这个问题与https://math.stackexchange.com/questions/3449433/selecting-cards-to-form-a-fair-game?noredirect=1#comment7090962_3449433。我试图直观地说明如何从一堆N+M
球中抽出 3 个球,其中N
是蓝色球的数量,M
是绿色球的数量。
下面是我在 tikz 中尝试执行此操作的示例。在这里,我首先按顺序排列球,然后展示移除一个球后它们的外观。两个球之间的一条线表示这两个球被选中,加上虚线球。
然而我希望做一些改变。
- 因此,我们选择 3 个球,两个球通过线连接,另一个球通过虚线连接。如果这三个球颜色相同,我希望连接它们的线为 X 样式(虚线、红色等)
- 如果三个球有不同的颜色,我希望连接它们的线是 Y 样式(实线、橙色等)
- 为什么我的代码在特定情况下会失败
2
? - 有没有办法让代码适用于不同数量的蓝球和绿球?
- 我如何选择中心多边形的旋转,使其与外部多边形的角相匹配?现在我改变了内部多边形以手动适应。
下面的代码不起作用有什么特殊原因吗?
\foreach \x in {1,...,\n}{ \begin{scope}[shift={(b.corner \x)}] \missingPentagonTemp{2}{c}; \end{scope} }
这是我想要的效果(由油漆提供)。
平均能量损失
\documentclass[a4paper,11pt,margin=5pt]{standalone}
\usepackage{tikz,xcolor}
\usetikzlibrary{arrows}
\usetikzlibrary{trees}
\usetikzlibrary{shapes.geometric}
\usetikzlibrary{decorations}
\usepackage{ifthenx}
\newcommand{\missingPentagon}[2]{
% draw edges
\node[draw=none, minimum size=3cm, regular polygon, regular polygon sides = \n] (#2) {};
\def\n{6}
\pgfmathsetmacro{\start}{int(min(#1+1, \n))};
\pgfmathsetmacro{\stop}{int(max(#1-1, 2))};
\ifthenelse{\equal{#1}{1}}{
\foreach\x in{2,...,\n}{
\foreach\y in{\x,...,\n}{
\draw[color = black, dashed](#2.corner \x)--(#2.corner \y);
}
}
\foreach \x in {1}{
\draw[fill=green!20,dashed](#2.corner \x) circle[radius=1em] node {\x};
}
\foreach \x in {2,...,\n}{
\draw[fill=blue!20](#2.corner \x) circle[radius=1em] node {\x};
}
}{
\foreach\x in {1,...,\stop}{
\foreach\y in {\x,...,\n}{
\ifthenelse{\equal{#1}{\y}}{}{
\draw[color = black, dashed](#2.corner \x) -- (#2.corner \y)};
}
}
\foreach\x in {\start,...,\n}{
\foreach\y in {\x,...,\n}{
\ifthenelse{\equal{#1}{\y}}{}{
\draw[color = black, dashed](#2.corner \x) -- (#2.corner \y)};
}
}
\foreach \x in {1}{
\draw[fill=green!20](#2.corner \x) circle[radius=1em] node {\x};
}
\foreach \x in {2,...,\n}{
\ifthenelse{\equal{#1}{\x}}{
\draw[fill=blue!20,dashed](#2.corner \x) circle[radius=1em] node {\x};
}{
\draw[fill=blue!20](#2.corner \x) circle[radius=1em] node {\x};
};
}
}
}
\begin{document}
\begin{tikzpicture}
\def\n{6}
\node[draw=none, minimum size=3cm, regular polygon, regular polygon sides = \n] (a) {};
\node[draw=none, minimum size=10cm, regular polygon, regular polygon sides = \n] (b) {};
\foreach \x in {4}{
\draw[fill=green!20](a.corner \x) circle[radius=1em] node {1};
}
\foreach \x in {1,...,3}{
\pgfmathparse{int(\x+3)}
\draw[fill=blue!20](a.corner \x) circle[radius=1em] node {\pgfmathresult};
}
\foreach \x in {5,...,6}{
\pgfmathparse{int(\x-3)}
\draw[fill=blue!20](a.corner \x) circle[radius=1em] node {\pgfmathresult};
}
%\foreach \x in {1,...,\n}{
% \begin{scope}[shift={(b.corner \x)}]
% \missingPentagonTemp{2}{c};
% \end{scope}
%}
\begin{scope}[shift={(b.corner 1)}]
\missingPentagon{4}{c};
\end{scope}
\begin{scope}[shift={(b.corner 2)}]
\missingPentagon{5}{c};
