Symbol inside a circle in a table

Question

I need to have a sequence of lines, in which some symbols are circled. Right now I am using a table, the output is shown in the picture:

However as you can see the circles are overlapping with the border of the table, moreover the circles on SN are bigger than the one for S[0-9]. I would like to have an equal distance from the border and an equal circle size for every symbol. These should also include having the same dimension for each row. The code I am using is for the circle the following:

\usepackage{mathtools}% superior to amsmath
\usepackage{tikz}
\makeatletter
\newcommand\mathcircled[1]{%
    \mathpalette\@mathcircled{#1}%
}
\newcommand\@mathcircled[2]{%
    \tikz[baseline=(math.base)] \node[draw,circle,inner sep=1pt] (math) {$\m@th#1#2$};%
}
\makeatother

The code for the table:

\begin{center}
        \begin{table}[]
            \centering
            \label{my-label}
            \begin{tabular}{ccccc}
                \hline
                S1  & S1  & $\mathcircled{S1}$  & ... & S1  \\ \hline
                $\mathcircled{S2}$  & $\mathcircled{S2}$  & S2  & ... & S2  \\ \hline
                ... & ... & ... & ... & ... \\ \hline
                SN  & $\mathcircled{SN}$  & $\mathcircled{SN}$  & ... & SN  \\ \hline
            \end{tabular}
        \end{table}
    \end{center}

I have taken the code from a previous question: How do I draw a circle around a term in an align* equation?

Thank you.

MWE:

\documentclass[12pt]{article}

\usepackage{amsthm}
\usepackage{mathtools}
\usepackage{multirow}
\usepackage{mathtools}% superior to amsmath
\usepackage{tikz}
\makeatletter
\newcommand\mathcircled[1]{%
    \mathpalette\@mathcircled{#1}%
}
\newcommand\@mathcircled[2]{%
    \tikz[baseline=(math.base)] \node[draw,circle,inner sep=1pt] (math) {$\m@th#1#2$};%
}
\makeatother
\usepackage{tikz-cd}
\newtheorem{definition}{Definition}[section]    %% this does it

\newcommand{\cmmnt}[1]{\ignorespaces}


\begin{document}

    \newpage


    \begin{center}
        \begin{table}[]
            \centering
            \label{my-label}
            \begin{tabular}{ccccc}
                \hline
                S1  & S1  & $\mathcircled{S1}$  & ... & S1  \\ \hline
                $\mathcircled{S2}$  & $\mathcircled{S2}$  & S2  & ... & S2  \\ \hline
                ... & ... & ... & ... & ... \\ \hline
                SN  & $\mathcircled{SN}$  & $\mathcircled{SN}$  & ... & SN  \\ \hline
            \end{tabular}
        \end{table}
    \end{center}

can you make a minimal working example (MWE)? – samcarter_is_at_topanswers.xyz Apr 24 '18 at 23:15 — samcarter_is_at_topanswers.xyz, Apr 24 '18 at 23:15

score 4 · Accepted Answer · 2018-04-24T23:35:18.590

4

Here is a proposal. Note that you need to reorder table and center.

\documentclass{report}
\usepackage{mathtools}% superior to amsmath
\usepackage{tikz}
\makeatletter
\newcommand\mathcircled[1]{%
    \mathpalette\@mathcircled{#1}%
}
\newcommand\@mathcircled[2]{%
    \tikz[baseline=(math.base),outer sep=auto]{\node[draw,circle,inner
    sep=1pt]
    (math) {$#1#2$};
    \path (math.north)--++(0,1pt);
    \path (math.south)--++(0,-1pt);}%
}
\makeatother
\begin{document}
\begin{table}[ht]
            \centering

            \begin{tabular}{ccccc}
                \hline
                S1  & S1  & $\mathcircled{S1}$  & ... & S1  \\ \hline
                $\mathcircled{S2}$  & $\mathcircled{S2}$  & S2  & ... & S2  \\ \hline
                ... & ... & ... & ... & ... \\ \hline
                SN  & $\mathcircled{SN}$  & $\mathcircled{SN}$  & ... & SN  \\ \hline
            \end{tabular}
\caption{A table with circles.}\label{my-label}
\end{table}

\end{document}

UPDATE: Because you asked me to do that. @samcarter was faster anyway, but personally I'd not go that way.

\documentclass{report}
\usepackage{mathtools}% superior to amsmath
\usepackage{tikz}
\makeatletter
\newcommand\mathcircled[1]{%
    \mathpalette\@mathcircled{#1}%
}
\newcommand\@mathcircled[2]{%
    \tikz[baseline=(math.base),outer sep=auto]{\node[draw,circle,inner
    sep=1pt,minimum width=0.8cm]
    (math) {$#1#2$};
    \path (math.north)--++(0,1pt);
    \path (math.south)--++(0,-1pt);}%
}
\makeatother
\begin{document}
\begin{table}[ht]
            \centering

            \begin{tabular}{ccccc}
                \hline
                S1  & S1  & $\mathcircled{S1}$  & ... & S1  \\ \hline
                $\mathcircled{S2}$  & $\mathcircled{S2}$  & S2  & ... & S2  \\ \hline
                ... & ... & ... & ... & ... \\ \hline
                SN  & $\mathcircled{SN}$  & $\mathcircled{SN}$  & ... & SN  \\ \hline
            \end{tabular}
\caption{A table with circles.}\label{my-label}
\end{table}

