3

I have a set of equations summarized in this way:

enter image description here

The code that produces this is the following:

\documentclass[12pt]{article}    
\usepackage{empheq}
\usepackage[left=2.5cm,top=2.5cm,right=2.5cm,bottom=2.5cm]{geometry}
\usepackage{amsmath}

\begin{document}

\newcommand*\widefbox[1]{\fbox{\hspace{2em}#1\hspace{2em}}}
\begin{subequations}
\begin{empheq}[box=\widefbox]{align}
%
\Aboxed{a = & b + c + d}
 \nonumber \\
& d = f + g
 \nonumber \\
& \qquad
f = m/4
 \nonumber \\
&
d = j + k
 \nonumber \\
& \qquad
j = n\cdot3/4
 \nonumber \\
& d = l + o
 \nonumber \\
\Aboxed{& d = m/4 + q + n\cdot 3/4 + k + l + 0 }
 \nonumber \\
\Aboxed{&c = p + q}
 \nonumber \\
& \qquad
p = h/2 + h/2
 \nonumber \\
\Aboxed{&b = r + s}
 \nonumber \\
\Aboxed{&t = u + w}
 \nonumber \\  \nonumber\\
\Aboxed{a = & b + c + d}
 \nonumber \\ \nonumber\\
\Aboxed{a = & zz + z'}
 \nonumber \\ \nonumber\\
\end{empheq}
\end{subequations}

\end{document}

I would like to produce something similar to this:

enter image description here

Is there a way to achieve those arrows, red boxes and circled numbers 1, 2 and 3 ?

Update:

Following @Ignasi's approach, when I try to apply this to the real example, I encounter quite a lot of difficulties, this is the nearest result I could achieve:

enter image description here

where:

1) The circled numbers and the arrows appear in the following page, instead of next to the equations (see image)

2) I could not manage to align the J[p] equations.

3) I could not manage to box the last equation,

4) Is there a way to caption this scheme?

Could you please help me to obtain this result?

This is the code to where I reached so far:

\documentclass[12pt]{article}    
\usepackage{empheq}
\usepackage[left=2.5cm,top=2.5cm,right=2.5cm,bottom=2.5cm]{geometry}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{tikzmark, positioning}

