diff --git a/papers/dual_decomposition_minimal_counterexamples/paper.aux b/papers/dual_decomposition_minimal_counterexamples/paper.aux index 5de24eb..67bb3a4 100644 --- a/papers/dual_decomposition_minimal_counterexamples/paper.aux +++ b/papers/dual_decomposition_minimal_counterexamples/paper.aux @@ -3,13 +3,16 @@ \newlabel{lem:triangulate}{{1.1}{1}} \newlabel{def:minimal}{{1.2}{1}} \newlabel{lem:mindeg}{{1.4}{1}} +\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{The reduced dual}}{2}{}\protected@file@percent } +\newlabel{def:reduced-dual}{{2.1}{2}} +\newlabel{def:edge-names}{{2.3}{2}} +\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces The four steps of Definition\nonbreakingspace 2.1\hbox {}, illustrated on $G' = $ the dodecahedron (dual of the icosahedron) with $F_v$ the inner pentagon and $i = 0$. Top left: delete the five boundary vertices of $F_v$, leaving five degree-$2$ vertices on a new face $F$. Top right: order them clockwise as $A_0,\dots ,A_4$. Bottom left: add $v_n$ joined to $A_0, A_1, A_2$. Bottom right: add the chord $A_3 A_4$, giving the cubic plane graph $\setbox \z@ \hbox {\mathsurround \z@ $\textstyle G$}\mathaccent "0362{G}'_{v,0}$.}}{3}{}\protected@file@percent } +\newlabel{fig:reduced-dual-steps}{{1}{3}} +\newlabel{lem:pentagonal-externals}{{2.4}{3}} \newlabel{tocindent-1}{0pt} \newlabel{tocindent0}{0pt} \newlabel{tocindent1}{17.77782pt} \newlabel{tocindent2}{0pt} \newlabel{tocindent3}{0pt} -\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{The reduced dual}}{2}{}\protected@file@percent } -\newlabel{def:reduced-dual}{{2.1}{2}} -\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces The four steps of Definition\nonbreakingspace 2.1\hbox {}, illustrated on $G' = $ the dodecahedron (dual of the icosahedron) with $F_v$ the inner pentagon and $i = 0$. Top left: delete the five boundary vertices of $F_v$, leaving five degree-$2$ vertices on a new face $F$. Top right: order them clockwise as $A_0,\dots ,A_4$. Bottom left: add $v_n$ joined to $A_0, A_1, A_2$. 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The chord $A_{i+3}A_{i+4}$ is the \emph{merged edge}; the +edge $A_{i+1}v_n$ is the \emph{spike edge}; and the edges $A_iv_n$ and +$A_{i+2}v_n$ are the \emph{side edges}. In the $i = 0$ case of +Figure~\ref{fig:reduced-dual-steps} these are $\{A_3, A_4\}$, $\{A_1, v_n\}$, +and $\{A_0, v_n\}, \{A_2, v_n\}$ respectively. +\end{definition} + +We will use the following structural fact about proper $3$-edge-colourings near +a pentagonal face of a cubic plane graph; it is stated for a generic such graph +$H$, not specifically for the reduced dual. + +\begin{lemma}[Pentagonal externals] +\label{lem:pentagonal-externals} +Let $H$ be a cubic plane graph and $F$ a pentagonal face of $H$, with +$\partial F$ traversed clockwise as $u_0, u_1, u_2, u_3, u_4$. For each $i$ +let $f_i$ be the unique edge of $H$ incident to $u_i$ that does not lie on +$\partial F$. An assignment $\varphi$ of colours from $\{1, 2, 3\}$ to the ten +edges incident to $\{u_0, \dots, u_4\}$ is proper at every $u_i$ if and only if +there is some index $j$ such that +\[ + \varphi(f_j) = \varphi(f_{j+1}) = \varphi(f_{j+2}) + \quad\text{and}\quad + \{\varphi(f_{j+3}), \varphi(f_{j+4})\} + = \{1, 2, 3\} \setminus \{\varphi(f_j)\}, +\] +indices mod $5$. +\end{lemma} + +\begin{proof} +Write $e_i = u_i u_{i+1}$ for the boundary edges of $\partial F$ (indices mod +$5$). A colouring $\varphi$ is proper at every $u_i$ if and only if at each +$u_i$ the three incident edges $e_{i-1}, e_i, f_i$ receive three distinct +colours; whenever this holds, $\varphi(f_i)$ is forced to be the unique colour +in $\{1, 2, 3\} \setminus \{\varphi(e_{i-1}), \varphi(e_i)\}$, and $\varphi$ +restricts to a proper $3$-edge-colouring of the cycle $\partial F$. + +\textbf{($\Rightarrow$)} The line graph of $\partial F$ is $C_5$, whose +maximum independent set has size $2$, so no colour