我有一个特殊的练习,我希望我创建的图表tikzpicture
居中,但我还希望\item
包含它的标签保持不变,并与同一级别的其他枚举器保持一致。我已经成功找到了无数其他问题的解决方案(这是我的第一个 LaTeX 文档),但无论我尝试了什么答案,标签\item
(在本例中为 a (b))似乎都会被拖着走,如下所示:
以下是我尝试过的几个例子:
\begin{center}
\begin{tikzpicture}
....
\end{tikzpicture}
\end{center}
{\centering
\begin{tikzpicture}
....
\end{tikzpicture}
}
\begin{figure}
\begin{center}
\begin{tikzpicture}
....
\end{tikzpicture}
\end{center}
\end{figure}
\begin{tikzpicture}
\centering
....
\end{tikzpicture}
...等等。我用前两个图得到了“最佳”结果,它们在我看来是相同的。(b)被拉到中心,并且tikzpicture
和它下面的文本之间没有适当的垂直间距。但是,后续图表(它们都与\item
第一个图表相同)没有这个间距问题。在我看来,\begin{center}
拉入一些应该超出其范围的东西似乎很奇怪,但我确信有一个简单的解释。这是一个最小(ish)的工作示例(我打算重构它以尽量减少重复):
\documentclass[12pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb,bm,enumitem,tikz}
\usetikzlibrary{calc,positioning,shapes}
\tikzset{
between/.style args={#1 and #2}{
at = ($(#1)!0.5!(#2)$)
}
}
\tikzstyle{mydot}=[circle,fill,inner sep=2pt]
\tikzstyle{myellipse}=[ellipse,draw=cyan,thick,minimum width=60pt,minimum height=110pt,inner sep=0pt]
\tikzstyle{every edge}=[draw=cyan,thick]
\tikzstyle{relarrow}=[->, shorten >= 0pt, shorten <= 0pt, out=25, in=155]
\begin{document}
\begin{enumerate}
\item
\begin{enumerate}
\item $R=\{(5,5),(6,5),(6,6)\}$,\\
$S=\{(4,6),(5,5),(5,7),(6,6)\}$
\item % Relation R
\begin{center}
\begin{tikzpicture}[
>=latex,
shorten >= 3pt,
shorten <= 3pt,
baseline={([yshift=-1em] current bounding box.north)}
]
\node[mydot,label={left:4}] (a1) {};
\node[mydot,below=of a1,label={left:5}] (a2) {};
\node[mydot,below=of a2,label={left:6}] (a3) {};
\node[mydot,right=5cm of a1,label={right:5}] (b1) {};
\node[mydot,below=of b1,label={right:6}] (b2) {};
\node[mydot,below=of b2,label={right:7}] (b3) {};
\path[->] (a2) edge (b1);
\path[->] (a3) edge (b1)
edge(b2);
\node[myellipse,label={above:A}] at (a2.center) (c1) {};
\node[myellipse,label={above:B}] at (b2.center) (c2) {};
\node at (2.5, 0.75) (r1) {$R$};
\path[relarrow] (2.1, 0.4) edge (2.95, 0.35);
\end{tikzpicture}
\end{center}
$R$ is not a function because it fails both
properties (1) and (2) of the definition of a function.
$4\in A$, but there is no $(4,y)\in R$ for any
$y\in B$.
Also, \mbox{$(6,5)\in R$} and $(6,6)\in R\text{, but }
5\neq6$.
% Relation S
\begin{center}
\begin{tikzpicture}[
>=latex,
shorten >= 3pt,
shorten <= 3pt,
baseline={([yshift=-1em] current bounding box.north)}
]
\node[mydot,label={left:4}] (a1) {};
\node[mydot,below=of a1,label={left:5}] (a2) {};
\node[mydot,below=of a2,label={left:6}] (a3) {};
\node[mydot,right=5cm of a1,label={right:5}] (b1) {};
\node[mydot,below=of b1,label={right:6}] (b2) {};
\node[mydot,below=of b2,label={right:7}] (b3) {};
\path[->] (a1) edge (b2);
\path[->] (a2) edge (b1)
edge (b3);
\path[->] (a3) edge (b2);
\node[myellipse,label={above:A}] at (a2.center) (c1) {};
\node[myellipse,label={above:B}] at (b2.center) (c2) {};
\node at (2.5, 0.75) (r1) {$S$};
\path[relarrow] (2.1, 0.4) edge (2.95, 0.35);
\end{tikzpicture}
\end{center}
$S$ is not a function because it fails property
(2) of the definition of a function.
