\documentclass[11pt]{article}
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\begin{document}
\assigntitle{31}{Graph Theory II}
Last week, you learned the basic definition of a \term{graph}. But
there are several ways that this basic definition can be usefully
generalized in order to model various sorts of structures. This week
we'll look at several of these generalizations.
\section{Weighted graphs and minimum spanning trees}
\label{sec:weighted}
\topic{weighted graphs}
A \term{weighted}, or \term{labelled} graph is one where each edge has
some kind of \term{weight} or \term{label} associated with it. This
is useful when you want to represent not only the fact that things are
connected in certain ways, but that there is some sort of cost,
length, capacity, or other sort of measurement associated with each
connection. Let's look at an example.
\topic{oooooOOOOOooo}
It is the year 2055. Humans have established several colonies on Mars.
\pref{fig:mars-colonies} shows a graph representing the colonies
(labeled $a$ through $h$) and the roads between them. (There used to
be a road between $b$ and $c$, but as you can see it has been taken
over by Martians.)
\begin{problem}
Is it possible to walk along every road exactly once without
retracing your steps or walking along the same road twice? If not,
can you do it if you leave out one road? If you leave out two roads?
\end{problem}
\begin{problem}
Is it possible to walk along the roads in such a way that you visit
every colony exactly once?
\end{problem}
\topic{zoom!}
The humans decide to link some of the colonies with special high-speed
trains, which need to be built along existing roads. However,
high-speed trains are really expensive. The number next to each road
shows the cost of turning that road into a high-speed train (in
zillions of zlqrts, the currency used by the colonists).
\diagramg{pdf}{mars-colonies}{Mars colonies, with the cost of turning
each road into a high-speed train}
\begin{problem}
Suppose the colonists want to link $a$ and $f$ with high-speed
trains. Which roads should be turned into trains to link $a$ and
$f$ at minimum cost? What is the minimum cost?
\end{problem}
As it turns out, the humans decide they want to link \emph{all} the
colonies with high-speed trains; that is, it should be possible to get
from any colony to any other just by riding trains. However, building
a high-speed train along \emph{every} road would cost too much money,
not to mention that it would be unnecessary. For example, there would
be no need to build a high-speed train from, say, $a$ to $d$ if there
were already high-speed trains from $a$ to $c$ and $c$ to $d$: anyone
wanting to get from $a$ to $d$ could first take the train to $c$ and
from there to $d$.
\topic{minimum spanning trees}
So, the problem is this: which roads should be turned into trains so
that all the colonies are connected for the \emph{least total amount
of money}? Such a subset of edges in a weighted graph which
connects all the vertices and has minimum total weight is called a
\term{minimum spanning tree}.
\topic{Kruskal's algorithm}
It turns out that there is a simple algorithm for finding minimum
spanning trees in a weighted graph, called \term{Kruskal's
algorithm}. It works by building up the minimal spanning tree one
edge at a time:
\begin{itemize}
\item Look at the edges in order of weight, from smallest to largest.
\item For each edge, if it would form a cycle with other edges already
in the minimum spanning tree, throw it out. Otherwise, add it to
the tree.
\end{itemize}
Let's see how this works, using \pref{fig:mars-colonies} as an
example. First, we look at the edge with the smallest weight, which
is the edge from $g$ to $h$ (weight $2$). It doesn't make a cycle
with any of the edges already in the minimum spanning tree (there
\emph{aren't} any other edges in the minimum spanning tree yet!) so we
add it to the minimum spanning tree, which I've indicated in
\pref{fig:mars-colonies1} by coloring it red.
\diagramg{pdf}{mars-colonies1}{The first step of Kruskal's
algorithm. The minimum spanning tree so far is colored red.}
Now we look at the edge with the next smallest weight, which is the
edge from $g$ to $f$ with weight $4$. Again, there are no cycles in
sight yet, so we add it to the minimum spanning tree
(\pref{fig:mars-colonies2}).
\diagramg{pdf}{mars-colonies2}{The second step of Kruskal's algorithm}
The next smallest edge is the one from $f$ to $h$, with weight
$5$. (Actually, there are two edges with weight $5$; we are free to
consider them in either order we like.) However, edges $fg$ and $gh$
are already in the minimum spanning tree, so adding $fh$ would create
the cycle $fgh$. We don't need to build a train from $f$ to $h$ if
there are already trains from $f$ to $g$ and $g$ to $h$! So we throw
away edge $fh$ (shown in \pref{fig:mars-colonies3} by drawing it with
a dotted line).
\diagramg{pdf}{mars-colonies3}{The third step of Kruskal's algorithm}
I'll show one more step: the next smallest edge is $ac$; it wouldn't
form any cycles, so we add it (\pref{fig:mars-colonies4}).
\diagramg{pdf}{mars-colonies4}{The fourth step of Kruskal's algorithm}
\begin{problem}
Complete Kruskal's algorithm for the given graph. Which other roads
should be turned into high-speed trains? What is the total
(minimum) cost of connecting all the colonies with trains?
\end{problem}
\begin{problem}
\topic{the claw has chosen!}
It turns out that the Martians are actually friendly, and offer to
help build a high-speed train from $b$ to $c$. With the aliens'
help, this would only cost one zillion zlqrts. How does the minimum
spanning tree change? What is the new minimum total cost to connect
all the colonies?
\end{problem}
\section{Directed graphs}
\label{sec:directed}
\topic{directional edges}
In all the graphs we've seen so far, the edges have been
\term{undirected}---that is, they just connect two vertices, with no
inherent directionality. For some applications, however, the
\term{direction} of the edges matters. For example, suppose we are
making a graph to represent the streets in downtown DC, with a vertex
for each intersection. As anyone who has driven around in downtown DC
knows, any map of downtown DC is useless if it doesn't show you
\term{which roads are one-way}! So the \emph{direction} of the
connections matters.
