我希望能够用 LaTeX 绘制此图像,但不知道如何操作。我对 LaTeX 中的图表还不熟悉,所以任何帮助我都会非常感激。
红色、绿色和蓝色圆圈以及注释不是图像的一部分;它们用于解释长度要求。此外,最终生成的图表不需要遵循相同的格式。只要画出所有线条,我就能确定定位。
它也是一个真子集符号,而不是 C。提前谢谢您。
编辑:在建议使用 Tikz 之后我明白了(感谢@SebGlav)。
\begin{center}
\begin{tikzpicture}[scale=1.5]
\node (a) at (0,0) [fill,circle,inner sep=1.2pt] {};
\node (b) at (1,0) [fill,circle,inner sep=1.2pt] {};
\node (c) at (4,0) [fill,circle,inner sep=1.2pt] {};
\node (d) at (5,0) [fill,circle,inner sep=1.2pt] {};
\draw (0,0) -- (1,0) -- (4,0) -- (5,0);
\draw [gray,decorate,decoration={brace,amplitude=5pt},
xshift=0pt,yshift=-4pt]
(1,0) -- (0,0)
node [black,midway,below=4pt,xshift=0pt]
{\footnotesize $I_1$};
\draw [gray,decorate,decoration={brace,amplitude=5pt},
xshift=0pt,yshift=-4pt]
(4,0) -- (1,0)
node [black,midway,below=4pt,xshift=0pt]
{\footnotesize $I_2$};
\draw [gray,decorate,decoration={brace,amplitude=5pt},
xshift=0pt,yshift=-4pt]
(5,0) -- (4,0)
node [black,midway,below=4pt,xshift=0pt]
{\footnotesize $I_3$};
\end{tikzpicture}
\bigskip
\begin{tikzpicture}[scale=1.5]
\node (a) at (0,0) [fill,circle,inner sep=1.2pt] {};
\node (b) at (2,0) [fill,circle,inner sep=1.2pt] {};
\node (c) at (3,0) [fill,circle,inner sep=1.2pt] {};
\node (d) at (5,0) [fill,circle,inner sep=1.2pt] {};
\draw (0,0) -- (1,0) -- (4,0) -- (5,0);
\draw [gray,decorate,decoration={brace,amplitude=5pt},
xshift=0pt,yshift=-4pt]
(2,0) -- (0,0)
node [black,midway,below=4pt,xshift=0pt]
{\footnotesize $I'_1$};
\draw [gray,decorate,decoration={brace,amplitude=5pt},
xshift=0pt,yshift=-4pt]
(3,0) -- (2,0)
node [black,midway,below=4pt,xshift=0pt]
{\footnotesize $I'_2$};
\draw [gray,decorate,decoration={brace,amplitude=5pt},
xshift=0pt,yshift=-4pt]
(5,0) -- (3,0)
node [black,midway,below=4pt,xshift=0pt]
{\footnotesize $I'_3$};
\end{tikzpicture}
\end{center}
答案1
您可以使用\rule
s 和tabular
s 绘制这些结构布局:
\documentclass{article}
\begin{document}
Consider the following partition:
\begin{center}
\begin{tabular}{ c @{\quad} c @{\quad} l }
\makebox[2em]{$I_1$}\makebox[6em]{$I_2$} \\[-.4\normalbaselineskip]
\rule[.5ex]{2em}{.4pt}\rule{.4pt}{1ex}\hspace{-.4pt}\rule[.5ex]{6em}{.4pt} &
$\rightarrow$ & partition 1 \\
\makebox[5.5em]{$I'_1$}\makebox[2.5em]{$I'_2$} \\[-.4\normalbaselineskip]
\rule[.5ex]{5.5em}{.4pt}\rule{.4pt}{1ex}\hspace{-.4pt}\rule[.5ex]{2.5em}{.4pt} &
$\rightarrow$ & partition 2
\end{tabular}
\end{center}
We note that $I_1 \subset I'_1$ and $I'_2 \subset I_2$. Thus,
\begin{center}
\begin{tabular}{ l @{\quad} c @{\quad} c }
& & \makebox[2em]{$I_1$}\makebox[3.5em]{$I^B_2$}\makebox[2.5em]{$I^A_2$} \\[-.4\normalbaselineskip]
partition 1 becomes & $\rightarrow$ &
\rule[.5ex]{2em}{.4pt}\rule{.4pt}{1ex}\hspace{-.4pt}\rule[.5ex]{3.5em}{.4pt}\rule{.4pt}{1ex}\hspace{-.4pt}\rule[.5ex]{2.5em}{.4pt} \\
& & \makebox[2em]{$I^{\prime A}_1$}\makebox[3.5em]{$I^{\prime C}_1$}\makebox[2.5em]{$I'_2$} \\[-.4\normalbaselineskip]
partition 2 becomes & $\rightarrow$ &
\rule[.5ex]{2em}{.4pt}\rule{.4pt}{1ex}\hspace{-.4pt}\rule[.5ex]{3.5em}{.4pt}\rule{.4pt}{1ex}\hspace{-.4pt}\rule[.5ex]{2.5em}{.4pt}
\end{tabular}
\end{center}
\end{document}