\newcounter{lecture} \setcounter{lecture}{11}
\def\thetoday{February 12, 2009}
\def\thetitle{Randomized Algorithms}

\input{lec_hdr.tex}

\topic {Randomized algorithm for finding a perfect matching}

Today we extend the randomized algorithm from last lecture that
determined the existence of a perfect matching in a graph $G(V,E)$, to
actually finding a perfect matching in $G$.  The key computational step
will be a matrix inversion.

First we establish a key (and surprising) property of subsets of random
numbers:

\begin{definition} 
A \emph{set system} $(S,F)$ consists of a finite set $S$ of elements, 
$S = \{x_1,x_2,\ldots,x_n\}$ and a family $F$ of subsets of $S$, $F =
\{S_1,\ldots,S_k\}, S_j \subseteq S$.  
\end{definition}

For each element of $S$, assign weight $w_i$ to $x_i$, where $w_i$ is
chosed uniformly and independently from $\{1,\ldots,2n\}$.  Denote the
weight of a set $S_j$ to be $\sum_{x_i \in S_j} w_i$.

\begin{lemma}[Isolating lemma]
The probability that there is a unique minimum weight set
is at least $1/2$.
\end{lemma}

\begin{proof}
Fix the weight of all elements except $x_i$. Define the \emph{threshold} for element $x_i$ to be the number $\alpha_i$ such that
if $w_i > \alpha_i$ then $x_i$ is in no set with minimum weight, and if
$w_i \leq \alpha_i$ the $x_i$ is in some set with minimum weight.

Clearly, if $w_i < \alpha_i$, then $x_i$ is in \emph{every} set with
minimum weight.  The only ambiguity occurs when $w_i = \alpha_i$. In
this case we say that $x_i$ is \emph{singular}.  

A key observation is that the weight of element $x_i$ is
\emph{independent} of the threshold value $\alpha_i$. Since $w_i$ is
chosen uniformly at random from $\{1,2,\ldots,2n\}$, the probability
that an element is singular is at most $1/2n$.

\begin{obs} If no element is singular, then the subset with minimum
weight is unique.
\end{obs}

Since $S$ contains $n$ elements, the probability that $S$ containts a
singular element is at most $n \times \frac{1}{2n} = \frac{1}{2}$.
Thus, with probability at least $1/2$, there exists no singular element,
implying the lemma.  
\end{proof}

\topic{Matchings in bipartite graphs}

Given a bipartite graph $G(U,V,E)$, to each edge $(i,j) \in E$ assign a
weight $w_{i,j}$  chosen uniformly and independently from $[1,2m]$,
where $m = |E|$.  By the isolating lemma, the minimum weight perfect
matching in $G$ will be unique with probability at least $1/2$.

As in the previous lecture, we
define the matrix $B$ such that
\[b_{i,j} = \left\{ 
\begin{array}{ll} 
x_{i,j} & \mbox{if } i \sim j, \  i \in U,\  j \in V\\
0 & \mbox{otherwise.}
\end{array}
\right.
\]

Recall that the determinant of this matrix is equivalent to zero iff
$G$ has no perfect matchings.  Set $x_{i,j} = 2^{w_{i,j}}$.

\begin{lemma} Suppose that the minimum weight perfect matching $M$ is
unique with weight $W$.  Then $|B| \neq 0$ and the highest power of 2
that divides $|B|$ is $2^W$.
\end{lemma}

\begin{proof}
A permutation of length $n$ corresponds to a perfect matching in $M$.
Recall that \[|B| = \sum_\pi sign(\pi) value(\pi),\] where $value(\pi) 
= \prod_{i=1}^n b_{i,\pi(i)}$.  Let $\pi_M$ be the permutation
corresponding to $M$, with $value(\pi_M) = 2^W$.  Then the value of
every other permutation is either $0$ or a greater power of $2$ than
$W$.  This implies the lemma.
\end{proof}

Thus, we can recover $W$ by calculating the determinant of $B$.  The
following lemma gives us a way of recovering which edges belong to
$M$.  Denote $B_{i,j}$ as the matrix derived from $B$ by removing the
$i$th row and $j$th column.

\begin{lemma}
The edge $(u_i,v_i)$ belongs to $M$ iff
$\frac{|B_{i,j}|2^{w_{i,j}}}{2^W}$ is odd.
\end{lemma}
\begin{proof}

