weight truncation!

main.tex

@@ -1,5 +1,25 @@
 \documentclass[12pt]{article}
 \usepackage{chao}
+\usepackage[breakable, theorems, skins]{tcolorbox}
+\tcbset{enhanced}
+\DeclareRobustCommand{\note}[2][blue]{%
+\begin{tcolorbox}[
+        breakable,
+        left=0pt,
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+        width=\dimexpr\textwidth\relax, 
+        enlarge left by=0mm,
+        boxsep=5pt,
+        arc=0pt,outer arc=0pt,
+        ]
+        #2
+\end{tcolorbox}
+}
+
 \title{connectivity interdiction}
 \author{}
 \date{}
@@ -13,7 +33,7 @@
 \section{``Cut-free'' Proof}
 \begin{problem}[b-free knapsack]\label{bfreeknap}
     Consider a set of elements $E$ and two weights $w:E\to \Z_+$ and 
-    $c:E\to Z_+$ and a budget $b\in \Z_+$. Given a feasible set $\mathcal 
+    $c:E\to \Z_+$ and a budget $b\in \Z_+$. Given a feasible set $\mathcal 
     F\subset 2^E$, find $\min_{X
     \in \mathcal F, F\subset E} w(X\setminus F)$ such that $c(F)\leq b$.
 \end{problem}
@@ -131,7 +151,7 @@ How to derive normalized min cut for connectivity interdiction?
 \begin{aligned}
 \max&   &                   z&          &   & \\
 s.t.&   &   \sum_{e} y_e c(e) &\leq B   &   &\text{(budget for $F$)}\\
-&       &   \sum_{e\in T} x_e&\geq 1   &   &\forall T\quad \text{($x$ forms a cut)}\\
+&       &   \sum_{e\in T} x_e&\geq 1   &   &\forall T\quad \text{($x$ is a cut)}\\
 &       &   \sum_{e} \min(0,x_e-y_e) w(e)&\geq z   &   &\\
 &       &   y_e,x_e&\in\{0,1\}      &   &\forall e
 \end{aligned}
@@ -143,7 +163,7 @@ we can assume that $y_e\leq x_e$.
 \begin{aligned}
 \min&   &                   \sum_{e} (x_e&-y_e) w(e)          &   & \\
 % s.t.&   &   \sum_{e} (x_e-y_e) w(e)&\geq z   &   &\\
-s.t.&       &   \sum_{e\in T} x_e&\geq 1   &   &\forall T\quad \text{($x$ forms a cut)}\\
+s.t.&       &   \sum_{e\in T} x_e&\geq 1   &   &\forall T\quad \text{($x$ is a cut)}\\
 &       &   \sum_{e} y_e c(e) &\leq B   &   &\text{(budget for $F$)}\\
 &       &   x_e&\geq y_e   &   &\forall e\quad(F\subset C)\\
 &       &   y_e,x_e&\in\{0,1\}      &   &\forall e
@@ -154,15 +174,15 @@ Now this LP looks similar to the normalized min-cut problem.
 
 A further reformulation (the new $x$ is $x-y$) gives us the following,
 
-\begin{equation*}
+\begin{equation}\label{lp:cutinterdict}
 \begin{aligned}
 \min&   &                   \sum_{e} x_e w(e)          &   & \\
-s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T\quad \text{($x$ forms a cut)}\\
+s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T\quad \text{($x+y$ is a cut)}\\
 &       &   \sum_{e} y_e c(e) &\leq B   &   &\text{(budget for $F$)}\\
 % &       &   x_e&\geq y_e   &   &\forall e\quad(F\subset C)\\
 &       &   y_e,x_e&\in\{0,1\}      &   &\forall e
 \end{aligned}
-\end{equation*}
+\end{equation}
 
