266 lines
13 KiB
TeX
266 lines
13 KiB
TeX
%% filename: amsart-template.tex
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%% date: 2014/07/24
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%% Copyright 2008-2010, 2014 American Mathematical Society.
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%% ====================================================================
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% AMS-LaTeX v.2 template for use with amsart
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%
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% Remove any commented or uncommented macros you do not use.
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\documentclass{amsart}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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\theoremstyle{definition}
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\newtheorem{definition}[theorem]{Definition}
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\newtheorem{example}[theorem]{Example}
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\theoremstyle{remark}
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\newtheorem{remark}[theorem]{Remark}
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\numberwithin{equation}{section}
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\begin{document}
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\title{Humans Suffice: A Novel Proof of the Four Color Theorem}
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% Remove any unused author tags.
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% author one information
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\author{Eric Bauerfeld}
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\address{}
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\curraddr{}
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\email{eric@dideric.is}
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\thanks{Dr. Joe Fields}
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\thanks{Holepunch}
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% author two information
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% \author{}
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% \address{}
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% \curraddr{}
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% \email{}
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% \thanks{}
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\subjclass[2020]{Primary }
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\keywords{Discrete , Four Color Theorem,}
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\date{}
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\dedicatory{Dedicated to all who value more than machines}
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\begin{abstract}
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\end{abstract}
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\maketitle
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\section{Kempe's Proof (Valid Portion)}
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\subsection{Setup}
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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.
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\subsection{Every Planar Graph Has a Vertex of Degree at Most 5}
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\begin{lemma}
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Every planar graph has at least one vertex of degree $\leq 5$.
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\end{lemma}
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\begin{proof}
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Let $G$ be a connected planar graph with $V$ vertices, $E$ edges, and $F$ faces. By Euler's formula,
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\[
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V - E + F = 2.
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\]
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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:
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\[
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V - E + \frac{2E}{3} \geq 2 \implies E \leq 3V - 6.
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\]
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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 $\leq 5$.
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\end{proof}
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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 $v$.
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\subsection{Cases of Degree at Most 3}
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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.
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\subsection{Case of Degree 4}
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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.
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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.
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Consider the Kempe chain $K_{13}$ containing $a$ (using colors 1 and 3).
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\begin{itemize}
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\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 $v$.
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\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 $v$.
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\end{itemize}
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In both cases we obtain a valid 4-coloring of $G$, a contradiction.
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\section{Resolution of Degree 5 Case}
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\subsection{At Least 12 Vertices of Degree 5}
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Since we have already handled every vertex of degree $\leq 4$, we may assume without loss of generality that the minimal counterexample $G$ has minimum degree 5. We now show that $G$ must contain at least 12 vertices of degree exactly 5.
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\begin{lemma}
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If $G$ is a planar graph with minimum degree 5, then $G$ contains at least 12 vertices of degree exactly 5.
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\end{lemma}
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\begin{proof}
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For each $k \geq 5$, let $n_k$ denote the number of vertices of degree exactly $k$ in $G$. Since every vertex has degree $\geq 5$,
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\[
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V = \sum_{k \geq 5} n_k, \qquad 2E = \sum_{k \geq 5} k\, n_k.
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\]
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Using $E \leq 3V - 6$, we obtain $2E \leq 6V - 12$, so
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\[
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\sum_{k \geq 5} k\, n_k \leq 6\sum_{k \geq 5} n_k - 12.
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\]
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Moving all terms to the right-hand side gives
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\[
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12 \leq 6\sum_{k \geq 5} n_k - \sum_{k \geq 5} k\, n_k = \sum_{k \geq 5} (6 - k)\, n_k.
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\]
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We now expand this sum by separating the $k = 5$, $k = 6$, and $k \geq 7$ terms:
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\[
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\sum_{k \geq 5}(6-k)\,n_k
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= \underbrace{(6-5)\,n_5}_{=\,n_5}
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+ \underbrace{(6-6)\,n_6}_{=\,0}
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+ \sum_{k \geq 7}(6-k)\,n_k
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= n_5 - \sum_{k \geq 7}(k-6)\,n_k.
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\]
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Therefore $12 \leq n_5 - \sum_{k \geq 7}(k-6)\,n_k$, which rearranges to
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\[
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n_5 \geq 12 + \sum_{k \geq 7}(k-6)\,n_k.
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\]
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Since $k - 6 \geq 1 > 0$ and $n_k \geq 0$ for all $k \geq 7$, each term in the sum is non-negative, so $\sum_{k \geq 7}(k-6)\,n_k \geq 0$ and thus $n_5 \geq 12$.
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\end{proof}
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\subsection{Two Non-Adjacent Vertices of Degree 5}
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\begin{lemma}
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If $G$ is a planar graph with minimum degree 5, then $G$ contains two non-adjacent vertices of degree exactly 5.
