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\begin{document}
\title{Graph Theory Notes \\\bigskip\textrm{\large\today}}
\author{H. A. Kierstead}
\maketitle
\chapter{Introduction}
This is an on going project. Before each lecture there will likely be changes to
the material not yet covered, especially before each lecture. There may also be changes
immediately after a lecture based on class questions, etc. Later corrections to previous
material will be marked in red. For these reasons, you should continuously download
the current version.
Use at your own risk. There are bound to be typos. Corrections are appreciated, and
rewarded with small amounts of extra credit, especially when they indicate mathematical
understanding. But you are also encouraged to ask questions in emails about things
you do not understand. I will try to answer questions quickly, but if they are complicated,
or there are many, I may need to wait until the next class.
These notes are meant to enhance, not replace, the lectures and class discussions.
They are intended to be concise records of proofs, to free students from the need
to take careful notes during class. However the motivation for these proofs is left
to the lectures and discussions in class.
\section{Graphs}
Formally a \emph{graph} is an ordered pair $G=(V,E)$ where $E$ is an irreflexive,
symmetric, binary relation on $V$. Since $E$ is symmetric there is no need to keep
track of the order of pairs $(x,y)\in E$; since it is irreflexive there are no ordered
singletons $(x,x)$ in $E$. This leads to a more intuitive formulation. We take
$E$ to be a set of unordered pairs of elements from $V$. Elements of $V$ are called
\emph{vertices}; elements of $E$ are called \emph{edges}. If $x,y\in V$ are vertices
and $\{x,y\}\in E$ is an edge we usually (but not always) denote $\{x,y\}$ by the
shorthand notation $xy$. So $xy=\{x,y\}=\{y,x\}=yx$. The vertices $x$ and $y$
are called \emph{ends}, or \emph{endpoints}, of the edge $xy$.\emph{ }The ends $x$
and $y$ of an edge $xy$ are said to be \emph{adjacent} and the end $x$ is said
to be \emph{incident} to the edge $xy$. We also say that $x$ and $y$ are \emph{joined}
or \emph{linked} (not connected) by the edge $xy$. Two edges are said to be \emph{adjacent}
if they have a common end. In this course, all graphs have a finite number of vertices,
unless it is explicitly stated that they have infinitely many. Graphs are illustrated
by drawing dots for vertices and joining adjacent vertices by lines or curves.
Our definition of graph is what West calls a \emph{simple graph}. Most of the time
we will only be interested in simple graphs, and so we begin with the simplest definition.
When necessary, we will introduce the more complicated notions of \emph{directed
graphs }and \emph{multigraphs, }but here is a quick hint\emph{. }A directed graph
$G=(V,E)$ is any binary relation (not necessarily irreflexive or symmetric) on $V$.
In other words $E$ is any set of \emph{ordered} pairs of vertices. A \emph{multigraph}
is obtained by letting\emph{ }by letting $E$ be a multiset; then two vertices can
have more than one edge between them. A \emph{hypergraph} is obtained by letting
$E$ be a set of subsets of $V$, where the elements of $E$ can have any size. If
they all have size $k$ then we get a $k$-\emph{uniform hypergraph}, also called
a $k$-graph. So ordinary graphs are $2$-uniform hypergraphs.
The study of graph theory involves a huge number of of parameters---see the front
and back inside covers of West or the last two pages of Diestel. This can be quite
daunting. My strategy is to introduce these parameters as they are needed. Please
feel free to interrupt lectures to be reminded of their meanings. Most of the time
my notation will agree with West, and I will try to emphasize differences. Next we
introduce some very basic notation.
Given a graph $G$, $V(G)$ denotes the set of vertices of $G$ and $E(G)$ denotes
the set of edges of $G$. Set $|G|:=|V(G)|$ and $\left\Vert G\right\Vert :=|E(G)|$;
this is not standard, and instead the book uses $v(G)=|V(G)|$ and $e(G)=|E(G)|$.
Suppose $v\in V(G)$ is a vertex of $G$. Define
\begin{align*}
N_{G}(v) & =\{w\in V(G):vw\in E(G)\}; & E_{G}(v) & =\{e\in E(G):v\mbox{ is an end of }e\}.
\end{align*}
The set $N_{G}(v)$ is called the (open) \emph{neighborhood} of $v$, and its elements
are called \emph{neighbors} of $v$. So a vertex $w$ is a neighbor of $v$ iff it
is adjacent to $v$. When there is no confusion with other graphs the subscript $G$
is often dropped. The closed neighborhood of $v$ is $N[v]=N(v)\cup\{v\}$---we dropped
the subscript. The set $E_{G}(v)$ is the set of edges incident to $v$; again, we
may drop the subscript $G$. The text does not provide notation for this set. For
simple graphs $|N(v)|=|E(v)|$. However for multigraphs this may not hold, since
two vertices might be joined by several edges. With this in mind, define the degree
of a vertex $v$ to be $d_{G}(v):=|E_{G}(v)|$, but note that for simple graphs $d_{G}(v)=|N_{G}(v)|$.
A graph is $k$\emph{-regular} if every vertex has degree $k$.
We use the following set theoretic notation. The sets of natural numbers, integers
and positive integers are denoted, respectively, by $\mathbb{N},$ $\mathbb{Z}$
and $\mathbb{Z}^{+}$. \textcolor{red}{For integers $a$ and $b$ let $\{a,\dots,b\}$
indicate the set $\{i\in\mathbb{N}:a\leq i\leq b\}$. Then $\{0,\dots,-1\}=\emptyset$.}
For $n\in\mathbb{N}$ set $[n]:=\{1,\dots,n\}$; in particular $[0]=\emptyset$.
For a set $X$ and an element $y$, set $X+y:=X\cup\{y\}$ and $X-y:=X\smallsetminus\{y\}$.
Finally, $\binom{X}{n}$ is the set of all $n$-element subsets of $X$.
\textcolor{red}{In this course all graphs $G$ satisfy $1\leq|G|<\infty$, unless
a specific exception is stated.}
\section{Proofs by Mathematical Induction}
Most proofs in graph theory involve mathematical induction, or at least the Least
Element Axiom. Here we quickly review this technique. Also see the discussion in
West on pages 19--20, and especially the \emph{induction trap} on page 42.
Here we present mathematical induction in terms of the Least Element Axion (LEA).
First we give a careful definition of ``least element''.
\begin{defn}
Let $B\subseteq\mathbb{N}$ be a set of natural numbers. A number $l\in\mathbb{N}$
is a least element of $B$ if
\begin{enumerate}
\item [(L1)]\label{L1} $\{0,\dots,l-1\}\subseteq\mathbb{N}\smallsetminus B$, and
\item [(L2)]\label{L2}$l\in B$.
\end{enumerate}
\end{defn}
The following axiom (LEA) is fundamental.
\begin{namedthm}
[Axiom][LEA]Every nonempty set of natural numbers has a least element.
\end{namedthm}
Consider a set of ``good''\emph{ }natural numbers $S$, and let $B=\mathbb{N}\smallsetminus S$
be the set of ``bad'' numbers. We would like to prove that every natural number
is ``good'', i.e., $S=\mathbb{N}$. Here is a way to organize the argument.
\begin{thm}[Principle of Induction]
\label{PMI-th}Suppose $S\subseteq\mathbb{N}$. Then $S=\mathbb{N}$, if
\begin{equation}
\forall n\in\mathbb{N}~(\{0,\dots,n-1\}\subseteq S\rightarrow n\in S).\label{induction}
\end{equation}
\end{thm}
\begin{proof}
Suppose $S\subseteq\mathbb{N}$ satisfies \eqref{induction}. As $S\subseteq\mathbb{N}$
we only need prove $\mathbb{N}\subseteq S$. Arguing by contradiction, assume $\mathbb{N}\nsubseteq S$.
Then $B:=\mathbb{N}\smallsetminus S\ne\emptyset$. By LEA, $B$ has a least element
$l$. By (L1) applied to $l$, $\{0,\dots,l-1\}\subseteq\mathbb{N}\smallsetminus B=S$.
By \eqref{induction}, $l\in S$, so $l\notin\mathbb{N}\smallsetminus S=B$. This
contradicts (L2).
\end{proof}
Using the Principle of Induction to prove that $S=\mathbb{N}$, we prove \eqref{induction}.
For this we consider \emph{any} natural number $n$. If $\{0,\dots,n-1\}\nsubseteq S$
then we are done, so we suppose $\{0,\dots,n-1\}\subseteq S$. We use this ``induction
hypothesis'' to prove that $n\in S.$ If $n=0$ then there is no natural number
$kf(G')$, that \textbf{also satisfies}
$P$; you use the induction hypothesis to claim that $G'$ satisfies $Q$; finally,
you use that $G'$ satisfies $Q$ to prove that $G$ satisfies $Q$. \textbf{You
do not start with a small graph, and then make it bigger!}
\section{Ramsey's Theorem for Graphs}
Ramsey's Theorem is an important generalization of the Pigeonhole Principle. Here
we only consider its simplest version applied to graphs; the general version is a
statement about $k$-uniform hypergraphs. In the past it was presented as part of
MAT 415, but it has been moved to MAT 416 because its presentation benefits from
the language of graph theory.\emph{ }
Let $G=(V,E)$ be a graph, and suppose $X\subseteq V$. The set $X$ is a \emph{clique
in $G$} if $xy\in E$\emph{ }for all distinct vertices $x,y\in X$. It is an \emph{independent
set}, or \emph{coclique}, in $G$ if $xy\notin E$ for all vertices $x,y\in X$.
A clique (coclique) $X$ is a $b$-clique ($b$-coclique) if $|X|=b$. Let\emph{
$\omega(G):=\max\{|X|:X$ is a clique in $G\}$, and $\alpha(G):=\max\{|X|:X$ is
a coclique in $G\}$.}
A graph\emph{ $H$ }is a \emph{subgraph} of\emph{ $G$, }denoted $H\subseteq G$,
if $V(H)\subseteq V(G)$ and $E(H)\subseteq E(G)$. It is an \emph{induced subgraph}
of $G$ if $H\subseteq G$ and \emph{$E(H)=\{xy\in E(G):x\in V(H)\mbox{ and }y\in V(H)\}$.
}\textcolor{black}{For $X\subseteq V$, $G[X]$ is the induced subgraph of $G$ that
has vertex set $X$.}\textcolor{red}{{} }The\emph{ complement }of $G$ is the graph,
$\overline{G}:=(V(G),\overline{E}(G))$, where $\overline{E}(G):=\binom{V(G)}{2}\smallsetminus E(G)$.
\begin{thm}[Ramsey's Theorem 8.3.7,11]
\label{Ramsey}For all graphs $G$ and $a,b\in\mathbb{Z}^{+}$, if $|G|\geq2^{a+b-2}$
then $\omega(G)\geq a$ or $\alpha(G)\geq b$.\end{thm}
\begin{proof}
Argue by induction on $n=a+b$. (That is, let $S$ be the set of natural numbers
$n$ such that for all positive integers $a,b$ if $n=a+b$, and $G$ is a graph
with $|G|\geq2^{a+b-2}$ then $\omega(G)\geq a$ or $\alpha(G)\geq b$. Show that
for all $n\in\mathbb{N}$ if $\{0,\dots,n-1\}\subseteq S$ then $n\in S$.) Consider
any $n=a+b$ with $a,b\in\mathbb{Z}^{+}$, and any graph $G$ with $|G|\geq2^{a+b-2}$.
\noindent \emph{Base step:} $\min\{a,b\}=1$. Since $|G|\geq1$, $G$ has a vertex
$v$. Since $\{v\}$ is both a clique and an independent set, both $\omega(G)\geq1$
and $\alpha(G)\geq1$. So we are done regardless of whether $a=1$ or $b=1$.
\noindent \emph{Induction Step:}\textbf{ $\min\{a,b\}\geq2$} (so $a-1,b-1\in\mathbb{Z}^{+})$.
(We assume the induction hypothesis: the theorem holds for all $a',b'\in\mathbb{Z}^{+}$
with $a'+b'1$ then $G$ is hamiltonian.\end{cor}
\begin{proof}
By Theorem~\ref{DO}, it suffices to prove $G$ is connected. Consider $x,y\in V(G)$.
Then
\begin{gather*}
|G|\geq|N[x]\cup N[y]|=|N[x]|+|N[y]|-|N[x]\cap N[y]|\geq|G|+2-|N[x]\cap N[y]|\\
|N[x]\cap N[y]|\geq2.
\end{gather*}
So $x$ is connected to $y$ by a path of length at most $2$.
\end{proof}
Here is a weaker, but slightly less local condition that ensures a graph is hamiltonian.
\begin{cor}[Ore's Theorem 1960 7.2.9]
\label{Ore}If $G$ is a graph with $|G|\geq3$ and $d(x)+d(y)\geq|G|$ for all
distinct nonadjacent vertices $x$ and $y$ then $G$ is hamiltonian.\end{cor}
\begin{proof}
As in the proof of Corollary~\ref{DiracC}, $G$ is connected, since for distinct
vertices $x$ and $y$ either $xy\in E$ or $|N[x]|+|N[y]|\geq|G|+2$. So we are
done by Theorem~\ref{DO}.
\end{proof}
Pósa (when he was in high school) posed a conjecture extending Dirac's Theorem:
\begin{conjecture}[Pósa 1963]
\label{Posa-con}If $\delta(G)\geq\frac{2}{3}|G|$ then $G$ contains the square
(2-power) of a hamiltonian cycle.
\end{conjecture}
The next Theorem answers a related question.
\begin{thm}[Fan \& Kierstead 1996 \cite{FK-SP}]
If $\delta(G)\geq\frac{2|G|-1}{3}$ then $G$ contains the square of a hamiltonian
path. The degree condition is best possible.
\end{thm}
The next conjecture generalizes Corollary~\ref{DiracC} and Conjecture~\ref{Posa-con}.
\begin{conjecture}[Seymour 1974]
\label{SeyK-pow-con}If $\delta(G)\geq\frac{k}{k+1}|G|$ then $G$ contains the
$k$-th power of a hamiltonian cycle.
\end{conjecture}
Seymour's Conjecture was proved for sufficiently large graphs (even more vertices
than the number of electrons in the known universe when $k\geq2$).
\begin{thm}[Komlós, Sárközy \& Szemerédi 1998 \cite{KSS-Sey}]
For every integer $k$ there exists an integer $n$ such that Conjecture~\ref{SeyK-pow-con}
is true for graphs $G$ with $|G|\geq n$.
\end{thm}
The next theorem improves the bound on $n$ when $k=2$.
\begin{thm}[Ch\^au, DeBiasio \& Kierstead 2011 \cite{100}]
Conjecture~\ref{Posa-con} is true for graphs $G$ with $|G|\geq2\times10^{8}$.
\end{thm}
\begin{x}Prove that $K_{a,a-1}$ is not hamiltonian for any integer $a>1$. More
generally, prove that if $\alpha(G)>\frac{1}{2}|G|$ then G is not hamiltonian. Determine
$\delta(K_{a,a-1})$.\end{x}
\begin{x}For disjoint sets $A,B,\{v\}$, let $G=\overline{K}(A,B)+K(v,A\cup B)$.
Prove that $G$ is not hamiltonian. Determine $\delta(G)$ when $|A|=|B|$.\end{x}
\begin{x}Let $P=v_{1}\dots v_{t}\subseteq G$ be a maximum path and $x\in V(G-P)$.
Prove that $\left\Vert \{v_{t},x\},P\right\Vert \leq t-1$.
\end{x}
\begin{x}Prove that if $\delta(G)\geq\frac{|G|-1}{2}$ then $G$ has a hamiltonian
path.\end{x}
\begin{x} (+) Let $G$ be an $X,Y$-bigraph with $|X|=|Y|=k$ and $\delta(G)\geq\frac{k+1}{2}{\color{red}{\normalcolor \geq2}}$.
Prove that $G$ contains a hamiltonian cycle. {[}Hint: First prove that if $G$ contains
a maximal path $P=P_{t}$ then $G[P]$ contains a cycle of length at least $2\lfloor\frac{t}{2}\rfloor$.
Then show that $G$ contains a hamiltonian path.{]} For all $k$ give an example
to show that the bound on the minimum degree cannot be lowered. \end{x}
\section{Even graphs and Euler's Theorem}
Sometimes when proving a statement by induction it is easier to prove a stronger
statement. This phenomenon is called the \emph{inventors paradox.} The reason this
is possible is that while more must be proved, the induction hypothesis provides
more to base an argument on. In this section we see an elementary example of this.
It is easier to prove our result for multigraphs. Proposition~\ref{trail} is a
minor example of this; we will see more serious examples later.
\begin{defn}
A multigraph is \emph{eulerian} if it has a closed trail containing all edges. (Note
that $T=v$ is closed, since its only vertex is its first and its last.) Such a trail
is said to be an eulerian trail. A multigraph is \emph{even if every vertex has even
degree}.
\end{defn}
For $H\subseteq G$ and $v\in V(G)\smallsetminus V(H)$, set $d_{H}(v):=0$. If $T$
is a trail in a graph $G$ then let $d_{T}(v)=d_{G(T)}(v)$.
\begin{fact}
\noindent If $H$ and $G$ are even graphs with $H\subseteq G$ then $H':=G-E(H)$
is even.\end{fact}
\begin{proof}
Since $G$ and $H$ are even, every $v\in V(G)$ satisfies
\[
d_{H'}(v)=d_{G}(v)-d_{H}(v)\equiv0\mod2.
\]
\end{proof}
\begin{prop}
\label{trail}Let $T=v_{1}...v_{n}$ be a trail in a multigraph $G=(V,E)$. Then
$d_{T}(v)$ is even for every vertex $v$, except that if $T$ is open then $d_{T}(v_{1})$
and $d(v_{n})$ are odd.\end{prop}
\begin{proof}
Let $T':=T+v_{n}v_{1}$ and $G'=G+v_{n}v_{1}$ ($v_{n}v_{1}$ may be a loop of multiple
edge). Then $T'$ is closed. It suffices to show that $d_{T'}(v_{i})$ is even for
all $v\in V$, since $d_{T}(v_{i})\not\equiv d_{T'}(v_{i})\mod2$ if and only if
$v_{i}\in\{v_{1},v_{n}\}$ and $v_{1}\ne v_{n}$. Argue by induction on $n$.
If $T'$ is a cycle, or $t=1$, then $d_{T'}(v_{i})\in\{0,2\}$ for every $v\in V(G)$.
Otherwise, there exist $1*0$ then $G$ contains a cycle.\end{cor}
\begin{proof}
Some component $H$ of $G$ contains an edge. Since $H$ is connected, $\delta(H)\geq1$.
Since $G$ is even this can be strengthened to $\delta(H)\geq2$. So by Lemma~\ref{delta2},
$H\subseteq G$ contains a cycle.
\end{proof}
\begin{proof}
[Second proof of Theorem~\ref{Eulerian} (Sufficiency)] Suppose $G$ is even and
has at most one nontrivial component $G'$. We argue by induction on $\left\Vert G'\right\Vert $.
If $G'$ is a cycle (or $\left\Vert G'\right\Vert =0$), then the cycle in its natural
order (or any vertex) is the Eulerian trail.
Otherwise, by Corollary~\ref{cycle}, $G'$ contains a cycle $C$. Let $H$ be a
nontrivial component of $G'-E(C)$ (maybe $H=G'-E(C)$), and set $H'=G'-E(H)$. Both
$H$ and $H'$ are even. Also $H'$ is connected, since all components of $G'-E(C)$
that are contained in $H'$ are connected to each other in $H'$ by edges of $C$.
Moreover, $\left\Vert H\right\Vert \leq\left\Vert G'\right\Vert -\left\Vert C\right\Vert <\left\Vert G'\right\Vert $
and $\left\Vert H'\right\Vert =\left\Vert G'\right\Vert -\left\Vert H\right\Vert <\left\Vert G'\right\Vert $.
So $H$ and $H'$ are nonempty even connected graphs with $\left\Vert G'\right\Vert =\left\Vert H\right\Vert +\left\Vert H'\right\Vert $.
By the induction hypothesis $H$ and $H'$ contain Eulerian trails $T$ and $T'$.
Moreover, T contains a vertex $v_{1}\in C$ and $T'$ contains all vertices of $C$.
Choose notation so that $T=v_{1}\dots v_{n}v_{1}$ and $T'=v_{1}u_{2}\dots u_{m}v_{1}$.
Then $v_{1}Tv_{n}v_{1}T'u_{m}v_{1}$ is an Eulerian trail in $G'$, and $G$.\end{proof}
\begin{thm}[1.2.33]
\label{EulPaths}A connected graph $G$ with exactly $q$ vertices of odd degree
decomposes into $\max\{1,\frac{q}{2}\}$ trails.\end{thm}
\begin{proof}
By Lemma~\ref{HandSh}, $q$ is even. Let $G^{+}$ be the result of adding a new
vertex $v^{+}$ to $G$ so that $N(v^{+})$ is the set of vertices with odd degree
in $G$. Since $q$ is even, and every $v\in V(G)$ satisfies $d_{G^{+}}(v)\equiv d_{G}(v)+1\mod2$
if and only if $d_{G}(v)$ is odd, $G^{+}$ is even. By Theorem~\ref{Eulerian},
$G^{+}$ has an Eulerian Trail $T$. Removing $v^{+}$ partitions $T$ into $\frac{q}{2}$
trails that decompose $G$.
\end{proof}
Alternatively, we could have proved Theorem~\ref{EulPaths} by adding $q$ edges
connecting disjoint pairs of odd degree vertices.
\chapter{Cut-vertices, -edges and trees}
\begin{defn}[1.2.12.]
A \emph{cut-vertex} is a vertex in a graph $G$ is a vertex such that $G-v$ has
more components than $G$. Similarly, a \emph{cut-edge} is an edge of $G$ such that
$G-e$ has more components than $G$.
Notice that a vertex $v$ is a cut-vertex of $G$ if and only if $H-v$ is neither
empty nor connected, where $H$ is the component of $G$ (maybe $H=G$) containing
$v$. Similarly an edge $e$ is a cut-edge of $G$ if and only if $H-e$ is not connected,
where $H$ is the component of $G$ containing $e$. Thus $G$ is not a cut-edge
if its ends are connected in $G-e$.\end{defn}
\begin{thm}[1.2.14.]
\label{cutedge}An edge $e=xy$ in $G$ is not a cut-edge iff it belongs to a cycle.\end{thm}
\begin{proof}
First suppose $e$ is not a cut edge. Then there exists an $x,y$-path $P$ in $G-e$.
So $xPyx$ is a cycle in $G$. Now suppose $e$ is on a cycle $C$ in $G$. Then
$x(C-e)y$ is a path connecting the ends of $e$, and so $e$ is not a cut-edge.\end{proof}
\begin{thm}
\label{n-c-v}The ends of a maximal path $P=x\dots y\subseteq G$ are not cut-vertices
of $G$.\end{thm}
\begin{proof}
Let $H$ be the component of $G$ containing $x$. Suppose $x$ is a cut-vertex.
Then $|H|\geq2$, $y\ne x$. and some $u\in V(H-x)$ is not connected to $y$. As
$H$ is connected, there is a $u,y$ walk $W$ in $H$. If $x\notin V(W)$ then set
$W'=W$; else the predecessor $v$ of $x$ on $W$ is in $N(x)\subseteq P-x$; set
$W'=uWvPy$. Anyway $W'$ is a $u,y$-walk, a contradiction.\end{proof}
\begin{defn}
A graph is \emph{acyclic }if it contains no cycle. Acyclic graphs are also called
\emph{forests}. A connected acyclic graph is called a \emph{tree}. A \emph{leaf}
is a vertex $v$ with $d(v)=1$. We say that a graph $G$ satisfies (A) if it is
acyclic, (C) if it is connected, and (E), if $|G|=\|G\|+1$. \end{defn}
\begin{lem}
\label{ELeaf}A graph $G$ with $\|G\|\geq1$ has at least two leaves if it satisfies
(A) or both (C) and (E).\end{lem}
\begin{proof}
First suppose that $G$ is acyclic. Let $P=v_{1}\dots v_{t}$ be a maximum path in
$G$. Since $G$ has an edge, $v_{1}\ne v_{t}$. Since $P$ is maximum and acyclic
$N(v_{1})=\{v_{2}\}$ and $N(v_{t})=\{v_{t-1}\}$. So $v_{1}$ and $v_{t}$ are distinct
leaves.
Now suppose that $G$ satisfies (C) and (E). Let $L$ be the set of leaves in $G$.