 Note that now this is almost a positive covering LP. Let $L(\lambda)= \min \{ w(C\setminus F)-\lambda(b-c(F)) | \forall \text{cut $C$}\;\forall F\subset C 
 % \land c(F)\leq b
@@ -183,24 +203,62 @@ We have shown that the budget $B$ in normalized min-cut does not really matter a
 Given a graph $G=(V,E)$ and a cost function $c:V\to \Z_+$, the strength $\sigma(G)$ is defined as $\sigma(G)=\min_{\Pi}\frac{c(\delta(\Pi))}{|\Pi|-1}$, where $\Pi$ is any partition of $V$, $|\Pi|$ is the number of parts in the partition and $\delta(\Pi)$ is the set of edges between parts. Note that an alternative formulation of strength (using graphic matroid rank) is $\sigma(G)=\min_{F\subset E} \frac{|E-F|}{r(E)-r(F)}$, which in general is the fractional optimum of matroid base packing.
 
 The principal sequence of partitions of $G$ is a piecewise linear concave curve $L(\lambda)=\min_\Pi c(\delta(\Pi))-\lambda |\Pi|$. ($L(\lambda)=\min_{F\in E}c(E\setminus F)-\lambda(r(E)-r(F)+1)$) Cunningham used principal partition to computed graph strength\cite{cunningham_optimal_1985}. There is a list of good properties mentioned in \cite[Section 6]{chekuri_lp_2020}(implicated stated in \cite{cunningham_optimal_1985}).
-\begin{itemize}
+\begin{enumerate}
 \item We can assume $G$ is connected and deal with the smallest strength component. One can see this by fractional base packing on the direct sum of matroids. Note that on disconnected graphs we should use the edge set definition instead of partitions.
 \item $L(\lambda)$ is piecewise linear concave since it is the lower envelope of some line arrangement.
 \item For each line segment on $L(\lambda)$ there is a corresponding partition $\Pi$. If $\lambda^*$ is a breakpoint on $L(\lambda)$, then there are two optimal solution (say partitions $P_1$ and $P_2$, assume $|P_1|\leq|P_2|$) to $\min_\Pi c(\delta(\Pi))-\lambda^* |\Pi|$.  Then $P_2$ is a refinement of $P_1$.
     \begin{proof}[sketch]
-        Suppose that $P_2$ is not a refinement of $P_1$. We claim that the meet of $P_1$ and $P_2$ achieves a objective value at least no larger than $P_1$ or $P_2$ does. The correspondence between graphic matroid rank function and partitions of $V$ gives us a reformulation $L(\lambda^*)=\min_{F\subset E}c(E-F)-\lambda^*(r(E)-r(F)+1)$. Here $F$ is the set of edges in each part of $\Pi$. Let $g(F)=c(E-F)+\lambda^*r(F)-\lambda^* n$. Then the claim is equivalent to the fact that for two optimal solutions $F_1,F_2$ to $L(\lambda^*)$, $g(F_1\cap F_2)\leq g(F_1)=g(F_2)\leq g(F_1\cup F_2)$, which can be seen by the submodularity of $g$.
