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Discrete Math, 3rd day, 6/23/04
REU 2004

Instructor: László Babai
Scribes: Justin Sinz and Travis Schedler


\begin{exercise}
% latex2html id marker 834Let $R$ be an equivalence relation...
...on
induces the original equivalence relation in the natural way).
\end{exercise}


\begin{exercise}
Recall: Every connected graph contains a spanning tree. Prove t...
...the (finite) number of
cycles (cycles have no repeated vertices).
\end{exercise}


\begin{exercise}
% latex2html id marker 838(A)
\begin{description}
\item[(a)]I...
...em[(c)$^\ast$]Disprove (a) for connected graphs.
\end{description}\end{exercise}


\begin{exercise}
% latex2html id marker 840(B)\\
(Helly-type Theorem) Let $R_...
...\neq \emptyset$.
Then $V(R_1)\cap \ldots\cap V(R_k)\neq\emptyset.$\end{exercise}


\begin{exercise}
% latex2html id marker 842(C)\\
Let $G$ be a connected grap...
...st paths in $G$. Prove
that $ V(P_1) \cap V(P_2) \neq \emptyset. $\end{exercise}

Note: (C) and (B) imply (A).

Definition 3.1   A legal coloring of a graph $ G$ is an assignment of a ``color'' to each vertex such that adjacent vertices have different ``colors''.
The chromatic number $ \chi (G) $ of $ G$ is defined to be the minimum number of colors necessary to color $ G$ legally.


\begin{exercise}
% latex2html id marker 844The graph $G$ is bipartite if and only if $\chi (G) \leq 2$.
\end{exercise}


\begin{exercise}
% latex2html id marker 846Construct a graph $G$ such that $K...
... \chi (G)=4.$ \\
{\sc Hint:} $11$ vertices, $5$-fold symmetry.
\end{exercise}

For a set of $ n$ positive real numbers, $ x_1, \ldots, x_n$ we can define several means:

(1)
the arithmetic mean $ \displaystyle{A(x_1, \ldots, x_n) =
\frac{x_1
+ x_2 + \cdots + x_n}{n}}$;
(2)
the quadratic mean $ \displaystyle{Q(x_1, \ldots,
x_n) = \sqrt{\frac{x_1^2 + \cdots + x_n^2}{n}} = \sqrt{A(x_1^2,
\ldots, x_n^2)}}$;
(3)
the geometric mean $ \displaystyle{G(x_1, \ldots, x_n) =
\sqrt[n]{x_1 x_2 \cdots x_n}}$;
(4)
the harmonic mean $ \displaystyle{H(x_1,
\ldots, x_n) = \frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}} =
\frac{1}{A(\frac{1}{x_1}, \ldots, \frac{1}{x_n})}}$.


\begin{exercise}
% latex2html id marker 848Prove, for any $x_1, \ldots, x_n$, ...
...$ is difficult; $G \geq H$ follows
immediately from $A \geq G$.
\end{exercise}


\begin{exercise}
% latex2html id marker 850
[JENSEN'S INEQUALITY]
There is a gen...
...(x_1, \ldots, x_n)) \leq A(f(x_1), \ldots, f(x_n)).
\end{equation}\end{exercise}

Note: A continuous function satisfying inequality ([*]) is concave and a continuous function satisfying inequality ([*]) is convex. However, not every function satisfying either (or even both) inequalities is continuous. (Prove!$ ^\ast$)


\begin{exercise}
% latex2html id marker 852Prove Jensen's inequality. Hint: Fi...
... any $n < 2^k$ as well,
thus proving the inequality for all $n$.
\end{exercise}


\begin{exercise}
% latex2html id marker 854Deduce Exercise \ref{mean} from Exercise \ref{jensen}.
\end{exercise}

We proved in class the

Theorem 3.2 (cf. Exercise 6.1.22 in the text)   There exists $ c > 0$ such that, if $ \cal G$ is a graph with $ m$ edges and $ n$ vertices, and $ \cal G$ has no $ 4$-cycles, i.e.  $ {\cal G} \not \supseteq C_4$, then $ m \leq c
n^{3/2}$. In other words, $ m = O(n^{3/2})$.

In fact we showed that, as $ m$ tends to infinity, $ m \lesssim \frac{1}{2} n^{3/2}$.

The following exercise strengthens this result:
\begin{exercise}
% latex2html id marker 856Show that, in the above situation o...
...uation}
\frac{n-1}{2} \geq {\frac{2m}{n} \choose 2}
\end{equation}\end{exercise}

The following exercise shows the tightness of the bound:
\begin{exercise}
% latex2html id marker 858Show that there exists $c' > 0$ su...
..._4$ with $m$edges and $n$ vertices, such that $m > c' n^{3/2}$.
\end{exercise}

\begin{exercise}
% latex2html id marker 860Show that $(\forall k) (\exists {\c...
...t
contain triangles which have arbitrarily high chromatic number.
\end{exercise}

\begin{exercise}
% latex2html id marker 862Show that $(\forall k,\ell) (\exist...
...ich do not contain any odd cycles of length $3, 5,
\ldots, \ell$.
\end{exercise}

Remark 3.3   Exercise [*] remains valid if we drop the word "odd":


\begin{theorem}
% latex2html id marker 864
[Erd\H os 1960]
$(\forall k,\ell) (\e...
... which do not contain any ycles of length $3, 4, 5,
\ldots, \ell$.
\end{theorem}

Erdos proved this in 1960 using the probabilistic method but did not exhibit any specific examples of such graphs. No explicit construction was known of such graphs until 30 years later, around 1990.


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Laszlo Babai 2004-06-27