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LaTeX Error: Unicode character ↔ (U+2194) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.91 through the in/out-spoke ↔ + face-boundary bijection to some +You may provide a definition with +\DeclareUnicodeCharacter + +[2] [3] (./loose_conjecture_counterexamples.aux) ) +Here is how much of TeX's memory you used: + 3250 strings out of 478268 + 48531 string characters out of 5846347 + 353625 words of memory out of 5000000 + 21437 multiletter control sequences out of 15000+600000 + 479526 words of font info for 68 fonts, out of 8000000 for 9000 + 1141 hyphenation exceptions out of 8191 + 55i,8n,62p,249b,244s stack positions out of 10000i,1000n,20000p,200000b,200000s +{/usr/local/texlive/2022/texm +f-dist/fonts/enc/dvips/cm-super/cm-super-ts1.enc} +Output written on loose_conjecture_counterexamples.pdf (3 pages, 158789 bytes). + +PDF statistics: + 80 PDF objects out of 1000 (max. 8388607) + 48 compressed objects within 1 object stream + 0 named destinations out of 1000 (max. 500000) + 1 words of extra memory for PDF output out of 10000 (max. 10000000) + diff --git a/papers/coloring_nested_tire_graphs/notes/loose_conjecture_counterexamples.pdf b/papers/coloring_nested_tire_graphs/notes/loose_conjecture_counterexamples.pdf new file mode 100644 index 0000000..1dbf7ae Binary files /dev/null and b/papers/coloring_nested_tire_graphs/notes/loose_conjecture_counterexamples.pdf differ diff --git a/papers/coloring_nested_tire_graphs/notes/loose_conjecture_counterexamples.tex b/papers/coloring_nested_tire_graphs/notes/loose_conjecture_counterexamples.tex new file mode 100644 index 0000000..8c08f70 --- /dev/null +++ b/papers/coloring_nested_tire_graphs/notes/loose_conjecture_counterexamples.tex @@ -0,0 +1,161 @@ +\documentclass[11pt]{article} +\usepackage{amsmath,amssymb,amsthm} +\usepackage{graphicx} +\usepackage{geometry} +\usepackage{booktabs} +\geometry{margin=1in} + +\title{Counterexamples to the loose chain pigeonhole conjecture\\ +(and what they suggest about the right statement)} +\author{} +\date{} + +\newtheorem*{obs}{Observation} + +\begin{document} +\maketitle + +\section*{The conjecture as stated} + +From \texttt{cut\_depth\_label.tex}, the loose chain pigeonhole +conjecture says: + +\begin{quote} +For every non-trivial cut tire $T$ (with $\ge 1$ in/out spoke), the +joint projection $\pi(T)$ is non-empty, $S_3$-closed, and contains +at least one full $S_3$-orbit (i.e.\ $|\pi(T)| \ge 6$). +\end{quote} + +\section*{Counterexamples already in the data} + +Re-examining the cut tire tests on $G'_1$ of Holton-McKay \#0 +(\texttt{cut\_tire\_conjecture\_tests.tex}): + +\begin{center} +\small +\begin{tabular}{lr|rrr|r|cl} +\toprule +$d$ & face & out & in & total spokes & $|\pi(T)|$ & orbit sizes & \\ +\midrule +$1$ & $0$ & $5$ & $0$ & $5$ & $93$ & $[3, 6^{15}]$ & ✓\\ +$1$ & $1$ & $1$ & $0$ & $1$ & $3$ & $[3]$ & \textbf{✗}\\ +$2$ & $0$ & $4$ & $3$ & $7$ & $126$ & $[6^{21}]$ & ✓\\ +$2$ & $1$ & $4$ & $3$ & $7$ & $126$ & $[6^{21}]$ & ✓\\ +$4$ & $0$ & $1$ & $0$ & $1$ & $3$ & $[3]$ & \textbf{✗}\\ +$4$ & $1$ & $2$ & $1$ & $3$ & $21$ & $[3, 6^3]$ & ✓\\ +$6$ & $0$ & $3$ & $2$ & $5$ & $93$ & $[3, 6^{15}]$ & ✓\\ +\bottomrule +\end{tabular} +\end{center} + +Two non-trivial cut tires have $|\pi(T)| = 3 < 6$: +$(d, \text{face}) = (1, 1)$ and $(4, 0)$. Both have exactly $1$ +spoke. In each case, $\pi(T)$ is the $3$-element $S_3$-orbit of a +single-color spoke configuration --- which has $S_3$-stabilizer of +order $2$ (the two permutations fixing the chosen color and swapping +the others), so the orbit has size $|S_3| / 2 = 3$. + +\section*{Why this happens (and is general)} + +For a cut tire with exactly $k$ in/out spokes, the projection +$\pi(T) \subseteq \{1, 2, 3\}^k$. $S_3$ acts by permuting colors. +The $S_3$-orbit of $\sigma \in \{1,2,3\}^k$ has size: +\begin{itemize} + \item $3$ if $\sigma$ uses exactly $1$ color (stabilizer of order $2$); + \item $6$ if $\sigma$ uses $\geq 2$ distinct colors (trivial stabilizer). +\end{itemize} + +For $k = 1$: every $\sigma \in \{1, 