刪除演算法中的索引和行

Question

我想請問您是否可以：

(1) 刪除左側索引和最後一行索引（全部為 0，黃色突出顯示）？

(2) 刪除演算法內的所有行（綠色突出顯示）？

這是代碼：

\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} 

先感謝您。

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

擺脫無用的重新定義，這些重新定義本質上是重申預設值。

如果您不需要縮排參考線，請指定相關選項。

但是，我得到的不是零，而是行號，因此您的文件中可能有其他內容導致編號錯誤。

如果您不需要行號，請不要指定[1]after \begin{algorithmic}，如第二個範例所示。

\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}