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\title{Notes on ideal base packing}
\author{}
\date{}

\DeclareMathOperator*{\opt}{OPT}
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\begin{document}
% \maketitle

\section{Ideal base packing}
Try to generalize Thorup's ideal tree packing \cite{Thorup2008} 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{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 $B_{F^*}$.
\end{remark}

\begin{comment}     % principal sequence of partition
\section{Principal sequence of partition on graphs}
For a graph $G=(V,E)$ with edge capacity $c:V\to \Z_+$, the strength $\sigma(G)$ is defined as $\sigma(G)=\min_{\Pi}\frac{c(\delta(\Pi))}{|\Pi|-1}$, where $\Pi$ is any partition of $V$, $|\Pi|$ is the number of parts in the partition and $\delta(\Pi)$ is the set of edges between parts. Note that an alternative formulation of strength (using graphic matroid rank function) is $\sigma(G)=\min_{F\subset E} \frac{c(E-F)}{r(E)-r(F)}$, which in general is the fractional optimum of matroid base packing.

The principal sequence of partitions of $G$ is a pwl concave curve $L(\lambda)= \min_\Pi c(\delta(\Pi))-\lambda |\Pi|$. (alternatively, $L(\lambda)=\min_{F\in E}c(E\setminus F)-\lambda(r(E)-r(F)+1)$) Cunningham used principal partition to computed graph strength\cite{cunningham_optimal_1985}. There is a list of good properties mentioned in \cite[Section 6]{chekuri_lp_2020}(implicated stated in \cite{cunningham_optimal_1985}).
\begin{enumerate}
\item We can assume $G$ is connected and deal with the smallest strength component. One can see this by fractional base packing on the direct sum of matroids. Note that on disconnected graphs we should use the edge set definition instead of partitions.
\item $L(\lambda)$ is piecewise linear concave since it is the lower envelope of some line arrangement.
\item For each line segment on $L(\lambda)$ there is a corresponding partition $\Pi$. If $\lambda^*$ is a breakpoint on $L(\lambda)$, then there are two optimal solution (say partitions $P_1$ and $P_2$, assume $|P_1|\leq|P_2|$) to $\min_\Pi c(\delta(\Pi))-\lambda^* |\Pi|$.  Then $P_2$ is a refinement of $P_1$.
    \begin{proof}[sketch]
        Suppose that $P_2$ is not a refinement of $P_1$. We claim that the meet of $P_1$ and $P_2$ achieves a objective value at least no larger than $P_1$ or $P_2$ does. The correspondence between graphic matroid rank function and partitions of $V$ gives us a reformulation $L(\lambda^*)=\min_{F\subset E}c(E-F)-\lambda^*(r(E)-r(F)+1)$. Here $F$ is the set of edges in each part of $\Pi$. Let $g(F)=c(E-F)+\lambda^*r(F)-\lambda^* n$. Then the claim is equivalent to the fact that for two optimal solutions $F_1,F_2$ to $L(\lambda^*)$, $g(F_1\cap F_2)\leq g(F_1)=g(F_2)\leq g(F_1\cup F_2)$, which can be seen by the submodularity of $g$ and the optimality of $F_1,F_2$.
    \end{proof}
The number of breakpoints on $L(\lambda)$ is at most $n-1$.
\item Let $\lambda^*$ be a breakpoint on $L(\lambda)$ induced by edge set $F$. The next breakpoint is induced by the edge set $F'\subset F$ and $F'$ is the solution to strength problem on the smallest strength component of $F$. $\lambda^*$ is the strength of the smallest strength component in $F$. These claims can be seen by the following arguments. From the previous bullet we have $\min_{\Delta F} c(E-F+\Delta F)-\lambda^*(r(E)-r(F-\Delta F)+1)=L(\lambda^*)$. Consider the largest $\lambda^*$ which allows $\Delta F=\emptyset$ to be an optimal solution. Such $\lambda^*$ would be the next breakpoint. For any $\Delta F$, $c(E-F+\Delta F)-\lambda^*(r(E)-r(F-\Delta F)+1)\geq c(E-F)-\lambda^*(r(E)-r(F)+1)$. Thus we have $\lambda^*\leq \frac{c(\Delta F)}{r(F)-r(F-\Delta F)}$.
\item Consider $\lambda\in [0,\e]$ for some small enough $\e$. The Lagrangian dual $\min_F c(E\setminus F)-\lambda (r(E)-r(F)+1)$ gets the optimum at $F=E$. 
That is $c(E\setminus F')-\lambda(r(E)-r(F')+1)>-\lambda$ for all $F'\subsetneq E$.
We are interested in the upperbound $\e$ of $\lambda$ such that the optimal $F$ is a proper subset of $E$ when $\lambda >\e$. Therefore, the upperbound is $\e=\min_{F\subsetneq E}\frac{c(E\setminus F)}{r(E)-r(F)}$, which is exactly the strength.
\end{enumerate}
\end{comment}

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