From e6cadf07c24514db8a5daea391a53ac26f2b2726 Mon Sep 17 00:00:00 2001 From: Yu Cong Date: Fri, 21 Nov 2025 17:11:26 +0800 Subject: [PATCH] z --- main.tex | 29 +++++++++++++++++++++++------ ref.bib | 17 +++++++++++++++++ 2 files changed, 40 insertions(+), 6 deletions(-) diff --git a/main.tex b/main.tex index 0159724..34a58fc 100644 --- a/main.tex +++ b/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} diff --git a/ref.bib b/ref.bib index 9e918f4..78b2a40 100644 --- a/ref.bib +++ b/ref.bib @@ -44,3 +44,20 @@ keywords = {k-way cuts, tree packing}, pages = {159--165}, } + +@article{thorup_fully-dynamic_2007, + title = {Fully-{Dynamic} {Min}-{Cut}*}, + volume = {27}, + issn = {0209-9683, 1439-6912}, + url = {http://link.springer.com/10.1007/s00493-007-0045-2}, + doi = {10.1007/s00493-007-0045-2}, + language = {en}, + number = {1}, + urldate = {2023-02-05}, + journal = {Combinatorica}, + author = {Thorup, Mikkel}, + month = feb, + year = {2007}, + pages = {91--127}, + file = {Thorup_2007_Fully-Dynamic Min-Cut.pdf:/Users/congyu/Zotero/storage/Q329VHH3/Thorup_2007_Fully-Dynamic Min-Cut.pdf:application/pdf}, +}