我想将 tikzcd 图表与文本并排放置，但正如我们在下面的示例中看到的那样，有时图表（或文本）会转到下一页。
抱歉，文件太大了。
先感谢您。
\documentclass[a4paper,12pt,reqno,twoside]{book}
\usepackage{amsmath}
\usepackage[latin1]{inputenc}
\usepackage[brazil]{babel}
\usepackage[adobe-utopia]{mathdesign}
\usepackage{helvet}
\renewcommand{\familydefault}{\sfdefault}
\usepackage{setspace}
\usepackage{amsthm}
\usepackage{paralist}
\usepackage{needspace}
\usepackage{paracol}
\usepackage[pdftex]{graphicx}
\usepackage[dvips,a4paper,top=2.54cm,bottom=2.0cm,left=2.0cm,right=2.54cm]{geometry}
\onehalfspacing
\parskip8pt
\vfuzz3pt
\hfuzz3pt
\usepackage{tikz-cd}
\tikzset{
every picture/.prefix style={
execute at begin picture=\shorthandoff{"}
}
}
\renewenvironment{proof}{{\bfseries Demonstração}}{\qed}
\begin{document}
\begin{proof}
\begin{sloppypar}
\begin{paracol}{2}
\begin{inparaenum}[(1)]
\noindent
\item
\begin{tikzcd}
& \square \arrow[dl, swap, "\alpha_1"] & \\
\bullet \arrow[rr, swap, "\alpha_2"] & & \bullet \arrow[ul, swap, "\alpha_3"] \arrow[d,"\alpha_4"] \\
\bullet \arrow[u, "\alpha_5"] & & \bullet \arrow[ll, "\alpha_6"]
\end{tikzcd}
\switchcolumn
\needspace{4\baselineskip}
As relações do tipo $2$ são: $r1=\alpha_1 \alpha_2 \alpha_4$ e $r2=\alpha_5 \alpha_2 \alpha_3$. Os ciclos elementares são: $C_1 = \alpha_2 \alpha_3 \alpha_1$ e $C_2 = \alpha_2 \alpha_4 \alpha_6 \alpha_5$. Analogamente ao anterior, temos a solução $47$.
\switchcolumn*
\noindent
\item
\begin{tikzcd}
&& \bullet \arrow[dl, swap, "\alpha_1"] &&\\
\bullet \arrow[r, bend left, "\alpha_2"] & \bullet \arrow[rr, swap, "\alpha_4"] \arrow[l, bend left, "\alpha_3"] && \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d,"\alpha_6"] & \\
&\bullet \arrow[u, "\alpha_7"] && \bullet \arrow[ll, "\alpha_8"] \arrow[r, bend left, "\alpha_9"]& \bullet \arrow[l, bend left, "\alpha_{10}"]
\end{tikzcd}
\switchcolumn
\needspace{4\baselineskip}
As relações do tipo $2$ são: $r1=\alpha_1 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_2 \alpha_4$, $r4=\alpha_1 \alpha_3$, $r5=\alpha_7 \alpha_3$, $r6=\alpha_6 \alpha_9$ e $r7=\alpha_{10} \alpha_8$. Os ciclos elementares são: $C_1 = \alpha_1 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_7$, $C_3=\alpha_2 \alpha_3$ e $C_4=\alpha_9 \alpha_{10}$. Logo, o programa mostra nenhuma solução.
\switchcolumn*
\noindent
\item
\begin{tikzcd}
\bullet \arrow[dr, bend left, "\alpha_2"] && \bullet \arrow[dl, swap, "\alpha_1"] &\\
& \bullet \arrow[rr, swap, "\alpha_4"] \arrow[ul, bend left, "\alpha_3"] && \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d,"\alpha_6"] \\
\bullet \arrow[r, bend left, "\alpha_{10}"] &\bullet \arrow[l, bend left, "\alpha_9"] \arrow[u, swap, "\alpha_7"] && \bullet \arrow[ll, "\alpha_8"]
\end{tikzcd}
\switchcolumn
\needspace{4\baselineskip}
As relações do tipo $2$ são: $r1=\alpha_1 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_2 \alpha_4$, $r4=\alpha_1 \alpha_3$, $r5=\alpha_7 \alpha_3$, $r6=\alpha_8 \alpha_9$ e $r7=\alpha_{10} \alpha_7$. Os ciclos elementares são: $C_1 = \alpha_1 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_7$, $C_3=\alpha_2 \alpha_3$ e $C_4=\alpha_9 \alpha_{10}$. Não obtemos nenhuma álgebra de incidência não hereditária.
