[Tex/LaTex] Highlighting specific parts of a TikZ drawing

highlightingtikz-pgf

Friends, sometimes during an automata keynote, I have to present a string recognition step-by-step example. Please consider the following TikZ code:

\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,/tikz/initial text=] 
   \node[state,initial] (q_0)   {$q_0$}; 
   \node[state] (q_1) [right=of q_0] {$q_1$}; 
   \node[state,accepting] (q_2) [right=of q_1] {$q_2$}; 
    \path[->] 
    (q_0) edge node {a} (q_1)
    (q_1) edge [loop above] node {a} ()
    (q_1) edge node {b} (q_2);
\end{tikzpicture}

which outputs the following automaton:

Automaton 1

So far so good. Then, for an example, I take a string and show how the execution will happen. Let's say I want to check aab:

I achieve these slides by manually highlighting the specific parts of my automaton. The full code is as follows:

\documentclass{beamer}

\usepackage{tikz}
\usetikzlibrary{automata,positioning}

\begin{document}

\begin{frame}{String recognition}

\begin{itemize}
    \item \textbf{String:} \texttt{aab}
\end{itemize}

\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,/tikz/initial text=] 
   \node[state,initial] (q_0)   {$q_0$}; 
   \node[state] (q_1) [right=of q_0] {$q_1$}; 
   \node[state,accepting] (q_2) [right=of q_1] {$q_2$}; 
    \path[->] 
    (q_0) edge node {a} (q_1)
    (q_1) edge [loop above] node {a} ()
    (q_1) edge node {b} (q_2);
\end{tikzpicture}

\end{frame}

\begin{frame}{String recognition}

\begin{itemize}
    \item \textbf{String:} \texttt{aab}
\end{itemize}

\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,every initial by arrow/.style={draw=blue},/tikz/initial text=] 
   \node[state,initial,draw=blue] (q_0)   {\color{blue} $q_0$}; 
   \node[state] (q_1) [right=of q_0] {$q_1$}; 
   \node[state,accepting] (q_2) [right=of q_1] {$q_2$}; 
    \path[->] 
    (q_0) edge node {a} (q_1)
    (q_1) edge [loop above] node {a} ()
    (q_1) edge node {b} (q_2);
\end{tikzpicture}

\end{frame}

\begin{frame}{String recognition}

\begin{itemize}
    \item \textbf{String:} \texttt{aab}
\end{itemize}

\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,/tikz/initial text=] 
   \node[state,initial] (q_0)   {$q_0$}; 
   \node[state] (q_1) [right=of q_0] {$q_1$}; 
   \node[state,accepting] (q_2) [right=of q_1] {$q_2$}; 
    \path[->] 
    (q_0) edge [draw=blue] node {\color{blue} a} (q_1)
    (q_1) edge [loop above] node {a} ()
    (q_1) edge node {b} (q_2);
\end{tikzpicture}

\end{frame}

\begin{frame}{String recognition}

\begin{itemize}
    \item \textbf{String:} \texttt{aab}
\end{itemize}

\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,/tikz/initial text=] 
   \node[state,initial] (q_0)   {$q_0$}; 
   \node[state] (q_1) [draw=blue,right=of q_0] {\color{blue} $q_1$}; 
   \node[state,accepting] (q_2) [right=of q_1] {$q_2$}; 
    \path[->] 
    (q_0) edge node {a} (q_1)
    (q_1) edge [loop above] node {a} ()
    (q_1) edge node {b} (q_2);
\end{tikzpicture}

\end{frame}

\begin{frame}{String recognition}

\begin{itemize}
    \item \textbf{String:} \texttt{aab}
\end{itemize}

\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,/tikz/initial text=] 
   \node[state,initial] (q_0)   {$q_0$}; 
   \node[state] (q_1) [right=of q_0] {$q_1$}; 
   \node[state,accepting] (q_2) [right=of q_1] {$q_2$}; 
    \path[->] 
    (q_0) edge node {a} (q_1)
    (q_1) edge [draw=blue,loop above] node {\color{blue} a} ()
    (q_1) edge node {b} (q_2);
\end{tikzpicture}

