Commit f4d59d15 by Conor McCoid

### Thesis: nearly finished proof for section on adjacent intersections

parent 177a5331
 ... @@ -45,8 +45,8 @@ idk something ... @@ -45,8 +45,8 @@ idk something \begin{proof} \begin{proof} Consider four nodes of $H(\Gamma)$, indexed by $J_0$, $J_1$, $J_2$ and $J_3$. Consider four nodes of $H(\Gamma)$, indexed by $J_0$, $J_1$, $J_2$ and $J_3$. Suppose $J_0$ and $J_1$ are adjacent, as are $J_2$ and $J_3$. Suppose $J_0$ and $J_1$ are adjacent, as are $J_2$ and $J_3$. These intersections may be configured as seen in Figure \ref{fig:config four}, where there are $n$ edges between nodes $J_1$ and $J_2$ and $m$ edges between $J_0$ and $J_3$. These intersections may be configured as seen in Figure \ref{fig:config four}, where there are at least $n$ edges between nodes $J_1$ and $J_2$ and at least $m$ edges between $J_0$ and $J_3$. Without loss of generality suppose $n+2 \leq m$. Without loss of generality $n \leq m$. \begin{figure} \begin{figure} \centering \centering \begin{tikzpicture} \begin{tikzpicture} ... @@ -64,7 +64,91 @@ Suppose that the edge between $J_0$ and $J_1$ belongs to $A_b(\Gamma)$, as does ... @@ -64,7 +64,91 @@ Suppose that the edge between $J_0$ and $J_1$ belongs to $A_b(\Gamma)$, as does Then nodes of $H(\Gamma \cup \set{\eta})$ will be calculated for $J_0 \cup J_1$ and $J_2 \cup J_3$. Then nodes of $H(\Gamma \cup \set{\eta})$ will be calculated for $J_0 \cup J_1$ and $J_2 \cup J_3$. These two nodes will be adjacent if $|(J_0 \cup J_1) \cap (J_2 \cup J_3)| = |J_0 \cup J_1|-1$. These two nodes will be adjacent if $|(J_0 \cup J_1) \cap (J_2 \cup J_3)| = |J_0 \cup J_1|-1$. Note that if $0 < n = |J_1|$ then $J_2$ has no indices in common with $J_1$. Likewise, $J_0$ has no indices in common with $J_3$. The two nodes of the next generation therefore cannot be adjacent. Since $J_1$ and $J_2$ are separated by $n$ edges $|J_1 \cap J_2| = |J_1| - n$. The nodes $J_0$ and $J_2$ are separated by $n+1$ edges, and so $|J_0 \cap J_2| = |J_1| - (n+1)$. Therefore, $J_2$ contains $J_1 \setminus J_0$ but does not contain any of $J_0 \setminus J_1$ and $(J_0 \cup J_1) \cap J_2 = J_1 \cap J_2$. Likewise, $|J_1 \cap J_3| = |J_1| - (n+1)$ and $J_1$ contains none of $J_3 \setminus J_2$. The nodes $J_0$ and $J_3$ are separated by the smaller of $n+2$ and $m$ edges. Since $m \geq n$, this means that $|J_1| - (n+2) \leq |J_0 \cap J_3| \leq |J_1| - n$. We consider each of the three possibilities in turn. \newcommand{\PolyConfig}[1]{ \foreach \x in {0,1,...,#1} { \coordinate (P\x) at (90-\x*360/#1+180/#1:1);} \foreach \x in {0,1} {\filldraw[black] (P\x) circle (1pt) node[above] {$J_\x$};} \foreach \x in {2,3} {\filldraw[black] (P\x) circle (1pt) node[below] {$J_\x$};} \foreach \x in {4,...,#1} {\filldraw[black] (P\x) circle (1pt);} \draw[thick,black] (P0) -- (P1); \draw[thick,black] (P2) -- (P3); \draw[thick,dotted] (P1) -- node[right] {$n$} (P2); \draw[thick,dotted] (P4) -- node[left] {$n$} (P3); } \begin{figure} \centering \begin{tikzpicture} \matrix[row sep=1em,column sep=1em]{ \PolyConfig{4} & \PolyConfig{5} \draw[thick,dashed] (P0) -- (P4); & \PolyConfig{6} \draw[thick,dashed] (P0) -- (P5) -- (P4); \\ }; \end{tikzpicture} \caption{The three choices of $m$ in Figure \ref{fig:config four}: $m=n$ (left); $m=n+1$ (centre) and; $m \geq n+2$ (right).