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PDF statistics: - 185 PDF objects out of 1000 (max. 8388607) - 52 compressed objects within 1 object stream + 195 PDF objects out of 1000 (max. 8388607) + 58 compressed objects within 1 object stream 0 named destinations out of 1000 (max. 500000) 18 words of extra memory for PDF output out of 10000 (max. 10000000) diff --git a/papers/coloring_nested_tire_graphs/notes/birkhoff_internally_6_connected.pdf b/papers/coloring_nested_tire_graphs/notes/birkhoff_internally_6_connected.pdf index d11f16d..13f8b84 100644 Binary files a/papers/coloring_nested_tire_graphs/notes/birkhoff_internally_6_connected.pdf and b/papers/coloring_nested_tire_graphs/notes/birkhoff_internally_6_connected.pdf differ diff --git a/papers/coloring_nested_tire_graphs/notes/birkhoff_internally_6_connected.tex b/papers/coloring_nested_tire_graphs/notes/birkhoff_internally_6_connected.tex index c1539e8..fec25dd 100644 --- a/papers/coloring_nested_tire_graphs/notes/birkhoff_internally_6_connected.tex +++ b/papers/coloring_nested_tire_graphs/notes/birkhoff_internally_6_connected.tex @@ -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}; + \node[red, anchor=east] at (11, 2.2) {FORBIDDEN}; \node[anchor=west] at (0, 1.5) {\textbullet\, separating $5$-cycle isolating $1$ vertex}; - \node[green!50!black, anchor=west] at (5.0, 1.5) {ALLOWED}; - % 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) {}; \foreach \i in {0,...,4} { \pgfmathsetmacro{\ang}{72*\i+90} @@ -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}; + \node[red] at (0, -1.95) {\small allowed case}; \end{scope} \end{tikzpicture} \end{center} @@ -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 diff --git a/papers/coloring_nested_tire_graphs/notes/even_separating_cycle.aux b/papers/coloring_nested_tire_graphs/notes/even_separating_cycle.aux new file mode 100644 index 0000000..4064ba6 --- /dev/null +++ b/papers/coloring_nested_tire_graphs/notes/even_separating_cycle.aux @@ -0,0 +1,10 @@ +\relax +\@writefile{toc}{\contentsline {paragraph}{Statement.}{1}{}\protected@file@percent } +\@writefile{toc}{\contentsline {paragraph}{Short answer.}{1}{}\protected@file@percent } +\newlabel{lem:cut-parity}{{}{1}} +\@writefile{toc}{\contentsline {paragraph}{Translating to primal cycles.}{1}{}\protected@file@percent } +\@writefile{toc}{\contentsline {paragraph}{Is this realisable?}{2}{}\protected@file@percent } +\@writefile{toc}{\contentsline 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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}