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通过哪个 pkg 来可视化超立方体图？

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通过哪个 pkg 来可视化超立方体图？

我正在尝试将超立方体可视化为下图。我已寻求帮助这里但是将节点与包放在一起。第 N 个超立方体有 2^n 个节点，每个顶点的度数为 n。我正在尝试找到一种方法来可视化度数更大的超立方体，这对 pkg 来说是一个很大的挑战。我的目标是可视化立方体的遍历。您将如何进行这种可视化？

关于第 3 个超立方体的示例（8 个节点，每个顶点的度为 3）

在此处输入图片描述

答案1

我将把“超立方体”解释为指定图形的一种方式。

我使用 LaTeX3 进行整数和二进制之间的转换，并使各种递归更容易。然后 TikZ 进行实际渲染。

（它可以对临时变量进行清理，并且一些样式选项将是一个合理的补充。）

\documentclass{article}
%\url{http://tex.stackexchange.com/q/129613/86}
\usepackage[landscape]{geometry}
\usepackage{tikz}
\usepackage{expl3}
\usepackage{xparse}

\ExplSyntaxOn

\int_new:N \l__binp_int
\int_set:Nn \l__binp_int {4}
\tl_new:N \l__bina_tl
\tl_new:N \l__binb_tl
\tl_new:N \l__binhw_tl
\int_new:N \l__bina_int
\int_new:N \l__binb_int
\int_new:N \l__binc_int
\int_new:N \l__bind_int
\fp_new:N \l__bina_fp

\DeclareDocumentCommand \BinaryPrecision {m}
{
  \int_set:Nn \l__binp_int {#1}
}

\cs_new_nopar:Npn \int_to_bin:Nn #1#2
{
  \tl_clear:N #1
  \int_set:Nn \l__bina_int {#2}
  \prg_replicate:nn {\l__binp_int}
  {
    \int_set:Nn \l__binb_int {\int_mod:nn {\l__bina_int} {2}}
    \tl_put_left:Nx #1 {\int_use:N \l__binb_int}
    \int_set:Nn \l__bina_int {\int_div_truncate:nn {\l__bina_int} {2}}
  }
}

\cs_new_nopar:Npn \bin_hamming_weight:NN #1#2
{
  \tl_set_eq:NN \l__binhw_tl #2
  \tl_replace_all:Nnn \l__binhw_tl {0} {}
  \int_set:Nn #1 {\tl_count:N \l__binhw_tl}
}

\cs_new_nopar:Npn \bin_flip_one:NNn #1#2#3
{
  \tl_set_eq:NN #2#1
  \prg_replicate:nn {#3 - 1}
  {
    \tl_replace_once:Nnn #2 {1} {a}
  }
  \tl_replace_once:Nnn #2 {1} {0}
  \tl_replace_all:Nnn #2 {a} {1}
}

\DeclareDocumentCommand \HammingWeight {m m}
{
  \int_to_bin:Nn \l__binhw_tl {#2}
  \bin_hamming_weight:NN #1 \l__binhw_tl
}