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\end{lemma}
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\begin{proof}
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By the previous lemma, $G$ contains at least 12 vertices of degree exactly 5. Let $S$ denote the set of all degree-5 vertices, so $|S| \geq 12$. Suppose for contradiction that every two vertices in $S$ are adjacent, i.e., $S$ induces a clique in $G$. Then every vertex $v \in S$ is adjacent to all other $|S| - 1 \geq 11$ vertices of $S$. But $\deg(v) = 5$, so $v$ has at most 5 neighbors in total. Since $11 > 5$, this is a contradiction. Therefore $S$ does not form a clique, and there exist two vertices $v_1, v_2 \in S$ that are non-adjacent.
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\end{proof}
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\subsection{The Reduced Subgraph $G'$}
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By the previous lemma, we may fix two non-adjacent vertices $v_0, v_1 \in G$ each of degree 5. Define
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\[
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G' = G - \{v_0, v_1\},
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\]
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as the subgraph of $G$ obtained by deleting $v_0$ and $v_1$ together with all edges incident to either vertex. Since $G'$ has strictly fewer vertices than $G$, the minimality of $G$ guarantees that $G'$ admits a proper 4-coloring $\phi : V(G') \to \{1,2,3,4\}$. Because $v_0$ and $v_1$ are non-adjacent in $G$, neither is a neighbor of the other, so each retains all 5 of its neighbors in $G'$. Let $N(v_i)$ denote the set of neighbors of $v_i$ in $G$; then $N(v_0)$ and $N(v_1)$ are each sets of 5 vertices, all of which lie in $G'$ and are assigned colors by $\phi$. Note also that $N(v_0)$ and $N(v_1)$ may overlap. Let $\Phi$ be the set of all possible 4-colorings of $G'$.
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\subsection{Not All Colorings in $\Phi$ Saturate Both Neighborhoods}
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\begin{lemma}
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It is not possible that every $\phi \in \Phi$ satisfies both $|\phi(N(v_0))| = 4$ and $|\phi(N(v_1))| = 4$.
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\end{lemma}
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\begin{proof}
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We show the stronger fact that there exists $\phi \in \Phi$ with $|\phi(N(v_0))| \leq 3$.
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Since $v_1$ is a single vertex, $G - \{v_1\}$ is a planar graph on $V - 1$ vertices. Because $V - 1 < V$ and $G$ is a minimal counterexample, $G - \{v_1\}$ is 4-colorable. Fix a proper 4-coloring $\psi$ of $G - \{v_1\}$. This coloring assigns a color to every vertex of $G'$ and also to $v_0$.
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Let $\phi = \psi|_{V(G')}$ be the restriction of $\psi$ to $G'$. Since $\psi$ is a proper coloring of $G - \{v_1\} \supseteq G'$, the restriction $\phi$ is a proper 4-coloring of $G'$, so $\phi \in \Phi$.
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Because $\psi$ is a proper coloring of $G - \{v_1\}$ and $v_0 \in V(G - \{v_1\})$, the color $\psi(v_0)$ differs from the color of every neighbor of $v_0$. In particular, $\psi(v_0) \notin \phi(N(v_0))$. Since $\psi(v_0) \in \{1,2,3,4\}$ and $\psi(v_0)$ does not appear in $\phi(N(v_0))$, it follows that $|\phi(N(v_0))| \leq 3$. Therefore not every coloring in $\Phi$ uses all 4 colors on $N(v_0)$, and in particular it is impossible that every $\phi \in \Phi$ saturates both $N(v_0)$ and $N(v_1)$ with all 4 colors.
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\end{proof}
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\section{Merged Subgraphs of $G'$}
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Let $N(v_0) = \{a_0, a_1, a_2, a_3, a_4\}$ be the five neighbors of $v_0$ in $G'$. For each pair of non-adjacent vertices $a_i, a_j \in N(v_0)$ (i.e., $\{a_i, a_j\} \notin E(G')$), define the graph $G'_{ij}$ to be the result of identifying $a_i$ and $a_j$ in $G'$: formally, $G'_{ij}$ is obtained from $G'$ by removing $a_i$ and $a_j$ and adding a new vertex $a_{ij}$ adjacent to every vertex in $N_{G'}(a_i) \cup N_{G'}(a_j)$. Define the set of all such merged subgraphs by
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\[
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\mathcal{M} = \bigl\{\, G'_{ij} : a_i, a_j \in N(v_0),\ \{a_i, a_j\} \notin E(G') \,\bigr\}.
