[Tex/LaTex] Option in the algorithm with Latex

algorithmicalgorithms

This is my algorithm:

 \usepackage{algorithm}
 \usepackage{algorithmic}
 \begin{document}
 \begin{algorithm}[H]
 \caption{My algo}
 \begin{algorithmic}[1]
 \STATE instruction 1
 \STATE instruction 2
 \end{algorithmic}
 \end{algorithm}
 \end{document}

How can I specify which instruction have applied this option: [1] ?
enter image description here

I'd like to delete the number before "Début" and before "Fin", I'd like also add spacing before "ch" and "po"

Best Answer

The optional argument to algorithmic ([1] in this case), provides the line numbered at every line. A numbering modulo 2 would be obtained using [2], and so forth.

Here's a slight variation of Is it possible to have connecting loop lines (like algorithm2e) in algorithmic? - using algpseudocode (from algorithmicx) - to insert Début and Fin as keywords. Others still need to be translated to French:

enter image description here

\documentclass{article}
\usepackage{algorithm}% http://ctan.org/pkg/algorithm
\usepackage{algpseudocode}% http://ctan.org/pkg/algorithmicx

\makeatletter
\renewcommand{\ALG@name}{Algorithme}% Algorithm in French
\algnewcommand{\Debut}{\Statex \textbf{D\'ebut}\pushindent}
\algnewcommand{\Fin}{\Statex \textbf{Fin}\popindent}

% This is the vertical rule that is inserted
\def\therule{\makebox[\algorithmicindent][l]{\hspace*{.5em}\vrule height .75\baselineskip depth .25\baselineskip}}%

\newtoks\therules% Contains rules
\therules={}% Start with empty token list
\def\appendto#1#2{\expandafter#1\expandafter{\the#1#2}}% Append to token list
\def\gobblefirst#1{% Remove (first) from token list
  #1\expandafter\expandafter\expandafter{\expandafter\@gobble\the#1}}%
\def\LState{\State\unskip\the\therules}% New line-state
\def\pushindent{\appendto\therules\therule}%
\def\popindent{\gobblefirst\therules}%
\def\printindent{\unskip\the\therules}%
\def\printandpush{\printindent\pushindent}%
\def\popandprint{\popindent\printindent}%

%      ***      DECLARED LOOPS      ***
% (from algpseudocode.sty)
\algdef{SE}[WHILE]{While}{EndWhile}[1]
  {\printandpush\textbf{Tant que} #1}
  {\popandprint\textbf{Fin Tant que}}%
\algdef{SE}[FOR]{For}{EndFor}[1]
  {\printandpush\algorithmicfor\ #1\ \algorithmicdo}
  {\popandprint\algorithmicend\ \algorithmicfor}%
\algdef{S}[FOR]{ForAll}[1]
  {\printindent\algorithmicforall\ #1\ \algorithmicdo}%
\algdef{SE}[LOOP]{Loop}{EndLoop}
  {\printandpush\algorithmicloop}
  {\popandprint\algorithmicend\ \algorithmicloop}%
\algdef{SE}[REPEAT]{Repeat}{Until}
  {\printandpush\algorithmicrepeat}[1]
  {\popandprint\algorithmicuntil\ #1}%
\algdef{SE}[IF]{If}{EndIf}[1]
  {\printandpush\algorithmicif\ #1\ \algorithmicthen}
  {\popandprint\algorithmicend\ \algorithmicif}%
\algdef{C}[IF]{IF}{ElsIf}[1]
  {\popandprint\pushindent\algorithmicelse\ \algorithmicif\ #1\ \algorithmicthen}%
\algdef{Ce}[ELSE]{IF}{Else}{EndIf}
  {\popandprint\pushindent\algorithmicelse}%
\algdef{SE}[PROCEDURE]{Procedure}{EndProcedure}[2]
   {\printandpush\algorithmicprocedure\ \textproc{#1}\ifthenelse{\equal{#2}{}}{}{(#2)}}%
   {\popandprint\algorithmicend\ \algorithmicprocedure}%
\algdef{SE}[FUNCTION]{Function}{EndFunction}[2]
   {\printandpush\algorithmicfunction\ \textproc{#1}\ifthenelse{\equal{#2}{}}{}{(#2)}}%
   {\popandprint\algorithmicend\ \algorithmicfunction}%
\makeatother

\begin{document}

\begin{algorithm}
  \caption{My algorithme}
  \begin{algorithmic}[1]
    \Debut
      \LState $r\gets a\bmod b$
      \While{$r\not=0$}\Comment{We have the answer if r is 0}
        \LState $a\gets b$
        \LState $b\gets r$
        \LState $r\gets a\bmod b$
      \EndWhile\label{euclidendwhile}
      \LState \textbf{Retour} $b$\Comment{The gcd is b}
    \Fin
  \end{algorithmic}
\end{algorithm}
\end{document}

\State provides a "regular line" (with number), while \Statex provides an unnumbered line. As such, I've defined both Début and Fin to be of \Statex.


Adding the macro

\def\LStatex{\Statex\unskip\the\therules}% New line-state

allows you to print an unnumbered line. Now you can use it as follows:

enter image description here

\begin{algorithm}
  \caption{My algorithme}
  \begin{algorithmic}[1]
    \Debut
      \LState $r\gets a\bmod b$
      \While{$r\not=0$}\Comment{We have the answer if r is 0}
        \LState $a\gets b$
        \LState Here is a very long expression
        \LStatex \expandafter\hskip\algorithmicindent\relax that has to go to a new line.
        \LState $r\gets a\bmod b$
      \EndWhile\label{euclidendwhile}
      \LState \textbf{Retour} $b$\Comment{The gcd is b}
    \Fin
  \end{algorithmic}
\end{algorithm}