update proof for graphic matroid
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@@ -15,7 +15,7 @@
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\maketitle
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\section{Ideal base packing}
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Try to generalize Thorup's ideal tree packing \cite{Thorup_2008} to matroids.
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Try to generalize Thorup's ideal tree packing \cite{Thorup_2008} to matroids.
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We cannot expect it to work on all matroids.
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The goal is to figure out some sufficient conditions and their relations with basepacking($\lambda\leq c \sigma$) and random contraction($\lambda \leq c \frac{|E|}{r(E)}$).
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@@ -26,114 +26,114 @@ One can select all bases and see that for every $k$-cocycle there is a base that
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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.
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\begin{algo}
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\textsc{\underline{Ideal Utilization}}($M$)\\
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If the groundset $E$ is empty, stop\\
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Let $F^*\in \argmin_{F\subset E} \frac{c(E-F)}{r(E)-r(F)}$ and let $\sigma$ be the strength\\
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for $e\in E-F^*$:\\
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\quad $u^*(e)=1/\sigma$\\
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\textsc{Ideal Utilization}($M|F^*$)
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\textsc{\underline{Ideal Utilization}}($M$)\\
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If the groundset $E$ is empty, stop\\
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Let $F^*\in \argmin_{F\subset E} \frac{c(E-F)}{r(E)-r(F)}$ and let $\sigma$ be the strength\\
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for $e\in E-F^*$:\\
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\quad $u^*(e)=1/\sigma$\\
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\textsc{Ideal Utilization}($M|F^*$)
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\end{algo}
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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 $(M/A)\setminus (E-B)$.
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\begin{lemma}
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Consider the ideal utilizations $u^*(e)$ assigned by the above algorithm.
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\begin{enumerate}
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\item $\sigma(M)\leq \sigma(M|F^*)$
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\item $u^*(e)$ is unique even though the $F^*$ may not be unique.
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\item There is a fractional base packing $y$ such that $\sum_{B:e\in B}y(B)=u^*(e)$.
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\item Each base in the above base packing is a minimum base with respect to the ideal utilizations $u^*(e)$.
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\end{enumerate}
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Consider the ideal utilizations $u^*(e)$ assigned by the above algorithm.
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\begin{enumerate}
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\item $\sigma(M)\leq \sigma(M|F^*)$
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\item $u^*(e)$ is unique even though the $F^*$ may not be unique.
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\item There is a fractional base packing $y$ such that $\sum_{B:e\in B}y(B)=u^*(e)$.
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\item Each base in the above base packing is a minimum base with respect to the ideal utilizations $u^*(e)$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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The proofs are similar to those in \cite{thorup_fully-dynamic_2007}.
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\begin{enumerate}
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\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
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\[
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\sigma(M|F^*)=\frac{c(F^*-F')}{r(F^*)-r(F')}
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=\frac{c(E-F')-c(E-F^*)}{r(E)-r(F')-(r(E)-r(F^*))}
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\geq \frac{c(E-F^*)}{r(E)-r(F^*)},
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\]
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the last inequality follows from $\frac{c(E-F')}{r(E)-r(F')}\geq \frac{c(E-F^*)}{r(E)-r(F^*)}$.
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The proofs are similar to those in \cite{thorup_fully-dynamic_2007}.
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\begin{enumerate}
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\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
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\[
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\sigma(M|F^*)=\frac{c(F^*-F')}{r(F^*)-r(F')}
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=\frac{c(E-F')-c(E-F^*)}{r(E)-r(F')-(r(E)-r(F^*))}
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\geq \frac{c(E-F^*)}{r(E)-r(F^*)},
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\]
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the last inequality follows from $\frac{c(E-F')}{r(E)-r(F')}\geq \frac{c(E-F^*)}{r(E)-r(F^*)}$.
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\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
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\[
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\frac{c(E-F)}{r(E)-r(F)}\leq
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\frac{c(E)-c(F_1)-c(F_2)}{r(E)-r(F_1)-r(F_2)}
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\leq \sigma
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\]
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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 $1/\sigma$.
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\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
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\[
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\frac{c(E-F)}{r(E)-r(F)}\leq
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\frac{c(E)-c(F_1)-c(F_2)}{r(E)-r(F_1)-r(F_2)}
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\leq \sigma
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\]
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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 $1/\sigma$.