Since $G$ is connected and has an edge, $\delta(G)\geq1$. Since $G$ satisfies
(E),
\begin{eqnarray*}
2|G|-|L| & \leq & \sum_{v\in V(G)}d(v)=2\|G\|=_{(E)}2|G|-2\\
2 & \leq & |L|.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\qedhere
\end{eqnarray*}
\end{proof}
\begin{lem}
\label{leaf}Suppose $G$ is a graph with a leaf $l$ and $G'=G-l$. Then each condition
(A), (C), (E) is satisfied by $G$ iff it is satisfied by $G'$.\end{lem}
\begin{proof}
Suppose $G$ is acyclic. Since removing a vertex cannot create a cycle, $G'$ is
acyclic. Now suppose $G'$ is acyclic. Since every vertex in a cycle has degree 2,
adding a leaf $l$ cannot create a cycle, and so $G$ is acyclic.
Suppose $G'$ is connected. Since $l$ has a neighbor in $V(G'$), $G$ is connected.
Now suppose $G$ is connected. Since $d(l)=1$, there is a maximal path $P$ with
an end $l$. Thus $l$ is not a cut-vertex, and so $G'$ is connected.
Since $|G|=|G'|+1$ and $\left\Vert G\right\Vert =\left\Vert G'\right\Vert +1$,
$G$ satisfies (E) iff $G'$ does.\end{proof}
\begin{thm}[2.1.4]
\label{Tree}If a graph $G$ satisfies at least two of the conditions (A), (C),
and (E) then it satisfies all three.\end{thm}
\begin{proof}
Argue by induction on $|G|$.
\noindent \emph{Base Step:} $|G|=1$. By inspection, $G$ satisfies all of (A), (C)
and (E).
\noindent \emph{Induction Step:}\textbf{ $|G|\geq2$}. Since $G$ satisfies (C) or
(E), it has an edge. Since $G$ satisfies (A) or both (C) and (E), it has a leaf
$l$ by Lemma~\ref{ELeaf}. Let $G'=G-l$. By Lemma~\ref{leaf}, $G'$ satisfies
the same two conditions that $G$ does. By the induction hypothesis $G'$ satisfies
all three conditions. By Lemma~\ref{leaf}, $G$ does also.
\end{proof}
For emphasis, we state that Theorem~\ref{Tree} implies every tree $T$ satisfies
$|T|=\left\Vert T\right\Vert +1$.
\begin{thm}[2.1.4]
\label{cor:uniquepath}G is a tree iff there is exactly one path between any two
vertices.\end{thm}
\begin{proof}
Both statements imply $G$ is connected. So it suffices to show that if $G$ is connected,
then $G$ has a cycle $C$ if and only if there are vertices $x$ and $y$ and distinct
$x,y$-paths $P$ and $Q$. If $C$ is a cycle with edge $xy$ then let $P=xy$ and
$Q=x(C-xy)y.$ If $P$ and $Q$ are distinct $x,y$-paths with (say) $uv\notin E$
then $uP^{*}xQyP^{*}v$ is a $u,v$-walk in $G-uv$, so $uv$ is not a cut-edge;
by Theorem~\ref{cutedge}, $G$ has a cycle containing $xy$.
\end{proof}
A \emph{spanning tree} of a graph $G$ is a tree $T\subseteq G$ with $V(T)=V(G)$.
\begin{cor}[2.1.5]
\label{cTree}Let $T=(V,E)$ be a tree. Then
\begin{enumerate}
\item Removing an edge disconnects $T$.
\item Adding an edge $xy\notin E$ with $x,y\in V$ to $T$ creates a unique cycle.
\end{enumerate}
\end{cor}
\begin{proof}
(1) Since $T$ is acyclic, every edge is a cut edge by Theorem~\ref{cutedge}.
(2) By Theorem~\ref{cor:uniquepath} there is a \emph{unique} $x,y$-path $P\subseteq T$.
So $P+xy$ is a cycle in $T':=T+xy$. Consider any cycle $D\subseteq T'$. Since
$T$ is acyclic, $xy\in E(D)$. By the uniqueness of $P,$ $D-xy=P$; so $D=P+xy$.\end{proof}
\begin{cor}
\label{Maxtreee-cor}Let $G=(V,E)$ be a connected graph.
\begin{enumerate}
\item Every minimally connected spanning subgraph $F\subseteq G$ is acyclic.
\item Every maximally acyclic spanning subgraph $F\subseteq G$ is connected.
\item Every maximal subtree $F\subseteq G$ is spanning.
\end{enumerate}
\end{cor}
\begin{proof}
(1) Let $F$ be a minimal connected, spanning subgraph of $G$. Then every edge of
$F$ is a cut-edge. By Theorem~\ref{cutedge}, no edge of $F$ is on a cycle, so
$F$ is acyclic.
(2) Let $F$ be a maximal acyclic spanning subgraph of $G$. First consider any edge
$xy\in E$. If $xy\in E(F)$ then $x$ and $y$ are connected in $F$; else by maximality
there is a cycle \textcolor{red}{$C\subseteq F+xy$} with $xy\in E(C)$, and so $x$
and $y$ are connected by the path $x(C-xy)y\subseteq F$. Now consider any $u,v\in V$.
We show by induction on $t:=\dist_{G}(u,v)$ that $u$ is connected to $v$ in $F$.
We have already seen this if $t\leq1$. If $t\geq2$ then let $(u=)v_{1}\dots v_{t-1}v_{t}(=v)$
be a $u,v$-path in $G$. By induction $u$ is connected to $v_{t-1}$ and $v_{t-1}$
is connected to $v$ in $F$, so we are done.
(3) Let $F\subseteq G$ be a subtree with $|F|$ maximum. Consider any $x\in V(G)$.
Since $G$ is connected, there is an $x,F$-path $(x=)x_{1}\dots x_{t}\subseteq G$.
As $|F|$ is maximum, $t=1$, and $x\in V(F)$.\end{proof}
\begin{prop}[2.1.5]
\label{T-T'-P}Suppose $T$ and $T'$ are spanning trees of a graph $G$. Then for
every $e=ab\in E(T)\smallsetminus E(T')$
\begin{enumerate}
\item there exists $e'\in E(T')\smallsetminus E(T)$ such that $T-e+e'$ is a spanning
tree of $G$; and
\item there exists $e'\in E(T')\smallsetminus E(T)$ such that $T'+e-e'$ is a spanning
tree of $G$.
\end{enumerate}
\end{prop}
\begin{proof}
(1) Let $H=T\cup T'-e$. Then $H$ is a connected, spanning subgraph of $G$, $T^{-}:=T-e$
is acyclic, and using Theorem~\ref{Tree}, $|T^{-}|=|T|=\left\Vert T\right\Vert +1=\left\Vert T^{-}\right\Vert +2$
and $T^{-}$ is not a tree. By Corollary~\ref{Maxtreee-cor}(2), $T^{-}$ is not
maximally acyclic in $H$. So there is $e'\in E(H-T^{-})=E(T')$ with $T^{*}:=T^{-}+e'$
acyclic. As $|T^{*}|=\left\Vert T^{*}\right\Vert +1$, Theorem~\ref{Tree} implies
$T^{*}=T-e+e'$ is a tree, spanning $G$.
(2) By Corollary~\ref{cTree}(2), $T'+e$ contains a unique cycle $C$. Since $T$
is acyclic, $C\nsubseteq T$; let $e'\in E(C-T)\subseteq E(T')$. Then $T^{*}:=T'+e-e'$
is acyclic, since $e'$ is an edge of the unique cycle in $T'+e$. By Theorem~\ref{Tree},
$T^{*}$ is a tree, since $|T^{*}|=|T|=\left\Vert T\right\Vert +1=\left\Vert T^{*}\right\Vert +1$.\end{proof}
\begin{defn}
If $G=(V,E)$ is a graph with a function $w:E\rightarrow\mathbb{N}$ then $w$ is
called a \emph{weight function}. For $H\subseteq G$ set
\[
w(H)=\sum_{e\in E(H)}w(e).
\]
A \emph{minimal spanning tree} of $G$ (with respect to $w$) is a spanning tree
$T\subseteq G$ such that $w(T)\leq w(T')$ for all spanning trees $T'\subseteq G$.
Consider the following algorithm.\end{defn}
\begin{description}
\item [{Minimal~Spanning~Tree~Algorithm}]~
\begin{enumerate}
\item input a connected graph $G=(V,E)$ with a weight function $w$;
\item let $e'_{1},\dots,e'_{|G|}$ be an enumeration of $E$ that is increasing with $w$;
\item for $i$ from $1$ to $|G|-1$ do
\begin{enumerate}
\item choose $e_{i}\in E\smallsetminus\{e{}_{1},e{}_{2}\dots,e_{i-1}\}$ such that $(V,\{e_{1},\dots,e_{i}\})$
is acyclic and subject to this $w(e_{i})$ is minimum
\item end do;
\end{enumerate}
\item output $T=(V,\{e_{1},\dots,e_{|G|-1}\})$;
\end{enumerate}
\end{description}
\begin{thm}
The Minimal Spanning Tree algorithm computes a minimal spanning tree of the input
graph $G$.\end{thm}
\begin{proof}
It suffices to prove by induction on $i$ that there is a minimal spanning tree $T^{*}$
such that $e_{1},\dots,e_{i}\in E(T^{*})$. The base step $i=0$ is trivial, so suppose
$i\geq1$. If $e_{i}\in E(T')$ we are done, so suppose not. By induction there is
a minimal spanning tree $T'$ with $\{e_{j}:1\leq j\leq i-1\}\subseteq E(T')$. By
Proposition~\ref{T-T'-P}, there is $e'\in E(T')\smallsetminus E(T)$ such that
$T'-e'+e_{i}$ is a spanning tree. By hypothesis, $e_{1},\dots,e_{i-1}\in T\cap T'$.
So $(V,\{e_{1},\dots,e_{i-1},e'\})$ is acyclic. Thus $e'$ was a candidate when
we chose $e_{i}$ for $T$. So $w(e_{i})\leq w(e')$. Thus:
\[
w(T')\leq w(T^{*}):=w(T')-w(e')+w(e_{i})\leq w(T').
\]
Thus $w(T^{*})=w(W')$. So $T^{*}$ is a minimal spanning tree with $e_{1},\dots e_{i}\in E(T^{*})$.\end{proof}
\begin{prop}
If $T$ is a tree with $k$ edges and $G$ is a nontrivial graph with $\delta(G)\geq k$
then $G$ contains a copy of $T$, i.e., a subgraph isomorphic to $T$.\end{prop}
\begin{proof}
Argue by induction on $k$.
\noindent \emph{Base Step:} $k=0$. Then $T\cong K_{1}$, so $T\cong G[\{v\}]$ for
any vertex $v$.
\noindent \emph{Induction Step:}\textbf{ }$k>1$. Let $l$ be a leaf of $T$. Then
$T':=T-l$ is a tree with $\left\Vert T'\right\Vert =k-1$. By the induction hypothesis
there exists $H'\subseteq G$ with $H'\cong T'$. Let $p$ be the unique neighbor
of $l$ in $T$, and let $x$ be the image of $p$ in $H'$. Since $|H'|=\left\Vert H'\right\Vert +1=k$
and $x$ is not adjacent to itself, $x$ has at most $k-1$ neighbors in $H'$. Since
$\delta(G)\geq k$, there exists $y\in N_{G}(x)\smallsetminus V(H')$. Set $H=H'+y+xy$.
Then $H\subseteq G$ and we can extend the isomorphism between $T'$ and $H'$ to
an isomorphism between $T$ and $H$ by mapping $l$ to $y$.
\end{proof}
\begin{x}
(+) Let $d_{1},\dots,d_{n}$ be positive integers with $n\geq2$. Prove that there
exists a tree with vertex degrees $d_{1},\dots,d_{n}$ iff $\sum d_{i}=2n-2$.
\end{x}
\begin{x}
({*}) Every tree is bipartite. Prove that every tree has a leaf in its larger partite
set, and in both partite sets if they have equal size.
\end{x}
\begin{x}
({*}) Let $T$ be a tree such that $2k$ of its vertices have odd degree. Prove that
${\color{red}T}$ decomposes into $k$ paths. Is it easier to prove this for forests?
\end{x}
\begin{x}
\textcolor{black}{(+) }Let $T$ be a tree with even order. Prove that $T$ has exactly
one spanning subgraph such that every vertex has odd degree.
\end{x}
\begin{x}(+) A \emph{root }of a graph $G$ is a special vertex $r$. A spanning
tree $T$ of a graph with root $r$ is normal if every edge $xy\in E(G)$ satisfies
either $x\in rTy$ or $y\in rTx$. Prove that every connected graph with root $r$
has a normal spanning tree {[}Hint: Prove the stronger statement that every path
$P$ with end $r$ is contained in a normal spanning tree.{]} \end{x}
\begin{x}(+) Let $\mathcal{T}$ be a set of subtrees of a tree $G$ such that $T\cap T'\ne\emptyset$
for all $T,T'\in\mathcal{T}$. Prove that $\bigcap\mathcal{T}\ne\emptyset$. \end{x}
\begin{x}
({*}) A \emph{directed graph }is a binary relation $G=(V,E)$. So the edges are \emph{ordered}
pairs. We still write $xy$ for the directed edge $(x,y)$. Let $E_{G}^{+}(x):=\{xy\in E\}$,
$d_{G}^{+}(x):=|E_{G}^{+}(x)|$, and $\delta^{+}(G):=\min_{v\in V}d_{G}^{+}(v)\}$.
A \emph{directed cycle} $C=(V,E)$ is a digraph with the form $V=\{v_{1},\dots,v_{s}\}$
and $E=\{v_{i}v_{i\oplus1}:i\in[s]\}$. This directed cycle is also denoted by $v_{1}\dots v_{s}v_{1}$
when it is clear form the context that the edges are directed.
Prove that a directed graph $G$ has a directed cycle if $\delta^{+}(G)\geq1$. {[}Hint:
You might want to define the notion of a directed path.{]}
\end{x}
\chapter{Networks, Matching and Connectivity}
In this Chapter we use the Max-Flow, Min-Cut Theorem to prove theorems about matchings
and connectivity of graphs. First we introduce more notation for multigraphs and
oriented edges.
Let $G=(V,E)$ be a multigraph. We call an edge with two distinct ends a \emph{link
}and an edge with two identical ends a \emph{loop. }Since many edges may have the
same ends, we need notation to denote the ends of an edge $e\in E$. Formally we
write $e\in E(x,y)$ to mean that one end of $e$ is $x$ and the other end is $y$.
This allows for the possibility that $e$ is a loop and both its ends are $x$, i.e.
$y=x$. Informally, we may write $e=xy$ to say that $e=xy$, but this is an abuse
of the equality symbol.
It is natural to extend this notation to sets of vertices. For sets $A,B\subseteq V$
let
\[
E(A,B)=\bigcup\{E(a,b):(a,b)\in A\times B\}.
\]
When one of the sets, say $A$, is a singleton $\{x\}$, we write $E(x,B)$. Thus
our earlier notation has the form $E(x)=E(x,V)$. We set
\[
\left\Vert A,B\right\Vert _{G}:=\sum_{a\in A}|E(a,B)|=|E(A,B)|+\left\Vert G[A\cap B]\right\Vert =\sum_{b\in B}|E(A,b)|.
\]
These definitions also apply to subsets $F\subseteq E$.
In this chapter every link $e\in E(x,y)$ will have two opposite \emph{orientations};
loops have only one orientation. The set of all orientations of edges of $E$ is
denoted by $\vec{E}$. We refer to an arbitrary orientation of $e$ by $\vec{e}$,
and its opposite orientation by $\cev e$; if $e$ is a loop then $\vec{e}=\cev e$.
We may identify which orientation $\vec{e}$ is by writing $\vec{e}\in\vec{E}(x,y)$
if $e$ is oriented from $x$ to $y$; else $\cev e\in\vec{E}(x,y)=\cev E(y,x)$.
Let $\vec{F}\subseteq\vec{E}$ denote a subset of orientations. Then $\cev F=\{\cev e:\vec{e}\in\vec{F}\}$.
Let $A,B\subseteq V$. Set $\overline{A}=V\smallsetminus A$ be the complement of
$A$. Define
\[
\vec{F}(A,B)=\{\vec{e}\in\vec{F}:\vec{e}\in\vec{E}(x,y)~\mbox{and}~(x,y)\in A\times B~\mbox{and}~a\ne b\},
\]
abbreviate $\vec{F}(x,B)=\vec{F}(\{x\},B)$, etc., and set $\vec{F}(x)=\vec{F}(x,\overline{\{x\}})$.
Consider a function $f:\vec{E}\rightarrow\mathbb{Z}$. Put
\[
f(A,B)=\sum_{\vec{e}\in\vec{E}(A,B)}f(\vec{e}),
\]
and as usual $f(x,B)=f(\{x\},B)$, etc. Also $f(x):=f(x,V)$.
\section{Max-Flow, Min-Cut Theorem}
A \emph{network} is a tuple $N=(V,E,s,t,c)$, where $G:=(V,E)$ is the multigraph,
$s\in V$ is the source, $t\in V$ is the sink, $s\ne t$, and $c:\vec{E}\rightarrow\mathbb{N}$
is the capacity function. A function $f:\vec{E}\rightarrow\mathbb{Z}$ is a \emph{flow}
in $N$ if it satisfies:
\begin{enumerate}
\item[(F1)] $f(\vec{e})=-f(\cev e)$ for all links $\vec{e}\in\vec{E}$;
\item[(F2)] $f(v)=0$ for all $v\in V\smallsetminus\{s,t\}$; and
\item[(F3)] $f(\vec{e})\leq c(\vec{e})$ for all $\vec{e}\in\vec{E}$.
\end{enumerate}
Notice that (F1) and (F3) imply $-c(\cev e)\leq f(\vec{e})\le c(\vec{e})$.
The \emph{value of a flow $f$ is $v(f)=f(s)$. }Let $\mathcal{F}(N)$ be the set
of all flows in $N$. It is nonempty because the $0$-flow is in $\mathcal{F}(N)$.
A \emph{cut} in the network $N$ is a pair $(S,\overline{S})$ with $s\in S\subseteq V$
and $t\in\overline{S}$. The \emph{capacity of a cut $(S,\overline{S})$} is $c(S,\overline{S})$.
Let $\mathcal{S}(N)$ be the set of all cuts in $N$. It is nonempty as $(s,\overline{\{s\}})\in\mathcal{S}$.
\begin{thm}
\label{FlowOverCuts-th}Let $(S,\overline{S})$ be a cut in a network $N=(V,E,s,t,c)$
with flow $f$. Then
\[
v(f)=f(S,\overline{S}).
\]
\end{thm}
\begin{proof}
\[
f(S,\overline{S})=f(S,V)-f(S,S)=_{(F1)}f(s)+\sum_{v\in S-s}f(v)-0=_{(F2)}f(s)=v(f).\qedhere
\]
\end{proof}
\begin{thm}[Max-Flow, Min-Cut]
\label{MaxFMinC-th}Every network $N=(V,E,x,y,c)$ satisfies
\[
\max_{f\in\mathcal{F}}v(f)=\min_{(S,\overline{S})\in\mathcal{S}}c(S,\overline{S}).
\]
\end{thm}
\begin{proof}
By Theorem~\ref{FlowOverCuts-th}, and the capacity constraint every flow $f$ and
cut $(S,\overline{S})$ satisfies
\begin{equation}
v(f)=f(S,\overline{S})\leq_{(F3)}c(S,\overline{S}).\label{MFMCpr-eq}
\end{equation}
Now let $f$ be a maximum flow in $N$. It suffices to construct a cut $(X,\overline{X})$
such that $v(f)=c(X,\overline{X})$. Let $\vec{D}=\{\vec{e}\in\vec{E}:f(\vec{e})v(f)$, contradicting the
maximality of $f$. So $\vec{H}$ contains no $s,t$-path.
Let $S=\{v\in V:\exists~\mbox{a directed \ensuremath{s,v}-path in \ensuremath{H}}\}$.
Then $s\in S$ and $t\notin S$. So $(S,\overline{S})$ is a cut. Consider an oriented
edge $\vec{e}\in(S,\overline{S})$; say $\vec{e}\in\vec{E}(x,y)$. As there is a
directed $s,x$-path in $\vec{H}$, but no directed $s,y$-path in $\vec{H}$, $c(\vec{e})=f(\vec{e})$.
Thus $v(f)=f(S,\overline{S})=c(S,\overline{S})$.
\end{proof}
\section{Matchings}
A \emph{matching} is a set of \emph{links} with no common ends. A \emph{maximal }matching
is a matching that cannot be enlarged by adding an edge. A \emph{maximum} matching
is matching with maximum size among all matchings in the graph. A vertex is said
to be $M$-\emph{saturated} if and only if it is the end of an edge in $M$; otherwise
it is $M$-\emph{free}, and a set of vertices $X$ is said to be $M$-\emph{saturated}
if every $x\in X$ is $M$-saturated. The matching $M$ is \emph{perfect} if every
vertex is $M$-saturated.
\begin{defn}
Given a matching $M$ in a graph $G$, an $M$-\emph{alternating} path is a path
$P$ such that each vertex $v\in V(P)$ is incident to at most one edge in $E(P)\smallsetminus M$.
Such a path is $M$-\emph{augmenting} if its ends are $M$-free.
\end{defn}
We will need the following easy proposition several times.
\begin{prop}
\label{Dle2}Every connected multigraph $G$ with $\Delta(G)\leq2$ is a path or
cycle.\end{prop}
\begin{proof}
Let $P=v_{1}\dots v_{t}$ be a maximum path in $G$. As $\Delta(G)\leq2$, no other
edge is incident to the internal vertices of $P$. As $P$ is maximum, no edge joins
an end of $P$ with a vertex of $V\smallsetminus V(P)$. So $G$ is the path $P$
or $G$ is the cycle $P+v_{1}v_{t}$. \end{proof}
\begin{thm}[3.1.10 Berge]
\label{Berge}A matching $M$ in a graph $G$ is not maximum iff $G$ has an $M$-augmenting
path.\end{thm}
\begin{proof}
Suppose $P$ is an $M$-augmenting path. Then
\[
M'=M\bigtriangleup E(P)=_{def}(M\smallsetminus E(P))\cup(E(P)\smallsetminus M)
\]
is a larger matching.
Now suppose $M$ is not maximum. Choose a matching $M'$ with $|M'|>|M|$. Let $H$
be the spanning submultigraph with edge set\textcolor{black}{{} $M\cup M'$, where
each $e\in M\cap M'$ is duplicated in $H$. Since each vertex is incident to at
most one edge of each matching, $\Delta(H)\leq2$. Using Proposition~\ref{Dle2},
the components of $H$ are either alternating paths or cycles with the same number
of edges from $M$ as $M'$. Since $|M|<|M'|$, }some component $Q$ of $H$ has
more edges from $M'$ than $M$. Such a component must be an $M$-augmenting path.
\end{proof}
\begin{x}({*}) Two players Alice and Bob play a game on a graph $G$. Alice begins
the game by choosing any vertex. All other plays consist of the player, whose turn
it is, choosing an unchosen vertex that is joined to the last chosen vertex. The
winner is the last player to play legally. Prove that Alice has a winning strategy
if $G$ has no perfect matching, and Bob has a winning strategy if it does. \end{x}
\section{Bipartite matching}
A bipartite $G$ with bipartition $\{X,Y\}$ is called an $X,Y$-bigraph. For $S\subseteq X$
set $N(S):=\bigcup_{v\in S}N(v)$. For a function $f:A\rightarrow B$ and $S\subseteq A$,
let $f(S):=\{y\in B:\exists x\in S(f(x)=y)\}$ be the range of $f$ restricted to
$S$.
\begin{defn}
A\emph{ cover} of a graph $G$ is a subset $Q\subseteq V(G)$ that contains at least
one end of every edge.
\end{defn}
Let $C$ be an odd cycle with $\left\Vert C\right\Vert =2k+1$. Since $C$ is $2$-regular,
every set $Q\subseteq V(C)$ covers at most $2|Q|$ edges. Thus every vertex cover
of $C$ has at least $k+1$ vertices. On the other hand, every matching $M$ in $C$
has $2|M|$ ends; so $|M|\leq k\leq k+1\leq|C|$.
\begin{thm}[{König, Egerváry {[}1931{]} 3.1.16}]
\label{KE-1}If $G$ is a graph with a maximum matching $M$ and a minimum cover
$W$ then $|M|\leq|W|$; if $G$ is bipartite then $|M|=|W|$.\end{thm}
\begin{proof}
Order $V$ as $v_{1}\prec\dots\prec v_{|G|}$. Since $W$ is a cover, every edge
is incident to some vertex of $W$ (possibly two). Define a function $g:M\rightarrow W$
by $g(e)$ is the least $w\in e\cap W$. Since $M$ is a matching, no vertex of $W$
can be incident to two edges of $M$. So $g$ is an injection. Thus $|M|\leq|W|$.
