Equations using Tree Diagrams

Question

I would like to write an equation over binary trees. Here is the code I have so far: `

\[
\begin{array}{lcr}
\sum(-1)^{p+qr} 
\begin{tikzpicture}[scale = 0.8]
        \node[circle] at (1,-1) {$\gamma_q$};
        \node[circle] at (1,-2.5) {$f_k$};
        \draw[thick] (0.5,0) -- (1,-0.7);
        \draw[thick] (1,0) -- (1,-0.7);
        \draw[thick] (1.5,0) -- (1,-0.7);
        \draw[thick] (0,-1.3) -- (1,-2.2);
        \draw[thick] (0.5,-1.3) -- (1,-2.2);
        \draw[thick] (1,-1.3) -- (1,-2.2);
        \draw[thick] (1.5,-1.3) -- (1,-2.2);
        \draw[thick] (2,-1.3) -- (1,-2.2);
        \draw[thick] (1,-2.75) -- (1,-3.4);
    \end{tikzpicture}
    &
    -\sum(-1)^{\varepsilon}
    &
\begin{tikzpicture}[scale = 0.8]
        \node[circle] at (0,-1) {$f_{j_1}$};
        \node[circle] at (1.35,-1) {$f_{j_i}$};
        \node[circle] at (2.55,-1) {$f_{j_k}$};
        \node[circle] at (1,-2.5) {$\rho_k$};
        \node[circle] at (0.6,-1) {$\ldots$};
        \node[circle] at (2.0,-1) {$\dots$};
        \draw[thick] (-0.6,0) -- (0,-0.7);
        \draw[thick] (-0.2,0) -- (0,-0.7);
        \draw[thick] (0.25,0) -- (0,-0.7);
        \draw[thick] (1.0,0) -- (1.35,-0.7);
        \draw[thick] (1.4,0) -- (1.35,-0.7); 
        \draw[thick] (2.9,0) -- (2.4,-0.7);
        \draw[thick] (2,0) -- (2.4,-0.7);
        \draw[thick] (1.8,0) -- (2.4,-0.7);
        \draw[thick] (3.1,0) -- (2.4,-0.7);
        \draw[thick] (0,-1.3) -- (1,-2.2);
        \draw[thick] (0.5,-1.3) -- (1,-2.2);
        \draw[thick] (1,-1.3) -- (1,-2.2);
        \draw[thick] (1.5,-1.3) -- (1,-2.2);
        \draw[thick] (2.2,-1.3) -- (1,-2.2);
        \draw[thick] (1,-2.75) -- (1,-3.4);
    \end{tikzpicture}
\end{array}
\]

` While the trees in the diagram are correct (although probably poor from a coding point of view), the positioning of the summation signs is problematic: I would like them to appear at about the midpoint of each tree.

Is there a way to write equations in TikZ, or is there some other way to improve the placement of the summation symbols?

Off-topic: If you highlight your code and press the button with {} the code will be placed in it's environment without need of spaces before every line. (Welcome!) — koleygr, Jan 16 '18 at 19:52
You could put the two tikzpicture environments into \raiseboxes. — Jasper Habicht, Jan 16 '18 at 19:53

Zarko · Answer 1 · 2018-01-16T20:55:03.547

5

replace

\begin{tikzpicture}[scale=0.8]

with

\begin{tikzpicture}[scale=0.8, baseline=(current bounding box.center)]

addedndum: off-topic: trees drawn as tree. also is corrected array:

