我怎样才能轻松地设计这个格子?

我怎样才能轻松地设计这个格子?

我有一个问题:我该如何使用乳胶以便它可以复制这种结构(乳胶)?

在此处输入图片描述

不幸的是,我不太了解这些库。了解如何绘制某些图表对我以后绘制与此非常相似的其他格子很有帮助。

答案1

使用纯粹的tikz、相当基本的库calcchains并且positioning

\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{calc, chains, 
                positioning}

\begin{document}
\begin{tikzpicture}[
node distance = 8mm and 2mm,
  start chain = going right,
     N/.style = {inner sep=1pt, on chain}
                    ]
\node (n21) [N] {$\{1,2\}$};
\node (n22) [N] {$\{1,3\}$};
\node (n23) [N] {$\{1,4\}$};
\node (n24) [N] {$\{2,3\}$};
\node (n25) [N] {$\{2,4\}$};
\node (n26) [N] {$\{3,4\}$};
%
\node (n11) [N, below=of n21]                           {$\{1\}$};
\node (n12) [N, below=of $(n22.south)!0.5!(n23.south)$] {$\{2\}$};
\node (n13) [N, below=of $(n24.south)!0.5!(n25.south)$] {$\{3\}$};
\node (n14) [N, below=of n26]                           {$\{4\}$};
%
\draw (n11) -- (n21)    (n11) -- (n22)
      (n12) -- (n21)    (n12) -- (n24)  (n12) -- (n25)
      (n13) -- (n22)    (n13) -- (n24)  (n13) -- (n26)
      (n14) -- (n23)    (n14) -- (n25)  (n14) -- (n26);
\draw[red, very thick]  (n11) -- (n23);
%
\node (n31) [N, above=of n21]               {$\{1,2,3\}$};
\node (n32) [N, above=of n21.north -| n12]  {$\{1,2,4\}$};
\node (n33) [N, above=of n21.north -| n13]  {$\{1,3,4\}$};
\node (n34) [N, above=of n26]               {$\{2,3,4\}$};
%
\draw (n31) -- (n21)    (n31) -- (n22)  (n31) -- (n23)
      (n32) -- (n21)    (n32) -- (n23)  (n32) -- (n25)
      (n33) -- (n23)    (n33) -- (n26)
      (n34) -- (n24)    (n34) -- (n25)  (n34) -- (n26);
\draw[red, very thick]  (n33) -- (n23);
%
\node (n41) [N, above=of $(n32.north)!0.5!(n33.north)$] {$\{1,2,3,4\}$};
%
\draw (n41) -- (n31)    (n41) -- (n32)  (n41) -- (n34);
\draw[red, very thick]  (n41) -- (n33);
\end{tikzpicture}
\end{document}

在此处输入图片描述

答案2

该图似乎具有以下逻辑:首先找到{1,2,3,4}通过删除一个条目而出现的该列表的所有子集,然后将它们放在下一行。在下一行中,放入通过删除另一个元素而出现的所有子集。继续这样做,在最后一行中得到一个元素列表。