2, 3\}$ uses exactly $1$ color, +so every orbit has size $3$, and $|\pi(T)| \in \{0, 3\}$. The +conjecture's ``$|\pi(T)| \ge 6$'' is automatically false for any +non-empty $\pi(T)$ on a $1$-spoke tire. + +For $k \ge 2$, $\sigma$ can use $1$ or $\geq 2$ colors: +\begin{itemize} + \item Single-color $\sigma$ contributes a size-$3$ orbit. + \item Multi-color $\sigma$ contributes size-$6$ orbits. +\end{itemize} +$|\pi(T)| \ge 6$ requires at least one multi-color $\sigma$, which +is not automatic but typically holds when $k$ is large or the tire +allows enough flexibility. + +\section*{Why the conjecture's chain half can still fail} + +Even if we restrict to $k \ge 2$ cut tires (avoiding the trivial +$k = 1$ counterexample), the chain pigeonhole conclusion is not +immediate: + +\begin{itemize} + \item Chain composition involves \emph{compatibility} between + adjacent cut tires. A full $S_3$-orbit at one layer projects + through the in/out-spoke ↔ face-boundary bijection to some + subset at the adjacent layer; this projection might land in + a size-$3$ orbit even if the source was size-$6$. + \item The chain at $d_{\max}$ terminates (no in spokes); the + innermost cut tire's structure constrains everything outward. + \item Sufficiency for $\mathcal{R}_0 \cap \mathcal{R}_1 \neq + \emptyset$ requires the chain to preserve the $S_3$-orbit + structure all the way to the cut, on both sides, and have a + common orbit there. +\end{itemize} + +The counterexamples at $1$-spoke tires don't break the chain at +those depths because they're at non-essential ``side'' faces +(face $1$ at $d = 1$, face $0$ at $d = 4$; the main chain runs +through the larger faces). But they show the per-tire claim +is not universal. + +\section*{Refined conjectures} + +In view of these counterexamples, three natural refinements: + +\begin{enumerate} +\item \textbf{Restrict to $\ge 2$ spokes.} ``For every cut tire + $T$ with $\geq 2$ in/out spokes total, $|\pi(T)| \ge 6$ and + $\pi(T)$ contains a full $S_3$-orbit.'' This avoids the + $k = 1$ trivial case. Still empirically open whether + $k = 2$ cut tires always have $\ge 6$ projections. + +\item \textbf{Weaken to non-emptiness.} ``For every non-trivial + cut tire, $\pi(T)$ is non-empty and $S_3$-closed.'' This + is much weaker but might still suffice for chain pigeonhole + \emph{provided} chain composition preserves non-emptiness --- + which is itself the hard step. + +\item \textbf{Allow size-$3$ orbits.} ``For every non-trivial cut + tire, $\pi(T)$ is non-empty and $S_3$-closed; the orbits in + $\pi(T)$ have size $3$ or $6$.'' This is just a description + of the empirical findings, not a useful conjecture --- chain + pigeonhole at a size-$3$-orbit layer is more restrictive than + at a size-$6$ one. +\end{enumerate} + +\section*{Where to look for harder counterexamples} + +The two counterexamples found are at low-spoke tires. More +substantive counterexamples would look like: + +\begin{itemize} + \item A $k \ge 2$ spoke cut tire whose face boundary is structured + such that only single-color $\sigma$'s extend, leaving + $|\pi(T)| = 3$. This would require the face boundary to + force all spoke colors equal --- an unusual constraint. + \item A chain in which adjacent compatibility kills all size-$6$ + orbits, leaving only size-$3$ at the cut layer. This would + be a chain-level counterexample to the bottom-line + pigeonhole. +\end{itemize} + +Neither is observed in the present data. But the data is from a +single $6$-cut on a single Holton-McKay graph. A broader empirical +sweep (multiple cuts, multiple Holton-McKay graphs, larger $G'$) +might surface harder counterexamples. + +\paragraph{Suggested next step.} Restrict the conjecture to +$k \ge 2$ tires and re-run the empirical test. If still empirically +true after broader sweeps, attempt to prove the restricted form via +the existing Prop 1.13 lower bound (which gives $\ge 2^n + 2(-1)^n$ +colorings for spoke-only cut tires of cycle length $n$) plus a +combinatorial argument that the multi-color $\sigma$'s are non-empty. + +\end{document}