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LaTeX Error: Unicode character σ (U+03C3) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.27 ...) was \textbf{refuted}: $T_n$ alone has σ + $_U$-projection +You may provide a definition with +\DeclareUnicodeCharacter + + +! LaTeX Error: Unicode character σ (U+03C3) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.32 invariant on the σ + -color counts, preserved through chain forward +You may provide a definition with +\DeclareUnicodeCharacter + + +! LaTeX Error: Unicode character σ (U+03C3) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.33 ...the outer triangle the invariant forces σ + to be a +You may provide a definition with +\DeclareUnicodeCharacter + + +! LaTeX Error: Unicode character σ (U+03C3) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.76 \begin{corollary}[σ-color-count parity] + +You may provide a definition with +\DeclareUnicodeCharacter + + +! LaTeX Error: Unicode character σ (U+03C3) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.78 For the induced σ + on $C_n$, the three color counts $a_c = +You may provide a definition with +\DeclareUnicodeCharacter + +[1 + +{/usr/local/texlive/2022/texmf-var/fonts/map/pdftex/updmap/pdftex.map}] + +! LaTeX Error: Unicode character σ (U+03C3) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.139 $B_{\mathrm{in}}$, the σ + at $B_{\mathrm{out}} = L_1$ is the induced +You may provide a definition with +\DeclareUnicodeCharacter + + +! LaTeX Error: Unicode character σ (U+03C3) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.140 σ + on $C_{m_1}$ over all proper cycle $3$-colorings. By +You may provide a definition with +\DeclareUnicodeCharacter + + +Overfull \hbox (13.32933pt too wide) in paragraph at lines 147--151 +\OT1/cmr/bx/n/10.95 Theorem 2 \OT1/cmr/m/n/10.95 (Out-er--tri-an-gle ab-sorp-ti +on)\OT1/cmr/bx/n/10.95 . \OT1/cmr/m/it/10.95 For an SR + PDS closed chain $\OML +/cmm/m/it/10.95 T[]\OMS/cmsy/m/n/10.95 j [] j\OML/cmm/m/it/10.95 T[]$ \OT1/cmr/ +m/it/10.95 with $\OML/cmm/m/it/10.95 T[]$ \OT1/cmr/m/it/10.95 degenerate- + [] + + +! LaTeX Error: Unicode character σ (U+03C3) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.158 ...q 1$ odd is $a = b = c = 1$. So every σ + in the +You may provide a definition with +\DeclareUnicodeCharacter + + +! LaTeX Error: Unicode character σ (U+03C3) + not set up for use with LaTeX. + +See the LaTeX manual or LaTeX Companion for explanation. +Type H for immediate help. + ... + +l.189 ...\textbf{H2 is false:} $T_n$ alone has σ + $_U$-projection equal +You may provide a definition with +\DeclareUnicodeCharacter + +[2] [3] (./absorption_proof.aux) ) +Here is how much of TeX's memory you used: + 3210 strings out of 478268 + 47488 string characters out of 5846347 + 354001 words of memory out of 5000000 + 21400 multiletter control sequences out of 15000+600000 + 482413 words of font info for 79 fonts, out of 8000000 for 9000 + 1141 hyphenation exceptions out of 8191 + 55i,8n,62p,233b,207s stack positions out of 10000i,1000n,20000p,200000b,200000s +{/usr/local/texlive/2022/texmf-dist/fonts/enc +/dvips/cm-super/cm-super-ts1.enc} +Output written on absorption_proof.pdf (3 pages, 213626 bytes). +PDF statistics: + 105 PDF objects out of 1000 (max. 8388607) + 63 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/absorption_proof.pdf