z

main.tex

@@ -12,10 +12,12 @@
 \DeclareMathOperator*{\cl}{span}
 
 \begin{document}
-\maketitle
+% \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.
 
@@ -28,23 +30,24 @@ for $e\in E-F^*$:\\
 \textsc{Ideal Utilization}($M|F^*$)
 \end{algo}
 
-Consider this process on the lattice of flats. The set of flats of $M$ forms a geometric lattice. Take two flats $A,B$ that $A\subset B$ in the lattice and consider the sublattice between $A$ and $B$. This sublattice is exactly the lattice of flats of matroid $(M/A)\setminus (E-B)$.
+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 $\sigma(M)\geq \sigma(M|F^*)$
+\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 
 \[
-\frac{c(F^*-F')}{r(F^*)-r(F')}
+\sigma(M|F^*)=\frac{c(F^*-F')}{r(F^*)-r(F')}
 =\frac{c(E-F')-c(E-F^*)}{r(E)-r(F')-(r(E)-r(F^*))}
-\leq \frac{c(E-F^*)}{r(E)-r(F^*)},
+\geq \frac{c(E-F^*)}{r(E)-r(F^*)},
 \]
 the last inequality follows from $\frac{c(E-F')}{r(E)-r(F')}\geq \frac{c(E-F^*)}{r(E)-r(F^*)}$.
 
@@ -54,7 +57,21 @@ the last inequality follows from $\frac{c(E-F')}{r(E)-r(F')}\geq \frac{c(E-F^*)}
 \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 fractional base packing $y$ recursively. Suppose that we have the desired fraction packing $y'$ on $M|F^*$...
 \end{enumerate}
 \end{proof}