Now suppose that $G$ is an $X,Y$-bigraph. Define a network $N=(X\cup Y+s+t,E',s,t,c)$,
where $E'$ is the set of edges $E'=E(K(s,X))\cup E_{G}(X,Y)\cup E(K(Y,t))$, and
\begin{eqnarray*}
c(\vec{e}) & = & \begin{cases}
1 & \mbox{if \ensuremath{\vec{e}\in\vec{E}'(s,X)\cup\vec{E}'(Y,t)} }\\
\infty:=|X|+1 & \mbox{if \ensuremath{\vec{e}\in\vec{E}'(X,Y)}}\\
0 & \mbox{else}
\end{cases}.
\end{eqnarray*}
Let $f$ be a maximum flow. By (F1) and (F3)
\begin{eqnarray*}
-0 & \leq & -f(\cev e)=f(\vec{e})\leq1~\mbox{for}~\vec{e}\in\vec{E}'(s,X)\cup\vec{E}'(Y,t)~\mbox{and}\\
-0 & \leq & -f(\cev e)=f(\vec{e})\leq\infty~\mbox{for}~\vec{e}\in\vec{E}'(X,Y).
\end{eqnarray*}
By (F2), $f(x,Y)\in\{0,1\}$ for all $x\in X$ and $f(X,y)\in\{0,1\}$ for all $y\in Y$.
It follows that $M':=\{xy\in E(X,Y):f({\color{red}x,y})=1\}$ is a matching with
$|M'|=f(X+s,Y+t)=v(f)$.
Let $(P,\overline{P})$ be a minimum cut. Now $(\{s\},\overline{\{s\}})$ is a cut
with $c(P,\overline{P})\leq c(\{{\color{red}s}\},\overline{\{{\color{red}s}\}})=|X|<\infty$,
and $c(\vec{e})=\infty$ for all $\vec{e}\in\vec{E}'(X,Y)$, so $\vec{E}'(P,\overline{P})\subseteq\vec{E}{\color{red}'}(s,\overline{P}\cap X)\cup\vec{E}'(P\cap Y,t)$.
Thus
\begin{equation}
N_{G}(P\cap X)\subseteq P\cap Y.\label{N(X)inP}
\end{equation}
Thus $C:=(P\cap Y)\cup(\overline{P}\cap X)$ is a cover: for any $xy\in E(X,Y)$
either $x\in\overline{P}\cap X\subseteq C$ or $x\in P$; in the latter case, $y\in N(x)\subseteq P\cap Y\subseteq C$
by \eqref{N(X)inP}. Using Theorem~\ref{MaxFMinC-th},
\[
|W|\leq|P\cap Y|+|\overline{P}\cap X|=c(P,t)+c({\color{red}s},\overline{P})\leq c(P,\overline{P})=v(f)=|M'|\leq{\color{red}|M|}\leq|W|.
\]
Thus equality holds throughout, and $|M|=|W|$.\end{proof}
\begin{cor}[Hall]
\label{Hall-th}Every $X,Y$-bigraph has a matching saturating $X$ if and only
if
\begin{equation}
|S|\leq|N(S)|~\mbox{for all}~\mbox{every set}~S\subseteq X.\label{eq:Hall}
\end{equation}
\end{cor}
\begin{proof}
Let $G$ be an $X,Y$-bigraph with a maximum matching $M$. If $X$ is saturated
then \eqref{eq:Hall} holds since the vertices of any $S\subseteq X$ are matched
to distinct elements of $N(S)$.
Otherwise, let $W$ be a minimum cover. Then $\left\Vert X\smallsetminus W,Y\smallsetminus W\right\Vert =0$.
So $N(S)\subseteq Y\cap W$, where $S=X\smallsetminus W$. By Theorem~\ref{KE-1},
$|W|=|M|<|X|.$ Thus
\[
|X\cap W|+|X\smallsetminus W|=|X|>|W|=|X\cap W|+|Y\cap W|.
\]
So \eqref{eq:Hall} fails for $S$, since canceling $|X\cap W|$ yields
\[
|S|>|Y\cap W|\geq|N(S)|.\qedhere
\]
\end{proof}
A multigraph is $k$-regular if every vertex has degree $k$. A $k$\emph{-factor
of a graph} is a $k$-regular, spanning subgraph. So the edge set of a $1$-factor
is a perfect matching.
\begin{cor}[3.1.13]
\label{HallC}Every $k$-regular bipartite multigraph $G$ has a perfect matching.\end{cor}
\begin{proof}
Suppose $G$ is an $k$-regular $X,Y$-bimultigraph. Then
\[
k|X|=\left\Vert X,Y\right\Vert =k|Y|.
\]
It follows that $|X|=|Y|$. Thus it suffices to show that $G$ has a matching that
saturates $X$. By Hall's Theorem, it suffices to check \eqref{eq:Hall}. Consider
any subset $S\subseteq X$. Then
\[
k|S|=\left\Vert S,Y\right\Vert {\color{black}{\color{black}{\color{red}{\color{black}=}}\left\Vert S,N(S)\right\Vert {\color{black}{\color{red}{\color{black}\leq}}\left\Vert X,N(S)\right\Vert {\color{red}{\color{black}=}}k}}}|N(S)|.
\]
So $|S|\leq|N(S)|$.
\end{proof}
\begin{x}({*}) Let $\mathcal{S}=\{S_{i}:i\in I\}$ be a family of sets. A \emph{system
of distinct representatives} (sdr) for $\mathcal{S}$ is a sequence $(a_{i}:i\in I)$
of distinct elements such that $a_{i}\in S_{i}$ for all $i\in I$. Prove that if
$\mathcal{S}$ is finite then $\mathcal{S}$ has an sdr if and only if $|J|\leq|\bigcup_{j\in J}S_{j}|$
for all $J\subseteq I$. \end{x}
\begin{x}({*}) Prove that there is an injection $f:\binom{[2k+1]}{k+1}\rightarrow\binom{[2k+1]}{k}$
such that $f(S)\subseteq S$ for all $S\in\binom{[2k+1]}{k+1}$.\end{x}
\begin{x}\textcolor{black}{({*})}\textcolor{red}{{} }Let $\mathcal{P}=\{P_{1},\dots,P_{t}\}$
and \emph{$\mathcal{Q}=\{Q_{1},\dots,Q_{t}\}$ }be two partitions of a set $S$ into
$t$ subsets of size $k$. Prove that\emph{ }$\mathcal{P}$ and $\mathcal{Q}$ have
a common sdr, that is there exist an sdr $(a_{i}:i\in[t])$ of $\mathcal{P}$ and
a permutation $\sigma:[t]\rightarrow[t]$\emph{ }such that $(a_{\sigma(i)}:i\in[t])$
is an sdr of $\mathcal{Q}$.\end{x}
\begin{x}(+) Suppose $G$ is an $X,Y$-bigraph with $\delta(G)\geq1$ such that
every edge $xy$ with $x\in X$, satisfies $d(x)\geq d(y)$. Prove that $G$ has
a matching that saturates every vertex in $X$. {[}Hint: Consider assigning each
edge $xy$ with $x\in X$ the weight $w(xy)=\frac{1}{d(x)}$.{]}\end{x}
\begin{x}(+) For $k,n\in\mathbb{N},$ let $G$ be an $A,B$-bigraph with $|A|=n=|B|$
such that $\delta(G)\geq k$, and for all $X\subseteq A,Y\subseteq B$, if $|X|,|Y|\geq k$
then $|E(X,Y)|\ne\emptyset$. Prove: $G$ has a perfect matching.\end{x}
\begin{x}({*}) Let $G$ be an $X,Y$-bigraph with $|X|=n=|Y|$. Prove that $G$
a perfect matching if $\delta(G)\geq\frac{n}{2}$.
\end{x}
\begin{x}(++) Prove that for all partitions $\mathcal{P}=\{P_{i}:i\in[n]\}$ and\emph{
$\mathcal{Q}=\{Q_{i}:i\in[n]\}$} of a region $S$ of area $n$\emph{ }into parts
of area\textcolor{red}{{} }$1$ there exists a permutation $\sigma:[n]\rightarrow[n]$
such that $\mbox{area }(P_{i}\cap Q_{\sigma(i)})\geq f(n)$ for all $i\in[n]$, where
\[
f(n):=\begin{cases}
\frac{4}{(n+1)^{2}} & \mbox{if }n\mbox{ is odd}\\
\frac{4}{n(n+2)} & \mbox{if }n\mbox{ is even}
\end{cases}.
\]
Also show that the function $f$ is optimal.\emph{ }\end{x}
\begin{x}({*}) Prove that every bipartite graph has a matching of size $\frac{\left\Vert G\right\Vert }{\Delta(G)}$.\end{x}
\section{General matching}
Notice that if $H$ is a component of a graph $G$ and $|H|$ is odd then $G$ does
not have a perfect matching.
\begin{defn}
Let $\mathcal{C}_{G}$ be the set of components of the graph $G$. A component with
an odd number of vertices is said to be an \emph{odd component}. Let $\mathcal{O_{G}}$
be the set of odd components of $G$ and $o(G)=|\mathcal{O}_{G}|$. \end{defn}
\begin{thm}[{Tutte {[}1947{]} 3.3.3}]
\label{TutteE}A graph $G=(V,E)$ has a perfect matching iff
\begin{equation}
o(G-S)\leq|S|\mbox{ for all }S\subseteq V.\label{tutte}
\end{equation}
\end{thm}
\begin{proof}
If $G$ has a perfect matching then \eqref{tutte} holds, since for any $S\subseteq V$,
the odd components of $G-S$ must have vertices matched to distinct vertices in $S$.
Now we assume \eqref{tutte}, and show that $G$ has a perfect matching by induction
on $\left\Vert \overline{G}\right\Vert $.
\noindent \textbf{Base Step: }$G-U$ consists of disjoint complete subgraphs. By
\eqref{tutte}, $|U|\geq o(G-U);$ so one vertex in each odd component can be matched
to a vertex in $U$. The remaining vertices of $G-U$ are in disjoint even cliques.
By \eqref{tutte}, $o(G\smallsetminus\emptyset)=|\emptyset|=0$; so $|G|$ is even,
and the remaining vertices of $U$ form an even clique. Each of these even cliques
has a perfect matching. Combining all these matchings yields a perfect matching of
$G$.
\noindent \textbf{Induction Step: }$G-U$ has an incomplete component. Now $G$ has
a path $aba'$ with $aa'\notin E$. As $b\notin U$, there is $b'\in V$ with $bb'\notin E$.
Adding $aa'$ or $bb'$ does not increase $o(G-S)$ for any $S\subseteq G$, so by
induction $G+aa'$ and $G+bb'$ have perfect matchings $M^{+}=M+aa'$ and $L^{+}=L+bb'$,
respectively. In $G$, the $M$-free vertices are $a$ and $a'$, and the $L$-free
vertices are $b$ and $b'$. The component of $G[M\cup L]$ containing $a$ is an
alternating path $P=a\dots x$ with $x\in\{a',b,b'\}$. If $x\in\{a',b\}$ then $aPx$
or $aPxa'$ is $M$-augmenting; else $baPb'$ is $L$-augmenting. Anyway, $G$ has
a perfect matching.
\end{proof}
The next theorem is a generalization of Tutte's Theoerm. We will give two proofs.
The first builds on the previous proof. The second starts from scratch.
\begin{thm}[{Tutte {[}1947{]} 3.3.3+3.3.7}]
\label{TutteG}Let $G=(V,E)$ be a graph with a maximum matching $M$. Then the
number of $M$-free vertices of $G$ is equal to
\begin{equation}
d:=\max_{S\subseteq V}o(G-S)-|S|.\label{tutteg}
\end{equation}
\end{thm}
\begin{proof}
[First Proof]If $d=0$ then we are done by Theorem~\ref{TutteE}. So assume $d>0$.
Choose $S\subseteq V$ so that $d=o(G-S)-|S|$. Every matching has at least $d$
free vertices, since each odd component has a vertex that is free or is matched to
a vertex of $S$.
Let $G^{+}:=G\vee K_{d}(Q)$, and consider $T\subseteq V(G^{+})$. As $d\geq1$,
$G^{+}$ is connected. Also
\[
|G^{+}|=|G|+d\equiv o(G-S)+|S|+d=2d+2|S|\equiv0\mod2.
\]
Thus if $T=\emptyset$ then $o(G-T)-|T|=0$. If $o(G^{+}-T)\geq2$ , then $Q\subseteq T$,
and so
\[
o(G^{+}-T)-|T|\leq(o(G-(T\smallsetminus Q))-|T\smallsetminus Q|)-|Q|\leq d-d\leq0.
\]
Otherwise, $o(G^{+}-T)-|T|\leq1-1=0$. By Theorem~\ref{TutteE}, $G^{+}$ has a
perfect matching $M^{+}$. Then $M:=M^{+}\smallsetminus E(Q,V)$ is a matching in
$G$ with at most $|Q|=d$ unsaturated vertices.
\end{proof}
\begin{defn}
A graph G is \emph{factor critical} if $G-v$ has a perfect matching for every vertex
$v\in V(G)$.\textcolor{black}{{} A set $S$ is }\textcolor{black}{\emph{matchable}}\textcolor{black}{{}
into $\mathcal{O}_{G-S}$ if there exists a matching $M$ }\textcolor{red}{saturating
$S$}\textcolor{black}{{} such that each edge $e\in M$ has one end in $S$ and one
end in an odd component, and at most one vertex of each odd component is saturated.}\end{defn}
\begin{proof}
[Second Proof]For any set $S\subseteq V$ and matching $M$, there are at least
$o(G-S)-|S|$ unsaturated vertices: each odd component $H\subseteq G-S$ has an $M$-unsaturated
vertex, unless $M\cap E(S,V(H))\ne\emptyset$, and there are at most $|S|$ such
edges in $M$. So it suffices to show that there exists a set $S\subseteq V$ and
a matching $M$ with exactly $o(G-S)-|S|$ unsaturated vertices.
Argue by induction on $|G|$. For the base step $|G|=1$, let $S=\emptyset$. Then
$o(G-S)-|S|=1$ and the only vertex of $G$ is unsaturated by any matching. Now consider
the induction step.
Choose a set $S\subseteq V$ so that $o(G-S)-|S|$ is maximum, and subject to this,
$|S|$ is also maximum. We first prove the following three claims:
\begin{claim*}[1]
\label{odd}Every component of $G-S$ is odd.\end{claim*}
\begin{proof}
Suppose $H\in\mathcal{C}_{G-S}$ with $|H|$ even. Choose a non-cut vertex $x$ (end
of a maximal path) of $H$, and set $S'=S+x$. Then
\[
\mathcal{O}_{G-S'}=\mathcal{O}_{G-S}+(H-x)\mbox{ and }|S'|=|S|+1.
\]
Thus $o(G-S)-|S|=o(G-S')-|S'|$, contradicting the choice of $S$, since $|S|<|S'|$.\end{proof}
\begin{claim*}[2]
\label{fc}Every odd component of $G-S$ is factor critical. \end{claim*}
\begin{proof}
Consider any $H\in\mathcal{O}_{G-S}$ and any vertex $x\in V(H)$. We must show that
$H'=H-x$ has a perfect matching. By the induction hypothesis, it suffices to show
that $o(H'-T)-|T|\leq0$ for all $T\subseteq V(H')$. So consider any such $T$,
and set $S'=S\cup T+x$. Then $|S'|=|S|+|T|+1>|S|$, and so by the choice of $S$
\[
o(G-S)-|S|>o(G-S')-|S'|.
\]
Since $T+x\subseteq V(H)$,
\[
\mathcal{O}_{G-S'}=(\mathcal{O}_{G-S}-H)\cup\mathcal{O}_{H'-T}.
\]
So
\begin{gather*}
o(G-S)-|S|>o(G-S')-|S'|=o(G-S)-1+o(H'-T)-|S|-|T|-1\\
2>o(H'-T)-|T|.
\end{gather*}
Moreover, by Claim (1), $H$ is an odd component, and so $|H'|$ is even. Thus
\[
o(H'-T)-|T|\equiv|H'-T|+|T|\equiv|H'|\equiv0\mod2.
\]
Hence $1\not\not\ne o(H'-T)-|T|$, and so $0\geq o(H'-T)-|T|$.\end{proof}
\begin{claim*}[3]
\label{mi}$S$ is matchable into $\mathcal{O}_{G-S}$.\end{claim*}
\begin{proof}
Let $H$ be the $S,\mathcal{O}_{G-S}$-bigraph with edge set
\[
F:=\{xD:x\in S,D\in\mathcal{O}_{G-S}\mbox{ and }N(x)\cap V(D)\ne\emptyset\}.
\]
We will show $H$ has a matching saturating $S$ by checking \eqref{eq:Hall} of
Corollary~\ref{Hall-th}. Consider any $T\subseteq S$, and set $S'=S-T$. Then
$\mathcal{O}_{G-S}\smallsetminus N_{H}(T)\subseteq\mathcal{O}_{G-S'}$. By the choice
of $S$
\begin{gather*}
o(G-S)-|S|\geq o(G-S')-|S'|\geq o(G-S)-N_{H}(T)-|S|+|T|\\
|N_{{\color{red}{\normalcolor H}}}(T)|\geq|T|\qedhere.
\end{gather*}
\end{proof}
Finally, we obtain a matching $M$ as follows. By Claim~(3) there is a matching
$M_{0}$ that saturates $S$ and one vertex of $|S|$ odd components. For each $H\in\mathcal{O}_{G-S}$
choose a vertex $v_{H}$, and if possible, choose $v_{H}$ so that it is $M_{0}$-saturated.
Next use Claim~(2) to obtain matchings $M_{H}$ of $H-v_{H}$ for every odd component
$H\in\mathcal{O}_{G-S}$. Then
\[
M:=M_{0}\cup\bigcup_{H\in\mathcal{O}_{G-S}}M_{H}
\]
is matching of $G$. Using Claim~(1), it saturates every vertex of $G$ except
those $o(G-S)-|S|$ vertices $v_{H}$ that are not saturated by $M_{0}$.
\end{proof}
We have actually proved a stronger statement.
\begin{thm}
Let $G=(V,E)$ be a graph and $S^{*}\subseteq V$ be a set of vertices such that
for all $S\subseteq V$
\begin{enumerate}
\item $o(G-S)-|S|\leq o(G-S^{*})-|S^{*}|;$ and
\item if equality holds in (1) then $|S|\leq|S^{*}|$.
\end{enumerate}
Then every component of $G-S^{*}$ is factor-critical; and every maximum matching
saturates $S^{*}$, matches $S^{*}$ into $\mathcal{O}_{G-S^{*}}$, and leaves at
most one vertex of each component of $G-S^{*}$ free.
\end{thm}
A graph $G=(V,E)$ is transitive if for all $x,y\in V$ there is an automorphism
$\varphi$ of $G$ with $\varphi(x)=y$. For example $C_{n}$, $K_{n,n}$ and the
Petersen graph, but not $P_{n}$, are transitive.
\begin{x}
({*}) Prove that a transitive graph does not have a cut vertex.
\end{x}
\begin{x}(+) Prove that if $G$ is a \textcolor{black}{connected,} transitive graph
with $|G|$ even then $G$ has a perfect matching. (Lovasz has conjectured that every
connected transitive graph has a hamiltonian path; there are only four known examples
of such graphs that are not hamiltonian.) {[}Hint: Use Theorem~\ref{TutteG} to
show that if $G$ does not have a perfect matching then some, but not all, of its
vertices have the property that they are saturated in every maximum matching.{]}\end{x}
\section{Applications of Matching Theorems}
A cut-edge is also called a \emph{bridge}. A \emph{bridgeless }graph is a graph without
cut-edges. It need not be connected.
\begin{thm}[{Petersen {[}1891{]} 3.3.8}]
\label{Pete1}Every bridgeless cubic graph $G=(V,E)$ contains a $1$-factor.\end{thm}
\begin{proof}
By Tutte's Theorem, it suffices to show that $o(G-S)\leq|S|$ for every subset $S\subseteq V$.
Fix any such $S$ and consider any \textcolor{red}{${\normalcolor H\in\mathcal{O}_{G-S}}$}.
Since $G$ is cubic and $|H|$ is odd,
\[
3|H|=\left\Vert H,V\right\Vert =2\|H\|+\left\Vert H,S\right\Vert \equiv1\mod2.
\]
It follows that $\left\Vert H,S\right\Vert $ is odd, and since $G$ is bridgeless,
$\left\Vert H,S\right\Vert \geq3$. Thus
\[
3o(G-S)\leq\left\Vert S,V\smallsetminus S\right\Vert \leq3|S|,
\]
and so $o(G-S)\leq|S|$.\end{proof}
\begin{thm}[{Petersen {[}1891{]} 3.3.9}]
\label{Pete2}Every regular multigraph with positive even degree has a $2$-factor.\end{thm}
\begin{proof}
Suppose $G'$ is $2k$-regular with $k\in\mathbb{Z}^{+}$. It suffices to show that
each component $G=(V,E)$ of $G'$ has a $2$-factor. By Euler's Theorem~\ref{Eulerian},
$G$ has an Eulerian trail $T=v_{1}\dots v_{n}(=v_{1})$. Let $V'=\{v':v\in V\}$
and $V''=\{v'':v\in V\}$ be sets of new vertices, disjoint from $V$ and each other,
where $v\mapsto v'$ and $v\mapsto v''$ are injections. Define a $V',V''$-bigraph
by $E(H)=\{e_{i}^{*}:i\in[n]\}$, where $e_{i}^{*}\in E_{H}(v'_{i},v''_{i\oplus1})$.
Since each vertex $v$ of $G$ is incident to $2k$ edges, it appears $k$ times
in $T$. Say $v=v_{i_{1}}=\dots=v_{i_{k}}$. Then
\[
E(v)=\{v_{i_{1}-1}v_{i_{1}},v_{i_{1}}v_{i_{1}+1},\dots,v_{i_{k}-1}v_{i_{k}},v_{i_{k}}v_{i_{k}+1}\}
\]
and
\[
E_{H}(v')=\{v'v''_{i_{1}+1},\dots,v'v''_{i_{k}+1}\}~\mbox{and}~E_{H}(v'')=\{v''v'_{i_{1}-1},\dots,v''v'_{i_{k}-1}\}.
\]
So $H$ is $k$-regular. By the Corollary~\ref{HallC}, $H$ has a perfect matching
$M$. Let $F=\{xy\in E:x'y''\in M\}$. Then $(V,F)$ is a $2$-factor of $G$: for
each $y\in V$ there exists a unique $x$ such that $x'y''\in M$ and a unique $z$
such that $y'z''\in M$. Since $T$ is a trail, and $xy,yz\in T$, we have $xy\ne yz$
(maybe $xy$ and $yz$ were parallel edges if $G$ is a multi graph).
\end{proof}
\begin{x}({*}) Prove that a $3$-regular graph has a $1$-factor iff it decomposes
into copies of $P_{4}$.\end{x}
\begin{x}
(+) Suppose $G$ is a graph on $2k$ vertices with $k\geq3$, whose complement $\overline{G}$
does not have a $1$-factor. Let $S$ be the set whose existence is guaranteed by
Tutte's Theorem (applied to $\overline{G}$). Prove that
\begin{enumerate}
\item If $|S|=0$ then $G$ contains $K_{c,2k-c}$ for some odd $c$.
\item If $|S|\geq k-1$ then $G$ contains $K_{k+1}$.
\item If $1\leq|S|\leq k-2$ then $\Delta(G)\geq k+1$. {[}Hint: $\frac{x+1}{x+2}(2k-x)>k$
when $1\leq x\leq k-2$.{]}
\end{enumerate}
\end{x}
\begin{x}(+) Let $M$ be a matching in a graph $G$ with an $M$-unsaturated vertex
$u$. Prove that if $G$ has no $M$-augmenting path starting at $u$ then $G$ has
a maximum matching $L$ such that $u$ is $L$-unsaturated. \end{x}
\begin{x}(+) A graph is \emph{claw-free} if it does not contain an induced $K_{1,3}$.
Prove that a \textcolor{black}{connected} claw-free graph of even order has a $1$-factor.
