53 lines
1.9 KiB
TeX
53 lines
1.9 KiB
TeX
\documentclass[12pt, a4paper]{article}
|
|
|
|
%packages
|
|
\usepackage[ngerman]{babel}
|
|
\usepackage[utf8x]{inputenc}
|
|
%Formel packages
|
|
\usepackage{amsmath}
|
|
\usepackage{amsthm}
|
|
\usepackage{amsbsy}
|
|
\usepackage{amssymb}
|
|
|
|
|
|
\usepackage{enumerate}
|
|
\usepackage{geometry}
|
|
\geometry{a4paper, top=25mm, left=18mm, right=15mm, bottom=30mm,
|
|
headsep=10mm, footskip=12mm}
|
|
|
|
|
|
\begin{document}
|
|
|
|
\section*{Computer Graphics - Excercise 2}
|
|
\subsection*{3.1.1}
|
|
\subsection*{(a)}
|
|
The more obtuse the triangle, the more acute the angle of the corner of the voronoi area gets. Because the both sides of this corner has to be in a 90° angle to the sides of the traingle.
|
|
\subsection*{(b)}
|
|
The formula of triangle is: $\frac{1}{2}*h*b$
|
|
As for the red triangle: $\frac{1}{2}*h*b$
|
|
with: $h=\frac{||p_i - p_j||}{2}$ and $b=$
|
|
|
|
\subsection*{(c)}
|
|
|
|
\subsection*{3.1.2}
|
|
\subsection*{(a)}
|
|
We know every vertex($v$) has three edges($e$) and 2 faces($f$). In addition every edge($e$) can separated into two half edges ($e_h$), this means $e_h = 2*e = 6*v$.\\
|
|
If we add the memory for every part it results in a formula:\\
|
|
\begin{center}
|
|
memory $= v*16\text{bytes} + e*4\text{bytes} + e_h*16\text{bytes} + f*4\text{bytes}$\\
|
|
$\rightarrow \text{through assumptions:}$\\
|
|
memory $= v*(16+3*4+6*16+2*4)\text{bytes}=v*132\text{bytes}$
|
|
\end{center}
|
|
\subsection*{(b)}
|
|
Because in a quad mash two triangles are combined into one quad, the resulting faces will be reduced by a half. The resulting ratio will be 1:3:1 (for v : e : f)
|
|
\subsection*{(c)}
|
|
We know every vertex($v$) has three edges($e$) and one face($f$). In addition every edge($e$) can separated into two half edges ($e_h$), this means $e_h = 2*e = 6*v$.\\
|
|
If we add the memory for every part it results in a formula:\\
|
|
\begin{center}
|
|
memory $= v*16\text{bytes} + e*4\text{bytes} + e_h*16\text{bytes} + f*4\text{bytes}$\\
|
|
$\rightarrow \text{through assumptions:}$\\
|
|
memory $= v*(16+3*4+6*16+1*4)\text{bytes}=v*132\text{bytes}$
|
|
\end{center}
|
|
\end{document}
|
|
|