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%\renewcommand{\coursecopyrightnames}{Prof.~Albert~R.~Meyer}

\begin{document} 
\inclassproblems{5, Fri.}                    % !! start the paper

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\begin{problem}
Suppose that one domino can cover exactly two squares on a chessboard,
either vertically or horizontally.

\bparts

\ppart Can you tile an $8 \times 8$ chessboard with 32 dominos?

\mfigure{!}{2in}{chess1}

\solution{Yes.  Place 4 vertical dominos in each column.}

\ppart Can you tile an $8 \times 8$ chessboard with 31 dominos if
opposite corners are removed?

\mfigure{!}{2in}{chess2}

\solution{No!  Opposing corners are the same color.  Therefore,
removing opposite corners leaves an unequal number of white and black
squares.  Since every domino covers one black square and one white
square, no tiling is possible.}

\instatements{\newpage}

\problempart Now suppose that an assortment of squares are removed
from a chessboard.  An example is shown below.

\mfigure{!}{2in}{chess3}

Given a truncated chessboard, show how to construct a bipartite graph
$G$ that has a perfect matching if and only if the chessboard can be
tiled with dominos.

\solution[\vspace{2in}]{Create a vertex for every white square and a
vertex for every black square.  Put an edge between squares that share
an edge.  (This graph is bipartite, since the coloring of the squares
defines a valid 2-coloring of the vertices.)

If a perfect matching exists in this graph, then a tiling exists: put
a domino over each pair of matched vertices.  On the other hand, if a
tiling exists, then a perfect matching exists: match squares covered
by the same domino.}

\ppart Based on this construction and Hall's theorem, can you state a
necessary and sufficient condition for a truncated chessboard to be
tilable with dominos?  Try not to mention graphs or matchings!

\solution{A board can be tiled with dominos if and only if every set
of white squares is adjacent to at least as many black squares and
vice versa.}

\end{problemparts}

\end{problem}

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\instatements{\newpage}

\begin{problem}
At Guiness brewery in the eary 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 $G_n$.

\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}

\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*}

(You may also find this corollary useful on the problem set!)

\solution{We use Hall's Theorem.  Let $L$ denote the set of vertices
on the left and $R$ denote the set of vertices on the right.  We must
show that every set of vertices $L' \subseteq L$ on the left is
adjacent to a set of vertices $R' \subseteq R$ such that $\card{L'}
\leq \card{R'}$.  If this condition holds, then Hall's theorem implies
that a matching for the vertices on the left exists.

Let $L' \subseteq L$ be a subset of vertices on the left, and let $R'
\subseteq R$ be the set of all vertices on the right adjacent to some
vertex in $L$.  Thus, every edge incident to a vertex in $L'$ is also
incident to a vertex in $R'$.  Therefore, the sum of the degrees of
the vertices in $L'$ is at most the sum of the degrees of vertices in
$R'$.  Thus, we have:

\begin{eqnarray*}
c \cdot \card{L'} & \leq & d \cdot \card{R'}
\end{eqnarray*}

Since $c \geq d > 0$ by assumption, this implies $\card{L'} \leq
\card{R'}$.  Thus, the desired condition holds.}
\end{problem}

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

\newtheorem*{conjecture*}{Conjecture}

\begin{problem}
Some simple questions in graph theory still await resolution by an
ingenious 6.042 student!  Prove or disprove:

\begin{conjecture*}[Erd\"{o}s-Gyarfas]
Every graph with mimimum degree at least 3 has a cycle whose length is
a power of 2.
\end{conjecture*}
\end{problem}

Now that doesn't seem so hard...

\solution{This is an open problem.  Reportedly, Paul Erd\"{o}s
offered \$100 for a proof and \$50 for a counterexample.}

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\end{document}

