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+++ b/main.tex
@@ -21,9 +21,10 @@ Always remember that $\mathcal F$ is usually not explicitly given.
 
 \begin{problem}[Normalized knapsack]\label{nknap}
     Given the same input as \autoref{bfreeknap}, find $\min \limits_{X
-    \in \mathcal F, F\subset E} \frac{w(X\setminus F)}{b+1-c(F)}$ such that 
+    \in \mathcal F, F\subset E} \frac{w(X\setminus F)}{B-c(F)}$ such that 
     $c(F)\leq b$.
 \end{problem}
+In \cite{vygen_fptas_2024} the normalized min-cut problem use $B=b+1$. Here we use any integer $B>b$ and see how their method works.
 
 Denote by $\tau$ the optimum of \autoref{nknap}. Define a new weight 
 $w_\tau:\E\to \R$,
@@ -39,12 +40,14 @@ w_\tau(e)=\begin{cases}
     Let $(X^N,F^N)$ be the optimal solution to \autoref{nknap}. 
     Every element in $F^N$ is heavy.
 \end{lemma}
-proof is exactly the same as lemma 1 in \cite{vygen_fptas_2024}.
+The proof is exactly the same as \cite[Lemma 1]{vygen_fptas_2024}.
+
+The following two lemmas show (a general version of) that the optimal cut $C^N$ to normalized min-cut is exactly the minimum cut under weights $w_\tau$.
 
 \begin{lemma}\label{lem:lb}
-    For any $X\in \mathcal F$, $w_\tau(X)\ge \tau(1+b)$.
+    For any $X\in \mathcal F$, $w_\tau(X)\ge \tau B$.
 \end{lemma}
-proof is the same.
+
 
 \begin{lemma}
     % $w_\tau (X^N)\le \tau(b+1)$.
@@ -53,23 +56,32 @@ proof is the same.
 \begin{proof}
     \begin{align*}
         w_\tau (X^N) & \le w(X^N\setminus F^N) + w_\tau(F^N)\\
-                    & = \tau \cdot(b+1-c(F^N)) + \tau\cdot c(F^N)\\
-                    & = \tau(b+1)
+                    & = \tau \cdot(B-c(F^N)) + \tau\cdot c(F^N)\\
+                    & = \tau B
     \end{align*}
     Thus by \autoref{lem:lb}, $X^N$ gets the minimum.
 \end{proof}
 
-Let $(X^*,F^*)$ be the optimal solution to \autoref{bfreeknap}.
+Now we show the counter part of \cite[Theorem 5]{vygen_fptas_2024}, which states the optimal solution to \autoref{bfreeknap} is a $\alpha$-approximate solution to $\min_{F\in \mathcal{F}} w_\tau(F)$. 
 
-\begin{lemma}
+\begin{lemma}[Lemma 4 in \cite{vygen_fptas_2024}]\label{lem:conditionalLB}
+    Let $(X^*,F^*)$ be the optimal solution to \autoref{bfreeknap}.
     $X^*$ is either an $\alpha$-approximate solution to $\min_{X\in\mathcal F} 
     w_\tau(X)$ for some $\alpha>1$, or $w(X^*\setminus F^*)\geq 
-    \tau(\alpha+(\alpha-1)b)$.
+    \tau(\alpha B-b)$.
 \end{lemma}
-The proof is the same.
 
-In fact, corollary 1 and theorem 5 are also the same as those in 
-\cite{vygen_fptas_2024}. Finally we get a knapsack version of Theorem 4:
+% In fact, corollary 1 and theorem 5 are also the same as those in 
+% \cite{vygen_fptas_2024}. 
+Then following the argument of Corollary 1 in \cite{vygen_fptas_2024}, assume that $X^*$ is not an $\alpha$-approximate solution to $\min_{X\in\mathcal F} 
+    w_\tau(X)$ for some $\alpha>1$. We have
+\[
+\frac{w(C^N\setminus F^N)}{w(C^*\setminus F^*)}\leq \frac{\tau(B-c(F^N))}{\tau(\alpha B-b)}\leq \frac{B}{\alpha B-b},
+\]
+where the second inequality uses \autoref{lem:conditionalLB}.
+One can see that if $\alpha>2$, $\frac{w(C^N\setminus F^N)}{w(C^*\setminus F^*)}\leq \frac{B}{\alpha B-b} <1$ which implies $(C^*,F^*)$ is not optimal. Thus for $\alpha >2$, $X^*$ must be a $2$-approximate solution to $\min_{X\in\mathcal F} w_\tau(X)$.
+
+Finally we get a knapsack version of Theorem 4:
 \begin{theorem}[Theorem 4 in \cite{vygen_fptas_2024}]
     Let $X^{\min}$ be the optimal solution to $\min_{X\in\mathcal F} w_\tau(X)$.
     The optimal set $X^*$ in \autoref{bfreeknap} is a