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    \makebox[\textwidth][l]{\textbf{MATH 6404: Applied [Combinatorics and] Graph Theory} \hfill CU Denver} \\
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    \makebox[\textwidth][l]{Spring 2026 \hfill Instructor: Carlos Mart\'inez}
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    \textbf{Lecture #1:} #2 \\
    \textbf{Date:} #3 \hfill     \textbf{Scribe:} #4
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\begin{document}

\thispagestyle{firstpage}
\scribebox{7}{Counting Lattice Paths}{February 11, 2026}{Baihan Liu}

\section{Definition of Lattice Paths}

Let $P$ be a lattice path:
\[
P:\ (x_0,y_0),(x_1,y_1),\dots,(x_n,y_n),
\]
where $x_i,y_i\in\mathbb{Z}$ for all $i\in [n]_0$.


A step is the vector
\[
[x_i-x_{i-1},\,y_i-y_{i-1}].
\]
A north step, denoted $N$, is the vector $[0,1]$. 
An east step, denoted $E$, is the vector $[1,0]$.

We will focus on lattice paths that involve only $N$ and $E$ steps.


\vspace{0.5cm}



\begin{figure}[H]
\center
\begin{tikzpicture}[scale=0.85, every node/.style={font=\small}];

  % Grid (black): fits the endpoint of NEEENENNE, which is (5,4)
  \draw[step=1, black, line width=0.4pt] (0,0) grid (5,4);

  % Optional lattice points (comment out if you don't want dots)
  \foreach \x in {0,...,5}
    \foreach \y in {0,...,4}
      \fill[black] (\x,\y) circle (1.1pt);

  % Path P in blue following NEEENENNE, starting at (0,0)
  \draw[blue, line width=1.8pt, -{Latex[length=2.2mm]}]
    (0,0) -- (0,1) -- (1,1) -- (2,1) -- (3,1) -- (3,2) -- (4,2) -- (4,3) -- (4,4) -- (5,4);

  % Optional start/end labels
  \node[below left] at (0,0) {$(0,0)$};
  \node[above right] at (5,4) {$(5,4)$};
\end{tikzpicture}
\caption{A lattice path $P:\ \texttt{NEEENENNE}$.}
\label{fig:lattice-path-P}
\end{figure}

\section{Catalan Numbers and Dyck Paths}
A Dyck path of semilength n is a lattice path from $(0,0)$ to $(n,n)$ such that it never goes below the main diagonal $y=x$. 

\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.0]

  % Grid (3 by 3)
  \draw[step=1, black, line width=0.4pt] (0,0) grid (3,3);

  % Diagonal y=x (red)
  \draw[red, line width=1.2pt] (0,0) -- (3,3);

  % Blue path: NENNEE  (from (0,0) to (3,3))
  \draw[blue, line width=1.8pt, -{Latex[length=2.2mm]}]
    (0,0) -- (0,1) -- (1,1) -- (1,2) -- (1,3) -- (2,3) -- (3,3);

  % Green path: NNEEEN (from (0,0) to (3,3))
  \draw[green!70!black, line width=1.8pt, -{Latex[length=2.2mm]}]
    (0,0) -- (0,1) -- (0,2) -- (1,2) -- (2,2) -- (3,2) -- (3,3);
  \node[below left] at (0,0) {$(0,0)$};
  \node[above right] at (3,3) {$(3,3)$};

\end{tikzpicture}
\caption{Dyck path (Blue, NENNEE), a path that is not a Dyck path (Green, NNEEEN), diagonal $y=x$ (red).}
\label{fig:dyck-vs-not}
\end{figure}

\noindent
\textbf{Question.} Let $D(n)$ be the set of Dyck paths of semilength $n$. What is $|D(n)|$?

\noindent
We let $C_n := |D(n)|$, where $C_n$ is the $n$th Catalan number, with the convention that $C_0=1$.