\begin{document}

\newcommand*\widefbox[1]{\fbox{\hspace{2em}#1\hspace{2em}}}

\begin{subequations}
\begin{empheq}[box=\widefbox]{align}
%
%
\tikzmark{1}\Aboxed{
E [\rho ] = & \underbrace{ T[\rho ] + V_{ee}[\rho ] }_{=\,\, F[\rho ]} + V_{ne}[\rho]
}
 \nonumber \\
& %
\tikzmark{2}
V_{ne}[\rho ] = \int \rho \left ( \mathbf{r} \right ) v_{\text{ext}} \left ( \mathbf{r} \right ) \mathrm{d}\mathbf {r}
 \nonumber \\
& \qquad \qquad
v_{\text{ext}}\left ( \mathbf{r}_{i} \right ) = - \sum_{A=1}^{M} \frac{Z_{A}}{r_{iA}}
\nonumber \\
&
\tikzmark{3}
V_{ne}[\rho ] =  \int -  \sum_{A=1}^{M} \frac{Z_{A}}{r_{1A}} \rho \left ( \mathbf{r}_{1} \right )  \mathrm{d}\mathbf {r}_{1}
 \nonumber \\
& \qquad \qquad
\rho (\mathbf{r}) \equiv \rho_{\text{S}}(\mathbf{r}) = \sum_{i=1}^{N} \left | \varphi_{i} \left ( \mathbf{x} \right ) \right |^{2} = \rho_{0}(\mathbf{r})
\nonumber \\
& %
\tikzmark{4}
V_{ne}[\rho ] =  \int -  \sum_{A=1}^{M} \frac{Z_{A}}{r_{1A}}  \sum_{i=1}^{N} \left | \varphi_{i} \left ( \mathbf{x}_{1} \right ) \right |^{2}  \mathrm{d}\mathbf {r}_{1}
 \nonumber \\
\tikzmark{5}
\Aboxed{
&V_{ne}[\rho ] =  - \sum_{i=1}^{N} \int   \sum_{A=1}^{M} \frac{Z_{A}}{r_{1A}}  \left | \varphi_{i} \left ( \mathbf{x}_{1} \right ) \right |^{2}  \mathrm{d}\mathbf {r}_{1}
}
 \nonumber \\
\tikzmark{6}
\Aboxed{
&V_{ee}[\rho ] = J[\rho ] + V_{\text{non-classical}}[\rho ]
}
 \nonumber \\
& \qquad \qquad J[\rho]  = \frac{1}{2} \int \int \frac{\rho(\mathbf{r}_{1})\rho(\mathbf{r}_{2})}{r_{12}} \mathrm{d}\mathbf {r}_{1} \mathrm{d}\mathbf {r}_{2}  \nonumber \\
& \qquad \qquad \qquad  \rho (\mathbf{r}) \equiv \rho_{\text{S}}(\mathbf{r}) = \sum_{i=1}^{N} \left | \varphi_{i} \left ( \mathbf{x} \right ) \right |^{2} = \rho_{0}(\mathbf{r})  \nonumber \\
&
\Aboxed{
J[\rho]  = \frac{1}{2} \sum_{i=1}^{N} \sum_{j=1}^{N} \int \int \left | \varphi_{i} (\mathbf{r}_{1}) \right |^{2} \frac{1}{r_{12}} \left | \varphi_{j} (\mathbf{r}_{2}) \right |^{2} \mathrm{d}\mathbf {r}_{1}  \mathrm{d}\mathbf {r}_{2}  }
\nonumber \\
\tikzmark{7}
\Aboxed{
&T[\rho ]  = T_{\text{S}}[\rho ] + T_{\text{C}}[\rho ]
}
 \nonumber \\
& \qquad \qquad T_{\text{S}}[\rho ] = -\frac{1}{2}\sum_{i=1}^{N} \expval{\nabla^{2}}{\varphi _{i}}
 \nonumber \\
\tikzmark{8}
\Aboxed{
E_{\text{XC}} [\rho] &= \left ( T[\rho] - T_{\text{S}}[\rho] \right ) + \left ( E_{ee}[\rho] - J[\rho] \right ) = T_{\text{C}}[\rho] + V_{\text{non-classsical}}[\rho]
}
\nonumber \\
\tikzmark{9}
\Aboxed{
E [\rho ] = & T_{\text{S}}[\rho ] + J[\rho] +  V_{ne}[\rho ] + E_{\text{XC}} [\rho]
}
\nonumber \\
%\begin{empheq}[box=\fbox]{align}
%\end{subequations}
%
\tikzmark{10}
E [\rho ] = &  -\frac{1}{2}\sum_{i=1}^{N} \expval{\nabla^{2}}{\varphi _{i}} +  \frac{1}{2} \sum_{i=1}^{N} \sum_{j=1}^{N} \int \int \left | \varphi_{i} (\mathbf{r}_{1}) \right |^{2} \frac{1}{r_{12}} \left | \varphi_{j} (\mathbf{r}_{2}) \right |^{2} \mathrm{d}\mathbf {r}_{1}  \mathrm{d}\mathbf {r}_{2}
\nonumber \\
&  + E_{\text{XC}} [\rho] - \sum_{i=1}^{N} \int   \sum_{A=1}^{M} \frac{Z_{A}}{r_{1A}}  \left | \varphi_{i} \left ( \mathbf{x}_{1} \right ) \right |^{2}  \mathrm{d}\mathbf {r}_{1}\\
%}
%\end{empheq}
\end{empheq}
\end{subequations}

\begin{tikzpicture}[remember picture, overlay]
\foreach \i [count=\ni] in {1,9,10}
    \node[draw, circle, inner sep=2pt, left=2mm of pic cs:\i, yshift=.5ex] (c\ni) {\ni};