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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,
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\[
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\begin{aligned}
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\sigma &< \frac{c(F_1)-c(F_1\cap F_2)}{r(F_1)-r(F_1\cap F_2)} & & \\
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&\leq \frac{c(F_1\cup F_2)-c(F_2)}{r(F_1\cup F_2)-r(F_2)} & &\text{submodularity of $r$}\\
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&=\frac{c(E)-c(F_2)-(c(E)-c(F_1\cup F_2))}{r(E)-r(F_2)-(r(E)-r(F_1\cup F_2))}
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\end{aligned}
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\]
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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 $M|F_1$.
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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,
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\[
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\begin{aligned}
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\sigma & < \frac{c(F_1)-c(F_1\cap F_2)}{r(F_1)-r(F_1\cap F_2)} & & \\
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& \leq \frac{c(F_1\cup F_2)-c(F_2)}{r(F_1\cup F_2)-r(F_2)} & & \text{submodularity of $r$} \\
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& =\frac{c(E)-c(F_2)-(c(E)-c(F_1\cup F_2))}{r(E)-r(F_2)-(r(E)-r(F_1\cup F_2))}
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\end{aligned}
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\]
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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 $M|F_1$.
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\item
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% I don't think this is true on general matroids.
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We define the ideal base packing $y$ by induction. Suppose that the ideal base packing $y'$ on $M|F^*$ is known.
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Let $y^*$ be the optimal fractional base packing of $M$ with capacity $c$.
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(Following Thorup's notation, $y^*(B)$ is a probability on the set of bases and ``optimal'' means that $1/\sigma = \max_e \frac{\sum_{B:e\in B}y^*(B)}{c(e)}$.)
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\item
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% I don't think this is true on general matroids.
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We define the ideal base packing $y$ by induction. Suppose that the ideal base packing $y'$ on $M|F^*$ is known.
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Let $y^*$ be the optimal fractional base packing of $M$ with capacity $c$.
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(Following Thorup's notation, $y^*(B)$ is a probability on the set of bases and ``optimal'' means that $1/\sigma = \max_e \frac{\sum_{B:e\in B}y^*(B)}{c(e)}$.)
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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 $S=B_{F^*}\cup (B\setminus F^*)$.
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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 $e\in E\setminus F^*$.
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We choose an edge $e$ with probability proportional to its capacity $c(e)$.
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\[
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\sum_{e\in E\setminus F^*} \frac{c(e)}{c(E\setminus F^*)}\frac{\sum_{B:e\in B} \Pr[B]}{c(e)}
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\geq \frac{r(E)-r(F^*)}{c(E\setminus F^*)}=\frac{1}{\sigma}
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\]
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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 $E\setminus F^*$.
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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 $S=B_{F^*}\cup (B\setminus F^*)$.
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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 $e\in E\setminus F^*$.
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We choose an edge $e$ with probability proportional to its capacity $c(e)$.
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\[
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\sum_{e\in E\setminus F^*} \frac{c(e)}{c(E\setminus F^*)}\frac{\sum_{B:e\in B} \Pr[B]}{c(e)}
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\geq \frac{r(E)-r(F^*)}{c(E\setminus F^*)}=\frac{1}{\sigma}
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\]
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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 $E\setminus F^*$.
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In graphic matroids it follows easily that $S$ is a spanning tree.
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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 $F^*$.}
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$S$ 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 $r(E)-r(F^*)$.
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Characterization of matroids where $S$ is a base is another interesting problem.
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\note{From now on we assume $S$ is a base. This should holds in all $(k,2k-1)$-sparsity matroids.}
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Then the lemma follows by induction.
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In graphic matroids it follows easily that $S$ is a spanning tree.
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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 $F^*$.}
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$S$ 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 $r(E)-r(F^*)$.
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Characterization of matroids where $S$ is a base is another interesting problem.
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\note{From now on we assume $S$ is a base. This should holds in all $(k,2k-1)$-sparsity matroids.}
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Then the lemma follows by induction.
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\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.
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Now suppose the claim holds for $M|F^*$.
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In the previous bullet point we have already shown that every element in $E\setminus F^*$ have the same utilization.
