From 6199b825b3fc0ddf2776f2dd1c2ea2226ed60b18 Mon Sep 17 00:00:00 2001 From: Yu Cong Date: Mon, 12 Jan 2026 14:14:46 +0800 Subject: [PATCH] update proof for graphic matroid --- .zed/tasks.json | 25 ------ main.tex | 229 +++++++++++++++++++++++++----------------------- ref.bib | 2 +- 3 files changed, 118 insertions(+), 138 deletions(-) delete mode 100644 .zed/tasks.json diff --git a/.zed/tasks.json b/.zed/tasks.json deleted file mode 100644 index 7500c78..0000000 --- a/.zed/tasks.json +++ /dev/null @@ -1,25 +0,0 @@ -// Static tasks configuration. -// -[ - { - "label": "forward_search", - "command": "/Applications/Skim.app/Contents/SharedSupport/displayline -r -z -b $ZED_ROW $ZED_DIRNAME/$ZED_STEM.pdf", - "allow_concurrent_runs": false, - "reveal": "never", - "hide": "always" - }, - { - "label": "pdflatex_view", - "command": "cd \"$ZED_DIRNAME\" && pdflatex -shell-escape -synctex=-1 \"$ZED_STEM\" && /Applications/Skim.app/Contents/SharedSupport/displayline -r -z -b $ZED_ROW \"$ZED_STEM\".pdf", - "allow_concurrent_runs": false, - "reveal": "no_focus", - "hide": "on_success" - }, - { - "label": "latexmk_view", - "command": "cd \"$ZED_DIRNAME\" && latexmk -pdf \"$ZED_STEM\" && /Applications/Skim.app/Contents/SharedSupport/displayline -r -z -b $ZED_ROW \"$ZED_STEM\".pdf", - "allow_concurrent_runs": false, - "reveal": "no_focus", - "hide": "on_success" - } -] \ No newline at end of file diff --git a/main.tex b/main.tex index 95be47d..d65c923 100644 --- a/main.tex +++ b/main.tex @@ -15,7 +15,7 @@ \maketitle \section{Ideal base packing} -Try to generalize Thorup's ideal tree packing \cite{Thorup_2008} to matroids. +Try to generalize Thorup's ideal tree packing \cite{Thorup_2008} to matroids. We cannot expect it to work on all matroids. The goal is to figure out some sufficient conditions and their relations with basepacking($\lambda\leq c \sigma$) and random contraction($\lambda \leq c \frac{|E|}{r(E)}$). @@ -26,114 +26,114 @@ One can select all bases and see that for every $k$-cocycle there is a base that Let $M=(E,\mathcal I)$ be a matroid with capacity $c:E\to \R_{\geq 0}$ on elements and let $\sigma=\min_{F\subset E} \frac{c(E-F)}{r(E)-r(F)}$ be its weighted strength. \begin{algo} -\textsc{\underline{Ideal Utilization}}($M$)\\ -If the groundset $E$ is empty, stop\\ -Let $F^*\in \argmin_{F\subset E} \frac{c(E-F)}{r(E)-r(F)}$ and let $\sigma$ be the strength\\ -for $e\in E-F^*$:\\ -\quad $u^*(e)=1/\sigma$\\ -\textsc{Ideal Utilization}($M|F^*$) + \textsc{\underline{Ideal Utilization}}($M$)\\ + If the groundset $E$ is empty, stop\\ + Let $F^*\in \argmin_{F\subset E} \frac{c(E-F)}{r(E)-r(F)}$ and let $\sigma$ be the strength\\ + for $e\in E-F^*$:\\ + \quad $u^*(e)=1/\sigma$\\ + \textsc{Ideal Utilization}($M|F^*$) \end{algo} We will work on the lattice of flats. The set of flats of $M$ forms a geometric lattice. Take two flats $A,B$ that $A\subset B$ and consider the sublattice between $A$ and $B$. This sublattice is exactly the lattice of flats of matroid $(M/A)\setminus (E-B)$. \begin{lemma} -Consider the ideal utilizations $u^*(e)$ assigned by the above