\foreach \i in {2,...,4}
    \draw[->] ([shift={(2mm,-4.pt)}]pic cs:1) |- ([shift={(-4pt,.5ex)}]pic cs:\i);
\foreach \i in {5,...,8}
    \draw[->] ([shift={(2mm,-4.pt)}]pic cs:1) |- ([yshift=.5ex]pic cs:\i);
\end{tikzpicture}
  • 1. Circled numbers in next page because your equations doesn't fit in one page. Reduce top and bottom margins. 2. \Aboxed needs one and only one & inside. Change \Aboxed by \fbox{$\displaymath.... 3. Same problem with \Aboxed, it's only valid for one line equations. I don't know if a nested boxed empheq will work. 4. Use \captionof command to caption non floating environments. – Ignasi Oct 3 '17 at 10:27
2

A simple solution with pstricks. I define an \Acolorboxed command, mimicked on the \Aboxed command from mathtools:

\documentclass[12pt]{article}
\usepackage{empheq}
\usepackage[margin=2.5cm]{geometry}
\usepackage[svgnames]{xcolor}
\usepackage{pst-node, multido}
\usepackage{auto-pst-pdf} % to compile with pdflatex
\makeatletter
\colorlet{framecolor}{Tomato}
\colorlet{bgcolor}{white}
\newcommand{\fcolorboxed}[3]{\fcolorbox{#1}{#2}{\m@th$\displaystyle#3$}}
\newcommand\Acolorboxed[1]{\let\bgroup{\romannumeral-`}\@Acolorboxed#1&&\ENDDNE}
\def\@Acolorboxed#1&#2&#3\ENDDNE{%
  \ifnum0=`{}\fi \setbox \z@
    \hbox{$\displaystyle#1{}\m@th$\kern\fboxsep \kern\fboxrule }%
    \edef\@tempa {\kern \wd\z@ &\kern -\the\wd\z@ \fboxsep
        \the\fboxsep \fboxrule \the\fboxrule }\@tempa \fcolorboxed{framecolor}{bgcolor}{#1#2}%
}
\makeatother
\def\ma-ht{\fontdimen22\textfont2}
\newcommand{\myeqlabel}[1]{\cput[linecolor=Tomato](-2em,0.7ex){\color{Tomato}#1}}
\begin{document}

\newcommand*\widefbox[1]{\hspace{2em}#1\hspace{2em}}
\begin{postscript}
\begin{subequations}
\begin{empheq}[box=\widefbox]{align}
\myeqlabel{1}\Acolorboxed{\rnode[b]{I}{a} = & b + c + d}
 \nonumber \\
\pnode[0,\ma-ht]{E1}& d = \begin{aligned}[t]
& f + g \\
& f = m/4
 \end{aligned} \nonumber\\
\pnode[0,\ma-ht]{E2}& d = \begin{aligned}[t]
 & j + kr \\
& j = n\cdot3/4
 \end{aligned} \nonumber \\
\pnode[0,\ma-ht]{E3} & d = l + o
 \nonumber \\
\pnode[0,\ma-ht]{E4}\Aboxed{& d = m/4 + q + n\cdot 3/4 + k + l + 0 }
 \nonumber \\
\pnode[0,\ma-ht]{E5}\Aboxed{&c = p + q}
 \nonumber \\
& \phantom{c ={}}
p = h/2 + h/2
 \nonumber \\
\pnode[0,\ma-ht]{E6}\Aboxed{&b = r + s}
 \nonumber \\
\pnode[0,\ma-ht]{E7}\Aboxed{&t = u + w}
 \nonumber \\[\baselineskip]
\myeqlabel{2}\Acolorboxed{a = & b + c + d}
 \nonumber \\[\baselineskip]
\myeqlabel{3}\Acolorboxed{a = & zz + z'}
 \nonumber \\ \nonumber\\
\end{empheq}
\end{subequations}
\psset{linewidth =0.4pt, linejoin=1, arrowinset=0.12, angleA=-90, angleB =-180, arrows=->, nodesepB=0.4em, nodesepA=1.44\fboxsep}
\multido{\i =1 + 1}{7}{\ncangle{I}{E\i}}
\end{postscript}

\end{document} 

enter image description here

Edit: A code for the real situation.

