\newcounter{lecture} \setcounter{lecture}{12}
\def\thetoday{February 19, 2009}
\def\thetitle{The Probabilistic Method}

\input{lec_hdr.tex}

The probabilistic method is an elegant and powerful technique for
analysis of algorithms and combinatorial structures.  It is based on a
simple premise: in order to prove the existance of a structure, we
merely need to show that there is a positive probability that the
structure exists.

This method was first used by Szele in 1943, but Erd\"os was the first
to fully realize the power of this method and apply it to a wide
variety of problems.

\subsection*{Chromatic Number of a Graph}

A fundamental question in graph theory is the relationship between the
subgraphs of a graph and its chromatic number.  For example, can we
show that triangle free graphs must have a chromatic number bounded by
some constant $k$?

Recall that a \emph{proper vertex $k$-coloring} of a graph $G(V,E)$ is
an assignment of $k$ colors to vertices such that no two vertices of
the same color share an edge.  The \emph{chromatic number}
$\chi (G)$ is equal to the smallest number of
colors needed to have a proper vertex coloring.

\begin{definition} The \emph{girth} of a graph $G$, $g(G)$, is the
length of the smallest cycle in $G$.
\end{definition}

In 1954, B. Descartes was the first to show that triangle-free graphs
can have arbitrarily high chromatic numbers.  But this construction
contained many short cycles.  Is there a dependence on short cycles
and an arbitrarily high chromatic number? 

In 1959, Paul Erd\"os used the probabilistic method to prove the
existence of graphs with arbitrarily high girth and chromatic number.

\begin{thm}[Erd\"os, 1959] For every $g,k > 0$, there exists a graph
$G$ with $\chi(G) \geq k$ and $g(G) \geq g$.
\end{thm}
\begin{proof}
Consider a random graph on $n$ vertices $G(n,p)$ such that there is an
edge between every pair or vertices with probability $p$ independent
of every other pair.  The idea is that if a graph has chromatic number
$k$, then there must exist at least one independent set of size at
least $\frac{n}{k}$, since a coloring corresponds to an independent set.

Therefore, in order to show that $\chi(G) \geq k$, it suffices to
prove that with high probability the size of any independent set in
$G$ is at most $\frac{n}{k}$.  We will prove that with high
probability, the graph doesn't have any independent set of size
$\frac{n}{2k}$.

Recall from basic probability the ``union bound'' or ``subadditivity
property'' of probabilities.  That is, for any (arbitrary, not necessarily
disjoint) events $E_1, E_2,\ldots, E_j$, 
\[Pr(\cup_{i=1}^j E_i) \leq
\sum_{i=1}^j Pr(E_i)\]

The probability that any set of $n/2k$ vertices is an independent set
is $(1-p)^{\binom{n/2k}{2}}$.  There are $\binom{n}{n/2k}$
possibilities for vertex sets of size $n/2k$.
By the union bound, the probability of $G_{n,p}$ having such an independent
set is therefore at most
\begin{eqnarray*}
Pr\left[G_{n,p} \mbox{ has an independent set of size } \frac{n}{2k}\right] &\leq
&\binom{n}{n/2k} (1-p)^{\binom{n/2k}{2}}\\ 
&\leq &2^n e^{-pn^2/8k} \\
&\leq &e^{\log 2n - pn^2/8k} \\
&\leq &e^{n \log 2 - n^{\epsilon + 1}/8k}
\end{eqnarray*}
where we set $p = n^{\epsilon - 1}$ for some $\epsilon < 1$.  The
above expression tends to zero as $n \rightarrow \infty$, so is
therefore smaller than $1/4$ for $n$ large enough.

Let $X$ be the random variable counting the number of cycles of length
$g$ and smaller.  By linearity of expectation,
\begin{eqnarray*}
\E[X] &=& \sum_{i=1}^g \binom{n}{i}\frac{(i-1)!}{2}p^i \\
&\leq &g(np)^g = gn^{\epsilon g}
\end{eqnarray*}
if $0 < \epsilon < 1/g$, the above expression is $o(n)$.  Thus, for
sufficiently large $n$, $\E(X) < 1/4$.  By Markov's
inequality, 
\[Pr (X > n/2) \leq Pr(X > 2\E(X)) < 1/2\]


Therefore, if we choose $n$ sufficiently large, and $p = n^{\epsilon -
1}$ for $0 < \epsilon < 1/g$, the probability that $G_{n,p}$ has a
independent set of size $\frac{n}{2k}$ or that the number of cycles of length
at most $g$ is $\frac{n}{2}$ is less than $1$ by the union bound. 

Now, we can construct a graph $G'$ by removing a vertex from each
short cycle.  The number of vertices in $G'$ is at least $\frac{n}{2}$, the
size of the maximum independent set in $G'$ is no more that $\frac{n}{2k}$,
and there are no cycles of length less than $g$.  This implies that
$\chi(G') > k$ and $g(G') > g$.
\end{proof}

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