enumeration of the final solution.
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@@ -180,6 +180,17 @@ For general matroids, we want to show the following.
Given a positive integer $k'<k$, then there exists a distribution on $k'$-cocycles such that for any base $B$ of $M'$, the expected size of intersection is $O(1)$ for fixed $k'$.
\end{conjecture}
\begin{theorem}
If \autoref{conj:dist} is true for matroid $M$, then one can compute minimum $k$-cocycle in polynomial time.
\end{theorem}
\begin{proof}
It follows from \autoref{conj:dist} that the minimum $k$-cocycle $C^*_k$ shares at most $h=O(1)$ elements with some base in the ideal base packing. The number of bases we need in the ideal base packing is polynomial (see next subsection).
We enumerate all bases in this set and for each base $B$ enumerate all subsets with size in range $[r-h,r-k]$.
Such a subset $I$ must be indenpendent and be contained in the flat $\overline{C^*_k}$. Then for such subset we further enumerate another subset $X$ such that $I\cup X$ is a rank-$r-k$ independent set. Thus, $\overline{\cl(I\cup X)}$ is a $k$-cocycle and we take the minimum one among all enumerations.
One can check that each part of the enumeration can be done in polynomial time.
\end{proof}
Certainly \autoref{conj:dist} does not hold on any matroid.
A counterexample would be uniform matroids $U_{2n,n}$.
The size of every $k'$-cocycle is $k'+n$, and the size of expected intersection should be $O(n)$.
@@ -237,19 +248,22 @@ We randomly pick one set in the span-partition $\mathcal S_F$ and merge it into
For graphic matroid, the desired bound is equivalent to the following conjecture.
\begin{conjecture}
Let $G=(V,E)$ be a connected graph with $n$ vertices. Contract edges uniformly random (ignore parallel edges) and remove loops until the remaining graph $H$ has 2 vertices.
Then for any spanning tree $T$ of $G$, the expected number of edges in $H\cap T$ is at most 2.
Let $G=(V,E)$ be a connected graph with $n$ vertices. Contract edges uniformly random (ignore parallel edges) and remove loops until the remaining graph $H$ has 2 vertices.
Then for any spanning tree $T$ of $G$, the expected number of edges in $H\cap T$ is at most 2.
\end{conjecture}
However, this is not the case.
Consider an edge $(u,v)$ and one round.
\[
\Pr[\text{$(u,v)$ is not contracted}]\leq 1-\frac{1}{|E|} \leq
\frac{n(n-1)-2}{n(n-1)}=\frac{(n+1)(n-2)}{n(n-1)}
\Pr[\text{$(u,v)$ is not contracted}]\leq 1-\frac{1}{|E|} \leq
\frac{n(n-1)-2}{n(n-1)}=\frac{(n+1)(n-2)}{n(n-1)}
\]
Then the probability that edge $(u,v)$ survives in the end is at most $\prod_{k=3}^n \frac{(n+1)(n-2)}{n(n-1)}=\frac{n+1}{3(n-1)}$.
Then the number of remaining edges in any spanning tree is at most $(n+1)/3$.
\subsection{Hypergraphic matroid}
Hypergraphic matroids are not closed under contraction (cf. Tamás Király's thesis).
\section{Greedy base packing}
\section{Principal sequence + KT contraction}