fix bug

main.tex

@@ -330,20 +330,20 @@ How is $C^*$ related to the optimal solution to IP?
 One can see that the Lagrangian dual (denoted by LD) is at most the optimum of IP. So we have $\opt(LD) \leq \opt(IP)$.
 
 We can assume WLOG that the optimal $\lambda^*$ in LD is the intersection of two lines with positive and negative slopes. Then there exists an optimal solution $(\lambda^*, C^{LD},F^{LD})$ to LD such that $c(F^{LD})\leq b$. Then we have 
-\[
-\opt(LD)+\lambda^* b -\lambda c(F^{LD}) = w(C^{LD}-F^{LD}) \geq \opt(IP),
-\]
+\begin{equation}\label{eq:ub}
+L(\lambda^*)\geq \opt(LD)+\lambda^* b -\lambda c(F^{LD}) = w(C^{LD}-F^{LD}) \geq \opt(IP)=w(C^*-F^*),    
+\end{equation}
 since $\opt(IP)$ is the smallest $b$-free min cut.
-Thus we can approximate the optimum of IP within an additive error if $\lambda^*$ is known.
 
-We can also recover IPCO alg's 2-approximation.
-\begin{itemize}
-\item $L(\lambda^*)=\opt(LD)+\lambda^* b\geq \opt(IP)$
-\item $L(\lambda^*)-\lambda^* b=LD \leq IP$ and 
-\item $L(\lambda^*)\geq \lambda^* b$ (since $LD \geq 0$)
-\end{itemize}
+We have $L(\lambda^*)\leq w_{\lambda^*}*(C^*)$ since $L(\lambda^*)$ is the value of the minimum cut in $(G,w_{\lambda^*})$. Now we prove $L(\lambda^*)+b\lambda \geq w_{\lambda^*}*(C^*)$.
 
-Then $\opt(IP)\in [L(\lambda^*),2L(\lambda^*)]$.
+\begin{equation*}
+\begin{aligned}
+L(\lambda^*)+b\lambda^* &\geq \opt(IP)+\lambda^* c(F^*)\\ &=w(C^*-F^*)+\lambda^* c(F^*)\\ &\geq w_{\lambda^*}(C^*) 
+\end{aligned}
+\end{equation*}
+The first line uses \autoref{eq:ub} and the fact that $c(F^*)\leq b$. The last line follows from the definition of $w_{\lambda}$.
+Finally the approximation guarantee $w_{\lambda^*}(C^*)\in [L(\lambda^*),2L(\lambda^*))$ easily follows since $L(\lambda^*)-\lambda^* b = LD > 0$.
 
 \subsection{complexity}
 
@@ -359,7 +359,7 @@ return the optimal $(C,F)$
 \paragraph{time for $\lambda^*$} $L(\lambda)-b\lambda$ is pwl concave. The number of segments is at most the number of lines which has a trivial upperbound of $2^m 2^m$. We need almost linear time to find the solution to a fixed $\lambda$. So parametric seach gives complexity $m^{1+o(1)} O(\log 4^m)$.
 
 \paragraph{time for the rest parts} Reweighting takes linear time.
-Finding 2-approx mincut takes $\tilde O(n^4)$. FPTAS for knapsack takes $O(\frac{1}{\e}m^2)$. The total complexity is $O(\frac{1}{\e}m^2n^4)$.
+Finding $<2$-approx mincut takes $\tilde O(n^3)$. FPTAS for knapsack takes $O(\frac{1}{\e}m^2)$. The total complexity is $O(\frac{1}{\e}m^2n^3)$.
 
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