Note double integrals are not obtained with two \int commands (which results in a very bad spacing), but with \iint. We have a tighter spacing loading the esint package.

I also simplified the code for the maths part defining \dd for the differential symbol in integrals (with a better spacing), an \abs command for the absolute value and had to define \expval (which is not a standard LaTeX command) with the \DeclarePairedDelimiterX command from mathtools. Final remark: you don't have to load amsmath when you load empheq since it loads mathtools, which loads the former.

\documentclass[11pt]{article}
\usepackage{empheq, nccmath}
\usepackage[margin=2.5cm]{geometry}
\usepackage{caption}
\usepackage[svgnames]{xcolor}
\usepackage{pst-node, multido}
\usepackage{auto-pst-pdf} % to compile with pdflatex
\makeatletter
\colorlet{framecolor}{Tomato}
\colorlet{bgcolor}{white}
\newcommand{\fcolorboxed}[3]{\fcolorbox{#1}{#2}{\m@th$\displaystyle#3$}}
\newcommand\Acolorboxed[1]{\setlength{\fboxrule}{0.8pt}\let\bgroup{\romannumeral-`}\@Acolorboxed#1&&\ENDDNE}
\def\@Acolorboxed#1&#2&#3\ENDDNE{%
\ifnum0=`{}\fi \setbox \z@
\hbox{$\displaystyle#1{}\m@th$\kern\fboxsep \kern\fboxrule }%
\edef\@tempa {\kern \wd\z@ &\kern -\the\wd\z@ \fboxsep
    \the\fboxsep \fboxrule \the\fboxrule }\@tempa \fcolorboxed{framecolor}{bgcolor}{#1#2}%
}
\makeatother
\def\ma-ht{\fontdimen22\textfont2}
\newcommand{\myeqlabel}[1]{\cput[linecolor=Tomato](-1.5em,0.7ex){\color{Tomato}#1}}
\newcommand*{\dd}{\mathop{}\!\mathrm{d}} %
\usepackage{esint}
\DeclarePairedDelimiter\abs{\lvert}{\rvert}
\DeclarePairedDelimiterX\expval[2]{\langle}{\rangle}%
{#1\,\delimsize\vert\,\mathopen{}#2\,\delimsize\vert\,\mathopen{}#1}
\usepackage{tikz}
\usetikzlibrary{tikzmark, positioning}