\DeclareDocumentCommand \HyperCube {m}
{
  \int_set:Nn \l__binp_int {#1}
  \int_zero_new:c {l__hyper_0_int}
  \int_set:cn {l__hyper_0_int} {-1}
  \int_step_inline:nnnn {1} {1} {#1}
  {
    \int_zero_new:c {l__hyper_##1_int}
    \int_set:cn  {l__hyper_##1_int} {\int_use:c {l__hyper_\int_eval:n {##1 - 1}_int} * (#1 - ##1 + 1) / ##1}
  }
  \fp_set:Nn \l__bina_fp {2^{#1}-1}
  \int_set:Nn \l__binc_int {\fp_to_int:N \l__bina_fp}
  \int_step_inline:nnnn {0} {1} {\l__binc_int}
  {
    \int_to_bin:Nn \l__bina_tl {##1}
    \bin_hamming_weight:NN \l__binb_int \l__bina_tl
    \node[every~ hypercube~ label/.try] (hg- \l__bina_tl) at (\int_use:c {l__hyper_\int_use:N \l__binb_int _int},\int_use:N \l__binb_int) {##1};
    \int_incr:c {l__hyper_\int_use:N \l__binb_int _int}
    \int_incr:c {l__hyper_\int_use:N \l__binb_int _int}
  }
  \int_step_inline:nnnn {0} {1} {\l__binc_int}
  {
    \int_to_bin:Nn \l__bina_tl {##1}
    \bin_hamming_weight:NN \l__binb_int \l__bina_tl
    \int_step_inline:nnnn {1} {1} {\l__binb_int}
    {
      \bin_flip_one:NNn \l__bina_tl \l__binb_tl {####1}
      \draw[every~ hypercube~ edge/.try] (hg- \l__bina_tl) -- (hg- \l__binb_tl);
    }
  }
}

\ExplSyntaxOff


\begin{document}
\begin{tikzpicture}
\HyperCube{3}
\end{tikzpicture}

\begin{tikzpicture}
\HyperCube{4}
\end{tikzpicture}

\begin{tikzpicture}
\HyperCube{5}
\end{tikzpicture}

\end{document}

3、4、5 生成器上的超立方体

答案2

\matrix这也可以使用 TikZ 和 a （或）来实现，tikz-cd尽管对于三维以上的空间来说会变得复杂。

这是一种将每个已经放置的节点移动到维度的方法n在维度中n+ 1，n+ 2，…，n 总计并将其与其父节点（->）连接起来。

宏\currentTransform的设置方式是，它包含每个节点的所有应用转换（来自0-0），这用于连接节点在其自己的维度中（<->）。（分组\foreach循环的一个很好的理由。）

代码

\documentclass[tikz,convert=false]{standalone}
% a little support for all the keys
\def\hyperset{\pgfqkeys{/tikz/hyper}}
\tikzset{hypher/.code=\hyperset{#1}}

% Dimension setup
\hyperset{
  set color/.code 2 args={\colorlet{tikz@hyper@dimen@#1}{#2}},
  set color={0}{black},
  set color={1}{blue!50!black},
  set color={2}{red},
  set color={3}{green},
  set color={4}{yellow!80!black}}
\hyperset{
  set dimens/.style args={#1:(#2)}{
    dimen #1/.style={/tikz/shift={(#2)}}},
  set dimens=0:(0:0),
  set dimens=1:(right:1),
  set dimens=2:(up:1),
  set dimens=3:(30:.75),
  set dimens=4:(180+70:.5),
  every hyper node/.style 2 args={%
    shape=circle,
    inner sep=+0pt,
    fill,
    draw,
    minimum size=+3pt,
    color=tikz@hyper@dimen@#1,
    label={[hyper/label #1/.try, hyper/dimen #1 style/.try]#1-#2}
  },
  every hyper edge/.style={draw},
  every hyper shift edge/.style={->,tikz@hyper@dimen@#1!80},
  every normal hyper edge/.style={<->,tikz@hyper@dimen@#1!40},
}
\newcommand*{\hyper}[1]{% #3 = max level
  \def\currentTransform{}
  \node[hyper/every hyper node/.try={0}{0}, hyper/dimen 0 node/.try] (0-0) {};
  \hyperhyper{0}{0}{#1}}
\newcommand*{\hyperhyper}[3]{% #1 = current level
                             % #2 = current number
                             % #3 = maxlevel
  \foreach \dimension in {#3,...,\the\numexpr#1+1\relax} {
    \edef\newNumber{\the\numexpr#2+\dimension\relax}
    \node[hyper/every hyper node/.try={\dimension}{\newNumber}, hyper/dimen \dimension node/.try, hyper/dimen \dimension\space style/.try] at ([hyper/dimen \dimension] #1-#2) (\dimension-\newNumber) {};
    \path (#1-#2) edge[hyper/every hyper edge/.try=\dimension, hyper/every hyper shift edge/.try=\dimension, hyper/dimen \dimension\space style/.try] (\dimension-\newNumber);
    \ifnum\newNumber>\dimension\relax
      \foreach \oldShift in \currentTransform {
        \if\relax\detokenize\expandafter{\oldShift}\relax\else
          \path (\dimension-\newNumber) edge[hyper/every hyper edge/.try=\dimension, hyper/every normal hyper edge/.try=\dimension, hyper/dimen \dimension\space style/.try] (\dimension-\the\numexpr\newNumber-\oldShift\relax);
        \fi
      }
    \fi
    \edef\currentTransform{\dimension,\currentTransform}%
    \ifnum\dimension<#3\relax
      \edef\temp{{\dimension}{\the\numexpr#2+\dimension\relax}{#3}}
      \expandafter\hyperhyper\temp
    \fi
  }
}
\tikzset{
  @only for the animation/.style={
    hyper/dimen #1 style/.style={opacity=0}}}
\begin{document}
\foreach \DIM in {0,...,4,4,3,...,0} {
\begin{tikzpicture}[
  >=stealth,
  scale=3,
  every label/.append style={font=\tiny,inner sep=+0pt}, label position=above left,
  @only for the animation/.list={\the\numexpr\DIM+1\relax,...,5}
  ]
  \hyper{4}
\end{tikzpicture}}
\end{document}

