%% filename: amsart-template.tex
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%     AMS-LaTeX v.2 template for use with amsart
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%     Remove any commented or uncommented macros you do not use.

\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{hobby}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{xca}[theorem]{Exercise}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\numberwithin{equation}{section}

\begin{document}

\title{Humans Suffice: A Novel Proof of the Four Color Theorem}

%    Remove any unused author tags.
%    author one information
\author{Eric Bauerfeld}
\address{}
\curraddr{}
\email{eric@dideric.is}
\thanks{Dr. Joe Fields}
\thanks{Holepunch}

%    author two information
% \author{}
% \address{}
% \curraddr{}
% \email{}
% \thanks{}

\subjclass[2020]{Primary }

\keywords{Discrete , Four Color Theorem,}

\date{}

\dedicatory{Dedicated to those who doubt the machine}

\begin{abstract}
\end{abstract}

\maketitle

\section{Kempe's Proof (Valid Portion)}

\subsection{Setup}

Kempe's strategy, published in 1879, follows a \emph{minimal counterexample} argument. Suppose, for contradiction, that there exists a planar graph requiring 5 colors. Among all such graphs, let $G$ be one with the fewest vertices. Then every planar graph with fewer vertices than $G$ is 4-colorable, but $G$ itself is not.

\subsection{Every Planar Graph Has a Vertex of Degree at Most 5}

\begin{lemma}
Every planar graph has at least one vertex of degree $\leq 5$.
\end{lemma}

\begin{proof}
Let $G$ be a connected planar graph with $V$ vertices, $E$ edges, and $F$ faces. By Euler's formula,
\[
  V - E + F = 2.
\]
Since every face is bounded by at least 3 edges and each edge borders at most 2 faces, we have $2E \geq 3F$, hence $F \leq \frac{2E}{3}$. Substituting into Euler's formula:
\[
  V - E + \frac{2E}{3} \geq 2 \implies E \leq 3V - 6.
\]
If every vertex had degree $\geq 6$, then $2E \geq 6V$, so $E \geq 3V$, contradicting $E \leq 3V - 6$. Therefore at least one vertex has degree $\leq 5$.
\end{proof}

Since $G$ is a minimal counterexample, it must contain a vertex $v$ of degree at most 5. Kempe argued by cases on the degree of $v$.

\subsection{Cases of Degree at Most 3}

Suppose $v$ has degree $\leq 3$. Remove $v$ from $G$ to obtain the graph $G - v$. Since $G - v$ has fewer vertices than $G$, it is 4-colorable by minimality. Fix such a 4-coloring. Now reinsert $v$: its at most 3 neighbors occupy at most 3 of the 4 colors, so at least one color remains available for $v$. This yields a valid 4-coloring of $G$, a contradiction.

\subsection{Case of Degree 4}

Suppose $v$ has degree exactly 4 with neighbors $a$, $b$, $c$, $d$ appearing in cyclic order around $v$ in the planar embedding. Remove $v$ and 4-color $G - v$ by minimality. If the four neighbors do not all receive distinct colors, then at least one color is unused among them, and we may assign that color to $v$, giving a contradiction.

So assume $a$, $b$, $c$, $d$ receive all four distinct colors; call them $1$, $2$, $3$, $4$ respectively. Define a \emph{Kempe chain} to be a maximal connected subgraph whose vertices are colored with exactly two specified colors.

Consider the Kempe chain $K_{13}$ containing $a$ (using colors 1 and 3).

\begin{itemize}
  \item \textbf{Case 1}: $c$ is not in $K_{13}$. Swap colors 1 and 3 throughout $K_{13}$. This is still a valid coloring of $G - v$, and now $a$ receives color 3, so color 1 is free for $v$.

  \item \textbf{Case 2}: $c$ is in $K_{13}$. Then there is a path of alternating colors 1 and 3 from $a$ to $c$ in the planar embedding. Because $a$ and $c$ alternate around $v$ with $b$ and $d$, this path separates $b$ from $d$ in the plane. Therefore $b$ and $d$ lie in different Kempe chains for colors 2 and 4. Swap colors 2 and 4 in the chain containing $b$; now $b$ receives color 4, freeing color 2 for $v$.
\end{itemize}

In both cases we obtain a valid 4-coloring of $G$, a contradiction.