\begin{document}

\newcommand*\widefbox[1]{\fbox{\hspace{2em}#1\hspace{2em}}}
\begin{postscript}
  \begin{subequations}
    \begin{empheq}[box=\widefbox]{align}
      %
      \myeqlabel{1}\Acolorboxed{\pnode[1em, -4.27ex]{I}%
        E [\rho ]= & \underbrace{T[\rho ] + V_{ee}[\rho ] }_{=\,\, F[\rho ]} + V_{ne}[\rho]
      }
      \nonumber \\
      & \begin{alignedat}{2}
        \pnode[0,\ma-ht]{E1}V_{ne}[\rho ] & = & & \int \rho \left ( \mathbf{r} \right ) v_{\text{ext}} \left ( \mathbf{r} \right ) \dd\mathbf {r} \\
        & & &\,v_{\text{ext}}\left ( \mathbf{r}_{i} \right ) = - \sum_{A=1}^{M} \frac{Z_{A}}{r_{iA}} \\[-1ex]
        \pnode[0,\ma-ht]{E2}
        V_{ne}[\rho ] & = & & \int - \sum_{A=1}^{M} \frac{Z_{A}}{r_{1A}} \rho \left ( \mathbf{r}_{1} \right ) \dd\mathbf {r}_{1} \\
        & & & \, \rho (\mathbf{r}) \equiv \rho_{\text{S}}(\mathbf{r}) = \sum_{i=1}^{N}\abs{\varphi_{i} \left ( \mathbf{x} \right )}^{2} = \rho_{0}(\mathbf{r}) \\
        \pnode[0,\ma-ht]{E3}
        V_{ne}[\rho ] & = & & \int - \sum_{A=1}^{M} \frac{Z_{A}}{r_{1A}} \sum_{i=1}^{N} \abs{\varphi_{i} \left (\mathbf{x}_{1} \right) }^{2} \mathrm{d}\mathbf {r}_{1}
      \end{alignedat}\nonumber \\
      \pnode[0,\ma-ht]{E4} \Aboxed{
      &V_{ne}[\rho ]= - \sum_{i=1}^{N} \int \sum_{A=1}^{M} \frac{Z_{A}}{r_{1A}} \abs{\varphi_{i} \left ( \mathbf{x}_{1} \right )}^{2} \dd\mathbf {r}_{1}
      }
      \nonumber\\
      \pnode[0,\ma-ht]{E5}
      \Aboxed{%
        & V_{ee}[\rho ] = J[\rho ] + V_{\text{non-classical}}[\rho ]
      } \nonumber \\
      & \begin{aligned}\phantom{V_{ee}[\rho ] ={}} J[\rho] = {} & \mfrac{1}{2} \iint \frac{\rho (\mathbf{r}_{1})\rho (\mathbf{r}_{2})}{r_{12}} \dd\mathbf {r}_{1} \dd\mathbf {r}_{2} \\
                                                 & \rho (\mathbf{r}) \equiv \rho_{\text{S}}(\mathbf{r}) = \sum_{i=1}^{N} \abs{\varphi_{i} \left ( \mathbf{x} \right )}^{2} = \rho_{0}(\mathbf{r}) r \\
        \Aboxed{
        J[\rho]= & \mfrac{1}{2} \sum_{i=1}^{N} \sum_{j=1}^{N} \iint \abs*{\varphi_{i} (\mathbf{r}_{1})}^{2} \frac{1}{r_{12}} \abs*{\varphi_{j} (\mathbf{r}_{2})}^{2} \dd\mathbf {r}_{1} \dd\mathbf {r}_{2}}
      \end{aligned}
      \nonumber \\
      \pnode[0,\ma-ht]{E6}
      \Aboxed{
      &T[\rho ] = T_{\text{S}}[\rho ] + T_{\text{C}}[\rho ]
      }
      \nonumber \\
      %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
      &\phantom{T[\rho ] ={}} T_{\text{S}}[\rho ] = -\mfrac{1}{2}\sum_{i=1}^{N}\expval*{\nabla^{2}}{\varphi _{i}}%
      \nonumber \\
      \pnode[0,\ma-ht]{E7}
      \Aboxed{
        & E_{\text{XC}} [\rho] = \left ( T[\rho] - T_{\text{S}}[\rho] \right ) + \left ( E_{ee}[\rho] - J[\rho] \right ) = T_{\text{C}}[\rho] + V_{\text{non-classical}}[\rho]
      }
      \nonumber \\
      \myeqlabel{2}
      \Acolorboxed{
        E [\rho ] &= T_{\text{S}}[\rho ] + J[\rho] + V_{ne}[\rho ] + E_{\text{XC}} [\rho]
      }
      \nonumber \\
      \myeqlabel{3}
      \Acolorboxed{E [\rho ] &=\begin{aligned}[t]-\mfrac{1}{2}\sum_{i=1}^{N} %
          \expval*{\nabla^{2}}{\varphi _{i}} + \mfrac{1}{2} \sum_{i=1}^{N} \sum_{j=1}^{N} \iint \abs{\varphi_{i} (\mathbf{r}_{1})}^{2} \frac{1}{r_{12}} \abs*{\varphi_{j} (\mathbf{r}_{2})}^{2} \dd\mathbf {r}_{1} \dd\mathbf {r}_{2}
          \\[-1.5ex]
          + E_{\text{XC}} [\rho] - \sum_{i=1}^{N} \int \sum_{A=1}^{M} \frac{Z_{A}}{r_{1A}} \abs{\varphi_{i} \left ( \mathbf{x}_{1} \right )}^{2} \dd\mathbf {r}_{1}
        \end{aligned}\nonumber} \\
    \end{empheq}
  \end{subequations}
  %
  \psset{linewidth=0.6pt, linejoin=1, linecolor=LightSteelBlue, linearc=0.08, arrowinset=0, angleA=-90, angleB=-180, arrows=->, nodesepB=0.4em, nodesepA=1.44\fboxsep}
  \multido{\i=1+1}{3}{\ncangle{I}{E\i}}
  \psset{nodesepB=0em}
  \multido{\i=4+1}{4}{\ncangle{I}{E\i}}
  \captionof{figure}{Some nice equations}
\end{postscript}

