coloring_nested_tire_graphs: redo second-link analysis for maximal planar

Previous version had loose formulas and overstated what second-link
length forces.  Replaced with cleaner version that:

- States the maximal-planar constraints explicitly
  (E = 3V-6, F = 2V-4, sum of deg = 6V-12).
- Notes the FORCED 12 degree-5 vertices when all degrees ∈ {5,6}.
- Gives the correct second-link length formula:
    L_2(v) = d + sum_{u in link(v)} (deg(u) - 5)
  Earlier version had this wrong.
- Concretely: pentakis dodecahedron has L_2 = 10 around every
  vertex, but its dual (Buckyball) STILL has 6-edge cyclic cuts
  arising from non-second-link constructions.

So second-link length being large doesn't prevent small non-facial
cyclic cuts via other separators.  The min cut size is not pinned
down by local link structure alone.

Bottom line unchanged: min non-facial cyclic cut for a min 4CT
counterexample could be 6, 7, 8, ... and Birkhoff alone doesn't
distinguish.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>

papers/coloring_nested_tire_graphs/notes/even_separating_cycle.aux

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papers/coloring_nested_tire_graphs/notes/even_separating_cycle.tex

@@ -106,41 +106,51 @@ possible in principle.
 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}
+\section*{What the maximal-planar constraint forces}
 
-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 $G$).
+Maximal planar (= triangulation) gives strong constraints:
+$E = 3V - 6$, $F = 2V - 4$, $\sum \deg = 6V - 12$.  Average degree
+approaches $6$ from below.  Internally $6$-connected forces min
+degree $\ge 5$.
 
-\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.
+\paragraph{Forced $12$ degree-$5$ vertices.}  If all degrees are in
+$\{5, 6\}$:
+\[
+5 n_5 + 6 n_6 = 6V - 12, \quad n_5 + n_6 = V \implies n_5 = 12.
+\]
+So exactly $12$ degree-$5$ vertices, with the remaining $V - 12$
+of degree $6$.  For $V > 12$, at least some degree-$5$ vertex has a
+degree-$6$ link neighbour.
 
-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.
+\paragraph{Second-link length.}  For vertex $v$ of degree $d$ in a
+triangulation, the second link (cycle of vertices at distance
+exactly $2$) has length
+\[
+L_2(v) = d + \sum_{u \in \mathrm{link}(v)} (\deg u - 5).
+\]
+(Each link vertex contributes one ``shared'' boundary vertex with
+each link-cycle neighbour, plus $\deg u - 5$ private boundary
+vertices inside $u$'s fan.)  For the icosahedron ($d = 5$, all
+link degrees $5$): $L_2 = 5$.
 
-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).
+\paragraph{Second-link length does \emph{not} pin down min cyclic
+cut.}  The pentakis dodecahedron ($V = 32$): every degree-$5$
+vertex has all $5$ link vertices of degree $6$, so $L_2 = 5 + 5 =
+10$ around degree-$5$ vertices.  Around degree-$6$ vertices,
+$L_2 = 6 + 0 \cdot 2 + 1 \cdot 4 = 10$.  So no second-link
+separator has length $6$.
 
-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.
+\emph{But} the pentakis dodecahedron's dual (Buckminsterfullerene
+graph) \emph{does} have $6$-edge cyclic cuts --- they arise as
+separations \emph{not} surrounding a single vertex.  So
+second-link length is just one source of cyclic cuts; longer
+constructions can yield smaller cuts.
+
+\paragraph{Conclusion of this section.}  The maximal-planar
+constraint forces some structural relations but does not pin down
+the minimum non-facial cyclic cut to any specific value.  A
+minimum $4$CT counterexample could in principle have any min cut
+size $\ge 6$.
 
 \section*{What I can conclude}