\section{Resolution of Degree 5 Case}

\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, fill=black, inner sep=2pt, minimum size=5pt}]
  \node[label=below:{$v_0$}] (v0) at (0,0) {};
  \node[label=above:{$a_0$}] (a0) at (90:1.8) {};
  \node[label=left:{$a_1$}]  (a1) at (162:1.8) {};
  \node[label=left:{$a_2$}]  (a2) at (234:1.8) {};
  \node[label=right:{$a_3$}] (a3) at (306:1.8) {};
  \node[label=right:{$v_1$}] (v1) at (18:1.8) {};
  \draw (a0) -- (a1) -- (a2) -- (a3) -- (v1) -- (a0);
  \draw (v0) -- (a0);
  \draw (v0) -- (a1);
  \draw (v0) -- (a2);
  \draw (v0) -- (a3);
  \draw (v0) -- (v1);
  \draw[dotted] (a0) -- ++(75:0.7)  (a0) -- ++(105:0.7);
  \draw[dotted] (a1) -- ++(147:0.7) (a1) -- ++(177:0.7);
  \draw[dotted] (a2) -- ++(219:0.7) (a2) -- ++(249:0.7);
  \draw[dotted] (a3) -- ++(291:0.7) (a3) -- ++(321:0.7);
  \draw[dotted] (v1) -- ++(3:0.7)   (v1) -- ++(33:0.7);
\end{tikzpicture}
\caption{The neighborhood of a degree-5 vertex $v_0$: a 5-cycle on $a_0,a_1,a_2,a_3,v_1$ with $v_0$ adjacent to each.}
\end{figure}

Let $G$ be a minimum example of a maximal planar graph requiring 5 colors, and let $v_0$ be a graph in $G$ of degree 5, and let $v_1$ be a neighbor of $v_0$.

\begin{lemma}
Let $G^*$ be the subgraph of $G$ obtained by deleting the edge $\{v_0, v_1\}$, and let $G'$ be the graph obtained by merging vertices $a_0$ and $a_3$ in $G^*$. Then there is at least one proper 4-coloring of $G*$ which assigns $v_0$ and $v_1$ the same color and $a_0$ and $a_3$ the same color.
\end{lemma}
\begin{proof}
Since $G'$ has fewer vertices than $G$, it admits a proper 4-coloring $\phi$ by minimality. Any such $\phi$ must assign $v_0$ and $v_1$ the same color: if they received distinct colors, $\phi$ would induce a proper 4-coloring of $G^*$ in which $v_0 \neq v_1$, and re-inserting edge $\{v_0,v_1\}$ would yield a proper 4-coloring of $G$, contradicting minimality. The induced coloring $\Phi$ of $G^*$ assigns $a_0$ and $a_3$ the common color $\phi(a_{03})$ by construction, so $\Phi$ is the required coloring.
\end{proof}

\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, fill=black, inner sep=2pt, minimum size=5pt}]
  \node[label=below:{$v_0$}] (v0) at (0,0) {};
  \node[label=above:{$a_0$}] (a0) at (90:1.8) {};
  \node[label=left:{$a_1$}]  (a1) at (162:1.8) {};
  \node[label=left:{$a_2$}]  (a2) at (234:1.8) {};
  \node[label=right:{$a_3$}] (a3) at (306:1.8) {};
  \node[label=right:{$v_1$}] (v1) at (18:1.8) {};
  \draw (a0) -- (a1) -- (a2) -- (a3) -- (v1) -- (a0);
  \draw (v0) -- (a0);
  \draw (v0) -- (a1);
  \draw (v0) -- (a2);
  \draw (v0) -- (a3);
  \draw[dotted] (a0) -- ++(75:0.7)  (a0) -- ++(105:0.7);
  \draw[dotted] (a1) -- ++(147:0.7) (a1) -- ++(177:0.7);
  \draw[dotted] (a2) -- ++(219:0.7) (a2) -- ++(249:0.7);
  \draw[dotted] (a3) -- ++(291:0.7) (a3) -- ++(321:0.7);
  \draw[dotted] (v1) -- ++(3:0.7)   (v1) -- ++(33:0.7);
\end{tikzpicture}
\caption{The neighborhood of $v_0$ and $v_1$ in $G^*$: $G$ with the edge $\{v_0, v_1\}$ deleted, so the spoke to $v_1$ is absent.}
\end{figure}

