Commit 3db4a39a authored by Conor McCoid's avatar Conor McCoid
Browse files

Tetra: removed more general pairs proof, now only refers to specific case of two groups

parent 16c4ed76
......@@ -61,11 +61,11 @@ Note that the $\sign(p)$ function used here is defined as
\item[Step 1: Change of coordinates.] Find an affine transformation such that the three vertices of $V$ are mapped to (0,0), (1,0) and (0,1), the vertices of a reference triangle $Y$.
Use this transformation to map $U$ to the triangle $X$.
\item[Step 2: Select reference line.] Choose a reference line of the reference triangle $Y$.
Apply another affine transformation (usually trivial) to the vertices of $X$ such that the edge of $Y$ lies on $\Set{p,q}{q \in [0,1]}$ and $Y \in \Set{p,q}{p \geq 0}$.
Apply another affine transformation (usually trivial) to the vertices of $X$ such that the edge of $Y$ lies on $\Set{p,q}{q \in [0,1]}$ and $Y \subset \Set{p,q}{p \geq 0}$.
The $i$--th vertex of $X$ has coordinates $(p_i,q_i)$.
\begin{description}
\item[2a: Intersections.] Test if $\sign(p_i) \neq \sign(p_j)$.
If so, calculate the intersection with the reference line and test if it lies on the reference triangle.
If so, calculate the intersection with the reference line and test if it lies on the edge of the reference triangle.
Repeat this step for all three pairs of vertices of $X$.
At most two intersections are found for each reference line, $q_0^1$ and $q_0^2$.
One may remove duplicates at this stage but it is not necessary.
......@@ -146,6 +146,7 @@ For example, the edge between the $i$--th and $j$--th vertices of $X$ intersects
The case where $\pxi{i}=0$ is considered in (nb: self-cite) and will be briefly summarized here.
Moving $\vec{x}_i$ an imperceptible distance into $Y$ does not change the shape of the polyhedron of intersection.
Thus, the degenerate case where $\pxi{i}=0$ can be treated as the non-degenerate case where $\pxi{i}=\epsilon/2$.
It is therefore practical to use the binary-valued sign function previously defined.
\newcommand{\pairs}{\text{pairs}}
......@@ -157,12 +158,11 @@ Only 0, 3 or 4 intersections may occur between the edges of $X$ and the plane $P
For an intersection to exist, $\spi{i}$ and $\spi{j}$ must disagree.
There are four $\pxi{i}$ ($i=1,...,4$), and $\spi{i}$ may take one of two values.
There are only three ways to partition four objects ($\pxi{i}$) into two groups (either 0 or 1), which may be proven by the partition function.
These partitionings are listed in Table \ref{tab:partition}.
These partitionings are listed in Table \ref{tab:partition}, where $m(a)$ and $m(b)$ are the multiplicities of elements labelled $a$ and $b$, respectively.
\begin{table}
\centering
\begin{tabular}{c|c|c}
$m_A(a)$ & $m_A(b)$ & $\pairs(A)$ \\ \hline
$m(a)$ & $m(b)$ & pairs \\ \hline
4 & 0 & 0 \\
3 & 1 & 3 \\
2 & 2 & 4 \\
......@@ -170,15 +170,8 @@ These partitionings are listed in Table \ref{tab:partition}.
\caption{Ways to partition four elements into two parts.}
\label{tab:partition}
\end{table}
The number of pairs of distinct elements of a multiset is equal to the sum of the products of the multiplicities of two of the elements of the multiset.
That is, if $A=\set{a_1,...,a_1,a_2,...,a_n}$ then the number of pairs of distinct elements of $A$ is equal to:
\begin{equation} \label{eq:pairs}
\pairs(A) = \sum_{i<j}^n m_A(a_i) m_A(a_j).
\end{equation}
(nb: prove in appendix?)
Since there are only two types of objects (whether $\spi{i}=0$ or 1) this reduces to multiplying the numbers in each group together.
The result is listed in the last column of Table \ref{tab:partition}, and represents the number of intersections calculated.
A pair is formed by taking one element of each group.
The number of pairs is then the product of the two multiplicities.
\end{proof}
This proposition tells us that the part of $X$ that intersects the plane of $Y$ is a triangle, a quadrilateral, or does not exist.
......@@ -995,12 +988,12 @@ Let the hyperplane $P$ be defined by $p=0$ for some linear function $p$.
An $n$--simplex has $n+1$ vertices.
Each vertex has a value of $\sign(p_i)$ equal to either 0 or 1.
There are $\lceil{(n+1)/2}\rceil$ ways to partition $n+1$ objects into two groups.
Those partitionings with the largest and smallest value of $\pairs(A)$ are listed in Table \ref{tab:partition all}.
Those partitionings with the largest and smallest number of pairs are listed in Table \ref{tab:partition all}.
As it is assumed that $X$ intersects $P$ it must be that the number of edges that intersect $P$ is between $n$ and $\lceil{(n+1)/2}\rceil \lfloor{(n+1)/2}\rfloor$.
\begin{table}
\centering
\begin{tabular}{c|c|c}
$m_A(a)$ & $m_A(b)$ & $\pairs(A)$ \\ \hline
$m(a)$ & $m(b)$ & pairs \\ \hline
$n+1$ & 0 & 0 \\
$n$ & 1 & $n$ \\
$\vdots$ & $\vdots$ & $\vdots$ \\
......
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