Find (easy) a small counter example if the graph is not connected.\end{x}
\begin{x}(+) Let $G$ be a $k$-regular graph with $|G|$ even that remains connected
when any $k-2$ edges are removed. Prove that $G$ has a $1$-factor. \end{x}
\section{Introduction to Connectivity}
\begin{defn}
A\emph{ separating set} or \emph{vertex cut} of a graph $G$ is a set $S\subseteq V(G)$
such that $G-S$ has more than one component. The \emph{connectivity} $\kappa(G)$
is the minimum size of a vertex set $S$ such that $G-S$ has more than one component
or only one vertex. A graph $G$ is $k$-connected if $k\leq\kappa(G)$. Two vertices
$x$ and $y$ are \emph{separated} by $S$ if they are in different components of
$G-S$.
\end{defn}
Note that it is not possible to disconnect a complete graph by removing vertices.
The definition sets the connectivity of a complete graph $K$ equal to $|K|-1$.
\begin{defn}
A \emph{disconnecting set} of edges in a graph $G$ is a set $F\subseteq E(G)$ such
that $G-F$ has more than one component. The \emph{edge-connectivity} $\kappa'(G)$
of $G$ is the minimum size of a disconnecting set of edges. It is $k$-edge-connected
if $k\leq\kappa'(G)$. Two vertices $x$ and $y$ are \emph{separated} by $F$ if
they are in different components of $G-F$.
\end{defn}
An \emph{edge cut} in $G$ is a set of edges of the form $E(S,\overline{S})$, where
$\emptyset\ne S\ne V(G)$.
\begin{thm}[{Whitney {[}1932{]} 4.1.9}]
Every graph $G=(V,E)$ satisfies
\[
\kappa(G)\leq\kappa'(G)\leq\delta(G).
\]
\end{thm}
\begin{proof}
Choose a vertex $v\in V$ with $d(v)=\delta(G)$. Then $E(v)$ is a disconnecting
set of edges of size $\delta(G)$, and so $\kappa'(G)\leq\delta(G)$.
For the first inequality, consider a minimum edge cut $E(S,\overline{S})$; so $\left\Vert S,\overline{S}\right\Vert =\kappa'(G)$.
Note that $\kappa(G)\leq|G|-1$. First suppose that every vertex in $S$ is adjacent
to every vertex in $\overline{S}$. Using calculus,
\[
\kappa'(G)=\left\Vert S,\overline{S}\right\Vert =|S||\overline{S}|\geq|G|-1\geq\kappa(G).
\]
Else there are $x\in S$ and $y\in\overline{S}$ with $xy\notin E$. Define $f:E(S,\overline{S})\rightarrow V$
by $f(e)=z$ if ${\normalcolor e=xz}$; else $f(e)\in e\cap S$. So $f(e)\in e\smallsetminus\{x,y\}$.
Every $x,y$-path $P$ contains an edge $e\in E(S,\overline{S})$, and $f(e)$ is
an interior vertex of $P$. So $\mbox{range}(f)$ separates $x$ from $y$. Thus
\[
\kappa'(G)=\left\Vert S,\overline{S}\right\Vert \geq|\mbox{range}(f)|\geq\kappa(G).\qedhere
\]
\end{proof}
\begin{thm}
Every cubic graph $G=(V,E)$ satisfies $\kappa(G)=\kappa'(G)$.\end{thm}
\begin{proof}
Put $\kappa:=\kappa(G)$ and $\kappa':=\kappa'(G)$. If $\kappa=0$ then $G$ is
disconnected, and so $\kappa'=0$ also. If $\kappa=3$ then $3=\kappa\leq\kappa'\leq\delta(G)=3$,
and again $\kappa=\kappa'$. So assume $\kappa\in[2]$. Let $S$ be a separating
set with $|S|=\kappa$, and let $H_{1},H_{2}$ be two components of $G-S$. Since
$S$ is minimum, every vertex $v\in S$ has a neighbor in each $H_{i}$. Since $G$
is cubic, there exists $i$ such that $v$ has a unique neighbor $w_{v}$ in $H_{i}$.
Choose such a $w_{v}$, preferring $w_{v}\in H_{1}$, and set $F:=\{vw_{v}:v\in S\}$.
Then $|F|=|S|$. Moreover $F$ is a disconnecting set of edges: for each $i\in[2]$,
F disconnects $H_{i}$ from $S$ unless $w_{v_{i}}\notin H_{i}$ for some $v_{i}\in S$.
In this case, $S=\{v_{1},v_{2}\}$, and two neighbors in $H_{1}$ by the preference
for $H_{1}$. Since $\delta(G)=3$ and $v_{1}w_{v_{1}}\in H_{2}$, $v_{1}v_{2}\notin E$.
Thus $F$ separates $v_{1}$ from $v_{2}$. So $\kappa'\leq|F|=|S|=\kappa\leq\kappa'$.\end{proof}
\begin{lem}[Expansion Lemma 4.2.3]
\label{exp-lem}If $G$ is $k$-connected and $G'$ is obtained from $G$ by adding
a new vertex $x$ with at least $k$ neighbors in $G$ then $G'$ is $k$-connected.\end{lem}
\begin{proof}
Since $|G'|=|G|+1$, it suffices to show that $G'$ does not have a $(k-1)$-separating
set. Consider any $(k-1)$-set $S.$ Then $G-S$ is connected and $x$ has a neighbor
in $G-S$, so $G'-S$ is connected.
\end{proof}
\begin{x}Prove that an $r$-connected graph on an even number of vertices with no
induced subgraph isomorphic to \textcolor{red}{${\normalcolor K_{1,r+1}}$} has a
$1$-factor.\end{x}
\section{Low Connectivity}
For graphs $H\subseteq G$, a path $v_{1}\dots v_{n}\subseteq G$ is an $H$-path
if $v_{1},v_{n}\in H$ and $v_{2}\dots v_{n-1}\subseteq G-H$.
\begin{defn}
A sequence of graphs $G_{0},\dots,G_{t}$ is a $2$-witness for a graph $G$ iff\end{defn}
\begin{enumerate}
\item $G_{0}$ is a cycle and $G_{t}=G$; and
\item for all $i\in[t]$ there is a $G_{i-1}$-path $P_{i}$ with $G_{i}=G_{i-1}+P_{i}$.\end{enumerate}
\begin{thm}[{Whitney {[}1932{]} 4.2.8}]
\label{Whitney}A graph $G=(V,E)$ is $2$-connected iff it has a $2$-witness
set.\end{thm}
\begin{proof}
First suppose $G$ is $2$-connected. Then $\delta(G)\geq2$, and so $G$ contains
a cycle $C$. Let $H\subseteq G$ be a maximal subgraph such that $H$ has a $2$-witness
$G_{0},\dots,G_{t}$. It exists because $C$ is a candidate. It suffices to show
that $H=G$.
Suppose $v_{0}\in V(G-H)$. Since $G$ is connected, there exists a $v_{0},H$-path
$Q=v_{0}\dots v_{s}$. Since $G$ is $2$-connected, $G-v_{s}$ is connected. So
there exists a $v_{s-1},H$-path $P$ in $G-v_{s}$. Then $P_{t+1}:=v_{s}v_{s-1}P$
is an $H$-path in $G_{t+1}:=H+P_{t+1}$, contradicting the maximality of $H$. We
conclude that $H$ is a spanning subgraph of $G$.
Now suppose $xy\in E(G-H)$. Then $xy$ is an $H$-path of $G_{t+1}:=H+xy$, contradicting
the maximality of $H$. So $H$ is an induced, spanning subgraph of $G$. Thus $G=H.$
Now suppose $G$ has a $2$-witness $G_{0},\dots,G_{t}$. Argue by induction on $t$
that $G$ is $2$-connected. The base step $t=0$ is easy since the cycle $G_{0}$
is $2$-connected. So consider the induction step $t\geq1$. By the induction hypothesis,
$G_{t-1}$ is $2$-connected. Say $G=G_{t-1}+P_{t}$, where $P_{t}:=v_{1}\dots v_{s}$.
Consider any $x\in V(G)$. We must show that $G-x$ is connected. Since $G_{t-1}$
is $2$-connected, $G_{t-1}-x$ is connected. Also every vertex in $P_{t}-x$ is
connected to a vertex $v\in\{v_{1},v_{s}\}\subseteq G_{t-1}-x$ in $G$, even if
$x\in V(P_{t})$. It follows that $G$ is $2$-connected.\end{proof}
\begin{defn}
Let $e=xy$ be an edge in a graph $G$, and fix a new vertex $v_{e}$. The graph
$G\cdot e$ obtained by \emph{contracting} $e$ is defined by
\[
G\cdot e:=(G-x-y)\cup K(v_{e},N_{G}(\{x,y\})-x-y).
\]
Note that if $P'$ is a path in $G\cdot e$ then either $P'$ is a path in $G$ or
$v_{e}\in V(P')$. In the latter case we can obtain a path in $G$ by replacing $v_{e}$
by one of $x,y,xy,yx$. If $P$ is a path in $G$ then either $P$ is a path in $G\cdot e$
or one or both of $x,y$ are in $V(P)$. In the latter case we can obtain a path
$P'$ in $G\cdot e$ by replacing one of $x,y,xPy,yPx$ by $v_{e}$.\end{defn}
\begin{lem}
\label{contrac-L}Let $G=(V,E)$ be a graph and $xy\in E$. If $S'$ is a separating
set of $G':=G\cdot xy$ then either $S'\subseteq V$ and $S'$ is a separating
set of $G$ or $v_{xy}\in S'$ and $S:=S'-v_{xy}+x+y$ is a separating set of $G$.\end{lem}
\begin{proof}
Suppose $v_{xy}\notin S'$. Then $S'\subseteq V(G)$ and $v_{xy}$ is in a component
$H$ of $G'-S'$. Choose a vertex $w$ in another component of $G'-S'$; then $w\in G$.
Consider any $w,\{x,y\}$-path $P=w\dots u'u$ in $G$. Then $P':=wPu'v_{xy}$ is
a $w,v_{xy}$ path in $G'$. Since $S'\subseteq V$ is a separating set in $G'$,
there is $v\in S'\cap V(P)\subseteq V$. Thus $S'$ separates $x$ from $w$ in $G$.
Else $v_{xy}\in S$. Let $u$ and $v$ be vertices in distinct components of $G'-S'$.
If $P$ is a $u,v$-path in $G$ then there is a $u,v$-path $P'$ in $G'$ such
that $V(P')\subseteq V(P)+v_{xy}$, and $v_{xy}\in V(P')$ only if $x$ or $y$ is
in $V(P)$. As $S'$ separates $u$ and $v$ in $G'$, there is $w\in V(P')\cap S'$.
If $w\ne v_{xy}$ then $w\in S$. If $w=v_{xy}$ then $x$ or $y$ is in $V(P)\cap S$.
So $S$ separates $u$ and $v$ in $G$.\end{proof}
\begin{lem}[{Thomassen {[}1980{]} 6.2.9}]
\label{Thomassen3}Every $3$-connected graph $G$ with $|G|\geq5$ has an edge
$e$ such that $G\cdot e$ is $3$-connected. \end{lem}
\begin{proof}
Suppose not. Consider any edge $xy$. Since $G':=G\cdot xy$ is not $3$-connected
and $|G'|\geq4$, $G'$ has a separating $2$-set $S'$. By Lemma~\ref{contrac-L},
$S'$ has the form $S'=\{v_{xy},z\}$ and $S:=\{x,y,z\}$ is a separating $3$-set
in $G$.
So far the edge $xy\in E$ and the $2$-separating set $S'$ are arbitrary. Now
choose $xy$ and $S'=\{v_{xy},z\}$, and set $S=\{x,y,z\}$ as above, so that $G-S$
has a component $H$ that is as large as possible among all possible choices of $xy$
and $S'$. Let $H'$ be another component of $G-S$. Since $S$ is a minimal separating
set, each of $x,y,z$ has a neighbor in each of $H$ and $H'$. Let $u$ be a neighbor
of $z$ in $H'$. Then $G\cdot uz$ has a separating set $\{v_{uz},v\}$, and $\{u,v,z\}$
is a separating set for $G$.
Put $H^{+}=G[H+x+y]$. Then $H^{+}$ is connected, and $u,z\notin V(H^{+})$. Thus
$H^{*}:=H^{+}-v$ is disconnected, since otherwise $H^{*}$ is contained in a component
$C$ of $G-\{u,v,z\}$ with $|C|\geq|H|+|\{x,y\}-v|\geq|H|+1$, contradicting the
choice of $xy,z,H$. As $x$ and $y$ are adjacent, they are in the same component
of $H^{*}$; let $U$ be a different component of $H^{*}$. Suppose $ab\in E(U,V\smallsetminus U)$
with $a\in U\subseteq H$. Then $b\in V(H)\cup S=V(H^{+})+z$. If $b\in H^{+}$ then
$b=v$; else $b=z$. Thus $\{v,z\}$ separates $U$ from $H'$, contradicting
$\kappa(G)\geq3$.\end{proof}
\begin{defn}
A sequence of graphs $G_{0},\dots,G_{s}$ is a $3$\emph{-witness for a graph $G$}
iff\end{defn}
\begin{enumerate}
\item $G_{0}=K_{4}$ and $G_{s}=G$; and
\item for all $i\in[s]$ there is $xy\in E(G_{i})$ with $G_{i-1}=G\cdot xy$ and $d_{G_{i}}(x),d_{G_{i}}(y)\geq3$.\end{enumerate}
\begin{thm}
A graph $G$ is $3$-connected iff it has a $3$-witness.\end{thm}
\begin{proof}
First suppose that $G$ is $3$-connected. Then $|G|\geq4$. We show by induction
on $|G|$ that $G$ has a $3$-witness. Suppose $|G|=4$. If $xy\notin E(G)$ then
$V(G)\smallsetminus\{x,y\}$ is a $2$-set that separates $x$ from $y$, a contradiction.
So $G=K_{4}$, and $G_{0}=K_{4}=G$ is a $3$-witness for $G$. Otherwise, $|G|\geq5$.
By Lemma~\ref{Thomassen3}, there exists an edge $xy\in E(G)$ such that $G\cdot xy$
is $3$-connected. Since $G$ is $3$-connected, $d(x),d(y)\geq3$. By induction,
$G\cdot xy$ has a $3$-witness $G_{0},\dots,G_{s}$. So $G_{0},\dots,G_{s},G$ is
a $3$-witness for $G$.
Now let $G_{0},\dots,G_{s}$ be a $3$-witness for $G$. Argue by induction on $s$
that $G$ is $3$-connected. If $s=0$ then $K_{4}=G_{0}=G$ is $3$-connected. Otherwise,
for some edge $xy\in E(G)$, both $G_{s-1}=G\cdot xy$ and $d_{G}(x),d_{G}(y)\geq3$.
By induction $G\cdot xy$ is $3$-connected. Suppose for a contradiction that $S$
is a $2$-separator in $G$. If $S=\{x,y\}$ then $v_{xy}$ is a cut vertex of $G\cdot xy$,
a contradiction. So there is a component $H$ of $G-S$ containing at least one,
say $x$, of $x$ and $y$, and another component $H'$ containing neither $x$ nor
$y$. As $d_{G}(x)\geq3$ and $N(x)\subseteq V(H)\cup S$, $x$ has a neighbor $v$
in $H$. If $y\in H$ then $S$ separates $v_{xy}$ from $H'$ in $G\cdot xy$; else
$y\in S$, and $S':=S-y+v_{xy}$ separates $v$ from $H'$ in $G\cdot xy$. Anyway,
$\kappa(G\cdot xy)\leq2$, a contradiction.
\end{proof}
The last paragraph of the above proof is subtle. If $d_{G}(x)<3$ then we could have
$S=N(x)$, and $V(H)=\{x\}$. Then $H-x$ is not a component of $G\cdot xy-S'$ since
$H-x$ has no vertices.
\begin{conjecture}
[Lovasz]\label{LovPart}There exists a function $f:\mathbb{Z}^{+}\rightarrow\mathbb{Z}^{+}$
such that for all $k\in\mathbb{Z}^{+}$ and $f(k)$-connected graphs $G$ and all
vertices $x,y\in V(G)$, there exists a partition $\{V_{1},V_{2}\}$ of $V(G)$ such
that $G[V_{1}]$ is an $x,y$-path and $G[V_{2}]$ is $k$-connected.
\begin{x}Show that Conjecture~\ref{LovPart} is true in the case $k=1$ with $f(1):=3$.\end{x}
\end{conjecture}
\section{Menger's Theorems}
\begin{defn}
A \emph{directed $x,y$-path }is a directed graph $\vec{P}=(\{v_{1}\dots v_{t}\},\{\vec{e}_{1},\dots,\vec{e}_{t-1}\})$,
where $\vec{e}_{i}\in E(v_{i},v_{i+1})$.
\end{defn}
\begin{defn}
Two $x,y$-paths (directed or not) are edge disjoint if they have no common edges.
(They may have common vertices.) They are \emph{independent} or \emph{internally
disjoint} if they have no common internal vertices. Define the following related
parameters:
\begin{enumerate}
\item $\vec{\lambda}'(x,y)$ is the maximum size of a set of edge-disjoint, directed $x,y$-paths;
\item $\lambda'(x,y)$ is the maximum size of a set of edge-disjoint $x,y$-paths;
\item $\lambda(x,y)$ is the maximum size of a set of independent $x,y$-paths;
\item $\vec{\lambda}(x,y)$ is the maximum size of a set of independent, directed $x,y$-paths.
\end{enumerate}
\end{defn}
\begin{defn}
Consider a graph $G=(V,E)$ and an orientation $\vec{G}$ of $G$. Let $x,y\in V$
be distinct. A set $\vec{F}\subseteq\vec{E}$ \emph{is a directed} $x,y$\emph{-disconnecting
set} if there are no directed $x,y$-paths in $\vec{G}-\vec{F}$; a set $F\subseteq E$\emph{
}is an\emph{ $x,y$}-\emph{disconnecting set} if there are no $x,y$-paths in $G-F$.
Now suppose $xy\notin E$. A subset $S\subseteq V(G)-\{x,y\}$ is an \emph{$x,y$-separating
set} if there is no $x,y$-path in $G-S$. Now assume $\vec{xy}\notin\vec{E}$. A
subset $S\subseteq V(G)-\{x,y\}$ is is a \emph{directed $x,y$-separating set} if
there is no directed $x,y$-path in $G-S$. Define the following related parameters:
\begin{enumerate}
\item $\vec{\kappa}'(x,y)$ is the minimum size of a directed $x,y$-disconnecting set;
\item $\kappa'(x,y)$ is the minimum size of an $x,y$-disconnecting set;
\item if $xy\notin E$ then $\kappa(x,y)$ is the minimum size of an $x,y$- separating
set;
\item if $\vec{xy}\notin E$ then $\vec{\kappa}(x,y)$ is the minimum size of a directed
$x,y$-separating set.
\end{enumerate}
If $ab\in E(G)$ then no $S\subseteq V$ separates $a$ and $b$. This is why $\kappa(a,b)$
is only defined when $ab\notin E$.
\textcolor{red}{Suppose $\vec{P}=(V,\vec{E})$ is a directed graph. Then $P=(V,\{e:\vec{e}\in E\})$.
For a (undirected) graph $G=(V,E)$ let $\overleftrightarrow{G}=(V,\{\vec{e},\cev e:e\in E\})$.
Notice that if $\vec{G}$ is a directed graph then it is possible that $\overleftrightarrow{G}\ne\vec{G}$;
for instance $\vec{G}$ could be a directed path. For an $x,y$-path $P$ let $\overrightarrow{P}$
be the directed $x,y$-path obtained by directing the edges of $P$ toward $y$.
Now, if $\vec{P}$ is a directed $x,y$-path then $\overrightarrow{P}=\vec{P}$. }\end{defn}
\begin{thm}[Menger 1927, directed edge]
\label{MengerDE-thm}Every directed graph $G(V,\vec{E})$ satisfies \linebreak{}
$\vec{\lambda}'(x,y)=\vec{\kappa}'(x,y)$ for all distinct $x,y\in V$.\end{thm}
\begin{proof}
Consider the network $N=(G,x,y,c)$, where $c(\vec{e})=1$ for all $\vec{e}\in\vec{E}$.
For a flow $f$ let $H(f)$ be the spanning graph of $G$ whose edge set is $\vec{E}(H(f))=\{\vec{e}\in\vec{E}:f(\vec{e})=1\}$.
Among all maximum flows choose a flow $f$ such that $\left\Vert H(f)\right\Vert $
is minimal. Then $H(f)$ has no directed cycle: Suppose $\vec{C}$ is a directed
cycle in $H(f)$. Each $\vec{e}\in E(\vec{C})$ satisfies $f(\vec{e})=1=-f(\cev e)$.
Set
\[
f'(\vec{e})=\begin{cases}
f(\vec{e}) & \mbox{if}~\vec{e}\notin\vec{E}(\overleftrightarrow{C})\\
0 & \mbox{if}~e\in E(\overleftrightarrow{C})
\end{cases}.
\]
Then $f'$ is a flow (check (F1--F3)) and $v(f')=v(f)$, but $\left\Vert H(f')\right\Vert =\left\Vert H(f)\right\Vert -\|\vec{C}\|$,
contradicting the choice of $f$.
Let $(U,\overline{U})$ be a minimum cut in $N$. By the Max-flow, Min-Cut Theorem,
$c(U,U')=v(f)$. Also every directed $x,y$-path contains an edge of $\vec{F}:=\vec{E}(U,\overline{U})$.
Thus $\vec{F}$ is a directed $x,y$-disconnecting set. So $\vec{\kappa}'(x,y)\leq c(U,\overline{U})$.
Let $\mathcal{P}$ be a set of $\vec{\lambda}'(x,y)$ edge-disjoint directed $x,y$-paths
and $\vec{S}$ be a directed $x,y$-disconnecting of size $\vec{\kappa}'(x,y)$.
Then every path $P\in\mathcal{P}$ contains an edge of $\vec{S}$. So $\vec{\lambda}'(x,y)\leq\vec{\kappa}'(x,y)$.
Now $\vec{\lambda}'(x,y)\leq\vec{\kappa}'(x,y)\leq c(U,\overline{U})=v(f)$. Thus
it suffices to show that $v(f)\leq\vec{\lambda}'(x,y)$. Argue by induction on $l:=\vec{\lambda}'(x,y)$.
For the base step $l=0$, arguing by contraposition, suppose $v(f)\geq1$. For
each $v\in G$ with $d_{H(f)}^{+}(v)\geq1$ fix one edge $\vec{e}_{v}\in\vec{E}(v,V-v)$.
By (F2), if $\vec{e}_{v}=\vec{vw}$ then $\vec{e}_{w}$ exists unless $w=y$. Consider
the maximum directed walk $\vec{W}=v_{1}\dots v_{t}$, where $v_{1}=x$ and $\vec{v_{i}v}_{i+1}=\vec{e}_{v_{i}}$
for all $i\in[t-1]$. As $v(f)\geq1$, it has at least one edge $\vec{xv}_{2}$.
Since $H(f)$ does not contain a directed cycle $\vec{W}$ is a directed path. Since
$G$ is finite, $\vec{W}$ must end. Since it does not end before $y$, it is a directed
$x,y$-path. So $\vec{\lambda}'(x,y)\geq1$.
Now suppose $l\geq1$ and let $P=xx_{1}\dots y\in\mathcal{P}$. Set $G'=G-\vec{E}(P)$
and $N'=(G',x,y,c)$. Then $v_{N}(f)=v_{N'}(f)+f(\vec{xx_{1}})=v_{N'}(f)+1$ since
$\vec{E}_{N}(x,V-x)=\vec{E}_{N'}(x,V-x)+\vec{xx_{1}}$. If $\mathcal{P}'$ is a set
of directed, edge-disjoint $x,y$-paths in $G'$ then $\mathcal{P}'+P$ is a set
of directed, edge-disjoint $x,y$-paths in $G$. Thus $\vec{\lambda}'_{G'}(x,y)+1\leq\vec{\lambda}_{G}'(x,y)$.
Let $f'$ be a maximum flow in $N'$. Then $v_{N'}(f)\leq v_{N'}(f')$. Now, using
induction,
\[
v_{N}(f)=v_{N'}(f)+1\leq v_{N'}(f')+1=_{i.h.}\vec{\lambda}'_{G'}(x,y)+1\leq\vec{\lambda}_{G}'(x,y).\qedhere
\]
\end{proof}
\begin{thm}[Menger 1927, edge]
\label{MengerE-thm}Every graph $G(V,E)$ satisfies $\lambda'(x,y)=\kappa'(x,y)$
for all distinct $x,y\in V$.\end{thm}
\begin{proof}
It suffices to prove
\[
\vec{\kappa}_{\overleftrightarrow{G}}(x,y)=_{(0)}\vec{\lambda}_{\overleftrightarrow{G}}(x,y)\leq_{(1)}\lambda'_{G}(x,y)\leq_{(2)}\kappa'_{G}(x,y)\leq_{(3)}\vec{\kappa}'_{\overleftrightarrow{G}}(x,y).
