[Tex/LaTex] How to connect nodes with an arrow TikZ

nodestikz-pgf

The arrow connecting the bottom blue rectangle "Update E from H" to the red diamond "Done" has two discontinuities. How can I fix my code below to connect these lines at the empty nodes?

\usepackage{tikz}
\usetikzlibrary{shapes.geometric, arrows, shadows}
\begin{document}

\begin{figure}
\begin{center}
\tikzstyle{greenrect} = [rectangle, rounded corners, minimum width=3cm, minimum height=1cm,text justified, draw=black, fill=green!60]
\tikzstyle{bluerect} = [rectangle, rounded corners, minimum width=3cm, minimum height=1cm,text justified, draw=black, fill=blue!50]
\tikzstyle{whiterect} = [rectangle, minimum width=3cm, minimum height=1cm, text centered, draw=black]
\tikzstyle{reddiamond} = [diamond, minimum width=1cm, minimum height=1cm, text centered, draw=black, fill=red!80]
\tikzstyle{arrow} = [thick,->,>=stealth, line width=1.5pt]
\tikzstyle{line} = [thick,-,>=stealth, line width=1.5pt]

\begin{tikzpicture}[auto, node distance=2cm,>=latex']
\node(Initialize) [greenrect] {
\begin{tabular}{c}Initialize Fields to Zero\\$\vec{E} = \vec{H} = 0$ \end{tabular}
};
\node (Done) [reddiamond, below of=Initialize] {Done};

\node(E2H) [bluerect, below of=Done, node distance=3cm] {
\begin{tabular}{c}Update H from E\\
$\displaystyle{\vec{H}\Bigr\lvert_{t+\triangle t/2} = \vec{H}\Bigr\lvert_{t-\triangle t/2} - \frac{\triangle t}{\mu} (\nabla \times \vec{E}\Bigr\lvert_{t})}$\\
\end{tabular}
};

\node (H2E) [bluerect, below of=E2H, node distance=2.5cm] {
\begin{tabular}{c} Update E from H \\
$\displaystyle{\vec{E}\Bigr\lvert_{t+\triangle t/2} = \vec{E}\Bigr\lvert_{t-\triangle t/2} - \frac{\triangle t}{\epsilon} (\nabla \times \vec{H}\Bigr\lvert_{t})}$\\
\end{tabular}
};

\node (Finished) [right of=Done, node distance=6cm] {
\begin{tabular}{c}
Finished\\ 
\end{tabular}
};

\node (Empty1) [below of=H2E, node distance=2cm] {};
\node (Empty2) [left of=Empty1, node distance=5cm] {};
\node (Empty3) [left of=Done, node distance=5cm] {};

\draw [arrow] (Initialize) -- (Done);
\draw [arrow] (Done) -- node[name=no] {No} (E2H);
\draw [arrow] (E2H) -- (H2E);
\draw [arrow] (Done) -- node[name=yes] {Yes} (Finished);
\draw [line] (H2E) -- (Empty1);
\draw [line] (Empty1) -| (Empty2);
\draw [line] (Empty2) |- (Empty3);
\draw [arrow] (Empty3) -- (Done);

\end{tikzpicture}
\end{center}
\caption{Block diagram describing a simple FDTD Algorithm}
\end{figure}

\end{document}

Best Answer

Use \coordinates instead of \nodes:

\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric, arrows, shadows,positioning}


\begin{document}

\begin{figure}
\begin{center}
\tikzstyle{greenrect} = [rectangle, rounded corners, minimum width=3cm, minimum height=1cm,text justified, draw=black, fill=green!60]
\tikzstyle{bluerect} = [rectangle, rounded corners, minimum width=3cm, minimum height=1cm,text justified, draw=black, fill=blue!50]
\tikzstyle{whiterect} = [rectangle, minimum width=3cm, minimum height=1cm, text centered, draw=black]
\tikzstyle{reddiamond} = [diamond, minimum width=1cm, minimum height=1cm, text centered, draw=black, fill=red!80]
\tikzstyle{arrow} = [thick,->,>=stealth, line width=1.5pt]
\tikzstyle{line} = [thick,-,>=stealth, line width=1.5pt]

\begin{tikzpicture}[auto, node distance=2cm,>=latex']
\node(Initialize) [greenrect] {
\begin{tabular}{c}Initialize Fields to Zero\\$\vec{E} = \vec{H} = 0$ \end{tabular}
};
\node (Done) [reddiamond, below of=Initialize] {Done};

\node(E2H) [bluerect, below of=Done, node distance=3cm] {
\begin{tabular}{c}Update H from E\\
$\displaystyle{\vec{H}\Bigr\lvert_{t+\triangle t/2} = \vec{H}\Bigr\lvert_{t-\triangle t/2} - \frac{\triangle t}{\mu} (\nabla \times \vec{E}\Bigr\lvert_{t})}$\\
\end{tabular}
};

\node (H2E) [bluerect, below of=E2H, node distance=2.5cm] {
\begin{tabular}{c} Update E from H \\
$\displaystyle{\vec{E}\Bigr\lvert_{t+\triangle t/2} = \vec{E}\Bigr\lvert_{t-\triangle t/2} - \frac{\triangle t}{\epsilon} (\nabla \times \vec{H}\Bigr\lvert_{t})}$\\
\end{tabular}
};

\node (Finished) [right of=Done, node distance=6cm] {
\begin{tabular}{c}
Finished\\ 
\end{tabular}
};

\coordinate[below = 2cm of H2E] (Empty1);
\coordinate[left=5cm of Empty1] (Empty2);
\coordinate[left = 5cm of Done] (Empty3);

\draw [arrow] (Initialize) -- (Done);
\draw [arrow] (Done) -- node[name=no] {No} (E2H);
\draw [arrow] (E2H) -- (H2E);
\draw [arrow] (Done) -- node[name=yes] {Yes} (Finished);
\draw [arrow] 
  (H2E) -- (Empty1) -- (Empty2) |- (Done);

\end{tikzpicture}
\end{center}
\caption{Block diagram describing a simple FDTD Algorithm}
\end{figure}

\end{document}

The result:

Remarks:

Instead of the old deprecated of= syntax use the positioning library and =of (see an example in the code I used for palcing the coordinates).
Instead of the old \tikzstyle use \tikzset (refer to the package documentation).

You can easily draw the arrow using simply

\draw [arrow] 
  (H2E) -- (Empty1) -- (Empty2) |- (Done);

Related Solutions

[Tex/LaTex] Tikz – How to connect nodes

One way would be to add a stage at the start of the arrows that goes horizontally for a bit:

\draw[-latex] (A.east) -- ++(1,0) |- (B.west);

(adjust the 1 to get the length you want).

This does mean that the first part of the line would get drawn 4 times. It may not show (I don't know). To avoid that, you could draw it only the once; thus the later lines would be:

\draw[-latex] (A.east) ++(1,0) |- (C.west);

An alternative would be to put a coordinate node at that point:

\draw (A.east) -- ++(1,0) coordinate (div);
\draw[-latex] (div) |- (B.west);
\draw[-latex] (div) |- (C.west);

[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)

Best Answer

Remarks:

Related Solutions

[Tex/LaTex] Tikz – How to connect nodes

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

Related Question