Tetra and thesis: moved sign of denominator to thesis; finsihed lemma on...

Tetra and thesis: moved sign of denominator to thesis; finsihed lemma on adjacency of next gen, but doesnt consider possibility of smaller cycles containing three of four vertex sets

Research/Edge Intersection/Tetrahedra/mccoid2020tetrahedra.tex

@@ -1082,82 +1082,6 @@ For each of these the numerator of $\vec{h}(J | \Gamma_j) \cdot \vec{e}_j$ is th
 By this corollary, if there is a change in sign of $\vec{h}(J | \Gamma) \cdot \vec{e}_\eta$ then the entire $m$--face of $X$ defined by the indices $J$ ends up on the other side of the $(n-m)$--face of $Y$ defined by $\Gamma \cup \set{\eta}$.
 If the $J$--th $m$--face of $X$ does not have $m+1$ intersections then the signs of all existing intersections can be found using the intersections of the $(m-1)$--faces of $X$ with indices that are subsets of $J$.
 
-For the sake of notation let
-\begin{align*}
-X_J = & \begin{bmatrix} \vec{x}_{i_0} & \dots & \vec{x}_{i_m} \end{bmatrix}, \quad &
-I_\Gamma = & \begin{bmatrix} \vec{e}_{\gamma_1} & \dots & \vec{e}_{\gamma_m} \end{bmatrix}
-\end{align*}
-for $J = \set{i_j}_{j=0}^m$ and $\Gamma = \set{\gamma_j}_{j=1}^m$.
-Then $$\vec{h}(J|\Gamma) \cdot \vec{e}_\eta = \frac{\begin{vmatrix} X_J^\top \vec{e}_\eta & X_J^\top I_\Gamma \end{vmatrix} }{ \begin{vmatrix} \vec{1} & X_J^\top I_\Gamma \end{vmatrix}}.$$
-
-\begin{lemma}
-Suppose $\sign(\vec{h}(J \setminus \set{i} | \Gamma) \cdot \vec{e}_\eta) \neq \sign(\vec{h}(J \setminus \set{j} | \Gamma) \cdot \vec{e}_\eta)$ and an intersection $\vec{h}(J \setminus \set{i,j} | \Gamma \setminus \set{\gamma})$ was calculated for some $\gamma \in \Gamma$, then
-\begin{equation*}
-\sign \left ( \begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \right ) = 
-\sign \left ( \begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
-\begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} 
-\begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix} \right ).
-\end{equation*}
-\end{lemma}
-
-\begin{proof}
-Without loss of generality, suppose $\vec{x}_i$ is the first column of $X_{J \setminus \set{j}}$ and likewise $\vec{x}_j$ is the first column of $X_{J \setminus \set{i}}$.
-Furthermore, to adhere to the notation already established the first column of $I_{\Gamma \cup \set{\eta}}$ must be $\vec{e}_\eta$.
-
-By assumption $\vec{h}(J \setminus \set{i} | \Gamma) \cdot \vec{e}_\eta - \vec{h}(J \setminus \set{j} | \Gamma) \cdot \vec{e}_\eta$ has the same sign as $\vec{h}(J \setminus \set{i} | \Gamma) \cdot \vec{e}_\eta$.
-We begin by simplifying this expression:
-\begin{align*}
-\vec{h}(J \setminus \set{i} | \Gamma)\cdot \vec{e}_\eta & - \vec{h}(J \setminus \set{j} | \Gamma) \cdot \vec{e}_\eta =
-\frac{ \begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
-}{ \begin{vmatrix} \vec{1} & X_{J \setminus \set{i}}^\top I_\Gamma \end{vmatrix}} - 
-\frac{ \begin{vmatrix} X_{J \setminus \set{j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
-}{ \begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix}} \\
-= & \frac{ \begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix}
-	\begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} -
-	\begin{vmatrix} \vec{1} & X_{J \setminus \set{i}}^\top I_\Gamma \end{vmatrix}
-	\begin{vmatrix} X_{J \setminus \set{j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} }{
-	\begin{vmatrix} \vec{1} & X_{J \setminus \set{i}}^\top I_\Gamma \end{vmatrix}
-	\begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix} }.
-\end{align*}
-We expand the numerator of this expression:
-\begin{align*}