} \end{figure} If $|J_0 \cap J_3| = |J_1| - (n+2)$ then $J_3$ contains no indices of $J_0 \setminus J_1$ and $(J_0 \cup J_1) \cap J_3 = J_1 \cap J_3$. Moreover, $J_1 \cap (J_2 \cup J_3) = J_1 \cap J_2$ and $|(J_0 \cup J_1) \cap (J_2 \cup J_3)| = |J_1| - n = |J_0 \cup J_1| - n - 1$. The two children must then be siblings if they are to be adjacent. If $|J_0 \cap J_3| = |J_1| - n$ then, by symmetry, $J_3$ contains $J_0 \setminus J_1$ but not $J_1 \setminus J_0$. Therefore, $J_3 \setminus J_2 = J_0 \setminus J_1$. Put another way, the transition from $J_2$ to $J_3$ is the inverse of the transition from $J_0$ to $J_1$. This means that \begin{align*} (J_0 \cup J_1) \cap (J_2 \cup J_3) = & ( (J_0 \setminus J_1) \cup J_1 ) \cap (J_2 \cup (J_0 \setminus J_1) ) \\ = & ( (J_0 \setminus J_1) \cap J_2 ) \cup (J_1 \cap J_2) \cup (J_1 \cap (J_0 \setminus J_1) ) \cup (J_0 \setminus J_1) \\ = & (J_1 \cap J_2) \cup (J_0 \setminus J_1) \end{align*} which has cardinality $|J_1| - n + 1 = |J_0 \cup J_1| - n$. The two children will then be adjacent for $n=1$, which will be referred to as cousins. If $|J_0 \cap J_3| = |J_1| - (n+1)$ then $J_3$ contains as many indices of $J_0$ as it does $J_1$. This means that either $J_3$ contains neither $J_0 \setminus J_1$ nor $J_1 \setminus J_0$, or it contains both. The former case has cardinality identical to when $|J_0 \cap J_3| = |J_1| - (n+2)$, while the latter case to when it equals $|J_1| - n$. This latter case therefore allows two children to be adjacent for $n=1$, which will be referred to as second cousins. \newcommand{\PentaConfig}{ \foreach \x in {1,2,3,4,5} { \coordinate (P\x) at (198-\x*360/5:1); \filldraw (P\x) circle (1pt);} \draw[thick,black] (P1) \foreach \x in {2,3,4,5} {-- (P\x)} -- cycle; } \begin{figure} \centering \begin{tikzpicture} \matrix[row sep=1em,column sep=1em]{ \PentaConfig \node[above] at (P1) {$ij$}; \node[above] at (P2) {$jk$}; \node[right] at (P3) {$kl$}; \node[below] at (P4) {$lm$}; \node[left] at (P5) {$mi$}; & \PentaConfig \node[above] at (P1) {$ijk$}; \node[above] at (P2) {$jkl$}; \node[left] at (P5) {$ijm$}; \node[right] at (P3) {$klm$}; \node[below] at (P4) {$ilm$}; \\}; \end{tikzpicture} \caption{The two possible cases of second cousins. Left: only siblings are adjacent. Right: all children are adjacent.} \end{figure} \end{proof} \end{proof} Take the graph of the set of intersections $\set{\vec{h}(J | \Gamma)}_J$ with edges defined by its adjacency matrix $A$. Take the graph of the set of intersections $\set{\vec{h}(J | \Gamma)}_J$ with edges defined by its adjacency matrix $A$. ... ...
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