algorithm. -\begin{enumerate} -\item $\sigma(M)\leq \sigma(M|F^*)$ -\item $u^*(e)$ is unique even though the $F^*$ may not be unique. -\item There is a fractional base packing $y$ such that $\sum_{B:e\in B}y(B)=u^*(e)$. -\item Each base in the above base packing is a minimum base with respect to the ideal utilizations $u^*(e)$. -\end{enumerate} + Consider the ideal utilizations $u^*(e)$ assigned by the above algorithm. + \begin{enumerate} + \item $\sigma(M)\leq \sigma(M|F^*)$ + \item $u^*(e)$ is unique even though the $F^*$ may not be unique. + \item There is a fractional base packing $y$ such that $\sum_{B:e\in B}y(B)=u^*(e)$. + \item Each base in the above base packing is a minimum base with respect to the ideal utilizations $u^*(e)$. + \end{enumerate} \end{lemma} \begin{proof} -The proofs are similar to those in \cite{thorup_fully-dynamic_2007}. -\begin{enumerate} -\item Let $F'\subset F^*$ be the optimal flat in $M|F^*$. Note that the lattice of flats is the same as the sublattice between $F^*$ and $\emptyset$ in $\mathcal L(M)$. Thus $F'$ is also a flat in $M$. Then we have -\[ -\sigma(M|F^*)=\frac{c(F^*-F')}{r(F^*)-r(F')} -=\frac{c(E-F')-c(E-F^*)}{r(E)-r(F')-(r(E)-r(F^*))} -\geq \frac{c(E-F^*)}{r(E)-r(F^*)}, -\] -the last inequality follows from $\frac{c(E-F')}{r(E)-r(F')}\geq \frac{c(E-F^*)}{r(E)-r(F^*)}$. + The proofs are similar to those in \cite{thorup_fully-dynamic_2007}. + \begin{enumerate} + \item Let $F'\subset F^*$ be the optimal flat in $M|F^*$. Note that the lattice of flats is the same as the sublattice between $F^*$ and $\emptyset$ in $\mathcal L(M)$. Thus $F'$ is also a flat in $M$. Then we have + \[ + \sigma(M|F^*)=\frac{c(F^*-F')}{r(F^*)-r(F')} + =\frac{c(E-F')-c(E-F^*)}{r(E)-r(F')-(r(E)-r(F^*))} + \geq \frac{c(E-F^*)}{r(E)-r(F^*)}, + \] + the last inequality follows from $\frac{c(E-F')}{r(E)-r(F')}\geq \frac{c(E-F^*)}{r(E)-r(F^*)}$. -\item If there are two disjoint flats $F_1,F_2$ achieving the same optimal strength. Consider the span of their union $F=\cl(F_1\cup F_2)$. It is not hard to see that -\[ -\frac{c(E-F)}{r(E)-r(F)}\leq -\frac{c(E)-c(F_1)-c(F_2)}{r(E)-r(F_1)-r(F_2)} -\leq \sigma -\] -Thus, both inequalities are tight. This fact implies that $\sigma(M)=\frac{c(F_1)}{r(F_1)}=\frac{c(F_2)}{r(F_2)}=\frac{c(E)}{r(E)}$. Now suppose that in the first step we choose $F_1$. Then $\empty$ should be the optimal flat in $M|F_1$ since $\sigma(M|F_1)\geq \sigma(M)=\frac{c(F_1)}{r(F_1)}$. Then the ideal utilization for any element is $1/\sigma$. + \item If there are two disjoint flats $F_1,F_2$ achieving the same optimal strength. Consider the span of their union $F=\cl(F_1\cup F_2)$. It is not hard to see that + \[ + \frac{c(E-F)}{r(E)-r(F)}\leq + \frac{c(E)-c(F_1)-c(F_2)}{r(E)-r(F_1)-r(F_2)} + \leq \sigma + \] + Thus, both inequalities are tight. This fact implies that $\sigma(M)=\frac{c(F_1)}{r(F_1)}=\frac{c(F_2)}{r(F_2)}=\frac{c(E)}{r(E)}$. Now suppose that in the first step we choose $F_1$. Then $\empty$ should be the optimal flat in $M|F_1$ since $\sigma(M|F_1)\geq \sigma(M)=\frac{c(F_1)}{r(F_1)}$. Then the ideal utilization for any element is $1/\sigma$. -Now suppose that $F_1$ and $F_2$ are not disjoint\footnote{In fact, the analysis still works if $F_1$ and $F_2$ are disjoint. The disjoint case can be removed.