-& \left ( \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} + \sum\limits_{k=0}^{m-1} (-1)^k \vec{x}_i^\top \vec{e}_{\gamma_k} \begin{vmatrix} \vec{1} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \setminus \set{\gamma_k}} \end{vmatrix} \right ) \begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \\
-& - \left ( \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} + \sum\limits_{k=0}^{m-1} (-1)^k \vec{x}_j^\top \vec{e}_{\gamma_k} \begin{vmatrix} \vec{1} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \setminus \set{\gamma_k}} \end{vmatrix} \right ) \begin{vmatrix} X_{J \setminus \set{j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \\
-& = \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix}
-	\begin{vmatrix} 1 & \vec{x}_i^\top I_{\Gamma \cup \set{\eta}} \\ 1 & \vec{x}_j^\top I_{\Gamma \cup \set{\eta}} \\ \vec{0} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
-	+ \begin{vmatrix} \vec{x}_i^\top I_\Gamma \vec{w} & \vec{x}_i^\top I_{\Gamma \cup \set{\eta}} \\ \vec{x}_j^\top I_\Gamma \vec{w} & \vec{x}_j^\top I_{\Gamma \cup \set{\eta}} \\ \vec{0} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \\
-& = \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix}
-	\begin{vmatrix} 1 & \vec{x}_i^\top I_{\Gamma \cup \set{\eta}} \\ 1 & \vec{x}_j^\top I_{\Gamma \cup \set{\eta}} \\ \vec{0} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
-	+ \begin{vmatrix} 0 & \vec{x}_i^\top I_{\Gamma \cup \set{\eta}} \\ 0 & \vec{x}_j^\top I_{\Gamma \cup \set{\eta}} \\ -X_{J \setminus \set{i,j}}^\top I_\Gamma \vec{w} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \\
-& = \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} \begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}.
-\end{align*}
-where $w_k = (-1)^k \begin{vmatrix} \vec{1} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \setminus \set{\gamma_k}} \end{vmatrix}.$
-To prove the last equality note that
-each element of $X_{J \setminus \set{i,j}}^\top I_\Gamma \vec{w}$ is equal to the same value:
-\begin{align*}
-\left ( X_{J \setminus \set{i,j}}^\top I_\Gamma \vec{w} \right )_l = & 
-\sum_{k=0}^{m-1} (-1)^k \vec{x}_l^\top \vec{e}_{\gamma_k} \begin{vmatrix} \vec{1} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \setminus \set{\gamma_k}} \end{vmatrix} \\
-= & \begin{vmatrix} 0 & \vec{x}_l^\top I_\Gamma \\ \vec{1} & X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} \\
-= & \begin{vmatrix} -1 & \vec{0}^\top \\ \vec{1} & X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} \\
-= & - \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix}.
-\end{align*}
-The simplified expression is then
-\begin{equation*}
-\vec{h}(J \setminus \set{i} | \Gamma)\cdot \vec{e}_\eta - \vec{h}(J \setminus \set{j} | \Gamma) \cdot \vec{e}_\eta =
-	\frac{ \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix}
-	\begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} }{
-	\begin{vmatrix} \vec{1} & X_{J \setminus \set{i}}^\top I_\Gamma \end{vmatrix}
-	\begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix}} .
-\end{equation*}
-Therefore, the sign of $\begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}$ is
-\begin{equation*}
-\sign \left ( \begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \right ) = 
-\sign \left ( \begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
-\begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} 
-\begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix} \right ).
-\end{equation*}
-\end{proof}
-
 \subsection{Algorithm for the intersection of n-dimensional simplices}
 
 Let $[a..b]$ represent the set of integers between $a$ and $b$.