\end{document} 

enter image description here

  • Thanks a lot for your approach. I am trying to apply this to the real example (see below), but I cannot manage to obtain the result. For the moment, we can forget about the coloured scheme, it is more important to get the scheme right. I would appreciate if you could have a look. Million Thanks – DavidC. Oct 2 '17 at 21:14
  • Did you compile with pdflatex -shell-escape? – Bernard Oct 2 '17 at 21:20
  • Yes, exactly, I compiled with pdflatex -shell-escape – DavidC. Oct 2 '17 at 21:29
  • 1
    @DavidC.: I've added a code for your real situation. Please check if it compiles for you. – Bernard Oct 3 '17 at 13:09
3

You can use tikzmarks inside the empheq environments and later on, draw the numbers and connections:

\documentclass[12pt]{article}    
\usepackage{empheq}
\usepackage[left=2.5cm,top=2.5cm,right=2.5cm,bottom=2.5cm]{geometry}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{tikzmark, positioning}

\begin{document}

\newcommand*\widefbox[1]{\fbox{\hspace{2em}#1\hspace{2em}}}

\begin{subequations}
\begin{empheq}[box=\widefbox]{align}
%
\tikzmark{1}\Aboxed{a = & b + c + d}
 \nonumber \\
& \tikzmark{2} d = f + g
 \nonumber \\
& \qquad
f = m/4
 \nonumber \\
&
\tikzmark{3} d = j + k
 \nonumber \\
& \qquad
j = n\cdot3/4
 \nonumber \\
&\tikzmark{4} d = l + o
 \nonumber \\
\tikzmark{5}\Aboxed{& d = m/4 + q + n\cdot 3/4 + k + l + 0 }
 \nonumber \\
\tikzmark{6}\Aboxed{&c = p + q}
 \nonumber \\
& \qquad
p = h/2 + h/2
 \nonumber \\
\tikzmark{7}\Aboxed{&b = r + s}
 \nonumber \\
\tikzmark{8}\Aboxed{&t = u + w}
 \nonumber \\  \nonumber\\
\tikzmark{9}\Aboxed{a = & b + c + d}
 \nonumber \\ \nonumber\\
\tikzmark{10}\Aboxed{a = & zz + z'}
 \nonumber \\ \nonumber
\end{empheq}
\end{subequations}

\begin{tikzpicture}[remember picture, overlay]
\foreach \i [count=\ni] in {1,9,10}
    \node[draw, circle, inner sep=2pt, left=2mm of pic cs:\i, yshift=.5ex] (c\ni) {\ni};

\foreach \i in {2,...,4}
    \draw[->] ([shift={(2mm,-4.pt)}]pic cs:1) |- ([shift={(-4pt,.5ex)}]pic cs:\i);
\foreach \i in {5,...,8}
    \draw[->] ([shift={(2mm,-4.pt)}]pic cs:1) |- ([yshift=.5ex]pic cs:\i);
\end{tikzpicture}

\end{document}

enter image description here

Update: red color addition.