\switchcolumn*
\noindent
\item
\begin{tikzcd}
\bullet \arrow[dr, bend left, "\alpha_2"] && \bullet \arrow[dl, swap, "\alpha_1"] && \bullet \arrow[dl, bend left, "\alpha_{10}"]\\
& \bullet \arrow[rr, swap, "\alpha_4"] \arrow[ul, bend left, "\alpha_3"] && \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d,"\alpha_6"] \arrow[ur, bend left, "\alpha_9"] &\\
&\bullet \arrow[u, swap, "\alpha_7"] && \bullet \arrow[ll, "\alpha_8"] &
\end{tikzcd}
\switchcolumn
\needspace{4\baselineskip}
As relações do tipo $2$ são: $r1=\alpha_1 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_2 \alpha_4$, $r4=\alpha_1 \alpha_3$, $r5=\alpha_7 \alpha_3$, $r6=\alpha_4 \alpha_9$, $r7=\alpha_{10} \alpha_5$ e $r8=\alpha_{6} \alpha_7$. Os ciclos elementares são: $C_1 = \alpha_1 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_7$, $C_3=\alpha_2 \alpha_3$ e $C_4=\alpha_9 \alpha_{10}$. Logo, o programa mostra a álgebra $48$.
\switchcolumn*
\noindent
\item
\begin{tikzcd}
& \bullet \arrow[d, bend left, "\alpha_2"] &&\\
& \bullet \arrow[dl, swap, "\alpha_1"] \arrow[u, bend left, "\alpha_3"] && \\
\bullet \arrow[rr, swap, "\alpha_4"] && \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d,"\alpha_6"] &\\
\bullet \arrow[u, "\alpha_7"] && \bullet \arrow[ll, "\alpha_8"] \arrow[r, bend left, "\alpha_9"] & \bullet \arrow[l, bend left, "\alpha_{10}"]
\end{tikzcd}
\switchcolumn
\needspace{4\baselineskip}
As relações do tipo $2$ são: $r1=\alpha_1 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_5 \alpha_3$, $r4=\alpha_2 \alpha_1$, $r5=\alpha_{10} \alpha_8$ e $r6=\alpha_6 \alpha_9$. Os ciclos elementares são: $C_1 = \alpha_1 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_7$, $C_3=\alpha_2 \alpha_3$ e $C_4=\alpha_9 \alpha_{10}$. Portanto não tem nenhuma álgebra de incidência não hereditária.
\switchcolumn*
\noindent
\item
\begin{tikzcd}
& \bullet \arrow[d, bend left, "\alpha_1"] & \\
& \bullet \arrow[dl, swap, "\alpha_3"] \arrow[u, bend left, "\alpha_2"] & \\
\bullet \arrow[rr, swap, "\alpha_4"] & & \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d, "\alpha_6"] \\
\bullet \arrow[u, "\alpha_7"] & & \bullet \arrow[dl, "\alpha_8"] \\
& \bullet \arrow[ul, "\alpha_9"] &
\end{tikzcd}
\switchcolumn
\needspace{4\baselineskip}
As relações do tipo $2$ são: $r1=\alpha_3 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_5 \alpha_2$ e $r4=\alpha_1 \alpha_3$. Os ciclos elementares são: $C_1 = \alpha_3 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_9 \alpha_7$ e $C_3=\alpha_1 \alpha_2$. Portanto, não temos nenhuma solução.
