coloring_nested_tire_graphs: cyclic-edge-conn distinction + even-cycle question

UPDATED: birkhoff_internally_6_connected.tex now adds the distinction between "internally 6-connected" (= cyclic edge conn ≥ 6 in dual) and the framework's needed condition (= cyclic edge conn EXACTLY 6, so 6-edge cuts exist). Notes that this is a real a priori restriction not provided by Birkhoff alone. NEW NOTE: even_separating_cycle.tex (3 pages) Addresses: "must a min 4CT counterexample have a separating n-cycle with n even and n ≥ 6?" Honest answer: I don't know of a proof either way. Key contributions: - Lemma (cut-parity in cubic graphs): |C| ≡ |S| ≡ |T| (mod 2). So even-length cycles in primal G ↔ cuts with even-sized sides in dual G^*. - |V(G^*)| = 2|V(G)| - 4 is always even, so both sides have matching parity. - Birkhoff doesn't rule out odd-length separating cycles ≥ 7. - Second-link heuristic: in internally 6-conn triangulations, the "second link" of any vertex is typically a 6-cycle, giving abundant separating 6-cycles in practice. But this is heuristic, not proven for all such triangulations. Conjecture (stated, not proven): every internally 6-conn planar triangulation with ≥ 12 vertices has a separating even n-cycle with n ≥ 6. Equivalent: every planar cubic graph with cyclic edge connectivity ≥ 6 and ≥ 20 vertices has a cyclic edge cut of size exactly 6. This is a structural question; I don't know a planar cubic counterexample. Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
2026-05-27 00:16:22 -04:00
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@@ -50,17 +50,19 @@ $1$ vertex is irreducible by this argument.
 \begin{center}
 \begin{tikzpicture}[scale=1.1]
  \node[anchor=west] at (0, 4.3) {\textbf{Internally $6$-connected $=$ all four below:}};
-  % case row
+  % All status labels right-aligned at the same column (x = 11), well
+  % clear of the longest bullet text.
  \node[anchor=west] at (0, 3.6) {\textbullet\, no separating $3$-cycle};
-  \node[red, anchor=west] at (5.0, 3.6) {FORBIDDEN};
+  \node[red, anchor=east] at (11, 3.6) {FORBIDDEN};
  \node[anchor=west] at (0, 2.9) {\textbullet\, no separating $4$-cycle};
-  \node[red, anchor=west] at (5.0, 2.9) {FORBIDDEN};
+  \node[red, anchor=east] at (11, 2.9) {FORBIDDEN};
  \node[anchor=west] at (0, 2.2) {\textbullet\, no separating $5$-cycle isolating $\ge 2$ on either side};
-  \node[red, anchor=west] at (5.0, 2.2) {FORBIDDEN};
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-  % Schematic: a 5-cycle with 1 isolated vertex on one side
-  \begin{scope}[shift={(7, 2.7)}, scale=0.5]
+  \node[green!50!black, anchor=east] at (11, 1.5) {ALLOWED};
+  % Schematic: a 5-cycle with 1 isolated vertex on one side --- placed
+  % below the bullet list so it doesn't collide with the case labels.
+  \begin{scope}[shift={(5.5, -0.6)}, scale=0.55]
    \node[circle, fill=red!80!black, inner sep=2pt] (c) at (0, 0) {};
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@@ -71,7 +73,7 @@ $1$ vertex is irreducible by this argument.
      \pgfmathtruncatemacro{\j}{mod(\i+1,5)}
      \draw[blue, very thick] (n\i) -- (n\j);
    }
-    \node[red] at (0, -1.95) {\small allowed};
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@@ -100,6 +102,47 @@ blue $5$-cycle is the separator; on one side is the isolated red
 vertex, on the other are the remaining $6$ vertices.  This is the
 allowed Birkhoff configuration.

+\section*{Internally $6$-connected vs.\ cyclic edge connectivity}
+
+Birkhoff's condition rules out small \emph{cuts} but does not
+force any specific separating \emph{cycle} to exist.  In the cubic
+dual $G^*$, the dictionary is:
+\begin{itemize}
+\item $k$-vertex cut in $G$ $\leftrightarrow$ $k$-edge cut in $G^*$.
+\item separating $k$-cycle in $G$ (with $\ge 2$ vertices on each
+      side) $\leftrightarrow$ $k$-edge cut in $G^*$ separating $G^*$
+      into two pieces \emph{each containing a cycle}
+      (= \emph{cyclic} edge cut).
+\end{itemize}
+
+So $G$ internally $6$-connected $\iff$ $G^*$ has \emph{cyclic edge
+connectivity} $\ge 6$.  Whether it is \emph{exactly} $6$ (= some
+separating $6$-cycle in $G$ exists with both sides $\ge 2$
+vertices) is an additional question.
+
+\paragraph{Hidden assumption in the cut-tire framework.}  Our
+chain-DP framework
+(\texttt{chain\_half\_analysis.tex}, \texttt{boundary\_cut\_tire.tex})
+operates on $6$-edge cuts of the cubic dual.  It therefore has
+nontrivial content on $G$ only when $G^*$ has cyclic edge
+connectivity \emph{exactly} $6$ --- i.e.\ when $G$ has a
+separating $6$-cycle with $\ge 2$ vertices on each side.  Birkhoff
+gives ``$\ge 6$''; we need ``$= 6$''.