输出

在此处输入图片描述

算法（beamer，逐步）

在此处输入图片描述

答案3

这里可以使用 PSTricks 来可视化 3 超立方体：

% arara: latex
% arara: dvips
% arara: ps2pdf
\documentclass[preview,border=3pt]{standalone}
\usepackage{pst-node}
\begin{document}
\begin{psmatrix}[mnode=circle,colsep=0.8,rowsep=0.8]
& [name=8] 8 \\[-0.4\psyunit]
[name=5] 5 & [name=6] 6 & [name=7] 7\\
[name=2] 2 & [name=3] 3 & [name=4] 4\\[-0.4\psyunit]
& [name=1] 1
\ncline{8}{5}\ncline{8}{6}\ncline{8}{7}
\ncline{5}{3}\ncline{5}{2}
\ncline{6}{2}\ncline{6}{4}
\ncline{7}{3}\ncline{7}{4}
\ncline{1}{2}\ncline{1}{3}\ncline{1}{4}
\ncline[offset=5pt, linewidth=0.5\pslinewidth, arrows=->, nodesep=5pt]{1}{2}
\ncarc[offset=3pt, linewidth=0.5\pslinewidth, arrows=->, nodesep=4pt]{2}{5}
\end{psmatrix}
\end{document}

得出：

在此处输入图片描述

但我不知道有哪个包能够自动完成第 N 个超立方体。

答案4

作为替代方案，这里是tikz-cd原始图像/Op 的真实请求的版本。

代码

\documentclass[tikz,convert=false]{standalone}
\usepackage{tikz-cd}
\tikzset{
  shorten/.style={/tikz/shorten >={#1},/tikz/shorten <={#1}}}
\begin{document}
\begin{tikzcd}[
  every arrow/.append style={-,thick},
  matrix of math nodes maybe/.append style={/tikz/cells={/tikz/nodes={/tikz/draw,/tikz/shape=circle,align=center,text width=\widthof{$x_1x_2x_3$}}}},
  row sep=large,
  column sep=large,
  thick,
  ]
        & x_1x_2x_3 \dar \dlar \drar \\
 x_1x_2 & x_1   x_3      \dlar \drar & x_2x_3 \\
 x_1 \uar
     \uar[bend left=20, shorten=.1cm, -stealth]
        &    x_2         \ular \urar &    x_3 \uar\\
        & \emptyset \uar \ular \urar \ular[shift left=.25cm, shorten=.25cm, -stealth]
\end{tikzcd}
\end{document}

输出

在此处输入图片描述

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