\documentclass[a4paper,12pt]{article}
\usepackage{chao}
\usepackage{algo}
\geometry{margin=2cm}

\title{Ideal and greedy base packing}
\author{}
\date{}

\DeclareMathOperator*{\opt}{OPT}
\DeclareMathOperator*{\len}{len}
\DeclareMathOperator*{\cl}{span}

\begin{document}
\maketitle

\section{Ideal base packing}
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($\lambda \leq c \frac{|E|}{r(E)}$).

The idea is that we want to find a small set of bases such that for any minimum $k$-cocycle there is a base that the cocycle hits the base $O(k)$ times.
The number of bases should be as small as possible.
One can select all bases and see that for every $k$-cocycle there is a base that only gets hit exactly $k$ times.

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}($M|F^*$)
\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}
\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 $\frac{c(E-F')}{r(E)-r(F')}\geq \frac{c(E-F^*)}{r(E)-r(F^*)}$.

		\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 $M|F_1$.

		\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 $E\setminus F^*$.

		      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}
\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.
\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 $1/u(f)$.

	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
	\[
		\frac{1}{u'(f)}\leq\frac{c'(A'-C_2)}{r(A')-r(A'\setminus C_2)}
		\leq \frac{c(C_2)}{r(A')-r(A'\setminus C_2)}\leq \frac{1}{u(f)}.
	\]
\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 $0$ ($\infty$).
\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.
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.
Thorup proved that in \textbf{graphic matroids} there is a fixed distribution of $k$-cocycles such that for any base from the ideal base packing, the expected size of intersection is at most $O(k)$ (Lemma~7 in \cite{Thorup_2008}).
Then it follows that if we take a random spanning tree from the ideal tree packing, there is a fixed $(k+1)$-cut $C$ such that the expected size of intersection is at most $O(k)$, which implies $\sum_{e\in C}c(e)u^*(e)\in O(k)$.

How is this fixed $(k+1)$-cut (or $k$-cocycle) related to the minimum $(k+1)$-cut (minimum $k$-cocycle) ?
The minimum $k$-cocycle has smaller capacity than $C$.
Note that Thorup used a greedy way to construct the cocycle $C$. Elements in $C$ always has the largest possible utilization.
These facts implies that the minimum $k$-cocycle has a smaller value $\sum c(e)u^*(e)$ than $C$.

However, Lemma~7 in \cite{Thorup_2008} does not generalize to all matroids and we need to take a close look at the construction of $C$.
Let $F$ be the optimal flat for strength and assume $k=r(E)-r(F)>k'$. We want to find the minimum $k'$-cocycle.
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]\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
	\[
		\E_{F'}[|T\setminus F'|]\leq 2k \frac{k'}{k+1}\leq 2k'.
	\]
\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 $k'<k$, 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 $O(n)$.

\begin{lemma}\label{lem:partition}
	Let $M=(E,\mathcal I)$ be a matroids and let $F$ be a flat of $M$.
	Then $\cl(F+e)\setminus F$ is a partition of $E\setminus F$.
\end{lemma}
\begin{proof}
	Suppose for contradiction that there are two elements $x,y\in E-F$ such that $\cl(F+x)\setminus F$ and $\cl(F+y)\setminus F$ have non-empty intersection. Let $z$ be an element in the intersection.
	Then we find a circuit $C_{x,z}$ in $\cl(F+e)$ such that $C_{x,z}\setminus F= \set{x,z}$.
	Note that this circuit exists since $z$ is in the span of $F+e$.
	Let $C_{y,z}$ denote the analogous circuit for $y$.
	Then it follows from the circuit axiom that there is another circuit $C\subset C_{x,z}\cup C_{y,z}\setminus \set{z}$, which implies $y\in \cl(F+x)$ and thus contradicts the assumption.
\end{proof}

\subsection{Support size}

Recall that we construct the ideal base packing recursively. Suppose that the ideal base packing for $M|F^*$ contains $n$ bases and let $m$ be the size of support of the optimal base packing of $M$. Then the number of bases in the ideal base packing of $M$ is $nm$. Note that $m$ is upperbounded by the $|E|$ since the number of constraints is at most $|E|$ in the tree packing LP.
The support size can be exponential. Consider a path with $n$ points and parallel edges. The depth of recursion can be $n-1$.

Do we need all bases in the packing? Say we are interested in the minimum $k$-cocycle and want to show that we can find a set of bases such that for any minimum $k$-cocycle there is a base whose intersection with the cocyle is at most $O(k)$ elements.
The strategy is to first find a $\geq k$-cocycle using the utilization algorithm, then randomly delete edges in the $\geq k$-cocycle to make the rank defieciency exactly $k$. Notice that only elements in the $\geq k$-cocycle matter. Thus we only need constant recursion depth and $O(m^k)$ bases where $m$ is the number of elements.

Note that LP gives an ideal tree packing with $O(m)$ support size.

