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main.tex

@@ -208,7 +208,7 @@ The principal sequence of partitions of $G$ is a piecewise linear concave curve
 \item $L(\lambda)$ is piecewise linear concave since it is the lower envelope of some line arrangement.
 \item For each line segment on $L(\lambda)$ there is a corresponding partition $\Pi$. If $\lambda^*$ is a breakpoint on $L(\lambda)$, then there are two optimal solution (say partitions $P_1$ and $P_2$, assume $|P_1|\leq|P_2|$) to $\min_\Pi c(\delta(\Pi))-\lambda^* |\Pi|$.  Then $P_2$ is a refinement of $P_1$.
     \begin{proof}[sketch]
-        Suppose that $P_2$ is not a refinement of $P_1$. We claim that the meet of $P_1$ and $P_2$ achieves a objective value at least no larger than $P_1$ or $P_2$ does. The correspondence between graphic matroid rank function and partitions of $V$ gives us a reformulation $L(\lambda^*)=\min_{F\subset E}c(E-F)-\lambda^*(r(E)-r(F)+1)$. Here $F$ is the set of edges in each part of $\Pi$. Let $g(F)=c(E-F)+\lambda^*r(F)-\lambda^* n$. Then the claim is equivalent to the fact that for two optimal solutions $F_1,F_2$ to $L(\lambda^*)$, $g(F_1\cap F_2)\leq g(F_1)=g(F_2)\leq g(F_1\cup F_2)$, which can be seen by the submodularity of $g$.
+        Suppose that $P_2$ is not a refinement of $P_1$. We claim that the meet of $P_1$ and $P_2$ achieves a objective value at least no larger than $P_1$ or $P_2$ does. The correspondence between graphic matroid rank function and partitions of $V$ gives us a reformulation $L(\lambda^*)=\min_{F\subset E}c(E-F)-\lambda^*(r(E)-r(F)+1)$. Here $F$ is the set of edges in each part of $\Pi$. Let $g(F)=c(E-F)+\lambda^*r(F)-\lambda^* n$. Then the claim is equivalent to the fact that for two optimal solutions $F_1,F_2$ to $L(\lambda^*)$, $g(F_1\cap F_2)\leq g(F_1)=g(F_2)\leq g(F_1\cup F_2)$, which can be seen by the submodularity of $g$ and the optimality of $F_1,F_2$.
     \end{proof}
 The number of breakpoints on $L(\lambda)$ is at most $n-1$.
 \item Let $\lambda^*$ be a breakpoint on $L(\lambda)$ induced by edge set $F$. The next breakpoint is induced by the edge set $F'\subset F$ and $F'$ is the solution to strength problem on the smallest strength component of $F$. $\lambda^*$ is the strength of the smallest strength component in $F$. These claims can be seen by the following arguments. From the previous bullet we have $\min_{\Delta F} c(E-F+\Delta F)-\lambda^*(r(E)-r(F-\Delta F)+1)=L(\lambda^*)$. Consider the largest $\lambda^*$ which allows $\Delta F=\emptyset$ to be an optimal solution. Such $\lambda^*$ would be the next breakpoint. For any $\Delta F$, $c(E-F+\Delta F)-\lambda^*(r(E)-r(F-\Delta F)+1)\geq c(E-F)-\lambda^*(r(E)-r(F)+1)$. Thus we have $\lambda^*\leq \frac{c(\Delta F)}{r(F)-r(F-\Delta F)}$.