删除算法中的索引和行

Question

我想问您是否可以：

（1）删除左侧的索引和最后一行的索引（全0，黄色突出显示）？

（2）删除算法中的所有行（绿色突出显示）？

以下是代码：

\documentclass[journal]{IEEEtran}
\usepackage[commentColor=black]{algpseudocodex}

\floatplacement{algorithm}{tbp}
\makeatletter
\newcommand{\algorithmname}{\ALG@name}
\renewcommand{\floatc@ruled}[2]{{\@fs@cfont #1} #2\par}
\makeatother
\tikzset{algpxIndentLine/.style={draw=black}}
\algrenewcommand{\alglinenumber}[1]{\bfseries\footnotesize #1}
\algrenewcommand{\textproc}{\bfseries}

\begin{algorithm}[h]
\caption{Sequence}
\begin{algorithmic}[1]
\Statex \textbf{Step I}
\State Do $U_{LL}$
\If{$n$ is odd number}
\State Do $U_n$
\Else
\State Do $L_n$
\EndIf
\Statex \textbf{Step II}
\For{$i=1,\ldots,n$}
\If{$i\bmod{2}==1$}
\State Do $L_i$
\Else
\State Do $U_i$
\EndIf
\EndFor
\Statex \textbf{Step III}
\For{$i=1,\ldots,n-1$}
\If{$i\bmod{2}==1$}
\State Do $U_i$
\Else
\State Do $L_i$
\EndIf
\EndFor
\State Do $L_{LL}$
\end{algorithmic}
\end{algorithm} 

先感谢您。

Answer 1

\documentclass[journal]{IEEEtran}

\usepackage{algorithm}
\floatplacement{algorithm}{tbp}
\makeatletter
\renewcommand{\floatc@ruled}[2]{{\@fs@cfont #1} #2\par}
\makeatother

\usepackage[commentColor=black,noEnd=false,indLines=false]{algpseudocodex}

\begin{document}

\begin{algorithm}
\caption{Sequence}
\begin{algorithmic}
\Statex \textbf{Step I}
\State Do $U_{LL}$
\If{$n$ is odd number}
\State Do $U_n$
\Else
\State Do $L_n$
\EndIf
\Statex \textbf{Step II}
\For{$i=1,\ldots,n$}
\If{$i\bmod{2}==1$}
\State Do $L_i$
\Else
\State Do $U_i$
\EndIf
\EndFor
\Statex \textbf{Step III}
\For{$i=1,\ldots,n-1$}
\If{$i\bmod{2}==1$}
\State Do $U_i$
\Else
\State Do $L_i$
\EndIf
\EndFor
\State Do $L_{LL}$
\end{algorithmic}
\end{algorithm}

\end{document}

Answer 2

摆脱无用的重新定义，它们本质上是重申默认设置。

如果您不想要缩进指南，请指定相关选项。

但是，我得到的不是零，而是行号，因此您的文档中的其他内容可能会导致编号错误。

如果您不想要行号，请不要[1]在之后指定\begin{algorithmic}，如第二个示例所示。

\documentclass[journal]{IEEEtran}

\usepackage{algorithm}
\usepackage[
  commentColor=black,
  indLines=false,
]{algpseudocodex}

\algrenewcommand{\alglinenumber}[1]{\bfseries\footnotesize #1}
\algrenewcommand{\textproc}{\bfseries}

\begin{document}

\begin{algorithm}
\caption{Sequence (with line numbers)}
\begin{algorithmic}[1]
\Statex \textbf{Step I}
\State Do $U_{LL}$
\If{$n$ is odd number}
\State Do $U_n$
\Else
\State Do $L_n$
\EndIf
\Statex \textbf{Step II}
\For{$i=1,\ldots,n$}
\If{$i\bmod{2}==1$}
\State Do $L_i$
\Else
\State Do $U_i$
\EndIf
\EndFor
\Statex \textbf{Step III}
\For{$i=1,\ldots,n-1$}
\If{$i\bmod 2==1$}
\State Do $U_i$
\Else
\State Do $L_i$
\EndIf
\EndFor
\State Do $L_{LL}$
\end{algorithmic}
\end{algorithm}

\newpage

\begin{algorithm}
\caption{Sequence (without line numbers)}
\begin{algorithmic}
\Statex \textbf{Step I}
\State Do $U_{LL}$
\If{$n$ is odd number}
\State Do $U_n$
\Else
\State Do $L_n$
\EndIf
\Statex \textbf{Step II}
\For{$i=1,\ldots,n$}
\If{$i\bmod 2==1$}
\State Do $L_i$
\Else
\State Do $U_i$
\EndIf
\EndFor
\Statex \textbf{Step III}
\For{$i=1,\ldots,n-1$}
\If{$i\bmod{2}==1$}
\State Do $U_i$
\Else
\State Do $L_i$
\EndIf
\EndFor
\State Do $L_{LL}$
\end{algorithmic}
\end{algorithm} 

\end{document}