doWhile 循环缩进问题

Question

我一直在尝试使用 do-while 循环为 [1] 中所示的流程图编写多个循环语句。但是，管理缩进却很困难。有人能帮忙解决这个问题吗？

\documentclass{article}
\usepackage{algorithmicx}
\usepackage{algpseudocode}

\algdef{SE}[DOWHILE]{Do}{doWhile}{\algorithmicdo}[1]{\algorithmicwhile\ #1}%

\begin{document}
\begin{algorithmic}
  \Do
    \State a,b,c,d,e,m,n
  \doWhile{$!f$} % <--- use \doWhile for the "while" at the end
  
  $g$
  
    \Do
    \State abc
    \doWhile{$!h$}
    
    $i$
    \Do
    \State Pv
    \doWhile{$!j$}
    
    $r s t $
  %  \doWhile{$!$}
  %   \State l
\end{algorithmic}
\end{document}

问题在于缩进。

Answer 1

尽管编程语言通常会忽略程序文本中的空格，但根据惯例，缩进可用于提高程序的可读性。为了实现此目的，缩进通常应遵循某种一致的方式，以反映程序的结构。

在您的代码中，我无法理解程序的结构。不幸的是，该algorithmicx包（参见其文档) 也无法遵循您的结构并且以不符合您预期的方式缩进您的代码。

尤其是，我（algorithmicx也）看不到：

1. 是什么$g$。如果是一些抽象语句，则应将其写为\State $g$以便正确缩进。这同样适用于$i$和$r s t$。

2. 为什么你希望下面的行$g$不是直接缩进$g$，而是更靠右。这只能解释为，当$g$表示某种形式的块语句（例如while）时，其主体延伸到其下面的行。

获得的结果（在添加缺失的\State宏和一些\medskip用于添加垂直空间的宏之后）反映了程序的结构algorithmicx：

另一方面，如果您确实想在块头后面引入任意缩进，您可以为此目的定义一种新类型的块（我将其命名为\Arbitrary{header} ... \endArbitrary）：

\documentclass{article}
\usepackage{algorithmicx}
\usepackage{algpseudocode}

\algdef{SE}[DOWHILE]{Do}{doWhile}{\algorithmicdo}[1]{\algorithmicwhile\ #1}%
\algblockdefx[ARBITRARY]{Arbitrary}{endArbitrary}[1]{#1}

\begin{document}
\begin{algorithmic}
  \Do
    \State a,b,c,d,e,m,n
  \doWhile{$!f$} % <--- use \doWhile for the "while" at the end
  \medskip
  \Arbitrary{$g$}
    \medskip
    \Do
    \State abc
    \doWhile{$!h$}
    \medskip
    \State $i$
    \Do
    \State Pv
    \doWhile{$!j$}
    \medskip
    \State $r s t $
  \endArbitrary
\end{algorithmic}
\end{document}

其结果是：

Answer 2

下面是我在 LaTex 上编写的基于嵌套 Do-While 的伪代码。有人可以检查一下，并可以供将来遇到同样问题的人参考。

\documentclass[11pt]{article}
\usepackage[fleqn]{amsmath}
\usepackage[ruled,vlined]{algorithm2e}
\PassOptionsToPackage{noend}{algpseudocode}
\usepackage{algpseudocode}

    %%%%%%%%%%%%%%%%%%
    %% DoWhile algorithm macro definition 
    \makeatletter
    \newcommand*{\algrule}[1][\algorithmicindent]{\makebox[#1][l]{\hspace*{.5em}\vrule height .75\baselineskip depth .25\baselineskip}}
    
    \newcount\ALG@printindent@tempcnta
    \def\ALG@printindent{%
        \ifnum \theALG@nested>0% is there anything to print
        \ifx\ALG@text\ALG@x@notext% is this an end group without any text?
        % do nothing
        \addvspace{-3pt}% FUDGE for cases where no text is shown, to make the rules line up
        \else
        \unskip
        % draw a rule for each indent level
        \ALG@printindent@tempcnta=1
        \loop
        \algrule[\csname ALG@ind@\the\ALG@printindent@tempcnta\endcsname]%
        \advance \ALG@printindent@tempcnta 1
        \ifnum \ALG@printindent@tempcnta<\numexpr\theALG@nested+1\relax% can't do <=, so add one to RHS and use < instead
        \repeat
        \fi
        \fi
    }
\usepackage{etoolbox}
% the following line injects our new indent handling code in place of the default spacing
\patchcmd{\ALG@doentity}{\noindent\hskip\ALG@tlm}{\ALG@printindent}{}{\errmessage{failed to patch}}
\makeatother
% end vertical rule patch for algorithmicx
\algdef{SE}[DOWHILE]{Do}{doWhile}{\algorithmicdo}[1]{\algorithmicwhile\ #1}%

\begin{algorithm}%\captionsetup{labelfont={sc,bf}, labelsep=newline}
    \caption{Workspace boundary determination}
    \label{alg:PoEG}    
    \begin{algorithmic}
        \State Input $z_0,~z_f,~n,~m$
        \State \quad \quad ~ $\Delta_z = \dfrac{z_0-z_f}{n}, \textit{tol}$
        \State \quad \quad ~ $\varepsilon_f= 2\pi~, \Delta\varepsilon = \dfrac{\varepsilon_f}{m}$
        \State \quad \quad ~ $k_{\textup{max}}, \Delta\alpha_0, i = j=1$
        \medskip
        \State Output $\boldsymbol{\theta}, \boldsymbol{\psi}, \boldsymbol{z} $
        \medskip
        \State initialize ~$z \gets z_0,~ \psi=\theta =\gets \emptyset$
        \Do
            \Do 
                \Do
                    \Do 
                        \State $\Delta\psi = \psi + \Delta\alpha c\varepsilon$
                        \State $\Delta\theta = \theta + \Delta\alpha s\varepsilon $ 
                        \State $\psi = \psi +\Delta\varepsilon$
                        \State $\theta = \psi + \Delta\varepsilon$ 
                        \State Compute $\boldsymbol{J}_{dh}$ 
                        \State $k = cond(\boldsymbol{J}_{dh})$
        
                    \doWhile{$k<k_{max}$}
        
                    \State $\psi = \psi -\Delta\varepsilon$
                    \State $\theta = \psi - \Delta\varepsilon$ 
        
                    \State $\Delta\alpha = \Delta\alpha/2$
        
                \doWhile{$\Delta\alpha > tol $}
                \State $\psi(i,j) = \psi $
                \State $\theta(i,j) = \theta $
        
                \State $ j =j+1$
                \State $\Delta\alpha = \Delta\alpha_0$
                \State $\varepsilon = \varepsilon + \Delta\varepsilon$
            \doWhile{$\varepsilon < 2\pi$}
        
            \State $ \theta =0$
            \State $\psi =0$
            \State $\varepsilon =0$
            \State $\Delta\alpha=\Delta\alpha_0$
            \State $i=i+1$
            \State $z_i=z+\Delta_z$

        \doWhile{$z<z_f$} \\ % <--- use \doWhile for the "while" at the end
        plot$(\boldsymbol{\psi},\boldsymbol{\theta},\boldsymbol{z})$\\ \medskip Note: $s$ and $c$ stands for sine and cosine, respectively.
        %\endArbitrary
    \end{algorithmic}
\end{algorithm}

\end{document}