Commit f4d59d15 authored by Conor McCoid's avatar Conor McCoid
Browse files

Thesis: nearly finished proof for section on adjacent intersections

parent 177a5331
......@@ -45,8 +45,8 @@ idk something
\begin{proof}
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 $J_3$.
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 $J_3$.
Without loss of generality suppose $n+2 \leq m$.
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 $n \leq m$.
\begin{figure}
\centering
\begin{tikzpicture}
......@@ -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 $J_2 \cup J_3$.
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}
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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