连接多面体的顶点

Question

我需要将 5-gone 的每个顶点与 10-gone 当前连接的顶点的对立面连接起来。参见图片（颜色必须是黑色）：

到目前为止，我有以下内容，但是两个顶点没有正确连接。

\documentclass[10pt]{article}
\usepackage{amstext}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{array}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\usepackage[margin=1.5in]{geometry}    % For reducing margin
\usepackage{graphicx}
\usetikzlibrary{shapes,positioning}
\usepackage[utf8]{inputenc}



\begin{document}

\begin{tikzpicture}
\node[regular polygon, regular polygon sides=10, minimum size=4cm, draw=black, shape border rotate=20, name=x] at (6,6) {};
\node[regular polygon, regular polygon sides=5, minimum size=1.5cm, draw=black, name=y] at (6,6) {};

\foreach \corner in {1,2,...,10}
\node[circle,ball color=black,inner sep=0pt,minimum size=5pt] at (x.corner \corner){};

\foreach \corner in {1,2,...,5}
\node[circle,ball color=black,inner sep=0pt,minimum size=5pt] at (y.corner \corner){};

\draw foreach[
  evaluate={\cornerA=int(2*\c);
            \cornerB=int(\cornerA-1);}] \c in {1,...,5}{
              (x.corner \cornerB) -- (y.corner \c)
              };
              
\draw foreach[
  evaluate={\cornerA=int(2*\c);
            \cornerB=int(\cornerA+4);}] \c in {1,...,5}{
              (x.corner \cornerB) -- (y.corner \c)
              };

\end{tikzpicture}

Answer 1

对原始代码的更改：

• 将外部多边形旋转 18°（180°/10），而不是 20°。
• 用于outer sep = 0pt多边形，以便角实际上位于指定的直径处。
• 命名球节点以便它们可以这样使用，这将使边缘停止在其边界处而不是在它们上面绘制。
• 该函数会计算对角的值mod，通常如果角的编号是从 0 到 9 的话就会这样<corner opposite> = mod(<corner>+5, 10)，但是由于它们的编号是从 1 到 5，所以我们只在 Mod 函数内部添加 4，在外部添加 1。

我还通过库添加了类似的绘图graphs，这非常接近我最近的另一个回答，我使用相同的 foreach 评估来绘制对角线边缘。

我用红色和黑色的双重函数绘制了它们，以表明它们不会穿过整个图表，而是停止在内球上。

代码

\documentclass[tikz]{standalone}
\usetikzlibrary{graphs.standard, shapes.geometric}
\begin{document}
\begin{tikzpicture}[
  rp/.style={regular polygon, regular polygon sides={#1}, outer sep=+0pt},
  ball/.style={
    circle, outer sep=auto, ball color=black, inner sep=0pt, minimum size=5pt}
]
\node[rp=10, minimum size=4cm,   draw=black, rotate=18, name=outer] at (6,6) {};
\node[rp=5,  minimum size=1.5cm, draw=black, name=inner] at (6,6) {};

\foreach \corner in {1,2,...,10}
  \node[ball] (outer-corner-\corner) at (outer.corner \corner){};
\foreach \corner in {1,2,...,5}
  \node[ball] (inner-corner-\corner) at (inner.corner \corner){};

\foreach[evaluate={
  \cOuter=int(2*\cInner-1);
  \cOpposite=int(mod(\cOuter+4,10)+1);
}]\cInner in {1,...,5}
  \draw (outer-corner-\cOuter)
     -- (inner-corner-\cInner)
     -- (outer-corner-\cOpposite);
\end{tikzpicture}

\tikz\graph[
  nodes={
    circle, outer sep=+0pt, ball color=black, inner sep=+0pt, minimum size=+5pt},
  counterclockwise, phase = 90, typeset =,
]{
  subgraph C_n[name=inner, n= 5] -!-
  subgraph C_n[name=outer, n=10],
  { [edge={double=black, red, thick}]
    \foreach[evaluate={
      \cOuter=int(2*\cInner-1);
      \cOpposite=int(mod(\cOuter+4,10)+1);
      }] \cInner in {1,...,5} {
        inner \cInner -- {outer \cOuter, outer \cOpposite}},
  }
};
\end{document}

输出

Answer 2

虽然我们找到了更自动化的解决方案，但这里有一个手动的解决方案：

\documentclass[10pt]{article}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\usepackage[margin=1.5in]{geometry}    % For reducing margin
\usetikzlibrary{shapes,positioning}


\begin{document}

\begin{tikzpicture}
\node[regular polygon, regular polygon sides=10, minimum size=4cm, draw=black, shape border rotate=20, name=x] at (6,6) {};
\node[regular polygon, regular polygon sides=5, minimum size=1.5cm, draw=black, name=y] at (6,6) {};

\foreach \corner in {1,2,...,10}
\node[circle,ball color=black,inner sep=0pt,minimum size=5pt] at (x.corner \corner){};

\foreach \corner in {1,2,...,5}
\node[circle,ball color=black,inner sep=0pt,minimum size=5pt] at (y.corner \corner){};

%\draw foreach[
%  evaluate={\cornerA=int(2*\c);
%            \cornerB=int(\cornerA-1);
%            \cornerC=int(mod(\cornerB+5,10);}] \c in {1,...,5}{
%              (x.corner \cornerB) -- (y.corner \c) -- (x.corner \cornerC)
%              };

\draw foreach \a/\b/\c in {1/1/6, 2/3/8, 3/5/10, 4/7/2, 5/9/4}
    {(x.corner \b)--(y.corner \a)--(x.corner \c)};
\end{tikzpicture}
\end{document}

更新：

此代码应用了 Qrrbrbirlbel 提出的旋转校正和一种计算顶点连接的自动方法。此连接仍然绘制在五边形顶点处停止。

\documentclass[10pt]{article}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\usepackage[margin=1.5in]{geometry}    % For reducing margin
\usetikzlibrary{shapes,positioning}


\begin{document}

\begin{tikzpicture}
\node[regular polygon, regular polygon sides=10, minimum size=4cm, draw=black, shape border rotate=18, name=x] at (6,6) {};
\node[regular polygon, regular polygon sides=5, minimum size=1.5cm, draw=black, name=y] at (6,6) {};

\foreach \corner in {1,2,...,10}
\node[circle,ball color=black,inner sep=0pt,minimum size=5pt] at (x.corner \corner){};

\foreach \corner in {1,2,...,5}
\node[circle,ball color=black,inner sep=0pt,minimum size=5pt] at (y.corner \corner){};

\draw foreach[
  evaluate={\cornerA=int(2*\c);
            \cornerB=int(\cornerA-1);
            \cornerC=ifthenelse(\cornerB+5>10, int(mod(\cornerB+5,10)), int(\cornerB+5));}] \c in {1,...,5}{
              (x.corner \cornerB) -- (y.corner \c) -- (x.corner \cornerC)
              };
\end{tikzpicture}
\end{document}