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  \makebox[\textwidth][l]{\textbf{MATH 6404: Applied [Combinatorics and] Graph
  Theory} \hfill CU Denver} \
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 % Add semester and instuctor
    \makebox[\textwidth][l]{Spring 2026 \hfill Instructor: Carlos Mart\'inez}
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    \textbf{Date:} #3 \hfill     \textbf{Scribe:} #4
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%%% -+-+-+-+-+-+- BEGIN HERE -+-+-+-+-+-+- %%%
\newcommand{\lecturenumber}{$4$}
\newcommand{\lecturetitle}{Combinatorial Arguments (Continued)}
\newcommand{\scribename}{Jon Terry}
\newcommand{\lecturedate}{February 2, 2026}

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The purpose of this lecture is to demonstrate that bijections can be quite
useful in combinatorial proofs. 
But in order to conduct the demonstration, we
need some definitions. 

\section{Trees}
A \textit{tree} $T=(V,E)$ is an undirected graph that is connected and acyclic.
For example, the following graph is a tree.

\begin{figure}[ht]
  \centering
  \resizebox{.4\linewidth}{!}{%
        \begin{tikzpicture}
            \Vertex[x=0,  y=0,  label=1]{1}
            \Vertex[x=-2, y=-2, label=2]{2}
            \Vertex[x=-3, y=-4, label=4]{4}
            \Vertex[x=-1, y=-4, label=5]{5}
            \Vertex[x=2,  y=-2, label=3]{3}
            \Vertex[x=2,  y=-4, label=6]{6}
            \Edge(1)(2)
            \Edge(2)(4)
            \Edge(2)(5)
            \Edge(1)(3)
            \Edge(3)(6)
        \end{tikzpicture}
  }
  \caption{A tree.}
  \label{fig:a tree}
\end{figure}
The vertices 4, 5, and 6 are each called \textit{leaves} of the tree. Formally,
if $T=(V,E)$ is a tree, then $u\in V$ is a leaf of $T$ if and only if $\left| \{ v\in V :
\{u,v\}\in E \} \right|=1 $.

\subsection{Facts about Trees}
Here are some facts about trees, presented without proof.
For any tree $T=(V,E)$, the following statements are true: 
\begin{itemize}
  \item 
  If $\left| V \right| \ge 2$, then $T$ has at least two leaves.
  \item 
  If $u\in V$ is a leaf of $T$, then $T'=T-u$ is a also a tree. 
  In other
  words, removing a leaf from a tree yields a new tree (but with one less vertex
  than the tree we started with).
  \item 
  $\left| V \right|=\left| E \right| + 1 $. However the converse is not
  always true. 
  In other words if $G=(V,E)$ is any graph with $\left| V \right|
  =\left| E \right| +1$, $G$ is not necessarily a tree.
  \item For all $u,v\in V$, there is a unique $\{u,v\}$-path in $T$. This fact
  follows almost immediately from the fact that $T$ is connected and acyclic.
  Also, be aware that we often write $\,\exists!\,$ to stand for the phrase
  ``there exists a unique \textellipsis.''
\end{itemize}

\section{Bijections}
First, recall that given two sets $X$ and $Y$, the function $f:X\to Y$ is a
\textit{bijection} between $X$ and $Y$ if $f$ is \textit{onto}, meaning that for
all $y\in Y$ there is an $x\in X$ such that $f(x)=y$, and $f$ is
\textit{one-to-one}, meaning that if $f(x_1)=f(x_2)$, then $x_1=x_2$. 
Now, if
$f$ is a bijection, then we can define another function $f^{-1}=
g:Y\to X$ given by $y=f(x)\mapsto x$. 
So another way to think about a bijection
is to say that $f:X\to Y$ is a bijection if there exists a function $g:Y\to X$
such that $f\circ g= \text{id}_Y$ and $g\circ f=\text{id}_X$. 

Now, here's what's neat about bijections and why they are useful in combinatorial proofs. 
Suppose we have a set $A$ whose size (i.e.  \textit{cardinality}) we want to know, but this quantity is difficult to
calculate. On the other hand, suppose we have a second set, $B$, whose
cardinality is known (or at least easy to find).  If we suspect that the cardinality of $B$ and the cardinality of $A$ are the same, then notice that
if we can find a bijection between $A$ and $B$, then $\left| A \right| = \left|
B \right|$ and we're done!