Research/PhD thesis/CHAP_Simplices.tex

@@ -7,6 +7,93 @@ investigates methodology for various aspects of the algorithm, including how to
 
 There are a number of practical concerns when it comes to implementing the algorithm.
 
+\subsection{Connectivity of signs}
+
+As has been explained in detail in Section (nb: reference) those intersections of a given $m$--face of $X$ and a given collection of hyperplanes of $Y$ are connected in the signs of their components.
+This connectivity is what lends the algorithm its consistency of shape.
+This connectivity was presented above through what is effectively Cramer's rule used to solve the intersections.
+The numerators and denominators then have specific connections with one another.
+However, this represents a particularly inaccurate, unstable and inefficient manner in calculating the intersections, as will be discussed in more detail further on (nb: ref?).
+It is then desirable to maintain the connectivity of the Cramer's rule representation while using better subroutines to calculate the intersections.
+
+There exists connections between the denominators of one generation of intersections and the numerators of the previous generations.
+These connections will be important to know for the total connectivity of the signs.
+For the sake of notation let
+\begin{align*}
+X_J = & \begin{bmatrix} \vec{x}_{i_0} & \dots & \vec{x}_{i_m} \end{bmatrix}, \quad &
+I_\Gamma = & \begin{bmatrix} \vec{e}_{\gamma_1} & \dots & \vec{e}_{\gamma_m} \end{bmatrix}
+\end{align*}
+for $J = \set{i_j}_{j=0}^m$ and $\Gamma = \set{\gamma_j}_{j=1}^m$.
+Then $$\vec{h}(J|\Gamma) \cdot \vec{e}_\eta = \frac{\begin{vmatrix} X_J^\top \vec{e}_\eta & X_J^\top I_\Gamma \end{vmatrix} }{ \begin{vmatrix} \vec{1} & X_J^\top I_\Gamma \end{vmatrix}}.$$
+
+\begin{lemma}
+Suppose $\sign(\vec{h}(J \setminus \set{i} | \Gamma) \cdot \vec{e}_\eta) \neq \sign(\vec{h}(J \setminus \set{j} | \Gamma) \cdot \vec{e}_\eta)$ and an intersection $\vec{h}(J \setminus \set{i,j} | \Gamma \setminus \set{\gamma})$ was calculated for some $\gamma \in \Gamma$, then
+\begin{equation*}
+\sign \left ( \begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \right ) = 
+\sign \left ( \begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
+\begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} 
+\begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix} \right ).
+\end{equation*}
+\end{lemma}
+
+\begin{proof}
+Without loss of generality, suppose $\vec{x}_i$ is the first column of $X_{J \setminus \set{j}}$ and likewise $\vec{x}_j$ is the first column of $X_{J \setminus \set{i}}$.
+Furthermore, to adhere to the notation already established the first column of $I_{\Gamma \cup \set{\eta}}$ must be $\vec{e}_\eta$.
+
+By assumption $\vec{h}(J \setminus \set{i} | \Gamma) \cdot \vec{e}_\eta - \vec{h}(J \setminus \set{j} | \Gamma) \cdot \vec{e}_\eta$ has the same sign as $\vec{h}(J \setminus \set{i} | \Gamma) \cdot \vec{e}_\eta$.
+We begin by simplifying this expression:
+\begin{align*}
+\vec{h}(J \setminus \set{i} | \Gamma)\cdot \vec{e}_\eta & - \vec{h}(J \setminus \set{j} | \Gamma) \cdot \vec{e}_\eta =
+\frac{ \begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
+}{ \begin{vmatrix} \vec{1} & X_{J \setminus \set{i}}^\top I_\Gamma \end{vmatrix}} - 
+\frac{ \begin{vmatrix} X_{J \setminus \set{j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
+}{ \begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix}} \\
+= & \frac{ \begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix}
+	\begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} -
+	\begin{vmatrix} \vec{1} & X_{J \setminus \set{i}}^\top I_\Gamma \end{vmatrix}
+	\begin{vmatrix} X_{J \setminus \set{j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} }{
