From a6e28b598fef9ffe96c00d67200d6144bdfe7be4 Mon Sep 17 00:00:00 2001
From: Yu Cong <sxlxcsxlxc@gmail.com>
Date: Tue, 10 Mar 2026 14:05:34 +0800
Subject: [PATCH] enumeration of the final solution.

---
 main.tex | 22 ++++++++++++++++++----
 1 file changed, 18 insertions(+), 4 deletions(-)

diff --git a/main.tex b/main.tex
index e2567bd..21ede3b 100644
--- a/main.tex
+++ b/main.tex
@@ -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}