如何將此 ntheorem 解決方案移植到 amsthm + thmtools？

Question

是否可以重現這個愉快的定理環境使用amsthm和thmtools?

Answer 1

您也可以使用該套件mdframed來繪製框架。定理的設定可以透過您最喜歡的套件來完成。設定定理環境後，您可以用 包圍該環境\surroundwithmdframed。對於您所需的樣式，您可以使用以下定義，其中短線的長度為 0.25 公分。當然這可以做得更優雅。

_{為了簡化程式碼，我使用calcTikZ 函式庫}

\usepackage[framemethod=TikZ]{mdframed}
\usetikzlibrary{calc}
\makeatletter
\newrobustcmd*\mdf@tikzbox@tfl@spare[1]{%three or four borders
    \clip(0,0)rectangle(\mdfboundingboxwidth,\mdfboundingboxheight);%
    \begin{scope}[mdfcorners]%
       \clip[preaction=mdfouterline]%
            [postaction=mdfbackground]%
            [postaction=mdfinnerline]#1;%
    \end{scope}%
    \path[mdfmiddleline,mdfcorners]($(O|-P)-(0,0.25cm)$)--(O|-P)--(P)--($(P)-(0,0.25cm)$);
    \path[mdfmiddleline,mdfcorners]($(P|-O)+(0,0.25cm)$)--(P|-O)--(O)--($(O)+(0,0.25cm)$);
  }%
\newrobustcmd*\changelinestyle{\let\mdf@tikzbox@tfl\mdf@tikzbox@tfl@spare}
\makeatother
\surroundwithmdframed[settings=\changelinestyle]{theorem}

透過這個定義，所有選項mdframed都是允許的。

\documentclass{article}
\usepackage{amsmath,amsthm}
\usepackage{thmtools}
\declaretheorem[]{theorem}

\usepackage[framemethod=TikZ]{mdframed}
\usetikzlibrary{calc}
\makeatletter
\newrobustcmd*\mdf@tikzbox@tfl@spare[1]{%three or four borders
    \clip(0,0)rectangle(\mdfboundingboxwidth,\mdfboundingboxheight);%
    \begin{scope}[mdfcorners]%
       \clip[preaction=mdfouterline]%
            [postaction=mdfbackground]%
            [postaction=mdfinnerline]#1;%
    \end{scope}%
    \path[mdfmiddleline,mdfcorners]($(O|-P)-(0,0.25cm)$)--(O|-P)--(P)--($(P)-(0,0.25cm)$);
    \path[mdfmiddleline,mdfcorners]($(P|-O)+(0,0.25cm)$)--(P|-O)--(O)--($(O)+(0,0.25cm)$);
  }%
\newrobustcmd*\changelinestyle{\let\mdf@tikzbox@tfl\mdf@tikzbox@tfl@spare}
\makeatother
\surroundwithmdframed[settings=\changelinestyle,middlelinecolor=blue,roundcorner=3pt,middlelinewidth=1.2pt]{theorem}

\usepackage{lipsum}
\begin{document}
\lipsum[2]

\begin{theorem}%
If~$G$ is a connected graph of order $n\geq 3$ and size~$m$, then
\[
    g(G)\geq \frac{m}{6}-\frac{n}{2}+1.%
\]
\end{theorem}%

\lipsum[2]

\end{document}