+	\begin{vmatrix} \vec{1} & X_{J \setminus \set{i}}^\top I_\Gamma \end{vmatrix}
+	\begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix} }.
+\end{align*}
+We expand the numerator of this expression:
+\begin{align*}
+& \left ( \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} + \sum\limits_{k=0}^{m-1} (-1)^k \vec{x}_i^\top \vec{e}_{\gamma_k} \begin{vmatrix} \vec{1} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \setminus \set{\gamma_k}} \end{vmatrix} \right ) \begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \\
+& - \left ( \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} + \sum\limits_{k=0}^{m-1} (-1)^k \vec{x}_j^\top \vec{e}_{\gamma_k} \begin{vmatrix} \vec{1} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \setminus \set{\gamma_k}} \end{vmatrix} \right ) \begin{vmatrix} X_{J \setminus \set{j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \\
+& = \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix}
+	\begin{vmatrix} 1 & \vec{x}_i^\top I_{\Gamma \cup \set{\eta}} \\ 1 & \vec{x}_j^\top I_{\Gamma \cup \set{\eta}} \\ \vec{0} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
+	+ \begin{vmatrix} \vec{x}_i^\top I_\Gamma \vec{w} & \vec{x}_i^\top I_{\Gamma \cup \set{\eta}} \\ \vec{x}_j^\top I_\Gamma \vec{w} & \vec{x}_j^\top I_{\Gamma \cup \set{\eta}} \\ \vec{0} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \\
+& = \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix}
+	\begin{vmatrix} 1 & \vec{x}_i^\top I_{\Gamma \cup \set{\eta}} \\ 1 & \vec{x}_j^\top I_{\Gamma \cup \set{\eta}} \\ \vec{0} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
+	+ \begin{vmatrix} 0 & \vec{x}_i^\top I_{\Gamma \cup \set{\eta}} \\ 0 & \vec{x}_j^\top I_{\Gamma \cup \set{\eta}} \\ -X_{J \setminus \set{i,j}}^\top I_\Gamma \vec{w} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \\
+& = \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} \begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}.
+\end{align*}
+where $w_k = (-1)^k \begin{vmatrix} \vec{1} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \setminus \set{\gamma_k}} \end{vmatrix}.$
+To prove the last equality note that
+each element of $X_{J \setminus \set{i,j}}^\top I_\Gamma \vec{w}$ is equal to the same value:
+\begin{align*}
+\left ( X_{J \setminus \set{i,j}}^\top I_\Gamma \vec{w} \right )_l = & 
+\sum_{k=0}^{m-1} (-1)^k \vec{x}_l^\top \vec{e}_{\gamma_k} \begin{vmatrix} \vec{1} & X_{J \setminus \set{i,j}}^\top I_{\Gamma \setminus \set{\gamma_k}} \end{vmatrix} \\
+= & \begin{vmatrix} 0 & \vec{x}_l^\top I_\Gamma \\ \vec{1} & X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} \\
+= & \begin{vmatrix} -1 & \vec{0}^\top \\ \vec{1} & X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} \\
+= & - \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix}.
+\end{align*}
+The simplified expression is then
+\begin{equation*}
+\vec{h}(J \setminus \set{i} | \Gamma)\cdot \vec{e}_\eta - \vec{h}(J \setminus \set{j} | \Gamma) \cdot \vec{e}_\eta =
+	\frac{ \begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix}
+	\begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} }{
+	\begin{vmatrix} \vec{1} & X_{J \setminus \set{i}}^\top I_\Gamma \end{vmatrix}
+	\begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix}} .
+\end{equation*}
+Therefore, the sign of $\begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}$ is
+\begin{equation*}
+\sign \left ( \begin{vmatrix} \vec{1} & X_J^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix} \right ) = 
+\sign \left ( \begin{vmatrix} X_{J \setminus \set{i}}^\top I_{\Gamma \cup \set{\eta}} \end{vmatrix}
+\begin{vmatrix} X_{J \setminus \set{i,j}}^\top I_\Gamma \end{vmatrix} 
+\begin{vmatrix} \vec{1} & X_{J \setminus \set{j}}^\top I_\Gamma \end{vmatrix} \right ).
+\end{equation*}
+\end{proof}
+
 \subsection{Adjacent intersections}
 
