blob: 366a7a6292b8c0b62135b3c801400245dc298890 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
|
%
% euklidmatrix.tex
%
% (c) 2021 Prof Dr Andreas Müller, OST Ostscheizer Fachhochschule
%
\bgroup
\definecolor{darkgreen}{rgb}{0,0.6,0}
\begin{frame}[t]
\frametitle{Euklidischer Algorithmus: Beispiel}
\setlength{\abovedisplayskip}{0pt}
\setlength{\belowdisplayskip}{0pt}
\vspace{-0pt}
\begin{block}{Finde $\operatorname{ggT}(25,15)$}
\vspace{-12pt}
\begin{align*}
a_0&=25 & b_0 &= 15 &\uncover<2->{25&=15 \cdot {\color{orange} 1} + 10 &q_0 &= {\color{orange}1} & r_0 &= 10}\\
\uncover<3->{a_1&=15 & b_1 &= 10}&\uncover<4->{15&=10 \cdot {\color{darkgreen}1} + \phantom{0}5 &q_1 &= {\color{darkgreen}1} & r_1 &= \phantom{0}5}\\
\uncover<5->{a_2&=10 & b_2 &= \phantom{0}5}&\uncover<6->{10&=\phantom{0}5 \cdot {\color{blue} 2} + \phantom{0}0 &q_2 &= {\color{blue}2} & r_2 &= \phantom{0}0 }
\end{align*}
\end{block}
\vspace{-5pt}
\uncover<7->{%
\begin{block}{Matrix-Operationen}
\begin{align*}
Q
&=
\uncover<9->{Q({\color{blue}2})}
\uncover<8->{Q({\color{darkgreen}1})}
Q({\color{orange}1})
=
\uncover<9->{
\begin{pmatrix*}[r]0&1\\1&-{\color{blue}2}\end{pmatrix*}
}
\uncover<8->{
\begin{pmatrix*}[r]0&1\\1&-{\color{darkgreen}1}\end{pmatrix*}
}
\begin{pmatrix*}[r]0&1\\1&-{\color{orange}1}\end{pmatrix*}
=
\ifthenelse{\boolean{presentation}}{
\only<7>{
\begin{pmatrix*}[r]\phantom{-}0&1\\1&-1\end{pmatrix*}
}
\only<8>{
\begin{pmatrix*}[r]
1&-1\\-1&2
\end{pmatrix*}
}
}{}
\only<9->{
\begin{pmatrix*}[r]
{\color{red}-1}&{\color{red}2}\\3&-5
\end{pmatrix*}}
\end{align*}
\end{block}}
\vspace{-5pt}
\uncover<10->{%
\begin{block}{Relationen ablesen}
\[
\begin{pmatrix}
\operatorname{ggT}(a,b)\\0
\end{pmatrix}
=
Q
\begin{pmatrix}a\\b\end{pmatrix}
\uncover<11->{%
\quad
\Rightarrow\quad
\left\{
\begin{aligned}
\operatorname{ggT}({\usebeamercolor[fg]{title}25},{\usebeamercolor[fg]{title}15}) &= 5 =
{\color{red}-1}\cdot {\usebeamercolor[fg]{title}25} + {\color{red}2}\cdot {\usebeamercolor[fg]{title}15} \\
0 &= \phantom{5=-}3\cdot {\usebeamercolor[fg]{title}25} -5\cdot {\usebeamercolor[fg]{title}15}
\end{aligned}
\right.}
\]
\end{block}}
\end{frame}
|
