1

I'd like to put a tikzcd diagram side by side with a text but, as we can see in the example below, sometimes diagram (or text) goes to the next page.

Sorry about the very large file.

Thank you in advance.

example

\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}
  • What is the question? – sergej Aug 8 '15 at 18:46
  • What could I do to keep the tikzcd diagram in the first page? – Vasco Aug 8 '15 at 21:33
1

Sorry, can't find brazil for my babel but this seems to work for me.

enter image description here

\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}

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