}. It suffices to prove that their meet $F_1\cap F_2$ is the optimal flat in $M|F_1$. First note that $F_1\cap F_2$ is a flat in $M$ and $M|F_1$. We claim that $\frac{c(F_1)-c(F_1\cap F_2)}{r(F_1)-r(F_1\cap F_2)}\leq \sigma$. Suppose this is not true. We have, -\[ -\begin{aligned} -\sigma &< \frac{c(F_1)-c(F_1\cap F_2)}{r(F_1)-r(F_1\cap F_2)} & & \\ - &\leq \frac{c(F_1\cup F_2)-c(F_2)}{r(F_1\cup F_2)-r(F_2)} & &\text{submodularity of $r$}\\ - &=\frac{c(E)-c(F_2)-(c(E)-c(F_1\cup F_2))}{r(E)-r(F_2)-(r(E)-r(F_1\cup F_2))} -\end{aligned} -\] -which shows $c(E)-c(\cl(F_1\cup F_2))\leq c(E)-c(F_1\cup F_2)<\sigma(r(E)-r(\cl(F_1\cup F_2)))$ and contradicts to the fact that $F_1,F_2$ are the optimal flats. Thus $\frac{c(F_1)-c(F_1\cap F_2)}{r(F_1)-r(F_1\cap F_2)}\leq \sigma$ holds and $F_1\cap F_2$ is the optimal flat in $M|F_1$. + Now suppose that $F_1$ and $F_2$ are not disjoint\footnote{In fact, the analysis still works if $F_1$ and $F_2$ are disjoint. The disjoint case can be removed.}. It suffices to prove that their meet $F_1\cap F_2$ is the optimal flat in $M|F_1$. First note that $F_1\cap F_2$ is a flat in $M$ and $M|F_1$. We claim that $\frac{c(F_1)-c(F_1\cap F_2)}{r(F_1)-r(F_1\cap F_2)}\leq \sigma$. Suppose this is not true. We have, + \[ + \begin{aligned} + \sigma & < \frac{c(F_1)-c(F_1\cap F_2)}{r(F_1)-r(F_1\cap F_2)} & & \\ + & \leq \frac{c(F_1\cup F_2)-c(F_2)}{r(F_1\cup F_2)-r(F_2)} & & \text{submodularity of $r$} \\ + & =\frac{c(E)-c(F_2)-(c(E)-c(F_1\cup F_2))}{r(E)-r(F_2)-(r(E)-r(F_1\cup F_2))} + \end{aligned} + \] + which shows $c(E)-c(\cl(F_1\cup F_2))\leq c(E)-c(F_1\cup F_2)<\sigma(r(E)-r(\cl(F_1\cup F_2)))$ and contradicts to the fact that $F_1,F_2$ are the optimal flats. Thus $\frac{c(F_1)-c(F_1\cap F_2)}{r(F_1)-r(F_1\cap F_2)}\leq \sigma$ holds and $F_1\cap F_2$ is the optimal flat in $M|F_1$. -\item -% I don't think this is true on general matroids. -We define the ideal base packing $y$ by induction. Suppose that the ideal base packing $y'$ on $M|F^*$ is known. -Let $y^*$ be the optimal fractional base packing of $M$ with capacity $c$. -(Following Thorup's notation, $y^*(B)$ is a probability on the set of bases and ``optimal'' means that $1/\sigma = \max_e \frac{\sum_{B:e\in B}y^*(B)}{c(e)}$.) + \item + % I don't think this is true on general matroids. + We define the ideal base packing $y$ by induction. Suppose that the ideal base packing $y'$ on $M|F^*$ is known. + Let $y^*$ be the optimal fractional base packing of $M$ with capacity $c$. + (Following Thorup's notation, $y^*(B)$ is a probability on the set of bases and ``optimal'' means that $1/\sigma = \max_e \frac{\sum_{B:e\in B}y^*(B)}{c(e)}$.) -We uniformly and independently choose a base $B_{F^*}$ with $y'(B_{F^*})>0$ and a base $B$ with $y^*(B)>0$ and construct a new set $S=B_{F^*}\cup (B\setminus F^*)$. -For any base $B$, the size of $S$ is $r(F^*)+r(E)-|B\cap F^*|$. However, if $B$ is in the support