53 lines
1.9 KiB
TeX
53 lines
1.9 KiB
TeX
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\documentclass[12pt, a4paper]{article}
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%packages
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\usepackage[ngerman]{babel}
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\usepackage[utf8x]{inputenc}
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%Formel packages
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\usepackage{amsmath}
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\usepackage{amsthm}
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\usepackage{amsbsy}
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\usepackage{amssymb}
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\usepackage{enumerate}
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\usepackage{geometry}
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\geometry{a4paper, top=25mm, left=18mm, right=15mm, bottom=30mm,
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headsep=10mm, footskip=12mm}
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\begin{document}
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\section*{Computer Graphics - Excercise 2}
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\subsection*{3.1.1}
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\subsection*{(a)}
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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.
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\subsection*{(b)}
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The formula of triangle is: $\frac{1}{2}*h*b$
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As for the red triangle: $\frac{1}{2}*h*b$
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with: $h=\frac{||p_i - p_j||}{2}$ and $b=$
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\subsection*{(c)}
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\subsection*{3.1.2}
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\subsection*{(a)}
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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 $e_h = 2*e = 6*v$.\\
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If we add the memory for every part it results in a formula:\\
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\begin{center}
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memory $= v*16\text{bytes} + e*4\text{bytes} + e_h*16\text{bytes} + f*4\text{bytes}$\\
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$\rightarrow \text{through assumptions:}$\\
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memory $= v*(16+3*4+6*16+2*4)\text{bytes}=v*132\text{bytes}$
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\end{center}
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\subsection*{(b)}
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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)
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\subsection*{(c)}
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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 $e_h = 2*e = 6*v$.\\
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If we add the memory for every part it results in a formula:\\
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\begin{center}
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memory $= v*16\text{bytes} + e*4\text{bytes} + e_h*16\text{bytes} + f*4\text{bytes}$\\
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$\rightarrow \text{through assumptions:}$\\
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memory $= v*(16+3*4+6*16+1*4)\text{bytes}=v*132\text{bytes}$
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\end{center}
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\end{document}
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