diff --git a/main.tex b/main.tex
index 232cf4e..24208bf 100644
--- a/main.tex
+++ b/main.tex
@@ -283,10 +283,10 @@ s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T\quad \text{($x+y$
  
 We write its Lagrangian dual.
 \[
-\max_{\lambda\geq 0} \min_{\text{cut $C$ and $F\subseteq C$}} w(C-F)-\lambda(b-c(F))
+LD = \max_{\lambda\geq 0} \min_{\substack{\text{cut $C$} \\ F\subseteq C}} w(C-F)-\lambda(b-c(F))
 \]
 
-For fixed $\lambda$, we define $L(\lambda)=\min\limits_{\text{cut $C$ and $F\subseteq C$}} w(C)-w(F)+\lambda c(F)$. Note that the Lagrangian dual is $\max_\lambda L(\lambda) - \lambda b$. 
+For fixed $\lambda$, we define $L(\lambda)=\min \set{ w(C)-w(F)+\lambda c(F),\text{cut $C$ and $F\subseteq C$} }$. Note that the Lagrangian dual is $\max_{\lambda\geq 0} L(\lambda) - \lambda b$. 
 
 If the optimal $C^*$ of $L(\lambda)$ is known, any edge in $C^*$ with $w(e)\geq c(e)\lambda$ will be added to $F^*$. So now the hard part is to find $C^*$. We reweight edges.
 
@@ -300,7 +300,7 @@ w_\lambda(e)=\begin{cases}
 Then clearly edges in $F^*$ are heavy and edges in $C^*-F^*$ are light.
 
 Now consider the min cut under capacity $w_\lambda$. For any cut $C$ in $G$, its capacity $w_\lambda(C)$ can be write as the sum of heavy elements ($\lambda c(F)$) and light elements ($w(C-F)$) which is at least $L(\lambda)$. 
-On the other hand, $C^*$ is a feasible cut and the capacity of $C^*$ is exactly $L(\lambda)$. Thus $C^*$ is a mincut in the reweighted graph and we can easily find it.
+On the other hand, $C^*$ is a feasible cut and the capacity of $C^*$ is exactly $L(\lambda)$. Thus $C^*$ is a mincut in the reweighted graph and we can find it in near-linear time.
 
 \subsection{Approximation}
 
@@ -326,6 +326,22 @@ L(\lambda^*)+b\lambda^* &\geq \opt(IP)+\lambda^* c(F^*)\\ &=w(C^*-F^*)+\lambda^*
 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 $L(\lambda^*)-\lambda^* b = LD > 0$.
 
+\begin{remark}
+Let $F'\subset C^*$ be the set of light elements in $C^*$ under weights $w_{\lambda^*}$.
+The real gap between $w_\lambda^*(C^*)$ and $L(\lambda^*)$ is
+\[
+\begin{aligned}
+    &\lambda^*(c(F^*)-\textcolor{blue}{c(F^{LD})})\\
+-   &\textcolor{blue}{[w(C^{LD}\setminus F^{LD})-w(C^*\setminus F^*)]}\\
+-   &\textcolor{blue}{[\lambda^*(c(F^*)-c(F'))+w(F')-w(F^*)]}
+\end{aligned}
+\]
+Note that everything in blue is non-negative.
+And we get that upperbound of $\lambda^*b$ by throwing away all blue terms and use $c(F^*)\leq b$.
+
+Can we show that the gap is 0 or much smaller than 2?
+\end{remark}
+
 \subsection{complexity}
 
 \begin{algo}
@@ -348,8 +364,8 @@ Finding $<2$-approx mincut takes $\tilde O(n^3)$.
 The total complexity is $\tilde O(m^2+n^3 T)$, where $T$ is the running time of FPTAS for 0-1 knapsack.\footnote{An $1+\e$ approximate solution to knapsack can be found in time $\tilde O(m+\frac{1}{\e^2})$ \cite{10.1145/3618260.3649730}.}
 
 \subsection{Avoid the parametric search}
-The $m^2$ term in the complexity is hard to improve.
-So we consider using binary search to find an $(1+\e)$-approximate $\lambda^*$.
+The $\tilde O(m^2)$ term in the complexity is hard to improve.
+We want to find an $(1+\e)$-approximate $\lambda^*$ in near linear time.
 
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