+
+This is a real \emph{a priori} restriction: if a minimum $4$CT
+counterexample turned out to have $G^*$ with cyclic edge
+connectivity $\ge 7$ (= no separating $6$-cycle in $G$), our
+framework would be empty on it.
+
+\paragraph{Empirically.}  All graphs in the test suite
+(icosahedron / dodecahedron, pentakis dodecahedron / Buckyball,
+Holton--McKay $\#0$ through $\#5$) have many $6$-edge cuts in
+their cubic duals --- i.e.\ all known internally $6$-connected
+triangulations of moderate size have plenty of separating
+$6$-cycles.  Whether a hypothetical minimum $4$CT counterexample
+\emph{must} have one is a structural question discussed in the
+next note.
+
 \section*{Why this matters for the framework}

 For our cut-tire chain DP framework, we test on graphs whose primal
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+\documentclass[11pt]{article}
+\usepackage{amsmath,amssymb,amsthm}
+\usepackage{graphicx}
+\usepackage{geometry}
+\usepackage{booktabs}
+\geometry{margin=1in}
+
+\title{Must a minimum $4$CT counterexample have a separating
+$n$-cycle with $n$ even and $n \ge 6$?}
+\author{}
+\date{}
+
+\newtheorem*{thm}{Theorem}
+\newtheorem*{lem}{Lemma}
+\newtheorem*{prop}{Proposition}
+\newtheorem*{conj}{Conjecture}
+\newtheorem*{rem}{Remark}
+
+\begin{document}
+\maketitle
+
+\section*{The question}
+
+\paragraph{Statement.}  Let $G$ be a hypothetical minimum
+$4$-colour counterexample (a minimum planar triangulation
+requiring $\ge 5$ colours).  Must $G$ contain a separating
+$n$-cycle with $n$ \emph{even} and $n \ge 6$?
+
+\paragraph{Short answer.}  I do not know of a proof either way.
+The question is subtle: a parity computation (below) shows that
+cuts in the cubic dual $G^*$ have a definite parity tied to the
+side sizes, but I do not see this parity forcing the existence of
+even-length separating cycles.
+
+\section*{Parity computation in the cubic dual}
+
+Let $G$ be a planar triangulation and $G^*$ its cubic planar dual.
+$G$ has $V$ vertices, $3V - 6$ edges, $2V - 4$ faces; $G^*$ has
+$2V - 4$ vertices, $3V - 6$ edges, $V$ faces.  In particular,
+$|V(G^*)| = 2V - 4$ is \emph{always even}.
+
+\begin{lem}[Cut size parity in cubic graphs]
+\label{lem:cut-parity}
+Let $G^*$ be a cubic graph and let $C$ be an edge cut separating
+$V(G^*)$ into sides $S$ and $T = V(G^*) \setminus S$.  Then
+\[
+|C| \equiv |S| \equiv |T| \pmod{2}.
+\]
+\end{lem}
+
+\begin{proof}
+Counting degree on side $S$: $3|S| = 2 e_S + |C|$, where $e_S$ is
+the number of edges with both endpoints in $S$.  Hence
+$|C| = 3|S| - 2e_S \equiv |S| \pmod{2}$.  Since
+$|S| + |T| = |V(G^*)|$ is even, $|S| \equiv |T| \pmod 2$.
+\end{proof}
+
+\paragraph{Translating to primal cycles.}  Via the $G \leftrightarrow
+G^*$ duality, an $n$-cycle in $G$ corresponds to an $n$-edge cut in
+$G^*$.  Lemma~\ref{lem:cut-parity} says: an \emph{even} $n$-cycle
+in $G$ ($n \ge 6$) corresponds to a $G^*$ cut with both sides
+having an \emph{even} number of vertices.  An \emph{odd}
+$n$-cycle in $G$ ($n \ge 7$) corresponds to both sides having an
+\emph{odd} number of vertices.
+
+\section*{What Birkhoff gives us}
+
+For a minimum $4$CT counterexample $G$:
+\begin{itemize}
+\item $G$ is internally $6$-connected (no separating $3$-cycle, no
+      separating $4$-cycle, no separating $5$-cycle with $\ge 2$
+      vertices on each side).
+\item Equivalently in $G^*$: cyclic edge connectivity $\ge 6$.
+\end{itemize}
+
+So all sufficiently small cuts in $G^*$ are excluded.  The smallest
+non-trivial cuts can be $6$-edge ($n = 6$, even) or $7$-edge
+($n = 7$, odd), and Birkhoff alone permits both.
+
+\section*{A potential ``no'' --- can $G^*$ have only odd cuts $\ge 7$?}
+
+In principle, the minimum non-trivial cut could be $7$ (and all
+non-trivial cuts could be odd-length).  In this case the
+minimum primal separating cycle has length $7$ (odd), and the
+question's answer is ``no.''