\]
Theorem~\ref{MengerDE-thm} implies (0). As every $x,y$-path meets every $x,y$-disconnecting
set, (2) holds.
For (1), let $\vec{\mathcal{P}}$ be a set of $\vec{\lambda}_{\overleftrightarrow{G}}'(x,y)$
edge-disjoint, directed $x,y$-paths in $\overleftrightarrow{G}$ with $\|\bigcup\{\vec{E}(\vec{P}):\vec{P}\in\mathcal{\vec{P}}\|$
minimum, and set $\mathcal{P}=\{P:\vec{P}\in\vec{\mathcal{P}}\}$. Suppose some $P,Q\in\mathcal{P}$
have a common edge $vw$. Since $\overrightarrow{P}$ and $\overrightarrow{Q}$ are
edge-disjoint, (say) $\vec{vw}\in\vec{E}(\overline{P})$ and $\vec{wv}\in\vec{E}(\overrightarrow{Q})$.
Then $\overrightarrow{\mathcal{P}}+x\overrightarrow{P}v\overrightarrow{Q}y+x\overrightarrow{Q}w\overrightarrow{P}y-\overrightarrow{P}-\overrightarrow{Q}$
is a set of $\vec{\lambda}_{\overleftrightarrow{G}}(x,y)$ edge-disjoint, directed
$x,y$-paths in $\overleftrightarrow{G}$ with two less edges, contradicting the
choice of $\vec{\mathcal{P}}$. So $\vec{\lambda}'_{\overleftrightarrow{G}}(x,y)=|\mathcal{P}|\leq\lambda'_{G}(x,y)$.
For (3), let $\vec{F}$ be a directed $x,y$-disconnecting set of size $\vec{\kappa}'(x,y)$,
and consider any $x,y$-path $P$ in $G$. Then there is an edge $\vec{vw}$ in $\vec{E}(\vec{P})\cap\vec{F}$.
Thus $vw\in F:=\{e:\vec{e}\in\vec{F}\}$ is an $x,y$-disconnecting set. So $\kappa'_{G}(x,y)\leq|F|\leq|\vec{F}|=\vec{\kappa}'_{\overleftrightarrow{G}}(x,y).$\end{proof}
\begin{thm}[Menger 1927, directed-vertex]
\label{MengerVD-thm}Every directed graph $\vec{G}(V,\vec{E})$ satisfies $\vec{\lambda}(x,y)=\vec{\kappa}(x,y)$
for all distinct $x,y\in V$ with $\vec{xy}\notin\vec{E}$.\end{thm}
\begin{proof}
Let $\vec{H}=(V'\cup V'',\{\vec{v'w}'':\vec{vw}\in\vec{E}\}\cup\{\vec{v''v'}:v\in V\})$,
where $V'=\{v':v\in V\}$, $V''=\{v'':v\in V\}$, and $V'\cap V''=\emptyset$. For
a directed path $\vec{P}=v_{1}v_{2}\dots v_{s}$ in $\vec{G}$ let $\vec{P}'={\color{red}v_{1}''}v_{1}'v_{2}''v_{1}'\dots v_{{\color{red}s}-1}''v_{{\color{red}s}-1}'v_{{\color{red}s}}''{\color{red}v_{s}'}$.
Then $\vec{P}'$ is a directed path in $\vec{H}$. Every directed ${\color{red}v''_{1},v_{s}'}$-path
in $\vec{H}$ has the form of $\vec{P}'$ \textcolor{red}{for} a unique path $\vec{P}=v_{1}\dots v_{s}$.
So ${\color{red}v''v'}\in E(\vec{P}')$ if and only if $v\in V(\vec{P})$. It suffices
to prove
\[
\vec{\kappa}'_{\vec{H}}({\color{red}x'',y'})=_{(0)}\vec{\lambda}'_{\vec{H}}({\color{red}x'',y'})\leq_{(1)}\vec{\lambda}{}_{\vec{G}}(x,y)\leq_{(2)}\vec{\kappa}{}_{\vec{G}}(x,y)\leq_{(3)}\vec{\kappa}'_{\vec{H}}({\color{red}x'',y'}).
\]
Theorem~\ref{MengerDE-thm} implies (0). As every directed $x,y$-path meets every
directed $x,y$-separating set, (2) holds.
For (1), let $\vec{\mathcal{P}'}=\{\vec{P}'_{i}:i\in[l]\}$ be a set of $l:=\vec{\lambda}_{\vec{H}}'({\color{red}x'',y'})$
edge-disjoint, directed $x,y$-paths in $\vec{H}$. Then $\vec{\mathcal{P}}=\{\vec{P}_{i}:i\in[l]\}$
is a set of $\vec{\lambda}_{\vec{G}}'(x,y)$ independent, directed $x,y$-paths in
$\vec{G}$. So (1) hods.
For (3), let $\vec{F}$ be a directed, ${\color{red}x'',y'}$-disconnecting set in
$\vec{H}$ of size $\vec{\kappa}'_{\vec{H}}({\color{red}x'',y'})$. Suppose $\vec{e}\in\vec{F}$.
As $\vec{xy}\notin\vec{E}$, if $\vec{e}=v'w''$ then either $v'\ne x'$ or $w''\ne y''$.
In the first case replace $\vec{e}$ by $\vec{e}':=\vec{v''v'}$; else replace $\vec{e}$
by $\vec{e}':=\vec{w''w'}$. Call the resulting set $\vec{F}'$. Then $|\vec{F}'|\leq|\vec{F}|$,
and $\vec{F}'$ is also a directed $x',y''$-disconnecting set in $\vec{H}$. Set
$S=\{v\in V:\vec{v''v'}\in\vec{F}'\}$. Suppose $\vec{P}$ is a directed $x,y$-path
in $G$. Then there is $\vec{v''v'}\in\vec{E}(\vec{P}')\cap\vec{F}'$, and $v\in V(\vec{P})\cap S$.
So $S$ is a directed $x,y$-separating set in $\vec{G}$. Now $\vec{\kappa}_{\vec{G}}(x,y)\leq|S|\leq|F'|\leq|F|=\vec{\kappa}'_{\vec{H}}({\color{red}x'',y'})$,
so (2) holds. \end{proof}
\begin{thm}[Menger 1927, vertex 4.2.17]
\label{MengerV-thm}Every graph $G(V,E)$ satisfies $\lambda(x,y)=\kappa(x,y)$
for all distinct $x,y\in V$ with $xy\notin E$.\end{thm}
\begin{proof}
It suffices to prove
\[
\vec{\kappa}{}_{\overleftrightarrow{G}}(x,y)=_{(0)}\vec{\lambda}{}_{\overleftrightarrow{G}}(x,y)\leq_{(1)}\lambda_{G}(x,y)\leq_{(2)}\kappa{}_{G}(x,y)\leq_{(3)}\vec{\kappa}{}_{\overleftrightarrow{G}}(x,y).
\]
Theorem~\ref{MengerVD-thm} implies (0). As every $x,y$-path meets every $x,y$-separating
set, (2) holds. For (1) note that if $\vec{P}$ and $\vec{Q}$ are independent, directed
$x,y$-paths then $P$ and $Q$ are independent $x,y$-paths. For (3) suppose $S$
is a directed $x,y$-separating set in $\overleftrightarrow{G}$, and consider an
$x,y$-path $P\subseteq G$. As $\overrightarrow{P}$ is a directed $x,y$-path,
$V(\overrightarrow{P})\cap S\ne\emptyset$, so $S$ is an $x,y$-separating set and
$(3)$ holds. \end{proof}
\begin{defn}
Let $G=(V,E)$ be a graph and $A$ and $B$ be subsets of $V$. An $A,B$\emph{-path}
is a path with exactly one end in $A$ and exactly one end in $B,$ and whose internal
vertices are in neither $A$ nor $B$. Let $l(A,B)$ be the maximum size of a set
of disjoint $A,B$-paths. An $A,B$\emph{-separating set} is a set of vertices $S$
such that $G-S$ has no $A,B$-paths (possibly $S\cap A\ne\emptyset$ or $S\cap B\ne\emptyset$).
Let $k(A,B)$ be the minimum cardinality of an $A,B$-separating set.\end{defn}
\begin{thm}[Menger 1927 4.2.17]
\label{menger}Let $G=(V,E)$ be a graph, and suppose $A,B\subseteq V$. Then $l(A,B)=k(A,B)$. \end{thm}
\begin{proof}
For new vertices $a,b\notin V$, let $G^{+}=G\cup K(a,A)\cup K(b,B)$. Then $l_{G}(A,B)=\lambda_{G^{+}}(a,b)=\kappa_{G^{+}}(a,b)=k_{G}(A,B)$.\end{proof}
\begin{thm}[4.2.21]
\label{locMeng-th-1}Every graph $G=(V,E)$ satisfies $\kappa(G)=t:=\min_{a,b\in V,a\ne b}\lambda(a,b)$.\end{thm}
\begin{proof}
Choose $a,b$ with $t=\lambda(a,b)$. If $G$ is complete then
\[
t=\lambda(a,b)=1+(|G|-2)=|G|-1=\kappa(G),
\]
since $ab$ is an $a,b$-path, and $acb$ is also an $a,b$-path for all $c\in V-a-b$.
Otherwise $G$ has a separating set $S$ with $|S|=\kappa(G)$. Let $x,y$ be vertices
in distinct components of $G-S$. Then
\begin{equation}
t=\lambda(a,b)\leq\lambda(x,y)\leq|S|=\kappa(G).\label{t1$ then $W+a$ is a separator of $G$ with size $\kappa(G')+1\geq\kappa(G)$.
Else $|G|=|W|+2$, and so $\kappa(G)\leq|G|-1=\kappa(G')+1$. \end{proof}
\begin{defn}[4.2.22]
Let $G=(V,E)$ be a graph with $x\in V$ and $U\subseteq V$. An $x,U$-fan is a
set $\mathcal{F}$ of $x,U$-paths such that $|\mathcal{F}|=|U|$ and $F\cap F'=\{x\}$
for all $F,F'\in\mathcal{F}$.\end{defn}
\begin{thm}[4.2.23]
\label{Menger-Fan} A graph $G=(V,E)$ is $k$-connected if and only if $|G|\geq k+1$
and $G$ has an $x,U$-fan for all $x\in V$ and all $k$-sets $U\subseteq V-x$
.\end{thm}
\begin{proof}
(Sketch) Suppose $G$ is $k$-connected, and consider $x\in V$ and $U\in\binom{V-x}{k}$.
Let $u\notin V$ be a new vertex. By Lemma~\ref{exp-lem}, $G^{+}:=G\cup K(u,U)$
is $k$-connected. Thus there exist $k$ independent $u,x$-paths in $G^{+}$. Deleting
$u$ yields an $x,U$-fan in $G$.
Conversely, the hypothesis implies $\delta(G)\geq k$, and for all $x$ and $y$,
there exist $k$ disjoint $N(x),N(y)$-paths. Thus by Theorem~\ref{locMeng-th-1}
\[
\kappa(G)=\min_{x,y\in V,x\ne y}\lambda(x,y)\geq k.
\]
\end{proof}
\begin{thm}[HW 4.2.24]
Let $G=(V,E)$ be a $k$-connected graph with $k\geq2$. Then for any $k$-set $S\subseteq V$
there is a cycle $C\subseteq G$ with $S\subseteq V(C)$. \end{thm}
\begin{proof}
Let $C\subseteq G$ be a cycle containing as many vertices of $S$ as possible. It
exists because $\delta(G)\geq\kappa(G)\geq2$. We claim that $S\subseteq V(C)$.
Otherwise, let $v\in S\smallsetminus V(C)$. Then $|S\cap V(C)||S\cap V(C)|$. So $D=x_{i+1}\overrightarrow{C}x_{i}FvFx_{i+1}$
is a cycle containing $S\cap V(C)+v$.
\end{proof}
\begin{x}({*}) Every $2$-connected graph $G$ has a cycle of length at least\textcolor{red}{{}
${\normalcolor \min\{|G|,2\kappa(G)\}}$}.\end{x}
\begin{x}({*}) If $G$ is a $2$-connected graph with $\alpha(G)\leq\kappa(G)$
then $G$ is hamiltonian.\end{x}
\begin{x}(+) Let $G$ be a $2$-connected graph that does not induce $K_{1,3}$.
Then $G$ has a cycle of length at least\textcolor{red}{{} ${\normalcolor \min\{|G|,4\kappa(G)\}}$}.\end{x}
\begin{x}(++) The Conjecture~\ref{LovPart} is known to be true in the case $k=2$
with $f(2)\geq5$. Prove this with $f(2)\geq8$. {[}Hint: Choose $P$ so that $G-P$
contains the biggest $2$-connected subgraph possible.{]}\end{x}
\begin{x}({*}) Let $G=(V,E)$ is a graph with $x\in V$ and $Y,Z\subseteq V$, $|Y|=4$,
and $|Z|=5$. Suppose $\mathcal{Q}=\{Q_{y}:y\in Y\}$ is an $x,Y$-fan in $G$, where
each $Q_{y}$ is an $x,y$-path. Similarly, suppose $\mathcal{R}=\{R_{z}:z\in Z\}$
is an $x,Z$-fan in $G$, where each $R_{z}$ is an $x,z$-path. Prove: There exists
a $x,(Y+z)$-fan in $G$ for some $z\in Z$. \textcolor{red}{{[}Hint: Add new vertices
$w$ and $x'$ with neighborhoods $N(w)=Z$ and $N(x')=Y+{\color{blue}w}$; then
apply a theorem.{]}}\end{x}
\chapter{Graph coloring}
\begin{defn}
Let $G=(V,E)$ be a graph and $C$ be a set (of colors). A \emph{proper $C$-coloring}
of $G$ is a function $f:V\rightarrow C$ such that for all vertices $x,y\in V$
if $xy\in E(G)$ then $f(x)\ne f(y)$. If $k$ is a positive integer, we say that
$f$ is a proper $k$-coloring if it is a proper $[k]$-coloring. The chromatic number
$\chi(G)$ is the least $k$ such that $G$ has a proper $k$-coloring. In this case
$G$ is said to be $k$-chromatic. If $G$ has a $k$-coloring then it is said to
be $k$-colorable. In this chapter we will assume that all colorings are proper unless
otherwise stated. For $i\in C$, $f^{-1}(i)$ is called a color class.\end{defn}
\begin{prop}[HW]
\label{colP}Every graph $G$ satisfies $\omega(G),\frac{|G|}{\alpha(G)}\leq\chi(G)\leq\Delta(G)+1$.
\end{prop}
\begin{x}({*}) Prove Proposition~\ref{colP}.\end{x}
\section{Examples}
\begin{example}[{5.2.3 Mycielski {[}1955{]}}]
\label{Mycielski}For every positive integer $k$ there exists a graph $G_{k}$
with $\omega(G_{k})\leq2$ and $\chi(G_{k})=k$.\end{example}
\begin{proof}
We argue by induction on $k$. For $k\leq2$ let $G_{k}=K_{k}$. Now suppose $k\geq3$
and we have constructed $G_{k-1}=(V_{k-1},E_{k-1})$ as required. We first construct
$G_{k}=(V_{k},E_{k})$ as follows: Let $V_{k-1}'=\{v':v\in V_{k-1}\}$ be a set of
new vertices, $x_{k}$ be a new vertex, and put
\[
V_{k}=V_{k-1}\cup V_{k-1}'+x_{k}\mbox{ and},
\]
\[
E_{k}=E_{k-1}\cup\{uv':uv\in E_{k-1}\}\cup\{x_{k}v':v'\in V_{k-1}'\}.
\]
So $N(v')\cap V_{k-1}=N(v)\cap V_{k-1}$ for all $v\in V_{k-1}$.
Suppose $\omega(G_{k})\geq3$, and choose $Q=K_{3}\subseteq G_{k}$. Then $k\geq3$.
Since $N(x_{k})=V_{k-1}'$ is independent, $x_{k}\notin Q$, and $|V_{k-1}'\cap Q|\leq1$.
As $\omega(G_{k-1})=_{i.h.}2$, there is at least one $v'\in V{}_{k-1}'\cap Q$.
Hence $N(v')\cap Q=N(v)\cap Q$, and $Q-v'+v$ is a $K_{3}$ in $G_{k-1}$, a contradiction.
Now $\chi(G_{k})\leq k$: If $k\leq2$ this is obvious; else $G_{k-1}$ has a $(k-1)$-coloring
$f'$ by induction. Extend $f'$ to a $k$-coloring $f$ of $G_{k}$ by setting $f(v')=k$
(new color), and $f(x_{k})=1$.
Finally, $\chi(G_{k})\geq k$: If $k\leq2$ this is obvious. If $k\geq3$ then it
suffices to show that every $(k-1)$-coloring $g$ of $G_{k}-x_{k}$ satisfies $g(V'_{k-1})=[k-1]$,
since $x_{k}$ will require a new color. Suppose not. After possibly renaming color
classes, let $k-1\in[k-1]\smallsetminus g(V_{k-1}')$. For a contradiction, we construct
a $(k-2)$-coloring $h$ of $G_{k-1}$. Define $h:V_{k-1}\rightarrow[k-2]$ by:
\[
h(v)=\begin{cases}
g(v) & \mbox{ if }g(v)\ne k-1\\
g(v') & \mbox{ if }g(v)=k-1
\end{cases}.
\]
If $uv\in E_{k-1}$ then $g(u)\ne g(v)$ since $g$ is proper; if $g(u)\ne k-1\ne g(v)$
then
\[
h(u)=g(u)\ne g(v)=h(v);
\]
else exactly one of $u,v$ is colored with $k-1$ by $g$; say $g(v)=k-1$. Since
$uv'\in E_{k}$,
\[
h(u)=g(u)\ne g(v')=h(v).\qedhere
\]
\end{proof}
\begin{x}({*}) Let $G_{k}$ be the graph in Example\textcolor{red}{~}\textcolor{black}{\ref{Mycielski}.}\textcolor{red}{{}
}Prove that $G_{k}$ is critical, i.e., $\chi(G_{k}-e)<\chi(G_{k})\text{ }(=k)$
for all $e\in E_{k}$. {[}Hint: There are three types of edges to consider.{]}\end{x}
\begin{x}({*}) Let $G$ be graph, and suppose any two odd cycles $C,C'\subseteq G$
have a common vertex. Prove that $\chi(G)\leq5$.\end{x}
\begin{x}({*}) Let $P=\{v_{1},\dots,v_{n}\}$ be a path, and suppose $G=(V,E)$
is a graph\textcolor{red}{{} }such that $V$ is a subset of the set of subpaths of
$P$, and $E=\{RQ\in V:R\cap Q\ne\emptyset\}$. Prove that $\chi(G)=\omega(G)$.
{[}Hint: Order V so that if $Q,R\in V$ (subpaths of $P$), the first vertex of $Q$
is $v_{i}$, the first vertex of $R$ is $v_{j}$, and $i1$.
Let $P$ be a maximum path; as $G$ is connected and $|G|\geq2$, $|P|\geq2$. Pick
an end $u$ of $P$ with $u\ne v$. As $u$ is not a cut-vertex, $G-u$ is connected.
By induction, there exists an ordering $v_{2},\dots,v_{n}$ of $V-u$ satisfying
({*}) for $G-u$. So $v_{1}:=u,v_{2},\dots,v_{n}$ satisfies ({*}) for $G$.
\end{proof}
Define a $b$\emph{-obstruction} to be $K_{b}$, or, if $b=3$, an odd cycle, and
let (non-standard) $\omega^{*}(G)$ be the largest integer $b$ such that $G$ contains
a $b$-obstruction. Then $\omega^{*}(G)\leq\chi(G)\leq\Delta(G)+1$.
The following notation will be useful. Recall that for a graph $G=(V,E)$ and subsets
$A,B\subseteq V$, $\left\Vert A,B\right\Vert =\left\Vert A,B\right\Vert _{G}=|E_{G}(A,B)|$.
If $A=\{a\}$ or $H\subseteq G$ we may also write $\left\Vert a,B\right\Vert =\left\Vert \{a\},B\right\Vert $
or $\left\Vert A,H\right\Vert =\left\Vert A,V(H)\right\Vert $.
\begin{thm}
[Brooks (1941)]\label{Brooks}Every graph satisfies $\chi(G)\leq\max\{\omega^{*}(G),\Delta(G)\}$.\end{thm}
\begin{proof}
Set $\Delta:=\Delta(G)$, $\chi:=\chi(G)$, $\omega^{*}:=\omega^{*}(G)$, and argue
by induction on $|G|$. Since $\chi\leq\Delta+1$, it suffices to show that $\omega^{*}=\Delta+1$
or $\chi\leq\Delta$. If $\Delta\leq1$ then $\omega^{*}=\Delta+1$. If $\Delta=2$
then $2\leq\omega^{*}\leq\chi\leq3$; if $\omega^{*}=2$ then $\chi\leq2$ as $G$
has no odd cycle. So assume $3\leq\omega^{*}\leq\Delta$.
By Proposition~\ref{Sep}, assume $G$ is $2$-connected. Let $S$ be a maximal
independent set, and put $G':=G-S$. Then $\Delta(G')<\Delta$, since every vertex
of $G'$ has a neighbor in $S$. If $\omega^{*}(G')<\Delta$ then $(\Delta-1)$-color
$G'$ by induction, and use a new color for $S$. Else, consider a $\Delta$-obstruction
$Q\subseteq G'$. Choose $y\in S$ with $\left\Vert y,Q\right\Vert \geq1$. Since
either $Q=K_{\Delta}$ and $\omega(G)\leq\Delta$ or $\Delta=3$ and $Q$ is an odd
cycle on more than $\Delta$ vertices, $V(Q)\nsubseteq N(y)$. As $Q$ is connected,
there is an edge $wx\in E(Q)$ with $x\in N(y)$ and $w\notin N(y)$; let $w'\in N_{Q}(x)-w$.
As $G$ is $2$-connected, there is a minimum $y,Q$-path $P:=y\dots z$ in $G-x.$
If $z\ne w$ then set $w^{*}=w$. If $z=w$ then $\left\Vert P\right\Vert \geq2$,
and so $yw'\notin E$; set $w^{*}=w'$. Anyway $z\ne w^{*}$, $w^{*}x\in E$, and
$w^{*}y\notin E$. Since $Q-w^{*}$ and $P-y$ are connected through $z$, Lemma~\ref{order-lem}
implies $H:=G[Q\cup P]$ has an ordering $L:=w^{*},y,v_{1},\dots,v_{t},x$, where
each $v_{i}$ has a neighbor to its right. Using induction, $\Delta$-color $H':=G-H$
by $f$. Since
\[
\left\Vert w^{*},H'\right\Vert +\left\Vert y,H'\right\Vert \leq2\Delta-\left\Vert w^{*},H\right\Vert -\left\Vert y,H\right\Vert \leq2\Delta-(\Delta-1)-2\leq\Delta-1,
\]
some color $\beta$ is not used on $N(w^{*})\cup N(y)$. Extend $f$ to $G$ by setting
$f(w^{*})=\beta=f(y)$ and coloring the remaining vertices in the order $L$. This
is possible, since each $v_{i}$ has at most $\Delta-1$ colored neighbors when colored,
and $x$ has two neighbors $w^{*},y$ colored the same.\end{proof}
\begin{conjecture}
[Borodin \& Kostochka 1977]If a graph $G$ satisfies $8,\omega(G)<\Delta(G)$ then
$\chi(G)<\Delta(G)$.
\end{conjecture}
Reed used sophisticated methods to prove the conjecture for $\Delta(G)>10^{14}$.
\begin{x}({*}) For a graph $G$ let $\theta(G)=\max_{{\normalcolor {\color{red}{\normalcolor uv}}}\in E(G)}(d(u)+d(v))$.