\section{Bijective Proofs}
We can use this technique of constructing a bijection between sets to prove that
there are $n^{n-2}$ labeled trees on $n$ vertices. Before we formally state the
theorem, let's define our target set, $B$. Let $B$ be the set of all strings of
length $n-2$ we can form from an alphabet of $[n] \coloneqq \{1,2,\ldots,n\}$ symbols. The
claim is that there are $n^{n-2}$ such strings.  Why? Well, notice there are $n$
choices for the first symbol of such a string, and $n$ choices for the second
symbol, and so on, up to the last ($n-2$) symbol. Since these choices are
independent of each other, then by the product principle discussed in Lecture 3,
we have $\left| B \right| = n^{n-2}$. 


\begin{theorem}
\label{theorem: main}
For any $n \geq 1$, the number of labeled trees on $n$ vertices is $n^{n-2}$.
\end{theorem}

\begin{proof}
  If $n=1$, then the theorem is true, since $1^{1-2}=1^{-1}=1/1=1$ and a graph
  with one vertex and no edges is a tree trivially. 
  So assume $n\ge 2$. Let
  $f:\mathcal{T} \to P([n], n-2)$, where $\mathcal{T}$ is the set of all trees
  on $n$ vertices and $P([n],n-2)$ is the set of all strings of length $n-2$
  that we can obtain from an alphabet of $n$ symbols. 
  We define $f$ as Algorithm~\ref{alg: prufer}.
  
\begin{algorithm}[ht]
\DontPrintSemicolon
\While{$|V(T)|>2$}{
  Let $\ell_i'$ be the leaf of $T$ with the largest label\;
  Let $w_i$ be the unique neighbor of $\ell_i'$\;
  Remove $\ell_i'$ from $T$\;
}
\Return $(w_1,w_2,\ldots,w_{n-2})$\;
\caption{Pr\"ufer Code Algorithm.}
\label{alg: prufer}
\end{algorithm}
This called Pr\"ufer's algorithm, and it gives us the map we want from the set
of trees to the set of $n-2$-length strings.
For example, consider the tree

\begin{figure}[ht]
  \centering
  \resizebox{.4\linewidth}{!}{%
        \begin{tikzpicture}
            \Vertex[x=0,  y=0,  label=1]{1}
            \Vertex[x=4, y=-1, label=2]{2}
            \Vertex[x=3.5, y=-2.5, label=3]{3}
            \Vertex[x=2, y=-3, label=4]{4}
            \Vertex[x=-1,  y=-3, label=5]{5}
            \Vertex[x=-2,  y=-1, label=6]{6}
            \Edge(1)(4)
            \Edge(2)(3)
            \Edge(3)(4)
            \Edge(5)(6)
            \Edge(6)(3)
        \end{tikzpicture}
  }
  \caption{A tree on six nodes.}
  \label{fig:a tree on six nodes}
\end{figure}

Applying Pr\"ufer's algorithm to this tree produces the following output:

\begin{table}[ht]
  \centering
  \caption{Pr\"ufer's algorithm on a six-node tree.}
  \label{tab:prufers algorithm}
  \begin{tabular}{c | l | l}
   Step & Largest leaf ($\ell_i'$)  to delete & Neighbor ($w_i$)\\
   \hline
   1 & 5 & 6\\
   2 & 6 & 3\\
   3 & 2 & 3\\
   4 & 3 & 4\\
   5 & Stop (only two vertices left). & -- \\
  \end{tabular}
\end{table}

\noindent Thus, the Pr\"ufer sequence for the tree given above is
$6\,3\,3\,4$.\\