\documentclass{article}
\usepackage{tikz}
%\usetikzlibrary{trees}

\begin{document}

\[
\begin{array}{cccc}
\sum(-1)^{p+qr}
    &
\begin{tikzpicture}[baseline=(current bounding box.center),
                    grow'=up, anchor=south, scale=0.5,
                    sibling distance=5mm]
\draw (0,0) -- + (0,1) node {$f_k$}
    child
    child
    child { node{$\gamma_q$}
        child
        child
        child
          }
    child
    child
    ;
\end{tikzpicture}
    &
    -\sum(-1)^{\varepsilon}
    &
\begin{tikzpicture}[baseline=(current bounding box.center),
                    grow'=up, anchor=south, scale=0.5,
                    sibling distance=7mm]
\draw (0,0) -- + (0,1) node {$\rho_k$}
    child { node (a) {$f_{j1}$}
        child
        child
        child
          }
    child {node {}}
    child
    child {  node (b) {$f_{j1}$}
        child
        child
        child
          }
    child
    child {node {}}
   child {  node (c) {$f_{jk}$}
        child
        child
        child
          }
    ;
   \draw[dotted, very thick] (a) -- (b)  (b) -- (c);
\end{tikzpicture}
\end{array}
\]
\end{document}

edited Jan 16 '18 at 20:55

answered Jan 16 '18 at 20:06

Zarko

296,517

@An Coileanach, i add code for drawing trees. see, if it is useful to you. – Zarko Jan 16 '18 at 20:58
Thanks for this - I will definitely use this approach for drawing trees in future. – An Coileanach Jan 16 '18 at 21:00
@AnCoileanach, complete solution is useful and very consistent :-) – Zarko Jan 16 '18 at 22:45
@Zarko Very nice solution! (+1) What would be the magic line to make the picture vertically centered at the center of the equation (rather than the baseline)? – Jan 16 '18 at 23:30
baseline=(current bounding box.center) :-). this set baseline of picture to it center. – Zarko Jan 17 '18 at 00:03
@marmot A bit of yshift would do, cf. https://tex.stackexchange.com/a/75199/ – Torbjørn T. Jan 17 '18 at 10:03

score 2 · Accepted Answer · answered Jan 16 '18 at 19:58

One of several possibilities is

\documentclass{article}
\usepackage{tikz}
\begin{document}
\[
\begin{array}{lcr}
\sum(-1)^{p+qr} \vcenter{\hbox{
\begin{tikzpicture}[scale = 0.8]
        \node[circle] at (1,-1) {$\gamma_q$};
        \node[circle] at (1,-2.5) {$f_k$};
        \draw[thick] (0.5,0) -- (1,-0.7);
        \draw[thick] (1,0) -- (1,-0.7);
        \draw[thick] (1.5,0) -- (1,-0.7);
        \draw[thick] (0,-1.3) -- (1,-2.2);
        \draw[thick] (0.5,-1.3) -- (1,-2.2);
        \draw[thick] (1,-1.3) -- (1,-2.2);
        \draw[thick] (1.5,-1.3) -- (1,-2.2);
        \draw[thick] (2,-1.3) -- (1,-2.2);
        \draw[thick] (1,-2.75) -- (1,-3.4);
    \end{tikzpicture}}}
    &
    -\sum(-1)^{\varepsilon}
    &
\vcenter{\hbox{\begin{tikzpicture}[scale = 0.8]
        \node[circle] at (0,-1) {$f_{j_1}$};
        \node[circle] at (1.35,-1) {$f_{j_i}$};
        \node[circle] at (2.55,-1) {$f_{j_k}$};
        \node[circle] at (1,-2.5) {$\rho_k$};
        \node[circle] at (0.6,-1) {$\ldots$};
        \node[circle] at (2.0,-1) {$\dots$};
        \draw[thick] (-0.6,0) -- (0,-0.7);
        \draw[thick] (-0.2,0) -- (0,-0.7);
        \draw[thick] (0.25,0) -- (0,-0.7);
        \draw[thick] (1.0,0) -- (1.35,-0.7);
        \draw[thick] (1.4,0) -- (1.35,-0.7); 
        \draw[thick] (2.9,0) -- (2.4,-0.7);
        \draw[thick] (2,0) -- (2.4,-0.7);
        \draw[thick] (1.8,0) -- (2.4,-0.7);
        \draw[thick] (3.1,0) -- (2.4,-0.7);
        \draw[thick] (0,-1.3) -- (1,-2.2);
        \draw[thick] (0.5,-1.3) -- (1,-2.2);
        \draw[thick] (1,-1.3) -- (1,-2.2);
        \draw[thick] (1.5,-1.3) -- (1,-2.2);
        \draw[thick] (2.2,-1.3) -- (1,-2.2);
        \draw[thick] (1,-2.75) -- (1,-3.4);
    \end{tikzpicture}}}
\end{array}
\]
\end{document}

