random merging does not work

main.tex

@@ -226,9 +226,30 @@ Note that for graphic matroids the number of spans is $O((r(E)-r(F))^2)$.
 For any spanning tree $T$, the number of spans hitting $T$ is exactly $r(E)-r(F)$ and these spans have a nice structure. If we contract each component in $G[F]$ to a vertex and consider spans a set of parallel edges, then the set of spans hitting $T$ is a tree (with parallel edges) in $G[F]$.
 For rigidity matroids, the number of rigid components in $F$ cannot be bounded by $r(E)-r(F)$.\footnote{Since the 1-thin cover inducing a cocircuit can have any number of rigid components.}
 
-Let $\mathcal S_F$ be the span partition of $E-F$ described in \autoref{lem:partition}. Let $B$ be a fixed base. We say a part $S\in \mathcal S_F$ is good if $S\cap B$ is non-empty. Note that if we merge a part $S$ into $F$, then the resulting span partition is a coarsening of $\mathcal S_F$. For general matroids, the set of good parts in $\mathcal S_F$ never merge when merging a good part into $F$.
+Let $\mathcal S_F$ be the span partition of $E-F$ described in \autoref{lem:partition}. Let $B$ be a fixed base.
+Note that if we merge a part $S$ into $F$, then the resulting span partition is a coarsening of $\mathcal S_F-S$.
+% We say a part $S\in \mathcal S_F$ is good if $S\cap B$ is non-empty.
+% For general matroids, the set of good parts in $\mathcal S_F$ never merge when merging a good part into $F$.
 It would be nice if we can characterize good parts in rigidity matroids with 1-thin cover.
 
+\paragraph{Randomly merge span-partition}
+We randomly pick one set in the span-partition $\mathcal S_F$ and merge it into $F$ until there is only one part.
+
+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.
+\end{conjecture}
+
+However, this is not the case.
+Consider an edge $(u,v)$ and one round.
+\[
+    \Pr[\text{$(u,v)$ is not contracted}]\leq 1-\frac{1}{|E|} \leq
+    \frac{n(n-1)-2}{n(n-1)}=\frac{(n+1)(n-2)}{n(n-1)}
+\]
+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 $(n+1)/3$.
+
 \section{Greedy base packing}
 
 \section{Principal sequence + KT contraction}