+        Suppose that $P_2$ is not a refinement of $P_1$. We claim that the meet of $P_1$ and $P_2$ achieves a objective value at least no larger than $P_1$ or $P_2$ does. The correspondence between graphic matroid rank function and partitions of $V$ gives us a reformulation $L(\lambda^*)=\min_{F\subset E}c(E-F)-\lambda^*(r(E)-r(F)+1)$. Here $F$ is the set of edges in each part of $\Pi$. Let $g(F)=c(E-F)+\lambda^*r(F)-\lambda^* n$. Then the claim is equivalent to the fact that for two optimal solutions $F_1,F_2$ to $L(\lambda^*)$, $g(F_1\cap F_2)\leq g(F_1)=g(F_2)\leq g(F_1\cup F_2)$, which can be seen by the submodularity of $g$ and the optimality of $F_1,F_2$.
     \end{proof}
 The number of breakpoints on $L(\lambda)$ is at most $n-1$.
-\item Let $\lambda^*$ be a breakpoint on $L(\lambda)$ induced by edge set $F$. The next breakpoint is induced by the edge set $F'\subset F$ and $F'$ is the solution to strength problem on the smallest strength component of $F$. $\lambda^*$ is the strength of the smallest strength component in $F$. These claims can be seen by the following arguments. From the previous bullet we can see that $\min_{\Delta F} c(E-F+\Delta F)-\lambda^*(r(E)-r(F-\Delta F)+1)=L(\lambda^*)$. Consider the largest $\lambda^*$ which allows $\Delta F=\emptyset$ to be an optimal solution. Such $\lambda^*$ would be the next breakpoint. For any $\Delta F$, $c(E-F+\Delta F)-\lambda^*(r(E)-r(F-\Delta F)+1)\geq c(E-F)-\lambda^*(r(E)-r(F)+1)$. Thus we have $\lambda^*\leq \frac{c(\Delta F)}{r(F)-r(F-\Delta F)}$.
-\end{itemize}
+\item Let $\lambda^*$ be a breakpoint on $L(\lambda)$ induced by edge set $F$. The next breakpoint is induced by the edge set $F'\subset F$ and $F'$ is the solution to strength problem on the smallest strength component of $F$. $\lambda^*$ is the strength of the smallest strength component in $F$. These claims can be seen by the following arguments. From the previous bullet we have $\min_{\Delta F} c(E-F+\Delta F)-\lambda^*(r(E)-r(F-\Delta F)+1)=L(\lambda^*)$. Consider the largest $\lambda^*$ which allows $\Delta F=\emptyset$ to be an optimal solution. Such $\lambda^*$ would be the next breakpoint. For any $\Delta F$, $c(E-F+\Delta F)-\lambda^*(r(E)-r(F-\Delta F)+1)\geq c(E-F)-\lambda^*(r(E)-r(F)+1)$. Thus we have $\lambda^*\leq \frac{c(\Delta F)}{r(F)-r(F-\Delta F)}$.
+\item Consider $\lambda\in [0,\e]$ for some small enough $\e$. The Lagrangian dual $\min_F c(E\setminus F)-\lambda (r(E)-r(F)+1)$ gets the optimum at $F=E$. 
+That is $c(E\setminus F')-\lambda(r(E)-r(F')+1)>-\lambda$ for all $F'\subsetneq E$.
+We are interested in the upperbound $\e$ of $\lambda$ such that the optimal $F$ is a proper subset of $E$ when $\lambda >\e$. Therefore, the upperbound is $\e=\min_{F\subsetneq E}\frac{c(E\setminus F)}{r(E)-r(F)}$, which is exactly the strength.
+\end{enumerate}
 