如果给定列表完全包含在上一级列表中,则该列表将与上一级列表连接。有人可能想知道是否可以让 LaTeX 决定在哪里绘制这些连接线。答案是

是的

当然,这需要一些准备,比如会员测试,这是例如这里,以及一个查找两个列表交集的函数,我在这里添加了它。结果是

\documentclass[tikz,border=3mm]{standalone}
\newcounter{iloop}
\makeatletter
\pgfmathdeclarefunction{memberQ}{2}{%
  \begingroup%
    \edef\pgfutil@tmpb{0}%memberQ({\lstPast},\inow)
    \edef\pgfutil@tmpa{#2}%
    \expandafter\pgfmath@member@i#1\pgfmath@token@stop%
    \edef\pgfmathresult{\pgfutil@tmpb}%
    \pgfmath@smuggleone\pgfmathresult%
  \endgroup}
\def\pgfmath@member@i#1{%
    \ifx\pgfmath@token@stop#1%
    \else
      \edef\pgfutil@tmpc{#1}%
      \ifx\pgfutil@tmpc\pgfutil@tmpa\relax%
      \gdef\pgfutil@tmpb{1}%
      \fi%
      \expandafter\pgfmath@member@i%
    \fi}  
\pgfmathdeclarefunction{intersection}{2}{%
  \begingroup%
    \pgfmathparse{int(dim(#1)-1)}%
    \pgfutil@tempcnta=\pgfmathresult%
    \pgfutil@tempcntb=0%
    \edef\pgfutil@tmpc{}%
    \edef\pgfutil@tmpd{}%
    \loop%
    \pgfmathsetmacro{\pgfutil@tmpe}{{#1}[\the\pgfutil@tempcntb]}%
    \pgfmathtruncatemacro{\pgfutil@tmpa}{memberQ("#2",\pgfutil@tmpe)}%
    \ifnum\pgfutil@tmpa=1%
     \ifx\pgfutil@tmpc\pgfutil@tmpd%
     \edef\pgfutil@tmpc{\pgfutil@tmpe}%
     \else%
     \edef\pgfutil@tmpc{\pgfutil@tmpc,\pgfutil@tmpe}%
     \fi%
    \fi% 
    \advance\pgfutil@tempcntb1%
    \ifnum\the\pgfutil@tempcntb<\the\pgfutil@tempcnta\repeat%
    \edef\pgfmathresult{\pgfutil@tmpc}%
    \pgfmathsmuggle\pgfmathresult%
  \endgroup}          
\makeatother
\begin{document}
\begin{tikzpicture}
 \def\myn{0}
 \foreach \X [count=\Y,remember=\X as \LastX,remember=\myn as \mylastn] in {{"1,2,3,4"},{"1,2,3","1,2,4","1,3,4","2,3,4"},%
 {"1,2","1,3","1,4","2,3","2,4","3,4"},{"1","2","3","4"}}
 {\pgfmathtruncatemacro{\myn}{dim({\X})-1}
  \pgfmathsetmacro{\myfirst}{{\X}[0]}
  \pgfmathsetmacro{\mylast}{{\X}[\myn]}
  \ifnum\Y=1
    \pgfmathsetmacro{\myfirst}{{{\X}}[0]}
    \node (L-1-1) at (0,0) {$\{\myfirst\}$};
  \else 
   \node (L-\Y-1) at (-4,1.5-\Y*1.5) {$\{\myfirst\}$};
   \node (L-\Y-\the\numexpr\myn+1) at (4,1.5-\Y*1.5) {$\{\mylast\}$};
   \path (L-\Y-1.center) -- (L-\Y-\the\numexpr\myn+1\relax.center) 
    foreach \Z in {1,...,\the\numexpr\myn-1}
     {[/utils/exec=\pgfmathsetmacro{\myentry}{{\X}[\Z]}] 
        node[pos=\Z/\myn] (L-\Y-\the\numexpr\Z+1) {$\{\myentry\}$} };
   \ifnum\Y=2
     \foreach \Z in {0,...,\the\numexpr\myn}
     {\draw (L-\Y-\the\numexpr\Z+1) -- (L-1-1);}
   \else
     \foreach \Z in {0,...,\the\numexpr\myn}        
     {\pgfmathsetmacro{\CurrentItem}{{\X}[\Z]}%
      \setcounter{iloop}{0}%        
      \loop\pgfmathsetmacro{\LastItem}{{\LastX}[\value{iloop}]}%
      \stepcounter{iloop}%
      \pgfmathsetmacro{\myintersection}{intersection("\CurrentItem","\LastItem")}%
      \pgfmathtruncatemacro{\nint}{dim(\myintersection)-dim(\CurrentItem)}%
      \ifnum\nint=0
       \draw (L-\Y-\the\numexpr\Z+1) -- 
        (L-\the\numexpr\Y-1\relax-\the\numexpr\value{iloop});
      \fi
      \ifnum\value{iloop}<\the\numexpr\mylastn+1\relax%
      \repeat
      }
   \fi
  \fi
 }
 \draw[red,very thick]  (L-1-1) -- (L-2-3) -- (L-3-3) -- (L-4-1);
\end{tikzpicture}

\end{document}

在此处输入图片描述

唯一硬编码的是红线,我没有看到任何模式。不过,画这条线很简单,

\draw[red,very thick]  (L-1-1) -- (L-2-3) -- (L-3-3) -- (L-4-1);

优点是这适用于其他类似的图表,而且不必手工绘制。(例如,到目前为止,在另一个答案中,{1,3,4}和之间的联系{1,3}缺失了。如果我必须用爪子画这些,我很可能会错过更多联系。)

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