----
title: Matroid Rainbow Base Cover
----

This problem is proposed in [@emlektabla16_2024].

Let $M=(E,\mathcal B)$  be a rank-$r$ matroid whose ground set decomposes into two disjoint bases. Furthermore,
assume that $E$ is colored by $r$ colors, each color appearing exactly twice. A basis of $M$ is called rainbow if
it does not contain two elements of the same color.

::: Problem
What is the minimum number of rainbow bases needed to cover $E$?
:::

Kristóf Bérczi conjectured the minimum number is 3.

Currently known bounds:

- upperbound: $\floor{\log_2 |E|}+1$ by matroid intersection;
- lowerbound: $3$ on graphic matroid of $K_4$.

::: {.Conjecture #conj-doublecover}
Let $M_1$ and $M_2$ be two matroid on the same ground set $E$. Assume that $E$ decomposes into two bases in both of them.
Then $M_1$ and $M_2$ has four common bases that cover each element exactly twice.
:::

Let $M_1$ be the partition matroid of colors and let $M_2$ be $M$.
If [@conj-doublecover] is true, then 3 common bases will be enough to cover $E$.

# Known results

Let $\beta(M)$ be the covering number of matroid $M$. Given two matroids $M_1,M_2$ on the same ground set, $\beta(M_1\cap M_2)$ is the minimum number of common independent sets needed to cover the ground set.
[@conj-doublecover] is basically asking for $\beta(M_1\cap M_2)$ when the ground set $E$ is partitioned into 2 common bases of $M_1$ and $M_2$.

## Results and conjectures on $\beta(M_1\cap M_2)$

One can see that $\beta(M_1\cap M_2)\geq \min \set{\beta(M_1),\beta(M_2)}$. Aharoni and Berger showed that $\beta(M_1\cap M_2)\leq 2 \max \set{\beta(M_1),\beta(M_2)}$

::: {.Conjecture title="Aharoni and Berger"}
Let $M_1$ and $M_2$ be two matroids on the same ground set.

1. If $\beta(M_1)\neq \beta(M_2)$, then $\beta(M_1\cap M_2)=\max \set{\beta(M_1),\beta(M_2)}$.
2. If $\beta(M_1)= \beta(M_2)$, then $\beta(M_1\cap M_2)\leq \max \set{\beta(M_1),\beta(M_2)}+1$.
:::

::: {.Theorem title="Davies and McDiarmid"}
Let $M_1$ and $M_2$ be strongly base orderable matroids on the same ground set.
Then $\beta(M_1\cap M_2)= \max \set{\beta(M_1),\beta(M_2)}$.
:::

See [@berczi_partitioning_2024] for refs.