$(5,5)\in S\text{ and }(5,7)\in S\text{, but }
5\neq7$.
\end{enumerate}
\end{enumerate}
\end{document}
答案1
我会将此类图表视为数学显示,但添加一些介绍性文字会更好。
\documentclass[12pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb,bm,enumitem,tikz}
\usetikzlibrary{calc,positioning,shapes}
\tikzset{
between/.style args={#1 and #2}{at = ($(#1)!0.5!(#2)$)},
mydot/.style={circle,fill,inner sep=2pt},
myellipse/.style={
ellipse,
draw=cyan,
thick,
minimum width=60pt,
minimum height=110pt,
inner sep=0pt
},
every edge/.style={draw=cyan,thick},
relarrow/.style={->, shorten >= 0pt, shorten <= 0pt, out=25, in=155},
}
\begin{document}
\begin{enumerate}
\item
\begin{enumerate}
\item $R=\{(5,5),(6,5),(6,6)\}$,\\
$S=\{(4,6),(5,5),(5,7),(6,6)\}$
\item % Relation R
\[
\begin{tikzpicture}[
>=latex,
shorten >= 3pt,
shorten <= 3pt,
baseline={([yshift=-1em] current bounding box.north)}
]
\node[mydot,label={left:4}] (a1) {};
\node[mydot,below=of a1,label={left:5}] (a2) {};
\node[mydot,below=of a2,label={left:6}] (a3) {};
\node[mydot,right=5cm of a1,label={right:5}] (b1) {};
\node[mydot,below=of b1,label={right:6}] (b2) {};
\node[mydot,below=of b2,label={right:7}] (b3) {};
\path[->] (a2) edge (b1);
\path[->] (a3) edge (b1) edge(b2);
\node[myellipse,label={above:$A$}] at (a2.center) (c1) {};
\node[myellipse,label={above:$B$}] at (b2.center) (c2) {};
\node at (2.5, 0.75) (r1) {$R$};
\path[relarrow] (2.1, 0.4) edge (2.95, 0.35);
\end{tikzpicture}
\]
$R$ is not a function because it fails both
properties (1) and (2) of the definition of a function.
$4\in A$, but there is no $(4,y)\in R$ for any $y\in B$.
Also, $(6,5)\in R$ and $(6,6)\in R$, but~$5\neq6$.
\item % Relation S
\[
\begin{tikzpicture}[
>=latex,
shorten >= 3pt,
shorten <= 3pt,
baseline={([yshift=-1em] current bounding box.north)}
]
\node[mydot,label={left:4}] (a1) {};
\node[mydot,below=of a1,label={left:5}] (a2) {};
\node[mydot,below=of a2,label={left:6}] (a3) {};
\node[mydot,right=5cm of a1,label={right:5}] (b1) {};
\node[mydot,below=of b1,label={right:6}] (b2) {};
\node[mydot,below=of b2,label={right:7}] (b3) {};
\path[->] (a1) edge (b2);
\path[->] (a2) edge (b1) edge (b3);
\path[->] (a3) edge (b2);
\node[myellipse,label={above:$A$}] at (a2.center) (c1) {};
\node[myellipse,label={above:$B$}] at (b2.center) (c2) {};
\node at (2.5, 0.75) (r1) {$S$};
\path[relarrow] (2.1, 0.4) edge (2.95, 0.35);
\end{tikzpicture}
\]
$S$ is not a function because it fails property (2) of
the definition of a function.
$(5,5)\in S$ and $(5,7)\in S$, but~$5\neq7$.