\end{frame}

\begin{frame}{String recognition}

\begin{itemize}
    \item \textbf{String:} \texttt{aab}
\end{itemize}

\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,/tikz/initial text=] 
   \node[state,initial] (q_0)   {$q_0$}; 
   \node[state] (q_1) [draw=blue,right=of q_0] {\color{blue} $q_1$}; 
   \node[state,accepting] (q_2) [right=of q_1] {$q_2$}; 
    \path[->] 
    (q_0) edge node {a} (q_1)
    (q_1) edge [loop above] node {a} ()
    (q_1) edge node {b} (q_2);
\end{tikzpicture}

\end{frame}

\begin{frame}{String recognition}

\begin{itemize}
    \item \textbf{String:} \texttt{aab}
\end{itemize}

\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,/tikz/initial text=] 
   \node[state,initial] (q_0)   {$q_0$}; 
   \node[state] (q_1) [right=of q_0] {$q_1$}; 
   \node[state,accepting] (q_2) [right=of q_1] {$q_2$}; 
    \path[->] 
    (q_0) edge node {a} (q_1)
    (q_1) edge [loop above] node {a} ()
    (q_1) edge [draw=blue] node {\color{blue} b} (q_2);
\end{tikzpicture}

\end{frame}

\begin{frame}{String recognition}

\begin{itemize}
    \item \textbf{String:} \texttt{aab}
\end{itemize}

\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,/tikz/initial text=] 
   \node[state,initial] (q_0)   {$q_0$}; 
   \node[state] (q_1) [right=of q_0] {$q_1$}; 
   \node[state,accepting,draw=blue] (q_2) [right=of q_1] {\color{blue} $q_2$}; 
    \path[->] 
    (q_0) edge node {a} (q_1)
    (q_1) edge [loop above] node {a} ()
    (q_1) edge node {b} (q_2);
\end{tikzpicture}

\end{frame}

\end{document}

As you can see, the drawing is basically the same for every step, apart from the highlighting. I thought of creating a new command called \myautomaton which takes an integer as parameter, representing the current step. In the drawing definition itself, I tried to use the \ifthenelse command from the xifthen package to check if, e.g., state q_0 should be highlighted. Unfortunately, I had some problems, as I tried to insert them like:

\node[state,initial\ifthenelse{#1=2}{,draw=blue}{} ] (q_0) {\ifthenelse{#1=2}{\color{blue}}{}$q_0$};

Anyway, I was wondering if there's any technique to highlight specific parts of a TikZ drawing without so much code repetition. Any ideas?

UPDATE: I'd like to say that all answers to this question are equally awesome! The accepted answer is the one that better suits my needs.

Usually, my automata representations tend to become more and more complex. Besides, the string complexity also follows the model tendency.

When I have several states and transitions, I find the \automaton{} approach better, as it gives me the flexibility of setting which highlight step I want based on a state number and not on the input string. That way, I can map all the possible highlights in my automaton and call them when needed.

For conventional automata representations, all answers are feasible. Thanks to all the answerers.

Best Answer

fabikw's answer is excellent if you're working with beamer. In order to get this working in a non-beamer application, I would use a /.code key to check whether the id of a part of the drawing is equal to the current step number, and set the color accordingly.

Here's one way of doing that:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{positioning,automata}

\begin{document}

\newif\ifhighlight
\tikzset{
    highlight/.is if=highlight,
    highlight=false,
    step number/.initial=0
}

\pgfkeys{
    /tikz/id/.code={
        \pgfmathparse{ifthenelse(#1==\pgfkeysvalueof{/tikz/step number},"true","false")}
        \pgfkeys{/tikz/highlight=\pgfmathresult}
        \ifhighlight
            \tikzset{
                orange!60!red,
            }
        \fi
    }
}


