support size

main.tex

@@ -79,17 +79,17 @@ We define the ideal base packing $y$ by induction. Suppose that the ideal base p
 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 $S=B_{F^*}\cup (B\setminus F^*)$.
+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 $S=B_{F^*}\cup (B\setminus F^*)$.
 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 $e\in E\setminus F^*$.
 We choose an edge $e$ with probability proportional to its capacity $c(e)$.
 \[
 \sum_{e\in E\setminus F^*} \frac{c(e)}{c(E\setminus F^*)}\frac{\sum_{B:e\in B} \Pr[B]}{c(e)}
 \geq \frac{r(E)-r(F^*)}{c(E\setminus F^*)}=\frac{1}{\sigma}
 \]
-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 $E\setminus F^*$.
+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 $E\setminus F^*$.
 
 In graphic matroids it follows easily that $S$ is a spanning tree.
-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 $F^*$.}
+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 $F^*$.}
 $S$ 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 $r(E)-r(F^*)$. 
 Characterization of matroids where $S$ is a base is another interesting problem.
 \note{From now on we assume $S$ is a base. This should holds in all $(k,2k-1)$-sparsity matroids.}
@@ -149,6 +149,9 @@ However, Lemma~7 in \cite{Thorup_2008} does not generalize to all matroids and w
 
 \subsection{Support size}
 
+Recall that we construct the ideal tree packing recursively. Suppose that the ideal base packing for $M|F^*$ is has $n$ bases and let $m$ be the size of support of the optimal base packing of $M$. Then the number of bases in the ideal base packing of $M$ is $nm$. Note that $m$ is upperbounded by the size of the groundset.
+So the support size can be exponential.
+
 \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 $F^*=\argmin_{F\subset E}\frac{c(E\setminus F)}{r(E)-r(F)}$.