\subsection{Rigidity matroids}
\begin{conjecture}
	Let $M$ be a connected 2D rigidity matroid on graph $G=(V,E)$. Let $F^*$ be the optimal flat for strength $F^*=\argmin_{F\subset E}\frac{c(E\setminus F)}{r(E)-r(F)}$.
	Let $X\subset E\setminus F^*$ be a independent set with rank $r(E)-r(F^*)$. Then for any maximal independent set $B_{F^*}\subset F^*$, $X\cup B_{F^*}$ is a base of $M$.
\end{conjecture}
\begin{remark}
	The intuition is that rigidity of $F^*\cup X$ only depends on the 1-thin cover of $F^*$ but not the base $B_{F^*}$.
	Consider a non-proper 1-thin cover where the rigid components come from those of 1-thin cover of $F^*$ and singleton elements of $X$. A proper 1-thin cover can be computed through coarsening.
	For a subset of rigid components $\mathcal P$, let $t=|\bigcup_{P\in \mathcal P} V[P]|$ be the number of vertices. If the number of edges $\sum_{P\in \mathcal P} 2|P|-3$ is at least $2t-3$ then we merge these components into a new one.
	One can see that in this process we do not care the actual base $B_{F^*}$ and only the 1-thin cover matters.
\end{remark}

\begin{conjecture}
	\autoref{conj:dist} is true when $M$ is a 2D rigidity matroid.
\end{conjecture}

Try to minic Thorup's proof for graphic matroids. It follows from \autoref{lem:partition} that spans form a partition on $E-F$.
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.
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 $(n+1)/3$.

\subsection{Hypergraphic matroid cocycle and hypergraph k-cut}

Let $H=(V,E)$ be a hypergraph and let $M=(E,\mathcal I)$ be a hypergraphic matroid on the hyperedge set $E$. A subset $I$ of hyperedges is independent in $M$ if the union of any subset $I'\subseteq I$ has at least $|I'|+1$ vertices.
One can see that hypergraphic matroid is a count matroid induced by $|E[V]|\leq |V|-1$.
The rank of a hypergraphic matroid is given by $\min\set{|V|-|\mathcal P|+e_H(\mathcal P):\text{$\mathcal P$ is a partition of $V$}}$, where $e_H(\mathcal P)$ is the number of inter-component hyperedges.
Hypergraphic matroids are not closed under contraction.\footnote{see Tamás Király's thesis \url{https://tkiraly.web.elte.hu/pub/tkiraly_thesis.pdf}}

Let $\mathcal P=\set{V_1,\ldots,V_k}$ be a non-empty $k$-partition of $V$.
Then the $k$-cut of a hypergraph is the set of hyperedges intersecting at least 2 parts of $\mathcal P$.
We say $H$ is partition connected if $e_H(\mathcal P)\geq |\mathcal P|-1$ for any partition $\mathcal P$.
It follows from the rank function that $\rank(M(H))=|V|-1$ iff $H$ is partition-connected.
Given a hypergraph with $<k$ components, we can add at most $O(|V|)$ dummy hyperedges with zero cost to make it partition connected. We always assume the input hypergraph is partition connected since adding zero-cost hyperedges does not affect the $k$-cut cost.

\begin{theorem}\label{thm:kcut}
	Let $H$ be a partition-connected hypergraph and let $M$ be the hypergraphic matroid on $H$.
	The minimum $k$-cut of $H$ is the same as the minimum $(k-1)$-cocycle of $M$.
\end{theorem}
\begin{proof}
	First consider any $k$-cut $\delta(\mathcal P)$ induced by partition $\mathcal P=\set{V_1,\ldots,V_k}$. Let $X=E\setminus \delta(\mathcal P)$. The rank of $X$ is
	\[
	    \rank(X)= \min_{\mathcal P'} \set{|V|-|\mathcal P'|+e_X(\mathcal P')}\leq |V|-|\mathcal P|+e_X(\mathcal P)\leq |V|-k.
	\]
	So $\delta(\mathcal P)$ must contain a $(k-1)$-cocycle.
	
	On the other hand, let $C$ be any $(k-1)$-cocycle of $M$. Let $F$ be the complement of $C$.
	$F$ is a flat of $M$ with rank $r-k+1$.
	Take any maximal independent set $I_F\subset F$. It is known that one can idenfity two vertices as a graph edge in each hyperedge of an independent set, such that the resulting graph is a forest. Let $T=(V[H],E)$ be such a forest of $I_F$. Note that we include isolated vertices.
	So the number of components of $I_F$ is exactly $k$. Let $\mathcal Q$ be the partition of $V[H]$ into components of $T$.
	It follows from the definition that $\delta(\mathcal Q)$ is a $k$-cut.
\end{proof}

\autoref{thm:kcut} reduces minimum $k$-cut problem on hypergraphs to minimum $k$-cocycle on hypergraphic matroids. Now we try to apply the ideal base packing framework on hypergraphic matroids.

\begin{conjecture}
	Let $H$ be a parititon-connected hypergraph and let $M$ be the hypergraphic matroid on $H$.
	There is a distribution $\mathcal D$ on $k$-cocycles such that for any base in the ideal base packing, the expected size of their intersection is at most $\max\set{\gamma,2k}$.
\end{conjecture}
This is not true. Consider a hypergraph with $\gamma\in[h-k-1,h+1]$ and an ideal base that every inter-component hyperedge has size $\gamma$. One can see that $e_B(\mathcal Q)$ has an upperbound $h$.
I think in general one cannot do better than $\gamma k$.

\section{Greedy base packing}

\section{Principal sequence + KT contraction}

Recently, Mohit Daga \cite{Daga_2025} combines the principal sequence of partitions and Kawarabayashi-Thorup contractions to get a sub-$n^k$ deterministic algorithm for $k$-cut in simple unweighted graphs.

Recall that in the ideal tree packing framework \cite{Thorup_2008}, one finds the first partition with $\geq k$ parts in the principal sequence and then merge parts to get a bound. However, the idea in \cite{Daga_2025} is that, instead of finding the first partition with $\geq k$ parts, one finds the last partition $P$ with $<k$ parts and show that the optimal $k$-cut can be expressed as the $E[G/P]$ together with some internal cuts and some singleton isolations inside parts of $P$.

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