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%%%  EDIT THE FOLLOWING TO UPDATE DATE AND LECTURE NUMBER

\head{CS G399}{Algorithmic Power Tools I}{Fall 2007}{Scribe: Hooman
Javaheri}{19 October 2007}


\newcommand{\p}{\mathcal{P}}
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\newcommand{\xpy}{x + \lambda y}
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\newcommand{\eps}{\varepsilon}


\noindent {\bf Lecture Outline:}


%%% Itemized outline of the lecture
\begin{itemize}
  \item Complete Steiner network analysis

\end{itemize}


In this lecture, we will complete the analysis for Jain's algorithm
as a 2-approximation algorithm for the Generalized Steinr Network
problem.

\section{Review of the problem and the algorithm}
The Generalized Steiner Network problem is:

{\bf Problem 1.}{ Given a graph $G(V,E)$, a cost function $c:E \to
Z^ + $, and a connectivity requirement $r_{uv}$ for each pair
$(u,v)$, find a min-cost subgraph of $G$ satisfying the connectivity
requirement for all $(u,v)$, where $u_{e}$ is maximum number of
copies you can pick for edge $e$}

and the corresponding LP can be written this way

\[ \min\sum c_{e}x_{e} \]
  \[ s.t. \sum_{e \in \delta(S)} x_e \geq f(S) \]
  \[ 0 \leq x_{e} \leq u_{e} \]
  where,
\[
f(S) = \mathop {\max }\limits_{\mathop {u,v}\limits_{u \in S,v \in
\bar S} } r_{uv}
\]

 $\delta(S)$ is the set of edges connecting $S$ to
  $\overline{S}$.

Here is the Jain's algorithm for the problem:

\begin{algorithm}\label{alg:jain}
\SetAlFnt{\small} \dontprintsemicolon
 $1.$ $F \leftarrow {\O}$ , Define $f$ according to $r_{uv}$s.
 $f' \leftarrow f$.\;
 $2.$  \hspace{0.1in}\Repeat{$f'(S) = 0$} {$2.1.$\hspace{0.1in} Solve LP for $f'$ to obtain a solution $x$ with desired property:
       \hspace{0.1in}$\exists e \hspace{0.1in}s.t. \hspace{0.1in}x_e \geq 1/2$\;
       $2.2.$\hspace{0.1in} Add $\lceil x_e \rceil$ copies of all $e$ s.t. $x_e \geq 1/2$ to $F$ \;
       $2.3.$\hspace{0.1in} Remove the above edges $e$ from $G$. \;
       $2.4.$\hspace{0.1in} $f'(S) \leftarrow \max(0,f(S) - \delta_F(S))$
       where, $\delta_F(S)$ is set of edges of $F$ crossing $S$. \;}
 $3.$ Return $F$

\caption{Jain's algorithm for generalized steiner network problem}
\end{algorithm}


In the previous lecture, we started a 3 step approach to analyze the
algorithm:
\begin{enumerate}
 \item LP can be solved in Poly-Time ,in fact, we will find an optimal BFS in each iteration.
 \item The algorithm is a 2-approximation assuming the desired property in
 state 2.1 is TRUE.
 \item \begin{Theorem} Desired Property: for all BFSs,\hspace{0.1in} $\exists e \hspace{0.1in}s.t. \hspace{0.1in}x_e \geq
                1/2$ \hspace{0.1in} Actually, we will prove it for 1/3 rather than
                1/2. The approach for 1/2 is the similar, but requires some
                 complicated case analysis for which we refer to the original paper.
        \end{Theorem}
\end{enumerate}

\section{Completing the analysis: Proving desired property theorem}

Having proved first two steps, we have got one step to go, the last
but definitely not the least step.

As we mentioned before, we are going to prove part 3 in two steps:

\begin{enumerate}
\item    \begin{Theorem}
         Suppose $x$ is a BFS of Generalized Steiner Network's LP

         Further assume $x_e \in (0,1)$ not including 0 and 1.
         Suppose there are $m$ such edges.
         There exist a set of $m$ "TIGHT" constraints  that are
         independent and form a laminar family.
        \end{Theorem}
\item   if 1 then, $\exists e \hspace{0.1in}s.t. \hspace{0.1in}x_e \geq
 1/2$.

\end{enumerate}



Let's define terminology here in more detail.
\begin{itemize}
\item { \bf Independent:} for every $S$ if $e$ cross the cut the
coefficient of $x_e$ in LP is 1, otherwise is zero, therefore we can
define a vector $A_S$ of coefficients of each edge. Note that WLOG
we can number the edges in arbitrary order and define $A_S$ as a m
dimension 0-1 vector, where $m$ is the numbers of edges .

Collection of sets is independent if their corresponding $A_S$'s are
independent.

\item { \bf Cross: } Two sets $S$ and
$T$ are said to cross if $ S-T , T-S , S \cap T $ are all non-empty.

\item { \bf Laminar family: } A laminar family is a collection of sets no
two of which cross.

\end{itemize}

\subsection{Proof of Theorem 1, assuming Theorem 2.}

Assuming that Theorem 2 is true, we are going to prove Theorem 1.
First, let's take a close look at a laminar family.

Let $L$ denote the laminar family in Figure \ref{fig:lam2}. As we
mentioned before laminar family can be represented with a tree
hierarchy. Figure \ref{fig:tree} shows the corresponding tree for
laminar family $L$.

\begin{figure}
\center \epsfig{file=laminar2.eps, width =300px} \caption{Laminar
Family L} \label{fig:lam2}
\end{figure}

In order to prove that $\exists e \hspace{0.1in}s.t.
\hspace{0.1in}x_e \geq
 1/3$, we only need to show $\exists$ set $S$ in this Laminar family
 that has at most 3 1's in its $A_S$ (i. e. there are at most 3
 edges crossing this set).

