我正在尝试用文本写出一组方程式来推断下一组方程式。我想做的事情如下:
\documentclass{article}
\begin{document}
And so,
\(\left. \begin{array}[t]{l}
a \\
b \\
c
\end{array} \right\}
\Rightarrow \left. \begin{array}[t]{l}
d \\
e \\
f
\end{array} \right\}\)
\end{document}
但它并没有产生期望的结果(底部),而是产生了顶部的结果。
除 之外我还可以使用其他符号吗\}
?
答案1
虽然有点麻烦,但还是可以的。使用abraces
你可以对齐整个结构垂直然后将其旋转/移动到位:
\documentclass{article}
\usepackage{abraces,graphicx}
\newcommand{\mrot}[1]{\rotatebox{90}{$#1\mathstrut$}}
\begin{document}
And so,
\raisebox{.8\baselineskip}{\rotatebox{-90}{$\setlength{\arraycolsep}{.5\arraycolsep}
\begin{array}{@{}c@{}}
\aunderbrace[lD1r]{\begin{array}{rrr}
\mrot{d} & \mrot{e} & \mrot{f}
\end{array}} \\
\mrot{\Rightarrow} \hspace{2.3\normalbaselineskip} \\
\aoverbrace[LU1R]{\begin{array}{rrr}
\mrot{a} & \mrot{b} & \mrot{c}
\end{array}}
\end{array}$}}
\end{document}
答案2
我不喜欢不对称的括号,但这是一个使用 TikZ 的机会:
代码:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.pathreplacing,matrix,positioning,calc}
\tikzset{
mybrace/.style={
decorate,
decoration={brace,aspect=#1},
line width=1pt
}
}
\begin{document}
And so,
\(
\left.
\begin{array}{l}
a \\
b \\
c
\end{array}
\right\}
\Rightarrow
\left.
\begin{array}{l}
d \\
e \\
f
\end{array}
\right\}
\)
And so,
\begin{tikzpicture}[baseline=2.5ex,every node/.style={text depth=0.2ex,text height=1.3ex}]
\matrix[matrix of math nodes]
(mat1)
{
a \\
b \\
c \\
};
\matrix[matrix of math nodes,right=0.8cm of mat1]
(mat2)
{
d \\
e \\
f \\
};
\node[yshift=-0.8ex]
at ( $ (mat1-1-1.east)!0.5!(mat2-1-1.west) $ )
{$\Rightarrow$};
\foreach \Valor in {1,2}
\draw[mybrace=0.25]
(mat\Valor-1-1.north east) -- (mat\Valor-1-1.north east|-mat\Valor-3-1.south east);
\end{tikzpicture}
\end{document}
答案3
我不会使用不对称的括号,而只是用数组偏移括号,这样做delarray
的目的是为了方便:
\documentclass{article}
\usepackage{delarray}
\begin{document}
And so,
\(\begin{array}[t].{l}\}
a \\
b \\
c
\end{array}
\begin{array}[t]{@{}c@{}}\\{}\Rightarrow {}\\{}\end{array}
\begin{array}[t].{l}\}
d \\
e \\
f
\end{array}\)
\end{document}
答案4
我相信你能逃脱
\documentclass{article}
\begin{document}
And so, $(a,b,c)\Rightarrow (d,e,f)$
\end{document}