[Tex/LaTex] Tikz: Division and Addition for Node Names

tikz-pgf

I am trying to draw a graph that is essentially just a long line with a triangle at one end and a square at the other end. I need there to be a chain of length 26 between the triangle and the square. When I tried with the code below, I got the error "Missing Number, treated as 0". Other attempts have told me that there are no nodes named "P0+1". How do I do division, addition, etc. on node names to iterate through them as I'd like?

    \begin{tikzpicture}
    \foreach \x in {0,1,2,3,4,5,6,7,8,9,10,11,12}{
    \node (P\x) at (-\x/2,0){};
    }
    \foreach \x in {13,14,15,16,17,18,19,20,21,22,23,24,25,26}{
    \node (P\x) at ({\x-13}/2,0){};
    }

    \node (P27) at (14,1){};
    \node (P28) at (14,-1){};
    \node (P29) at (-14,1){};
    \node (P30) at (-14,-1){};
    \node (P31) at (-15,0){};


    \foreach \x in {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31}{
    \fill (P\x) circle (2pt);
    }

    \foreach \x in {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25}{
    \draw[line width=1 pt] (P\x)--(P\x+1);
    }

    \draw[line width=1 pt] (P26)--(P27);
    \draw[line width=1 pt] (P27)--(P28);
    \draw[line width=1 pt] (P28) -- (P26);
    \draw[line width=1 pt] (P13)--(P29);
    \draw[line width=1 pt] (P29)--(P30);
    \draw[line width=1 pt] (P30) -- (P31);
    \draw[line width=1 pt] (P31) -- (P13);


    \draw (0,-3) node[below]{Figure 3.4};
    \end{tikzpicture}

Best Answer

There are certain places where TikZ don't parse the math such as node names and a few other places. Inside a loop you can instead use the evaluate=<var> as <resulting var> using <formula> syntax. You could have used a counter that starts from 1 in this specific case too.

Moreover you need to use (...) to nest the math and {...} to hide it from the parser. Hence the code becomes

\begin{tikzpicture}
    \foreach \x in {0,1,2,3,4,5,6,7,8,9,10,11,12}{
    \node (P\x) at (-\x/2,0){};
    }
    \foreach \x in {13,14,15,16,17,18,19,20,21,22,23,24,25,26}{
    \node (P\x) at ({(\x-13)/2)},0){};
    }

    \node (P27) at (14,1){};
    \node (P28) at (14,-1){};
    \node (P29) at (-14,1){};
    \node (P30) at (-14,-1){};
    \node (P31) at (-15,0){};


    \foreach \x in {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31}{
    \fill (P\x) circle (2pt);
    }

    \foreach \x[evaluate=\x as \evalx using int(\x+1)] in {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25}{
    \draw[line width=1 pt] (P\x)--(P\evalx);
    }

    \draw[line width=1 pt] (P26)--(P27);
    \draw[line width=1 pt] (P27)--(P28);
    \draw[line width=1 pt] (P28) -- (P26);
    \draw[line width=1 pt] (P13)--(P29);
    \draw[line width=1 pt] (P29)--(P30);
    \draw[line width=1 pt] (P30) -- (P31);
    \draw[line width=1 pt] (P31) -- (P13);


    \draw (0,-3) node[below]{Figure 3.4};
    \end{tikzpicture}

enter image description here

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] How to get access to the size parameters of a TikZ node

Here is one possibility with not-exactly-failproof node construction. The triangles are also accesible with (<nodename>-leftri) and (<nodename>-rightri). The possible failure reason would be rotating the node. It's possible to include that but it's unnecessarily complicated at this stage and I don't know if you ever need it.

\documentclass[tikz]{standalone}
\usetikzlibrary{shapes.geometric}

\begin{document}

\begin{tikzpicture}[mynode/.style args={#1 and #2}{
draw,
minimum height=#1,
minimum width=#2,
rounded corners=0.5*#1,
append after command={
\pgfextra{%                 
    \begin{pgfinterruptpath}
    \node[isosceles triangle,
          isosceles triangle stretches,
          draw=black,inner sep=0,
          anchor=west,
          minimum width=0.5*#1,
          minimum height=0.2*#1,
          fill=black] 
          (\tikzlastnode-leftri) at ([xshift=0.5*#1]\tikzlastnode.west) {};
    \node[isosceles triangle,
          isosceles triangle stretches,
          draw=black,inner sep=0,
          anchor=west,
          minimum width=0.5*#1,
          minimum height=0.2*#1,
          isosceles triangle apex angle=90]
          (\tikzlastnode-rightri) at ([xshift=-0.5*#1]\tikzlastnode.east) {};
\end{pgfinterruptpath}
        }   
    }
}
]

\node[mynode=1cm and 4cm] (a) {};
\node[mynode=3cm and 8cm] (b) at (0,3) {};
\draw (a-leftri) -- (b-rightri);
\end{tikzpicture}

\end{document}

enter image description here

Best Answer

Related Solutions

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

[Tex/LaTex] How to get access to the size parameters of a TikZ node

Related Question