In a \term{directed graph}, each edge has a \term{direction}, which we
indicate in drawings of the graph by putting an arrowhead on one end.
Note that \emph{every} edge has a direction---if we want to indicate a
two-way connection, we just draw two edges, one going in each direction.
\diagramg{pdf}{choose}{The plot thickens}
\topic{write your own adventure}
\pref{fig:choose} shows the plot outline of a choose-your-own
adventure story, where $1$ is the beginning of the story, and $8$ and
$9$ are the endings.
\begin{problem}
Please write a story corresponding to the given plot outline. Here
is an example of the beginning of such a story:
\begin{enumerate}
\item It is a lazy Saturday afternoon, and you are bored. You
suddenly have a great idea! If you decide to skateboard down the
middle of the highway wearing a pretzel costume, go to 2. If you
decide to fill your swimming pool with blue Gatorade, go to 3.
\end{enumerate}
This corresponds to the vertex labeled $1$ on the graph since there
are edges going from $1$ to $2$ and $3$. You can use this as \#1 if
you want, or you can make up your own.
\end{problem}
\topic{DAGs}
The graph in \pref{fig:choose} is called a \term{directed acyclic
graph}, or DAG, since there are no cycles in the graph: there is
no way to start at some node, follow edges in the proper directions,
and end up back at the node where you started. (Obviously there
\emph{would} be cycles if the edges were undirected, but you are not
allowed to go backwards along the edges.)
\begin{problem}
\topic{topological sort}
List the vertices in order so that whenever there is an edge from
vertex $m$ to vertex $n$, $m$ comes before $n$ in the list of
vertices. For example, $1$, $2$, $3$, $6$, \dots would \emph{not}
be a correct order, since there is an edge from $6$ to $2$, so $6$
has to come before $2$ in the list.
(Note: such a list is called a \term{topological sort}, and it turns
out that it is always possible to topologically sort a DAG.)
\end{problem}
\begin{problem}
How many different possible stories are there? In other words, how
many different paths are there from vertex $1$ to either vertex $8$
or vertex $9$? (\emph{Hint}: consider the vertices in order of your
answer to the previous problem; for each vertex, figure out how many
paths there are to it from vertex $1$.)
\end{problem}
\subsection{Directed graphs and degree}
\label{sec:dir-and-degree}
\topic{in- and outdegree}
In an undirected graph, the \term{degree} of each vertex was just the
number of edges connected to it. But in a directed graph, this isn't
enough: it matters whether each edge is coming \emph{into} or going
\emph{out of} the vertex. So, in a directed graph, we talk about two
different degrees: the \term{indegree} and the \term{outdegree}. The
indegree of a vertex is the number of edges pointing \emph{towards}
the vertex (coming \emph{in}) and the outdegree is the number of edges
pointing \emph{away from} the vertex (going \emph{out}).
\begin{problem}
List the indegree and outdegree of each vertex in the graph in
\pref{fig:choose}.
\end{problem}
\begin{problem}
\topic{Eulerian cycles in directed graphs}
A directed graph has an Eulerian cycle if every vertex has equal
indegree and outdegree---every time you come in, you have to be able
to go out. (Note, the indegree and outdegree of each individual
vertex have to be the same as each other; this doesn't mean they
have to be equal to the indegree and outdegree of other vertices.)
Draw a directed graph which does have an Eulerian cycle when
considered as an undirected graph (that is, by ignoring all the
arrows) but does \emph{not} have an Eulerian cycle when considered
as a directed graph.
\end{problem}
\section{Self-loops}
\label{sec:self-loops}
Another way we can classify different types of graphs is by asking
whether \term{self-loops} are allowed. A self-loop is an edge from a
vertex to itself. For example, the graph in \pref{fig:self-loops} has
two edges, one from $A$ to $B$ and the other from $B$ to itself.
\diagramg{pdf}{self-loops}{A graph with a self-loop}
\topic{state transition diagrams}
When might self-loops be useful? One common situation is if a graph
(often a directed graph) is used to represent transitions in some sort
of system; a self-loop would indicate that the system stays in the
same state. For example, the directed, weighted graph in
\pref{fig:nuclear} is a (completely made up) model of a nuclear
reactor.
\diagramg{pdf}{nuclear}{The state model of a nuclear reactor}
State $A$ represents normal operation; state $B$ is a danger state in
which there is excess energy in the system; and state !! represents a
meltdown/explosion/general catastrophic failure. The edges represent
transitions between the states, and are labeled with the probability
of each transition in any given minute. For example, if the reactor
is in state $A$, then nine out of ten times, on average, it will still
be in state $A$ after one minute (since the self-loop from $A$ to $A$
is labeled with $0.9$); one time out of ten, on average, it will be in
state $B$ after one minute. From state $B$, after one minute there is
a $90\%$ chance the reactor will be back in state $A$, a $9.9\%$
chance it will still be in state $B$, and an $0.1\%$ chance that it
will blow up.
\begin{problem}
If the reactor is in state $A$, what is the probability that it will
blow up after\dots
\begin{subproblems}
\item exactly one minute?
\item exactly two minutes?
\item exactly three minutes?
\end{subproblems}
\emph{Hint:} part (c) is a bit tricky. Start by finding all the
paths of length $3$ from $A$ to !!.
\end{problem}
The information in \pref{fig:nuclear} doesn't have to be represented
by a graph---for example, it could be represented in a table as
well---but a graph is a nice visual way to represent the information.
\end{document}