<2-approx for general set functions

main.tex

@@ -340,8 +340,61 @@ Note that everything in blue is non-negative.
 And we get that upperbound of $\lambda^*b$ by throwing away all blue terms and using $c(F^*)\leq b$.
 
 Can we show that the gap is 0 or much smaller than 2?
+
+No. There are examples (a 4-vertex path with parallel edges) where the gap is almost $b\lambda^*$.\footnote{see \url{https://gitea.talldoor.uk/sxlxc/edge_conn_interdiction/src/branch/master/gap.py}}
+One cannot do better than $b\lambda^*$.
 \end{remark}
 
+\subsection{general objective function}
+Consider the following optimization problem.
+\begin{problem}\label{prob:generalf}
+Let $S$ be a set of $n$ elements and let $f:2^S\to \Z_+$ be a set function on $S$.
+Given a cost function $c:S\to \Z_+$ and a budget $b\in Z_+$,
+find $\min_{F\subset X\subset S} \set{f(X\setminus F)| c(F)\leq b}$.
+\end{problem}
+
+Now we show that the $<2$-approximation holds for general set function.
+First we need to generalize the definition of Lagrangian dual and $L(\lambda)$ to general set function $f$.
+Let the Lagrangian dual be
+\[
+LD = \max_{\lambda\geq 0} \min_{F\subset X\subset S} f(X\setminus F)+ \lambda(c(F)-b)
+\]
+and define $L(\lambda)=\min_{F\subset X\subset S} f(X\setminus F)+ \lambda c(F)$.
+Let $\lambda^{LD},X^{LD},F^{LD}$ be the optimal solution to $LD$.
+We have shown that $LD$ is the maximum of the lower envelope of some linear functions when $f$ is modular.
+Similarly, for general set function $LD$ is the maximum of a pwl-concave function and we can assume $c(F^{LD})\leq b$.
+
+Let $f_\lambda (X)=\min_{F\subset X} f(X\setminus F)+\lambda c(F)$ be the generalized ``reweighted-mincut''.
+
+\begin{lemma}
+Let $(X^*,F^*)$ be the optimal solution to \autoref{prob:generalf}.
+We have $L(\lambda^{LD})\leq f_{\lambda^{LD}}(X^*)\leq L(\lambda^{LD})+b\lambda^{LD}$.
+\end{lemma}
+\begin{proof}
+For the first inequality, we have
+\begin{equation*}
+\begin{aligned}
+L(\lambda^{LD}) & = \min_{F\subset X\subset S} f(X\setminus F)+ \lambda^{LD} c(F)\\
+                & \leq \min_{F\subset X^*} f(X^*\setminus F)+\lambda^{LD} c(F)\\
+                & = f_{\lambda^{LD}}(X^*).
+\end{aligned}
+\end{equation*}
+
+To see the upperbound:
+\begin{equation*}
+\begin{aligned}
+L(\lambda^{LD})+b\lambda^{LD} &= f(X^{LD}\setminus F^{LD})+ \lambda^{LD} c(F^{LD}) + b\lambda^{LD}\\
+&\geq f(X^*\setminus F^*) + b\lambda^{LD}\\
+&\geq f(X^*\setminus F^*) + c(F^*)\lambda^{LD}\\
+&\geq f_{\lambda^{LD}}(X^*)
+\end{aligned}
+\end{equation*}
+\end{proof}
+
+Now the $<2$-approximation easily follows from the fact that $L(\lambda^{LD})-b\lambda^{LD}=LD > 0$
+
+Note that if $f$ is submodular, then one can solve $f_\lambda(X)$ in polynomial time. However, the framework doesn't seem very useful if one cannot enumerate all $<2$-approximate solutions to $L(\lambda^{LD})$.
+
 \subsection{complexity}
 
 \begin{algo}