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Discrete Math, 4th day, 6/24/04
REU 2004

Instructor: László Babai
Scribes: Eric Patterson and Travis Schedler


\begin{exercise}
% latex2html id marker 815
If $\cal G$\ is connected with $n \g...
...
the center of a star (which leaves a highly disconnected graph).
\end{exercise}

Definition 4.1   A graph $ G$ is $ k$-edge-connected between vertices $ x$ and $ y$ if there are $ k$ edge-disjoint paths between $ x$ and $ y.$

Definition 4.2   A graph $ G$ is $ k$-vertex-connected between vertices $ x$ and $ y$ if there are $ k$ vertex-disjoint paths between $ x$ and $ y.$

Suppose we found $ 57$-edge-disjoint paths between $ x$ and $ y,$ and we think $ 57$ is optimal. How can we demonstrate that $ 58$ such paths cannot be found?

Ben's Conjecture: If there are no more than $ 57$ edge-disjoint paths from $ x$ to $ y,$ then there is a partition $ V=A\dot{\cup} B$ such that $ x\in
A,$ $ y\in B,$ and the maximum number of edges between $ A$ and $ B$ is less than or equal to $ 57.$

Definition 4.3   An $ \left(x,y\right)$-cut of a graph $ G=\left(V,E\right)$ is a partition $ V=A\dot{\cup} B$ such that $ x\in A$ and $ y\in B.$

Theorem 4.4 (Menger's Theorem)   The maximum number of edge-disjoint paths between $ x$ and $ y$ equals the minimum number of edges between the two parts of an $ \left(x,y\right)$-cut.

Puzzles


\begin{exercise}
% latex2html id marker 817If we have a $10^{10} \times 10^{10...
...s (10^{10} + 0.1)$-square then we can fit
$10^{20} + 1$\ squares.
\end{exercise}


\begin{exercise}
% latex2html id marker 819
Suppose we have a big rectangle whic...
..._n$\ is an integer. Show that
either $a$\ and $b$\ is an integer.
\end{exercise}

Set Game. The Set game is played with a deck of cards. Each card has four properties with three possible values: a shape (oval,diamond,squiggle), a color (red, green, purple), a shading (blank, solid, shaded), and a number (1,2,3). Each card is distinct, so there are $ 3^4=81$ cards. Twelve cards are placed face up on a table. The objective is to find a ``set'' among the twelve cards. A ``set'' is formed by three cards so that each property is either the same on all cards or different on all cards. The first player to find a ``set'' gets that ``set,'' after which three new cards from the deck are placed face up to replace them. The player who finds the most ``sets'' wins.

We can generalize the Set game so that there are $ n$ properties on each card, and each property has three possible values. In this $ n$-dimensional Set game we have $ 3^n$ cards. We are interested in how many cards we can draw from the deck without finding a ``set.'' Let $ \alpha(n)$ denote the maximum number.

Let $ \mathbb{Z}_3$ be the set $ \{0,1,2\}$ with addition mod $ 3,$ so $ 2+2=1$ for example. Let $ \mathbb{Z}_3^n$ be the set $ \{(x_1,\dots,x_n):\ x_i\in
\mathbb{Z}_3\}.$ For $ \underline{x}=(x_1,\dots,x_n)$ and $ \underline{y}=(y_1,\dots,y_n)$ in $ \mathbb{Z}_3^n,$ we define addition by $ \underline{x}+\underline{y}=(x_1+y_1,\dots,x_n+y_n).$ In the Set game ($ n=4$), the cards correspond to the quadruples in $ \mathbb{Z}_3^4;$ for instance [oval, green, blank, 3] $ \leftrightarrow \left[1,1,0,2\right].$

Observation 4.5   Three distinct elements $ \underline{x}, \underline{y},\underline{z}
\in \mathbb{Z}_3^4$ form a set if and only if $ \underline{x}+\underline{y}+
\underline{z}=\underline{0}.$

Similarly, we can work in $ \mathbb{Z}_3^n$ for the $ n$-dimensional Set game. So an alternative definition for $ \alpha(n)$ is the maximum number of elements of $ \mathbb{Z}_3^n$ without a solution to $ \underline{x}+\underline{y}+\underline{z}=\underline{0}$ where $ \underline{x},\underline{y},\underline{z}$ are distinct.

Definition 4.6   A function $ f:\mathbb{N} \to \mathbb{R}$ is supermultiplicative if $ f(n+m)\geq f(n)\cdot f(m).$


\begin{exercise}
% latex2html id marker 821Prove: if $f$\ is supermultiplicati...
... \sqrt[n]{f(n)}$\ exists, and $L\geq \sqrt[n]{f(n)}$\ for
all $n.$\end{exercise}


\begin{exercise}
% latex2html id marker 823Prove $\alpha(n)$\ is supermultiplicative.
\end{exercise}

Observe that $ 2^n \leq \alpha(n) \leq 3^n,$ so $ 2\leq L \leq 3$ by the previous two exercises, where $ L=\lim_{n\to \infty}\sqrt[n]{\alpha(n)}.$


\begin{exercise}
% latex2html id marker 825Prove: $\alpha(4) \geq 20.$\ Infer: $L \geq \sqrt[4]{20}\approx 2.115.$\end{exercise}


\begin{exercise}
% latex2html id marker 827Prove: $\alpha(2)=4.$\ Infer $\alpha(4)\leq 36.$\end{exercise}


\begin{exercise}
% latex2html id marker 829Prove: $\alpha(3)=9.$\ Infer $\alpha(4)\leq 27.$\end{exercise}


\begin{exercise}
% latex2html id marker 831Prove: $\alpha(4)\leq 24.$\end{exercise}


\begin{exercise}
% latex2html id marker 833$^\ast$\ Prove $\alpha(4)=20.$\end{exercise}

Conjecture 4.7 (Open Problem)   $ \lim_{n\to \infty} \sqrt[n]{\alpha(n)}=3.$


\begin{exercise}
% latex2html id marker 835Color the integers red and blue. Is...
...bers. What is the largest $n$\ for which
you can find an example?
\end{exercise}

Theorem 4.8 (Van der Waerden, 1928)   If we color the positive integers with $ k$ colors, then for all $ \ell$ there is an $ \ell$-term arithmetic progression in one color.

Theorem 4.9 (Szemerédi, 1975)   For all $ \epsilon > 0$ and $ \ell,$ there is an $ N$ such that if $ A\subseteq \{1,\dots,N\}$ and $ \vert A\vert\geq \epsilon N,$then $ A$ contains an $ \ell$-term arithmetic progression.


\begin{exercise}
Prove that Szemer\'edi's Theorem implies van der Waerden's Theorem.
\end{exercise}

History:

Theorem 4.10   $ \alpha(n)=o(3^n).$

On April 8, 2004, a major breakthrough was announced in a very old problem that has fascinated mathematicians for ages:

Theorem 4.11 (Green-Tao, 2004)   For all $ \ell,$ there is an $ \ell$-term arithmetic progression among prime numbers.

Ben Green and Terence Tao announced this result in their paper ``The Primes Contain Arbitrarily Long Arithmetic Progressions.'' The proof uses Szemerédi's Theorem and the ideas of Furstenberg's proof. It is posted on arXiv.


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Laszlo Babai 2004-06-27