Note that \[|B_{i,j}| 2^{w_{i,j}} = \sum_{\pi:\pi(i)=j} sign(\pi)
value(\pi).\] Let $\pi_M$ be the permutation corresponding to $M$. If
$(u_i,v_j)$ is an edge in $M$, then one permutation, $\pi_M$, will have
weight $2^W$.  All other permutations will either have weight $0$ or a
higher power of $2$ than $W$.  Therefore, $2^W$ will be the highest
power of $2$ which divides the righthand side iff $(u_i,v_j) \in M$.
\end{proof}

$|B_{i,j}|$ is the $(i,j)$ entry of the cofactor matrix, which we can
compute from the following relationship: \[T^{-1} = \frac{1}{det(T)}
(cofactor(T))^T\] We can therefore find the perfect matching with
minimum weight $M$ with a single matrix inversion.  Since $M$ will be
unique with probability at least $1/2$, we only need to run this
algorithm a constant number of times to have an arbitrarily high
probability of success.


\topic{Matchings in general graphs}

Recall from last lecture the definition of the Tutte matrix $T$ 
for a graph $G(V,E)$:
\[t_{i,j} = \left\{ 
\begin{array}{ll} 
x_{i,j} & \mbox{if } i \sim j, \  i > j \\
-x_{i,j} & \mbox{if } i \sim j, \  i < j \\
0 & \mbox{otherwise.}
\end{array}
\right.
\]

Recall that we showed that the determinant of $T$ is equivalent to $0$
iff $T$ has a perfect matching.  

To find this perfect matching, for each $i,j$ set $x_{i,j}$ to
$2^{w_{i,j}}$, where $w_{i,j}$ is chosen uniformly and independently
from $\{1,\ldots,2m\}$ with $m = |E|$.  

\begin{lemma} 
Suppose that the minimum weight perfect matching in $G(V,E)$ is unique.
Let this matching be $M$ and its weight be $W$.  Then $|T| \neq 0$, and
the highest power of $2$ which divides $|T|$ is $2^{2W}$.
\end{lemma}

\begin{proof}
For a permutation $\pi$, we define $value(\pi)$ as in the bipartite
case.  In addition, for a cycle decomposition $\{c_1, c_2, \ldots, c_k\}
= \pi$, \[value(\pi) = \prod_{i=1}^k value(c_i).\]

\begin{obs}
Terms corresponding to permutations with odd cycles cancel each
other.  
\end{obs}

\begin{proof}
To see this let $\pi = \{c_1, c_2, \ldots, c_k\}$ be a permutation with
an odd cycle.  WLOG assume that this cycle is $c_1$.  Note that if we
reverse the direction of all the edges to get cycle $c'_1$, $value(c_1)
= -value(c_1)$.  Let $\pi' = \{c'_1, c_2, c_3, \ldots, c_k\}$.  Then
$value(\pi) = -value(\pi')$.  Since reversing the edges of a cycle does
not change the sign of a permutation, $\pi$ and $\pi'$ cancel each other
in the determinant calculation, which implies the observation.
\end{proof}

Therefore, we only need to consider even cycles.

Each perfect matching corresponds to a permutation with $n/2$ cycles of
length $2$.  For a given even cycle $x_1,x_2,...,x_{2k}$ for some $1
\leq k \leq n/2 $, note that there are two corresponding perfect
matchings of these vertices.

\begin{obs} 
The value of an even cycle of $c$ length greater than $2$ is at least 
as much as the average value of its component perfect matchings.
\end{obs}

\begin{proof}
The contribution from the even cycle is just the product of the weights
of its edges.  Denote the value of the $(x_{2i},x_{2i+1})$ perfect
matching as $a$, and denote the value of the $(x_{2i+1},x_{2(i+1)})$
matching as $b$.  Then the value cycle of cycle $c$ is $ab$, and the
weight of the two perfect matchings are $a^2$ and $b^2$ respectively.
Therefore,
\begin{eqnarray}
(a-b)^2 &\geq& 0 \\
a^2+b^2 &\geq& 2ab\\
\frac{a^2+b^2}{2} &\geq& ab
\end{eqnarray}
\end{proof}

Since $M$ is unique, it follows that the weight of any permutation with
even cycles of length greater than $2$ will be strictly greater than
$2^{2W}$.  This means that all of the terms in the determinant will be
composed of sums of powers of $2$ with exponents greater than $2W$ plus
$2^{2W}$, which implies the lemma.  
\end{proof}


\begin{lemma} 
If $M$ is the unique minimum weight matching in $G$ with weight $W$,
then $\frac{|T_{i,j}|2^{2w_{i,j}}}{2^{2W}}$ is odd iff the edge $(i,j)
\in M$.
\end{lemma}

The proof of this lemma for general graphs is identical to the
corresponding lemma for bipartite graphs.  So by the same algorithm as
for the bipartite case, we can determine the minimum weight perfect
matching of a general graph $G$ with a single matrix inversion.