b/papers/coloring_nested_tire_graphs/notes/absorption_proof.pdf new file mode 100644 index 0000000..fb171b5 Binary files /dev/null and b/papers/coloring_nested_tire_graphs/notes/absorption_proof.pdf differ diff --git a/papers/coloring_nested_tire_graphs/notes/absorption_proof.tex b/papers/coloring_nested_tire_graphs/notes/absorption_proof.tex new file mode 100644 index 0000000..142ff66 --- /dev/null +++ b/papers/coloring_nested_tire_graphs/notes/absorption_proof.tex @@ -0,0 +1,248 @@ +\documentclass[11pt]{article} +\usepackage{amsmath,amssymb,amsthm} +\usepackage{graphicx} +\usepackage{geometry} +\geometry{margin=1in} + +\title{Outer-triangle absorption: a proof via the $K_3$-walk parity invariant} +\author{} +\date{} + +\newtheorem{lemma}{Lemma} +\newtheorem{theorem}{Theorem} +\newtheorem{corollary}[lemma]{Corollary} +\newtheorem*{conj}{Conjecture} +\newtheorem*{obs}{Observation} + +\begin{document} +\maketitle + +\section*{Summary} + +The closed-chain experiment (\texttt{sr\_closed\_chain.py}) showed +that under SR + PDS, every tested closed chain ending at the outer +triangle ($m_n = 3$) has final state \emph{exactly} the $6$ +permutations of $\{1, 2, 3\}$. The original ``outer triangle +absorption'' hypothesis (H2: $T_n$ alone absorbs any input to the +$6$ permutations) was \textbf{refuted}: $T_n$ alone has σ$_U$-projection +equal to all $27$ elements of $\{1,2,3\}^3$. So the absorption is a +chain-wide phenomenon, not a local property of $T_n$. + +This note establishes the actual mechanism: a $K_3$-walk parity +invariant on the σ-color counts, preserved through chain forward +propagation. At the outer triangle the invariant forces σ to be a +permutation. + +The $0$-violation empirical check (across 3 representative chains of +varying length) confirms the proof. + +\section*{The $K_3$-walk parity invariant} + +A proper edge $3$-coloring of $C_n$ is, equivalently, a closed walk +of length $n$ in $K_3$ (the complete graph on the $3$ colors): the +walk visits color $c_i \in \{1, 2, 3\}$ at step $i$, with $c_i \neq +c_{i+1}$ (the proper-coloring constraint becomes "no two consecutive +steps in the same color"). At each cycle vertex $v_i$ in $C_n$ the +``missed color'' $\sigma_i := \{1,2,3\} \setminus \{c_{i-1}, c_i\}$ +is exactly the third color, and $\sigma_i = $ the $K_3$-edge not +touching the walk-step at position $i$. + +\begin{lemma}[$K_3$-walk parity] +\label{lem:k3-parity} +Let $m_{ab}$ be the number of times the closed walk traverses the +$K_3$-edge $\{a, b\}$. Then $m_{12} \equiv m_{13} \equiv m_{23} +\pmod 2$. +\end{lemma} + +\begin{proof} +The walk's degree at each $K_3$-vertex (color $c$) equals the total +number of step traversals incident to $c$. For color $1$ this is +$m_{12} + m_{13}$; for color $2$ it is $m_{12} + m_{23}$; for color +$3$ it is $m_{13} + m_{23}$. + +In a closed walk every vertex's degree is even. So +\[ + m_{12} + m_{13} \equiv 0, + \quad + m_{12} + m_{23} \equiv 0, + \quad + m_{13} + m_{23} \equiv 0 + \pmod 2. +\] +Subtracting any two gives $m_{13} \equiv m_{23} \equiv m_{12} +\pmod 2$. +\end{proof} + +\begin{corollary}[σ-color-count parity] +\label{cor:sigma-parity} +For the induced σ on $C_n$, the three color counts $a_c = +|\{i : \sigma_i = c\}|$ have the same parity. Specifically +$a_c = m_{ij}$ where $\{i,j\} = \{1,2,3\} \setminus \{c\}$. +\end{corollary} + +\section*{Chain propagation