3.4 KiB
title
| title |
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| 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 #prob-rainbowbasecover}
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|}+1by matroid intersection; - lowerbound:
3on graphic matroid ofK_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 (allow repetition) 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.
- If
\beta(M_1)\neq \beta(M_2), then\beta(M_1\cap M_2)=\max \set{\beta(M_1),\beta(M_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. :::
Cases that have been verified:
- If
\beta(M_1)=\beta(M_2)=2then\beta(M_1\cap M_2)=3.^[Note that this case does not give an answer to [@prob-rainbowbasecover] since the covering in\beta(M_1\cap M_2)uses common independent sets instead of common bases.] - If
\beta(M_1)=2,\beta(M_2)=3, then\beta(M_1\cap M_2)\leq 4 - ...
::: {.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. Bérczi and Schwarcz generalized SBO using the following definition.
::: {.Definition title="strongly base reorderable"}
A matroid is called strongly base reorderable(SBRO) if for any pair A,B of bases, there exists bases A',B' and a bijection \phi:A'\setminus B'\to B'\setminus A' such that A'\cap B'=A\cap B,A'\cup B'=A\cup B, and (A'\setminus X)\cup \phi(X) is a basis for every X\subset A'\setminus B'.
:::
::: Lemma The class of SRBO matroids is complete(closed under taking minors, duals, direct sums, truncations and induction by directed graphs). :::
They also showed that SBRO is sufficient for \beta(M_1\cap M_2)=\max \set{\beta(M_1),\beta(M_2)}.
::: Theorem
Let M_1 and M_2 be strongly base reorderable matroids on the same ground set.
Then \beta(M_1\cap M_2)= \max \set{\beta(M_1),\beta(M_2)}.
:::
Circuit cover
Let M=(E,\mathcal B) be a matroid and let A,B\in \mathcal B be two bases of M.
Define a graph G=(A\Delta B, F).