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Note that we also have shown in the first bullet point that the ideal utilization is larger in $M$ than in $M|F^*$.
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Thus, the construction conincides with the greedy algorithm for minimum matroid base.
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\end{enumerate}
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\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.
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Now suppose the claim holds for $M|F^*$.
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In the previous bullet point we have already shown that every element in $E\setminus F^*$ have the same utilization.
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Note that we also have shown in the first bullet point that the ideal utilization is larger in $M$ than in $M|F^*$.
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Thus, the construction conincides with the greedy algorithm for minimum matroid base.
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\end{enumerate}
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\end{proof}
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The following lemma does not seem related to the ideal tree packing.
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\begin{lemma}
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If we decrease the capacity of an edge, no ideal edge utilization decreases.
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If we decrease the capacity of an edge, no ideal edge utilization decreases.
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\end{lemma}
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\begin{proof}
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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 $u(f)$.
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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 $1/u(f)$.
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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 $u(f)$.
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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 $1/u(f)$.
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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 $A'\subseteq A$.
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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 $A'\subseteq A$.
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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
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\[
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\frac{1}{u(f)}=\frac{c(C)}{r(A)-r(A\setminus C)}
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\geq \frac{c(C_1)+c(C_2)}{(r(A)-r(A\setminus C_1))+(r(A')-r(A'\setminus C_2))}.
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\]
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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
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\[
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\frac{1}{u'(f)}\leq\frac{c'(A'-C_2)}{r(A')-r(A'\setminus C_2)}
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\leq \frac{c(C_2)}{r(A')-r(A'\setminus C_2)}\leq \frac{1}{u(f)}.
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\]
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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
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\[
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\frac{1}{u(f)}=\frac{c(C)}{r(A)-r(A\setminus C)}
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\geq \frac{c(C_1)+c(C_2)}{(r(A)-r(A\setminus C_1))+(r(A')-r(A'\setminus C_2))}.
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\]
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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
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\[
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\frac{1}{u'(f)}\leq\frac{c'(A'-C_2)}{r(A')-r(A'\setminus C_2)}
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\leq \frac{c(C_2)}{r(A')-r(A'\setminus C_2)}\leq \frac{1}{u(f)}.
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\]
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\end{proof}
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\begin{remark}
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If we increase the capacity, no ideal edge utilization increases. The proof is similar.
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Removing (contracting) edges has the same effect on ideal utilization as setting the capacity to $0$ ($\infty$).
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If we increase the capacity, no ideal edge utilization increases. The proof is similar.
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Removing (contracting) edges has the same effect on ideal utilization as setting the capacity to $0$ ($\infty$).
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\end{remark}
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\subsection{Counting}
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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.
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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.
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The expectation is exactly $\sum_{e\in D} c(e)u^*(e)$.
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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.
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@@ -150,44 +150,49 @@ Let $F$ be the optimal flat for strength and assume $k=r(E)-r(F)>k'$. We want to
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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.
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Then we do $k-k'$ random contractions in $M/F$ to get a random $k'$-cocycle $C_{k'}$.
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\begin{lemma}[Lemma~7 in \cite{Thorup_2008}, restated]
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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 $2k'$.
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\begin{lemma}[Lemma~7 in \cite{Thorup_2008}, restated]\label{lem:idealload}
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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 $2k'$.
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\end{lemma}
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\begin{proof}
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We can assume the matroid is connected. (Otherwise we can remove loops and coloops and add dummy elements.)
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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 $G[F]$.
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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
|
||||
\[
|
||||
\E_{F'}[|T\setminus F'|]\leq 2k \frac{k'}{k+1}\leq 2k'.
|
||||
\]
|
||||
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 $k'$.
|
||||
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}
|
||||
|
||||
A counterexample would be uniform matroids $U_{2n,n}$.
|
||||
The size of every $k'$-cocycle is $k'+n$, and for any $k'$-cocycle there are bases using $O(n)$ elements in the cocycle.
|
||||
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 $E\setminus F$.
|
||||
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.
|
||||
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}
|
||||
@@ -202,18 +207,18 @@ 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 $M$.
|
||||
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.
|
||||
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.
|
||||
\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 $E-F$.
|
||||
|
||||
Reference in New Issue
Block a user