Prove that if $\theta(G)\leq2r+1$ then $\chi(G)\leq r+1$. Also prove that if $\theta(G)\leq2r$
and $\omega^{*}(G)\leq r$ then $\chi(G)\leq r$.\end{x}
\section{Turán's Theorem}
Let $n,s\in\mathbb{Z}^{+}$. In this section we determine the number of edges a graph
on $n$ vertices must have to ensure it contains $K_{s}$. In other words, how many
edges can we put into a graph on $n$ vertices without getting $K_{s}$.
\begin{defn}
A graph is said to be \emph{$r$-partite} if it is $r$-colorable. Saying $r$-partite
instead of $r$-colorable tends to emphasize the partition into $r$ independent
sets provided by the $r$-coloring. These independent sets are called \emph{parts.
}The \emph{complete $r$-partite} $K_{n_{1},\dots n_{r}}$ graph is the $r$-partite
graph with $r$ parts of sizes $n_{1},\dots,n_{r}$ such that any two vertices in
different parts are adjacent. The \emph{Turán} graph $T_{n,r}$ is the complete $r$-partite
graph on $n$ vertices such that any two parts differ in size by at most one.\end{defn}
\begin{lem}[5.2.8]
\label{r-part}Among all $r$-partite graphs on $n$ vertices, $T_{n,r}$ has the
most edges.\end{lem}
\begin{proof}
Let $G$ be an $r$-partite graph on $n$ vertices with as many edges as possible;
say $\mathcal{X}$ is an $r$-partition of $G$. Clearly, $G$ is a complete $r$-partite
graph. So, if $G\ne T_{n,r}$ then there exist parts $X,Y\in\mathcal{X}$ with $|X|-|Y|\geq2$
and $x\in X$. Let $G'$ be the complete $r$-partite graph with $r$-partition $\mathcal{X}':=\mathcal{X}-X-Y+(X-x)+(Y+x)$.
Then
\[
E(G')\supseteq E(G)-\{xy:y\in Y\}+\{xx':x'\in X-x\}.
\]
Thus
\[
\|G'\|\geq\|G\|-|Y|+|X|-1\geq\|G\|+1,
\]
a contradiction. So $G\cong T_{n,r}$.\end{proof}
\begin{thm}[{5.2.9 Turán {[}1941{]}}]
\label{Turan}Among all graphs $G=(V,E)$ on $n$ vertices with $\omega(G)\leq r$,
the one with the most edges is $T_{n,r}$. \end{thm}
\begin{proof}
Evidently $T_{n,r}$ is a candidate. Argue by induction on $r$ that if $G$ satisfies
$|G|=n$, $\omega(G)\leq r$, and $\left\Vert G\right\Vert \geq\left\Vert T_{n,r}\right\Vert $
then $G\cong T_{n,r}$. If $\omega(G)=1$ then $G\cong T_{n,1}$; so suppose $r>1$.
Choose $v\in V$ with $d(v)=\Delta:=\Delta(G)$. Set $N:=N(v)$, $G':=G[N]$, $S:=V-N(v)$
and $G'':=G[S]$. Then $|G'|=\Delta$, and $\omega(G')\leq r-1$, since $K+v$ is
a clique in $G$ for every clique $K$ in $G'$. Set $H:=T_{\Delta,r-1}\vee\overline{K}(S)$.
Then $H$ is an $r$-partite graph on $n$ vertices, and $\omega(H)\leq r$, since
any clique in $H$ has at most $r-1$ vertices in $T_{\Delta,r-1}$ and one vertex
in $S$. So
\begin{align}
\|G\| & =\|G'\|+\|G''\|+|E(N,S)|\nonumber \\
& =\|G'\|+\sum_{v\in S}d_{G}(v)-\|G''\|\nonumber \\
& \leq\|T_{\Delta,r-1}\|+\sum_{v\in S}d_{G}(v)-\|G''\| & \mbox{(induction)}\label{eq:ih}\\
& \leq\|T_{\Delta,r-1}\|+\Delta|S| & \mbox{(maximum degree)}\label{eq:S}\\
& =\|H\|\nonumber \\
& \leq_{\mbox{}}\|T_{n,r}\| & \mbox{(Lemma\text{ \ref{r-part}})}\label{eq:Lem}
\end{align}
Inequality \eqref{eq:ih} is strict unless $G'\cong T_{\Delta,r-1}$. Inequality
\eqref{eq:S} is strict unless $G''=\overline{K}(S)$ and $G=G'\vee G''$. Inequality
\eqref{eq:Lem} is strict unless $H\cong T_{n,r}$. If $\|G\|\geq\|T_{n,r}\|$ then
all three inequalities are tight, and so
\[
G\cong T_{\Delta,r-1}\vee\overline{K}(S)\cong H\cong T_{n,r}.\qedhere
\]
\end{proof}
\begin{x}({*}) Prove that if $\omega(G)\leq r$ then $\left\Vert G\right\Vert \leq(1-1/r)|G|^{2}/2$.\end{x}
\section{Edge Coloring}
\begin{defn}
Let $G=(V,E)$ be a graph. A \emph{proper $k$-edge-coloring }of $G$ is a function
$f:E\rightarrow[k]$ such that $f(e)=f(e')$ implies that $e$ and $e'$ are not
adjacent ($e\cap e'=0$). The \emph{chromatic index }$\chi'(G)$ of $G$ is the least
$k$ such that $G$ has a proper $k$-edge-coloring. In this section we will assume
that all edge colorings are proper. Note that this is not the case when we consider
Ramsey Theory.
\begin{x}({*}) Let $P$ be the Petersen graph and $v\in V(P)$. Determine $\chi'(P-v)$.\end{x}\end{defn}
\begin{thm}[{7.1.17 König {[}1916{]}}]
\label{KVT}Every bipartite graph G satisfies $\chi'(G)=\Delta(G)$.\end{thm}
\begin{proof}
Argue by induction on $\Delta=\Delta(G)$. The base step $\Delta=1$ is trivial since
$G$ has no adjacent edges, and so all edges can receive the same color. So consider
the induction step $\Delta>1$. It suffices to find a $\Delta$-regular bipartite
multigraph $H$ with $G\subseteq H$: By Theorem \ref{HallC}, $H$ has a perfect
matching $M$. Color all edges in $M\cap E(G)$ with color $\Delta$, and set \textcolor{black}{${\color{black}G'=}G-M$}.
Then $\Delta(G')\leq\Delta(H-M)\leq\Delta-1$, and so by induction, $G'$ has a $(\Delta-1)$-edge-coloring.
Using another color on $M$ yields a $\Delta$-edge-coloring of $G$.
It remains to construct $H$. Suppose $G$ is an $A,B$-bigraph with $|A|\leq|B|$.
Let $A\subseteq A'$, where $A'\cap B=\emptyset$ and $|A'|=|B|$. Choose an $A',B$-bipartite
multigraph $H\supseteq G$ with $\left\Vert H\right\Vert $ maximal subject to $\Delta(H)\leq\Delta$.
It exists because $G$ is a candidate. Now
\[
\|H\|=\sum_{v\in A'}d_{G'}(v)=\sum_{v\in B}d_{G'}(v)\leq\Delta|B|.
\]
If $\left\Vert H\right\Vert =\Delta|B|$ then $H$ is $\Delta$-regular. Else there
exist vertices $a\in A'$ and $b\in B$ with $d_{H}(a),d_{H}(b)<\Delta$. Then $H':=H+e$,
where $e\in E(a,b)$ is a new, possibly parallel edge, contradicts the maximality
of $H$.
\end{proof}
Now we consider edge coloring of general graphs. The fundamental result is Theorem~\ref{Vizing}
due to Vizing. The following lemma does most of the work in its proof.
\begin{lem}
\label{VL}Suppose $G=(V,E)$ is a simple graph with $\Delta(G)\leq k\in\mathbb{N}$,
and $v\in V$. If $\chi'(G-v)\leq k$ and $d(x)=k$ for at most one $x\in N(v)$
then $\chi'(G)\leq k$.\end{lem}
\begin{proof}
Argue by induction on $k$. If $k=1$ then $E$ is a matching, and so $\chi'(G)\leq1$.
Now suppose $k>1$. For \textcolor{black}{a function $f:E\rightarrow[k]$, $x\in V$
and} $\alpha\in[k]$, set
\[
f(x):=[k]\smallsetminus\{f(e):e\in E(x)\}\mbox{ and }f_{\alpha}:=\{u\in N(v):\alpha\in f(u)\}.
\]
By adding edges and vertices to $G$, we may assume $k-1\leq d(x)\leq k=d(v)$ for
all $x\in N(v)$, and $d(y)=k$ for exactly one $y\in N(v)$. So $|f(x)|=2$ for
all $x\in N(v)-y$ and $|f(y)|=1$. Pick a $k$-edge-coloring $f$ of $G':=G-v$
with $T(f):=\{\beta\in[k]:1\leq|f_{\beta}|\leq2\}$ maximum.
Suppose $|f_{\alpha}|\ne1$ for all $\alpha\in[k]$. Since $\sum_{\alpha\in[k]}|f_{\alpha}|=\sum_{x\in N(v)}|f(x)|=2k-1$,
there exist $\beta,\gamma\in[k]$ with $|f_{\beta}|=0$ and $|f_{\gamma}|\geq3$;
say $w\in f_{\gamma}$. Set $G_{\beta,\gamma}=(V-v,E_{\beta,\gamma})$, where $E_{\beta,\gamma}=\{e\in E:f(e)\in\{\beta,\gamma\}\}$.
Then the component of $G_{\beta,\gamma}$ containing $w$ is a path $P$ with ends
$w$ and (say) $z$, where $f(z)\cap\{\beta,\gamma\}\ne\emptyset$. Switching colors
$\gamma$ and $\beta$ on the edges of $P$ yields a new coloring $f'$ of $G'$.
Then $f'(u)=f(u)$ for $u\in V(G')\smallsetminus\{w,z\}$, and $f'(w)=f(w)-\gamma+\beta$.
So $f'_{\gamma}\ne\emptyset$ and $w\in f'_{\beta}\subseteq\{w,z\}$. Thus ${\normalcolor T(f)\subsetneqq T(f')+\beta}$,
contradicting the choice of $f$.
So $f_{\alpha}=\{z\}$ for some $\alpha\in[k]$ and $z\in N(v)$; say $\alpha=k$.
Set $M=f^{-1}(k)+vz$. Since neither $z$ nor $v$ are incident to any edges colored
$k$, $M$ is a matching. Put $H=G-M$. Since $f_{k}=\{z\}$ and $vz\in M$, every
vertex of $N[v]$ is $M$-saturated. So $d_{H}(x)\leq k-1$ for every $x\in N_{H}(v)$,
and equality holds at most once. Since $f^{-1}(k)\subseteq M$, $f$ is a $(k-1)$-coloring
of $H-v$, and $\Delta(H-v)\leq k-1$. So $\Delta(H)\leq k-1$. By induction, $\chi'(G)\leq\chi'(H)+1\leq_{i.h.}k.$\end{proof}
\begin{thm}[7.1.10 Vizing (1964)]
\label{Vizing}Every graph $G=(V,E)$ satisfies $\chi'(G)\leq\Delta(G)+1$.\end{thm}
\begin{proof}
Set $k:=\Delta(G)+1$ and argue by induction on $|G|$. If $|G|=1$ then $\chi'(G)\leq1=k$.
Otherwise choose $v\in V$. By induction, $\chi'(G-v)\leq k$, and so by Lemma~\ref{VL},
$\chi'(G)\leq k$.
\end{proof}
\begin{thm}
[Full Vizing (1964)]\label{VTM}Every multigraph $M$ satisfies $\chi'(M)\leq\Delta(M)+\mu(M)$.
\end{thm}
\begin{x}[*]Let $G$ be a graph with $\Delta(G)=k$. Put $X=\{v\in V(G):d(v)=k\}$.
Prove that if $G[X]$ is acyclic then $\chi'(G)\leq k$. {[}Hint: Use Lemma~\ref{VL}.{]}\end{x}
\begin{x}[+]Prove Theorem~\ref{VTM}.\end{x}
\begin{conjecture}
[Goldberg (1973), Seymour (1979)]Every multigraph $M$ with $\chi'(M)\geq\Delta(M)+2$
satisfies $\chi'(M)=\max_{H\subseteq M}\lceil\frac{\left\Vert H\right\Vert }{\lfloor|H|/2\rfloor}\rceil$.
\end{conjecture}
\begin{x}[*]Prove that $\chi'(M)\geq\max_{H\subseteq M}\lceil\frac{\left\Vert H\right\Vert }{\lfloor|H|/2\rfloor}\rceil$.\end{x}
\begin{defn}[4.2.18]
The line graph $H=L(G)$ of a graph $G=(V,E)$ is defined by
\[
V(H)=E\mbox{ and }E(H)=\{ee':e\cap e'\ne\emptyset\}.
\]
\end{defn}
If $H$ is the line graph of a simple graph $G$ then $H$ contains neither an induced
copy of $K_{1,3}$ nor an induced copy of $K_{5}-e$ (a $K_{5}$ missing one edge).
Also, $\chi(H)=\chi'(G)$ and $\omega(H)=\Delta(G)$, unless $\Delta(G)=2$ and $\omega(G)=3$.
So the following theorem (with an extra observation for the case $\Delta(G)=2<\omega(G)$)
extends Vizing's Theorem for simple graphs.
\begin{thm}
[Kierstead \& Schmerl 1983]Every graph $H$ that contains neither an induced copy
of $K_{1,3}$ nor an induced copy of $K_{5}-e$ satisfies $\chi(H)\leq\omega(H)+1$.
\end{thm}
\section{List Coloring}
\begin{defn}
Let $G=(V,E)$ be a graph and $C$ a set of colors. We write $2^{C}$ for the power
set of $C$. A \emph{list assignment} for $G$ is a function $f:V\rightarrow2^{C}$.
One should think of $f(v)\subseteq C$ as the set of colors that are available for
coloring the vertex $v$. A $k$\emph{-list assignment} is a list assignment $f$
such that $|f(v)|=k$ for all $v\in V$. Given a list assignment $f$, an $f$-coloring
is a proper coloring $g$ such that $g(v)\in f(v)$ for all $v\in V$. In this case
$G$ is $f$-colorable. The graph $G$ is $k$\emph{-list-colorable} (also \emph{k-choosable})
if for every $k$-list assignment $f$ it is $f$-colorable. The \emph{list-chromatic
number }(also \emph{choosability}, also \emph{choice number})\emph{ $\chi_{l}(G)$
of }$G$ is the least $k$ such that it is $k$-list colorable.
\end{defn}
\begin{example}
Let $G=K_{t,t^{t}}$. Then $\chi(G)=2$, but $\chi_{l}(G)\geq t+1$. \end{example}
\begin{proof}
Let $X,Y$ be a bipartition of $G$ with $|X|=t$. Let $f$ be a $t$-list assignment
for $G$ such that the vertices of $X$ have disjoint lists of size $t$, and for
each $\sigma\in\prod_{x\in X}f(x)$ there exists $y_{\sigma}\in Y$ with $f(y)=\mbox{range}(\sigma)$.
Then for any $f$-coloring $\sigma$ of $G[X]$, the vertex $v_{\sigma}$ cannot
be colored from the list $f(y_{\sigma})=\sigma$. \end{proof}
\begin{defn}
An \emph{edge-list assignment} for $G$ is a function $f:E\rightarrow2^{C}$. One
should think of $f(e)\subseteq C$ as the set of colors that are available for coloring
the edge $e$. A $k$\emph{-edge-list assignment} is a list assignment $f$ such
that $|f(e)|=k$ for all $e\in E$. Given an edge-list assignment $f$, an $f$-coloring
is a proper edge-coloring $g$ such that $g(e)\in f(e)$ for all $e\in E$. In this
case, $G$ is $f$-list-colorable. The graph $G$ is $k$\emph{-edge-list-colorable}
(also \emph{k-edge-choosable}) if for every $k$-edge-list assignment\textcolor{red}{{}
${\color{black}f}$}\textcolor{black}{, it is $f$-colorable}. The \emph{list-chromatic
index }(also \emph{edge-choosability,} \emph{edge-choice number})\emph{ $\chi'_{l}(G)$
of }$G$ is the least $k$ such that it is $k$-edge-list colorable.\end{defn}
\begin{conjecture}
Every graph $G$ satisfies $\chi'_{l}(G)=\chi'(G)$.\end{conjecture}
\begin{defn}
A \emph{kernel} of a digraph $D=(V,A)$ is an independent set $S\subseteq V$ such
that for every $x\in V\smallsetminus S$ there exists $y\in S$ with $xy\in A$.
The $5$-cycle $G=(V,E)=v_{1}v_{2}v_{3}v_{4}v_{5}v_{1}$ has several orientations.\emph{
\[
E_{1}=\{\vec{v_{1}v_{2}},\vec{v_{2}v_{3}},\vec{v_{3}v_{4}},\vec{v_{4}v_{5}},\vec{v_{5}v_{1}}\}~and~E_{2}=\{\vec{v_{1}v_{2}},\vec{v_{3}v_{2}},\vec{v_{3}v_{4}},\vec{v_{5}v_{4}},\vec{v_{5}v_{1}}\}.
\]
}$\vec{G}_{1}:=(V,\vec{E}_{1})$ does not have a kernel since $\Delta^{+}(\vec{G}_{1})=1$,
$\alpha(G)=2$, and $|\vec{G}|=5>\alpha(G)\cdot(\Delta^{+}(\vec{G}_{1})+1)$, but
$\{v_{2},v_{4}\}$ is a kernel of $\vec{G}_{2}:=(V,\vec{E}_{2})$.\end{defn}
\begin{lem}
[8.4.29 Bondy \& Boppana \& Siegel]\label{kern}Let $D=(V,A)$ be a digraph all
of whose induced subgraphs have kernels. If $f$ is a list assignment for $D$ satisfying
$d^{+}(v)<|f(v)|$ for all $v\in V$ then $D$ has an $f$-coloring. \end{lem}
\begin{proof}
Argue by induction on $|D|$. Let $v_{0}\in V$. Since $|f(v_{0})|>d^{+}(v_{0})\geq0$,
there is $\alpha\in f(v_{0})$. Set $W=\{v\in V:\alpha\in f(v)\}$. Then $v_{0}\in W$.
By hypothesis $D[W]$ has a (nonempty) kernel $S$. Color every vertex in $S$ with
$\alpha$. This is possible because $S$ is independent and $S\subseteq W$.
Now it suffices to $f$-color $D'=D-S$ so that no vertex in $D-S$ is colored $\alpha$.
For this purpose, let $f'$ be the list assignment for $D'$ defined by $f'(v)=f(v)-\alpha$.
Since $|D'|<|D|$, using induction, it suffices to show $|f'(v)|>d_{D'}^{+}(v)\mbox{ for all }v\in V\smallsetminus S.$
If $v\notin W$ then $\alpha\notin f(v)$, and so
\[
|f'(v)|=|f(v)|>d_{D}^{+}(v)\geq d_{D'}^{+}(v).
\]
Else $v\in W$. Since $S$ is a kernel of $D[W]$, there exists $w\in S=V\smallsetminus V(D')$
with $vw\in A$. So
\[
|f'(v)|=|f(v)-\alpha|>d_{D}^{+}(v)-1\geq d_{D'}^{+}(v).\qedhere
\]
\end{proof}
\begin{thm}
[8.4.30 Galvin 1995]\label{Galvin}Every $X,Y$-bigraph $G$ satisfies $\chi_{l}'(G)=\Delta(G)$. \end{thm}
\begin{proof}
Let $\Delta:=\Delta(G)$ and set $H:=L(G)$. Then $\chi'_{l}(G)=\chi_{l}(H)$ and
$\chi'(G)=\Delta$ (Theorem~\ref{KVT}). Fix a $\Delta$-edge coloring $c:E(G)\rightarrow[\Delta]$.
\textcolor{black}{Let $L$ be a $\Delta$-edge-list assignment for $G$; so $L$
is a $\Delta$-list assignment for $H$. Our plan is to apply Lemma~\ref{kern}
to $H$ to show that it has an $L$-coloring $f$; then $f$ is an $L$-edge-coloring
of $G$. So it suffices to show $H$ has an orientation $D:=(\vec{E}(G),A)$ such
that (i) $\Delta^{+}(D)\leq\Delta-1$ and (ii) every induced subgraph $D'\subseteq D$
has a kernel.}
\textcolor{black}{Each $\vec{ee}'\in E(H)$ sa}tisfies $e\cap e'\subseteq X$ or
$e\cap e'\subseteq Y$. Define $D$ by putting
\[
\vec{ee}'\in A\mbox{ iff (}e\cap e'\subseteq X\wedge c(e)>c(e'))\vee(e\cap e'\in Y\wedge c(e)c(e_{x'})>c(e_{x})$; so $\vec{e_{x}e^{*}}\in A$.
Hence $S$ is a kernel for $D'$.
\end{proof}
\chapter{Planar graphs}
We have been informally drawing graphs in the Euclidean plane $\mathbb{R}^{2}$ since
the start of the semester. Now we formalize the definition of a drawing of a graph
in $\mathbb{R}^{2}$.
\section{Very Basic Topology of the Euclidean Plane}
For $x\in\mathbb{R}^{2}$ the \emph{open ball around} $x$ \emph{with radius} $r$
is the set $B_{r}(x):=\{y\in\mathbb{R}^{2}:\left\Vert x,y\right\Vert 0$ such that $B_{r}(x)\subseteq U$. In particular, $\mathbb{R}^{2}$ and $\emptyset$
are open. The complement of an open set is a \emph{closed }set\emph{.} The \emph{frontier
}of a set $X$ is the set of all points $x\in\mathbb{R}^{2}$ such that $B_{r}(x)\cap X\ne\emptyset$
and $B_{r}(x)\smallsetminus X\ne\emptyset$ for all $r>0$. Note that if $X$ is
open, then its frontier lies in $\mathbb{R}^{2}\smallsetminus X$.
Let $p,q\in\mathbb{R}^{2}$. The $p,q$-line segment $L_{p,q}$ is the subset of
$\mathbb{R}^{2}$ defined by $L(p,q):=\{p+\lambda(q-p):0\le\lambda\leq1\}$ and $\mathring{L}(p,q):=L(p,q)\smallsetminus\{p,q\}$.
For distinct points $p_{0},\dots,p_{k}\in\mathbb{R}^{2}$, the union $A(p_{0},\dots,p_{k}):=\bigcup_{i\in[k]}L(p_{i-1},p_{i})$
is a (polygonal) $p_{0},p_{k}$-arc provided $L(p_{i-1},p_{i})\cap\mathring{L}(p_{j-1},p_{j})=\emptyset$
for all distinct $i,j\in[k]$. We say that $p_{0}$ and $p_{k}$ are linked by $A(p_{0},\dots,p_{k})$.
If $A(p_{0},\dots,p_{k})$ is an arc and $L(p_{0},\dots,p_{k})\cap\mathring{L}(p_{k},p_{0})=\emptyset$
then $P(p_{0},\dots,p_{k},p_{0}):=L(p_{0},\dots,p_{k})\cup L(p_{k},p_{0})$ is a
polygon. Note that arcs and polygons are closed in $\mathbb{R}^{2}$.
Let $U$ be an open set. Two points $x,y\in U$ are linked in $U$ if there exists
an $x,y$-arc contained in $U$. The relation of being linked is an equivalence relation
on $U$. Its equivalence classes are called \emph{regions}. Regions are open: Suppose
$R\subseteq U$ is a region and $x\in R$. Then there exists a $r>0$ such that $B_{r}(x)\subseteq U$.
Clearly every $y\in B_{r}(x)$ is linked to $x$ in $U$, since $L(x,y)\subseteq B_{r}(x)$.
So $B_{r}(x)\subseteq R$. A closed set $X$ separates a region $R$ if $R\smallsetminus X$
has more than one region.
\begin{thm}
[Jordan Curve Theorem for Polygons]For every polygon $P\subseteq\mathbb{R}^{2}$,
the set $\mathbb{R}^{2}\smallsetminus P$ has exactly two regions. Each of these
regions has the entire polygon as its frontier. \end{thm}
\begin{proof}
\textit{\textcolor{red}{To be continued ...}}
\end{proof}
\section{Graph Drawings}
Recall that in this class all graphs are finite. Let $G=(V,E)$ be a (multi)graph.
A \emph{drawing} of $G$ is a graph $\widetilde{G}:=(\widetilde{V},\widetilde{E})$
such that $G\cong\widetilde{G}$ , $\widetilde{V}\subseteq\mathbb{R}^{2}$, and each
edge $\tilde{e}\in\widetilde{E}$ is an arc linking its ends. So edges are no longer
just pairs of vertices, but have their own identity and structure (we need this anyway
to formally deal with different edges linking the same two vertices). It should be
clear that every finite graph has a drawing. Moreover, by moving vertices slightly
and readjusting edges, we can (and do) require the following additional properties
for drawings, without restricting the set of (finite) graphs that can be drawn.
\begin{enumerate}
\item No three edges have a common internal point.
