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fix typo

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Yu Cong
2025-11-26 16:55:44 +08:00
parent 029ca788de
commit 5ef046ea74
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notes.tex
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@@ -160,14 +160,14 @@ Now we prove \autoref{contraction_gap}.
\begin{proof} \begin{proof}
We follow the same framework as the graphic case. Let $c$ be the 2-hitting set gap and let $c'$ be the cogirth-packing gap of $M$. We have We follow the same framework as the graphic case. Let $c$ be the 2-hitting set gap and let $c'$ be the cogirth-packing gap of $M$. We have
\begin{equation}\label{eq1} \begin{equation}\label{eq1}
\lambda(M'/t)\leq \lambda_2(G)\leq c\sigma_2(G)\leq c(|J|+\sigma(M'/T)|_{E\setminus J}) \lambda(M'/\tau)\leq \lambda_2(M)\leq c\sigma_2(M)\leq c(|J|+\sigma(M'/T)|_{E\setminus J})
\end{equation} \end{equation}
where $J$ is the minimum hitting set of $\{C(B,f)\setminus f|\forall B\}$. where $J$ is the minimum hitting set of $\{C(B,f)\setminus f|\forall B\}$ and $\lambda_2(M)$ ($\sigma_2(M)$) is the integral (fractional) 2-hitting set of bases of $M$.
We then apply \autoref{TcutTjoin}. We then apply \autoref{TcutTjoin}.
If $\lambda(M'/t)=\lambda(M)$, then we have $\lambda(M'/t)= \lambda(M)\leq c \sigma(M)\leq c\sigma(M'/t)$, since the optimal solution to $\sigma(M'/t)$ is feasible to $\sigma(M)$; If $\lambda(M'/\tau)=\lambda(M)$, then we have $\lambda(M'/\tau)= \lambda(M)\leq c \sigma(M)\leq c\sigma(M'/\tau)$, since the optimal solution to $\sigma(M'/\tau)$ is feasible to $\sigma(M)$;
Otherwise, we extend \autoref{eq1} and get $\lambda(M'/t)\leq c(|J|+\sigma(M'/T))=c(\lambda(M)+\sigma(M'/T))\leq c(c'+1)\sigma(M'/T)$. Otherwise, we extend \autoref{eq1} and get $\lambda(M'/\tau)\leq c(|J|+\sigma(M'/\tau))=c(\lambda(M)+\sigma(M'/\tau))\leq c(c'+1)\sigma(M'/\tau)$.
\end{proof} \end{proof}
To finish the proof of constant gap of $M/T$ we note that $(M/T_t)|E(G)=M/T$ and $(M/T_{i+1})|E(G)=[((M/T_{i})|E(G))+e_{i+1}]/e_{i+1}$ and the constant gap for deletion minors can be proven by wrapping the entire proof into an induction on the size of $E(G)$. To finish the proof of constant gap of $M/T$ we note that $(M/T_t)|E(G)=M/T$ and $(M/T_{i+1})|E(G)=[((M/T_{i})|E(G))+e_{i+1}]/e_{i+1}$ and the constant gap for deletion minors can be proven by wrapping the entire proof into an induction on the size of $E(G)$.