\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, fill=black, inner sep=2pt, minimum size=5pt}]
  \node[label=below:{$v_0$}]    (v0)  at (0,0) {};
  \node[label=above:{$a_{03}$}] (a03) at (18:0.9) {};
  \node[label=left:{$a_1$}]     (a1)  at (162:1.8) {};
  \node[label=left:{$a_2$}]     (a2)  at (234:1.8) {};
  \node[label=right:{$v_1$}]    (v1)  at (18:1.8) {};
  \draw (a03) -- (a1) -- (a2) -- (a03);
  \draw (a03) -- (v1);
  \draw (v0) -- (a03);
  \draw (v0) -- (a1);
  \draw (v0) -- (a2);
  \draw[dotted] (a03) -- ++(75:0.7)  (a03) -- ++(105:0.7);
  \draw[dotted] (a03) -- ++(255:0.7) (a03) -- ++(285:0.7);
  \draw[dotted] (a1)  -- ++(147:0.7) (a1)  -- ++(177:0.7);
  \draw[dotted] (a2)  -- ++(219:0.7) (a2)  -- ++(249:0.7);
  \draw[dotted] (v1)  -- ++(3:0.7)   (v1)  -- ++(33:0.7);
\end{tikzpicture}
\caption{The neighborhood of $v_0$ and $v_1$ in $G'$: $G^*$ with $a_0$ and $a_3$ merged into $a_{03}$.}
\end{figure}

\begin{lemma}
  Let $\Phi$ be a proper 4-coloring of $G*$ induced by a proper 4-coloring of $G'$, let $\alpha = \Phi(v_0) = \Phi(v_1)$, and let $\beta = \Phi(a_{0}) = \Phi(a_{3})$. Then for each of the two remaining colors $\gamma$ and $\delta$, there exists a Kempe chain in $G'$ connecting $v_0$ to $v_1$ using colors $\{\alpha, \gamma\}$ and $\{\alpha, \delta\}$ respectively.
\end{lemma}
\begin{proof}
We show $v_0$ and $v_1$ lie in the same $\{\alpha, \gamma\}$-Kempe chain; the argument for $\{\alpha, \delta\}$ is identical.

By the previous lemma, $\Phi$ must exist. Suppose for contradiction that $v_0$ and $v_1$ lie in \emph{different} $\{\alpha, \gamma\}$-Kempe chains. Swapping colors $\alpha$ and $\gamma$ throughout $v_0$'s chain yields a new proper 4-coloring $\Phi'$ of $G*$ in which $\Phi'(v_0) = \gamma$ while $\Phi'(v_1) = \alpha$. Since $\gamma \neq \alpha$, the coloring $\Phi'$ assigns $v_0$ and $v_1$ distinct colors, which allows us to properly color $G$. Therefore $v_0$ and $v_1$ lie in the same $\{\alpha, \gamma\}$-Kempe chain, and by the same argument they lie in the same $\{\alpha, \delta\}$-Kempe chain.
\end{proof}


