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Diffstat (limited to 'vorlesungen/slides/a/ecc/quadrieren.tex')
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diff --git a/vorlesungen/slides/a/ecc/quadrieren.tex b/vorlesungen/slides/a/ecc/quadrieren.tex new file mode 100644 index 0000000..942c73b --- /dev/null +++ b/vorlesungen/slides/a/ecc/quadrieren.tex @@ -0,0 +1,59 @@ +% +% quadrieren.tex -- Quadrieren +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\bgroup +\begin{frame}[t] +\setlength{\abovedisplayskip}{5pt} +\setlength{\belowdisplayskip}{5pt} +\frametitle{Quadrieren} +\vspace{-20pt} +\begin{columns}[t,onlytextwidth] +\begin{column}{0.40\textwidth} +\begin{block}{Problem} +\( g = g_1 = g_2 \) +$\Rightarrow$ +Tangente +\\ +\uncover<2->{{\color{red}ohne Analysis!}} +\end{block} +\begin{center} +\includegraphics[width=\textwidth]{../../buch/chapters/90-crypto/images/elliptic.pdf} +\end{center} +\end{column} +\begin{column}{0.56\textwidth} +\uncover<3->{% +\begin{block}{Lösung} +Finde $h\in G$ derart, dass +\begin{align*} +g(t) +&= +tg + (1-t)h +\\ +\uncover<4->{% +\begin{pmatrix}X(t)\\Y(t)\end{pmatrix} +&= +t\begin{pmatrix}x_g\\y_g\end{pmatrix} ++(1-t) \begin{pmatrix}x_h\\y_h\end{pmatrix} +} +\end{align*} +\uncover<5->{eingesetzt +\[ +p(t) += +Y(t)^2+X(t)Y(t)-X(t)^3-aX(t)-b += +0 +\]}% +\uncover<6->{% +Nullstellen $0$ (doppelt) und $1$ hat:} +\[ +\uncover<7->{p(t) = c(t^3-t)} +\] +\uncover<8->{Koeffizientenvergleich: einfachere Gleichungen für $x_h$ und $y_h$} +\end{block}} +\end{column} +\end{columns} +\end{frame} +\egroup |