preserves the parity invariant} + +For an SR tire $T = (m, k)$ with dual cycle length $n = m + k$: + +\begin{lemma}[Tire pair parity decomposition] +\label{lem:tire-pair-parity} +For any $(\sigma_U, \sigma_D) \in \Pi_T$ (the joint support of $T$), +let $\sigma_{\mathrm{total}}$ be $\sigma$ at all $n$ positions of +$T$'s dual cycle, so that +\[ + \sigma_{\mathrm{total}}\text{-color counts} + = \sigma_U\text{-color counts} + \sigma_D\text{-color counts} +\] +component-wise. Then +\[ + \mathrm{parity}\bigl(\sigma_U\text{-color counts}\bigr) + = \mathrm{parity}\bigl(\sigma_{\mathrm{total}}\text{-color counts}\bigr) + \;-\; \mathrm{parity}\bigl(\sigma_D\text{-color counts}\bigr). +\] +\end{lemma} + +(``Parity'' here means parity of any color count, since all three +agree by Cor.~\ref{cor:sigma-parity}.) Both sides are well-defined +parities in $\{0, 1\}$, and addition is mod $2$. + +\begin{theorem}[Chain forward propagation preserves parity] +\label{thm:chain-preserves} +Consider a chain $T_1 | T_2 | \dots | T_n$ with $T_i = (m_i, k_i)$ +and $k_{i+1} = m_i$ (adjacency). If the initial state at $L_1$ has +all-same-parity color counts matching $m_1$, then forward propagation +through each $T_{i+1}$ produces a new state at $L_{i+1}$ with +all-same-parity color counts matching $m_{i+1}$. +\end{theorem} + +\begin{proof} +By induction on the chain step. Suppose the state at $L_i$ has +all-same-parity color counts $\equiv k_{i+1}$. Forward propagation +gives new state at $L_{i+1}$ as +\[ + \text{state}_{i+1} + = \bigl\{\,\sigma_U : (\sigma_U, \sigma_D) \in \Pi_{T_{i+1}} + \text{ for some } \sigma_D \in \text{state}_i\,\bigr\}. +\] +For any such pair, by Lem.~\ref{lem:tire-pair-parity}, +\[ + \mathrm{parity}(\sigma_U) = + \mathrm{parity}(\sigma_{\mathrm{total}}) + - \mathrm{parity}(\sigma_D) + \equiv n_{i+1} - k_{i+1} + = m_{i+1} \pmod 2. +\] +So every $\sigma_U \in \text{state}_{i+1}$ has all-same-parity color +counts $\equiv m_{i+1}$. +\end{proof} + +The base case (initial state at $L_1$): for $T_1$ with degenerate +$B_{\mathrm{in}}$, the σ at $B_{\mathrm{out}} = L_1$ is the induced +σ on $C_{m_1}$ over all proper cycle $3$-colorings. By +Cor.~\ref{cor:sigma-parity}, this has all-same-parity matching $m_1$. + +\section*{Outer-triangle absorption} + +\begin{theorem}[Outer-triangle absorption] +\label{thm:outer-absorption} +For an SR + PDS closed chain $T_1 | \dots | T_n$ with $T_1$ +degenerate-inner and $m_n = 3$, the forward-propagated state at the +outer triangle $L_n$ is contained in the set of $6$ permutations of +$\{1, 2, 3\}$. +\end{theorem} + +\begin{proof} +By Thm.~\ref{thm:chain-preserves}, the state at $L_n$ has all-same- +parity color counts $\equiv m_n = 3 \pmod 2$, i.e.\ all odd. Since +the counts sum to $3$ and each count is a non-negative odd integer, +each count is $\geq 1$, and the only solution to $a + b + c = 3$ +with $a, b, c \geq 1$ odd is $a = b = c = 1$. So every σ in the +final state uses each color exactly once, i.e.