\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. 
Certainly it won't 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($\lambda \leq c \frac{|E|}{r(E)}$).

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 $B_{F^*}$ in the support of $y'$ and a base $B$ in the support of $y^*$ 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 $e\in E\setminus F^*$.
\[
\sum_{e\in E\setminus F^*} \frac{c(e)}{c(E\setminus F^*)} \left[\sum_{B:e\in B} \frac{\Pr[B]}{c(e)}\right] 
\geq \frac{r(E)-r(F^*)}{c(E\setminus F^*)}=\frac{1}{\sigma}
\]
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 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. In general matroids $S$ may not be independent.
\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 preceeding 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}

\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{Hard part}
Lemma~7 in \cite{Thorup_2008} does not generalize to all matroids.

Let $F^*$ be the optimal flat.
Choose an random element $f\in E-F^*$ and construct a new flat $F=\cl(F^*+f)$ and repeat this process until $r(F)=r(E)-k$. Let $X$ be a independent set with rank $r(E)-r(F^*)$ inside the cocycle $E-F^*$.
We generate another flat $F\supset F^*$ in the following way. 
Initially we set $F=F^*$.
Randomly choose an edge $f$ in $E-F$, update $F$ to $\cl(F+f)$. Repeat this operation $t$ times for a fixed $t< r(E)-r(F^*)$.
We want to upperbound the expected size of $X\setminus F$ using $c(r(E)-r(F))$.
\note{This seems stronger than constant gap...}

In graphs this seems like a random contraction on $G/\mathcal F^*$. However, instead of the probability certain mincut is preserved, we are interested the expected number of remaining edges of a spanning tree.

\subsection{Rigidity matroids}
\begin{conjecture}\label{conj:idealrigidbase}
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}
Assume \autoref{conj:idealrigidbase} is true.
???
\end{conjecture}

\section{Greedy base packing}


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