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\title{Testing the rainbow and K\"onig-lift conjectures on cut tires}
\author{}
\date{}

\newtheorem*{obs}{Observation}

\begin{document}
\maketitle

\section*{What was tested}

For the cut tires arising on $G'_1$ of Holton-McKay graph \#0 (with
the matching $6$-edge cut from \texttt{cut\_depth\_label.tex}), under
the redefined cut tire definition (face boundary $+$ labelled
pendants at degree-$2$ vertices), I:
\begin{enumerate}
  \item Built each cut tire as a graph (face boundary plus pendants).
  \item Brute-force enumerated proper edge $3$-colorings.
  \item Computed the projection support $\{(\sigma_{\mathrm{out}},
        \sigma_{\mathrm{in}}) : \chi \text{ proper}\}$.
  \item Checked $S_3$-closure and orbit decomposition.
\end{enumerate}

\section*{Results}

\begin{center}
\small
\begin{tabular}{lr|rrrrr|r|cl}
\toprule
$d$ & face & $|f|$ & out & in & $|E|$ & $\#$col. & $|\pi|$
    & $S_3$-cl.\ & orbit sizes\\
\midrule
$1$ & $0$ & $12$ & $5$ & $0$ & $17$ & $96$  & $93$  & Yes & $[3, 6^{15}]$\\
$1$ & $1$ & $4$  & $1$ & $0$ & $5$  & $6$   & $3$   & Yes & $[3]$\\
$2$ & $0$ & $7$  & $4$ & $3$ & $14$ & $126$ & $126$ & Yes & $[6^{21}]$\\
$2$ & $1$ & $7$  & $4$ & $3$ & $14$ & $126$ & $126$ & Yes & $[6^{21}]$\\
$3$ & $0$--$2$ & $2$ & $0$ & $0$ & $2$ & $3$ & $1$ & Yes & $[1]$\\
$4$ & $0$ & $4$  & $1$ & $0$ & $5$  & $6$   & $3$   & Yes & $[3]$\\
$4$ & $1$ & $8$  & $2$ & $1$ & $11$ & $24$  & $21$  & Yes & $[3, 6^3]$\\
$5$ & $0$ & $14$ & $4$ & $2$ & $20$ & --- (too big) & --- & --- & ---\\
$5$ & $1$ & $2$  & $0$ & $0$ & $2$  & $3$   & $1$   & Yes & $[1]$\\
$6$ & $0$ & $12$ & $3$ & $2$ & $17$ & $96$  & $93$  & Yes & $[3, 6^{15}]$\\
$7$ & $0$ & $2$  & $0$ & $0$ & $2$  & $3$   & $1$   & Yes & $[1]$\\
\bottomrule
\end{tabular}
\end{center}

\section*{Observations}

\begin{obs}[Universal $S_3$-closure]
Every cut tire's projection $\pi(\sigma_{\mathrm{out}}, \sigma_{\mathrm{in}})$
is closed under the diagonal $S_3$ action on the $3$ colors.  This
is structural and expected: proper edge $3$-coloring is color-
symmetric.
\end{obs}

\begin{obs}[Orbit sizes are $3$ or $6$ only]
Every $S_3$-orbit in every cut tire projection has size $3$ (the
constant-color orbit, when present) or $6$ (the generic orbit using
all $3$ colors).  No size-$2$ orbits, which would correspond to
$\sigma$'s with non-trivial $S_3$ stabilizer; these don't occur.
\end{obs}

\begin{obs}[Non-trivial cut tires have substantial projection support]
The two ``main'' cut tires at depth $2$ (face length $7$, $4$ out
$+\ 3$ in spokes) each have $126$ proper edge $3$-colorings, all
$126$ distinct in their joint $(\sigma_{\mathrm{out}}, \sigma_{\mathrm{in}})$
projection.  This is $21$ full $S_3$-orbits of size $6$.

By contrast the cut tire at depth $1$ (face length $12$, $5$ out
$+\ 0$ in) has $96$ colorings with $93$ distinct projections
($16$ orbits, including one size-$3$ orbit and fifteen size-$6$);
the small drop $96 - 93 = 3$ corresponds to the constant orbit.
\end{obs}

\section*{Why the rainbow conjecture is not directly testable here}

The rainbow conjecture from \texttt{rainbow\_proof.tex} states:
for an antipodal-chord SP tire $T = (m_1, (0, m/2), \mathrm{SP})$
with $m \in \{4, 6\}$ even and $m_1 \ge m - 1$, the inner-spoke
projection $\pi_D(\mathcal{C}(T))$ equals the perms-per-face set
$\mathcal{P}_m$ (size $36$).

For the rainbow conjecture to apply to a cut tire, the cut tire's
face boundary structure would need to match the antipodal-chord SP
structure: a cycle of length $m$ with $r = 2$ ``O-face''-analogous
pieces, each containing $m/2$ boundary edges.

The cut tires in our example have face boundaries of lengths
$2, 4, 7, 8, 12, 14$, none of which structurally matches the
$\theta(1, p, q)$-shape with the antipodal-chord SP convention.
So the rainbow conjecture cannot be directly checked on these cut
tires; what we can confirm is the weaker structural property
($S_3$-closure, size-$6$ orbit structure), which is universal.

