I want to draw the following tree in a question. But, I don't want to upload a picture. Can anyone tell me how to do it using MathJax?The Tree

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    $\begingroup$ Why don't you want to upload a picutre? I don't believe it's possible to replicate this with MathJax... $\endgroup$ Commented Apr 20, 2016 at 10:19
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    $\begingroup$ See How could I add a graph to a post at MSE? and some other questions on meta tagged (images). $\endgroup$ Commented Apr 20, 2016 at 11:28
  • $\begingroup$ As far as I know, diagonal lines are not possible with any graph-drawing environment (namely amscd) available for use on the website. $\endgroup$
    – pjs36
    Commented Apr 20, 2016 at 12:46
  • $\begingroup$ It would be nice if TikZ were supported by MathJax, but it's not. Uploading a picture may be your only option. $\endgroup$ Commented Apr 20, 2016 at 13:26
  • $\begingroup$ Som older questions related to TikZ: meta.math.stackexchange.com/questions/9545/… and meta.math.stackexchange.com/questions/3190/… $\endgroup$ Commented Apr 20, 2016 at 13:29
  • $\begingroup$ Hmm... Seems like it's the only way. Sad. Anyways, thank you for the input. $\endgroup$ Commented Apr 20, 2016 at 15:43
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    $\begingroup$ You can find an image of (probably) that tree in this answer of mine. Feel free to use it if you wish/need. $\endgroup$ Commented Apr 30, 2016 at 19:54

1 Answer 1


It can be done in MathJax, but it is painful and a complete hack. Here is one way:

\hskip 1em

which produces

$$ \require{enclose} \def\uline#1#2{\enclose{updiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} \def\dline#1#2{\enclose{downdiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} % \def\place#1#2#3{\smash{\rlap{\hskip{#1em}\raise{#2em}{#3}}}} % \hskip 1em % \place{0}{12}{\bullet} \place{2}{0}{\bullet} \place{4}{4}{\bullet} \place{6}{0}{\bullet} \place{8}{8}{\bullet} \place{10}{0}{\bullet} \place{12}{4}{\bullet} \place{14}{0}{\bullet} \place{16}{12}{\bullet} % \place{.3}{4.4}{\dline{3.6}{7.6}} \place{2.3}{.5}{\uline{1.6}{3.6}} \place{4.3}{.4}{\dline{1.6}{3.6}} \place{4.3}{4.4}{\uline{3.6}{3.7}} \place{8.3}{4.4}{\dline{3.6}{3.6}} \place{10.3}{.5}{\uline{1.6}{3.6}} \place{12.3}{.4}{\dline{1.6}{3.6}} \place{12.2}{4.4}{\uline{3.7}{7.6}} % \place{-1}{12.5}{\frac01} \place{1}{-.5}{\frac13} \place{2.75}{4}{\frac12} \place{6.5}{-.5}{\frac23} \place{7.75}{9.5}{\frac11} \place{9}{-.5}{\frac32} \place{12.75}{4}{\frac21} \place{14.5}{-.5}{\frac31} \place{16.5}{12.5}{\frac10} % \hskip18em\Rule{0em}{14em}{1.5em} $$

It allows you to define an abstract grid on which you place the items you want (bullets for dots, fractions, and diagonal lines created using the enclose package). I also use \hskip and \Rule to define the size of the complete diagram, and an \hskip at the front to compensate for the use of -1 for the left-most labels.

Anyway, you can make these kinds of diagrams with some effort.

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    $\begingroup$ I'd like to mention this doesn't work on the iOS app $\endgroup$
    – KoA
    Commented Apr 22, 2016 at 5:28
  • $\begingroup$ The numbers after place, for example \place{2}{0} are relative coordinates with respect to previous position? Or are they absolute coordinates? $\endgroup$ Commented Apr 22, 2016 at 7:54
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    $\begingroup$ @shieldbug1 And I can confirm that it does not render correctly on the mobile version of the site, using iOS's Safari (but switching to the "full site" in Safari still, it does work). Very interesting answer, either way! $\endgroup$
    – pjs36
    Commented Apr 22, 2016 at 12:29
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    $\begingroup$ @MartinSleziak, they are absolute with respect to the lower left-hand corner of the graph. $\endgroup$ Commented Apr 22, 2016 at 13:10
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    $\begingroup$ Thx for this great hack. $+1$ The diagram flooted on the code out of the screen though. $\endgroup$ Commented Dec 4, 2018 at 5:05
  • $\begingroup$ Here's what it looks like on an Android phone; viz., a ZTE Blade V8. $\endgroup$
    – Shaun
    Commented Dec 9, 2018 at 2:13
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    $\begingroup$ Here's what it looks like on the same phone but using the desktop site. $\endgroup$
    – Shaun
    Commented Dec 9, 2018 at 2:15

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