\topic {Randomized algorithm for global min-cut}

In the \emph{global min-cut} problem, we wish to find a minimum set of
edges such that a graph becomes disconnected.  Formally, 

\begin{definition}
Given an undirected graph $G(V,E)$, a {\bf global min-cut} is a
partition of $V$ into two subsets $(A,B)$ such that the number of
edges between $A$ and $B$ is minimized.
\end{definition}

To derive an algorithm for this, note that the global min-cut is the
minimum over all possible $s-t$ cuts.

{\bf Solution 1:} \emph{Compute the min $s-t$ cut for all pairs $s, t
\in V$.}

This algorithm requires $O(n^2)$ calls to a min s-t cut solver.  We can
reduce this by noting that any node $s$ must appear in one of $A$ or
$B$, meaning we can reduce the number of min s-t cut solves by a factor
of $n$.


{\bf Solution 2:} \emph{Fix $s$ and find min $s-t$ cut for all $t \in
V$.}

For a long time, the intuition was that global min-cut was harder than
$s-t$ cut.  David Karger was able to show that this is not the case with
a randomized algorithm.

\begin{algorithm}

\caption{Karger's randomized global min-cut}
\begin{algorithmic}
\REPEAT
\STATE Choose an edge $\{u,v\}$ uniformly at random from $E$.
\STATE Contract the vertices $u$ and $v$ to a super-vector $w$. 
\STATE Keep parallel edges but remove self-loops.
\UNTIL {$G$ has only 2 vertices.}
\STATE Report the corresponding cut.
\end{algorithmic}
\end{algorithm}

\begin{thm}  The probability that the algorithm finds the minimum cut
in $G$ is at least $2/n^2$.
\end{thm}

\begin{proof}
Let $F$ be the set of edges in a global min-cut and supposes $|F|=k$.
Let $E$ by the even that the algorithms does not contract an edge from
$F_in$ step $1$. Since the minimum cut has value $k$, the degree of each
vertex must be at least $k$, which implies the following claim:

\begin{claim} $|E| \geq \frac{k|V|}{2} = \frac{kn}{2}$.
\end{claim}

Denote $E_i$ be the event that the $i$th edge picked belongs to the
min-cut.  The probability that the first edge picked belongs to the
min-cut is therefore:

\begin{eqnarray*}
Pr (E_1) &=& 1 - \frac{k}{|E|} \\
&\geq & 1 = \frac{2}{n}
\end{eqnarray*}

Similarly, 
\begin{eqnarray*}
Pr (E_2 | E_1) &\geq& 1 - \frac{2}{n-1} \\
Pr (E_i | E_1 \cap E_2 \cap \ldots) &\geq & 1 - \frac{2}{n-i+1}
\end{eqnarray*}

Combining these to get the total probability of success gives
\begin{eqnarray*}
Pr(\mbox{success}) &=& Pr (E_1 \cap E_2 \cap \ldots \cap E_{n-2})  \\
&\geq& Pr(E_1)Pr(E_2|E_1)\ldots Pr(E_{n-2}|E_1\cap\ldots E_{n-3})\\
&\geq& (1-\frac{2}{n})(1 - \frac{2}{n-1})\ldots (1-\frac{2}{3}) \\
&\geq& 2 \frac{(n-2)!}{n!} \geq \frac{2}{n^2},
\end{eqnarray*}

So the algorithm will succeed at with probability at least $2/{n^2}$.
\end{proof}

The probability of finding a min-cut seems rather low, and goes to zero
as $n \rightarrow \infty$ .  However, what happens to the probability of
success if we run the algorithm $t$ times? The probability of failure
will be at most \[(1-\frac{2}{n^2})^t.\] So if we set $t = cn^2$ for
some constant $c$, the probability of failure will be at most $(1 - 
\frac{2}{n^2})^{cn^2} \leq e^{-2c}$. To make the probability of 
failure as small as, say $e^{-14}$, it is only necessary to run the 
algorithm $7n^2$ times.

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