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\lecturehead{CS G399}{Algorithmic Power Tools I}{Fall 2007}{18
September 2007}{San Tan}

\newcommand{\p}{\mathcal{P}}
\newcommand{\U}{\mathcal{U}}
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\noindent {\bf Lecture Outline:}
\begin{itemize}
\item
Rounding of LP relaxations
\item
Randomized rounding for set cover
\end{itemize}

This lecture introduces the technique of rounding LP relaxations for
designing approximation algorithms.  We introduce the randomized
rounding approach using the set cover problem.  The rounding algorithm
produces a set cover solution that, with probability almost $1$,
covers all elements with cost $O(\ln n)$ times that of optimal cost.

\section{Rounding for set cover}
Recall the definition of the set cover problem.
\begin{problem}
\label{prob:setcover} Given a universe $\U = \{e_1, \ldots, e_n\}$
of $n$ elements, a collection $\coll$ of $m$ subsets of $\U$, and a
cost function $c: \coll \rightarrow Q^+$, the set cover problem
seeks a minimum-cost subset of $\coll$ that covers all elements of
$\U$.
\end{problem}

Last class we formulated a linear programming relaxation for the set
cover problem.
\begin{eqnarray*}
& min & \sum_{S \in \coll}c_{S}x_{S}\\
& s.t. & \sum_{e\in S}x_{S}\ge 1;  \forall e\in \U \\
&      & x_{S}\ge 0; \forall S\in \coll \\
&      & x_{S}\in [0,1], \forall S\in \coll
\end{eqnarray*}

Suppose $\{x_{S}^{*}\}$ is an optimal solution for above LP
problem. Notice $x_{S}^{*}$ could be a fractional value between $0$
and $1$.  In order to apply the LP solution back to the set cover
problem, we need to round $\{x_{s}^{*}\}$.  Let us consider two
natural rounding methods.
\begin{itemize}
\item {\bf Nearest rounding:}
\[x_{S}^{*} \ge \frac{1}{2} \to x_{S}=1\]
\[x_{S}^{*} < \frac{1}{2} \to x_{S}=1\]

\item {\bf Randomized rounding:} Interpret $x_{s}$ as a probability that set $s$ is selected.
\end{itemize}

To analyze the rounding approaches, we must ask two questions:
\begin{itemize}
\item {\bf q1:} What's the cost of rounding algorithm (compared with optimal cost given by LP)?
\item {\bf q2:} Are all elements covered?
\end{itemize}

For nearest rounding, consider this scenario: for every set $S$ that
contains a given element $e$, we have $x_{S}^{*}<\frac{1}{2}$.  In
this case, $e$ will definitely not be covered after rounding.  It is
easy to construct examples where this is the case for all
elements. This scenario is surely undesirable.

Let's consider randomized rounding.
\begin{itemize}
\item {\bf q1:} Suppose $X={\text{collection of selected sets after randomized rounding}}$.
Then the total cost is
\[C(X)=\sum_{S\in X}c(S). \]

The expected total cost is
\begin{eqnarray*}
E[ C(X) ] & = & \sum_{S \in X}\Pr[ S \in X ] \cdot c(S) ] \\
& = & \sum_{S \in X} x_{S}^{*}c(S) \\
& = & OPT_{LP} \\
& \le & OPT
\end{eqnarray*}
So the expected cost of randomized rounding is as good as the LP cost,
which we know is a lower bound on the cost of an optimal set cover.

\item {\bf q2:} For an element $e$, assume that it occurs in sets
$s_{1}, s_{2}, ..., s_{k}$. The probability that $e$ is not covered is
\[ p_{\bar{e}} = \prod_{i = {1,\cdots,k} }(1-x_{s_{i}}^{*})  \]

Since $x_{s_{1}}^* + x_{s_{2}}^* + \cdots + x_{s_{k}}^*  \ge 1$, we
have
\begin{eqnarray*}
p_{\bar{e}} & = & (1-x_{s_{1}})(1-x_{s_{2}})\cdots (1-x_{s_{k}}) \\
& \le & \left(1 - \frac{1}{k}\right)^{k}\\
& \le & \frac{1}{e}
\end{eqnarray*}

So the probability that $e$ is covered by $X$ is
\[ p_{e} = 1 - p_{\bar{e}} \ge 1- \frac{1}{e}. \]

\end{itemize}

The expected cost of randomized rounding is good, but we have not
guaranteed that all elements are covered.  In fact, there is a very
good chance that all the elements are not covered.  On the other hand,
any given element is covered with at least a constant non-zero
probability.  An improvement to the simple randomized rounding is to
repeat randomized rounding for $t$ times and combine the result
\[ X= \bigcup_{i=1}^{t}X_{i}, \]
in which $X_{i}$ is the result of $i^{th}$ randomized rounding. The
improved randomized rounding has
\[ E[ C(X) ] \le t\cdot OPT, \]
\[ \Pr[\text{given element is not covered}] \le \frac{1}{e^{t}}. \]

Then the probability that at least one element is not covered is
\begin{eqnarray*}
\Pr[\text{some element is not covered}] & \le & \sum_{e_i \in {\cal U}} \Pr[e_i
\text{is not covered}] \\
& =\frac{n}{e^{t}}
\end{eqnarray*}
(Note that $\Pr[\text{some element is not covered}]$ is not equal to
$1-(1-\frac{1}{e^{t}})^n$, since the events defined by the coverage of
each element are not independent.) \\

Given a threshold $\varepsilon$, we now compute how many rounds of
randomized rounding are needed to assure that the probability that
some element is not covered is lower than $\varepsilon$.
\[ \frac{n}{e^{t}} \le \varepsilon \Rightarrow t \ge
\ln\frac{n}{\varepsilon}. \]

So a solution that repeats randomized rounding for
$\frac{n}{\varepsilon}$ times and return $X= \bigcup_{i=1}^{t}X_{i}$
satisfies
\[ E[ C(X) ] \le OPT \cdot \ln(\frac{n}{\varepsilon}), \]
\[ p[ \text{X is not a valid solution} ] \le \varepsilon. \] \\

From Markov's inequality $p_{r}[ Y \ge \alpha ] \le \frac{E[ Y
]}{\alpha}$, we have
\[p_{s}[ C(X) \ge 4 \cdot OPT \cdot \ln(\frac{n}{\varepsilon}) ] \le \frac{ E[ C(X) ] }{4 \cdot OPT \cdot \ln(\frac{n}{\varepsilon} )} \le
\frac{1}{4}. \]

If we set $\varepsilon=\frac{1}{4}$,
\[ p[ X \text{ is valid and } C(X) \le 4 \cdot OPT \cdot \ln(4n)] \ge
\frac{1}{2}, \] and corresponding algorithm is
\begin{enumerate}
\item Solve LP;
\item Repeat randomized rounding for $\ln(4n)$ times and get a
solution $X= \bigcup_{i=1}^{t}X_{i}$;
\item If $C(X) \ge 4 \cdot OPT \cdot \ln(4n)$ or X is not valid, go
back to step 2; otherwise return X as a final solution.
\end{enumerate}

The expected number of iterations of step 2 of the above algorithm
before the algorithm terminates is at most $2$.  In fact, the
probability that the algorithm terminates in $T$ iterations is at
least $1 - 1/2^T$. 

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