+
+\paragraph{Is this realisable?}  We need a planar cubic graph $G^*$
+satisfying:
+\begin{itemize}
+\item cyclic edge connectivity $\ge 7$;
+\item all non-trivial cyclic edge cuts have odd size.
+\end{itemize}
+
+Lemma~\ref{lem:cut-parity} gives a constraint: if a $G^*$ has
+\emph{any} cut at all of size $\ge 6$, its parity is fixed by side
+size.  For ``all cuts odd'' to hold, no even-sized cut would
+separate.  In a cubic graph this corresponds to the side sizes
+being \emph{odd}.  Since $|V(G^*)|$ is even (for any planar
+triangulation dual), the sides $(|S|, |V(G^*)| - |S|)$ have
+matching parity --- both even or both odd.  ``All cuts odd''
+means ``all cyclic separations have odd side sizes,'' which is
+possible in principle.
+
+\paragraph{However:} I don't know of a planar cubic graph that has
+\emph{only} odd cyclic cuts.  The known internally $6$-connected
+examples (icosahedron's dual = dodecahedron, etc.) all have
+many even cuts of size $6$ as well as odd cuts.
+
+\section*{A heuristic suggesting ``yes'' --- vertex links}
+
+Every vertex $v$ in a planar triangulation $G$ has a \emph{link}:
+the cycle formed by its neighbours.  By Birkhoff, this link is a
+$5$-cycle isolating $v$.  In $G^*$, this corresponds to a
+$5$-edge cut isolating a single triangle face of $G^*$ (= one
+vertex of $G$).
+
+\paragraph{Next layer.}  Consider the ``second link'' of $v$: the
+set of vertices at $G$-distance exactly $2$ from $v$.  This forms
+a cycle around the link, of length depending on the degrees of
+link vertices.
+
+If all link vertices have degree $5$ (= minimum for internally
+$6$-connected): the link's $5$ vertices contribute $5 \cdot 5 = 25$
+incidences, of which $5$ are to $v$ (the centre) and $2 \cdot 5
+= 10$ are between link vertices (the $5$-cycle).  The remaining
+$25 - 5 - 10 = 10$ incidences go to second-link vertices.  But the
+$5$ link vertices share their second-link vertices in pairs (each
+triangle face containing $v$, a link vertex, and a second-link
+vertex), so the second link has $\le 5$ distinct vertices.
+
+Working through Euler more carefully: if all degrees are $\ge 5$
+and the link of $v$ has length $5$, the second link has length
+exactly $5 \cdot (5 - 4) = 5$ (in the icosahedron, second link =
+link of antipodal vertex).  But in larger triangulations
+(degrees of link vertices $\ge 5$, some higher), the second link
+is generically a cycle of length $\sum_{u \in \text{link}}(\deg(u)
+- 4) = 5 + \sum_u(\deg u - 5) \ge 5$ with equality only when all
+link degrees are exactly $5$ (icosahedron case).
+
+So in larger internally $6$-connected triangulations, second-link
+length $\ge 6$, often equal to $6$, often even (especially in
+``vertex-transitive enough'' graphs).  This is a heuristic for why
+$6$-cycle separators are abundant, but it's not a proof.
+
+\section*{What I can conclude}
+
+\begin{itemize}
+\item \textbf{Parity is determined by side size}
+      (Lemma~\ref{lem:cut-parity}).  Even $n$-cycle separators in $G$
+      correspond to even-sided cuts in $G^*$.
+\item \textbf{Birkhoff doesn't rule out odd cuts.}  Minimum
+      non-trivial cyclic cut in $G^*$ could in principle be of any
+      size $\ge 6$.
+\item \textbf{No known proof that even $n \ge 6$ separators must
+      exist} in min $4$CT counterexamples.
+\item \textbf{Empirically}, in all tested internally $6$-connected
+      planar triangulations (icosahedron, pentakis dodecahedron,
+      Holton--McKay duals), even $6$-cycle separators with
+      both sides $\ge 2$ vertices exist in abundance.
+\end{itemize}
+
+\section*{Conjecture}
+
+\begin{conj}
+Every internally $6$-connected planar triangulation $G$ with
+$|V(G)| \ge 12$ has a separating $n$-cycle with $n$ even and
+$n \ge 6$.
+\end{conj}
+
+Equivalently: every planar cubic graph with cyclic edge
+connectivity $\ge 6$ and $|V| \ge 20$ has a cyclic edge cut of
+size $6$.
+
+This conjecture seems plausible based on the second-link
+heuristic, but I don't have a proof.  A planar cubic graph that
+violates it would be a structural curiosity worth a name --- a
+``cyclically $7$-edge-connected planar cubic graph'' --- and I do
+not know an example.
+
+\paragraph{Relevance to the cut-tire framework.}  If the
+conjecture holds, our cut-tire framework's domain assumption (=
+cyclic edge connectivity exactly $6$ in $G^*$) is automatically
+satisfied by every minimum $4$CT counterexample.  If it doesn't
+hold, we'd need to either prove that the counterexample is not
+of the violating type, or extend the framework to higher-size
+cuts.
+
+\end{document}