecture for cographic matroids and several other classes of matroids.

Projective planes

Moving beyond binary matroids, one could ask if Tuza’s Conjecture holds for other classes of matroids. Here, we ask if it holds for projective planes. We’ve already seen that the conjecture fails for the Fano plane. It fails spectacularly for all other finite projective planes as well. This result is closely related to classical results in projective geometry.

A finite projective plane is a set system $\Pi = (E,\mathcal{L})$ — the elements of $E$ are called points, and the elements of $\mathcal{L}$ are called lines — with the following properties:

1. Any pair of distinct points is contained in exactly one line;
2. Any pair of lines intersects in exactly one point; and
3. There are four points such that no line contains more than two of them.

For each finite projective plane $\Pi$ there is a number $q \ge 2$, the order of $\Pi$, such that every point is contained in $q+1$ lines, and every line contains $q+1$ points. It follows that a projective plane of order $q$ contains exactly $q^2 + q + 1$ points.

Well-known examples are the Desarguesian planes $\mathrm{PG}(2,q)$, whose points and lines are the 1- and 2-dimensional subspaces of $\mathbb{F}_q^3$, respectively, but there are other examples as well.

Arcs

The lines in $\mathcal{L}$ form the collection of hyperplanes of a rank-3 matroid on $E$, so it makes sense to talk of $\tau(\Pi)$ and $\nu(\Pi)$ for finite projective planes $\Pi$.

An arc in $\Pi$ is a set $A$ of points such that no line intersects $A$ in more than two points. In our terminology, $A$ is a triangle-free restriction of $\Pi$. (Confusingly, the term triangle is also used for an arc of size 3, which is the opposite of our use of the term.)

The following result is well known in projective geometry.

Lemma: Let $\Pi$ be a projective plane of order $q$ and let $A$ be an arc. Then $|A| \le q+2$. If $q$ is odd, then $|A| \le q+1$.

Proof. Let $p \in A$ be any point, and let $\ell_1, \ell_2, \ldots, \ell_{q+1}$ be an enumeration of the lines incident with $p$. Each point in $\Pi$ is on one of those $q+1$ lines, and each line contains at most one point from $A$ other than $p$, so $|A| \le 1 + (q+1)$.

To prove the second bound, suppose for the sake of contradiction that $q$ is odd and $|A| = q+2$. Every line of $\Pi$ intersects $A$ in either $0$ or $2$ points; this is in particular true for all lines through a fixed point $p \not\in A$. It follows that $|A|$ is twice the number of lines incident with $p$ that intersect $A$, so $|A|$ must be even: a contradiction, so $|A| < q+2$. $\Box$

Arcs of size $q+1$ are called ovals, while arcs of size $q+2$ are called hyperovals. Constructions for ovals and hyperovals exist for the planes $\mathrm{PG}(2,q)$; Segre proved that in $\mathrm{PG}(2,q)$ of odd order, all ovals are nondegenerate conic sections.

By the lemma, sets containing more than $q+2$ points must contain a triangle, so $\tau(\Pi)$ is large.

Lemma: Let $\Pi$ be a projective plane of order $q$. If $q$ is even, then $\tau(\Pi) \ge q^2 – 1$. If $q$ is odd, then $\tau(\Pi) \ge q^2$.

Trivially, $\nu(\Pi) \le \frac{1}{3}(q^2+q+1)$. This is sufficient to show that Tuza’s conjecture fails for projective planes of sufficiently large order, as $\tau(\Pi) \sim 3\nu(\Pi)$ as $q \to \infty$.

The trivial upper bound on $\nu$ is correct, as can be shown by a simple construction.

Lemma: Let $\Pi$ be a projective plane of order $q \ge 3$. Then $\Pi$ can be decomposed into disjoint triangles and at most one point. Consequently, $\nu(\Pi) = \lfloor\frac{1}{3}(q^2+q+1)\rfloor$.

Proof. The exact construction depends on the remainder of $q$ modulo 3. Here, we only give the construction for the case $q = 3k+1$; the case $q = 3k+2$ allows a similar construction, while the case $q=3k$ is easier. See the figure following the proof for an illustration in the case $q=4$.

Let $p \in \Pi$ be any point, and let $\ell_0, \ell_1, \ldots, \ell_q$ be an enumeration of the lines incident with $p$. Let $e \in \ell_0$ be a point distinct from $p$, and let $\ell’ = {e, e’_1, \ldots, e’_q}$ and $\ell” = {e, e”_1, \ldots, e”_q}$ be two distinct lines incident with $e$ such that $e’_i, e”_i \in \ell_i$ for $i = 1, 2, \ldots, q$.

Consider the $(q+2)/3$ disjoint triangles

$\{e’_2, e’_3, e’_4\}, \{e’_5, e’_6, e’_7\}, \ldots, \{e’_{q-2}, e’_{q-1}, e’_q\},\quad\text{and}\quad\{e,e”_1, e”_2\}$,

and let $Z$ be the set of all points in those triangles. $Z$ contains exactly one point on each of the lines $\ell_i$, except that it contains two points of $\ell_2$. It follows that each of the sets $\ell_i\setminus (Z \cup {e})$ (for $i \neq 2$) and $\ell_2\setminus Z$ can be partitioned into triangles. $\Box$

It follows from the above results that

$\frac{\tau(\Pi)}{\nu(\Pi)} \ge \frac{q^2-1}{\frac{1}{3}(q^2+q+1)} \ge \frac{15}{7} > 2$

whenever the order $q$ of $\Pi$ is at least 4. For the (unique) projective geometry $\Pi = \mathrm{PG}(2,3)$ of order $q = 3$, we have $\tau(\Pi) = 9$ and $\nu(\Pi) = 4$, while for the (unique) projective geometry $\Pi = \mathrm{PG}(2,2)$ or order $q=2$ we have $\tau(\Pi)=3$ and $\nu(\Pi) = 1$, so Tuza’s Conjecture fails for all finite projective planes.

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.