of $y^*$ then $|S|$ is exactly $r(E)$. To see this, consider the average relative load of $y^*$ on $e\in E\setminus F^*$. -We choose an edge $e$ with probability proportional to its capacity $c(e)$. -\[ -\sum_{e\in E\setminus F^*} \frac{c(e)}{c(E\setminus F^*)}\frac{\sum_{B:e\in B} \Pr[B]}{c(e)} -\geq \frac{r(E)-r(F^*)}{c(E\setminus F^*)}=\frac{1}{\sigma} -\] -Note that $B$ is taken from the support of the optimal base packing, then $\sum_{B:e\in B} \frac{\Pr[B]}{c(e)}$ are the same for all elements in $E\setminus F^*$ and every $B$ contains $r(E)-r(F^*)$ edges in $E\setminus F^*$. + We uniformly and independently choose a base $B_{F^*}$ with $y'(B_{F^*})>0$ and a base $B$ with $y^*(B)>0$ and construct a new set $S=B_{F^*}\cup (B\setminus F^*)$. + For any base $B$, the size of $S$ is $r(F^*)+r(E)-|B\cap F^*|$. However, if $B$ is in the support of $y^*$ then $|S|$ is exactly $r(E)$. To see this, consider the average relative load of $y^*$ on $e\in E\setminus F^*$. + We choose an edge $e$ with probability proportional to its capacity $c(e)$. + \[ + \sum_{e\in E\setminus F^*} \frac{c(e)}{c(E\setminus F^*)}\frac{\sum_{B:e\in B} \Pr[B]}{c(e)} + \geq \frac{r(E)-r(F^*)}{c(E\setminus F^*)}=\frac{1}{\sigma} + \] + Note that $B$ is taken from the support of the optimal base packing, then $\sum_{B:e\in B} \frac{\Pr[B]}{c(e)}$ are the same for all elements in $E\setminus F^*$ and every $B$ contains $r(E)-r(F^*)$ edges in $E\setminus F^*$. -In graphic matroids it follows easily that $S$ is a spanning tree. -However, $S$ may not be independent in general matroids.\footnote{Here is a systematical way to construct counterexamples. Let $B$ be a base of $M|F^*$ and let $X\subset E-F^*$ be an independent set such that $B\sqcup X$ is a base of $M$. Let $B'$ be the counterexample with largest intersection with $B$. We can assume the circuit $C\subset B'\sqcup X$ contains $B'\setminus B$ since otherwise we can do multiple symmetric exchange with $B$ and $B'$. Then we can divide $C$ into 3 parts, $R=B'\setminus B, S=C\cap B, T=C\cap X$. One can set $R=\set{(0,0,1)^T},S=\set{(1,1,1)^T}$ and $T=\set{(1,0,0)^T,(0,1,0)^T}$. Then it is easy to add more dimensions and get the desired $F^*$.} -$S$ is independent does not imply that $M$ is a direct sum of $M|F^*$ and $M\setminus F^*$ since the rank of $M\setminus F^*$ can be larger than $r(E)-r(F^*)$. -Characterization of matroids where $S$ is a base is another interesting problem. -\note{From now on we assume $S$ is a base. This should holds in all $(k,2k-1)$-sparsity matroids.} -Then the lemma follows by induction. + In graphic matroids it follows easily that $S$ is a spanning tree. + However, $S$ may not be independent in general matroids.\footnote{Here is a systematical way to construct counterexamples. Let $B$ be a base of $M|F^*$ and let $X\subset E-F^*$ be an independent set such that $B\sqcup X$ is a base of $M$. Let $B'$ be the counterexample with largest intersection with $B$. We can assume the circuit $C\subset B'\sqcup X$ contains $B'\setminus B$ since otherwise we can do multiple symmetric exchange with $B$ and $B'$. Then we can divide $C$ into 3 parts, $R=B'\setminus B, S=C\cap B, T=C\cap X$. One can set $R=\set{(0,0,1)^T},S=\set{(1,1,1)^T}$ and $T=\set{(1,0,0)^T,(0,1,0)^T}$. Then it is easy to add more dimensions and get the desired $F^*$.