Having established that $f:\mathcal{T}\to P([n],n-2)$, we define 
$g:P([n],n-2)\to \mathcal{T}$ as the inverse of the Pr\"ufer algorithm. So,
starting with a Pr\"ufer sequence, $g$ constructs a tree $T'$ whose first $n-2$
edges are $\{\ell_i, w_i\}$ for all $i=\{1,\ldots,n-2\}$, where 
\[
  \ell_i = \max \{[n]\setminus
  \{\ell_1,\ell_2,\ldots,\ell_{i-1},w_i,w_{i+1},\ldots,w_{n-2}\}\},
\]
and $w_i$ is the $i$th vertex in the Pr\"ufer sequence. The last edge of $T'$
is found by connecting the two remaining vertices found in $\{[n]\setminus
\{\ell_1,\ldots,\ell_{n-2}\}\}$. In other words, $g$ forms a tree from a
Pr\"ufer sequence by sequentially picking the largest \textit{available} vertex
(found by taking the list of all vertices and excluding from consideration all
the vertices already used and those remaining in the sequence) and paring it
with the current vertex in the Pr\"ufer sequence. The final edge consists of the
two vertices that were never deleted as leaves by $f$ in the original process.

We can see how this works by applying the reverse algorithm to the Pr\"ufer
sequence from our example:

\begin{table}[h]
  \centering
  \caption{Inverse Pr\"ufer algorithm on the sequence 6\,3\,3\,4 }
  \label{tab:inverse prufer algorithm}
  \begin{tabular}{c | l | l | l}
    Step & Largest leaf $(\ell_i)$ to pick & $w_i$ & Edge\\
    \hline
    1 & $\ell_1 = \{[6]\setminus \{6,3,3,4\}\} = 5$ & 6 & $\{5,6\}$\\
    2 & $\ell_2 = \{[6]\setminus \{5,3,3,4\}\} = 6$ & 3 & $\{6,3\}$\\
    3 & $\ell_3 = \{[6]\setminus \{5,6,3,4\}\} = 2$ & 3 & $\{2,3\}$\\
    4 & $\ell_4 = \{[6]\setminus \{5,6,2,4\}\} = 3$ & 4 & $\{3,4\}$\\
  \end{tabular}
\end{table}

Continuing with the proof, we claim that $g=f^{-1}$. Notice that $f$ enumerated
the edges of $T$ to be
\[
  \{\{\ell_1',w_1\},\{\ell_2',w_2\},\ldots,\{\ell_{n-2}',w_{n-2}\}\}\cup
  \{[n]\setminus \{\ell_1',\ell_2',\ldots,\ell_{n-2}'\}\},
\]
while $g$ enumerated the edges of $T'$ as 
\[
  \{\{\ell_1,w_1\},\{\ell_2,w_2\},\ldots,\{\ell_{n-2},w_{n-2}\}\}\cup
  \{[n]\setminus \{\ell_1,\ell_2,\ldots,\ell_{n-2}\}\}.
\]
Therefore, we need to show that $\ell_i'=\ell_i$ for all
$i=\{1,2,\ldots,n-2\}$. To see why this is so, recognize that when we go to pick
$\ell_i'$ we start with original set of $[n]$ vertices. But the vertices
$\ell_1',\ell_2',\ldots,\ell_{i-1}'$ have already been picked (so the algorithm
cannot pick them again), and furthermore, we can't pick $w_i,w_{i+1},\ldots,w_{n-2}$
since they have degree  $\ge 2$ because they are neighbors of ``future''
leaves. Thus we have, 
\[
  \ell_i' = \max \{[n]\setminus \{\ell_1',\ell_2',\ldots,\ell_{i-1},w_i,w_{i+1},\ldots,w_{n-2}\}\}.
\]
Since this is exactly the set that defines $\ell_i$ for all $i=\{1,\ldots,n-2\}$, the $\ell_i'$'s and $\ell_i$'s match. 
This establishes that $f$ is bijection between the set of all trees on $n$ vertices $\mathcal{T}$ and the set $P([n],n-2)$ of all strings of length $n-2$ that can be formed from an alphabet of $n$ symbols. 
Therefore, since  $|P([n],n-2)|=n^{n-2}$, then $|\mathcal{T}|=n^{n-2}$ which proves the claim.
\end{proof}
The count in Theorem~\ref{theorem: main} is Cayley's formula (check out \href{https://oeis.org/A000272}{OEIS A000272}).


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