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everything in english

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Martin Thoma 2013-09-01 11:43:30 +02:00
parent 98e1d1955c
commit 93cd7986b9
2 changed files with 16 additions and 27 deletions
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source-code/Pseudocode/Calculate-Legendre/Calculate-Legendre.png
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source-code/Pseudocode/Calculate-Legendre/Calculate-Legendre.tex
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@ -9,49 +9,38 @@
\usepackage{braket} % needed for \Set
\usepackage{algorithm,algpseudocode}
\usepackage{tikz}
\usetikzlibrary{decorations.pathreplacing,calc}
\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};}
\newcommand*{\AddNote}[4]{%
\begin{tikzpicture}[overlay, remember picture]
\draw [decoration={brace,amplitude=0.5em},decorate,very thick]
($(#3)!(#1.north)!($(#3)-(0,1)$)$) --
($(#3)!(#2.south)!($(#3)-(0,1)$)$)
node [align=center, text width=2.5cm, pos=0.5, anchor=west] {#4};
\end{tikzpicture}
}%
\begin{document}
\begin{preview}
\begin{algorithm}[H]
\begin{algorithmic}
\Require $p \in \mathbb{P}, a \in \mathbb{Z}, p \geq 3$
\If{$a \geq p$ or $a < 0$}\Comment{Regel (III)}
\State \Return $\Call{CalculateLegendre}{a \mod p, p}$ \Comment{nun: $a \in [0, \dots, p-1]$}
\Require $p \in \mathbb{P}, a \in \mathbb{Z}, p \geq 3$
\Procedure{CalculateLegendre}{$a$, $p$}
\If{$a \geq p$ or $a < 0$}\Comment{rule (III)}
\State \Return $\Call{CalculateLegendre}{a \mod p, p}$ \Comment{now: $a \in [0, \dots, p-1]$}
\ElsIf{$a == 0$ or $a == 1$}
\State \Return $a$ \Comment{nun: $a \in [2, \dots, p-1]$}
\ElsIf{$a == 2$} \Comment{Regel (VII)}
\State \Return $a$ \Comment{now: $a \in [2, \dots, p-1]$}
\ElsIf{$a == 2$} \Comment{rule (VII)}
\If{$a \equiv \pm 1 \mod 8$}
\State \Return 1
\Else
\State \Return -1
\EndIf \Comment{nun: $a \in [3, \dots, p-1]$}
\ElsIf{$a == p-1$} \Comment{Regel (VI)}
\EndIf \Comment{now: $a \in [3, \dots, p-1]$}
\ElsIf{$a == p-1$} \Comment{rule (VI)}
\If{$p \equiv 1 \mod 4$}
\State \Return 1
\Else
\State \Return -1
\EndIf \Comment{nun: $a \in [3, \dots, p-2]$}
\ElsIf{!$\Call{isPrime}{a}$} \Comment{Regel (II)}
\State $p_1, p_2, \dots, p_n \gets \Call{Faktorisiere}{a}$
\State \Return $\prod_{i=1}^n \Call{CalculateLegendre}{p_i, p}$ \Comment{nun: $a \in \mathbb{P}, \sqrt{p-2} \geq a \geq 3$}
\Else
\EndIf \Comment{now: $a \in [3, \dots, p-2]$}
\ElsIf{!$\Call{isPrime}{a}$} \Comment{rule (II)}
\State $p_1, p_2, \dots, p_n \gets \Call{Factorize}{a}$
\State \Return $\prod_{i=1}^n \Call{CalculateLegendre}{p_i, p}$
\Else \Comment{now: $a \in \mathbb{P}, \sqrt{p-2} \geq a \geq 3$}
\State \Return $(-1) \cdot \Call{CalculateLegendre}{p, a}$
\EndIf
\EndProcedure
\end{algorithmic}
\caption{Calculate Legendre-Symbol}
%\AddNote{top}{bottom}{right}{calclulate $p$ such that: $b^p \leq Z < b^{p+1}$} %\tikzmark{top},\tikzmark{right},\tikzmark{bottom}
\label{alg:euclidBaseTransformation}
\caption{Calculate Legendre symbol}
\label{alg:calculateLegendreSymbol}
\end{algorithm}
\end{preview}
\end{document}