} + $S$ is independent does not imply that $M$ is a direct sum of $M|F^*$ and $M\setminus F^*$ since the rank of $M\setminus F^*$ can be larger than $r(E)-r(F^*)$. + Characterization of matroids where $S$ is a base is another interesting problem. + \note{From now on we assume $S$ is a base. This should holds in all $(k,2k-1)$-sparsity matroids.} + Then the lemma follows by induction. -\item If $F^*=\emptyset$ or the size of the groundset is 1, then one can easily see the claim holds since every element have the same ideal utilitization. -Now suppose the claim holds for $M|F^*$. -In the previous bullet point we have already shown that every element in $E\setminus F^*$ have the same utilization. -Note that we also have shown in the first bullet point that the ideal utilization is larger in $M$ than in $M|F^*$. -Thus, the construction conincides with the greedy algorithm for minimum matroid base. -\end{enumerate} + \item If $F^*=\emptyset$ or the size of the groundset is 1, then one can easily see the claim holds since every element have the same ideal utilitization. + Now suppose the claim holds for $M|F^*$. + In the previous bullet point we have already shown that every element in $E\setminus F^*$ have the same utilization. + Note that we also have shown in the first bullet point that the ideal utilization is larger in $M$ than in $M|F^*$. + Thus, the construction conincides with the greedy algorithm for minimum matroid base. + \end{enumerate} \end{proof} The following lemma does not seem related to the ideal tree packing. \begin{lemma} -If we decrease the capacity of an edge, no ideal edge utilization decreases. + If we decrease the capacity of an edge, no ideal edge utilization decreases. \end{lemma} \begin{proof} -Let $c'$ be the decreased capacity. Suppose for a contradiction that there is a element $f$ with decreased utilization $u'(f)< u(f)$. Among all such edges, let $f$ maximize $u(f)$. Consider edges with $u(e)>u$. We have $u'(e)\geq u(e)$ for such edges since $f$ is the counterexample with largest $u(f)$. -Let $M|E_{\leq u(f)}$ be the matroid restricted to the edge set with $u(e)\leq u(f)$. For simplicity, in this proof we write $A$ for $E_{\leq u(f)}$ and $A'$ for $E_{\leq u'(f)}$. Note that $f$ is in $A-F^*$ where $F^*$ is the smallest optimal flat and the strength of $M|A$ is $1/u(f)$. + Let $c'$ be the decreased capacity. Suppose for a contradiction that there is a element $f$ with decreased utilization $u'(f)< u(f)$. Among all such edges, let $f$ maximize $u(f)$. Consider edges with $u(e)>u$. We have $u'(e)\geq u(e)$ for such edges since $f$ is the counterexample with largest $u(f)$. + Let $M|E_{\leq u(f)}$ be the matroid restricted to the edge set with $u(e)\leq u(f)$. For simplicity, in this proof we write $A$ for $E_{\leq u(f)}$ and $A'$ for $E_{\leq u'(f)}$. Note that $f$ is in $A-F^*$ where $F^*$ is the smallest