\end{document}

And thanks @percusse, you're right, as always. ;-)

edited Apr 24 '18 at 23:35

answered Apr 24 '18 at 23:27

Thank you, however, I need also the circle to be of the same dimension. – Guido Muscioni Apr 24 '18 at 23:29
@GuidoMuscioni Well, I can write something assuming that $SN$ are the largest objects you will encircle. But is that really what you want? What if you encounter even larger object later? – Apr 24 '18 at 23:31
Related : when-should-we-use-begincenter-instead-of-centering – percusse Apr 24 '18 at 23:32
I mean I would like to have all circles to have the same dimension, $SN$ is a simplification, I have also more complex symbols. @marmot – Guido Muscioni Apr 24 '18 at 23:32
1

@GuidoMuscioni Without scanning twice you will not know what your widest element would be. You need to set a maximum width to have a one shot solution – percusse Apr 24 '18 at 23:34
@percusse Yet another point on which I agree with you. ;-) – Apr 24 '18 at 23:36
@percusse Do you know why outer sep does not work here? – Apr 24 '18 at 23:37
@GuidoMuscioni Thanks for accepting my answer, but consider accepting samcarter's, who was faster. – Apr 24 '18 at 23:40
I have written why I do not consider his as a solution. @marmot – Guido Muscioni Apr 24 '18 at 23:41
@GuidoMuscioni Please keep marmots answer as the accepted! – samcarter_is_at_topanswers.xyz Apr 24 '18 at 23:42
@GuidoMuscioni Please don't call me "his" :) – samcarter_is_at_topanswers.xyz Apr 24 '18 at 23:44
@marmot Outer sep is used when trying to find a point on the shape border. It doesn't actually pad a node. Test: \begin{tikzpicture}\node[outer sep=10cm] (a) at (0,0) {Z};\draw (2,2) -- (a);\end{tikzpicture} Remove and add the \draw command for different picture sizes. – percusse Apr 24 '18 at 23:46
@percusse Thanks!! Yes, I agree. I was just too naive when I thought it does the trick. (But I learned a new thing: outer sep=auto ;-) – Apr 24 '18 at 23:58

score 4 · Answer 2 · answered Apr 25 '18 at 00:27

You can simply use the rules from booktabs which have some vertical padding around them:

\documentclass[12pt]{article}

\usepackage{tikz}

\makeatletter
\newcommand\mathcircled[1]{%
    \mathpalette\@mathcircled{#1}%
}
\newcommand\@mathcircled[2]{%
    \tikz[baseline=(math.base)] \node[draw,circle,inner sep=1pt] (math) {$\m@th#1#2$};%
}
\makeatother

\usepackage{booktabs}

\begin{document}

        \begin{table}[]
            \centering
            \label{my-label}
            \begin{tabular}{ccccc}
                \toprule
                S1 & S1 & $\mathcircled{S1}$ & ... & S1 \\ \midrule
                $\mathcircled{S2}$ & $\mathcircled{S2}$ & S2 & ... & S2 \\ \midrule
                ... & ... & ... & ... & ... \\ \midrule
                SN & $\mathcircled{SN}$ & $\mathcircled{SN}$ & ... & SN \\ \bottomrule
            \end{tabular}
        \end{table}

    \end{document}

samcarter_is_at_topanswers.xyz · Answer 3 · 2018-04-24T23:56:38.283

To make the circles the same size, you can specify a minimum size

\documentclass[12pt]{article}

\usepackage{tikz}
\makeatletter
\newcommand\mathcircled[1]{%
    \mathpalette\@mathcircled{#1}%
}
\newcommand\@mathcircled[2]{%
    \tikz[baseline=(math.base)] \node[draw,circle,minimum size=30pt] (math) {$\m@th#1#2$};%
}
\makeatother

\usepackage{booktabs}


\begin{document}


\begin{table}[htbp]
    \centering
    \renewcommand{\arraystretch}{1.8}
    \belowrulesep=1ex
    \aboverulesep=1ex
    \begin{tabular}{ccccc}
        \midrule
        S1  & S1  & $\mathcircled{S1}$  & $\dots$ & S1  \\ \midrule
        $\mathcircled{S2}$  & $\mathcircled{S2}$  & S2  & $\dots$ & S2  \\ \midrule
        $\dots$ & $\dots$ & $\dots$ & $\dots$ & $\dots$ \\ \midrule
        SN  & $\mathcircled{SN}$  & $\mathcircled{SN}$  & $\dots$ & SN  \\ \midrule
    \end{tabular}
\end{table}

\end{document}

Thank you, however this solution overlaps circle with the borders. @samcarter — Guido Muscioni, Apr 24 '18 at 23:41
@GuidoMuscioni They don't overlap, I adjusted the row height to make the circles touch the lines. You can choose another array stretch. — samcarter_is_at_topanswers.xyz, Apr 24 '18 at 23:43

Symbol inside a circle in a table

3 Answers3