 Our proof will be based on case analysis and contradiction. In
 other words we will show that if all sets have more than 3 crossing
 edges, then we obtain a contradiction.

 \begin{itemize}
 \item {\bf CASE 1:} Let suppose all sets are disjoint. We have $m$
 edges correspond to $2m$ end-points. In this scenario, each edge
 crosses two sets. Considering we have only $m$ sets, it is trivial to
 see that Theorem 1 holds true in this situation.

 \item {\bf CASE 2:} Let's look at the tree corresponding to first 10
 sets of the laminar family $L$ (Figure \ref{fig:tree}.). If we
 assume that each set has at least 4 crossing edges, adding up the
 number of the edges for leaves of the tree, we will have 24
 end-points which is not possible since we at most have 20 endpoints
 for 10 sets.

 \item {\bf CASE 3:} ???

 \end{itemize}

\begin{figure}
\center \epsfig{file=tree.eps, width =300px} \caption{Corresponding
tree for L} \label{fig:tree}
\end{figure}
%% Section 1

\subsection{Proof of Theorem 2.}

{\bf Definition:} Function $\delta$ is SUBMODULAR if for all $S$ and
$T$ both of the following conditions hold,
\[ | \delta(S)| + | \delta(T) | \geq | \delta(S-T) | + | \delta(T-S)
| \] \center and,
\[ | \delta(S)| + | \delta(T) | \geq | \delta(S \cap T) | + | \delta(T \cup S)
| \]


We can easily show that $\delta(S)$ in our problem is submodular.
Figure \ref{fig:submodular} depicts $S$ and $T$ crossing. We can
define 6 types of edges corresponds to their origins and
destinations. Using the counting argument it's easy to see that all
types of edges but Type $VI$ counts in both right-hand-side and
let-hand-side equally and type $VI$ edges only count toward the
left-hand side.

\begin{figure}
\center \epsfig{file=cross2.eps, width =200px}
\caption{Corresponding tree for L} \label{fig:submodular}
\end{figure}


\begin{Lemma}
If $S$ and $T$ are crossing and tight, then\\
either $S-T$ and $T-S$ are tight and \[ A_{S-T} + A_{T-S} = A_S +
A_T \] or $S \cap T$ and $S \cup T$ are tight and \[ A_{S \cap T} +
A_{T \cup S} = A_S + A_T
\]
\end{Lemma}
\begin{proof}
Given solution $x$, $S$ and $T$, if $x$ is tight at $S$ and $T$ we
have \[ \delta_x(S) = \sum_{e \in \delta(S)} { x_e} = f(S) \]
\[ \delta_x(T) = \sum_{e \in \delta(T)}  x_e = f(T) \]

According to submodularity of cut functions, we have\\
 either $\delta_x(S-T) = f(S-T)$ and $\delta_x(T-S) = f(T-S)$ \\
or $\delta_x(S \cap T) = f(S \cap T)$ and $\delta_x(T \cup S) = f(T
\cup S)$

    \begin{Lemma}
    $f(S)$ is weakly supermodular. i. e.\\
    \hspace{2in} Either
    \[ f(S) + f(T) \leq f(S-T) + f(T-S) \]
    \hspace{2in}or
    \[ f(S) + f(T) \leq f(S \cap T) + f(S \cup T) \]

    \end{Lemma}
    \begin{proof}
    Proof is similar to proof of submodularity of cut
    functions.
    \end{proof}

Let us suppose $f(S) + f(T) \leq f(S-T) + f(T-S)$, Therefore
\[\delta_x(S-T) + \delta_x(T-S) \leq \delta_x(S) + \delta_x(T) \]
 \[ = f(S) + f(T) \]
 \[ \leq f(S-T) + f(T-S) \]

 Since $x$ is feasible, $\delta(S_T) \geq f(S-T) $ and $  \delta_x(T-S) \geq
 f(S-T)$.
 Therefore,
 $\delta_x(S \cap T) = f(S \cap T)$ and $\delta_x(T \cup S) = f(T
\cup S)$ (by definition of $\delta(S)$ and $f(S)$). We then also
have
\[\delta_x(S-T) + \delta_x(T-S) = \delta_x(S) + \delta_x(T)\]
 This leads to
 \[ A_{S-T} + A_{T-S} = A_S + A_T \]

Note: assuming second part of Lemma 2 and same approach we can get
$A_{S \cap T} + A_{T \cup S} = A_S + A_T$.

\end{proof}

\begin{Lemma}
Let $L$ be a laminar family and $S$ be a set not in $L$ and also
independent of $L$. Suppose $S$ crosses $T$ in $L$. Then each of $ S
\cap T , S \cup T , S-T , T-S$ cross fewer sets in $L$ than $S$.
\end{Lemma}
\begin{proof}
Figure \ref{fig:lemma3} depicts $S$ crossing $T$. Imagine how the
other sets in $L$ can cross $S$. There are in total 3 types of
crossing sets. S may cross all 3 types, while $S-T$ ans $S \cup T$
only cross Type I and II, also $T-S$ and $S \cap T$ only cross Type
III. None of these sets cross $T$. Therefore, each of those sets
cross fewer sets
than $S$ itself.\\

We will wrap up the proof in the next class.
\end{proof}
\begin{figure}
\center \epsfig{file=cross3.eps, width =250px}
\caption{Corresponding tree for L} \label{fig:lemma3}
\end{figure}



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