\end{inparaenum}
\end{paracol}
\end{sloppypar}
\end{proof}
\end{document}
答案1
抱歉，找不到，brazil但这babel似乎对我有用。
\documentclass[a4paper,12pt,reqno,twoside]{book}
\usepackage{amsmath,paralist}
\usepackage[latin1]{inputenc}
\usepackage[brazil]{babel}
\usepackage[adobe-utopia]{mathdesign}
\usepackage{helvet}
\renewcommand{\familydefault}{\sfdefault}
\usepackage{setspace}
\usepackage{amsthm}
\usepackage{paralist}
\usepackage{needspace}
\usepackage{paracol}
\usepackage[pdftex]{graphicx}
\usepackage[dvips,a4paper,top=2.54cm,bottom=2.0cm,left=2.0cm,right=2.54cm]{geometry}
\onehalfspacing
\parskip8pt
\vfuzz3pt
\hfuzz3pt
\usepackage{tikz-cd}
\tikzset{
every picture/.prefix style={
execute at begin picture=\shorthandoff{"}
}
}
\renewenvironment{proof}{{\bfseries Demonstração}}{\qed}
\begin{document}
\begin{proof}
\begin{sloppypar}
\begin{compactenum}[(1)]
\item \begin{tikzcd}
& \square \arrow[dl, swap, "\alpha_1"] & \\
\bullet \arrow[rr, swap, "\alpha_2"] & & \bullet \arrow[ul, swap, "\alpha_3"] \arrow[d,"\alpha_4"] \\
\bullet \arrow[u, "\alpha_5"] & & \bullet \arrow[ll, "\alpha_6"]
\end{tikzcd}\hfill%
\begin{minipage}{.55\linewidth}
\needspace{4\baselineskip}
\noindent As relações do tipo $2$ são: $r1=\alpha_1 \alpha_2 \alpha_4$ e $r2=\alpha_5 \alpha_2 \alpha_3$. Os ciclos elementares são: $C_1 = \alpha_2 \alpha_3 \alpha_1$ e $C_2 = \alpha_2 \alpha_4 \alpha_6 \alpha_5$. Analogamente ao anterior, temos a solução $47$.
\end{minipage}
\item\begin{tikzcd}
&& \bullet \arrow[dl, swap, "\alpha_1"] &&\\
\bullet \arrow[r, bend left, "\alpha_2"] & \bullet \arrow[rr, swap, "\alpha_4"] \arrow[l, bend left, "\alpha_3"] && \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d,"\alpha_6"] & \\
&\bullet \arrow[u, "\alpha_7"] && \bullet \arrow[ll, "\alpha_8"] \arrow[r, bend left, "\alpha_9"]& \bullet \arrow[l, bend left, "\alpha_{10}"]
\end{tikzcd}\hfill%
\begin{minipage}{.55\linewidth}
As relações do tipo $2$ são: $r1=\alpha_1 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_2 \alpha_4$, $r4=\alpha_1 \alpha_3$, $r5=\alpha_7 \alpha_3$, $r6=\alpha_6 \alpha_9$ e $r7=\alpha_{10} \alpha_8$. Os ciclos elementares são: $C_1 = \alpha_1 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_7$, $C_3=\alpha_2 \alpha_3$ e $C_4=\alpha_9 \alpha_{10}$. Logo, o programa mostra nenhuma solução.
\end{minipage}
\item\begin{tikzcd}
\bullet \arrow[dr, bend left, "\alpha_2"] && \bullet \arrow[dl, swap, "\alpha_1"] &\\
& \bullet \arrow[rr, swap, "\alpha_4"] \arrow[ul, bend left, "\alpha_3"] && \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d,"\alpha_6"] \\
\bullet \arrow[r, bend left, "\alpha_{10}"] &\bullet \arrow[l, bend left, "\alpha_9"] \arrow[u, swap, "\alpha_7"] && \bullet \arrow[ll, "\alpha_8"]
\end{tikzcd}\hfill%
\begin{minipage}{.55\linewidth}
As relações do tipo $2$ são: $r1=\alpha_1 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_2 \alpha_4$, $r4=\alpha_1 \alpha_3$, $r5=\alpha_7 \alpha_3$, $r6=\alpha_8 \alpha_9$ e $r7=\alpha_{10} \alpha_7$. Os ciclos elementares são: $C_1 = \alpha_1 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_7$, $C_3=\alpha_2 \alpha_3$ e $C_4=\alpha_9 \alpha_{10}$. Não obtemos nenhuma álgebra de incidência não hereditária.
