TikZ Colored Equations – Using \boxed{}, TikZ, and Colored Equation Backgrounds

amsmathhighlightingmath-modetikz-pgf

I wonder if someone could provide me with a simple MWE for using TikZ together with \boxed to produce a colored equation background. If possible with rounded corners.

I am using \boxed{} inside the split environment from amsmath.

Best Answer

There are some packages that can help you, for example:

and along this site some answers could be a good starting point:

Here, I provide you a MWE using the hf-tikz package. Actually \boxed{} is not used at all, thus I don't know if this will meet your requirements. Notice that, in the following example, it is shown how to highlight the whole equation or just a part of it, that is the major potentiality of the package. You should compile twice to get the right result.

\documentclass{article}
\usepackage{amsmath}
\usepackage[customcolors]{hf-tikz}

\begin{document}
\begin{equation}\label{e:barwq}\begin{split}\tikzmarkin{a}(0.2,-0.5)(-0.2,0.65)
H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}
\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\
&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot
\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\tikzmarkend{a}
\end{split}\end{equation}

\hfsetfillcolor{blue!10}
\hfsetbordercolor{blue}
\begin{equation}\label{e:barwq2}\begin{split}
H_c&=\tikzmarkin{b}(0,-0.6)(0,0.65)\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}
\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\tikzmarkend{b}\\
&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot
\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].
\end{split}\end{equation}

\hfsetfillcolor{green!10}
\hfsetbordercolor{green!50!black}
\begin{equation}\label{e:barwq3}\begin{split}
H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}
\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\
&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot
\tikzmarkin{c}(0.05,-0.6)(-0.05,0.65)\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\tikzmarkend{c}
\end{split}\end{equation}

\end{document}

Result:

enter image description here

According to the request in the comments, here are the two possibilities to get rigid corners. It is needed the version 0.2 of the package.

The option norndcorners to have always rigid corners: just loading the package with \usepackage[customcolors,norndcorners]{hf-tikz} the previous document becomes:

enter image description here

The second possibility allows to have rigid corners singularly; load the package with the shade option and then use the key disable rounded corners=true in the tikzmarkin command.

In this example, the second equation is highlighted with rigid corners while the other two with rounded corners:

\documentclass{article}
\usepackage{amsmath}
\usepackage[customcolors,shade]{hf-tikz}

\begin{document}
\begin{equation}\label{e:barwq}\begin{split}\tikzmarkin{a}(0.2,-0.5)(-0.2,0.65)
H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}
\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\
&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot
\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\tikzmarkend{a}
\end{split}\end{equation}

\hfsetfillcolor{blue!10}
\hfsetbordercolor{blue}
\begin{equation}\label{e:barwq2}\begin{split}
H_c&=\tikzmarkin[disable rounded corners=true]{b}(0,-0.6)(0,0.65)\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}
\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\tikzmarkend{b}\\
&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot
\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].
\end{split}\end{equation}

\hfsetfillcolor{green!10}
\hfsetbordercolor{green!50!black}
\begin{equation}\label{e:barwq3}\begin{split}
H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}
\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\
&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot
\tikzmarkin{c}(0.05,-0.6)(-0.05,0.65)\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\tikzmarkend{c}
\end{split}\end{equation}

\end{document}

The result:

enter image description here

Related Solutions

[Tex/LaTex] Drawing a triangle with rounded corners in TikZ

Make the radius of the rounded corners smaller. Remove =10mm and you will see the instant effect. Also, in your case, I find no need to load the xcolor package since it is used internally by tikz, unless, as noted by Alain Matthes, you use some of its options for naming your colors. You can uncomment it if you like. The libraries are not needed here, too, so I commented them out.

Here is your figure when the 10mm is dropped.

enter image description here

And here is one with scaling (not really necessary, just for demonstration purposes).

%\documentclass{article}

\documentclass[border=5]{standalone}

\usepackage{tikz}
%\usepackage{xcolor}
%\usetikzlibrary{arrows,shapes,snakes,automata,backgrounds,petri}

\begin{document}
\begin{tikzpicture}[scale=3]
    \draw [rounded corners,fill=gray!20] (0,0)--(1,0)--(0.5,1)--cycle;
\end{tikzpicture}
\end{document}

enter image description here

Here I have scaled the figure up to 3 since your figure is too small.

To make a point, see what happens when I scale the figure up by 5 and set rounded corners=10mm.

enter image description here

[Tex/LaTex] tikz: Draw rectangle with rounded corners only north/south

Thank you for your suggestions! I liked the method proposed in the answer Torbjørn T. linked to the most, so I'm answering here myself.

Code:

\documentclass{article}
\usepackage{tikz}

\begin{document}

\begin{figure}[h]
\centering

\begin{tikzpicture}
\fill [blue,draw]
  (0,0) --
  ++(5,0) {[rounded corners=10] --
  ++(0,5) --
  ++(-5,0)} --
  cycle
  {};
\fill [red,draw]
  (0,0) --
  ++(5,0) {[rounded corners=10] --
  ++(0,-5) --
  ++(-5,0)} --
  cycle
  {};
\end{tikzpicture}

\end{figure}

\end{document}

Output:

Best Answer

Related Solutions

[Tex/LaTex] Drawing a triangle with rounded corners in TikZ

[Tex/LaTex] tikz: Draw rectangle with rounded corners only north/south

Related Question