optimal flat and the strength of $M|A$ is $1/u(f)$. -Similarly, let $M|A'$ be the corresponding matroid under capacity $c'$. Note that if $e\notin A$, then $u'(e)\geq u(e)> u(f)> u'(f)$, so $e\notin A'$. It follows that $A'\subseteq A$. + Similarly, let $M|A'$ be the corresponding matroid under capacity $c'$. Note that if $e\notin A$, then $u'(e)\geq u(e)> u(f)> u'(f)$, so $e\notin A'$. It follows that $A'\subseteq A$. -Now consider the optimal cocycle $C=A-F^*$. We divide $C$ into 2 parts, $C_2=(A-F^*)\cap A'$ and $C_1=(A-F^*)- A'$. Note that by submodularity of rank function, we have -\[ -\frac{1}{u(f)}=\frac{c(C)}{r(A)-r(A\setminus C)} -\geq \frac{c(C_1)+c(C_2)}{(r(A)-r(A\setminus C_1))+(r(A')-r(A'\setminus C_2))}. -\] -We also know that $\frac{1}{u(f)}\leq \frac{c(C_1)}{r(A)-r(A\setminus C_1)}$. Then it follows that $\frac{c(C_2)}{r(A')-r(A'\setminus C_2)}\leq \frac{1}{u(f)}$. Hence, we get a contradiction -\[ -\frac{1}{u'(f)}\leq\frac{c'(A'-C_2)}{r(A')-r(A'\setminus C_2)} -\leq \frac{c(C_2)}{r(A')-r(A'\setminus C_2)}\leq \frac{1}{u(f)}. -\] + Now consider the optimal cocycle $C=A-F^*$. We divide $C$ into 2 parts, $C_2=(A-F^*)\cap A'$ and $C_1=(A-F^*)- A'$. Note that by submodularity of rank function, we have + \[ + \frac{1}{u(f)}=\frac{c(C)}{r(A)-r(A\setminus C)} + \geq \frac{c(C_1)+c(C_2)}{(r(A)-r(A\setminus C_1))+(r(A')-r(A'\setminus C_2))}. + \] + We also know that $\frac{1}{u(f)}\leq \frac{c(C_1)}{r(A)-r(A\setminus C_1)}$. Then it follows that $\frac{c(C_2)}{r(A')-r(A'\setminus C_2)}\leq \frac{1}{u(f)}$. Hence, we get a contradiction + \[ + \frac{1}{u'(f)}\leq\frac{c'(A'-C_2)}{r(A')-r(A'\setminus C_2)} + \leq \frac{c(C_2)}{r(A')-r(A'\setminus C_2)}\leq \frac{1}{u(f)}. + \] \end{proof} \begin{remark} -If we increase the capacity, no ideal edge utilization increases. The proof is similar. -Removing (contracting) edges has the same effect on ideal utilization as setting the capacity to $0$ ($\infty$). + If we increase the capacity, no ideal edge utilization increases. The proof is similar. + Removing (contracting) edges has the same effect on ideal utilization as setting the capacity to $0$ ($\infty$). \end{remark} \subsection{Counting} -Ideal base packing is a distribution on some bases. Given a subset $D\subset E$, consider the expected size of intersection with a base sampled from the ideal distribution. +Ideal base packing is a distribution on some bases. Given a subset $D\subset E$, consider the expected size of intersection with a base sampled from the ideal distribution. The expectation is exactly $\sum_{e\in D} c(e)u^*(e)$. Recall that our goal is to show that for any minimum $k$-cocycle a random base in the ideal base packing uses $O(k)$ elements in the cocycle in expectation. @@ -150,44 +150,49 @@ Let $F$ be the optimal flat for strength and assume $k=r(E)-r(F)>k'$. We want to Basically we need to do random contraction on $M/ F$. Let $\mathcal X$ be the set $\set{X|X=B\setminus F \land r(X)=k}$. That is, we consider all bases that hitten by the $k$-cocycle exactly $k$ times and for each of them we collect the intersection with the $k$-cocycle. Then we do $k-k'$ random contractions in $M/F$ to get a random $k'$-cocycle $C_{k'}$. -\begin{lemma}[Lemma~7 in \cite{Thorup_2008}, restated] -For graphic matroids, there is a distribution on $C_{k'}$ such that for any base in the ideal base packing, the expected size of its intersection with $C_{k'}$ is at most $2k'$. +\begin{lemma}[Lemma~7 in \cite{Thorup_2008}, restated]\label{lem:idealload} + For graphic matroids, there is a distribution on $C_{k'}$ such that for any base in the ideal base packing, the expected size of its intersection with $C_{k'}$ is at most $2k'$. \end{lemma} \begin{proof} -We can assume the matroid is connected. (Otherwise we can remove loops and coloops and add dummy elements.) -In graphic matroids, $F$ corresponds to a partition $\mathcal P_F$ with $k+1$ parts, where each part is the vertex set of a component in $G[F]$. -One can carefully design the distributions for contractions so instead of contracting edges, we consider randomly merging parts in $\mathcal P_F$. -We uniformly and randomly choose $k-k'+1$ parts in $\mathcal P_F$ and merge them into a big part. -Denote the resulting partition by $\mathcal P_{F'}$. -Let $T$ be a spanning tree in the support of ideal tree packing. -Recall that the number of inter-component edges of $T$ in $\mathcal P_F$ is $k$. -Then we have -\[ -\E_{F'}[|T\setminus F'|]\leq 2k \frac{k'}{k+1}\leq 2k'. -\] + We can assume the matroid is connected. (Otherwise we can remove loops and coloops and add dummy elements.) + In graphic matroids, $F$ corresponds to a partition $\mathcal P_F$ with $k+1$ parts, where each part is the vertex set of a component in $G[F]$. + One can carefully design the distributions for contractions so instead of contracting edges, we consider randomly merging parts in $\mathcal P_F$. + We uniformly and randomly choose $k-k'+1$ parts in $\mathcal P_F$ and merge them into a big part. + Denote the resulting partition by $\mathcal P_{F'}$. + Let $T$ be a spanning tree in the support of ideal tree packing. + Recall that the number of inter-component edges of $T$ in $\mathcal P_F$ is $k$. + Then we have + \[ + \E_{F'}[|T\setminus F'|]\leq 2k \frac{k'}{k+1}\leq 2k'. + \] \end{proof} +Finally, we want to upperbound the ideal load of the minimum $k$-cocycle ($(k+1)$-cut). +By \autoref{lem:idealload}, there is a distribution of some $k$-cocycles that use at most $2k$ edges in any ideal spanning tree. +Then there is a special $k$-cocycle $C$ that use (expectedly) at most $2k$ edges in a random ideal spanning tree. +Note that for edge set $D$ the ideal load $\sum_{e\in D} u^*(e)c(e)$ can be interpreted as the expected number of edges of $D$ in a random ideal spanning tree. +Now we consider the minimum $k$-cocycle $\mathcal K$. Its ideal load $\sum_{e\in \mathcal K} u^*(e)c(e)$ must be at most that of $C$, since $C$ has the largest the ideal load per edge capacity and $c(\mathcal K)\leq c(C)$. + For general matroids, we want to show the following. \begin{conjecture}\label{conj:dist} -Let $M'$ be the contraction minor $M/F^*$. The rank of $M'$ is $k$. -Given a positive integer $k'