Great answer! I also tried the raisebox solution proposed by Jasper Habicht, and both did the trick, but this way is slightly nicer looking. Thanks for your help. For some reason, I'm not allowed to accept your answer just yet, but I will attend to that shortly. — An Coileanach, Jan 16 '18 at 20:02
@AnCoileanach Note that you don't really need the array at all here, you are after all just writing a few things after one another on the same line. — Torbjørn T., Jan 16 '18 at 20:18

score 1 · Answer 3 · answered Jan 16 '18 at 20:03

A tabular solution:

\documentclass{article}
\usepackage{lipsum}
\usepackage{tikz}
\usepackage{array}


\begin{document}

\[
\begin{tabular}{m{1.6cm} m{2cm} m{1.2cm} m{2.5cm}}
$\sum(-1)^{p+qr}$ &
\begin{tikzpicture}[scale = 0.8]
        \node[circle] at (1,-1) {$\gamma_q$};
        \node[circle] at (1,-2.5) {$f_k$};
        \draw[thick] (0.5,0) -- (1,-0.7);
        \draw[thick] (1,0) -- (1,-0.7);
        \draw[thick] (1.5,0) -- (1,-0.7);
        \draw[thick] (0,-1.3) -- (1,-2.2);
        \draw[thick] (0.5,-1.3) -- (1,-2.2);
        \draw[thick] (1,-1.3) -- (1,-2.2);
        \draw[thick] (1.5,-1.3) -- (1,-2.2);
        \draw[thick] (2,-1.3) -- (1,-2.2);
        \draw[thick] (1,-2.75) -- (1,-3.4);
    \end{tikzpicture}
    &$
    -\sum(-1)^{\varepsilon}
    $&
\begin{tikzpicture}[scale = 0.8]
        \node[circle] at (0,-1) {$f_{j_1}$};
        \node[circle] at (1.35,-1) {$f_{j_i}$};
        \node[circle] at (2.55,-1) {$f_{j_k}$};
        \node[circle] at (1,-2.5) {$\rho_k$};
        \node[circle] at (0.6,-1) {$\ldots$};
        \node[circle] at (2.0,-1) {$\dots$};
        \draw[thick] (-0.6,0) -- (0,-0.7);
        \draw[thick] (-0.2,0) -- (0,-0.7);
        \draw[thick] (0.25,0) -- (0,-0.7);
        \draw[thick] (1.0,0) -- (1.35,-0.7);
        \draw[thick] (1.4,0) -- (1.35,-0.7); 
        \draw[thick] (2.9,0) -- (2.4,-0.7);
        \draw[thick] (2,0) -- (2.4,-0.7);
        \draw[thick] (1.8,0) -- (2.4,-0.7);
        \draw[thick] (3.1,0) -- (2.4,-0.7);
        \draw[thick] (0,-1.3) -- (1,-2.2);
        \draw[thick] (0.5,-1.3) -- (1,-2.2);
        \draw[thick] (1,-1.3) -- (1,-2.2);
        \draw[thick] (1.5,-1.3) -- (1,-2.2);
        \draw[thick] (2.2,-1.3) -- (1,-2.2);
        \draw[thick] (1,-2.75) -- (1,-3.4);
    \end{tikzpicture}
\end{tabular}
\]
\end{document}

Output:

Welcome (since you already tried to accept -no matter which answer- and could not, I gave you some points to have this advantage... :P). Welcome and happy TeXing. — koleygr, Jan 16 '18 at 20:06
@koleygr Questioners can always accept answers. Upvoting is restricted, though, so people need some points to reward multiple answers to the same question or answers to other people's questions. — cfr, Jan 17 '18 at 02:11
@cfr thanks for the info... I thought that he needed at least 10 points... Could not imagine something else... Good to know because it would be really bad idea not to be like this... — koleygr, Jan 17 '18 at 02:22

Equations using Tree Diagrams

3 Answers3