\newcommand{\automaton}[1]{
\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto,/tikz/initial text=,step number=#1,] 
   \begin{scope}[id=1]  % Necessary to colour the initial arrow
   \node[initial,state] (q_0)   {$q_0$};
   \end{scope}
   \node[state,id=3] (q_1) [right=of q_0] {$q_1$}; 
   \node[state,accepting,id=6] (q_2) [right=of q_1] {$q_2$}; 
    \path[->] 
    (q_0) edge [id=2] node [id=2] {a} (q_1)
    (q_1) edge [loop above,id=4] node {a} ()
    (q_1) edge [id=5] node {b} (q_2);
\end{tikzpicture}
}

\automaton{1}

\automaton{2}

\automaton{3}

\automaton{4}

\automaton{3}

\automaton{5}

\automaton{6}
\end{document}

Related Solutions

[Tex/LaTex] Rotate a node but not its content: the case of the ellipse decoration

The source of the difficulty is that ellipses are constructed in a particular way in TikZ. They are paths that start from the x-axis and proceed counter-clockwise around their centre. The vast majority of the time, the exact parametrisation doesn't matter. You appear to have found the one situation where it does!

In the actual question, you only want to be able to mirror the ellipse, and so draw it starting from the negative x-axis (the title of the question suggests a more flexible approach). That's actually not too hard since we can exploit the symmetry of the ellipse. The key is to provide it with a negative x-radius, since then it will start from the negative x-axis (and proceed clockwise, but we could correct for that by negating the y-radius as well). To do this, we interrupt the call from the node shape to the drawing command and flip the sign of the x-radius. The simplest way to do this is to redefine the \pgfpathellipse macro to do the negation and then call the original macro. The following code does this.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations,shapes,decorations.markings}
\makeatletter
\let\origpgfpathellipse=\pgfpathellipse
\def\revpgfpathellipse#1#2#3{%
  #2%
  \pgf@xa=-\pgf@x
  \origpgfpathellipse{#1}{\pgfqpoint{\pgf@xa}{0pt}}{#3}}
\makeatother

\tikzset{
  reversed ellipse/.style={
    ellipse,
    reverse the ellipse%
  },
  reverse the ellipse/.code={
    \let\pgfpathellipse=\revpgfpathellipse
  }
}
\begin{document}
\begin{tikzpicture}
\node[ellipse,
  draw,
  postaction={
    decorate,
    decoration={
      markings,
      mark=at position 1 with {
        \arrow[line width=5pt,blue]{>}
      }
    }
  }
] at (0,0) {hello world};

\node[reversed ellipse,
  draw,
  postaction={
    decorate,
    decoration={
      markings,
      mark=at position 1 with {
        \arrow[line width=5pt,blue]{>}
      }
    }
  }
] at (0,-2) {hello world};
\end{tikzpicture}

\end{document}

Here's the result:

rotated ellipse

(the arrow got clipped, but you can see where it lies)

[Tex/LaTex] Array cell highlighting tikz

This is the best I can think: use a tikz matrix to create a matrix of math nodes (which you can include inside a math environment and delimit with brackets if you want), and then use the implicit naming of nodes to refer to individual cells of the matrix, as for example: m-1-1.north east to refer to the north east corner of the first element.

In order to avoid alignment problems, you have to ensure that all the nodes of that matrix have the same dimensions, by giving a minimum width and minimum height option. I'm not very satisfied with this solution, because it requires you to know the dimensions of the larger cell. However, appropiate values are not difficult to find by trial and error.

After some tries, my code is the following:

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{inputenc}
\usepackage{xcolor}
\usepackage{tikz}
\begin{document}
\thispagestyle{empty}
\usetikzlibrary{matrix}
\usetikzlibrary{calc,fit}
\tikzset{%
  highlight1/.style={rectangle,rounded corners,color=red!,fill=red!15,draw,fill opacity=0.5,thick,inner sep=0pt}
}
\tikzset{%
  highlight2/.style={rectangle,rounded corners,color=green!,fill=green!15,draw,fill opacity=0.5,thick,inner sep=0pt}
}
\begin{equation}
\renewcommand{\arraystretch}{1.5}
 A_{L}=
\begin{tikzpicture}[baseline=(m.center)]
    \matrix (m) [matrix of math nodes, left delimiter={[}, right delimiter={]},
     row sep=1mm, nodes={minimum width=3em, minimum height=1.6em}] {
      -T^{1}_{11} & 0 &  0 & -T^{1}_{12} \\
      -T^{2}_{12} &  -T^{2}_{11} & 0 & 0  \\
        0 & -T^{3}_{12} & |(r)| T^{3}_{11} & 0 \\
        0 & 0 & 0 & 0 \\
      }; 
      \node[highlight2, fit=(m-1-1.north west) (m-2-2.south east)] {};
      \node[highlight1, fit=(m-3-1.north west) (m-4-4.south east)] {};
\end{tikzpicture} 
\left[\begin{array}{c}
\phi_{A} \\
\phi_{B} \\
\phi_{C} \\
\phi_{D}
\end{array}\right]
\label{eq:ALphif}
\end{equation}

\begin{equation}\renewcommand{\arraystretch}{1.5}
B_{L}= 
\begin{tikzpicture}[baseline=(m.center)]
    \matrix (m) [matrix of math nodes, left delimiter={[}, right delimiter={]},
       row sep=1mm, nodes={minimum width=5.5em, minimum height=1.6em}] {
        (T^{1}_{11}+T^{1}_{12}) & 0 & 0 \\
        0 & (T^{2}_{11}+T^{2}_{12}) & 0 \\ 
        0 & 0 & (T^{3}_{11}+T^{3}_{12}) \\
        0 & 0 & 0 \\
        };
    \node[highlight2, fit=(m-1-1.north west) (m-2-2.south east)] {};
    \node[highlight1, fit=(m-3-1.north west) (m-4-3.south east)] {};
\end{tikzpicture}
\left[\begin{array}{c}
\phi_{1} \\
\phi_{2} \\
\phi_{3}
\end{array}\right]
\label{eq:BLphii}
\end{equation}


\begin{equation}
\renewcommand{\arraystretch}{1.5}
 A_{R}= 
\begin{tikzpicture}[baseline=(m.center)]
    \matrix (m) [matrix of math nodes, left delimiter={[}, right delimiter={]},
      row sep=1mm, nodes={minimum width=2.5em, minimum height=1.6em}] {
        T^{2}_{22} & T^{2}_{21} & 0 & 0 \\
        0 & T^{3}_{22} & T^{3}_{21} & 0 \\ 
        0 &  0 &  0 &  0 \\
        T^{1}_{21} & 0 & 0 & T^{1}_{22}\\
        };
    \node[highlight2, fit=(m-1-1.north west) (m-2-2.south east)] {};
    \node[highlight1, fit=(m-3-1.north west) (m-4-4.south east)] {};
\end{tikzpicture}
\left[\begin{array}{c}
\phi_{A} \\
\phi_{B} \\
\phi_{C} \\
\phi_{D}
\end{array}\right]
\label{eq:ARphif}
\end{equation}

\begin{equation}
\renewcommand{\arraystretch}{1.5}
B_{R}= 
\begin{tikzpicture}[baseline=(m.center)]
    \matrix (m) [matrix of math nodes, left delimiter={[}, right delimiter={]},
       row sep=1mm, nodes={minimum width=6.5em, minimum height=1.6em}] {
        0 & -(T^{2}_{21}+T^{2}_{22}) & 0 \\
        0 &  0 & -(T^{3}_{21}+T^{3}_{22})\\ 
        0 & 0 & 0 \\
        -(T^{1}_{21}+T^{1}_{22}) & 0 & 0 \\
        };
    \node[highlight2, fit=(m-1-1.north west) (m-2-3.south east)] {};
    \node[highlight1, fit=(m-3-1.north west) (m-4-3.south east)] {};
\end{tikzpicture}
\left[\begin{array}{c}
\phi_{1} \\
\phi_{2} \\
\phi_{3}
\end{array}\right]
\label{eq:BRphii}
\end{equation}
\end{document}

Which produces the following output:

Output

Best Answer

Related Solutions

[Tex/LaTex] Rotate a node but not its content: the case of the ellipse decoration

[Tex/LaTex] Array cell highlighting tikz

Related Question