\begin{figure}[h]
\centering
\begin{tikzpicture}
  \coordinate (v0pos)  at (0,0);
  \coordinate (a03pos) at (18:0.9);
  \coordinate (a1pos)  at (162:1.8);
  \coordinate (a2pos)  at (234:1.8);
  \coordinate (v1pos)  at (18:1.8);
  % Kempe chain highlights beneath nodes
  \draw[line width=4pt, green!50,  dashed] (a1pos) to[bend left=40] (v1pos);
  \draw[line width=4pt, green!80, dashed] (a1pos) to (v0pos);
  \draw[line width=4pt, yellow!80, dashed] (a2pos) to[bend right=40] (v1pos);
  \draw[line width=4pt, yellow!80, dashed] (a2pos) to (v0pos);
  % Regular edges
  \draw (a03pos) -- (a1pos) -- (a2pos) -- (a03pos);
  \draw (a03pos) -- (v1pos);
  \draw (v0pos) -- (a03pos);
  \draw (v0pos) -- (a1pos);
  \draw (v0pos) -- (a2pos);
  % Dotted lines
  \draw[dotted] (a03pos) -- ++(75:0.7)  (a03pos) -- ++(105:0.7);
  \draw[dotted] (a03pos) -- ++(255:0.7) (a03pos) -- ++(285:0.7);
  \draw[dotted] (a1pos)  -- ++(147:0.7) (a1pos)  -- ++(177:0.7);
  \draw[dotted] (a2pos)  -- ++(219:0.7) (a2pos)  -- ++(249:0.7);
  \draw[dotted] (v1pos)  -- ++(3:0.7)   (v1pos)  -- ++(33:0.7);
  % Colored nodes
  \node[circle, draw, fill=red!60,    inner sep=2pt, minimum size=10pt, label=below:{$v_0$}]    at (v0pos)  {};
  \node[circle, draw, fill=blue!60,   inner sep=2pt, minimum size=10pt, label=above:{$a_{03}$}] at (a03pos) {};
  \node[circle, draw, fill=green!60,  inner sep=2pt, minimum size=10pt, label=left:{$a_1$}]     at (a1pos)  {};
  \node[circle, draw, fill=yellow!80, inner sep=2pt, minimum size=10pt, label=left:{$a_2$}]     at (a2pos)  {};
  \node[circle, draw, fill=red!60,    inner sep=2pt, minimum size=10pt, label=right:{$v_1$}]    at (v1pos)  {};
\end{tikzpicture}
\caption{A proper 4-coloring of $G'$ with $\phi(v_0)=\phi(v_1)=\text{red}$, $\phi(a_{03})=\text{blue}$, $\phi(a_1)=\text{green}$, $\phi(a_2)=\text{yellow}$. By the previous lemma, $v_0$ and $v_1$ must lie in both the $\{\text{red},\text{green}\}$-Kempe chain (dashed green, passing through $a_1$) and the $\{\text{red},\text{yellow}\}$-Kempe chain (dashed yellow, passing through $a_2$).}
\end{figure}

\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, inner sep=2pt, minimum size=10pt}]
  \coordinate (v0pos)  at (0,0);
  \coordinate (a0pos)  at (90:1.8);
  \coordinate (a1pos)  at (162:1.8);
  \coordinate (a2pos)  at (234:1.8);
  \coordinate (a3pos)  at (306:1.8);
  \coordinate (v1pos)  at (18:1.8);

  \coordinate (green_mid)  at (0, 2.4);
  \coordinate (yellow_mid) at (306:2.3);
  % Kempe chain highlights beneath nodes
  \draw[line width=4pt, green!50,  dashed] (a1pos) to[curve through={(green_mid)}]  (v1pos);
  \draw[line width=4pt, green!80,  dashed] (a1pos) to (v0pos);
  \draw[line width=4pt, yellow!80, dashed] (a2pos) to[curve through={(yellow_mid)}] (v1pos);
  \draw[line width=4pt, yellow!80, dashed] (a2pos) to (v0pos);

  \node[fill=red!60,    label=below:{$v_0$}] (v0) at (v0pos) {};
  \node[fill=blue!60,   label=above:{$a_0$}] (a0) at (a0pos) {};
  \node[fill=green!60,  label=left:{$a_1$}]  (a1) at (a1pos) {};
  \node[fill=yellow!80, label=left:{$a_2$}]  (a2) at (a2pos) {};
  \node[fill=blue!60,   label=right:{$a_3$}] (a3) at (a3pos) {};
  \node[fill=red!60,    label=right:{$v_1$}] (v1) at (v1pos) {};

  \draw (a0) -- (a1) -- (a2) -- (a3) -- (v1) -- (a0);
  \draw (v0) -- (a0);
  \draw (v0) -- (a1);
  \draw (v0) -- (a2);
  \draw (v0) -- (a3);
  \draw[dotted] (a0) -- ++(75:0.7)  (a0) -- ++(105:0.7);
  \draw[dotted] (a1) -- ++(147:0.7) (a1) -- ++(177:0.7);
  \draw[dotted] (a2) -- ++(219:0.7) (a2) -- ++(249:0.7);
  \draw[dotted] (a3) -- ++(291:0.7) (a3) -- ++(321:0.7);
  \draw[dotted] (v1) -- ++(3:0.7)   (v1) -- ++(33:0.7);
\end{tikzpicture}
\caption{The same coloring lifted to $G^*$: $\phi(v_0)=\phi(v_1)=\text{red}$, $\phi(a_0)=\phi(a_3)=\text{blue}$, $\phi(a_1)=\text{green}$, $\phi(a_2)=\text{yellow}$. The $\{\text{red},\text{green}\}$- and $\{\text{red},\text{yellow}\}$-Kempe chains connecting $v_0$ to $v_1$ pass through $a_1$ and $a_2$ respectively.}
\end{figure}

