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main.tex

@@ -81,16 +81,25 @@ Let $y^*$ be the optimal fractional base packing of $M$ with capacity $c$.
 
 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 $e\in E\setminus F^*$.
+We choose an edge $e$ with probability proportional to its capacity $c(e)$.
 \[
-\sum_{e\in E\setminus F^*} \frac{c(e)}{c(E\setminus F^*)} \left[\sum_{B:e\in B} \frac{\Pr[B]}{c(e)}\right] 
+\sum_{e\in E\setminus F^*} \frac{c(e)}{c(E\setminus F^*)}\frac{\sum_{B:e\in B} \Pr[B]}{c(e)}
 \geq \frac{r(E)-r(F^*)}{c(E\setminus F^*)}=\frac{1}{\sigma}
 \]
 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 $E\setminus F^*$.
 
-In graphic matroids it follows easily that $S$ is a spanning tree. In general matroids $S$ may not be independent.
+In graphic matroids it follows easily that $S$ is a spanning tree.
+However, $S$ may not be independent in general matroids. (can we find an example or prove that $S$ is a base in all matroids?)
+$S$ 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.}
 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.
+
+\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 previous 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}
 
@@ -128,15 +137,15 @@ Ideal base packing is a distribution on some bases. Given a subset $D\subset E$,
 The expectation is exactly $\sum_{e\in D} c(e)u^*(e)$.
 
 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.
-Thorup proves in \textbf{graphic matroids} that there is a fixed distribution of $k$-cocycles such that for any base from the ideal base packing, the expected size of intersection is at most $O(k)$ (Lemma~7 in \cite{Thorup_2008}).
+Thorup proved that in \textbf{graphic matroids} there is a fixed distribution of $k$-cocycles such that for any base from the ideal base packing, the expected size of intersection is at most $O(k)$ (Lemma~7 in \cite{Thorup_2008}).
 Then it follows that if we take a random spanning tree from the ideal tree packing, there is a fixed $(k+1)$-cut $C$ such that the expected size of intersection is at most $O(k)$, which implies $\sum_{e\in C}c(e)u^*(e)\in O(k)$.
 
 How is this fixed $(k+1)$-cut (or $k$-cocycle) related to the minimum $(k+1)$-cut (minimum $k$-cocycle) ?
 The minimum $k$-cocycle has smaller capacity than $C$.
-Note that Thorup uses a greedy way to construct the cocycle $C$. Elements in $C$ always has the largest possible utilization.
+Note that Thorup used a greedy way to construct the cocycle $C$. Elements in $C$ always has the largest possible utilization.
 These facts implies that the minimum $k$-cocycle has a smaller value $\sum c(e)u^*(e)$ than $C$.
 
-However, Lemma~7 in \cite{Thorup_2008} does not generalize to all matroids and we need to dive into the construction of $C$.
+However, Lemma~7 in \cite{Thorup_2008} does not generalize to all matroids and we need to take a close look at the construction of $C$.
 
 \subsection{Support size}