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\begin{document}

\recitation{7}{October 1, 2004}

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%[\stamp]

\section*{The Chemist and The Maltster}

\begin{problem}

At Guiness brewery in the early 1900's, W. S. Gosset (a chemist) and
E. S. Beaven (a ``maltster'') were working to improve barley.  Gosset
and Beavan planned to grow several varieties of barley in a field and
compare the yields.  However, local variations in the fertility of the
field might skew the results.  Their solution was to divide the field
into many small plots and grow each crop in several different places.

Similar thinking led to the use of {\em Latin squares} in experiment
design.  A Latin square is an $n \times n$ array of numbers such that
each row and each column contains every number from $1$ to $n$.  For
example, here is a $4 \times 4$ Latin square:

{\Large
\[
\begin{array}{|c|c|c|c|}
\hline
1 & 2 & 3 & 4 \\ \hline
3 & 4 & 2 & 1 \\ \hline
2 & 1 & 4 & 3 \\ \hline
4 & 3 & 1 & 2 \\ \hline
\end{array}
\]
}

You can imagine that this array is an agricultural field where each
square is a small plot, and the number inside indicates the variety of
barley planted there.

There are some nice connections between Latin squares and graph
theory.  For example, see if you can construct a graph $G_n$ such that
there is a one-to-one correspondence between $n \times n$ Latin
squares and valid $n$-colorings of $G_n$.

(Recall that an \term{$n$-coloring} of a graph is a way of assigning
one of $n$ colors to each vertex so that vertices joined by an edge
are assigned different colors.)

\solution{Create a vertex in $G$ for each entry in the Latin Square.
Then connect each vertex to every other vertex in the same row and to
every other vertex in the same column.  Now color the graph with $n$
colors, each corresponding to a number between 1 and $n$.

Notice that every pair of vertices in the same row are connected, so
no two vertices in the same row can get the same color.  Similarly,
since every pair of vertices in the same column are connected, no two
vertices in the same column can get same color either.  These coloring
constraint match the constraints on Latin squares, so there is a
one-to-one correspondence between colorings of $G$ and $n \times n$
Latin squares.}

\end{problem}

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\instatements{\newpage}

\section*{Perfect Matching from Hall's Theorem}

\begin{problem}
The next question about Latin squares will require a corollary of
Hall's Theorem.  A bipartite graph is {\em regular} if every vertex on
the left has degree $c$ and every vertex on the right has degree $d$.
Prove the following:

\begin{corollary*}
A regular bipartite graph has a matching for the vertices on the left
if $c \geq d > 0$.
\end{corollary*}

For your reference, here is the statement of Hall's Theorem.

\textbox{

If $S$ is a set of vertices, then $N(S)$ is the set of neighboring
vertices; that is, $N(S)$ consists of all vertices that are adjacent
to a vertex in $S$, but not actually in $S$, as shown in the figure
below.

\mfigure{!}{2in}{rec7-matching}

Let $G = (L \cup R, E)$ be a bipartite graph such that every edge has one
endpoint in $L$ and one endpoint in $R$.  A \term{matching} for $L$ is a
subset of the edges such that exactly one edge is incident to each vertex
in $L$ and at most one edge is incident to each vertex in $R$.  The edges
shown in bold in the example above form such a matching.

\begin{theorem}[Hall's Theorem]
There is a matching for the $L$ vertices iff $\size{N(L')} \geq \size{L'}$
for every set $L' \subseteq L$.
\end{theorem}
}

\solution{We use Hall's Theorem.  Let $L$ denote the set of vertices on
the left of the regular bipartite graph, and $R$ denote the set of
vertices on the right.

Now let $L' \subseteq L$ be any subset of vertices on the left, and let
$R'$ be $N(L')$, namely, the set of all vertices on the right adjacent to
some vertex in $L'$.

Since every edge incident to a vertex in $L'$ is also incident to a vertex
in $R'$, the number of edges incident to vertices in $L'$ is at most as
large as the number of vertices incident to $R'$.  But the number of edges
incident to $L'$ is $c \cdot \size{L'}$ and the number of edges incident
to $R'$ is $d \cdot \size{R'}$.  Thus, we have:
\begin{eqnarray*}
c \cdot \size{L'} & \leq & d \cdot \size{R'}
\end{eqnarray*}
Since $c \geq d > 0$ by assumption, this implies $\size{L'} \leq
\size{R'}$.  Thus, the desired condition holds.}
\end{problem}

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\instatements{\newpage}

\section*{Completing Latin Squares}

\begin{problem}
Now we return to Quadra\footnote{Get it?  That's Latin for
``squares''.  Latin -- squares?  HAHAHA! Haha...  Heh.  Okay, just
shut up and do the problem.}.

\bparts

\ppart Suppose your teammate wrote only three lines of a $5 \times 5$
Latin square before deciding to chuck it all and become a maltster:

{\Large
\[
\begin{array}{|c|c|c|c|c|}
\hline
2 & 4 & 5 & 3 & 1 \\ \hline
4 & 1 & 3 & 2 & 5 \\ \hline
3 & 2 & 1 & 5 & 4 \\ \hline
  &   &   &   &   \\ \hline
  &   &   &   &   \\ \hline
\end{array}
\]
}

Can you fill in the last two lines to extend this ``Latin rectangle''
to a complete Latin square?

\solution{
Here is one possible solution:

{\Large
\[
\begin{array}{|c|c|c|c|c|}
\hline
2 & 4 & 5 & 3 & 1 \\ \hline
4 & 1 & 3 & 2 & 5 \\ \hline
3 & 2 & 1 & 5 & 4 \\ \hline
1 & 5 & 2 & 4 & 3 \\ \hline
5 & 3 & 4 & 1 & 2 \\ \hline
\end{array}
\]
}
}

\ppart Show that filling in the next row of a Latin rectangle is
equivalent to finding a matching in some bipartite graph.

\solution[\vspace{2in}]{Construct a bipartite graph as follows.  The
vertices on the left are the columns of the Latin rectangle, and the
vertices on the right are the numbers $1$ to $n$.  Put an edge between
a column and a number if the number has {\em not yet appeared} in the
column.  Thus, a matching in this graph would associate each column
with a distinct number that has not yet appeared in that column.
These numbers would form the next row of the Latin rectangle.}

\ppart Prove that a matching must exist in this bipartite graph and,
consequently, a Latin rectangle can always be extended to a Latin
square.

\solution{First, we show that the bipartite graph described above has
a matching.  Each column-vertex on the left has degree $n - k$ and
each number-vertex on the right has degree $n - k$ as well.
Therefore, a matching for the columns exists by the corollary in the
preceding problem.  This implies that we can add rows to the Latin
rectangle by the procedure described above as long as $k < n$.  At
that point, we have a Latin square.}

\eparts

\end{problem}

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\end{document}