\subsection{Recurrence Relation}

\begin{lemma}
    For $n \ge 1, C_n = \displaystyle\sum_{j = 1}^n C_{j-1} C_{n-j}.$
\end{lemma}

\begin{proof}
Let \(P: v_0, v_1, \dots, v_{2n}\in D(n)\), where $v_0=(0,0)$ and $v_{2n}=(n,n).$ While the endpoint is fixed on the diagonal, \(P\) may also touch the diagonal \(y=x\) several times along the way. Let \(j\) be the first return of \(P\) to the diagonal, i.e., $j$ is the smallest $j>0$ such that $v_{2j} = (j,j)$. This is an example of extremality. For example, in Figure 3, $j = 3$ is the smallest $j>0$ with $v_6 = (3,3)$.


\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=0.85, every node/.style={font=\small}]

  % Grid for NNENEENENE ends at (5,5)
  \draw[step=1, black, line width=0.4pt] (0,0) grid (5,5);

  % Diagonal (red): y=x
  \draw[red, line width=1.2pt] (0,0) -- (5,5);

  % Optional lattice points
  \foreach \x in {0,...,5}
    \foreach \y in {0,...,5}
      \fill[black] (\x,\y) circle (1.1pt);

  % Path P in blue: NNENEENENE
  \draw[blue, line width=1.8pt, -{Latex[length=2.2mm]}]
    (0,0) -- (0,1) -- (0,2) -- (1,2) -- (1,3) -- (2,3) -- (3,3) -- (3,4) -- (4,4) -- (4,5) -- (5,5);

  % Start/end labels
  \node[below left] at (0,0) {$(0,0)$};
  \node[above right] at (5,5) {$(n,n)$};

  % Mark points
  \fill[orange] (0,1) circle (2.2pt);
  \fill[orange] (2,3) circle (2.2pt);
  \fill[orange] (3,3) circle (2.2pt);

  % Smaller point labels
  \node[left=2pt, fill=white, inner sep=1pt, font=\scriptsize] at (0,1) {$(0,1)$};
  \node[above right=2pt, fill=white, inner sep=1pt, font=\scriptsize] at (2,3) {$(2,3)$};
  \node[below right=2pt, fill=white, inner sep=1pt, font=\scriptsize] at (3,3) {$(3,3)$};

\end{tikzpicture}
\caption{A Dyck path $P:\ \texttt{NNENEENENE}$ (blue).}
\label{fig:lattice-path-NNENEENENE-marked}
\end{figure}


P can be decomposed as $P = P_1+P_2$, where $P_1$ is a Dyck path from $(0,1)$ to $(j-1,j)$ and $P_2$ is a Dyck path from $(j,j)$ to $(n,n)$.

To see why \(P_1\) is a Dyck path, observe that the first step in \(P\) must be \((0,1)\), and the last step before the first return is \((1,0)\).
Moreover, by our extreme choice of \(j\), from \((0,1)\) to $(j-1,j)$ the path does not touch the line \(y=x\).

Thus, $P_1 \in D(j-1)$, and $P_2 \in D(n-j)$.
For a fixed $j$, there are $|D(j-1)||D(n-j)| = C_{j-1} C_{n-j}$ different ways of building $P$.

$j$ can be anywhere in $j \in \{1,2,\dots, n\}$.
These options for first return are mutually exclusive and collectively exhaustive, so we can use sum principle over choice of $j$.
Thus,
\[
C_n = \sum_{j = 1}^n C_{j-1} C_{n-j}, \qquad n \ge 1.
\]
\end{proof}

This is \href{https://oeis.org/A000108}{OEIS A000108}.



\subsection{Closed-form Formula}
\begin{lemma}
For $n \ge 1, C_n = \displaystyle\frac{1}{n+1}$ $\displaystyle\binom{2n}{n}$.
\end{lemma}


\begin{proof}
There are $\displaystyle\binom{2n}{n}$ different lattice paths from $(0,0)$ to $(n,n)$.
Some of these are Dyck paths, which we called ``good paths''.
Some of these are not Dyck paths, which we called ``bad paths''.

The total number of paths is equal to the number of good paths plus the number of bad paths:

\[\binom{2n}{n} = |G| + |B|.\]
 \[C_n = |G|= \binom{2n}{n} - |B|.\]

Thus, to find the number of ``good paths'', we only need to find the number of ``bad paths''.