-(there is a $\pm1$ difference between principal partition and graph strength... but we dont care those $c\lambda$ terms since the difficult part is minimize $L(\lambda)$ for fixed $\lambda$)
+% (there is a $\pm1$ difference between principal partition and graph strength... but we dont care those $c\lambda$ terms since the difficult part is minimize $L(\lambda)$ for fixed $\lambda$)
 
 \subsection{principal sequence of partitions for cut interdiction}
-Now we focus One $L(\lambda)=\min \{w(C\setminus F)-\lambda(b-c(F)) | \forall \text{cut } C\;\forall F\subset C\}$. We can still assume that $G$ is connected and see that $L(\lambda)$ is pwl concave. Let $\lambda^*$ be a breakpoint on $L$. Suppose that there are two optimal solutions $(C_1,F_1)$ and $(C_2,F_2)$ at $\lambda^*$. For fixed $C$ ($C_1=C_2$), the same argument for principal partition still works. However, the difficult part is that $C$ might not be the same.
+Now we focus on $L(\lambda)=\min \{w(C\setminus F)-\lambda(b-c(F)) | \forall \text{cut } C\;\forall F\subset C\}$. We can still assume that $G$ is connected and see that $L(\lambda)$ is pwl concave (1 and 2 still hold). Let $\lambda^*$ be a breakpoint on $L$. Suppose that there are two optimal solutions $(C_1,F_1)$ and $(C_2,F_2)$ at $\lambda^*$. For fixed $C$ ($C_1=C_2$), the same argument for principal partition still works. However, the difficult part is that $C$ might not be the same. So it's unlikely that 3 and 4 hold. For cut interdiction problem, 5 shows connections between normalized mincut and the original interdiction problem. Recall that we observe the denominator in normalized min-cut can be relaxed (that is, we can use $\frac{w(C\setminus F)}{B-c(F)}$ for any $B>b$, instead of restricting to $B=b+1$) and the analysis still works. Now following the previous argument for 5, we assume $\lambda\in [0,\e]$ for small enough positive $\e$. For any $C$, we have $F=C$ since $w(C\setminus F)$ is dominating. For the remaining term $-\lambda(b-c(F))$ we are selecting a cut $F$ with smallest cose with respect to $c$. Note that we can assume that any cut in $G$ has larger cost than $b$ since otherwise the optimum is simply 0. 
+% Now we can see that $B$ in the denominator $B-c(F)$ should be the cost of mincut in $G$.
+Let $B$ be the minimum cost of cuts in $G$. 
+We have $-\lambda(b-B)\leq w(C\setminus F)-\lambda(b-c(F))$ for any cut $C$ and $F\subsetneq C$. Thus the upperbound is $\e=\min \frac{w(C\setminus F)}{B-c(F)}$.
+
+\note{It remains to show that the optimal solution at $\e$ guarantees $c(F)\leq b$? or maybe we don't need this for normalized mincut. I think normalized min-cut should not require $c(F)\leq b$. Further checks are needed. What we can guarantee is that $c(F)\leq B$.}
+
+\subsection{a fundamental difference}
+Consider $L(\lambda)$ for cut problem. One can see that the optimal $\lambda$ is clearly 0 since $L(\lambda)$ is pwl concave and the slope is negative at $\lambda=0$. What we are really interested in is the first segment on $L$. At the left end, $L(0)$ is exactly the weight of minimum cut. (the complementary slackness condition is satisfied.) At the right end, as we have shown in the previous paragraph, $\lambda$ equals to the value of the strength (which is the optimum of the linear relaxation of the cut IP). However, for cut interdiction problems $L(0)$ is not the optimum. Need to understand this better...
 
 \subsection{integrality gap}
-I guess the 2 approximation cut enumeration algorithm implies a integrality gap of 2 for cut interdiction problem.
+I guess the 2-approximate min-cut enumeration algorithm implies an integrality gap of 2 for cut interdiction problem.
+
+First consider the dual of linear relaxation of \autoref{lp:cutinterdict}.
+
+\begin{equation}\label{lp:dualcutint}
+\begin{aligned}
+\max&       &       \sum_T z_T &- b\lambda          &       &\\
+s.t.&       &       \sum_{T\ni e} z_T &\leq w(e)    &   &\forall e\in E\\
+    &       &       \sum_{T\ni e} z_T &\leq c(e)\lambda &   &\forall e \in E\\
+    &       &                           z_T,\lambda &\geq 0 &   &
+\end{aligned}
+\end{equation}
+We want to prove something like tree packing theorem for \autoref{lp:dualcutint}.
+\begin{conjecture}
+    The optimum of \autoref{lp:dualcutint} is $\min \set{\frac{w(C\setminus F)}{B-c(F)}| \forall \text{cut $C$}, c(F)\leq b}$, where $B$ is the cost of mincut in $G$ and $b$ is the budget.
+\end{conjecture}
+
+I believe the previous conjecture is not likely to be true. This one seems better,
+
+\begin{conjecture}\label{conj:optimaldual}
+    $\lambda=\min \frac{w(C\setminus F)}{B-c(F)}$ is optimal for \autoref{lp:dualcutint}.
+\end{conjecture}
+
+Assuming \autoref{conj:optimaldual} is true, \autoref{lp:dualcutint} in fact gives the idea of weight truncation. The capacity of each edge $e$ in the ``tree packing'' is $\min\{c(e)\lambda,w(e)\}$.
+
+\begin{conjecture}
+    \autoref{lp:cutinterdict} has an integrality gap of 2.
+\end{conjecture}
 
 \section{Random Stuff}