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Commit 4997461b authored by Vincent Nivoliers's avatar Vincent Nivoliers
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Presentation/presentation.tex
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......@@ -225,4 +225,87 @@
}
}
\frame{
\frametitle{Calcul exact}
\begin{center}
\begin{myblock}{0.8\li}
\centering
Il peut arriver que le déterminant ne soit pas nul\\
mais que l'arrondi en virgule flottante le soit.
\end{myblock}
\end{center}
\visible<2>{
\vspace{8mm}
\textbf{Calculer en précision exacte :}
\vspace{3mm}
\begin{itemize}
\item entiers sans limite de taille ;
\item nombre rationels ;
\item opérations dont le résultat est rationnel (pas de racine carrée !) ;
\item le prédicat ne doit prendre en entrée que des nombres initiaux.
\end{itemize}
}
}
\frame{
\frametitle{Généralisation de l'orientation}
\centering
\includegraphics[width=0.3\li]{Figures/generalized_orient}
\vspace{8mm}
\begin{myblock}{0.8\li}
\centering
examiner le signe de :
$
\det{
\begin{pmatrix}
x_0 & y_0 & 1 \\
x_1 & y_1 & 1 \\
x_2 & y_2 & 1
\end{pmatrix}
}
$
\end{myblock}
}
\frame{
\frametitle{Simulation de simplicité (Edelsbrnner \& Mücke 90)}
\begin{center}
\begin{myblock}{0.8\li}
Utiliser des \eps{} avec une puissance \hl{exponentielle} en $i$ et $j$ :
\begin{equation*}
\eps_{i,j} = \eps^{2^{i*d - j}}
\end{equation*}
\end{myblock}
\end{center}
\vspace{5mm}
\textbf{Principe :}
\begin{itemize}
\item $\sum_{k=0}^n 2^k = 2^{k+1} - 1$
\item $\prod_{k=0}^n\eps^{2^k} = \eps^{\sum_{k=0}^n2^k} = \eps^{2^{k+1} - 1} >> \eps^{2^{k+1}}$
\end{itemize}
}
\frame{
\frametitle{Extraire les termes}
\begin{equation*}
\begin{pmatrix}
p_{0,0} + \eps{0,0} & p_{0,1} + \eps{0,1} & \cdots & p_{0,d} + \eps{0,d-1} & 1 \\
p_{1,0} + \eps{1,0} & p_{1,1} + \eps{1,1} & \cdots & p_{1,d} + \eps{1,d-1} & 1 \\
\vdots & \vdots & \ddots & \vdots \\
p_{d,0} + \eps{d,0} & p_{d,1} + \eps{d,1} & \cdots & p_{d,d-1} + \eps{d,d-1} & 1
\end{pmatrix}
\end{equation*}
\vspace{5mm}
\begin{itemize}
\item pour chaque point aligner la perturbation voulue sur la diagonale ;
\item développement par les cofacteurs.
\end{itemize}
}
\end{document}
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