Square curly bracket symbol

bracketsdelimitersmath-modemath-operators

I would like to create "square-curly bracket" delimiters, somehow looking as in this question. I came up with the code below, which produces some good results, but has also some inconvenients. For instance we get these examples:


You can see that the fraction bar is not really centered (because of the exponent 'j'), and the formatting under the summation symbols is too big.

My question is:

How would you proceed to have nice-looking "square-curly" brackets?

MWE:

\documentclass{article}

\usepackage{tikz, mathtools, titletoc}

\DeclareRobustCommand{\fp}[1]{%
    \let\mybox\relax%
    \newsavebox\mybox%
    \sbox{\mybox}{$#1$}%
    \def\WIDTH{\the\dimexpr\wd\mybox + 6pt \relax}%
    \def\HEIGHT{\the\dimexpr\ht\mybox * 10 / 20 +  \dp\mybox * 10/20 + 2pt\relax}%
    \def\SHIFT{2pt}%
    \begin{tikzpicture}[baseline = -0.57ex]
    \draw  [line width=0.6]  (\SHIFT, \HEIGHT) -- (0, \HEIGHT) -- (0, 1.0pt) -- (-1.5pt, 0) -- (0, -1.0pt) -- (0, -\HEIGHT) -- (\SHIFT, -\HEIGHT);%
    \node[anchor = west] at (-0.01, 0) {\copy\mybox};%
    \draw  [line width=0.6]  (-\SHIFT + \WIDTH, \HEIGHT) -- (0 + \WIDTH, \HEIGHT) -- (0 + \WIDTH, 1.0pt) -- (1.5pt + \WIDTH, 0) -- (0 + \WIDTH, -1.0pt) -- (0 + \WIDTH, -\HEIGHT) -- (-\SHIFT + \WIDTH, -\HEIGHT);%
    \end{tikzpicture}%
}


\begin{document}

$\fp{x}         \qquad          \fp{ \frac{x a}{d} } = \left\{ \frac{x a}{d} \right\}    \leq \frac{1}{2}$

\[      \fp{ -x } = \fp{ \dfrac{a x}{d} } = \fp{ \frac{x}{2} }   = 
\fp{ \dfrac{- t \cdot a^j }{ R } }      =       \fp{ \dfrac{- t  a_j }{ R } }   \]

\[      \sum_{ \fp{r/m} \in S } r       =       \sum_{ \fp{ \frac{r}{m} } \in S } r.        \]

\end{document}

Best Answer

Here is a different approach. First we define a tikz style called sqbrace that draws the new brace between points. It has an optional argument to shift it perpendicular to its direction. For example,

\begin{tikzpicture}
\draw(0,0)--(2,1);
\draw[sqbrace=2](0,0)--(2,1);
\end{tikzpicture}

enter image description here

The 2 shifts the brace by 2 units, where a unit is the amount of overhang in the brace (set to 1.5pt.

Then we define a macro \mybrace{<contents>} that draws the braces around a node containing <contents>. The node is vcentered and contains vphantom contents to get the baseline in the right place.

enter image description here

\documentclass{article}

\usepackage{tikz, amsmath}
\usetikzlibrary{decorations.pathreplacing, calc}
\tikzset{
  sqbrace/.style={decorate, decoration={show path construction,
    lineto code={
      \path (\tikzinputsegmentfirst); \pgfgetlastxy{\xstart}{\ystart}
      \path (\tikzinputsegmentlast); \pgfgetlastxy{\xend}{\yend}
      \path ($(0,0)!1.5pt!(\ystart-\yend,\xend-\xstart)$); \pgfgetlastxy{\xperp}{\yperp}
      \path ($(0,0)!1.5pt!(\xend-\xstart, \yend-\ystart)$); \pgfgetlastxy{\xpar}{\ypar}
      \draw[line width=.5pt, shift={(#1*\xperp,#1*\yperp)}] (\xstart-\xperp,\ystart-\yperp)--(\xstart,\ystart)--
        ([shift={(-.5*\xpar,-.5*\ypar)}]$.5*(\xstart,\ystart)+.5*(\xend,\yend)$)--
        ([shift={(.866*\xperp,.866*\yperp)}]$.5*(\xstart,\ystart)+.5*(\xend,\yend)$)--
        ([shift={(.5*\xpar,.5*\ypar)}]$.5*(\xstart,\ystart)+.5*(\xend,\yend)$)--
        (\xend,\yend)--(\xend-\xperp,\yend-\yperp);
    }
  }},
  sqbrace/.default={0}
}
\newcommand{\mybrace}[1]{{}\mathbin{\vcenter{\hbox{\tikz{
  \node[inner ysep=.0pt, inner xsep=2pt](M){$#1\vphantom{\left(#1\right)}$};
  \draw[sqbrace](M.north east)--(M.south east);
  \draw[sqbrace](M.south west)--(M.north west);
}}}}{}}

\begin{document}
\[
\sum_{j=0}^N\mybrace{\dfrac{-t\cdot a^j}{R}}\qquad\sum_{\mybrace{\scriptstyle r/m}\in S}r\qquad\sum_{\mybrace{\frac{r}{m}}\in S}r
\]

\begin{tikzpicture}
\draw(0,0)--(2,1);
\draw[sqbrace=2](0,0)--(2,1);
\end{tikzpicture}
\end{document}
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