A path $P$ is bad, i.e., $P \in B$, if it crosses below the diagonal $y=x$ at least once. Thus, it touches the off-diagonal $y = x-1$ at least once, as shown in Figure 4. 


\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=0.85, every node/.style={font=\small}]

  % Grid 6 by 6
  \draw[step=1, black, line width=0.4pt] (0,0) grid (6,6);

  % Lattice points (optional)
  \foreach \x in {0,...,6}
    \foreach \y in {0,...,6}
      \fill[black] (\x,\y) circle (1.1pt);

  % Diagonal y=x (red), extended a bit outside the grid
  \draw[red, line width=1.2pt] (-0.4,-0.4) -- (6.4,6.4);

  % Dotted line y=x-1 (orange), extended a bit outside the grid
  \draw[orange, dotted, line width=1.4pt] (-0.4,-1.4) -- (7.4,6.4);

  % ------------------------------------------------------------
  % Shared part of both paths (draw once in a mixed color)
  % (0,0) -> (4,3)
  \draw[blue!55!green!60!black, line width=2.2pt]
    (0,0) -- (0,1) -- (0,2) -- (1,2) -- (2,2) -- (3,2) -- (4,2) -- (4,3);

  % Blue tail only: (4,3) -> (6,6)
  \draw[blue, line width=1.8pt, -{Latex[length=2.2mm]}]
    (4,3) -- (4,4) -- (4,5) -- (5,5) -- (5,6) -- (6,6);

  % Green tail only: (4,3) -> (7,5)
  \draw[green!60!black, line width=1.8pt, -{Latex[length=2.2mm]}]
    (4,3) -- (5,3) -- (6,3) -- (6,4) -- (7,4) -- (7,5);
  % ------------------------------------------------------------

  % Corner labels
  \node[below left] at (0,0) {$(0,0)$};
  \node[above right] at (6,6) {$(6,6)$};

  % Mark and label the outside point (7,5)
  \fill[black] (7,5) circle (1.3pt);
  \node[above right=1pt, font=\scriptsize, fill=white, inner sep=1pt] at (7,5) {$(7,5)$};

\end{tikzpicture}
\caption{Bad path $P$(blue, NNEEEENNNENE), $\phi(P)$ (green, NNEEEENEENEN),  diagonal $y=x$ (red), offset line $y=x-1$ (orange dotted).}
\label{fig:lattice-paths-6x6}
\end{figure}


The reflection of $(n,n)$ across $y = x-1$ is $(n+1, n-1)$. Let $(i+1, i)$ be the last lattice point of $P$ on $y = x-1$, where $i \in \{0,1,2,\dots, n\}$. The subpath of $P$ from $(i+1, i)$ to $(n,n)$ is the final portion of $P$. 

Let $\phi: B \longrightarrow \mathcal{P}((0,0), (n+1, n-1))$ by reflecting the final portion of $P \in B$ after final intersection with $y = x-1$. For example, in Figure 4, the green path is the image of the blue path under $\phi$. In the step encoding, the final portion 
NNENE is reflected to EENEN.

$\phi$ turns $P \in B$, which touches $y = x-1$, into a $(0,0)$ to $(n+1, n-1)$ path that touches $y=x-1$ at the same points. 

Conversely, every path from $(0,0)$ to $(n+1, n-1)$ crosses the off-diagonal $y = x-1$, since $(0,0)$ and $(n+1,n-1)$ are on different sides of it. 
Reflecting the final portion of the path produces a path from $(0,0)$ to $(n,n)$ that intersects $y=x-1$, hence lies in $B$.