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\]
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\subsection{Colorings of Merged Subgraphs Extend to $G'$}
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For each $G'_{ij} \in \mathcal{M}$, let $\Phi(G'_{ij})$ denote the set of proper 4-colorings of $G'_{ij}$. Each such coloring can be extended to a coloring of $G'$ by assigning the color of the merged vertex $a_{ij}$ back to both $a_i$ and $a_j$. Formally, for $\psi \in \Phi(G'_{ij})$, define $\widetilde{\psi} : V(G') \to \{1,2,3,4\}$ by
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\[
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\widetilde{\psi}(v) = \begin{cases} \psi(a_{ij}) & \text{if } v = a_i \text{ or } v = a_j, \\ \psi(v) & \text{otherwise.} \end{cases}
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\]
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Let $\widetilde{\Phi}_{ij} = \{\,\widetilde{\psi} : \psi \in \Phi(G'_{ij})\,\}$ denote the set of all such extensions.
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\begin{lemma}
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For each $G'_{ij} \in \mathcal{M}$, we have $\widetilde{\Phi}_{ij} \subseteq \Phi$.
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\end{lemma}
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\begin{proof}
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Let $\psi \in \Phi(G'_{ij})$ and let $\widetilde{\psi}$ be its extension to $G'$. We verify that $\widetilde{\psi}$ is a proper coloring of $G'$ by checking every edge $\{u, w\} \in E(G')$.
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\textbf{Case 1: neither $u$ nor $w$ is $a_i$ or $a_j$.} Then $\{u, w\}$ is also an edge of $G'_{ij}$, and $\widetilde{\psi}(u) = \psi(u) \neq \psi(w) = \widetilde{\psi}(w)$.
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\textbf{Case 2: $u \in \{a_i, a_j\}$ and $w \notin \{a_i, a_j\}$ (or vice versa).} Then $w \in N_{G'}(a_i) \cup N_{G'}(a_j)$, so $w$ is adjacent to $a_{ij}$ in $G'_{ij}$. Since $\psi$ is proper, $\psi(w) \neq \psi(a_{ij}) = \widetilde{\psi}(u)$.
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\textbf{Case 3: $u = a_i$ and $w = a_j$.} This case cannot occur, since $a_i$ and $a_j$ are non-adjacent in $G'$ by the definition of $\mathcal{M}$.
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In all cases the endpoints of every edge receive distinct colors, so $\widetilde{\psi} \in \Phi$.
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\end{proof}
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\subsection{4-Saturating Merged Colorings Extend to 4-Saturating Colorings of $G'$}
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For each $G'_{ij} \in \mathcal{M}$, the vertices of $G'_{ij}$ that would be adjacent to $v_0$ if $v_0$ were reinserted are exactly $\{a_{ij}\} \cup (N(v_0) \setminus \{a_i, a_j\})$; call this set $N_{ij}(v_0)$. Define
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\[
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\Phi_4 = \{\, \phi \in \Phi : |\phi(N(v_0))| = 4 \,\}
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\]
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to be the set of colorings of $G'$ that use all 4 colors on $N(v_0)$, and for each $G'_{ij} \in \mathcal{M}$ define
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\[
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\Phi_4(G'_{ij}) = \{\, \psi \in \Phi(G'_{ij}) : |\psi(N_{ij}(v_0))| = 4 \,\}
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\]
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to be the colorings of $G'_{ij}$ that use all 4 colors on the effective neighborhood of $v_0$.
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\begin{lemma}
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For each $G'_{ij} \in \mathcal{M}$, the extensions of $\Phi_4(G'_{ij})$ lie in $\Phi_4$, i.e., $\widetilde{\Phi_4(G'_{ij})} \subseteq \Phi_4$.
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\end{lemma}
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\begin{proof}
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Let $\psi \in \Phi_4(G'_{ij})$, so $|\psi(N_{ij}(v_0))| = 4$. By definition of the extension,
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\[
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\widetilde{\psi}(N(v_0))
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= \widetilde{\psi}(\{a_i, a_j\} \cup (N(v_0) \setminus \{a_i, a_j\}))
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= \{\psi(a_{ij})\} \cup \psi(N(v_0) \setminus \{a_i, a_j\})
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= \psi(N_{ij}(v_0)).
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\]
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The first equality expands $N(v_0)$; the second uses $\widetilde{\psi}(a_i) = \widetilde{\psi}(a_j) = \psi(a_{ij})$ and $\widetilde{\psi}(u) = \psi(u)$ for $u \notin \{a_i, a_j\}$; the third uses $N_{ij}(v_0) = \{a_{ij}\} \cup (N(v_0) \setminus \{a_i,a_j\})$. Therefore $|\widetilde{\psi}(N(v_0))| = |\psi(N_{ij}(v_0))| = 4$. Since also $\widetilde{\psi} \in \Phi$ by the previous lemma, we conclude $\widetilde{\psi} \in \Phi_4$.
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\end{proof}
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\end{document}
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