appears more than twice on +$\partial F$; and since $\partial F$ is an odd cycle, all three colours appear. +The colour multiset on $(\varphi(e_0), \dots, \varphi(e_4))$ is therefore +$(2, 2, 1)$, with the singleton at a unique position. Cyclically shifting +indices we may place this position at $0$; let $c$ be the singleton colour. +The remaining four edges form the path $e_1 e_2 e_3 e_4$, which by propriety +alternates between the other two colours, so for some labelling +$\{a, b, c\} = \{1, 2, 3\}$, +\[ + (\varphi(e_0), \varphi(e_1), \varphi(e_2), \varphi(e_3), \varphi(e_4)) + = (c, a, b, a, b). +\] +Reading off the forced values of $\varphi(f_i)$, +\[ + \varphi(f_0) = a, \quad + \varphi(f_1) = b, \quad + \varphi(f_2) = \varphi(f_3) = \varphi(f_4) = c, +\] +which is the lemma's pattern at $j = 2$ (the cyclic shift maps this back to +the corresponding $j$ in the original indexing). This case is the unique +proper $3$-edge-colouring of $\partial F$ up to cyclic shift and permutation of +$\{1, 2, 3\}$ (since $5 \cdot 3! = 30 = P(C_5, 3)$, the chromatic polynomial of +$C_5$ at $3$), so it exhausts every proper $\varphi$. + +\textbf{($\Leftarrow$)} The lemma's hypothesis is invariant under cyclic +shifts of indices and under permutations of $\{1, 2, 3\}$, so we may assume +$j = 2$, $\varphi(f_2) = \varphi(f_3) = \varphi(f_4) = c$, $\varphi(f_0) = a$, +and $\varphi(f_1) = b$, with $\{a, b, c\} = \{1, 2, 3\}$. Propriety at $u_i$ +and $u_{i+1}$ requires $\varphi(e_i) \notin \{\varphi(f_i), \varphi(f_{i+1})\}$, +which gives +\[ + \varphi(e_0) = c, \quad + \varphi(e_1) = a, \quad + \varphi(e_2) \in \{a, b\}, \quad + \varphi(e_3) \in \{a, b\}, \quad + \varphi(e_4) = b. +\] +The remaining propriety condition $\varphi(e_{i-1}) \neq \varphi(e_i)$ holds +automatically at $u_0, u_1, u_4$, forces $\varphi(e_2) = b$ at $u_2$, and then +forces $\varphi(e_3) = a$ at $u_3$. The resulting triples +$(\varphi(e_{i-1}), \varphi(e_i), \varphi(f_i))$ at $u_0, u_1, u_2, u_3, u_4$ +are +\[ + (b, c, a), \quad (c, a, b), \quad (a, b, c), \quad (b, a, c), \quad (a, b, c), +\] +each a permutation of $\{1, 2, 3\}$, so $\varphi$ is proper at every $u_i$. +\end{proof} + +\begin{remark} +The two-element condition $\{\varphi(f_{j+3}), \varphi(f_{j+4})\} += \{1,2,3\}\setminus\{\varphi(f_j)\}$ cannot be dropped: a 3-colouring +satisfying $\varphi(f_j) = \varphi(f_{j+1}) = \varphi(f_{j+2})$ alone need not +extend, e.g.\ $(1,1,1,1,2)$. +\end{remark} + +Since $\widehat{G}'_{v,i}$ is the dual of a triangulation on fewer vertices than +$G$, it is $3$-edge-colourable by the minimality of $G$. The following lemma +constrains every such colouring. + +\begin{lemma} +\label{lem:chord-apex} +Let $G$ be a minimal counterexample, and let $\widehat{G}'_{v,i}$ be a reduced +dual of its dual $G'$. Then in every proper $3$-edge-colouring of +$\widehat{G}'_{v,i}$, the merged edge and the spike edge receive the same +colour. +\end{lemma} + +\begin{proof} +% TODO. Intended argument: given a proper 3-edge-colouring of +% \widehat{G}'_{v,i}, the colour at each A_k of the new edge there (one of the +% three v_n-edges for k in {i, i+1, i+2}, or the chord for k in {i+3, i+4}) +% determines the colour of f_k in any lift to G' --- it is the unique colour +% not used by the two original edges at A_k. Because the chord has a single +% colour, f_{i+3} and f_{i+4} agree. If the merged edge and the spike edge +% receive distinct colours, the three v_n-edges contribute three distinct +% colours to f_i, f_{i+1}, f_{i+2}, and the resulting f-vector has shape +% (X, Y, Z, W, W) with X, Y, Z distinct and W \neq Y. By +% Lemma~\ref{lem:pentagonal-externals} applied to G' at the face F_v, this +% extends to a proper 3-edge-colouring of G' --- contradicting that G' (the +% dual of the minimal counterexample G) is not 3-edge-colourable. Hence the +% merged edge and the spike edge must share a colour. +\end{proof} + \end{document}