\begin{lemma}
  Let $\Phi$ be a proper 4-coloring of $G*$ induced by a proper 4-coloring of $G'$, let $\alpha = \Phi(v_0) = \Phi(v_1)$, let $\beta = \Phi(a_{0}) = \Phi(a_{3})$, let $\delta = \Phi(a_{2})$, and let $\gamma = \Phi(a_{1})$. Then the coloring $\Phi'$ obtained by doing a Kempe switch on the $\{\beta, \delta\}$-Kempe chain incident to $a_0$ must assign $\alpha = \Phi'(v_0) = \Phi'(v_1)$, $\beta = \Phi'(a_3)$, $\delta = \Phi'(a_0) = \Phi'(a_2)$, and $\gamma = \Phi'(a_1)$ (IE the coloring for all vertices except $a_0$ remains the same)
\end{lemma}
\begin{proof}
  The kempe switch will assign $\Phi'(a_0) = \delta$ by definition. By the previous lemma, there must be a $\{\alpha, \gamma\}$ kempe chain incident to $v_0$, $v_1$, and $a_1$. Therefore the $\{\beta, \delta\}$ component incident to $a_2$ and $a_3$ cannot be connected to the $\{\beta, \delta\}$ component incident to $a_0$, so the kempe switch we do to obtain $\Phi'$ will not recolor vertices $a_2$ and $a_3$. Since the $\{\beta, \delta\}$ kempe switch also maintains vertices assigned $\alpha$ and $\gamma$ in $\Phi$, $\Phi'$ will not recolor vertices $a_1$, $v_0$, and $v_1$. Therefore $\alpha = \Phi'(v_0) = \Phi'(v_1)$, $\beta = \Phi'(a_3)$, $\delta = \Phi'(a_0) = \Phi'(a_2)$, and $\gamma = \Phi'(a_1)$.
\end{proof}

\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, inner sep=2pt, minimum size=10pt}]
  \coordinate (v0pos)  at (0,0);
  \coordinate (a0pos)  at (90:1.8);
  \coordinate (a1pos)  at (162:1.8);
  \coordinate (a2pos)  at (234:1.8);
  \coordinate (a3pos)  at (306:1.8);
  \coordinate (v1pos)  at (18:1.8);

  \coordinate (green_mid)  at (0, 2.4);
  \coordinate (yellow_mid) at (306:2.3);
  % Kempe chain highlights beneath nodes
  \draw[line width=4pt, green!50,  dashed] (a1pos) to[curve through={(green_mid)}]  (v1pos);
  \draw[line width=4pt, green!80,  dashed] (a1pos) to (v0pos);
  % \draw[line width=4pt, yellow!80, dashed] (a2pos) to[curve through={(yellow_mid)}] (v1pos);
  % \draw[line width=4pt, yellow!80, dashed] (a2pos) to (v0pos);

  \node[fill=red!60,    label=below:{$v_0$}] (v0) at (v0pos) {};
  \node[fill=yellow!60,   label=above:{$a_0$}] (a0) at (a0pos) {};
  \node[fill=green!60,  label=left:{$a_1$}]  (a1) at (a1pos) {};
  \node[fill=yellow!80, label=left:{$a_2$}]  (a2) at (a2pos) {};
  \node[fill=blue!60,   label=right:{$a_3$}] (a3) at (a3pos) {};
  \node[fill=red!60,    label=right:{$v_1$}] (v1) at (v1pos) {};

  \draw (a0) -- (a1) -- (a2) -- (a3) -- (v1) -- (a0);
  \draw (v0) -- (a0);
  \draw (v0) -- (a1);
  \draw (v0) -- (a2);
  \draw (v0) -- (a3);
  \draw[dotted] (a0) -- ++(75:0.7)  (a0) -- ++(105:0.7);
  \draw[dotted] (a1) -- ++(147:0.7) (a1) -- ++(177:0.7);
  \draw[dotted] (a2) -- ++(219:0.7) (a2) -- ++(249:0.7);
  \draw[dotted] (a3) -- ++(291:0.7) (a3) -- ++(321:0.7);
  \draw[dotted] (v1) -- ++(3:0.7)   (v1) -- ++(33:0.7);
\end{tikzpicture}
\end{figure}

\end{document}

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