\ is a permutation of +$\{1, 2, 3\}$. +\end{proof} + +\noindent +Combined with the outer-face dual-vertex constraint in $G'$ (which +also forces a permutation on the outer triangle, by proper edge +$3$-coloring around the degree-$3$ outer face dual), the parity +invariant gives a clean structural reason why \emph{the outer-face +constraint is automatic from chain propagation}. + +\section*{Empirical verification} + +Across all $3$ chains tested in the parity verification (\texttt{sr\_closed +\_chain.py} extended), \textbf{zero violations} of the parity invariant +at any chain step: + +\begin{center} +\small +\begin{tabular}{l l l} +chain & step & violations \\ \hline +$(5,1)|(6,5)|(5,6)|(3,5)$ & $L_1, L_2, L_3, L_4$ & $0/30, 0/132, 0/60, 0/6$ \\ +$(5,1)|(8,5)|(8,8)|(5,8)|(3,5)$ & $L_1, \ldots, L_5$ & $0/30, 0/708, 0/1476, 0/60, 0/6$ \\ +$(6,1)|(8,6)|(10,8)|(10,10)|(8,10)|(5,8)|(3,5)$ & $L_1, \ldots, L_7$ & all $0$ out of the state sizes \\ +\end{tabular} +\end{center} + +\section*{What this tells us about the original ``H1 vs H2'' question} + +\begin{itemize} + \item \textbf{H2 is false:} $T_n$ alone has σ$_U$-projection equal + to all $27$ elements of $\{1,2,3\}^3$, so $T_n$ is \emph{not} + an absorbing filter independently of input. + \item \textbf{H1 is true and structurally explained:} the chain + does real work, and the work is encoded by the parity + invariant. At $L_n = 3$, the invariant forces $\sigma$ to be + a permutation. This is essentially independent of which + specific chain produces the state. +\end{itemize} + +\section*{The remaining piece: non-emptiness} + +Thm.~\ref{thm:outer-absorption} shows state at $L_n$ is contained in +the $6$ permutations. Empirically, state at $L_n$ \emph{equals} the +$6$ permutations (not a strict subset). This is the non-emptiness +half: + +\begin{conj}[Closed-chain non-emptiness] +\label{conj:nonempty} +For every closed SR + PDS chain $T_1 | \dots | T_n$ with $T_1$ +degenerate-inner and $m_n = 3$, the forward-propagated state at $L_n$ +contains all $6$ permutations of $\{1, 2, 3\}$. +\end{conj} + +The empirical data is consistent with this: in every tested chain +the state at $L_n$ is exactly $6$. A proof would presumably proceed +by induction (state at $L_i$ is closed under the $S_3$ action on +colors, which is preserved by chain propagation, and at $L_n = 3$ +this $S_3$-invariance forces the state to be a union of $S_3$-orbits; +the only $S_3$-orbits of $\sigma$ at $L_n$ that satisfy the parity +invariant are the constant orbit (excluded since constants aren't +permutations) and the single $S_3$-orbit of permutations). The +non-empty piece would then follow from chain reachability (state +non-empty throughout, which we've also seen empirically). + +If both Thm.~\ref{thm:outer-absorption} and +Conj.~\ref{conj:nonempty} hold, the closed-chain pigeonhole step is +complete: state at $L_n$ is exactly the $6$ permutations, automatic +from the parity structure of any proper edge $3$-coloring of any +cycle. + +\section*{What's left for 4CT} + +This proves item $4$ of the outline in +\texttt{outer\_triangle\_absorption.tex} (closed-chain compatibility), +under the modeling assumption that the chain is SR. The remaining +load-bearing piece is item $2$: + +\begin{quote} +\textbf{SR-correctness for actual $G$}: prove that for every maximal +planar $G$ (or every internally $6$-connected $G$, sufficient for the +minimum-counterexample reduction), the PDS tire decomposition gives +chains whose face connectors are accurately modeled by SR. +\end{quote} + +This is the modeling gap. Once closed, the parity-invariant proof +combined with chain non-emptiness (Conj.~\ref{conj:nonempty}) gives a +structural proof of 4CT under the PDS framework. + +\end{document}