\paragraph{What would test the rainbow conjecture properly.}
Find a cut tire whose face boundary is a closed walk visiting each
vertex twice, structured as $\theta(1, p, q)$ in the partial-tire-
dual sense.  Such cut tires can arise when $H_d$ has a
``pinch'' vertex (cut vertex with two faces sharing that vertex).
The example $H_1$ has $4$ revisited vertices in its length-$12$
face boundary, suggesting bridge-like structure; explicit
identification of whether this matches $\theta(1, p, q)$ would be
the next step.

\section*{Why the K\"onig-lift conjecture is not directly testable here}

The K\"onig-lift conjecture from \texttt{worst\_case\_proof\_sketch.tex}
applies to pairs of adjacent SP tires $(T_1, T_2)$ sharing a cycle
$\gamma$ where \emph{both} sides give direct $\gamma$-face partitions
(both have chord(s) on $\gamma$).

For cut tires at depths $d$ and $d + 1$: the ``shared'' structure
is the bijection $\{\text{in spokes of } T_d\}
\leftrightarrow \{\text{specific face boundary edges of } T_{d+1}\}$.
This is not the same as ``both tires give $\gamma$-face partitions''
because $T_{d+1}$'s face boundary is not (in general) a $\gamma$-cycle
that $T_d$ also borders --- the cut tires sit in $H_d$ and $H_{d+1}$
respectively, with different vertex sets.

So the K\"onig-lift conjecture would need restatement for the
cut-tire chain.  A correct restatement might say:

\paragraph{Cut-tire K\"onig-lift analog (tentative).}
For each pair of adjacent cut tires $(T_d, T_{d+1})$ in a chain, the
bijection $\beta : \{\text{in spokes of } T_d\} \to
\{\text{face-boundary edges of } T_{d+1} \text{ adjacent to } V(f_d)\}$
preserves a Latin structure: any Latin $\sigma$ on the shared
positions in $T_{d+1}$'s face boundary can be lifted to a proper
edge $3$-coloring of $T_d$ via $\beta$, and vice versa.

\paragraph{What this would require to test.}
\begin{enumerate}
  \item Explicit identification of the bijection $\beta$ for each
        pair $(T_d, T_{d+1})$.  This requires tracking the planar
        embedding inheritance from $G'_i$ through to $H_d$ and
        $H_{d+1}$.
  \item Computing the Latin structure on each $T_{d+1}$'s face
        boundary edges (analogous to the perms-per-face structure
        on $\gamma$).
  \item Comparing the projected supports under $\beta$.
\end{enumerate}

None of this is automated in the present code.

\section*{What can be concluded from the empirical data}

\begin{enumerate}
  \item \textbf{$S_3$-closure holds universally.}  Every cut tire
        projection is $S_3$-symmetric, with orbits of size $3$ or $6$.
        This matches the partial-tire-dual data
        (\texttt{orbit\_decomposition.tex}, Obs.\ ``S3-closed'').
  \item \textbf{Non-trivial cut tires have rich projections.}  At
        depth $2$, the projection is $126 = 21 \cdot 6$ ($21$ full
        $S_3$-orbits).  At depth $6$, $93 = 1 \cdot 3 + 15 \cdot 6$
        ($16$ orbits).  This is substantially more than the
        rainbow's $36 = 6 \cdot 6$ specific orbit, so cut tires here
        are \emph{looser} than the rainbow case --- chain pigeonhole
        should be \emph{easier} when projections are large.
  \item \textbf{Trivial cut tires (length-$2$ faces) contribute
        nothing.}  Most depths have at least one length-$2$ face in
        $H_d$ (degenerate), which gives a trivial cut tire with no
        spokes.  These are not informative for chain pigeonhole.
  \item \textbf{Neither conjecture is directly applicable.}  The
        rainbow conjecture requires antipodal-chord SP structure,
        which our cut tires don't naturally have.  The K\"onig-lift
        conjecture requires both sides give $\gamma$-face
        partitions, which the cut-tire chain doesn't naturally
        produce.
\end{enumerate}

\paragraph{What's actually closer to provable.}
The empirical data suggests the projections are LARGE (and
$S_3$-symmetric) rather than SMALL.  A natural chain pigeonhole
statement is:

\begin{quote}
\textbf{Conjecture (loose).}  For each cut tire $T$ arising in the
chain, the projection $\pi(T)$ has size $\ge c \cdot 6$ for some
absolute constant $c$ depending only on $|E(T)|$, with $\pi(T)$
$S_3$-closed.  Chain composition through $T_1, T_2, \ldots$ yields
$|\mathcal{R}_i| \ge c'$ uniform in chain length.
\end{quote}

The constant $c$ might be derivable from Prop 1.13 of \texttt{paper.tex}
($2^n + 2(-1)^n$ colorings for spoke-only $D(T)$).  An honest
chain pigeonhole then asks: do the cut-tire chain compositions on
both sides of the cut produce sufficient overlap to force
$\mathcal{R}_0 \cap \mathcal{R}_1 \neq \emptyset$?

This requires the full chain machinery, not just per-tire analysis.

\end{document}
