Add edge-deletion subgraph 4-colorability for a minimal counterexample
Defines D(G) as the family of single-edge-deletion spanning subgraphs of a maximal planar graph G, and shows that when G_0 is a minimum-order 5-chromatic maximal planar graph every member of D(G_0) is 4-colorable, via a coloring pulled back from the smaller minor G_0/uv. Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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@@ -289,6 +289,35 @@ to identifying further necessary properties of a minimum-order
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$5$-chromatic maximal planar graph, of which flip-asymmetry is the
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first.
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\section{Edge-deletion subgraphs}
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\begin{definition}[Edge-deletion subgraph]\label{def:edge-deletion}
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Let $G$ be a maximal planar graph and $uv \in E(G)$. The
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\emph{edge-deletion subgraph at $uv$} is the spanning subgraph
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$G - uv = (V(G),\,E(G) \setminus \{uv\})$. Write
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$\mathcal{D}(G) = \{G - uv : uv \in E(G)\}$.
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\end{definition}
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\begin{theorem}\label{thm:edge-deletion-4colorable}
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Let $G_0$ be a maximal planar graph of minimum order with
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$\chi(G_0) \geq 5$. Then every $H \in \mathcal{D}(G_0)$ is
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$4$-colorable.
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\end{theorem}
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\begin{proof}
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Fix $uv \in E(G_0)$ and let $G_0 / uv$ denote the simple planar graph
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obtained by contracting $uv$ and discarding parallel edges. Since
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$|V(G_0/uv)| = |V(G_0)| - 1$, the minimality of $G_0$ supplies a
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proper $4$-coloring $c$ of $G_0 / uv$. Let $z$ be the contracted
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vertex and define $c'\colon V(G_0) \to \{1,2,3,4\}$ by
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$c'(u) = c'(v) = c(z)$ and $c'(y) = c(y)$ for $y \notin \{u, v\}$.
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Every edge of $G_0 - uv$ is either disjoint from $\{u, v\}$ or
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incident to exactly one of them; in either case the corresponding
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edge of $G_0 / uv$ has distinct endpoints under $c$, so $c'$ assigns
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its endpoints distinct colors. The edge $uv$ itself is absent from
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$G_0 - uv$, so $c'$ is a proper $4$-coloring of $G_0 - uv$.
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\end{proof}
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\end{document}
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%-----------------------------------------------------------------------
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