\item The only vertices contained in an edge are its endpoints.
\item No two edges are tangent.
\item No two edges have more than one common internal point.
\end{enumerate}
A crossing in a drawing of a graph is a point that is in the interior of two edges.
A \emph{plane} (multi)graph is a drawing of a (multi)graph that has no crossing.
A \emph{planar} (multi)graph is a (multi)graph that has a plane drawing.
Let $\widetilde{G}:=(\widetilde{V},\widetilde{E})$ be a plane (multi)graph. The
faces of $\widetilde{G}$ are the regions of $\mathbb{R}^{2}\smallsetminus(\widetilde{V}\cup\bigcup\widetilde{E})$.
The frontier of a face is called its \emph{boundary}. A boundary is the union of
some edges and vertices of $G$---so it is a subgraph of $\tilde{G}$. The edges
of a face $f$ can be ordered to form its \emph{boundary} \emph{walk(s)}; this is
a closed walk(s) that contains every edge once or twice. Let $l(f)$ be the length
of its boundary walk. Let $F(\widetilde{G})$ be the set of faces of $\widetilde{G}$.
Note that a plane cycle is a polygon. By the Jordan Curve Theorem we have:
\begin{prop}
[\textcolor{red}{Top.}]A plane cycle is the boundary of exactly two faces.
\end{prop}
\textit{\textcolor{red}{To be continued ...}}
\section{Basic facts}
\begin{fact}
\textcolor{red}{[\textcolor{red}{Top.}]\label{af}Let $\widetilde{G}$ be a plane
multigraph with $e\in E$. If $e$ is a cut-edge then $|F(\widetilde{G})-e)|=|F(\widetilde{G})|$;
if $e$ is not a cut-edge then $|F(\widetilde{G}-e)|=|F(\widetilde{G})|-1$.}
\end{fact}
\begin{fact}
\label{contr-fact}If $G$ is a planar multigraph and $e\in E$ then $G\cdot e$
is a planar graph.
\end{fact}
\begin{fact}
\textcolor{red}{[\textcolor{red}{Top.}]\label{cl}Let $\widetilde{G}$ be a plane
graph with link (non-loop-edge) $e=xy$. Let $\widetilde{G}\cdot e$ be the plane
graph formed from $\widetilde{G}$ by deleting $y$ and adding edges $xv$ in the
face of $\widetilde{G}$-y containing $y$ for all $v\in N(y)\smallsetminus N(x)$.
Then $|F(\widetilde{G}\cdot e)|=|F(\widetilde{G})|$. Deleting a loop of $\widetilde{G}$,
reduces the number of faces by $1$.}
\end{fact}
\textcolor{red}{For an edge $e$ and a face $f$, let $\iota(e,f)$ be the number
of times $e$ appears on the boundary walk of $f$.}
\begin{fact}
\textcolor{red}{[\textcolor{red}{Top.}]\label{sfl}Let $\widetilde{G}$ be a plane
multigraph. Every non-cut-edge appears exactly once on the bounding walk of exactly
two faces. Every cut-edge appears exactly twice on the bounding walk of exactly one
face. In particular,
\[
\sum_{f\in F(\widetilde{G})}l(f)=\sum_{f\in F(\widetilde{G})}\sum_{e\in E(\widetilde{G})}\iota(e,f)=\sum_{e\in E(\widetilde{G})}\sum_{f\in F(\widetilde{G})}\iota(e,f)=2\left\Vert \widetilde{G}\right\Vert .
\]
}\end{fact}
\begin{thm}[6.1.21 Euler's Formula (1758)]
\label{EF}All connected, planar multigraphs $G$ satisfy
\[
|G|-\left\Vert G\right\Vert +|F(G)|=2.
\]
\end{thm}
\begin{proof}
Argue by induction on $|G|$.
\noindent \textbf{Base Step: $|G|=1$}. In this case all edges of $G$ are loops.
Argue by secondary induction on $\left\Vert G\right\Vert $. For the base step $\left\Vert G\right\Vert =0$,
note that $G$ has one vertex and one face, and so
\[
|G|-\left\Vert G\right\Vert +|F(G)|=1-0+1=2.
\]
For the induction step, set $G':=G-l$ for some loop $l$. Then using Fact~\ref{cl}:
\[
|G|-\left\Vert G\right\Vert +|F(G)|=|G'|-(\left\Vert G'\right\Vert +1)+(|F(G')|+1)=_{i.h.}2.
\]
\noindent \textbf{Induction Step: }$|G|>1$. Since $G$ is connected, $G$ has a
non-loop edge $e$. Set $G'=G\cdot e$. By Fact~\ref{cl},
\[
|G|-\left\Vert G\right\Vert +|F(G)|=(|G'|+1)-(\left\Vert G'\right\Vert +1)+|F(G')|=2.
\]
\end{proof}
\begin{thm}
\label{esg}If $G$ is a simple\textcolor{black}{{} planar} graph with at least three
vertices then $\left\Vert G\right\Vert \leq3|G|-6$. Moreover, if $G$ has girth
greater than $3$ then $\left\Vert G\right\Vert \leq2|G|-4$.\end{thm}
\begin{proof}
We may assume that $G$ is a maximal planar graph, i.e., it is not a spanning subgraph
of any planar graph with more edges. Then $G$ is connected, since otherwise we could
add an edge between two components of $G$ while maintaining planarity. Since $G$
is connected and has at least three vertices, $\|G\|\ge2$; since it is simple it
has no parallel edges. Thus the length of every face is at least $3$. By Fact~\ref{sfl},
\[
2\left\Vert G\right\Vert =\sum_{f\in F(G)}l(f)\geq3|F(G)|.
\]
So $|F(G)|\leq\frac{2}{3}\left\Vert G\right\Vert $. By Theorem~\ref{EF}
\[
2=|G|-\left\Vert G\right\Vert +|F(G)|\leq|G|-\frac{1}{3}\left\Vert G\right\Vert ,
\]
and so $3|G|-6\geq\left\Vert G\right\Vert $.
Now suppose that the girth of $G$ is greater than $3$. Then every face boundary
has length at least 4. So
\[
2\left\Vert G\right\Vert =\sum_{f\in F(G)}l(f)\geq4|F(G)|.
\]
Thus $|F(G)|\leq\frac{1}{2}\left\Vert G\right\Vert $. By Theorem~\ref{EF}
\[
2=|G|-\left\Vert G\right\Vert +|F(G)|\leq|G|-\frac{1}{2}\left\Vert G\right\Vert ,
\]
and so $2|G|-4\geq\left\Vert G\right\Vert $.
\end{proof}
A graph $H$ is a subdivision of a graph $G$ if $H$ is formed by replacing some
of the edges $xy\in E(G)$ by an $x,y$-path $P$ whose internal vertices have degree
$2$.
\begin{cor}
\label{Knp}Neither $K_{5}$ nor $K_{3,3}$ is planar.\end{cor}
\begin{proof}
If $K_{5}$ is planar then Theorem~\ref{esg} yields the following contradiction:
\[
10=\left\Vert K_{5}\right\Vert \leq3|K_{5}|-6=9.
\]
If $K_{3,3}$ is planar, then since it is bipartite, and so has girth greater than
$3$, Theorem~\ref{esg} yields the contradiction:
\[
9=\left\Vert K_{3,3}\right\Vert \leq2|K_{3,3}|-4=8.
\]
\end{proof}
A graph $H$ is a subdivision of a graph $G$ if $H$ is formed by replacing some
of the edges $xy\in E(G)$ by an $x,y$-path $P$ whose internal vertices are not
vertices of $G$ and have degree $2$ in $H$. The vertices of $G$ are called branch
vertices and the new vertices are called subdivision vertices. Notice $G$ is not
a subgraph of a proper subdivision of itself.
\begin{thm}[6.2.2 Kuratowski (1930)]
\label{KT}A graph is planar iff it contains neither a subdivision of $K_{5}$ nor
a subdivision of $K_{3,3}$.
\end{thm}
We will break the proof of Kuratowski's Theorem into smaller pieces. Call a subdivision
of $K_{5}$ or $K_{3,3}$ a $K$\emph{-graph} (for Kuratowski).
\begin{lem}
\label{csd}Let $e=x_{1}x_{2}$ be an edge of a graph $G$. If $G$ contains no $K$-graph
then $G\cdot e$ contains no $K$-graph.\end{lem}
\begin{proof}
We prove the contrapositive. Suppose $G\cdot e$ contains a $K$-graph $Q$. If $Q\subseteq G$
we are done; else $v_{e}\in V(Q)$. Then $3\leq d_{Q}(v_{e})\leq4$ and $N_{Q}(v_{e})\subseteq N_{G}(x_{1})\cup N_{G}(x_{2})$.
If there is $i\in[2]$ with $|N_{Q}(v_{e})\smallsetminus N_{G}(x_{i})|\leq1$ then
replacing $v_{e}$ by $x_{i}$ or $x_{1},x_{2},e$ yields a subdivision of $Q$.
Otherwise, $d_{Q}(v_{e})=4$, $|N_{Q}(v_{e})\smallsetminus N_{G}(x_{1})|=2=|N_{Q}(v_{e})\smallsetminus N_{G}(x_{2})|$,
and $Q$ is a subdivision of $K_{5}(\{v_{e},a,b,c,d\})$. We claim $G$ contains
a subdivision of $K_{3,3}$! Let $Q'=Q-v_{e}+x_{1}+x_{2}+e$. Then $Q'$ is a subdivision
of $Q'=K_{3,3}\{\{x_{1},a,b\},\{x_{2},c,d\}\}$. \end{proof}
\begin{cor}
\label{noK}If $G$ is planar then $G$ does not contain a $K$-graph. \end{cor}
\begin{proof}
Suppose $Q\subseteq G$ is a $K$-graph, and show that $G$ is not planar by induction
on the number $h$ of subdivision vertices in $Q$. The base step $h=0$ is Corollary~\ref{Knp}.
For the induction step $h>0$, consider a subdivision vertex $x$ and one of its
two neighbors y. Contracting $xy$ yields a $K$-graph with one less subdivision
vertex. Thus by the induction $G\cdot xy$ is nonplanar, and by Fact~\ref{contr-fact},
$G$ is nonplanar.
\end{proof}
Let $G=(V,E)$ be a graph, and $S\subseteq V$. An $S$-lobe is a subgraph of the
form $G[S\cup V(H)]$, where $H$ is a component of $G-S$.
\begin{lem}
[\textcolor{red}{Top.}]\label{2cp}Let $G$ be a planar $2$\textup{-connected}
planar graph with plane drawing $\widetilde{G}$ . Then the boundary walk of every
face of $\widetilde{G}$ is a cycle.\end{lem}
\begin{proof}
By Theorem \ref{Whitney}, $G$ has an a $2$-witness $P_{0},\dots,P_{h}$. Argue
by induction on $h$. For the base step $h=0$, note that $P_{0}$ is a cycle that
bounds the only two faces of $\widetilde{G}$ (Jordan Curve Theorem). Now consider
the induction step. Let $P_{h}$ have ends $x,y$, and note that $\mathring{P}_{h}$
is contained in some face $f$ of $\widetilde{H}=\widetilde{G}$$-\mathring{P_{h}}$.
By the induction hypothesis, every face of $\widetilde{H}$ is bounded by a cycle.
Let $C=xv_{1}\dots v_{a}yv_{a+2}\dots v_{b}x$ be the cycle that bounds $f$. Let
$f_{1}$ and $f_{2}$ be the faces of $\widetilde{G}$ that are contained in $f$
and bounded by $xv_{1}CyP^{*}x$ and $yv_{a+2}CxPy$. Then $F(\widetilde{G})=F(\widetilde{H})-f+f_{1}+f_{2}$,
and all are bounded by cycles.\end{proof}
\begin{thm}
\label{3KT}If $G$ is a $3$-connected graph that does not contain any $K$-graph
then $G$ is planar.\end{thm}
\begin{proof}
We argue by induction on $|G|$. If $|G|\le4$ then $G$ is planar, since $G\subseteq K_{4}$,
and $K_{4}$ is planar. So suppose $|G|\ge5$. By Theorem \ref{Thomassen3}, $G$
has an edge $e=xy$ such that $H:=G\cdot e$ is $3$-connected. By Corollary \ref{csd},
$H$ does not contain a $K$-graph. Thus by the induction hypothesis, $H$ is planar.
Let $\widetilde{H}$ be a drawing of $H$.
Let $H'=H-v_{e}$, and let $f$ be the face of $\widetilde{H}-v_{e}$ that contains
$v_{e}.$ Since $H'$ is $2$-connected, Lemma~\ref{2cp} implies that $f$ is bounded
by a cycle $C$, say with orientation $\overrightarrow{C}$, and $(N_{G}(x)\cup N_{G}(y))\smallsetminus\{x,y\}=N_{H}(v_{e})\subseteq V(C)$.
Choose the notation so that $d_{G}(x)\leq d_{G}(y)$. Obtain a drawing $\widetilde{G'}$
of $G':=G-y$ from $\widetilde{H}'$ by drawing $x$ at the point corresponding to
$v_{e}$ in $\widetilde{H}$ and deleting the edges $v_{e}z$ with \textcolor{red}{${\normalcolor z\in N_{G}(y)\smallsetminus N_{G}(x)}$}.
Our goal is to extend $\widetilde{G}'$ to a drawing $\widetilde{G}$ of $G$ by
adding $y$ and the edges in $E(y)$ to $\widetilde{G}'$.
Let $x_{1},\dots,x_{k}$ be the neighbors of $x$ in $G'$ arranged in cyclic order
around $\overrightarrow{C}$, and \textcolor{red}{${\normalcolor U:=N_{G}(y)-x}$}.
If there exists an index $i$ such that $U\subseteq C_{i}:=V(x_{i}\overrightarrow{C}x_{i+1})$
then we can extend $\widetilde{G'}$ to $\widetilde{G}$ by drawing $y$ in the face
$f'$ of $\widetilde{G'}$ bounded by $xx_{i}Cx_{i+1}x$, and then drawing edges
from $y$ to each vertex of $U+x$. This is possible, because all vertices of $U+x$
appear on the boundary of $f'$. Otherwise for every index $i$ there exists $u_{i}\in U$
such that $u_{i}\notin C_{i}$.
If there exists an index $i$ and a vertex $v_{i}\in N(y)\cap V(C_{i}-x_{i}-x_{i+1})$
then the vertices
\[
v_{i},u_{i},x,x_{i},x_{i+1},y
\]
are the branch vertices of a subdivision of $K_{3,3}$ with bipartition $\{\{v_{i},u_{i},x\},\{x_{i},x_{i+1},y\}\}$,
where the edges of the cycle $x_{i}v_{i}x_{i+1}u_{i}x_{i}$ are represented by the
paths
\[
x_{i}\overrightarrow{C}v_{i},v_{i}\overrightarrow{C}x_{i+1},x_{i+1}\overrightarrow{C}u_{i},u_{i}\overrightarrow{C}x_{i}.
\]
(Draw the picture.) This is a contradiction.
Otherwise, \textcolor{red}{${\normalcolor U\subseteq N_{G}(x)}$}. Since $d_{G}(x)\leq d_{G}(y)$,
we have $U=N_{G}(x)-y$. Thus $U$ separates $xy$ from the rest of $G$. Since $G$
is $3$-connected, $|U|\geq3$. Say $a,b,c\in U$. Then $x,y,a,b,c$ are the branch
vertices of a subdivision of a $K_{5}$, where the edges of the cycle $abca$ are
represented by the paths $a\overrightarrow{C}b,b\overrightarrow{C}c,c\overrightarrow{C}a$.
(Draw the picture.) This is a contradiction.
\end{proof}
\begin{lem}
\label{edge-max-lem}Let $G=(V,E)$ be a graph with $V=V_{1}\cup V_{2}$ such that
$S:=V_{1}\cap V_{2}$ is a minimum separating set with $|S|\leq2$. If $G$ is edge-maximal
among graphs without $K$-graphs then so are $G_{1}:=G[V_{1}]$ and $G_{2}:=G[V_{2}]$,
and $G[S]=K_{2}$.\end{lem}
\begin{proof}
Let $S=\{u^{i}:i\in[|S|]\}$. As $S$ is a minimum separating set, there are $v_{j}^{i}\in N(u^{i})\cap(V_{j}\smallsetminus S)$
for all $u^{i}\in S$ and $j\in[2]$, and $E(V_{1}\smallsetminus S,V_{2}\smallsetminus S)=\emptyset$.
Let $w_{i}\in V_{i}\smallsetminus S$ for $i\in[2]$. By maximality $G+e$ contains
a $K$-graph $H$ with $e\in E(H)$ for all $e\in E(\overline{G})$.
If $S=\emptyset$ then $e:=w_{1}w_{2}$ is a cut-edge in $G+e$. As $K$-graphs contain
no cut-edges, this is a contradiction. Now suppose $|S|=1$. Then $G':=G+v_{1}^{1}v_{1}^{2}$
contains a $K$-graph $H$. As any two branch vertices of $H$ are linked by $3$
disjoint paths in $H$, all the branch vertices are contained in, say, $G_{1}$.
Thus there is a path $P=v_{1}^{1}v_{2}^{1}\dots u^{1}$ with $G'[V_{2}+v_{1}^{1}]=P$.
Replacing $P$ by $v_{1}^{1}u^{1}$ yields an $K$-graph in $G_{1}\subseteq G$ a
contradiction. So $|S|=2$. Suppose $e:=u^{1}u^{2}\notin E$. Then $G+e$ contains
a $K$-graph $H$ with $e\in E(H)$. Again, the branch vertices of $H$ are contained
in (say) $G_{1}$. Then $H\subseteq G_{1}+e$. Replacing $e$ by a $u^{1},u^{2}$-path
in $G_{2}$ yields an $K$-graph in $G$, a contradiction. So $G[S]=K_{2}$.
Finally, suppose that $e\in E(\overline{G}_{i})$; say $i=1$. Then $G+e$ contains
an $K$-graph $H$ with $e\in E(H)$. If the branch vertices of $H$ are contained
in $G_{2}$, then $G_{1}\cap H$ is a $u^{1},u^{2}$-path $P$. Replacing $P$ with
$u^{1}u^{2}$ yields a $K$-graph in $G_{2}\subseteq G$, a contradiction. So the
branch vertices of $H$ are contained in $G_{1}$. If $H\cap G_{2}-S\ne\emptyset$
then $H\cap G_{2}$ is a $u^{1},u^{2}$ path $P$. Replacing $P$ by $u^{1}u^{2}$
yields an $H$ graph in $G_{1}+e$, proving that $G_{i}$ is maximum. \end{proof}
\begin{thm}
\label{max-3-con-thm}Let $G=(V,E)$ be a graph with $|G|\geq4$. If $G$ is edge-maximal
among graphs with no $K$-graph then $G$ is $3$-connected.\end{thm}
\begin{proof}
Argue by induction on $|G|$. Suppose $S=\{x,y\}$ is a minimum separating set. If
$|S|\geq3$ we are done, so suppose not. Choose sets $V_{1}$ and $V_{2}$ such that
$V=V_{1}\cup V_{2}$ and $S=V_{1}\cap V_{2}$. By Lemma~\ref{edge-max-lem}, $G[S]=K_{2}$
and each $G_{i}:=G[V_{i}]$ is edge-maximal among graphs with no $K$-graph. By induction
each $G_{i}$ is either $K_{3}$ or $3$-connected, so Theorem~\ref{3KT} implies
each $G_{i}$ is planar. For each $G_{i}$ let $\widetilde{G}_{i}$ be a plane drawing
of $G_{i}$. \textcolor{red}{Each face $f$ of $\widetilde{G}_{i}$ is a triangle,
as otherwise we could draw a new edge in $f$ without creating a $K$-graph. Let
$xyz_{i}x$ be the boundary of a face $f_{i}$ }\textcolor{blue}{of $G_{i}$}\textcolor{red}{.
Let $v_{i}$ be a new vertex and $G_{{\color{blue}i}}^{+}:=G_{{\color{blue}i}}\cup K(\{v_{i}\},\{x,y,z_{i}\})$.
Drawing $v$ and its incident edges in $f_{i}$ shows that $G_{i}^{+}$ is planar.}
As $e:=z_{1}z_{2}\in E(\overline{G})$, $G':=G+e$ contains a $K$-graph $H$ with
$e\in H$. If all the branch vertices of $H$ are contained in the same $G_{i}$
then there is a path $P=z_{i}z_{3-i}\dots{\color{blue}u}$ in $G'$, where $u\in S$.
Replacing $P$ by ${\color{blue}u}z_{i}$ yields a $K$-graph in ${\color{blue}G_{i}}$,
contradicting Corollary~\ref{noK}. Thus $H$ has branch vertices in each $G_{i}-S$.
In $G'$ there are at most three ${\color{blue}V_{1}\smallsetminus S,V_{2}\smallsetminus S}$-paths.
Thus the only possibility is that $H$ is a subdivision of $K_{3,3}$ with, say,
exactly one branch vertex $v$ in $G_{2}-S$. Then $H\cap G'[V_{2}+z_{1}]$ is a
$v,\{x,y,z_{1}\}$-fan $F$. \textcolor{red}{Thus $H$ with $F$ replaced by $K(v_{1},\{x,y,z_{1}\})$
is a subdivision of $K_{3,3}$ in $G_{i}^{+}$, contradicting Corollary~\ref{noK}. }
\end{proof}
\begin{proof}[Proof of Kuratowski's Theorem \ref{3KT}]
First suppose $G$ contains a $K$-graph. Then by Corollary \ref{noK}, $G$ is
not planar. Now suppose $G$ contains no $K$-graph. Then $G$ is a spanning subgraph
of a graph $G'$ that is edge-maximal among graphs with no $K$-graph. By Theorem~\ref{max-3-con-thm},
$G'$ is $3$-connected. By Theorem~\ref{3KT}, $G'$ is planar, so $G$ is planar.\end{proof}
\begin{thm}[8.4.32 Thomassen (1994)]
\label{5C}Every planar graph $G$ is $5$-list colorable.\end{thm}
\begin{proof}
It suffices to prove the following more technical statement by induction on $G$.
\begin{claim*}
Suppose $\widetilde{G}$ is a drawing of a simple planar graph $G$ such that every
inner face has length three, and the boundary of the outer face is a cycle $C=v_{1}v_{2}\dots v_{s}v_{1}$
with $x=v_{1}$ and $y=v_{2}$. If $L$ is a list assignment for $G$ such that
\begin{enumerate}
\item $L(x)=\{\alpha\}$, $L(y)=\{\beta\}$, and $\alpha\ne\beta$,
\item $|L(v)|=3$ for all vertices $v\in\{v_{3},\dots,v_{s}\}$, and
\item $|L(v)|=5$ for all vertices $v\in V(G-C)$,
\end{enumerate}
\noindent then $G$ has an $L$-coloring.
\end{claim*}
The claim implies the theorem since adding edges and vertices to $G$, and deleting
colors from some lists of $L$ does not make it easier to $L$-color $G$. Moreover,
every face of an edge-maximal planar graph is bounded by a $C_{3}$. So it suffices
to prove the claim.
\begin{proof}[Proof of Claim]
Argue by induction on $|G|$. Note that $|G|\geq|C|\geq3$. First consider the base
step $|G|=3$. Color $x$ with $\alpha$ and $y$ with $\beta$. The last vertex
$z$ has three colors in its list, so it can be colored with a color distinct from
$\alpha$ and $\beta$.
Now consider the induction step $|G|>3$.
\noindent \emph{Case 1:}\textbf{ }$C$ has a chord $v_{i}v_{j}$ with $i>j$. Let
$C_{1}=v_{i}v_{i\oplus1}\dots v_{j}v_{i}$ and $C_{2}=v_{j}v_{j\oplus1}\dots v_{i}v_{j}$
be the two nonspanning cycles contained in $C+v_{i}v_{j}$. Let $\widetilde{G}_{i}$
be the plane graph formed by $C_{i}$ and its interior. Then $\widetilde{G}_{1}\cup\widetilde{G}_{2}=G$,
$\widetilde{G}_{1}\cap\widetilde{G}_{2}=\widetilde{G}[\{v_{i}v_{j}\}]$, and $x,y\in V(C_{1})$.