\end{scope}
\begin{scope}[shift={(b.corner 3)}]
\missingPentagon{6}{c};
\end{scope}
\begin{scope}[shift={(b.corner 4)}]
\missingPentagon{1}{c};
\end{scope}
\begin{scope}[shift={(b.corner 5)}]
\missingPentagon{2}{c};
\end{scope}
\begin{scope}[shift={(b.corner 6)}]
\missingPentagon{3}{c};
\end{scope}
\end{tikzpicture}
\end{document}
答案1
这是尝试用 来回答问题pic
。此代码检查顶点是否在缺失或突出显示的顶点列表中(是的,您现在可以使用列表)并相应地绘制边。
至于你的观点:
- 完毕。
- 完成。此时样式已经固定。可以更改它。
- 说实话,我并没有仔细思考代码。抱歉。
- 是的。这些是清单。
- 已更改。也可以只旋转多边形,但我喜欢 mod 条件。;-)
- 再说一遍,我没有尝试。抱歉。
完整代码:
\documentclass[a4paper,11pt,margin=5pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\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}
\makeatother
\tikzset{circ/.style={circle,minimum size=2em,draw},
pics/missing polygon/.style={code={
\tikzset{missing polygon/.cd,#1}
\def\pv##1{\pgfkeysvalueof{/tikz/missing polygon/##1}}
\node[draw=none, minimum size=\pv{size},
regular polygon, regular polygon sides =\pv{n}] (-poly) {};
% test if the highlighted node is in the missin nodes
\edef\lsthigh{\pv{highlight}}
\foreach \XX in \lsthigh
{\pgfmathtruncatemacro{\ktest}{memberQ(\pv{miss},\XX)}
\xdef\pgfmathresult{\ktest}}
\edef\ktest{\pgfmathresult}
\foreach \XX in {1,...,\the\numexpr\pv{n}-1}
{
\foreach \YY in {2,...,\pv{n}}
{
\pgfmathtruncatemacro{\itest}{memberQ(\pv{miss},\XX)+memberQ(\pv{miss},\YY)}
\ifnum\itest=0
\pgfmathtruncatemacro{\jtest}{memberQ(\pv{highlight},\XX)+memberQ(\pv{highlight},\YY)}
\draw \ifnum\jtest=0 [dashed]\fi \ifnum\ktest=1 [solid]\fi
(-poly.corner \XX) -- (-poly.corner \YY);
\fi
}
}
\foreach \XX in {1,...,\pv{n}}
{\pgfmathtruncatemacro{\itest}{memberQ(\pv{highlight},\XX)}
\ifnum\itest=1
\node[missing polygon/highlighted] (-poly-vertex-\XX) at (-poly.corner \XX){\XX};
\else
\pgfmathtruncatemacro{\jtest}{memberQ(\pv{miss},\XX)}
\ifnum\jtest=1
\node[missing polygon/missing] (-poly-vertex-\XX) at (-poly.corner \XX){\XX};
\else
\node[missing polygon/regular] (-poly-vertex-\XX) at (-poly.corner \XX){\XX};
\fi
\fi
}
}},missing polygon/.cd,size/.initial=3cm,n/.initial=6,
miss/.initial={2},highlight/.initial={1},
highlighted/.style={circ,fill=green!20},
missing/.style={circ,fill=blue!20,dashed},
regular/.style={circ,fill=blue!20}
}
\begin{document}
\begin{tikzpicture}
\def\n{6}
\node[draw=none, minimum size=3cm, regular polygon, regular polygon sides = \n] (a) {};
\node[draw=none, minimum size=10cm, regular polygon, regular polygon sides = \n] (b) {};
\foreach \X in {1,...,\n}{
\pgfmathparse{int(1+Mod(\X-4,6))}
\draw \ifnum\X=4 [fill=green!20] \else [fill=blue!20]\fi
(a.corner \X) circle[radius=1em] node {\pgfmathresult};
}
\path foreach \X in {1,...,\n}
{ [/utils/exec=\pgfmathtruncatemacro{\mymiss}{int(1+Mod(\X-4,6))}]
(b.corner \X) pic(m\X){missing polygon={miss={\mymiss}}}};
\end{tikzpicture}
\end{document}