\end{enumerate}
\end{enumerate}
\end{document}
需要注意的几点:
\tikzstyle
已弃用集合名称应采用数学模式:
label={above:$A$}
适用于此文本
Also, \mbox{$(6,5)\in R$} and $(6,6)\in R\text{, but } 5\neq6$.
应该
Also, $(6,5)\in R$ and $(6,6)\in R$, but~$5\neq6$.
同样地,文本
$(5,5)\in S\text{ and }(5,7)\in S\text{, but } 5\neq7$.
应该
$(5,5)\in S$ and $(5,7)\in S$, but~$5\neq7$.
\]
和以下文本之间不要留空行。
答案2
答案3
像这样:
解决方案相当简单。在 之后,\item
只需b
添加~
或\mbox{}
。完整的 MWE 是(经过微小更改后)的:
\documentclass[12pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb,bm}
\usepackage{enumitem,tikz}
\usepackage{tikz}
\usetikzlibrary{calc,positioning,shapes}
\tikzset{ >=latex,
shorten >=3pt,
shorten <=3pt,
between/.style args ={#1 and #2}{at = ($(#1)!0.5!(#2)$)},
mydot/.style ={circle,fill,inner sep=2pt},
myellipse/.style ={ellipse,draw=cyan,thick,
minimum width=60pt,minimum height=110pt,
inner sep=0pt},
every edge/.style ={draw=cyan,thick},
relarrow/.style ={->, out=25, in=155}
}
\begin{document}
\begin{enumerate}
\item
\begin{enumerate}
\item $R=\{(5,5),(6,5),(6,6)\}$,\\
$S=\{(4,6),(5,5),(5,7),(6,6)\}$
\item \mbox{}% <-- added (Relation R )
\begin{center}
\begin{tikzpicture}[
baseline={([yshift=-1em] current bounding box.north)}
]
\node[mydot,label={left:4}] (a1) {};
\node[mydot,below=of a1,label={left:5}] (a2) {};
\node[mydot,below=of a2,label={left:6}] (a3) {};
\node[mydot,right=5cm of a1,label={right:5}] (b1) {};
\node[mydot,below=of b1,label={right:6}] (b2) {};
\node[mydot,below=of b2,label={right:7}] (b3) {};
\path[->] (a2) edge (b1);
\path[->] (a3) edge (b1) edge (b2);
\node[myellipse,label={above:A}] at (a2.center) (c1) {};
\node[myellipse,label={above:B}] at (b2.center) (c2) {};
\node at (2.5, 0.75) (r1) {$R$};
\path[relarrow] (2.1, 0.4) edge (2.95, 0.35);
\end{tikzpicture}
\end{center}
$R$ is not a function because it fails both properties (1) and (2) of the definition of a function. $4\in A$, but there is no $(4,y)\in R$ for any $y\in B$. Also, $(6,5)\in R$ and $(6,6)\in R$, but $5\neq6$.
% Relation S
\begin{center}
\begin{tikzpicture}[
baseline={([yshift=-1em] current bounding box.north)}
]
\node[mydot,label=left:4] (a1) {};
\node[mydot,below=of a1,label=left:5] (a2) {};
\node[mydot,below=of a2,label=left:6] (a3) {};
\node[mydot,right=5cm of a1,label=right:5] (b1) {};
\node[mydot,below=of b1,label=right:6] (b2) {};
\node[mydot,below=of b2,label=right:7] (b3) {};
\path[->] (a1) edge (b2);
\path[->] (a2) edge (b1)
edge (b3);
\path[->] (a3) edge (b2);
\node[myellipse,label=above:A] at (a2.center) (c1) {};
\node[myellipse,label=above:B] at (b2.center) (c2) {};
\node at (2.5, 0.75) (r1) {$S$};
\path[relarrow] (2.1, 0.4) edge (2.95, 0.35);
\end{tikzpicture}
\end{center}
$S$ is not a function because it fails property (2) of the definition of a function. $(5,5)\in S$ and $(5,7)\in S$, but $5\neq7$.
\end{enumerate}
\end{enumerate}
\end{document}