By the induction hypothesis, there exists an $L$-coloring ${\color{red}g_{1}}$
of $\widetilde{G}_{1}$. Set $x'=v_{i}$, $\alpha'={\color{red}g_{1}}(x')$, $y'=v_{j}$,
$\beta'={\color{red}g_{1}}(y')$, $L'(x')=\{\alpha'\}$, $L'(y')=\{\beta'\}$ and
$L'(v)=L(v)$ for all vertices of $\widetilde{G}_{2}-x'-y'$. Then by the induction
hypothesis there exists an $L'$-coloring $g_{2}$ of $\widetilde{G}_{2}$. It follows
that $f=g_{1}\cup g_{2}$ is an $L$-coloring of $G$.
\noindent \emph{Case 2:}\textbf{ }$C$ does not have a chord. Since every interior
face is bounded by a $C_{3}$, $G[N(v_{s})]$ contains a hamiltonian $x,v_{s-1}$-path
$P$. Moreover, since $C$ has no chords, the outer face of $\widetilde{G'}=\widetilde{G}-v_{s}$
is bounded by the cycle $C'=xPv_{s-1}C^{*}v_{1}\text{ }(=x)$. Of course, the interior
faces of $\widetilde{G'}$ have length three. Let $\gamma,\delta\in L(v_{s})$ be
distinct colors not equal to $\alpha$. Define a list assignment $L'$ for $L$ by
\[
L'(v)=\begin{cases}
L(v)-\gamma-\delta & \mbox{if }v\in V(P)-x-v_{s-1}\\
L(v) & \mbox{else}
\end{cases}
\]
(and shrinking oversized lists). By induction $\widetilde{G'}$ has an $L'$-coloring
$f'$. Pick $\varepsilon\in\{\gamma,\delta\}$ with $\varepsilon\ne f'(v_{s-1})$.
Finally, extend $f'$ to an $L$-coloring $f$ of $G$ by setting $f(v_{s})=\varepsilon.$
\end{proof}
This completes the proof of the claim and the Theorem.
\end{proof}
\begin{x}[*]Let $G$ be a simple planar graph with girth (length of the shortest
cycle) $k$. Prove that $\left\Vert G\right\Vert \leq\frac{k}{k-2}(|G|-2)$.\end{x}
\begin{x}[*]Prove that every simple planar graph $G$ with $|G|\geq4$ has at least
four vertices with degree less than six. \end{x}
\begin{x}[+]Prove that every simple planar graph $G$ with $\delta(G)=5$ has a
matching with at most $\frac{1}{5}|G|$ unsaturated vertices. \end{x}
\begin{x}[*]Prove that the vertices of a simple planar graph can be ordered so that
every vertex is preceded by at most five of its neighbors. Similarly, prove that
the vertices of every planar bipartite graph can be ordered so that each vertex is
preceded by at most three of its neighbors.\end{x}
\begin{x}[*]Prove that every planar bipartite graph satisfies $\chi_{l}(G)\leq4$.
{[}Hint: Use the previous problem.\end{x}
\begin{x}[+]Prove that every orientation of every $X,Y$-bigraph has a kernel. {[}Hint:
When is $X$ a kernel?{]}\end{x}
\begin{x}[*]Prove that every bipartite planar graph $G$ satisfies $\chi_{l}(G)\leq3$.
{[}Hint: Use the previous problem.{]}\end{x}
\chapter{Extras}
\section{Lower Bounds on Ramsey's Theorem}
\begin{thm}
\label{rlb}For every integer $k\geq2$ there exists a graph $G$ such that $\omega(G)N(1-2\frac{n^{k}}{k!}2^{-(k-1)k/2}) & (b)\\
& \geq N(1-(\frac{n}{2^{k/2-1/2}})^{k})\geq0. & (c) & \qedhere
\end{align*}
\end{proof}
\section{Equitable Coloring}
\begin{defn}
An equitable $k$-coloring of a graph $G=(V,E)$ is a proper coloring $f:V\rightarrow[k]$
such that difference $||f^{-1}(i)|-|f^{-1}(j)||$ in the sizes of the the $i$-th
and $j$-th color classes is at most $1$ for all $i,j\in[k]$. In particular, every
color is used if $|G|\leq k$.\end{defn}
\begin{thm}
[Hajnal \& Szemer\'edi Theorem (1976)]\label{hsz}Every graph $G$ with maximum
degree at most $r$ has an equitable $(r+1)$-coloring.
\end{thm}
\bigskip{}
The proof was long and sophisticated, and does not provide a polynomial time algorithm.
Kierstead and Kostochka found a much simpler and shorter proof. This better understanding
has led to many new results, several of which are stated below.
Let $\theta(G)=\max\{d(x)=d(y):xy\in E(G)\}$.
\begin{thm}
[Kiestead \& Kostochka (2008)] \label{ore} For every $r\geq3$, each graph $G$
with $\theta(G)\leq2r+1$ has an equitable $(r+1)$-coloring.
\end{thm}
\begin{thm}
[Kierstead, Kostochka, Mydlarz \& Szemer\'edi]There is an algorithm that constructs
an equitable $k$-coloring of any graph $G$ with $\Delta(G)+1\leq k$, using time
$O(r|G|^{2})$.\end{thm}
\begin{problem}
Find a polynomial time algorithm for constructing the coloring in Theorem~\eqref{ore}.
\end{problem}
One might hope to prove an equitable version of Brooks' Theorem, but the following
example shows that the statement would require special care: For $r$ is odd, $K_{r,r}$
satisfies $\Delta(K_{r,r})=r$ and $\omega(G)=2$, but has no $r$-equitable coloring.
Chen, Lih and Wu~\cite{CLW} proposed the following common strengthening of Theorem~\ref{hsz}
and Brooks' Theorem.
\begin{conjecture}
\label{main} Let $G$ be a connected graph with $\Delta(G)\leq r$. Then $G$ has
no equitable $r$-coloring if and only if either (a) $G=K_{r+1}$, or (b) $r=2$
and $G$ is an odd cycle, or (c) $r$ is odd and $G=K_{r,r}$.
\end{conjecture}
Kierstead and Kostochka have proved the conjecture for $r\leq4$, and also for $r\geq\frac{1}{4}|G|$.
\bigskip{}
\textbf{Proof of Theorem~\ref{hsz}.} Let $G$ be a graph with $\Delta(G)\leq r$.
We may assume that $\left\vert G\right\vert $ is divisible by $r+1$: If $\left\vert G\right\vert =s(r+1)-p$,
where $p\in\lbrack r]$ then set $G^{\prime}:=G+K_{p}$. Then $\left\vert G^{\prime}\right\vert $
is divisible by $r+1$ and $\Delta(G^{\prime})\leq r$. Moreover, the restriction
of any equitable $(r+1)$-coloring of $G^{\prime}$ to $G$ is an equitable $(r+1)$-coloring
of $G$. So we may assume $\left\vert G\right\vert =(r+1)s$.
We argue by induction on $\left\Vert G\right\Vert $. The base step $\left\Vert G\right\Vert =0$
is trivial, so consider the induction step. Let $u$ be a non-isolated vertex. By
the induction hypothesis, there exists an equitable $(r+1)$-coloring of $G-E(u)$.
We are done unless some color class $V$ contains an edge $uv$. Since $\Delta(G)\leq r$,
some color class $W$ contains no neighbors of $u$. Moving $u$ to $W$ yields an
$(r+1)$-coloring of $G$ with all classes of size $s$, except for one \emph{small}
class $V^{-}:=V-u$ of size $s-1$ and one \emph{large} class $V^{+}:=W+u$ of size
$s+1$. Such a coloring is called \emph{nearly equitable}.
Given a nearly equitable $(r+1)$-coloring, define an auxiliary digraph $\mathcal{H}$,
whose vertices are the color classes, so that $UW$ is a directed edge if and only
if some vertex $y\in U$ has no neighbors in $W$. In this case we say that $y$
\emph{witnesses} $UW$. Let $\mathcal{A}$ be the set of classes from which $V^{-}$
can be reached in $\mathcal{H}$, $\mathcal{B}$ be the set of classes not in $\mathcal{A}$
and $\mathcal{B}^{\prime}$ be the set of classes reachable from $V^{+}$ in $\mathcal{H}[\mathcal{B}]$.
Set $a:=\left\vert \mathcal{A}\right\vert $, $b:=\left\vert \mathcal{B}\right\vert $,
$b^{\prime}:=\left\vert \mathcal{B}^{\prime}\right\vert $, $A:=\bigcup\mathcal{A}$,
$B:=\bigcup\mathcal{B}$ and $B^{\prime}:=\bigcup\mathcal{B}^{\prime}$. Then $r+1=a+b$.
Since every vertex $y\in B$ has a neighbor in every class of $\mathcal{A}$ and
every vertex $z\in B^{\prime}$ also has a neighbor in every class of $\mathcal{B}-\mathcal{B}^{\prime}$,
\begin{equation}
d_{A}(y)\geq a\mbox{ for all }y\in B\mbox{ and }d_{A\cup B\setminus B^{\prime}}(z)\geq a+b-b^{\prime}\mbox{ for all }z\in B^{\prime}\mbox{.}\tag{*}\label{y}
\end{equation}
\textbf{Case 0}: $V^{+}\in\mathcal{A}$. Then there exists a $V^{+},V^{-}$-path
$\mathcal{P}=V_{1},\ldots,V_{k}$ in $\mathcal{H}$. Moving each witness $y_{j}$
of $V_{j}V_{j+1}$ to $V_{j+1}$ yields an equitable $(r+1)$-coloring of $G$.\qquad{}$\Box$
We now argue by a secondary induction on $b$, whose base step $b=0$ holds by Case
0. Also $\left\vert A\right\vert =as-1$ and $\left\vert B\right\vert =bs+1$. Now
consider the secondary induction step.
A class $W\in\mathcal{A}$ is \emph{terminal}, if every $U\in\mathcal{A}-W$ can
reach $V^{-}$ in $\mathcal{H}-W$. Let $\mathcal{A}^{\prime}$ be the set of terminal
classes, $a':=\left\vert \mathcal{A}^{\prime}\right\vert $ and $A^{\prime}:=\bigcup\mathcal{A}^{\prime}$.
An edge $wz$ is \emph{solo} if $w\in W\in\mathcal{A}^{\prime}$, $z\in B$ and $N_{W}(z)=\{w\}$.
Ends of solo edges are \emph{solo} vertices and \emph{solo neighbors} of each other.
Order $\mathcal{A}$ as $V^{-},X_{1},\dots,X_{a-1}$ so that each $X_{i}$ has a
previous out-neighbor.
\textbf{Case 1:} For some $a-b\leq i\leq a-1$, class $X_{i}$ is not terminal. Then
some $X_{j}\in\mathcal{A}^{\prime}$ cannot reach $V^{-}$ in $\mathcal{H}-X_{i}$.
So $j>i$ and $X_{j}$ has no out-neighbors before $X_{i}$. In particular, $d_{\mathcal{A}}^{+}(X_{j})***2\left\vert B-N_{B}(S)\right\vert $. Using
$\left\vert S\right\vert +\left\vert D\right\vert =s$ and $r\left\vert S\right\vert \geq\left\vert E(S,A)\right\vert +\left\vert N_{B}(S)\right\vert $,
\[
bs+(a-1)\left\vert S\right\vert =b\left\vert D\right\vert +r\left\vert S\right\vert >\left\vert B-N_{B}(S)\right\vert +\left\vert E(S,A)\right\vert +\left\vert N_{B}(S)\right\vert >bs+\left\vert E(S,A)\right\vert .
\]
Thus $(a-1)\left\vert S\right\vert >\left\vert E(S,A)\right\vert $, and so there
exists $w\in S$ with $d_{A}(w)\leq a-2$. Thus $w$ witnesses some edge $X_{j}X\in E(\mathcal{H}[A])$.
Since $w\in S$, it has a solo neighbor $y\in B$.
Move $w$ to $X$ and $y$ to $X_{j}$. This yields nearly equitable colorings of
$G[A+y]$ and $G[B-y]$. Since $X_{j}$ is terminal, $X+w$ can still reach $V^{-}$.
Thus by Case 0, $G[A+y]$ has an equitable $a$-coloring. By (\ref{y}), $\Delta(G[B-y])\leq b-1$.
So by the primary induction hypothesis $G[B-y]$ has an equitable $b$-coloring.
After combining these equitable colorings we are done.
\textbf{Case 2}: All the last $b$ classes $X_{a-b},\ldots,X_{a-1}$ are terminal.
Then $a^{\prime}\geq b$. For $y\in B^{\prime}$, let $\sigma(y)$ be the number
of solo neighbors of $y$. Similarly to (\ref{y}),
\[
r\geq d(y)\geq a+b-b^{\prime}+d_{B^{\prime}}(y)+a^{\prime}-\sigma(y)\geq r+1+d_{B^{\prime}}(y)+a^{\prime}-b^{\prime}-\sigma(y).
\]
So $\sigma(y)\geq a^{\prime}-b^{\prime}+d_{B^{\prime}}(y)+1$. Let $I$ be a maximal
independent set with $V^{+}\subseteq I\subseteq B^{\prime}$. Then $\sum_{y\in I}(d_{B^{\prime}}(y)+1)\geq\left\vert B^{\prime}\right\vert =b^{\prime}s+1$.
Since $a^{\prime}\geq b$,
\[
\sum_{y\in I}\sigma(y)\geq\sum_{y\in I}(a^{\prime}-b^{\prime}+d_{B^{\prime}}(y)+1)\geq s(a^{\prime}-b^{\prime})+b^{\prime}s+1>a^{\prime}s=\left\vert A^{\prime}\right\vert .
\]
So some vertex $w\in W\in\mathcal{A}^{\prime}$ has two solo neighbors $y_{1}$
and $y_{2}$ in the independent set $I$.
Since the class $Y$ of $y_{1}$ is reachable from $V^{+}$, we can equitably $b$-color
$G[B-y_{1}]$. Let $Y^{\prime}$ be the new class of $y_{2}$. If $w$ witnesses
an edge $WX$ of $G[A]$ then we are done by Case 1; otherwise we can move $w$ to
some class $U\subseteq B^{\prime}-y_{1}$. Replacing $w$ with $y_{1}$ in $W$ to
get $W^{\ast}$ and moving $w$ to $U$ yields a new nearly equitable $(r+1)$-coloring
of $G$. If $U\in\mathcal{A}$ then we are done by Case 0; otherwise at least $a+1$
classes, $W^{\ast},Z^{\prime}$, and all $X\in\mathcal{A}^{\prime}-W$, can reach
$V^{-}$. In this case we are done by the secondary induction hypothesis.
\appendix
\chapter{Exceptional notation}
\noindent $|G|=|V(G)|=n(G)$, number
\noindent $\left\Vert G\right\Vert =|E(G)|=e(G)$, number
\noindent $E(v)=\{vu:vu\in E\}$, edge set
\noindent $K(A,B)=(A\cup B,\{ab:a\ne b\wedge(a,b)\in A\times B\})$, graph
\noindent $K(A)=K(A,A)$, graph
\noindent $E_{G}(A,B)=\{ab\in E(G):a\ne b\wedge(a,b)\in A\times B\}$, edge set
\chapter{Standards }
\subsection{\textcolor{red}{MAT 416--Level C}}
\begin{enumerate}
\item \textcolor{red}{Theorem~\ref{OddCycle} (Characterization of bipartite graphs).}
\item \textcolor{red}{Theorem~\ref{Eulerian} (Euler's Theorem). You may use Lemma~\ref{trail}.}
\item \textcolor{red}{Theorem~\ref{Tree} (Tree Theorem). Provide proofs for required
parts of Lemma~\ref{ELeaf} and Lemma~\ref{leaf}. }
\item \textcolor{red}{Theorem~\ref{Berge} (Berge's theorem on maximum matchings).}
\item \textcolor{red}{Corollary~\ref{HallC}. You may use Hall's Theorem.}
\item \textcolor{blue}{Theorem~\ref{Pete1} Petersens's Matching Theorem.}
\item \textcolor{blue}{Theorem~\ref{Whitney} (Whitney's Theorem).}
\item \textcolor{blue}{Lemma~\ref{kern} (Kernel Lemma).}
\item \textcolor{blue}{Euler's Formula Theorem~\ref{EF} and Theorem~\ref{esg}.}
\end{enumerate}
\subsection{\textcolor{red}{MAT 416--Level A}}
\textcolor{red}{All MAT 416--Level C and HW \#1--3, }\textcolor{blue}{HW4,5, }HW
6\textcolor{red}{{} except (+) and:}
\begin{enumerate}
\item \textcolor{red}{Theorem~\ref{Ramsey} (Ramsey's Theorem for graphs).}
\item \textcolor{red}{Corollary~\ref{DiracC} (Dirac's Theorem). Provide proofs for required
parts of Theorem~\ref{DO}.}
\item \textcolor{red}{Theorem~\ref{MaxFMinC-th} (Max-Flow, Min-Cut Theorem). You may
use Theoerm~\ref{FlowOverCuts-th}.}
\item \textcolor{red}{Theorem~\ref{KE-1} (König--Egerváry Theorem) using Max-Flow, Min-Cut
Theorem.}
\item \textcolor{red}{Corollary~\ref{Hall-th} (Hall's Theorem) using König, Egerváry
Theorem.}
\item \textcolor{red}{Theorem~\ref{TutteG} (Tutte's Theorem).}
\item \textcolor{red}{Theorem~\ref{Pete2} (Petersen's $2$-Factor Theorem).}
\item \textcolor{blue}{Theorem~\ref{Thomassen3} (Thomassen's Contraction Theorem).}
\item \textcolor{blue}{Theorems~\ref{MengerDE-thm}, \ref{MengerE-thm}, \ref{MengerVD-thm},
\ref{MengerV-thm}, \ref{menger}, \ref{locMeng-th-1}, and \ref{Menger-Fan} (Various
versions of Menger's Theorem). You only need to be able to prove Theorem~\ref{MengerDE-thm},
but you should know and be able to use the others, especially \ref{MengerV-thm},
\ref{menger}, \ref{locMeng-th-1}, and \ref{Menger-Fan}.}
\item \textcolor{blue}{Example~\ref{Mycielski} Mycielski's Construction.}
\item \textcolor{blue}{Theorem~\ref{Brooks} (Brooks' Theorem). You may assume Proposition~\ref{Sep}
and Lemma~\ref{order-lem}, but be prepared to prove them if asked.}
\item \textcolor{blue}{Theorem~\ref{Turan} (Turan's Theorem). You may assume Lemma \ref{r-part},
but be prepared to prove it if asked.}
\item \textcolor{blue}{Theorem~\ref{Vizing} (Vizing's Theorem). This includes proving
Lemma \ref{VL}.}
\item \textcolor{blue}{Theorem~\ref{Galvin} (Galvin's Theorem).}
\item \textcolor{blue}{Kuratowski's Theorem \ref{KT}. You may assume:\ref{csd}, \ref{noK}
and \ref{2cp}. You should be able to prove Theorems \ref{3KT} and \ref{max-3-con-thm},
and Theorem \ref{KT}, using Theorems \ref{3KT} and \ref{max-3-con-thm} (easy).}
\item Thomassen's 5-Choosability Theorem~\ref{5C}.
\item Lower bound on $\Ram(k,k)$ Theorem~\ref{rlb}.
\end{enumerate}
\subsection{MAT--513}
All MAT 416--Level A, \textcolor{red}{HW \#1--3, }\textcolor{blue}{HW \#4,5, }HW\#6.
\medskip{}
\section{{\LARGE{}Sample MAT 416/513 Midterm 1}}
\medskip{}
\noindent \textbf{Directions:} Use one sheet per problem. Order the sheets by problem
before submitting.
MAT\textbf{ 513 Students:} Do the \textbf{last} \textbf{4} problems.
MAT\textbf{ 416 Students: }Do \textbf{any} \textbf{6} problems; one will be treated
as extra credit. The first four problems are intended to be somewhat easier. Only
the six chosen problems should be turned in.
\begin{enumerate}
\item Prove: If an acyclic graph $G$ satisfies $|G|=\left\Vert G\right\Vert +1$ then
it is connected.
\item Prove: Every $k$-regular bigraph $G$ has a perfect matching.
\item Let $P\subseteq G$ be an $x,y$-path. Prove: $G[P]$ contains an $x,y$-path $Q$
with $Q=G[Q]$.
\item Let $T$ be a forest (acyclic graph) such that $2k$ of its vertices have odd degree.
Prove that $T$ decomposes into $k$ paths.
{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}{*}
\item Prove: A graph is bipartite if and only if it contains no odd cycle.
\item Prove: A graph $G=(V,E)$ has a perfect matching if $o(G-S)\leq|S|$ for all $S\subseteq V$.
\item Let $n\geq2$ and $d_{1},\dots,d_{n}\in\mathbb{Z}^{+}$. Prove: If $\sum_{i=1}^{n}d_{i}=2n-2$
then there is a tree with vertices $v_{1},\dots,v_{n}$ such that $d(v_{i})=d_{i}$
for all $i\in[n]$.
\item For $k,n\in\mathbb{N},$ let $G$ be an $A,B$-bigraph with $|A|=n=|B|$ such that
$\delta(G)\geq k$, and for all $X\subseteq A,Y\subseteq B$, if $|X|,|Y|\geq k$
then $|E(X,Y)|\ne\emptyset$. Prove: $G$ has a perfect matching.
\end{enumerate}
\chapter{Matching card trick}
Consider a deck of $2k+1$ cards numbered $1,\dots,2k+1$, and denoted by $[2k+1]$.
The class chooses a \emph{hand} $H$ consisting of\emph{ }$k+1$\emph{ }of these
cards, and gives them to Professor A. Professor A looks at them, puts one of them
in his pocket, and then has a student spread the remaining $k$ cards face-up on
a table. Professor B, who has observed none of this transaction, now enters the room,
looks at the cards on the table and identifies the one in Professor A's pocket. How
is this done?
\textbf{Solution. \noindent}Our arithmetic is done modulo $k+1$, and we use $k+1$
instead of $0$ for the representative of its equivalence class. Arrange the cards
of $H$ in order as $c_{1}<\dots\dots>\overline{d}_{k+1}$.
Then $c_{x}=\overline{d}_{k+2-(1-y)}=\overline{d}_{y}$.
\chapter{Alternative proofs}
\section{Hall's Theoerm}
\begin{proof}[Proof of Theorem~\ref{Hall}]
If $M$ is a matching saturating $X$ and $S\subseteq X$ then $|S|=|E(S,Y){\color{red}{\normalcolor \cap M}}|\leq|N(S)|$;
so \eqref{eq:Hall} holds.
Suppose \eqref{eq:Hall} holds for some $X,Y$-bigraph with no matching saturating
$X$; among such counterexamples choose $G$ with $|G|$ minimal, and subject to
this $\left\Vert G\right\Vert $ maximal. By minimality, (1) $N(X)=Y$: if $y\in Y\smallsetminus N(X)$
then $G-y$ is a smaller counterexample. Also (2) all $a\in X$ satisfy $N(a)\ne Y$:
else, since $G-a$ has a matching saturating $X-a$ and there is an unsaturated vertex
$b\in Y$ by \eqref{eq:Hall}, $G$ has a matching saturating $X$.
Let $a\in X$; by (2) there is $b\in Y$ with $ab\notin E$. Since $G+ab$ satisfies
\eqref{eq:Hall}, maximality implies $G+ab$ has a matching $M^{+}$ saturating $X$.
By (1) there is $a'\in X$ with $a'b\in E$; by (2) there is $b'\in Y$ with $a'b'\notin E$.
Again by maximality, $G+a'b'$ has a matching $L^{+}$ saturating $X$. Set $M:=M^{+}-ab$
and $L:=L^{+}-a'b'$.
Let $H$ be the spanning submultigraph of $G$ with edge set $M\cup L+a'b$, where
edges in $M\cap L$ have multiplicity $2$. (Figure~\ref{Fig:HallT-1}). Then $\Delta(H)\leq2$.
As $G$ is bipartite, Proposition~\ref{Dle2} implies the components of $H$ are
paths and even cycles. Each cycle has a perfect matching. As $d_{H}(a)=1$ and $d_{H}(x)=2$
for all $x\in X-a$, each path $P$ has an end $y\in Y$ and a matching saturating
$V(P)-y$. Combining these matchings yields a matching saturating $X$.
Here is an alternative way through the last paragraph.
Let $H$ be the spanning subgraph of $G$ with $E(H)=M\triangle L$. (Figure~\ref{Fig:HallT-1}).
Then $\Delta(H)\leq2$ and $d_{H}(b),d_{H}(b')\leq1=d_{H}(a)=d_{H}(a')0$, and fix a set $\mathcal{P}'$ of disjoint $A,B$-paths with $|\mathcal{P}'|**