With Bernard's Acolorboxed definition and a little change in round labels, it's easy to get the desired result:

enter image description here

\documentclass[12pt]{article}    
\usepackage{empheq}
\usepackage[left=2.5cm,top=2.5cm,right=2.5cm,bottom=2.5cm]{geometry}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{tikzmark, positioning}

\makeatletter
\colorlet{framecolor}{red}
\colorlet{bgcolor}{white}
\newcommand{\fcolorboxed}[3]{\fcolorbox{#1}{#2}{\m@th$\displaystyle#3$}}
\newcommand\Acolorboxed[1]{\let\bgroup{\romannumeral-`}\@Acolorboxed#1&&\ENDDNE}
\def\@Acolorboxed#1&#2&#3\ENDDNE{%
  \ifnum0=`{}\fi \setbox \z@
    \hbox{$\displaystyle#1{}\m@th$\kern\fboxsep \kern\fboxrule }%
    \edef\@tempa {\kern \wd\z@ &\kern -\the\wd\z@ \fboxsep
        \the\fboxsep \fboxrule \the\fboxrule }\@tempa \fcolorboxed{framecolor}{bgcolor}{#1#2}%
}
\makeatother

\begin{document}

\newcommand*\widefbox[1]{\fbox{\hspace{2em}#1\hspace{2em}}}

\begin{subequations}
\begin{empheq}[box=\widefbox]{align}
%
\tikzmark{1}\Acolorboxed{a = & b + c + d}
 \nonumber \\
& \tikzmark{2} d = f + g
 \nonumber \\
& \qquad
f = m/4
 \nonumber \\
&
\tikzmark{3} d = j + k
 \nonumber \\
& \qquad
j = n\cdot3/4
 \nonumber \\
&\tikzmark{4} d = l + o
 \nonumber \\
\tikzmark{5}\Aboxed{& d = m/4 + q + n\cdot 3/4 + k + l + 0 }
 \nonumber \\
\tikzmark{6}\Aboxed{&c = p + q}
 \nonumber \\
& \qquad
p = h/2 + h/2
 \nonumber \\
\tikzmark{7}\Aboxed{&b = r + s}
 \nonumber \\
\tikzmark{8}\Aboxed{&t = u + w}
 \nonumber \\  \nonumber\\
\tikzmark{9}\Acolorboxed{a = & b + c + d}
 \nonumber \\ \nonumber\\
\tikzmark{10}\Acolorboxed{a = & zz + z'}
 \nonumber \\ \nonumber
\end{empheq}
\end{subequations}

\begin{tikzpicture}[remember picture, overlay]
\foreach \i [count=\ni] in {1,9,10}
    \node[draw, red, circle, inner sep=2pt, left=2mm of pic cs:\i, yshift=.5ex] (c\ni) {\ni};

\foreach \i in {2,...,4}
    \draw[->] ([shift={(2mm,-4.pt)}]pic cs:1) |- ([shift={(-4pt,.5ex)}]pic cs:\i);
\foreach \i in {5,...,8}
    \draw[->] ([shift={(2mm,-4.pt)}]pic cs:1) |- ([yshift=.5ex]pic cs:\i);
\end{tikzpicture}

\end{document}
  • Thanks a lot for all your effort. I have been quite a long time trying to apply this to the real example, but I have encountered the difficulties explained below. Could you please help me to achieve this? Again, thank you very much – DavidC. Oct 2 '17 at 21:10
  • 1
    @DavidC. Answer updated with red labels and boxes. – Ignasi Oct 3 '17 at 7:32
  • Thank you for your update. Please see updated post: I am finding difficulties to apply this code to the real example. Thanks again – DavidC. Oct 3 '17 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.