# How to write this formula?

How to write this formula in Latex？

• Welcome to TeX.SX! We are no do-it-for-me-service. Please provide a minimal example of what you've tried. If you don't know how to write a specific symbol, try Detexify. – TeXnician Aug 7 '17 at 13:57
• – JPi Aug 7 '17 at 14:03

Another solution using tikz to draw the symbols (although I like egreg's solution better):

\documentclass[12pt]{article}
\usepackage{tikz}

\newlength{\lineheight}
\newcommand{\updateheight}{\settoheight{\lineheight}{A}}
\tikzset{ThreePoints/.style={circle, fill}}
\newcommand{\drawthreepoints}{%
\node [ThreePoints] (first) at (0,0) {};
\node [ThreePoints] (second) at (0.5,0.8) {};
\node [ThreePoints] (third) at (-0.5,0.8) {};
}
\newcommand{\resizepoints}[1]{\updateheight\resizebox{!}{\lineheight}{#1}}
\newcommand{\threepoints}{\resizepoints{\begin{tikzpicture}
\drawthreepoints
\end{tikzpicture}}}
\newcommand{\threepointsone}{\resizepoints{\begin{tikzpicture}
\drawthreepoints
\draw [line width=3pt] (third) -- (second);
\end{tikzpicture}}}
\newcommand{\threepointstwo}{\resizepoints{\begin{tikzpicture}
\drawthreepoints
\draw [line width=3pt] (third) -- (second) -- (first);
\end{tikzpicture}}}

\begin{document}
$$d(G)=d(\threepoints)d(\threepoints,G)+d(\threepointsone)d(\threepointsone,G)+d(\threepointstwo)d(\threepointstwo,G)$$
\end{document}

It's rather easy with pict2e. The argument to \TR is a three digit binary number, which sets draws the lines joining the dots; the first digit represents the top, then we proceed in clockwise order.

\documentclass{article}
\usepackage{pict2e}

\makeatletter
\newcommand{\TR}[1]{\TR@aux#1000\@nil}
\def\TR@aux#1#2#3#4\@nil{%
\,\vcenter{\hbox{\TR@aux@i{#1}{#2}{#3}}}\,%
}
\def\TR@aux@i#1#2#3{%
\setlength\unitlength{.5em}%
\begin{picture}(1,0.866)
\put(0.5,0){\circle*{.4}}
\put(0,0.866){\circle*{.4}}
\put(1,0.866){\circle*{.4}}
\ifnum#1=1 \polyline(0,0.866)(1,0.866) \fi
\ifnum#2=1 \polyline(1,0.866)(0.5,0) \fi
\ifnum#3=1 \polyline(0.5,0)(0,0.866) \fi
\end{picture}%
}
\makeatother

\begin{document}

$d(G)= d(\TR{000})d(\TR{000},G)+ d(\TR{100})d(\TR{100},G)+ d(\TR{110})d(\TR{110},G)$

$\TR{000}\quad\TR{001}\quad\TR{010}\quad\TR{011}$

$\TR{100}\quad\TR{101}\quad\TR{110}\quad\TR{111}$

\end{document}

• You could actually use \Line, here, instead of \polyline. And what about a standard \mathpalette wrapper? – GuM Aug 7 '17 at 22:15
• @GuM You looked in the manual, didn't you? :-) The \mathpalette thing is left as an exercise. – egreg Aug 8 '17 at 8:13

Like this?

\documentclass{article}\usepackage[show frame]{geometry}

\newcommand{\threedots}{\!\begin{smallmatrix} \bullet\mkern10mu\bullet \\ \bullet \end{smallmatrix}\!}

\newcommand{\threedotsl}{\!\begin{smallmatrix}\bullet\mkern-4mu\mathord{\relbar\mkern-9mu\relbar}\mkern-4mu\bullet \\
\bullet \end{smallmatrix}\!}

\newcommand{\threedotsll}{\!\stackMath\stackinset{r}{9.7pt}{c}{0pt}{\adjustbox{rotate=-4, origin=c, scale={1}{0.8}}{$\not$}}%
{\begin{smallmatrix}\bullet\mkern-4mu\mathord{\relbar\mkern-9mu\relbar}\mkern-4mu\bullet \\
\bullet \end{smallmatrix}}\!}

\begin{document}

\begin{equation*}
d(G) = d\bigl(\threedots\bigr)d\bigl(\threedots, G\bigr) + d\bigl(\threedotsl\bigr)d\bigl(\threedotsl, G\bigr) + d\bigl(\threedotsll)d\bigl(\threedotsll, G)
\end{equation*}

\end{document}