 Assuming that a set of intersections for a given collection $\Gamma$ of hyperplanes forms a convex object,
@@ -39,13 +126,19 @@ Therefore, any other child of either parent, called a sibling, will be adjacent
 It remains to determine if any other intersections of this generation are adjacent.
 
 \begin{lemma}
-idk something
+Let a father be a given parent of a child node, and let the mother of this child be the other parent.
+Two children of a given generation of intersections are adjacent under one of three circumstances:
+\begin{description}
+\item[siblings] the children share exactly one parent;
+\item[cousins] the two fathers are adjacent as are the two mothers, ie. the four parents form a 4-cycle;
+\item[second cousins] the two fathers are adjacent while the two mothers are both adjacent to a fifth intersection, ie. the four parents are part of a 5-cycle, and all four parents share all but exactly three vertices.
+\end{description}
 \end{lemma}
 
 \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 at least $n$ edges between nodes $J_1$ and $J_2$ and at least $m$ 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 $J_2$ that do not pass through $J_3$ and at least $m$ edges between $J_0$ and $J_3$ that do not pass through $J_2$.
 Without loss of generality $n \leq m$.
 \begin{figure}
 \centering
@@ -122,11 +215,21 @@ This means that either $J_3$ contains neither $J_0 \setminus J_1$ nor $J_1 \setm
 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.
 
+The configuration with $n=1$ and $m=2$ necessarily places the four nodes into a 5-cycle.
+The arrangements of vertex sets that adhere to this condition are limited.
+If the five vertex sets share all but four or more vertices then all five are adjacent.
+The same is true if they share all but one.
+The two cases are then when the five sets share all but two or three vertices.
+
 \newcommand{\PentaConfig}{
-	\foreach \x in {1,2,3,4,5} {
+	\foreach \x in {0,1,...,4} {
 		\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;
+	\draw[thick,black] (P1) \foreach \x in {2,3,4,0} {-- (P\x)} -- cycle;
+	\foreach \x in {1,2,3,4,0} {
+		\tikzmath{\y = int(mod(\x+1,5));}
+		\coordinate (E\x) at ($ (P\x) !.5! (P\y) $);}
+	\draw[thick,dotted] (E0) \foreach \x in {1,2,...,4} { -- (E\x)} -- cycle;
 	}
 \begin{figure}
 \centering
@@ -137,20 +240,68 @@ This latter case therefore allows two children to be adjacent for $n=1$, which w
 			\node[above] at (P2) {$jk$};
 			\node[right] at (P3) {$kl$};
 			\node[below] at (P4) {$lm$};
-			\node[left] at (P5) {$mi$}; &
+			\node[left] at (P0) {$mi$}; &
 			\PentaConfig 
+			\foreach \x in {0,1,...,4} {
+				\tikzmath{\y = int(mod(\x+2,5));}
+				\draw[thick,dotted] (E\x) -- (E\y);}
 			\node[above] at (P1) {$ijk$};
 			\node[above] at (P2) {$jkl$};
-			\node[left] at (P5) {$ijm$};
 			\node[right] at (P3) {$klm$};
-			\node[below] at (P4) {$ilm$}; \\};
+			\node[below] at (P4) {$ilm$};
+			\node[left] at (P0) {$ijm$}; \\};
 	\end{tikzpicture}
 	\caption{The two possible cases of second cousins.
 		Left: only siblings are adjacent.
 		Right: all children are adjacent.}
+	\label{fig:second cousins}
 \end{figure}
+Figure \ref{fig:second cousins} shows these two scenarios.
+To construct them, one may begin at any node with two vertices.
+Travel along one edge to the second node by changing one of the vertices.
+Travel along the other edge off of the first node by changing a different vertex so that this third node is not adjacent to the second.
+For example, in the case where the vertex sets share all but two vertices one can start with the set $\set{i,j}$.
+The adjacent sets are then $\set{j,k}$ and $\set{m,i}$.
+The final two nodes are adjacent to one another as well as the second and third nodes, but not the first.
+This uniquely determines the sets.
+
+As shown in Figure \ref{fig:second cousins} the children of the case where all but two vertices are shared are adjacent only if they are siblings.
+Take $J_0 = \set{i,j}$ and $J_3 = \set{l,m}$.
+The set $J_3$ clearly does not contain $i$ or $k$, the vertices of $J_0 \setminus J_1$ and $J_1 \setminus J_0$, respectively.
+The children $\set{k,l,m}$ and $\set{i,j,k}$ differ by two vertices and therefore are not adjacent.
+
+Meanwhile, all children are adjacent when the vertex sets share all but three vertices.
+Using $J_0 = \set{i,j,k}$ and $J_3 = \set{i,l,m}$ both $i$ and $l$ are within $J_3$.
+The children $\set{i,k,l,m}$ and $\set{i,j,k,l}$ differ by only one vertex and are adjacent.
+By symmetry the same is true of the other children around this cycle.
+
+
 \end{proof}
 
