Shifted hexagons in TikZ picture

Question

Recently there was a great answer in How to draw a 3D hexagonal structure with TikZ?

But I noticed a small "bug" which I can't seem to be able to fix. I stripped the code down that shows the problem:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes}

\begin{document}
\begin{tikzpicture}
  \begin{scope}[%
      every node/.style={anchor=west,regular polygon, regular polygon sides=6,draw,inner sep=0.5cm},
      transform shape]
    \node (A) {A};
    \node (B) at (A.corner 1) {B};
    \node (C) at (B.corner 5) {C};
    \node (D) at (A.corner 5) {D};
    \node (E) at (D.corner 5) {E};
    \foreach \hex in {A,...,E}
    {
      \foreach \corn in {1,...,6}
        \draw[fill=white] (\hex.corner \corn) circle (2pt); 
    }
  \end{scope}
\end{tikzpicture}
\end{document}

This code draws five nodes with hexagonal shape, but the hexagons are a few, but always a different "number of few", pixels "off", as shown in this picture:

• Which value is responsible for this?
• Why is the "shifting" happening with different strength?
• How to fix it?

Answer 1

You would need outer sep=0 and also inner sep is dependent on the node contents and the node size change accordingly. Instead you can use minimum height/width keys.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes}

\begin{document}
\begin{tikzpicture}
  \begin{scope}[%
every node/.style={anchor=west,
regular polygon, 
regular polygon sides=6,
draw,
minimum width=2cm,
outer sep=0,
},
      transform shape]
    \node (A) {A};
    \node (B) at (A.corner 1) {B};
    \node (C) at (B.corner 5) {C};
    \node (D) at (A.corner 5) {DECF};
    \node (E) at (D.corner 5) {E};
    \foreach \hex in {A,...,E}
    {
      \foreach \corn in {1,...,6}
        \draw[fill=white] (\hex.corner \corn) circle (2pt); 
    }
  \end{scope}
\end{tikzpicture}
\end{document}