Reflecting the final portion twice returns it to its original position. Thus, $B$ and paths from $(0,0)$ to $(n+1, n-1)$ are in bijection.
In fact, $\phi$ is an involution on $B \cup \mathcal{P}((0,0), (n+1, n-1))$.
Therefore,

\[|B| = |\mathcal{P}((0,0), (n+1, n-1))| = \binom{n+1+n-1}{n+1} = \binom{2n}{n+1}\]

\[C_n = \binom{2n}{n} -|B| = \binom{2n}{n} -\binom{2n}{n+1} = \frac{(2n)!}{n!n!} - \frac{(2n)!}{(n+1)!(n-1)!} = (1-\frac{n}{n+1}) \binom{2n}{n} = \frac{1}{n+1} \binom{2n}{n}\]

\end{proof}


\subsection{Final Project}
One final project idea is to explore more Catalan objects. 
Stanley’s book is a useful reference.
\begin{table}[H]
\centering
\renewcommand{\arraystretch}{1.3} % taller rows
\setlength{\tabcolsep}{10pt}      % wider columns


\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
n & 0 & 1 & 2 & 3 & 4 & 5 & \dots \\
\hline
$C_n$ & 1 & 1 & 2 & 5 & 14 & 42 & \dots \\
\hline
\end{tabular}

\caption{Catalan numbers}
\label{tab:two_by_seven}
\end{table}

\section{Fibonacci Numbers}
Let $F_0 = F_1 = 1$, and $F_n = F_{n-1} + F_{n-2}$, for all $n \ge 2$. 

\begin{table}[H]
\centering
\renewcommand{\arraystretch}{1.3} % taller rows
\setlength{\tabcolsep}{10pt}      % wider columns
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
n & 0 & 1 & 2 & 3 & 4 & \dots \\
\hline
$F_n$ & 1 & 1 & 2 & 3 & 5 & \dots \\
\hline
\end{tabular}

\caption{Fibonacci Number}
\label{tab:two_by_seven}
\end{table}
This is \href{https://oeis.org/A000045}{OEIS A000045}.

\noindent
A combinatorial interpretation: Let $P_n$ be a path with $n$ nodes. Consider the set of all possible ``matchings" (ways of pairing/not pairing adjacent nodes) in $P_n$.


\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.00, transform shape]
\tikzset{
  VertexStyle/.style={circle,draw,inner sep=0.6pt,minimum size=4mm},
  EdgeStyle/.style={line width=0.6pt}
}

% common x-coordinates
\def\xA{0}\def\xB{1.2}\def\xC{2.4}\def\xD{3.6}\def\xE{4.8}

% (1)
\node[font=\normalsize] at (-0.6, 0) {(1)};
\node[VertexStyle] (a1) at (\xA,0) {};
\node[VertexStyle] (a2) at (\xB,0) {};
\node[VertexStyle] (a3) at (\xC,0) {};
\node[VertexStyle] (a4) at (\xD,0) {};
\draw[EdgeStyle] (a1) -- (a2);
\draw[EdgeStyle] (a3) -- (a4);

% (2)
\node[font=\normalsize] at (-0.6, -1.2) {(2)};
\node[VertexStyle] (b1) at (\xA,-1.2) {};
\node[VertexStyle] (b2) at (\xB,-1.2) {};
\node[VertexStyle] (b3) at (\xC,-1.2) {};
\node[VertexStyle] (b4) at (\xD,-1.2) {};


% (3)
\node[font=\normalsize] at (-0.6, -2.4) {(3)};
\node[VertexStyle] (c1) at (\xA,-2.4) {};
\node[VertexStyle] (c2) at (\xB,-2.4) {};
\node[VertexStyle] (c3) at (\xC,-2.4) {};
\node[VertexStyle] (c4) at (\xD,-2.4) {};
\draw[EdgeStyle] (c3) -- (c4);

\end{tikzpicture}
\caption{Some possible ``matchings'' in $P_4$.}
\end{figure}


\noindent
\textbf{Claim}: $F_n$ is the number of different matchings in $P_n$.

\begin{proof}







For $n \ge 2$,

1. If we match $\{n-1,n\}$, then $n-1$ cannot be matched again. There are $F_{n-2}$ matching options for nodes $1,2,\dots, n-2$. 

