tikz 中用于访问标签的循环

Question

我想使用 Tikz 创建类似这样的绘图：

我找到了一个很好的 tikz 示例，它创建了类似的东西：

\documentclass[border=10pt]{standalone}%
\usepackage{tikz}

\begin{document}


\begin{tikzpicture}
\foreach \n in {0,...,4} {
  \foreach \k in {0,...,\n} {
    \node at (\k-\n/2,-\n) {${\n \choose \k}$};
  }
}
\end{tikzpicture}

\end{document}

我以为通过访问列表上的标签来更改节点上的标签很简单，例如lst[\n]，其中lst包含单项式，但结果比我想象的要复杂得多。此外，我找不到以编程方式添加指数的方法。使用循环的 tikz 可以做到这一点吗，还是我必须创建一个临时解决方案？

Answer 1

扩展你的例子：

\documentclass[border=10pt]{standalone}%
\usepackage{ifthen}
\usepackage{tikz}
\usetikzlibrary{math}

\begin{document}

\newcommand{\exponent}[2]{\ifthenelse{#2=0}{}{\ifthenelse{#2=1}{$#1$}{$#1^{#2}$}}}

\begin{tikzpicture}
\foreach \x in {0,...,7} {
  \foreach \n in {0,...,\x} {

    \pgfmathparse{int(subtract(\x,\n))}
    \edef\xexp{\pgfmathresult}
    \edef\nexp{\n}

    \node[inner sep = 8pt] (X\xexp-N\nexp) at (2*\n-\x,-\x) {\ifthenelse{\xexp=0 \AND \nexp=0}{$1$}{\exponent{\xi}{\xexp}\exponent{\eta}{\nexp}}};
    \node[inner sep = 16pt] (dX\xexp-N\nexp) at (2*\n-\x,-\x) {}; % d: dummy
  }
}

\draw[blue, thick, dashed] (X0-N0.north) -- (X1-N0.west) -- (X1-N1.south) -- (X0-N1.east) node[inner sep = 0] (S1) {} -- cycle;

\draw[red, thick, dashed] (dX0-N0.north) -- (dX4-N0.west) -- (dX4-N1.south) -- (dX3-N1.south) --
(dX1-N3.south) -- (dX1-N4.south) -- (dX0-N4.east) node[inner sep = 0] (S4) {} -- cycle;

\node[right = 0.5cm, above = 0.5cm] (S1t) at (S1.west) {$S_1(\Omega_{\rm st}^{(q)})$};
\draw[blue] (S1t) edge[thick, out=270, in=90, ->] (S1);

\node[right = 0.5cm, above = 0.5cm] (S4t) at (S4.west) {$S_4(\Omega_{\rm st}^{(q)})$};
\draw[red] (S4t) edge[thick, out=270, in=90, ->] (S4);

\end{tikzpicture}

\end{document}