\end{minipage}
\item\begin{tikzcd}
\bullet \arrow[dr, bend left, "\alpha_2"] && \bullet \arrow[dl, swap, "\alpha_1"] && \bullet \arrow[dl, bend left, "\alpha_{10}"]\\
& \bullet \arrow[rr, swap, "\alpha_4"] \arrow[ul, bend left, "\alpha_3"] && \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d,"\alpha_6"] \arrow[ur, bend left, "\alpha_9"] &\\
&\bullet \arrow[u, swap, "\alpha_7"] && \bullet \arrow[ll, "\alpha_8"] &
\end{tikzcd}\hfill%
\begin{minipage}{.55\linewidth}
As relações do tipo $2$ são: $r1=\alpha_1 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_2 \alpha_4$, $r4=\alpha_1 \alpha_3$, $r5=\alpha_7 \alpha_3$, $r6=\alpha_4 \alpha_9$, $r7=\alpha_{10} \alpha_5$ e $r8=\alpha_{6} \alpha_7$. Os ciclos elementares são: $C_1 = \alpha_1 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_7$, $C_3=\alpha_2 \alpha_3$ e $C_4=\alpha_9 \alpha_{10}$. Logo, o programa mostra a álgebra $48$.
\end{minipage}
\item\begin{tikzcd}
& \bullet \arrow[d, bend left, "\alpha_2"] &&\\
& \bullet \arrow[dl, swap, "\alpha_1"] \arrow[u, bend left, "\alpha_3"] && \\
\bullet \arrow[rr, swap, "\alpha_4"] && \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d,"\alpha_6"] &\\
\bullet \arrow[u, "\alpha_7"] && \bullet \arrow[ll, "\alpha_8"] \arrow[r, bend left, "\alpha_9"] & \bullet \arrow[l, bend left, "\alpha_{10}"]
\end{tikzcd}\hfill%
\begin{minipage}{.55\linewidth}
As relações do tipo $2$ são: $r1=\alpha_1 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_5 \alpha_3$, $r4=\alpha_2 \alpha_1$, $r5=\alpha_{10} \alpha_8$ e $r6=\alpha_6 \alpha_9$. Os ciclos elementares são: $C_1 = \alpha_1 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_7$, $C_3=\alpha_2 \alpha_3$ e $C_4=\alpha_9 \alpha_{10}$. Portanto não tem nenhuma álgebra de incidência não hereditária.
\end{minipage}
\item \begin{tikzcd}
& \bullet \arrow[d, bend left, "\alpha_1"] & \\
& \bullet \arrow[dl, swap, "\alpha_3"] \arrow[u, bend left, "\alpha_2"] & \\
\bullet \arrow[rr, swap, "\alpha_4"] & & \bullet \arrow[ul, swap, "\alpha_5"] \arrow[d, "\alpha_6"] \\
\bullet \arrow[u, "\alpha_7"] & & \bullet \arrow[dl, "\alpha_8"] \\
& \bullet \arrow[ul, "\alpha_9"] &
\end{tikzcd}\hfill%
\begin{minipage}{.55\linewidth}
As relações do tipo $2$ são: $r1=\alpha_3 \alpha_4 \alpha_6$, $r2=\alpha_7 \alpha_4 \alpha_5$, $r3=\alpha_5 \alpha_2$ e $r4=\alpha_1 \alpha_3$. Os ciclos elementares são: $C_1 = \alpha_3 \alpha_4 \alpha_5$, $C_2 = \alpha_4 \alpha_6 \alpha_8 \alpha_9 \alpha_7$ e $C_3=\alpha_1 \alpha_2$. Portanto, não temos nenhuma solução.
\end{minipage}
\end{compactenum}
\end{sloppypar}
\end{proof}
\end{document}