\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.05, transform shape]
\usetikzlibrary{decorations.pathreplacing,calc}
\tikzset{
  VertexStyle/.style={circle,draw,inner sep=0.6pt,minimum size=4mm},
  EdgeStyle/.style={line width=0.6pt}
}

% x-positions for 6 displayed vertices
\def\xA{0}\def\xB{1.2}\def\xC{2.4}\def\xD{3.6}\def\xE{4.8}\def\xF{6.0}
\def\y{0}

% vertices (a path on n vertices)
\node[VertexStyle, label=above:$1$]      (v1) at (\xA,\y) {};
\node[VertexStyle, label=above:$2$]      (v2) at (\xB,\y) {};
\node[VertexStyle, label=above:$3$]      (v3) at (\xC,\y) {};
\node[VertexStyle, label=above:$\cdots$] (v4) at (\xD,\y) {};

% put n-2 and n-1 labels OUTSIDE the circles
\node[VertexStyle] (v5) at (\xE,\y) {};
\node[above=2pt] at (v5.north) {$n-2$};

\node[VertexStyle] (v6) at (\xF,\y) {};
\node[above=2pt] at (v6.north) {$n-1$};

% (optional) show the final n as text to the right, if you want to indicate continuation
\node[VertexStyle] (v7) at (\xF+1.2,\y) {};
\node[above=2pt] at (v7.north) {$n$};

% edges
\draw[EdgeStyle] (v1) -- (v2) -- (v3) -- (v4) -- (v5);

\draw[EdgeStyle, blue, line width=1.2pt] (v6) -- (v7)
  node[midway,below=5pt,font=\scriptsize] {matched};

% brace for the first n-2 positions: 1 through n-2
\draw[decorate,decoration={brace,mirror,raise=3pt}]
  (v1.south) -- (v5.south)
  node[midway,below=8pt] {$n-2$ nodes};

\end{tikzpicture}
\end{figure}








2. If we do not match $n$ with $n-1$. $n-1$ is free to be matched with $n-2$. There are $F_{n-1}$ matchings options for nodes $1,2,\dots, n-1$.



\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.05, transform shape]
%\usetikzlibrary{decorations.pathreplacing,calc}
\tikzset{
  VertexStyle/.style={circle,draw,inner sep=0.6pt,minimum size=4mm},
  EdgeStyle/.style={line width=0.6pt}
}

% x-positions for 6 displayed vertices
\def\xA{0}\def\xB{1.2}\def\xC{2.4}\def\xD{3.6}\def\xE{4.8}\def\xF{6.0}
\def\y{0}

% vertices (a path on n vertices)
\node[VertexStyle, label=above:$1$]      (v1) at (\xA,\y) {};
\node[VertexStyle, label=above:$2$]      (v2) at (\xB,\y) {};
\node[VertexStyle, label=above:$3$]      (v3) at (\xC,\y) {};
\node[VertexStyle, label=above:$\cdots$] (v4) at (\xD,\y) {};

% put n-2 and n-1 labels OUTSIDE the circles
\node[VertexStyle] (v5) at (\xE,\y) {};
\node[above=2pt] at (v5.north) {$n-2$};

\node[VertexStyle] (v6) at (\xF,\y) {};
\node[above=2pt] at (v6.north) {$n-1$};

% (optional) show the final n as text to the right, if you want to indicate continuation
\node[VertexStyle] (v7) at (\xF+1.2,\y) {};
\node[above=2pt] at (v7.north) {$n$};

% edges
\draw[EdgeStyle] (v1) -- (v2) -- (v3) -- (v4) -- (v5) -- (v6);

\draw[EdgeStyle, white, line width=1.2pt] (v6) -- (v7)
  node[midway,below=5pt,font=\scriptsize] {unmatched};

% brace for the first n-2 positions: 1 through n-2
\draw[decorate,decoration={brace,mirror,raise=3pt}]
  (v1.south) -- (v6.south)
  node[midway,below=8pt] {$n-1$ nodes};

\end{tikzpicture}
\end{figure}



By assumptions, $F_n = F_{n-1} + F_{n-2}$. Proof holds inductively.

\end{proof}

\begin{remark}
    The answer so far for $F_n$ is a recursion formula, which is less nice than our best answer for $C_n$, which is a closed-form formula.
\end{remark} 




\end{document}