+As a corollary to this lemma it is impossible to determine $A(\Gamma \cup \set{\eta})$ from $A(\Gamma)$ alone.
+This is because the case of second cousins requires knowledge of the shared vertices of the vertex sets of the parents.
+Therefore, we cannot completely eliminate the comparison of vertex sets when computing intersections of the next generation.
+
+For each intersection in $H(\Gamma)$ compare its sign in the $\vec{e}_\eta$ direction with each adjacent intersection as identified in $A(\Gamma)$.
+If the sign is different calculate an intersection in $H(\Gamma \cup \set{\eta})$.
+Add a column to $B(\Gamma \cup \set{\eta})$, the adjacency matrix between $H(\Gamma)$ and $H(\Gamma \cup \set{\eta})$.
+There are two non-zero entries in this column, one for each parent.
+The matrix $B(\Gamma \cup \set{\eta})$ must be formed before forming $A(\Gamma \cup \set{\eta})$.
+
+The matrix $A(\Gamma \cup \set{\eta})$ must be formed in three parts, one part each for siblings, cousins and second cousins.
+The sibling part is formed easily.
+Each column of $B(\Gamma \cup \set{\eta})$ is associated with a child in $H(\Gamma \cup \set{\eta})$.
+For each of these children the associated row of $A(\Gamma \cup \set{\eta})$ is the `xor' combination of the two rows of $B(\Gamma \cup \set{\eta})$ associated with the child's parents.
+Each parent has a limited number of children (nb: prove exactly how much?) and this `xor' combination need only be done for twice this number, with all other entries being zero.
+
+
+
+%---Efficiency comparison---%
+Based on Proposition (nb: check ref on this) the number of intersections can grow quadratically with the generations.
+For $n$--simplices there are up to $n-1$ generations, meaning at the last generation there can, in theory, be up to $(n/2)^{2(n-1)}$ interesctions.
+If using a brute force method to determine adjacency between intersections this will require $(n/2)^{2(n-1)} * (n+1)$ comparisons.
+%---end---%
+
 Take the graph of the set of intersections $\set{\vec{h}(J | \Gamma)}_J$ with edges defined by its adjacency matrix $A$.
 Place the nodes of this graph into two groups, one with non-negative values along the $\vec{e}_\gamma$ direction and one with negative values.
 Remove the intragroup edges.

Research/PhD thesis/OUT_mccoid_v1_20220102.tex

@@ -8,7 +8,7 @@
 \usepackage{subcaption}
 \usepackage{float}
 \usepackage{tikz}
-\usetikzlibrary{decorations.pathreplacing,positioning,calc,intersections,3d,shapes.geometric}
+\usetikzlibrary{decorations.pathreplacing,positioning,calc,intersections,3d,shapes.geometric,math}
 
 \newcommand{\dxdy}[2]{\frac{d #1}{d #2}}
 \newcommand{\dxdyk}[3]{\frac{d^{#3} #1}{d {#2}^{#3}}}